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Tyn Myint-U Lokenath Debnath Linear Partial Differential Equations for Scientists and Engineers Fourth Edition Birkhauser ¨ Boston • Basel • Berlin Tyn Myint-U 5 Sue Terrace Westport, CT 06880 USA Lokenath Debnath Department of Mathematics University of Texas-Pan American 1201 W. University Drive Edinburgh, TX 78539 USA Cover design by Alex Gerasev. Mathematics Subject Classification (2000): 00A06, 00A69, 34B05, 34B24, 34B27, 34G20, 35-01, 35-02, 35A15, 35A22, 35A25, 35C05, 35C15, 35Dxx, 35E05, 35E15, 35Fxx, 35F05, 35F10, 35F15, 35F20, 35F25, 35G10, 35G20, 35G25, 35J05, 35J10, 35J20, 35K05, 35K10, 35K15, 35K55, 35K60, 35L05, 35L10, 35L15, 35L20, 35L25, 35L30, 35L60, 35L65, 35L67, 35L70, 35Q30, 35Q35, 35Q40, 35Q51, 35Q53, 35Q55, 35Q58, 35Q60, 35Q80, 42A38, 44A10, 44A35 49J40, 58E30, 58E50, 65L15, 65M25, 65M30, 65R10, 70H05, 70H20, 70H25, 70H30, 76Bxx, 76B15, 76B25, 76D05, 76D33, 76E30, 76M30, 76R50, 78M30, 81Q05 Library of Congress Control Number: 2006935807 ISBN-10: 0-8176-4393-1 e-ISBN-10: 0-8176-4560-8 ISBN-13: 978-0-8176-4393-5 e-ISBN-13: 978-0-8176-4560-1 Printed on acid-free paper.
c 2007 Birkhauser Boston ¨ All rights reserved. This work may not be translated or copied in whole or in part without the written permission of the publisher (Birkhauser Boston, c/o Springer Science ¨ +Business Media LLC, 233 Spring Street, New York, NY 10013, USA) and the author, except for brief excerpts in connection with reviews or scholarly analysis. Use in connection with any form of information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed is forbidden. The use in this publication of trade names, trademarks, service marks and similar terms, even if they are not identified as such, is not to be taken as an expression of opinion as to whether or not they are subject to proprietary rights. 987654321 www.birkhauser.com (SB) To the Memory of U and Mrs. Hla Din U and Mrs. Thant Tyn Myint-U In Loving Memory of My Mother and Father Lokenath Debnath “True Laws of Nature cannot be linear.” “The search for truth is more precious than its possession.” “Everything should be made as simple as possible, but not a bit simpler.” Albert Einstein “No human investigation can be called real science if it cannot be demonstrated mathematically.” Leonardo Da Vinci “First causes are not known to us, but they are subjected to simple and constant laws that can be studied by observation and whose study is the goal of Natural Philosophy ... Heat penetrates, as does gravity, all the substances of the universe; its rays occupy all regions of space. The aim of our work is to expose the mathematical laws that this element follows ... The differential equations for the propagation of heat express the most general conditions and reduce physical questions to problems in pure Analysis that is properly the object of the theory.” James Clerk Maxwell “One of the properties inherent in mathematics is that any real progress is accompanied by the discovery and development of new methods and simplifications of previous procedures ... The unified character of mathematics lies in its very nature; indeed, mathematics is the foundation of all exact natural sciences.” David Hilbert “ ... partial differential equations are the basis of all physical theorems. In the theory of sound in gases, liquid and solids, in the investigations of elasticity, in optics, everywhere partial differential equations formulate basic laws of nature which can be checked against experiments.” Bernhard Riemann “The effective numerical treatment of partial differential equations is not a handicraft, but an art.” Folklore “The advantage of the principle of least action is that in one and the same equation it relates the quantities that are immediately relevant not only to mechanics but also to electrodynamics and thermodynamics; these quantities are space, time and potential.” Max Planck “The thorough study of nature is the most ground for mathematical discoveries.” “The equations for the flow of heat as well as those for the oscillations of acoustic bodies and of fluids belong to an area of analysis which has recently been opened, and which is worth examining in the greatest detail.” Joseph Fourier “Of all the mathematical disciplines, the theory of differential equation is the most important. All branches of physics pose problems which can be reduced to the integration of differential equations. More generally, the way of explaining all natural phenomena which depend on time is given by the theory of differential equations.” Sophus Lie “Differential equations form the basis for the scientific view of the world.” V.I. Arnold “What we know is not much. What we do not know is immense.” “The algebraic analysis soon makes us forget the main object [of our research] by focusing our attention on abstract combinations and it is only at the end that we return to the original objective. But in abandoning oneself to the operations of analysis, one is led to the generality of this method and the inestimable advantage of transforming the reasoning by mechanical procedures to results often inaccessible by geometry ... No other language has the capacity for the elegance that arises from a long sequence of expressions linked one to the other and all stemming from one fundamental idea.” “It is India that gave us the ingenious method of expressing all numbers by ten symbols, each symbol receiving a value of position, as well as an absolute value. We shall appreciate the grandeur of the achievement when we remember that it escaped the genius of Archimedes and Appolonius.” P.S. Laplace “The mathematician’s best work is art, a high perfect art, as daring as the most secret dreams of imagination, clear and limpid. Mathematical genius and artistic genius touch one another.” G˙osta Mittag-Leffler Contents Preface to the Fourth Edition . . . . . . . . . . . . . . . . . xv Preface to the Third Edition . . . . . . . . . . . . . . . . . xix 1 Introduction 1 1.1 Brief Historical Comments . . . . . . . . . . . . . . . . . 1 1.2 Basic Concepts and Definitions . . . . . . . . . . . . . . . 12 1.3 Mathematical Problems . . . . . . . . . . . . . . . . . . . 15 1.4 Linear Operators . . . . . . . . . . . . . . . . . . . . . . . 16 1.5 Superposition Principle . . . . . . . . . . . . . . . . . . . 20 1.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 2 First-Order, Quasi-Linear Equations and Method of Characteristics 27 2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 27 2.2 Classification of First-Order Equations . . . . . . . . . . . 27 2.3 Construction of a First-Order Equation . . . . . . . . . . 29 2.4 Geometrical Interpretation of a First-Order Equation . . 33 2.5 Method of Characteristics and General Solutions . . . . . 35 2.6 Canonical Forms of First-Order Linear Equations . . . . 49 2.7 Method of Separation of Variables . . . . . . . . . . . . . 51 2.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 55 3 Mathematical Models 63 3.1 Classical Equations . . . . . . . . . . . . . . . . . . . . . 63 3.2 The Vibrating String . . . . . . . . . . . . . . . . . . . . 65 3.3 The Vibrating Membrane . . . . . . . . . . . . . . . . . . 67 3.4 Waves in an Elastic Medium . . . . . . . . . . . . . . . . 69 3.5 Conduction of Heat in Solids . . . . . . . . . . . . . . . . 75 3.6 The Gravitational Potential . . . . . . . . . . . . . . . . . 76 3.7 Conservation Laws and The Burgers Equation . . . . . . 79 3.8 The Schr¨odinger and the Korteweg–de Vries Equations . 81 3.9 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 83 4 Classification of Second-Order Linear Equations 91 4.1 Second-Order Equations in Two Independent Variables . 91 x Contents 4.2 Canonical Forms . . . . . . . . . . . . . . . . . . . . . . . 93 4.3 Equations with Constant Coefficients . . . . . . . . . . . 99 4.4 General Solutions . . . . . . . . . . . . . . . . . . . . . . 107 4.5 Summary and Further Simplification . . . . . . . . . . . . 111 4.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 113 5 The Cauchy Problem and Wave Equations 117 5.1 The Cauchy Problem . . . . . . . . . . . . . . . . . . . . 117 5.2 The Cauchy–Kowalewskaya Theorem . . . . . . . . . . . 120 5.3 Homogeneous Wave Equations . . . . . . . . . . . . . . . 121 5.4 Initial Boundary-Value Problems . . . . . . . . . . . . . . 130 5.5 Equations with Nonhomogeneous Boundary Conditions . 134 5.6 Vibration of Finite String with Fixed Ends . . . . . . . . 136 5.7 Nonhomogeneous Wave Equations . . . . . . . . . . . . . 139 5.8 The Riemann Method . . . . . . . . . . . . . . . . . . . . 142 5.9 Solution of the Goursat Problem . . . . . . . . . . . . . . 149 5.10 Spherical Wave Equation . . . . . . . . . . . . . . . . . . 153 5.11 Cylindrical Wave Equation . . . . . . . . . . . . . . . . . 155 5.12 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 158 6 Fourier Series and Integrals with Applications 167 6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 167 6.2 Piecewise Continuous Functions and Periodic Functions . 168 6.3 Systems of Orthogonal Functions . . . . . . . . . . . . . . 170 6.4 Fourier Series . . . . . . . . . . . . . . . . . . . . . . . . . 171 6.5 Convergence of Fourier Series . . . . . . . . . . . . . . . . 173 6.6 Examples and Applications of Fourier Series . . . . . . . 177 6.7 Examples and Applications of Cosine and Sine Fourier Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183 6.8 Complex Fourier Series . . . . . . . . . . . . . . . . . . . 194 6.9 Fourier Series on an Arbitrary Interval . . . . . . . . . . 196 6.10 The Riemann–Lebesgue Lemma and Pointwise Convergence Theorem . . . . . . . . . . . . . . . . . . . . 201 6.11 Uniform Convergence, Differentiation, and Integration . . 208 6.12 Double Fourier Series . . . . . . . . . . . . . . . . . . . . 212 6.13 Fourier Integrals . . . . . . . . . . . . . . . . . . . . . . . 214 6.14 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 220 7 Method of Separation of Variables 231 7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 231 7.2 Separation of Variables . . . . . . . . . . . . . . . . . . . 232 7.3 The Vibrating String Problem . . . . . . . . . . . . . . . 235 7.4 Existence and Uniqueness of Solution of the Vibrating String Problem . . . . . . . . . . . . . . . . . . . . . . . . 243 7.5 The Heat Conduction Problem . . . . . . . . . . . . . . . 248 Contents xi 7.6 Existence and Uniqueness of Solution of the Heat Conduction Problem . . . . . . . . . . . . . . . . . . . . . 251 7.7 The Laplace and Beam Equations . . . . . . . . . . . . . 254 7.8 Nonhomogeneous Problems . . . . . . . . . . . . . . . . . 258 7.9 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 265 8 Eigenvalue Problems and Special Functions 273 8.1 Sturm–Liouville Systems . . . . . . . . . . . . . . . . . . 273 8.2 Eigenvalues and Eigenfunctions . . . . . . . . . . . . . . . 277 8.3 Eigenfunction Expansions . . . . . . . . . . . . . . . . . . 283 8.4 Convergence in the Mean . . . . . . . . . . . . . . . . . . 284 8.5 Completeness and Parseval’s Equality . . . . . . . . . . . 286 8.6 Bessel’s Equation and Bessel’s Function . . . . . . . . . . 289 8.7 Adjoint Forms and Lagrange Identity . . . . . . . . . . . 295 8.8 Singular Sturm–Liouville Systems . . . . . . . . . . . . . 297 8.9 Legendre’s Equation and Legendre’s Function . . . . . . . 302 8.10 Boundary-Value Problems Involving Ordinary Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 308 8.11 Green’s Functions for Ordinary Differential Equations . . 310 8.12 Construction of Green’s Functions . . . . . . . . . . . . . 315 8.13 The Schr¨odinger Equation and Linear Harmonic Oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . 317 8.14 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 321 9 Boundary-Value Problems and Applications 329 9.1 Boundary-Value Problems . . . . . . . . . . . . . . . . . . 329 9.2 Maximum and Minimum Principles . . . . . . . . . . . . 332 9.3 Uniqueness and Continuity Theorems . . . . . . . . . . . 333 9.4 Dirichlet Problem for a Circle . . . . . . . . . . . . . . . . 334 9.5 Dirichlet Problem for a Circular Annulus . . . . . . . . . 340 9.6 Neumann Problem for a Circle . . . . . . . . . . . . . . . 341 9.7 Dirichlet Problem for a Rectangle . . . . . . . . . . . . . 343 9.8 Dirichlet Problem Involving the Poisson Equation . . . . 346 9.9 The Neumann Problem for a Rectangle . . . . . . . . . . 348 9.10 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 351 10 Higher-Dimensional Boundary-Value Problems 361 10.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 361 10.2 Dirichlet Problem for a Cube . . . . . . . . . . . . . . . . 361 10.3 Dirichlet Problem for a Cylinder . . . . . . . . . . . . . . 363 10.4 Dirichlet Problem for a Sphere . . . . . . . . . . . . . . . 367 10.5 Three-Dimensional Wave and Heat Equations . . . . . . . 372 10.6 Vibrating Membrane . . . . . . . . . . . . . . . . . . . . . 372 10.7 Heat Flow in a Rectangular Plate . . . . . . . . . . . . . 375 10.8 Waves in Three Dimensions . . . . . . . . . . . . . . . . . 379 xii Contents 10.9 Heat Conduction in a Rectangular Volume . . . . . . . . 381 10.10 The Schr¨odinger Equation and the Hydrogen Atom . . . 382 10.11 Method of Eigenfunctions and Vibration of Membrane . . 392 10.12 Time-Dependent Boundary-Value Problems . . . . . . . . 395 10.13 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 398 11 Green’s Functions and Boundary-Value Problems 407 11.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 407 11.2 The Dirac Delta Function . . . . . . . . . . . . . . . . . . 409 11.3 Properties of Green’s Functions . . . . . . . . . . . . . . . 412 11.4 Method of Green’s Functions . . . . . . . . . . . . . . . . 414 11.5 Dirichlet’s Problem for the Laplace Operator . . . . . . . 416 11.6 Dirichlet’s Problem for the Helmholtz Operator . . . . . . 418 11.7 Method of Images . . . . . . . . . . . . . . . . . . . . . . 420 11.8 Method of Eigenfunctions . . . . . . . . . . . . . . . . . . 423 11.9 Higher-Dimensional Problems . . . . . . . . . . . . . . . . 425 11.10 Neumann Problem . . . . . . . . . . . . . . . . . . . . . . 430 11.11 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 433 12 Integral Transform Methods with Applications 439 12.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 439 12.2 Fourier Transforms . . . . . . . . . . . . . . . . . . . . . . 440 12.3 Properties of Fourier Transforms . . . . . . . . . . . . . . 444 12.4 Convolution Theorem of the Fourier Transform . . . . . . 448 12.5 The Fourier Transforms of Step and Impulse Functions . 453 12.6 Fourier Sine and Cosine Transforms . . . . . . . . . . . . 456 12.7 Asymptotic Approximation of Integrals by Stationary Phase Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . 458 12.8 Laplace Transforms . . . . . . . . . . . . . . . . . . . . . 460 12.9 Properties of Laplace Transforms . . . . . . . . . . . . . . 463 12.10 Convolution Theorem of the Laplace Transform . . . . . 467 12.11 Laplace Transforms of the Heaviside and Dirac Delta Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 470 12.12 Hankel Transforms . . . . . . . . . . . . . . . . . . . . . . 488 12.13 Properties of Hankel Transforms and Applications . . . . 491 12.14 Mellin Transforms and their Operational Properties . . . 495 12.15 Finite Fourier Transforms and Applications . . . . . . . . 499 12.16 Finite Hankel Transforms and Applications . . . . . . . . 504 12.17 Solution of Fractional Partial Differential Equations . . . 510 12.18 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 521 13 Nonlinear Partial Differential Equations with Applications 535 13.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 535 Contents xiii 13.2 One-Dimensional Wave Equation and Method of Characteristics . . . . . . . . . . . . . . . . . . . . . . . . 536 13.3 Linear Dispersive Waves . . . . . . . . . . . . . . . . . . . 540 13.4 Nonlinear Dispersive Waves and Whitham’s Equations . . 545 13.5 Nonlinear Instability . . . . . . . . . . . . . . . . . . . . . 548 13.6 The Traffic Flow Model . . . . . . . . . . . . . . . . . . . 549 13.7 Flood Waves in Rivers . . . . . . . . . . . . . . . . . . . . 552 13.8 Riemann’s Simple Waves of Finite Amplitude . . . . . . . 553 13.9 Discontinuous Solutions and Shock Waves . . . . . . . . . 561 13.10 Structure of Shock Waves and Burgers’ Equation . . . . . 563 13.11 The Korteweg–de Vries Equation and Solitons . . . . . . 573 13.12 The Nonlinear Schr¨odinger Equation and Solitary Waves. 581 13.13 The Lax Pair and the Zakharov and Shabat Scheme . . . 590 13.14 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 595 14 Numerical and Approximation Methods 601 14.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 601 14.2 Finite Difference Approximations, Convergence, and Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . 602 14.3 Lax–Wendroff Explicit Method . . . . . . . . . . . . . . . 605 14.4 Explicit Finite Difference Methods . . . . . . . . . . . . . 608 14.5 Implicit Finite Difference Methods . . . . . . . . . . . . . 624 14.6 Variational Methods and the Euler–Lagrange Equations . 629 14.7 The Rayleigh–Ritz Approximation Method . . . . . . . . 647 14.8 The Galerkin Approximation Method . . . . . . . . . . . 655 14.9 The Kantorovich Method . . . . . . . . . . . . . . . . . . 659 14.10 The Finite Element Method . . . . . . . . . . . . . . . . . 663 14.11 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 668 15 Tables of Integral Transforms 681 15.1 Fourier Transforms . . . . . . . . . . . . . . . . . . . . . . 681 15.2 Fourier Sine Transforms . . . . . . . . . . . . . . . . . . . 683 15.3 Fourier Cosine Transforms . . . . . . . . . . . . . . . . . 685 15.4 Laplace Transforms . . . . . . . . . . . . . . . . . . . . . 687 15.5 Hankel Transforms . . . . . . . . . . . . . . . . . . . . . . 691 15.6 Finite Hankel Transforms . . . . . . . . . . . . . . . . . . 695 Answers and Hints to Selected Exercises 697 1.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 697 2.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 698 3.9 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 704 4.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 707 5.12 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 712 6.14 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 715 7.9 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 724 xiv Contents 8.14 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 726 9.10 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 727 10.13 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 731 11.11 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 739 12.18 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 740 14.11 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 745 Appendix: Some Special Functions and Their Properties 749 A-1 Gamma, Beta, Error, and Airy Functions . . . . . . . . . 749 A-2 Hermite Polynomials and Weber–Hermite Functions . . . 757 Bibliography 761 Index 771 Preface to the Fourth Edition “A teacher can never truly teach unless he is still learning himself. A lamp can never light another lamp unless it continues to burn its own flame. The teacher who has come to the end of his subject, who has no living traffic with his knowledge but merely repeats his lessons to his students, can only load their minds; he cannot quicken them.” Rabindranath Tagore An Indian Poet 1913 Nobel Prize Winner for Literature The previous three editions of our book were very well received and used as a senior undergraduate or graduate-level text and research reference in the United States and abroad for many years. We received many comments and suggestions from many students, faculty and researchers around the world. These comments and criticisms have been very helpful, beneficial, and encouraging. This fourth edition is the result of the input. Another reason for adding this fourth edition to the literature is the fact that there have been major discoveries of new ideas, results and methods for the solution of linear and nonlinear partial differential equations in the second half of the twentieth century. It is becoming even more desirable for mathematicians, scientists and engineers to pursue study and research on these topics. So what has changed, and will continue to change is the nature of the topics that are of interest in mathematics, applied mathematics, physics and engineering, the evolution of books such is this one is a history of these shifting concerns. This new and revised edition preserves the basic content and style of the third edition published in 1989. As with the previous editions, this book has been revised primarily as a comprehensive text for senior undergraduates or beginning graduate students and a research reference for professionals in mathematics, science and engineering, and other applied sciences. The main goal of the book is to develop required analytical skills on the part of the xvi Preface to the Fourth Edition reader, rather than to focus on the importance of more abstract formulation, with full mathematical rigor. Indeed, our major emphasis is to provide an accessible working knowledge of the analytical and numerical methods with proofs required in mathematics, applied mathematics, physics, and engineering. The revised edition was greatly influenced by the statements that Lord Rayleigh and Richard Feynman made as follows: “In the mathematical investigation I have usually employed such methods as present themselves naturally to a physicist. The pure mathematician will complain, and (it must be confessed) sometimes with justice, of defi- cient rigor. But to this question there are two sides. For, however important it may be to maintain a uniformly high standard in pure mathematics, the physicist may occasionally do well to rest content with arguments, which are fairly satisfactory and conclusive from his point of view. To his mind, exercised in a different order of ideas, the more severe procedure of the pure mathematician may appear not more but less demonstrative. And further, in many cases of difficulty to insist upon highest standard would mean the exclusion of the subject altogether in view of the space that would be required.” Lord Rayleigh “... However, the emphasis should be somewhat more on how to do the mathematics quickly and easily, and what formulas are true, rather than the mathematicians’ interest in methods of rigorous proof.” Richard P. Feynman We have made many additions and changes in order to modernize the contents and to improve the clarity of the previous edition. We have also taken advantage of this new edition to correct typographical errors, and to update the bibliography, to include additional topics, examples of applications, exercises, comments and observations, and in some cases, to entirely rewrite and reorganize many sections. This is plenty of material in the book for a year-long course. Some of the material need not be covered in a course work and can be left for the readers to study on their own in order to prepare them for further study and research. This edition contains a collection of over 900 worked examples and exercises with answers and hints to selected exercises. Some of the major changes and additions include the following: 1. Chapter 1 on Introduction has been completely revised and a new section on historical comments was added to provide information about the historical developments of the subject. These changes have been made to provide the reader to see the direction in which the subject has developed and find those contributed to its developments. 2. A new Chapter 2 on first-order, quasi-linear, and linear partial differential equations, and method of characteristics has been added with many new examples and exercises. Preface to the Fourth Edition xvii 3. Two sections on conservation laws, Burgers’ equation, the Schr¨odinger and the Korteweg-de Vries equations have been included in Chapter 3. 4. Chapter 6 on Fourier series and integrals with applications has been completely revised and new material added, including a proof of the pointwise convergence theorem. 5. A new section on fractional partial differential equations has been added to Chapter 12 with many new examples of applications. 6. A new section on the Lax pair and the Zakharov and Shabat Scheme has been added to Chapter 13 to modernize its contents. 7. Some sections of Chapter 14 have been revised and a new short section on the finite element method has been added to this chapter. 8. A new Chapter 15 on tables of integral transforms has been added in order to make the book self-contained. 9. The whole section on Answers and Hints to Selected Exercises has been expanded to provide additional help to students. All figures have been redrawn and many new figures have been added for a clear understanding of physical explanations. 10. An Appendix on special functions and their properties has been expanded. Some of the highlights in this edition include the following: • The book offers a detailed and clear explanation of every concept and method that is introduced, accompanied by carefully selected worked examples, with special emphasis given to those topics in which students experience difficulty. • A wide variety of modern examples of applications has been selected from areas of integral and ordinary differential equations, generalized functions and partial differential equations, quantum mechanics, fluid dynamics and solid mechanics, calculus of variations, linear and nonlinear stability analysis. • The book is organized with sufficient flexibility to enable instructors to select chapters appropriate for courses of differing lengths, emphases, and levels of difficulty. • A wide spectrum of exercises has been carefully chosen and included at the end of each chapter so the reader may further develop both rigorous skills in the theory and applications of partial differential equations and a deeper insight into the subject. • Many new research papers and standard books have been added to the bibliography to stimulate new interest in future study and research. Index of the book has also been completely revised in order to include a wide variety of topics. • The book provides information that puts the reader at the forefront of current research. With the improvements and many challenging worked-out problems and exercises, we hope this edition will continue to be a useful textbook for xviii Preface to the Fourth Edition students as well as a research reference for professionals in mathematics, applied mathematics, physics and engineering. It is our pleasure to express our grateful thanks to many friends, colleagues, and students around the world who offered their suggestions and help at various stages of the preparation of the book. We offer special thanks to Dr. Andras Balogh, Mr. Kanadpriya Basu, and Dr. Dambaru Bhatta for drawing all figures, and to Mrs. Veronica Martinez for typing the manuscript with constant changes and revisions. In spite of the best efforts of everyone involved, some typographical errors doubtless remain. Finally, we wish to express our special thanks to Tom Grasso and the staff of Birkh¨auser Boston for their help and cooperation. Tyn Myint-U Lokenath Debnath Preface to the Third Edition The theory of partial differential equations has long been one of the most important fields in mathematics. This is essentially due to the frequent occurrence and the wide range of applications of partial differential equations in many branches of physics, engineering, and other sciences. With much interest and great demand for theory and applications in diverse areas of science and engineering, several excellent books on PDEs have been published. This book is written to present an approach based mainly on the mathematics, physics, and engineering problems and their solutions, and also to construct a course appropriate for all students of mathematical, physical, and engineering sciences. Our primary objective, therefore, is not concerned with an elegant exposition of general theory, but rather to provide students with the fundamental concepts, the underlying principles, a wide range of applications, and various methods of solution of partial differential equations. This book, a revised and expanded version of the second edition published in 1980, was written for a one-semester course in the theory and applications of partial differential equations. It has been used by advanced undergraduate or beginning graduate students in applied mathematics, physics, engineering, and other applied sciences. The prerequisite for its study is a standard calculus sequence with elementary ordinary differential equations. This revised edition is in part based on lectures given by Tyn Myint-U at Manhattan College and by Lokenath Debnath at the University of Central Florida. This revision preserves the basic content and style of the earlier editions, which were written by Tyn Myint-U alone. However, the authors have made some major additions and changes in this third edition in order to modernize the contents and to improve clarity. Two new chapters added are on nonlinear PDEs, and on numerical and approximation methods. New material emphasizing applications has been inserted. New examples and exercises have been provided. Many physical interpretations of mathematical solutions have been added. Also, the authors have improved the exposition by reorganizing some material and by making examples, exercises, and ap- xx Preface to the Third Edition plications more prominent in the text. These additions and changes have been made with the student uppermost in mind. The first chapter gives an introduction to partial differential equations. The second chapter deals with the mathematical models representing physical and engineering problems that yield the three basic types of PDEs. Included are only important equations of most common interest in physics and engineering. The third chapter constitutes an account of the classifi- cation of linear PDEs of second order in two independent variables into hyperbolic, parabolic, and elliptic types and, in addition, illustrates the determination of the general solution for a class of relatively simple equations. Cauchy’s problem, the Goursat problem, and the initial boundary-value problems involving hyperbolic equations of the second order are presented in Chapter 4. Special attention is given to the physical significance of solutions and the methods of solution of the wave equation in Cartesian, spherical polar, and cylindrical polar coordinates. The fifth chapter contains a fuller treatment of Fourier series and integrals essential for the study of PDEs. Also included are proofs of several important theorems concerning Fourier series and integrals. Separation of variables is one of the simplest methods, and the most widely used method, for solving PDEs. The basic concept and separability conditions necessary for its application are discussed in the sixth chapter. This is followed by some well-known problems of applied mathematics, mathematical physics, and engineering sciences along with a detailed analysis of each problem. Special emphasis is also given to the existence and uniqueness of the solutions and to the fundamental similarities and differences in the properties of the solutions to the various PDEs. In Chapter 7, self-adjoint eigenvalue problems are treated in depth, building on their introduction in the preceding chapter. In addition, Green’s function and its applications to eigenvalue problems and boundary-value problems for ordinary differential equations are presented. Following the general theory of eigenvalues and eigenfunctions, the most common special functions, including the Bessel, Legendre, and Hermite functions, are discussed as examples of the major role of special functions in the physical and engineering sciences. Applications to heat conduction problems and the Schr¨odinger equation for the linear harmonic oscillator are also included. Boundary-value problems and the maximum principle are described in Chapter 8, and emphasis is placed on the existence, uniqueness, and wellposedness of solutions. Higher-dimensional boundary-value problems and the method of eigenfunction expansion are treated in the ninth chapter, which also includes several applications to the vibrating membrane, waves in three dimensions, heat conduction in a rectangular volume, the threedimensional Schr¨odinger equation in a central field of force, and the hydrogen atom. Chapter 10 deals with the basic concepts and construction of Green’s function and its application to boundary-value problems. Preface to the Third Edition xxi Chapter 11 provides an introduction to the use of integral transform methods and their applications to numerous problems in applied mathematics, mathematical physics, and engineering sciences. The fundamental properties and the techniques of Fourier, Laplace, Hankel, and Mellin transforms are discussed in some detail. Applications to problems concerning heat flows, fluid flows, elastic waves, current and potential electric transmission lines are included in this chapter. Chapters 12 and 13 are entirely new. First-order and second-order nonlinear PDEs are covered in Chapter 12. Most of the contents of this chapter have been developed during the last twenty-five years. Several new nonlinear PDEs including the one-dimensional nonlinear wave equation, Whitham’s equation, Burgers’ equation, the Korteweg–de Vries equation, and the nonlinear Schr¨odinger equation are solved. The solutions of these equations are then discussed with physical significance. Special emphasis is given to the fundamental similarities and differences in the properties of the solutions to the corresponding linear and nonlinear equations under consideration. The final chapter is devoted to the major numerical and approximation methods for finding solutions of PDEs. A fairly detailed treatment of explicit and implicit finite difference methods is given with applications The variational method and the Euler–Lagrange equations are described with many applications. Also included are the Rayleigh–Ritz, the Galerkin, and the Kantorovich methods of approximation with many illustrations and applications. This new edition contains almost four hundred examples and exercises, which are either directly associated with applications or phrased in terms of the physical and engineering contexts in which they arise. The exercises truly complement the text, and answers to most exercises are provided at the end of the book. The Appendix has been expanded to include some basic properties of the Gamma function and the tables of Fourier, Laplace, and Hankel transforms. For students wishing to know more about the subject or to have further insight into the subject matter, important references are listed in the Bibliography. The chapters on mathematical models, Fourier series and integrals, and eigenvalue problems are self-contained, so these chapters can be omitted for those students who have prior knowledge of the subject. An attempt has been made to present a clear and concise exposition of the mathematics used in analyzing a variety of problems. With this in mind, the chapters are carefully organized to enable students to view the material in an orderly perspective. For example, the results and theorems in the chapters on Fourier series and integrals and on eigenvalue problems are explicitly mentioned, whenever necessary, to avoid confusion with their use in the development of PDEs. A wide range of problems subject to various boundary conditions has been included to improve the student’s understanding. In this third edition, specific changes and additions include the following: xxii Preface to the Third Edition 1. Chapter 2 on mathematical models has been revised by adding a list of the most common linear PDEs in applied mathematics, mathematical physics, and engineering science. 2. The chapter on the Cauchy problem has been expanded by including the wave equations in spherical and cylindrical polar coordinates. Examples and exercises on these wave equations and the energy equation have been added. 3. Eigenvalue problems have been revised with an emphasis on Green’s functions and applications. A section on the Schr¨odinger equation for the linear harmonic oscillator has been added. Higher-dimensional boundary-value problems with an emphasis on applications, and a section on the hydrogen atom and on the three-dimensional Schr¨odinger equation in a central field of force have been added to Chapter 9. 4. Chapter 11 has been extensively reorganized and revised in order to include Hankel and Mellin transforms and their applications, and has new sections on the asymptotic approximation method and the finite Hankel transform with applications. Many new examples and exercises, some new material with applications, and physical interpretations of mathematical solutions have also been included. 5. A new chapter on nonlinear PDEs of current interest and their applications has been added with considerable emphasis on the fundamental similarities and the distinguishing differences in the properties of the solutions to the nonlinear and corresponding linear equations. 6. Chapter 13 is also new. It contains a fairly detailed treatment of explicit and implicit finite difference methods with their stability analysis. A large section on the variational methods and the Euler–Lagrange equations has been included with many applications. Also included are the Rayleigh–Ritz, the Galerkin, and the Kantorovich methods of approximation with illustrations and applications. 7. Many new applications, examples, and exercises have been added to deepen the reader’s understanding. Expanded versions of the tables of Fourier, Laplace, and Hankel transforms are included. The bibliography has been updated with more recent and important references. As a text on partial differential equations for students in applied mathematics, physics, engineering, and applied sciences, this edition provides the student with the art of combining mathematics with intuitive and physical thinking to develop the most effective approach to solving problems. In preparing this edition, the authors wish to express their sincere thanks to those who have read the manuscript and offered many valuable suggestions and comments. The authors also wish to express their thanks to the editor and the staff of Elsevier–North Holland, Inc. for their kind help and cooperation. Tyn Myint-U Lokenath Debnath 1 Introduction “If you wish to foresee the future of mathematics, our proper course is to study the history and present condition of the science.” Henri Poincar´e “However varied may be the imagination of man, nature is a thousand times richer, ... Each of the theories of physics ... presents (partial differential) equations under a new aspect ... without the theories, we should not know partial differential equations.” Henri Poincar´e 1.1 Brief Historical Comments Historically, partial differential equations originated from the study of surfaces in geometry and a wide variety of problems in mechanics. During the second half of the nineteenth century, a large number of famous mathematicians became actively involved in the investigation of numerous problems presented by partial differential equations. The primary reason for this research was that partial differential equations both express many fundamental laws of nature and frequently arise in the mathematical analysis of diverse problems in science and engineering. The next phase of the development of linear partial differential equations was characterized by efforts to develop the general theory and various methods of solution of linear equations. In fact, partial differential equations have been found to be essential to the theory of surfaces on the one hand and to the solution of physical problems on the other. These two areas of mathematics can be seen as linked by the bridge of the calculus of variations. With the discovery of the basic concepts and properties of distributions, the modern theory of linear partial differential equations is now 2 1 Introduction well established. The subject plays a central role in modern mathematics, especially in physics, geometry, and analysis. Almost all physical phenomena obey mathematical laws that can be formulated by differential equations. This striking fact was first discovered by Isaac Newton (1642–1727) when he formulated the laws of mechanics and applied them to describe the motion of the planets. During the three centuries since Newton’s fundamental discoveries, many partial differential equations that govern physical, chemical, and biological phenomena have been found and successfully solved by numerous methods. These equations include Euler’s equations for the dynamics of rigid bodies and for the motion of an ideal fluid, Lagrange’s equations of motion, Hamilton’s equations of motion in analytical mechanics, Fourier’s equation for the diffusion of heat, Cauchy’s equation of motion and Navier’s equation of motion in elasticity, the Navier–Stokes equations for the motion of viscous fluids, the Cauchy–Riemann equations in complex function theory, the Cauchy–Green equations for the static and dynamic behavior of elastic solids, Kirchhoff’s equations for electrical circuits, Maxwell’s equations for electromagnetic fields, and the Schr¨odinger equation and the Dirac equation in quantum mechanics. This is only a sampling, and the recent mathematical and scientific literature reveals an almost unlimited number of differential equations that have been discovered to model physical, chemical and biological systems and processes. From the very beginning of the study, considerable attention has been given to the geometric approach to the solution of differential equations. The fact that families of curves and surfaces can be defined by a differential equation means that the equation can be studied geometrically in terms of these curves and surfaces. The curves involved, known as characteristic curves, are very useful in determining whether it is or is not possible to find a surface containing a given curve and satisfying a given differential equation. This geometric approach to differential equations was begun by Joseph-Louis Lagrange (1736–1813) and Gaspard Monge (1746–1818). Indeed, Monge first introduced the ideas of characteristic surfaces and characteristic cones (or Monge cones). He also did some work on second-order linear, homogeneous partial differential equations. The study of first-order partial differential equations began to receive some serious attention as early as 1739, when Alex-Claude Clairaut (1713– 1765) encountered these equations in his work on the shape of the earth. On the other hand, in the 1770s Lagrange first initiated a systematic study of the first-order nonlinear partial differential equations in the form f (x, y, u, ux, uy)=0, (1.1.1) where u = u (x, y) is a function of two independent variables. Motivated by research on gravitational effects on bodies of different shapes and mass distributions, another major impetus for work in partial differential equations originated from potential theory. Perhaps the most 1.1 Brief Historical Comments 3 important partial differential equation in applied mathematics is the potential equation, also known as the Laplace equation uxx +uyy = 0, where subscripts denote partial derivatives. This equation arose in steady state heat conduction problems involving homogeneous solids. James Clerk Maxwell (1831–1879) also gave a new initiative to potential theory through his famous equations, known as Maxwell’s equations for electromagnetic fields. Lagrange developed analytical mechanics as the application of partial differential equations to the motion of rigid bodies. He also described the geometrical content of a first-order partial differential equation and developed the method of characteristics for finding the general solution of quasi-linear equations. At the same time, the specific solution of physical interest was obtained by formulating an initial-value problem (or a Cauchy Problem) that satisfies certain supplementary conditions. The solution of an initial-value problem still plays an important role in applied mathematics, science and engineering. The fundamental role of characteristics was soon recognized in the study of quasi-linear and nonlinear partial differential equations. Physically, the first-order, quasi-linear equations often represent conservation laws which describe the conservation of some physical quantities of a system. In its early stages of development, the theory of second-order linear partial differential equations was concentrated on applications to mechanics and physics. All such equations can be classified into three basic categories: the wave equation, the heat equation, and the Laplace equation (or potential equation). Thus, a study of these three different kinds of equations yields much information about more general second-order linear partial differential equations. Jean d’Alembert (1717–1783) first derived the onedimensional wave equation for vibration of an elastic string and solved this equation in 1746. His solution is now known as the d’Alembert solution. The wave equation is one of the oldest equations in mathematical physics. Some form of this equation, or its various generalizations, almost inevitably arises in any mathematical analysis of phenomena involving the propagation of waves in a continuous medium. In fact, the studies of water waves, acoustic waves, elastic waves in solids, and electromagnetic waves are all based on this equation. A technique known as the method of separation of variables is perhaps one of the oldest systematic methods for solving partial differential equations including the wave equation. The wave equation and its methods of solution attracted the attention of many famous mathematicians including Leonhard Euler (1707–1783), James Bernoulli (1667–1748), Daniel Bernoulli (1700–1782), J.L. Lagrange (1736–1813), and Jacques Hadamard (1865–1963). They discovered solutions in several different forms, and the merit of their solutions and relations among these solutions were argued in a series of papers extending over more than twenty-five years; most concerned the nature of the kinds of functions that can be represented by trigonometric (or Fourier) series. These controversial problems were finally resolved during the nineteenth century. 4 1 Introduction It was Joseph Fourier (1768–1830) who made the first major step toward developing a general method of solutions of the equation describing the conduction of heat in a solid body in the early 1800s. Although Fourier is most celebrated for his work on the conduction of heat, the mathematical methods involved, particularly trigonometric series, are important and very useful in many other situations. He created a coherent mathematical method by which the different components of an equation and its solution in series were neatly identified with the different aspects of the physical solution being analyzed. In spite of the striking success of Fourier analysis as one of the most useful mathematical methods, J.L. Lagrange and S.D. Poisson (1781–1840) hardly recognized Fourier’s work because of its lack of rigor. Nonetheless, Fourier was eventually recognized for his pioneering work after publication of his monumental treatise entitled La Th´eorie Auatytique de la Chaleur in 1822. It is generally believed that the concept of an integral transform originated from the Integral Theorem as stated by Fourier in his 1822 treatise. It was the work of Augustin Cauchy (1789–1857) that contained the exponential form of the Fourier Integral Theorem as f (x) = 1 2π
∞ −∞ e ikx
∞ −∞ e −ikξ f (ξ) dξ dk. (1.1.2) This theorem has been expressed in several slightly different forms to better adapt it for particular applications. It has been recognized, almost from the start, however, that the form which best combines mathematical simplicity and complete generality makes use of the exponential oscillating function exp (ikx). Indeed, the Fourier integral formula (1.1.2) is regarded as one of the most fundamental results of modern mathematical analysis, and it has widespread physical and engineering applications. The generality and importance of the theorem is well expressed by Kelvin and Tait who said: “ ... Fourier’s Theorem, which is not only one of the most beautiful results of modern analysis, but may be said to furnish an indispensable instrument in the treatment of nearly every recondite question in modern physics. To mention only sonorous vibrations, the propagation of electric signals along a telegraph wire, and the conduction of heat by the earth’s crust, as subjects in their generality intractable without it, is to give but a feeble idea of its importance.” This integral formula (1.1.2) is usually used to define the classical Fourier transform of a function and the inverse Fourier transform. No doubt, the scientific achievements of Joseph Fourier have not only provided the fundamental basis for the study of heat equation, Fourier series, and Fourier integrals, but for the modern developments of the theory and applications of the partial differential equations. One of the most important of all the partial differential equations involved in applied mathematics and mathematical physics is that associated with the name of Pierre-Simon Laplace (1749–1827). This equation was first discovered by Laplace while he was involved in an extensive study of 1.1 Brief Historical Comments 5 gravitational attraction of arbitrary bodies in space. Although the main field of Laplace’s research was celestial mechanics, he also made important contributions to the theory of probability and its applications. This work introduced the method known later as the Laplace transform, a simple and elegant method of solving differential and integral equations. Laplace first introduced the concept of potential, which is invaluable in a wide range of subjects, such as gravitation, electromagnetism, hydrodynamics, and acoustics. Consequently, the Laplace equation is often referred to as the potential equation. This equation is also an important special case of both the wave equation and the heat equation in two or three dimensions. It arises in the study of many physical phenomena including electrostatic or gravitational potential, the velocity potential for an imcompossible fluid flows, the steady state heat equation, and the equilibrium (time independent) displacement field of a two- or three-dimensional elastic membrane. The Laplace equation also occurs in other branches of applied mathematics and mathematical physics. Since there is no time dependence in any of the mathematical problems stated above, there are no initial data to be satisfied by the solutions of the Laplace equation. They must, however, satisfy certain boundary conditions on the boundary curve or surface of a region in which the Laplace equation is to be solved. The problem of finding a solution of Laplace’s equation that takes on the given boundary values is known as the Dirichlet boundary-value problem, after Peter Gustav Lejeune Dirichlet (1805–1859). On the other hand, if the values of the normal derivative are prescribed on the boundary, the problem is known as Neumann boundary-value problem, in honor of Karl Gottfried Neumann (1832–1925). Despite great efforts by many mathematicians including Gaspard Monge (1746–1818), AdrienMarie Legendre (1752–1833), Carl Friedrich Gauss (1777–1855), SimeonDenis Poisson (1781–1840), and Jean Victor Poncelet (1788–1867), very little was known about the general properties of the solutions of Laplace’s equation until 1828, when George Green (1793–1841) and Mikhail Ostrogradsky (1801–1861) independently investigated properties of a class of solutions known as harmonic functions. On the other hand, Augustin Cauchy (1789–1857) and Bernhard Riemann (1826–1866) derived a set of first-order partial differential equations, known as the Cauchy–Riemann equations, in their independent work on functions of complex variables. These equations led to the Laplace equation, and functions satisfying this equation in a domain are called harmonic functions in that domain. Both Cauchy and Riemann occupy a special place in the history of mathematics. Riemann made enormous contributions to almost all areas of pure and applied mathematics. His extraordinary achievements stimulated further developments, not only in mathematics, but also in mechanics, physics, and the natural sciences as a whole. Augustin Cauchy is universally recognized for his fundamental contributions to complex analysis. He also provided the first systematic and rigorous 6 1 Introduction investigation of differential equations and gave a rigorous proof for the existence of power series solutions of a differential equation in the 1820s. In 1841 Cauchy developed what is known as the method of majorants for proving that a solution of a partial differential equation exists in the form of a power series in the independent variables. The method of majorants was also introduced independently by Karl Weierstrass (1815–1896) in that same year in application to a system of differential equations. Subsequently, Weierstrass’s student Sophie Kowalewskaya (1850–1891) used the method of majorants and a normalization theorem of Carl Gustav Jacobi (1804–1851) to prove an exceedingly elegant theorem, known as the Cauchy–Kowalewskaya theorem. This theorem quite generally asserts the local existence of solutions of a system of partial differential equations with initial conditions on a noncharacteristic surface. This theorem seems to have little practical importance because it does not distinguish between well-posed and ill-posed problems; it covers situations where a small change in the initial data leads to a large change in the solution. Historically, however, it is the first existence theorem for a general class of partial differential equations. The general theory of partial differential equations was initiated by A.R. Forsyth (1858–1942) in the fifth and sixth volumes of his Theory of Differential Equations and by E.J.B. Goursat (1858–1936) in his book entitled Cours d’ analyse mathematiques (1918) and his Lecons sur l’ integration des equations aux d´eriv´ees, volume 1 (1891) and volume 2 (1896). Another notable contribution to this subject was made by E. Cartan’s book, Lecons sur les invariants int´egraux, published in 1922. Joseph Liouville (1809– 1882) formulated a more tractable partial differential equation in the form uxx + uyy = k exp (au), (1.1.3) and obtained a general solution of it. This equation has a large number of applications. It is a special case of the equation derived by J.L. Lagrange for the stream function ψ in the case of two-dimensional steady vortex motion in an incompossible fluid, that is, ψxx + ψyy = F (ψ), (1.1.4) where F (ψ) is an arbitrary function of ψ. When ψ = u and F (u) = keau , equation (1.1.4) reduces to the Liouville equation (1.1.3). In view of the special mathematical interest in the nonhomogeneous nonlinear equation of the type (1.1.4), a number of famous mathematicians including Henri Poincar´e, E. Picard (1856–1941), Cauchy (1789–1857), Sophus Lie (1842– 1899), L.M.H. Navier (1785–1836), and G.G. Stokes (1819–1903) made many major contributions to partial differential equations. Historically, Euler first solved the eigenvalue problem when he developed a simple mathematical model for describing the the ‘buckling’ modes of a vertical elastic beam. The general theory of eigenvalue problems for second-order differential equations, now known as the Sturm–Liouville Theory, originated from the study of a class of boundary-value problems due to 1.1 Brief Historical Comments 7 Charles Sturm (1803–1855) and Joseph Liouville (1809–1882). They showed that, in general, there is an infinite set of eigenvalues satisfying the given equation and the associated boundary conditions, and that these eigenvalues increase to infinity. Corresponding to these eigenvalues, there is an infinite set of orthogonal eigenfunctions so that the linear superposition principle can be applied to find the convergent infinite series solution of the given problem. Indeed, the Sturm–Liouville theory is a natural generalization of the theory of Fourier series that greatly extends the scope of the method of separation of variables. In 1926, the WKB approximation method was developed by Gregor Wentzel, Hendrik Kramers, and MarcelLouis Brillouin for finding the approximate eigenvalues and eigenfunctions of the one-dimensional Schr¨odinger equation in quantum mechanics. This method is now known as the short-wave approximation or the geometrical optics approximation in wave propagation theory. At the end of the seventeenth century, many important questions and problems in geometry and mechanics involved minimizing or maximizing of certain integrals for two reasons. The first of these were several existence problems, such as, Newton’s problem of missile of least resistance, Bernoulli’s isoperimetric problem, Bernoulli’s problem of the brachistochrone (brachistos means shortest, chronos means time), the problem of minimal surfaces due to Joseph Plateau (1801–1883), and Fermat’s principle of least time. Indeed, the variational principle as applied to the propagation and reflection of light in a medium was first enunciated in 1662 by one of the greatest mathematicians of the seventeenth century, Pierre Fermat (1601–1665). According to his principle, a ray of light travels in a homogeneous medium from one point to another along a path in a minimum time. The second reason is somewhat philosophical, that is, how to discover a minimizing principle in nature. The following 1744 statement of Euler is characteristic of the philosophical origin of what is known as the principle of least action: “As the construction of the universe is the most perfect possible, being the handiwork of all-wise Maker, nothing can be met with in the world in which some maximal or minimal property is not displayed. There is, consequently, no doubt but all the effects of the world can be derived by the method of maxima and minima from their final causes as well as from their efficient ones.” In the middle of the eighteenth century, Pierre de Maupertius (1698–1759) stated a fundamental principle, known as the principle of least action, as a guide to the nature of the universe. A still more precise and general formulation of Maupertius’ principle of least action was given by Lagrange in his Analytical Mechanics published in 1788. He formulated it as δS = δ
t2 t1 (2T) dt = 0, (1.1.5) where T is the kinematic energy of a dynamical system with the constraint that the total energy, (T + V ), is constant along the trajectories, and V is 8 1 Introduction the potential energy of the system. He also derived the celebrated equation of motion for a holonomic dynamical system d dt  ∂T ∂q˙i  − ∂T ∂qi = Qi , (1.1.6) where qi are the generalized coordinates, ˙qi is the velocity, and Qi is the force. For a conservative dynamical system, Qi = − ∂V ∂qi , V = V (qi), ∂V ∂q˙i = 0, then (1.1.6) can be expressed in terms of the Lagrangian, L = T − V , as d dt  ∂L ∂q˙i  − ∂L ∂qi = 0. (1.1.7) This principle was then reformulated by Euler in a way that made it useful in mathematics and physics. The work of Lagrange remained unchanged for about half a century until William R. Hamilton (1805–1865) published his research on the general method in analytical dynamics which gave a new and very appealing form to the Lagrange equations. Hamilton’s work also included his own variational principle. In his work on optics during 1834–1835, Hamilton elaborated a new principle of mechanics, known as Hamilton’s principle, describing the stationary action for a conservative dynamical system in the form δA = δ
t1 t0 (T − V ) dt = δ
t1 t0 L dt = 0. (1.1.8) Hamilton’s principle (1.1.8) readily led to the Lagrange equation (1.1.6). In terms of time t, the generalized coordinates qi , and the generalized momenta pi = (∂L/q˙i) which characterize the state of a dynamical system, Hamilton introduced the function H (qi , pi , t) = piq˙i − L(qi , pi , t), (1.1.9) and then used it to represent the equation of motion (1.1.6) as a system of first order partial differential equations q˙i = ∂H ∂pi , p˙i = − ∂H ∂q˙i . (1.1.10) These equations are known as the celebrated Hamilton canonical equations of motion, and the function H (qi , pi , t) is referred to as the Hamiltonian which is equal to the total energy of the system. Following the work of Hamilton, Karl Jacobi, Mikhail Ostrogradsky (1801–1862), and Henri Poincar´e (1854–1912) put forth new modifications of the variational principle. Indeed, the action integral S can be regarded as a function of generalized coordinates and time provided the terminal point is not fixed. In 1842, Jacobi showed that S satisfies the first-order partial differential equation 1.1 Brief Historical Comments 9 ∂S ∂t + H  qi , ∂S ∂qi , t = 0, (1.1.11) which is known as the Hamilton–Jacobi equation. In 1892, Poincar´e defined the action integral on the trajectories in phase space of the variable qi and pi as S =
t1 t0 [piq˙i − H (pi , qi)] dt, (1.1.12) and then formulated another modification of the Hamilton variational principle which also yields the Hamilton canonical equations (1.1.10). From (1.1.12) also follows the celebrated Poincar´e–Cartan invariant I =  C (piδqi − Hδt), (1.1.13) where C is an arbitrary closed contour in the phase space. Indeed, the discovery of the calculus of variations in a modern sense began with the independent work of Euler and Lagrange. The first necessary condition for the existence of an extremum of a functional in a domain leads to the celebrated Euler–Lagrange equation. This equation in its various forms now assumes primary importance, and more emphasis is given to the first variation, mainly due to its power to produce significant equations, than to the second variation, which is of fundamental importance in answering the question of whether or not an extremal actually provides a minimum (or a maximum). Thus, the fundamental concepts of the calculus of variations were developed in the eighteenth century in order to obtain the differential equations of applied mathematics and mathematical physics. During its early development, the problems of the calculus of variations were reduced to questions of the existence of differential equations problems until David Hilbert developed a new method in which the existence of a minimizing function was established directly as the limit of a sequence of approximations. Considerable attention has been given to the problem of finding a necessary and sufficient condition for the existence of a function which extremized the given functional. Although the problem of finding a sufficient condition is a difficult one, Legendre and C.G.J. Jacobi (1804–1851) discovered a second necessary condition and a third necessary condition respectively. Finally, it was Weierstrass who first provided a satisfactory foundation to the theory of calculus of variations in his lectures at Berlin between 1856 and 1870. His lectures were essentially concerned with a complete review of the work of Legendre and Jacobi. At the same time, he reexamined the concepts of the first and second variations and looked for a sufficient condition associated with the problem. In contrast to the work of his predecessors, Weierstrass introduced the ideas of ‘strong variations’ and ‘the excess function’ which led him to discover a fourth necessary condition 10 1 Introduction and a satisfactory sufficient condition. Some of his outstanding discoveries announced in his lectures were published in his collected work. At the conclusion of his famous lecture on ‘Mathematical Problems’ at the Paris International Congress of Mathematicians in 1900, David Hilbert (1862–1943), perhaps the most brilliant mathematician of the late nineteenth century, gave a new method for the discussion of the minimum value of a functional. He obtained another derivation of Weierstrass’s excess function and a new approach to Jacobi’s problem of determining necessary and sufficient conditions for the existence of a minimum of a functional; all this without the use of the second variation. Finally, the calculus of variations entered the new and wider field of ‘global’ problems with the original work of George D. Birkhoff (1884–1944) and his associates. They succeeded in liberating the theory of calculus of variations from the limitations imposed by the restriction to ‘small variations’, and gave a general treatment of the global theory of the subject with large variations. In 1880, George Fitzgerald (1851–1901) probably first employed the variational principle in electromagnetic theory to derive Maxwell’s equations for an electromagnetic field in a vacuum. Moreover, the variational principle received considerable attention in electromagnetic theory after the work of Karl Schwarzchild in 1903 as well as the work of Max Born (1882–1970) who formulated the principle of stationary action in electrodynamics in a symmetric four-dimensional form. On the other hand, Poincar´e showed in 1905 that the action integral is invariant under the Lorentz transformations. With the development of the special theory of relativity and the relativistic theory of gravitation in the beginning of the twentieth century, the variational principles received tremendous attention from many great mathematicians and physicists including Albert Einstein (1879–1955), Hendrix Lorentz (1853–1928), Hermann Weyl (1885–1955), Felix Klein (1849– 1925), Amalie Noether (1882–1935), and David Hilbert. Even before the use of variational principles in electrodynamics, Lord Rayleigh (1842–1919) employed variational methods in his famous book, The Theory of Sound, for the derivation of equations for oscillations in plates and rods in order to calculate frequencies of natural oscillations of elastic systems. In his pioneering work in the 1960’s, Gerald Whitham first developed a general approach to linear and nonlinear dispersive waves using a Lagrangian. He successfully formulated the averaged variational principle, which is now known as the Whitham averaged variational principle, which was employed to derive the basic equations for linear and nonlinear dispersive wave propagation problems. In 1967, Luke first explicitly formulated a variational principle for nonlinear water waves. In 1968, Bretherton and Garret generalized the Whitham averaged variational principle to describe the conservation law for the wave action in a moving medium. Subsequently, Ostrovsky and Pelinovsky (1972) also generalized the Whitham averaged variational principle to nonconservative systems. 1.1 Brief Historical Comments 11 With the rapid development of the theory and applications of differential equations, the closed form analytical solutions of many different types of equations were hardly possible. However, it is extremely important and absolutely necessary to provide some insight into the qualitative and quantitative nature of solutions subject to initial and boundary conditions. This insight usually takes the form of numerical and graphical representatives of the solutions. It was E. Picard (1856–1941) who first developed the method of successive approximations for the solutions of differential equations in most general form and later made it an essential part of his treatment of differential equations in the second volume of his Trait´e d’Analyse published in 1896. During the last two centuries, the calculus of finite differences in various forms played a significant role in finding the numerical solutions of differential equations. Historically, many well known integration formulas and numerical methods including the Euler–Maclaurin formula, Gregory integration formula, the Gregory–Newton formula, Simpson’s rule, Adam– Bashforth’s method, the Jacobi iteration, the Gauss–Seidel method, and the Runge–Kutta method have been developed and then generalized in various forms. With the development of modern calculators and high-speed electronic computers, there has been an increasing trend in research toward the numerical solution of ordinary and partial differential equations during the twentieth century. Special attention has also given to in depth studies of convergence, stability, error analysis, and accuracy of numerical solutions. Many well-known numerical methods including the Crank–Nicolson methods, the Lax–Wendroff method, Richtmyer’s method, and Stone’s implicit iterative technique have been developed in the second half of the twentieth century. All finite difference methods reduce differential equations to discrete forms. In recent years, more modern and powerful computational methods such as the finite element method and the boundary element method have been developed in order to handle curved or irregularly shaped domains. These methods are distinguished by their more general character, which makes them more capable of dealing with complex geometries, allows them to use non-structured grid systems, and allows more natural imposition of the boundary conditions. During the second half of the nineteenth century, considerable attention was given to problems concerning the existence, uniqueness, and stability of solutions of partial differential equations. These studies involved not only the Laplace equation, but the wave and diffusion equations as well, and were eventually extended to partial differential equations with variable coefficients. Through the years, tremendous progress has been made on the general theory of ordinary and partial differential equations. With the advent of new ideas and methods, new results and applications, both analytical and numerical studies are continually being added to this subject. Partial differential equations have been the subject of vigorous mathematical research for over three centuries and remain so today. This is an active 12 1 Introduction area of research for mathematicians and scientists. In part, this is motivated by the large number of problems in partial differential equations that mathematicians, scientists, and engineers are faced with that are seemingly intractable. Many of these equations are nonlinear and come from such areas of applications as fluid mechanics, plasma physics, nonlinear optics, solid mechanics, biomathematics, and quantum field theory. Owing to the ever increasing need in mathematics, science, and engineering to solve more and more complicated real world problems, it seems quite likely that partial differential equations will remain a major area of research for many years to come. 1.2 Basic Concepts and Definitions A differential equation that contains, in addition to the dependent variable and the independent variables, one or more partial derivatives of the dependent variable is called a partial differential equation. In general, it may be written in the form f (x, y, . . . , u, ux, uy,...,uxx, uxy,...)=0, (1.2.1) involving several independent variables x, y, ..., an unknown function u of these variables, and the partial derivatives ux, uy, ..., uxx, uxy, ..., of the function. Subscripts on dependent variables denote differentiations, e.g., ux = ∂u/∂x, uxy = ∂ 2 /∂y ∂x. Here equation (1.2.1) is considered in a suitable domain D of the ndimensional space Rn in the independent variables x, y, .... We seek functions u = u (x, y, . . .) which satisfy equation (1.2.1) identically in D. Such functions, if they exist, are called solutions of equation (1.2.1). From these many possible solutions we attempt to select a particular one by introducing suitable additional conditions. For instance, uuxy + ux = y, uxx + 2yuxy + 3xuyy = 4 sin x, (1.2.2) (ux) 2 + (uy) 2 = 1, uxx − uyy = 0, are partial differential equations. The functions u (x, y)=(x + y) 3 , u (x, y) = sin (x − y), are solutions of the last equation of (1.2.2), as can easily be verified. 1.2 Basic Concepts and Definitions 13 The order of a partial differential equation is the order of the highestordered partial derivative appearing in the equation. For example uxx + 2xuxy + uyy = e y is a second-order partial differential equation, and uxxy + xuyy + 8u = 7y is a third-order partial differential equation. A partial differential equation is said to be linear if it is linear in the unknown function and all its derivatives with coefficients depending only on the independent variables; it is said to be quasi-linear if it is linear in the highest-ordered derivative of the unknown function. For example, the equation yuxx + 2xyuyy + u = 1 is a second-order linear partial differential equation, whereas uxuxx + xuuy = sin y is a second-order quasi-linear partial differential equation. The equation which is not linear is called a nonlinear equation. We shall be primarily concerned with linear second-order partial differential equations, which frequently arise in problems of mathematical physics. The most general second-order linear partial differential equation in n independent variables has the form n i,j=1 Aijuxixj + n i=1 Biuxi + F u = G, (1.2.3) where we assume without loss of generality that Aij = Aji. We also assume that Bi , F, and G are functions of the n independent variables xi . If G is identically zero, the equation is said to be homogeneous; otherwise it is nonhomogeneous. The general solution of a linear ordinary differential equation of nth order is a family of functions depending on n independent arbitrary constants. In the case of partial differential equations, the general solution depends on arbitrary functions rather than on arbitrary constants. To illustrate this, consider the equation uxy = 0. If we integrate this equation with respect to y, we obtain ux (x, y) = f (x). 14 1 Introduction A second integration with respect to x yields u (x, y) = g (x) + h (y), where g (x) and h (y) are arbitrary functions. Suppose u is a function of three variables, x, y, and z. Then, for the equation uyy = 2, one finds the general solution u (x, y, z) = y 2 + yf (x, z) + g (x, z), where f and g are arbitrary functions of two variables x and z. We recall that in the case of ordinary differential equations, the first task is to find the general solution, and then a particular solution is determined by finding the values of arbitrary constants from the prescribed conditions. But, for partial differential equations, selecting a particular solution satisfying the supplementary conditions from the general solution of a partial differential equation may be as difficult as, or even more difficult than, the problem of finding the general solution itself. This is so because the general solution of a partial differential equation involves arbitrary functions; the specialization of such a solution to the particular form which satisfies supplementary conditions requires the determination of these arbitrary functions, rather than merely the determination of constants. For linear homogeneous ordinary differential equations of order n, a linear combination of n linearly independent solutions is a solution. Unfortunately, this is not true, in general, in the case of partial differential equations. This is due to the fact that the solution space of every homogeneous linear partial differential equation is infinite dimensional. For example, the partial differential equation ux − uy = 0 (1.2.4) can be transformed into the equation 2uη = 0 by the transformation of variables ξ = x + y, η = x − y. The general solution is u (x, y) = f (x + y), where f (x + y) is an arbitrary function. Thus, we see that each of the functions 1.3 Mathematical Problems 15 (x + y) n , sin n (x + y), cos n (x + y), exp n (x + y), n = 1, 2, 3,... is a solution of equation (1.2.4). The fact that a simple equation such as (1.2.4) yields infinitely many solutions is an indication of an added difficulty which must be overcome in the study of partial differential equations. Thus, we generally prefer to directly determine the particular solution of a partial differential equation satisfying prescribed supplementary conditions. 1.3 Mathematical Problems A problem consists of finding an unknown function of a partial differential equation satisfying appropriate supplementary conditions. These conditions may be initial conditions (I.C.) and/or boundary conditions (B.C.). For example, the partial differential equation (PDE) ut − uxx = 0, 0 <x<l, t=""> 0, with I.C. u (x, 0) = sin x, 0 ≤ x ≤ l, t > 0, B.C. u (0, t) = 0, t ≥ 0, B.C. u (l, t) = 0, t ≥ 0, constitutes a problem which consists of a partial differential equation and three supplementary conditions. The equation describes the heat conduction in a rod of length l. The last two conditions are called the boundary conditions which describe the function at two prescribed boundary points. The first condition is known as the initial condition which prescribes the unknown function u (x, t) throughout the given region at some initial time t, in this case t = 0. This problem is known as the initial boundary-value problem. Mathematically speaking, the time and the space coordinates are regarded as independent variables. In this respect, the initial condition is merely a point prescribed on the t-axis and the boundary conditions are prescribed, in this case, as two points on the x-axis. Initial conditions are usually prescribed at a certain time t = t0 or t = 0, but it is not customary to consider the other end point of a given time interval. In many cases, in addition to prescribing the unknown function, other conditions such as their derivatives are specified on the boundary and/or at time t0. In considering the problem of unbounded domain, the solution can be determined uniquely by prescribing initial conditions only. The corresponding problem is called the initial-value problem or the Cauchy problem. The mathematical definition is given in Chapter 5. The solution of such a prob- 16 1 Introduction lem may be interpreted physically as the solution unaffected by the boundary conditions at infinity. For problems affected by the boundary at infinity, boundedness conditions on the behavior of solutions at infinity must be prescribed. A mathematical problem is said to be well-posed if it satisfies the following requirements: 1. Existence: There is at least one solution. 2. Uniqueness: There is at most one solution. 3. Continuity: The solution depends continuously on the data. The first requirement is an obvious logical condition, but we must keep in mind that we cannot simply state that the mathematical problem has a solution just because the physical problem has a solution. We may well be erroneously developing a mathematical model, say, consisting of a partial differential equation whose solution may not exist at all. The same can be said about the uniqueness requirement. In order to really reflect the physical problem that has a unique solution, the mathematical problem must have a unique solution. For physical problems, it is not sufficient to know that the problem has a unique solution. Hence the last requirement is not only useful but also essential. If the solution is to have physical significance, a small change in the initial data must produce a small change in the solution. The data in a physical problem are normally obtained from experiment, and are approximated in order to solve the problem by numerical or approximate methods. It is essential to know that the process of making an approximation to the data produces only a small change in the solution. 1.4 Linear Operators An operator is a mathematical rule which, when applied to a function, produces another function. For example, in the expressions L[u] = ∂ 2u ∂x2 + ∂ 2u ∂y2 , M [u] = ∂ 2u ∂x2 − ∂u ∂x + x ∂u ∂y , L =  ∂ 2/∂x2 + ∂ 2/∂y2 and M =  ∂ 2/∂x2 − ∂/∂x + x (∂/∂y) are called the differential operators. An operator is said to be linear if it satisfies the following: 1. A constant c may be taken outside the operator: L[cu] = cL[u] . (1.4.1) 1.4 Linear Operators 17 2. The operator operating on the sum of two functions gives the sum of the operator operating on the individual functions: L[u1 + u2] = L[u1] + L[u2] . (1.4.2) We may combine (1.4.1) and (1.4.2) as L[c1u1 + c2u2] = c1L[u1] + c2L[u2] , (1.4.3) where c1 and c2 are any constants. This can be extended to a finite number of functions. If u1, u2, ..., uk are k functions and c1, c2, ..., ck are k constants, then by repeated application of equation (1.4.3) L ⎡ ⎣  k j=1 cjuj ⎤ ⎦ =  k j=1 cjL[uj ] . (1.4.4) We may now define the sum of two linear differential operators formally. If L and M are two linear operators, then the sum of L and M is defined as (L + M) [u] = L[u] + M [u] , (1.4.5) where u is a sufficiently differentiable function. It can be readily shown that L + M is also a linear operator. The product of two linear differential operators L and M is the operator which produces the same result as is obtained by the successive operations of the operators L and M on u, that is, LM [u] = L(M [u]), (1.4.6) in which we assume that M [u] and L(M [u]) are defined. It can be readily shown that LM is also a linear operator. In general, linear differential operators satisfy the following: 1. L + M = M + L (commutative) (1.4.7) 2. (L + M) + N = L + (M + N) (associative) (1.4.8) 3. (LM) N = L(MN) (associative) (1.4.9) 4. L(c1M + c2N) = c1LM + c2LN (distributive). (1.4.10) For linear differential operators with constant coefficients, 5. LM = ML (commutative). (1.4.11) Example 1.4.1. Let L = ∂ 2 ∂x2 + x ∂ ∂y and M = ∂ 2 ∂y2 − y ∂ ∂y . 18 1 Introduction LM [u] =  ∂ 2 ∂x2 + x ∂ ∂y ∂ 2u ∂y2 − y ∂u ∂y  = ∂ 4u ∂x2∂y2 − y ∂ 3u ∂x2∂y + x ∂ 3u ∂y3 − xy ∂ 2u ∂y2 , ML[u] =  ∂ 2 ∂y2 − y ∂ ∂y ∂ 2u ∂x2 + x ∂u ∂y  = ∂ 4u ∂y2∂x2 + x ∂ 3u ∂y3 − y ∂ 3u ∂y∂x2 − xy ∂ 2u ∂y2 , which shows that LM = ML. Now let us consider a linear second-order partial differential equation. In the case of two independent variables, such an equation takes the form A (x, y) uxx + B (x, y) uxy + C (x, y) uyy +D (x, y) ux + E (x, y) uy + F (x, y) u = G (x, y), (1.4.12) where A, B, C, D, E, and F are the coefficients, and G is the nonhomogeneous term. If we denote L = A ∂ 2 ∂x2 + B ∂ 2 ∂x∂y + C ∂ 2 ∂y2 + D ∂ ∂x + E ∂ ∂y + F, then equation (1.4.12) may be written in the form L[u] = G. (1.4.13) Very often the square bracket is omitted and we simply write Lu = G. Let v1, v2, ..., vn be n functions which satisfy L[vj ] = Gj , j = 1, 2,...,n and let w1, w2, ..., wn be n functions which satisfy L[wj ]=0, j = 1, 2, . . . , n. If we let uj = vj + wj then, the function u = n j=1 uj 1.4 Linear Operators 19 satisfies the equation L[u] = n j=1 Gj . This is called the principle of linear superposition. In particular, if v is a particular solution of equation (1.4.13), that is, L[v] = G, and w is a solution of the associated homogeneous equation, that is, L[w] = 0, then u = v + w is a solution of L[u] = G. The principle of linear superposition is of fundamental importance in the study of partial differential equations. This principle is used extensively in solving linear partial differential equations by the method of separation of variables. Suppose that there are infinitely many solutions u1 (x, y), u2 (x, y), ... un (x, y), ... of a linear homogeneous partial differential equation Lu = 0. Can we say that every infinite linear combination c1u1 +c2u2 +···+cnun + ··· of these solutions, where c1, c2, ..., cn, ... are any constants, is again a solution of the equation? Of course, by an infinite linear combination, we mean an infinite series and we must require that the infinite series ∞ k=0 ck uk = limn→∞ n k=0 ck uk (1.4.14) must be convergent to u. In general, we state that the infinite series is a solution of the homogeneous equation. There is another kind of infinite linear combination which is also used to find the solution of a given linear equation. This is concerned with a family of solutions u (x, y; k) of the linear equation, where k is any real number, not just the values 1, 2, 3,.... If ck = c (k) is any function of the real parameter k such that
b a c (k) u (x, y; k) dk or
∞ −∞ c (k) u (x, y; k) dk (1.4.15) is convergent, then, under suitable conditions, the integral (1.4.15), again, is a solution. This may be called the linear integral superposition principle. To illustrate these ideas, we consider the equation Lu = ux + 2uy = 0. (1.4.16) It is easy to verify that, for every real k, the function u (x, y; k) = e k(2x−y) (1.4.17) is a solution of (1.4.16). Multiplying (1.4.17) by e −k and integrating with respect to k over −1 ≤ k ≤ 1 gives 20 1 Introduction u (x, y) =
1 −1 e −k e k(2x−y) dk = e 2x−y−1 2x − y − 1 (1.4.18) It is easy to verify that u (x, y) given by (1.4.18) is also a solution of (1.4.16). It is also easy to verify that u (x, y; k) = e −ky cos (k x), k ∈ R is a one-parameter family of solutions of the Laplace equation ∇2u ≡ uxx + uyy = 0 (1.4.19) It is also easy to check that v (x, y; k) = ∂ ∂k u (x, y; k) (1.4.20) is also a one-parameter family of solutions of (1.4.19), k ∈ R. Further, for any (x, y) in the upper half-plane y > 0, the integral v (x, y) ≡
∞ 0 u (x, y, k) dk =
∞ 0 e −ky cos (k x) dk, (1.4.21) is convergent, and v (x, y) is a solution of (1.4.19) for x ∈ R and y > 0. This follows from direct computation of vxx and vyy. The solution (1.4.21) is another example of the linear integral superposition principle. 1.5 Superposition Principle We may express supplementary conditions using the operator notation. For instance, the initial boundary-value problem utt − c 2uxx = G (x, t) 0 < x < l, t > 0, u (x, 0) = g1 (x) 0 ≤ x ≤ l, ut (x, 0) = g2 (x) 0 ≤ x ≤ l, (1.5.1) u (0, t) = g3 (t) t ≥ 0, u (l, t) = g4 (t) t ≥ 0, may be written in the form L[u] = G, M1 [u] = g1, M2 [u] = g2, (1.5.2) M3 [u] = g3, M4 [u] = g4, where gi are the prescribed functions and the subscripts on operators are assigned arbitrarily. 1.5 Superposition Principle 21 Now let us consider the problem L[u] = G, M1 [u] = g1, M2 [u] = g2, (1.5.3) . . . Mn [u] = gn. By virtue of the linearity of the equation and the supplementary conditions, we may divide problem (1.5.3) into a series of problems as follows: L[u1] = G, M1 [u1]=0, M2 [u1]=0, (1.5.4) . . . Mn [u1]=0, L[u2]=0, M1 [u2] = g1, M2 [u2]=0, (1.5.5) . . . Mn [u2]=0, L[un]=0, M1 [un]=0, M2 [un]=0, (1.5.6) . . . Mn [un] = gn. Then the solution of problem (1.5.3) is given by u = n i=1 ui . (1.5.7) Let us consider one of the subproblems, say, (1.5.5). Suppose we find a sequence of functions φ1,φ2, ..., which may be finite or infinite, satisfying the homogeneous system 22 1 Introduction L[φi ]=0, M2 [φi ]=0, (1.5.8) . . . Mn [φi ]=0, i = 1, 2, 3,... and suppose we can express g1 in terms of the series g1 = c1M1 [φ1] + c2M1 [φ2] + .... (1.5.9) Then the linear combination u2 = c1φ1 + c2φ2 + ..., (1.5.10) is the solution of problem (1.5.5). In the case of an infinite number of terms in the linear combination (1.5.10), we require that the infinite series be uniformly convergent and sufficiently differentiable, and that all the series Nk (ui) where N0 = L, Nj = Mj for j = 1, 2, ..., n convergence uniformly. 1.6 Exercises 1. For each of the following, state whether the partial differential equation is linear, quasi-linear or nonlinear. If it is linear, state whether it is homogeneous or nonhomogeneous, and gives its order. (a) uxx + xuy = y, (b) uux − 2xyuy = 0, (c) u 2 x + uuy = 1, (d) uxxxx + 2uxxyy + uyyyy = 0, (e) uxx + 2uxy + uyy = sin x, (f) uxxx + uxyy + log u = 0, (g) u 2 xx + u 2 x + sin u = e y , (h) ut + uux + uxxx = 0. 2. Verify that the functions u (x, y) = x 2 − y 2 u (x, y) = e x sin y u (x, y)=2xy are the solutions of the equation uxx + uyy = 0. 3. Show that u = f (xy), where f is an arbitrary differentiable function satisfies 1.6 Exercises 23 xux − yuy = 0 and verify that the functions sin (xy), cos (xy), log (xy), e xy, and (xy) 3 are solutions. 4. Show that u = f (x) g (y) where f and g are arbitrary twice differentiable functions satisfies uuxy − uxuy = 0. 5. Determine the general solution of the differential equation uyy + u = 0. 6. Find the general solution of uxx + ux = 0, by setting ux = v. 7. Find the general solution of uxx − 4uxy + 3uyy = 0, by assuming the solution to be in the form u (x, y) = f (λx + y), where λ is an unknown parameter. 8. Find the general solution of uxx − uyy = 0. 9. Show that the general solution of ∂ 2u ∂t2 − c 2 ∂ 2u ∂x2 = 0, is u (x, t) = f (x − ct) + g (x + ct), where f and g are arbitrary twice differentiable functions. 10. Verify that the function u = φ (xy) + x ψ
4y x
5 , is the general solution of the equation x 2uxx − y 2uyy = 0. 11. If ux = vy and vx = −uy, show that both u and v satisfy the Laplace equations ∇2u = 0 and ∇2 v = 0. 24 1 Introduction 12. If u (x, y) is a homogeneous function of degree n, show that u satisfies the first-order equation xux + yuy = nu. 13. Verify that u (x, y, t) = A cos (kx) cos (ly) cos (nct) + B sin (kx) sin (ly) sin (nct), where k 2 + l 2 = n 2 , is a solution of the equation utt = c 2 (uxx + uyy). 14. Show that u (x, y; k) = e −ky sin (kx), x ∈ R, y> 0, is a solution of the equation ∇2u ≡ uxx + uyy = 0 for any real parameter k. Verify that u (x, y) =
∞ 0 c (k) e −ky sin (kx) dk is also a solution of the above equation. 15. Show, by differentiation that, u (x, t) = 1 √ 4πkt exp  − x 2 4kt , x ∈ R, t> 0, is a solution of the diffusion equation ut = k uxx, where k is a constant. 16. (a) Verify that u (x, y) = log
4 x 2 + y 2
5 , satisfies the equation uxx + uyy = 0 for all (x, y) = (0, 0). (b) Show that 1.6 Exercises 25 u (x, y, z) =  x 2 + y 2 + z 2 − 1 2 is a solution of the Laplace equation uxx + uyy + uzz = 0 except at the origin. (c) Show that u (r) = a rn satisfies the equation r 2u ′′ + 2ru′ − n (n + 1) u = 0. 17. Show that un (r, θ) = r n cos (nθ) and un (r, θ) = r n sin (nθ), n = 0, 1, 2, 3, ··· are solutions of the Laplace equation ∇2u ≡ urr + 1 r ur + 1 r 2 uθθ = 0. 18. Verify by differentiation that u (x, y) = cos x cosh y satisfies the Laplace equation uxx + uyy = 0. 19. Show that u (x, y) = f  2y + x 2 + g  2y − x 2 is a general solution of the equation uxx − 1 x ux − x 2uyy = 0. 20. If u satisfies the Laplace equation ∇2u ≡ uxx+uyy = 0, show that both xu and yu satisfy the biharmonic equation ∇4 ⎛ ⎝ xu yu ⎞ ⎠ = 0, but xu and yu will not satisfy the Laplace equation. 21. Show that u (x, y, t) = f (x + iky − iωt) + g (x − iky − iωt) is a general solution of the wave equation utt = c 2 (uxx + uyy), where f and g are arbitrary twice differentiable functions, and ω 2 = c 2  k 2 − 1 , k, ω, c are constants. 26 1 Introduction 22. Verify that u (x, y) = x 3 + y 2 + e x (cos x sin y cosh y − sin x cos y sinh y) is a classical solution of the Poisson equation uxx + uyy = (6x + 2). 23. Show that u (x, y) = exp
4 − x b
5 f (ax − by) satisfies the equation b ux + a uy + u = 0. 24. Show that utt − c 2uxx + 2b ut = 0 has solutions of the form u (x, t)=(A cos kx + B sin kx) V (t), where c, b, A and B are constants. 25. Show that c 2  urr + 1 r ur  − utt = 0 has solutions of the form u (r, t) = V (r) r cos (nct), n = 0, 1, 2,.... Find a differential equation for V (r). 2 First-Order, Quasi-Linear Equations and Method of Characteristics “As long as a branch of knowledge offers an abundance of problems, it is full of vitality.” David Hilbert “Since a general solution must be judged impossible from want of analysis, we must be content with the knowledge of some special cases, and that all the more, since the development of various cases seems to be the only way to bringing us at last to a more perfect knowledge.” Leonhard Euler 2.1 Introduction Many problems in mathematical, physical, and engineering sciences deal with the formulation and the solution of first-order partial differential equations. From a mathematical point of view, first-order equations have the advantage of providing a conceptual basis that can be utilized for second-, third-, and higher-order equations. This chapter is concerned with first-order, quasi-linear and linear partial differential equations and their solution by using the Lagrange method of characteristics and its generalizations. 2.2 Classification of First-Order Equations The most general, first-order, partial differential equation in two independent variables x and y is of the form 28 2 First-Order, Quasi-Linear Equations and Method of Characteristics F (x, y, u, ux, uy)=0, (x, y) ∈ D ⊂ R 2 , (2.2.1) where F is a given function of its arguments, and u = u (x, y) is an unknown function of the independent variables x and y which lie in some given domain D in R2 , ux = ∂u ∂x and uy = ∂u ∂y . Equation (2.2.1) is often written in terms of standard notation p = ux and q = uy so that (2.2.1) takes the form F (x, y, u, p, q)=0. (2.2.2) Similarly, the most general, first-order, partial differential equation in three independent variables x, y, z can be written as F (x, y, z, u, ux, uy, uz)=0. (2.2.3) Equation (2.2.1) or (2.2.2) is called a quasi-linear partial differential equation if it is linear in first-partial derivatives of the unknown function u (x, y). So, the most general quasi-linear equation must be of the form a (x, y, u) ux + b (x, y, u) uy = c (x, y, u), (2.2.4) where its coefficients a, b, and c are functions of x, y, and u. The following are examples of quasi-linear equations: x  y 2 + u ux − y  x 2 + u uy =  x 2 − y 2 u, (2.2.5) uux + ut + nu2 = 0, (2.2.6)  y 2 − u 2 ux − xy uy = xu. (2.2.7) Equation (2.2.4) is called a semilinear partial differential equation if its coefficients a and b are independent of u, and hence, the semilinear equation can be expressed in the form a (x, y) ux + b (x, y) uy = c (x, y, u). (2.2.8) Examples of semilinear equations are xux + yuy = u 2 + x 2 , (2.2.9) (x + 1)2 ux + (y − 1)2 uy = (x + y) u 2 , (2.2.10) ut + aux + u 2 = 0, (2.2.11) where a is a constant. Equation (2.2.1) is said to be linear if F is linear in each of the variables u, ux, and uy, and the coefficients of these variables are functions only of the independent variables x and y. The most general, first-order, linear partial differential equation has the form a (x, y) ux + b (x, y) uy + c (x, y) u = d (x, y), (2.2.12) 2.3 Construction of a First-Order Equation 29 where the coefficients a, b, and c, in general, are functions of x and y and d (x, y) is a given function. Unless stated otherwise, these functions are assumed to be continuously differentiable. Equations of the form (2.2.12) are called homogeneous if d (x, y) ≡ 0 or nonhomogeneous if d (x, y) = 0. Obviously, linear equations are a special kind of the quasi-linear equation (2.2.4) if a, b are independent of u and c is a linear function in u. Similarly, semilinear equation (2.2.8) reduces to a linear equation if c is linear in u. Examples of linear equations are xux + yuy − nu = 0, (2.2.13) nux + (x + y) uy − u = e x , (2.2.14) yux + xuy = xy, (2.2.15) (y − z) ux + (z − x) uy + (x − y) uz = 0. (2.2.16) An equation which is not linear is often called a nonlinear equation. So, first-order equations are often classified as linear and nonlinear. 2.3 Construction of a First-Order Equation We consider a system of geometrical surfaces described by the equation f (x, y, z, a, b)=0, (2.3.1) where a and b are arbitrary parameters. We differentiate (2.3.1) with respect to x and y to obtain fx + p fz = 0, fy + q fz = 0, (2.3.2) where p = ∂z ∂x and q = ∂z ∂y . The set of three equations (2.3.1) and (2.3.2) involves two arbitrary parameters a and b. In general, these two parameters can be eliminated from this set to obtain a first-order equation of the form F (x, y, z, p, q)=0. (2.3.3) Thus the system of surfaces (2.3.1) gives rise to a first-order partial differential equation (2.3.3). In other words, an equation of the form (2.3.1) containing two arbitrary parameters is called a complete solution or a complete integral of equation (2.3.3). Its role is somewhat similar to that of a general solution for the case of an ordinary differential equation. On the other hand, any relationship of the form f (φ, ψ)=0, (2.3.4) 30 2 First-Order, Quasi-Linear Equations and Method of Characteristics which involves an arbitrary function f of two known functions φ = φ (x, y, z) and ψ = ψ (x, y, z) and provides a solution of a first-order partial differential equation is called a general solution or general integral of this equation. Clearly, the general solution of a first-order partial differential equation depends on an arbitrary function. This is in striking contrast to the situation for ordinary differential equations where the general solution of a firstorder ordinary differential equation depends on one arbitrary constant. The general solution of a partial differential equation can be obtained from its complete integral. We obtain the general solution of (2.3.3) from its complete integral (2.3.1) as follows. First, we prescribe the second parameter b as an arbitrary function of the first parameter a in the complete solution (2.3.1) of (2.3.3), that is, b = b (a). We then consider the envelope of the one-parameter family of solutions so defined. This envelope is represented by the two simultaneous equations f (x, y, z, a, b (a)) = 0, (2.3.5) fa (x, y, z, a, b (a)) + fb (x, y, z, b (a)) b ′ (a)=0, (2.3.6) where the second equation (2.3.6) is obtained from the first equation (2.3.5) by partial differentiation with respect to a. In principle, equation (2.3.5) can be solved for a = a (x, y, z) as a function of x, y, and z. We substitute this result back in (2.3.5) to obtain f {x, y, z, a (x, y, z), b (a (x, y, z))} = 0, (2.3.7) where b is an arbitrary function. Indeed, the two equations (2.3.5) and (2.3.6) together define the general solution of (2.3.3). When a definite b (a) is prescribed, we obtain a particular solution from the general solution. Since the general solution depends on an arbitrary function, there are infinitely many solutions. In practice, only one solution satisfying prescribed conditions is required for a physical problem. Such a solution may be called a particular solution. In addition to the general and particular solutions of (2.3.3), if the envelope of the two-parameter system (2.3.1) of surfaces exists, it also represents a solution of the given equation (2.3.3); the envelope is called the singular solution of equation (2.3.3). The singular solution can easily be constructed from the complete solution (2.3.1) representing a two-parameter family of surfaces. The envelope of this family is given by the system of three equations f (x, y, z, a, b)=0, fa (x, y, z, a, b)=0, fb (x, y, z, a, b)=0. (2.3.8) In general, it is possible to eliminate a and b from (2.3.8) to obtain the equation of the envelope which gives the singular solution. It may be pointed out that the singular solution cannot be obtained from the general 2.3 Construction of a First-Order Equation 31 solution. Its nature is similar to that of the singular solution of a first-order ordinary differential equation. Finally, it is important to note that solutions of a partial differential equation are expected to be represented by smooth functions. A function is called smooth if all of its derivatives exist and are continuous. However, in general, solutions are not always smooth. A solution which is not everywhere differentiable is called a weak solution. The most common weak solution is the one that has discontinuities in its first partial derivatives across a curve, so that the solution can be represented by shock waves as surfaces of discontinuity. In the case of a first-order partial differential equation, there are discontinuous solutions where z itself and not merely p = ∂z ∂x and q = ∂z ∂y are discontinuous. In fact, this kind of discontinuity is usually known as a shock wave. An important feature of quasi-linear and nonlinear partial differential equations is that their solutions may develop discontinuities as they move away from the initial state. We close this section by considering some examples. Example 2.3.1. Show that a family of spheres x 2 + y 2 + (z − c) 2 = r 2 , (2.3.9) satisfies the first-order linear partial differential equation yp − xq = 0. (2.3.10) Differentiating the equation (2.3.9) with respect to x and y gives x + p (z − c) = 0 and y + q (z − c)=0. Eliminating the arbitrary constant c from these equations, we obtain the first-order, partial differential equation yp − xq = 0. Example 2.3.2. Show that the family of spheres (x − a) 2 + (y − b) 2 + z 2 = r 2 (2.3.11) satisfies the first-order, nonlinear, partial differential equation z 2  p 2 + q 2 + 1 = r 2 . (2.3.12) We differentiate the equation of the family of spheres with respect to x and y to obtain (x − a) + z p = 0, (y − b) + z q = 0. Eliminating the two arbitrary constants a and b, we find the nonlinear partial differential equation 32 2 First-Order, Quasi-Linear Equations and Method of Characteristics z 2  p 2 + q 2 + 1 = r 2 . All surfaces of revolution with the z-axis as the axis of symmetry satisfy the equation z = f  x 2 + y 2 , (2.3.13) where f is an arbitrary function. Writing u = x 2 + y 2 and differentiating (2.3.13) with respect to x and y, respectively, we obtain p = 2x f′ (u), q = 2y f′ (u). Eliminating the arbitrary function f (u) from these results, we find the equation yp − xq = 0. Theorem 2.3.1. If φ = φ (x, y, z) and ψ = ψ (x, y, z) are two given functions of x, y, and z and if f (φ, ψ) = 0, where f is an arbitrary function of φ and ψ, then z = z (x, y) satisfies a first-order, partial differential equation p ∂ (φ, ψ) ∂ (y, z) + q ∂ (φ, ψ) ∂ (z, x) = ∂ (φ, ψ) ∂ (x, y) , (2.3.14) where ∂ (φ, ψ) ∂ (x, y) =     φx φy ψx ψy     . (2.3.15) Proof. We differentiate f (φ, ψ) = 0 with respect to x and y respectively to obtain the following equations: ∂f ∂φ  ∂φ ∂x + p ∂φ ∂z  + ∂f ∂ψ  ∂ψ ∂x + p ∂ψ ∂z  = 0, (2.3.16) ∂f ∂φ  ∂φ ∂y + q ∂φ ∂z  + ∂f ∂ψ  ∂ψ ∂y + q ∂ψ ∂z  = 0. (2.3.17) Nontrivial solutions for ∂f ∂φ and ∂f ∂ψ can be found if the determinant of the coefficients of these equations vanishes, that is,       φx + pφz ψx + pψz φy + qφz ψy + qψz       = 0. (2.3.18) Expanding this determinant gives the first-order, quasi-linear equation (2.3.14). 2.4 Geometrical Interpretation of a First-Order Equation 33 2.4 Geometrical Interpretation of a First-Order Equation To investigate the geometrical content of a first-order, partial differential equation, we begin with a general, quasi-linear equation a (x, y, u) ux + b (x, y, u) uy − c (x, y, u)=0. (2.4.1) We assume that the possible solution of (2.4.1) in the form u = u (x, y) or in an implicit form f (x, y, u) ≡ u (x, y) − u = 0 (2.4.2) represents a possible solution surface in (x, y, u) space. This is often called an integral surface of the equation (2.4.1). At any point (x, y, u) on the solution surface, the gradient vector ∇f = (fx, fy, fu)=(ux, uy, −1) is normal to the solution surface. Clearly, equation (2.4.1) can be written as the dot product of two vectors a ux + b uy − c = (a, b, c) · (ux, uy − 1) = 0. (2.4.3) This clearly shows that the vector (a, b, c) must be a tangent vector of the integral surface (2.4.2) at the point (x, y, u), and hence, it determines a direction field called the the characteristic direction or Monge axis. This direction is of fundamental importance in determining a solution of equation (2.4.1). To summarize, we have shown that f (x, y, u) = u (x, y) − u = 0, as a surface in the (x, y, u)-space, is a solution of (2.4.1) if and only if the direction vector field (a, b, c) lies in the tangent plane of the integral surface f (x, y, u) = 0 at each point (x, y, u), where ∇f = 0, as shown in Figure 2.4.1. A curve in (x, y, u)-space, whose tangent at every point coincides with the characteristic direction field (a, b, c), is called a characteristic curve. If the parametric equations of this characteristic curve are x = x (t), y = y (t), u = u (t), (2.4.4) then the tangent vector to this curve is
4 dx dt , dy dt , du dt
5 which must be equal to (a, b, c). Therefore, the system of ordinary differential equations of the characteristic curve is given by dx dt = a (x, y, u), dy dt = b (x, y, u), du dt = c (x, y, u). (2.4.5) These are called the characteristic equations of the quasi-linear equation (2.4.1). 34 2 First-Order, Quasi-Linear Equations and Method of Characteristics Figure 2.4.1 Tangent and normal vector fields of solution surface at a point (x, y, u). In fact, there are only two independent ordinary differential equations in the system (2.4.5); therefore, its solutions consist of a two-parameter family of curves in (x, y, u)-space. The projection on u = 0 of a characteristic curve on the (x, t)-plane is called a characteristic base curve or simply characteristic. Equivalently, the characteristic equations (2.4.5) in the nonparametric form are dx a = dy b = du c . (2.4.6) The typical problem of solving equation (2.4.1) with a prescribed u on a given plane curve C is equivalent to finding an integral surface in (x, y, u) space, satisfying the equation (2.4.1) and containing the three-dimensional space curve Γ defined by the values of u on C, which is the projection on u = 0 of Γ. Remark 1. The above geometrical interpretation can be generalized for higher-order partial differential equations. However, it is not easy to visualize geometrical arguments that have been described for the case of three space dimensions. Remark 2. The geometrical interpretation is more complicated for the case of nonlinear partial differential equations, because the normals to possible 2.5 Method of Characteristics and General Solutions 35 solution surfaces through a point do not lie in a plane. The tangent planes no longer intersect along one straight line, but instead, they envelope along a curved surface known as the Monge cone. Any further discussion is beyond the scope of this book. We conclude this section by adding an important observation regarding the nature of the characteristics in the (x, t)-plane. For a quasi-linear equation, characteristics are determined by the first two equations in (2.4.5) with their slopes dy dx = b (x, y, u) a (x, y, u) . (2.4.7) If (2.4.1) is a linear equation, then a and b are independent of u, and the characteristics of (2.4.1) are plane curves with slopes dy dx = b (x, y) a (x, y) . (2.4.8) By integrating this equation, we can determine the characteristics which represent a one-parameter family of curves in the (x, t)-plane. However, if a and b are constant, the characteristics of equation (2.4.1) are straight lines. 2.5 Method of Characteristics and General Solutions We can use the geometrical interpretation of first-order, partial differential equations and the properties of characteristic curves to develop a method for finding the general solution of quasi-linear equations. This is usually referred to as the method of characteristics due to Lagrange. This method of solution of quasi-linear equations can be described by the following result. Theorem 2.5.1. The general solution of a first-order, quasi-linear partial differential equation a (x, y, u) ux + b (x, y, u) uy = c (x, y, u) (2.5.1) is f (φ, ψ)=0, (2.5.2) where f is an arbitrary function of φ (x, y, u) and ψ (x, y, u), and φ = constant = c1 and ψ = constant = c2 are solution curves of the characteristic equations dx a = dy b = du c . (2.5.3) The solution curves defined by φ (x, y, u) = c1 and ψ (x, y, u) = c2 are called the families of characteristic curves of equation (2.5.1). 36 2 First-Order, Quasi-Linear Equations and Method of Characteristics Proof. Since φ (x, y, u) = c1 and ψ (x, y, u) = c2 satisfy equations (2.5.3), these equations must be compatible with the equation dφ = φxdx + φydy + φudu = 0. (2.5.4) This is equivalent to the equation a φx + b φy + c φu = 0. (2.5.5) Similarly, equation (2.5.3) is also compatible with a ψx + b ψy + c ψu = 0. (2.5.6) We now solve (2.5.5), (2.5.6) for a, b, and c to obtain a ∂(φ,ψ) ∂(y,u) = b ∂(φ,ψ) ∂(u,x) = c ∂(φ,ψ) ∂(x,y) . (2.5.7) It has been shown earlier that f (φ, ψ) = 0 satisfies an equation similar to (2.3.14), that is, p ∂ (φ, ψ) ∂ (y, u) + q ∂ (φ, ψ) ∂ (u, x) = ∂ (φ, ψ) ∂ (x, y) . (2.5.8) Substituting, (2.5.7) in (2.5.8), we find that f (φ, ψ) = 0 is a solution of (2.5.1). This completes the proof. Note that an analytical method has been used to prove Theorem 2.5.1. Alternatively, a geometrical argument can be used to prove this theorem. The geometrical method of proof is left to the reader as an exercise. Many problems in applied mathematics, science, and engineering involve partial differential equations. We seldom try to find or discuss the properties of a solution to these equations in its most general form. In most cases of interest, we deal with those solutions of partial differential equations which satisfy certain supplementary conditions. In the case of a first-order partial differential equation, we determine the specific solution by formulating an initial-value problem or a Cauchy problem. Theorem 2.5.2. (The Cauchy Problem for a First-Order Partial Differential Equation). Suppose that C is a given curve in the (x, y)-plane with its parametric equations x = x0 (t), y = y0 (t), (2.5.9) where t belongs to an interval I ⊂ R, and the derivatives x ′ 0 (t) and y ′ 0 (t) are piecewise continuous functions, such that (x ′ 0 ) 2 + (y ′ 0 ) 2 = 0. Also, suppose that u = u0 (t) is a given function on the curve C. Then, there exists a solution u = u (x, y) of the equation 2.5 Method of Characteristics and General Solutions 37 F (x, y, u, ux, uy) = 0 (2.5.10) in a domain D of R2 containing the curve C for all t ∈ I, and the solution u (x, y) satisfies the given initial data, that is, u (x0 (t), y0 (t)) = u0 (t) (2.5.11) for all values of t ∈ I. In short, the Cauchy problem is to determine a solution of equation (2.5.10) in a neighborhood of C, such that the solution u = u (x, y) takes a prescribed value u0 (t) on C. The curve C is called the initial curve of the problem, and u0 (t) is called the initial data. Equation (2.5.11) is called the initial condition of the problem. The solution of the Cauchy problem also deals with such questions as the conditions on the functions F, x0 (t), y0 (t), and u0 (t) under which a solution exists and is unique. We next discuss a method for solving a Cauchy problem for the firstorder, quasi-linear equation (2.5.1). We first observe that geometrically x = x0 (t), y = y0 (t), and u = u0 (t) represent an initial curve Γ in (x, y, u)-space. The curve C, on which the Cauchy data is prescribed, is the projection of Γ on the (x, y)-plane. We now present a precise formulation of the Cauchy problem for the first-order, quasi-linear equation (2.5.1). Theorem 2.5.3. (The Cauchy Problem for a Quasi-linear Equation). Suppose that x0 (t), y0 (t), and u0 (t) are continuously differentiable functions of t in a closed interval, 0 ≤ t ≤ 1, and that a, b, and c are functions of x, y, and u with continuous first-order partial derivatives with respect to their arguments in some domain D of (x, y, u)-space containing the initial curve Γ : x = x0 (t), y = y0 (t), u = u0 (t), (2.5.12) where 0 ≤ t ≤ 1, and satisfying the condition y ′ 0 (t) a (x0 (t), y0 (t), u0 (t)) − x ′ 0 (t) b (x0 (t), y0 (t), u0 (t)) = 0. (2.5.13) Then there exists a unique solution u = u (x, y) of the quasi-linear equation (2.5.1) in the neighborhood of C : x = x0 (t), y = y0 (t), and the solution satisfies the initial condition u0 (t) = u (x0 (t), y0 (t)), for 0 ≤ t ≤ 1. (2.5.14) Note: The condition (2.5.13) excludes the possibility that C could be a characteristic. Example 2.5.1. Find the general solution of the first-order linear partial differential equation. 38 2 First-Order, Quasi-Linear Equations and Method of Characteristics x ux + y uy = u. (2.5.15) The characteristic curves of this equation are the solutions of the characteristic equations dx x = dy y = du u . (2.5.16) This system of equations gives the integral surfaces φ = y x = C1 and ψ = u x = C2, where C1 and C2 are arbitrary constants. Thus, the general solution of (2.5.15) is f
4y x , u x
5 = 0, (2.5.17) where f is an arbitrary function. This general solution can also be written as u (x, y) = x g
4y x
5 , (2.5.18) where g is an arbitrary function. Example 2.5.2. Obtain the general solution of the linear Euler equation x ux + y uy = nu. (2.5.19) The integral surfaces are the solutions of the characteristic equations dx x = dy y = du nu . (2.5.20) From these equations, we get y x = C1, u x n = C2, where C1 and C2 are arbitrary constants. Hence, the general solution of (2.5.19) is f
4y x , u x n
5 = 0. (2.5.21) This can also be written as u x n = g
4y x
5 or u (x, y) = x n g
4y x
5 . (2.5.22) This shows that the solution u (x, y) is a homogeneous function of x and y of degree n. 2.5 Method of Characteristics and General Solutions 39 Example 2.5.3. Find the general solution of the linear equation x 2 ux + y 2 uy = (x + y) u. (2.5.23) The characteristic equations associated with (2.5.23) are dx x 2 = dy y 2 = du (x + y) u . (2.5.24) From the first two of these equations, we find x −1 − y −1 = C1, (2.5.25) where C1 is an arbitrary constant. It follows from (2.5.24) that dx − dy x 2 − y 2 = du (x + y) u or d (x − y) x − y = du u . This gives x − y u = C2, (2.5.26) where C2 is a constant. Furthermore, (2.5.25) and (2.5.26) also give xy u = C3, (2.5.27) where C3 is a constant. Thus, the general solution (2.5.23) is given by f  xy u , x − y u  = 0, (2.5.28) where f is an arbitrary function. This general solution representing the integral surface can also be written as u (x, y) = xy g  x − y u  , (2.5.29) where g is an arbitrary function, or, equivalently, u (x, y) = xy h  x − y xy  , (2.5.30) where h is an arbitrary function. 40 2 First-Order, Quasi-Linear Equations and Method of Characteristics Example 2.5.4. Show that the general solution of the linear equation (y − z) ux + (z − x) uy + (x − y) uz = 0 (2.5.31) is u (x, y, z) = f  x + y + z, x2 + y 2 + z 2 , (2.5.32) where f is an arbitrary function. The characteristic curves satisfy the characteristic equations dx y − z = dy z − x = dz x − y = du 0 (2.5.33) or du = 0, dx + dy + dz = 0, xdx + ydy + zdz = 0. Integration of these equations gives u = C1, x + y + z = C2, and x 2 + y 2 + z 2 = C3, where C1, C2 and C3 are arbitrary constants. Thus, the general solution can be written in terms of an arbitrary function f in the form u (x, y, z) = f  x + y + z, x2 + y 2 + z 2 . We next verify that this is a general solution by introducing three independent variables ξ, η, ζ defined in terms of x, y, and z as ξ = x + y + z, η = x 2 + y 2 + z 2 , and ζ = y + z, (2.5.34) where ζ is an arbitrary combination of y and z. Clearly the general solution becomes u = f (ξ,η), and hence, uζ = ux ∂x ∂ζ + uy ∂y ∂ζ + uz ∂z ∂ζ . (2.5.35) It follows from (2.5.34) that 0 = ∂x ∂ζ + ∂y ∂ζ + ∂z ∂ζ , 0=2  x ∂x ∂ζ + y ∂y ∂ζ + z ∂z ∂ζ  , ∂y ∂ζ + ∂z ∂ζ = 1. It follows from the first and the third results that ∂x ∂ζ = −1 and, therefore, x = y ∂y ∂ζ + z ∂z ∂ζ , y = y ∂y ∂ζ + y ∂z ∂ζ , z = z ∂y ∂ζ + z ∂z ∂ζ . 2.5 Method of Characteristics and General Solutions 41 Clearly, it follows by subtracting that x − y = (z − y) ∂z ∂ζ , x − z = (y − z) ∂y ∂ζ . Using the values for ∂x ∂ζ , ∂z ∂ζ , and ∂y ∂ζ in (2.5.35), we obtain (z − y) ∂u ∂ζ = (y − z) ∂u ∂x + (z − x) ∂u ∂y + (x − y) ∂u ∂z . (2.5.36) If u = u (ξ,η) satisfies (2.5.31), then ∂u ∂ζ = 0 and, hence, (2.5.36) reduces to (2.5.31). This shows that the general solution (2.5.32) satisfies equation (2.5.31). Example 2.5.5. Find the solution of the equation u (x + y) ux + u (x − y) uy = x 2 + y 2 , (2.5.37) with the Cauchy data u = 0 on y = 2x. The characteristic equations are dx u (x + y) = dy u (x − y) = du x 2 + y 2 = ydx + xdy − udu 0 = xdx − ydy − udu 0 . Consequently, d xy − 1 2 u 2  = 0 and d 1 2  x 2 − y 2 − u 2 = 0. (2.5.38) These give two integrals u 2 − x 2 + y 2 = C1 and 2xy − u 2 = C2, (2.5.39) where C1 and C2 are constants. Hence, the general solution is f  x 2 − y 2 − u 2 , 2xy − u 2 = 0, where f is an arbitrary function. Using the Cauchy data in (2.5.39), we obtain 4C1 = 3C2. Therefore 4  u 2 − x 2 + y 2 = 3  2xy − u 2 . Thus, the solution of equation (2.5.37) is given by 7u 2 = 6xy + 4  x 2 − y 2 . (2.5.40) Example 2.5.6. Obtain the solution of the linear equation ux − uy = 1, (2.5.41) 42 2 First-Order, Quasi-Linear Equations and Method of Characteristics with the Cauchy data u (x, 0) = x 2 . The characteristic equations are dx 1 = dy −1 = du 1 . (2.5.42) Obviously, dy dx = −1 and du dx = 1. Clearly, x + y = constant = C1 and u − x = constant = C2. Thus, the general solution is given by u − x = f (x + y), (2.5.43) where f is an arbitrary function. We now use the Cauchy data to find f (x) = x 2 − x, and hence, the solution is u (x, y)=(x + y) 2 − y. (2.5.44) The characteristics x + y = C1 are drawn in Figure 2.5.1. The value of u must be given at one point on each characteristic which intersects the line y = 0 only at one point, as shown in Figure 2.5.1. Figure 2.5.1 Characteristics of equation (2.5.41). 2.5 Method of Characteristics and General Solutions 43 Example 2.5.7. Obtain the solution of the equation (y − u) ux + (u − x) uy = x − y, (2.5.45) with the condition u = 0 on xy = 1. The characteristic equations for equation (2.5.45) are dx y − u = dy u − x = du x − y . (2.5.46) The parametric forms of these equations are dx dt = y − u, dy dt = u − x, du dt = x − y. These lead to the following equations: x˙ + ˙y + ˙u = 0 and xx˙ + yy˙ + uu˙ = 0, (2.5.47) where the dot denotes the derivative with respect to t. Integrating (2.5.47), we obtain x + y + u = const. = C1 and x 2 + y 2 + u 2 = const. = C2. (2.5.48) These equations represent circles. Using the Cauchy data, we find that C 2 1 = (x + y) 2 = x 2 + y 2 + 2xy = C2 + 2. Thus, the integral surface is described by (x + y + u) 2 = x 2 + y 2 + u 2 + 2. Hence, the solution is given by u (x, y) = 1 − xy x + y . (2.5.49) Example 2.5.8. Solve the linear equation y ux + x uy = u, (2.5.50) with the Cauchy data u (x, 0) = x 3 and u (0, y) = y 3 . (2.5.51) The characteristic equations are dx y = dy x = du u 44 2 First-Order, Quasi-Linear Equations and Method of Characteristics or du u = dx − dy y − x = dx + dy y + x . Solving these equations, we obtain u = C1 x − y = C2 (x + y) or u = C2 (x + y), x2 − y 2 = C1 C2 = constant = C. So the characteristics are rectangular hyperbolas for C > 0 or C < 0. Thus, the general solution is given by f  u x + y , x2 − y 2  = 0 or, equivalently, u (x, y)=(x + y) g  x 2 − y 2 . (2.5.52) Using the Cauchy data, we find that g  x 2 = x 2 , that is, g (x) = x. Consequently, the solution becomes u (x, y)=(x + y)  x 2 − y 2 on x 2 − y 2 = C > 0. Similarly, u (x, y)=(x + y)  y 2 − x 2 on y 2 − x 2 = C > 0. It follows from these results that u → 0 in all regions, as x → ± y (or y → ± x), and hence, u is continuous across y = ± x which represent asymptotes of the rectangular hyperbolas x 2 − y 2 = C. However, ux and uy are not continuous, as y → ± x. For x 2 − y 2 = C > 0, ux = 3x 2 + 2xy − y 2 = (x + y) (3x − y) → 0, as y → −x. uy = −3y 2 − 2xy + x 2 = (x + y) (x − 3y) → 0, as y → −x. Hence, both ux and uy are continuous as y → −x. On the other hand, ux → 4x 2 , uy → −4x 2 as y → x. This implies that ux and uy are discontinuous across y = x. Combining all these results, we conclude that u (x, y) is continuous everywhere in the (x, t)-plane, and ux, uy are continuous everywhere in the (x, t)-plane except on the line y = x. Hence, the partial derivatives ux, uy are discontinuous on y = x. Thus, the development of discontinuities across characteristics is a significant feature of the solutions of partial differential equations. 2.5 Method of Characteristics and General Solutions 45 Example 2.5.9. Determine the integral surfaces of the equation x  y 2 + u ux − y  x 2 + u uy =  x 2 − y 2 u, (2.5.53) with the data x + y = 0, u = 1. The characteristic equations are dx x (y 2 + u) = dy −y (x 2 + u) = du (x 2 − y 2) u (2.5.54) or dx x (y 2 + u) = dy y − (x 2 + u) = du u (x 2 − y 2) = dx x + dy y + du u 0 . Consequently, log (xyu) = log C1 or xyu = C1. From (2.5.54), we obtain xdx x 2 (y 2 + u) = ydy −y 2 (x 2 + u) = du (x 2 − y 2) u = xdx + ydy − du 0 , whence we find that x 2 + y 2 − 2u = C2. Using the given data, we obtain C1 = −x 2 and C2 = 2x 2 − 2, so that C2 = −2 (C1 + 1). Thus the integral surface is given by x 2 + y 2 − 2u = −2 − 2xyu or 2xyu + x 2 + y 2 − 2u +2=0. (2.5.55) 46 2 First-Order, Quasi-Linear Equations and Method of Characteristics Example 2.5.10. Obtain the solution of the equation x ux + y uy = x exp (−u) (2.5.56) with the data u = 0 on y = x 2 . The characteristic equations are dx x = dy y = du x exp (−u) (2.5.57) or y x = C1. We also obtain from (2.5.57) that dx = e udu which can be integrated to find e u = x + C2. Thus, the general solution is given by f
4 e u − x, y x
5 = 0 or, equivalently, e u = x + g
4y x
5 . (2.5.58) Applying the Cauchy data, we obtain g (x)=1 − x. Thus, the solution of (2.5.56) is given by e u = x + 1 − y x or u = log
4 x + 1 − y x
5 . (2.5.59) Example 2.5.11. Solve the initial-value problem ut + u ux = x, u (x, 0) = f (x), (2.5.60) where (a) f (x) = 1 and (b) f (x) = x. The characteristic equations are dt 1 = dx u = du x = d (x + u) x + u . (2.5.61) 2.5 Method of Characteristics and General Solutions 47 Integration gives t = log (x + u) − log C1 or (u + x) e −t = C1. Similarly, we get u 2 − x 2 = C2. For case (a), we obtain 1 + x = C1 and 1 − x 2 = C2, and hence C2 = 2C1 − C 2 1 . Thus,  u 2 − x 2 = 2(u + x) e −t − (u + x) 2 e −2t or u − x = 2e −t − (u + x) e −2t . A simple manipulation gives the solution u (x, t) = x tanh t + sech t. (2.5.62) Case (b) is left to the reader as an exercise. Example 2.5.12. Find the integral surface of the equation u ux + uy = 1, (2.5.63) so that the surface passes through an initial curve represented parametrically by x = x0 (s), y = y0 (s), u = u0 (s), (2.5.64) where s is a parameter. The characteristic equations for the given equations are dx u = dy 1 = du 1 , which are, in the parametric form, dx dτ = u, dy dτ = 1, du dτ = 1, (2.5.65) 48 2 First-Order, Quasi-Linear Equations and Method of Characteristics where τ is a parameter. Thus the solutions of this parametric system in general depend on two parameters s and τ . We solve this system (2.5.65) with the initial data x (s, 0) = x0 (s), y (s, 0) = y0 (s), u (s, 0) = u0 (s). The solutions of (2.5.65) with the given initial data are x (s, τ ) = τ 2 2 + τ u0 (s) + x0 (s) y (s, τ ) = τ + y0 (s) u (s, τ ) = τ + u0 (s) ⎫ ⎬ ⎭ . (2.5.66) We choose a particular set of values for the initial data as x (s, 0) = 2s 2 , y (s, 0) = 2s, u (s, 0) = 0, s> 0. Therefore, the solutions are given by x = 1 2 τ 2 + 2s 2 , y = τ + 2s, u = τ. (2.5.67) Eliminating τ and s from (2.5.67) gives the integral surface (u − y) 2 + u 2 = 2x or 2u = y ±  4x − y 2 1 2 . (2.5.68) The solution surface satisfying the data u = 0 on y 2 = 2x is given by 2u = y −  4x − y 2 1 2 . (2.5.69) This represents the solution surface only when y 2 < 4x. Thus, the solution does not exist for y 2 > 4x and is not differentiable when y 2 = 4x. We verify that y 2 = 4x represents the envelope of the family of characteristics in the (x, t)-plane given by the τ -eliminant of the first two equations in (2.5.67), that is, F (x, y, s)=2x − (y − 2s) 2 − 4s 2 = 0. (2.5.70) This represents a family of parabolas for different values of the parameter s. Thus, the envelope is obtained by eliminating s from equations ∂F ∂s = 0 and F = 0. This gives y 2 = 4x, which is the envelope of the characteristics for different s, as shown in Figure 2.5.2. 2.6 Canonical Forms of First-Order Linear Equations 49 Figure 2.5.2 Dotted curve is the envelope of the characteristics. 2.6 Canonical Forms of First-Order Linear Equations It is often convenient to transform the more general first-order linear partial differential equation (2.2.12) a (x, y) ux + b (x, y) uy + c (x, y) u = d (x, y), (2.6.1) into a canonical (or standard) form which can be easily integrated to find the general solution of (2.6.1). We use the characteristics of this equation (2.6.1) to introduce the new transformation by equations ξ = ξ (x, y), η = η (x, y), (2.6.2) where ξ and η are once continuously differentiable and their Jacobian J (x, y) ≡ ξxηy − ξyηx is nonzero in a domain of interest so that x and y can be determined uniquely from the system of equations (2.6.2). Thus, by chain rule, ux = uξξx + uηηx, uy = uξξy + uηηy, (2.6.3) we substitute these partial derivatives (2.6.3) into (2.6.1) to obtain the equation A uξ + B uη + cu = d, (2.6.4) where A = uξx + bξy, B = aηx + bηy. (2.6.5) From (2.6.5) we see that B = 0 if η is a solution of the first-order equation aηx + bηy = 0. (2.6.6) This equation has infinitely many solutions. We can obtain one of them by assigning initial condition on a non-characteristic initial curve and solving the resulting initial-value problem according to the method described 50 2 First-Order, Quasi-Linear Equations and Method of Characteristics earlier. Since η (x, y) satisfies equation (2.6.6), the level curves η (x, y) = constant are always characteristic curves of equation (2.6.1). Thus, one set of the new transformations are the characteristic curves of (2.6.1). The second set, ξ (x, y) = constant, can be chosen to be any one parameter family of smooth curves which are nowhere tangent to the family of the characteristic curves. We next assert that A = 0 in a neighborhood of some point in the domain D in which η (x, y) is defined and J = 0. For, if A = 0 at some point of D, then B = 0 at the same point. Consequently, equations (2.6.5) would form a system of linear homogeneous equations in a and b, where the Jacobian J is the determinant of its coefficient matrix. Since J = 0, both a and b must be zero at that point which contradicts the original assumption that a and b do not vanish simultaneously. Finally, since B = 0 and A = 0 in D, we can divide (2.6.4) by A to obtain the canonical form uξ + α (ξ,η) u = β (ξ,η), (2.6.7) where α (ξ,η) = c A and β (ξ,η) = d A . Equation (2.6.7) represents an ordinary differential equation with ξ as the independent variable and η as a parameter which may be treated as constant. This equation (2.6.7) is called the canonical form of equation (2.6.1) in terms of the coordinates (ξ,η). Generally, the canonical equation (2.6.7) can easily be integrated and the general solution of (2.6.1) can be obtained after replacing ξ and η by the original variables x and y. We close this section by considering some examples that illustrate this procedure. In practice, it is convenient to choose ξ = ξ (x, y) and η (x, y) = y or ξ = x and η = η (x, y) so that J = 0. Example 2.6.1. Reduce each of the following equations ux − uy = u, (2.6.8) yux + uy = x, (2.6.9) to canonical form, and obtain the general solution. In (2.6.8), a = 1, b = −1, c = −1 and d = 0. The characteristic equations are dx 1 = dy −1 = du u . The characteristic curves are ξ = x + y = c1, and we choose η = y = c2 where c1 and c2 are constants. Consequently, ux = uξ and uy = uξ + uη, and hence, equation (2.6.8) becomes uη = u. Integrating this equation gives 2.7 Method of Separation of Variables 51 ln u (ξ,η) = −η + ln f (ξ), where f (ξ) is an arbitrary function of ξ only. Equivalently, u (ξ,η) = f (ξ) e −η . In terms of the original variables x and y, the general solution of equation (2.6.8) is u (x, y) = f (x + y) e −y , (2.6.10) where f is an arbitrary function. The characteristic equations of (2.6.9) are dx y = dy 1 = du x . It follows from the first two equations that ξ (x, y) = x− y 2 2 = c1; we choose η (x, y) = y = c2. Consequently, ux = uξ and uy = −y uξ + uη and hence, equation (2.6.9) reduces to uη = ξ + 1 2 η 2 . Integrating this equation gives the general solution u (ξ,η) = ξη + 1 6 η 3 + f (ξ), where f is an arbitrary function. Thus, the general solution of (2.6.9) in terms of x and y is u (x, y) = xy − 1 3 y 3 + f  x − y 2 2  . 2.7 Method of Separation of Variables During the last two centuries several methods have been developed for solving partial differential equations. Among these, a technique known as the method of separation of variables is perhaps the oldest systematic method for solving partial differential equations. Its essential feature is to transform the partial differential equations by a set of ordinary differential equations. The required solution of the partial differential equations is then exposed as a product u (x, y) = X (x) Y (y) = 0, or as a sum u (x, y) = X (x) + Y (y), where X (x) and Y (y) are functions of x and y, respectively. Many signifi- cant problems in partial differential equations can be solved by the method 52 2 First-Order, Quasi-Linear Equations and Method of Characteristics of separation of variables. This method has been considerably refined and generalized over the last two centuries and is one of the classical techniques of applied mathematics, mathematical physics and engineering science. Usually, the first-order partial differential equation can be solved by separation of variables without the need for Fourier series. The main purpose of this section is to illustrate the method by examples. Example 2.7.1. Solve the initial-value problem ux + 2uy = 0, u (0, y)=4 e −2y . (2.7.1ab) We seek a separable solution u (x, y) = X (x) Y (y) = 0 and substitute into the equation to obtain X′ (x) Y (y)+2X (x) Y ′ (y)=0. This can also be expressed in the form X′ (x) 2X (x) = − Y ′ (y) Y (y) . (2.7.2) Since the left-hand side of this equation is a function of x only and the right-hand is a function of y only, it follows that (2.7.2) can be true if both sides are equal to the same constant value λ which is called an arbitrary separation constant. Consequently, (2.7.2) gives two ordinary differential equations X′ (x) − 2λX (x)=0, Y ′ (y) + λY (y)=0. (2.7.3) These equations have solutions given, respectively, by X (x) = A e2λx and Y (y) = B e−λy , (2.7.4) where A and B are arbitrary integrating constants. Consequently, the general solution is given by u (x, y) = AB exp (2λx − λy) = C exp (2λx − λy), (2.7.5) where C = AB is an arbitrary constant. Using the condition (2.7.1b), we find 4 e −2y = u (0, y) = Ce−λy , and hence, we deduce that C = 4 and λ = 2. Therefore, the final solution is u (x, y) = 4 exp (4x − 2y). (2.7.6) 2.7 Method of Separation of Variables 53 Example 2.7.2. Solve the equation y 2u 2 x + x 2u 2 y = (xyu) 2 . (2.7.7) We assume u (x, y) = f (x) g (y) = 0 is a separable solution of (2.7.7), and substitute into the equation. Consequently, we obtain y 2 {f ′ (x) g (y)} 2 + x 2 {f (x) g ′ (y)} 2 = x 2 y 2 {f (x) g (y)} 2 , or, equivalently, 1 x 2  f ′ (x) f (x) 02 + 1 y 2  g ′ (y) g (y) 02 = 1, or 1 x 2  f ′ (x) f (x) 02 = 1 − 1 y 2  g ′ (y) g (y) 02 = λ 2 , where λ 2 is a separation constant. Thus, 1 x f ′ (x) f (x) = λ and g ′ (y) y g (y) =  1 − λ2 . (2.7.8) Solving these ordinary differential equations, we find f (x) = A exp  λ 2 x 2  and g (y) = B exp  1 2 y  1 − λ2  , where A and B are arbitrary constant. Thus, the general solution is u (x, y) = C exp  λ 2 x 2 + 1 2 y 2  1 − λ2  , (2.7.9) where C = AB is an arbitrary constant. Using the condition u (x, 0) = 3 exp  x 2/4 , we can determine both C and λ in (2.7.9). It turns out that C = 3 and λ = (1/2), and the solution becomes u (x, y) = 3 exp 1 4
4 x 2 + y 2 √ 3
5 . (2.7.10) Example 2.7.3. Use the separation of variables u (x, y) = f (x) + g (y) to solve the equation u 2 x + u 2 y = 1. (2.7.11) Obviously, 54 2 First-Order, Quasi-Linear Equations and Method of Characteristics {f ′ (x)} 2 = 1 − {g ′ (y)} 2 = λ 2 , where λ 2 is a separation constant. Thus, we obtain f ′ (x) = λ and g ′ (y) =  1 − λ2 . Solving these ordinary differential equations, we find f (x) = λx + A and g (y) = y  1 − λ2 + B, where A and B are constants of integration. Finally, the solution of (2.7.11) is given by u (x, y) = λx + y  1 − λ2 + C, (2.7.12) where C = A + B is an arbitrary constant. Example 2.7.4. Use u (x, y) = f (x) + g (y) to solve the equation u 2 x + uy + x 2 = 0. (2.7.13) Obviously, equation (2.7.13) has the separable form {f ′ (x)} 2 + x 2 = −g ′ (y) = λ 2 , where λ 2 is a separation constant. Consequently, f ′ (x) =  λ2 − x 2 and g ′ (y) = −λ 2 . These can be integrated to obtain f (x) =
 λ2 − x 2 dx + A = λ 2
cos2 θ dθ + A, (x = λ sin θ) = 1 2 λ 2 1 sin−1
4x λ
5 + x λ 2 1 − x 2 λ2 3 + A and g (y) = −λ 2 y + B. Finally, the general solution is given by u (x, y) = 1 2 λ 2 sin−1
4x λ
5 + x 2  λ2 − x 2 − λ 2 y + C, (2.7.14) where C = A + B is an arbitrary constant. 2.8 Exercises 55 Example 2.7.5. Use v = ln u and v = f (x) + g (y) to solve the equation x 2u 2 x + y 2u 2 y = u 2 . (2.7.15) In view of v = ln u, vx = 1 u ux and vy = 1 u uy, and hence, equation (2.7.15) becomes x 2 v 2 x + y 2 v 2 y = 1. (2.7.16) Substitute v (x, y) = f (x) + g (y) into (2.7.16) to obtain x 2 {f ′ (x)} 2 + y 2 {g ′ (y)} 2 = 1 or x 2 {f ′ (x)} 2 = 1 − y 2 {g ′ (y)} 2 = λ 2 , where λ 2 is a separation constant. Thus, we obtain f ′ (x) = λ x and g ′ (y) = 1 y  1 − λ2 . Integrating these equations gives f (x) = λ ln x + A and g (y) =  1 − λ2 ln y + B, where A and B are integrating constants. Therefore, the general solution of (2.7.16) is given by v (x, y) = λ ln x +  1 − λ2 ln y + ln C = ln
4 x λ · y √ 1−λ2 · C
5 , (2.7.17) where ln C = A + B. The final solution is u (x, y) = e v = C xλ · y √ 1−λ2 , (2.7.18) where C is an integrating constant. 2.8 Exercises 1. (a) Show that the family of right circular cones whose axes coincide with the z-axis x 2 + y 2 = (z − c) 2 tan2 α satisfies the first-order, partial differential equation 56 2 First-Order, Quasi-Linear Equations and Method of Characteristics yp − xq = 0. (b) Show that all the surfaces of revolution, z = f  x 2 + y 2 with the z-axis as the axis of symmetry, where f is an arbitrary function, satisfy the partial differential equation yp − xq = 0. (c) Show that the two-parameter family of curves u − ax − by − ab = 0 satisfies the nonlinear equation xp + yq + pq = u. 2. Find the partial differential equation arising from each of the following surfaces: (a) z = x + y + f (xy), (b) z = f (x − y), (c) z = xy + f  x 2 + y 2 , (d) 2z = (αx + y) 2 + β. 3. Find the general solution of each of the following equations: (a) ux = 0, (b) a ux + b uy = 0; a, b, are constant, (c) ux + y uy = 0, (d)  1 + x 2 ux + uy = 0, (e) 2xy ux +  x 2 + y 2 uy = 0, (f) (y + u) ux + y uy = x − y, (g) y 2ux − xy uy = x (u − 2y), (h) yuy − xux = 1, (i) y 2up + u 2xq = −xy2 , (j) (y − xu) p + (x + yu) q = x 2 + y 2 . 4. Find the general solution of the equation ux + 2xy2uy = 0. 5. Find the solution of the following Cauchy problems: (a) 3ux + 2uy = 0, with u (x, 0) = sin x, (b) y ux + x uy = 0, with u (0, y) = exp  −y 2 , (c) x ux + y uy = 2xy, with u = 2 on y = x 2 , (d) ux + x uy = 0, with u (0, y) = sin y, (e) y ux + x uy = xy, x ≥ 0, y ≥ 0, with u (0, y) = exp  −y 2 for y > 0, and u (x, 0) = exp  −x 2 for x > 0, 2.8 Exercises 57 (f) ux + x uy =  y − 1 2 x 2 2 , with u (0, y) = exp (y), (g) x ux + y uy = u + 1, with u (x, y) = x 2 on y = x 2 , (h) u ux − u uy = u 2 + (x + y) 2 , with u = 1 on y = 0, (i) x ux + (x + y) uy = u + 1, with u (x, y) = x 2 on y = 0. (j) √ x ux + u uy + u 2 = 0, u (x, 0) = 1, 0 <x< ∞.="" (k)="" u="" x2ux="" +="" e="" −yuy="" 2="0," (x,="" 0)="1," 0="" <x<="" 6.="" solve="" the="" initial-value="" problem="" ut="" ux="0" with="" initial="" curve="" x="1" τ="" ,="" t="τ,u" =="" τ.="" 7.="" find="" solution="" of="" cauchy="" 2xy="" ="" y="" uy="0," ="" −="" ="" on="" 8.="" following="" equations:="" (a)="" z="" uz="0," (b)="" (x="" y)="" (c)="" (y="" z)="" (z="" x)="" (d)="" yz="" xz="" xy="" (e)="" 9.="" equation="" data="" (0,="" (1,="" 58="" first-order,="" quasi-linear="" equations="" and="" method="" characteristics="" 10.="" show="" that="" u1="e" u2="e" −y="" are="" solutions="" nonlinear="" (ux="" uy)="" but="" their="" sum="" (e="" )="" is="" not="" a="" equation.="" 11.="" u)="" y),="" 12.="" integral="" surfaces="" for="" each="" data:="" (s,="" draw="" in="" case.="" 13.="" containing="" does="" exist.="" 14.="" problems:="" 2ux="" 2uy="0," →="" as="" ∞,="" −x="" <x<y,="" 2x="" 1)=""> 0, u = 2y on x = 1, (e) x ux − 2y uy = x 2 + y 2 for x > 0, y > 0, u = x 2 on y = 1, (f) ux + 2 uy =1+ u, u (x, y) = sin x on y = 3x + 1, (g) ux + 3uy = u, u (x, y) = cos x on y = x, (h) ux + 2x uy = 2xu, u (x, 0) = x 2 , (i) u ux + uy = u, u (x, 0) = 2x, 1 ≤ x ≤ 2, (j) ux + uy = u 2 , u (x, 0) = tanh x. 2.8 Exercises 59 Show that the solution of (j) is unbounded on the critical curve y tanh (x − y) = 1. 15. Find the solution surface of the equation  u 2 − y 2 ux + xy uy + xu = 0, with u = y = x, x > 0. 16. (a) Solve the Cauchy problem ux + uuy = 1, u (0, y) = ay, where a is a constant. (b) Find the solution of the equation in (a) with the data x (s, 0) = 2s, y (s, 0) = s 2 , u  0, s2 = s. 17. Solve the following equations: (a) (y + u) ux + (x + u) uy = x + y, (b) x u  u 2 + xy ux − y u  u 2 + xy uy = x 4 , (c) (x + y) ux + (x − y) uy = 0, (d) y ux + x uy = xy  x 2 − y 2 , (e) (cy − bz) zx + (az − cx) zy = bx − ay. 18. Solve the equation x zx + y zy = z, and find the curves which satisfy the associated characteristic equations and intersect the helix x + y 2 = a 2 , z = b tan−1  y x . 19. Obtain the family of curves which represent the general solution of the partial differential equation (2x − 4y + 3u) ux + (x − 2y − 3u) uy = −3 (x − 2y). Determine the particular member of the family which contains the line u = x and y = 0. 20. Find the solution of the equation y ux − 2xy uy = 2xu with the condition u (0, y) = y 3 . 60 2 First-Order, Quasi-Linear Equations and Method of Characteristics 21. Obtain the general solution of the equation (x + y + 5z) p + 4zq + (x + y + z)=0 (p = zx, q = zy), and find the particular solution which passes through the circle z = 0, x2 + y 2 = a 2 . 22. Obtain the general solution of the equation  z 2 − 2yz − y 2 p + x (y + z) q = x (y − z) (p = zx, q = zy). Find the integral surfaces of this equation passing through (a) the x-axis, (b) the y-axis, and (c) the z-axis. 23. Solve the Cauchy problem (x + y) ux + (x − y) uy = 1, u (1, y) = 1 √ 2 . 24. Solve the following Cauchy problems: (a) 3 ux + 2 uy = 0, u (x, 0) = f (x), (b) a ux + b uy = c u, u (x, 0) = f (x), where a, b, c are constants, (c) x ux + y uy = c u, u (x, 0) = f (x), (d) u ux + uy = 1, u (s, 0) = αs, x (s, 0) = s, y (s, 0) = 0. 25. Apply the method of separation of variables u (x, y) = f (x) g (y) to solve the following equations: (a) ux + u = uy, u (x, 0) = 4e −3x , (b) uxuy = u 2 , (c) ux + 2uy = 0, u (0, y)=3e −2y , (d) y 2u 2 x + x 2u 2 y = (xyu) 2 , (e) x 2uxy + 9y 2u = 0, u (x, 0) = exp  1 x , (f) y ux − x uy = 0, (g) ut = c 2 (uxx + uyy), (h) uxx + uyy = 0. 26. Use a separable solution u (x, y) = f (x) + g (y) to solve the following equations: (a) u 2 x + u 2 y = 1, (b) u 2 x + u 2 y = u, (c) u 2 x + uy + x 2 = 0, (d) x 2u 2 x + y 2u 2 y = 1, (e) y ux + x uy = 0, u (0, y) = y 2 . 2.8 Exercises 61 27. Apply v = ln u and then v (x, y) = f (x) + g (y) to solve the following equations: (a) x 2u 2 x + y 2u 2 y = u 2 , (b) x 2u 2 x + y 2u 2 y = (xyu) 2 . 28. Apply √ u = v and v (x, y) = f (x) + g (y) to solve the equation x 4u 2 x + y 2u 2 y = 4u. 29. Using v = ln u and v = f (x) + g (y), show that the solution of the Cauchy problem y 2u 2 x + x 2u 2 y = (xyu) 2 , u (x, 0) = e x 2 is u (x, y) = exp 4 x 2 + i √ 3 2 y 2 5 . 30. Reduce each of the following equations into canonical form and find the general solution: (a) ux + uy = u, (b) ux + x uy = y, (c) ux + 2xy uy = x, (d) ux − y uy − u = 1. 31. Find the solution of each of the following equations by the method of separation of variables: (a) ux − uy = 0, u (0, y)=2e 3y , (b) ux − uy = u, u (x, 0) = 4e −3x , (c) a ux + b uy = 0, u (x, 0) = αeβx , where a, b, α and β are constants. 32. Find the solution of the following initial-value systems (a) ut + 3uux = v − x, vt − cvx = 0 with u (x, 0) = x and v (x, 0) = x. (b) ut + 2uux = v − x, vt − cvx = 0 with u (x, 0) = x and v (x, 0) = x. 62 2 First-Order, Quasi-Linear Equations and Method of Characteristics 33. Solve the following initial-value systems (a) ut + uux = e −xv, vt − avx = 0 with u (x, 0) = x and v (x, 0) = e x . (b) ut − 2uux = v − x, vt + cvx = 0 with u (x, 0) = x and v (x, 0) = x. 34. Consider the Fokker–Planck equation (See Reif (1965)) in statistical mechanics to describe the evolution of the probability distribution function in the form ut = uxx + (x u)x , u (x, 0) = f (x). Neglecting the term uxx, solve the first-order linear equation ut − x ux = u with u (x, 0) = f (x). 3 Mathematical Models “Physics can’t exist without mathematics which provides it with the only language in which it can speak. Thus, services are continuously exchanged between pure mathematical analysis and physics. It is really remarkable that among works of analysis most useful for physics were those cultivated for their own beauty. In exchange, physics, exposing new problems, is as useful for mathematics as it is a model for an artist.” Henri Poincar´e “It is no paradox to say in our most theoretical models we may be nearest to our most practical applications.” A. N. Whitehead “... builds models based on data from all levels: gene expression, protein location in the cell, models of cell function, and computer representations of organs and organisms.” E. Pennisi 3.1 Classical Equations Partial differential equations arise frequently in formulating fundamental laws of nature and in the study of a wide variety of physical, chemical, and biological models. We start with a special type of second-order linear partial differential equation for the following reasons. First, second-order linear equations arise more frequently in a wide variety of applications. Second, their mathematical treatment is simpler and easier to understand than that of first-order equations in general. Usually, in almost all physical 64 3 Mathematical Models phenomena (or physical processes), the dependent variable u = u (x, y, z, t) is a function of three space variables, x, y, z and time variable t. The three basic types of second-order partial differential equations are: (a) The wave equation utt − c 2 (uxx + uyy + uzz)=0. (3.1.1) (b) The heat equation ut − k (uxx + uyy + uzz)=0. (3.1.2) (c) The Laplace equation uxx + uyy + uzz = 0. (3.1.3) In this section, we list a few more common linear partial differential equations of importance in applied mathematics, mathematical physics, and engineering science. Such a list naturally cannot ever be complete. Included are only equations of most common interest: (d) The Poisson equation ∇2u = f (x, y, z). (3.1.4) (e) The Helmholtz equation ∇2u + λu = 0. (3.1.5) (f) The biharmonic equation ∇4u = ∇2  ∇2u = 0. (3.1.6) (g) The biharmonic wave equation utt + c 2∇4u = 0. (3.1.7) (h) The telegraph equation utt + aut + bu = c 2uxx. (3.1.8) (i) The Schr¨odinger equations in quantum physics i
ψt = −
2 2m  ∇2 + V (x, y, z) ψ, (3.1.9) ∇2Ψ + 2m
2 [E − V (x, y, z)] Ψ = 0. (3.1.10) (j) The Klein–Gordon equation
u + λ 2u = 0, (3.1.11) 3.2 The Vibrating String 65 where ∇2 ≡ ∂ 2 ∂x2 + ∂ 2 ∂y2 + ∂ 2 ∂z2 , (3.1.12) is the Laplace operator in rectangular Cartesian coordinates (x, y, z),
≡ ∇2 − 1 c 2 ∂ 2 ∂t2 , (3.1.13) is the d’Alembertian, and in all equations λ, a, b, c, m, E are constants and h = 2π
is the Planck constant. (k) For a compressible fluid flow, Euler’s equations ut + (u · ∇) u = − 1 ρ ∇p, ρt + div (ρu)=0, (3.1.14) where u = (u, v, w) is the fluid velocity vector, ρ is the fluid density, and p = p (ρ) is the pressure that relates p and ρ (the constitutive equation or equation of state). Many problems in mathematical physics reduce to the solving of partial differential equations, in particular, the partial differential equations listed above. We will begin our study of these equations by first examining in detail the mathematical models representing physical problems. 3.2 The Vibrating String One of the most important problems in mathematical physics is the vibration of a stretched string. Simplicity and frequent occurrence in many branches of mathematical physics make it a classic example in the theory of partial differential equations. Let us consider a stretched string of length l fixed at the end points. The problem here is to determine the equation of motion which characterizes the position u (x, t) of the string at time t after an initial disturbance is given. In order to obtain a simple equation, we make the following assumptions: 1. The string is flexible and elastic, that is the string cannot resist bending moment and thus the tension in the string is always in the direction of the tangent to the existing profile of the string. 2. There is no elongation of a single segment of the string and hence, by Hooke’s law, the tension is constant. 3. The weight of the string is small compared with the tension in the string. 4. The deflection is small compared with the length of the string. 5. The slope of the displaced string at any point is small compared with unity. 66 3 Mathematical Models 6. There is only pure transverse vibration. We consider a differential element of the string. Let T be the tension at the end points as shown in Figure 3.2.1. The forces acting on the element of the string in the vertical direction are T sin β − T sin α. By Newton’s second law of motion, the resultant force is equal to the mass times the acceleration. Hence, T sin β − T sin α = ρ δs utt (3.2.1) where ρ is the line density and δs is the smaller arc length of the string. Since the slope of the displaced string is small, we have δs ≃ δx. Since the angles α and β are small sin α ≃ tan α, sin β ≃ tan β. Figure 3.2.1 An Element of a vertically displaced string. 3.3 The Vibrating Membrane 67 Thus, equation (3.2.1) becomes tan β − tan α = ρ δx T utt. (3.2.2) But, from calculus we know that tan α and tan β are the slopes of the string at x and x + δx: tan α = ux (x, t) and tan β = ux (x + δx, t) at time t. Equation (3.2.2) may thus be written as 1 δx (ux)x+δx − (ux)x ! = ρ T utt, 1 δx [ux (x + δx, t) − ux (x, t)] = ρ T utt. In the limit as δx approaches zero, we find utt = c 2uxx (3.2.3) where c 2 = T /ρ. This is called the one-dimensional wave equation. If there is an external force f per unit length acting on the string. Equation (3.2.3) assumes the form utt = c 2uxx + F, F = f /ρ, (3.2.4) where f may be pressure, gravitation, resistance, and so on. 3.3 The Vibrating Membrane The equation of the vibrating membrane occurs in a large number of problems in applied mathematics and mathematical physics. Before we derive the equation for the vibrating membrane we make certain simplifying assumptions as in the case of the vibrating string: 1. The membrane is flexible and elastic, that is, the membrane cannot resist bending moment and the tension in the membrane is always in the direction of the tangent to the existing profile of the membrane. 2. There is no elongation of a single segment of the membrane and hence, by Hooke’s law, the tension is constant. 3. The weight of the membrane is small compared with the tension in the membrane. 4. The deflection is small compared with the minimal diameter of the membrane. 68 3 Mathematical Models 5. The slope of the displayed membrane at any point is small compared with unity. 6. There is only pure transverse vibration. We consider a small element of the membrane. Since the deflection and slope are small, the area of the element is approximately equal to δxδy. If T is the tensile force per unit length, then the forces acting on the sides of the element are T δx and T δy, as shown in Figure 3.3.1. The forces acting on the element of the membrane in the vertical direction are T δx sin β − T δx sin α + T δy sin δ − T δy sin γ. Since the slopes are small, sines of the angles are approximately equal to their tangents. Thus, the resultant force becomes T δx (tan β − tan α) + T δy (tan δ − tan γ). By Newton’s second law of motion, the resultant force is equal to the mass times the acceleration. Hence, T δx (tan β − tan α) + T δy (tan δ − tan γ) = ρ δA utt (3.3.1) where ρ is the mass per unit area, δA ≃ δxδy is the area of this element, and utt is computed at some point in the region under consideration. But from calculus, we have Figure 3.3.1 An element of vertically displaced membrane. 3.4 Waves in an Elastic Medium 69 tan α = uy (x1, y) tan β = uy (x2, y + δy) tan γ = ux (x, y1) tan δ = ux (x + δx, y2) where x1 and x2 are the values of x between x and x+δx, and y1 and y2 are the values of y between y and y + δy. Substituting these values in (3.3.1), we obtain T δx [uy (x2, y + δy) − uy (x1, y)] + T δy [ux (x + δx, y2) − ux (x, y1)] = ρ δxδy utt. Division by ρ δxδy yields T ρ uy (x2, y + δy) − uy (x1, y) δy + ux (x + δx, y2) − ux (x, y1) δx = utt. (3.3.2) In the limit as δx approaches zero and δy approaches zero, we obtain utt = c 2 (uxx + uyy), (3.3.3) where c 2 = T /ρ. This equation is called the two-dimensional wave equation. If there is an external force f per unit area acting on the membrane. Equation (3.3.3) takes the form utt = c 2 (uxx + uyy) + F, (3.3.4) where F = f /ρ. 3.4 Waves in an Elastic Medium If a small disturbance is originated at a point in an elastic medium, neighboring particles are set into motion, and the medium is put under a state of strain. We consider such states of motion to extend in all directions. We assume that the displacements of the medium are small and that we are not concerned with translation or rotation of the medium as a whole. Let the body under investigation be homogeneous and isotropic. Let δV be a differential volume of the body, and let the stresses acting on the faces of the volume be τxx, τyy, τzz, τxy, τxz, τyx, τyz, τzx, τzy. The first three stresses are called the normal stresses and the rest are called the shear stresses. (See Figure 3.4.1). We shall assume that the stress tensor τij is symmetric describing the condition of the rotational equilibrium of the volume element, that is, 70 3 Mathematical Models Figure 3.4.1 Volume element of an elastic body. τij = τji, i = j, i, j = x, y, z. (3.4.1) Neglecting the body forces, the sum of all the forces acting on the volume element in the x-direction is (τxx)x+δx − (τxx)x ! δyδz + " (τxy) y+δy − (τxy) y # δzδx + (τxz) z+δz − (τxz) z ! δxδy. By Newton’s law of motion this resultant force is equal to the mass times the acceleration. Thus, we obtain (τxx)x+δx − (τxx)x ! δyδz + " (τxy) y+δy − (τxy) y # δzδx + (τxz) z+δz − (τxz) z ! δxδy = ρ δxδyδz utt (3.4.2) where ρ is the density of the body and u is the displacement component in the x-direction. Hence, in the limit as δV approaches zero, we obtain ∂τxx ∂x + ∂τxy ∂y + ∂τxz ∂z = ρ ∂ 2u ∂t2 . (3.4.3) Similarly, the following two equations corresponding to y and z directions are obtained: 3.4 Waves in an Elastic Medium 71 ∂τyx ∂x + ∂τyy ∂y + ∂τyz ∂z = ρ ∂ 2v ∂t2 , (3.4.4) ∂τzx ∂x + ∂τzy ∂y + ∂τzz ∂z = ρ ∂ 2w ∂t2 , (3.4.5) where v and w are the displacement components in the y and z directions respectively. We may now define linear strains [see Sokolnikoff (1956)] as εxx = ∂u ∂x, εyz = 1 2  ∂w ∂y + ∂v ∂z  , εyy = ∂v ∂y , εzx = 1 2  ∂u ∂z + ∂w ∂x  , (3.4.6) εzz = ∂w ∂z , εxy = 1 2  ∂v ∂x + ∂u ∂y  , in which εxx, εyy, εzz represent unit elongations and εyz, εzx, εxy represent unit shearing strains. In the case of an isotropic body, generalized Hooke’s law takes the form τxx = λθ + 2µεxx, τyz = 2µεyz, τyy = λθ + 2µεyy, τzx = 2µεzx, (3.4.7) τzz = λθ + 2µεzz, τxy = 2µεxy, where θ = εxx + εyy + εzz is called the dilatation, and λ and µ are Lame’s constants. Expressing stresses in terms of displacements, we obtain τxx = λθ + 2µ ∂u ∂x, τxy = µ  ∂v ∂x + ∂u ∂y  , (3.4.8) τxz = µ  ∂w ∂x + ∂u ∂z  . By differentiating equations (3.4.8), we obtain ∂τxx ∂x = λ ∂θ ∂x + 2µ ∂ 2u ∂x2 , ∂τxy ∂y = µ ∂ 2v ∂x∂y + µ ∂ 2u ∂y2 , (3.4.9) ∂τxz ∂z = µ ∂ 2w ∂x∂z + µ ∂ 2u ∂z2 . Substituting equation (3.4.9) into equation (3.4.3) yields 72 3 Mathematical Models λ ∂θ ∂x + µ  ∂ 2u ∂x2 + ∂ 2v ∂x∂y + ∂ 2w ∂x∂z  + µ  ∂ 2u ∂x2 + ∂ 2u ∂y2 + ∂ 2u ∂z2  = ρ ∂ 2u ∂t2 . (3.4.10) We note that ∂ 2u ∂x2 + ∂ 2v ∂x∂y + ∂ 2w ∂x∂z = ∂ ∂x  ∂u ∂x + ∂v ∂y + ∂w ∂z  = ∂θ ∂x, and introduce the notation △ = ∇2 = ∂ 2 ∂x2 + ∂ 2 ∂y2 + ∂ 2 ∂z2 . The symbol △ or ∇2 is called the Laplace operator. Hence, equation (3.4.10) becomes (λ + µ) ∂θ ∂x + µ∇2u = ρ ∂ 2u ∂t2 . (3.4.11) In a similar manner, we obtain the other two equations which are (λ + µ) ∂θ ∂y + µ∇2 v = ρ ∂ 2v ∂t2 . (3.4.12) (λ + µ) ∂θ ∂z + µ∇2w = ρ ∂ 2w ∂t2 . (3.4.13) The set of equations (3.4.11)–(3.4.13) is called the Navier equations of motion. In vector form, the Navier equations of motion assume the form (λ + µ) grad div u + µ∇2u = ρ utt, (3.4.14) where u = ui + vj + wk and θ = div u. (i) If div u = 0, the general equation becomes µ∇2u = ρ utt, or utt = c 2 T ∇2u, (3.4.15) where cT is called the transverse wave velocity given by cT =  µ/ρ. This is the case of an equivoluminal wave propagation, since the volume expansion θ is zero for waves moving with this velocity. Sometimes these waves are called waves of distortion because the velocity of propagation depends on µ and ρ; the shear modulus µ characterizes the distortion and rotation of the volume element. 3.4 Waves in an Elastic Medium 73 (ii) When curl u = 0, the vector identity curl curl u = grad div u − ∇2u, gives grad div u = ∇2u, Then the general equation becomes (λ + 2µ) ∇2u = ρ utt, or utt = c 2 L∇2u, (3.4.16) where cL is called the longitudinal wave velocity given by cL = $ λ + 2µ ρ . This is the case of irrotational or dilatational wave propagation, since curl u = 0 describes irrotational motion. Equations (3.4.15) and (3.4.16) are called the three-dimensional wave equations. In general, the wave equation may be written as utt = c 2∇2u, (3.4.17) where the Laplace operator may be one, two, or three dimensional. The importance of the wave equation stems from the facts that this type of equation arises in many physical problems; for example, sound waves in space, electrical vibration in a conductor, torsional oscillation of a rod, shallow water waves, linearized supersonic flow in a gas, waves in an electric transmission line, waves in magnetohydrodynamics, and longitudinal vibrations of a bar. To give a more general method of decomposing elastic waves into transverse and longitudinal wave forms, we write the Navier equations of motion in the form c 2 T ∇2u +  c 2 L − c 2 T grad (div u) = utt. (3.4.18) We now decompose this equation into two vector equations by defining u = uT + uL, where uT and uL satisfy the equations div uT = 0 and curl uL = 0. (3.4.19ab) Since uT is defined by (3.4.19a) that is divergenceless, it follows from vector analysis that there exists a rotation vector ψ such that 74 3 Mathematical Models uT = curl ψ, (3.4.20) where ψ is called the vector potential. On the other hand, uL is irrotational as given by (3.4.19b), so there exists a scalar function φ (x, t), called the scalar potential such that uL = grad φ. (3.4.21) Using (3.4.20) and (3.4.21), we can write u = curl ψ + grad φ. (3.4.22) This means that the displacement vector field is decomposed into a divergenceless vector and irrotational vector. Inserting u = uT + uL into (3.4.18), taking the divergence of each term of the resulting equation, and then using (3.4.19a) gives div c 2 L∇2uL − (uL) tt! = 0. (3.4.23) It is noted that the curl of the square bracket in (3.4.23) is also zero. Clearly, any vector whose divergence and curl both vanish is identically a zero vector. Consequently, c 2 L∇2uL = (uL) tt . (3.4.24) This shows that uL satisfies the vector wave equation with the wave velocity cL. Since uL = grad φ, it is clear that the scalar potential φ also satisfies the wave equation with the same wave speed. All solutions of (3.4.24) represent longitudinal waves that are irrotational (since ψ = 0). Similarly, we substitute u = uL + uT into (3.4.18), take the curl of the resulting equation, and use the fact that curl uL = 0 to obtain curl c 2 T ∇2uT − (uT ) tt! = 0. (3.4.25) Since the divergence of the expression inside the square bracket is also zero, it follows that c 2 T ∇2uT = (uT ) tt . (3.4.26) This is a vector wave equation for uT whose solutions represent transverse waves that are irrotational but are accompanied by no change in volume (equivoluminal, transverse, rotational waves). These waves propagate with a wave velocity cT . We close this section by seeking time-harmonic solutions of (3.4.18) in the form u = Re U (x, y, z) e iωt! . (3.4.27) 3.5 Conduction of Heat in Solids 75 Invoking (3.4.27) into equation (3.4.18) gives the following equation for the function U cT ∇2U + (cL − cT ) grad (div U) + ω 2U = 0. (3.4.28) Inserting, u = uT + uL, and using the above method of taking the divergence and curl of (3.4.28) respectively leads to equation for UL and UT as follows ∇2UL + k 2 L ∇2UL = 0, ∇2UT + k 2 T UT = 0, (3.4.29) where k 2 L = ω 2 c 2 L and k 2 T = ω 2 c 2 T . (3.4.30) Equations (3.4.29) are called the reduced wave equations (or the Helmholtz equations) for UL and UT . Obviously, equations (3.4.29) can also be derived by assuming time-harmonic solutions for uL and uT in the form ⎛ ⎝ uL uT ⎞ ⎠ = e iωt ⎛ ⎝ UL UT ⎞ ⎠ , (3.4.31) and substituting these results into (3.4.24) and (3.4.26) respectively. 3.5 Conduction of Heat in Solids We consider a domain D∗ bounded by a closed surface B∗ . Let u (x, y, z, t) be the temperature at a point (x, y, z) at time t. If the temperature is not constant, heat flows from places of higher temperature to places of lower temperature. Fourier’s law states that the rate of flow is proportional to the gradient of the temperature. Thus the velocity of the heat flow in an isotropic body is v = −Kgradu, (3.5.1) where K is a constant, called the thermal conductivity of the body. Let D be an arbitrary domain bounded by a closed surface B in D∗ . Then the amount of heat leaving D per unit time is

B vnds, where vn = v · n is the component of v in the direction of the outer unit normal n of B. Thus, by Gauss’ theorem (Divergence theorem) 76 3 Mathematical Models

B vnds =


D div (−Kgradu) dx dy dz = −K


D ∇2u dx dy dz. (3.5.2) But the amount of heat in D is given by


D σρu dx dy dz, (3.5.3) where ρ is the density of the material of the body and σ is its specific heat. Assuming that integration and differentiation are interchangeable, the rate of decrease of heat in D is −


D σρ ∂u ∂t dx dy dz. (3.5.4) Since the rate of decrease of heat in D must be equal to the amount of heat leaving D per unit time, we have −


D σρut dx dy dz = −K


D ∇2u dx dy dz, or −


D σρut − K∇2u ! dx dy dz = 0, (3.5.5) for an arbitrary D in D∗ . We assume that the integrand is continuous. If we suppose that the integrand is not zero at a point (x0, y0, z0) in D, then, by continuity, the integrand is not zero in a small region surrounding the point (x0, y0, z0). Continuing in this fashion we extend the region encompassing D. Hence the integral must be nonzero. This contradicts (3.5.5). Thus, the integrand is zero everywhere, that is, ut = κ∇2u, (3.5.6) where κ = K/σρ. This is known as the heat equation. This type of equation appears in a great variety of problems in mathematical physics, for example the concentration of diffusing material, the motion of a tidal wave in a long channel, transmission in electrical cables, and unsteady boundary layers in viscous fluid flows. 3.6 The Gravitational Potential In this section, we shall derive one of the most well-known equations in the theory of partial differential equations, the Laplace equation. 3.6 The Gravitational Potential 77 Figure 3.6.1 Two particles at P and Q. We consider two particles of masses m and M, at P and Q as shown in Figure 3.6.1. Let r be the distance between them. Then, according to Newton’s law of gravitation, a force proportional to the product of their masses, and inversely proportional to the square of the distance between them, is given in the form F = G mM r 2 , (3.6.1) where G is the gravitational constant. It is customary in potential theory to choose the unit of force so that G = 1. Thus, F becomes F = mM r 2 . (3.6.2) If r represents the vector P Q, the force per unit mass at Q due to the mass at P may be written as F = −mr r 3 = ∇
4m r
5 , (3.6.3) which is called the intensity of the gravitational field of force. We suppose that a particle of unit mass moves under the attraction of the particle of mass m at P from infinity up to Q. The work done by the force F is 78 3 Mathematical Models
r ∞ Fdr =
r ∞ ∇
4m r
5 dr = m r . (3.6.4) This is called the potential at Q due to the particle at P. We denote this by V = − m r , (3.6.5) so that the intensity of force at P is F = ∇
4m r
5 = −∇V. (3.6.6) We shall now consider a number of masses m1, m2, ..., mn, whose distances from Q are r1, r2, ..., rn, respectively. Then the force of attraction per unit mass at Q due to the system is F = n k=1 ∇ mk rk = ∇ n k=1 mk rk . (3.6.7) The work done by the forces acting on a particle of unit mass is
r ∞ F · dr = n k=1 mk rk = −V. (3.6.8) Then the potential satisfies the equation ∇2V = −∇2n k=1 mk rk = − n k=1 ∇2  mk rk  = 0, rk = 0. (3.6.9) In the case of a continuous distribution of mass in some volume R, we have, as in Figure 3.6.2. V (x, y, z) =


R ρ (ξ, η, ζ) r dR, (3.6.10) where r = % (x − ξ) 2 + (y − η) 2 + (z − ζ) 2 and Q is outside the body. It immediately follows that ∇2V = 0. (3.6.11) This equation is called the Laplace equation, also known as the potential equation. It appears in many physical problems, such as those of electrostatic potentials, potentials in hydrodynamics, and harmonic potentials in the theory of elasticity. We observe that the Laplace equation can be viewed as the special case of the heat and the wave equations when the dependent variables involved are independent of time. 3.7 Conservation Laws and The Burgers Equation 79 Figure 3.6.2 Continuous Mass Distribution. 3.7 Conservation Laws and The Burgers Equation A conservation law states that the rate of change of the total amount of material contained in a fixed domain of volume V is equal to the flux of that material across the closed bounding surface S of the domain. If we denote the density of the material by ρ (x, t) and the flux vector by q (x, t), then the conservation law is given by d dt
V ρ dV = −
S (q · n) dS, (3.7.1) where dV is the volume element and dS is the surface element of the boundary surface S, n denotes the outward unit normal vector to S as shown in Figure 3.7.1, and the right-hand side measures the outward flux — hence, the minus sign is used. Applying the Gauss divergence theorem and taking d dt inside the integral sign, we obtain
V  ∂ρ ∂t + div q  dV = 0. (3.7.2) This result is true for any arbitrary volume V , and, if the integrand is continuous, it must vanish everywhere in the domain. Thus, we obtain the differential form of the conservation law ρt + div q = 0. (3.7.3) 80 3 Mathematical Models Figure 3.7.1 Volume V of a closed domain bounded by a surface S with surface element dS and outward normal vector n. The one-dimensional version of the conservation law (3.7.3) is ∂ρ ∂t + ∂q ∂x = 0. (3.7.4) To investigate the nature of the discontinuous solution or shock waves, we assume a functional relation q = Q (ρ) and allow a jump discontinuity for ρ and q. In many physical problems of interest, it would be a better approximation to assume that q is a function of the density gradient ρx as well as ρ. A simple model is to take q = Q (ρ) − νρx, (3.7.5) where ν is a positive constant. Substituting (3.7.5) into (3.7.4), we obtain the nonlinear diffusion equation ρt + c (ρ) ρx = νρxx, (3.7.6) where c (ρ) = Q′ (ρ). We multiply (3.7.6) by c ′ (ρ) to obtain ct + c cx = ν c′ (ρ) ρxx, = ν & cxx − c ′′ (ρ) ρ 2 x ' . (3.7.7) If Q (ρ) is a quadratic function in ρ, then c (ρ) is linear in ρ, and c ′′ (ρ) = 0. Consequently, (3.7.7) becomes ct + c cx = ν cxx. (3.7.8) 3.8 The Schr¨odinger and the Korteweg–de Vries Equations 81 As a simple model of turbulence, c is replaced by the fluid velocity field u (x, t) to obtain the well-known Burgers equation ut + u ux = ν uxx, (3.7.9) where ν is the kinematic viscosity. Thus the Burgers equation is a balance between time evolution, nonlinearity, and diffusion. This is the simplest nonlinear model equation for diffusive waves in fluid dynamics. Burgers (1948) first developed this equation primarily to shed light on the study of turbulence described by the interaction of the two opposite effects of convection and diffusion. However, turbulence is more complex in the sense that it is both three dimensional and statistically random in nature. Equation (3.7.9) arises in many physical problems including one-dimensional turbulence (where this equation had its origin), sound waves in a viscous medium, shock waves in a viscous medium, waves in fluid-filled viscous elastic tubes, and magnetohydrodynamic waves in a medium with finite electrical conductivity. We note that (3.7.9) is parabolic provided the coefficient of ux is constant, whereas the resulting (3.7.9) with ν = 0 is hyperbolic. More importantly, the properties of the solution of the parabolic equation are significantly different from those of the hyperbolic equation. 3.8 The Schr¨odinger and the Korteweg–de Vries Equations We consider the following Fourier integral representation of a quasi-monochromatic plane wave solution u (x, t) =
∞ −∞ F (k) exp [i {kx − ω (k)t}] dk, (3.8.1) where the spectrum function F (k) is determined from the given initial or boundary conditions and has the property F (−k) = F ∗ (k), and ω = ω (k) is the dispersion relation. We assume that the initial wave is slowly modulated as it propagates in a dispersive medium. For such a quasimonochromatic wave, most of the energy is confined in a neighborhood of a specified wave number k = k0, so that spectrum function F (k) has a sharp peak around the point k = k0 with a narrow wave number width k − k0 = δk = O (ε), and the dispersion relation ω (k) can be expanded about k0 in the form ω = ω0 + ω ′ 0 (δk) + 1 2! ω ′′ 0 (δk) 2 + 1 3! ω ′′′ 0 (δk) 3 + ··· , (3.8.2) where ω0 = ω (k0), ω ′ 0 = ω ′ (k0), ω ′′ 0 = ω ′′ (k0), and ω ′′′ 0 = ω ′′′ (k0). 82 3 Mathematical Models Substituting (3.8.2) into (3.8.1) gives a new form u (x, t) = A (x, t) exp [i(k0x − ω0t)] + c.c., (3.8.3) where c.c. stands for the complex conjugate and A (x, t) is the complex wave amplitude given by A (x, t) =
∞ 0 F (k0 + δk) exp i  (x − ω ′ 0 t) (δk) − 1 2 ω ′′ 0 (δk) 2 t − 1 3 ω ′′′ 0 (δk) 3 t 0 d (δk), (3.8.4) where it has been assumed that ω (−k) = − ω (k). Since (3.8.4) depends on (x − ω ′ 0 t) δk, (δk) 2 t, (δk) 3 t where δk = O (ε) is small, the wave amplitude A (x, t) is a slowly varying function of x ∗ = (x − ω ′ 0 t) and time t. We keep only the term with (δk) in (3.8.4) and neglect all terms with (δk) n , n = 2, 3, ··· , so that (3.8.4) becomes A (x, t) =
∞ 0 F (k0 + δk) exp [i {(x − ω ′ 0 t)} (δk)] d (δk). (3.8.5) A simple calculation reveals that A (x, t) satisfies the evolution equation ∂A ∂t + cg ∂A ∂x = 0, (3.8.6) where cg = ω ′ 0 is the group velocity. In the next step, we retain only terms with (δk) and (δk) 2 in (3.8.4) to obtain A (x, t) =
∞ 0 F (k0 + δk) exp i  (x − ω ′ 0 t) (δk) − 1 2 ω ′′ 0 (δk) 2 0 d (δk). (3.8.7) A simple calculation shows that A (x, t) satisfies the linear Schr¨odinger equation i  ∂A ∂t + ω ′ 0 ∂A ∂x  + 1 2 ω ′′ 0 ∂ 2A ∂x2 = 0. (3.8.8) Using the slow variables ξ = ε (x − ω ′ 0 t), τ = ε 2 t, (3.8.9) the modulated wave amplitude A (ξ, τ ) satisfies the linear Schr¨odinger equation i Aτ + 1 2 ω ′′ 0Aξξ = 0. (3.8.10) 3.9 Exercises 83 On the other hand, for the frequencies at which the group velocity ω ′ 0 reaches an extremum, ω ′′ 0 = 0. In this case, the cubic term in the dispersion relation (3.7.2) plays an important role. Consequently, equation (3.8.4) reduces to a form similar to (3.8.7) with ω ′′ 0 = 0 in the exponential factor. Once again, a simple calculation from the resulting integral (3.8.4) reveals that A (x, t) satisfies the linearized Korteweg–de Vries (KdV) equation ∂A ∂t + ω ′ 0 ∂A ∂x + 1 6 ω ′′′ 0 ∂ 3A ∂x3 = 0. (3.8.11) By transferring to the new variables ξ = x−ω ′ 0 t and τ = t which correspond to a reference system moving with the group velocity ω ′ 0 , we obtain the linearized KdV equation ∂A ∂τ + 1 6 ω ′′′ 0 ∂ 3A ∂ξ3 = 0. (3.8.12) This describes waves in a dispersive medium with a weak high frequency dispersion. One of the remarkable nonlinear model equations is the Korteweg–de Vries (KdV) equation in the form ut + αu ux + βuxxx = 0, −∞ <x< ∞,="" t=""> 0. (3.8.13) This equation arises in many physical problems including water waves, ion acoustic waves in a plasma, and longitudinal dispersive waves in elastic rods. The exact solution of this equation is called the soliton which is remarkably stable. We shall discuss the soliton solution in Chapter 13. Another remarkable nonlinear model equation describing solitary waves is known as the nonlinear Schr¨odinger (NLS) equation written in the standard form i ut + 1 2 ω ′′ 0 uxx + γ |u| 2 u = 0, −∞ <x< ∞,="" t=""> 0. (3.8.14) This equation admits a solution called the solitary waves and describes the evolution of the water waves; it arises in many other physical systems that include nonlinear optics, hydromagnetic and plasma waves, propagation of heat pulse in a solid, and nonlinear instability problems. The solution of this equation will be discussed in Chapter 13. 3.9 Exercises 1. Show that the equation of motion of a long string is utt = c 2uxx − g, where g is the gravitational acceleration. 84 3 Mathematical Models 2. Derive the damped wave equation of a string utt + a ut = c 2uxx, where the damping force is proportional to the velocity and a is a constant. Considering a restoring force proportional to the displacement of a string, show that the resulting equation is utt + aut + bu = c 2uxx, where b is a constant. This equation is called the telegraph equation. 3. Consider the transverse vibration of a uniform beam. Adopting Euler’s beam theory, the moment M at a point can be written as M = −EI uxx, where EI is called the flexural rigidity, E is the elastic modulus, and I is the moment of inertia of the cross section of the beam. Show that the transverse motion of the beam may be described by utt + c 2uxxxx = 0, where c 2 = EI/ρA, ρ is the density, and A is the cross-sectional area of the beam. 4. Derive the deflection equation of a thin elastic plate ∇4u = q/D, where q is the uniform load per unit area, D is the flexural rigidity of the plate, and ∇4u = uxxxx + 2uxxyy + uyyyy. 5. Derive the one-dimensional heat equation ut = κuxx, where κ is a constant. Assuming that heat is also lost by radioactive exponential decay of the material in the bar, show that the above equation becomes ut = κuxx + he−αx , where h and α are constants. 6. Starting from Maxwell’s equations in electrodynamics, show that in a conducting medium electric intensity E, magnetic intensity H, and current density J satisfy ∇2X = µεXtt + µσXt, where X represents E, H, and J, µ is the magnetic inductive capacity, ε is the electric inductive capacity, and σ is the electrical conductivity. 3.9 Exercises 85 7. Derive the continuity equation ρt + div (ρu)=0, and Euler’s equation of motion ρ [ut + (u · grad) u] + grad p = 0, in fluid dynamics. 8. In the derivation of the Laplace equation (3.6.11), the potential at Q which is outside the body is ascertained. Now determine the potential at Q when it is inside the body, and show that it satisfies the Poisson equation ∇2u = −4πρ, where ρ is the density of the body. 9. Setting U = e iktu in the wave equation Utt = ∇2U and setting U = e −k 2 tu in the heat equation Ut = ∇2U, show that u (x, y, z) satisfies the Helmholtz equation ∇2u + k 2u = 0. 10. The Maxwell equations in vacuum are ∇ × E = − ∂B ∂t , ∇ × B = µε ∂E ∂t , ∇ · E = 0, ∇ · B = 0, where µ and ε are universal constants. Show that the magnetic field B = (0, By (x, t), 0) and the electric field E = (0, 0, Ez (x, t)) satisfy the wave equation ∂ 2u ∂t2 = c 2 ∂ 2u ∂x2 , where u = By or Ez and c = (µε) − 1 2 is the speed of light. 11. The equations of gas dynamics are linearized for small perturbations about a constant state u = 0, ρ = ρ0, and p0 = p (ρ0) with c 2 0 = p ′ (ρ0). In terms of velocity potential φ defined by u = ∇φ, the perturbation equations are ρt + ρ0 div u = 0, p − p0 = −ρ0φt = c 2 0 (ρ − ρ0), ρ − ρ0 = − ρ0 c 2 0 φt. 86 3 Mathematical Models Show that f and u satisfy the three dimensional wave equations ftt = c 2 0 ∇2 f, and utt = c 2 0 ∇2u, where f = p, ρ, or φ and ∇2 ≡ ∂ 2 ∂x2 + ∂ 2 ∂y2 + ∂ 2 ∂z2 . 12. Consider a slender body moving in a gas with arbitrary constant velocity U, and suppose (x1, x2, x3) represents the frame of reference in which the motion of the gas is small and described by the equations of problem 11. The body moves in the negative x1 direction, and (x, y, z) denotes the coordinates fixed with respect to the body so that the coordinate transformation is (x, y, z)=(x1 + U t, x2, x3). Show that the wave equation φtt = c 2 0∇2φ reduces to the form  M2 − 1 Φxx = Φyy + Φzz, where M ≡ U/c0 is the Mach number and Φ is the potential in the new frame of reference (x, y, z). 13. Consider the motion of a gas in a taper tube of cross section A (x). Show that the equation of continuity and the equation of motion are ρ = ρ0  1 − ∂ξ ∂x − ξ A ∂A ∂x  = ρ0 1 − 1 A ∂ ∂x (Aξ) , and ρ0 ∂ 2 ξ ∂t2 = − ∂p ∂x, where x is the distance along the length of the tube, ξ (x) is the displacement function, p = p (ρ) is the pressure-density relation, ρ0 is the average density, and ρ is the local density of the gas. Hence derive the equation of motion ξtt = c 2 ∂ ∂x 1 A ∂ ∂x (Aξ) , c2 = ∂p ∂ρ. Find the equation of motion when A is constant. If A (x) = a0 exp (2αx) where a0 and α are constants, show that the above equation takes the form ξtt = c 2 (ξxx + 2αξx). 14. Consider the current I (x, t) and the potential V (x, t) at a point x and time t of a uniform electric transmission line with resistance R, 3.9 Exercises 87 inductance L, capacity C, and leakage conductance G per unit length. (a) Show that both I and V satisfy the system of equations LIt + RI = −Vx, CVt + GV = −Ix. Derive the telegraph equation uxx − c −2 utt − a ut − bu = 0, for u = I or V, where c 2 = (LC) −1 , a = RC + LG and b = RG. (b) Show that the telegraph equation can be written in the form utt − c 2uxx + (p + q) ut + pq u = 0, where p = G C and q = R L . (c) Apply the transformation u = v exp − 1 2 (p + q)t to transform the above equation into the form vtt − c 2 vxx = 1 4 (p − q) 2 v. (d) When p = q, show that there exists an undisturbed wave solution in the form u (x, t) = e −ptf (x + ct), which propagates in either direction, where f is an arbitrary twice differentiable function of its argument. If u (x, t) = A exp [i(kx − ωt)] is a solution of the telegraph equation utt − c 2uxx − αut − βu = 0, α = p + q, β = pq, show that the dispersion relation holds ω 2 + iαω −  c 2 k 2 + β 2 = 0. Solve the dispersion relation to show that u (x, t) = exp  − 1 2 p t exp i  kx − t 2  4c 2k 2 + (4q − p 2)  . When p 2 = 4q, show that the solution represents attenuated nondispersive waves. 88 3 Mathematical Models (e) Find the equations for I and V in the following cases: (i) Lossless transmission line (R = G = 0), (ii) Ideal submarine cable (L = G = 0), (iii) Heaviside’s distortionless line (R/L = G/C = constant = k). 15. The Fermi–Pasta–Ulam model is used to describe waves in an anharmonic lattice of length l consisting of a row of n identical masses m, each connected to the next by nonlinear springs of constant κ. The masses are at a distance h = l/n apart, and the springs when extended or compressed by an amount d exert a force F = κ  d + α d2 where α measures the strength of nonlinearity. The equation of motion of the ith mass is my¨i = κ " (yi+1 − yi) − (yi − yi−1) + α ( (yi+1 − yi) 2 − (yi − yi−1) 2 )# , where i = 1, 2, 3...n, yi is the displacement of the ith mass from its equilibrium position, and κ, α are constants with y0 = yn = 0. Assume a continuum approximation of this discrete system so that the Taylor expansions yi+1 − yi = hyx + h 2 2! yxx + h 3 3! yxxx + h 4 4! yxxxx + o  h 5 , yi − yi−1 = hyx − h 2 2! yxx + h 3 3! yxxx − h 4 4! yxxxx + o  h 5 , can be used to derive the nonlinear differential equation ytt = c 2 [1 + 2αhyx] yxx + o  h 4 , ytt = c 2 [1 + 2αhyx] yxx + c 2h 2 12 yxxxx + o  h 5 , where c 2 = κh2 m . Using a change of variables ξ = x − ct, τ = cαht, show that u = yξ satisfies the Korteweg–de Vries (KdV) equation uτ + uuξ + βuξξξ = o  ε 2 , ε = αh, β = h 24α . 16. The one-dimensional isentropic fluid flow is obtained from Euler’s equations (3.1.14) in the form ut + u ux = − 1 ρ px, ρt + (ρu)x = 0, p = p (ρ). 3.9 Exercises 89 (a) Show that u and ρ satisfy the one-dimensional wave equation  u ρ  tt − c 2  u ρ  xx = 0, where c 2 = dp dρ is the velocity of sound. (b) For a compressible adiabatic gas, the equation of state is p = Aργ , where A and γ are constants; show that c 2 = γp ρ . 17. (a) Obtain the two-dimensional unsteady fluid flow equations from (3.1.14). (b) Find the two-dimensional steady fluid flow equations from (3.1.14). Hence or otherwise, show that  c 2 − u 2 ux − u v (uy + vx) +  c 2 − v 2 vy = 0, where c 2 = p ′ (ρ). (c) Show that, for an irrotational fluid flow (u = ∇φ), the above equation reduces to the quasi-linear partial differential equations  c 2 − φ 2 x φxx − 2φxφyφxy +  c 2 − φ 2 y φyy = 0. (d) Show that the slope of the characteristic C satisfies the quadratic equation  c 2 − u 2  dy dx2 + 2uv  dy dx +  c 2 − v 2 = 0. Hence or otherwise derive  c 2 − v 2  dv du2 − 2uv  dv du +  c 2 − u 2 = 0. 18. For an inviscid incompressible fluid flow under the body force, F = −∇Φ, the Euler equations are ∂u ∂t + u · ∇u = −∇Φ − 1 ρ ∇p, divu = 0. (a) Show that the vorticity ω = ∇ × u satisfies the vorticity equation Dω Dt = ∂ω ∂t + u · ∇ω = ω · ∇u. (b) Give the interpretation of this vorticity equation. (c) In two dimensions, show that Dω Dt = 0 (conservation of vorticity). 90 3 Mathematical Models 19. The evolution of the probability distribution function u (x, t) in nonequilibrium statistical mechanics is described by the Fokker–Planck equation (See Reif (1965)) ∂u ∂t = ∂ ∂x  ∂u ∂t + x  u. (a) Use the change of variables ξ = x et and v = u e−t to show that the Fokker–Planck equation assumes the form with u (x, t) = e t v (ξ, τ ) vt = e 2t vξξ. (b) Make a suitable change of variable t to τ (t), and transform the above equation into the standard diffusion equation vt = vξξ. 20. The electric field E (x) and the electromagnetic field H (x) in free space (a vacuum) satisfy the Maxwell equations Et = c curl H, Ht = −c curl H, divE = 0 = divH, where c is the constant speed of light in a vacuum. Show that both E and H the three-dimensional wave equations Ett = c 2∇2E and Htt = c 2∇2H, where x = (x, y, z) and ∇2 is the three-dimensional Laplacian. 21. Consider longitudinal vibrations of a free elastic rod with a variable cross section A (x) with x measured along the axis of the rod from the origin. Assuming that the material of the rod satisfies Hooke’s law, show that the displacement function u (x, t) satisfies the generalized wave equation utt = c 2uxx + c 2 A (x)  dA dx  ux, where c 2 = (λ/ρ), λ is a constant that describes the elastic nature of the material, and ρ is the line density of the rod. When A (x) is constant, the above equation reduces to one-dimensional wave equation. 4 Classification of Second-Order Linear Equations “When we have a good understanding of the problem, we are able to clear it of all auxiliary notions and to reduce it to simplest element.” Ren´e Descartes “The first process ... in the effectual study of sciences must be one of simplification and reduction of the results of previous investigations to a form in which the mind can grasp them.” James Clerk Maxwell 4.1 Second-Order Equations in Two Independent Variables The general linear second-order partial differential equation in one dependent variable u may be written as n i,j=1 Aijuxixj + n i=1 Biuxi + F u = G, (4.1.1) in which we assume Aij = Aji and Aij , Bi , F, and G are real-valued functions defined in some region of the space (x1, x2,...,xn). Here we shall be concerned with second-order equations in the dependent variable u and the independent variables x, y. Hence equation (4.1.1) can be put in the form Auxx + Buxy + Cuyy + Dux + Euy + F u = G, (4.1.2) where the coefficients are functions of x and y and do not vanish simultaneously. We shall assume that the function u and the coefficients are twice continuously differentiable in some domain in R 2 . 92 4 Classification of Second-Order Linear Equations The classification of partial differential equations is suggested by the classification of the quadratic equation of conic sections in analytic geometry. The equation Ax2 + Bxy + Cy2 + Dx + Ey + F = 0, represents hyperbola, parabola, or ellipse accordingly as B2 − 4AC is positive, zero, or negative. The classification of second-order equations is based upon the possibility of reducing equation (4.1.2) by coordinate transformation to canonical or standard form at a point. An equation is said to be hyperbolic, parabolic, or elliptic at a point (x0, y0) accordingly as B 2 (x0, y0) − 4A (x0, y0) C (x0, y0) (4.1.3) is positive, zero, or negative. If this is true at all points, then the equation is said to be hyperbolic, parabolic, or elliptic in a domain. In the case of two independent variables, a transformation can always be found to reduce the given equation to canonical form in a given domain. However, in the case of several independent variables, it is not, in general, possible to find such a transformation. To transform equation (4.1.2) to a canonical form we make a change of independent variables. Let the new variables be ξ = ξ (x, y), η = η (x, y). (4.1.4) Assuming that ξ and η are twice continuously differentiable and that the Jacobian J =       ξx ξy ηx ηy       , (4.1.5) is nonzero in the region under consideration, then x and y can be determined uniquely from the system (4.1.4). Let x and y be twice continuously differentiable functions of ξ and η. Then we have ux = uξξx + uηηx, uy = uξξy + uηηy, uxx = uξξξ 2 x + 2uξηξxηx + uηηη 2 x + uξξxx + uηηxx, (4.1.6) uxy = uξξξxξy + uξη (ξxηy + ξyηx) + uηηηxηy + uξξxy + uηηxy, uyy = uξξξ 2 y + 2uξηξyηy + uηηη 2 y + uξξyy + uηηyy. Substituting these values in equation (4.1.2) we obtain A ∗uξξ + B ∗uξη + C ∗uηη + D∗uξ + E ∗uη + F ∗u = G ∗ , (4.1.7) where 4.2 Canonical Forms 93 A ∗ = Aξ2 x + Bξxξy + Cξ2 y , B ∗ = 2Aξxηx + B (ξxηy + ξyηx)+2Cξyηy, C ∗ = Aη2 x + Bηxηy + Cη2 y , D∗ = Aξxx + Bξxy + Cξyy + Dξx + Eξy, (4.1.8) E ∗ = Aηxx + Bηxy + Cηyy + Dηx + Eηy, F ∗ = F, G∗ = G. The resulting equation (4.1.7) is in the same form as the original equation (4.1.2) under the general transformation (4.1.4). The nature of the equation remains invariant under such a transformation if the Jacobian does not vanish. This can be seen from the fact that the sign of the discriminant does not alter under the transformation, that is, B ∗2 − 4A ∗C ∗ = J 2  B 2 − 4AC , (4.1.9) which can be easily verified. It should be noted here that the equation can be of a different type at different points of the domain, but for our purpose we shall assume that the equation under consideration is of the single type in a given domain. The classification of equation (4.1.2) depends on the coefficients A (x, y), B (x, y), and C (x, y) at a given point (x, y). We shall, therefore, rewrite equation (4.1.2) as Auxx + Buxy + Cuyy = H (x, y, u, ux, uy), (4.1.10) and equation (4.1.7) as A ∗uξξ + B ∗uξη + C ∗uηη = H∗ (ξ, η, u, uξ, uη). (4.1.11) 4.2 Canonical Forms In this section we shall consider the problem of reducing equation (4.1.10) to canonical form. We suppose first that none of A, B, C, is zero. Let ξ and η be new variables such that the coefficients A∗ and C ∗ in equation (4.1.11) vanish. Thus, from (4.1.8), we have A ∗ = Aξ2 x + Bξxξy + Cξ2 y = 0, C ∗ = Aη2 x + Bηxηy + Cη2 y = 0. These two equations are of the same type and hence we may write them in the form Aζ2 x + Bζxζy + Cζ2 y = 0, (4.2.1) 94 4 Classification of Second-Order Linear Equations in which ζ stand for either of the functions ξ or η. Dividing through by ζ 2 y , equation (4.2.1) becomes A  ζx ζy 2 + B  ζx ζy  + C = 0. (4.2.2) Along the curve ζ = constant, we have dζ = ζxdx + ζydy = 0. Thus, dy dx = − ζx ζy , (4.2.3) and therefore, equation (4.2.2) may be written in the form A  dy dx2 − B  dy dx + C = 0, (4.2.4) the roots of which are dy dx =
4 B +  B2 − 4AC
5 /2A, (4.2.5) dy dx =
4 B −  B2 − 4AC
5 /2A. (4.2.6) These equations, which are known as the characteristic equations, are ordinary differential equations for families of curves in the xy-plane along which ξ = constant and η = constant. The integrals of equations (4.2.5) and (4.2.6) are called the characteristic curves. Since the equations are first-order ordinary differential equations, the solutions may be written as φ1 (x, y) = c1, c1 = constant, φ2 (x, y) = c2, c2 = constant. Hence the transformations ξ = φ1 (x, y), η = φ2 (x, y), will transform equation (4.1.10) to a canonical form. (A) Hyperbolic Type If B2 − 4AC > 0, then integration of equations (4.2.5) and (4.2.6) yield two real and distinct families of characteristics. Equation (4.1.11) reduces to uξη = H1, (4.2.7) 4.2 Canonical Forms 95 where H1 = H∗/B∗ . It can be easily shown that B∗ = 0. This form is called the first canonical form of the hyperbolic equation. Now if new independent variables α = ξ + η, β = ξ − η, (4.2.8) are introduced, then equation (4.2.7) is transformed into uαα − uββ = H2 (α, β, u, uα, uβ). (4.2.9) This form is called the second canonical form of the hyperbolic equation. (B) Parabolic Type In this case, we have B2 − 4AC = 0, and equations (4.2.5) and (4.2.6) coincide. Thus, there exists one real family of characteristics, and we obtain only a single integral ξ = constant (or η = constant). Since B2 = 4AC and A∗ = 0, we find that A ∗ = Aξ2 x + Bξxξy + Cξ2 y =
4√ A ξx + √ C ξy
52 = 0. From this it follows that A ∗ = 2Aξxηx + B (ξxηy + ξyηx)+2Cξyηy = 2
4√ A ξx + √ C ξy
5
4√ A ηx + √ C ηy
5 = 0, for arbitrary values of η (x, y) which is functionally independent of ξ (x, y); for instance, if η = y, the Jacobian does not vanish in the domain of parabolicity. Division of equation (4.1.11) by C ∗ yields uηη = H3 (ξ, η, u, uξ, uη), C∗ = 0. (4.2.10) This is called the canonical form of the parabolic equation. Equation (4.1.11) may also assume the form uξξ = H∗ 3 (ξ, η, u, uξ, uη), (4.2.11) if we choose η = constant as the integral of equation (4.2.5). (C) Elliptic Type For an equation of elliptic type, we have B2 − 4AC < 0. Consequently, the quadratic equation (4.2.4) has no real solutions, but it has two complex conjugate solutions which are continuous complex-valued functions of the real variables x and y. Thus, in this case, there are no real characteristic curves. However, if the coefficients A, B, and C are analytic functions of x and y, then one can consider equation (4.2.4) for complex x and y. A function of two real variables x and y is said to be analytic in a certain domain if in some neighborhood of every point (x0, y0) of this domain, the 96 4 Classification of Second-Order Linear Equations function can be represented as a Taylor series in the variables (x − x0) and (y − y0). Since ξ and η are complex, we introduce new real variables α = 1 2 (ξ + η), β = 1 2i (ξ − η), (4.2.12) so that ξ = α + iβ, η = α − iβ. (4.2.13) First, we transform equations (4.1.10). We then have A ∗∗ (α, β) uαα + B ∗∗ (α, β) uαβ + C ∗∗ (α, β) uββ = H4 (α, β, u, uα, uβ), (4.2.14) in which the coefficients assume the same form as the coefficients in equation (4.1.11). With the use of (4.2.13), the equations A∗ = C ∗ = 0 become  Aα2 x + Bαxαy + Cα2 y −  Aβ2 x + Bβxβy + Cβ2 y +i[2Aαxβx + B (αxβy + αyβx)+2Cαyβy]=0,  Aα2 x + Bαxαy + Cα2 y −  Aβ2 x + Bβxβy + Cβ2 y −i[2Aαxβx + B (αxβy + αyβx)+2Cαyβy]=0, or, (A ∗∗ − C ∗∗) + iB∗∗ = 0, (A ∗∗ − C ∗∗) − iB∗∗ = 0. These equations are satisfied if and only if A ∗∗ = C ∗∗ and B ∗∗ = 0. Hence, equation (4.2.14) transforms into the form A ∗∗uαα + A ∗∗uββ = H4 (α, β, u, uα, uβ). Dividing through by A∗∗, we obtain uαα + uββ = H5 (α, β, u, uα, uβ), (4.2.15) where H5 = (H4/A∗∗). This is called the canonical form of the elliptic equation. We close this discussion of canonical forms by adding an important comment. From mathematical and physical points of view, characteristics or characteristic coordinates play a very important physical role in hyperbolic equations. However, they do not play a particularly physical role in parabolic and elliptic equations, but their role is somewhat mathematical 4.2 Canonical Forms 97 in solving these equations. In general, first-order partial differential equations such as advection-reaction equations are regarded as hyperbolic because they describe propagation of waves like the wave equation. On the other hand, second-order linear partial differential equations with constant coefficients are sometimes classified by the associated dispersion relation ω = ω (κ) as defined in Section 13.3. In one-dimensional case, ω = ω (k). If ω (k) is real and ω ′′ (k) = 0, the equation is called dispersive. The word dispersive simply means that the phase velocity cp = (ω/k) of a plane wave solution, u (x, t) = A exp [i(kx − ωt)] depends on the wavenumber k. This means that waves of different wavelength propagate with different phase velocities and hence, disperse in the medium. If ω = ω (k) = σ (k) + iµ (k) is complex, the associated partial differential equation is called diffusive. From a physical point of view, such a classification of equations is particularly useful. Both dispersive and diffusive equations are physically important, and such equations will be discussed in Chapter 13. Example 4.2.1. Consider the equation y 2uxx − x 2uyy = 0. Here A = y 2 , B = 0, C = −x 2 . Thus, B 2 − 4AC = 4x 2 y 2 > 0. The equation is hyperbolic everywhere except on the coordinate axes x = 0 and y = 0. From the characteristic equations (4.2.5) and (4.2.6), we have dy dx = x y , dy dx = − x y . After integration of these equations, we obtain 1 2 y 2 − 1 2 x 2 = c1, 1 2 y 2 + 1 2 x 2 = c2. The first of these curves is a family of hyperbolas 1 2 y 2 − 1 2 x 2 = c1, and the second is a family of circles 1 2 y 2 + 1 2 x 2 = c2. To transform the given equation to canonical form, we consider 98 4 Classification of Second-Order Linear Equations ξ = 1 2 y 2 − 1 2 x 2 , η = 1 2 y 2 + 1 2 x 2 . From the relations (4.1.6), we have ux = uξξx + uηηx = −xuξ + xuη, uy = uξξy + uηηy = yuξ + yuη, uxx = uξξξ 2 x + 2uξηξxηx + uηηη 2 x + uξξxx + uηηxx = x 2uξξ − 2x 2uξη + x 2uηη − uξ + uη. uyy = uξξξ 2 y + 2uξηξyηy + uηηη 2 y + uξξyy + uηηyy = y 2uξξ + 2y 2uξη + y 2uηη + uξ + uη. Thus, the given equation assumes the canonical form uξη = η 2 (ξ 2 − η 2) uξ − ξ 2 (ξ 2 − η 2) uη. Example 4.2.2. Consider the partial differential equation x 2uxx + 2xy uxy + y 2uyy = 0. In this case, the discriminant is B 2 − 4AC = 4x 2 y 2 − 4x 2 y 2 = 0. The equation is therefore parabolic everywhere. The characteristic equation is dy dx = y x , and hence, the characteristics are y x = c, which is the equation of a family of straight lines. Consider the transformation ξ = y x , η = y, where η is chosen arbitrarily. The given equation is then reduced to the canonical form y 2uηη = 0. Thus, uηη = 0 for y = 0. 4.3 Equations with Constant Coefficients 99 Example 4.2.3. The equation uxx + x 2uyy = 0, is elliptic everywhere except on the coordinate axis x = 0 because B 2 − 4AC = −4x 2 < 0, x = 0. The characteristic equations are dy dx = ix, dy dx = −ix. Integration yields 2y − ix2 = c1, 2y + ix2 = c2. Thus, if we write ξ = 2y − ix2 , η = 2y + ix2 , and hence, α = 1 2 (ξ + η)=2y, β = 1 2i (ξ − η) = −x 2 , we obtain the canonical form uαα + uββ = − 1 2β uβ. It should be remarked here that a given partial differential equation may be of a different type in a different domain. Thus, for example, Tricomi’s equation uxx + xuyy = 0, (4.2.16) is elliptic for x > 0 and hyperbolic for x < 0, since B2 − 4AC = −4x. For a detailed treatment, see Hellwig (1964). 4.3 Equations with Constant Coefficients In this case of an equation with real constant coefficients, the equation is of a single type at all points in the domain. This is because the discriminant B2 − 4AC is a constant. From the characteristic equations dy dx =
4 B +  B2 − 4AC
5 /2A, (4.3.1) 100 4 Classification of Second-Order Linear Equations we can see that the characteristics y = 4 B + √ B2 − 4AC 2A 5 x + c1, y = 4 B − √ B2 − 4AC 2A 5 x + c2, (4.3.2) are two families of straight lines. Consequently, the characteristic coordinates take the form ξ = y − λ1x, η = y − λ2x, (4.3.3) where λ1,2 = B+ √ B2 − 4AC 2A . (4.3.4) The linear second-order partial differential equation with constant coeffi- cients may be written in the general form as Auxx + Buxy + Cuyy + Dux + Euy + F u = G (x, y). (4.3.5) In particular, the equation Auxx + Buyy + Cuyy = 0, (4.3.6) is called the Euler equation. (A) Hyperbolic Type If B2 − 4AC > 0, the equation is of hyperbolic type, in which case the characteristics form two distinct families. Using (4.3.3), equation (4.3.5) becomes uξη = D1uξ + E1uη + F1u + G1 (ξ,η), (4.3.7) where D1, E1, and F1 are constants. Here, since the coefficients are constants, the lower order terms are expressed explicitly. When A = 0, equation (4.3.1) does not hold. In this case, the characteristic equation may be put in the form −B (dx/dy) + C (dx/dy) 2 = 0, which may again be rewritten as dx/dy = 0, and − B + C (dx/dy)=0. Integration gives x = c1, x = (B/C) y + c2, where c1 and c2 are integration constants. Thus, the characteristic coordinates are 4.3 Equations with Constant Coefficients 101 ξ = x, η = x − (B/C) y. (4.3.8) Under this transformation, equation (4.3.5) reduces to the canonical form uξη = D∗ 1uξ + E ∗ 1uη + F ∗ 1 u + G ∗ 1 (ξ,η), (4.3.9) where D∗ 1 , E∗ 1 , and F ∗ 1 are constants. The canonical form of the Euler equation (4.3.6) is uξη = 0. (4.3.10) Integrating this equation gives the general solution u = φ (ξ) + ψ (η) = φ (y − λ1, x) + ψ (y − λ2, x), (4.3.11) where φ and ψ are arbitrary functions, and λ1 and λ2 are given by (4.3.3). (B) Parabolic Type When B2 − 4AC = 0, the equation is of parabolic type, in which case only one real family of characteristics exists. From equation (4.3.4), we find that λ1 = λ2 = (B/2A), so that the single family of characteristics is given by y = (B/2A) x + c1, where c1 is an integration constant. Thus, we have ξ = y − (B/2A) x, η = hy + kx, (4.3.12) where η is chosen arbitrarily such that the Jacobian of the transformation is not zero, and h and k are constants. With the proper choice of the constants h and k in the transformation (4.3.12), equation (4.3.5) reduces to uηη = D2uξ + E2uη + F2u + G2 (ξ,η), (4.3.13) where D2, E2, and F2 are constants. If B = 0, we can see at once from the relation B 2 − 4AC = 0, that C or A vanishes. The given equation is then already in the canonical form. Similarly, in the other cases when A or C vanishes, B vanishes. The given equation is is then also in canonical form. The canonical form of the Euler equation (4.3.6) is uηη = 0. (4.3.14) 102 4 Classification of Second-Order Linear Equations Integrating twice gives the general solution u = φ (ξ) + η ψ (ξ), (4.3.15) where ξ and η are given by (4.3.12). Choosing h = 1, k = 0 and λ =  B 2A for simplicity, the general solution of the Euler equation in the parabolic case is u = φ (y − λx) + y ψ (y − λx). (4.3.16) (C) Elliptic Type When B2 − 4AC < 0, the equation is of elliptic type. In this case, the characteristics are complex conjugates. The characteristic equations yield y = λ1x + c1, y = λ2x + c2, (4.3.17) where λ1 and λ2 are complex numbers. Accordingly, c1 and c2 are allowed to take on complex values. Thus, ξ = y − (a + ib) x, η = y − (a − ib) x, (4.3.18) where λ1,2 = a + ib in which a and b are real constants, and a = B 2A , and b = 1 2A  4AC − B2 . Introduce the new variables α = 1 2 (ξ + η) = y − ax, β = 1 2i (ξ − η) = −bx. (4.3.19) Application of this transformation readily reduces equation (4.3.5) to the canonical form uαα + uββ = D3uα + E3uβ + F3u + G3 (α, β), (4.3.20) where D3, E3, F3 are constants. We note that B2 − AC < 0, so neither A nor C is zero. In this elliptic case, the Euler equation (4.3.6) gives the complex characteristics (4.3.18) which are ξ = (y − ax) − ibx, η = (y − ax) + ibx = ξ. (4.3.21) Consequently, the Euler equation becomes uξξ = 0, (4.3.22) with the general solution u = φ (ξ) + ψ  ξ . (4.3.23) The appearance of complex arguments in the general solution (4.3.23) is a general feature of elliptic equations. 4.3 Equations with Constant Coefficients 103 Example 4.3.1. Consider the equation 4 uxx + 5 uxy + uyy + ux + uy = 2. Since A = 4, B = 5, C = 1, and B2 − 4AC = 9 > 0, the equation is hyperbolic. Thus, the characteristic equations take the form dy dx = 1, dy dx = 1 4 , and hence, the characteristics are y = x + c1, y = (x/4) + c2. The linear transformation ξ = y − x, η = y − (x/4), therefore reduces the given equation to the canonical form uξη = 1 3 uη − 8 9 . This is the first canonical form. The second canonical form may be obtained by the transformation α = ξ + η, β = ξ − η, in the form uαα − uββ = 1 3 uα − 1 3 uβ − 8 9 . Example 4.3.2. The equation uxx − 4 uxy + 4 uyy = e y , is parabolic since A = 1, B = −4, C = 4, and B2 − 4AC = 0. Thus, we have from equation (4.3.12) ξ = y + 2x, η = y, in which η is chosen arbitrarily. By means of this mapping, the equation transforms into uηη = 1 4 e η . Example 4.3.3. Consider the equation uxx + uxy + uyy + ux = 0. 104 4 Classification of Second-Order Linear Equations Since A = 1, B = 1, C = 1, and B2 − 4AC = −3 < 0, the equation is elliptic. We have λ1,2 = B + √ B2 − 4AC 2A = 1 2 + i √ 3 2 , and hence, ξ = y − 4 1 2 + i √ 3 2 5 x, η = y − 4 1 2 − i √ 3 2 5 x. Introducing the new variables α = 1 2 (ξ + η) = y − 1 2 x, β = 1 2i (ξ − η) = − √ 3 2 x, the given equation is then transformed into canonical form uαα + uββ = 2 3 uα + 2 √ 3 uβ. Example 4.3.4. Consider the wave equation utt − c 2uxx = 0, c is constant. Since A = −c 2 , B = 0, C = 1, and B2 − 4AC = 4c 2 > 0, the wave equation is hyperbolic everywhere. According to (4.2.4), the equation of characteristics is −c 2  dt dx2 +1=0, or dx2 − c 2 dt2 = 0. Therefore, x + ct = ξ = constant, x − ct = η = constant. Thus, the characteristics are straight lines, which are shown in Figure 4.3.1. The characteristics form a natural set of coordinates for the hyperbolic equation. In terms of new coordinates ξ and η defined above, we obtain uxx = uξξ + 2uξη + uηη, utt = c 2 (uξξ − 2uξη + uηη), so that the wave equation becomes 4.3 Equations with Constant Coefficients 105 Figure 4.3.1 Characteristics for the wave equation. −4c 2uξη = 0. Since c = 0, we have uξη = 0. Integrating with respect to ξ, we obtain uη = ψ1 (η). where ψ1 is the arbitrary function of η. Integrating with respect to η, we obtain u (ξ,η) =
ψ1 (η) dη + φ (ξ). If we set ψ (η) = * ψ1 (η) dη, the general solution becomes u (ξ,η) = φ (ξ) + ψ (η), which is, in terms of the original variables x and t, u (x, t) = φ (x + ct) + ψ (x − ct), provided φ and ψ are arbitrary but twice differentiable functions. 106 4 Classification of Second-Order Linear Equations Note that φ is constant on “wavefronts” x = −ct + ξ that travel toward decreasing x as t increases, whereas ψ is constant on wavefronts x = ct + η that travel toward increasing x as t increases. Thus, any general solution can be expressed as the sum of two waves, one traveling to the right with constant velocity c and the other traveling to the left with the same velocity c. Example 4.3.5. Find the characteristic equations and characteristics, and then reduce the equations uxx +  sech4 x uyy = 0, (4.3.24ab) to the canonical forms. In equation (4.3.24a), A = 1, B = 0 and C = −sech4 x. Hence, B 2 − 4AC = 4 sech4 x > 0. Hence, the equation is hyperbolic. The characteristic equations are dy dx = B + √ B2 − 4AC 2A = + sech2 x. Integration gives y + tanh x = constant. Hence, ξ = y + tanh x, η = y − tanh x. Using these characteristic coordinates, the given equation can be transformed into the canonical form uξη = (η − ξ) " 4 − (ξ − η) 2 # (uξ − uη). (4.3.25) In equation (4.3.24b), A = 1, B = 0 and C = sech4 x. Hence, B 2 − 4AC = + isech2 x. Integrating gives y + i tanh x = constant. Thus, ξ = y + i tanh x, η = y − i tanh x. The new real variables α and β are 4.4 General Solutions 107 α = 1 2 (ξ + η) = y, β = 1 2i (ξ − η) = tanh x. In terms of these new variables, equation (4.3.24b) can be transformed into the canonical form uαα + uββ = 2β 1 − β 2 uβ, |β| < 1. (4.3.26) Example 4.3.6. Consider the equation uxx + (2 cosecy) uxy +  cosec2 y uyy = 0. (4.3.27) In this case, A = 1, B = 2 cosecy and C = cosec2y. Hence, B2 − 4AC = 0, and dy dx = B 2A = cosec y. The characteristic curves are therefore given by ξ = x + cos y and η = y. Using these variables, the canonical form of (4.3.27) is uηη =  sin2 η cos η uξ. (4.3.28) 4.4 General Solutions In general, it is not so simple to determine the general solution of a given equation. Sometimes further simplification of the canonical form of an equation may yield the general solution. If the canonical form of the equation is simple, then the general solution can be immediately ascertained. Example 4.4.1. Find the general solution of x 2uxx + 2xy uxy + y 2uyy = 0. In Example 4.2.2, using the transformation ξ = y/x, η = y, this equation was reduced to the canonical form uηη = 0, for y = 0. Integrating twice with respect to η, we obtain u (ξ,η) = ηf (ξ) + g (ξ), where f (ξ) and g (ξ) are arbitrary functions. In terms of the independent variables x and y, we have u (x, y) = y f
4y x
5 + g
4y x
5 . 108 4 Classification of Second-Order Linear Equations Example 4.4.2. Determine the general solution of 4 uxx + 5 uxy + uyy + ux + uy = 2. Using the transformation ξ = y − x, η = y − (x/4), the canonical form of this equation is (see Example 4.3.1) uξη = 1 3 uη − 8 9 . By means of the substitution v = uη, the preceding equation reduces to vξ = 1 3 v − 8 9 . This can be easily integrated by separating the variables. Integrating with respect to ξ, we have v = 8 3 + 1 3 e (ξ/3)F (η). Integrating with respect to η, we obtain u (ξ,η) = 8 3 η + 1 3 g (η) e ξ/3 + f (ξ), where f (ξ) and g (η) are arbitrary functions. The general solution of the given equation becomes u (x, y) = 8 3  y − 1 4  + 1 3 g
4 y − x 4
5 e 1 3 (y−x) + f (y − x). Example 4.4.3. Obtain the general solution of 3 uxx + 10 uxy + 3 uyy = 0. Since B2 − 4AC = 64 > 0, the equation is hyperbolic. Thus, from equation (4.3.2), the characteristics are y = 3x + c1, y = 1 3 x + c2. Using the transformations ξ = y − 3x, η = y − 1 3 x, the given equation can be reduced to the form  64 3  uξη = 0. 4.4 General Solutions 109 Hence, we obtain uξη = 0. Integration yields u (ξ,η) = f (ξ) + g (η). In terms of the original variables, the general solution is u (x, y) = f (y − 3x) + g
4 y − x 3
5 . Example 4.4.4. Find the general solution of the following equations y uxx + 3 y uxy + 3 ux = 0, y = 0, (4.4.1) uxx + 2 uxy + uyy = 0, (4.4.2) uxx + 2 uxy + 5 uyy + ux = 0. (4.4.3) In equation (4.4.1), A = y, B = 3y, C = 0, D = 3, E = F = G = 0. Hence B2 − 4AC = 9y 2 > 0 and the equation is hyperbolic for all points (x, y) with y = 0. Consequently, the characteristic equations are dy dx = B + √ B2 − 4AC 2A = 3y + 3y 2y = 3, 0. Integrating gives y = c1 and y = 3x + c2. The characteristic curves are ξ = y and η = y − 3x. In terms of these variables, the canonical form of (4.4.1) is ξ uξη + uη = 0. Writing v = uη and using the integrating factor gives v = uη = 1 ξ C (η), where C (η) is an arbitrary function. Integrating again with respect to η gives u (ξ,η) = 1 ξ
C (η) dη + g (ξ) = 1 ξ f (η) + g (ξ), where f and g are arbitrary functions. Finally, in terms of the original variables, the general solution is 110 4 Classification of Second-Order Linear Equations u (x, y) = 1 y f (y − 3x) + g (y). (4.4.4) Equation (4.4.2) has coefficients A = 1, B = 2, C = 1, D = E = F = G = 0. Hence, B2 − 4AC = 0, the equation is parabolic. The characteristic equation is dy dx = 1, and the characteristics are ξ = y − x = c1 and η = y. Using these variables, equation (4.4.2) takes the canonical form uηη = 0. Integrating twice gives the general solution u (ξ,η) = η f (ξ) + g (ξ), where f and g are arbitrary functions. In terms of x and y, this solution becomes u (x, y) = y f (y − x) + g (y − x). (4.4.5) The coefficients of equation (4.4.3) are A = 1, B = 2, C = 5, E = 1, F = G = 0 and hence B2 − 4AC = −16 < 0, equation (4.4.3) is elliptic. The characteristic equations are dy dx = (1 + 2i). The characteristics are y = (1 − 2i) x + c1, y = (1 + 2i) x + c2, and hence, ξ = y − (1 − 2i) x, η = y − (1 + 2i) x, and new real variables α and β are α = 1 2 (ξ + η) = y − x, η = 1 2i (ξ − η)=2x. The canonical form is given by (uαα + uββ) = 1 4 (uα − 2 uβ). (4.4.6) It is not easy to find a general solution of (4.4.6). 4.5 Summary and Further Simplification 111 Example 4.4.5. Use u = f (ξ), ξ = √x 4κt to solve the parabolic system ut = κ uxx, −∞ <x< ∞,="" t=""> 0, (4.4.7) u (x, 0) = 0, x< 0; u (x, 0) = u0, x> 0, (4.4.8) where κ and u0 are constant. We use the given transformations to obtain ut = f ′ (ξ) ξt = − 1 2 x √ 4κt3 f ′ (ξ), uxx = ∂ ∂x (ux) = ∂ ∂x (f ′ (ξ) · ξx) = 1 4κt f ′′ (ξ). Consequently, equation (4.4.7) becomes f ′′ (ξ)+2 ξf′ (ξ)=0. The solution of this equation is f ′ (ξ) = A exp  −ξ 2 , where A is a constant of integration. Integrating again gives f (ξ) = A
ξ 0 e −α 2 dα + B, where B is an integrating constant. Using the given conditions yields 0 = A
−∞ 0 e −α 2 dα + B, u0 = A
∞ 0 e −α 2 dα + B, which give A = u0 √ π and B = 1 2 u0. Thus, the final solution is u (x, t) = u0 1 1 √ π
√x 4κt 0 e −α 2 dα + 1 2 3 . 4.5 Summary and Further Simplification We summarize the classification of linear second-order partial differential equations with constant coefficients in two independent variables. 112 4 Classification of Second-Order Linear Equations hyperbolic: urs = a1ur + a2us + a3u + f1, (4.5.1) urr − uss = a ∗ 1ur + a ∗ 2us + a ∗ 3u + f ∗ 1 , (4.5.2) parabolic: urs = b1ur + b2us + b3u + f2, (4.5.3) elliptic: urr + uss = c1ur + c2us + c3u + f3, (4.5.4) where r and s represent the new independent variables in the linear transformations r = r (x, y), s = s (x, y), (4.5.5) and the Jacobian J = 0. To simplify equation (4.5.1) further, we introduce the new dependent variable v = u e−(ar+bs) , (4.5.6) where a and b are undetermined coefficients. Finding the derivatives, we obtain ur = (vr + av) e ar+bs , us = (vs + bv) e ar+bs , urr =  vrr + 2avr + a 2 v e ar+bs , urs = (vrs + avs + bvr + abv) e ar+bs , uss =  vss + 2bvs + b 2 v e ar+bs . Substitution of these equation (4.5.1) yields vrs + (b − a1) vr + (a − a2) vs + (ab − a1a − a2b − a3) v = f1 e −(ar+bs) . In order that the first derivatives vanish, we set b = a1 and a = a2. Thus, the above equation becomes vrs = (a1a2 + a3) v + g1, where g1 = f1 e −(a2r+a1s) . In a similar manner, we can transform equations (4.5.2)–(4.5.4). Thus, we have the following transformed equations corresponding to equations (4.5.1)–(4.5.4). hyperbolic: vrs = h1v + g1, vrr − vss = h ∗ 1 v + g ∗ 1 , (4.5.7) parabolic: vss = h2v + g2, elliptic: vrr + vss = h3v + g3. In the case of partial differential equations in several independent variables or in higher order, the classification is considerably more complex. For further reading, see Courant and Hilbert (1953, 1962). 4.6 Exercises 113 4.6 Exercises 1. Determine the region in which the given equation is hyperbolic, parabolic, or elliptic, and transform the equation in the respective region to canonical form. (a) xuxx + uyy = x 2 , (b) uxx + y 2uyy = y, (c) uxx + xyuyy = 0, (d) x 2uxx − 2xyuxy + y 2uyy = e x , (e) uxx + uxy − xuyy = 0, (f) e xuxx + e yuyy = u, (g) uxx − √y uxy +  x 4 uyy + 2x ux − 3y uy + 2u = exp  x 2 − 2y , y ≥ 0, (h) uxx − √y uxy + xuyy = cos  x 2 − 2y , y ≥ 0, (i) uxx − yuxy + xux + yuy + u = 0, (j) sin2 x uxx + sin 2x uxy + cos2 x uyy = x, 2. Obtain the general solution of the following equations: (i) x 2uxx + 2xyuxy + y 2uyy + xyux + y 2uy = 0, (ii) rutt − c 2 rurr − 2c 2ur = 0, c = constant, (iii) 4ux + 12uxy + 9uyy − 9u = 9, (iv) uxx + uxy − 2uyy − 3ux − 6uy = 9 (2x − y), (v) yux + 3y uxy + 3ux = 0, y = 0. (vi) uxx + uyy = 0, (vii) 4 uxx + uyy = 0, (viii) uxx − 2 uxy + uyy = 0, (ix) 2 uxx + uyy = 0, (x) uxx + 4 uxy + 4 uyy = 0, (xi) 3 uxx + 4 uxy − 3 4 uyy = 0. 114 4 Classification of Second-Order Linear Equations 3. Find the characteristics and characteristic coordinates, and reduce the following equations to canonical form: (a) uxx + 2uxy + 3uyy + 4ux + 5uy + u = e x , (b) 2uxx − 4uxy + 2uyy + 3u = 0, (c) uxx + 5uxy + 4uyy + 7uy = sin x, (d) uxx + uyy + 2ux + 8uy + u = 0, (e) uxy + 2uyy + 9ux + uy = 2, (f) 6uxx − uxy + u = y 2 , (g) uxy + ux + uy = 3x, (h) uyy − 9ux + 7uy = cos y, (i) x 2uxx − y 2uyy − ux =1+2y 2 , (j) uxx + yuyy + 1 2 uy + 4yux = 0, (k) x 2y 2uxx + 2xyuxy + uyy = 0, (l) uxx + yuyy = 0. 4. Determine the general solutions of the following equations: (i) uxx − 1 c 2 uyy = 0, c = constant, (ii) uxx + uyy = 0, (iii) uxxxx + 2uxxyy + uyyyy = 0, (iv) uxx − 3uxy + 2uyy = 0, (v) uxx + uxy = 0, (vi) uxx + 10uxy + 9uyy = y. 5. Transform the following equations to the form vξη = cv, c = constant, (i) uxx − uyy + 3ux − 2uy + u = 0, (ii) 3uxx + 7uxy + 2uyy + uy + u = 0, by introducing the new variables v = u e−(aξ+bη) , where a and b are undetermined coefficients. 6. Given the parabolic equation uxx = aut + bux + cu + f, where the coefficients are constants, by the substitution u = v e 1 2 bx, for the case c = −  b 2/4 , show that the given equation is reduced to the heat equation vxx = avt + g, g = fe−bx/2 . 7. Reduce the Tricomi equation uxx + xuyy = 0, 4.6 Exercises 115 to the canonical form (i) uξη − [6 (ξ − η)]−1 (uξ − uη)=0, for x < 0, (ii) uαα + uββ + 1 3β = 0, x > 0. Show that the characteristic curves for x < 0 are cubic parabolas. 8. Use the polar coordinates r and θ (x = r cos θ, y = r sin θ) to transform the Laplace equation uxx + uyy = 0 into the polar form ∇2u = urr + 1 r ur + 1 r 2 uθθ = 0. 9. (a) Using the cylindrical polar coordinates x = r cos θ, y = r sin θ, z = z, transform the three-dimensional Laplace equation uxx + uyy + uzz = 0 into the form urr + 1 r ur + 1 r 2 uθθ + uzz = 0. (b) Use the spherical polar coordinates (r, θ, φ) so that x = r sin φ cos θ, y = r sin φ sin θ, z = r cos φ to transform the three-dimensional Laplace equation uxx + uyy + uzz = 0 into the form urr + 2 r ur + 1 r 2 sin φ (sin φ uφ)φ + 1 r 2 sin2 φ uθθ = 0. (c) Transform the diffusion equation ut = κ (uxx + uyy), into the axisymmetric form ut = κ  urr + 1 r ur  . 10. (a) Apply a linear transformation ξ = ax + by and η = cx + dy, to transform the Euler equation A uxx + 2B uxy + C uyy = 0 into canonical form, where a, b, c, d, A, B and C are constants . (b) Show that the same transformation as in (a) can be used to transform the nonhomogeneous Euler equation A uxx + 2B uxy + C uyy = F (x, y, u, ux, uy) into canonical form. 116 4 Classification of Second-Order Linear Equations 11. Obtain the solution of the Cauchy problem uxx + uyy = 0, u (x, 0) = f (x) and uy (x, 0) = g (x). 12. Classify each of the following equations and reduce it to canonical form: (a) y uxx − x uyy = 0, x> 0, y> 0; (b) uxx +  sech4 x uyy = 0, (c) y 2uxx + x 2uyy = 0, (d) uxx −  sech4 x uyy = 0, (e) uxx + 6uxy + 9uyy + 3y uy = 0, (f) y 2uxx + 2xy uxy + 2x 2uyy + xux = 0, (g) uxx − (2 cos x) uxy +  1 + cos2 x uyy + u = 0, (h) uxx + (2 cosec y) uxy +  cosec2y uyy = 0. (i) uxx − 2 uxy + uyy + 3 ux − u + 1 = 0, (j) uxx − y 2uyy + ux − u + x 2 = 0, (k) uxx + y uyy − x uy + y = 0. 13. Transform the equation uxy + y uyy + sin (x + y)=0 into the canonical form. Use the canonical form to find the general solution. 14. Classify each of the following equations for u (x, t): (a) ut = (p ux)x , (b) utt − c 2uxx + αu = 0, (c) (a ux)x + (a ut) t = 0, (d) uxt − a ut = 0, where p (x), c (x, t), a (x, t), and α (x) are given functions that take only positive values in the (x, t) plane. Find the general solution of the equation in (d). 5 The Cauchy Problem and Wave Equations “Since a general solution must be judged impossible from want of analysis, we must be content with the knowledge of some special cases, and that all the more, since the development of various cases seems to be the only way to bringing us at last to a more perfect knowledge.” Leonhard Euler “What would geometry be without Gauss, mathematical logic without Boole, algebra without Hamilton, analysis without Cauchy?” George Temple 5.1 The Cauchy Problem In the theory of ordinary differential equations, by the initial-value problem we mean the problem of finding the solutions of a given differential equation with the appropriate number of initial conditions prescribed at an initial point. For example, the second-order ordinary differential equation d 2u dt2 = f  t, u, du dt  and the initial conditions u (t0) = α,  du dt  (t0) = β, constitute an initial-value problem. An analogous problem can be defined in the case of partial differential equations. Here we shall state the problem involving second-order partial differential equations in two independent variables. 118 5 The Cauchy Problem and Wave Equations We consider a second-order partial differential equation for the function u in the independent variables x and y, and suppose that this equation can be solved explicitly for uyy, and hence, can be represented in the from uyy = F (x, y, u, ux, uy, uxx, uxy). (5.1.1) For some value y = y0, we prescribe the initial values of the unknown function and of the derivative with respect to y u (x, y0) = f (x), uy (x, y0) = g (x). (5.1.2) The problem of determining the solution of equation (5.1.1) satisfying the initial conditions (5.1.2) is known as the initial-value problem. For instance, the initial-value problem of a vibrating string is the problem of finding the solution of the wave equation utt = c 2uxx, satisfying the initial conditions u (x, t0) = u0 (x), ut (x, t0) = v0 (x), where u0 (x) is the initial displacement and v0 (x) is the initial velocity. In initial-value problems, the initial values usually refer to the data assigned at y = y0. It is not essential that these values be given along the line y = y0; they may very well be prescribed along some curve L0 in the xy plane. In such a context, the problem is called the Cauchy problem instead of the initial-value problem, although the two names are actually synonymous. We consider the Euler equation Auxx + Buxy + Cuyy = F (x, y, u, ux, uy), (5.1.3) where A, B, C are functions of x and y. Let (x0, y0) denote points on a smooth curve L0 in the xy plane. Also let the parametric equations of this curve L0 be x0 = x0 (λ), y0 = y0 (λ), (5.1.4) where λ is a parameter. We suppose that two functions f (λ) and g (λ) are prescribed along the curve L0. The Cauchy problem is now one of determining the solution u (x, y) of equation (5.1.3) in the neighborhood of the curve L0 satisfying the Cauchy conditions u = f (λ), (5.1.5a) ∂u ∂n = g (λ), (5.1.5b) 5.1 The Cauchy Problem 119 on the curve L0 where n is the direction of the normal to L0 which lies to the left of L0 in the counterclockwise direction of increasing arc length. The function f (λ) and g (λ) are called the Cauchy data. For every point on L0, the value of u is specified by equation (5.1.5a). Thus, the curve L0 represented by equation (5.1.4) with the condition (5.1.5a) yields a twisted curve L in (x, y, u) space whose projection on the xy plane is the curve L0. Thus, the solution of the Cauchy problem is a surface, called an integral surface, in the (x, y, u) space passing through L and satisfying the condition (5.1.5b), which represents a tangent plane to the integral surface along L. If the function f (λ) is differentiable, then along the curve L0, we have du dλ = ∂u ∂x dx dλ + ∂u ∂y dy dλ = df dλ, (5.1.6) and ∂u ∂n = ∂u ∂x dx dn + ∂u ∂y dy dn = g, (5.1.7) but dx dn = − dy ds and dy dn = dx ds . (5.1.8) Equation (5.1.7) may be written as ∂u ∂n = − ∂u ∂x dy ds + ∂u ∂y dx ds = g. (5.1.9) Since       dx dλ dy dλ −dy ds dx ds       = (dx) 2 + (dy) 2 ds dλ = 0, (5.1.10) it is possible to find ux and uy on L0 from the system of equations (5.1.6) and (5.1.9). Since ux and uy are known on L0, we find the higher derivatives by first differentiating ux and uy with respect to λ. Thus, we have ∂ 2u ∂x2 dx dλ + ∂ 2u ∂x ∂y dy dλ = d dλ  ∂u ∂x , (5.1.11) ∂ 2u ∂x ∂y dx dλ + ∂ 2u ∂y2 dy dλ = d dλ  ∂u ∂y  . (5.1.12) From equation (5.1.3), we have A ∂ 2u ∂x2 + B ∂ 2u ∂x ∂y + C ∂ 2u ∂y2 = F, (5.1.13) 120 5 The Cauchy Problem and Wave Equations where F is known since ux and uy have been found. The system of equations can be solved for uxx, uxy, and uyy, if           dx dλ dy dλ 0 0 dx dλ dy dλ ABC           = C  dx dλ2 − B  dx dλ dy dλ + A  dy dλ2 = 0. (5.1.14) The equation A  dy dx2 − B  dy dx + C = 0, (5.1.15) is called the characteristic equation. It is then evident that the necessary condition for obtaining the second derivatives is that the curve L0 must not be a characteristic curve. If the coefficients of equation (5.1.3) and the function (5.1.5) are analytic, then all the derivatives of higher orders can be computed by the above process. The solution can then be represented in the form of a Taylor series: u (x, y) = ∞ n=0 ∞ k=0 1 k! (n − k)! ∂ nu0 ∂xk 0 ∂yn−k 0 (x − x0) k (y − y0) n−k , (5.1.16) which can be shown to converge in the neighborhood of the curve L0. Thus, we may state the famous Cauchy–Kowalewskaya theorem. 5.2 The Cauchy–Kowalewskaya Theorem Let the partial differential equation be given in the form uyy = F (y, x1, x2,...,xn, u, uy, ux1 , ux2 ...,uxn , ux1y, ux2y,...,uxny, ux1x1 , ux2x2 ,...,uxnxn ),(5.2.1) and let the initial conditions u = f (x1, x2,...,xn), (5.2.2) uy = g (x1, x2,...,xn), (5.2.3) be given on the noncharacteristic manifold y = y0. If the function F is analytic in some neighborhood of the point  y 0 , x0 1 , x0 2 ,...,x0 n , u0 , u0 y ,... and if the functions f and g are analytic in some neighborhood of the point  x 0 1 , x0 2 ,...,x0 n , then the Cauchy problem has a unique analytic solution in some neighborhood of the point  y 0 , x0 1 , x0 2 ,...,x0 n . 5.3 Homogeneous Wave Equations 121 For the proof, see Petrovsky (1954). The preceding statement seems equally applicable to hyperbolic, parabolic, or elliptic equations. However, we shall see that difficulties arise in formulating the Cauchy problem for nonhyperbolic equations. Consider, for instance, the famous Hadamard (1952) example. The problem consists of the elliptic (or Laplace) equation uxx + uyy = 0, and the initial conditions on y = 0 u (x, 0) = 0, uy (x, 0) = n −1 sin nx. The solution of this problem is u (x, y) = n −2 sinh ny sin nx, which can be easily verified. It can be seen that, when n tends to infinity, the function n −1 sin nx tends uniformly to zero. But the solution n −2 sinh ny sin nx does not become small, as n increases for any nonzero y. Physically, the solution represents an oscillation with unbounded amplitude  n −2 sinh ny as y → ∞ for any fixed x. Even if n is a fixed number, this solution is unstable in the sense that u → ∞ as y → ∞ for any fixed x for which sin nx = 0. It is obvious then that the solution does not depend continuously on the data. Thus, it is not a properly posed problem. In addition to existence and uniqueness, the question of continuous dependence of the solution on the initial data arises in connection with the Cauchy–Kowalewskaya theorem. It is well known that any continuous function can accurately be approximated by polynomials. We can apply the Cauchy–Kowalewskaya theorem with continuous data by using polynomial approximations only if a small variation in the initial data leads to a small change in the solution. 5.3 Homogeneous Wave Equations To study Cauchy problems for hyperbolic partial differential equations, it is quite natural to begin investigating the simplest and yet most important equation, the one-dimensional wave equation, by the method of characteristics. The essential characteristic of the solution of the general wave equation is preserved in this simplified case. We shall consider the following Cauchy problem of an infinite string with the initial condition 122 5 The Cauchy Problem and Wave Equations utt − c 2uxx = 0, x ∈ R, t> 0, (5.3.1) u (x, 0) = f (x), x ∈ R, (5.3.2) ut (x, 0) = g (x), x ∈ R. (5.3.3) By the method of characteristics described in Chapter 4, the characteristic equation according to equation (4.2.4) is dx2 − c 2 dt2 = 0, which reduces to dx + c dt = 0, dx − c dt = 0. The integrals are the straight lines x + ct = c1, x − ct = c2. Introducing the characteristic coordinates ξ = x + ct, η = x − ct, we obtain uxx = uξξ + 2 uξη + uηη, utt = c 2 (uξξ − 2 uξη + uηη). Substitution of these in equation (5.3.1) yields −4c 2uξη = 0. Since c = 0, we have uξη = 0. Integrating with respect to ξ, we obtain uη = ψ ∗ (η), where ψ ∗ (η) is an arbitrary function of η. Integrating again with respect to η, we obtain u (ξ,η) =
ψ ∗ (η) dη + φ (ξ). If we set ψ (η) = * ψ ∗ (η) dη, we have u (ξ,η) = φ (ξ) + ψ (η), where φ and ψ are arbitrary functions. Transforming to the original variables x and t, we find the general solution of the wave equation 5.3 Homogeneous Wave Equations 123 u (x, t) = φ (x + ct) + ψ (x − ct), (5.3.4) provided φ and ψ are twice differentiable functions. Now applying the initial conditions (5.3.2) and (5.3.3), we obtain u (x, 0) = f (x) = φ (x) + ψ (x), (5.3.5) ut (x, 0) = g (x) = c φ′ (x) − c ψ′ (x). (5.3.6) Integration of equation (5.3.6) gives φ (x) − ψ (x) = 1 c
x x0 g (τ ) dτ + K, (5.3.7) where x0 and K are arbitrary constants. Solving for φ and ψ from equations (5.3.5) and (5.3.7), we obtain φ (x) = 1 2 f (x) + 1 2c
x x0 g (τ ) dτ + K 2 , ψ (x) = 1 2 f (x) − 1 2c
x x0 g (τ ) dτ − K 2 . The solution is thus given by u (x, t) = 1 2 [f (x + ct) + f (x − ct)] + 1 2c
x+ct x0 g (τ ) dτ −
x−ct x0 g (τ ) dτ = 1 2 [f (x + ct) + f (x − ct)] + 1 2c
x+ct x−ct g (τ ) dτ. (5.3.8) This is called the celebrated d’Alembert solution of the Cauchy problem for the one-dimensional wave equation. It is easy to verify by direct substitution that u (x, t), represented by (5.3.8), is the unique solution of the wave equation (5.3.1) provided f (x) is twice continuously differentiable and g (x) is continuously differentiable. This essentially proves the existence of the d’Alembert solution. By direct substitution, it can also be shown that the solution (5.3.8) is uniquely determined by the initial conditions (5.3.2) and (5.3.3). It is important to note that the solution u (x, t) depends only on the initial values of f at points x − ct and x + ct and values of g between these two points. In other words, the solution does not depend at all on initial values outside this interval, x − ct ≤ x ≤ x + ct. This interval is called the domain of dependence of the variables (x, t). Moreover, the solution depends continuously on the initial data, that is, the problem is well posed. In other words, a small change in either f or g results in a correspondingly small change in the solution u (x, t). Mathematically, this can be stated as follows: For every ε > 0 and for each time interval 0 ≤ t ≤ t0, there exists a number δ (ε, t0) such that 124 5 The Cauchy Problem and Wave Equations |u (x, t) − u ∗ (x, t)| < ε, whenever |f (x) − f ∗ (x)| < δ, |g (x) − g ∗ (x)| < δ. The proof follows immediately from equation (5.3.8). We have |u (x, t) − u ∗ (x, t)| ≤ 1 2 |f (x + ct) − f ∗ (x + ct)| + 1 2 |f (x − ct) − f ∗ (x − ct)| + 1 2c
x+ct x−ct |g (τ ) − g ∗ (τ )| dτ < ε, where ε = δ (1 + t0). For any finite time interval 0 <t<t0, a="" small="" change="" in="" the="" initial="" data="" only="" produces="" solution.="" this="" shows="" that="" problem="" is="" well="" posed.="" example="" 5.3.1.="" find="" solution="" of="" initial-value="" utt="c" 2uxx,="" x="" ∈="" r,="" t=""> 0, u (x, 0) = sin x, ut (x, 0) = cos x. From (5.3.8), we have u (x, t) = 1 2 [sin (x + ct) + sin (x − ct)] + 1 2c
x+ct x−ct cos τ dτ = sin x cos ct + 1 2c [sin (x + ct) − sin (x − ct)] = sin x cos ct + 1 c cos x sin ct. It follows from the d’Alembert solution that, if an initial displacement or an initial velocity is located in a small neighborhood of some point (x0, t0), it can influence only the area t>t0 bounded by two characteristics x−ct = constant and x+ct = constant with slope ± (1/c) passing through the point (x0, t0), as shown in Figure 5.3.1. This means that the initial displacement propagates with the speed dx dt = c, whereas the effect of the initial velocity propagates at all speeds up to c. This infinite sector R in this figure is called the range of influence of the point (x0, t0). According to (5.3.8), the value of u (x0, t0) depends on the initial data f and g in the interval [x0 − ct0, x0 + ct0] which is cut out of the initial line by the two characteristics x−ct = constant and x+ct = constant with slope ± (1/c) passing through the point (x0, t0). The interval [x0 − ct0, x0 + ct0] 5.3 Homogeneous Wave Equations 125 Figure 5.3.1 Range of influence on the line t = 0 is called the domain of dependence of the solution at the point (x0, t0), as shown in Figure 5.3.2. Figure 5.3.2 Domain of dependence 126 5 The Cauchy Problem and Wave Equations Since the solution u (x, t) at every point (x, t) inside the triangular region D in this figure is completely determined by the Cauchy data on the interval [x0 − ct0, x0 + ct0], the region D is called the region of determinancy of the solution. We will now investigate the physical significance of the d’Alembert solution (5.3.8) in greater detail. We rewrite the solution in the form u (x, t) = 1 2 f (x + ct) + 1 2c
x+ct 0 g (τ ) dτ + 1 2 f (x − ct) − 1 2c
x−ct 0 g (τ ) dτ. (5.3.9) Or, equivalently, u (x, t) = φ (x + ct) + ψ (x − ct), (5.3.10) where φ (ξ) = 1 2 f (ξ) + 1 2c
ξ 0 g (τ ) dτ, (5.3.11) ψ (η) = 1 2 f (η) − 1 2c
η 0 g (τ ) dτ. (5.3.12) Evidently, φ (x + ct) represents a progressive wave traveling in the negative x-direction with speed c without change of shape. Similarly, ψ (x − ct) is also a progressive wave propagating in the positive x-direction with the same speed c without change of shape. We shall examine this point in greater detail. Treat ψ (x − ct) as a function of x for a sequence of times t. At t = 0, the shape of this function of u = ψ (x). At a subsequent time, its shape is given by u = ψ (x − ct) or u = ψ (ξ), where ξ = x − ct is the new coordinate obtained by translating the origin a distance ct to the right. Thus, the shape of the curve remains the same as time progresses, but moves to the right with velocity c as shown in Figure 5.3.3. This shows that ψ (x − ct) represents a progressive wave traveling in the positive xdirection with velocity c without change of shape. Similarly, φ (x + ct) is also a progressive wave propagating in the negative x-direction with the same speed c without change of shape. For instance, u (x, t) = sin (x + ct) (5.3.13) represent sinusoidal waves traveling with speed c in the positive and negative directions respectively without change of shape. The propagation of waves without change of shape is common to all linear wave equations. To interpret the d’Alembert formula we consider two cases: Case 1. We first consider the case when the initial velocity is zero, that is, g (x)=0. 5.3 Homogeneous Wave Equations 127 Figure 5.3.3 Progressive Waves. Then, the d’Alembert solution has the form u (x, t) = 1 2 [f (x + ct) + f (x − ct)] . Now suppose that the initial displacement f (x) is different from zero in an interval (−b, b). Then, in this case the forward and the backward waves are represented by u = 1 2 f (x). The waves are initially superimposed, and then they separate and travel in opposite directions. We consider f (x) which has the form of a triangle. We draw a triangle with the ordinate x = 0 one-half that of the given function at that point, as shown in Figure 5.3.4. If we displace these graphs and then take the sum of the ordinates of the displaced graphs, we obtain the shape of the string at any time t. As can be seen from the figure, the waves travel in opposite directions away from each other. After both waves have passed the region of initial disturbance, the string returns to its rest position. Case 2. We consider the case when the initial displacement is zero, that is, f (x)=0, 128 5 The Cauchy Problem and Wave Equations Figure 5.3.4 Triangular Waves. and the d’Alembert solution assumes the form u (x, t) = 1 2
x+ct x−ct g (τ ) dτ = 1 2 [G (x + ct) − G (x − ct)] , where G (x) = 1 c
x x0 g (τ ) dτ. If we take for the initial velocity g (x) = ⎧ ⎨ ⎩ 0 |x| > b g0 |x| ≤ b, then, the function G (x) is equal to zero for values of x in the interval x ≤ −b, and G (x) = ⎧ ⎪⎪⎪⎪⎨ ⎪⎪⎪⎪⎩ 1 c
x −b g0 dτ = g0 c (x + b) for −b ≤ x ≤ b, 1 c
x −b g0 dτ = 2bg0 c for x > b. 5.3 Homogeneous Wave Equations 129 Figure 5.3.5 Graph of u (x, t) at time t. As in the previous case, the two waves which differ in sign travel in opposite directions on the x-axis. After some time t the two functions (1/2) G (x) and − (1/2) G (x) move a distance ct. Thus, the graph of u at time t is obtained by summing the ordinates of the displaced graphs as shown in Figure 5.3.5. As t approaches infinity, the string will reach a state of rest, but it will not, in general, assume its original position. This displacement is known as the residual displacement. In the preceding examples, we note that f (x) is continuous, but not continuously differentiable and g (x) is discontinuous. To these initial data, there corresponds a generalized solution. By a generalized solution we mean the following: Let us suppose that the function u (x, t) satisfies the initial conditions (5.3.2) and (5.3.3). Let u (x, t) be the limit of a uniformly convergent sequence of solutions un (x, t) which satisfy the wave equation (5.3.1) and the initial conditions un (x, 0) = fn (x),  ∂un ∂t  (x, 0) = gn (x). Let fn (x) be a continuously differentiable function, and let the sequence converge uniformly to f (x); let gn (x) be a continuously differentiable function, and * x x0 gn (τ ) dτ approach uniformly to * x x0 g (τ ) dτ . Then, the function u (x, t) is called the generalized solution of the problem (5.3.1)–(5.3.3). In general, it is interesting to discuss the effect of discontinuity of the function f (x) at a point x = x0, assuming that g (x) is a smooth function. Clearly, it follows from (5.3.8) that u (x, t) will be discontinuous at each 130 5 The Cauchy Problem and Wave Equations point (x, t) such that x+ct = x0 or x−ct = x0, that is, at each point of the two characteristic lines intersecting at the point (x0, 0). This means that discontinuities are propagated along the characteristic lines. At each point of the characteristic lines, the partial derivatives of the function u (x, t) fail to exist, and hence, u can no longer be a solution of the Cauchy problem in the usual sense. However, such a function may be called a generalized solution of the Cauchy problem. Similarly, if f (x) is continuous, but either f ′ (x) or f ′′ (x) has a discontinuity at some point x = x0, the first- or second-order partial derivatives of the solution u (x, t) will be discontinuous along the characteristic lines through (x0, 0). Finally, a discontinuity in g (x) at x = x0 would lead to a discontinuity in the first- or second-order partial derivatives of u along the characteristic lines through (x0, 0), and a discontinuity in g ′ (x) at x0 will imply a discontinuity in the second-order partial derivatives of u along the characteristic lines through (x0, 0). The solution given by (5.3.8) with f, f ′ , f ′′ , g, and g ′ piecewise continuous on −∞ <x< ∞="" is="" usually="" called="" the="" generalized="" solution="" of="" cauchy="" problem.="" 5.4="" initial="" boundary-value="" problems="" we="" have="" just="" determined="" initial-value="" problem="" for="" infinite="" vibrating="" string.="" will="" now="" study="" effect="" a="" boundary="" on="" solution.="" (a)="" semi-infinite="" string="" with="" fixed="" end="" let="" us="" first="" consider="" end,="" that="" is,="" utt="c" 2uxx,="" 0="" <x<="" ∞,="" t=""> 0, u (x, 0) = f (x), 0 ≤ x < ∞, (5.4.1) ut (x, 0) = g (x), 0 ≤ x < ∞, u (0, t)=0, 0 ≤ t < ∞. It is evident here that the boundary condition at x = 0 produces a wave moving to the right with the velocity c. Thus, for x > ct, the solution is the same as that of the infinite string, and the displacement is influenced only by the initial data on the interval [x − ct, x + ct], as shown in Figure 5.4.1. When x < ct, the interval [x − ct, x + ct] extends onto the negative x-axis where f and g are not prescribed. But from the d’Alembert formula u (x, t) = φ (x + ct) + ψ (x − ct), (5.4.2) where 5.4 Initial Boundary-Value Problems 131 Figure 5.4.1 Displacement influenced by the initial data on [x − ct, x + ct]. φ (ξ) = 1 2 f (ξ) + 1 2c
ξ 0 g (τ ) dτ + K 2 , (5.4.3) ψ (η) = 1 2 f (η) − 1 2c
η 0 g (τ ) dτ − K 2 , (5.4.4) we see that u (0, t) = φ (ct) + ψ (−ct)=0. Hence, ψ (−ct) = −φ (ct). If we let α = −ct, then ψ (α) = −φ (−α). Replacing α by x − ct, we obtain for x < ct, ψ (x − ct) = −φ (ct − x), and hence, ψ (x − ct) = − 1 2 f (ct − x) − 1 2c
ct−x 0 g (τ ) dτ − K 2 . The solution of the initial boundary-value problem, therefore, is given by 132 5 The Cauchy Problem and Wave Equations u (x, t) = 1 2 [f (x + ct) + f (x − ct)] + 1 2c
x+ct x−ct g (τ ) dτ for x > ct, (5.4.5) u (x, t) = 1 2 [f (x + ct) − f (ct − x)] + 1 2c
x+ct ct−x g (τ ) dτ for x < ct. (5.4.6) In order for this solution to exist, f must be twice continuously differentiable and g must be continuously differentiable, and in addition f (0) = f ′′ (0) = g (0) = 0. Solution (5.4.6) has an interesting physical interpretation. If we draw the characteristics through the point (x0, t0) in the region x > ct, we see, as pointed out earlier, that the displacement at (x0, t0) is determined by the initial values on [x0 − ct0, x0 + ct0]. If the point (x0, t0) lies in the region x > ct as shown in Figure 5.4.1, we see that the characteristic x + ct = x0 + ct0 intersects the x-axis at (x0 + ct0, 0). However, the characteristic x − ct = x0 − ct0 intersects the t-axis at (0, t0 − x0/c), and the characteristic x + ct = ct0 − x0 intersects the x-axis at (ct0 − x0, 0). Thus, the disturbance at (ct0 − x0, 0) travels along the backward characteristic x + ct = ct0 − x0, and is reflected at (0, t0 − x0/c) as a forward moving wave represented by −φ (ct0 − x0). Example 5.4.1. Determine the solution of the initial boundary-value problem utt = 4 uxx, x > 0, t > 0, u (x, 0) = |sin x| , x > 0, ut (x, 0) = 0, x ≥ 0, u (x, 0) = 0, t ≥ 0. For x > 2t, u (x, t) = 1 2 [f (x + 2t) + f (x − 2t)] = 1 2 [|sin (x + 2t)|−|sin (x − 2t)|] , and for x < 2t, u (x, t) = 1 2 [f (x + 2t) − f (2t − x)] = 1 2 [|sin (x + 2t)|−|sin (2t − x)|] . Notice that u (0, t) = 0 is satisfied by u (x, t) for x < 2t (that is, t > 0). 5.4 Initial Boundary-Value Problems 133 (B) Semi-infinite String with a Free End We consider a semi-infinite string with a free end at x = 0. We will determine the solution of utt = c 2uxx, 0 <x< ∞,="" t=""> 0, u (x, 0) = f (x), 0 ≤ x < ∞, (5.4.7) ut (x, 0) = g (x), 0 ≤ x < ∞, ux (0, t)=0, 0 ≤ t < ∞. As in the case of the fixed end, for x > ct the solution is the same as that of the infinite string. For x < ct, from the d’Alembert solution (5.4.2) u (x, t) = φ (x + ct) + ψ (x − ct), we have ux (x, t) = φ ′ (x + ct) + ψ ′ (x − ct). Thus, ux (0, t) = φ ′ (ct) + ψ ′ (−ct)=0. Integration yields φ (ct) − ψ (−ct) = K, where K is a constant. Now, if we let α = −ct, we obtain ψ (α) = φ (−α) − K. Replacing α by x − ct, we have ψ (x − ct) = φ (ct − x) − K, and hence, ψ (x − ct) = 1 2 f (ct − x) + 1 2c
ct−x 0 g (τ ) dτ − K 2 . The solution of the initial boundary-value problem, therefore, is given by u (x, t) = 1 2 [f (x + ct) + f (x − ct)] + 1 2c
x+ct x−ct g (τ ) dτ for x > ct. (5.4.8) u (x, t) = 1 2 [f (x + ct) + f (ct − x)] + 1 2c
x+ct 0 g (τ ) dτ +
ct−x 0 g (τ ) dτ for x < ct. (5.4.9) We note that for this solution to exist, f must be twice continuously differentiable and g must be continuously differentiable, and in addition, f ′ (0) = g ′ (0) = 0. 134 5 The Cauchy Problem and Wave Equations Example 5.4.2. Find the solution of the initial boundary-value problem utt = uxx, 0 <x< ∞,="" t=""> 0, u (x, 0) = cos
4πx 2
5 , 0 ≤ x < ∞, ut (x, 0) = 0, 0 ≤ x < ∞, ux (x, 0) = 0, t ≥ 0. For x>t u (x, t) = 1 2 " cos π 2 (x + t) + cos π 2 (x − t) # = cos
4π 2 x
5 cos
4π 2 t
5 , and for x 0, t > 0, u (x, 0) = f (x), x ≥ 0, (5.5.1) ut (x, 0) = g (x), x ≥ 0, u (0, t) = p (t), t ≥ 0, we proceed in a manner similar to the case of homogeneous boundary conditions. Using equation (5.4.2), we apply the boundary condition to obtain u (0, t) = φ (ct) + ψ (−ct) = p (t). If we let α = −ct, we have ψ (α) = p
4 − α c
5 − φ (−α). Replacing α by x − ct, the preceding relation becomes ψ (x − ct) = p
4 t − x c
5 − φ (ct − x). 5.5 Equations with Nonhomogeneous Boundary Conditions 135 Thus, for 0 ≤ x < ct, u (x, t) = p
4 t − x c
5 + 1 2 [f (x + ct) − f (ct − x)] + 1 2c
x+ct ct−x g (τ ) dτ = p
4 t − x c
5 + φ (x + ct) − ψ (ct − x), (5.5.2) where φ (x + ct = ξ) is given by (5.3.11), and ψ (η) is given by ψ (η) = 1 2 f (η) + 1 2c
η 0 g (τ ) dτ. (5.5.3) The solution for x > ct is given by the solution (5.4.5) of the infinite string. In this case, in addition to the differentiability conditions satisfied by f and g, as in the case of the problem with the homogeneous boundary conditions, p must be twice continuously differentiable in t and p (0) = f (0), p′ (0) = g (0), p′′ (0) = c 2 f ′′ (0). We next consider the initial boundary-value problem utt = c 2uxx, x > 0, t > 0, u (x, 0) = f (x), x ≥ 0, ut (x, 0) = g (x), x ≥ 0, ux (0, t) = q (t), t ≥ 0. Using (5.4.2), we apply the boundary condition to obtain ux (0, t) = φ ′ (ct) + ψ ′ (−ct) = q (t). Then, integrating yields φ (ct) − ψ (−ct) = c
t 0 q (τ ) dτ + K. If we let α = −ct, then ψ (α) = φ (−α) − c
−α/c 0 q (τ ) dτ − K. Replacing α by x − ct, we obtain ψ (x − ct) = φ (ct − x) − c
t−x/c 0 q (τ ) dτ − K. The solution of the initial boundary-value problem for x < ct, therefore, is given by 136 5 The Cauchy Problem and Wave Equations u (x, t) = 1 2 [f (x + ct) + f (ct − x)] + 1 2c
x+ct 0 g (τ ) dτ +
ct−x 0 g (τ ) dτ −c
t−x/c 0 q (τ ) dτ. (5.5.4) Here f and g must satisfy the differentiability conditions, as in the case of the problem with the homogeneous boundary conditions. In addition f ′ (0) = q (0), g′ (0) = q ′ (0). The solution for the initial boundary-value problem involving the boundary condition ux (0, t) + h u (0, t)=0, h = constant can also be constructed in a similar manner from the d’Alembert solution. 5.6 Vibration of Finite String with Fixed Ends The problem of the finite string is more complicated than that of the infinite string due to the repeated reflection of waves from the boundaries We first consider the vibration of the string of length l fixed at both ends. The problem is that of finding the solution of utt = c 2uxx, 0 < x < l, t > 0, u (x, 0) = f (x), 0 ≤ x ≤ l, ut (x, 0) = g (x), 0 ≤ x ≤ l, (5.6.1) u (0, t)=0, u (l, t)=0, t ≥ 0, From the previous results, we know that the solution of the wave equation is u (x, t) = φ (x + ct) + ψ (x − ct). Applying the initial conditions, we have u (x, 0) = φ (x) + ψ (x) = f (x), 0 ≤ x ≤ l, ut (x, 0) = c φ′ (x) − c ψ′ (x) = g (x), 0 ≤ x ≤ l. Solving for φ and ψ, we find φ (ξ) = 1 2 f (ξ) + 1 2c
ξ 0 g (τ ) dτ + K 2 , 0 ≤ ξ ≤ l, (5.6.2) ψ (η) = 1 2 f (η) − 1 2c
η 0 g (τ ) dτ − K 2 , 0 ≤ η ≤ l. (5.6.3) 5.6 Vibration of Finite String with Fixed Ends 137 Hence, u (x, t) = 1 2 [f (x + ct) + f (x − ct)] + 1 2c
x+ct x−ct g (τ ) dτ, (5.6.4) for 0 ≤ x + ct ≤ l and 0 ≤ x − ct ≤ l. The solution is thus uniquely determined by the initial data in the region t ≤ x c , t ≤ l − x c , t ≥ 0. For larger times, the solution depends on the boundary conditions. Applying the boundary conditions, we obtain u (0, t) = φ (ct) + ψ (−ct)=0, t ≥ 0, (5.6.5) u (l, t) = φ (l + ct) + ψ (l − ct)=0, t ≥ 0. (5.6.6) If we set α = −ct, equation (5.6.5) becomes ψ (α) = −φ (−α), α ≤ 0, (5.6.7) and if we set α = l + ct, equation (5.6.6) takes the form φ (α) = −ψ (2l − α), α ≥ l. (5.6.8) With ξ = −η, we may write equation (5.6.2) as φ (−η) = 1 2 f (−η) + 1 2c
−η 0 g (τ ) dτ + K 2 , 0 ≤ −η ≤ l. (5.6.9) Thus, from (5.6.7) and (5.6.9), we have ψ (η) = − 1 2 f (−η) − 1 2c
−η 0 g (τ ) dτ − K 2 , −l ≤ η ≤ 0. (5.6.10) We see that the range of ψ (η) is extended to −l ≤ η ≤ l. If we put α = ξ in equation (5.6.8), we obtain φ (ξ) = −ψ (2l − ξ), ξ ≥ l. (5.6.11) Then, by putting η = 2l − ξ in equation (5.6.3), we obtain ψ (2l − ξ) = 1 2 f (2l − ξ) − 1 2c
2l−ξ 0 g (τ ) dτ − K 2 , 0 ≤ 2l − ξ ≤ l. (5.6.12) Substitution of this in equation (5.6.11) yields 138 5 The Cauchy Problem and Wave Equations φ (ξ) = − 1 2 f (2l − ξ) + 1 2c
2l−ξ 0 g (τ ) dτ + K 2 , l ≤ ξ ≤ 2l. (5.6.13) The range of φ (ξ) is thus extended to 0 ≤ ξ ≤ 2l. Continuing in this manner, we obtain φ (ξ) for all ξ ≥ 0 and ψ (η) for all η ≤ l. Hence, the solution is determined for all 0 ≤ x ≤ l and t ≥ 0. In order to observe the effect of the boundaries on the propagation of waves, the characteristics are drawn through the end point until they meet the boundaries and then continue inward as shown in Figure 5.6.1. It can be seen from the figure that only direct waves propagate in region 1. In regions 2 and 3, both direct and reflected waves propagate. In regions, 4,5,6, ... , several waves propagate along the characteristics reflected from both of the boundaries x = 0 and x = l. Example 5.6.1. Determine the solution of the following problem utt = c 2uxx, 0 < x < l, t > 0, u (x, 0) = sin (πx/l), 0 ≤ x ≤ l, ut (x, 0) = 0, 0 ≤ x ≤ l, u (0, t)=0, u (l, t)=0, t ≥ 0. From equations (5.6.2) and (5.6.3), we have Figure 5.6.1 Regions of wave propagation. 5.7 Nonhomogeneous Wave Equations 139 φ (ξ) = 1 2 sin  πξ l  + K 2 , 0 ≤ ξ ≤ l. ψ (η) = 1 2 sin
4πη l
5 − K 2 , 0 ≤ η ≤ l. Using equation (5.6.10), we obtain ψ (η) = − 1 2 sin
4 − πη l
5 − K 2 , −l ≤ η ≤ 0 = 1 2 sin
4πη l
5 − K 2 . From equation (5.6.13), we find φ (ξ) = − 1 2 sin (π l (2l − ξ) ) + K 2 , l ≤ ξ ≤ 2l. Again by equation (5.6.7) and from the preceding φ (ξ), we have φ (η) = 1 2 sin
4πη l
5 − K 2 , −2l ≤ η ≤ −l. Proceeding in this manner, we determine the solution u (x, t) = φ (ξ) + ψ (η) = 1 2 " sin π l (x + ct) + sin π l (x − ct) # for all x in (0, l) and for all t > 0. Similarly, the solution of the finite initial boundary-value problem utt = c 2uxx, 0 < x < l, t > 0, u (x, 0) = f (x), 0 ≤ x ≤ l, ut (x, 0) = g (x), 0 ≤ x ≤ l, u (0, t) = p (t), u (l, t) = q (t), t ≥ 0, can be determined by the same method. 5.7 Nonhomogeneous Wave Equations We shall consider next the Cauchy problem for the nonhomogeneous wave equation utt = c 2uxx + h ∗ (x, t), (5.7.1) with the initial conditions u (x, 0) = f (x), ut (x, 0) = g ∗ (x). (5.7.2) 140 5 The Cauchy Problem and Wave Equations By the coordinate transformation y = ct, (5.7.3) the problem is reduced to uxx − uyy = h (x, y), (5.7.4) u (x, 0) = f (x), (5.7.5) uy (x, 0) = g (x), (5.7.6) where h (x, y) = −h ∗/c2 and g (x) = g ∗/c. Let P0 (x0, y0) be a point of the plane, and let Q0 be the point (x0, 0) on the initial line y = 0. Then the characteristics, x + y = constant, of equation (5.7.4) are two straight lines drawn through the point P0 with slopes + 1. Obviously, they intersect the x-axis at the points P1 (x0 − y0, 0) and P2 (x0 + y0, 0), as shown in Figure 5.7.1. Let the sides of the triangle P0P1P2 be designated by B0, B1, and B2, and let D be the region representing the interior of the triangle and its boundaries B. Integrating both sides of equation (5.7.4), we obtain

R (uxx − uyy) dR =

R h (x, y) dR. (5.7.7) Now we apply Green’s theorem to obtain Figure 5.7.1 Triangular Region. 5.7 Nonhomogeneous Wave Equations 141

R (uxx − uyy) dR =  B (uxdy + uydx). (5.7.8) Since B is composed of B0, B1, and B2, we note that
B0 (ux dy + uy dx) =
x0+y0 x0−y0 uy dx,
B1 (ux dy + uy dx) =
B1 (−ux dx − uy dy), = u (x0 + y0, 0) − u (x0, y0),
B2 (ux dy + uy dx) =
B2 (ux dx + uy dy), = u (x0 − y0, 0) − u (x0, y0). Hence,  B (ux dy + uy dx) = −2 u (x0, y0) + u (x0 − y0, 0) +u (x0 + y0, 0) +
x0+y0 x0−y0 uy dx. (5.7.9) Combining equations (5.7.7), (5.7.8) and (5.7.9), we obtain u (x0, y0) = 1 2 [u (x0 + y0, 0) + u (x0 − y0, 0)] + 1 2
x0+y0 x0−y0 uy dx − 1 2

R h (x, y) dR. (5.7.10) We have chosen x0, y0 arbitrarily, and as a consequence, we replace x0 by x and y0 by y. Equation (5.7.10) thus becomes u (x, y) = 1 2 [f (x + y) + f (x − y)] + 1 2
x+y x−y g (τ ) dτ − 1 2

R h (x, y) dR. In terms of the original variables u (x, t) = 1 2 [f (x + ct) + f (x − ct)] + 1 2c
x+ct x−ct g ∗ (τ ) dτ − 1 2

R h (x, t) dR. (5.7.11) Example 5.7.1. Determine the solution of uxx − uyy = 1, u (x, 0) = sin x, uy (x, 0) = x. 142 5 The Cauchy Problem and Wave Equations Figure 5.7.2 Triangular Region. It is easy to see that the characteristics are x + y = constant = x0 + y0 and x − y = constant = x0 − y0, as shown in Figure 5.7.2. Thus, u (x0, y0) = 1 2 [sin (x0 + y0) + sin (x0 − y0)] + 1 2
x0+y0 x0−y0 τ dτ − 1 2
y0 0
−y+x0+y0 y+x0−y0 dx dy = 1 2 [sin (x0 + y0) + sin (x0 − y0)] + x0y0 − 1 2 y 2 0 . Now dropping the subscript zero, we obtain the solution u (x, y) = 1 2 [sin (x + y) + sin (x − y)] + xy − 1 2 y 2 . 5.8 The Riemann Method We shall discuss Riemann’s method of integrating the linear hyperbolic equation L[u] ≡ uxy + aux + buy + cu = f (x, y), (5.8.1) 5.8 The Riemann Method 143 where L denotes the linear operator, and a (x, y), b (x, y), c (x, y), and f (x, y) are differentiable functions in some domain D∗ . The method consists essentially of the derivation of an integral formula which represents the solution of the Cauchy problem. Let v (x, y) be a function having continuous second-order partial derivatives. Then, we may write vuxy − uvxy = (vux)y − (vuy)x , vaux = (avu)x − u (av)x , (5.8.2) vbuy = (bvu)y − u (bv)y , so that vL[u] − uM [v] = Ux + Vy, (5.8.3) where M is the operator represented by M [v] = vxy − (av)x − (bv)y + cv, (5.8.4) and U = auv − uvy, V = buv + vux. (5.8.5) The operator M is called the adjoint operator of L. If M = L, then the operator L is said to be self-adjoint. Now applying Green’s theorem, we have

D (Ux + Vy) dx dy =  C (U dy − V dx), (5.8.6) where C is the closed curve bounding the region of integration D which is in D∗ . Let Λ be a smooth initial curve which is continuous, as shown in Figure 5.8.1. Since equation (5.8.1) is in first canonical form, x and y are the characteristic coordinates. We assume that the tangent to Λ is nowhere parallel to the x or y axis. Let P (α, β) be a point at which the solution to the Cauchy problem is sought. Line P Q parallel to the x axis intersects the initial curve Λ at Q, and line P R parallel to the y axis intersects the curve Λ at R. We suppose that u and ux or uy are prescribed along Λ. Let C be the closed contour P QRP bounding D. Since dy = 0 on P Q and dx = 0 on P R, it follows immediately from equations (5.8.3) and (5.8.6) that

D (vL[u] − uM [v]) dx dy =
R Q (U dy − V dx) +
P R U dy −
Q P V dx. (5.8.7) 144 5 The Cauchy Problem and Wave Equations Figure 5.8.1 Smooth initial curve. From equation (5.8.5), we find
Q P V dx =
Q P bvu dx +
Q P vux dx. Integrating by parts, we obtain
Q P vuxdx = [uv] Q P −
Q P uvxdx. Hence, we may write
Q P V dx = [uv] Q P +
Q P u (bv − vx) dx. Substitution of this integral in equation (5.8.7) yields [uv]P = [uv]Q +
Q P u (bv − vx) dx −
P R u (av − vy) dy −
R Q (U dy − V dx) +

D (vL[u] − uM [v]) dx dy. (5.8.8) Suppose we can choose the function v (x, y; α, β) to be the solution of the adjoint equation 5.8 The Riemann Method 145 M [v]=0, (5.8.9) satisfying the conditions vx = bv when y = β, vy = av when x = α, (5.8.10) v = 1 when x = α and y = β. The function v (x, y; α, β) is called the Riemann function. Since L[u] = f, equation (5.8.8) reduces to, [u]P = [uv]Q −
R Q uv (a dy − b dx) +
R Q (uvydy + vuxdx) +

D vf dx dy. (5.8.11) This gives us the value of u at the point P when u and ux are prescribed along the curve Λ. When u and uy are prescribed, the identity [uv]R − [uv]Q =
R Q ( (uv)x dx + (uv)y dy) , may be used to put equation (5.8.8) in the form [u]P = [uv]R −
R Q uv (a dy − b dx) −
R Q (uvxdx + vuydy) +

D vf dx dy. (5.8.12) By adding equations (5.8.11) and (5.8.12), the value of u at P is given by [u]P = 1 2
4 [uv]Q + [uv]R
5 −
R Q uv (a dy − b dx) − 1 2
R Q u (vxdx − vydy) + 1 2
R Q v (uxdx − uydy) +

D vf dx dy (5.8.13) which is the solution of the Cauchy problem in terms of the Cauchy data given along the curve Λ. It is easy to see that the solution at the point (α, β) depends only on the Cauchy data along the arc QR on Λ. If the initial data were to change outside this arc QR, the solution would change only outside the triangle P QR. Thus, from Figure 5.8.2, we can see that each characteristic separates the region in which the solution remains unchanged from the region in which it varies. Because of this fact, the unique continuation of the solution across any characteristic is not possible. This is evident from Figure 5.8.2. The solution on the right of the characteristic P1R1 is determined by the initial data given in Q1R2, whereas the solution 146 5 The Cauchy Problem and Wave Equations Figure 5.8.2 Solution on the right and left of the characteristic. on the left is determined by the initial data given on Q1R1. If the initial data on R1R2 were changed, the solution on the right of P1R1 only will be affected. It should be remarked here that the initial curve can intersect each characteristic at only one point. Suppose, for example, the initial curve Λ intersects the characteristic at two points, as shown in Figure 5.8.3. Then, the solution at P obtained from the initial data on QR will be different from the solution obtained from the initial data on RS. Hence, the Cauchy problem, in this case, is not solvable. Figure 5.8.3 Initial curve intersects the characteristic at two points. 5.8 The Riemann Method 147 Example 5.8.1. The telegraph equation wtt + a ∗wt + b ∗w = c 2wxx, may be transformed into canonical form L[u] = uξη + ku = 0, by the successive transformations w = u e−a ∗ t/2 , and ξ = x + ct, η = x − ct, where k =  a ∗2 − 4b ∗ /16c 2 . We apply Riemann’s method to determine the solution satisfying the initial conditions u (x, 0) = f (x), ut (x, 0) = g (x). Since t = 1 2c (ξ − η), the line t = 0 corresponds to the straight line ξ = η in the ξ − η plane. The initial conditions may thus be transformed into [u] ξ=η = f (ξ), (5.8.14) [uξ − uη] ξ=η = c −1 g (ξ). (5.8.15) We next determine the Riemann function v (ξ,η; α, β) which satisfies vξη + kv = 0, (5.8.16) vξ (ξ, β; α, β)=0, (5.8.17) vη (α, η; α, β)=0, (5.8.18) v (α, β; α, β)=1. (5.8.19) The differential equation (5.8.16) is self-adjoint, that is, L[v] = M [v] = vξη + kv. We assume that the Riemann function is of the form v (ξ,η; α, β) = F (s), with the argument s = (ξ − α) (η − β). Substituting this value in equation (5.8.16), we obtain 148 5 The Cauchy Problem and Wave Equations sFss + Fs + kF = 0. If we let λ = √ 4ks, the above equation becomes F ′′ (λ) + 1 λ F ′ (λ) + F (λ)=0. This is the Bessel equation of order zero, and the solution is F (λ) = J0 (λ), disregarding Y0 (λ) which is unbounded at λ = 0. Thus, the Riemann function is v (ξ,η; α, β) = J0
4 4k (ξ − α) (η − β)
5 which satisfies equation (5.8.16) and is equal to one on the characteristics ξ = α and η = β. Since J ′ 0 (0) = 0, equations (5.8.17) and (5.8.18) are satisfied. From this, it immediately follows that [vξ] ξ=η = √ k (ξ − β)  (ξ − α) (η − β) [J ′ 0 (λ)]ξ=η , [vη] ξ=η = √ k (ξ − α)  (ξ − α) (η − β) [J ′ 0 (λ)]ξ=η . Thus, we have [vξ − uη] ξ=η = √ k (α − β)  (ξ − α) (ξ − β) [J ′ 0 (λ)]ξ=η . (5.8.20) From the initial condition u (Q) = f (β) and u (R) = f (α), (5.8.21) and substituting equations (5.8.15), (5.8.19), and (5.8.20) into equation (5.8.13), we obtain u (α, β) = 1 2 [f (α) + f (β)] − 1 2
α β √ k (α − β)  (τ − α) (τ − β) J ′ 0
4 4k (τ − α) (τ − β)
5 f (τ ) dτ + 1 2c
α β J0
4 4k (τ − α) (τ − β)
5 g (τ ) dτ. (5.8.22) Replacing α and β by ξ and η, and substituting the original variables x and t, we obtain 5.9 Solution of the Goursat Problem 149 u (x, t) = 1 2 [f (x + ct) + f (x − ct)] + 1 2
x+ct x−ct G (x, t, τ ) dτ, (5.8.23) where G (x, t, τ ) = / −2 √ k ctf (τ ) J0 42 4k " (τ − x) 2 − c 2t 2 # 501% (τ − x) 2 − c 2t 2 + c −1 g (τ ) J0 42 4k " (τ − x) 2 − c 2t 2 # 5 . If we set k = 0, we arrive at the d’Alembert solution for the wave equation u (x, t) = 1 2 [f (x + ct) + f (x − ct)] + 1 2c
x+ct x−ct g (τ ) dτ. 5.9 Solution of the Goursat Problem The Goursat problem is that of finding the solution of a linear hyperbolic equation uxy = a1 (x, y) ux + a2 (x, y) uy + a3 (x, y) u + h (x, y), (5.9.1) satisfying the prescribed conditions u (x, y) = f (x), (5.9.2) on a characteristic, say, y = 0, and u (x, y) = g (x) (5.9.3) on a monotonic increasing curve y = y (x) which, for simplicity, is assumed to intersect the characteristic at the origin. The solution in the region between the x-axis and the monotonic curve in the first quadrant can be determined by the method of successive approximations. The proof is given in Garabedian (1964). Example 5.9.1. Determine the solution of the Goursat problem utt = c 2uxx, (5.9.4) u (x, t) = f (x), on x − ct = 0, (5.9.5) u (x, t) = g (x), on t = t(x), (5.9.6) where f (0) = g (0). 150 5 The Cauchy Problem and Wave Equations The general solution of the wave equation is u (x, t) = φ (x + ct) + ψ (x − ct). Applying the prescribed conditions, we obtain f (x) = φ (2x) + ψ (0), (5.9.7) g (x) = φ (x + c t(x)) + ψ (x − c t(x)). (5.9.8) It is evident that f (0) = φ (0) + ψ (0) = g (0). Now, if s = x − c t(x), the inverse of it is x = α (s). Thus, equation (5.9.8) may be written as g (α (s)) = φ (x + c t(x)) + ψ (s). (5.9.9) Replacing x by (x + c t(x)) /2 in equation (5.9.7), we obtain f  x + c t(x) 2  = φ (x + c t(x)) + ψ (0). (5.9.10) Thus, using (5.9.10), equation (5.9.9) becomes ψ (s) = g (α (s)) − f  α (s) + c t(α (s)) 2  + ψ (0). Replacing s by x − ct , we have ψ (x − ct) = g (α (x − ct)) − f  α (x − c t) + c t(α (x − c t)) 2  + ψ (0). Hence, the solution is given by u (x, t) = f  x + c t 2  − f  α (x − c t) + c t(α (x − c t)) 2  + g (α (x − c t)). (5.9.11) Let us consider a special case when the curve t = t(x) is a straight line represented by t − kx = 0 with a constant k > 0. Then s = x − ckx and hence x = s/ (1 − ck). Using these values in (5.9.11), we obtain u (x, t) = f  x + c t 2  − f  (1 + c k) (x − c t) 2 (1 − c k)  + g  x − c t 1 − c k  . (5.9.12) When the values of u are prescribed on both characteristics, the problem of finding u of a linear hyperbolic equation is called a characteristic initialvalue problem. This is a degenerate case of the Goursat problem. 5.9 Solution of the Goursat Problem 151 Consider the characteristic initial-value problem uxy = h (x, y), (5.9.13) u (x, 0) = f (x), (5.9.14) u (0, y) = g (y), (5.9.15) where f and g are continuously differentiable, and f (0) = g (0). Integrating equation (5.9.13), we obtain u (x, y) =
x 0
y 0 h (ξ,η) dη dξ + φ (x) + ψ (y), (5.9.16) where φ and ψ are arbitrary functions. Applying the prescribed conditions (5.9.14) and (5.9.15), we have u (x, 0) = φ (x) + ψ (0) = f (x), (5.9.17) u (0, y) = φ (0) + ψ (y) = g (y). (5.9.18) Thus, φ (x) + ψ (y) = f (x) + g (y) − φ (0) − ψ (0). (5.9.19) But from (5.9.17), we have φ (0) + ψ (0) = f (0). (5.9.20) Hence, from (5.9.16), (5.9.19) and (5.9.20), we obtain u (x, y) = f (x) + g (y) − f (0) +
x 0
y 0 h (ξ,η) dη dξ. (5.9.21) Example 5.9.2. Determine the solution of the characteristic initial-value problem utt = c 2uxx, u (x, t) = f (x) on x + ct = 0, u (x, t) = g (x) on x − ct = 0, where f (0) = g (0). Here it is not necessary to reduce the given equation to canonical form. The general solution of the wave equation is u (x, t) = φ (x + ct) + ψ (x − ct). The characteristics are x + ct = 0, x − ct = 0. 152 5 The Cauchy Problem and Wave Equations Applying the prescribed conditions, we have u (x, t) = φ (2x) + ψ (0) = f (x) on x + ct = 0, (5.9.22) u (x, t) = φ (0) + ψ (2x) = g (x) on x − ct = 0. (5.9.23) We observe that these equations are compatible, since f (0) = g (0). Now, replacing x by (x + ct) /2 in equation (5.9.22) and replacing x by (x − ct) /2 in equation (5.9.23), we have φ (x + ct) = f  x + ct 2  − ψ (0), φ (x − ct) = g  x − ct 2  − φ (0). Hence, the solution is given by u (x, t) = f  x + ct 2  + g  x − ct 2  − f (0). (5.9.24) We note that this solution can be obtained by substituting k = −1/c into (5.9.12). Example 5.9.3. Find the solution of the characteristic initial-value problem y 3uxx − yuyy + uy = 0, (5.9.25) u (x, y) = f (x) on x + y 2 2 = 4 for 2 ≤ x ≤ 4, u (x, y) = g (x) on x − y 2 2 = 0 for 0 ≤ x ≤ 2, with f (2) = g (2). Since the equation is hyperbolic except for y = 0, we reduce it to the canonical form uξη = 0, where ξ = x +  y 2/2 and η = x −  y 2/2 . Thus, the general solution is u (x, y) = φ  x + y 2 2  + ψ  x − y 2 2  . (5.9.26) Applying the prescribed conditions, we have f (x) = φ (4) + ψ (2x − 4), (5.9.27) g (x) = φ (2x) + ψ (0). (5.9.28) Now, if we replace (2x − 4) by  x − y 2/2 in (5.9.27) and (2x) by  x + y 2/2 in (5.9.28), we obtain 5.10 Spherical Wave Equation 153 ψ  x − y 2 2  = f  x 2 − y 2 4 + 2 − φ (4), φ  x + y 2 2  = g  x 2 + y 2 4  − ψ (0). Thus, u (x, y) = f  x 2 − y 2 4 + 2 + g  x 2 + y 2 4  − φ (4) − ψ (0). But from (5.9.27) and (5.9.28), we see that f (2) = φ (4) + ψ (0) = g (2). Hence, u (x, y) = f  x 2 − y 2 4 + 2 + g  x 2 + y 2 4  − f (2). 5.10 Spherical Wave Equation In spherical polar coordinates (r, θ, φ), the wave equation (3.1.1) takes the form 1 r 2 ∂ ∂r  r 2 ∂u ∂r  + 1 r 2 sin θ ∂ ∂θ  sin θ ∂u ∂θ  + 1 r 2 sin2 θ ∂ 2u ∂φ2 = 1 c 2 ∂ 2u ∂t2 .(5.10.1) Solutions of this equation are called spherical symmetric waves if u depends on r and t only. Thus, the solution u = u (r, t) which satisfies the wave equation with spherical symmetry in three-dimensional space is 1 r 2 ∂ ∂r  r 2 ∂u ∂r  = 1 c 2 ∂ 2u ∂t2 . (5.10.2) Introducing a new dependent variable U = ru (r, t), this equation reduces to a simple form Utt = c 2Urr. (5.10.3) This is identical with the one-dimensional wave equation (5.3.1) and has the general solution in the form U (r, t) = φ (r + ct) + ψ (r − ct), (5.10.4) or, equivalently, u (r, t) = 1 r [φ (r + ct) + ψ (r − ct)] . (5.10.5) 154 5 The Cauchy Problem and Wave Equations This solution consists of two progressive spherical waves traveling with constant velocity c. The terms involving φ and ψ represent the incoming waves to the origin and the outgoing waves from the origin respectively. Physically, the solution for only outgoing waves generated by a source is of most interest, and has the form u (r, t) = 1 r ψ (r − ct), (5.10.6) where the explicit form of ψ is to be determined from the properties of the source. In the context of fluid flows, u represents the velocity potential so that the limiting total flux through a sphere of center at the origin and radius r is Q (t) = limr→0 4πr2ur (r, t) = −4π ψ (−ct). (5.10.7) In physical terms, we say that there is a simple (or monopole) point source of strength Q (t) located at the origin. Thus, the solution (5.10.6) can be expressed in terms of Q as u (r, t) = − 1 4πr Q
4 t − r c
5 . (5.10.8) This represents the velocity potential of the point source, and ur is called the radial velocity. In fluid flows, the difference between the pressure at any time t and the equilibrium value is given by p − p0 = ρ ut = − ρ 4πr Q˙
4 t − r c
5 , (5.10.9) where ρ is the density of the fluid. Following an analysis similar to Section 5.3, the solution of the initialvalue problem with the initial data u (r, 0) = f (r), ut (r, 0) = g (r), r ≥ 0, (5.10.10) where f and g are continuously differentiable, is given by u (r, t) = 1 2r (r + ct) f (r + ct)+(r − ct) f (r − ct) + 1 c
r+ct r−ct τg (τ ) dτ , (5.10.11) provided r ≥ ct. However, when r < ct, this solution fails because f and g are not defined for r < 0. This initial data at t = 0, r ≥ 0 determine the solution u (r, t) only up to the characteristic r = ct in the r-t plane. To find u for r < ct, we require u to be finite at r = 0 for all t ≥ 0, that is, U = 0 at r = 0. Thus, the solution for U (r, t) is 5.11 Cylindrical Wave Equation 155 U (r, t) = 1 2 (r + ct) f (r + ct)+(r − ct) f (r − ct) + 1 c
r+ct r−ct τg (τ ) dτ , (5.10.12) provided r ≥ ct ≥ 0, and U (r, t) = 1 2 [φ (ct + r) + ψ (ct − r)] , ct ≥ r ≥ 0, (5.10.13) where φ (ct) + ψ (ct)=0, for ct ≥ 0. (5.10.14) In view of the fact that Ur + 1 c Ut is constant on each characteristic r + ct = constant, it turns out that φ ′ (ct + r)=(r + ct) f ′ (r + ct) + f (r + ct) + 1 c (r + ct) g (r + ct), or φ ′ (ct) = ctf′ (ct) + f (ct) + t g (ct). Integration gives φ (t) = tf (t) + 1 c
t 0 τg (τ ) dτ + φ (0), so that ψ (t) = −tf (t) − 1 c
t 0 τg (τ ) dτ − φ (0). Substituting these values into (5.10.13) and using U (r, t) = ru (r, t), we obtain, for ct > r, u (r, t) = 1 2r (ct + r) f (ct + r) − (ct − r) f (ct − r) + 1 c
ct+r ct−r τg (τ ) dτ . (5.10.15) 5.11 Cylindrical Wave Equation In cylindrical polar coordinates (R, θ, z), the wave equation (3.1.1) assumes the form uRR + 1 R uR + 1 R2 uθθ + uzz = 1 c 2 utt. (5.11.1) If u depends only on R and t, this equation becomes 156 5 The Cauchy Problem and Wave Equations uRR + 1 R uR = 1 c 2 utt. (5.11.2) Solutions of (5.11.2) are called cylindrical waves. In general, it is not easy to find the solution of (5.11.1). However, we shall solve this equation by using the method of separation of variables in Chapter 7. Here we derive the solution for outgoing cylindrical waves from the spherical wave solution (5.10.8). We assume that sources of constant strength Q (t) per unit length are distributed uniformly on the z-axis. The solution for the cylindrical waves produced by the line source is given by the total disturbance u (R, t) = − 1 4π
∞ −∞ 1 r Q
4 t − r c
5 dz = − 1 2π
∞ 0 1 r Q
4 t − r c
5 dz, (5.11.3) where R is the distance from the z-axis so that R2 =  r 2 − z 2 . Substitution of z = R sinh ξ and r = R cosh ξ in (5.11.3) gives u (R, t) = − 1 2π
∞ 0 Q  t − R c cosh ξ  dξ. (5.11.4) This is usually considered as the cylindrical wave function due to a source of strength Q (t) at R = 0. It follows from (5.11.4) that utt = − 1 2π
∞ 0 Q ′′  t − R c cosh ξ  dξ, (5.11.5) uR = 1 2πc
∞ 0 cosh ξ Q′  t − R c cosh ξ  dξ, (5.11.6) uRR = − 1 2πc2
∞ 0 cosh2 ξ Q′′  t − R c cosh ξ  dξ, (5.11.7) which give c 2  uRR + 1 R uR  − utt = 1 2π
∞ 0 d dξ c R Q ′  t − R c cosh ξ  sinh ξ dξ = lim ξ→∞ c 2πR Q ′  t − R c cosh ξ  sinh ξ = 0, provided the differentiation under the sign of integration is justified and the above limit is zero. This means that u (R, t) satisfies the cylindrical wave equation (5.11.2). In order to find the asymptotic behavior of the solution as R → 0, we substitute cosh ξ = c(t−ζ) R into (5.11.4) and (5.11.6) to obtain u = − 1 2π
t−R/c −∞ Q (ζ) dζ " (t − ζ) 2 − R2 c 2 # 1 2 , (5.11.8) uR = 1 2π
t−R/c −∞  t − ζ R  Q′ (ζ) dζ " (t − ζ) 2 − R2 c 2 # 1 2 , (5.11.9) 5.11 Cylindrical Wave Equation 157 which, in the limit R → 0, give uR ∼ 1 2πR
t −∞ Q ′ (ζ) dζ = 1 2πR Q (t). (5.11.10) This leads to the result lim R→0 2πR uR = Q (t), (5.11.11) or u (R, t) ∼ 1 2π Q (t) log R as R → 0. (5.11.12) We next investigate the nature of the cylindrical wave solution near the waterfront (R = ct) and in the far field (R → ∞). We assume Q (t) = 0 for t < 0 so that the lower limit of integration in (5.11.8) may be taken to be zero, and the solution is non-zero for τ = t − R c > 0, where τ is the time passed after the arrival of the wavefront. Consequently, (5.11.8) becomes u (R, t) = − 1 2π
τ 0 Q (ζ) dζ (t − ζ)  t − ζ + 2R c ! 1 2 . (5.11.13) Since 0 <ζ<τ , 2R c > R c >τ>τ − ζ > 0, so that the second factor under the radical is approximately equal to 2R c when R ≫ cτ , and hence, u (R, t) ∼ − 1 2π
4 c 2R
5 1 2
τ 0 Q (ζ) dζ (t − ζ) 1 2 = −
4 c 2R
5 1 2 q (τ ) = −
4 c 2R
5 1 2 q  t − R c  , R ≫ ct 2 , (5.11.14) where q (τ ) = 1 2π
τ 0 Q (ζ) dζ √ τ − ζ . (5.11.15) Evidently, the amplitude involved in the solution (5.11.14) decays like R− 1 2 for large R (R → ∞). Example 5.11.1. Determine the asymptotic form of the solution (5.11.4) for a harmonically oscillating source of frequency ω. We take the source in the form Q (t) = q0 exp [−i(ω + iε)t], where ε is positive and small so that Q (t) → 0 as t → −∞. The small imaginary part ε of ω will make insignificant contributions to the solution at finite time as ε → 0. Thus, the solution (5.11.4) becomes u (R, t) = −
4 q0 2π
5 e −iωt
∞ 0 exp  iωR c cosh ξ  dξ = −  iq0 4  e −iωtH (1) 0  ωR c  , (5.11.16) 158 5 The Cauchy Problem and Wave Equations where H (1) 0 (z) is the Hankel function given by H (1) 0 (z) = 2 πi
∞ 0 exp (iz cosh ξ) dξ. (5.11.17) In view of the asymptotic expansion of H (1) 0 (z) in the form H (1) 0 (z) ∼  2 πz 1 2 exp " i
4 z − π 4
5# , z → ∞, (5.11.18) the asymptotic solution for u (R, t) in the limit  ωR c → ∞ is u (R, t) ∼ −  iq0 4  2c πωR1 2 exp −i  ωt − ωR c − π 4  . This represents the cylindrical wave propagating with constant velocity c. The amplitude of the wave decays like R− 1 2 as R → ∞. Example 5.11.2. For a supersonic flow (M > 1) past a solid body of revolution, the perturbation potential Φ satisfies the cylindrical wave equation ΦRR + 1 R ΦR = N 2Φxx, N2 = M2 − 1, where R is the distance from the path of the moving body and x is the distance from the nose of the body. It follows from problem 12 in 3.9 Exercises that Φ satisfies the equation Φyy + Φzz = N 2 Φxx. This represents a two-dimensional wave equation with x ↔ t and N2 ↔ 1 c 2 . For a body of revolution with (y, z) ↔ (R, θ), ∂ ∂θ ≡ 0, the above equation reduces to the cylindrical wave equation ΦRR + 1 R ΦR = 1 c 2 Φtt. 5.12 Exercises 1. Determine the solution of each of the following initial-value problems: (a) utt − c 2uxx = 0, u (x, 0) = 0, ut (x, 0) = 1. (b) utt − c 2uxx = 0, u (x, 0) = sin x, ut (x, 0) = x 2 . (c) utt − c 2uxx = 0, u (x, 0) = x 3 , ut (x, 0) = x. 5.12 Exercises 159 (d) utt − c 2uxx = 0, u (x, 0) = cos x, ut (x, 0) = e −1 . (e) utt − c 2uxx = 0, u (x, 0) = log  1 + x 2 , ut (x, 0) = 2. (f) utt − c 2uxx = 0, u (x, 0) = x, ut (x, 0) = sin x. 2. Determine the solution of each of the following initial-value problems: (a) utt − c 2uxx = x, u (x, 0) = 0, ut (x, 0) = 3. (b) utt − c 2uxx = x + ct, u (x, 0) = x, ut (x, 0) = sin x. (c) utt − c 2uxx = e x , u (x, 0) = 5, ut (x, 0) = x 2 . (d) utt − c 2uxx = sin x, u (x, 0) = cos x, ut (x, 0) = 1 + x. (e) utt − c 2uxx = xet , u (x, 0) = sin x, ut (x, 0) = 0. (f) utt − c 2uxx = 2, u (x, 0) = x 2 , ut (x, 0) = cos x. 3. A gas which is contained in a sphere of radius R is at rest initially, and the initial condensation is given by s0 inside the sphere and zero outside the sphere. The condensation is related to the velocity potential by s (t) =  1/c2 ut, at all times, and the velocity potential satisfies the wave equation utt = ∇2u. Determine the condensation s (t) for all t > 0. 4. Solve the initial-value problem uxx + 2uxy − 3uyy = 0, u (x, 0) = sin x, uy (x, 0) = x. 5. Find the longitudinal oscillation of a rod subject to the initial conditions u (x, 0) = sin x, ut (x, 0) = x. 6. By using the Riemann method, solve the following problems: (a) sin2 µ φxx − cos2 µ φyy −  λ 2 sin2 µ cos2 µ φ = 0, φ (0, y) = f1 (y), φ (x, 0) = g1 (x), φx (0, y) = f2 (y), φy (x, 0) = g2 (x). 160 5 The Cauchy Problem and Wave Equations (b) x 2uxx − t 2utt = 0, u (x, t1) = f (x), ut (x, t2) = g (x). 7. Determine the solution of the initial boundary-value problem utt = 4 uxx, 0 <x< ∞,="" t=""> 0, u (x, 0) = x 4 , 0 ≤ x < ∞, ut (x, 0) = 0, 0 ≤ x < ∞, u (0, t)=0, t ≥ 0. 8. Determine the solution of the initial boundary-value problem utt = 9 uxx, 0 <x< ∞,="" t=""> 0, u (x, 0) = 0, 0 ≤ x < ∞, ut (x, 0) = x 3 , 0 ≤ x < ∞, ux (0, t)=0, t ≥ 0. 9. Determine the solution of the initial boundary-value problem utt = 16 uxx, 0 <x< ∞,="" t=""> 0, u (x, 0) = sin x, 0 ≤ x < ∞, ut (x, 0) = x 2 , 0 ≤ x < ∞, u (0, t)=0, t ≥ 0. 10. In the initial boundary-value problem utt = c 2uxx, 0 < x < l, t > 0, u (x, 0) = f (x), 0 ≤ x ≤ l, ut (x, 0) = g (x), 0 ≤ x ≤ l, u (0, t)=0, t ≥ 0, if f and g are extended as odd functions, show that u (x, t) is given by the solution (5.4.5) for x > ct and solution (5.4.6) for x < ct. 11. In the initial boundary-value problem utt = c 2uxx, 0 < x < l, t > 0, u (x, 0) = f (x), 0 ≤ x ≤ l, ut (x, 0) = g (x), 0 ≤ x ≤ l, ux (0, t)=0, t ≥ 0, if f and g are extended as even functions, show that u (x, t) is given by solution (5.4.8) for x > ct, and solution (5.4.9) for x < ct. 5.12 Exercises 161 12. Determine the solution of the initial boundary-value problem utt = c 2uxx, 0 <x< ∞,="" t=""> 0, u (x, 0) = f (x), 0 ≤ x < ∞, ut (x, 0) = 0, 0 ≤ x < ∞, ux (0, t) + h u (0, t)=0, t ≥ 0, h = constant. State the compatibility condition of f. 13. Find the solution of the problem utt = c 2uxx, at < x < ∞, t> 0, u (x, 0) = f (x), 0 <x< ∞,="" ut="" (x,="" 0)="0," 0="" <x<="" u="" (at,="" t)="0," t=""> 0, where f (0) = 0 and a is constant. 14. Find the solution of the initial boundary-value problem utt = uxx, 0 <x< 2,="" t=""> 0, u (x, 0) = sin (πx/2), 0 ≤ x ≤ 2, ut (x, 0) = 0, 0 ≤ x ≤ 2, u (0, t)=0, u (2, t)=0, t ≥ 0. 15. Find the solution of the initial boundary-value problem utt = 4 uxx, 0 <x< 1,="" t=""> 0, u (x, 0) = 0, 0 ≤ x ≤ 1, ut (x, 0) = x (1 − x), 0 ≤ x ≤ 1, u (0, t)=0, u (1, t)=0, t ≥ 0. 16. Determine the solution of the initial boundary-value problem utt = c 2uxx, 0 < x < l, t > 0, u (x, 0) = f (x), 0 ≤ x ≤ l, ut (x, 0) = g (x), 0 ≤ x ≤ l, ux (0, t)=0, ux (l, t)=0, t ≥ 0, by extending f and g as even functions about x = 0 and x = l. 17. Determine the solution of the initial boundary-value problem utt = c 2uxx, 0 < x < l, t > 0, u (x, 0) = f (x), 0 ≤ x ≤ l, ut (x, 0) = g (x), 0 ≤ x ≤ l, u (0, t) = p (t), u (l, t) = q (t), t ≥ 0. 162 5 The Cauchy Problem and Wave Equations 18. Determine the solution of the initial boundary-value problem utt = c 2uxx, 0 < x < l, t > 0, u (x, 0) = f (x), 0 ≤ x ≤ l, ut (x, 0) = g (x), 0 ≤ x ≤ l, ux (0, t) = p (t), ux (l, t) = q (t), t ≥ 0. 19. Solve the characteristic initial-value problem xy3uxx − x 3 y uyy − y 3ux + x 3uy = 0, u (x, y) = f (x) on y 2 − x 2 = 8 for 0 ≤ x ≤ 2, u (x, y) = g (x) on y 2 + x 2 = 16 for 2 ≤ x ≤ 4, with f (2) = g (2). 20. Solve the Goursat problem xy3uxx − x 3 y uyy − y 3ux + x 3uy = 0, u (x, y) = f (x) on y 2 + x 2 = 16 for 0 ≤ x ≤ 4, u (x, y) = g (y) on x = 0 for 0 ≤ y ≤ 4, where f (0) = g (4). 21. Solve utt = c 2uxx, u (x, t) = f (x) on t = t(x), u (x, t) = g (x) on x + ct = 0, where f (0) = g (0). 22. Solve the characteristic initial-value problem xuxx − x 3uyy − ux = 0, x = 0, u (x, y) = f (y) on y − x 2 2 = 0 for 0 ≤ y ≤ 2, u (x, y) = g (y) on y + x 2 2 = 4 for 2 ≤ y ≤ 4, where f (2) = g (2). 23. Solve uxx + 10 uxy + 9 uyy = 0, u (x, 0) = f (x), uy (x, 0) = g (x). 5.12 Exercises 163 24. Solve 4 uxx + 5 uxy + uyy + ux + uy = 2, u (x, 0) = f (x), uy (x, 0) = g (x). 25. Solve 3 uxx + 10 uxy + 3 uyy = 0, u (x, 0) = f (x), uy (x, 0) = g (x). 26. Solve uxx − 3 uxy + 2 uyy = 0, u (x, 0) = f (x), uy (x, 0) = g (x). 27. Solve x 2uxx − t 2utt = 0 x > 0, t> 0, u (x, 1) = f (x), ut (x, 1) = g (x). 28. Consider the initial boundary-value problem for a string of length l under the action of an external force q (x, t) per unit length. The displacement u (x, t) satisfies the wave equation ρ utt = T uxx + ρ q (x, t), where ρ is the line density of the string and T is the constant tension of the string. The initial and boundary conditions of the problem are u (x, 0) = f (x), ut (x, 0) = g (x), 0 ≤ x ≤ l, u (0, t) = u (l, t)=0, t > 0. Show that the energy equation is dE dt = [T uxut] l 0 +
l 0 ρqut dx, where E represents the energy integral E (t) = 1 2
l 0  ρ u2 t + T u2 x dx. Explain the physical significance of the energy equation. Hence or otherwise, derive the principle of conservation of energy, that is, that the total energy is constant for all t ≥ 0 provided that the string has free or fixed ends and there are no external forces. 164 5 The Cauchy Problem and Wave Equations 29. Show that the solution of the signaling problem governed by the wave equation utt = c 2uxx, x > 0, t > 0, u (x, 0) = ut (x, 0) = 0, x > 0, u (0, t) = U (t), t > 0, is u (x, t) = U
4 t − x c
5 H
4 t − x c
5 , where H is the Heaviside unit step function. 30. Obtain the solution of the initial-value problem of the homogeneous wave equation utt − c 2uxx = sin (kx − ωt), −∞ <x< ∞,="" t=""> 0, u (x, 0) = 0 = ut (x, 0), for all x ∈ R, where c, k and ω are constants. Discuss the non-resonance case, ω = ck and the resonance case, ω = ck. 31. In each of the following Cauchy problems, obtain the solution of the system utt − c 2uxx = 0, x ∈ R, t> 0, u (x, 0) = f (x) and ut (x, 0) = g (x) for x ∈ R, for the given c, f (x) and g (x): (a) c = 3, f (x) = cos x, g (x) = sin 2x. (b) c = 1, f (x) = sin 3x, g (x) = cos 3x. (c) c = 7, f (x) = cos 3x, g (x) = x. (d) c = 2, f (x) = cosh x, g (x)=2x. (e) c = 3, f (x) = x 3 , g (x) = x cos x. (f) c = 4, f (x) = cos x, g (x) = xe−x . 32. If u (x, t) is the solution of the nonhomogeneous Cauchy problem utt − c 2uxx = p (x, t), for x ∈ R, t> 0, u (x, 0) = 0 = ut (x, 0), for x ∈ R, 5.12 Exercises 165 and if v (x, t, τ ) is the solution of the nonhomogeneous Cauchy problem vtt − c 2 vxx = 0, for x ∈ R, t> 0, v (x, 0; τ )=0, vt (x, 0; τ ) = p (x, τ ), x ∈ R, show that u (x, t) =
t 0 v (x, t; τ ) dτ. This is known as the Duhamel principle for the wave equation. 33. Show that the solution of the nonhomogeneous diffusion equation with homogeneous boundary and initial data ut = κuxx + p (x, t), 0 < x < l, t > 0, u (0, t)=0= u (l, t), t > 0, u (x, 0) = 0, 0 < x < l, is u (x, t) =
t 0 v (x, t; τ ) dτ, where v = v (x, t; τ ) satisfies the homogeneous diffusion equation with nonhomogeneous boundary and initial data vtt = κvxx + p (x, t), 0 < x < l, t > 0, v (0, t; τ )=0= v (l, t; τ ), t > 0, v (x, τ ; τ ) = p (x, τ ). This is known as the Duhamel principle for the diffusion equation. 34. Use the Duhamel principle to solve the nonhomogeneous diffusion equation ut = κuxx + e −t sin πx, 0 < x < l, t > 0, with the homogeneous boundary and initial data u (0, t)=0, u (1, t)=0, t > 0, u (x, 0) = 0, 0 ≤ x ≤ 1. 35. (a) Verify that un (x, y) = exp  ny − √ n sin nx, is the solution of the Laplace equation 166 5 The Cauchy Problem and Wave Equations uxx + uyy = 0, x ∈ R, y > 0, u (x, 0) = 0, uy (x, 0) = n exp  − √ n sin nx, where n is a positive integer. (b) Show that this Cauchy problem is not well posed. 36. Show that the following Cauchy problems are not well posed: (a) ut = uxx, x ∈ R, t > 0, u (0, t) =  2 n sin  2n 2 t , ux (0, t)=0, t > 0. (b) uxx + uyy = 0, x ∈ R, t > 0, un (x, 0) → 0, (un)y (x, 0) → 0, as n → ∞. 6 Fourier Series and Integrals with Applications “The thorough study of nature is the most ground for mathematical discoveries.” Joseph Fourier “Nearly fifty years had passed without any progress on the question of analytic representation of an arbitrary function, when an assertion of Fourier threw new light on the subject. Thus a new era began for the development of this part of Mathematics and this was heralded in a stunning way by major developments in mathematical Physics.” Bernhard Riemann “Fourier created a coherent method by which the different components of an equation and its solution in series were neatly identified with different aspects of physical solution being analyzed. He also had a uniquely sure instinct for interpreting the asymptotic properties of the solutions of his equations for their physical meaning. So powerful was his approach that a full century passed before non-linear equations regained prominence in mathematical physics.” Ioan James 6.1 Introduction This chapter is devoted to the theory of Fourier series and integrals. Although the treatment can be extensive, the exposition of the theory here will be concise, but sufficient for its application to many problems of applied mathematics and mathematical physics. 168 6 Fourier Series and Integrals with Applications The Fourier theory of trigonometric series is of great practical importance because certain types of discontinuous functions which cannot be expanded in power series can be expanded in Fourier series. More importantly, a wide class of problems in physics and engineering possesses periodic phenomena and, as a consequence, Fourier’s trigonometric series become an indispensable tool in the analysis of these problems. We shall begin our study with the basic concepts and definitions of some properties of real-valued functions. 6.2 Piecewise Continuous Functions and Periodic Functions A single-valued function f is said to be piecewise continuous in an interval [a, b] if there exist finitely many points a = x1 < x2 < ... < xn = b, such that f is continuous in the intervals xj <x<xj+1 and="" the="" one-sided="" limits="" f="" (xj+)="" (xj+1−)="" exist="" for="" all="" j="1," 2,="" 3,...,n="" −="" 1.="" a="" piecewise="" continuous="" function="" is="" shown="" in="" figure="" 6.2.1.="" functions="" such="" as="" 1="" x="" sin="" (1="" x)="" fail="" to="" be="" closed="" interval="" [0,="" 1]="" because="" limit="" (0+)="" does="" not="" either="" case.="" if="" an="" [a,="" b],="" then="" it="" necessarily="" bounded="" integrable="" over="" that="" interval.="" also,="" follows="" immediately="" product="" of="" two="" on="" common="" b]="" if,="" addition,="" first="" derivative="" ′="" each="" intervals="" xj="" <x<xj+1,="" (xj−)="" exist,="" said="" smooth;="" second="" ′′="" 6.2.1="" graph="" function.="" 6.2="" periodic="" 169="" very="" smooth.="" (x)="" there="" exists="" real="" positive="" number="" p="" (x="" +="" p)="f" (x),="" (6.2.1)="" x,="" called="" period="" f,="" smallest="" value="" termed="" fundamental="" period.="" sample="" given="" 6.2.2.="" with="" p,="" 2p)="f" p),="" 3p)="f" 2p="" 2p),="" np)="f" (n="" 1)="" any="" integer="" n.="" hence,="" integral="" values="" n="" (x).="" (6.2.2)="" can="" readily="" f1,="" f2,="" ...,="" fk="" have="" ck="" are="" constants,="" c2f2="" ...="" ckfk,="" (6.2.3)="" has="" p.="" well="" known="" examples="" sine="" cosine="" functions.="" special="" case,="" constant="" also="" arbitrary="" thus,="" by="" relation="" (6.2.3),="" series="" a0="" a1="" cos="" a2="" 2x="" b1="" b2="" converges,="" obviously="" 2π.="" types="" series,="" which="" occur="" frequently="" problems="" applied="" mathematics="" mathematical="" physics,="" will="" treated="" later.="" 6.2.2="" 170="" 6="" fourier="" integrals="" applications="" 6.3="" systems="" orthogonal="" sequence="" {φn="" (x)}="" respect="" weight="" q=""
="" b="" φm="" φn="" dx="0," m="n." (6.3.1)="" we="" φn="1
" φ="" 2="" dx3="" (6.3.2)="" norm="" system="" (x)}.="" example="" 6.3.1.="" {sin="" mx},="" 2,...,="" form="" [−π,="" π],="" π="" −π="" mx="" nx="" ⎨="" ⎩="" 0,="" π,="" this="" notice="" equal="" unity,="" √="" π.="" φ1,="" φ2,="" φn,="" where="" may="" finite="" or="" infinite,="" satisfies="" relations="" 1,="" (6.3.3)="" orthonormal="" b].="" evident="" obtained="" from="" dividing="" its="" 6.3.2.="" x,sin="" .="" ,="" nx,sin="" forms="" π]="" since="" m,="" n,="" (6.3.4)="" 6.4="" 171="" integers="" normalize="" system,="" divide="" elements="" original="" their="" norms.="" 2π="" ,...,="" system.="" 2x,sin="" 2x,="" mutually="" other="" linearly="" independent.="" formally="" associate="" trigonometric="" write="" ∼="" ∞="" k="1" (ak="" kx="" bk="" kx),="" (6.4.1)="" symbol="" indicates="" association="" a0,="" ak,="" some="" unique="" manner.="" coefficients="" ak="" determined="" soon.="" coefficient="" (a0="" 2)="" instead="" used="" convenience="" representation.="" however,="" easy="" say="" right="" hand="" side="" itself="" converges="" represents="" indeed,="" converge="" diverge.="" let="" riemann="" defined="" π].="" suppose="" define="" nth="" partial="" sum="" sn="" n="" (6.4.2)="" represent="" shall="" seek="" best="" approximation="" sense="" least="" squares,="" is,="" minimize="" i="" (a0,="" bk)="
" [f="" (x)]2="" dx.="" (6.4.3)="" extremal="" problem.="" necessary="" condition="" bk,="" so="" minimum,="" derivatives="" these="" vanish.="" substituting="" equation="" into="" differentiating="" obtain="" 172="" ∂i="" ∂a0="−" ⎡="" ⎣f="" (aj="" jx="" bj="" jx)="" ⎤="" ⎦dx.="" (6.4.4)="" ∂ak="−2" ⎦cos="" dx.(6.4.5)="" ∂bk="−2" ⎦sin="" dx.(6.4.6)="" using="" orthogonality="" noting="" (6.4.7)="" integers,="" equations="" (6.4.4),="" (6.4.5),="" (6.4.6)="" become="" dx,="" (6.4.8)="" (6.4.9)="" (6.4.10)="" must="" vanish="" value.="" (6.4.11)="" (6.4.12)="" (6.4.13)="" note="" case="" reason="" writing="" rather="" than="" (6.4.1).="" (6.4.8),="" (6.4.9),="" ∂="" ∂a2="" 0="π," (6.4.14)="" ∂b2="" (6.4.15)="" mixed="" order="" remaining="" higher="" now="" expand="" taylor="" about="" a1,...,an,="" b1,...,bn),="" 6.5="" convergence="" 173="" ∆a0,...,bn="" ∆bn)="I" (a0,...,bn)="" ∆i,="" (6.4.16)="" ∆i="" stands="" terms.="" derivatives,="" vanish,="" 2!="" 1="" ∆a2="" ="" ∆b2="" 3="" (6.4.17)="" virtue="" (6.4.15),="" positive.="" minimum="" value,="" (6.4.11),="" (6.4.12),="" respectively.="" corresponding="" correspondence="" asserts="" nothing="" divergence="" constructed="" series.="" question="" arises="" whether="" possible="" investigation="" sufficient="" conditions="" representation="" turns="" out="" difficult="" remark="" possibility="" representing="" imply="" uniformly,="" matter="" fact,="" convergent="" need="" instance,="" log="" no="" introduce="" three="" kinds="" fouriers="" series:="" (i)="" pointwise="" convergence,="" (ii)="" uniform="" (iii)="" mean-square="" convergence.="" definition="" 6.5.1.="" (pointwise="" convergence).="" infinite="" 2∞="" fn="" a<x<b="" a<x<b.="" words,="" a<x<b,="" |f="" (x)|="" →="" ∞,="" 174="" 6.5.2.="" (="" uniformly="" ≤="" max="" a≤x≤b="" ∞.="" evidently,="" implies="" but="" converse="" true.="" 6.5.3.="" (or="" l="" )="" noted="" stronger="" both="" study="" long="" complex="" history.="" f.="" answer="" certainly="" obvious.="" 2π-="" function,="" leads="" questions="" local="" behavior="" near="" point="" global="" overall="" entire="" another="" deals="" (−π,="" π),="" theorem="" provide="" insight="" problem="" guarantee="" x.="" hand,="" 2π-periodic="" smooth="" r,="" every="" been="" 1876="" whose="" diverge="" at="" certain="" points.="" was="" open="" century="" point.="" 1966,="" lennart="" carleson="" (1966)="" provided="" affirmative="" deep="" states="" square="" almost="" obvious="" 175="" ≥="" (6.5.1)="" expanding="" gives="" [sn="" but,="" definitions="" (6.3.4),="" kx)="" 3="" ="" (6.5.2)="" s="" 32="" (6.5.3)="" consequently,="" πa2="" 0.="" (6.5.4)="" (6.5.5)="" independent="" (6.5.6)="" bessel’s="" inequality.="" 176="" see="" left="" nondecreasing="" above,="" therefore,="" (6.5.7)="" converges.="" lim="" k→∞="" (6.5.8)="" mean="" when="" limn→∞="" 4="" kx532="" (6.5.9)="" (6.5.10)="" parseval’s="" one="" central="" results="" theory="" derive="" many="" important="" numerical="" furthermore,="" holds="" true,="" set="" complete.="" parseval="" derived="" (6.5.11)="" respectively,="" multiply="" 6.4.11)="" integrate="" resulting="" expression="" .(6.5.12)="" replacing="" (6.5.12)="" (6.5.10).="" 6.6="" 177="" section="" different="" way.="" expansion="" kx).="" (6.6.1)="" assume="" term-by-term="" (we="" later="" this),="" (6.6.2)="" again,="" sides="" (6.6.3)="" similar="" manner,="" find="" (6.6.4)="" just="" found="" exactly="" same="" those="" 6.4.="" 6.6.1.="" <="" here="" 178="" 6.6.1="" 3="" ="" kπ="" (−1)k="" 3,....="" similarly,="" 179="" 4="" ....="" (6.6.5)="" 6.6.2.="" consider="" −π,="" −π<x<="" −πdx="" 6.6.2="" 180="" (cos="" "="" #="" kπ)="1" ="" kx0="" (6.6.6)="" 6.6.3.="" sawtooth="" wave="" −π<x<π,="" 2kπ)="" 2,....="" 6.6.3="" continuous.="" 181="" (−1)k+1="" 3x="" 4x="" ...="" (6.6.7)="" agree="" endpoints="" zero,="" endpoint.="" (π="" 0)="" 0)]="1" π)="" uniform.="" kx.="" (6.6.8)="" neighborhood="" difference="" between="" (x)and="" seems="" smaller="" increases,="" size="" region="" occurs="" decreases="" indicating="" nonuniform="" oscillatory="" nature="" close="" discontinuities="" gibbs="" phenomenon.="" what="" happens="" π?="" (6.6.8),="" put="" xn="π" approximate="" (xn)="n"
4=""
5="n" ="" rewritten="" πk="" ·=""
4π="" (6.6.9)="" 182="" identified="" definite="" subintervals="" (k−1)π="" obviously,="" subinterval="" length="" evaluate="" right-hand="" endpoint="" ≈="" 1.18π.="" xn,="" approaches="" left.="" tends="" jump="" (π−)−f="" (π+)="2π," and,="" sufficiently="" large="" (π−)="" 1.18π="" next="" draw="" graphs="" s7="" s10="" exhibit="" oscillations="" figures="" 6.6.4="" (a)="" (b).="" show="" so-called="" overshooting="" 3π,="" several="" 5="" 7="" well-known="" slowly="" (6.6.10)="" putting="" 8="1" 9="" 10="" 11√="" 13√="" 14="" 11="" view="" (6.6.10),="" (6.6.11)="" 6.7="" 183="" even="" odd="" (6.7.1)="" 184="" written="" kx,="" (6.7.2)="" formula="" (6.7.1).="" consequence,="" (6.7.3)="" (6.7.4)="" (6.7.5)="" (6.7.4).="" 6.7.1.="" −1,="" +1,="" (x+="" clearly,="" 3,...="" b2k="0" b2k−1="[(4/π)" (2k="" 1)].="" (6.7.6)="" consists="" only="" harmonics.="" loss="" harmonics="" due="" fact="" sgn="" (6.6.10).="" 185="" 6.7.1="" (6.7.7)="" examine="" manner="" terms="" tend="" 3,="" 6.7.2.="" investigate="" locate="" peak="" origin="" calculate="" height="" 6.7.2="" 186="" overshoot="" maximum="" [sin="" 2kx="" x]="2" (6.7.8)="" points="" 2n="" ,...,(2n="" so,="" becomes="" (1.852)="1.179." 1.179="" approximately.="" compared="" (1.179="" ×="" 100%="" 9%.="" onset="" phenomenon="" 6.7.3.="" historically,="" observed="" physicist="" a.="" michelson="" (1852–1931)="" end="" nineteenth="" century.="" graphically,="" he="" developed="" equipment="" harmonic="" analyser="" synthesizer.="" calculated="" sums,="" sn,="" graphically.="" graphical="" that,="" functions,="" were="" sums="" error="" origin,="" (the="" function)="" sums.="" j.w.="" (1839–1903)="" who="" explanation="" strikingly="" new="" showed="" errors="" associated="" computations.="" further="" 187="" 6.7.3="" discontinuity="" properties="" discontinuity.="" |sin="" x|="" 6.7.4,="" 2,...="" k)="" 1+(−1)k="" (2kx)="" 4k="" triangular="" 188="" 6.7.4="" rectified="" =="" ⎧="" −x,="" (6.7.9)="" 2nπ),="" 6.7.5.="" bn="0" 6.7.5="" 189="" |x|="" x="" dx="" integrating="" parts="2" πn2="" [cos="" nx]="" [(−1)n="" even,="" odd.="" 5x="" (6.7.10)="" yields="" following="" (2n="" 1)2="" (6.7.11)="" reciprocals="" squares="" n2="π" (6.7.12)="" vise="" versa.="" (2n)="" s.="" (6.7.12).="" zeta="" ζ="" (s)="∞" ns="" (6.7.13)="" 190="" complex.="" extended="" natural="" way="" extension="" analytic="" continuation="" include="" numbers="" except="" introduced="" bernhard="" 1841.="" proved="" made="" conjectures,="" still="" mathematics.="" zeros="" axis="" negative="" integers.="" conjectured="" lie="" line="" re="" hypothesis;="" unsolved="" line.="" fall="" 2002,="" fifty="" billion="" —="" them="" stated="" 6.7.4.="" extensions="" 6.7.6.="" ≡="" 6.7.6="" 191="" nxπ="" nπ="" "="" nx#π="" (−1)n="" 4∞="" (6.7.14)="" ···="π" 12="" (6.7.15)="" vice="" s,="" adding="" (6.7.16)="" then,="" subtracting="" 24="" (6.7.17)="" 192="" 6.7.7="" preceding="" sections,="" prescribed="" assumed="" (−∞,∞).="" practice,="" encounter="" π).="" simply="" extend="" periodically="" 2π,="" 6.7.7.="" way,="" able="" expansion,="" although="" interested="" (0,="" ways.="" denoted="" (see="" 6.7.8)="" fe="" (−x),="" while="" 6.7.9)="" f0="" −f="" 193="" 6.7.8="" expansions="" 6.7.9="" 194="" 6.8="" sometimes="" convenient="" form.="" easily="" euler’s="" formulas="" ix="" e="" −ix="" 2i="" ikx="" −ikx="" ak="" ibk="" c−k="" c0="a0" isin="" −ikxdx="" ikxdx.="" 195="" (6.8.1)="" −ikxdx.="" (6.8.2)="" |ck|="" (6.8.3)="" (6.8.1)–(6.8.2).="" multiplying="" −π<x<π="" ikxdx="∞" c¯k="∞" 6.8.1.="" ik)="" sinh="" eikx="" (6.8.4)="" apply="" |1="" ik|="" sinh2="" 4π="" −2π="1" simplifying="" result="" coth="" (6.8.5)="" 196="" 6.8.2.="" −1="" <a<="" 2a="" (6.8.6)="" (6.8.7)="" denote="" c="" nx)="∞" eix="" a="" eix="" ia="" sin2="" (6.8.8)="" equating="" imaginary="" part="" desired="" results.="" 6.9="" far="" concerned="" applications,="" restrictive,="" interest="" arbitrary,="" variable="" t="" transformation="" (b="" a)="" t,="" (6.9.1)="" [(b="" ((b="" 2π)t]="F" (t)="" kt="" kt),="" (6.9.2)="" dt,="" 197="" changing="" (2x="" ,(6.9.3)="" (6.9.4)="" (6.9.5)="" take="" [−l,l].="" once="" letting="" [−l,l]="" takes="" kπx="" (6.9.6)="" −l="" (6.9.7)="" (6.9.8)="" 2l,="" (6.9.6),="" determine="" (6.9.9)="" (6.9.10)="" (6.9.11)="" 198="" (6.9.12)="" finally,="" make="" change="" −l<x<l.="" 2l.="" lt="" (t).="" (6.9.13)="" smooth,="" expanded="" cke="" ikt,="" −iktdt.="" (6.9.14)="" exp="" ixπk="" 2l="" dx.(6.9.15)="" particular,="" time="" ω="" frequency,="" [an="" (nωt)="" (nωt)]="" (6.9.16)="" anand="" replaced="" (a1="" ωt="" ωt),="" (a2="" 2ωt="" 2ωt),="" (an="" nωt="" nωt),="" first,="" second,="" 6.9.1.="" −2="" <x<="" odd,="" 199="" 6.9.1="" 2.="" 6.9.2.="" (kπx)="" (−1)k−1="" πx.="" 6.9.3.="" (−l,l)="" 200="" 6.9.2="" directly="" πkx="" l.="" 6.9.4.="" l)="" know="" 2∞="" πt="" πkt="" 4l="" 6.10="" riemann–lebesgue="" lemma="" 201="" 6.9.5.="" solution="" 25="" 6.14="" exercises,="" −l<t<l.="" use="" exercise="" ⎫="" ⎬="" ⎭="1" ="" n="0" inlt="" earlier="" section,="" discuss="" proof="" lemma.="" 6.10.1.="" (riemann–lebesgue="" lemma)="" g="" λ→∞="" λx="" (6.10.1)="" proof.="" (λ)="
" (6.10.2)="" λ,="" λ="" (t="" λ)="−" λt,="" b−π="" a−π="" λt="" dt.="" (6.10.3)="" dummy="" variable,="" above="" (6.10.4)="" addition="" 202="" [g="" λ)]="" (6.10.5)="" bounded,="" |g="" m.="" a+π="" πm="" |i="" (λ)|="" λ)|="" (6.10.6)="" ε="" a),="" (6.10.7)="">Λ and all x in [a, b]. We now choose λ such that πM/λ < ε/2, whenever λ>Λ. Then |I (λ)| < ε 2 + ε 2 = ε. If g (x) is piecewise continuous in [a, b], then the proof consists of a repeated application of the preceding argument to every subinterval of [a, b] in which g (x) is continuous. Theorem 6.10.1. (Pointwise Convergence Theorem). If f (x) is piecewise smooth and periodic function with period 2π in [−π, π], then for any x a0 2 + ∞ k=1 (ak cos kx + bk sin kx) = 1 2 [f (x+) + f (x−)] , (6.10.8) where ak = 1 π
π −π f (t) cos kt dt, k = 0, 1, 2,..., (6.10.9) bk = 1 π
π −π f (t) sin kt dt, k = 1, 2, 3,.... (6.10.10) 6.10 The Riemann–Lebesgue Lemma and Pointwise Convergence Theorem 203 Proof. The nth partial sum sn (x) of the series (6.10.8) is sn (x) = 1 2 a0 + n k=1 (ak cos kx + bk sin kx). (6.10.11) We use integrals in (6.10.9)–(6.10.10) to replace ak and bk in (6.10.11) so that sn (x) = 1 2π
π −π 1 1+2n k=1 (cos kt cos kx + sin ktsin kx) 3 f (t) dt = 1 2π
π −π 1 1+2n k=1 cos k (x − t) 3 f (t) dt = 1 2π
π −π Dn (x − t) f (t) dt, (6.10.12) where Dn (θ) is called the Dirichlet kernel defined by Dn (θ)=1+2n k=1 cos kθ. (6.10.13) The next step is to study the properties of this kernel Dn (θ) which is an even function with period 2π and satisfies the condition 1 2π
π −π Dn (θ) dθ =1+0+0+ ... +0=1. (6.10.14) We find the value of the sum in (6.10.13) by Euler’s formula so that Dn (θ)=1+n k=1  e ikθ + e −ikθ = n k=−n e ikθ = e −inθ + ... +1+ ... + e inθ . This is a finite geometric series with the first term e −inθ, the ratio e iθ, and the last term e inθ, and hence, its sum is given by Dn (θ) = e −inθ − e i(n+1)θ 1 − e iθ = exp −  n + 1 2 iθ! − exp n + 1 2 iθ! exp  − 1 2 iθ − exp  + 1 2 iθ = sin  n + 1 2 θ sin 1 2 θ . (6.10.15) The graph of Dn (θ) is shown in Figure 6.10.1. It looks similar to that of the diffusion kernel as drawn in Figure 12.4.1 in Chapter 12 except for its symmetric oscillatory trail. 204 6 Fourier Series and Integrals with Applications Figure 6.10.1 Graph of Dn (θ) against θ. We next put t − x = θ in (6.10.12) to obtain sn (x) = 1 2π
x+π x−π Dn (θ) f (x + θ) dθ. (6.10.16) Since both Dn and f have period 2π, the limits of the integral can be taken from −π to π, and hence, (6.10.16) assumes the form sn (x) = 1 2π
π −π Dn (θ) f (x + θ) dθ. (6.10.17) We next use (6.10.14) to express the difference of sn (x) and 1 2 [f (x+) + f (x−)] in the form sn (x) − 1 2 [f (x+) + f (x−)] = 1 2π
0 −π Dn (θ) [f (x + θ) − f (x−)] dθ + 1 2π
π 0 Dn (θ) [f (x + θ) − f (x+)] dθ which is, by (6.10.15), = 1 2π
0 −π g− (θ) sin  n + 1 2  θ dθ + 1 2π
π 0 g+ (θ) sin  n + 1 2  θ dθ, (6.10.18) 6.10 The Riemann–Lebesgue Lemma and Pointwise Convergence Theorem 205 where g + (θ) =  sin θ 2 −1 [f (x + θ) − f (x + )] . (6.10.19) Since the denominators of the functions g + (θ) vanish at θ = 0, integrals in (6.10.18) may diverge at this point. However, by assumption, f is piecewise smooth, and hence, lim θ→0 + g + (θ) = lim θ→0 + f (x + θ) − f (x) θ 2 · 4 θ 2 sin θ 2 5 = 2f ′ (x +).(6.10.20) Evidently, the above limits exist, and g + (θ) are piecewise continuous elsewhere in the interval (−π, π). Therefore, by the Riemann–Lebesgue Lemma 6.10.1, both integrals in (6.10.18) vanish as n → ∞. Thus, limn→∞ sn (x) = 1 2 [f (x+) + f (x−)] . This proves that the Fourier series converges for each x in (−π, π). Remark 3. At a point of continuity the series converges to the function f (x). Remark 4. At a point of discontinuity, the series is equal to the arithmetic mean of the limits of the function on both sides of the discontinuity. Remark 5. The condition of piecewise smoothness under which the Fourier series converges pointwise is a sufficient condition. A large number of examples of applications is covered by this case. However, the pointwise convergence Theorem 6.10.1 can be proved under weaker conditions. Example 6.10.1. In Example 6.6.1, we obtained that the Fourier series expansion for  x + x 2 in [−π, π], as shown in Figure 6.10.2, is f (x) ∼ π 2 3 + ∞ k=1 4 k 2 (−1)k cos kx − 2 k (−1)k sin kx . Since f (x) = x + x 2 is piecewise smooth, the series converges, and hence, we write x + x 2 = π 2 3 + ∞ k=1 4 k 2 (−1)k cos kx − 2 k (−1)k sin kx , at points of continuity. At points of discontinuity, such as x = π, by virtue of the Pointwise Convergence Theorem, 206 6 Fourier Series and Integrals with Applications Figure 6.10.2 Graph of f (x). 1 2 π + π 2 +  −π + π 2 ! = π 2 3 + ∞ k=1 4 k 2 (−1)k cos kπ, (6.10.21) since f (π−) = π + π 2 and f (π+) = f (−π+) = −π + π 2 . Simplification of equation (6.10.21) gives π 2 = π 2 3 + ∞ k=1 4 k 2 (−1)2k , or π 2 6 = ∞ k=1 1 k 2 . The series can be used to obtain the sum of reciprocals of squares of odd positive integers, that is, ∞ n=1 1 (2n − 1)2 . We have π 2 6 = ∞ n=1 1 n2 = ∞ n=1 1 (2n) 2 + ∞ n=1 1 (2n − 1)2 = 1 4 · π 2 6 + ∞ n=1 1 (2n − 1)2 , 6.10 The Riemann–Lebesgue Lemma and Pointwise Convergence Theorem 207 or, ∞ n=1 1 (2n − 1)2 = π 2 6  1 − 1 4  = π 2 8 . Conversely, this series can be used to find the sum of reciprocals of squares of all positive integers. Example 6.10.2. Find the Fourier series of the following function f (x) = ⎧ ⎨ ⎩ 0, −2 ≤ x < 0 2 − x, 0 < x ≤ 2. This function is defined over the interval −2 ≤ x ≤ 2, where it is piecewise smooth with a finite discontinuity at x = 0. We use (6.9.7) and (6.9.8) to calculate the Fourier coefficients a0 = 1 2
2 0 (2 − x) dx = 1 ak = 1 2
2 0 (2 − x) cos  πkx 2  dx = 2 π 2k 2 " 1 − (−1)k # , k = 1, 2, 3,.... bk = 1 2
2 0 (2 − x) sin  πkx 2  dx = 2 πk , k = 1, 2, 3,.... Consequently, the Fourier series (6.9.6) becomes f (x) = 1 2 + 2 π ∞ k=1 ⎡ ⎣ ( 1 − (−1)k ) πk2 cos  πkx 2  + 1 k sin  πkx 2  ⎤ ⎦. (6.10.22) The function f (x) is continuous at x = 1 where f (1) = 1, so that the Fourier series (6.10.22) gives 1 = 1 2 + 2 π ∞ n=1 {1 − (−1)n } πn2 cos
4nπ 2
5 + 1 n sin
4nπ 2
5 . Since the factor 1 − (−1)n = 0 for even n, and cos nπ 2 = 0 when n is odd, every term of the cosine series vanishes for all n. Consequently, π 4 = ∞ n=1 1 n sin
4nπ 2
5 = ∞ n=1 (−1)n (2n + 1). (6.10.23) On the other hand, f (x) is discontinuous at x = 0 and the Fourier series must converge to 1 2 (0 + 2) = 1. Thus, 208 6 Fourier Series and Integrals with Applications π 2 4 = ∞ n=1 [1 − (−1)n ] n2 = 2∞ n=1 1 (2n − 1)2 , or ∞ n=1 1 (2n − 1)2 = π 2 8 . (6.10.24) 6.11 Uniform Convergence, Differentiation, and Integration In the preceding section, we have proved the pointwise convergence of the Fourier series for a piecewise smooth function. Here, we shall consider several theorems without proof concerning uniform convergence, term-by-term differentiation, and integration of Fourier series. Theorem 6.11.1. (Uniform and Absolute Convergence Theorem) Let f (x) be a continuous function with period 2π, and let f ′ (x) be piecewise continuous in the interval [−π, π]. If, in addition, f (−π) = f (π), then the Fourier series expansion for f (x) is uniformly and absolutely convergent. In the preceding theorem, we have assumed that f (x) is continuous and f ′ (x) is piecewise continuous. With less stringent conditions on f, the following theorem can be proved. Theorem 6.11.2. Let f (x) be piecewise smooth in the interval [−π, π]. If f (x) is periodic with period 2π, then the Fourier series for f converges uniformly to f in every closed interval containing no discontinuity. We note that the partial sums sn (x) of a Fourier series cannot approach the function f (x) uniformly over any interval containing a point of discontinuity of f. The behavior of the deviation of sn (x) from f (x) in such an interval is known as the Gibbs phenomenon. For instance, in the Example 6.7.1, the Fourier series of the function is given by f (x) = 4 π ∞ k=1 sin (2k − 1) x (2k − 1) . (6.11.1) From graphs of the partial sums sn (x) against the x-axis, as shown in Figures 6.7.2 and 6.7.3, we find that sn (x) oscillate above and below the value of f. It can be observed that, near the discontinuous points x = 0 and x = π, sn deviate from the function rather significantly. Although the magnitude of oscillation decreases at all points in the interval for large n, very near the points of discontinuity the amplitude remains practically independent of n as n increases. This illustrates the fact that the Fourier 6.11 Uniform Convergence, Differentiation, and Integration 209 series of a function f does not converge uniformly on any interval which contains a discontinuity. Termwise differentiation of Fourier series is, in general, not permissible. From Example 6.6.3, the Fourier series for f (x) = x is given by x = 2 sin x − sin 2x 2 + sin 3x 3 − ... , (6.11.2) which converges for all x, whereas the series after formal term-by-term differentiation, 1 ∼ 2 [cos x − cos 2x + cos 3x − ...] . This series is not the Fourier series of f ′ (x) = 1, since the Fourier series of f ′ (x) = 1 is the function 1. In fact, this series is not a Fourier series of any piecewise continuous function defined in [−π, π] as the coefficients do not tend to zero which contradicts the Riemann–Lebesque lemma. In fact, the series of f ′ (x) = 1 diverges for all x since the nth term, cos nx does not tend to zero as n → ∞. The difficulty arises from the fact that the given function f (x) = x in [−π, π] when extended periodically is discontinuous at the points + π, + 3π, .... We shall see below that the continuity of the periodic function is one of the conditions that must be met for the termwise differentiation of a Fourier series. Theorem 6.11.3. (Differentiation Theorem) Let f (x) be a continuous function in the interval [−π, π] with f (−π) = f (π), and let f ′ (x) be piecewise smooth in that interval. Then Fourier series for f ′ can be obtained by termwise differentiation of the series for f, and the differentiated series converges pointwise to f ′ at points of continuity and to [f ′ (x) + f ′ (−x)] /2 at discontinuous points. The termwise integration of Fourier series is possible under more general conditions than termwise differentiation. We recall that in calculus, the series of functions to be integrated must converge uniformly in order to assure the convergence of a termwise integrated series. However, in the case of Fourier series, this condition is not necessary. Theorem 6.11.4. (Integration Theorem) Let f (x) be piecewise continuous in [−π, π], and periodic with period 2π. Then the Fourier series of f (x) a0 2 + ∞ k=1 (ak cos kx + bk sin kx), whether convergent or not, can be integrated term by term between any limits. 210 6 Fourier Series and Integrals with Applications Example 6.11.1. In Example 6.7.2, we have found that f (x) = |sin x| is represented by the Fourier series sin x = 2 π + 4 π ∞ k=1 cos (2kx) (1 − 4k 2) , −π < x < π. (6.11.3) Since f (x) = |sin x| is continuous in the interval [−π, π] and f (−π) = f (π), we differentiate the series term by term, obtaining cos x = − 8 π ∞ k=1 k sin (2kx) (1 − 4k 2) , (6.11.4) by use of Theorem 6.11.3, since f ′ (x) is piecewise smooth in [−π, π]. In this way, we obtain the Fourier sine series expansion of the cosine function in (−π, π). Note that the reverse process is not permissible. Example 6.11.2. Consider the function f (x) = x in the interval −π 0, lim λ→∞
b 0 f (x) sin λx x dx = π 2 f (0+). Proof.
b 0 f (x) sin λx x dx =
b 0 f (0+) sin λx x dx +
b 0 f (x) − f (0+) x sin λx dx = f (0+)
λb 0 sin t t dt +
b 0 f (x) − f (0+) x sin λx dx. Since f is piecewise smooth, the integrand of the last integral is bounded as λ → ∞, and thus, by the Riemann–Lebesgue lemma 6.10.1, the last integral tends to zero as λ → ∞. Hence, lim λ→∞
b 0 f (x) sin λx x dx = π 2 f (0+), (6.13.6) since
∞ 0 sin t t dt = π 2 . Theorem 6.13.1. (Fourier Integral Theorem) If f is piecewise smooth in every finite interval, and absolutely integrable on (−∞,∞), then 1 π
∞ 0
∞ −∞ f (t) cos k (t − x) dt dk = 1 2 [f (x+) + f (x−)] . Proof. Noting that |cos k (t − x)| ≤ 1 and that by hypothesis
∞ −∞ f (t) dt < ∞, we see that the integral
∞ −∞ f (t) cos k (t − x) dt converges independently of k and x. It therefore follows that in the double integral I =
λ 0
∞ −∞ f (t) cos k (t − x) dt dk, the order of integration may be interchanged. We then have 218 6 Fourier Series and Integrals with Applications I =
∞ −∞ f (t) 1
λ 0 cos k (t − x) dk3 dt =
∞ −∞ f (t) sin λ (t − x) (t − x) dt = 1
−M −∞ +
x −M +
M x +
∞ M 3 f (t) sin λ (t − x) (t − x) dt. If we substitute u = t − x, we have
M x f (t) sin λ (t − x) (t − x) dt =
M−x 0 f (u + x)  sin λu u  du which is equal to πf (x+) /2 in the limit λ → ∞, by Lemma 6.13.1. Similarly, the second integral tends to πf (x−) /2 when λ → ∞. If we make M sufficiently large, the absolute values of the first and the last integrals are each less than ε/2. Consequently, as λ → ∞
∞ 0
∞ −∞ f (t) cos k (t − x) dt dk = π 2 [f (x+) + f (x−)] . (6.13.7) If f is continuous at the point x, then f (x+) = f (x−) = f (x) so that integral (6.13.7) reduces to the Fourier integral representation for f as f (x) = 1 π
∞ 0
∞ −∞ f (t) cos k (t − x) dt dk. (6.13.8) We may express the Fourier integral representation (6.13.8) in complex form. In this case, we substitute cos k (t − x) = cos k (x − t) = 1 2 " e ik(x−t) + e −ik(x−t) # into equation (6.13.8) and write it as the sum of two integrals f (x) = 1 2π
∞ 0
∞ −∞ f (t) e ik(x−t) dt dk + 1 2π
∞ 0
∞ −∞ f (t) e −ik(x−t) dt dk. Changing the integration variable from k to −k in the second integral, we obtain f (x) = 1 2π
∞ 0
∞ −∞ f (t) e ik(x−t) dt dk −
−∞ 0
∞ −∞ f (t) e ik(x−t) dt dk = 1 2π
∞ 0
∞ −∞ f (t) e ik(x−t) dt dk +
0 −∞
∞ −∞ f (t) e ik(x−t) dt dk = 1 2π
∞ −∞
∞ −∞ f (t) e ik(x−t) dt dk. (6.13.9) 6.13 Fourier Integrals 219 Or, equivalently, f (x) = 1 √ 2π
∞ −∞ e ikxdk 1 √ 2π
∞ −∞ e −iktf (t) dt = 1 √ 2π
∞ −∞ F (k) e ikxdk, (6.13.10) where F (k) = 1 √ 2π
∞ −∞ e −iktf (t) dt. (6.13.11) Either (6.13.9) or (6.13.10) with coefficient F (k) is called the complex form of the Fourier integral representation for f (x). Now we assume that f (x) is either an even or an odd function. Any function that is not even or odd can be expressed as a sum of two such functions. Expanding the cosine function in (6.13.8), we obtain the Fourier cosine formula f (x) = f (−x) = 2 π
∞ 0 cos kx dk
∞ 0 cos kt f (t) dt. (6.13.12) Similarly, for an odd function, we obtain the Fourier sine formula f (x) = −f (−x) = 2 π
∞ 0 sin kx dk
∞ 0 sin kt f (t) dt. (6.13.13) Example 6.13.1. The rectangular pulse can be expressed as a sum of Heaviside functions f (x) = H (x + 1) − H (x − 1). Find its Fourier integral representation. From (6.13.5) we find f (x) = 1 π
∞ 0
1 −1 cos [k (t − x)] dt dk = 1 π
∞ 0 cos kx
1 −1 cos kt dt + sin kx
1 −1 sin kt dt dk = 2 π
∞ 0  sin k k  cos kx dk. Example 6.13.2. Find the Fourier cosine integral representation of the function f (x) = ⎧ ⎨ ⎩ 1, 0 <x< 1,="" 0,="" x="" ≥="" 1.="" 220="" 6="" fourier="" series="" and="" integrals="" with="" applications="" we="" have,="" from="" (6.13.12),="" f="" (x)="2" π=""
="" ∞="" 0="" cos="" kx="" dk="" 1="" kt="" dt="2" ="" sin="" k="" ="" dk,="" or,="" dk.="" 6.14="" exercises="" find="" the="" of="" following="" functions:="" (a)="" ⎨="" ⎩="" h="" −π<x<="" <="" π,="" is="" a="" constant="" (b)="" 2="" (c)="" +="" −π<x<π,="" (d)="" (e)="" (f)="" −π<x<π.="" 2.="" determine="" sine="" −="" <x<π,="" 3="" <x<π.="" 221="" 3.="" obtain="" cosine="" representation="" for="" 3x="" 4.="" expand="" functions="" in="" series:="" −1="" <x<="" 6,="" (πx="" l)="" <x<l,="" −2="" 2,="" −x="" 5.="" complex="" 2x="" 2)="" 222="" 6.="" expansion="" function="" π.="" use="" 6(a),="" show="" that="" 8="1+" 5="" 7="" ....="" 7.="" ,="" −l="" l.="" 7(a),="" 12="1" 4="" 8.="" each="" by="" performing="" differentiation="" appropriate="" sin2="" cos2="" 9.="" represented="" new="" which="" are="" obtained="" termwise="" integration="" to="" x:="" ∞="" (−1)k+1="" 1−(−1)k="" ="" 223="" sin(2k+1)x="" (2k+1)3="π" 2x−πx2="" 2π,="" sin(2k−1)x="" (2k−1)="⎧" 10.="" double="" (x,="" y)="1" <x<π="" <y<π,="" 2y="" y="" −π<x<π="" −π<y<π,="" x+y="" (g)="" <y<="" (h)="" <x<a="" <y<b,="" (i)="" (j)="" (k)="" −π<y<π.="" 11.="" deduce="" general="" formula="" rectangle="" −a<x<a,="" −b<y<b.="" 12.="" prove="" weierstrass="" approximation="" theorem:="" if="" continuous="" on="" interval="" −π="" ≤="" (−π)="f" (π),="" then,="" any="" ε=""> 0, there exists a trigonometric polynomial T (x) = a0 2 + n k=1 (ak cos kx + bk sin kx) such that |f (x) − T (x)| < ε for all x in [−π, π]. 224 6 Fourier Series and Integrals with Applications 13. Use the Fourier cosine or sine integral formula to show that (a) e −αx = 2 π
∞ 0 α α2+β2 cos βx dβ, x ≥ 0, α > 0, (b) e −αx = 2 π
∞ 0 β α2+β2 sin βx dβ, x > 0, α > 0. 14. Show that the Fourier integral representation of the function f (x) = ⎧ ⎨ ⎩ x 2 , 0 <xa is f (x) = 2 π
∞ 0 a 2 − 2 k 2  sin ak + 2a k cos ak cos kx k dk. 15. Apply the Parseval relation (6.5.10) to Example 6.7.3 or Example 6.7.4 to show that (a) ∞ n=1 1 (2n − 1)4 = π 4 96 and (b) ∞ n=1 1 n4 = π 4 90 . 16. (a) Obtain the Fourier series for the 2π-periodic odd function f (x) = x (π − x) on [0, π]. (b) Use the Parseval relation (6.5.10) to show that ∞ n=1 1 (2n − 1)6 = π 6 960 and ∞ n=1 1 n6 = π 6 945 . 17. If the 2π-periodic even function is given by f (x) = |x| for −π ≤ x ≤ π, show that f (x) = π 2 − 4 π ∞ n=1 cos (2n − 1) x (2n − 1)2 . 18. Consider the sawtooth function defined by f (x) = π − x, 0 <x< 2π,="" and="" f="" (x="" +="" 2nπ)="f" (x)="" with="" (0)="0." (a)="" show="" that="" the="" fourier="" series="" for="" this="" function="" is="" k="0" 2="" sin="" kx,="" has="" a="" jump="" discontinuity="" at="" origin="" 6.14="" exercises="" 225="" (0+)="π" ,="" (0−)="−" π="" −="" (b)="" max="" 0≤x≤="" n="" sn="" 0="" θ="" dθ="" .="" (c)="" result="" manifestation="" of="" gibbs="" phenomenon,="" is,="" near="" discontinuity,="" overshoots="" (or="" undershoots)="" it="" by="" approximately="" 9%="" jump.="" (d)="" if="" dn="" in="" ≤="" x="" d="" ′="" (x),="" where="" given="" (6.10.15).="" (e)="" using="" e="" ikx,="" first="" critical="" point="" to="" right="" occurs="" xn="π/" ="" 1="" limn→∞="" (xn)="2"
="" π.="" (f)="" draw="" graph="" fortieth="" partial="" sum="" s40="" 40="" kx="" −2π<x<="" then="" examine="" phenomenon="" (x).="" 19.="" consider="" characteristic="" interval="" [a,="" b]="" ⊂="" [−π,="" π]="" defined="" ⎨="" ⎩="" 1,="" b="" 0,="" otherwise.="" ∼="" 2π="" (b="" a)="" +="" k="0" exp="" (−ika)="" (−ikb)="" 2πik="" ·="" ikx="" 20.="" obtain="" (π="" x),="" derive="" following="" numerical="" 3="" 5="" 7="" ...="π" 12="" 9="" 113="" √="" 128="" 226="" 6="" integrals="" applications="" 21.="" triangular="" vertices="" (0,="" 0),="" 2,="" 1)="" (π,="" 0)="" 2x="" ="" 3x="" 5x="" 7x="" ...0="" ∞="" (2n="" 1)2="π" 8="" 22.="" sine="" cosine="" functions:="" <x<a,="" <x<a.="" 23.="" find="" full="" functions="" −1="" x,="" <x<="" +1="" 1.="" π,="" 2π.="" 24.="" cos="" <="" 25.="" 2i="" ="" n="0" inx="" complex="" 2π-periodic="" sawtooth="" 227="" −π<x<="" 26.="" suppose="" g="" have="" expansion="" −π="" π:="" a0="" (ak="" bk="" kx),="" α0="" (αk="" βk="" together="" their="" two="" derivatives="" are="" continuous="" on="" (−π)="f" (π),="" (π)="" hold.="" prove="" general="" parseval="" relation="" holds:="" dx="1" a0α0="" (akαk="" bkβk).="" when="" (6.5.10)="" special="" case="" above="" result.="" 27.="" integral="" representation="" (a="" |x|)="⎧" |x|=""> a, (b) f (x) = ⎧ ⎨ ⎩ sin x, |x| < π 0, |x| > π. 28. If f (x), x ∈ R, is defined by f (x) = ⎧ ⎪⎪⎪⎪⎨ ⎪⎪⎪⎪⎩ −1, −a<x< 0,="" +1,="" 0="" ≤="" x="" <="" a,="" otherwise,="" show="" that="" f="" (x)="" has="" the="" fourier="" sine="" integral="" representation="" π=""
="" ∞="" 1="" k="" (1="" −="" cos="" ka)="" sin="" kx="" dk.="" 228="" 6="" series="" and="" integrals="" with="" applications="" 29.="" if="" −x="" ,="" <x<="" ∞,="" (a)="" is="" +="" 2="" dk,="" (b)="" cosine="" 30.="" defined="" by="" ⎨="" ⎩="" x<="" e=""> 0, show that (a) the Fourier integral representation of f (x) is f (x) = 1 π
∞ 0 (cos kx + k sin kx) (1 + k 2) dk, (b) the Fourier cosine integral representation of f (x) is f (x) = 1 2π
∞ −∞  1 − ik 1 + k 2  e ikxdk. 31. (a) Obtain both the complex Fourier series and the usual Fourier series of f (x) = exp [x (1 + 2πi)] on the interval [−1, 1]. (b) Find the sum of each of the series ∞ k=1 1 (1 + π 2k 2) and ∞ k=1 (−1)k (1 + π 2k 2) . 32. Use Example 6.7.2 to calculate the value of the following series: (a) ∞ k=1 1 (4k 2 − 1), (b) ∞ k=1 (−1)k (4k 2 − 1), (c) ∞ k=1 1 (4k 2 − 1)2 , and (d) ∞ k=1 k 2 (4k 2 − 1)2 . 33. Show that the complex Fourier series of f (x) = x is given by x ∼ ∞ k=1 (−1)k k  i eikx + −∞ k=−1 (−1)k k  i eikx . 6.14 Exercises 229 34. (a) Show that the Fourier series for f (x) is defined by f (x) = ⎧ ⎨ ⎩ sin 2x, 0 ≤ x ≤ π 2 0, π 2 ≤ x ≤ π, is f (x) = 1 π + 1 2 sin 2x −  2 π ∞ k=1 cos 4kx (4k 2 − 1). (b) Show that ∞ k=1 1 (4k 2 − 1)2 = 1 16  π 2 − 8 . (c) Find the sum of the infinite series sin (4x) 1.2.3 + sin (2.4x) 3.4.5 + sin (3.4x) 5.6.7 + ..., 0 ≤ x ≤ π. 35. (a) Obtain the complex Fourier series of f (x) = cos (ax), −π ≤ x ≤ π, where a is real but not an integer. (b) Hence, show that π cot πx = 1 x − ∞ k=1 2x (k 2 − x 2) . (c) Derive the product formula sin πx = πx 3∞ n=1  1 − x 2 n2  . (d) Show that π 2 = 3∞ n=1 2n (2n − 1) · 2n (2n + 1) =  2 1 · 2 3  ·  4 3 · 4 5  ·  6 5 · 6 7  ·  8 7 · 8 9  .... 36. Obtain the Fourier series of the following functions: (a) f (x) = e x , 0 ≤ x ≤ 2π, f (x + 2π) = f (x). (b) f (x) = ⎧ ⎪⎪⎪⎪⎨ ⎪⎪⎪⎪⎩ +1, −π<x< −="" π="" 2="" ,="" 0="" <x<="" −1,="" 0,="" <="" x="" π,="" n="0," +="" 1,="" 2,...="" 230="" 6="" fourier="" series="" and="" integrals="" with="" applications="" draw="" the="" graph="" of="" this="" function.="" (c)="" f="" (x)="x" [x],="" where="" [x]="" is="" greatest="" integer="" not="" exceeding="" x.="" 37.="" find="" for="" each="" functions="" in="" −l<x<l="" defined="" outside="" interval="" so="" that="" (x="" 2l)="f" all="" x:="" (a)="" ⎨="" ⎩="" −l<x<="" l,="" l.="" (b)="" −x,="" −l="" ≤="" x,="" l="" (d)="" examine="" gibbs="" phenomenon="" at="" points="" discontinuity="" function="" (a).="" 38.="" prove="" following="" identities:="" 1="" n="" k="1" cos="" kx="sin" ="" sin="" .="" sn=""
="" −π="" (ξ="" x)="" ξ="" dξ,="" nth="" partial="" sum="" a="" (−π,="" π).="" 7="" method="" separation="" variables="" “however,="" emphasis="" should="" be="" somewhat="" more="" on="" how="" to="" do="" mathematics="" quickly="" easily,="" what="" formulas="" are="" true,="" rather="" than="" mathematicians’="" interest="" methods="" rigorous="" proof.”="" richard="" feynman="" “as="" science,="" has="" been="" adapted="" description="" natural="" phenomena,="" great="" practitioners="" field,="" such="" as="" von="" k´arm´an,="" taylor="" lighthill,="" have="" never="" concerned="" themselves="" logical="" foundations="" mathematics,="" but="" boldly="" taken="" pragmatic="" view="" an="" intellectual="" machine="" which="" works="" successfully.="" verified="" by="" further="" observation,="" still="" strikingly="" prediction,="" ....="" ”="" george="" temple="" 7.1="" introduction="" combined="" principle="" superposition="" widely="" used="" solve="" initial="" boundary-value="" problems="" involving="" linear="" differential="" equations.="" usually,="" dependent="" variable="" u="" (x,="" y)="" expressed="" separable="" form="" y="" (y),="" respectively.="" many="" cases,="" equation="" reduces="" two="" ordinary="" equations="" similar="" treatment="" can="" applied="" three="" or="" independent="" variables.="" however,="" question="" separability="" into="" no="" means="" trivial="" one.="" spite="" question,="" finding="" solutions="" large="" class="" problems.="" 232="" solution="" also="" known="" (or="" eigenfunction="" expansion).="" thus,="" procedure="" outlined="" above="" leads="" important="" ideas="" eigenvalues,="" eigenfunctions,="" orthogonality,="" very="" general="" powerful="" dealing="" examples="" illustrate="" nature="" solution.="" 7.2="" section,="" we="" shall="" introduce="" one="" most="" common="" elementary="" methods,="" called="" variables,="" solving="" applicable="" contains="" wide="" range="" mathematical="" physics,="" engineering="" science.="" now="" describe="" conditions="" applicability="" involve="" second-order="" consider="" homogeneous="" ∗ux∗x∗="" b="" ∗ux∗y∗="" c="" ∗uy∗y∗="" d="" ∗ux∗="" e="" ∗uy∗="" ∗u="0" (7.2.1)="" ∗="" stated="" chapter="" 4="" transformation="" y∗="" ),="" (7.2.2)="" ∂="" ∗,="" y∗)="0," always="" transform="" canonical="" uxx="" uyy="" ux="" uy="" when="" (i)="" hyperbolic,="" (ii)="" parabolic,="" (iii)="" elliptic.="" assume="" (7.2.3)="" (y)="0," (7.2.4)="" are,="" respectively,="" alone,="" twice="" continuously="" differentiable.="" substituting="" (7.2.3),="" obtain="" 233="" x′′y="" cxy="" ′′="" x′y="" exy="" ′="" xy="0," (7.2.5)="" primes="" denote="" differentiation="" respect="" appropriate="" let="" there="" exist="" p="" y),="" that,="" if="" divide="" a1="" b1="" a2="" b2="" [a3="" b3="" (y)]="" (7.2.6)="" dividing="" again="" x′′="" x′="" a3="−" (7.2.7)="" left="" side="" only.="" right="" depends="" only="" upon="" y.="" differentiate="" dx="" (7.2.8)="" integration="" yields="" (7.2.9)="" λ="" constant.="" from="" (7.2.9),="" (7.2.10)="" may="" rewrite="" a1x′′="" a2x′="" (a3="" λ)="" (7.2.11)="" b1y="" b2y="" (b3="" (7.2.12)="" coefficients="" constant,="" then="" reduction="" longer="" necessary.="" this,="" auxx="" buxy="" cuyy="" dux="" euy="" (7.2.13)="" 234="" a,="" b,="" c,="" d,="" e,="" constants="" zero.="" before,="" (7.2.13),="" ax′′y="" bx′y="" dx′y="" fxy="0." (7.2.14)="" division="" axy="" (7.2.15)="" ="" ′="" (7.2.16)=""
4=""
5′="" (7.2.17)="" obviously="" separable,="" both="" sides="" must="" equal="" constant="" λ.="" therefore,="" λy="0," (7.2.18)="" ="" (7.2.19)="" integrating="" (7.2.20)="" β="" determined.="" original="" (7.2.15),="" (7.2.21)="" comparing="" (7.2.21),="" clearly="" satisfy="" 7.3="" vibrating="" string="" problem="" 235="" just="" described="" given="" equation.="" now,="" take="" look="" boundary="" involved.="" several="" types="" conditions.="" ones="" appear="" frequently="" physics="" include="" dirichlet="" condition:="" prescribed="" neumann="" (∂u="" ∂n)="" mixed="" hu="" boundary,="" directional="" derivative="" along="" outward="" normal="" h="" continuous="" boundary.="" details,="" see="" 9="" besides="" these="" conditions,="" as,="" first,="" second,="" third="" other="" robin="" condition;="" condition="" portion="" another="" remainder="" variety="" treat="" later.="" separate="" listed="" above,="" it="" best="" choose="" coordinate="" system="" suitable="" instance,="" cartesian="" rectangular="" region="" lines="" polar="" (r,="" θ)="" circular="" r="constant" θ="constant." imposed="" say="" contain="" derivatives="" only,="" their="" depend="" example,="" [u="" uy]x="x0" =="" cannot="" separated.="" needless="" say,="" condition,="" uy,="" axis.="" first="" tension="" t="" density="" ρ="" stretched="" xaxis="" fixed="" its="" end="" points.="" seen="" 5="" utt="" 2uxx="0,"> 0, (7.3.1) u (x, 0) = f (x), 0 ≤ x ≤ l, (7.3.2) ut (x, 0) = g (x), 0 ≤ x ≤ l, (7.3.3) u (0, t)=0, t ≥ 0, (7.3.4) u (l, t)=0, t ≥ 0, (7.3.5) 236 7 Method of Separation of Variables where f and g are the initial displacement and initial velocity respectively. By the method of separation of variables, we assume a solution in the form u (x, t) = X (x) T (t) = 0. (7.3.6) If we substitute equation (7.3.6) into equation (7.3.1), we obtain XT′′ = c 2X′′T, and hence, X′′ X = 1 c 2 T ′′ T , (7.3.7) whenever XT = 0. Since the left side of equation (7.3.7) is independent of t and the right side is independent of x, we must have X′′ X = 1 c 2 T ′′ T = λ, where λ is a separation constant. Thus, X′′ − λX = 0, (7.3.8) T ′′ − λc2T = 0. (7.3.9) We now separate the boundary conditions. From equations (7.3.4) and (7.3.6), we obtain u (0, t) = X (0) T (t)=0. We know that T (t) = 0 for all values of t, therefore, X (0) = 0. (7.3.10) In a similar manner, boundary condition (7.3.5) implies X (l)=0. (7.3.11) To determine X (x) we first solve the eigenvalue problem (eigenvalue problems are also treated in Chapter 8) X′′ − λX = 0, X (0) = 0, X (l)=0. (7.3.12) We look for values of λ which gives us nontrivial solutions. We consider three possible cases λ > 0, λ = 0, λ < 0. Case 1. λ > 0. The general solution in this case is of the form 7.3 The Vibrating String Problem 237 X (x) = Ae− √ λ x + Be √ λ x where A and B are arbitrary constants. To satisfy the boundary conditions, we must have A + B = 0, Ae − √ λ l + Be √ λ l = 0. (7.3.13) We see that the determinant of the system (7.3.13) is different from zero. Consequently, A and B must both be zero, and hence, the general solution X (x) is identically zero. The solution is trivial and hence, is no interest. Case 2. λ = 0. Here, the general solution is X (x) = A + Bx. Applying the boundary conditions, we have A = 0, A + Bl = 0. Hence A = B = 0. The solution is thus identically zero. Case 3. λ < 0. In this case, the general solution assumes the form X (x) = A cos √ −λ x + B sin √ −λ x. From the condition X (0) = 0, we obtain A = 0. The condition X (l)=0 gives B sin √ −λ l = 0. If B = 0, the solution is trivial. For nontrivial solutions, B = 0, hence, sin √ −λ l = 0. This equation is satisfied when √ −λ l = nπ for n = 1, 2, 3,..., or −λn = (nπ/l) 2 . (7.3.14) For this infinite set of discrete values of λ, the problem has a nontrivial solution. These values of λn are called the eigenvalues of the problem, and the functions sin (nπ/l) x, n = 1, 2, 3,... are the corresponding eigenfunctions. We note that it is not necessary to consider negative values of n since sin (−n) πx/l = − sin nπx/l. 238 7 Method of Separation of Variables No new solution is obtained in this way. The solutions of problems (7.3.12) are, therefore, Xn (x) = Bn sin (nπx/l). (7.3.15) For λ = λn, the general solution of equation (7.3.9) may be written in the form Tn (t) = Cn cos
4nπc l
5 t + Dn sin
4nπc l
5 t, (7.3.16) where Cn and Dn are arbitrary constants. Thus, the functions un (x, t) = Xn (x) Tn (t) =
4 an cos nπc l t + bn sin nπc l t
5 sin
4nπx l
5 (7.3.17) satisfy equation (7.3.1) and the boundary conditions (7.3.4) and (7.3.5), where an = BnCn and bn = BnDn. Since equation (7.3.1) is linear and homogeneous, by the superposition principle, the infinite series u (x, t) = ∞ n=1
4 an cos nπc l t + bn sin nπc l t
5 sin
4nπx l
5 (7.3.18) is also a solution, provided it converges and is twice continuously differentiable with respect to x and t. Since each term of the series satisfies the boundary conditions (7.3.4) and (7.3.5), the series satisfies these conditions. There remain two more initial conditions to be satisfied. From these conditions, we shall determine the constants an and bn. First we differentiate the series (7.3.18) with respect to t. We have ut = ∞ n=1 nπc l
4 −an sin nπc l t + bn cos nπc l t
5 sin
4nπx l
5 . (7.3.19) Then applying the initial conditions (7.3.2) and (7.3.3), we obtain u (x, 0) = f (x) = ∞ n=1 an sin
4nπx l
5 , (7.3.20) ut (x, 0) = g (x) = ∞ n=1 bn
4nπc l
5 sin
4nπx l
5 . (7.3.21) These equations will be satisfied if f (x) and g (x) can be represented by Fourier sine series. The coefficients are given by an = 2 l
l 0 f (x) sin
4nπx l
5 dx, bn = 2 nπc
l 0 g (x) sin
4nπx l
5 dx, (7.3.22ab) 7.3 The Vibrating String Problem 239 The solution of the vibrating string problem is therefore given by the series (7.3.18) where the coefficients an and bn are determined by the formulae (7.3.22ab). We examine the physical significance of the solution (7.3.17) in the context of the free vibration of a string of length l. The eigenfunctions un (x, t)=(an cos ωnt + bn sin ωnt) sin
4nπx l
5 , ωn = nπc l , (7.3.23) are called the nth normal modes of vibration or the nth harmonic, and ωn represent the discrete spectrum of circular (or radian) frequencies or νn = ωn 2π = nc 2l , which are called the angular frequencies. The first harmonic (n = 1) is called the fundamental harmonic and all other harmonics (n > 1) are called overtones. The frequency of the fundamental mode is given by ω1 = πc l , ν1 = 1 2l $ T ∗ ρ . (7.3.24) Result (7.3.24) is considered the fundamental law (or Mersenne law) of a stringed musical instrument. The angular frequency of the fundamental mode of transverse vibration of a string varies as the square root of the tension, inversely as length, and inversely as the square root of the density. The period of the fundamental mode is T1 = 2c ω1 = 2l c , which is called the fundamental period. Finally, the solution (7.3.18) describes the motion of a plucked string as a superposition of all normal modes of vibration with frequencies which are all integral multiples (ωn = nω1 or νn = nν1) of the fundamental frequency. This is the main reason that stringed instruments produce sweeter musical sounds (or tones) than drum instruments. In order to describe waves produced in the plucked string with zero initial velocity (ut (x, 0) = 0), we write the solution (7.3.23) in the form un (x, t) = an sin
4nπx l
5 cos  nπct l  , n = 1, 2, 3,.... (7.3.25) These solutions are called standing waves with amplitude an sin  nπx l , which vanishes at x = 0, l n , 2l n , . . . ,l. These are called the nodes of the nth harmonic. The string displays n loops separated by the nodes as shown in Figure 7.3.1. It follows from elementary trigonometry that (7.3.25) takes the form un (x, t) = 1 2 an " sin nπ l (x − ct) + sin nπ l (x + ct) # . (7.3.26) This shows that a standing wave is expressed as a sum of two progressive waves of equal amplitude traveling in opposite directions. This result is in agreement with the d’Alembert solution. 240 7 Method of Separation of Variables Figure 7.3.1 Several modes of vibration in a string. Finally, we can rewrite the solution (7.3.23) of the nth normal modes in the form un (x, t) = cn sin
4nπx l
5 cos  nπct l − εn  , (7.3.27) where cn =  a 2 n + b 2 n 1 2 and tan εn =
4 bn an
5 . This solution represents transverse vibrations of the string at any point x and at any time t with amplitude cn sin  nπx l and circular frequency ωn = nπc l . This form of the solution enables us to calculate the kinetic and potential energies of the transverse vibrations. The total kinetic energy (K.E.) is obtained by integrating with respect to x from 0 to l, that is, Kn = K.E. =
l 0 1 2 ρ  ∂un ∂t 2 dx, (7.3.28) where ρ is the line density of the string. Similarly, the total potential energy (P.E.) is given by Vn = P.E. = 1 2 T ∗
l 0  ∂un ∂x 2 dx. (7.3.29) Substituting (7.3.27) in (7.3.28) and (7.3.29) gives Kn = 1 2 ρ
4nπc l cn
52 sin2  nπct l − εn 
l 0 sin2
4nπx l
5 dx = ρc2π 2 4l (n cn) 2 sin2  nπct l − εn  = 1 4 ρlω2 n c 2 n sin2 (ωnt − εn),(7.3.30) where ωn = nπc l . 7.3 The Vibrating String Problem 241 Similarly, Vn = 1 2 T ∗
4nπcn l
52 cos2  nπct l − εn 
l 0 cos2
4nπx l
5 dx = π 2T ∗ 4l (n cn) 2 cos2  nπct l − εn  = 1 4 ρlω2 n c 2 n cos2 (ωnt − εn). (7.3.31) Thus, the total energy of the nth normal mode of vibrations is given by En = Kn + Vn = 1 4 ρl(ωncn) 2 = constant. (7.3.32) For a given string oscillating in a normal mode, the total energy is proportional to the square of the circular frequency and to the square of the amplitude. Finally, the total energy of the system is given by E = ∞ n=1 En = 1 4 ρl∞ n=1 ω 2 n c 2 n , (7.3.33) which is constant because En = constant. Example 7.3.1. The Plucked String of length l As a special case of the problem just treated, consider a stretched string fixed at both ends. Suppose the string is raised to a height h at x = a and then released. The string will oscillate freely. The initial conditions, as shown in Figure 7.3.2, may be written u (x, 0) = f (x) = ⎧ ⎨ ⎩ hx/a, 0 ≤ x ≤ a h (l − x) / (l − a), a ≤ x ≤ l. Since g (x) = 0, the coefficients bn are identically equal to zero. The coeffi- cients an, according to equation (7.3.22a), are given by an = 2 l
l 0 f (x) sin
4nπx l
5 dx = 2 l
a 0 hx a sin
4nπx l
5 dx + 2 l
l a h (l − x) (l − a) sin
4nπx l
5 dx. Integration and simplification yields an = 2hl2 π 2a (l − a) 1 n2 sin
4nπa l
5 . Thus, the displacement of the plucked string is u (x, t) = 2hl2 π 2a (l − a) ∞ n=1 1 n2 sin
4nπa l
5 sin
4nπx l
5 cos
4nπc l
5 t. 242 7 Method of Separation of Variables Figure 7.3.2 Plucked String Example 7.3.2. The struck string of length l Here, we consider the string with no initial displacement. Let the string be struck at x = a so that the initial velocity is given by ut (x, 0) = ⎧ ⎨ ⎩ v0 a x, 0 ≤ x ≤ a v0 (l − x) / (l − a), a ≤ x ≤ l . Since u (x, 0) = 0, we have an = 0. By applying equation (7.3.22b), we find that bn = 2 nπc
a 0 v0 a x sin
4nπx l
5 dx + 2 nπc
l a v0 (l − x) (l − a) sin
4nπx l
5 dx = 2v0l 3 π 3ca (l − a) 1 n3 sin
4nπa l
5 . Hence, the displacement of the struck string is u (x, t) = 2v0l 3 π 3ca (l − a) ∞ n=1 1 n3 sin
4nπa l
5 sin
4nπx l
5 cos
4nπc l
5 t. 7.4 Existence and Uniqueness of Solution of the Vibrating String Problem 243 7.4 Existence and Uniqueness of Solution of the Vibrating String Problem In the preceding section we found that the initial boundary-value problem (7.3.1)–(7.3.5) has a formal solution given by (7.3.18). We shall now show that the expression (7.3.18) is the solution of the problem under certain conditions. First we see that u1 (x, t) = ∞ n=1 an cos
4nπc l t
5 sin
4nπx l
5 (7.4.1) is the formal solution of the problem (7.3.1)–(7.3.5) with g (x) ≡ 0, and u2 (x, t) = ∞ n=1 bn sin
4nπc l t
5 sin
4nπx l
5 (7.4.2) is the formal solution of the above problem with f (x) ≡ 0. By linearity of the problem, the solution (7.3.18) may be considered as the sum of the two formal solutions (7.4.1) and (7.4.2). We first assume that f (x) and f ′ (x) are continuous on [0, l], and f (0) = f (l) = 0. Then by Theorem 6.10.1, the series for the function f (x) given by (7.3.20) converges absolutely and uniformly on the interval [0, l]. Using the trigonometric identity sin
4nπx l
5 cos
4nπc l t
5 = 1 2 sin nπ l (x − ct) + 1 2 sin nπ l (x + ct), (7.4.3) u1 (x, t) may be written as u1 (x, t) = 1 2 ∞ n=1 an sin nπ l (x − ct) + 1 2 ∞ n=1 an sin nπ l (x + ct). Define F (x) = ∞ n=1 an sin
4nπx l
5 (7.4.4) and assume that F (x) is the odd periodic extension of f (x), that is, F (x) = f (x) 0 ≤ x ≤ l F (−x) = −F (x) for all x F (x + 2l) = F (x). We can now rewrite u1 (x, t) in the form u1 (x, t) = 1 2 [F (x − ct) + F (x + ct)] . (7.4.5) 244 7 Method of Separation of Variables To show that the boundary conditions are satisfied, we note that u1 (0, t) = 1 2 [F (−ct) + F (ct)] = 1 2 [−F (ct) + F (ct)] = 0 u1 (l, t) = 1 2 [F (l − ct) + F (l + ct)] = 1 2 [F (−l − ct) + F (l + ct)] = 1 2 [−F (l + ct) + F (l + ct)] = 0. Since u1 (x, 0) = 1 2 [F (x) + F (x)] = F (x) = f (x), 0 ≤ x ≤ l, we see that the initial condition u1 (x, 0) = f (x) is satisfied. Thus, equation (7.3.1) and conditions (7.3.2)–(7.3.3) with g (x) ≡ 0 are satisfied. Since f ′ is continuous in [0, l], F ′ exists and is continuous for all x. Thus, if we differentiate u1 (x, t) with respect to t, we obtain ∂u1 ∂t = 1 2 [−c F′ (x − ct) + c F′ (x + ct)] , and ∂u1 ∂t (x, 0) = 1 2 [−c F′ (x) + c F′ (x)] = 0. We therefore see that initial condition (7.3.3) is also satisfied. In order to show that u1 (x, t) satisfies the differential equation (7.3.1), we impose additional restrictions on f. Let f ′′ be continuous on [0, l] and f ′′ (0) = f ′′ (l) = 0. Then, F ′′ exists and is continuous everywhere, and therefore, ∂ 2u1 ∂t2 = 1 2 c 2 [F ′′ (x − ct) + F ′′ (x + ct)] , ∂ 2u1 ∂x2 = 1 2 [F ′′ (x − ct) + F ′′ (x + ct)] . We find therefore that ∂ 2u1 ∂t2 = c 2 ∂ 2u1 ∂x2 . Next, we shall state the assumptions which must be imposed on g to make u2 (x, t) the solution of problem (7.3.1)–(7.3.5) with f (x) ≡ 0. Let g 7.4 Existence and Uniqueness of Solution of the Vibrating String Problem 245 and g ′ be continuous on [0, l] and let g (0) = g (l) = 0. Then the series for the function g (x) given by (7.3.21) converges absolutely and uniformly in the interval [0, l]. Introducing the new coefficients cn = (nπc/l) bn, we have u2 (x, t) =  l πc∞ n=1 cn n sin
4nπc l t
5 sin
4nπx l
5 . (7.4.6) We shall see that term-by-term differentiation with respect to t is permitted, and hence, ∂u2 ∂t = ∞ n=1 cn cos
4nπc l t
5 sin
4nπx l
5 . (7.4.7) Using the trigonometric identity (7.4.3), we obtain ∂u2 ∂t = 1 2 ∞ n=1 cn sin nπ l (x − ct) + 1 2 ∞ n=1 cn sin nπ l (x + ct). (7.4.8) These series are absolutely and uniformly convergent because of the assumptions on g, and hence, the series (7.4.6) and (7.4.7) converge absolutely and uniformly on [0, l]. Thus, the term-by-term differentiation is justified. Let G (x) = ∞ n=1 cn sin
4nπx l
5 be the odd periodic extension of the function g (x). Then, equation (7.4.8) can be written in the form ∂u2 ∂t = 1 2 [G (x − ct) + G (x + ct)] . Integration yields u2 (x, t) = 1 2
t 0 G (x − ct′ ) dt′ + 1 2
t 0 G (x + ct′ ) dt′ = 1 2c
x+ct x−ct G (τ ) dτ. (7.4.9) It immediately follows that u2 (x, 0) = 0, and ∂u2 ∂t (x, 0) = G (x) = g (x), 0 ≤ x ≤ l. Moreover, u2 (0, t) = 1 2
t 0 G (−ct′ ) dt′ + 1 2
t 0 G (ct′ ) dt′ = − 1 2
t 0 G (ct′ ) dt′ + 1 2
t 0 G (ct′ ) dt′ = 0 246 7 Method of Separation of Variables and u2 (l, t) = 1 2
t 0 G (l − ct′ ) dt′ + 1 2
t 0 G (l + ct′ ) dt′ = 1 2
t 0 G (−l − ct′ ) dt′ + 1 2
t 0 G (l + ct′ ) dt′ = − 1 2
t 0 G (l + ct′ ) dt′ + 1 2
t 0 G (l + ct′ ) dt′ = 0. Finally, u2 (x, t) must satisfy the differential equation. Since g ′ is continuous on [0, l], G′ exists so that ∂ 2u2 ∂t2 = c 2 [−G ′ (x − ct) + G ′ (x + ct)] . Differentiating u2 (x, t) represented by equation (7.4.6) with respect to x, we obtain ∂u2 ∂x = 1 c ∞ n=1 cn sin
4nπc l t
5 cos
4nπx l
5 = 1 2c ∞ n=1 cn " − sin nπ l (x − ct) + sin nπ l (x + ct) # = 1 2c [−G (x − ct) + G (x + ct)] . Differentiating again with respect to x, we obtain ∂ 2u2 ∂x2 = 1 2c [−G ′ (x − ct) + G ′ (x + ct)] . It is quite evident that ∂ 2u2 ∂t2 = c 2 ∂ 2u2 ∂x2 . Thus, the solution of the initial boundary-value problem (7.3.1)–(7.3.5) is established. Theorem 7.4.2. (Uniqueness Theorem) There exists at most one solution of the wave equation utt = c 2uxx, 0 < x < l, t > 0, satisfying the initial conditions u (x, 0) = f (x), ut (x, 0) = g (x), 0 ≤ x ≤ l, and the boundary conditions u (0, t)=0, u (l, t)=0, t ≥ 0, where u (x, t) is a twice continuously differentiable function with respect to both x and t. 7.4 Existence and Uniqueness of Solution of the Vibrating String Problem 247 Proof. Suppose that there are two solutions u1 and u2 and let v = u1−u2. It can readily be seen that v (x, t) is the solution of the problem vtt = c 2 vxx, 0 < x < l, t > 0, v (0, t)=0, t ≥ 0, v (l, t)=0, t ≥ 0, v (x, 0) = 0, 0 ≤ x ≤ l, vt (x, 0) = 0, 0 ≤ x ≤ l. We shall prove that the function v (x, t) is identically zero. To do so, consider the energy integral E (t) = 1 2
l 0  c 2 v 2 x + v 2 t dx (7.4.10) which physically represents the total energy of the vibrating string at time t. Since the function v (x, t) is twice continuously differentiable, we differentiate E (t) with respect to t. Thus, dE dt =
l 0  c 2 vxvxt + vtvtt dx. (7.4.11) Integrating the first integral in (7.4.11) by parts, we have
l 0 c 2 vxvxtdx = c 2 vxvt !l 0 −
l 0 c 2 vtvxxdx. But from the condition v (0, t) = 0 we have vt (0, t) = 0, and similarly, vt (l, t) = 0 for x = l. Hence, the expression in the square brackets vanishes, and equation (7.4.11) becomes dE dt =
l 0 vt  vtt − c 2 vxx dx. (7.4.12) Since vtt − c 2vxx = 0, equation (7.4.12) reduces to dE dt = 0 which means E (t) = constant = C. Since v (x, 0) = 0 we have vx (x, 0) = 0. Taking into account the condition vt (x, 0) = 0, we evaluate C to obtain 248 7 Method of Separation of Variables E (0) = C = 1 2
l 0 c 2 v 2 x + v 2 t ! t=0 dx = 0. This implies that E (t) = 0 which can happen only when vx = 0 and vt = 0 for t > 0. To satisfy both of these conditions, we must have v (x, t) = constant. Employing the condition v (x, 0) = 0, we then find v (x, t) = 0. Therefore, u1 (x, t) = u2 (x, t) and the solution u (x, t) is unique. 7.5 The Heat Conduction Problem We consider a homogeneous rod of length l. The rod is sufficiently thin so that the heat is distributed equally over the cross section at time t. The surface of the rod is insulated, and therefore, there is no heat loss through the boundary. The temperature distribution of the rod is given by the solution of the initial boundary-value problem ut = kuxx, 0 < x < l, t > 0, u (0, t)=0, t ≥ 0, u (l, t)=0, t ≥ 0, (7.5.1) u (x, 0) = f (x), 0 ≤ x ≤ l. If we assume a solution in the form u (x, t) = X (x) T (t) = 0. Equation (7.5.1) yields XT′ = kX′′T. Thus, we have X′′ X = T ′ kT = −α 2 , where α is a positive constant. Hence, X and T must satisfy X′′ + α 2X = 0, (7.5.2) T ′ + α 2 kT = 0. (7.5.3) From the boundary conditions, we have u (0, t) = X (0) T (t)=0, u (l, t) = X (l) T (t)=0. Thus, X (0) = 0, X (l)=0, 7.5 The Heat Conduction Problem 249 for an arbitrary function T (t). Hence, we must solve the eigenvalue problem X′′ + α 2X = 0, X (0) = 0, X (l)=0. The solution of equation (7.5.2) is X (x) = A cos αx + B sin αx. Since X (0) = 0, A = 0. To satisfy the second condition, we have X (l) = B sin αl = 0. Since B = 0 yields a trivial solution, we must have B = 0 and hence, sin αl = 0. Thus, α = nπ l for n = 1, 2, 3 .... Substituting these eigenvalues, we have Xn (x) = Bn sin
4nπx l
5 . Next, we consider equation (7.5.3), namely, T ′ + α 2 kT = 0, the solution of which is T (t) = Ce−α 2kt . Substituting α = (nπ/l), we have Tn (t) = Cne −(nπ/l) 2kt . Hence, the nontrivial solution of the heat equation which satisfies the two boundary conditions is un (x, t) = Xn (x) Tn (t) = an e −(nπ/l) 2kt sin
4nπx l
5 , n = 1, 2, 3 ..., where an = BnCn is an arbitrary constant. By the principle of superposition, we obtain a formal series solution as u (x, t) = ∞ n=1 un (x, t), = ∞ n=1 an e −(nπ/l) 2kt sin
4nπx l
5 , (7.5.4) 250 7 Method of Separation of Variables which satisfies the initial condition if u (x, 0) = f (x) = ∞ n=1 an sin
4nπx l
5 . This holds true if f (x) can be represented by a Fourier sine series with Fourier coefficients an = 2 l
l 0 f (x) sin
4nπx l
5 dx. (7.5.5) Hence, u (x, t) = ∞ n=1 1 2 l
l 0 f (τ ) sin
4nπτ l
5 dτ3 e −(nπ/l) 2kt sin
4nπx l
5 (7.5.6) is the formal series solution of the heat conduction problem. Example 7.5.1. (a) Suppose the initial temperature distribution is f (x) = x (l − x). Then, from equation (7.5.5), we have an = 8l 2 n3π 3 , n = 1, 3, 5,.... Thus, the solution is u (x, t) =  8l 2 π 3  ∞ n=1,3,5,... 1 n3 e −(nπ/l) 2kt sin
4nπx l
5 . (b) Suppose the temperature at one end of the rod is held constant, that is, u (l, t) = u0, t ≥ 0. The problem here is ut = k uxx, 0 < x < l, t > 0, u (0, t)=0, u (l, t) = u0, (7.5.7) u (x, 0) = f (x), 0 < x < l. Let u (x, t) = v (x, t) + u0x l . Substitution of u (x, t) in equations (7.5.7) yields vt = k vxx, 0 < x < l, t > 0, v (0, t)=0, v (l, t)=0, v (x, 0) = f (x) − u0x l , 0 < x < l. 7.6 Existence and Uniqueness of Solution of the Heat Conduction Problem 251 Hence, with the knowledge of solution (7.5.6), we obtain the solution u (x, t) = ∞ n=1 1 2 l
l 0
4 f (τ ) − u0τ l
5 sin
4nπτ l
5 dτ3 e −(nπ/l) 2kt sin
4nπx l
5 +
4u0x l
5 . (7.5.8) 7.6 Existence and Uniqueness of Solution of the Heat Conduction Problem In the preceding section, we found that (7.5.4) is the formal solution of the heat conduction problem (7.5.1), where an is given by (7.5.5). We shall prove the existence of this formal solution if f (x) is continuous in [0, l] and f (0) = f (l) = 0, and f ′ (x) is piecewise continuous in (0, l). Since f (x) is bounded, we have |an| = 2 l     
l 0 f (x) sin
4nπx l
5 dx      ≤ 2 l
l 0 |f (x)| dx ≤ C, where C is a positive constant. Thus, for any finite t0 > 0,    an e −(nπ/l) 2kt sin
4nπx l
5   ≤ C e−(nπ/l) 2kt0 when t ≥ t0. According to the ratio test, the series of terms exp " − (nπ/l) 2 kt0 # converges. Hence, by the Weierstrass M-test, the series (7.5.4) converges uniformly with respect to x and t whenever t ≥ t0 and 0 ≤ x ≤ l. Differentiating equation (7.5.4) termwise with respect to t, we obtain ut = − ∞ n=1 an
4nπ l
52 k e−(nπ/l) 2kt sin
4nπx l
5 . (7.6.1) We note that    −an
4nπ l
52 k e−(nπ/l) 2kt sin
4nπx l
5     ≤ C
4nπ l
52 k e−(nπ/l) 2kt0 when t ≥ t0, and the series of terms C (nπ/l) 2 k exp " − (nπ/l) 2 kt0 # converges by the ratio test. Hence, equation (7.6.1) is uniformly convergent in the region 0 ≤ x ≤ l, t ≥ t0. In a similar manner, the series (7.5.4) can be differentiated twice with respect to x, and as a result uxx = − ∞ n=1 an
4nπ l
52 e −(nπ/l) 2kt sin
4nπx l
5 . (7.6.2) 252 7 Method of Separation of Variables Evidently, from equations (7.6.1) and (7.6.2), ut = k uxx. Hence, equation (7.5.4) is a solution of the one-dimensional heat equation in the region 0 ≤ x ≤ l, t ≥ 0. Next, we show that the boundary conditions are satisfied. Here, we note that the series (7.5.4) representing the function u (x, t) converges uniformly in the region 0 ≤ x ≤ l, t ≥ 0. Since the function represented by a uniformly convergent series of continuous functions is continuous, u (x, t) is continuous at x = 0 and x = l. As a consequence, when x = 0 and x = l, solution (7.5.4) satisfies u (0, t)=0, u (l, t)=0, for all t > 0. It remains to show that u (x, t) satisfies the initial condition u (x, 0) = f (x), 0 ≤ x ≤ l. Under the assumptions stated earlier, the series for f (x) given by f (x) = ∞ n=1 an sin
4nπx l
5 is uniformly and absolutely convergent. By Abel’s test of convergence the series formed by the product of the terms of a uniformly convergent series ∞ n=1 an sin
4nπx l
5 and a uniformly bounded and monotone sequence exp " − (nπ/l) 2 kt# converges uniformly with respect to t. Hence, u (x, t) = ∞ n=1 an e −(nπ/l) 2kt sin
4nπx l
5 converges uniformly for 0 ≤ x ≤ l, t ≥ 0, and by the same reasoning as before, u (x, t) is continuous for 0 ≤ x ≤ l, t ≥ 0. Thus, the initial condition u (x, 0) = f (x), 0 ≤ x ≤ l is satisfied. The existence of solution is therefore established. In the above discussion the condition imposed on f (x) is stronger than necessary. The solution can be obtained with a less stringent condition on f (x) (see Weinberger (1965)). 7.6 Existence and Uniqueness of Solution of the Heat Conduction Problem 253 Theorem 7.6.1. (Uniqueness Theorem) Let u (x, t) be a continuously differentiable function. If u (x, t) satisfies the differential equation ut = k uxx, 0 < x < l, t > 0, the initial conditions u (x, 0) = f (x), 0 ≤ x ≤ l, and the boundary conditions u (0, t)=0, u (l, t)=0, t ≥ 0, then, the solution is unique. Proof. Suppose that there are two distinct solutions u1 (x, t) and u2 (x, t). Let v (x, t) = u1 (x, t) − u2 (x, t). Then, vt = k vxx, 0 < x < l, t > 0, v (0, t)=0, v (l, t)=0, t ≥ 0, (7.6.3) v (x, 0) = 0, 0 ≤ x ≤ l, Consider the function defined by the integral J (t) = 1 2k
l 0 v 2 dx. Differentiating with respect to t, we have J ′ (t) = 1 k
l 0 vvtdx =
l 0 vvxxdx, by virtue of equation (7.6.3). Integrating by parts, we have
l 0 vvxxdx = [vvx] l 0 −
l 0 v 2 xdx. Since v (0, t) = v (l, t) = 0, J ′ (t) = −
l 0 v 2 x dx ≤ 0. From the condition v (x, 0) = 0, we have J (0) = 0. This condition and J ′ (t) ≤ 0 implies that J (t) is a nonincreasing function of t. Thus, 254 7 Method of Separation of Variables J (t) ≤ 0. But by definition of J (t), J (t) ≥ 0. Hence, J (t)=0, for t ≥ 0. Since v (x, t) is continuous, J (t) = 0 implies v (x, t)=0 in 0 ≤ x ≤ l, t ≥ 0. Therefore, u1 = u2 and the solution is unique. 7.7 The Laplace and Beam Equations Example 7.7.1. Consider the steady state temperature distribution in a thin rectangular slab. Two sides are insulated, one side is maintained at zero temperature, and the temperature of the remaining side is prescribed to be f (x). Thus, we are required to solve ∇2u = 0, 0 < x < a, 0 < y < b, u (x, 0) = f (x), 0 ≤ x ≤ a, u (x, b)=0, 0 ≤ x ≤ a, ux (0, y)=0, ux (a, y)=0. Let u (x, y) = X (x) Y (y). Substitution of this into the Laplace equation yields X′′ − λX = 0, Y ′′ + λX = 0. Since the boundary conditions are homogeneous on x = 0 and x = a, we have λ = −α 2 with α ≥ 0 for nontrivial solutions of the eigenvalue problem X′′ + α 2X = 0, X′ (0) = X′ (a)=0. The solution is X (x) = A cos αx + B sin αx. Application of the boundary conditions then yields B = 0 and α = (nπ/a) with n = 0, 1, 2,.... Hence, Xn (x) = A cos
4nπx a
5 . 7.7 The Laplace and Beam Equations 255 The solution of the Y equation is clearly Y (y) = C cosh αy + D sinh αy which can be written in the form Y (y) = E sinh α (y + F), where E =  D2 − C 2 1 2 and F = tanh−1 (C/D) ! /α. Applying the homogeneous boundary condition Y (b) = 0, we obtain Y (b) = E sinh α (b + F)=0 which implies F = −b, E = 0 for nontrivial solutions. Hence, we have u (x, y) = (b − y) b a0 2 + ∞ n=1 an cos
4nπx a
5 sinh(nπ a (y − b) ) . Now we apply the remaining nonhomogeneous condition to obtain u (x, 0) = f (x) = a0 2 + ∞ n=1 an cos
4nπx a
5 sinh  − nπb a  . Since this is a Fourier cosine series, the coefficients are given by a0 = 2 a
a 0 f (x) dx, an = −2 a sinh  nπb a
a 0 f (x) cos
4nπx a
5 dx, n = 1, 2,.... Thus, the solution is u (x, y) =  b − y b  a0 2 + ∞ n=1 a ∗ n sinh nπ a (b − y) sinh nπb a cos
4nπx a
5 , where a ∗ n = 2 a
a 0 f (x) cos
4nπx a
5 dx. If, for example f (x) = x in 0 <x<π, 0="" <y<π,="" then="" we="" find="" (note="" that="" a="π)" a0="π," a∗="" n="2" πn2="" [(−1)n="" −="" 1]="" ,="" 2,...="" and="" hence,="" the="" solution="" has="" final="" form="" u="" (x,="" y)="1" 2="" (π="" +="" ∞="" sinh="" nπ="" cos="" nx.="" 256="" 7="" method="" of="" separation="" variables="" example="" 7.7.2.="" as="" another="" example,="" consider="" transverse="" vibration="" beam.="" equation="" motion="" is="" governed="" by="" utt="" 2uxxxx="0," <="" x="" l,="" t=""> 0, where u (x, t) is the displacement and a is the physical constant. Note that the equation is of the fourth order in x. Let the initial and boundary conditions be u (x, 0) = f (x), 0 ≤ x ≤ l, ut (x, 0) = g (x), 0 ≤ x ≤ l, u (0, t) = u (l, t)=0, t > 0, (7.7.1) uxx (0, t) = uxx (l, t)=0, t> 0. The boundary conditions represent the beam being simple supported, that is, the displacements and the bending moments at the ends are zero. Assume a nontrivial solution in the form u (x, t) = X (x) T (t), which transforms the equation of motion into the forms T ′′ + a 2α 4T = 0, X(iv) − α 4X = 0, α > 0. The equation for X (x) has the general solution X (x) = A cosh αx + B sinh αx + C cos αx + D sin αx. The boundary conditions require that X (0) = X (l)=0, X′′ (0) = X′′ (l)=0. Differentiating X twice with respect to x, we obtain X′′ (x) = Aα2 cosh αx + Bα2 sinh αx − Cα2 cos αx − Dα2 sin αx. Now applying the conditions X (0) = X′′ (0) = 0, we obtain A + C = 0, α2 (A − C)=0, and hence, A = C = 0. The conditions X (l) = X′′ (l) = 0 yield B sinh αl + D sin αl = 0, B sinh αl − D sin αl = 0. 7.7 The Laplace and Beam Equations 257 These equations are satisfied if B sinh αl = 0, D sin αl = 0. Since sinh αl = 0, B must vanish. For nontrivial solutions, D = 0, sin αl = 0, and hence, α =
4nπ l
5 , n = 1, 2, 3,.... We then obtain Xn (x) = Dn sin
4nπx l
5 . The general solution for T (t) is T (t) = E cos  aα2 t + F sin  aα2 t . Inserting the values of α 2 , we obtain Tn (t) = En cos  a
4nπ l
52 t 0 + Fn sin  a
4nπ l
52 t 0 . Thus, the general solution for the transverse vibrations of a beam is u (x, t) = ∞ n=1 an cos  a
4nπ l
52 t 0 + bn sin  a
4nπ l
52 t 0 sin
4nπx l
5 . (7.7.2) To satisfy the initial condition u (x, 0) = f (x), we must have u (x, 0) = f (x) = ∞ n=1 an sin
4nπx l
5 from which we find an = 2 l
l 0 f (x) sin
4nπx l
5 dx. (7.7.3) Now the application of the second initial condition gives ut (x, 0) = g (x) = ∞ n=1 bna
4nπ l
52 sin
4nπx l
5 and hence, bn = 2 al  l nπ 2
l 0 g (x) sin
4nπx l
5 dx. (7.7.4) Thus, the solution of the initial boundary-value problem is given by equations (7.7.2)–(7.7.4). 258 7 Method of Separation of Variables 7.8 Nonhomogeneous Problems The partial differential equations considered so far in this chapter are homogeneous. In practice, there is a very important class of problems involving nonhomogeneous equations. First, we shall illustrate a problem involving a time-independent nonhomogeneous equations. Example 7.8.1. Consider the initial boundary-value problem utt = c 2uxx + F (x), 0 < x < l, t > 0, u (x, 0) = f (x), 0 ≤ x ≤ l, ut (x, 0) = g (x), 0 ≤ x ≤ l, (7.8.1) u (0, t) = A, u (l, t) = B, t > 0. We assume a solution in the form u (x, t) = v (x, t) + U (x). Substitution of u (x, t) in equation (7.8.1) yields vtt = c 2 (vxx + Uxx) + F (x), and if U (x) satisfies the equation c 2Uxx + F (x)=0, then v (x, t) satisfies the wave equation vtt = c 2 vxx. In a similar manner, if u (x, t) is inserted in the initial and boundary conditions, we obtain u (x, 0) = v (x, 0) + U (x) = f (x), ut (x, 0) = vt (x, 0) = g (x), u (0, t) = v (0, t) + U (0) = A, u (l, t) = v (l, t) + U (l) = B . Thus, if U (x) is the solution of the problem c 2Uxx + F = 0, U (0) = A, U (l) = B, then v (x, t) must satisfy vtt = c 2 vxx, v (x, 0) = f (x) − U (x), vt (x, 0) = g (x), (7.8.2) v (0, t)=0, v (l, t)=0. 7.8 Nonhomogeneous Problems 259 Now v (x, t) can be solved easily since U (x) is known. It can be seen that U (x) = A + (B − A) x l + x l
l 0 1 c 2
η 0 F (ξ) dξ dη −
x 0 1 c 2
η 0 F (ξ) dξ dη. As a specific example, consider the problem utt = c 2uxx + h, h is a constant u (x, 0) = 0, ut (x, 0) = 0, (7.8.3) u (0, t)=0, u (l, t)=0. Then, the solution of the system c 2Uxx + h = 0, U (0) = 0, U (l)=0, is U (x) = h 2c 2  lx − x 2 . The function v (x, t) must satisfy vtt = c 2 vxx, v (x, 0) = − h 2c 2  lx − x 2 , vt (x, 0) = 0, v (0, t)=0, v (l, t)=0. The solution is given (see Section 7.3 with g (x) = 0) by v (x, t) = ∞ n=1 an cos
4nπc l t
5 sin
4nπx l
5 , and the coefficient is an = 2 l
l 0 − h 2c 2  lx − x 2 sin
4nπx l
5 dx an = − 4l 2h n3π 3c 2 for n odd an = 0 for n even. The solution of the given initial boundary-value problem is, therefore, given by 260 7 Method of Separation of Variables u (x, t) = v (x, t) + U (x) = hx 2c 2 (l − x) + ∞ n=1  − 4l 2h c 2π 3  cos (2n − 1) (πct/l) (2n − 1)3 × sin (2n − 1) (πx/l).(7.8.4) Let us now consider the problem of a finite string with an external force acting on it. If the ends are fixed, we have utt − c 2uxx = h (x, t), 0 < x < l, t > 0, u (x, 0) = f (x), 0 ≤ x ≤ l, ut (x, 0) = g (x), 0 ≤ x ≤ l, (7.8.5) u (0, t)=0, u (l, t)=0, t ≥ 0. We assume a solution involving the eigenfunctions, sin (nπx/l), of the associated eigenvalue problem in the form u (x, t) = ∞ n=1 un (t) sin
4nπx l
5 , (7.8.6) where the functions un (t) are to be determined. It is evident that the boundary conditions are satisfied. Let us also assume that h (x, t) = ∞ n=1 hn (t) sin
4nπx l
5 . (7.8.7) Thus, hn (t) = 2 l
l 0 h (x, t) sin
4nπx l
5 dx. (7.8.8) We assume that the series (7.8.6) is convergent. We then find utt and uxx from (7.8.6) and substitution of these values into (7.8.5) yields ∞ n=1 u ′′ n (t) + λ 2 n un (t) ! sin
4nπx l
5 = ∞ n=1 hn (t) sin
4nπx l
5 , where λn = (nπc/l). Multiplying both sides of this equation by sin (mπx/l), where m = 1, 2, 3,..., and integrating from x = 0 to x = l, we obtain u ′′ n (t) + λ 2 n un (t) = hn (t) the solution of which is given by un (t) = an cos λnt + bn sin λnt + 1 λn
t 0 hn (τ ) sin [λn (t − τ )] dτ. (7.8.9) 7.8 Nonhomogeneous Problems 261 Hence, the formal solution (7.8.6) takes the final form u (x, t) = ∞ n=1  an cos λnt + bn sin λnt + 1 λn
t 0 hn (τ ) sin [λn (t − τ )] dτ0 · sin
4nπx l
5 . (7.8.10) Applying the initial conditions, we have u (x, 0) = f (x) = ∞ n=1 an sin
4nπx l
5 . Thus, an = 2 l
l 0 f (x) sin
4nπx l
5 dx. (7.8.11) Similarly, ut (x, 0) = g (x) = ∞ n=1 bnλn sin
4nπx l
5 . Thus, bn =  2 lλn 
l 0 g (x) sin
4nπx l
5 dx. (7.8.12) Hence, the formal solution of the initial boundary-value problem (7.8.5) is given by (7.8.10) with an given by (7.8.11) and bn given by (7.8.12). Example 7.8.2. Determine the solution of the initial boundary-value problem utt − uxx = h, 0 <x< 1,="" t=""> 0, h = constant, u (x, 0) = x (1 − x), 0 ≤ x ≤ 1, ut (x, 0) = 0, 0 ≤ x ≤ 1, (7.8.13) u (0, t)=0, u (1, t)=0, t ≥ 0. In this case, c = 1, λn = nπ, bn = 0 and an is given by an = 2
1 0 x (1 − x) sin nπx dx = 4 (nπ) 3 [1 − (−1)n ] . We also have hn = 2
1 0 h sin
4nπx l
5 dx = 2h nπ [1 − (−1)n ] . 262 7 Method of Separation of Variables Hence, the integral term in (7.8.9) represents φn (t) given by φn (t) = 1 λn
t 0 hn (τ ) sin [λn (t − τ )] dτ = 2h nπλ2 n [1 − (−1)n ] (1 − cos λnt). The solution (7.8.10) is thus given by u (x, t) = ∞ n=1  4 n3π 3 [1 − (−1)n ] cos nπt + 2h n3π 3 [1 − (−1)n ] (1 − cos nπt) 0 · sin nπx. (7.8.14) We have treated the initial boundary-value problem with the fixed end conditions. Problems with other boundary conditions can also be solved in a similar manner. We will now consider the initial boundary-value problem with timedependent boundary conditions, namely, utt − uxx = h (x, t), 0 < x < l, t > 0, u (x, 0) = f (x), 0 ≤ x ≤ l, ut (x, 0) = g (x), 0 ≤ x ≤ l, (7.8.15) u (0, t) = p (t), u (l, t) = q (t), t ≥ 0. We assume a solution in the form u (x, t) = v (x, t) + U (x, t). (7.8.16) Substituting this into equation (7.8.15), we obtain vtt − c 2 vxx = h − Utt + c 2Uxx. For the initial and boundary conditions, we have v (x, 0) = f (x) − U (x, 0), vt (x, 0) = g (x) − Ut (x, 0), v (0, t) = p (t) − U (0, t), v (l, t) = q (t) − U (l, t). In order to make the boundary conditions homogeneous, we set U (0, t) = p (t), U (l, t) = q (t). Thus, U (x, t) must take the form U (x, t) = p (t) + x l [q (t) − p (t)] . (7.8.17) 7.8 Nonhomogeneous Problems 263 The problem now is to find the function v (x, t) which satisfies vtt − c 2 vxx = h − Utt = H (x, t), v (x, 0) = f (x) − U (x, 0) = F (x), vt (x, 0) = g (x) − Ut (x, 0) = G (x), (7.8.18) v (0, t)=0, v (l, t)=0. This is the same type of problem as the one with homogeneous boundary condition that has previously been treated. Example 7.8.3. Find the solution of the problem utt − uxx = h, 0 <x< 1,="" t=""> 0, h = constant, u (x, 0) = x (1 − x), 0 ≤ x ≤ 1, ut (x, 0) = 0, 0 ≤ x ≤ 1, (7.8.19) u (0, t) = t, u (1, t) = sin t, t ≥ 0. In this case, we use (7.8.16) and (7.8.17) with c = 1 and λn = nπ so that u (x, t) = v (x, t) + U (x, t), U (x, t) = t + x (sin t − t). (7.8.20) Then, v must satisfy vtt − vxx = h + x sin t, v (x, 0) = x (1 − x), vt (x, 0) = −1, (7.8.21) v (0, t)=0, v (1, t)=0. It follows from (7.8.8) that hn (t)=2
1 0 (h + x sin t) sin nπx dx = 2h nπ [1 − (−1)n ] + 2 (−1)n+1 nπ sin t = a + b sin t(say). (7.8.22) We also find an = 2
1 0 x (1 − x) sin nπx dx = 4 (nπ) 3 [1 − (−1)n ] , and bn = 2 nπ
1 0 sin nπx dx = 2 (nπ) 2 [1 − (−1)n ] . 264 7 Method of Separation of Variables Then, we determine the integral term in (7.8.9) so that φn (t) = 1 nπ
t 0 (a + b sin τ ) sin [nπ (t − τ )] dτ = 1 nπ  a nπ (1 − cos nπt) + b 4 [(sin 2t − 2t) cos nπt − (cos 2t − 1) sin nπt] 0 . (7.8.23) Hence, the solution of the problem (7.8.21) is v (x, t) = ∞ n=1 [an cos nπt + bn sin nπt + φn (t)] sin nπx. (7.8.24) Thus, the solution of problem (7.8.19) is given by u (x, t) = v (x, t) + U (x, t), where v (x, t) is given by (7.8.24) and U (x, t) is given by (7.8.20) Example 7.8.4. Use the method of separation of variables to derive the Hermite equation from the Fokker–Planck equation of nonequilibrium statistical mechanics ut − uxx = (x u)x . (7.8.25) We seek a nontrivial separable solution u (x, t) = X (x) T (t) so that equation (7.8.25) reduces to a pair of ordinary differential equations X′′ + xX′ + (1 + n) X = 0 and T ′ + n T = 0, (7.8.26ab) where (−n) is a separation constant. We next use X (x) = exp  − 1 2 x 2  f (x) (7.8.27) and rescale the independent variable to obtain the Hermite equation for f in the form d 2f dξ2 − 2ξ df dξ + 2nf = 0. The solution of (7.8.26b) gives T (t) = cn exp (−nt), (7.8.28) where the coefficients cn are constants. 7.9 Exercises 265 Thus, the solution of the Fokker–Planck equation is given by u (x, t) = ∞ n=1 an exp  −nt − 1 2 x 2  Hn  x √ 2  , (7.8.29) where Hn is the Hermite function and an are arbitrary constants to be determined from the given initial condition u (x, 0) = f (x). (7.8.30) We make the change of variables ξ = x et and u = e t v, (7.8.31) in equation (7.8.25). Consequently, equation (7.8.25) becomes ∂v ∂t = e 2t ∂ 2v ∂ξ2 . (7.8.32) Making another change of variable t to τ (t), we transform (7.8.32) into the linear diffusion equation ∂v ∂τ = ∂ 2v ∂ξ2 . (7.8.33) Finally, we note that the asymptotic behavior of the solution u (x, t) as t → ∞ is of special interest. The reader is referred to Reif (1965) for such behavior. 7.9 Exercises 1. Solve the following initial boundary-value problems: (a) utt = c 2uxx, 0 <x< 1,="" t=""> 0, u (x, 0) = x (1 − x), ut (x, 0) = 0, 0 ≤ x ≤ 1, u (0, t) = u (1, t) = 0, t > 0. (b) utt = c 2uxx, 0 <x<π, t=""> 0, u (x, 0) = 3 sin x, ut (x, 0) = 0, 0 ≤ x ≤ π, u (0, t) = u (1, t) = 0, t > 0. 266 7 Method of Separation of Variables 2. Determine the solutions of the following initial boundary-value problems: (a) utt = c 2uxx, 0 <x<π, t=""> 0, u (x, 0) = 0, ut (x, 0) = 8 sin2 x, 0 ≤ x ≤ π, u (0, t) = u (π, t) = 0, t > 0. (b) utt = c 2uxx = 0, 0 <x< 1,="" t=""> 0, u (x, 0) = 0, ut (x, 0) = x sin πx, 0 ≤ x ≤ 1, u (0, t) = u (1, t) = 0, t > 0. 3. Find the solution of each of the following problems: (a) utt = c 2uxx = 0, 0 <x< 1,="" t=""> 0, u (x, 0) = x (1 − x), ut (x, 0) = x − tan πx 4 , 0 ≤ x ≤ 1, u (0, t) = u (π, t) = 0, t > 0. (b) utt = c 2uxx = 0, 0 <x<π, t=""> 0, u (x, 0) = sin x, ut (x, 0) = x 2 − πx, 0 ≤ x ≤ π, u (0, t) = u (π, t) = 0, t > 0. 4. Solve the following problems: (a) utt = c 2uxx = 0, 0 <x<π, t=""> 0, u (x, 0) = x + sin x, ut (x, 0) = 0, 0 ≤ x ≤ π, u (0, t) = ux (π, t) = 0, t > 0. (b) utt = c 2uxx = 0, 0 <x<π, t=""> 0, u (x, 0) = cos x, ut (x, 0) = 0, 0 ≤ x ≤ π, ux (0, t) = 0, ux (π, t) = 0, t > 0. 7.9 Exercises 267 5. By the method of separation of variables, solve the telegraph equation: utt + aut + bu = c 2uxx, 0 < x < l, t > 0, u (x, 0) = f (x), ut (x, 0) = 0, u (0, t) = u (l, t)=0, t > 0. 6. Obtain the solution of the damped wave motion problem: utt + aut = c 2uxx, 0 < x < l, t > 0, u (x, 0) = 0, ut (x, 0) = g (x), u (0, t) = u (l, t)=0. 7. The torsional oscillation of a shaft of circular cross section is governed by the partial differential equation θtt = a 2 θxx, where θ (x, t) is the angular displacement of the cross section and a is a physical constant. The ends of the shaft are fixed elastically, that is, θx (0, t) − h θ (0, t)=0, θx (l, t) + h θ (l, t)=0. Determine the angular displacement if the initial angular displacement is f (x). 8. Solve the initial boundary-value problem of the longitudinal vibration of a truncated cone of length l and base of radius a. The equation of motion is given by
4 1 − x h
52 ∂ 2u ∂t2 = c 2 ∂ ∂x
4 1 − x h
52 ∂u ∂x , 0 < x < l, t > 0, where c 2 = (E/ρ), E is the elastic modulus, ρ is the density of the material and h = la/ (a − l). The two ends are rigidly fixed. If the initial displacement is f (x), that is, u (x, 0) = f (x), find u (x, t). 9. Establish the validity of the formal solution of the initial boundaryvalue problems: utt = c 2uxx, 0 < x < π, t > 0, u (x, 0) = f (x), ut (x, 0) = g (x), 0 ≤ x ≤ π, ux (0, t)=0, ux (π, t)=0, t > 0. 10. Prove the uniqueness of the solution of the initial boundary-value problem: utt = c 2uxx, 0 < x < π, t > 0, u (x, 0) = f (x), ut (x, 0) = g (x), 0 ≤ x ≤ π, ux (0, t)=0, ux (π, t)=0, t > 0. 268 7 Method of Separation of Variables 11. Determine the solution of utt = c 2uxx + A sinh x, 0 < x < l, t > 0, u (x, 0) = 0, ut (x, 0) = 0, 0 ≤ x ≤ l, u (0, t) = h, u (l, t) = k, t > 0, where h, k, and A are constants. 12. Solve the problem: utt = c 2uxx + Ax, 0 <x< 1,="" t=""> 0, A = constant, u (x, 0) = 0, ut (x, 0) = 0, 0 ≤ x ≤ 1, u (0, t)=0, u (1, t)=0, t > 0. 13. Solve the problem: utt = c 2uxx + x 2 , 0 <x< 1,="" t=""> 0, u (x, 0) = x, ut (x, 0) = 0, 0 ≤ x ≤ 1, u (0, t)=0, u (1, t)=1, t ≥ 0. 14. Find the solution of the following problems: (a) ut = kuxx + h, 0 <x< 1,="" t=""> 0, h = constant, u (x, 0) = u0 (1 − cos πx), 0 ≤ x ≤ 1, u0 = constant, u (0, t)=0, u (l, t)=2u0, t ≥ 0. (b) ut = kuxx − hu, 0 < x < l, t > 0, h = constant, u (x, 0) = f (x), 0 ≤ x ≤ l, ux (0, t) = ux (l, t)=0, t > 0. 15. Obtain the solution of each of the following initial boundary-value problems: (a) ut = 4 uxx, 0 <x< 1,="" t=""> 0, u (x, 0) = x 2 (1 − x), 0 ≤ x ≤ 1, u (0, t) = 0, u (l, t) = 0, t ≥ 0. (b) ut = k uxx, 0 <x<π, t=""> 0, u (x, 0) = sin2 x, 0 ≤ x ≤ π, u (0, t) = 0, u (π, t) = 0, t ≥ 0. 7.9 Exercises 269 (c) ut = uxx, 0 <x< 2,="" t=""> 0, u (x, 0) = x, 0 ≤ x ≤ 2, u (0, t) = 0, ux (2, t) = 1, t ≥ 0. (d) ut = k uxx, 0 <x<l, t=""> 0, u (x, 0) = sin (πx/2l), 0 ≤ x ≤ l, u (0, t) = 0, u (l, t) = 1, t ≥ 0. 16. Find the temperature distribution in a rod of length l. The faces are insulated, and the initial temperature distribution is given by x (l − x). 17. Find the temperature distribution in a rod of length π, one end of which is kept at zero temperature and the other end of which loses heat at a rate proportional to the temperature at that end x = π. The initial temperature distribution is given by f (x) = x. 18. The voltage distribution in an electric transmission line is given by vt = k vxx, 0 < x < l, t > 0. A voltage equal to zero is maintained at x = l, while at the end x = 0, the voltage varies according to the law v (0, t) = Ct, t > 0, where C is a constant. Find v (x, t) if the initial voltage distribution is zero. 19. Establish the validity of the formal solution of the initial boundaryvalue problem: ut = k uxx, 0 < x < l, t > 0, u (x, 0) = f (x), 0 ≤ x ≤ l, u (0, t)=0, ux (l, t)=0, t ≥ 0. 20. Prove the uniqueness of the solution of the problem: ut = k uxx, 0 < x < l, t > 0, u (x, 0) = f (x), 0 ≤ x ≤ l, ux (0, t)=0, ux (l, t)=0, t ≥ 0. 270 7 Method of Separation of Variables 21. Solve the radioactive decay problem: ut − k uxx = Ae−ax , 0 < x < π, t > 0, u (x, 0) = sin x, 0 ≤ x ≤ π, u (0, t)=0, u (π, t)=0, t ≥ 0. 22. Determine the solution of the initial boundary-value problem: ut − k uxx = h (x, t), 0 < x < l, t > 0, k = constant, u (x, 0) = f (x), 0 ≤ x ≤ l, u (0, t) = p (t), u (l, t) = q (t), t ≥ 0. 23. Determine the solution of the initial boundary-value problem: ut − k uxx = h (x, t), 0 < x < l, t > 0, u (x, 0) = f (x), 0 ≤ x ≤ l, u (0, t) = p (t), ux (l, t) = q (t), t ≥ 0. 24. Solve the problem: ut − k uxx = 0, 0 <x< 1,="" t=""> 0, u (x, 0) = x (1 − x), 0 ≤ x ≤ 1, u (0, t) = t, u (1, t) = sin t, t ≥ 0. 25. Solve the problem: ut − 4uxx = xt, 0 <x< 1,="" t="" ≥="" 0,="" u="" (x,="" 0)="sin" πx,="" 0="" ≤="" x="" (0,="" t)="t," (1,="" 2="" ,="" 0.="" 26.="" solve="" the="" problem:="" ut="" −="" k="" uxx="x" cost,="" <="" π,=""> 0, u (x, 0) = sin x, 0 ≤ x ≤ π, u (0, t) = t 2 , u (π, t)=2t, t ≥ 0. 27. Solve the problem: ut − uxx = 2x 2 t, 0 <x< 1,="" t=""> 0, u (x, 0) = cos (3πx/2), 0 ≤ x ≤ 1, u (0, t)=1, ux (1, t) = 3π 2 , t ≥ 0. 28. Solve the problem: ut − 2 uxx = h, 0 <x< 1,="" t=""> 0, h = constant, u (x, 0) = x, 0 ≤ x ≤ 1, u (0, t) = sin t, ux (1, t) + u (1, t)=2, t ≥ 0. 7.9 Exercises 271 29. Determine the solution of the initial boundary-value problem: utt − c 2uxx = h (x, t), 0 < x < l, t > 0, u (x, 0) = f (x), 0 ≤ x ≤ l, ut (x, 0) = g (x), 0 ≤ x ≤ l, u (0, t) = p (t), ux (l, t) = q (t), t ≥ 0. 30. Determine the solution of the initial boundary-value problem: utt − c 2uxx = h (x, t), 0 < x < l, t > 0, u (x, 0) = f (x), 0 ≤ x ≤ l, ut (x, 0) = g (x), 0 ≤ x ≤ l, ux (0, t) = p (t), ux (l, t) = q (t), t ≥ 0. 31. Solve the problem: utt − uxx = 0, 0 <x< 1,="" t=""> 0, u (x, 0) = x, ut (x, 0) = 0, 0 ≤ x ≤ 1, u (0, t) = t 2 , u (1, t) = cost, t ≥ 0. 32. Solve the problem: utt − 4 uxx = xt, 0 <x< 1,="" t=""> 0, u (x, 0) = x, ut (x, 0) = 0, 0 ≤ x ≤ 1, u (0, t)=0, ux (1, t)=1+ t, t ≥ 0. 33. Solve the problem: utt − 9 uxx = 0, 0 <x< 1,="" t=""> 0, u (x, 0) = sin
4πx 2
5 , ut (x, 0) = 1 + x, 0 ≤ x ≤ 1, ux (0, t) = π/2, ux (1, t)=0, t ≥ 0. 34. Find the solution of the problem: utt + 2k ut − c 2uxx = 0, 0 < x < l, t > 0, u (x, 0) = 0, ut (x, 0) = 0, 0 ≤ x ≤ l, ux (0, t)=0, u (l, t) = h, t ≥ 0, h = constant. 35. Solve the problem: ut − c 2uxx + hu = hu0, −π < x < π, t > 0, u (x, 0) = f (x), −π ≤ x ≤ π, u (−π, t) = u (π, t), ux (−π, t) = ux (π, t), t ≥ 0, where h and u0 are constants. 272 7 Method of Separation of Variables 36. Prove the uniqueness theorem for the boundary-value problem involving the Laplace equation: uxx + uyy = 0, 0 < x < a, 0 < y < b, u (x, 0) = f (x), u (x, b)=0, 0 ≤ x ≤ a, ux (0, y)=0= ux (a, y), 0 ≤ y ≤ b. 37. Consider the telegraph equation problem: utt − c 2uxx + aut + bu = 0, 0 < x < l, t > 0, u (x, 0) = f (x), ut (x, 0) = g (x) for 0 ≤ x ≤ l, u (0, t)=0= u (l, t) for t ≥ 0, where a and b are positive constants. (a) Show that, for any T > 0,
l 0  u 2 t + c 2u 2 x + bu2 t=T dx ≤
l 0  u 2 t + c 2u 2 x + bu2 t=0 dx. (b) Use the above integral inequality from (a) to show that the initial boundary-value problem for the telegraph equation can have only one solution. 8 Eigenvalue Problems and Special Functions “The tool which serves as intermediary between theory and practice, between thought and observation, is mathematics; it is mathematics which builds the linking bridges and gives the ever more reliable forms. From this it has come about that our entire contemporary culture, in as much as it is based on the intellectual penetration and the exploitation of nature, has its foundations in mathematics.” David Hilbert “In 1836/7, he published some important joint work with his friend Sturm on what became known as Sturm–Liouville theory, which became important in physics.” Ioan James 8.1 Sturm–Liouville Systems In the preceding chapter, we determined the solutions of partial differential equations by the method of separation of variables. In this chapter, we generalize the method of separation of variables and the associated eigenvalue problems. This generalization, usually known as the Sturm–Liouville theory, greatly extends the scope of the method of separation of variables. Under separable conditions we transformed the second-order homogeneous partial differential equation into two ordinary differential equations (7.2.11) and (7.2.12) which are of the form a1 (x) d 2y dx2 + a2 (x) dy dx + [a3 (x) + λ] y = 0. (8.1.1) If we introduce 274 8 Eigenvalue Problems and Special Functions p (x) = exp
x a2 (t) a1 (t) dt , q (x) = a3 (x) a1 (x) p (x), s (x) = p (x) a1 (x) , (8.1.2) into equation (8.1.1), we obtain d dx  p dy dx + (q + λ s) y = 0, (8.1.3) which is known as the Sturm–Liouville equation. In terms of the Sturm– Liouville operator L ≡ d dx  p d dx + q, equation (8.1.3) can be written as L[y] + λ s (x) y = 0, (8.1.4) where λ is a parameter independent of x, and p, q, and s are real-valued functions of x. To ensure the existence of solutions, we let q and s be continuous and p be continuously differentiable in a closed finite interval [a, b]. The Sturm–Liouville equation is called regular in the interval [a, b] if the functions p (x) and s (x) are positive in the interval [a, b]. Thus, for a given λ, there exist two linearly independent solutions of a regular Sturm– Liouville equation in the interval [a, b]. The Sturm–Liouville equation L[y] + λ s (x) y = 0, a ≤ x ≤ b, together with the separated end conditions a1y (a) + a2y ′ (a)=0, b1y (b) + b2y ′ (b)=0, (8.1.5) where a1 and a2, and likewise b1 and b2, are not both zero and are given real numbers, is called a regular Sturm–Liouville (RSL) system. The values of λ for which the Sturm–Liouville system has a nontrivial solution are called the eigenvalues, and the corresponding solutions are called the eigenfunctions. For a regular Sturm–Liouville problem, we denote the domain of L by D (L), that is, D (L) is the space of all complex-valued functions y defined on [a, b] for which y ′′ ∈ L 2 ([a, b]) and which satisfy boundary conditions (8.1.5). Example 8.1.1. Consider the regular Sturm–Liouville problem y ′′ + λy = 0, 0 ≤ x ≤ π, y (0) = 0, y′ (π)=0. 8.1 Sturm–Liouville Systems 275 When λ ≤ 0, it can be readily shown that λ is not an eigenvalue. However, when λ > 0, the solution of the Sturm–Liouville equation is y (x) = A cos √ λ x + B sin √ λ x. Applying the condition y (0) = 0, we obtain A = 0. The condition y ′ (π)=0 yields B √ λ cos √ λ π = 0. Since λ = 0 and B = 0 yields a trivial solution, we must have cos √ λ π = 0, B = 0. This equation is satisfied if √ λ = 2n − 1 2 , n = 1, 2, 3,..., and hence, the eigenvalues are λn = (2n − 1)2 /4, and the corresponding eigenfunctions are sin  2n − 1 2  x, n = 1, 2, 3,.... Example 8.1.2. Consider the Euler equation x 2 y ′′ + xy′ + λu = 0, 1 ≤ x ≤ e with the end conditions y (1) = 0, y (e)=0. By using the transformation (8.1.2), the Euler equation can be put into the Sturm–Liouville form: d dx  x dy dx + 1 x λ y = 0. The solution of the Euler equation is y (x) = c1 x i √ λ + c2 x −i √ λ . Noting that x ia = e ia ln x = cos (a ln x) + isin (a ln x), the solution y (x) becomes y (x) = A cos
4√ λ ln x
5 + B sin
4√ λ ln x
5 , where A and B are constants related to c1 and c2. The end condition y (1) = 0 gives A = 0, and the end condition y (e) = 0 gives 276 8 Eigenvalue Problems and Special Functions sin √ λ = 0, B = 0, which in turn yields the eigenvalues λn = n 2π 2 , n = 1, 2, 3,..., and the corresponding eigenfunctions sin (nπ ln x), n = 1, 2, 3,.... Another type of problem that often occurs in practice is the periodic Sturm–Liouville system. The Sturm–Liouville equation d dx  p (x) dy dx + [q (x) + λs (x)] y = 0, a ≤ x ≤ b, in which p (a) = p (b), together with the periodic end conditions y (a) = y (b), y′ (a) = y ′ (b) is called a periodic Sturm–Liouville system. Example 8.1.3. Consider the periodic Sturm–Liouville system y ′′ + λy = 0, −π ≤ x ≤ π, y (−π) = y (π), y′ (−π) = y ′ (π). Here we note that p (x) = 1, hence p (−π) = p (π). When λ > 0, we see that the solution of the Sturm–Liouville equation is y (x) = A cos √ λ x + B sin √ λ x. Application of the periodic end conditions yields
4 2 sin √ λ π
5 B = 0,
4 2 √ λ sin √ λ π
5 A = 0. Thus, to obtain a nontrivial solution, we must have sin
4√ λ
5 π = 0, A = 0, B = 0. Consequently, λn = n 2 , n = 1, 2, 3,.... Since sin √ λ π = 0 is satisfied for arbitrary A and B, we obtain two linearly independent eigenfunctions cos nx, and sin nx corresponding to the same eigenvalue n 2 . It can be readily shown that if λ < 0, the solution of the Sturm–Liouville equation does not satisfy the periodic end conditions. However, when λ = 0 the corresponding eigenfunction is 1. Thus, the eigenvalues of the periodic Sturm–Liouville system are 0, & n 2 ' , and the corresponding eigenfunctions are 1, {cos nx}, {sin nx}, where n is a positive integer. 8.2 Eigenvalues and Eigenfunctions 277 8.2 Eigenvalues and Eigenfunctions In Examples 8.1.1 and 8.1.2 of the regular Sturm–Liouville systems in the preceding section, we see that there exists only one linearly independent eigenfunction corresponding to the eigenvalue λ, which is called an eigenvalue of multiplicity one (or a simple eigenvalue). An eigenvalue is said to be of multiplicity k if there exist k linearly independent eigenfunctions corresponding to the same eigenvalue. In Example 8.1.3 of the periodic Sturm– Liouville system, the eigenfunctions cos nx, sin nx correspond to the same eigenvalue n 2 . Thus, this eigenvalue is of multiplicity two. In the preceding examples, we see that the eigenfunctions are cos nx and sin nx for n = 1, 2, 3,.... It can be easily shown by using trigonometric identities that
π −π cos mx cos nx dx = 0, m = n,
π −π cos mx sin nx dx = 0, for all integers m, n,
π −π sin mx sin nx dx = 0, m = n. We say that these functions are orthogonal to each other in the interval [−π, π]. The orthogonality relation holds in general for the eigenfunctions of Sturm–Liouville systems Let φ (x) and ψ (x) be any real-valued integrable functions on an interval I. Then φ and ψ are said to be orthogonal on I with respect to a weight function ρ (x) > 0, if and only if, φ, ψ =
I φ (x) ψ (x) ρ (x) dx = 0. (8.2.1) The interval I may be of infinite extent, or it may be either open or closed at one or both ends of the finite interval. When φ = ψ in (8.2.1) we define the norm of φ by φ =
I φ 2 (x) ρ (x) dx 1 2 . (8.2.2) Theorem 8.2.1. Let the coefficients p, q, and s in the Sturm–Liouville system be continuous in [a, b]. Let the eigenfunctions φj and φk, corresponding to λj and λk, be continuously differentiable. Then φj and φk are orthogonal with respect to the weight function s (x) in [a, b]. Proof. Since φj corresponding to λj satisfies the Sturm–Liouville equation, we have d dx  p φ′ j + (q + λj s) φj = 0 (8.2.3) 278 8 Eigenvalue Problems and Special Functions and for the same reason d dx (p φ′ k )+(q + λks) φk = 0. (8.2.4) Multiplying equation (8.2.3) by φk and equation (8.2.4) by φj , and subtracting, we obtain (λj − λk) s φjφk = φj d dx (p φ′ k ) − φk d dx  p φ′ j = d dx (p φ′ k ) φj −  p φ′ j φk ! and integrating yields (λj − λk)
b a s φjφkdx = p  φjφ ′ k − φ ′ jφk !b a = p (b) φj (b) φ ′ k (b) − φ ′ j (b) φk (b) ! −p (a) φj (a) φ ′ k (a) − φ ′ j (a) φk (a) ! (8.2.5) the right side of which is called the boundary term of the Sturm–Liouville system. The end conditions for the eigenfunctions φj and φk are b1φj (b) + b2φ ′ j (b)=0, b1φk (b) + b2φ ′ k (b)=0. If b2 = 0, we multiply the first condition by φk (b) and the second condition by φj (b), and subtract to obtain φj (b) φ ′ k (b) − φ ′ j (b) φk (b) ! = 0. (8.2.6) In a similar manner, if a2 = 0, we obtain φj (a) φ ′ k (a) − φ ′ j (a) φk (a) ! = 0. (8.2.7) We see by virtue of (8.2.6) and (8.2.7) that (λj − λk)
b a s φj φk dx = 0. (8.2.8) If λj and λk are distinct eigenvalues, then
b a s φj φk dx = 0. (8.2.9) Theorem 8.2.2. The eigenfunctions of a periodic Sturm–Liouville system in [a, b] are orthogonal with respect to the weight function s (x) in [a, b]. 8.2 Eigenvalues and Eigenfunctions 279 Proof. The periodic conditions for the eigenfunctions φj and φk are φj (a) = φj (b), φ′ j (a) = φ ′ j (b), φk (a) = φk (b), φ′ k (a) = φ ′ k (b). Substitution of these into equation (8.2.5) yields (λj − λk)
b a s φj φk dx = [p (b) − p (a)] φj (a) φ ′ k (a) − φ ′ j (a) φk (a) ! . Since p (a) = p (b), we have (λj − λk)
b a s φj φk dx = 0. (8.2.10) For distinct eigenvalues λj = λk, (λj − λk) = 0 and thus,
b a s φj φk dx = 0. (8.2.11) Theorem 8.2.3. For any y, z ∈ D (L), we have the Lagrange identity yL[z] − zL[y] = d dx [p (yz′ − zy′ )] . (8.2.12) Proof. Using the definition of the Sturm–Liouville operator, we have yL[z] − zL[y] = y d dx  p dz dx + qyz − z d dx  p dy dx − qyz = d dx [p (yz′ − zy′ )] . Theorem 8.2.4. The Sturm–Liouville operator L is self-adjoint. In other words, for any y, z ∈ D (L), we have L[y] , z = y,L[z], (8.2.13) where < , > is the inner product in L 2 ([a, b]) defined by f,g =
b a f (x) g (x) dx. (8.2.14) Proof. Since all constants involved in the boundary conditions of Sturm– Liouville system are real, if z ∈ D (L), then z ∈ D (L). Also since p, q and s are real-valued, L[z] = L[z]. Consequently, we have L[y] , z−y,L[z] =
b a (z L[y] − y L[z]) dx = [p (z y′ − y z ′ )]b a , by (8.2.12). (8.2.15) 280 8 Eigenvalue Problems and Special Functions We next show that the right hand side of this equality vanishes for a regular RSL system. If p (a) = 0, the result follows immediately. If p (a) > 0, then y and z satisfy the boundary conditions of the form (8.1.5) at x = a. That is, ⎡ ⎣ y (a) y ′ (a) z (a) z ′ (a) ⎤ ⎦ ⎡ ⎣ a1 a2 ⎤ ⎦ = 0. Since a1 and a2 are not both zero, we have z (a) y ′ (a) − y (a) z ′ (a)=0. A similar argument can be used to the other end point x = b, so that the right-hand side of (8.2.15) vanishes. This proves the theorem. Theorem 8.2.5. All the eigenvalues of a regular Sturm–Liouville system with s (x) > 0 are real. Proof. Let λ be an eigenvalue of a RSL system and let y (x) be the corresponding eigenfunction. This means that y = 0 and L[y] = −λsy. Then 0 = L[y] , y−y,L[y] =  λ − λ
b a s (x)|y (x)| 2 dx. Since s (x) > 0 in [a, b] and y = 0, the integrand is a positive number. Thus, λ = λ, and hence, the eigenvalues are real. This completes the proof. Theorem 8.2.6. If φ1 (x) and φ2 (x) are any two solutions of the equation L[y] + λsy = 0 on [a, b], then p (x) W (x; φ1, φ2) = constant, where W is the Wronskian. Proof. Since φ1 and φ2 are solutions of L[y] + λsy = 0, we have d dx  p dφ1 dx  + (q + λs) φ1 = 0, d dx  p dφ2 dx  + (q + λs) φ2 = 0. Multiplying the first equation by φ2 and the second equation by φ1, and subtracting, we obtain φ1 d dx  p dφ2 dx  − φ2 d dx  p dφ1 dx  = 0. Integrating this equation from a to x, we obtain p (x) [φ1 (x) φ ′ 2 (x) − φ ′ 1 (x) φ2 (x)] = p (a) [φ1 (a) φ ′ 2 (a) − φ ′ 1 (a) φ2 (a)] p (x) W (x; φ, φ2) = constant. (8.2.16) This is called Abel’s formula where W is the Wronskian. 8.2 Eigenvalues and Eigenfunctions 281 Theorem 8.2.7. An eigenfunction of a regular Sturm–Liouville system is unique except for a constant factor. Proof. Let φ1 (x) and φ2 (x) be eigenfunctions corresponding to an eigenvalue λ. Then, according to Abel’s formula (8.2.16), we have p (x) W (x; φ1, φ2) = constant, p (x) > 0, where W is the Wronskian. Thus, if W vanishes at a point in [a, b], it must vanish for all x ∈ [a, b]. Since φ1 and φ2 satisfy the end condition at x = a, we have a1φ1 (a) + a2φ ′ 1 (a)=0, a1φ2 (a) + a2φ ′ 2 (a)=0. Since a1 and a2 are not both zero, we have       φ1 (a) φ ′ 1 (a) φ1 (a) φ ′ 2 (a)       = W (a; φ1, φ2)=0. Therefore, W (x; φ1φ2) = 0 for all x ∈ [a, b], which is a sufficient condition for the linear dependence of two functions φ1 and φ2. Hence, φ1 (x) differs from φ2 (x) only by a constant factor. Theorem 8.2.5 states that all eigenvalues of a regular Sturm–Liouville system are real, but it does not guarantee that any eigenvalue exists. However, it can be proved that a self-adjoint regular Sturm–Liouville system has a denumerably infinite number of eigenvalues. To illustrate this, we consider the following example. Example 8.2.1. Consider the Sturm–Liouville system y ′′ + λy = 0, 0 ≤ x ≤ 1, y (0) = 0, y (1) + hy′ (1) = 0, h> 0 a constant. Here p = 1, q = 0, s = 1. The solution of the Sturm–Liouville equation is y (x) = A cos √ λ x + B sin √ λ x. Since y (0) = 0, gives A = 0, we have y (x) = B sin √ λ x. Applying the second end condition, we have sin √ λ + h √ λ cos √ λ = 0, B = 0 282 8 Eigenvalue Problems and Special Functions which can be rewritten as tan √ λ = −h √ λ. If α = √ λ is introduced in this equation, we have tan α = −h α. This equation does not possess an explicit solution. Thus, we determine the solution graphically by plotting the functions ξ = tan α and ξ = −hα against α, as shown in Figure 8.2.1. The roots are given by the intersection of two curves, and as is evident from the graph, there are infinitely many roots αn for n = 1, 2, 3,.... To each root αn, there corresponds an eigenvalue λn = α 2 n , n = 1, 2, 3,.... Thus, there exists an ordered sequence of eigenvalues λ0 < λ1 < λ2 < λ3 <... with limn→∞ λn = ∞. The corresponding eigenfunctions are sin √ λn x . Figure 8.2.1 Intersection of ξ = tan α and ξ = −h α. 8.3 Eigenfunction Expansions 283 Theorem 8.2.8. A self-adjoint regular Sturm–Liouville system has an infinite sequence of real eigenvalues λ1 < λ2 < λ3 <... with limn→∞ λn = ∞. For each n the corresponding eigenfunction φn (x), uniquely determined up to a constant factor, has exactly n zeros in the interval (a, b). Proof of this theorem can be found in the book by Myint-U (1978). 8.3 Eigenfunction Expansions A real-valued function φ (x) is said to be square-integrable with respect to a weight function ρ (x) > 0, if, on an interval I,
I φ 2 (x) ρ (x) dx < +∞. (8.3.1) An immediate consequence of this definition is the Schwarz inequality    
I φ (x) ψ (x) ρ (x) dx     2 ≤
I φ 2 (x) ρ (x) dx
I ψ 2 (x) ρ (x) dx (8.3.2) for square-integrable functions φ (x) and ψ (x). Let {φn (x)}, for positive integers n, be an orthogonal set of squareintegrable functions with a positive weight function ρ (x) on an interval I. Let f (x) be a given function that can be represented by a uniformly convergent series of the form f (x) = ∞ n=1 cn φn (x), (8.3.3) where the coefficients cn are constants. Now multiplying both sides of (8.3.3) by φm (x) ρ (x) and integrating term by term over the interval I (uniform convergence of the series is a sufficient condition for this), we obtain
I f (x) φm (x) ρ (x) dx = ∞ n=1
I cnφn (x) φm (x) ρ (x) dx, and hence, for n = m,
I f (x) φn (x) ρ (x) dx = cn
I φ 2 n (x) ρ (x) dx. 284 8 Eigenvalue Problems and Special Functions Thus, cn = * I f φn ρ dx * I φ2 n ρ dx . (8.3.4) Hence, we have the following theorem: Theorem 8.3.1. If f is represented by a uniformly convergent series f (x) = ∞ n=1 cnφn (x) on an interval I, where φn are square-integrable functions orthogonal with respect to a positive weight function ρ (x), then cn are determined by cn = * I f φn ρ dx * I φ2 n ρ dx . Example 8.3.1. The Legendre polynomials Pn (x) are orthogonal with respect to the weight function ρ (x) = 1 on (−1, 1). If we assume that f (x) can be represented by the Fourier–Legendre series f (x) = ∞ n=1 cnPn (x) then, cn are given by cn = * 1 −1 f (x) Pn (x) dx * 1 −1 P2 n (x) dx =  2n + 1 2 
1 −1 f (x) Pn (x) dx. In the above discussion, we assumed that the given function f (x) is represented by a uniformly convergent series. This is rather restrictive, and we will show in the following section that f (x) can be represented by a mean-square convergent series. 8.4 Convergence in the Mean Let {φn} be the set of square-integrable functions orthogonal with respect to a weight function ρ (x) on [a, b]. Let sn (x) = n k=1 ckφk (x) 8.4 Convergence in the Mean 285 be the nth partial sum of the series ∞ k=1 ckφk (x). Let f be a square-integrable function. The sequence {sn} is said to converge in the mean to f (x) on the interval I with respect to the weight function ρ (x) if lim n→+∞
I [f (x) − sn (x)]2 ρ (x) dx = 0. (8.4.1) We shall now seek the coefficients ck such that sn (x) represents the best approximation to f (x) in the sense of least squares, that is, we seek to minimize the integral E (ck) =
I [f (x) − sn (x)]2 ρ (x) dx =
I f 2 ρ dx − 2 n k=1 ck
I f φk ρ dx + n k=1 c 2 k
I φ 2 k ρ dx. (8.4.2) This is an extremal problem. A necessary condition on the ck for E to be minimum is that the first partial derivatives of E with respect to these coefficients vanish. Thus, differentiating (8.4.2) with respect to ck, we obtain ∂E ∂ck = −2
I f φk ρ dx + 2ck
I φ 2 k ρ dx = 0 (8.4.3) and hence, ck = * I f φk ρ dx * I φ 2 k ρ dx . (8.4.4) Now if we complete the square, the right side of (8.4.2) becomes E =
I f 2 ρ dx + n k=1
I φ 2 k ρ dx ck − * I f φk ρ dx * I φ 2 k ρ dx 2 − n k=1 * I f φk ρ dx 2 * I φ 2 k ρ dx . The right side shows that E is a minimum if and only if ck is given by (8.4.4). Therefore, this choice of ck yields the best approximation to f (x) in the sense of least squares. For series convergent in the mean to f (x), we conventionally write f (x) ∼ ∞ k=1 ckφk (x), where the coefficients ck are the generalized Fourier coefficients and the series is the generalized Fourier series. This series may or may not be pointwise or uniformly convergent. 286 8 Eigenvalue Problems and Special Functions 8.5 Completeness and Parseval’s Equality Substituting the Fourier coefficients (8.4.4) into (8.4.2), we obtain
I 4 f (x) − n k=1 ckφk (x) 52 ρ (x) dx =
I f 2 ρ dx − n k=1 c 2 k
I φ 2 k ρ dx. Since the left side is nonnegative, we have n k=1 c 2 k
I φ 2 k ρ dx ≤
I f 2 ρ dx. (8.5.1) The integral on the right side is finite, and hence, the series on the left side is bounded above for any n. Thus, as n → ∞, the inequality (8.5.1) may be written as ∞ k=1 c 2 k
I φ 2 k ρ dx ≤
I f 2 ρ dx. (8.5.2) This is called Bessel’s inequality. If the series converges in the mean to f (x), that is, limn→∞
I 4 f (x) − n k=1 ckφk (x) 52 ρ (x) dx = 0, then, it follows from the above derivation that ∞ k=1 c 2 k
I φ 2 kρ dx =
I f 2 ρ dx which is called Parseval’s equality. Sometimes it is known as the completeness relation. Thus, when every continuous square-integrable function f (x) can be expanded into an infinite series f (x) = ∞ k=1 ckφk (x), the sequence of continuous square-integrable functions {φk} orthogonal with respect to the weight function ρ is said to be complete. Next we state the following theorem: Theorem 8.5.1. The eigenfunctions of any regular Sturm–Liouville system are complete in the space of functions that are piecewise continuous on the interval [a, b] with respect to the weight function s (x). Moreover, any piecewise smooth function on [a, b] that satisfies the end conditions of 8.5 Completeness and Parseval’s Equality 287 the regular Sturm–Liouville system can be expanded in an absolutely and uniformly convergent series f (x) = ∞ k=1 ckφk (x), where ck are given by ck =
b a f φk s (x) dx4
b a φ 2 k s (x) dx. Proof of a more general theorem can be found in Coddington and Levinson (1955). Example 8.5.1. Consider a cylindrical wire of length l whose surface is perfectly insulated against the flow of heat. The end l = 0 is maintained at the zero degree temperature, while the other end radiates freely into the surrounding medium of zero degree temperature. Let the initial temperature distribution in the wire be f (x). Find the temperature distribution u (x, t). The initial boundary-value problem is ut = k uxx, 0 < x < l, t > 0, (8.5.3) u (x, 0) = f (x), 0 < x ≤ l, (8.5.4) u (0, t)=0, t > 0, (8.5.5) h u (l, t) + u ′ (l, t)=0, t > 0, h > 0. (8.5.6) By the method of separation of variables, we assume a nontrivial solution in the form u (x, t) = X (x) T (t) and substituting it in the heat equation, we obtain X′′ + λX = 0, T′ + kλT = 0, where λ > 0 is a separation constant. The solution of the latter equation is T (t) = Ce−kλt (8.5.7) where C is an arbitrary constant. The former equation has to be solved subject to the boundary conditions X (0) = 0, h X (l) + X′ (l)=0. This is a Sturm–Liouville system which gives the solution with X (0) = 0 X (x) = B sin √ λ x, (8.5.8) 288 8 Eigenvalue Problems and Special Functions where B is a constant to be determined. Application of the second end condition (8.5.6) yields h sin √ λ l + √ λ cos √ λ l = 0 for B = 0 which can be rewritten as tan √ λ l = − √ λ/h. If α = √ λ l is introduced in the preceding equation, we have tan α = −a α, where a = (1/hl). As in Example 8.2.1, there exists a sequence of eigenvalues λ1 < λ2 < λ3 <... with limn→∞ λn = ∞. The corresponding eigenfunctions are sin √ λn x, and hence, Xn (x) = Bn sin  λn x. (8.5.9) Therefore, combining (8.5.7) with C = Cn and (8.5.9), the solution takes the form un (x, t) = an e −kλnt sin  λn x, an = BnCn which satisfies the heat equation and the boundary conditions. Since the heat equation is linear and homogeneous, we form the series solution u (x, t) = ∞ n=1 an e −kλnt sin  λn x, (8.5.10) which is also a solution, provided it converges and is twice differentiable with respect to x and once differentiable with respect to t. According to Theorem 8.2.1, the eigenfunctions sin √ λn x form an orthogonal system over the interval (0, l). Application of the initial condition yields u (x, 0) = f (x) ∼ ∞ n=1 an sin  λn x. If we assume that f is a piecewise smooth function on [a, b], then, by Theorem 8.5.1, we can expand f (x) in terms of the eigenfunctions, and formally write f (x) = ∞ n=1 an sin  λn x, where the coefficient an is given by an =
l 0 f (x) sin λn x dx4
l 0 sin2  λn x dx. With this value of an, the temperature distribution is given by (8.5.10). 8.6 Bessel’s Equation and Bessel’s Function 289 8.6 Bessel’s Equation and Bessel’s Function Bessel’s equation frequently occurs in problems of applied mathematics and mathematical physics involving cylindrical symmetry. The standard form of Bessel’s equation is given by x 2 y ′′ + xy′ +  x 2 − ν 2 y = 0, (8.6.1) where ν is a nonnegative real number. We shall first restrict our attention to x > 0. Since x = 0 is the regular singular point, a solution is taken in accordance with the Frobenius method to be y (x) = ∞ n=0 an x s+n , (8.6.2) where the index s is to be determined. Substitution of this series into equation (8.6.1) then yields  s 2 − ν 2 a0 x s + " (s + 1)2 − ν 2 # a1x s+1 + ∞ n=2 ("(s + n) 2 − ν 2 # an + an−2 ) x s+n = 0. (8.6.3) The requirement that the coefficient of x s vanish leads to the initial equation  s 2 − ν 2 a0 = 0, (8.6.4) from which it follows that s = + ν for arbitrary a0 = 0. Since the leading term in the series (8.6.2) is a0 x s , it is clear that for ν > 0 the solution of Bessel’s equation corresponding to the choice s = ν vanishes at the origin, whereas the solution corresponding to s = −ν is infinite at that point. We consider first the regular solution of Bessel’s equation, that is, the solution corresponding to the choice s = ν. The vanishing of the coefficient of x s+1 in equation (8.6.3) requires that (2ν + 1) a1 = 0, (8.6.5) which in turn implies that a1 = 0 (since ν ≥ 0). From the requirement that the coefficient of x s+n in equation (8.6.3) be zero, we obtain the two-term recurrence relation an = − an−2 n (2ν + n) . (8.6.6) Since a1 = 0, it is obvious that an = 0 for n = 3, 5, 7,.... The remaining coefficients are given by a2k = (−1)k a0 2 2kk! (ν + k) (ν + k − 1)...(ν + 1) (8.6.7) 290 8 Eigenvalue Problems and Special Functions for k = 1, 2, 3,.... This relation may also be written as a2k = (−1)k 2 νΓ (ν + 1) a0 2 2k+νk!Γ (ν + k + 1), k = 1, 2,..., (8.6.8) where Γ (α) is the gamma function, whose properties are described in the Appendix. Hence, the regular solution of Bessel’s equation takes the form y (x) = a0 ∞ k=0 (−1)k 2 νΓ (ν + 1) 2 2k+νk!Γ (ν + k + 1) x 2k+ν . (8.6.9) It is customary to choose a0 = 1 2 νΓ (ν + 1) (8.6.10) and to denote the corresponding solution by Jν (x). This solution, called the Bessel function of the first kind of order ν, is therefore given by Jν (x) = ∞ k=0 (−1)k x 2k+ν 2 2k+νk! Γ (ν + k + 1). (8.6.11) To determine the irregular solution of the Bessel equation for s = −ν, we proceed as above. In this way, we obtain as the analogue of equation (8.6.5) the relation (−2ν + 1) a1 = 0 from which it follows, without loss of generality, that a1 = 0. Using the recurrence relation an = − an−2 n (n − 2ν) , n ≥ 2 (8.6.12) we obtain the irregular solution of the Bessel function of the first kind of order −ν as J−ν (x) = ∞ k=0 (−1)k x 2k−ν 2 2k−νk! Γ (−ν + k + 1). (8.6.13) It can be easily proved that, if ν is not an integer, Jν and J−ν converge for all values of x, and are linearly independent. Thus, the general solution of the Bessel equation for nonintegral ν is y (x) = c1Jν (x) + c2J−ν (x). (8.6.14) If ν is an integer, say ν = n, then from equation (8.6.13), noting that, when gamma functions in the coefficients of the first n terms become infi- nite, the coefficients become zero, hence we have 8.6 Bessel’s Equation and Bessel’s Function 291 J−n (x) = ∞ k=n (−1)k x 2k−n 2 2k−nk! Γ (−n + k + 1). = (−1)n∞ k=0 (−1)k x 2k+n 2 2k+nk! Γ (n + k + 1). = (−1)n Jn (x). (8.6.15) This shows that J−n is not independent of Jn, and therefore, a second linearly independent solution is required. A number of distinct irregular solutions are discussed in the literature, but the one most commonly used, as defined by Watson (1966), is Yν (x) = (cos νπ) Jν (x) − J−ν (x) sin νπ . (8.6.16) For nonintegral ν, it is obvious that Yν (x), being a linear combination of Jν (x) and J−ν (x), is linearly independent of Jν (x). When ν is a nonnegative integer n, Yν (x) is indeterminate. But Yn (x) = limν→n Yν (x) exists and is a solution of the Bessel equation. Moreover, it is linearly independent of Jn (x). (For an extended treatment, see Watson (1966)). The function Yν (x) is called the Bessel function of the second kind of order ν. Thus, the general solution of the Bessel equation is y (x) = c1Jν (x) + c2Yν (x), for ν ≥ 0. (8.6.17) Like elementary functions, the Bessel functions are tabulated (see Jahnke et al. (1960)). For illustration, the functions J0, J1, Y0 and Y1 are plotted for small values of x in Figure 8.6.1. It should be noted that Jν (x) for ν ≥ 0 and J−ν (x) for a positive integer ν are finite at the origin, but J−ν (x) for nonintegral ν and Yν (x) for ν ≥ 0 approach infinity as x tends to zero. Some of the useful recurrence relations are Jν−1 (x) + Jν+1 (x) = 2ν x Jν (x), (8.6.18) νJν (x) + xJ′ ν (x) = xJν−1 (x), (8.6.19) Jν−1 (x) − Jν+1 (x)=2J ′ ν (x), (8.6.20) νJν (x) − xJ′ ν (x) = xJν+1 (x), (8.6.21) d dx [x νJν (x)] = x νJν−1 (x), (8.6.22) d dx x −νJν (x) ! = −x −νJν+1 (x). (8.6.23) All of these relations also hold true for Yν (x). 292 8 Eigenvalue Problems and Special Functions Figure 8.6.1 Graphs of Jν (x) and Yν (x). For |x| ≫ 1 and |x| ≫ ν, the asymptotic expansion of Jν (x) is Jν−1 (x) ∼ 2 2 πx 1/1 −  4ν 2 − 1 2 4ν 2 − 3 2 2! (8x) 2 +  4ν 2 − 1 2 4ν 2 − 3 2 4ν 2 − 5 2 4ν 2 − 7 2 4! (8x) 4 − ...0 cos φ − / 4ν 2 − 1 2 8x −  4ν 2 − 1 2 4ν 2 − 3 2 4ν 2 − 5 2 3! (8x) 3 + ...0 sin φ 3 (8.6.24) where φ = x −  ν + 1 2  π 2 . For |x| ≫ 1 and |x| ≫ ν, the asymptotic expansion of Yν (x) is 8.6 Bessel’s Equation and Bessel’s Function 293 Yν (x) ∼ 2 2 πx 1/1 −  4ν 2 − 1 2 4ν 2 − 3 2 2! (8x) 2 +  4ν 2 − 1 2 4ν 2 − 3 2 4ν 2 − 5 2 4ν 2 − 7 2 4! (8x) 4 − ...0 sin φ + / 4ν 2 − 1 2 8x −  4ν 2 − 1 2 4ν 2 − 3 2 4ν 2 − 5 2 3! (8x) 3 + ...0 cos φ 3 . (8.6.25) When ν = + (1/2), Bessel’s function may be expressed in the form J 1 2 (x) = 2 2 πx sin x, (8.6.26) J− 1 2 (x) = 2 2 πx cos x. (8.6.27) The Bessel functions which satisfy the condition Jν (akm) + hJ′ ν (akm)=0, h, a = constant, (8.6.28) are orthogonal to each other with respect to the weight function x, that is, for the nonnegative integer ν, the orthogonal relation is
a 0 xJν (xkn) Jν (xkm) dx = 0, n = m. (8.6.29) When n = m, we have the norm Jν (xkm) 2 =
a 0 x [Jν (xkm)]2 dx = 1 2k 2 m ( a 2 k 2 m [J ′ ν (akm)]2 +  a 2 k 2 m − ν 2 [Jν (akm)]2 ) , (8.6.30) where km are the roots of (8.6.28). We now give a particular example of the eigenfunction expansion theorem discussed in Sections 8.4 and 8.5. Assume a formal expansion of the function f (x) defined in 0 ≤ x ≤ a in the form f (x) = ∞ m=1 amJν (xkm), (8.6.31) where the summation is taken over all the positive roots k1, k2, k3, ..., of equation (8.6.28). Multiplying (8.6.31) by xJν (xkn), integrating, and utilizing (8.6.30), we obtain 294 8 Eigenvalue Problems and Special Functions
a 0 xf (x) Jν (xkm) dx = am
a 0 x [Jν (xkm)]2 dx = am 2k 2 m ( a 2 k 2 m [J ′ ν (akm)]2 +  a 2 k 2 m − ν 2 [Jν (akm)]2 ) . (8.6.32) Thus, we have the following theorem: Theorem 8.6.1. If bm =
a 0 xf (x) Jν (xkm) dx (8.6.33) then the expansion (8.6.31) of f (x) takes the form f (x) = ∞ m=1 2k 2 mbmJν (xkm) k 2 m [J ′ ν (akm)]2 + (a 2k 2 m − ν 2) [Jν (akm)]2 . (8.6.34) In particular, when h = 0 in (8.6.28), that is, when km are the positive roots of Jν (akm)=0, then (8.6.34) becomes f (x) = 2 a 2 ∞ m=1 bmJν (xkm) k 2 m [J ′ ν (akm)]2 = 2 a 2 ∞ m=1 bmJν (xkm) [Jν+1 (akm)]2 . (8.6.35) These expansions are known as the Bessel–Fourier series for f (x). They are generated by Sturm–Liouville problems involving the Bessel equation, and arise from problems associated with partial differential equations. Closely related to Bessel’s functions are Hankel’s functions of the first and second kind, defined by H(1) ν (x) = Jν (x) + iYν (x), H(2) ν (x) = Jν (x) − iYν (x), (8.6.36) respectively, where i = √ −1. Other closely related functions are the modified Bessel functions. Consider Bessel’s equation containing a parameter λ, namely, x 2 y ′′ + xy′ +  λ 2x 2 − ν 2 y = 0. (8.6.37) The general solution of this equation is y (x) = c1Jν (λx) + c2Yν (λx). If λ = i, then y (x) = c1Jν (ix) + c2Yν (ix). We write 8.7 Adjoint Forms and Lagrange Identity 295 Jν (ix) = ∞ k=0 (−1)k (ix) 2k+ν 2 2k+νk! Γ (ν + k + 1) = i ν Iν (x), where Iν (x) = ∞ k=0 x 2k+ν 2 2k+νk! Γ (ν + k + 1), (8.6.38) Iν (x) is called the modified Bessel function of the first kind of order ν. As in the case of Jν and J−ν, Iν and I−ν (which is defined in a similar manner) are linearly independent solutions except when ν is an integer. Consequently, we define the modified Bessel function of the second kind of order ν by Kν (x) = π 2  I−ν (x) − Iν (x) sin νπ  . (8.6.39) Thus, we obtain the general solution of the modified Bessel equation x 2 y ′′ + xy′ −  x 2 + ν 2 y = 0 (8.6.40) in the form y (x) = c1Iν (x) + c2Kν (x). (8.6.41) We should note that Iν (0) = ⎧ ⎨ ⎩ 1, ν = 0 0, ν> 0 (8.6.42) and that Kν approaches infinity as x → 0. For a detailed treatment of Bessel and related functions, refer to Watson’s (1966) Theory of Bessel Functions. The eigenvalue problems which involve Bessel’s functions will be described in Section 8.8 on singular Sturm–Liouville systems. 8.7 Adjoint Forms and Lagrange Identity Self-adjoint equations play a very important role in many areas of applied mathematics and mathematical physics. Here we will give a brief account of self-adjoint operators and the Lagrange identity. We consider the equation L[y] = a0 (x) y ′′ + a1 (x) y ′ + a2 (x) y = 0 296 8 Eigenvalue Problems and Special Functions defined on an interval I. Integrating z (x)L[y] by parts from a to x, we have
x a zL[y] dx = (za0) y ′ − (za0) ′ y + (za1) y !x a +
x a (za0) ′′ − (za1) ′ + (za2) ! y dx. (8.7.1) Now, if we define the second-order operator L ∗ by L ∗ [z]=(za0) ′′ − (za1) ′ + (za2) = a0 z ′′ (2a ′ 0 − a1) z ′ + (a ′′ 0 − a ′ 1 + a2) z the relation (8.7.1) takes the form
x a (zL[y] − yL∗ [z]) dx = [a0 (y ′ z − yz′ )+(a1 − a ′ 0 ) yz] x a . (8.7.2) The operator L ∗ is called the adjoint operator corresponding to the operator L. It can be readily verified that the adjoint of L ∗ is L itself. If L and L ∗ are the same, L is said to be self-adjoint. The necessary and sufficient condition for this is that a1 = 2a ′ 0 − a1, a2 = a ′′ 0 − a ′ 1 + a2, which is satisfied if a1 = a ′ 0 . Thus, if L is self-adjoint, we have L(y) = a0y ′′ + a ′ 0 y ′ + a2y = (a0y ′ ) ′ + a2 (x) y. (8.7.3) In general, L[y] is not self-adjoint. But if we let h (x) = 1 a0 exp 
x a1 (t) a0 (t) dt0 (8.7.4) then h (x)L[y] is self-adjoint. Thus, any second-order linear differential equation a0 (x) y ′′ + a1 (x) y ′ + a2 (x) y = 0 (8.7.5) can be made self-adjoint. Multiplying by h (x) given by equation (8.7.4), equation (8.7.5) is transformed into the self-adjoint form d dx p (x) dy dx + q (x) y = 0, (8.7.6) 8.8 Singular Sturm–Liouville Systems 297 where p (x) = exp 
x a1 (t) a0 (t) dt0 , q (x) =  a2 a0  exp 
x a1 (t) a0 (t) dt0 . (8.7.7) For example, the self-adjoint form of the Legendre equation  1 − x 2 y ′′ − 2xy′ + n (n + 1) y = 0 is d dx  1 − x 2 dy dx + n (n + 1) y = 0, (8.7.8) and the self-adjoint form of the Bessel equation x 2 y ′′ + xy′ +  x 2 − ν 2 y = 0 is d dx  x dy dx +  x − ν 2 x  y = 0. (8.7.9) Now, if we differentiate both sides of equation (8.7.2), we obtain zL[y] − yL∗ [z] = d dx [a0 (y ′ z − yz′ )+(a1 − a ′ 0 ) yz] (8.7.10) which is known as the Lagrange identity for the operator L. If we consider the integral from a to b of equation (8.7.2), we obtain Green’s identity
b a (zL[y] − yL∗ [z]) dx = [a0 (y ′ z − yz′ )+(a1 − a ′ 0 ) yz] b a . (8.7.11) When L is self-adjoint, this relation becomes
b a (zL[y] − yL[z]) dx = [a0 (y ′ z − yz′ )]b a . (8.7.12) 8.8 Singular Sturm–Liouville Systems A Sturm–Liouville equation is called singular when it is given on a semiinfinite or infinite interval, or when the coefficient p (x) or s (x) vanishes, or when one of the coefficients becomes infinite at one end or both ends of a finite interval. A singular Sturm–Liouville equation together with appropriate linear homogeneous end conditions is called a singular Sturm–Liouville (SSL) system. The conditions imposed in this case are not like the separated boundary end conditions in the regular Sturm–Liouville system. The 298 8 Eigenvalue Problems and Special Functions condition that is often necessary to prescribe is the boundedness of the function y (x) at the singular end point. To exhibit this, let us consider a problem with a singularity at the end point x = a. By the relation (8.7.12), for any twice continuously differentiable functions y (x) and z (x), we have on (a, b)
b a+ε {zL[y] − yL[z]} dx = p (b) [y ′ (b) z (b) − y (b) z ′ (b)] −p (a + ε) [y ′ (a + ε) z (a + ε) − y (a + ε) z ′ (a + ε)] , where ε is a small positive number. If the conditions lim x→a+ p (x) [y ′ (x) z (x) − y (x) z ′ (x)] = 0, (8.8.1) p (b) [y ′ (b) z (b) − y (b) z ′ (b)] = 0, (8.8.2) are imposed on y and z, it follows that
b a {zL[y] − yL[z]} dx = 0. (8.8.3) For example, when p (a) = 0, the relations (8.8.1) and (8.8.2) are replaced by the conditions 1. y (x) and y ′ (x) are finite as x → a 2. b1y (b) + b2y ′ (b) = 0. Thus, we say that the singular Sturm–Liouville system is self-adjoint, if any functions y (x) and z (x) that satisfy the end conditions satisfy
b a {zL[y] − yL[z]} dx = 0. Example 8.8.1. Consider the singular Sturm–Liouville system involving Legendre’s equation d dx  1 − x 2 dy dx + λy = 0, −1 <x< 1,="" with="" the="" conditions="" that="" y="" and="" ′="" are="" finite="" as="" x="" →="" +="" 1.="" in="" this="" case,="" p="" (x)="1" −="" 2="" s="" vanishes="" at="" legendre="" functions="" of="" first="" kind,="" pn="" (x),="" n="0," 2,...,="" eigenfunctions="" which="" corresponding="" eigenvalues="" λn="n" (n="" 1)="" for="" 2,....="" we="" observe="" here="" singular="" sturm–liouville="" system="" has="" infinitely="" many="" real="" eigenvalues,="" orthogonal="" to="" each="" other.="" example="" 8.8.2.="" another="" a="" is="" bessel="" equation="" fixed="" ν="" 8.8="" systems="" 299="" d="" dx="" ="" dy="" dx="" λ="" ="" 0="" <="" a,="" end="" (a)="0" 0+.="" q="" ,="" now="" (0)="0," becomes="" infinite="" 0+,="" therefore,="" singular.="" if="" kind="" order="" ν,="" namely="" jν="" (knx),="" 2,="" 3,...,="" where="" kna="" nth="" zero="" jν.="" function="" its="" derivative="" both="" .="" thus,="" other="" respect="" weight="" preceding="" examples,="" have="" seen="" (x).="" general,="" they="" square-integrable="" theorem="" 8.8.1.="" distinct="" proof.="" proceeding="" 8.2.1,="" arrive="" (λj="" λk)=""
="" b="" φj="" φk="" (b)="" φ="" k="" j="" !="" suppose="" boundary="" term="" vanishes,="" case="" mentioned="" earlier,="" b1y="" b2y="" then,="" integral="" exists="" by="" virtue="" (8.3.2).="" λj="λk," 8.8.3.="" consider="" involving="" hermite="" u="" ′′="" 2xu′="" λu="0," −∞="" <x<="" ∞,="" (8.8.4)="" not="" self-adjoint.="" let="" −x="" 2u="" takes="" self-adjoint="" form="" 1="" ∞.="" 300="" 8="" eigenvalue="" problems="" special="" nonnegative="" integers="" n,="" φn="" 2hn="" hn="" polynomials="" solutions="" (see="" magnus="" oberhettinger="" (1949)).="" now,="" impose="" tends="" +∞.="" satisfied="" because="" fact="" ne="" since="" square-integrable,="" ∞="" hm="" e="" m="n." 8.8.4.="" problem="" transverse="" vibration="" thin="" elastic="" circular="" membrane="" utt="c" urr="" 1="" r="" ur="" t=""> 0, u (r, 0) = f (r), ut (r, 0) = 0, 0 ≤ r ≤ 1, (8.8.5) u (1, t)=0, limr→0 u (r, t) < ∞, t ≥ 0. We seek a nontrivial separable solution in the form u (r, t) = R (r) T (t). Substituting this in the wave equation yields R′′ + (1/r) R′ R = 1 c 2 T ′′ T = −α 2 , where α is a positive constant. The negative sign in front of α 2 is chosen to obtain the solution periodic in time. Thus, we have rR′′ + R ′ + α 2 rR = 0, T′′ + α 2 c 2T = 0. The solution T (t) is therefore given by T (t) = A cos (αct) + B sin (αct). Next, it is required to determine the solution R (r) of the following singular Sturm–Liouville system d dr r dR dr + α 2 rR = 0, (8.8.6) R (1) = 0, limr→0 R (r) < ∞. (8.8.7) We note that in this case, p = r which vanishes at r = 0. The condition on the boundedness of the function R (r) is obtained from the fact that limr→0 u (r, t) = limr→0 R (r) T (t) < ∞ 8.8 Singular Sturm–Liouville Systems 301 which implies that limr→0 R (r) < ∞ (8.8.8) for arbitrary T (t). Equation (8.8.6) is Bessel’s equation of order zero, the solution of which is given by R (r) = CJ0 (αr) + DY0 (αr), (8.8.9) where J0 and Y0 are Bessel’s functions of the first and second kinds respectively of order zero. The condition (8.8.7) requires that D = 0 since Y0 (αr) → −∞ as r → 0. Hence, R (r) = CJ0 (αr). The remaining condition R (1) = 0 yields J0 (α) = 0. This transcendental equation has infinitely many positive zeros α1 < α2 < α3 <.... Thus, the solution of problem (8.8.5) is given by un (r, t) = J0 (αnr) (An cos αnct + Bn sin αnct), n = 1, 2, 3,.... Since the Bessel equation is linear and homogeneous, the linear superposition principle gives u (r, t) = ∞ n=1 J0 (αnr) (An cos αnct + Bn sin αnct), (8.8.10) is also a solution, provided the series converges and is sufficiently differentiable with respect to r and t. Differentiating (8.8.10) formally with respect to t, we obtain ut (r, t) = ∞ n=1 J0 (αnr) (−An αn c sin αnct + Bn αnc cos αnct). Application of the initial condition ut (r, 0) = 0 yields Bn = 0. Consequently, we have u (r, t) = ∞ n=1 AnJ0 (αnr) cos (αnct). (8.8.11) It now remains to show that u (r, t) satisfies the initial condition u (r, 0) = f (r). For this, we have u (r, 0) = f (r) ∼ ∞ n=1 AnJ0 (αnr). 302 8 Eigenvalue Problems and Special Functions If f (r) is piecewise smooth on [0, 1], then the eigenfunctions J0 (αnr) form a complete orthogonal system with respect to the weight function r over the interval (0, 1). Hence, we can formally expand f (r) in terms of the eigenfunctions. Thus, f (r) = ∞ n=1 AnJ0 (αnr), (8.8.12) where the coefficient An is represented by An =
1 0 rf (r) J0 (αnr) dr1
1 0 r [J0 (αnr)]2 dr. (8.8.13) The solution of the problem (8.8.5) is therefore given by (8.8.11) with the coefficients An given by (8.8.13). 8.9 Legendre’s Equation and Legendre’s Function The Legendre equation is  1 − x 2 y ′′ − 2xy′ + ν (ν + 1) y = 0, (8.9.1) where ν is a real number. This equation arises in problems with spherical symmetry in mathematical physics. Its coefficients are analytic at x = 0. Thus, if we expand near the point x = 0, the coefficients are p (x) = − 2x 1 − x 2 = −2x ∞ m=0 x 2m = ∞ m=0 (−2) x 2m+1 , and q (x) = ν (ν + 1) 1 − x 2 = ν (ν + 1) ∞ m=0 x 2m = ∞ m=0 ν (ν + 1) x 2m. We see that these series converge for |x| < 1. Thus, the Legendre equation on |x| < 1 has convergent power series solution at x = 0. Now to find the solution near the ordinary point x = 0, we assume y (x) = ∞ m=0 am x m. Substituting y, y ′ , and y ′′ in the Legendre equation, we obtain  1 − x 2 ∞ m=0 m (m − 1) am x m−2 − 2x ∞ m=0 mam x m−1 +ν (ν + 1) ∞ m=0 am x m = 0. 8.9 Legendre’s Equation and Legendre’s Function 303 Simplification gives ∞ m=0 [(m + 1) (m + 2) am+2 + (ν − m) (ν + m + 1) am] x m = 0. Therefore the coefficients in the power series must satisfy the recurrence relation am+2 = − (ν − m) (ν + m + 1) (m + 1) (m + 2) am, m ≥ 0. (8.9.2) This relation determines a2, a4, a6, ... in terms of a0, and a3, a5, a7, ... in terms of a1. It can easily be verified that a2k and a2k+1 can be expressed in terms of a0 and a1 respectively as a2k = (−1)k ν (ν − 2)...(ν − 2k + 2) (ν + 1) (ν + 3)...(ν + 2k − 1) (2k)! a0 and a2k+1 = (−1)k (ν − 1) (ν − 3)...(ν − 2k + 1) (ν + 2) (ν + 4)...(ν + 2k) (2k + 1)! a1. Hence, the solution of the Legendre equation is y (x) = a0 1 + ∞ k=1 (−1)k ν (ν − 2)...(ν − 2k + 2) (ν + 1) (ν + 3)...(ν + 2k − 1) x 2k (2k)! 3 +a1 x + ∞ k=1 (−1)k (ν − 1) (ν − 3)...(ν − 2k + 1) (ν + 2) (ν + 4)...(ν + 2k) x 2k+1 (2k + 1)! 3 = a0φν (x) + a1ψν (x). (8.9.3) It can easily be proved that the functions φν (x) and ψν (x) converge for x < 1 and are linearly independent. Now consider the case in which ν = n, with n a nonnegative integer. It is then evident from the recurrence relation (8.9.2) that, when m = n, an+2 = an+4 = ... = 0. Consequently, when n is even, the series φn (x) terminates with x n, whereas the series for ψn (x) does not terminate. When n is odd, it is the series for ψn (x) which terminates with x n, while that for φn (x) does not terminate. 304 8 Eigenvalue Problems and Special Functions In the first case (n even), φn (x) is a polynomial of degree n; the same is true for ψn (x) in the second case (n odd). Thus, for any nonnegative integer n, either φn (x) or ψn (x), but not both, is a polynomial of degree n. Consequently, the general solution of the Legendre equation contains a polynomial solution Pn (x) and an infinite series solution Qn (x) for a nonnegative integer n. To find the polynomial solution Pn (x), it is convenient to choose an so that Pn (1) = 1. Let this an be an = (2n)! 2 n (n!)2 . (8.9.4) Rewriting the recurrence relation (8.9.2), we have an−2 = − (n − 1) n 2 (2n − 1) an. Substituting an from (8.9.4) into this relation, we obtain an−2 = − (2n − 2)! 2 n (n − 1)! (n − 2)!, and an−4 = (2n − 4)! 2 n2! (n − 2)! (n − 4)!. It follows by induction that an−2k = (−1)k (2n − 2k)! 2 nk! (n − k)! (n − 2k)!. Hence, we may write Pn (x) in the form Pn (x) =  N k=0 (−1)k (2n − 2k)! 2 nk! (n − k)! (n − 2k)! x n−2k , (8.9.5) where N = (n/2) when n is even, and N = (n − 1) /2 when n is odd. This polynomial Pn (x) is called the Legendre function of the first kind of order n. It is also known as the Legendre polynomial of degree n. The first few Legendre polynomials are P0 (x)=1, P1 (x) = x, P2 (x) = 1 2  3x 2 − 1 , P3 (x) = 1 2  5x 3 − 3x , P4 (x) = 1 8  35x 4 − 30x 2 + 3 . 8.9 Legendre’s Equation and Legendre’s Function 305 Figure 8.9.1 The first four Legendre’s polynomials. These polynomials are plotted in Figure 8.9.1 for small values of x. Recall that for a given nonnegative integer n, only one of the two solutions φn (x) and ψn (x) of Legendre’s equation is a polynomial, while the other in an infinite series. This infinite series, when appropriately normalized, is called the Legendre function of the second kind. It is defined for |x| < 1 by Qn (x) = ⎧ ⎨ ⎩ φn (1) ψn (x) for n even −ψn (1) φn (x) for n odd . (8.9.6) Thus, when n is a nonnegative integer, the general solution of the Legendre equation is given by y (x) = c1Pn (x) + c2Qn (x). (8.9.7) The Legendre polynomial may also be expressed in the form Pn (x) = 1 2 nn! d n dxn  x 2 − 1 n . (8.9.8) This expression is known as the Rodriguez formula. Like Bessel’s functions, Legendre polynomials satisfy certain recurrence relations. Some of the important relations are 306 8 Eigenvalue Problems and Special Functions (n + 1) Pn+1 (x) − (2n + 1) xPn (x) + nPn−1 (x)=0, n ≥ 1, (8.9.9)  x 2 − 1 P ′ n (x) = nxPn (x) − nPn−1 (x), n ≥ 1, (8.9.10) nPn (x) + P ′ n−1 (x) − xP′ n (x)=0, n ≥ 1, (8.9.11) P ′ n+1 (x) = xP′ n (x)+(n + 1) Pn (x), n ≥ 0. (8.9.12) In addition, P2n (−x) = P2n (x), (8.9.13) P2n+1 (−x) = −P2n+1 (x). (8.9.14) These indicate that Pn (x) is an even function for even n, and an odd function for odd n. It can easily be shown that the Legendre polynomials form a sequence of orthogonal functions on the interval [−1, 1]. Thus, we have
1 −1 Pn (x) Pm (x) dx = 0, for n = m. (8.9.15) The norm of the function Pn (x) is given by Pn (x) 2 =
1 −1 P 2 n (x) dx = 2 2n + 1 . (8.9.16) Another important equation in mathematical physics, one which is closely related to the Legendre equation (8.9.1), is Legendre’s associated equation:  1 − x 2 y ′′ − 2xy′ + n (n + 1) − m2 1 − x 2 y = 0, (8.9.17) where m is an integer. Although this equation is independent of the algebraic sign of the integer m, it is often convenient to have the solutions for negative m differ somewhat from those for positive m. We consider first the case for a nonnegative integer m. Introducing the change of variable y =  1 − x 2 m/2 u, |x| < 1, Legendre’s associated equation becomes  1 − x 2 u ′′ − 2 (m + 1) xu′ + (n − m) (n + m + 1) u = 0. But this is the same as the equation obtained by differentiating the Legendre equation (8.9.1) m times. Thus, the general solution of (8.9.17) is given by y (x) =  1 − x 2 m/2 d mY (x) dxm , (8.9.18) 8.9 Legendre’s Equation and Legendre’s Function 307 and Y (x) = c1Pn (x) + c2Qn (x) (8.9.19) is the general solution of (8.9.1). Hence, we have the linearly independent solutions of (8.9.17), known as the associated Legendre functions of the first and second kind, respectively given by P m n (x) =  1 − x 2 m/2 d mPn (x) dxm , (8.9.20) and Q m n (x) =  1 − x 2 m/2 d mQn (x) dxm . (8.9.21) We observe that P 0 n (x) = Pn (x), Q0 n (x) = Qn (x), and that P m n (x) vanishes for m>n. The functions P −m n (x) and Q−m n (x) are defined by P −m n (x)=(−1)m (n − m)! (n + m)! P m n (x), m = 0, 1, 2, . . . , n, (8.9.22) Q −m n (x)=(−1)m (n − m)! (n + m)! Q m n (x), m = 0, 1, 2, . . . , n. (8.9.23) The first few associated Legendre functions are P 1 1 (x) =  1 − x 2 1 2 , P 1 2 (x)=3x  1 − x 2 1 2 , P 2 2 (x)=3  1 − x 2 . The associated Legendre functions of the first kind also form a sequence of orthogonal functions in the interval [−1, 1]. Their orthogonality, as well as their norm, is expressed by the equation
1 −1 P m n (x) P m k (x) dx = 2 (n − m)! (2n + 1) (n + m)! δnk. (8.9.24) Note that (8.9.15) and (8.9.16) are special cases of (8.9.24), corresponding to the choice m = 0. We finally observe that P m n (x) is bounded everywhere in the interval [−1, 1], whereas Qm n (x) is unbounded at the end points x = + 1. Problems in which Legendre’s polynomials arise will be treated in Chapter 10. 308 8 Eigenvalue Problems and Special Functions 8.10 Boundary-Value Problems Involving Ordinary Differential Equations A boundary-value problem consists in finding an unknown solution which satisfies an ordinary differential equation and appropriate boundary conditions at two or more points. This is in contrast to an initial-value problem for which a unique solution exists for an equation satisfying prescribed initial conditions at one point. The linear two-point boundary-value problem, in general, may be written in the form L[y] = f (x), a < x < b, (8.10.1) Ui [y] = αi , 1 ≤ i ≤ n, where L is a linear operator of order n and Ui is the boundary operator defined by Ui [y] = n j=1 aij y (j−1) (a) +n j=1 bij y (j−1) (b). (8.10.2) Here aij , bij , and αi are constants. The treatment of this problem can be found in Coddington and Levinson (1955). More complicated boundary conditions occur in practice. Treating a general differential system is rather complicated and difficult. A large class of boundary-value problems that occur often in the physical sciences consists of the second-order equations of the type y ′′ = f (x, y, y′ ), a<xξ . (8.12.1) Since G (x, ξ) is continuous at x = ξ, we have 316 8 Eigenvalue Problems and Special Functions c2 (ξ) y2 (ξ) − c1 (ξ) y1 (ξ)=0. (8.12.2) The discontinuity in the derivative of G at the point requires that dG dx (x, ξ)     x=ξ+ x=ξ− = c2 (ξ) y ′ 2 (ξ) − c1 (ξ) y ′ 1 (ξ) = − 1 p (ξ) . (8.12.3) Solving equations (8.12.2) and (8.12.3) for c1 and c2, we find c1 (ξ) = −y2 (ξ) p (ξ) W (y1, y2; ξ) , c2 (ξ) = −y1 (ξ) p (ξ) W (y1, y2; ξ) , (8.12.4) where W (y1, y2; ξ) is the Wronskian given by W (y1, y2; ξ) = y1 (ξ) y ′ 2 (ξ) − y2 (ξ) y ′ 1 (ξ). Since the two solutions are linearly independent, the Wronskian differs from zero. From Theorem 8.2.6, with λ = 0, we have p W = constant = C. (8.12.5) Hence, Green’s function is given by G (x, ξ) = ⎧ ⎨ ⎩ −y1 (x) y2 (ξ) /C, for x ≤ ξ −y2 (x) y1 (ξ) /C, for x ≥ ξ. (8.12.6) Thus, we state the following theorem: Theorem 8.12.1. If the associated homogeneous boundary-value problem of (8.11.1)–(8.11.3) has the trivial solution only, then Green’s function exists and is unique. Proof. The proof for uniqueness of Green’s function is left as an exercise for the reader. Example 8.12.1. Consider the problem y ′′ + y = −1, y (0) = 0, y
4π 2
5 = 0. (8.12.7) The solution of L[y]=(dy′/dx) + y = 0 satisfying y (0) = 0 is y1 (x) = sin x, 0 ≤ x<ξ and the solution of L[y] = 0 satisfying y (π/2) = 0 is y2 (x) = cos x, ξ < x ≤ π 2 . The Wronskian of y1 and y2 is then given by W (ξ) = y1 (ξ) y ′ 2 (ξ) − y2 (ξ) y ′ 1 (ξ) = −1. 8.13 The Schr¨odinger Equation and Linear Harmonic Oscillator 317 Since in this case p = 1, (8.12.6) becomes G (x, ξ) = ⎧ ⎨ ⎩ sin x cos ξ, for x ≤ ξ cos x sin ξ, for x ≥ ξ. Therefore, the solution of (8.12.7) is y (x) =
x 0 G (x, ξ) f (ξ) dξ +
π/2 x G (x, ξ) f (ξ) dξ =
x 0 cos x sin ξ dξ +
π/2 x sin x cos ξ dξ = −1 + sin x + cos x. It can be seen in the formula (8.12.6) that Green’s function is symmetric in x and ξ. Example 8.12.2. Construct the Green’s function for the two-point boundaryvalue problem y ′′ (x) + ω 2 y = f (x), y (a) = y (b)=0. This describes the forced oscillation of an elastic string with fixed ends at x = a and x = b. It is easy to check that sin ωx and cos ωx are two functions which satisfy the homogeneous equation y ′′ + ω 2y = 0. These are used to construct two functions y1 (x) and y2 (x) which satisfy the boundary conditions y1 (a) = y2 (b) = 0. Accordingly, y1 (x) = A sin ωx+B cos ωx and y2 (x) = C sin ωx+ D cos ωx, and the resulting functions are y1 (x) = sin ω (x − a), y2 (x) = sin ω (x − b). The corresponding Wronskian is W = −ω sin ω (a − b). Substituting these results into (8.11.10) yields G (x, ξ) = ⎧ ⎪⎨ ⎪⎩ sin ω(ξ−a) sin ω(x−b) −ω sin ω(a−b) , a ≤ ξ 0, the equation (8.13.1) takes the form d 2ψ dx2 +  β − α 2x 2 ψ = 0. (8.13.3) For small β and large x, β − α 2x 2 ∼ −α 2x 2 so that the equation becomes d 2ψ dx2 − α 2x 2ψ = 0. As |x|→∞, ψ (x) = x n exp
4 + αx2 2
5 satisfies (8.13.3) for a finite n so far as leading terms  ∼ −α 2x 2 are concerned. The positive exponential factor is unacceptable because of the boundary conditions, so the asymptotic solution ψ (x) = x n exp
4 − αx2 2
5 suggests the possibility of the exact solution in the form ψ (x) = v (x) exp
4 − αx2 2
5 where v (x) is to be determined. Substituting this result into (8.13.3), we obtain d 2v dx2 − 2αx dv dx + (β − α) v = 0. (8.13.4) In terms of a new independent variable ζ = x √ α, this equation reduces to the form d 2v dζ2 − 2ζ dv dζ +  β α − 1  v = 0. (8.13.5) We seek a power series solution v (ζ) = ∞ n=0 anζ n . (8.13.6) 8.13 The Schr¨odinger Equation and Linear Harmonic Oscillator 319 Substituting this series into equation (8.13.5) and equating the coefficients of ζ n to zero, we obtain the recurrence relation an+2 = (2n + 1 − β/α) (n + 1) (n + 2) an (8.13.7) which gives an+2 an ∼ 2 n as n → ∞. (8.13.8) This ratio is the same as that of the series for ζ n exp  ζ 2
4 ∼ x ne αx2
5 with finite n. This leads to the fact that ψ (x) = v (x) e −αx2/2 ∼ x ne αx2/2 which does not satisfy the basic requirement for |x|→∞. This unacceptable result can only be avoided if n is an integer and the series terminates so that it becomes a polynomial of degree n. This means that an+2 = 0 but an = 0 so that 2n + 1 − β α = 0, (8.13.9) or β α = (2n + 1). Substituting the values for α and β, it turns out that E ≡ En =  n + 1 2  ω
, n = 0, 1, 2,.... (8.13.10) This represents a discrete set of energies. Thus, in quantum mechanics, a stationary state of the harmonic oscillator can assume only one of the values from the set En. The energy is thus quantized, and forms a discrete spectrum. According to the classical theory, the energy forms a continuous spectrum, that is, all non-negative numbers are allowed for the energy of a harmonic oscillator. This shows a remarkable contrast between the results of the classical and quantum theory. The number n which characterizes the energy eigenvalues and eigenfunctions is called the quantum number. The value of n = 0 corresponds to the minimum value of the quantum number with the energy E0 = 1 2 ω
. (8.13.11) This is called the lowest (or ground) state energy which never vanishes as the lowest possible classical energy would. E0 is proportional to
, representing a quantum phenomenon. The discrete energy spectrum is in perfect agreement with the quantization rules of the quantum theory. 320 8 Eigenvalue Problems and Special Functions To determine the eigenfunctions for the harmonic oscillator associated with the eigenvalues En, we obtain the solution of equation (8.13.5) which has the form d 2v dζ2 − 2ζ dv dζ + 2nv = 0. (8.13.12) This is a well-known differential equation for the Hermite polynomials Hn (ζ) of degree n. Thus, the complete eigenfunctions can be expressed in terms of Hn (ζ) as ψn (x) = AnHn  x √ α exp  − αx2 2  , (8.13.13) where An are arbitrary constants. The Hermite polynomials Hn (x) are usually defined by Hn (x)=(−1)n e x 2 Dn
4 e −x 2
5 , D ≡ d dx. (8.13.14) They form an orthogonal system in (−∞,∞) with the weight function exp  −x 2 . The orthogonal relation for these polynomials is
∞ −∞ e −x 2 Hm (x) Hn (x) dx = ⎧ ⎨ ⎩ 0, n = m 2 nn! √ π, n = m. The Hermite polynomials Hn (x) for n = 0, 1, 2, 3, 4 are H0 (x)=1 H1 (x)=2x H2 (x) = −2+4x 2 H3 (x) = −12x + 8x 3 H4 (x) = 12 − 48x 2 + 16x 4 . Finally, the eigenfunctions ψn of the linear harmonic oscillator for the quantum number n = 0, 1, 2, 3 are given in Figure 8.13.1. 8.14 Exercises 321 Figure 8.13.1 Eigenfunctions ψn for n = 0, 1, 2, 3. 8.14 Exercises 1. Determine the eigenvalues and eigenfunctions of the following regular Sturm–Liouville systems: (a) y ′′ + λy = 0, y (0) = 0, y (π)=0. (b) y ′′ + λy = 0, y (0) = 0, y′ (1) = 0. (c) y ′′ + λy = 0, y ′ (0) = 0, y′ (π)=0. (d) y ′′ + λy = 0, y (1) = 0, y (0) + y ′ (0) = 0. 322 8 Eigenvalue Problems and Special Functions 2. Find the eigenvalues and eigenfunctions of the following periodic Sturm– Liouville systems: (a) y ′′ + λy = 0, y (−1) = y (−1), y′ (−1) = y ′ (1). (b) y ′′ + λy = 0, y (0) = y (2π), y′ (0) = y ′ (2π). (c) y ′′ + λy = 0, y (0) = y (π), y′ (0) = y ′ (π). 3. Obtain the eigenvalues and eigenfunctions of the following Sturm– Liouville systems: (a) y ′′ + y ′ + (1 + λ) y = 0, y (0) = 0, y (1) = 0. (b) y ′′ + 2y ′ + (1 − λ) y = 0, y (0) = 0, y′ (1) = 0. (c) y ′′ − 3y ′ + 3 (1 + λ) y = 0, y ′ (0) = 0, y′ (π) = 0. 4. Find the eigenvalues and eigenfunctions of the following regular Sturm– Liouville systems: (a) x 2y ′′ + 3xy′ + λy = 0, 1 ≤ x ≤ e, y (1) = 0, y (e) = 0. (b) d dx " (2 + x) 2 y ′ # + λy = 0, −1 ≤ x ≤ 1, y (−1) = 0, y (1) = 0. (c) (1 + x) 2 y ′′ + 2 (1 + x) y ′ + 3λy = 0, 0 ≤ x ≤ 1, y (0) = 0, y (1) = 0. 8.14 Exercises 323 5. Determine all eigenvalues and eigenfunctions of the Sturm–Liouville systems: (a) x 2y ′′ + xy′ + λy = 0, y (1) = 0, y, y′ are bounded at x = 0. (b) y ′′ + λy = 0, y (0) = 0, y, y′ are bounded at infinity. 6. Expand the function f (x) = sin x, 0 ≤ x ≤ π in terms of the eigenfunctions of the Sturm–Liouville problem y ′′ + λy = 0, y (0) = 0, y (π) + y ′ (π)=0. 7. Find the expansion of f (x) = x, 0 ≤ x ≤ π in a series of eigenfunctions of the Sturm–Liouville system y ′′ + λy = 0, y ′ (0) = 0, y′ (π)=0. 8. Transform each of the following equations into the equivalent selfadjoint form: (a) The Laguerre equation xy′′ + (1 − x) y ′ + ny = 0, n = 0, 1, 2,.... (b) The Hermite equation y ′′ − 2xy′ + 2ny = 0, n = 0, 1, 2,.... (c) The Tchebycheff equation  1 − x 2 y ′′ − xy′ + n 2 y = 0, n = 0, 1, 2,.... 9. If q (x) and s (x) are continuous and p (x) is twice continuously differentiable in [a, b], show that the solutions of the fourth-order Sturm– Liouville system 324 8 Eigenvalue Problems and Special Functions [p (x) y ′′] ′′ + [q (x) + λs (x)] y = 0, " a1y + a2 (py′′) ′ # x=a = 0, " b1y + b2 (py′′) ′ # x=b = 0, [c1y ′ + c2 (py′′)]x=a = 0, [d1y ′ + d2 (py′′)]x=b = 0, where a 2 1 + a 2 2 = 0, b 2 1 + b 2 2 = 0, c 2 1 + c 2 2 = 0, d 2 1 + d 2 2 = 0, are orthogonal with respect to s (x) in [a, b]. 10. If the eigenfunctions of the problem 1 r d dr (ry′ ) + λy = 0, 0 < r < a, c1y (a) + c2y ′ (a)=0, limr→0+ y (r) < ∞, satisfy limr→0+ ry′ (r)=0, show that all the eigenvalues are real for real c1 and c2. 11. Find the Green’s function for each of the following problems: (a) L[y] = y ′′ = 0, y (0) = 0, y′ (1) = 0. (b) L[y] =  1 − x 2 y ′′ − 2xy′ = 0, y (0) = 0, y′ (1) = 0. (c) L[y] = y ′′ + a 2y = 0, a = constant, y (0) = 0, y (1) = 0. 12. Determine the solution of each of the following boundary-value problems: (a) y ′′ + y = 1, y (0) = 0, y (1) = 0. 8.14 Exercises 325 (b) y ′′ + 4y = e x , y (0) = 0, y′ (1) = 0. (c) y ′′ = sin x, y (0) = 0, y (1) + 2y ′ (1) = 0. (d) y ′′ + 4y = −2, y (0) = 0, y  π 4 = 0. (e) y ′′ = −x, y (0) = 2, y (1) + y ′ (1) = 4. (f) y ′′ = −x 2 , y (0) + y ′ (0) = 4, y′ (1) = 2. (g) y ′′ = −x, y (0) = 1, y′ (1) = 2. 13. Determine the solution of the following boundary-value problems: (a) y ′′ = −f (x), y (0) = 0, y′ (1) = 0. (b) y ′′ = −f (x), y (−1) = 0, y (1) = 0. 14. Find the solution of the following boundary-value problems: (a) y ′′ − y = −f (x), y (0) = y (1) = 0. (b) y ′′ − y = −f (x), y′ (0) = y ′ (1) = 0. 15. Show that the Green’s function G (t, ξ) for the forced harmonic oscillator described by initial-value problem x¨ + ω 2x =  F m  sin Ωt, x (0) = a, x˙ (0) = 0, is 326 8 Eigenvalue Problems and Special Functions G (t, ξ) = 1 ω sin ω (t − ξ). Hence, the particular solution is xp (t) = F mω
t 0 sin ω (t − ξ) sin (Ωξ) dξ. 16. Determine the Green’s function for the boundary-value problem xy′′ + y ′ = −f (x), y (1) = 0, limx→0 |y (x)| < ∞. 17. Determine the Green’s function for the boundary-value problem xy′′ + y ′ − n 2 x y = −f (x), y (1) = 0, limx→0 |y (x)| < ∞. 18. Determine the Green’s function for the boundary-value problem 1 − x 2 y ′ !′ − h 2 (1 − x 2) y = −f (x), h = 1, 2, 3,..., lim r→+ 1 |y (x)| < ∞. 19. Prove the uniqueness of the Green’s function for the boundary-value problem L[y] = −f (x), a1y (a) + a2y ′ (a)=0, b1y (b) + b2y ′ (b)=0. 20. Find the Green’s function for the boundary-value problem L[y] = y (iv) = −f (x), y (0) = y (1) = y ′ (0) = y ′ (1) = 0. Prove that the homogeneous problem has a trivial solution only, and prove that the nonhomogeneous problem has a unique solution. 8.14 Exercises 327 21. Determine the Green’s function for the boundary-value problem y ′′ = −f (x), y (−1) = y (1), y′ (−1) = y ′ (1). 22. Consider the nonself-adjoint boundary-value problem L[y] = y ′′ + 3y ′ + 2y = −f (x), 2 y (0) − y (1) = 0, y′ (1) = 2. By direct integration of GL[y] from 0 to 1, show that y (x) = −2 G (1, x) −
1 0 G (x, ξ) f (ξ) dξ is the solution of the boundary-value problem, if G satisfies the system Gξξ − 3Gξ + 2G = 0, ξ = x, G (0, x)=0, 6 G (1, x) − 2 Gξ (1, x) + Gξ (0, x)=0. Find the Green’s function G (x, ξ). 23. Show that dG (x, ξ) dx     ξ=x+ ξ=x− = 1 p (x) is equivalent to dG (x, ξ) dx     x=ξ+ x=ξ− = − 1 p (ξ) . 24. (a) Apply the Pr¨ufer transformation R 2 = y 2 + p 2 (y ′ ) 2 , θ = tan−1  y py′  to transform the Sturm–Liouville equation (8.1.3) into the first order nonlinear equation in the form dR dx = 1 2 r  1 p − q − λr sin 2θ, dθ dx = (q + λr) sin2 θ + 1 p cos2 θ, where a<x 0. (9.2.2) But for v to be a maximum in D, vxx ≤ 0, vyy ≤ 0. Thus, vxx + vyy ≤ 0 which contradicts equation (9.2.2). Hence, the maximum of u must be attained on B. Theorem 9.2.2. (The Minimum Principle) If u (x, y) is harmonic in a bounded domain D and continuous in D = D ∪ B, then u attains its minimum on the boundary B of D. 9.3 Uniqueness and Continuity Theorems 333 Proof. The proof follows directly by applying the preceding theorem to the harmonic function −u (x, y). As a result of the above theorems, we see that u =constant which is evidently harmonic attains the same value in the domain D as on the boundary B. 9.3 Uniqueness and Continuity Theorems Theorem 9.3.1. (Uniqueness Theorem) The solution of the Dirichlet problem, if it exists, is unique. Proof. Let u1 (x, y) and u2 (x, y) be two solutions of the Dirichlet problem. Then u1 and u2 satisfy ▽2u1 = 0, ▽2u2 = 0 in D, u1 = f, u2 = f on B. Since u1 and u2 are harmonic in D, (u1 − u2) is also harmonic in D. But u1 − u2 = 0 on B. The maximum-minimum principle gives u1 − u2 = 0 at all interior points of D. Thus, we have u1 = u2. Therefore, the solution is unique. Theorem 9.3.2. (Continuity Theorem) The solution of the Dirichlet problem depends continuously on the boundary data. Proof. Let u1 and u2 be the solutions of ▽2u1 = 0 in D, u1 = f1 on B, and ▽2u2 = 0 in D, u2 = f2 on B. If v = u1 − u2, then v satisfies 334 9 Boundary-Value Problems and Applications ▽2 v = 0 in D, v = f1 − f2 on B. By maximum and minimum principles, f1 − f2 attains the maximum and minimum of v on B . Thus, if |f1 − f2| < ε, then −ε 0, there exists an integer N such that everywhere on B |fn − fm| < ε for n, m > N. It follows from the continuity theorem that for all n, m > N |un − um| < ε in D, and hence, the theorem is proved. 9.4 Dirichlet Problem for a Circle 1. Interior Problem We shall now establish the existence of the solution of the Dirichlet problem for a circle. The Dirichlet problem is ▽2u = urr + 1 r ur + 1 r 2 uθθ = 0, 0 ≤ r < a, 0 < θ ≤ 2π,(9.4.1) u (a, θ) = f (θ) for all θ in [0, 2π] . (9.4.2) By the method of separation of variables, we seek a solution in the form u (r, θ) = R (r) Θ (θ) = 0. Substitution of this in equation (9.4.1) yields 9.4 Dirichlet Problem for a Circle 335 r 2 R ′′ R + r R ′ R = − Θ ′′ Θ = λ. Hence, r 2R ′′ + rR ′ − λR = 0, (9.4.3) Θ ′′ + λΘ = 0. (9.4.4) Because of the periodicity conditions Θ (0) = Θ (2π) and Θ ′ (0) = Θ ′ (2π) which ensure that the function Θ is single-valued, the case λ < 0 does not yield an acceptable solution. When λ = 0, we have u (r, θ)=(A + B log r) (Cθ + D). Since log r → −∞ as r → 0+ (note that r = 0 is a singular point of equation (9.4.1)), B must vanish in order for u to be finite at r = 0. C must also vanish in order for u to be periodic with period 2π. Hence, the solution for λ = 0 is u = constant. When λ > 0, the solution of equation (9.4.4) is Θ (θ) = A cos √ λ θ + B sin √ λ θ. The periodicity conditions imply √ λ = n for n = 1, 2, 3,.... Equation (9.4.3) is the Euler equation and therefore, the general solution is R (r) = Crn + Dr−n . Since r −n → ∞ as r → 0, D must vanish for u to be continuous at r = 0. Thus, the solution is u (r, 0) = Crn (A cos n θ + B sin n θ) for n = 1, 2,.... Hence, the general solution of equation (9.4.1) may be written in the form u (r, θ) = a0 2 + ∞ n=1
4r a
5n (an cos nθ + bn sin nθ), (9.4.5) where the constant term (a0/2) represents the solution for λ = 0, and an and bn are constants. Letting ρ = r/a, we have u (ρ, θ) = a0 2 + ∞ n=1 ρ n (an cos nθ + bn sin nθ). (9.4.6) 336 9 Boundary-Value Problems and Applications Our next task is to show that u (r, θ) is harmonic in 0 ≤ r 0 such that |a0| < M, |an| < M, |bn| < M, n = 1, 2, 3,.... Thus, if we consider the sequence of functions {un} defined by un (ρ, θ) = ρ n (an cos nθ + bn sin nθ), (9.4.8) we see that |un| < 2ρ n 0M, 0 ≤ ρ ≤ ρ0 < 1. Hence, in any closed circular region, series (9.4.6) converges uniformly. Next, differentiate un with respect to r. Then, for 0 ≤ ρ ≤ ρ0 < 1,     ∂un ∂r     =    n a ρ n−1 (an cos nθ + bn sin nθ)    < 2
4n a
5 ρ n−1 0 M. Thus, the series obtained by differentiating series (9.4.6) term by term with respect to r converges uniformly. In a similar manner, we can prove that the series obtained by twice differentiating series (9.4.6) term by term with respect to r and θ converge uniformly. Consequently, ▽2u = urr + 1 r ur + 1 r 2 uθθ = ∞ n=1 ρ n−2 a 2 (an cos nθ + bn sin nθ) n (n − 1) + n − n 2 ! = 0, 0 ≤ ρ ≤ ρ0 < 1. Since each term of series (9.4.6) is a harmonic function, and since the series converges uniformly, u (r, θ) is harmonic at any interior point of the region 0 ≤ ρ < 1. It now remains to show that u satisfies the boundary data f (θ). Substitution of the Fourier coefficients an and bn into equation (9.4.6) yields 9.4 Dirichlet Problem for a Circle 337 u (ρ, θ) = 1 2π
2π 0 f (θ) dθ + 1 π ∞ n=1 ρ n
2π 0 f (τ ) × (cos nτ cos nθ + sin nτ sin nθ) dτ = 1 2π
2π 0 1 1+2∞ n=1 ρ n cos n (θ − τ ) 3 f (τ ) dτ. (9.4.9) The interchange of summation and integration is permitted due to the uniform convergence of the series. For 0 ≤ ρ ≤ 1 1+2∞ n=1 [ρ n cos n (θ − τ )] = 1 + ∞ n=1 " ρ n e in(θ−τ) + ρ n e −in(θ−τ) # =1+ ρ ei(θ−τ) 1 − ρ ei(θ−τ) + ρ e−i(θ−τ) 1 − ρ e−i(θ−τ) = 1 − ρ 2 1 − ρ ei(θ−τ) − ρ e−i(θ−τ) + ρ 2 = 1 − ρ 2 1 − 2ρ cos (θ − τ ) + ρ 2 . Hence, u (ρ, θ) = 1 2π
2π 0 1 − ρ 2 1 − 2ρ cos (θ − τ ) + ρ 2 f (τ ) dτ. (9.4.10) The integral on the right side of (9.4.10) is called the Poisson integral formula for a circle. Now if f (θ) = 1, then, according to series (9.4.9), u (r, θ) = 1 for 0 ≤ ρ ≤ 1. Thus, equation (9.4.10) gives 1 = 1 2π
2π 0 1 − ρ 2 1 − 2ρ cos (θ − τ ) + ρ 2 dτ. Hence, f (θ) = 1 2π
2π 0 1 − ρ 2 1 − 2ρ cos (θ − τ ) + ρ 2 f (θ) dτ, 0 ≤ ρ < 1. Therefore, u (ρ, θ) − f (θ) = 1 2π
2π 0  1 − ρ 2 [f (τ ) − f (θ)] 1 − 2ρ cos (θ − τ ) + ρ 2 dτ. (9.4.11) Since f (θ) is uniformly continuous on [0, 2π], for given ε > 0, there exists a positive number δ (ε) such that |θ − τ | < δ implies |f (θ) − f (τ )| < ε. If |θ − τ | ≥ δ so that θ − τ = 2nπ for n = 0, 1, 2,..., then 338 9 Boundary-Value Problems and Applications lim ρ→1− 1 − ρ 2 1 − 2ρ cos (θ − τ ) + ρ 2 = 0. In other words, there exists ρ0 such that if |θ − τ | ≥ δ, then 1 − ρ 2 1 − 2ρ cos (θ − τ ) + ρ 2 < ε, for 0 ≤ ρ ≤ ρ0 < 1. Hence, equation (9.4.10) yields |u (r, θ)| − f (θ)| ≤ 1 2π
2π |0−τ|≥δ  1 − ρ 2 |f (τ ) − f (θ)| 1 − 2ρ cos (θ − τ ) + ρ 2 dτ + 1 2π
2π |θ−τ|<δ  1 − ρ 2 |f (θ) − f (τ )| 1 − 2ρ cos (θ − τ ) + ρ 2 dτ ≤ 1 2π (2πε) 2 max 0≤θ≤2π |f (θ)| + ε 2π · 2π = ε 1+2  max 0≤θ≤2π |f (θ)|  which implies that lim ρ→1− u (r, θ) = f (θ) uniformly in θ. Therefore, we state the following theorem: Theorem 9.4.1. There exists one and only one harmonic function u (r, θ) which satisfies the continuous boundary data f (θ). This function is either given by u (r, θ) = 1 2π
2π 0 a 2 − r 2 a 2 − 2ar cos (θ − τ ) + r 2 f (τ ) dτ, (9.4.12) or u (r, θ) = a0 2 + ∞ n=1 r n a n (an cos nθ + bn sin nθ), (9.4.13) where an and bn are the Fourier coefficients of f (θ). For ρ = 0, the Poisson integral formula (9.4.10) becomes u (0, θ) = u (0) = 1 2π
2π 0 f (τ ) dτ. (9.4.14) This result may be stated as follows: Theorem 9.4.2. (Mean Value Theorem) If u is harmonic in a circle, then the value of u at the center is equal to the mean value of u on the boundary of the circle. 9.4 Dirichlet Problem for a Circle 339 Several comments are in order. First, the Continuity Theorem 9.3.2 for the Dirichlet problem for the Laplace equation is a special example of the general result that the Dirichlet problems for all elliptic equations are well-posed. Second, the formula (9.4.12) represents the unique continuous solution of the Laplace equation in 0 ≤ r 1.(9.4.18) 9.5 Dirichlet Problem for a Circular Annulus The natural extension of the Dirichlet problem for a circle is the Dirichlet problem for a circular annulus, that is ∇2u = 0, r2 <r<r1, (9.5.1)="" u="" (r1,="" θ)="f" (θ),="" (r2,="" (θ).="" (9.5.2)="" in="" addition,="" (r,="" must="" satisfy="" the="" periodicity="" condition.="" accordingly,="" f="" (θ)="" and="" g="" also="" be="" periodic="" with="" period="" 2π.="" proceeding="" as="" case="" of="" dirichlet="" problem="" for="" a="" circle,="" we="" obtain="" λ="0" +="" b="" log="" r)="" (cθ="" d).="" condition="" on="" requires="" that="" c="0." then,="" becomes="" 2="" b0="" r,="" where="" a0="2AD" solution=""> 0 is u (r, θ) =
4 Cr √ λ + Dr− √ λ
5
4A cos √ λ θ + B sin √ λ θ
5 , where √ λ = n = 1, 2, 3,.... Thus, the general solution is u (r, θ) = 1 2 (a0 + b0 log r) + ∞ n=1 anr n + bnr −n cos nθ +  cnr n + dnr −n sin nθ! , (9.5.3) where an, bn, cn, and dn are constants. Applying the boundary conditions (9.5.2), we find that the coefficients are given by 9.6 Neumann Problem for a Circle 341 a0 + b0 log r1 = 1 π
2π 0 f (τ ) dτ, anr n 1 + bnr −n 1 = 1 π
2π 0 f (τ ) cos nτ dτ, cnr n 1 + dnr −n 1 = 1 π
2π 0 f (τ ) sin nτ dτ, and a0 + b0 log r2 = 1 π
2π 0 g (τ ) dτ, anr n 2 + bnr −n 2 = 1 π
2π 0 g (τ ) cos nτ dτ, cnr n 2 + dnr −n 2 = 1 π
2π 0 g (τ ) sin nτ dτ. The constants a0, b0, an, bn, cn, dn for n = 1, 2, 3,... can then be determined. Hence, the solution of the Dirichlet problem for an annulus is given by (9.5.3). 9.6 Neumann Problem for a Circle Let u be a solution of the Neumann problem ∇2u = 0 in D, ∂u ∂n = f on B. It is evident that u + constant is also a solution. Thus, we see that the solution of the Neumann problem is not unique, and it differs from another by a constant. Consider the interior Neumann problem ∇2u = 0, r < R, (9.6.1) ∂u ∂n = ∂u ∂r = f (θ), r = R. (9.6.2) Before we determine a solution of the Neumann problem, a necessary condition for the existence of a solution will be established. In Green’s second formula
D
 v∇2u − u∇2 v dS =
B  v ∂u ∂n − u ∂v ∂n ds, (9.6.3) we put v = 1, so that ∇2v = 0 in D and ∂v/∂n = 0 on B. Then, the result is 342 9 Boundary-Value Problems and Applications
D
∇2u dS =
B ∂u ∂n ds. (9.6.4) Substituting of (9.6.1) and (9.6.2) into equation (9.6.4) yields
B f ds = 0 (9.6.5) which may also be written in the form R
2π 0 f (θ) dθ = 0. (9.6.6) As in the case of the interior Dirichlet problem for a circle, the solution of the Laplace equation is u (r, θ) = a0 2 + ∞ k=1 r k (ak cos kθ + bk sin kθ). (9.6.7) Differentiating this with respect to r and applying the boundary condition (9.6.2), we obtain ∂u ∂r (R, θ) = ∞ k=1 kRk−1 (ak cos kθ + bk sin kθ) = f (θ). (9.6.8) Hence, the coefficients are given by ak = 1 kπRk−1
2π 0 f (τ ) cos kτ dτ, k = 1, 2, 3,..., (9.6.9) bk = 1 kπRk−1
2π 0 f (τ ) sin kτ dτ, k = 1, 2, 3,.... Note that the expansion of f (θ) in a series of the form (9.6.8) is possible only by virtue of the compatibility condition (9.6.6) since a0 = 1 π
2π 0 f (τ ) dτ = 0. Inserting ak and bk in equation (9.6.7), we obtain u (r, θ) = a0 2 + R π
2π 0 1∞ k=1
4 r R
5k cos k (θ − τ ) 3 f (τ ) dτ. Using the identity − 1 2 log 1 + ρ 2 − 2ρ cos (θ − τ ) ! = ∞ k=1 1 k ρ k cos {k (θ − τ )} , 9.7 Dirichlet Problem for a Rectangle 343 with ρ = (r/R), we find that u (r, θ) = a0 2 − R 2π
2π 0 log R 2 − 2rR cos (θ − τ ) + r 2 ! f (τ ) dτ. (9.6.10) in which a constant factor R2 in the argument of the logarithm was eliminated by virtue of equation (9.6.6). In a similar manner, for the exterior Neumann problem, we can readily find the solution in the form u (r, θ) = a0 2 + R 2π
2π 0 log R 2 − 2rR cos (θ − τ ) + r 2 ! f (τ ) dτ. (9.6.11) 9.7 Dirichlet Problem for a Rectangle We first consider the boundary-value problem ∇2u = uxx + uyy = 0, 0 < x < a, 0 < y < b, (9.7.1) u (x, 0) = f (x), u (x, b)=0, 0 ≤ x ≤ a, (9.7.2) u (0, y)=0, u (a, y)=0, 0 ≤ y ≤ b. (9.7.3) We seek a nontrivial separable solution in the form u (x, y) = X (x) Y (y) Substituting u (x, y) in the Laplace equation, we obtain X′′ − λX = 0, (9.7.4) Y ′′ + λY = 0, (9.7.5) where λ is a separation constant. Since the boundary conditions are homogeneous for x = 0 and x = a, we choose λ = −α 2 with α > 0 in order to obtain nontrivial solutions of the eigenvalue problem X′′ + α 2X = 0, X (0) = X (a)=0. It is easily found that the eigenvalues are α = nπ a , n = 1, 2, 3,.... and the corresponding eigenfunctions are sin (nπx/a). Hence Xn (x) = Bn sin
4nπx a
5 . 344 9 Boundary-Value Problems and Applications The solution of equation (9.7.5) is Y (y) = C cosh αy +D sinh αy, which may also be written in the form Y (y) = E sinh α (y + F), where E =  D2 − C 2 1 2 and F = (1/α) tanh−1 (C/D). Applying the remaining homogeneous boundary condition u (x, b) = X (x) Y (b)=0, we obtain Y (b) = E sinh α (b + F)=0, and hence, F = −b, E = 0 for a nontrivial solution u (x, y). Thus, we have Yn (y) = En sinh(nπ a (y − b) ) . Because of linearity, the solution becomes u (x, y) = ∞ n=1 an sin
4nπx a
5 sinh(nπ a (y − b) ) , where an = BnEn. Now, we apply the nonhomogeneous boundary condition to obtain u (x, 0) = f (x) = ∞ n=1 an sinh  −nπb a  sin
4nπx a
5 . This is a Fourier sine series and hence, an = −2 a sinh  nπb a
a 0 f (x) sin
4nπx a
5 dx. Thus, the formal solution is given by u (x, y) = ∞ n=1 a ∗ n sinh & nπ a (b − y) ' sinh  nπb a sin
4nπx a
5 , (9.7.6) where a ∗ n = 2 a
a 0 f (x) sin
4nπx a
5 dx. 9.7 Dirichlet Problem for a Rectangle 345 To prove the existence of solution (9.7.6), we first note that sinh nπ a (b − y) sinh nπb a = e −nπy/a 1 − e −(2nπ/a)(b−y) 1 − e−2nπb/a ≤ C1e −nπy/a , where C1 is a constant. Since f (x) is bounded, we have |a ∗ n | ≤ 2 a
a 0 |f (x)| dx = C2. Thus, the series for u (x, y) is dominated by the series ∞ n=1 Me−nπy0/a for y ≥ y0 > 0, M = constant, and hence, u (x, y) converges uniformly in x and y whenever 0 ≤ x ≤ a, y ≥ y0 > 0. Consequently, u (x, y) is continuous in this region and satisfies the boundary values u (0, y) = u (a, y) = u (x, b) = 0. Now differentiating u twice with respect to x, we obtain uxx (x, y) = ∞ n=1 −a ∗ n
4nπ a
52 sinh nπ a (b − y) sinh nπb a sin
4nπx a
5 and differentiating u twice with respect to y, we obtain uyy (x, y) = ∞ n=1 a ∗ n
4nπ a
52 sinh nπ a (b − y) sinh nπb a sin
4nπx a
5 . It is evident that the series for uxx and uyy are both dominated by ∞ n=1 M∗n 2 e −nπy0/a and hence, converge uniformly for any 0 < y0 < b. It follows that uxxand uyy exist, and hence, u satisfies the Laplace equation. It now remains to show that u (x, 0) = f (x). Let f (x) be a continuous function and let f ′ (x) be piecewise continuous on [0, a]. If, in addition, f (0) = f (a) = 0, then, the Fourier series for f (x) converges uniformly. Putting y = 0 in the series for u (x, y), we obtain u (x, 0) = ∞ n=1 a ∗ n sin
4nπx a
5 . Since u (x, 0) converges uniformly to f (x), we write, for ε > 0, |sm (x, 0) − sn (x, 0)| < ε for m, n > Nε, 346 9 Boundary-Value Problems and Applications where sm (x, y) = ∞ n=1 a ∗ n sin
4nπx a
5 . We also know that sm (x, y) − sn (x, y) satisfies the Laplace equation and the boundary conditions at x = 0, x = a and y = b. Then, by the maximum principle, |sm (x, y) − sn (x, y)| < ε for m, n > Nε in the region 0 ≤ x ≤ a, 0 ≤ y ≤ b. Thus, the series for u (x, y) converges uniformly, and as a consequence, u (x, y) is continuous in the region 0 ≤ x ≤ a, 0 ≤ y ≤ b. Hence, we obtain u (x, 0) = ∞ n=1 a ∗ n sin
4nπx a
5 = f (x). Thus, the solution (9.7.6) is established. The general Dirichlet problem ∇2u = 0, 0 < x < a, 0 < y < b, u (x, 0) = f1 (x), u (x, a) = f2 (x), 0 ≤ x ≤ a, u (0, y) = f3 (y), u (b, y) = f4 (y), 0 ≤ y ≤ b can be solved by separating it into four problems, each of which has one nonhomogeneous boundary condition and the rest zero. Thus, determining each solution as in the preceding problem and then adding the four solutions, the solution of the Dirichlet problem for a rectangle can be obtained. 9.8 Dirichlet Problem Involving the Poisson Equation The solution of the Dirichlet problem involving the Poisson equation can be obtained for simple regions when the solution of the corresponding Dirichlet problem for the Laplace equation is known. Consider the Poisson equation ∇2u = uxx + uuu = f (x, y) in D, with the condition u = g (x, y) on B. Assume that the solution can be written in the form u = v + w, 9.8 Dirichlet Problem Involving the Poisson Equation 347 where v is a particular solution of the Poisson equation and w is the solution of the associated homogeneous equation, that is, ∇2w = 0, ∇2 v = f. As soon as v is ascertained, the solution of the Dirichlet problem ∇2w = 0 in D, w = −v + g (x, y) on B can be determined. The usual method of finding a particular solution for the case in which f (x, y) is a polynomial of degree n is to seek a solution in the form of a polynomial of degree (n + 2) with undetermined coefficients. As an example, we consider the torsion problem ∇2u = −2, 0 < x < a, 0 < y < b, u (0, y)=0, u (a, y) = 0; u (x, 0) = 0, u (x, b)=0. We let u = v + w. Now assume v to be the form v (x, y) = A + Bx + Cy + Dx2 + Exy + F y2 . Substituting this in the Poisson equation, we obtain 2D + 2F = −2. The simplest way of satisfying this equation is to choose D = −1 and F = 0. The remaining coefficients are arbitrary. Thus, we take v (x, y) = ax − x 2 so that v reduces to zero on the sides x = 0 and x = a. Next, we find w from ∇2w = 0, 0 < x < a, 0 < y < b, w (0, y) = −v (0, y)=0, w (a, y) = −v (a, 0) = 0, w (x, 0) = −v (x, 0) = −  ax − x 2 , w (x, b) = −v (x, b) = −  ax − x 2 . As in the Dirichlet problem (Section 9.7), the solution is found to be w (x, y) = ∞ n=1
4 an cosh nπy a + bn sinh nπy a
5 sin
4nπx a
5 . 348 9 Boundary-Value Problems and Applications Application of the nonhomogeneous boundary conditions yields w (x, 0) = −  ax − x 2 = ∞ n=1 an sin
4nπx a
5 , w (x, b) = −  ax − x 2 = ∞ n=1  an cosh nπb a + bn sinh nπb a  sin
4nπx a
5 , from which we find an = 2 a
a 0  x 2 − ax sin
4nπx a
5 dx =  0, if n is even −8a 2 π3n3 if n is odd and  an cosh nπb a + bn sinh nπb a  = 2 a
a 0  x 2 − ax sin
4nπx a
5 dx. Thus, we have bn =  1 − cosh nπb a an sinh  nπb a . Hence, the solution of the Dirichlet problem for the Poisson equation is given by u (x, y)=(a − x) x − 8a 2 π 3 ∞ n=1 " sinh (2n − 1) π(b−y) a + sinh (2n − 1) πy a # sinh (2n − 1) πb a sin (2n − 1) πx a (2n − 1)3 . 9.9 The Neumann Problem for a Rectangle Consider the Neumann problem ∇2u = 0, 0 < x < a, 0 < y < b, (9.9.1) ux (0, y) = f1 (y), ux (a, y) = f2 (y), 0 ≤ y ≤ b, (9.9.2) uy (x, 0) = g1 (x), uy (x, b) = g2 (x), 0 ≤ x ≤ a. (9.9.3) The compatibility condition that must be fulfilled in this case is
a 0 [g1 (x) − g2 (x)] dx +
b 0 [f1 (y) − f2 (y)] dy = 0. (9.9.4) We assume a solution in the form 9.9 The Neumann Problem for a Rectangle 349 u (x, y) = u1 (x, y) + u2 (x, y), (9.9.5) where u1 (x, y) is a solution of ∇2u1 = 0, ∂u1 ∂x (0, y)=0, ∂u1 ∂x (a, y)=0, (9.9.6) ∂u1 ∂x (x, 0) = g1 (x), ∂u1 ∂x (x, b) = g2 (x), and where g1 and g2 satisfy the compatibility condition
a 0 [g1 (x) − g2 (x)] dx = 0. (9.9.7) The function u2 (x, y) is a solution of ∇2u2 = 0, ∂u2 ∂x (0, y) = f1 (y) ∂u2 ∂x (a, y) = f2 (y) (9.9.8) ∂u2 ∂y (x, 0) = 0, ∂u2 ∂y (x, b)=0, where f1 and f2 satisfy the compatibility condition
b 0 [f1 (y) − f2 (y)] dy = 0. (9.9.9) Hence, u1 (x, y) and u2 (x, y) can be determined. Conditions (9.9.7) and (9.9.9) ensure that condition (9.9.4) is fulfilled. Thus, the problem is solved. However, the solution obtained in this manner is rather restrictive. In general, condition (9.9.4) does not imply conditions (9.9.7) and (9.9.9). Thus, generally speaking, it is not possible to obtain a solution of the Neumann problem for a rectangle by the method described above. To obtain a general solution, Grunberg (1946) proposed the following method. Suppose we assume a solution in the form u (x, y) = Y0 2 (y) + ∞ n=1 Xn (x) Yn (y), (9.9.10) 350 9 Boundary-Value Problems and Applications where Xn (x) = cos (nπx/a) is an eigenfunction of the eigenvalue problem X′′ + λX = 0, X′ (0) = X′ (a)=0, corresponding to the eigenvalue λn = (nπ/a) 2 . Then, from equation (9.9.10), we see that Yn (y) = 2 a
a 0 u (x, y) Xn (x) dx, = 2 a
a 0 u (x, y) cos
4nπx a
5 dx. (9.9.11) Multiplying both sides of equation (9.9.1) by 2 cos (nπx/a) and integrating with respect to x from 0 to a, we obtain 2 a
a 0 (uxx + uyy) cos
4nπx a
5 dx = 0, or, Y ′′ n + 2 a
a 0 uxx cos
4nπx a
5 dx = 0. Integrating the second term by parts and applying the boundary conditions (9.9.2), we obtain Y ′′ n (y) −
4nπ a
52 Yn (y) = Fn (y), (9.9.12) where Fn (y) = 2[f1 (y) − (−1)n f2 (y)] /a. This is an ordinary differential equation whose solution may be written in the form Yn (y) = An cosh
4nπy a
5 + Bn sinh
4nπy a
5 + 2 πn
y 0 Fn (τ ) sinh(nπ a (y − τ ) ) dτ. (9.9.13) The coefficients Anand Bn are determined from the boundary conditions Y ′ n (0) = 2 a
a 0 uy (x, 0) cos
4nπx a
5 dx = 2 a
a 0 g1 (x) cos
4nπx a
5 dx (9.9.14) and Y ′ n (b) = 2 a
a 0 g2 (x) cos
4nπx a
5 dx. (9.9.15) 9.10 Exercises 351 For n = 0, equation (9.9.12) takes the form Y ′′ 0 (y) = 2 a [f1 (y) − f2 (y)] and hence, Y ′ 0 (y) = 2 a
y 0 [f1 (τ ) − f2 (τ )] dτ + C, where C is an integration constant. Employing the condition (9.9.14) for n = 0, we find C = 2 a
a 0 g1 (x) dx. Thus, we have Y ′ 0 (y) = 2 a 
y 0 [f1 (τ ) − f2 (τ )] dτ +
a 0 g1 (x) dx0 . Consequently, Y ′ 0 (b) = 2 a /
b 0 [f1 (τ ) − f2 (τ )] dτ +
a 0 g1 (x) dx0 . Also from equation (9.9.14), we have Y ′ 0 (b) = 2 a
a 0 g2 (x) dx. It follows from these two expressions for Y ′ 0 (b) that
b 0 [f1 (y) − f2 (y)] dy +
a 0 [g1 (x) − g2 (x)] dx = 0. which is the necessary condition for the existence of a solution to the Neumann problem for a rectangle. 9.10 Exercises 1. Reduce the Neumann problem to the Dirichlet problem in the twodimensional case. 2. Reduce the wave equation un = c 2 (uxx + uyy + uzz) 352 9 Boundary-Value Problems and Applications to the Laplace equation uxx + uyy + uzz + uτ τ = 0 by letting τ = ict where i = √ −1. Obtain the solution of the wave equation in cylindrical coordinates via the solution of the Laplace equation. Assume that u (r, θ, z, τ ) is independent of z. 3. Prove that, if u (x, t) satisfies ut = k uxx for 0 ≤ x ≤ 1, 0 ≤ t ≤ t0, then the maximum value of u is attained either at t = 0 or at the end points x = 0 or x = 1 for 0 ≤ t ≤ t0. This is called the maximum principle for the heat equation. 4. Prove that a function which is harmonic everywhere on a plane and is bounded either above or below is a constant. This is called the Liouville theorem. 5. Show that the compatibility condition for the Neumann problem ∇2u = f in D ∂u ∂n = g on B is
D f dS +
B g ds = 0, where B is the boundary of domain D. 6. Show that the second degree polynomial P = Ax2 + Bxy + Cy2 + Dyz + F z2 + F xz is harmonic if E = − (A + C) and obtain P = A  x 2 − z 2 + Bxy + C  y 2 − z 2 + Dyz + F xz. 7. Prove that a solution of the Neumann problem ∇2u = f in D u = g on B differs from another solution by at most a constant. 9.10 Exercises 353 8. Determine the solution of each of the following problems: (a) ∇2u = 0, 1 <r< 2,="" 0="" <="" θ="" π,="" u="" (1,="" θ)="sin" θ,="" (2,="" ≤="" (r,="" 0)="0," π)="0," 1="" r="" 2.="" (b)="" ∇2u="0," <r<="" (θ="" −="" π),="" (c)="" 3,="" π="" (3,="" 1)="" (r="" 3),="" ="" r,="" 2="0," 3.="" (d)="" (r),="" 9.="" solve="" the="" boundary-value="" problem="" a="" b,="" α,="" (a,="" (θ),="" (b,="" α)="0," b.="" 10.="" verify="" directly="" that="" poisson="" integral="" is="" solution="" of="" laplace="" equation.="" 11.="" a,="" (π="" θ),="" (0,="" bounded.="" 354="" 9="" problems="" and="" applications="" 12.="" +="" 13.="" find="" dirichlet="" <θ<="" 2π,="" 14.="" following="" problems:="" (a)="" ur="" 2π.="" 15.="" where=""
="" f="" ds="" g="" 16.="" robin="" for="" semicircular="" disk="" 9.10="" exercises="" 355="" 17.="" 18.="" determine="" mixed="" r<r,="" hu="" h="constant." 19.="" (θ).="" 20.="" neumann="" sin="" 2θ,="" r1="" <r<r2,="" (r1,="" (r2,="" 21.="" 22.="" <x<="" 1,="" <y<="" (x,="" (x="" 1),="" x="" y)="0," y="" 1.="" (πx),="" 356="" cos="" πy="" ,="" πy,="" 23.="" ux="" (π,="" uy="" π.="" y,="" x,="" 24.="" steady-state="" temperature="" distribution="" in="" rectangular="" plate="" length="" width="" b="" described="" by="" at="" kept="" zero="" degrees,="" while="" insulated.="" prescribed="" 357="" heat="" allowed="" to="" radiate="" freely="" into="" surrounding="" medium="" degree="" temperature.="" is,="" boundary="" conditions="" are="" (x),="" b)="" a.="" distribution.="" 25.="" 26.="" harmonic="" function="" which="" vanishes="" on="" hypotenuse="" has="" values="" other="" two="" sides="" an="" isosceles="" right-angled="" triangle="" formed="" constant.="" 27.="" a)="0," 28.="" third="" 358="" 29.="" 30.="" obtain="" representation="" d,="" ∂u="" ∂n="g" d.="" 31.="" terms="" green’s="" 32.="" inside="" circular="" annular="" region="" governed="" urr="" uθθ="0," −π="" 33.="" consider="" radially="" symmetric="" solid="" homogeneous="" cylinder="" radius="" unity="" height="" h.="" z)="" satisfies="" equation="" uzz="0," z="" with="" conditions:="" (z),="" h)="0," 3πz="" =="" h).="" 359="" 34.="" show="" 33(c)="" given="" n="1" anj0="" (knr)="" sinh="" knz="" knh="" j0="" (kn)="0," 3,....="" 35.="" problem:="" <z<="" 0),="" <z<π,="" constant,="" uz="" <z<h,="" h),="" (h="" z),="" each="" above="" (a)–(d),="" uzz.="" 10="" higher-dimensional="" “as="" long="" as="" branch="" knowledge="" offers="" abundance="" problems,="" it="" full="" vitality.”="" david="" hilbert="" 10.1="" introduction="" treatment="" more="" than="" space="" variables="" much="" involved="" variables.="" here="" number="" multidimensional="" involving="" equation,="" wave="" equations="" various="" will="" be="" presented.="" included="" cube,="" sphere,="" three="" dimensional="" rectangular,="" cylindrical="" polar="" spherical="" coordinates.="" three-dimensional="" schr¨odinger="" central="" field="" hydrogen="" helium="" atoms="" discussed.="" we="" also="" forced="" vibration="" membrane="" three-dimensional,="" nonhomogeneous="" moving="" boundaries.="" 10.2="" cube="" uyy="" (10.2.1)="" <x<π,="" <y<π,="" <z<π.="" faces="" except="" face="" 362="" 0,="" y),="" assume="" nontrivial="" separable="" form="" (x)="" (y)z="" (z).="" substituting="" this="" x′′y="" xy="" ′′z="" z′′="0." division="" yields="" x′′="" ′′="" .="" since="" right="" side="" depends="" only="" left="" independent="" z,="" both="" must="" equal="" thus,="" have="" same="" reasoning,="" hence,="" ordinary="" differential="" µx="0," (λ="" µ)="" λz="0." using="" conditions,="" eigenvalue="" (0)="X" (π)="0," eigenvalues="" µ="−m2" m="1," 3,...="" corresponding="" eigenfunctions="" mx.="" similarly,="" gives="" λ−µ="−n" ny.="" 10.3="" 363="" λ="" m2="" follows="" ′′+λz="0" satisfying="" condition="" (z)="C" "="" n2="" #="" takes="" ∞="" amn=""
4=""
5="" mx="" applying="" condition,="" formally="" coefficient="" double="" fourier="" series="" thus="" ny="" dx="" dy.="" therefore,="" formal="" may="" written="" bmn="" √="" ny,="" (10.2.2)="" example="" 10.3.1.="" determining="" electric="" potential="" charge-free="" cylinder.="" coordinates="" (10.3.1)="" r<a,="" <z<l.="" let="" lateral="" surface="" top="" grounded,="" potential.="" base="" (10.3.2)="" 364="" (r)="" Θ="" (θ)z="" r′′="" rr′="" Θ′′="" 2r′′="" 2λ="−" 2r="" r2="" (10.3.3)="" µΘ="0," (10.3.4)="" (10.3.5)="" periodicity="" (2π),="" Θ′="" ′="" 2,...="" nθ,="" nθ.="" (θ)="A" nθ="" (10.3.6)="" suppose="" real="" negative="" β=""> 0. If the condition Z (l) = 0 is imposed, then the solution of equation (10.3.5) can be written in the form Z (z) = C sinh β (l − z). (10.3.7) Next we introduce the new independent variable ξ = βr. Equation (10.3.3) transforms into ξ 2 d 2R dξ2 + ξ dR dξ +  ξ 2 − n 2 R = 0 which is the Bessel equation of order n. The general solution is Rn (ξ) = DJn (ξ) + E Yn (ξ) where Jn and Yn are the Bessel functions of the first and second kind respectively. In terms of the original variable, we have 10.3 Dirichlet Problem for a Cylinder 365 Rn (r) = DJn (βr) + E Yn (βr). Since Yn (βr) is unbounded at r = 0, we choose E = 0. The condition R (a) = 0 requires that Jn (βa)=0. For each n ≥ 0, there exist positive zeros. Arranging these in an infinite increasing sequence, we have 0 < αn1 < αn2 <...<αnm <.... Thus, we obtain βnm = (αnm/a). Consequently, Rn (r) = DJn (αnmr/a). The solution u then finally takes the form u (r, θ, z) = ∞ n=0 ∞ m=1 Jn
4r a αnm
5 (anm cos nθ + bnm sin nθ) × sinh (l − z) a αnm . To satisfy the nonhomogeneous boundary condition, it is required that f (r, θ) = ∞ n=0 ∞ m=1 Jn
4r a αnm
5 (anm cos nθ + bnm sin nθ) sinh  l a αnm . The coefficients anm and bnm are given by a0m = 1 πa2 sinh  1 a α0m [J1 (α0m)]2
a 0
2π 0 f (r, θ) J0
4r a α0m
5 r dr dθ, anm = 2 πa2 sinh  1 a αnm [Jn+1 (αnm)]2
a 0
2π 0 f (r, θ) Jn
4r a αnm
5 × cos nθ r dr dθ, bnm = 2 πa2 sinh  1 a αnm [Jn+1 (αnm)]2
a 0
2π 0 f (r, θ) Jn
4r a αnm
5 × sin nθ r dr dθ. Example 10.3.2. We shall illustrate the same problem with different boundary conditions. Consider the problem 366 10 Higher-Dimensional Boundary-Value Problems ∇2u = 0, 0 ≤ r < a, 0 < z < π, u (r, θ, 0) = 0, u (r, θ, π)=0, u (a, θ, z) = f (θ, z). As before, by the separation of variables, we obtain r 2R ′′ + rR′ −  λr2 + µ R = 0, Θ ′′ + µΘ = 0, Z ′′ + λZ = 0. By the periodicity conditions, again as in the previous example, the Θ equation yields the eigenvalues µ = n 2 with n = 0, 1, 2,...; the corresponding eigenfunctions are sin nθ, cos nθ. Thus, we have Θ (θ) = An cos n cos θ + Bn sin nθ. Now let λ = β 2 with β > 0. Then, the boundary value problem Z ′′ + β 2Z = 0 Z (0) = 0, Z (π)=0, has the solution Z (z) = Cm sin mz, m = 1, 2, 3,.... Finally, we have r 2R ′′ + rR′ −  m2 r 2 + n 2 R = 0, or R ′′ + 1 r R ′ −  m2 + n 2 r 2  R = 0, the general solution of which is R (r) = DIn (mr) + EKn (mr), where In and Kn are the modified Bessel functions of the first and second kind, respectively. Since R must remain finite at r = 0, we set E = 0. Then R takes the form R (r) = DIn (mr). Applying the nonhomogeneous condition, we find the solution 10.4 Dirichlet Problem for a Sphere 367 u (r, θ, z) = ∞ m=1
4am0 2
5 I0 (mr) I0 (ma) sin mz + ∞ m=1 ∞ n=1 (amn cos nθ + bmn sin nθ) In (mr) In (ma) sin mz, where amn = 2 π 2
π 0
2π 0 f (θ, z) sin mz cos nθ dθ dz, bmn = 2 π 2
π 0
2π 0 f (θ, z) sin mz sin nθ dθ dz. 10.4 Dirichlet Problem for a Sphere Example 10.4.3. To determine the potential in a sphere, we transform the Laplace equation into spherical coordinates. It has the form ∇2u = urr + 2 r ur + 1 r 2 uθθ + cot θ r 2 uθ + 1 r 2 sin2 θ uϕϕ, (10.4.1) where 0 ≤ r<a, 0="" <θ<π,="" and="" <ϕ<="" 2π.="" let="" the="" prescribed="" potential="" on="" sphere="" be="" u="" (a,="" θ,="" ϕ)="f" (θ,="" ϕ).="" (10.4.2)="" we="" assume="" a="" nontrivial="" separable="" solution="" in="" form="" (r,="" (r)="" Θ="" (θ)Φ(ϕ).="" substitution="" of="" laplace="" equation="" yields="" r="" 2r="" ′′="" +="" 2rr′="" −="" λr="0," (10.4.3)="" sin2="" θ="" Θ′′="" sin="" cos="" Θ′="" ="" λ="" µ="" (10.4.4)="" Φ="" µΦ="0." (10.4.5)="" general="" is="" Φ(ϕ)="A" √="" ϕ="" bn="" ϕ.="" (10.4.6)="" periodicity="" condition="" requires="" that="" m="0," 1,="" 2,....="" since="" euler="" type,="" β="" .="" 368="" 10="" higher-dimensional="" boundary-value="" problems="" inserting="" this="" (10.4.3),="" obtain="" 2="" roots="" are="" −1="" 1+4λ="" (1="" β).="" hence,="" rβ="" d="" r−(1+β)="" (10.4.7)="" variable="" ξ="cos" transforms="" into="" 1="" 2ξΘ′="" (β="" 1)="" m2="" (10.4.8)="" which="" legendre’s="" associated="" equation.="" with="" for="" n="0," 2,...="" (θ)="E" p="" (cos="" θ)="" f="" qm="" θ).="" continuity="" at="" π="" corresponds="" to="" (ξ)="" 1.="" has="" logarithmic="" singularity="" choose="" thus,="" becomes="" consequently,="" spherical="" coordinates="" n="" np="" (anm="" mϕ="" bnm="" mϕ).="" order="" satisfy="" function="" boundary,="" it="" necessary="" mϕ)="" ≤="" π,="" by="" orthogonal="" properties="" functions="" mϕ,="" coefficients="" given="" anm="(2n" 2πan="" (n="" m)!=""
="" 2π="" dθ="" dϕ,="" 2,...,="" an0="(2n" 4πan="" pn="" 10.4="" dirichlet="" problem="" 369="" example="" 10.4.4.="" determine="" grounded="" conducting="" uniform="" field="" satisfies="" ∇2u="0," <="" a,="" <φ<="" 2π,="" →="" −e0="" as="" ∞.="" z="" direction="" so="" will="" independent="" φ.="" then,="" takes="" urr="" ur="" uθθ="" cot="" uθ="0." (θ).="" if="" set="" then="" second="" legendre="" qn="" θ),="" where="" first="" kind="" respectively.="" not="" singular="" r-equation="" obtained="" dn="" −(n+1)=""
4="" an="" −(n+1)
5="" infinity,="" must="" have="" a1="−E0," ≥="" ∞="" n+1="" 370="" using="" orthogonality="" functions,="" find="" e0="" n+2="" −π="" 3="" δn1,="" integral="" vanishes="" all="" except="" θ.="" 10.4.5.="" dielectric="" radius="" placed="" electric="" e0.="" potentials="" inside="" outside="" sphere.="" u1="" u2="" ∇2u1="∇2u2" =="" 0,="" k="" ∂u1="" ∂r="∂u2" ,="" −e0r="" ∞,="" sphere,="" respectively,="" constant.="" preceeding="" example,="" (10.4.9)="" finite="" origin,="" take="" anr="" npn="" a.="" (10.4.10)="" u2,="" approach="" infinity="" manner,="" −(n+1)pn="" (10.4.11)="" from="" two="" conditions="" b1="" ka1="−E0" 2b1="" 2.="" 371="" found="" 3e0="" (k="" 2).="" 2)="" −2="" 10.4.6.="" between="" concentric="" spheres="" held="" different="" constant="" potentials.="" here="" need="" solve="" b,="" case,="" depends="" only="" radial="" distance.="" ∂="" ="" ∂u="" ="0." elementary="" integration,="" c2="" c1="" arbitrary="" constants.="" applying="" boundary="" conditions,="" aa="" b="" b)="" ab="" (b="" a)="" (a="" bb="" r="" b="" 372="" 10.5="" three-dimensional="" wave="" heat="" equations="" three="" space="" variables="" may="" written="" utt="c" 2∇2u,="" (10.5.1)="" ∇2="" operator.="" (x,="" y,="" z,="" t)="U" z)="" t="" (t).="" substituting="" (10.5.1),="" λc2t="0," (10.5.2)="" λu="0," (10.5.3)="" −λ="" separation="" separated="" solutions="" determined.="" next="" consider="" ut="k∇2u." (10.5.4)="" before,="" seek="" (10.5.4),="" ′="" λkt="0," see="" here,="" previous="" essentially="" solving="" helmholtz="" 10.6="" vibrating="" membrane="" specific="" equation,="" us="" length="" width="" b.="" initial="" displacement="" (uxx="" uyy),="" x="" y=""> 0, (10.6.1) u (x, y, 0) = f (x, y), 0 ≤ x ≤ a, 0 ≤ y ≤ b, (10.6.2) ut (x, y, 0) = g (x, y), 0 ≤ x ≤ a, 0 ≤ y ≤ b, (10.6.3) u (0, y, t)=0, u (a, y, t)=0, (10.6.4) u (x, 0, t)=0, u (x, b, t)=0. (10.6.5) 10.6 Vibrating Membrane 373 We have just shown that the separated equations for the wave equation are T ′′ + λc2T = 0, (10.6.6) ∇2U + λU = 0, (10.6.7) where, in this case, ∇2U = Uxx + Uyy. Let λ = α 2 . Then the solution of equation (10.6.6) is T (t) = A cos αct + B sin αct. Now we look for a nontrivial solution of equation (10.6.7) in the form U (x, y) = X (x) Y (y). Substituting this into equation (10.6.7) yields X′′ − µX = 0, Y ′′ + (λ + µ) Y = 0. If we let µ = −β 2 , then the solutions of these equations take the form X (x) = C cos βx + D sin βx. Y (y) = E cos γy + F sin γy, where γ 2 = (λ + µ) = α 2 − β 2 . The homogeneous boundary conditions in x require that C = 0 and D sin βa = 0 which implies that β = (mπ/a) with D = 0. Similarly, the homogeneous boundary conditions in y require that E = 0 and F sin γb = 0 which implies that γ = (nπ/b) with F = 0. Noting that m and n are independent integers, we obtain the displacement function in the form u (x, y, t) = ∞ m=1 ∞ n=1 (amn cos αmn ct + bmn sin αmn ct) sin
4mπx a
5 sin
4nπy b
5 , (10.6.8) where αmn =  m2π 2/a2 +  n 2π 2/b2 , amn and bmn are constants. Now applying the nonhomogeneous initial conditions, we have 374 10 Higher-Dimensional Boundary-Value Problems u (x, y, 0) = f (x, y) = ∞ m=1 ∞ n=1 amn sin
4mπx a
5 sin
4nπy b
5 , and thus, amn = 4 ab
a 0
b 0 f (x, y) sin
4mπx a
5 sin
4nπy b
5 dx dy. (10.6.9) In a similar manner, the initial condition on ut implies ut (x, y, 0) = g (x, y) = ∞ m=1 ∞ n=1 bmn αmn c sin
4mπx a
5 sin
4nπy b
5 , from which it follows that bmn = 4 αmn abc
a 0
b 0 g (x, y) sin
4mπx a
5 sin
4nπy b
5 dx dy. (10.6.10) The solution of the rectangular membrane problem is, therefore, given by equation (10.6.8). Example 10.6.1. (Vibration of a Circular Membrane). For a circular elastic membrane that is stretched over a circular frame of radius a, the motion of the membrane can be described by a function u (r, θ, t) that satisfies the partial differential equation 1 c 2 utt = urr + 1 r ur + 1 r 2 uθθ, (10.6.11) where c 2 = (T /ρ), T is the tension in the membrane and ρ is its mass density. We consider the synchronous vibrations of the vibration of the membrane defined by the separable solution u (r, θ, t) = v (r, θ, t) cos (ωct). (10.6.12) Substituting (10.6.12) into (10.6.11) gives vrr + 1 r vr + 1 r 2 uθθ + ω 2 v = 0. (10.6.13) We seek a nontrivial separable solution v (r, θ) = R (r) Θ (θ) of equation (10.6.13) so that r 2R′′ + r R′ R + ω 2 r 2 = − Θ′′ Θ = λ 2 . (10.6.14) 10.7 Heat Flow in a Rectangular Plate 375 This must hold for all points of the membrane, 0 <r 0, (10.7.1) u (x, y, 0) = f (x, y), 0 ≤ x ≤ a, 0 ≤ y ≤ b, (10.7.2) 376 10 Higher-Dimensional Boundary-Value Problems ux (0, y, t)=0, ux (a, y, t)=0, (10.7.3ab) u (x, 0, t)=0, u (x, b, t)=0. (10.7.4ab) As shown earlier, the separated equations for this problem are found to be T ′ + λkT = 0, (10.7.5) ∇2U + λU = 0. (10.7.6) We assume a nontrivial separable solution in the form U (x, y) = X (x) Y (y). Inserting this in equation (10.7.6), we obtain X′′ − µX = 0, (10.7.7) Y ′′ + (λ + µ) Y = 0. (10.7.8) Because the conditions in x are homogeneous, we choose µ = −α 2 so that X (x) = A cos αx + B sin αx. Since X′ (0) = 0, B = 0 and since X′ (a) = 0, sin αa = 0, A = 0 which gives α = (mπ/a), m = 1, 2, 3,.... We note that µ = 0 is also an eigenvalue. Consequently, Xm (x) = Am cos (mπx/a), m = 0, 1, 2,.... Similarly, for nontrivial solution Y , we select β 2 = λ + µ = λ − α 2 so that the solution of equation (10.7.8) is Y (y) = C cos βy + D sin βy. Applying the homogeneous conditions, we find C = 0 and sin βb = 0, D = 0. Thus, we obtain β = (nπ/b) ; n = 1, 2, 3,..., and 10.7 Heat Flow in a Rectangular Plate 377 Yn (y) = Dn sin (nπy/b). Recalling that λ = α 2 + β 2 , the solution of equation (10.7.5) may be written in the form Tmn (t) = Emn e −(m2/a2+n 2/b2 )π 2kt . Thus, the solution of the heat equation satisfying the prescribed boundary conditions may be written as u (x, y, t) = ∞ m=0 ∞ n=1 amn e −(m2/a2+n 2/b2 )π 2kt cos
4mπx a
5 sin
4nπy b
5 , (10.7.9) where amn = AmDmEmn are arbitrary constants. Applying the initial condition, we obtain u (x, y, 0) = f (x, y) = ∞ m=0 ∞ n=1 amn cos
4mπx a
5 sin
4nπy b
5 . (10.7.10) This is a double Fourier series, and the coefficients are given by a0n =  2 ab
a 0
b 0 f (x, y) sin
4nπy b
5 dx dy, and for m ≥ 1 amn =  4 ab
a 0
b 0 f (x, y) cos
4mπx a
5 sin
4nπy b
5 dx dy. The solution of the heat equation is thus given by equation (10.7.9). Example 10.7.1. (Steady-state temperature in a Circular Disk). We next consider the steady-state temperature distribution u (r, θ) in a circular disk of radius r = a that satisfies the Laplace equation urr + 1 r ur + 1 r 2 uθθ = 0, 0 ≤ r ≤ a, 0 ≤ θ ≤ 2π, (10.7.11) u (r, θ) = f (θ), on r = a for all θ, (10.7.12) where f (θ) is a given function of θ. This is exactly the Dirichlet problem for a circle that was already solved in Section 9.4. We also consider the steady-state temperature distribution u (r, θ, φ) in a sphere of radius a where 0 ≤ r<a, 0="" <θ<π="" and="" <φ<="" 2π.="" for="" simplicity,="" we="" assume="" only="" steady="" temperature="" distribution="" which="" depends="" on="" r="" θ.="" thus,="" u="" is="" independent="" of="" the="" longitudinal="" coordinate="" 378="" 10="" higher-dimensional="" boundary-value="" problems="" φ,="" hence,="" steady-state="" (r,="" θ)="" satisfies="" laplace="" equation="" in="" spherical="" polar="" coordinates="" form="" ∂="" ∂r="" ="" 2="" ∂u="" ="" +="" 1="" sin="" θ="" ∂θ="" (10.7.13)="" seek="" a="" separable="" solution="" (r)="" Θ="" (θ)="" so="" that="" leads="" to="" d="" dr="" dθ="" dΘ="" (10.7.14)="" this="" must="" hold="" <r<a="" <θ<π.="" consequently,="" (10.7.15)="" or="" −="" λr="0," <="" a,="" (10.7.16)="" λ="" π.="" (10.7.17)="" can="" also="" be="" written="" as="" ′′="" ′="" (10.7.18)="" simplify="" by="" change="" variable.="" x="cos" θ,="" y="" (x)="Θ" (θ).="" using="" chain="" rule="" obtain="" dx="" (sin="" dy="" sin2="" dx="−" ="" .="" combining="" result="" with="" legendre="" 10.8="" waves="" three="" dimensions="" 379="" λy="0," −1="" ≤="" 1,="" (10.7.19)="" or,="" equivalently,="" 2y="" dx2="" 2x="" (10.7.20)="" was="" completely="" solved="" section="" 8.9.="" well-known="" sturm–liouville="" (−1)="" (+1)="" finite.="" results="" are="" =="" n="" (n="" 1),="" (x),="" 2,="" 3,...,="" where="" pn="" polynomial="" degree="" n.="" propagation="" due="" an="" initial="" disturbance="" rectangular="" volume="" best="" described="" problem="" utt="c" 2∇2u,="" b,="" z="" d,="" t=""> 0, (10.8.1) u (x, y, z, 0) = f (x, y, z), 0 ≤ x ≤ a, 0 ≤ y ≤ b, 0 ≤ z ≤ d, (10.8.2) ut (x, y, z, 0) = g (x, y, z), 0 ≤ x ≤ a, 0 ≤ y ≤ b, 0 ≤ z ≤ d, (10.8.3) u (0, y, z, t)=0, u (a, y, z, t)=0, (10.8.4) u (x, 0, z, t)=0, u (x, b, z, t)=0, (10.8.5) u (x, y, 0, t)=0, u (x, y, d, t)=0. (10.8.6) We assume a nontrivial separable solution in the form u (x, y, z, t) = U (x, y, z) T (t). The separated equations are given by T ′′ + λc2T = 0, (10.8.7) ∇2U + λU = 0. (10.8.8) We assume that U has the nontrivial separable solution in the form U (x, y, z) = X (x) Y (y)Z (z). Substitution of this into equation (10.8.8) yields X′′ − µX = 0, (10.8.9) Y ′′ − νY = 0, (10.8.10) Z ′′ + (λ + µ + ν)Z = 0. (10.8.11) 380 10 Higher-Dimensional Boundary-Value Problems Because of the homogeneous conditions in x, we let µ = −α 2 so that X (x) = A cos αx + B sin αx. As in the preceding examples, we obtain Xl (x) = Bl sin  lπx a  , l = 1, 2, 3,.... In a similar manner, we let ν = −β 2 to obtain Y (y) = C cos βy + D sin βy and accordingly, Ym (y) = Dm sin
4mπy b
5 , m = 1, 2, 3,.... We again choose γ 2 = λ + µ + ν = λ − α 2 − β 2 so that Z (z) = E cos (γz) + F sin (γz). Applying the homogeneous conditions in z, we obtain Zn (z) = Fn sin
4nπz d
5 . Since the solution of equation (10.8.7) is T (t) = G cos
4√ λ ct
5 + H sin
4√ λ ct
5 , the solution of the wave equation has the form u (x, y, z, t) = ∞ l=1 ∞ m=1 ∞ n=1
4 almn cos √ λ ct + blmn sin √ λ ct
5 × sin  lπx a  sin
4mπy b
5 sin
4nπz d
5 where almn and blmn are arbitrary constants. The coefficients almn are determined from the initial condition u (x, y, z, 0) = f (x, y, z) and are found to be almn = 8 abd
a 0
b 0
d 0 f (x, y, z) sin  lπx a  sin
4mπy b
5 sin
4nπz d
5 dx dy dz. Similarly the coefficients blmn are determined from the initial condition u (x, y, z, 0) = g (x, y, z) and are found to be blmn = 8 √ λ acbd
a 0
b 0
d 0 g (x, y, z) sin  lπx a  sin
4mπy b
5 sin
4nπz d
5 dx dy dz, where λ =  l 2 a 2 + m2 b 2 + n 2 d 2  π 2 . 10.9 Heat Conduction in a Rectangular Volume 381 10.9 Heat Conduction in a Rectangular Volume As in the case of the wave equation, the solution of the heat equation in three spaces variables can be determined. Consider the problem of heat distribution in a rectangular volume. The faces are maintained at zero degree temperature. The solid is initially heated so that the problem may be written as ut = k ∇2u, 0 < x < a, 0 < y < b, 0 < z < d, t > 0, u (x, y, z, 0) = f (x, y, z), 0 ≤ x ≤ a, 0 ≤ y ≤ b, 0 ≤ z ≤ d, u (0, y, z, t)=0, u (a, y, z, t)=0, u (x, 0, z, t)=0, u (x, b, z, t)=0, u (x, y, 0, t)=0, u (x, y, d, t)=0. As before, the separable equations are T ′ + λkT = 0, (10.9.1) ∇2U + λU = 0. (10.9.2) If we assume the solution U to be of the form U (x, y, z) = X (x) Y (y)Z (z), then the solution of the Helmholtz equation is Ulmn (x, y, z) = BlDmFn sin  lπx a  sin
4mπy b
5 sin
4nπz d
5 . Since the solution of equation (10.9.1) is T (t) = G e−λkt , the solution of the heat equation takes the form u (x, y, z, t) = ∞ l=1 ∞ m=1 ∞ n=1 almn e −λkt sin  lπx a  sin
4mπy b
5 sin
4nπz d
5 , where λ = l 2/a2 +  m2/b2 +  n 2/d2 ! π 2 and almn are constants. Application of the initial condition yields almn =  8 abd
a 0
b 0
d 0 f (x, y, z) sin  lπx a  sin
4mπy b
5 sin
4nπz d
5 dx dy dz. 382 10 Higher-Dimensional Boundary-Value Problems 10.10 The Schr¨odinger Equation and the Hydrogen Atom In quantum mechanics, the Hamiltonian (or energy operator) is usually denoted by H and is defined by H = p 2 2M + V (r) (10.10.1) where p = (
/i) ∇ = −i
∇ is the momentum of a particle of mass M, h = 2π
is the Planck constant, and V (r) is the potential energy. The physical state of a particle at time t is described as fully as possible by the wave function Ψ (r, t). The probability of finding the particle at position r = (x, y, z) within a finite volume dV = dx dy dz is


|Ψ| 2 dx dy dz. The particle must always be somewhere in the space, so the probability of finding the particle within the whole space is one, that is,
∞ −∞
∞ −∞
∞ −∞ |Ψ| 2 dx dy dz = 1. The time dependent Schr¨odinger equation for the function Ψ (r, t) is i
Ψt = HΨ, (10.10.2) where H is explicitly given by H = −
2 2M ∇2 + V (r). (10.10.3) Given the potential V (r), the fundamental problem of quantum mechanics is to obtain a solution of (10.10.2) which agrees with a given initial state Ψ (r, 0). For the stationary state solutions, we seek a solution of the form Ψ (r, t) = f (t) ψ (r). Substituting this into (10.10.2) gives df dt + iE
f = 0, (10.10.4) Hψ (r) = Eψ (r), (10.10.5) where E is a separation constant and has the dimension of energy. Integration of (10.10.4) gives 10.10 The Schr¨odinger Equation and the Hydrogen Atom 383 f (t) = A exp  − iEt
 , (10.10.6) where A is an arbitrary constant. Equation (10.10.5) is called the time independent Schr¨odinger equation. The great importance of this equation follows from the fact that the separation of variables gives not just some particular solution of (10.10.5), but generally yields all solutions of physical interest. If ψE (r) represents one particular solution of (10.10.5), then most general solutions of (10.10.2) can be obtained by the principle of superposition of such particular solutions. In fact, the general solution is given by ψ (r, t) =  E AE exp  − iEt
 ψE (r), (10.10.7) where the summation is taken over all admissible values of E, and AE is an arbitrary constant to be determined from the initial conditions. We now solve the eigenvalue problem for the Schr¨odinger equation for the spherically symmetric potential so that V (r) = V (r). The equation for the wave function ψ (r) is ∇2ψ + 2M
2 [E − V (r)] ψ = 0, (10.10.8) where ∇2 is the three-dimensional Laplacian. To determine the wave function ψ, it is convenient to introduce spherical polar coordinates (r, θ, φ) so that equation (10.10.8) takes the form 1 r 2 ∂ ∂r  r 2 ∂ψ ∂r  + 1 r 2 sin θ ∂ ∂θ  sin θ ∂ψ ∂θ  + 1 r 2 sin2 θ ∂ 2ψ ∂φ2 +K [E − V (r)] ψ = 0, (10.10.9) where K =  2M/
2 , ψ ≡ ψ (r, θ, φ), 0 ≤ r < ∞, 0 ≤ θ ≤ π, and 0 ≤ φ ≤ 2π. We seek a nontrivial separable solution of the form ψ = R (r) Y (θ, φ) and then substitute into (10.10.9) to obtain the following equations d dr  r 2 dR dr  + K (E − V ) r 2 − λ ! R = 0, (10.10.10) 1 sin θ ∂ ∂θ  sin θ ∂ ∂θ  + 1 sin2 θ ∂ 2 ∂φ2 Y + λY = 0, (10.10.11) where λ is a separation constant. 384 10 Higher-Dimensional Boundary-Value Problems We first solve (10.10.11) by separation of variables through Y = Θ (θ)Φ(φ) so that the equation becomes sin θ d dθ  sin θ dΘ dθ  +  λ sin2 θ − m2 Θ = 0, (10.10.12) d 2Φ dφ2 + m2Φ = 0, (10.10.13) where m2 is a separation constant. The general solution of (10.10.13) is Φ = A eimφ + B e−imφ , where A and B are arbitrary constants to be determined by the boundary conditions on ψ (r, θ, φ) = R (r) Θ (θ)Φ(φ) which will now be formulated. According to the fundamental postulate of quantum mechanics, the wave function for a particle without spin must have a definite value at every point in space. Hence, we assume that ψ is a single-valued function of position. In particular, ψ must have the same value whether the azimuthal coordinate φ is given by φ or φ + 2π, that is, Φ(φ) = Φ(φ + 2π). Consequently, the solution for Φ has the form Φ = C eimΦ, m = 0, + 1, + 2,..., (10.10.14) where C is an arbitrary constant. In order to solve (10.10.12), it is convenient to change the variable x = cos θ, Θ (θ) = u (x), −1 ≤ x ≤ 1 so that this equation becomes d dx  1 − x 2 du dx +  λ − m2 1 − x 2  u = 0. (10.10.15) For the particular case m = 0, this equation becomes d dx  1 − x 2 du dx + λ u = 0. (10.10.16) This is known as the Legendre equation, which gives the Legendre polynomials Pl (x) of degree l as solutions provided λ = l(l + 1) where l is a positive integer or zero. When m = 0, equation (10.10.15) with λ = l(l + 1) admits solutions which are well known as associated Legendre functions, P m l (x) of degree l and order m defined by P m l (x) =  1 − x 2 m/2 d m dxm P m l (x), x = cos θ. Clearly, P m l (x) vanishes when m>l. As for the negative integral values of m, it can be readily shown that 10.10 The Schr¨odinger Equation and the Hydrogen Atom 385 P −m l (x)=(−1)m (l − m)! (l + m)! P m l (x). Hence, the functions P −m l (x) differ from P m l (x) by a constant factor, and as a consequence, m is restricted to a positive integer or zero. Thus, the associated Legendre functions P m l (x) with |m| ≤ l are the only nonsingular and physically acceptable solutions of (10.10.15). Since |m| ≤ l, when l = 0, m = 0; when l = 1, m = −1, 0, +1; when l = 2, m = −2, −1, 0, 1, 2, etc. This means that, given l, there are exactly (2l + 1) different values of m = −l, ..., −1, 0, 1,..., l. The numbers l and m are called the orbital quantum member and the magnetic quantum number respectively. It is convenient to write down the solutions of (10.10.11) as functions which are normalized with respect to an integration over the whole solid angle. They are called spherical harmonics and are given by, for m ≥ 0, Y m l (θ, φ) = (2l + 1) 4π (l − m)! (l + m)! 1 2 (−1)m e imφP m l (cos θ). (10.10.17) Spherical harmonics with negative m and with |m| ≤ l are defined by Y m l (θ, φ)=(−1)m Y −m l (θ, φ). (10.10.18) We now return to a general discussion of the radial equation (10.10.10) which becomes, under the transformation R (r) = P (r) /r, d 2P dr2 + K (E − V ) − λ r 2 P (r)=0. (10.10.19) Almost all cases of physical interest require V (r) to be finite everywhere except at the origin r = 0. Also, V (r) → 0 as r → ∞. The Coulomb and square well potentials are typical examples of this kind. In the neighborhood of r = 0, V (r) can be neglected compared to the centrifugal term  ∼ 1/r2 so that equation (10.10.19) takes the form d 2P dr2 − l(l + 1) r 2 P (r) = 0 (10.10.20) for all states with l = 0. The general solution of this equation is P (r) = A rl+1 + B r−l , (10.10.21) where A and B are arbitrary constants. With the boundary condition P (0) = 0, B = 0 so that the solution is proportional to r l+1 . On the other hand, in view of the assumption that V (r) → 0 as r → ∞, the radial equation (10.10.19) reduces to d 2P dr2 + KE P (r)=0. (10.10.22) 386 10 Higher-Dimensional Boundary-Value Problems The general solution of this equation is P (r) = C eir√ KE + D e−ir√ KE. (10.10.23) The solution is oscillatory for E > 0, and exponential in nature for E < 0. The oscillatory solutions are not physically acceptable because the wave function does not tend to zero as r → ∞. When E < 0, the second term in (10.10.23) tends to infinity as r → ∞. Consequently, the only physically acceptable solutions for E > 0, have the asymptotic form P (r) = C e−αr/2 , (10.10.24) where KE = −  α 2/4 . Thus, the general solution of (10.10.19) can be written as P (r) = f (r) e −(α/2)r , so that f (r) satisfies the ordinary differential equation d 2f dr2 − α df dr − KV + l(l + 1) r 2 f = 0. (10.10.25) Note that this general solution is physically acceptable because the wave function tends to zero as r → 0 and as r → ∞. We now specify the form of the potential V (r). One of the most common potentials is the Coulomb potential V (r) = −Ze2/r representing the attraction between an atomic nucleus of charge +Ze and a moving electron of charge −e. For the hydrogen atom Z = 1. It is a two particle system consisting of a negatively charged electron interacting with a positively charged proton. On the other hand, a helium atom consists of two protons and two neutrons. There are two electrons in orbit around the nucleus of a helium atom. For the singly charged helium ion Z = 2, where Z represents the number of unit charges of the nucleus. Consequently, equation (10.10.25) reduces to d 2f dr2 − α df dr + KZe2 r − l(l + 1) r 2 f (r)=0. (10.10.26) We seek a power series solution of this equation in the form f (r) = r k∞ s=1 asr s , k = 0. (10.10.27) Substituting this series into (10.10.26), we obtain r k∞ s=1 [(s + k) (s + k − 1) − l(l + 1)] asr s+k−1 + ∞ s=1 Zke2 − α (s + k) ! asr s+k−1 = 0. 10.10 The Schr¨odinger Equation and the Hydrogen Atom 387 Clearly, the lowest power of r is (k − 1), so that [k (k + 1) − l(l + 1)] a1 = 0. This implies that k = l or − (l + 1) provided a1 = 0. The negative root of k is not acceptable because it leads to an unbounded solution. Equating the coefficient of r s+k−1 , we get the recurrence relation for the coefficients as as+1 = α (s + l) − ZKe2 s (s + 2l + 1) as, s = 1, 2, 3,.... (10.10.28) The asymptotic nature of this result is as+1 as ∼ α s as s → ∞. This ratio is the same as that of the series for e αr. This means that R (r) is unbounded as r → ∞, which is physically unacceptable. Hence, the series for f (r) must terminate, and f (r) must be a polynomial so that as+1 = 0, but as = 0. Hence α (s + l) − ZKe2 = 0, s = 1, 2, 3,..., or, α 2 4 = Z 2K2 e 4 4 (s + l) 2 = −KE. (10.10.29) Putting K =  2M/
2 , the energy levels are given by E = En = − Z 2K2 e 4 4n2K = − MZ2 e 4 2
2n2 , (10.10.30) where n = (s + l) is called the principal quantum number and n = 1, 2, 3,.... Thus, it turns out that the complete solution of the Schr¨odinger equation is given by ψn,l,m (r, θ, φ) = Rn,l (r) Y m l (θ, φ), where the radial part is the solution of the radial equation (10.10.10), and it depends on the principle quantum number n (energy levels) and the orbital quantum number l. However, it does not depend on the magnetic quantum number m. Of course, there are (2l + 1) states with the same l value but with different m values. Each of these states has the same energy, and therefore, such systems have a (2l + 1)-fold degeneracy, as a result of rotational symmetry. For the hydrogen atom, Z = 1, the discrete energy spectrum is En = − Me4 2
2n2 = − e 2 2an2 , (10.10.31) 388 10 Higher-Dimensional Boundary-Value Problems where a = 
2/e2M is called the Bohr radius of the hydrogen atom of mass M and charge of the electron, −e. This discrete energy spectrum depends only on the principle quantum number n (but not on m) and has an excellent agreement with experimental prediction of spectral lines. For a given n, there are n sets of l and s n = 1, {l = 0, s = 1} ; n = 2, ⎧ ⎨ ⎩ l = 0, s = 2 l = 1, s = 1 ⎫ ⎬ ⎭ ; n = 3, ⎧ ⎪⎪⎪⎪⎨ ⎪⎪⎪⎪⎩ l = 0, s = 3 l = 1, s = 2 l = 2, s = 1 ⎫ ⎪⎪⎪⎪⎬ ⎪⎪⎪⎪⎭ ; etc. Given n, there are exactly n values of l(l = 0, 1, 2,...,n − 1) and the highest value of l is n − 1. Thus, the three numbers n, l, m, determine a unique eigenfunction, ψn,l,m (r, θ, φ) = Rn,l (r) Y m l (θ, φ). Since the energy levels depend only on the principle quantum number n, there are, in general, several linearly independent eigenfunctions of the Schr¨odinger equation for the hydrogen atom corresponding to each energy level, so the energy levels are said to be degenerate. There are (2l + 1) different eigenfunctions of the same energy obtained by varying the magnetic quantum number m from −l to l. In general, the total number of degenerate energy states En for the hydrogen atom is then n−1 l=0 (2l + 1) = 2 n (n − 1) 2 + n = n 2 . (10.10.32) The energy levels of the hydrogen atom (10.10.31) can be expressed in terms the Rydberg, Ry, as En = − Ry n2 , (10.10.33) where Ry represents the Rydberg given by Ry = Me4 2
2 = M c2 e 4 2 (
c) 2 = M c2 2 ×  e 2
c 2 ≃ 5 × 105 2 eV ×  1 1372 ≃ 13.3 eV. Consequently, En = − 13.3 n2 eV. (10.10.34) 10.10 The Schr¨odinger Equation and the Hydrogen Atom 389 Thus, the ground state of the hydrogen atom, which is the most tightly bound, has an energy −13.3 eV (more accurately −13.6 eV ) and therefore, it would take 13.6 eV to release the electron from its ground state. Therefore, this is called the binding energy of the hydrogen atom. Finally, the electron is treated here as a nonrelativistic particle. However, in reality, small relativistic effects can be calculated. These are known as fine structure corrections. Thus, the nonrelativistic Schr¨odinger equation describes the hydrogen atom extremely well. Example 10.10.1. (Infinite Square well potential V0 → ∞). We consider the one-dimensional Schr¨odinger equation (10.10.8) in the form − 
2 2M  d 2 dx2 + V (x) ψ (x) = Eψ (x), (10.10.35) where the potential V (x) is given by V (x) = ⎧ ⎨ ⎩ V0, x ≤ −a, x ≥ a, 0, −a ≤ x ≤ a (10.10.36) and take the limit V0 → +∞ as shown in Figure 10.10.1. It is noted that the potential is zero inside the square well and E is the kinetic energy of the particle in this region (−a ≤ x ≤ a) which must be positive, E > 0. It is convenient to fix the origin at the center of the well so that V (x) is an even function of x. A case of special interest is that V0 > E ≥ 0 and eventually, V0 → ∞. Figure 10.10.1 Square well with potential V0 → ∞. 390 10 Higher-Dimensional Boundary-Value Problems The given potential is different in different regions, we solve (10.10.35) separately in three regions. Region 1. V = V0 in this region x ≤ −a. The Schr¨odinger equation (10.10.35) in this region is −
2 2M ψxx + V0ψ = Eψ. Or, equivalently, ψxx =  2M
2  (V0 − E) ψ, V0 >E> 0. (10.10.37) The general solution of (10.10.37) is ψ1 (x) = A ekx + B e−kx , (10.10.38) where A and B are constants and k = 2M
2 (V0 − E) 1 2 . (10.10.39) The wave function ψ1 (x) must be bounded as x → −∞ to retain its probabilistic interpretation, hence B = 0, and the solution in x ≤ −a is ψ1 (x) = A ekx . As V0 → ∞, k → ∞, and, in this limit, the solution must vanish, that is, ψ1 (x)=0, for x ≤ −a. (10.10.40) Region 2. V = 0 in −a ≤ x ≤ a. In this case, the equation takes the form ψxx + k 2ψ = 0, (10.10.41) where k 2 = 2ME
2 . (10.10.42) The general solution of (10.10.41) is given by ψ2 (x) = C sin kx + D cos kx, (10.10.43) where C and D are arbitrary constants. Region 3. V = V0 in this region x ≥ a. An argument similar to region 1 leads to zero solution, that is, 10.10 The Schr¨odinger Equation and the Hydrogen Atom 391 ψ3 (x)=0, for x ≥ a. (10.10.44) From a physical point of view, the solution of the Schr¨odinger equation must be continuous everywhere including at the boundaries. Thus, matching of solutions at x = + a is required so that ψ2 (a) = C sin ak + D cos ak =0= ψ3 (a), (10.10.45) ψ2 (−a) = −C sin ak + D cos ak =0= ψ1 (−a). (10.10.46) This system of linear homogeneous equations has nontrivial solutions for C and D only if the determinant of the coefficient matrix vanishes. This means that sin ak cos ak = 0. (10.10.47) There are two possible nontrivial solutions for the set of conditions (10.10.47). Case 1. Even solution: cos ak = 0. In this case, it follows from (10.10.45)–(10.10.46) that C = 0. Hence, ak = (2n + 1) π 2 , n is an integer, or, k 2 = k 2 n = " (2n + 1) π 2a #2 . (10.10.48) Consequently, (10.10.42) gives the energy levels E = En as En = (2n + 1)2 π 2
2 8M a2 . (10.10.49) In this case, the nontrivial solution in region 2 takes the form ψ2 (x) = D cos kx, −a ≤ x ≤ a. (10.10.50) Case 2. Odd solution: sin ak = 0. It follows from (10.10.45)–(10.10.46) that D = 0 and sin ak = 0 holds. Consequently, ak = nπ, n is an integer, n = 0, or, k 2 n =
4nπ a
52 . (10.10.51) 392 10 Higher-Dimensional Boundary-Value Problems Thus, the energy levels are given by En =
2k 2 n 2M = (nπ
) 2 2M a2 . (10.10.52) The nontrivial solution in region 2 is ψ2 (x) = C sin kx, −a ≤ x ≤ a. (10.10.53) Thus, it turns out that, corresponding to every value of En given by (10.10.49) or (10.10.52), there exists a physically acceptable solution. Hence, the general solution of the Schr¨odinger equation is obtained from (10.10.7) in the form ψ (x, t) =  n Cnψn (x) exp  − itEn
 , (10.10.54) where Cn are constants. In classical mechanics, the motion of the particle is allowed for E > 0. In quantum mechanics, it follows from (10.10.49) or (10.10.52) that particle motion is allowed for discrete values of energy, that is, the energy for this system is quantized. This is a remarkable contrast between the results of the classical mechanics and quantum mechanics. Finally, it follows from the above analysis is that lim |x|→∞ ψ (x)=0. (10.10.55) Such a system, where the wave function vanishes beyond range or asymptotically, is called a bound state, and energy is quantized. A very common example is the hydrogen atom which was discussed in this section. In the present system ψ (x) = 0 for x 2 ≥ a 2 . Therefore, this system is also referred to as a particle in a box of length 2a. The probability for finding the particle outside this region is zero. 10.11 Method of Eigenfunctions and Vibration of Membrane Consider the nonhomogeneous initial boundary-value problem L[u] = ρ utt − G in D (10.11.1) with prescribed homogeneous boundary conditions on the boundary B of D, and the initial conditions u (x1, x2,...,xn, 0) = f (x1, x2,...,xn), (10.11.2) ut (x1, x2,...,xn, 0) = g (x1, x2,...,xn). (10.11.3) 10.11 Method of Eigenfunctions and Vibration of Membrane 393 Here ρ ≡ ρ (x1, x2,...,xn) is a real-valued positive continuous function and G ≡ G (x1, x2,...,xn) is a real-valued continuous function. We assume that the only solution of the associated homogeneous problem L[u] = ρutt (10.11.4) with the prescribed boundary conditions is the trivial solution. Then, if there exists a solution of the given problem (10.11.1)–(10.11.3), it can be represented by a series of eigenfunctions of the associated eigenvalue problem L[ϕ] + λρϕ = 0 (10.11.5) with ϕ satisfying the boundary conditions given for u. For problems with one space variable, see Section 7.8. As a specific example, we shall determine the solution of the problem of forced vibration of a rectangular membrane of length a and width b. The problem is utt − c 2∇2u = F (x, y, t) in D (10.11.6) u (x, y, 0) = f (x, y), 0 ≤ x ≤ a, 0 ≤ y ≤ b, (10.11.7) ut (x, y, 0) = g (x, y), 0 ≤ x ≤ a, 0 ≤ y ≤ b, (10.11.8) u (0, y, t)=0, u (a, y, t)=0, (10.11.9) u (x, 0, t)=0, u (x, b, t)=0. (10.11.10) The associated eigenvalue problem is ∇2ϕ + λϕ = 0 in D, ϕ = 0 on the boundary B of D. The eigenvalues for this problem according to Section 10.6 are given by αmn =  m2π 2 a 2 + n 2π 2 b 2  , m, n = 1, 2, 3 ... and the corresponding eigenfunctions are ϕmn (x, y) = sin
4mπx a
5 sin
4nπy b
5 . Thus, we assume the solution u (x, y, t) = ∞ m=1 ∞ n=1 umn (t) sin
4mπx a
5 sin
4nπy b
5 and the forcing function 394 10 Higher-Dimensional Boundary-Value Problems F (x, y, t) = ∞ m=1 ∞ n=1 Fmn (t) sin
4mπx a
5 sin
4nπy b
5 . Here Fmn (t) are given by Fmn (t) = 4 ab
a 0
b 0 F (x, y, t) sin
4mπx a
5 sin
4nπy b
5 dx dy. Note that u automatically satisfies the homogeneous boundary conditions. Now inserting u (x, y, t) and F (x, y, t) in equation (10.11.6), we obtain u¨mn + c 2α 2 mnumn = Fmn, where α 2 mn = (mπ/a) 2 + (nπ/b) 2 . We have assumed that u is twice continuously differentiable with respect to t. Thus, the solution of the preceding ordinary differential equation takes the form umn (t) = Amn cos (αmnct) + Bmn sin (αmnct) + 1 (αmn c)
t 0 Fmn (τ ) sin [αmnc (t − τ )] dτ. The first initial condition gives u (x, y, 0) = f (x, y) = ∞ m=1 ∞ n=1 Amn sin
4mπx a
5 sin
4nπy b
5 . Assuming that f (x, y) is continuous in x and y, the coefficient Amn of the double Fourier series is given by Amn = 4 ab
a 0
b 0 f (x, y) sin
4mπx a
5 sin
4nπy b
5 dx dy. Similarly, from the remaining initial condition, we have ut (x, y, 0) = g (x, y) = ∞ m=1 ∞ n=1 Bmn (αmn c) sin
4mπx a
5 sin
4nπy b
5 , and hence, for continuous g (x, y), Bmn = 4 (ab αmnc)
a 0
b 0 g (x, y) sin
4mπx a
5 sin
4nπy b
5 dx dy. The solution of the given initial boundary-value problem is therefore given by u (x, y, t) = ∞ m=1 ∞ n=1 umn (t) sin
4mπx a
5 sin
4nπy b
5 , 10.12 Time-Dependent Boundary-Value Problems 395 provided the series for u and its first and second derivatives converge uniformly. If F (x, y, t) = e x+y cos ωt, then Fmn (t) = 4mnπ2 (m2π 2 + a 2) (n2π 2 + b 2) " 1+(−1)m+1 e a # × " 1+(−1)n+1 e b # cos ωt = Cmn cos ωt. Hence, we have umn (t) = 1 (αmnc)
t 0 Cmn cos ωtsin [αmnc (t − τ )] dτ = Cmn (α2 mnc 2 − ω2) (cos ωt − cos αmnct) provided ω = (αmnc). Thus, the solution may be written in the form u (x, y, t) = ∞ m=1 ∞ n=1 Cmn (α2 mnc 2 − ω2) (cos ωt − cos αmnct) × sin
4mπx a
5 sin
4nπy b
5 . 10.12 Time-Dependent Boundary-Value Problems The preceding chapters have been devoted to problems with homogeneous boundary conditions. Due to the frequent occurrence of problems with time dependent boundary conditions in practice, we consider the forced vibration of a rectangular membrane with moving boundaries. The problem here is to determine the displacement function u which satisfies utt − c 2∇2u = F (x, y, t), 0 < x < a, 0 < y < b, (10.12.1) u (x, y, 0) = f (x, y), 0 ≤ x ≤ a, 0 ≤ y ≤ b, (10.12.2) ut (x, y, 0) = g (x, y), 0 ≤ x ≤ a, 0 ≤ y ≤ b, (10.12.3) u (0, y, t) = p1 (y, t), 0 ≤ y ≤ b, t ≥ 0, (10.12.4) u (a, y, t) = p2 (y, t), 0 ≤ y ≤ b, t ≥ 0, (10.12.5) u (x, 0, t) = q1 (x, t), 0 ≤ x ≤ a, t ≥ 0, (10.12.6) u (x, b, t) = q2 (x, t), 0 ≤ x ≤ a, t ≥ 0. (10.12.7) For such problems, we seek a solution in the form u (x, y, t) = U (x, y, t) + v (x, y, t), (10.12.8) 396 10 Higher-Dimensional Boundary-Value Problems where v is the new dependent variable to be determined. Before finding v, we must first determine U. If we substitute equation (10.12.8) into equations (10.12.1)–(10.12.7), we respectively obtain vtt − c 2 (vxx + vyy) = F − Utt + c 2 (Uxx + Uyy) = F5 (x, y, t) and v (x, y, 0) = f (x, y) − U (x, y, 0) = f5(x, y), vt (x, y, 0) = g (x, y) − Ut (x, y, 0) = g5(x, y), v (0, y, t) = p1 (y, t) − U (0, y, t) = p51 (y, t), v (a, y, t) = p2 (y, t) − U (a, y, t) = p52 (y, t), v (x, 0, t) = q1 (x, t) − U (x, 0, t) = q51 (x, t), v (x, b, t) = q2 (x, t) − U (x, b, t) = q52 (x, t). In order to make the conditions on v homogeneous, we set p51 = p52 = q51 = q52 = 0, so that U (0, y, t) = p1 (y, t), U (a, y, t) = p2 (y, t), (10.12.9) U (x, 0, t) = q1 (x, t), U (x, b, t) = q2 (x, t). (10.12.10) In order that the boundary conditions be compatible, we assume that the prescribed functions take the forms p1 (y, t) = ϕ (y) p ∗ 1 (y, t), p2 (y, t) = ϕ (y) p ∗ 2 (y, t), q1 (x, t) = ψ (x) q ∗ 1 (x, t), q2 (x, t) = ψ (x) q ∗ 2 (x, t), where the function ϕ must vanish at the end points y = 0, y = b and the function ψ must vanish at x = 0, x = a. Thus, U (x, y, t) which satisfies equations (10.12.9)–(10.12.10) takes the form U (x, y, t) = ϕ (y) " p ∗ 1 + x a (p ∗ 2 + p ∗ 1 ) # + ψ (x) " q ∗ 1 + y b (q ∗ 2 + q ∗ 1 ) # . The problem then is to find the function v (x, y, t) which satisfies vtt − c 2 (vxx + vyy) = F5 (x, y, t), v (x, y, 0) = f5(x, y), vt (x, y, 0) = g5(x, y), v (0, y, t)=0, v (a, y, t)=0, v (x, 0, t)=0, v (x, b, t)=0. This is an initial boundary-value problem with homogeneous boundary conditions, which has already been solved. 10.12 Time-Dependent Boundary-Value Problems 397 As a particular case, consider the following problem utt − c 2 (uxx + uyy)=0, u (x, y, 0) = 0, ut (x, y, 0) = y b sin
4πx a
5 , u (0, y, t)=0, u (a, y, t)=0, u (x, 0, t)=0, u (x, b, t) = sin
4πx a
5 sin t. We assume a solution in the form u (x, y, t) = v (x, y, t) + U (x, y, t). The function U (x, y, t) which satisfies U (0, y, t)=0, U (a, y, t)=0, U (x, 0, t)=0, U (x, b, t) = sin
4πx a
5 sin t is U (x, y, t) = sin
4πx a
5
4y b sin t
5 . Thus, the new problem to be solved is vtt − c 2 (vxx + vyy) =  1 − c 2π 2 a 2  y b sin
4πx a
5 sin t, v (x, y, 0) = 0, vt (x, y, 0) = 0, v (0, y, t)=0, v (a, y, t)=0, v (x, 0, t)=0, v (x, b, t)=0. Then, we find Fmn from Fmn (t) = 4 ab
a 0
b 0 F (x, y, t) sin
4mπx a
5 sin
4nπy b
5 dx dy, where F (x, y, t) =  1 − c 2π 2 a 2  y b sin
4πx a
5 sin t, and obtain Fmn (t) = 2 (−1)n+1 an  1 − c 2π 2 a 2  sin t. Now we determine vmn (t) which are given by 398 10 Higher-Dimensional Boundary-Value Problems vmn (t) = Amn cos (αmnct) + Bmn sin (αmnct) + 1 αmnc
t 0 Fmn (τ ) sin [αmnc (t − τ )] dτ. Since v (x, y, 0) = 0, Amn = 0, but Bmn = 4 ab αmnc
a 0
b 0
4 − y b sin πx a
5 sin
4mπx a
5 sin
4nπy b
5 dx dy = 2 (−1)n αmnnac . Thus, we have vmn (t) = 2 (−1)n αmnnac sin (αmnct) + 2 (−1)n αmnca3n (1 − α2c 2)  a 2 − c 2π 2 (sin αmnct − αc sin t). The solution is therefore given by u (x, y, t) = y b sin
4πx a
5 sin t + ∞ m=1 ∞ n=1 vmn (t) sin
4mπx a
5 sin
4nπy b
5 . 10.13 Exercises 1. Solve the Dirichlet problem ∇2u = 0, 0 < x < a, 0 < y < b, 0 < z < c, u (0, y, z) = sin
4πy b
5 sin
4πz c
5 , u (a, y, z)=0, u (x, 0, z)=0, u (x, b, z)=0, u (x, y, 0) = 0, u (x, y, c)=0. 2. Solve the Neumann problem ∇2u = 0, 0 <x< 1,="" 0="" <y<="" <z<="" ux="" (0,="" y,="" z)="0," (1,="" uy="" (x,="" 0,="" uz="" 0)="cos" πx="" cos="" πy,="" 1)="0." 3.="" solve="" the="" robin="" boundary-value="" problem="" ∇2u="0," <="" x="" π,="" y="" z="" u="" (y,="" z),="" (π,="" +="" h="" π)="" ⎫="" ⎬="" ⎭="" 10.13="" exercises="" 399="" 4.="" determine="" solution="" of="" each="" following="" problems="" for="" a="" cylinder:="" (a)="" r<a,="" <θ<="" 2π,="" l,="" (a,="" θ,="" (r,="" l)="0," θ).="" (b)="" (θ,="" 5.="" find="" dirichlet="" sphere="" θ="" <ϕ<="" ϕ)="cos2" θ.="" 6.="" in="" region="" bounded="" by="" two="" concentric="" spheres="" r="" b,="" <φ<="" φ)="f" φ),="" (b,="" φ).="" 7.="" steady-state="" temperature="" distribution="" cylinder="" radius="" if="" constant="" flow="" heat="" t="" is="" supplied="" at="" end="" and="" surface="" are="" maintained="" zero="" temperature.="" 8.="" potential="" electrostatic="" field="" inside="" length="" l="" a,="" grounded,="" charged="" to="" u0.="" 9.="" electric="" upper="" half="" u1="" lower="" u2.="" 10.="" r<="" 11.="" neumann="" 400="" 10="" higher-dimensional="" ur="" ϕ),="" where=""
="" 2π="" π="" f="" sin="" dθ="" dϕ="0." 12.="" initial="" utt="c" 2∇2u,="" <x<=""> 0, u (x, y, 0) = sin2 πx sin πy, ut (x, y, 0) = 0, 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, u (0, y, t)=0, u (1, y, t)=0, 0 ≤ y ≤ 1, t> 0, u (x, 0, t)=0, u (x, 1, t)=0, 0 ≤ x ≤ 1, t> 0. 13. Obtain the solution of the problem utt = c 2∇2u, r < a, 0 <θ< 2π, t > 0, u (r, θ, 0) = f (r, θ), ut (r, θ, 0) = g (r, θ), u (a, θ, t)=0. 14. Determine the temperature distribution in a rectangular plate with radiation from its surface. The temperature distribution is described by ut = k (uxx + uyy) − h (u − u0), 0 < x < a, 0 < y < b, t > 0, u (x, y, 0) = f (x, y), u (0, y, t)=0, u (a, y, t)=0, u (x, 0, t)=0, u (x, b, t)=0, where k, h and u0 are constants. 15. Solve the heat conduction problem in a circular plate ut = k  urr + 1 r ur + 1 r 2 uθθ , r< 1, 0 <θ< 2π, t > 0, u (r, θ, 0) = f (r, θ), u (1, θ, t)=0. 16. Solve the initial boundary-value problem utt = c 2∇2u, 0 <x< 1,="" 0="" <y<="" <z<="" t=""> 0, u (x, y, z, 0) = sin πx sin πy sin πz, 10.13 Exercises 401 ut (x, y, z, 0) = 0, u (0, y, z, t) = u (1, y, z, t)=0, u (x, 0, z, t) = u (x, 1, z, t)=0, u (x, y, 0, t) = u (x, y, 1, t)=0. 17. Solve utt + k ut = c 2∇2u, 0 < x < a, 0 < y < b, 0 < z < d, t > 0, u (x, y, z, 0) = f (x, y, z), ut (x, y, z, 0) = g (x, y, z), u (0, y, z, t) = u (a, y, z, t)=0, u (x, 0, z, t) = u (x, b, z, t)=0, u (x, y, 0, t) = u (x, y, d, t)=0. 18. Obtain the solution of the problem for t > 0, utt = c 2  urr + 1 r ur + 1 r 2 uθθ + uzz , r < a, 0 <θ< 2π, 0 < z < l, u (r, θ, z, 0) = f (r, θ, z), ut (r, θ, z, 0) = g (r, θ, z), u (a, θ, z, t)=0, u (r, θ, 0, t) = u (r, θ, l, t)=0. 19. Determine the solution of the heat conduction problem ut = k ∇2u, 0 < x < a, 0 < y < b, 0 < z < c, t > 0, u (x, y, z, 0) = f (x, y, z), ux (0, y, z, t) = ux (a, y, z, t)=0, uy (x, 0, z, t) = uy (x, b, z, t)=0, uz (x, y, 0, t) = uz (x, y, c, t)=0. 402 10 Higher-Dimensional Boundary-Value Problems 20. Solve the problem ut = k ∇2u, r < a, 0 <θ< 2π, 0 < z < l, t > 0, u (r, θ, z, 0) = f (r, θ, z), ur (a, θ, z, t)=0, u (r, θ, 0, t) = u (r, θ, l, t)=0. 21. Find the temperature distribution in the section of a sphere cut out by the cone θ = θ0. The surface temperature is zero while the initial temperature is given by f (r, θ, ϕ). 22. Solve the initial boundary-value problem utt = c 2∇2u + F (x, y, t), 0 < x < a, 0 < y < b, t > 0, u (x, y, 0) = f (x, y), ut (x, y, 0) = g (x, y), ux (0, y, t) = ux (a, y, t) = 0 for all t > 0, uy (x, 0, t) = uy (x, b, t) = 0 for all t > 0. 23. Solve the problem utt = c 2∇2u + xy sin t, 0 < x < π, 0 < y < π, t > 0, u (x, y, 0) = 0, ut (x, y, 0) = 0, u (0, y, t) = u (π, y, t)=0, u (x, 0, t) = u (x, π, t)=0. 24. Solve ut = k∇2u + F (x, y, z, t), 0 < x < a, 0 < y < b, 0 < z < c, t > 0, u (x, y, z, 0) = f (x, y, z), u (0, y, z, t) = u (a, y, z, t)=0, 10.13 Exercises 403 u (x, 0, z, t) = u (x, b, z, t)=0, uz (x, y, 0, t) = uz (x, y, c, t)=0. 25. Solve the nonhomogeneous diffusion problem ut = k ∇2u + A, 0 < x < π, 0 < y < π, t > 0, u (x, y, 0) = 0, u (0, y, t) = u (π, y, t)=0, uy (x, 0, t) + u (x, 0, t)=0, uy (x, π, t) + u (x, π, t)=0, where k and A are constants. 26. Find the temperature distribution of the composite cylinder consisting of an inner cylinder 0 ≤ r ≤ r0 and an outer cylindrical tube r0 ≤ r ≤ a. The surface temperature is maintained at zero degrees, and the initial temperature distribution is given by f (r, θ, z). 27. Solve the initial boundary-value problem ut − c 2∇2u = 0, 0 < x < π, 0 < y < π, t > 0, u (x, y, 0) = 0, u (0, y, t) = u (π, y, t)=0, u (x, 0, t) = x (x − π) sin t, u (x, π, t)=0, 0 ≤ x ≤ π, t ≥ 0. 28. Solve the problem utt = c 2∇2u, r < a, 0 <θ< 2π, t > 0, u (r, θ, 0) = f (r, θ), ut (r, θ, 0) = g (r, θ), u (a, θ, t) = p (θ, t). 404 10 Higher-Dimensional Boundary-Value Problems 29. Solve ut = c 2∇2u, r < a, t > 0, u (r, θ, 0) = f (r, θ), ut (a, θ, t) = g (θ, t), 0 < θ < π. 30. Determine the solution of the biharmonic equation ∇4u = q/D with the boundary conditions u (x, 0) = u (x, b)=0, u
4 − a 2 , y
5 = u
4a 2 , y
5 = 0, uxx
4 − a 2 , y
5 = uxx
4a 2 , y
5 = 0, uyy (x, 0) = uyy (x, b)=0, where q is the load per unit area and D is the flexural rigidity of the plate. This is the problem of the deflection of a uniformly loaded plate, the sides of which are simply supported. 31. (a) Show that the solution of the one-dimensional Schr¨odinger equation for a free particle of mass M ψt =  i
2M  ψxx is ψ (x, t) =  N b  exp  − x 2 2b 2  , b =  a 2 + i
t M 1 2 , where a is an integrating constant that can be determined from the initial value of the wave function ψ (x, t), and N is also a constant that can be determined from the normalization of the probability (wave function) of finding the particle. (b) Show that the Gaussian probability density is |ψ| 2 = ψψ∗ = |N| 2 ac exp  − x 2 c 2  , and its mean width is δ = c √ 2 , C =  a 2 +
2 t 2 M2a 2 1 2 . 10.13 Exercises 405 32. Analogous to Example 10.10.1, solve the problem for a finite square well potential (see Figure 10.10.1) with a finite value for the height of the potential given as V (x) = ⎧ ⎨ ⎩ 0, for −a ≤ x ≤ a V0, for x ≤ −a, x ≥ a. 33. Consider the quantum mechanical problem described by the one-dimensional Schr¨odinger equation ψxx + k 2ψ = 0 where the wavenumber k = 1
 2M (E − V ) in the rectangular potential barrier of height V0 and width 2a, and V (x) = V0H (a − |x|), where H is the Heaviside unit step function. The particle is free for x < −a and x>a, and V (x) is an even function; the case V0 > E is of great interest here. Show that the general solution of the Schr¨odinger equation for V0 > E is ψ (x) = ⎧ ⎪⎪⎪⎪⎨ ⎪⎪⎪⎪⎩ A eikx + B e−ikx, x ≤ −a, C e−κx + D e+κx , −a ≤ x ≤ a, F eikx + G e−ikx, x ≥ a, where
k = √ 2ME and
κ =  2M (V0 − E). Matching the boundary conditions at x = −a, show that ⎡ ⎣ A B ⎤ ⎦ = 1 2 ⎡ ⎣  1 + iκ k exp (κa + ika)  1 − iκ k exp (−κa + ika)  1 − iκ k exp (κa − ika)  1 + iκ k exp (−κa − ika) ⎤ ⎦ ⎡ ⎣ C D ⎤ ⎦ where [ ] denotes a matrix. Using the matching conditions at x = a, show that ⎡ ⎣ C D ⎤ ⎦ = 1 2 ⎡ ⎣  1 − ik κ exp (aκ + iak)  1 + ik κ exp (aκ − iak)  1 + ik κ exp (−aκ + iak)  1 − ik κ exp (−aκ − ika) ⎤ ⎦ ⎡ ⎣ F G ⎤ ⎦. Hence, deduce ⎡ ⎣ A B ⎤ ⎦ = ⎡ ⎣  cosh 2aκ + iε 2 sinh 2aκ e 2iak 1 2 (iη) sinh 2aκ − 1 2 (iη) sinh 2aκ  cosh 2aκ − 1 2 iε sinh 2aκ e −2ika ⎤ ⎦ ⎡ ⎣ F G ⎤ ⎦ where ε =  κ k − k κ and η =  κ k + k κ . 11 Green’s Functions and Boundary-Value Problems “Potential theory has developed out of the vector analysis created by Gauss, Green, and Kelvin for the mathematical theories of gravitational attraction, of electrostatics and of the hydrodynamics of perfect fluids (i.e., incompressible and inviscid fluids). The first stage of abstraction was the study of harmonic functions, i.e., potential functions in space free from masses, charges, sources, or sinks. This led to the inspired intuition of Dirichlet and the early attempts to justify his ‘principle’.” George Temple 11.1 Introduction Boundary-value problems associated with either ordinary or partial differential equations arise most frequently in mathematics, mathematical physics and engineering science. The linear superposition principle is one of the most elegant and effective methods to represent solutions of boundaryvalue problems in terms of an auxiliary function known as Green’s function. Such a function was first introduced by George Green as early as 1828. Subsequently, the method of Green’s functions became a very useful analytical method in mathematics and in many of the applied sciences. In previous chapters, it has been shown that the eigenfunction method can effectively be used to express the solutions of differential equations as infinite series. On the other hand, solutions of differential equations can be obtained as an integral superposition in terms of Green’s functions. So the method of Green’s functions offers several advantages over eigenfunction expansions. First, an integral representation of solutions provides a direct way of describing the general analytical structure of a solution that may be obscured by an infinite series representation. Second, from an analytical point of view, the evaluation of a solution from an integral representation may prove simpler than finding the sum of an infinite series, particularly near 408 11 Green’s Functions and Boundary-Value Problems rapidly-varying features of a function, where the convergence of an eigenfunction expansion may be slow. Third, in view of the Gibbs phenomenon discussed in Chapter 6, the integral representation seems to impose less stringent requirements on the functions that describe the values that the solution must assume on a given boundary than the expansion based on eigenfunctions. Many physical problems are described by second-order nonhomogeneous differential equations with homogeneous boundary conditions or by secondorder homogeneous equations with nonhomogeneous boundary conditions. Such problems can be solved by the method of Green’s functions. We consider a nonhomogeneous partial differential equation of the form Lxu (x) = f (x), (11.1.1) where x = (x, y, z) is a vector in three (or higher) dimensions, Lx is a linear partial differential operator in three or more independent variables with constant coefficients, and u (x) and f (x) are functions of three or more independent variables. The Green’s function G (x, ξ) of this problem satisfies the equation LxG (x, ξ) = δ (x − ξ) (11.1.2) and represents the effect at the point x of the Dirac delta function source at the point ξ = (ξ, η, ζ). Multiplying (11.1.2) by f (ξ) and integrating over the volume V of the ξ space, so that dV = dξ dη dζ, we obtain
V LxG (x, ξ) f (ξ) dξ =
V δ (x − ξ) f (ξ) dξ = f (x). (11.1.3) Interchanging the order of the operator Lx and integral sign in (11.1.3) gives Lx
V G (x, ξ) f (ξ) dξ = f (x). (11.1.4) A simple comparison of (11.1.4) with (11.1.1) leads to the solution of (11.1.1) in the form u (x) =
V G (x, ξ) f (ξ) dξ. (11.1.5) Clearly, (11.1.5) is valid for any finite number of components of x. Accordingly, the Green’s function method can be applied, in general, to any linear, constant coefficient, nonhomogeneous partial differential equation in any number of independent variables. Another way to approach the problem is by looking for the inverse operator L −1 x . If it is possible to find L −1 x , then the solution of (11.1.1) can 11.2 The Dirac Delta Function 409 be obtained as u (x) = L −1 x (f (x)). It turns out that in many important cases it is possible, and the inverse operator can be expressed as an integral operator of the form u (x) = L −1 x (f (ξ)) =
V G (x, ξ) f (ξ) dξ. (11.1.6) The kernel G (x, ξ) is called the Green’s function which is, in fact, the characteristic of the operator Lx for any finite number of independent variables. In our study of partial differential equations with the aid of Green’s functions, special attention will be given to those three partial differential equations which occur most frequently in mathematics, mathematical physics and engineering science; the wave equation utt − c 2∇2u = f (x), (11.1.7) the heat or diffusion equation ut − κ∇2u = f (x), (11.1.8) and the potential or the Laplace equation ∇2u = f (x), (11.1.9) where the Laplacian ∇2 in an n-dimensional Euclidean space is given by ∇2 ≡ ∂ 2 ∂x2 1 + ∂ 2 ∂x2 2 + ... + ∂ 2 ∂x2 n , (11.1.10) and x = (x1, x2,...,xn). Clearly, the solutions of the wave and heat equations are functions of (n + 1) coordinates consisting of n space dimensions, x = (x1, x2,...,xn) and one time dimension t, whereas the solutions of the Laplace equation are functions of n space dimensions. This chapter deals with the basic idea and properties of Green’s functions and how to construct such functions for finding solutions of partial differential equations. Some examples of applications are provided in this chapter and in the next chapter. 11.2 The Dirac Delta Function The application of Green’s functions to boundary-value problems in ordinary differential equations was described earlier in Chapter 8. The Green’s function method is applied here to boundary-value problems in partial differential equations. The method provides solutions in integral form and is applicable to a wide class of problems in applied mathematics and mathematical physics. 410 11 Green’s Functions and Boundary-Value Problems Before developing the method of Green’s functions, we will first define the Dirac delta function δ (x − ξ,y − η) in two dimensions by (a) δ (x − ξ,y − η)=0, x = ξ, y = η, (11.2.1) (b)

Rε δ (x − ξ,y − η) dx dy = 1, Rε : (x − ξ) 2 + (y − η) 2 < ε2 , (11.2.2) (c)

R F (x, y) δ (x − ξ,y − η) dx dy = F (ξ,η), (11.2.3) for arbitrary continuous function F in the region Rε. The delta function is not a function in the ordinary sense. For an elegant treatment of the delta function as a generalized function, see L. Schwartz, Th´eorie des Distributions (1950, 1951). It is a symbolic function, and is often viewed as the limit of a distribution. If δ (x − ξ) and δ (y − η) are one-dimensional delta functions, we have

R F (x, y) δ (x − ξ) δ (y − η) dx dy = F (ξ,η). (11.2.4) Since (11.2.3) and (11.2.4) hold for an arbitrary continuous function F, we conclude that δ (x − ξ,y − η) = δ (x − ξ) δ (y − η). (11.2.5) Thus, we may state that the two-dimensional delta function is the product of two one-dimensional delta functions. Higher dimensional delta functions can be defined in a similar manner. δ (x1, x2,...,xn) = δ (x1) δ (x2)...δ (xn). (11.2.6) The expression for the δ-function become much more complicated when we introduce curvilinear coordinates. However, for simplicity, we transform the two-dimensional delta function from Cartesian coordinates x, y to curvilinear coordinates α, β by means of the transformation x = u (α, β) and y = v (α, β), (11.2.7) where u and v are single-valued continuous and differentiable functions of their arguments. We assume that under this transformation α = α1 and β = β1 correspond to x = ξ and y = η respectively. Changing the coordinates according to (11.2.7), we reduce equation (11.2.4) to

F (u, v) δ (u − ξ) δ (v − η)|J| dα dβ = F (ξ,η), (11.2.8) where J is the Jacobian of the transformation defined by 11.2 The Dirac Delta Function 411 J = ∂ (u, v) ∂ (α, β) =       uα uβ vα vβ       = 0. (11.2.9) Consequently, we can write δ (u − ξ) δ (v − η)|J| = δ (α − α1) δ (β − β1). (11.2.10) In particular, the transformation from rectangular Cartesian coordinates (x, y) to polar coordinates (r, θ) is defined by x = r cos θ, y = r sin θ, (11.2.11) so that the Jacobian J is J =       xr yr xθ yθ       = xryθ − yrrθ = r. (11.2.12) In this case, J vanishes at the origin and the transformation is singular at r = 0 for any θ. Hence, θ can be ignored and δ (x) δ (y) = δ (r) |J1| = 1 2π δ (r) r , (11.2.13) where J1 =
π 0 J dθ = 2πr. Similarly, the transformation from three-dimensional rectangular Cartesian coordinates (x, y, z) to spherical polar coordinates (r, θ, φ) is given by x = r sin θ cos φ, y = r sin θ sin φ, z = r cos θ, (11.2.14) where 0 ≤ r < ∞, 0 ≤ θ ≤ π, and 0 ≤ φ ≤ 2π. The Jacobian of the transformation is J = r 2 sin θ. This Jacobian vanishes for all points on the z-axis, that is, for θ = 0, and hence, the coordinate φ may be ignored. Also, J vanishes at the origin (r = 0) in which case both θ and φ may be ignored. Consequently, δ (x) δ (y) δ (z) = δ (r) |J2| = δ (r) 4πr2 , (11.2.15) where J2 =
π 0
2π 0 J dθ dφ =
π 0
2π 0 r 2 sin θ dφ = 4πr2 . 412 11 Green’s Functions and Boundary-Value Problems 11.3 Properties of Green’s Functions The solution of the Dirichlet problem in a domain D with boundary B ∇2u = h (x, y) in D u = f (x, y) on B (11.3.1) is given in Section 11.5 and has the form u (x, y) =

D G (x, y; ξ,η) h (ξ,η) dξ dη +
B f ∂G ∂n ds, (11.3.2) where G is the Green’s function and n denotes the outward normal to the boundary B of the region D. It is rather obvious then that the solution u (x, y) can be determined as soon as the Green’s function G is ascertained, so the problem in this technique is really to find the Green’s function. First, we shall define the Green’s function for the Dirichlet problem involving the Laplace operator. Then, the Green’s function for the Dirichlet problem involving the Helmholtz operator may be defined in a completely analogous manner. The Green’s function for the Dirichlet problem involving the Laplace operator is the function which satisfies (a) ∇2G = δ (x − ξ,y − η) in D, (11.3.3) G = 0 on B. (11.3.4) (b) G is symmetric, that is, G (x, y; ξ,η) = G (ξ,η; x, y), (11.3.5) (c) G is continuous in x, y, ξ, η, but (∂G/∂n) has a discontinuity at the point (ξ,η) which is specified by the equation limε→0
Cε ∂G ∂n ds = 1, (11.3.6) where n is the outward normal to the circle Cε : (x − ξ) 2 + (y − η) 2 = ε 2 . The Green’s function G may be interpreted as the response of the system at a field point (x, y) due to a δ function input at the source point (ξ,η). G is continuous everywhere in D, and its first and second derivatives are continuous in D except at (ξ,η). Thus, property (a) essentially states that ∇2G = 0 everywhere except at the source point (ξ,η). We will now prove property (b). Theorem 11.3.1. The Green’s function is symmetric. 11.3 Properties of Green’s Functions 413 Proof. Applying the Green’s second formula

D  φ∇2ψ − ψ∇2φ dS =
B  φ ∂ψ ∂n − ψ ∂φ ∂n ds, (11.3.7) to the functions φ = G (x, y; ξ,η) and ψ = G (x, y; ξ ∗ , η∗ ), we obtain

D G (x, y; ξ,η) ∇2G (x, y; ξ ∗ , η∗ ) − G (x, y; ξ ∗ , η∗ ) ∇2G (x, y; ξ,η) ! dx dy =
B G (x, y; ξ,η) ∂G ∂n (x, y; ξ ∗ , η∗ ) − G (x, y; ξ ∗ , η∗ ) ∂G ∂n (x, y; ξ,η) ds. Since G (x, y; ξ,η) and hence, G (x, y; ξ ∗ , η∗ ) must vanish on B, we have

D G (x, y; ξ,η) ∇2G (x, y; ξ ∗ , η∗ ) − G (x, y; ξ ∗ , η∗ ) ∇2G (x, y; ξ,η) ! dx dy = 0. But ∇2G (x, y; ξ,η) = δ (x − ξ,y − η), and ∇2G (x, y; ξ ∗ , η∗ ) = δ (x − ξ ∗ , y − η ∗ ). Since

D G (x, y; ξ,η) δ (x − ξ ∗ , y − η ∗ ) dx dy = G (ξ ∗ , η∗ ; ξ,η), and

D G (x, y; ξ ∗ , η∗ ) δ (x − ξ,y − η) dx dy = G (ξ,η; ξ ∗ , η∗ ), we obtain G (ξ,η; ξ ∗ , η∗ ) = G (ξ ∗ , η∗ ; ξ,η). Theorem 11.3.2. ∂G/∂n is discontinuous at (ξ,η); in particular limε→0
Cε ∂G ∂n ds = 1, Cε : (x − ξ) 2 + (y − η) 2 = ε 2 . Proof. Let Rε be the region bounded by Cε. Then, integrating both sides of equation (11.3.3), we obtain

Rε ∇2G dx dy =

R δ (x − ξ,y − η) dx dy = 1. 414 11 Green’s Functions and Boundary-Value Problems It therefore follows that limε→0

Rε ∇2G dx dy = 1. (11.3.8) Thus, by the Divergence theorem of calculus, limε→0

Cε ∂G ∂n ds = 1. 11.4 Method of Green’s Functions It is often convenient to seek G as the sum of a particular integral of the nonhomogeneous equation and the solution of the associated homogeneous equation. That is, G may assume the form G (ξ,η; x, y) = F (ξ,η; x, y) + g (ξ,η; x, y), (11.4.1) where F, known as the free-space Green’s function, satisfies ∇2F = δ (ξ − x, η − y) in D, (11.4.2) and g satisfies ∇2 g = 0 in D, (11.4.3) so that by superposition G = F + g satisfies equation (11.3.3). Also G = 0 on B requires that g = −F on B. (11.4.4) Note that F need not satisfy the boundary condition. Hereafter, (x, y) will denote the source point. Before we determine the solution of a particular problem, let us first find F for the Laplace and Helmholtz operators. (1) Laplace Operator In this case, F must satisfy the equation ∇2F = δ (ξ − x, η − y) in D. Then, for r = " (ξ − x) 2 + (η − y) 2 # 1 2 > 0, that is, for ξ = x, η = y, we have with (x, y) as the center ∇2F = 1 r ∂ ∂r  r ∂F ∂r  = 0, 11.4 Method of Green’s Functions 415 since F is independent of θ. Therefore, the solution is F = A + B log r. Applying condition (11.3.6), it follows directly from equation (11.3.8) with ∇2 g = 0, that limε→0
Cε ∂F ∂n ds = limε→0
2π 0 B r r dθ = 1. Thus, B = 1/2π and A is arbitrary. For simplicity, we choose A = 0. Then F takes the form F = 1 2π log r. (11.4.5) (2) Helmholtz Operator Here F is required to satisfy ∇2F + κ 2F = δ (x − ξ,y − η). Again for r > 0, we find 1 r ∂ ∂r  r ∂F ∂r  + κ 2F = 0, or, r 2Frr + rFr + κ 2 r 2F = 0. This is the Bessel equation of order zero, the solution of which is F (κr) = AJ0 (κr) + BY0 (κr). Since the behavior of J0 at r = 0 is not singular, we set A = 0. Thus, we have F (κr) = BY0 (κr). But, for very small r, Y0 (κr) ∼ 2 π log r. Applying condition (11.3.6), we obtain 1 = limε→0
Cε ∂F ∂n ds = limε→0
Cε B ∂Y0 ∂r ds = B · 2 πr · 2πr and hence, B = 1/4. Thus, F (κr) becomes 416 11 Green’s Functions and Boundary-Value Problems F (κr) = 1 4 Y0 (κr). (11.4.6) We may point out that, since  ∇2 + κ 2 approaches ∇2 as κ → 0, it should (and does) follow that 1 4 Y0 (κr) → 1 2 log r as κ → 0 + . 11.5 Dirichlet’s Problem for the Laplace Operator We are now in a position to determine the solution of the Dirichlet problem ∇2u = h in D, (11.5.1) u = f on B, by the method of Green’s function. By putting φ (ξ,η) = G (ξ,η; x, y) and ψ (ξ,η) = u (ξ,η) in equation (11.3.7), we obtain

D G (ξ,η; x, y) ∇2u − u (ξ,η) ∇2G ! dξ dη =
B G (ξ,η; x, y) ∂u ∂n − u (ξ,η) ∂G ∂n ds. But ∇2u = h (ξ,η) in D, and ∇2G = δ (ξ − x, η − y) in D. Thus, we have

D [G (ξ,η; x, y) h (ξ,η) − u (ξ,η) δ (ξ − x, η − y)] dξ dη =
B G (ξ,η; x, y) ∂u ∂n − u (ξ,η) ∂G ∂n ds. (11.5.2) Since G = 0 and u = f on B, and since G is symmetric, it follows that u (x, y) =

D G (x, y; ξ,η) h (ξ,η) dξ dη +
B f ∂G ∂n ds (11.5.3) 11.5 Dirichlet’s Problem for the Laplace Operator 417 which is the solution given by (11.3.2). As a specific example, consider the Dirichlet problem for a unit circle. Then ∇2 g = gξξ + gηη = 0 in D, (11.5.4) g = −F on B. But we already have from equation (11.4.5) that F = (1/2π) log r. If we introduce the polar coordinates (see Figure 11.5.1) ρ, θ, σ, β by means of the equations x = ρ cos θ, ξ = σ cos β, (11.5.5) y = ρ sin θ, η = σ sin β, then the solution of equation (11.5.4) is [see Section 9.4] g (σ, β) = a0 2 + ∞ n=1 σ n (an cos nβ + bn sin nβ), where g = − 1 4π log 1 + ρ 2 − 2ρ cos (β − θ) ! on B. Figure 11.5.1 Image point. 418 11 Green’s Functions and Boundary-Value Problems By using the relation log 1 + ρ 2 − 2ρ cos (β − θ) ! = −2 ∞ n=1 ρ n cos n (β − θ) n , and equating the coefficients of sin nβ and cos nβ to determine an and bn, we find an = ρ n 2πn cos nθ, bn = ρ n 2πn sin nθ. It therefore follows that g (ρ, θ; σ, β) = 1 2π ∞ n=1 (σρ) n n cos n (β − θ) = − 1 4π log " 1+(σρ) 2 − 2 (σρ) cos (β − θ) # . Hence, the Green’s function for the problem is G (ρ, θ; σ, β) = 1 4π log σ 2 + ρ 2 − 2σρ cos (β − θ) ! − 1 4π log " 1+(σρ) 2 − 2σρ cos (β − θ) # , (11.5.6) from which we find ∂G ∂n     on B =  ∂G ∂σ  σ=1 = 1 2π 1 − ρ 2 [1 + ρ 2 − 2ρ cos (β − θ)]. If h = 0, then solution (11.5.3) reduces to the Poisson integral formula similar to (9.4.10) and assumes the form u (ρ, θ) = 1 2π
2π 0 1 − ρ 2 1 + ρ 2 − 2ρ cos (β − θ) f (β) dβ. 11.6 Dirichlet’s Problem for the Helmholtz Operator We will now determine the Green’s function solution of the Dirichlet problem involving the Helmholtz operator, namely, ∇2u + κ 2u = h in D, (11.6.1) u = f on B, where D is a circular domain of unit radius with boundary B. Then, the Green’s function must satisfy 11.6 Dirichlet’s Problem for the Helmholtz Operator 419 ∇2G + κ 2G = δ (ξ − x, η − y) in D, (11.6.2) G = 0 on B. Again, we seek the solution in the form G (ξ,η; x, y) = F (ξ,η; x, y) + g (ξ,η; x, y). (11.6.3) From equation (11.4.6), we have F = 1 4 Y0 (κr), (11.6.4) where r = " (ξ − x) 2 + (η − y) 2 # 1 2 . The function g must satisfy ∇2 g + κ 2 g = 0 in D, (11.6.5) g = − 1 4 Y0 (κr) on B. This solution can be determined easily by the method of separation of variables. Thus, the solution in the polar coordinates defined by equation (11.5.5) may be written in the form g (ρ, θ, σ, β) = ∞ n=0 Jn (κσ) [an cos nβ + bn sin nβ] , (11.6.6) where a0 = − 1 8πJ0 (κ)
π −π Y0 " κ  1 + ρ 2 − 2ρ cos (β − θ) # dβ, an bn = = − 1 4πJn(κ) * π −π Y0 " κ  1 + ρ 2 − 2ρ cos (β − θ) # cos nβ dβ − 1 4πJn(κ) * π −π Y0 " κ  1 + ρ 2 − 2ρ cos (β − θ) # sin nβ dβ ⎫ ⎪⎪⎬ ⎪⎪⎭ n = 1, 2,.... To find the solution of the Dirichlet problem, we multiply both sides of the first equation of equation (11.6.1) by G and integrate. Thus, we have

D  ∇2u + κ 2u G (ξ,η; x, y) dξ dη =

D h (ξ,η) G (ξ,η; x, y) dξ dη. We then apply Green’s theorem on the left side of the preceding equation and obtain

D h (ξ,η) G (ξ,η; x, y) dξ dη −

D u  ∇2G + κ 2G dξ dη =
B (G un − u Gn) ds. 420 11 Green’s Functions and Boundary-Value Problems But ∇2G + κ 2G = δ (ξ − x, η − y) in D and G = 0 on B. We, therefore, have u (x, y) =

D h (ξ,η) G (ξ,η; x, y) dξ dη +
B f (ξ,η) Gnds, (11.6.7) where G is given by equation (11.6.3) with equations (11.6.4) and (11.6.6). 11.7 Method of Images We shall describe another method of obtaining Green’s functions. This method, called the method of images, is based essentially on the construction of Green’s function for a finite domain from that of an infinite domain. The disadvantage of this method is that it can be applied only to problems with simple boundary geometry. As an illustration, we consider the same Dirichlet problem solved in Section 11.5. Let P (ξ,η)be a point in the unit circle D, and let Q (x, y) be the source point also in D. The distance between P and Q is r. Let Q′ be the image which lies outside of D on the ray from the origin opposite to the source point Q (as shown in Figure 11.7.1) such that OQ/σ = σ/OQ′ , where σ is the radius of the circle passing through P centered at the origin. Figure 11.7.1 Image point. 11.7 Method of Images 421 Since the two triangles OPQ and OPQ′ are similar by virtue of the hypothesis (OQ) (OQ′ ) = σ 2 and by possessing a common angle at O, we have r ′ r = σ ρ , (11.7.1) where r ′ = P Q′ and ρ = OQ. If σ = 1, equation (11.7.1) becomes
4 r r ′
5 1 ρ = 1. Then, we clearly see that the quantity 1 2π log  r r ′ 1 ρ  = 1 2π log r − 1 2π log r ′ + 1 2π log 1 ρ (11.7.2) which vanishes on the boundary σ = 1, is harmonic in D except at Q, and satisfies equation (11.3.3). (Note the log r ′ is harmonic everywhere except at Q′ , which is outside the domain D.) This suggests that we should choose the Green’s function G = 1 2π log r − 1 2π log r ′ + 1 2π log 1 ρ . (11.7.3) Noting that Q′ is at (1/ρ, θ), the function G in polar coordinates takes the form G (ρ, θ, σ, β) = 1 4π log σ 2 + ρ 2 − 2σρ cos (β − θ) ! − 1 4π log 1 σ 2 + ρ 2 − 2 ρ σ cos (β − θ) + 1 2π log 1 σ (11.7.4) which is the same as G given by (11.5.6). It is quite interesting to observe the physical interpretation of the Green’s function (11.7.3) and (11.7.4). The first term represents the potential due to a unit line charge at the source point, whereas the second term represents the potential due to a negative unit charge at the image point. The third term represents a uniform potential. The sum of these potentials makes up the total potential field. Example 11.7.1. To illustrate an obvious and simple case, consider the semiinfinite plane η > 0. The problem is to solve ∇2u = h in η > 0, u = f on η = 0. 422 11 Green’s Functions and Boundary-Value Problems The image point should be obvious by inspection. Thus, if we construct G = 1 4π log " (ξ − x) 2 + (η − y) 2 # − 1 4π " (ξ − x) 2 + (η + y) 2 # , (11.7.5) the condition that G = 0 on η = 0 is clearly satisfied. It is also evident that G is harmonic in η > 0 except at the source point, and that G satisfies equation (11.3.3). With Gn|B = [−Gη] η=0, the solution (11.5.3) is given by u (x, y) = y π
∞ −∞ f (ξ) dξ (ξ − x) 2 + y 2 + 1 4π
∞ 0
∞ −∞ log 1 (ξ − x) 2 + (η − y) 2 (ξ − x) 2 + (η + y) 2 3 h (ξ,η) dξ dη. (11.7.6) Example 11.7.2. Another example that illustrates the method of images well is the Robin’s problem on the quarter infinite plane, that is, ∇2u = h (ξ,η) in ξ > 0, η> 0, u = f (η) on ξ = 0, (11.7.7) un = g (ξ) on η = 0. This illustrated in Figure 11.7.2. Figure 11.7.2 Images in the Robin problem. 11.8 Method of Eigenfunctions 423 Let (−x, y), (−x, −y), and (x, −y) be the three image points of the source point (x, y). Then, by inspection, we can immediately construct Green’s function G = 1 4π log " (ξ − x) 2 + (η − y) 2 #"(ξ − x) 2 + (η + y) 2 # " (ξ + x) 2 + (η − y) 2 #"(ξ + x) 2 + (η + y) 2 # . (11.7.8) This function satisfies ∇2G = 0 except at the source point, and G = 0 on ξ = 0 and Gη = 0 on η = 0. The solution from equation (11.3.3) is thus given by u (x, y) =

D G h dξ dη +
B (G un − u Gn) ds, =
∞ 0
∞ 0 G h dξ dη +
∞ 0 g (ξ) G (ξ, 0; x, y) dξ, +
∞ 0 f (η) Gξ (0, η; x, y) dξ. 11.8 Method of Eigenfunctions In this section, we will apply the method of eigenfunctions, described in Chapter 10, to obtain the Green’s function. We consider the boundary-value problem ∇2u = h in D, (11.8.1) u = f on B. For this problem, G must satisfy ∇2G = δ (ξ − x, η − y) in D, (11.8.2) G = 0 on B, and hence, the associated eigenvalue problem is ∇2φ + λφ = 0 in D, (11.8.3) φ = 0 on B. Let φmn be the eigenfunctions and λmn be the corresponding eigenvalues. We then expand G and δ in terms of the eigenfunctions φmn. Consequently, we write 424 11 Green’s Functions and Boundary-Value Problems G (ξ,η; x, y) =  m  n amn (x, y) φmn (ξ,η), (11.8.4) δ (ξ − x, η − y) =  m  n bmn (x, y) φmn (ξ,η), (11.8.5) where bmn = 1 φmn 2

D δ (ξ − x, η − y) φmn (ξ,η) dξ dη = φmn (x, y) φmn 2 (11.8.6) in which φmn 2 =

D φ 2 mndξ dη. Now substituting equations (11.8.4) and (11.8.5) into equation (11.8.2) and using the relation from equation (11.8.3) that ∇2φmn + λmnφmn = 0, we obtain −  m  n λmnamn (x, y) φmn (ξ,η) =  m  n φmn (x, y) φmn (ξ,η) φmn 2 . Hence, amn (x, y) = − φmn (x, y) λmn φmn 2 , (11.8.7) and the Green’s function is therefore given by G (ξ,η; x, y) = −  m  n φmn (x, y) φmn (ξ,η) λmn φmn 2 . (11.8.8) Example 11.8.1. As a particular example, consider the Dirichlet problem in a rectangular domain ∇2u = h in D {0 < x < a, 0 <y<b} ,="" u="0" on="" b.="" the="" eigenfunctions="" can="" be="" obtained="" explicitly="" by="" method="" of="" separation="" variables.="" we="" assume="" a="" nontrivial="" solution="" in="" form="" (ξ,η)="X" (ξ)="" y="" (η).="" substituting="" this="" following="" system="" 11.9="" higher-dimensional="" problems="" 425="" ∇2u="" +="" λu="0" d,="" b,="" yields,="" with="" α="" 2="" as="" constant,="" x′′="" 2x="0," ′′="" ="" λ="" −="" homogeneous="" boundary="" conditions="" x="" (0)="X" (a)="0" and="" (b)="0," functions="" are="" found="" to="" xm="" sin="" ="" mπξ="" ="" yn="" (η)="Bn"
4nπη="" b=""
5="" .="" then="" have="" λmn="π" m2="" n="" thus,="" obtain="" φmn="" knowing="" φmn,="" compute="" φmn="" 0=""
="" sin2="" dξ="" dη="" ab="" 4="" from="" equation="" (11.8.8)="" green’s="" function="" g="" (ξ,η;="" x,="" y)="−" 4ab="" π="" ∞="" m="1" mπx="" nπy=""
4="" nπη="" (m2b="" n2a="" 2)="" easily="" extended="" for="" applications="" three="" more="" dimensions.="" since="" most="" encountered="" physical="" sciences="" dimensions,="" will="" illustrate="" some="" examples="" suitable="" practical="" application.="" first="" extend="" our="" definition="" dirichlet="" problem="" involving="" laplace="" operator="" is="" that="" satisfies="" 426="" 11="" boundary-value="" ∇2g="δ" (x="" ξ,y="" η,="" z="" ζ)="" r,="" (11.9.1)="" s.="" (11.9.2)="" (x,="" y,="" z;="" ξ,="" (ξ,="" ζ;="" z).="" (11.9.3)="" (c)="" limε→0=""

="" sε="" ∂g="" ∂n="" ds="1," (11.9.4)="" where="" outward="" unit="" normal="" surface="" :="" ξ)="" (y="" η)="" (z="" proceeding="" two-dimensional="" case,="" (11.9.5)="" s,="" z)="


" r="" h="" dr="" s="" f="" gn="" ds.="" (11.9.6)="" again="" let="" z),="" ∇2f="δ" ∇2="" example="" 11.9.1.="" consider="" spherical="" domain="" radius="" a.="" must="" except="" at="" source="" point.="" τ="" 2ρ2="" a2="" 2τ="" ρ="" cos="" γ="" #="" 1="" (11.9.10)="" angle="" between="" ′="" now="" differentiating="" g,="" ∂τ="" =="" 4πa="" (a="" 2aρ="" γ)="" 428="" (ρ,="" θ,="" φ)="a" 4π="" 2π="" (α,="" ψ)="" dα="" dψ="" 3="" (11.9.11)="" θ="" (ψ="" φ).="" integral="" called="" three-dimensional="" poisson="" formula.="" exterior="" radially="" inward="" towards="" origin,="" simply="" replacing="" (11.9.11).="" 11.9.2.="" an="" helmholtz="" threedimensional="" radiation="" κ="" 2u="0," (11.9.12)="" limr→∞="" (ur="" iκu)="0," i="√" −1;="" limit="" condition="" condition,="" field="" point="" distance.="" satisfy="" 2g="δ" (ξ="" η="" ζ="" (11.9.13)="" dependent="" only="" write="" grr="" gr=""> 0. Note that the source point is taken as the origin. If we write the above equation in the form (G r) rr + κ 2 (G r) = 0 for r > 0 (11.9.14) then the solution can easily be seen to be G r = Aeiκr + Be−iκr , or, equivalently, G = A e iκr r + B e −iκr r . (11.9.15) In order for G to satisfy the radiation condition limr→∞ r (Gr + iκG)=0, the constant A = 0, and G then takes the form G = B e −iκr r . 11.9 Higher-Dimensional Problems 429 To determine B, we have limε→0

Sε ∂G ∂n dS = − limε→0

Sε B e −iκr r  1 r + iκ dS = 1 from which we obtain B = −1/4π, and consequently, G = − e −iκr 4πr . (11.9.16) Note that this reduces to (1/4πr) when κ = 0. Example 11.9.3. Show that the solution of the Poisson equation −∇2u = f (x, y, z), (11.9.17) is u (x, y, z) =


G (r) f (ξ, η, ζ) dξ dη dζ, (11.9.18) where the Green’s function G (r) is G (r) = 1 4πr = 1 4π ( (x − ξ) 2 + (y − η) 2 + (z − ζ) 2 )− 1 2 . (11.9.19) The Green’s function G satisfies the equation −∇2G = δ (x − ξ) δ (y − η) δ (z − ζ). (11.9.20) It is noted that everywhere except at (x, y, z)=(ξ, η, ζ), equation (11.9.20) is a homogeneous equation that can be solved by the method of separation of variables. However, at the point (ξ, η, ζ) this equation is no longer homogeneous. Usually, this point (ξ, η, ζ) represents a source point or a source point singularity in electrostatics or fluid mechanics. In order to solve (11.9.17), it is necessary to take into account the source point at (ξ, η, ζ). Without loss of generality, it is convenient to transform the frame of reference so that the source point is at the origin. This can be done by the transformation x1 = x − ξ, y1 = y − η, and z1 = z − ζ. Consequently, equation (11.9.20) becomes ∇2G = −δ (x1) δ (y1) δ (z1), (11.9.21) where ∇2 is the Laplacian in terms of x1, y1, and z1. Introducing the spherical polar coordinates x1 = r sin θ cos φ, y1 = r sin θ sin φ, z1 = r cos θ, equation (11.9.21) reduces to the form 430 11 Green’s Functions and Boundary-Value Problems ∇2G = − δ (r) 4πr2 , (11.9.22) where ∇2G ≡ 1 r 2 ∂ ∂r  r 2 ∂G ∂r  + 1 r 2 sin θ ∂ ∂θ  sin θ ∂G ∂θ  + 1 r 2 sin2 θ ∂ 2G ∂φ2 .(11.9.23) Since the right hand side of (11.9.22) is a function of r alone, and hence, G must be a function of r alone, we write (11.9.22) with (11.9.23) as 1 r 2 ∂ ∂r  r 2 ∂G ∂r  = − δ (r) 4πr2 . (11.9.24) We assume that G tends to zero as r → ∞. The solution of the corresponding homogeneous equation of (11.9.24) is G (r) = a r + b, (11.9.25) where a and b are constants of integration. Since G → 0 as r → ∞, b = 0 and we set a = 1 4π . Consequently, the solution for G is G (r) = 1 4πr . (11.9.26) This solution can be interpreted as the potential produced by a point charge at the point (ξ, η, ζ). Finally, the solution of (11.9.17) is then given by u (x, y, z) = −


∞ −∞ G (r) f (ξ, η, ζ) dξ dη dζ = 1 4π


∞ −∞ " (x − ξ) 2 + (y − η) 2 + (z − ζ) 2 #− 1 2 ×f (ξ, η, ζ) dξ dη dζ. (11.9.27) Physically, this solution of the Poisson equation represents the potential u (x, y, z) produced by a charge distribution of volume density f (x, y, z). 11.10 Neumann Problem We have noted in the chapter on boundary-value problems that the Neumann problem requires more attention than Dirichlet’s problem because an additional condition is necessary for the existence of a solution of the Neumann problem. We now consider the Neumann problem 11.10 Neumann Problem 431 ∇2u + κ 2u = h in R, ∂u ∂n = 0 on S. By the divergence theorem of calculus, we have


R ∇2u dR =

S ∂u ∂n dS. Thus, if we integrate the Helmholtz equation and use the preceding result, we obtain κ 2


R u dR =


R h dR. In the case of Poisson’s equation where κ = 0, this relation is satisfied only when


R h dR = 0. If we consider a heat conduction problem, this condition may be interpreted as the requirement that the net heat generation be zero. This is physically reasonable since the boundary is insulated in such a way that the net flux across it is zero. If we define Green’s function G, in this case, by ∇2G + κ 2G = δ (ξ − x, η − y, ζ − z) in R, ∂G ∂n = 0 on S. Then we must have κ 2


R G dR = 1 which cannot be satisfied for κ = 0. But, we know from a physical point of view that a solution exists if


R h dR = 0. Hence, we will modify the definition of Green’s function so that ∂G ∂n = C on S, where C is a constant. Integrating ∇2G = δ over R, we obtain C

S dS = 1. 432 11 Green’s Functions and Boundary-Value Problems It is not difficult to show that G remains symmetric if

S G dS = 0. Thus, under this condition, if we take C to be reciprocal of the surface area, the solution of the Neumann problem for Poisson’s equation is u (x, y, z) = C ∗ +


R G (x, y, z; ξ, η, ζ) h (ξ, η, ζ) dξ dη dζ, where C ∗ is a constant. We should remark here that the method of Green’s functions provides the solution in integral form. This is made possible by replacing a problem involving nonhomogeneous boundary conditions with a problem of finding Green’s function G with homogeneous boundary conditions. Regardless of method employed, the Green’s function of a problem with nonhomogeneous equation and homogeneous boundary conditions is the same as the Green’s function of a problem with homogeneous equation and nonhomogeneous boundary conditions, since one problem can be transferred to the other without difficulty. To illustrate, we consider the problem Lu = f in R, u = 0 on ∂R, where ∂R denotes the boundary of R. If we let v = w − u, where w satisfies Lw = f in R, then the problem becomes Lv = 0 in R, v = w on ∂R. Conversely, if we consider the problem Lu = 0 in R, u = g on ∂R, we can easily transform this problem into Lv = Lw ≡ w ∗ in R, v = 0 on ∂R, by putting v = w − u and finding w that satisfies w = g on ∂R. In fact, if we have Lu = f in R, u = g on ∂R, we can transform this problem into either one of the above problems. 11.11 Exercises 433 11.11 Exercises 1. If L denotes the partial differential operator Lu = Auxx + Buxy + Cuyy + Dux + Euy + F u, and if M denotes the adjoint operator Mv = (Av)xx + (Bv)xy + (Cv)yy − (Dv)x − (Ev)y + Fv, show that

R (vLu − uMv) dx dy =
∂R [U cos (n, x) + V cos (n, y)] ds, where U = Avux − u (Av)x − u (Bv)y + Duv, V = Bvux + Cvuy − u (Cv)y + Euv, and ∂R is the boundary of a region R. 2. Prove that the Green’s function for a problem, if it exists, is unique. 3. Determine the Green’s function for the exterior Dirichlet problem for a unit circle ∇2u = 0 in r > 1, u = f on r = 1. 4. Prove that for x = x (ξ,η) and y = y (ξ,η) δ (x − x0) δ (y − y0) = 1 |J| δ (ξ − ξ0) δ (η − η0), where J is the Jacobian and (x0, y0) corresponds to (ξ0, η0). Hence, show that for polar coordinates δ (x − x0) δ (y − y0) = 1 r δ (r − r0) δ (θ − θ0). 5. Determine, for an infinite wedge, the Green’s function that satisfies ∇2G + κ 2G = 1 r δ (r − r0, θ − θ0), G = 0, θ = 0, and θ = α. 6. Determine, for the Poisson’s equation, the Green’s function which vanishes on the boundary of a semicircular domain of radius R. 434 11 Green’s Functions and Boundary-Value Problems 7. Find the solution of the Dirichlet problem ∇2u = 0, 0 < x < a, 0 < y < b, u (0, y) = u (a, y) = u (x, b)=0, u (x, 0) = f (x). 8. Determine the solution of Dirichlet’s problem ∇2u = f (r, θ) in D, u = 0, on ∂D, where ∂D is the boundary of a circle D of radius R. 9. Determine the Green’s function for the semi-infinite region ζ > 0 for ∇2G + κ 2G = δ (ξ − x, η − y, ζ − z), G = 0, on ζ = 0. 10. Determine the Green’s function for the semi-infinite region ζ > 0 for ∇2G + κ 2G = δ (ξ − x, η − y, ζ − z), ∂G ∂n = 0, on ζ = 0. 11. Find the Green’s function in the quarter plane ξ > 0, η > 0 which satisfies ∇2G = δ (ξ − x, η − y), G = 0, on ξ = 0 and η = 0. 12. Find the Green’s function in the quarter plane ξ > 0, η > 0 which satisfies ∇2G = δ (ξ − x, η − y), Gξ (0, η)=0, G (ξ, 0) = 0. 13. Find the Green’s function in the half plane 0 <x< ∞,="" −∞="" <y<="" ∞="" for="" the="" problem="" ∇2u="f" in="" r,="" u="0," on="" x="0." 14.="" determine="" green’s="" function="" that="" satisfies="" ∇2g="δ" (x="" −="" ξ,y="" η)="" d="" :="" 0="" <="" a,="" g="0," ∂d="" y="0," is="" bounded="" at="" infinity.="" 11.11="" exercises="" 435="" 15.="" find="" r="" δ="" (r="" ρ,="" θ="" β),="" <θ<="" π="" 3="" ,="" <r<="" 1,="" and="" ∂g="" ∂n="0," 16.="" solve="" boundary-value="" 1="" ∂r="" ="" ∂u="" ="" +="" ∂="" 2u="" ∂z2="" κ="" ≥="" 0,="" z=""> 0, ∂u ∂z = ⎧ ⎨ ⎩ 0, r>a, z C, r < a, z = = 0, 0, C = constant. 17. Obtain the solution of the Laplace equation ∇2u = 0, 0 <r< ∞,="" 0="" <θ<="" 2π,="" u="" (r,="" 0+)="u" 2π−)="0." 18.="" determine="" the="" green’s="" function="" for="" equation="" ∇2u="" −="" κ="" 2u="0," vanishing="" on="" all="" sides="" of="" rectangle="" ≤="" x="" a,="" y="" b.="" 19.="" helmholtz="" +="" <="" −∞="" <y<="" and="" 20.="" solve="" exterior="" dirichlet="" problem="" in="" r=""> 1, u (1, θ, φ) = f (θ, φ). 21. By the method of images, determine the potential due to a point charge q near a conducting sphere of radius R with potential V . 22. By the method of images, show that the potential due to a conducting sphere of radius R in a uniform electric field E0 is given by U = −E0  r − R2 r 2  cos θ, where r, θ are polar coordinates with origin at the center of the sphere. 436 11 Green’s Functions and Boundary-Value Problems 23. Determine the potential in a cylinder of radius R and length l. The potential on the ends is zero, while the potential on the cylindrical surface is prescribed to be f (θ, z). 24. Consider the fundamental solution of the Fokker–Planck equation defined by ∂ ∂t − ∂ ∂x  ∂ ∂x + x  G (x, x′ ;t, t′ ) = δ (x − x ′ ) δ (t − t ′ ). Using the transformation of variables employed in Example 7.8.4, show that the above equation becomes ∂ ∂τ − ∂ ∂ξ2 G (ξ, ξ′ ; τ, τ ′ ) = δ (ξ − ξ ′ ) δ (τ − τ ′ ). Show that (a) the fundamental solution of the Fokker–Planck equation (see Reif (1965)) is G (x, x′ ;t, t′ ) = [2π {1 − exp [−2 (t − t ′ )]}] − 1 2 × exp 1 − 1 2 [x − x ′ exp {− (t − t ′ )}] 2 1 − exp {−2 (t − t ′)} 3 , (b) limt→∞ G (x, x′ ;t, t′ ) = 1 √ 2π exp  − 1 2 x 2  , (c) limt→∞ u (x, t) = 1 √ 2π exp  − 1 2 x 2 
∞ −∞ f (x ′ ) dx′ , where u (x, 0) = f (x). Give an interpretation of this asymptotic solution u (x, t) as t → ∞. 25. (a) Use the transformation u = ve−t to show that the telegraph equation utt − c 2uxx + 2ut = 0, can be reduced to the form vtt − c 2 vxx + v = 0. (b) Show that the fundamental solution of the transformed telegraph equation is given by G (x − x ′ , t − t ′ ) = 1 2 I0 % (t − t ′) 2 − (x − x ′) 2 ×H [(t − t ′ ) − (x − x ′ )] H [(t − t ′ )+(x − x ′ )] . 11.11 Exercises 437 (c) If the initial data for the telegraph equation are u (x, 0) = f (x), ut (x, 0) = g (x), show that the solution of the telegraph equation is given by u (x, t) =  1 2
x+t x−t g (ξ) I0 % t 2 − (x − ξ) 2 dξ + 1 2 ∂ ∂t
x+t x−t f (ξ) I0 % t 2 − (x − ξ) 2 dξ0 e −t , which is, by evaluating the second term, = e −t  1 2 [f (x − t) + f (x + t)] + 1 2
x+t x−t g (ξ) I0 % t 2 − (x − ξ) 2 dξ + t 2
x+t x−t f (ξ) " t 2 − (x − ξ) 2 #− 1 2 I1 % t 2 − (x − ξ) 2 dξ0 . 12 Integral Transform Methods with Applications “The theory of Fourier series and integrals has always had major difficulties and necessitated a large mathematical apparatus in dealing with questions of convergence. It engendered the development of methods of summation, although these did not lead to a completely satisfactory solution of the problem.... For the Fourier transform, the introduction of distribution (hence the space S) is inevitable either in an explicit or hidden form.... As a result one may obtain all that is desired from the point of view of the continuity and inversion of the Fourier transform.” L. Schwartz “In every mathematical investigation, the question will arise whether we can apply our mathematical results to the real world.” V. I. Arnold 12.1 Introduction The linear superposition principle is one of the most effective and elegant methods to represent solutions of partial differential equations in terms of eigenfunctions or Green’s functions. More precisely, the eigenfunction expansion method expresses the solution as an infinite series, whereas the integral solution can be obtained by integral superposition or by using Green’s functions with initial and boundary conditions. The latter offers several advantages over eigenfunction expansion. First, an integral representation provides a direct way of describing the general analytical structure of a solution that may be obscured by an infinite series representation. Second, from a practical point of view, the evaluation of a solution from an integral representation may prove simpler than finding the sum of an infinite series, 440 12 Integral Transform Methods with Applications particularly near rapidly-varying features of a function, where the convergence of an eigenfunction expansion is expected to be slow. Third, in view of the Gibbs phenomenon discussed in Chapter 6, the integral representation seems to be less stringent requirements on the functions that describe the initial conditions or the values of a solution are required to assume on a given boundary than expansions based on eigenfunctions. Integral transform methods are found to be very useful for finding solutions of initial and/or boundary-value problems governed by partial differential equations for the following reason. The differential equations can readily be replaced by algebraic equations that are inverted by the inverse transform so that the solution of the differential equations can then be obtained in terms of the original variables. The aim of this chapter is to provide an introduction to the use of integral transform methods for students of applied mathematics, physics, and engineering. Since our major interest is the application of integral transforms, no attempt will be made to discuss the basic results and theorems relating to transforms in their general forms. The present treatment is restricted to classes of functions which usually occur in physical and engineering applications. 12.2 Fourier Transforms We first give a formal definition of the Fourier transform by using the complex Fourier integral formula (6.13.10). Definition 12.2.1. If u (x, t) is a continuous, piecewise smooth, and absolutely integrable function, then the Fourier transform of u (x, t) with respect to x ∈ R is denoted by U (k, t) and is defined by F {u (x, t)} = U (k, t) = 1 √ 2π
∞ −∞ e −ikx u (x, t) dx, (12.2.1) where k is called the Fourier transform variable and exp (−ikx) is called the kernel of the transform. Then, for all x ∈ R, the inverse Fourier transform of U (k, t) is defined by F −1 {U (k, t)} = u (x, t) = 1 √ 2π
∞ −∞ e ikx U (k, t) dk. (12.2.2) We may note that the factor (1/2π) in the Fourier integral formula (6.13.9) has been split and placed in front of the integrals (12.2.1) and (12.2.2). Often the factor (1/2π) can be placed in only one of the relations (12.2.1) and (12.2.2). It is not uncommon to adopt the kernel exp (ikx) in (12.2.1) instead of exp (−ikx), and as a consequence, exp (−ikx) would be replaced by exp (ikx) in (12.2.2). 12.2 Fourier Transforms 441 Example 12.2.1. Show that (a) F & exp  −ax2 ' = 1 √ 2a exp  − k 2 4a  , a > 0, (12.2.3) (b) F {exp (−a |x|)} = 2 2 π a (a 2 + k 2) , a > 0, (12.2.4) (c) F & χ[−a,a] (x) ' = 2 2 π  sin ak k  , (12.2.5) where χ[−a,a] (x) = H (a − |x|) = ⎧ ⎨ ⎩ 1, |x| < a 0, |x| > a ⎫ ⎬ ⎭ . (12.2.6) Proof. We have, by definition (12.2.1), F & exp  −ax2 ' = 1 √ 2π
∞ −∞ e −ikx−ax2 dx = 1 √ 2π
∞ −∞ exp 1 −a  x + ik 2a 2 − k 2 4a 3 dx = 1 √ 2π exp  − k 2 4a 
∞ −∞ e −ay2 dy = 1 √ 2a exp  − k 2 4a  , in which the change of variable y =  x + ik 2a is used. The above result is correct, and the change of variable can be justified by methods of complex analysis because (ik/2a) is complex. If a = 1 2 , then F  exp  − 1 2 x 2 0 = exp  − 1 2 k 2  . (12.2.7) This shows that F {f (x)} = f (k). Such a function is said to be selfreciprocal under the Fourier transformation. The graphs of f (x) = e −ax2 and F (k) = F {f (x)} are shown in Figure 12.2.1 for a = 1. To prove (b), we write F {exp (−a |x|)} = 1 √ 2π
∞ −∞ exp (−a |x| − ikx) dx = 1 √ 2π
0 −∞ exp {(a − ik) x} dx +
∞ 0 exp {− (a + ik) x} dx = 1 √ 2π 1 a − ik + 1 a + ik = 2 2 π a a 2 + k 2 . 442 12 Integral Transform Methods with Applications Figure 12.2.1 Graphs of f (x) = exp(−ax2 ) and F (k). It is noted that f (x) = exp (−a |x|) decreases rapidly at infinity, and it is not differentiable at x = 0. The graphs of f (x) and its Fourier transform F (k) are shown in Figure 12.2.2. To prove (c), we have Fa (k) = F & χ[−a,a] (x) ' = 1 √ 2π
∞ −∞ e −ikxχ[−a,a] (x) dx = 1 √ 2π
a −a e −ikxdx = 2 2 π  sin ak a  . The graphs of χ[−a,a] (x) and Fa (k) are shown in Figure 12.2.3 with a = 1. Analogous to the Fourier cosine and sine series, there are Fourier cosine and sine integral transforms for odd and even functions respectively. Definition 12.2.2. Let f (x) be defined for 0 ≤ x < ∞, and extended as an even function in (−∞,∞) satisfying the conditions of Fourier InteFigure 12.2.2 Graphs of f (x) = exp (−a |x|) and F (k). 12.2 Fourier Transforms 443 Figure 12.2.3 Graphs of χ[−a,a] (x) and Fa (k). gral formula (6.13.9). Then, at the points of continuity, the Fourier cosine transform of f (x) and its inverse transform are defined by Fc {f (x)} = Fc (k) = 2 2 π
∞ 0 cos kx f (x) dx, (12.2.8) F −1 c {Fc (k)} = f (x) = 2 2 π
∞ 0 cos kx Fc (k) dk, (12.2.9) where Fc is the Fourier cosine transformation and F −1 c is its inverse transformation respectively. Definition 12.2.3. Similarly, the Fourier sine integral formula (6.13.3) leads to the Fourier sine transform and its inverse defined by Fs {f (x)} = Fs (k) = 2 2 π
∞ 0 sin kx f (x) dx, (12.2.10) F −1 s {Fs (k)} = f (x) = 2 2 π
∞ 0 sin kx Fs (k) dk, (12.2.11) where Fs is called the Fourier sine transformation and F −1 s is its inverse. Example 12.2.2. Show that (a) Fc & e −ax' = 2 2 π a (a 2 + k 2) , a > 0, (12.2.12) (b) Fs & e −ax' = 2 2 π k (a 2 + k 2) , a > 0, (12.2.13) (c) F −1 s  1 k e −sk0 = 2 2 π tan−1
4x s
5 . (12.2.14) 444 12 Integral Transform Methods with Applications We have, by definition, Fc & e −ax' = 2 2 π
∞ 0 e −ax cos kx dk, = 1 2 2 2 π
∞ 0 " e −(a−ik)x + e −(a+ik)x # dx, = 1 2 2 2 π 1 a − ik + 1 a + ik = 2 2 π a (a 2 + k 2) . The proof of (b) is similar and is left to the reader as an exercise. To prove (c), we use the standard definite integral 2 π 2 F −1 s & e −sk' =
∞ 0 e −sk sin kx dk = x s 2 + x 2 . Integrating both sides with respect to s from s to ∞ gives
∞ 0 e −sk k sin kx dk =
∞ s x ds x 2 + s 2 = " tan−1 s x # = (π/2) − tan−1
4 s x
5 . Consequently, F −1 s  1 k e −sk0 = 2 2 π
∞ 0 1 k e −sk sin kx dk = 2 2 π tan−1
4x s
5 . 12.3 Properties of Fourier Transforms Theorem 12.3.1. (Linearity). The Fourier transformation F is linear. Proof. We have F [f (x)] = 1 √ 2π
∞ −∞ e −ikxf (x) dx. Then, for any constants a and b, F [af (x) + bg (x)] = 1 √ 2π
∞ −∞ [af (x) + bg (x)] e −ikxdx, = a √ 2π
∞ −∞ f (x) e −ikxdx + b √ 2π
∞ −∞ g (x) e −ikxdx, = a F [f (x)] + b F [g (x)] . Theorem 12.3.2. (Shifting). Let F [f (x)] be a Fourier transform of f (x). Then F [f (x − c)] = e −ixcF [f (x)] , where c is a real constant. 12.3 Properties of Fourier Transforms 445 Proof. From the definition, we have, for c > 0, F [f (x − c)] = 1 √ 2π
∞ −∞ f (x − c) e −ikxdx, = 1 √ 2π
∞ −∞ f (ξ) e −ik(ξ+c) dξ, where ξ = x − c = e −ikcF [f (x)] . Theorem 12.3.3. (Scaling). If F is the Fourier transform of f, then F [f (cx)] = (1/ |c|) F (k/c), where c is a real nonzero constant. Proof. For c = 0, F [f (cx)] = 1 √ 2π
∞ −∞ f (cx) e −ikxdx. If we let ξ = cx, then F [f (cx)] = 1 |c| 1 √ 2π
∞ −∞ f (ξ) e −i(k/c)ξ dξ = (1/ |c|) F (k/c). Theorem 12.3.4. (Differentiation). Let f be continuous and piecewise smooth in (−∞,∞). Let f (x) approach zero as |x|→∞. If f and f ′ are absolutely integrable, then F [f ′ (x)] = ikF [f (x)] = ikF (k). Proof. F [f ′ (x)] = 1 √ 2π
∞ −∞ f ′ (x) e −ikxdx, = 1 √ 2π f (x) e −ikx  ∞ −∞ + ik √ 2π
∞ −∞ f (x) e −ikxdx , = ikF [f (x)] = ikF (k). This result can be easily extended. If f and its first (n − 1) derivatives are continuous, and if its nth derivative is piecewise continuous, then F " f (n) (x) # = (ik) n F [f (x)] = (ik) n F (k), n = 0, 1, 2,... (12.3.1) provided f and its derivatives are absolutely integrable. In addition, we assume that f and its first (n − 1) derivatives tend to zero as |x| tends to infinity. 446 12 Integral Transform Methods with Applications If u (x, t) → 0 as |x|→∞, then F  ∂u ∂x0 = 1 √ 2π
∞ −∞ e −ikx  ∂u ∂x dx, which is, integrating by parts, = 1 √ 2π e −ikxu (x, t) !∞ −∞ + ik √ 2π
∞ −∞ e −ikxu (x, t) dx, = ik F {u (x, t)} = ik U (k, t). (12.3.2) Similarly, if u (x, t) is continuously n times differentiable, and ∂mm ∂xm → 0 as |x|→∞ for m = 1, 2, 3,..., (n − 1) then F  ∂ nu ∂xn 0 = (ik) n F {u (x, t)} = (ik) n U (k, t). (12.3.3) It also follows from the definition (12.2.1) that F  ∂u ∂t 0 = dU dt , F  ∂ 2u ∂t2 0 = d 2U dt2 , ..., F  ∂ nu ∂tn 0 = d nU dtn . (12.3.4) The definition of the Fourier transform (12.2.1) shows that a sufficient condition for u (x, t) to have a Fourier transform is that u (x, t) is absolutely integrable in −∞ <x< ∞.="" this="" existence="" condition="" is="" too="" strong="" for="" many="" practical="" applications.="" simple="" functions,="" such="" as="" a="" constant="" function,="" sin="" ωx,="" and="" x="" nh="" (x),="" do="" not="" have="" fourier="" transforms="" even="" though="" they="" occur="" frequently="" in="" the="" above="" definition="" of="" transform="" has="" been="" extended="" more="" general="" class="" functions="" to="" include="" other="" functions.="" we="" simply="" state="" fact="" that="" there="" sense,="" useful="" applications,="" which="" stated="" others="" transforms.="" following="" are="" examples="" their="" (see="" lighthill,="" 1964):="" f="" {h="" (a="" −="" |x|)}="2" 2="" π="" ="" ak="" k="" ="" ,="" (12.3.5)="" where="" h="" (x)="" heaviside="" unit="" step="" {δ="" (x="" a)}="1" √="" 2π="" exp="" (−iak),="" (12.3.6)="" δ="" a)="" dirac="" delta="" 1="" iπk="" +="" (k)="" (−iak).="" (12.3.7)="" 12.3="" properties="" 447="" example="" 12.3.1.="" find="" solution="" dirichlet="" problem="" half-plane="" y=""> 0 uxx + uyy = 0, −∞ <x< ∞,="" y=""> 0, u (x, 0) = f (x), −∞ <x< ∞,="" u="" and="" ux="" vanish="" as="" |x|→∞,="" is="" bounded="" y="" →="" ∞.="" let="" (k,="" y)="" be="" the="" fourier="" transform="" of="" (x,="" with="" respect="" to="" x.="" then="" √="" 2π=""
="" ∞="" −∞="" e="" −ikxdx.="" application="" x="" gives="" uyy="" −="" k="" 2u="0," (12.3.8)="" 0)="F" (k)="" 0="" (12.3.9)="" solution="" this="" transformed="" system="" −|k|y="" .="" inverse="" in="" form="" 1="" f="" (ξ)="" −ikξdξ="" ikxdk,="1" dξ="" −k[i(ξ−x)]−|k|y="" dk.="" it="" follows="" from="" proof="" example="" 12.2.1="" (b)="" that="" dk="2y" (ξ="" x)="" 2="" +="" hence,="" dirichlet="" problem="" half-plane=""> 0 is u (x, y) = y π
∞ −∞ f (ξ) (ξ − x) 2 + y 2 dξ. From this solution, we can readily deduce a solution of the Neumann problem in the half-plane y > 0. Example 12.3.2. Find the solution of Neumann’s problem in the half-plane y > 0 uxx + uyy = 0, −∞ <x< ∞,="" y=""> 0, uy (x, 0) = g (x), −∞ <x< ∞,="" u="" is="" bounded="" as="" y="" →="" and="" ux="" vanish="" |x|→∞.="" 448="" 12="" integral="" transform="" methods="" with="" applications="" let="" v="" (x,="" y)="uy" y).="" then="" η)="" dη="" the="" neumann="" problem="" becomes="" ∂="" 2v="" ∂x2="" +="" ∂y2="∂" 2uy="" ∂y="" (uxx="" uyy)="0." 0)="uy" (x).="" this="" dirichlet="" for="" y),="" its="" solution="" given="" by="" π=""
="" ∞="" −∞="" g="" (ξ)="" dξ="" (ξ="" −="" x)="" 2="" .="" thus,="" we="" have="" η="" dη,="1" 2π="" 2η="" ,="1" log="" "="" (x="" ξ)="" #="" dξ,="" where="" an="" arbitrary="" constant="" can="" be="" added="" to="" solution.="" in="" other="" words,="" of="" any="" neumann’s="" uniquely="" determined="" up="" constant.="" 12.4="" convolution="" theorem="" fourier="" function="" (f="" ∗="" g)="" (x)="1" √="" f="" (12.4.1)="" called="" functions="" over="" interval="" (−∞,∞).="" 12.4.1.="" (convolution="" theorem).="" if="" (k)="" are="" transforms="" respectively,="" product="" (k).="" that="" is,="" {f="" (x)}="F" (12.4.2)="" or,="" equivalently,="" 449="" −1="" (k)}="f" (12.4.3)="" more="" explicitly,="" 1="" e="" ikxdk="(f" dξ.="" (12.4.4)="" proof.="" definition,="" [(f="" (x)]="1" −ikxdx="" −ikξdξ="" −ik(x−ξ)="" dx.="" change="" variable="" ξ,="" (η)="" −ikηdη="F" satisfies="" following="" properties:="" 1.="" (commutative).="" 2.="" (g="" h)="(f" h="" (associative).="" 3.="" (ag="" bh)="a" b="" h),="" (distributive),="" a="" constants.="" 12.4.2.="" (parseval’s="" formula).="" |f="" (x)|="" dx="
" (k)|="" dk.="" (12.4.5)="" formula="" gives="" ikξdk="" which="" putting="" ξ="0," (−x)="" (12.4.6)="" (x),="" −ikxdx,="1" e−ikxdx="F" (k),="" 450="" bar="" denotes="" complex="" conjugate.="" result="" dk,="" terms="" notation="" norm,="" f="F" physical="" systems,="" quantity="" |f|="" measure="" energy,="" represents="" power="" spectrum="" example="" 12.4.3.="" obtain="" initial-value="" heat="" conduction="" infinite="" rod="" ut="κ" uxx,="" <x<="" t=""> 0, (12.4.7) u (x, 0) = f (x), −∞ <x< ∞,="" (12.4.8)="" u="" (x,="" t)="" →="" 0,="" as="" |x|→∞,="" where="" represents="" the="" temperature="" distribution="" and="" is="" bounded,="" κ="" a="" constant="" of="" diffusivity.="" fourier="" transform="" with="" respect="" to="" x="" defined="" by="" (k,="" √="" 2π=""
="" ∞="" −∞="" e="" −ikxu="" dx.="" in="" view="" this="" transformation,="" equations="" (12.4.7)–(12.4.8)="" become="" ut="" +="" k2="" (12.4.9)="" 0)="F" (k).="" (12.4.10)="" solution="" transformed="" system="" (k)="" −k="" 2κ="" t="" .="" inverse="" transformation="" gives="" f="" 2κte="" ikxdk="" which="" is,="" convolution="" theorem="" 12.4.1,="" 12.4="" 451="1" (ξ)="" g="" (x="" −="" ξ)="" dξ,="" (x)="" 2κt="" has="" form="" −1="" (="" −κk2="" )="1" 2κt+ikxdk="1" −x="" 2="" 4κt="" consequently,="" final="" 4πκt="" exp="" 1="" 3="" (12.4.11)="
" ξ,="" (12.4.12)="" ,="" (12.4.13)="" called="" green’s="" function="" (or="" fundamental="" solution)="" diffusion="" equation.="" means="" that="" at="" any="" point="" time="" represented="" definite="" integral="" made="" up="" contribution="" due="" initial="" source="" t).="" response="" along="" rod="" an="" unit="" impulse="" heat="" physical="" meaning="" decomposed="" into="" spectrum="" impulses="" magnitude="" each="" resulting="" thus,="" integrated="" find="" (12.4.11).="" so-called="" principle="" superposition.="" using="" change="" variable="" ξ="" κt="ζ," dζ="dξ" we="" obtain="" π=""
4="" 2√="" ζ
5="" −ζ="" dζ.="" (12.4.14)="" or="" poisson="" representation="" distribution.="" convergent="" for=""> 0, and integrals obtained from it by differentiation under the integral sign with respect to x and t are uniformly convergent in the neighborhood of 452 12 Integral Transform Methods with Applications the point (x, t). Hence, u (x, t) and its derivatives of all orders exist for t > 0. In the limit t → 0+, solution (12.4.12) becomes formally u (x, 0) = f (x) =
∞ −∞ f (ξ) limt→0+ G (x − ξ, t) dξ. This limit represents the Dirac delta function δ (x − ξ) = limt→0+ 1 √ 4πκt e −(x−ξ) 2/4κt . (12.4.15) Consider a special case where f (x) = ⎧ ⎨ ⎩ 0, x< 0 a, x > 0 ⎫ ⎬ ⎭ = a H (x). Then, the solution (12.4.11) gives u (x, t) = a 2 √ πκt
∞ 0 exp 1 − (x − ξ) 2 4κt 3 dξ. If we introduce a change of variable η = ξ − x 2 √ κt then the above solution becomes u (x, t) = a √ π
∞ −x/2 √ κt e −η 2 dη = a √ π 1
0 −x/2 √ κt e −η 2 dη +
∞ 0 e −η 2 dη3 = a √ π 1
x/2 √ κt 0 e −η 2 dη + √ π 2 3 = a 2 1 + erf  x 2 √ κt , where erf (x) is called the error function and is defined by erf (x) = 2 √ π
x 0 e −η 2 dη. (12.4.16) This is a widely used and tabulated function. 12.5 The Fourier Transforms of Step and Impulse Functions 453 Figure 12.5.1 The Heaviside unit step function. 12.5 The Fourier Transforms of Step and Impulse Functions In this section, we shall determine the Fourier transforms of the step function and the impulse function, functions which occur frequently in applied mathematics and mathematical physics. The Heaviside unit step function is defined by H (x − a) = ⎧ ⎨ ⎩ 0, x</x<></x<></x<></x<></x<></x<></r<></x<></y<b}></x<></x<></a,></r</a,></r<></r<r1,> 0, and ε is a small positive constant, as shown in Figure 12.5.2. This type of function appears in practical applications; for instance, a force of large magnitude may act over a very short period of time. The Fourier transform of the impulse function is 12.5 The Fourier Transforms of Step and Impulse Functions 455 F [p (x)] = 1 √ 2π
∞ −∞ p (x) e −ikxdx = 1 √ 2π
a+ε a−ε h e−ikxdx = h √ 2π e −iak ik  e ikε − e −ikε = 2hε √ 2π e −iak  sin kε kε  . Now if we choose the value of h to be (1/2ε), then the impulse defined by I (ε) =
∞ −∞ p (x) dx becomes I (ε) =
a+ε a−ε 1 2ε dx = 1 which is a constant independent of ε. In the limit as ε → 0, this particular function pε (x) with h = (1/2ε) satisfies limε→0 pε (x)=0, x = a, limε→0 I (ε)=1. Thus, we arrive at the result δ (x − a)=0, x = a,
∞ −∞ δ (x − a) dx = 1. (12.5.4) This is the Dirac delta function which was defined earlier in Section 8.11. We now define the Fourier transform of δ (x) as the limit of the transform of pε (x). We then consider F [δ (x − a)] = limε→0 F [pε (x)] = limε→0 e −iak √ 2π  sin kε kε  = e −iak √ 2π (12.5.5) in which we note that, by L’Hospital’s rule, limε→0 (sin kε/kε) = 1. When a = 0, we obtain F [δ (x)] =
4 1/ √ 2π
5 . (12.5.6) 456 12 Integral Transform Methods with Applications Example 12.5.1. Slowing-down of Neutrons (see Sneddon (1951), p. 215). Consider the following physical problem ut = uxx + δ (x) δ (t), (12.5.7) u (x, 0) = δ (x), (12.5.8) lim |x|→∞ u (x, t)=0. (12.5.9) This is the problem of an infinite medium which slows neutrons, in which a source of neutrons is located. Here u (x, t) represents the number of neutrons per unit volume per unit time and δ (x) δ (t) represents the source function. Let U (k, t) be the Fourier transform of u (x, t). Then the Fourier transformation of equation (12.5.7) yields dU dt + k 2U = 1 √ 2π δ (t). The solution of this, after applying the condition U (k, 0) =  1/ √ 2π , is U (k, t) = 1 √ 2π e −k 2 t . Hence, the inverse Fourier transform gives the solution of the problem u (x, t) = 1 √ 2π
∞ −∞ e −k 2 t+ikxdk, = 1 √ 4πt e −x 2/4t . 12.6 Fourier Sine and Cosine Transforms For semi-infinite regions, the Fourier sine and cosine transforms determined in Section 12.2 are particularly appropriate in solving boundary-value problems. Before we illustrate their applications, we must first prove the differentiation theorem. Theorem 12.6.1. Let f (x) and its first derivative vanish as x → ∞. If Fc (k) is the Fourier cosine transform, then Fc [f ′′ (x)] = −k 2Fc (k) − 2 2 π f ′ (0). (12.6.1) Proof. Fc [f ′′ (x)] = 2 2 π
∞ 0 f ′′ (x) cos kx dx 12.6 Fourier Sine and Cosine Transforms 457 = 2 2 π [f ′ (x) cos kx] ∞ 0 + 2 2 π k
∞ 0 f ′ (x) sin kx dx = − 2 2 π f ′ (0) + 2 2 π k [f (x) sin kx] ∞ 0 − 2 2 π k 2
∞ 0 f (x) cos kx dx = − 2 2 π f ′ (0) − k 2Fc (k). In a similar manner, the Fourier cosine transforms of higher-order derivatives of f (x) can be obtained. Theorem 12.6.2. Let f (x) and its first derivative vanish as x → ∞. If Fs (k) is the Fourier sine transform, then Fs [f ′′ (x)] = 2 2 π k f (0) − k 2Fs (k). (12.6.2) The proof is left to the reader. Example 12.6.2. Find the temperature distribution in a semi-infinite rod for the following cases with zero initial temperature distribution: (a) The heat supplied at the end x = 0 at the rate g (t); (b) The end x = 0 is kept at a constant temperature T0. The problem here is to solve the heat conduction equation ut = κ uxx, x > 0, t> 0, u (x, 0) = 0, x > 0. (a) ux (0, t) = g (t) and (b) u (0, t) = T0, t ≥ 0. Here we assume that u (x, t) and ux (x, t) vanish as x → ∞. For case (a), let U (k, t) be the Fourier cosine transform of u (x, t). Then the transformation of the heat conduction equation yields Ut + κ k2 U = − 2 2 π g (t) κ. The solution of this equation with U (k, 0) = 0 is u (x, t) = 2 2 π
∞ 0 U (k, t) cos kx dk = − 2κ π
t 0 g (τ ) dτ
∞ 0 e −κk2 (t−τ) cos kx dk. 458 12 Integral Transform Methods with Applications The inner integral is given by (see Problem 6, Exercises 12.18)
∞ 0 e −k 2κ(t−τ) cos kx dk = 1 2 2 π κ (t − τ ) exp − x 2 4κ (t − τ ) . The solution, therefore, is u (x, t) = − 2 κ π
t 0 g (τ ) √ t − τ e −x 2/4κ(t−τ) dτ. (12.6.3) For case (b), we apply the Fourier sine transform U (k, t) of u (x, t) to obtain the transformed equation Ut + κ k2U = 2 2 π k T0 κ. The solution of this equation with zero initial condition is U (k, t) = T0 2 2 π
4 1 − e −κtk2
5 k . Then the inverse Fourier sine transformation gives u (x, t) = 2T0 π
∞ 0 sin kx k
4 1 − e −κtk2
5 dk. Making use of the integral
∞ 0 e −a 2x 2  sin kx k  dk = π 2 erf
4 x 2a
5 , the solution is found to be u (x, t) = 2T0 π π 2 − π 2 erf  x 2 √ κt = T0 erfc  x 2 √ κt , (12.6.4) where erfc (x)=1 − erf (x) is the complementary error function defined by erfc (x) = 2 √ π
∞ x e −α 2 dα. 12.7 Asymptotic Approximation of Integrals by Stationary Phase Method Although definite integrals represent exact solutions for many physical problems, the physical meaning of the solutions is often difficult to determine. In many cases the exact evaluation of the integrals is a formidable task. It is then necessary to resort to asymptotic methods. 12.7 Asymptotic Approximation of Integrals by Stationary Phase Method 459 We consider the typical integral solution u (x, t) =
b a F (k) e itθ(k) dk, (12.7.1) where F (k) is called the spectral function determined by the initial or boundary data in a<k<b, and="" θ="" (k),="" known="" as="" the="" phase="" function,="" is="" given="" by="" (k)="" ≡="" k="" x="" t="" −="" ω=""> 0. (12.7.2) We examine the asymptotic behavior of (12.7.1) for both large x and large t; one of the interesting limits is t → ∞ with (x/t) held fixed. Integral (12.7.1) can be evaluated by the Kelvin stationary phase method for large t. As t → ∞, the integrand of (12.7.1) oscillates very rapidly; consequently, the contributions to u (x, t) from adjacent parts of the integrand cancel one another except in the neighborhood of the points, if any, at which the phase function θ (k) is stationary, that is, θ ′ (k) = 0. Thus, the main contribution to the integral for large t comes from the neighborhood of the point k = k1 which determined by the solution of θ ′ (k1) = x t − ω ′ (k1)=0, a < k1 < b. (12.7.3) The point k = k1 known as the point of stationary phase, or simply, stationary point. We expand both F (k) and θ (k) in Taylor series about k = k1 so that u (x, t) =
b a F (k1)+(k − k1) F ′ (k1) + 1 2 (k − k1) 2 F ′′ (k1) + ... × exp  it θ (k1) + 1 2 (k − k1) 2 θ ′′ (k1) + 1 6 (k − k1) 3 θ ′′′ (k1) + ...0 dk (12.7.4) provided that θ ′′ (k1) = 0. Introducing the change of variable k − k1 = εα, where ε (t) =  2 t|θ ′′ (k1)| 01 2 , (12.7.5) we find that the significant contribution to integral (12.7.4) is u (x, t) ∼ ε
(b−k1)/ε −(k1−a)/ε F (k1) + εαF′ (k1) + 1 2 ε 2α 2F ′′ (k1) + ... × exp  i t θ (k1) + α 2 sgn θ ′′ (k1) + 1 3 ε  θ ′′′ (k1) |θ ′′ (k1)|  α 3 + ...0dα, (12.7.6) 460 12 Integral Transform Methods with Applications where sgn x denotes the signum function defined by sgn x = 1, x > 0 and sgn x = −1, x < 0. We then proceed to the limit as ε → 0 (t → ∞) and use the standard integral
∞ −∞ exp  ±iα2 dα = √ π exp  ± iπ 4  (12.7.7) to obtain the asymptotic approximation as t → ∞, u (x, t) ∼ F (k1) 2π t|θ ′′ (k1)| 1 2 exp ( i " t θ (k1) + π 4 sgn θ ′′ (k1) #) + O  ε 2 , (12.7.8) where O  ε 2 means that a function tends to zero like ε 2 (t) as t → ∞. If there is more than one stationary point, each one contributes a term similar to (12.7.8) and we obtain, for n stationary points k = kr, r = 1, 2,...n; u(x, t) ∼ n r=1 F(kr)  2π t|θ ′′ (kr)| 01 2 exp ( i " t θ (kr) + π 4 sgn θ ′′(kr) #), t → ∞. (12.7.9) If θ ′′ (k1) = 0, but θ ′′′ (k1) = 0, then asymptotic approximation (12.7.8) fails. This important special case can be handled in a similar fashion. The asymptotic approximation of (12.7.1) is then given by u (x, t) = F (k1) exp {itθ (k1)}
∞ −∞ exp i 6 t θ′′′ (k1) (k − k1) 3 dk ∼ Γ  4 3  6 t|θ ′′′ (k1)| 1 3 F (k1) exp itθ (k1) + πi 6 + O
4 t − 2 3
5 as t → ∞. (12.7.10) For an elaborate treatment of the stationary phase method, see Copson (1965). 12.8 Laplace Transforms Because of their simplicity, Laplace transforms are frequently used to solve a wide class of partial differential equations. Like other transforms, Laplace transforms are used to determine particular solutions. In solving partial differential equations, the general solutions are difficult, if not impossible, to obtain. The transform technique sometimes offers a useful tool for finding particular solutions. The Laplace transform is closely related to the complex Fourier transform, so the Fourier integral formula (6.13.10) can be used to define 12.8 Laplace Transforms 461 the Laplace transform and its inverse. We replace f (x) in (6.13.10) by H (x) e −cxf (x) for x > 0 to obtain f (x) H (x) e −cx = 1 2π
∞ −∞ e ikxdk
∞ 0 f (t) e −t(c+ik) dt or f (x) H (x) = 1 2π
∞ −∞ e x(c+ik) dk
∞ 0 f (t) e −t(c+ik) dt. Substituting s = c + ik so that ds = idk, we obtain, for x > 0, f (x) H (x) = 1 2πi
c+i∞ c−i∞ e xsds
∞ 0 f (t) e −stdt. (12.8.1) Thus, we give the following definition of the Laplace transform: If f (t) is defined for all values of t > 0, then the Laplace transform of f (t) is denoted by ¯f (s) or L {f (t)} and is defined by the integral ¯f (s) = L {f (t)} =
∞ 0 e −stf (t) dt, (12.8.2) where s is a positive real number or a complex number with a positive real part so that the integral is convergent. Hence, (12.8.1) gives f (x) = L −1 & ¯f (s) ' = 1 2πi
c+i∞ c−i∞ e xs ¯f (s) ds, c > 0, (12.8.3) for x > 0 and zero for x < 0. This complex integral is used to define the inverse Laplace transform which is denoted by L −1 & ¯f (s) ' = f (t). It can be verified easily that both L and L −1 are linear integral operators. We now find the Laplace transforms of some elementary functions. 1. Let f (t) = c, c is a constant. L[c] =
∞ 0 e −stc dt = − ce−st s ∞ 0 = c s . 2. Let f (t) = e at , a is a constant. L e at! =
∞ 0 e −ste atdt = − e −(s−a)t (s − a) ∞ 0 = 1 s − a , s ≥ a. 3. Let f (t) = t 2 . Then 462 12 Integral Transform Methods with Applications L t 2 ! =
∞ 0 e −st t 2 dt. Integration by parts yields L & t 2 ' = − t 2 e −st s ∞ 0 +
∞ 0 e −st s 2 t dt. Since t 2 e −st → 0 as t → ∞, we have, integrating by parts again, L t 2 ! = 2 s − e −st s t + 2 s
∞ 0 e −st s dt = 2 s 3 . 4. Let f (t) = sin ωt. ¯f (s) = L[sin ωt] =
∞ 0 e −st sin ωt dt = − e −st s sin ωt∞ 0 +
∞ 0 e −st s ω cos ωt dt = ω s − e −st s cos ωt − ω s
∞ 0 e −st s ω sin ωt dt ¯f (s) = ω s 2 − ω 2 s 2 ¯f (s). Thus, solving for ¯f (s), we obtain L[sin ωt] = ω/  s 2 + ω 2 . A function f (t) is said to be of exponential order as t → ∞ if there exist real constants M and a such that |f (t)| ≤ Meat for 0 ≤ t < ∞. Theorem 12.8.1. Let f be piecewise continuous in the interval [0, T] for every positive T, and let f be of exponential order, that is, f (t) = O (e at) as t → ∞ for some a > 0. Then, the Laplace transform of f (t) exists for Re s>a. Proof. Since f is piecewise continuous and of exponential order, we have |L(f (t))| =    
∞ 0 e −stf (t) dt     ≤
∞ 0 e −st |f (t)| dt ≤
∞ 0 e −stMeatdt = M
∞ 0 e −(s−a)t dt = M/ (s − a), Re s > a. Thus,
∞ 0 e −stf (t) dt exists for Re s>a. 12.9 Properties of Laplace Transforms 463 12.9 Properties of Laplace Transforms Theorem 12.9.1. (Linearity) If L[f (t)]and L[g (t)] are Laplace transforms of f (t) and g (t) respectively, then L[af (t) + bg (t)] = aL[f (t)] + bL[g (t)] where a and b are constants. Proof. L[af (t) + bg (t)] =
∞ 0 [af (t) + bg (t)] e −stdt = a
∞ 0 f (t) e −stdt + b
∞ 0 g (t) e −stdt = aL[f (t)] + bL[g (t)] . This shows that L is a linear operator. Theorem 12.9.2. (Shifting) If ¯f (s) is the Laplace transform of f (t), then the Laplace transform of e atf (t) is ¯f (s − a). Proof. By definition, we have L e atf (t) ! =
∞ 0 e −ste atf (t) dt =
∞ 0 e −(s−a)t f (t) dt = ¯f (s − a). Example 12.9.1. (a) If L t 2 ! = 2/s3 , then L[t s e t ]=2/ (s − 1)3 . (b) If L[sin ωt] = ω/  s 2 + ω 2 , then L[e at sin ωt] = ω/ " (s − 1)2 + ω 2 # . (c) If L {cos ωt} = s s 2+ω2 , then L {e at cos ωt} = s−a (s−a) 2+ω2 . (d) If L {t n} = n! sn+1 , then L {e att n} = n! (s−a) n+1 . Theorem 12.9.3. (Scaling) If the Laplace transform of f (t) is ¯f (s), then the Laplace transform of f (ct) with c > 0 is (1/c) ¯f (s/c). Proof. By definition, we have L[f (ct)] =
∞ 0 e −stf (ct) dt =
∞ 0 1 c e −(sξ/c) f (ξ) dξ (substituting ξ = ct) = (1/c) ¯f (s/c). 464 12 Integral Transform Methods with Applications Example 12.9.2. (a) If s s 2+1 = L[cost], then 1 ω s/ω (s/ω) 2 + 1 = s s 2 + ω2 = L[cos ωt] . (b) If 1 s−1 = L[e t ], then 1 a 1  s a − 1 = L e at! , or L e at! = 1 s − a . Theorem 12.9.4. (Differentiation) Let f be continuous and f ′ piecewise continuous, in 0 ≤ t ≤ T for all T > 0. Let f also be of exponential order as t → ∞. Then, the Laplace transform of f ′ (t) exists and is given by L[f ′ (t)] = sL[f (t)] − f (0) = s ¯f (s) − f (0). Proof. Consider the definite integral
T 0 e −stf ′ (t) dt = e −stf (t) !T 0 +
T 0 s e−stf (t) dt = e −sT f (T) − f (0) + s
T 0 e −stf (t) dt. Since |f (t)| ≤ Meat for large t, with a > 0 and M > 0,  e −sT f (T)   ≤ Me−(s−a)T . In the limit as T → ∞, e −sT f (T) → 0 whenever s>a. Hence, L[f ′ (t)] = sL[f (t)] − f (0) = s ¯f (s) − f (0). If f ′ and f ′′ satisfy the same conditions imposed on f and f ′ respectively, then, the Laplace transform of f ′′ (t) can be obtained immediately by applying the preceding theorem; that is L[f ′′ (t)] = sL[f ′ (t)] − f ′ (0) = s {sL[f (t)] − f (0)} − f ′ (0) = s 2L[f (t)] − sf (0) − f ′ (0) = s 2 ¯f (s) − sf (0) − f ′ (0). Clearly, the Laplace transform of f (n) (t) can be obtained in a similar manner by successive application of Theorem 12.9.4. The result may be written as L " f (n) (t) # = s nL[f (t)] − s n−1 f (0) − ... − s f(n−2) (0) − f (n−1) (0). 12.9 Properties of Laplace Transforms 465 Theorem 12.9.5. (Integration) If ¯f (s) is the Laplace transform of f (t), then L
t 0 f (τ ) dτ = ¯f (s) /s. Proof. L
t 0 f (τ ) dτ =
∞ 0
t 0 f (τ ) dτ e −stdt = − e −st s
t 0 f (τ ) dτ∞ 0 + 1 s
∞ 0 f (t) e −stdt = ¯f (s) /s since
t 0 f (τ ) dτ is of exponential order. In solving problems by the Laplace transform method, the difficulty arises in finding inverse transforms. Although the inversion formula exists, its evaluation requires a knowledge of functions of complex variables. However, for some problems of mathematical physics, we need not use this inversion formula. We can avoid its use by expanding a given transform by the method of partial fractions in terms of simple fractions in the transform variables. With these simple functions, we refer to the table of Laplace transforms given in the end of the book and obtain the inverse transforms. Here, we should note that we use the assumption that there is essentially a one-to-one correspondence between functions and their Laplace transforms. This may be stated as follows: Theorem 12.9.6. (Lerch) Let f and g be piecewise continuous functions of exponential order. If there exists a constant s0, such that L[f] = L[g] for all s>s0, then f (t) = g (t) for all t > 0 except possibly at the points of discontinuity. For a proof, the reader is referred to Kreider et al. (1966). In order to find a solution of linear partial differential equations, the following formulas and results are useful. If L {u (x, t)} = u (x, s), then L  ∂u ∂t 0 = s u (x, s) − u (x, 0), L  ∂ 2u ∂t2 0 = s 2 u (x, s) − s u (x, 0) − ut (x, 0), and so on. Similarly, it is easy to show that 466 12 Integral Transform Methods with Applications L  ∂u ∂x0 = du dx, L  ∂ 2u ∂x2 0 = d 2u dx2 ,...,L  ∂ nu ∂xn 0 = d nu dxn . The following results are useful for applications: L  erfc  a 2 √ t 0 = 1 s exp  −a √ s , a ≥ 0, (12.9.1) L ( exp (at) erf
4√ at
5) = √ a √ s (s − a) , a > 0. (12.9.2) Example 12.9.3. Consider the motion of a semi-infinite string with an external force f (t) acting on it. One end is kept fixed while the other end is allowed to move freely in the vertical direction. If the string is initially at rest, the motion of the string is governed by utt = c 2uxx + f (t), 0 <x< ∞,="" t=""> 0, u (x, 0) = 0, ut (x, 0) = 0, u (0, t)=0, ux (x, t) → 0, as x → ∞. Let u (x, s) be the Laplace transform of u (x, t). Transforming the equation of motion and using the initial conditions, we obtain uxx −  s 2 /c2 u = −f (s) /c2 . The solution of this ordinary differential equation is u (x, s) = Aesx/c + Be−sx/c + f (s) /s2 ! . The transformed boundary conditions are given by u (0, s)=0, and limx→∞ ux (x, s)=0. In view of the second condition, we have A = 0. Now applying the first condition, we obtain u (0, s) = B + f (s) /s2 ! = 0. Hence u (x, s) = f (s) /s2 ! " 1 − e −sx/c# . (a) When f (t) = f0, a constant, then u (x, s) = f0  1 s 3 − 1 s 3 e −sx/c . The inverse Laplace transform gives the solution 12.10 Convolution Theorem of the Laplace Transform 467 u (x, t) = f0 2 t 2 −
4 t − x c
52 when t ≥ x/c, = (f0/2)t 2 when t ≤ x/c. (b) When f (t) = cos ωt, where ω is a constant, then ¯f (s) =
∞ 0 e −st cos ωt dt = s/  ω 2 + s 2 . Thus, we have u (x, s) = 1 s (ω2 + s 2)
4 1 − e −sx/c
5 . (12.9.3) By the method of partial fractions, we write 1 s (s 2 + ω2) = 1 ω2 1 s − 1 s 2 + ω2 . Hence L −1 1 s (s 2 + ω2) = 1 ω2 (1 − cos ωt) = 2 ω2 sin2  ωt 2  . If we denote ψ (t) = sin2  ωt 2  , then the Laplace inverse of equation (12.9.3) may be written in the form u (x, t) = 2 ω2 " ψ (t) − ψ
4 t − x c
5# when t ≥ x/c, = 2 ω2 ψ (t) when t ≤ x/c. 12.10 Convolution Theorem of the Laplace Transform The function (f ∗ g) (t) =
t 0 f (t − ξ) g (ξ) dξ (12.10.1) is called the convolution of the functions f and g. Theorem 12.10.1. (Convolution) If ¯f (s) and g¯ (s) are the Laplace transforms of f (t) and g (t) respectively, then the Laplace transform of the convolution (f ∗ g) (t) is the product ¯f (s) ¯g (s). 468 12 Integral Transform Methods with Applications Figure 12.10.1 Region of integration. Proof. By definition, we have L[(f ∗ g) (t)] =
∞ 0 e −st
t 0 f (t − ξ) g (ξ) dξ dt =
∞ 0
t 0 e −stf (t − ξ) g (ξ) dξ dt. The region of integration is shown in Figure 12.10.1. By reversing the order of integration, we have L[(f ∗ g) (t)] =
∞ 0
∞ ξ e −stf (t − ξ) g (ξ) dt dξ =
∞ 0 g (ξ)
t ξ e −stf (t − ξ) dt dξ. If we introduce the new variable η = (t − ξ) in the inner integral, we obtain L[(f ∗ g) (t)] =
∞ 0 g (ξ)
∞ 0 e −s(ξ+η) f (η) dη dξ =
∞ 0 g (ξ) e −sξdξ
∞ 0 e −sηf (η) dη = ¯f (s) ¯g (s). (12.10.2) The convolution satisfies the following properties: 12.10 Convolution Theorem of the Laplace Transform 469 1. f ∗ g = g ∗ f (commutative). 2. f ∗ (g ∗ h)=(f ∗ g) ∗ h (associative). 3. f ∗ (αg + βh) = α (f ∗ g) + β (f ∗ h), (distributive), where α and β are constants. Example 12.10.1. Find the temperature distribution in a semi-infinite radiating rod. The temperature is kept constant at x = 0, while the other end is kept at zero temperature. If the initial temperature distribution is zero, the problem is governed by ut = kuxx − hu, 0 <x< ∞,="" t=""> 0, h = constant, u (x, 0) = 0, u (0, t) = u0, t> 0, u0 = constant, u (x, t) → 0, as x → ∞. Let u (x, s) be the Laplace transform of u (x, t). Then the transformation with respect to t yields uxx −  s + h k  u = 0, u (0, s) = u0/s, limx→∞ u (x, s)=0. The solution of this equation is u (x, s) = A ex √ (s+h)/k + B e−x √ (s+h)/k . The boundary condition at infinity requires that A = 0. Applying the other boundary condition gives u (0, s) = B = u0/s. Hence, the solution takes the form u (x, s)=(u0/s) exp " −x  (s + h) /k # . We find (by using the Table of Laplace Transforms) that L −1 " u0 s # = u0, and L −1 " exp ( −x  (s + h) /k )# = x exp −ht −  x 2/4kt ! 2 √ πkt3 . Thus, the inverse Laplace transform of u (x, s) is u (x, t) = L −1 " u0 s exp ( −x  (s + h) /k )# . 470 12 Integral Transform Methods with Applications By the Integration Theorem 12.9.5, we have u (x, t) =
t 0 u0 x exp −hτ −  x 2/4kτ ! 2 √ πk τ 3 2 dτ. Substituting the new variable η =
4 x/2 √ kτ
5 yields u (x, t) = 2 u0 √ π
∞ x/2 √ kt exp −η 2 +  hx2 /4kη2 ! dη. For the case h = 0, the solution u (x, t) becomes u (x, t) = 2 u0 √ π
∞ x/2 √ kt e −η 2 dη = 2 u0 √ π
∞ 0 e −η 2 dη − 2 u0 √ π
x/2 √ kt 0 e −η 2 dη = u0 1 − erf  x 2 √ kt = u0 erfc  x 2 √ kt . 12.11 Laplace Transforms of the Heaviside and Dirac Delta Functions We have defined the Heaviside unit step function. Now, we will find its Laplace transform. L[H (t − a)] =
∞ 0 e −stH (t − a) dt =
∞ a e −stdt =  1 s  e −as, s> 0. (12.11.1) Theorem 12.11.1. (Second Shifting) If ¯f (s) and g¯ (s) are the Laplace transforms of f (t) and g (t) respectively, then (a) L[H (t − a) f (t − a)] = e −as ¯f (s) = e −asL {f (t)} . (b) L {H (t − a) g (t)} = e −asL {g (t + a)} . Proof. (a) By definition L[H (t − a) f (t − a)] =
∞ 0 e −stH (t − a) f (t − a) dt =
∞ a e −stf (t − a) dt. Introducing the new variable ξ = t − a, we obtain 12.11 Laplace Transforms of the Heaviside and Dirac Delta Functions 471 L[H (t − a) f (t − a)] =
∞ 0 e −(ξ+a)s f (ξ) dξ = e −as
∞ 0 e −ξsf (ξ) dξ = e −as ¯f (s). To prove (b), we write L {H (t − a) g (t)} =
∞ a e −stg (t) dt (t − a = τ ) =
∞ 0 e −s(a+τ) g (a + τ ) dτ = e −saL {g (t + a)} . Example 12.11.1. (a) Given that f (t) = ⎧ ⎨ ⎩ 0, t < 2 t − 2, t ≥ 2 ⎫ ⎬ ⎭ = (t − 2) H (t − 2), find the Laplace transform of f (t). We have L[f (t)] = L[H (t − 2) (t − 2)] = e −2sL[t] =  1 s 2  e −2s . (b) Find the inverse Laplace transform of f (s) = 1 + e −2s s 2 . L −1 f (s) ! = L −1  1 s 2 + e −2s s 2  = L −1 1 s 2 + L −1 e −2s s 2 = t + H (t − 2) (t − 2) = ⎧ ⎨ ⎩ t, 0 ≤ t < 2, 2 (t − 1), t ≥ 2. The Laplace transform of the impulse function p (t) is given by L[p (t)] =
∞ 0 e −stp (t) dt =
a+ε a−ε h e−stdt = h − e −st s a+ε a−ε = h e−as s  e εs − e −εs = 2 h e−as s sinh (εs). (12.11.2) 472 12 Integral Transform Methods with Applications If we choose the value of h to be (1/2ε), then the impulse is given by I (ε) =
∞ −∞ p (t) dt =
a+ε a−ε 1 2ε dt = 1. Thus, in the limit as ε → 0, this particular impulse function satisfies limε→0 pε (t)=0, t = a, limε→0 I (ε)=1. From this result, we obtain the Dirac delta function which satisfies δ (t − a)=0, t = a,
∞ −∞ δ (t − a) dt = 1. (12.11.3) Thus, we may define the Laplace transform of δ (t) as the limit of the transform of pε (t). L[δ (t − a)] = limε→0 L[pε (t)] , = limε→0 e −as sinh (εs) εs (12.11.4) = e −as . If a = 0, we have L[δ (t)] = 1. (12.11.5) One very useful result that can be derived is the integral of the product of the delta function and any continuous function f (t).
∞ −∞ δ (t − a) f (t) dt = limε→0
∞ −∞ pε (t) f (t) dt, = limε→0
a+ε a−ε f (t) 2ε dt, = limε→0 1 2ε · 2ε f (t ∗ ), a − ε<t∗ <="" a="" +="" ε="f" (a).="" (12.11.6)="" suppose="" that="" f="" (t)="" is="" periodic="" function="" with="" period="" t.="" let="" be="" piecewise="" continuous="" on="" [0,="" t].="" then,="" the="" laplace="" transform="" of="" l[f="" (t)]="
" ∞="" 0="" e="" −stf="" dt,="∞" n="0"
="" (n+1)t="" nt="" dt.="" 12.11="" transforms="" heaviside="" and="" dirac="" delta="" functions="" 473="" if="" we="" introduce="" new="" variable="" ξ="t" −="" nt,="" then="" −nt="" s="" t="" −sξf="" (ξ)="" dξ,="∞" 1="" (s),="" where="" (s)="*" dξ="" over="" first="" period.="" since="" series="" geometric="" series,="" obtain="" for="" (1="" e−t="" s)="" .="" (12.11.7)="" example="" 12.11.2.="" find="" square="" wave="" 2c="" given="" by="" ⎨="" ⎩="" h,="" <t<c="" −h,="" c="" (t="" 2c)="f" (t),="" as="" shown="" in="" figure="" 12.11.1.="" 12.11.1="" function.="" 474="" 12="" integral="" methods="" applications="" −sξh="" −sξ="" (−h)="" ="" −cs="" 2="" thus,="" is,="" (12.11.7),="" e−2cs="h" −cs)="" e−2cs)="h" e−cs)="h" tanh=""
4cs=""
5="" 12.11.3.="" uniform="" bar="" length="" l="" fixed="" at="" one="" end.="" force="" f0,=""> 0 0, t< 0 be suddenly applied at the end x = l. If the bar is initially at rest, find the longitudinal displacement for t > 0. The motion of the bar is governed by the differential system utt = a 2uxx, 0 < x < l, t > 0, a = constant, u (x, 0) = 0, ut (x, 0) = 0, u (0, t)=0, ux (l, t)=(f0/E), where E is a constant and t > 0. Let u (x, s) be the Laplace transform of u (x, t). Then, u (x, s) satisfies the system uxx − s 2 a 2 u = 0, u (0, s)=0, ux (l, s)=(f0/Es). The solution of this differential equation is u (x, s) = Aexs/a + Be−xs/a . Applying the boundary conditions, we have A + B = 0,
4 s a e ls/a
5 A +
4 − s a e −ls/a
5 B = f0/Es. Solving for A and B, we obtain 12.11 Laplace Transforms of the Heaviside and Dirac Delta Functions 475 A = −B = af0 Es2  e ls/a + e−ls/a . Hence, the transform of the displacement function is given by u (x, s) = af0  e xs/a − e −xs/a Es2  e ls/a + e−ls/a . Before finding the inverse transform of u (x, s), multiply the numerator and denominator by  e −ls/a − e −3ls/a . Thus, we have u (x, s) =  af0 Es2  " e −(l−x)s/a − e −(l+x)s/a − e −(3l−x)s/a + e −(3l+x)s/a# × 1  1 − e−4ls/a . Since the denominator has the term  1 − e −4ls/a , the inverse transform u (x, t) is periodic with period (4l/a). Hence, the final solution may be written in the form u (x, t) = ⎧ ⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨ ⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩ 0, 0 <t< l−x="" a="" ,="" af0="" e="" ="" t="" −="" <t<="" l+x="" t="" !="" 3l−x="" 3l+x="" +="" 0,="" 4l="" which="" may="" be="" simplified="" to="" obtain="" u="" (x,="" t)="⎧" ⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨="" ⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩="" 0="" (l="" x)="" a,="" <="" 2x="" (3l="" −t="" a.="" this="" result="" can="" clearly="" seen="" in="" figure="" 12.11.2.="" 476="" 12="" integral="" transform="" methods="" with="" applications="" example="" 12.11.4.="" consider="" semi-infinite="" string="" fixed="" at="" the="" end="" x="0." is="" initially="" rest.="" let="" there="" an="" external="" force="" f="" δ=""
4="" v=""
5="" acting="" on="" string.="" concentrated="" f0="" point="" motion="" of="" governed="" by="" initial="" boundary-value="" problem="" utt="c" 2uxx="" 0)="0," ut="" (0,="" bounded="" as="" →="" ∞.="" s)="" laplace="" t).="" transforming="" wave="" equation="" and="" using="" conditions,="" we="" uxx="" s="" 2="" c="" exp="" (−xs="" v).="" solution="" aesx="" be−sx="" c
5="" ⎧="" ⎪⎨="" ⎪⎩="" f0v="" −sx="" (c="" 2−v="" 2)s="" for="" f0xe−sx="" 2cs="" 12.11.2="" graph="" 12.11="" transforms="" heaviside="" dirac="" delta="" functions="" 477="" condition="" that="" must="" infinity="" requires="" application="" yields="" b="⎧" ⎨="" ⎩="" −f0v="" hence,="" given="" (e="" −xs="" v−e="" c)="" f0xe−xs="" inverse="" therefore="" 2)="" f0x="" 2c="" 12.11.5.="" (the="" stokes="" rayleigh="" fluid="" dynamics).="" solve="" concerned="" unsteady="" boundary="" layer="" flows="" induced="" viscous="" infinite="" horizontal="" disk="" z="0" due="" oscillations="" its="" own="" plane="" frequency="" ω.="" velocity="" (z,="" uzz,=""> 0, t> 0, with the boundary and initial conditions u (z, t) = U0e iωt, z = 0, t> 0, u (z, t) → 0, as z → ∞, t> 0, u (z, 0) = 0, at t ≤ 0 for all z > 0, where u (z, t) is the velocity of fluid of kinematic viscosity ν and U0 is a constant. The Laplace transform solution of the equation with the transformed boundary conditions is u (z, s) = U0 (s − iω) exp  −z 2 s ν  . 478 12 Integral Transform Methods with Applications Using a standard table of inverse Laplace transforms, we obtain the solution u (z, t) = U0 2 e iωt " exp (−λz) erfc
4 ζ − √ iωt
5 + exp (λz) erfc
4 ζ + √ iωt
5# , where ζ =  z/2 √ νt is called the similarity variable of the viscous boundary layer theory, and λ = (iω/ν) 1 2 . This result describes the unsteady boundary layer flow. In view of the asymptotic formula for the complementary error function erfc
4 ζ + √ iωt
5 ∼ (2, 0) as t → ∞, the above solution for u (z, t) has the asymptotic representation u (z, t) ∼ U0 exp (iωt − λz) = U0 exp iωt −
4 ω 2ν
5 1 2 (1 + i) z . (12.11.8) This is called the Stokes steady-state solution. This represents the propagation of shear waves which spread out from the oscillating disk with velocity ω/k = √ 2νω
4 k = (ω/2ν) 1 2
5 and exponentially decaying amplitude. The boundary layer associated with the solution has thickness of the order (ν/ω) 1 2 in which the shear oscillations imposed by the disk decay exponentially with distance z from the disk. This boundary layer is called the Stokes layer. In other words, the thickness of the Stokes layer is equal to the depth of penetration of vorticity which is essentially confined to the immediate vicinity of the disk for high frequency ω. The Stokes problem with ω = 0 becomes the Rayleigh problem. In other words, the motion is generated in the fluid from rest by moving the disk impulsively in its own plane with constant velocity U0. In this case, the Laplace transform solution is u (z, s) = U0 s exp  −z 2 s ν  , so that the inversion gives the Rayleigh solution u (z, t) = U0 erfc  z 2 √ νt . (12.11.9) This describes the growth of a boundary layer adjacent to the disk. The associated boundary layer is called the Rayleigh layer of thickness of the order δ ∼ √ νt which grows with increasing time t. The rate of growth is of the order dδ/dt ∼  ν/t, which diminishes with increasing time. 12.11 Laplace Transforms of the Heaviside and Dirac Delta Functions 479 The vorticity of the unsteady flow is given by ∂u ∂z = U0 √ πνt exp  −ζ 2 (12.11.10) which decays exponentially to zero as z ≫ δ. Note that the vorticity is everywhere zero at t = 0 except at z = 0. This implies that it is generated at the disk and diffuses outward within the Rayleigh layer. The total viscous diffusion time is Td ∼ δ 2/ν. Another physical quantity related to the Stokes and Rayleigh problems is the skin friction on the disk defined by τ0 = µ  ∂u ∂z  z=0 , (12.11.11) where µ = νρ is the dynamic viscosity and ρ is the density of the fluid. The skin friction can readily be calculated from the flow field given by (12.11.8) or (12.11.9). Example 12.11.6. (The Nonhomogeneous Cauchy Problem for the Wave Equation). We consider the nonhomogeneous Cauchy problem utt − c 2uxx = q (x, t), x ∈ R, t > 0, (12.11.12) u (x, 0) = f (x), ut (x, 0) = g (x) for all x ∈ R, (12.11.13) where q (x, t) is a given function representing a source term. We use the joint Laplace and Fourier transform of u (x, t) U (k, s) = L[F {u (x, t)}] = 1 √ 2π
∞ −∞ e −ikxdx
∞ 0 e −stu (x, t) dt. (12.11.14) Application of the joint transform leads to the solution of the transformed Cauchy problem in the form U (k, s) = s F (k) + G (k) + Q (k, s) (s 2 + c 2k 2) . (12.11.15) The inverse Laplace transform of (12.11.15) gives U (k, t) = F (k) cos (ckt) + 1 ckG (k) sin (ckt) + 1 ckL −1  ck s 2 + c 2k 2 · Q (k, s) 0 = F (k) cos (ckt) + G (k) ck sin (ckt) + 1 ck
t 0 sin ck (t − τ ) Q (k, τ ) dτ. (12.11.16) The inverse Fourier transform leads to the exact integral solution 480 12 Integral Transform Methods with Applications u (x, t) = 1 2 √ 2π
∞ −∞  e ickt + e −ickt e ikxF (k) dk + 1 2 √ 2π
∞ −∞  e ickt − e −ickt e ikx · G (k) ick dk + 1 √ 2π · 1 2c
t 0 dτ
∞ −∞ Q (k, τ ) ik " e ick(t−τ) − e −ick(t−τ) # e ikxdk = 1 2 [f (x + ct) + f (x − ct)] + 1 2c
x+ct x−ct g (ξ) dξ + 1 2c
t 0 dτ 1 √ 2π
∞ −∞ Q (k, τ ) dk
x+c(t−τ) x−c(t−τ) e ikξdξ = 1 2 [f (x − ct) + f (x + ct)] + 1 2c
x+ct x−ct g (ξ) dξ + 1 2c
t 0 dτ
x+c(t−τ) x−c(t−τ) q (ξ, τ ) dξ. (12.11.17) In the case of the homogeneous Cauchy problem, q (x, t) ≡ 0, the solution of (12.11.17) reduces to the famous d’Alembert solution (5.3.8). Example 12.11.7. (The Heat Conduction Equation in a Semi-Infinite Medium and Fractional Derivatives). Solve the one-dimensional diffusion equation ut = κ uxx, x > 0, t> 0, (12.11.18) with the initial and boundary conditions u (x, 0) = 0, x > 0, (12.11.19) u (0, t) = f (t), t> 0, (12.11.20) u (x, t) → 0, as x → ∞, t> 0. (12.11.21) Application of the Laplace transform with respect to t to (12.11.18) gives d 2u dx2 − s κ u = 0. (12.11.22) The general solution of this equation is u (x, s) = A exp  −x 2 s κ  + B exp  x 2 s κ  , where A and B are integrating constants. For bounded solutions, B ≡ 0, and using u (0, s) = f (s), we obtain the solution u (x, s) = f (s) exp  −x 2 s κ  . (12.11.23) 12.11 Laplace Transforms of the Heaviside and Dirac Delta Functions 481 The Laplace inversion theorem gives the solution u (x, t) = x 2 √ πκ
t 0 f (t − τ ) τ − 3 2 exp  − x 2 4κτ  dτ, (12.11.24) which, by setting λ = x 2 √ κτ , or dλ = − x 4 √ κ τ − 3 2 dτ , = 2 √ π
∞ x 2 √κt f  t − x 2 4κλ2  e −λ 2 dλ. (12.11.25) This is the formal solution of the heat conduction problem. In particular, if f (t) = T0 = constant, solution (12.11.25) becomes u (x, t) = 2T0 √ π
∞ x 2 √κt e −λ 2 dλ = T0 erfc  x 2 √ κt . (12.11.26) Clearly, the temperature distribution tends asymptotically to the constant value T0, as t → ∞. Alternatively, solution (12.11.23) can be written as u (x, s) = f (s) s u0 (x, s), (12.11.27) where s u0 (x, s) = exp  −x 2 s κ  . (12.11.28) Consequently, the inversion of (12.11.27) gives a new representation u (x, t) =
t 0 f (t − τ )  ∂u0 ∂τ  dτ. (12.11.29) This is called the Duhamel formula for the diffusion equation. We consider another physical problem: determining the temperature distribution of a semi-infinite solid when the rate of flow of heat is prescribed at the end x = 0. Thus, the problem is to solve diffusion equation (12.11.18) subject to conditions (12.11.19), (12.11.21), and −k  ∂u ∂x = g (t) at x = 0, t> 0, (12.11.30) where k is a constant called thermal conductivity. Application of the Laplace transform gives the solution of the transformed problem u (x, s) = 1 k 2 κ s g (s) exp  −x 2 s κ  . (12.11.31) 482 12 Integral Transform Methods with Applications The inverse Laplace transform yields the solution u (x, t) = 1 k 2 κ π
t 0 g (t − τ ) τ − 1 2 exp  − x 2 4κτ  dτ, (12.11.32) which is, by the change of variable λ = x 2 √ κτ , = x k √ π
∞ x 2 √κt g  t − x 2 4κλ2  λ −2 e −λ 2 dλ. (12.11.33) In particular, if g (t) = T0 = constant, this solution becomes u (x, t) = T0 x k √ π
∞ √x 4κt λ −2 e −λ 2 dλ. Integrating this result by parts gives u (x, t) = T0 k 1 2 2 κt π exp  − x 2 4κt − x erfc  x 2 √ κt3 . (12.11.34) Alternatively, the heat conduction problem (12.11.18)–(12.11.21) can be solved by using fractional derivatives (Debnath 1995). We recall (12.11.23) and rewrite it as ∂u ∂x = − 2 s κ u. (12.11.35) This can be expressed in terms of a fractional derivative of order 1 2 as ∂u ∂x = − 1 √ κ L −1 &√ s u (x, s) ' = − 1 √ κ 0D 1 2 t u (x, t). (12.11.36) Thus, the heat flux is expressed in terms of the fractional derivative. In particular, when u (0, t) = constant = T0, then the heat flux at the surface is given by −k  ∂u ∂x x=0 = k √ κ D 1 2 t T0 = kT0 √ πκt . (12.11.37) Example 12.11.8. (Diffusion Equation in a Finite Medium). Solve the diffusion equation ut = κ uxx, 0 < x < a, t > 0, (12.11.38) with the initial and boundary conditions u (x, 0) = 0, 0 < x < a, (12.11.39) u (0, t) = U, t > 0, (12.11.40) ux (a, t)=0, t> 0, (12.11.41) 12.11 Laplace Transforms of the Heaviside and Dirac Delta Functions 483 where U is a constant. We introduce the Laplace transform of u (x, t) with respect to t to obtain d 2u dx2 − s κ u = 0, 0 < x < a, (12.11.42) u (0, s) = U s ,  du dx x=a = 0. (12.11.43) The general solution of (12.11.42) is u (x, s) = A cosh  x 2 s κ  + B sinh  x 2 s κ  , (12.11.44) where A and B are constants of integration. Using (12.11.43), we obtain the values of A and B, so that the solution (12.11.44) becomes u (x, s) = U s · cosh (a − x) s κ ! cosh  a s κ . (12.11.45) The inverse Laplace transform gives the solution u (x, t) = UL −1 / cosh (a − x) s κ s cosh  a s κ 0 . (12.11.46) The inversion can be carried out by the Cauchy residue theorem to obtain the solution u (x, t) = U 1 1 + 4 π ∞ n=1 (−1)n (2n − 1) cos  (2n − 1) (a − x) π 2a 0 × exp  − (2n − 1)2
4 π 2a
52 κt0 . (12.11.47) By expanding the cosine term, this becomes u (x, t) = U 1 1 − 4 π ∞ n=1 1 (2n − 1) sin 2n − 1 2a  πx0 × exp  − (2n − 1)2
4 π 2a
52 κt0 . (12.11.48) This result can be obtained by solving the problem by the method of separation of variables. Example 12.11.9. (Diffusion in a Finite Medium). Solve the one-dimensional diffusion equation in a finite medium 0 <z<a, where="" the="" concentration="" function="" c="" (z,="" t)="" satisfies="" equation="" 484="" 12="" integral="" transform="" methods="" with="" applications="" ct="κ" czz,="" 0="" <="" z="" a,="" t=""> 0, (12.11.49) and the initial and boundary data C (z, 0) = 0 for 0 < z < a, (12.11.50) C (z, t) = C0 for z = a, t > 0, (12.11.51) ∂C ∂z = 0 for z = 0, t> 0, (12.11.52) where C0 is a constant. Application of the Laplace transform of C (z, t) with respect to t gives d 2C dz2 −
4 s κ
5 C = 0, 0 < z < a, C (a, s) = C0 s ,  dC dz  z=0 = 0. The solution of this differential equation system is C (z, s) = C0 cosh  z s κ s cosh  a s κ , (12.11.53) which, by writing α = s κ , = C0 s (e αz + e −αz) (e αa + e−αa) = C0 s [exp {−α (a − z)} + exp {−α (a + z)}] ∞ n=0 (−1)n exp (−2nαa) = C0 s 1∞ n=0 (−1)n exp [−α {(2n + 1) a − z}] + ∞ n=0 (−1)n exp [−α {(2n + 1) a + z}] 3 .(12.11.54) Using the result (12.9.1), we obtain the final solution C (z, t) = C0 /∞ n=0 (−1)n erfc  (2n + 1) a − z 2 √ κt 0 + erfc  (2n + 1) a + z 2 √ κt 00 . (12.11.55) This solution represents an infinite series of complementary error functions. The successive terms of this series are, in fact, the concentrations at depth a − z, a + z, 3a − z, 3a + z, ... in the medium. The series converges rapidly for all except large values of  κt a2 . 12.11 Laplace Transforms of the Heaviside and Dirac Delta Functions 485 Example 12.11.10. (The Wave Equation for the Transverse Vibration of a Semi-Infinite String). Find the displacement of a semi-infinite string, which is initially at rest in its equilibrium position. At time t = 0, the end x = 0 is constrained to move so that the displacement is u (0, t) = A f (t) for t ≥ 0, where A is a constant. The problem is to solve the one-dimensional wave equation utt = c 2uxx, 0 ≤ x < ∞, t> 0, (12.11.56) with the boundary and initial conditions u (x, t) = A f (t) at x = 0, t ≥ 0, (12.11.57) u (x, t) → 0 as x → ∞, t ≥ 0, (12.11.58) u (x, t) = 0= ∂u ∂t at t = 0 for 0 <x< ∞.="" (12.11.59)="" application="" of="" the="" laplace="" transform="" u="" (x,="" t)="" with="" respect="" to="" t="" gives="" d="" 2u="" dx2="" −="" s="" 2="" c="" for="" 0="" ≤="" x="" <="" ∞,="" s)="A" f="" (s)="" at="" →="" as="" solution="" this="" differential="" equation="" system="" is="" exp=""
4="" xs=""
5="" .="" (12.11.60)="" inversion="" h="" (12.11.61)="" in="" other="" words,="" ⎨="" ⎩="" a="" ="" ,=""> x c 0, t < x c . (12.11.62) This solution represents a wave propagating at a velocity c with the characteristic x = ct. Example 12.11.11. (The Cauchy–Poisson Wave Problem in Fluid Dynamics). We consider the two-dimensional Cauchy–Poisson problem (Debnath 1994) for an inviscid liquid of infinite depth with a horizontal free surface. We assume that the liquid has constant density ρ and negligible surface tension. Waves are generated on the free surface of liquid initially at rest for time t < 0 by the prescribed free surface displacement at t = 0. 486 12 Integral Transform Methods with Applications In terms of the velocity potential φ (x, z, t) and the free surface elevation η (x, t), the linearized surface wave motion in Cartesian coordinates (x, y, z) is governed by the following equation and free surface and boundary conditions: ∇2φ = φxx + φzz = 0, −∞ < z ≤ 0, x ∈ R, t < 0, (12.11.63) φz − ηt = 0 φt + gη = 00 on z = 0, t> 0, (12.11.64) φz → 0 as z → −∞. (12.11.65) The initial conditions are φ (x, 0, 0) = 0 and η (x, 0) = η0 (x), (12.11.66) where η0 (x) is a given initial elevation with compact support. We introduce the Laplace transform with respect to t and the Fourier transform with respect to x defined by " ˜ φ (k, z, s), η˜ (k, s) # = 1 √ 2π
∞ −∞ e −ikxdx
∞ 0 e −st [φ, η] dt. (12.11.67) Application of the joint transform method to the above system gives ˜ φzz − k 2 ˜ φ = 0, −∞ < z ≤ 0, (12.11.68) ˜ φ = s η˜ − η˜0 (k) s ˜ φ + gη˜ = 0 ⎫ ⎪⎬ ⎪⎭ on z = 0, (12.11.69) ˜ φz → 0 as z → −∞, (12.11.70) where η˜0 (k) = F {η0 (x)} . The bounded solution of equation (12.11.68) is ˜ φ (k, s) = A exp (|k| z), (12.11.71) where A = A (s) is an arbitrary function of s. Substituting (12.11.71) into (12.11.69) and eliminating η˜ from the resulting equations gives A. Hence, the solutions for ˜ φ and η˜ are " ˜ φ, η˜ # = − g η˜0 exp (|k| z) s 2 + ω2 , s η˜0 s 2 + ω2 , (12.11.72) and the associated the dispersion relation is 12.11 Laplace Transforms of the Heaviside and Dirac Delta Functions 487 ω 2 = g |k| . (12.11.73) The inverse Laplace and Fourier transforms give the solutions φ (x, z, t) = − g √ 2π
∞ −∞ sin ωt ω exp (ikx + |k| z) ˜η0 (k) dk, (12.11.74) η (x, t) = 1 √ 2π
∞ −∞ η˜0 (k) cos ωt eikxdk, = 1 √ 2π
∞ 0 η˜0 (k) " e i(kx−ωt) + e i(kx+ωt) # dk, (12.11.75) in which ˜η0 (−k)=˜η0 (k) is assumed. Physically, the first and second integrals of (12.11.75) represent waves traveling in the positive and negative directions of x, respectively, with phase velocity ω k . These integrals describe superposition of all such waves over the wavenumber spectrum 0 <k< ∞.="" for="" the="" classical="" cauchy–poisson="" wave="" problem,="" η0="" (x)="aδ" (x),="" where="" δ="" is="" dirac="" delta="" function,="" so="" that="" ˜η0="" (k)="" a="" √="" 2π="" .="" thus,="" solution="" (12.11.75)="" becomes="" η="" (x,="" t)="a"
="" ∞="" 0="" "="" e="" i(kx−ωt)="" +="" i(kx+ωt)="" #="" dk.="" (12.11.76)="" integrals="" (12.11.74)="" and="" represent="" exact="" velocity="" potential="" φ="" free="" surface="" elevation="" all="" x="" t=""> 0. However, they do not lend any physical interpretations. In general, the exact evaluation of these integrals is a formidable task. So it is necessary to resort to asymptotic methods. It would be sufficient for the determination of the principal features of the wave motions to investigate (12.11.75) or (12.11.76) asymptotically for large time t and large distance x with (x, t) held fixed. The asymptotic solution for this kind of problem is available in many standard books; (for example, see Debnath 1994, p. 85). We use the stationary phase approximation of a typical wave integral (12.7.1), for t → ∞, given by (12.7.8) η(x, t) =
b a F(k) exp[itθ(k)]dk (12.11.77) ∼ f(k1) 2π t|θ ′′(k1)| 1 2 exp " i ( tθ(k1) + π 4 sgn θ ′′(k1) )#, (12.11.78) where θ (k) = kx t − ω (k), x > 0, and k = k1 is a stationary point that satisfies the equation θ ′ (k1) = x t − ω ′ (k1)=0, a < k1 < b. (12.11.79) 488 12 Integral Transform Methods with Applications Application of (12.11.78) to (12.11.75) shows that only the first integral in (12.11.75) has a stationary point for x > 0. Hence, the stationary phase approximation (12.11.78) gives the asymptotic solution, as t → ∞, x > 0, η(x, t) ∼ 1 t|ω′′(k1)| 1 2 η˜0(k1) exp i{k1x − tω(k1)} + iπ 4 sgn {−ω ′′(k1)} , (12.11.80) where k1 =  gt2/4x 2 is the root of the equation ω ′ (k) = x t . On the other hand, when x < 0, only the second integral of (12.11.75) has a stationary point k1 =  gt2/4x 2 , and hence, the same result (12.11.78) can be used to obtain the asymptotic solution for t → ∞ and x < 0 as η(x, t) ∼ 1 t|ω′′(k1)| 1 2 η˜0(k1) exp i {tω(k1) − k1 |x|} + iπ 4 sgn ω ′′(k1) . (12.11.81) In particular, for the classical Cauchy–Poisson solution (12.11.76), the asymptotic representation for η (x, t) follows from (12.11.81) in the form η (x, t) ∼ at 2 √ 2π √g x 3/2 cos  gt2 4x  , gt 2 ≫ 4x (12.11.82) and gives a similar result for η (x, t), when x < 0 and t → ∞. 12.12 Hankel Transforms We introduce the definition of the Hankel transform from the two-dimensional Fourier transform and its inverse given by F {f (x, y)} = F (k,l) = 1 2π
∞ −∞
∞ −∞ exp {−i(κ · r)} f (x, y) dx dy, (12.12.1) F −1 {F (k,l)} = f (x, y) = 1 2π
∞ −∞
∞ −∞ exp {i(κ · r)} F (k,l) dk dl, (12.12.2) where r = (x, y) and κ = (k,l). Introducing polar coordinates (x, y) = r (cos θ,sin θ) and (k,l) = κ (cos φ,sin φ), we find κ · r = κr cos (θ − φ) and then F (κ, φ) = 1 2π
∞ 0 rdr
2π 0 exp [−iκr cos (θ − φ)] f (r, θ) dθ. (12.12.3) 12.12 Hankel Transforms 489 We next assume f (r, θ) = exp (inθ) f (r), which is not a very severe restriction, and make a change of variable θ−φ = α− π 2 to reduce (12.12.3) to the form F (κ, φ) = 1 2π
∞ 0 rf (r) dr ×
2π+φ0 φ0 exp " in
4 φ − π 2
5 + i(nα − κr sin α) # dα, (12.12.4) where φ0 = π 2 − φ. We use the integral representation of the Bessel function of order n Jn (κr) = 1 2π
2π+φ0 φ0 exp [i(nα − κr sin α)] dα (12.12.5) so that integral (12.12.4) becomes F (κ, φ) = exp " in
4 φ − π 2
5#
∞ 0 rJn (κr) f (r) dr (12.12.6) = exp " in
4 φ − π 2
5# ˜fn (κ), (12.12.7) where ˜fn (κ) is called the Hankel transform of f (r) and is defined formally by Hn {f (r)} = ˜fn (κ) =
∞ 0 rJn (κr) f (r) dr. (12.12.8) Similarly, in terms of the polar variables with the assumption f (x, y) = f (r, θ) = e inθf (r) and with result (12.12.7), the inverse Fourier transform (12.12.2) becomes e inθf (r) = 1 2π
∞ 0 κ dκ
2π 0 exp [iκr cos (θ − φ)] F (κ, φ) dφ = 1 2π
∞ 0 κ ˜fn (κ) dκ
2π 0 exp " in
4 φ − π 2
5 + iκr cos (θ − φ) # dφ, which is, by the change of variables θ −φ = −  α + π 2 and θ0 = −  θ + π 2 , = 1 2π
∞ 0 κ ˜fn (κ) dκ
2π+θ0 θ0 exp [in (θ + α) − iκr sin α] dα = e inθ
∞ 0 κJn (κr) ˜fn (κ) dκ, by (12.12.5). (12.12.9) Thus, the inverse Hankel transform is defined by H−1 n " ˜fn (κ) # = f (r) =
∞ 0 κ Jn (κr) ˜fn (κ) dκ. (12.12.10) 490 12 Integral Transform Methods with Applications Instead of ˜fn (κ), we often simply write ˜f (κ) for the Hankel transform specifying the order. Integrals (12.12.8) and (12.12.10) exist for certain large classes of functions, which usually occur in physical applications. Alternatively, the famous Hankel integral formula (Watson, 1966, p 453) f (r) =
∞ 0 κJn (κr) dκ
∞ 0 p Jn (κp) f (p) dp, (12.12.11) can be used to define the Hankel transform (12.12.8) and its inverse (12.12.10). In particular, the Hankel transforms of zero order (n = 0) and of order one (n = 1) are often useful for the solution of problems involving Laplace’s equation in an axisymmetric cylindrical geometry. Example 12.12.1. Obtain the zero-order Hankel transforms of (a) r −1 exp (−ar), (b) δ(r) r , (c) H (a − r), where H (r) is the Heaviside unit step function. (a) f50 (κ) = H0 &1 r exp (−ar) ' =
∞ 0 exp (−ar) J0 (κr) dr = √ 1 κ2+a2 . (b) f50 (κ) = H0 ( δ(r) r ) =
∞ 0 δ (r) J0 (κr) dr = 1. (c) f50 (κ) = H0 {H (a − r)} =
a 0 rJ0 (κr) dr = 1 κ2
aκ 0 pJ0 (p) dp = 1 κ2 [pJ1 (p)]aκ 0 = a κ J1 (aκ). Example 12.12.2. Find the first-order Hankel transform of the following functions: (a) f (r) = e −ar , (b) f (r) = 1 r e −ar . (a) ˜f (κ) = H1 {e −ar} =
∞ 0 re−arJ1 (κr) dr = κ (a2+κ2) 3 2 . (b) ˜f (κ) = H1 ( e −ar r ) =
∞ 0 e −arJ1 (κr) dr = 1 κ " 1 − a  κ 2 + a 2 − 1 2 # . Example 12.12.3. Find the nth-order Hankel transforms of (a) f (r) = r nH (a − r), (b) f (r) = r n exp  −ar2 . 12.13 Properties of Hankel Transforms and Applications 491 (a) ˜f (κ) = Hn [r nH (a − r)] =
a 0 r n+1Jn (κr) dr = a n+1 κ Jn+1 (aκ). (b) ˜f (κ) = Hn r n exp  −ar2 ! =
∞ 0 r n+1Jn (κr) exp  −ar2 dr = κ n (2a) n+1 exp
4 − κ 2 4a
5 . 12.13 Properties of Hankel Transforms and Applications We state the following properties of the Hankel transforms: (i) The Hankel transform operator, Hn is a linear integral operator, that is, Hn {af (r) + bg (r)} = aHn {f (r)} + bHn {g (r)} for any constants a and b. (ii) The Hankel transform satisfies the Parseval relation
∞ 0 rf (r) g (r) dr =
∞ 0 k ˜f (k) ˜g (k) dk (12.13.1) where ˜f (k) and ˜g (k) are Hankel transforms of f (r) and g (r) respectively. To prove (12.13.1), we proceed formally to obtain
∞ 0 k ˜f (k) ˜g (k) dk =
∞ 0 k ˜f (k) dk
∞ 0 rJn (kr) g (r) dr =
∞ 0 rg (r) dr
∞ 0 kJn (kr) ˜f (k) dk =
∞ 0 rf (r) g (r) dr. (iii) Hn {f ′ (r)} = k 2n " (n − 1) ˜fn+1 (k) − (n + 1) ˜fn−1 (k) # provided rf (r) vanishes as r → 0 and as r → ∞. (iv) Hn  1 r d dr  r df dr  − n 2 r 2 f (r) 0 = −k 2 ˜fn (k) (12.13.2) provided both
4 r df dr
5 and rf (r) vanish as r → 0 and as r → ∞. 492 12 Integral Transform Methods with Applications We have, by definition, Hn  1 r d dr  r df dr  − n 2 r 2 f (r) 0 =
∞ 0 d dr  r df dr  Jn (kr) dr −
∞ 0 n 2 r 2 rf (r) Jn (kr) dr = r df dr Jn (kr) ∞ 0 −
∞ 0 kJ′ n (kr) r df dr dr −
∞ 0 n 2 r 2 [rf (r)] Jn (kr) dr, by partial integration = − [f (r) krJ′ n (kr)]∞ 0 +
∞ 0 d dr [k rJ′ n (kr)] f (r) dr −
∞ 0 n 2 r 2 rf (r) Jn (kr) dr, by partial integration which is, by the given assumption and Bessel’s differential equation (8.6.1), = −
∞ 0  k 2 − n 2 r 2  rf (r) Jn (kr) dr −
∞ 0 n 2 r 2 [rf (r)] Jn (kr) dr = −k 2
∞ 0 rf (r) Jn (kr) dr = −k 2Hn {f (r)} = −k 2 ˜fn (k). (v) (Scaling). If Hn {f (r)} = ˜fn (κ), then Hn {f (ar)} = 1 a 2 ˜fn
4κ a
5 , a > 0. (12.13.3) Proof. We have, by definition, Hn {f (ar)} =
∞ 0 rJn (κr) f (ar) dr = 1 a 2
∞ 0 s Jn
4κ a s
5 f (s) ds = 1 a 2 ˜fn
4κ a
5 . These results are used very widely in solving partial differential equations in the axisymmetric cylindrical configurations. We illustrate this point by considering the following examples of applications. Example 12.13.1. Obtain the solution of the free vibration of a large circular membrane governed by the initial-value problem ∂ 2u ∂r2 + 1 r ∂u ∂r = 1 c 2 ∂ 2u ∂t2 , 0 <r< ∞,="" t=""> 0, (12.13.4) u (r, 0) = f (r), ut (r, 0) = g (r), 0 ≤ r < ∞, (12.13.5) 12.13 Properties of Hankel Transforms and Applications 493 where c 2 = (T /ρ) = constant, T is the tension in the membrane, and ρ is the surface density of the membrane. Application of the Hankel transform of order zero u˜ (k, t) =
∞ 0 r u (r, t) J0 (kr) dr to the vibration problem gives d 2u˜ dt2 + k 2 c 2u˜ = 0 u˜ (k, 0) = ˜f (k), u˜t (k, 0) = ˜g (k). The general solution of this transformed system is u˜ (k, t) = ˜f (k) cos (ckt) + g˜ (k) ck sin (ckt). The inverse Hankel transformation gives u (r, t) =
∞ 0 k ˜f (k) cos (ckt) J0 (kr) dk + 1 c
∞ 0 g˜ (k) sin (ckt) J0 (kr) dr. (12.13.6) This is the desired solution. In particular, we consider the following initial conditions u (r, 0) = f (r) = A  1 + r 2 a2 1 2 , ut (r, 0) = g (r)=0 so that ˜g (k) = 0 and ˜f (k) = Aa
∞ 0 rJ0 (kr) dr (a 2 + r 2) 1 2 = Aa k e −ak by means of Example 12.12.1(a). Thus, solution (12.13.6) becomes u (r, t) = Aa
∞ 0 e −akJ0 (kr) cos (ckt) dk = Aa Re
∞ 0 e −k(a+ict)J0 (kr) dk = Aa Re ( r 2 + (a + ict) 2 )− 1 2 . (12.13.7) 494 12 Integral Transform Methods with Applications Example 12.13.2. Obtain the steady-state solution of the axisymmetric acoustic radiation problem governed by the wave equation in cylindrical polar coordinates (r, θ, z): c 2∇2u = utt, 0 <r< ∞,="" z=""> 0, t > 0 (12.13.8) uz = f (r, t) on z = 0, (12.13.9) where f (r, t) is a given function and c is a constant. We also assume that the solution is bounded and behaves as outgoing spherical waves. This is referred to as the Sommerfeld radiation condition. We seek a solution of the acoustic radiation potential u = e iωtφ (r, z) so that φ satisfies the Helmholtz equation φrr + 1 r φr + φzz + ω 2 c 2 φ = 0, 0 <r< ∞,="" z=""> 0 (12.13.10) with the boundary condition representing the normal velocity prescribed on the z = 0 plane φz = f (r) on z = 0, (12.13.11) where f (r) is a known function of r. We solve the problem by means of the zero-order Hankel transformation φ˜ (k, z) =
∞ 0 rJ0 (kr) φ (r, z) dr so that the given differential system becomes φ˜ zz = κ 2φ, z > ˜ 0, φ˜ z = ˜f (k) on z = 0 where κ = k 2 −  ω 2/c2 ! 1 2 . The solution of this system is φ˜ (k, z) = −κ −1 ˜f (k) e −κz , (12.13.12) where κ is real and positive for k > ω/c, and purely imaginary for k < ω/c. The inverse transformation yields the solution φ (r, z) = −
∞ 0 κ −1 ˜f (k) kJ0 (kr) e −κzdk. (12.13.13) Since the exact evaluation of this integral is difficult, we choose a simple form of f (r) as f (r) = A H (a − r), (12.13.14) 12.14 Mellin Transforms and their Operational Properties 495 where A is a constant and H (x) is the Heaviside unit step function so that ˜f (k) =
a 0 kJ0 (ak) dk = a k J1 (ak). Then the solution for this special case is given by φ (r, z) = −Aa
∞ 0 κ −1J1 (ak) J0 (kr) e −κzdk. (12.13.15) For an asymptotic evaluation of this integral, we express it in terms of the spherical polar coordinates (R, θ, φ), (x = R sin θ cos φ, y = R sin θ sin φ, z = R cos θ), combined with the asymptotic result J0 (kr) ∼  2 πkr1 2 cos
4 kr − π 4
5 as r → ∞ so that the acoustic potential u = e iωtφ is u ∼ − Aa√ 2 e iωt √ πR sin θ
∞ 0 J1 (ka) cos
4 kR sin θ − π 4
5 e −kzdk, where z = R cos θ. This integral can be evaluated asymptotically for R → ∞ by using the stationary phase approximation formula (12.7.8) to obtain u ∼ − Aac ωR sin θ J1 (k1a) e i(ωt−ωR/c) , (12.13.16) where k1 = ω/c sin θ is the stationary point. This solution represents the outgoing spherical waves with constant velocity c and decaying amplitude as R → ∞. 12.14 Mellin Transforms and their Operational Properties If f (t) is not necessarily zero for t < 0, it is possible to define the two-sided (or bilateral) Laplace transform f (p) =
∞ −∞ e −ptf (t) dt. (12.14.1) Then replacing f (x) with e −cxf (x) in Fourier integral formula (6.13.9), we obtain e −cxf (x) = 1 2π
∞ −∞ e −ikxdk
∞ −∞ f (t) e −t(c−ik) dt, 496 12 Integral Transform Methods with Applications or f (x) = 1 2π
∞ −∞ e x(c−ik) dk
∞ −∞ f (t) e −t(c−ik) dt. Making a change of variable p = c − ik and using definition (12.14.1), we obtain the formal inverse transform after replacing x by t as f (t) = 1 2πi
c+i∞ c−i∞ e pt f (p) dp, c > 0. (12.14.2) If we put e −t = x into (12.14.1) with f (− log x) = g (x) and f (p) ≡ G (p), then (12.14.1)–(12.14.2) become G (p) = M {g (x)} =
∞ 0 x p−1 g (x) dx, (12.14.3) g (x) = M−1 {G (p)} = 1 2πi
c+i∞ c−i∞ x −pG (p) dp. (12.14.4) The function G (p) is called the Mellin transform of g (x) defined by (12.14.3). The inverse Mellin transformation is given by (12.14.4). We state the following operational properties of the Mellin transforms: (i) Both M and M−1 are linear integral operators, (ii) M[f (ax)] = a −pF (p), (iii) M[x af (x)] = F (p + a), (iv) M[f ′ (x)] = − (p − 1) F (p − 1), provided that f (x) x p−1 !∞ 0 = 0, M[f ′′ (x)] = (p − 1) (p − 2) F (p − 2), ··· ··· ··· ··· ··· ··· , M f (n) (x) ! = (−1)nΓ(p) Γ(p−n) F (p − n), provided limx→0 x p−r−1f (r) (x) = 0, r = 0, 1, 2, ..., (n − 1), (v) M {xf′ (x)} = −pM {f (x)} = −pF (p), provided that [x pf (x)]∞ 0 = 0, M & x 2f ′′ (x) ' = (−1)2  p + p 2 F (p), ··· ··· ··· ··· ··· ··· , M & x nf (n) (x) ' = (−1)n Γ(p+n) Γ(p) F (p). (vi) M ( x d dx n f (x) ) = (−1)n p nF (p), n = 1, 2, .... 12.14 Mellin Transforms and their Operational Properties 497 (vii) Convolution Property M
∞ 0 f (xξ) g (ξ) dξ = F (p) G (1 − p), M
∞ 0 f
4 x ξ
5 g (ξ) dξ ξ = F (p) G (p). (viii) If F (p) = M(f (x)) and G (p) = M(g (x)), then, the following convolution result holds: M[f (x) g (x)] = 1 2πi
c+i∞ c−i∞ F (s) G (p − s) ds. In particular, when p = 1, we obtain the Parseval formula
∞ 0 f (x) g (x) dx = 1 2πi
c+i∞ c−i∞ F (s) G (1 − s) ds. The reader is referred to Debnath (1995) for other properties of the Mellin transform. Example 12.14.1. Show that the Mellin transform of (1 + x) −1 is π cosec πp, 0 < Re p < 1. We consider the standard definite integral
1 0 (1 − t) m−1 t p−1 dt = Γ (m) Γ (p) Γ (m + p) , Re p > 0, Re m > 0, and then change the variable t = x 1+x to obtain
∞ 0 x p−1dx (1 + x) m+p = Γ (m) Γ (p) Γ (m + p) . If we replace m + p by α, this gives M " (1 + x) −α # = Γ (p) Γ (α − p) Γ (α) . Setting α = 1 and using the result Γ (p) Γ (1 − p) = π cosec πp, 0 < Re p < 1, we obtain M " (1 + x) −1 # = π cosec πp, 0 < Re p < 1. 498 12 Integral Transform Methods with Applications Example 12.14.2. Obtain the solution of the boundary-value problem x 2uxx + xux + uyy = 0, 0 ≤ x < ∞, 0 <y< 1,="" u="" (x,="" 0)="0," and="" 1)="⎧" ⎨="" ⎩="" a,="" 0="" ≤="" x="" 1="" 0,=""> 1 , where A is constant. We apply the Mellin transform U (p, y) =
∞ 0 x p−1u (x, y) dx to reduce the system to the form Uyy + p 2U = 0, 0 <y< 1,="" u="" (p,="" 0)="0," and="" 1)="A"
="" 1="" 0="" x="" p−1="" dx="A" p="" .="" the="" solution="" of="" this="" differential="" system="" is="" y)="A" sin="" (py)="" ,="" <="" re="" 1.="" inverse="" mellin="" transform="" gives="" (x,="" 2πi="" c+i∞="" c−i∞="" −p="" dp,="" where="" analytic="" in="" a="" vertical="" strip="" p<π="" hence,="" <c<π.="" integrand="" has="" simple="" poles="" at="" r="1," 2,="" 3,="" ...="" which="" lie="" inside="" semi-circular="" contour="" right="" half-plane.="" application="" theory="" residues="" for=""> 1 u (x, y) = A π ∞ r=1 (−1)r x −rπ r sin (rπy). Example 12.14.3. Find the Mellin transform of the Weyl fractional integral ω (x, α) = Wα [f (ξ)] = 1 Γ (α)
∞ x f (ξ) (ξ − x) α−1 dξ. We rewrite the Weyl integral by setting k (x) = x α f (x), g (x) = 1 Γ (α) (1 − x) α−1 H (1 − x), so that 12.15 Finite Fourier Transforms and Applications 499 ω (x, α) =
∞ 0 k (ξ) g  x ξ  dξ ξ . The Mellin transform of this result is obtained by the convolution property (vii): Ω (p, a) = K (p) G (p), where K (p) = M[k (x)] = M[x αf (x)] = F (p + α) and G (p) = 1 Γ (α)
1 0 (1 − x) α−1 x p−1 dx = Γ (p) Γ (p + α) . Thus, Ω (p, a) = M[Wαf (ξ)] = Γ (p) Γ (p + α) F (p + α). (12.14.5) If α is complex with a positive real part such that n − 1 < Re α 0, u (x, 0) = 0, ut (x, 0) = 0, 0 < x < π, (12.15.7) u (0, t)=0, u (π, t)=0, t > 0. Applying the finite Fourier sine transform to the equation of motion with respect to x gives Fs utt − c 2uxx − f (x, t) ! = 0. Due to its linearity (see Problem 52, 12.18 Exercises), this can be written in the form Fs [utt] − c 2Fs [uxx] = Fs [f (x, t)] . (12.15.8) Let U (n, t) be the finite Fourier sine transform of u (x, t). Then we have Fs [utt] = 2 π
π 0 utt sin nx dx = d 2 dt2 2 π
π 0 u (x, t) sin nx dx = d 2Us (n, t) dt2 . We also have, from Theorem 12.15.1, Fs [uxx] = 2n π [u (0, t) − (−1)n u (π, t)] − n 2Us (n, t). Because of the boundary conditions u (0, t) = u (π, t)=0, Fs [uxx] becomes Fs [uxx] = −n 2Us (n, t). If we denote the finite Fourier sine transform of f (x, t) by Fs (n, t) = 2 π
π 0 f (x, t) sin nx dx, then, equation (12.15.8) takes the form d 2Us dt2 + n 2 c 2Us = Fs (n, t). This is a second-order ordinary differential equation, the solution of which is given by 502 12 Integral Transform Methods with Applications Us (n, t) = A cos nct + B sin nct + 1 nc
t 0 Fs (n, τ ) sin nc (t − τ ) dτ. Applying the initial conditions Fs [u (x, 0)] = 2 π
π 0 u (x, 0) sin nx dx = Us (n, 0) = 0, and Fs [ut (x, 0)] = d dtUs (n, 0) = 0, we have Us (n, t) = 1 nc
t 0 Fs (n, τ ) sin nc (t − τ ) dτ. Thus, the inverse transform of Us (n, t) is u (x, t) = ∞ n=1 Us (n, t) sin nx = ∞ n=1 1 nc
t 0 Fs (n, τ ) sin nc (t − τ ) dτ sin nx. In the case when f (x, t) = h which is a constant, then Fs [h] = 2 π
π 0 h sin nx dx = 2h nπ [1 − (−1)n ] . Now, we evaluate Us (n, t) = 1 nc
t 0 2h nπ [1 − (−1)n ] sin nc (t − τ ) dτ = 2h n3πc2 [1 − (−1)n ] (1 − cos nct). Hence, the solution is given by u (x, t) = 2h πc2 ∞ n=1 [1 − (−1)n ] n3 (1 − cos nct) sin nx. Example 12.15.2. Find the temperature distribution in a rod of length π. The heat is generated in the rod at the rate g (x, t) per unit time. The ends are insulated. The initial temperature distribution is given by f (x). The problem is to find the temperature function u (x, t) of the system ut = uxx + g (x, t), 0 < x < π, t > 0, u (x, 0) = f (x), 0 ≤ x ≤ π, ux (0, t)=0, ux (π, t)=0, t ≥ 0. 12.15 Finite Fourier Transforms and Applications 503 Let Us (n, t) be the finite Fourier cosine transform of u (x, t). As before, transformation of the heat equation with respect to x, using the boundary conditions, yields dUs dt = −n 2Us + Gs (n, t), where Gs (n, t) = 2 π
π 0 g (x, t) cos nx dx. Rewriting this equation, we obtain d dt
4 e n 2 t Us
5 = Gs e n 2 t . Thus, the solution is Us (n, t) =
t 0 e −n 2 (t−τ)Gs (n, τ ) dτ + A e−n 2 t . Transformation of the initial condition gives Us (n, 0) = 2 π
π 0 u (x, 0) cos nx dx = 2 π
π 0 f (x) cos nx dx. Hence, Us (n, t) takes the form Us (n, t) =
t 0 e −n 2 (t−τ)Gs (n, τ ) dτ + Us (n, 0) e −n 2 t . The solution u (x, t), therefore, is given by u (x, t) = Us (0, 0) 2 + ∞ n=1 Us (n, t) cos nx. Example 12.15.3. A rod with diffusion constant κ contains a fuel which produces neutrons by fission. The ends of the rod are perfectly reflecting. If the initial neutron distribution is f (x), find the neutron distribution u (x, t) at any subsequent time t. The problem is governed by ut = κuxx + bu, u (x, 0) = f (x), 0 < x < l, t > 0, ux (0, t) = ux (l, t)=0. If U (n, t) is the finite Fourier cosine transform of u (x, t), then by transforming the equation and using the boundary conditions, we obtain 504 12 Integral Transform Methods with Applications Ut +  κn2 − b U = 0. The solution of this equation is U (n, t) = C e−(κn2−b)t where C is a constant. Then applying the initial condition, we obtain U (n, t) = U (n, 0) e −(κn2−b)t , where U (n, 0) = 2 l
l 0 f (x) cos nx dx. Thus, the solution takes the form u (x, t) = U (0, 0) 2 + ∞ n=1 U (n, t) cos nx. If for instance f (x) = x in 0 <x<π, then,="" u="" (0,="" 0)="π," and="" (n,="" n2π="" [(−1)n="" −="" 1]="" ,="" n="1," 2,="" 3,...="" the="" solution="" is="" given="" by="" (x,="" t)="π" 2="" +="" ∞="" exp="" &="" ="" κn2="" b="" t="" '="" cos="" nx.="" 12.16="" finite="" hankel="" transforms="" applications="" fourier–bessel="" series="" representation="" of="" a="" function="" f="" (r)="" defined="" in="" 0="" ≤="" r="" can="" be="" stated="" following="" theorem:="" theorem="" 12.16.1.="" if="" fn="" (ki)="Hn" {f="" (r)}="
" rf="" jn="" (rki)="" dr,="" (12.16.1)="" then="" {fn="" (ki)}="2" i="1" j="" n+1="" (aki)="" (12.16.2)="" where="" ki="" (0="" <="" k1="" k2="" <...)="" are="" roots="" equation="" 505="" called="" nth-order="" transform="" (r),="" inverse="" (12.16.2).="" particular,="" when="" order="" zero="" its="" integral="" respectively="" j0="" (12.16.3)="" −1="" 1="" .="" (12.16.4)="" example="" find="" n.="" we="" have="" result="" for="" bessel=""
="" n+1jn="" (kir)="" dr="a" jn+1="" (aki),="" so="" that="" hn="" {r="" }="a" (aki).="" h0="" {1}="a" j1="" or,="" equivalently,="" ="" 0="2" kij1="" 12.16.2.="" &a="" '.="" definition="" rj0="" k="" 3="" 2a="" root="" (ax)="0." hence,="" state="" operational="" properties="" transform:="" 506="" 12="" methods="" with="" (i)="" df="" dr0="ki" 2n="" [(n="" 1)="" hn+1="" (n="" hn−1="" (r)}]="" (ii)="" h1="" (iii)="" d="" {rf′="" (a)="" ′="" (12.16.5)="" (iv)="" ′′="" (12.16.6)="" 12.16.3.="" axisymmetric="" heat="" conduction="" ut="κ" ="" urr="" ur="" ="" a,=""> 0, with the boundary and initial conditions u (r, t) = f (t) on r = a, t ≥ 0, u (r, 0) = 0, 0 ≤ r ≤ a, where u (r, t) represents the temperature distribution. We apply the finite Hankel transform defined by U (k, t) = H0 {u (r, t)} =
a 0 rJ0 (kir) u (r, t) dr so that the given equation with the boundary condition becomes dU dt + κk2 i U = κakiJ1 (aki) f (t). The solution of this equation with the transformed initial condition is U (k, t) = aκkiJ1 (aki)
t 0 f (τ ) e −κk2 i (t−τ) dτ. The inverse transformation gives the solution as u (r, t) =  2κ a ∞ i=1 kiJ0 (rki) J1 (aki)
t 0 f (τ ) e −κk2 i (t−τ) dτ. (12.16.7) In particular, if f (t) = T0 = constant, then this solution becomes u (r, t) =  2T0 a ∞ i=1 J0 (rki) kiJ1 (aki)
4 1 − e −κk2 i t
5 . (12.16.8) 12.16 Finite Hankel Transforms and Applications 507 In view of Example 12.16.1, result (12.16.8) becomes u (r, t) = T0 1 1 − 2 a ∞ i=1 J0 (rki) kiJ1 (aki) e −κk2 i t 3 . (12.16.9) This solution consists of the steady-state term, and the transient term which tends to zero as t → ∞. Consequently, the steady-state is attained in the limit t → ∞. Example 12.16.4. (Unsteady Viscous Flow in a Rotating Cylinder). The axisymmetric unsteady motion of a viscous fluid in an infinitely long circular cylinder of radius a is governed by vt = ν  vrr + 1 r vr − v r 2  , 0 ≤ r ≤ a, t > 0, where v = v (r, t) is the tangential fluid velocity and ν is the kinematic viscosity of the fluid. The cylinder is at rest until at t = 0+ it is caused to rotate, so that the boundary and initial conditions are v (r, t) = aΩf (t) H (t) on r = a, v (r, t)=0, at t = 0 for r < a, where f (t) is a physically realistic function of t. We solve the problem by using the joint Laplace and finite Hankel transforms of order one defined by V (km, s) =
a 0 rJ1 (rkm) dr
∞ 0 e −st v (r, t) dt, where V (km, s) is the Laplace transform of V (km, t), and km are the roots of equation J1 (akm) = 0. Application of the transform yields
4 s ν
5 V (km, s) = −akmV (a, s) J ′ 1 (akm) − k 2 mV (km, s), V (a, s) = a Ω f (s), where f (s) is the Laplace transform of f (t). The solution of this system is V (km, s) = − a 2νkmΩ f (s) J ′ 1 (akm) (s + νk2 m) . The joint inverse transformation gives 508 12 Integral Transform Methods with Applications v (r, t) = 2 a 2 ∞ m=1 J1 (rkm) [J ′ 1 (akm)]2 1 2πi
c+i∞ c−i∞ e st V (km, s) ds = −2νΩ ∞ m=1 kmJ1 (rkm) J ′ 1 (akm) 1 2πi
c+i∞ c−i∞ e st f (s) (s + νk2 m) ds = −2νΩ ∞ m=1 kmJ1 (rkm) J ′ 1 (akm)
t 0 f (τ ) exp −νk2 m (t − τ ) ! dτ, by the Convolution Theorem of the Laplace transform. In particular, when f (t) = cos ωt, the velocity field becomes v (r, t) = −2νΩ ∞ m=1 kmJ1 (rkm) J ′ 1 (akm)
t 0 cos ωτ exp −νk2 m (t − τ ) ! dτ = 2νΩ ∞ m=1 kmJ1 (rkm) J ′ 1 (akm) × 1 νk2 m exp  −νtk2 m −  ω sin ωt + νk2 m cos ωt (ω2 + ν 2k 4 m) 3 = vst (s, t) + vtr (r, t), (12.16.10) where the steady-state flow field vst and the transient flow field vtr are given by vst (r, t) = −2νΩ ∞ m=1 kmJ1 (rkm)  ω sin ωt + νk2 m cos ωt J ′ 1 (akm) (ω2 + ν 2k 4 m) , (12.16.11) vtr (r, t)=2ν 2Ω ∞ m=1 J1 (rkm) k 3 me −νtk2 m J ′ 1 (akm) (ω2 + ν 2k 4 m) . (12.16.12) Thus, the solution consists of the steady-state and transient components. In the limit t → ∞, the latter decays to zero, and the ultimate steady-state is attained and is given by (12.16.11), which has the form vst (r, t) = −2νΩ ∞ m=1 kmJ1 (rkm) cos (ωt − α) J ′ 1 (akm) (ω2 + ν 2k 4 m) 1 2 , (12.16.13) where tan α =  ω/νk2 m . Thus, we see that the steady solution suffers from a phase change of α+π. The amplitude of the motion remains bounded for all values of ω. The frictional couple exerted on the fluid by unit length of the cylinder of radius r = a is given by C =
2π 0 [Prθ] r=a a 2 dθ = 2πa2 [Prθ] r=a , 12.16 Finite Hankel Transforms and Applications 509 where Prθ = µ r (d/dr) (v/r) with µ = νρ calculated from (12.16.10). Thus, C = 4πµΩ 1 −a ∞ m=1 νkmJ1 (akm) νk2 m exp  −νtk2 m −  νk2 m cos ωt + ω sin ωt ! (ω2 + ν 2k 4 m) J ′ 1 (akm) + a 2 ∞ m=1 νk2 m νk2 m exp  −νtk2 m −  νk2 m cos ωt + ω sin ωt ! (ω2 + ν 2k 4 m) J ′ 1 (akm) 3 . (12.16.14) A particular case corresponding to ω = 0 is of special interest. The solution assumes the form v (r, t) = −2Ω ∞ m=1 J1 (rkm)
4 1 − e −νtk2 m
5 km J ′ 1 (akm) = vst + vtr,(12.16.15) where vst and vtr represent the steady-state and the transient flow fields respectively given by vst (r, t) = −2Ω ∞ m=1 J1 (rkm) kmJ ′ 1 (akm) , (12.16.16) vtr (r, t)=2Ω ∞ m=1 J1 (rkm) kmJ ′ 1 (akm) e −νtk2 m. (12.16.17) It follows from (12.16.16) that vst (r, t)=2Ω ∞ m=1 J1 (rkm) kmJ2 (akm) = 2Ω a 2 ∞ m=1 a 2 km J2 (akm) · J1 (rkm) J 2 2 (akm) = ΩH −1 1  a 2 km J2 (akm) 0 = Ωr, by Example 12.16.1. Thus, the steady-state solution has the closed form vst (r, t) = rΩ. (12.16.18) This represents the rigid body rotation of the fluid inside the cylinder. Thus, the final form of (12.16.15) is given by v (r, t) = rΩ − 2Ω ∞ m=1 J1 (rkm) e −νtk2 m kmJ2 (akm) . (12.16.19) In the limit t → ∞, the transients die out and the ultimate steady-state is attained as the rigid body rotation about the axis of the cylinder. 510 12 Integral Transform Methods with Applications 12.17 Solution of Fractional Partial Differential Equations (a) Fractional Diffusion Equation The fractional diffusion equation is given by ∂ αu ∂tα = κ ∂ 2u ∂x2 , x ∈ R, t > 0, (12.17.1) with the boundary and initial conditions u (x, t) → 0 as |x|→∞, (12.17.2) 0D α−1 t u (x, t) ! t=0 = f (x) for x ∈ R, (12.17.3) where κ is a diffusivity constant and 0 < α ≤ 1. Application of the Fourier transform to (12.17.1) with respect to x and using the boundary conditions (12.17.2) and (12.17.3) yields 0Dα t u˜ (k, t) = −κ k2 u, ˜ (12.17.4) 0D α−1 t u˜ (k, t) ! t=0 = ˜f (k), (12.17.5) where ˜u (k, t) is the Fourier transform of u (x, t) defined by (12.2.1). The Laplace transform solution of (12.17.4) and (12.17.5) yields u˜¯ (k, s) = ˜f (k) (s α + κ k2) . (12.17.6) The inverse Laplace transform of (12.17.6) gives u˜ (k, t) = ˜f (k)t α−1Eα,α  −κ k2 t α , (12.17.7) where Eα,β is the Mittag-Leffler function defined by Eα,β (z) = ∞ m=0 z m Γ (αm + β) , α > 0, β> 0. (12.17.8) Finally, the inverse Fourier transform leads to the solution u (x, t) =
∞ −∞ G (x − ξ, t) f (ξ) dξ, (12.17.9) where G (x, t) = 1 π
∞ −∞ t α−1Eα,α  −κ k2 t α cos kx dk. (12.17.10) 12.17 Solution of Fractional Partial Differential Equations 511 This integral can be evaluated by using the Laplace transform of G (x, t) as G (x, s) = 1 π
∞ −∞ cos kx dk s α + κk2 = 1 √ 4κ s −α/2 exp  − |x| √ κ s α/2  , (12.17.11) where L " t mα+β−1E (m) α,β (+atα ) # = m! s α−β (s α + a) m+1 , (12.17.12) and E (m) α,β (z) = d m dzm Eα,β (z). (12.17.13) The inverse Laplace transform of (12.17.11) gives the explicit solution G (x, t) = 1 √ 4κ t α 2 −1W
4 −ξ, − α 2 , α 2
5 , (12.17.14) where ξ = |x| √ κ tα/2 , and W (z, α, β) is the Wright function (see Erd´elyi 1953, formula 18.1 (27)) defined by W (z, α, β) = ∞ n=0 z n n! Γ (αn + β) . (12.17.15) It is important to note that when α = 1, the initial-value problem (12.17.1)–(12.17.3) reduces to the classical diffusion problem and solution (12.17.9) reduces to the classical solution because G (x, t) = 1 √ 4κt W  − x √ κt , − 1 2 , 1 2  = 1 √ 4πκt exp  − x 2 4κt . (12.17.16) The fractional diffusion equation (12.17.1) has also been solved by other authors including Schneider and Wyss (1989), Mainardi (1994, 1995), Debnath (2003) and Nigmatullin (1986) with a physical realistic initial condition u (x, 0) = f (x), x ∈ R. (12.17.17) The solutions obtained by these authors are in total agreement with (12.17.9). It is noted that the order α of the derivative with respect to time t in equation (12.17.1) can be of arbitrary real order including α = 2 so that equation (12.17.1) may be called the fractional diffusion-wave equation. For α = 2, it becomes the classical wave equation. The equation (12.17.1) with 1 < α ≤ 2 will be solved next in some detail. 512 12 Integral Transform Methods with Applications (b) Fractional Nonhomogeneous Wave Equation The fractional nonhomogeneous wave equation is given by ∂ αu ∂tα − c 2 ∂ 2u ∂x2 = q (x, t), x ∈ R, t > 0 (12.17.18) with the initial condition u (x, 0) = f (x), ut (x, 0) = g (x), x ∈ R, (12.17.19) where c is a constant and 1 < α ≤ 2. Application of the joint Laplace transform with respect to t and the Fourier transform with respect to x gives the transform solution u˜¯ (k, s) = ˜f (k) s α−1 s α + c 2k 2 + g˜ (k) s α−2 s α + c 2k 2 + q˜¯(k, s) s α + c 2k 2 , (12.17.20) where k is the Fourier transform variable and s is the Laplace transform variable. The inverse Laplace transform produces the following result: u˜ (k, t) = ˜f (k)L −1  s α−1 s α + c 2k 2 0 + ˜g (k)L −1  s α−2 s α + c 2k 2 0 +L −1  q˜¯(k, s) s α + c 2k 2 0 , (12.17.21) which, by (12.17.12), = ˜f (k) Eα,1  −c 2 k 2 t α + ˜g (k)tEα,2  −c 2 k 2 t α +
t 0 q˜(k, t − τ ) τ α−1Eα,α  −c 2 k 2 τ α dτ. (12.17.22) Finally, the inverse Fourier transform gives the formal solution u (x, t) = 1 √ 2π
∞ −∞ ˜f (k) Eα,1  −c 2 k 2 t α e ikxdk + 1 √ 2π
∞ −∞ t g˜ (k) Eα,2  −c 2 k 2 τ α e ikxdk + 1 √ 2π
t 0 τ α−1 dτ
∞ −∞ q˜(k, t − τ ) Eα,α  −c 2 k 2 τ α e ikxdk. (12.17.23) In particular, when α = 2, the fractional wave equation (12.17.18) reduces to the classical nonhomogeneous wave equation. In this particular case, we use 12.17 Solution of Fractional Partial Differential Equations 513 E2,1  −c 2 k 2 t 2 = cosh (ickt) = cos (ckt), (12.17.24) tE2,2  −c 2 k 2 t 2 = t · sinh (ickt) ickt = 1 ck sin ckt. (12.17.25) Consequently, solution (12.17.23) reduces to solution (12.11.17) for α = 2 as u (x, t) = 1 √ 2π
∞ −∞ ˜f (k) cos (ckt) e ikxdk + 1 √ 2π
∞ −∞ g˜ (k) sin (ckt) ck e ikxdk + 1 √ 2π c
t 0 dτ
∞ −∞ q˜(k, τ ) sin ck (t − τ ) k e ikxdk (12.17.26) = 1 2 [f (x − ct) + f (x + ct)] + 1 2c
x+ct x−ct g (ξ) dξ + 1 2c
t 0 dτ
x+c(t−τ) x−c(t−τ) q (ξ, τ ) dξ. (12.17.27) We now derive the solution of the nonhomogeneous fractional diffusion equation (12.17.18) with c 2 = κ and g (x) = 0. In this case, the joint transform solution (12.17.20) becomes u˜¯ (k, s) = ˜f (k) s α−1 (s α + κk2) + q˜¯(k, s) (s α + κk2) (12.17.28) which is inverted by using (12.17.12) to obtain ˜u (k, t) in the form u˜ (k, t) = ˜f (k) Eα,1  −κ k2 t α +
t 0 (t − τ ) α−1 Eα,α & −κ k2 (t − τ ) α ' q˜(k, τ ) dτ. (12.17.29) Finally, the inverse Fourier transform gives the exact solution u (x, t) = 1 √ 2π
∞ −∞ ˜f (k) Eα,1  −κ k2 t α e ikxdk + 1 √ 2π
t 0 dτ
∞ −∞ (t − τ ) α−1 Eα,α & −κ k2 (t − τ ) α ' q˜(k, τ ) e ikxdk. (12.17.30) Application of the Convolution Theorem of the Fourier transform gives the final solution in the form u (x, t) =
∞ −∞ G1 (x − ξ, t) f (ξ) dξ +
t 0 (t − τ ) α−1 dτ
∞ −∞ G2 (x − ξ, t − τ ) q (ξ, τ ) dξ, (12.17.31) 514 12 Integral Transform Methods with Applications where G1 (x, t) = 1 2π
∞ −∞ e ikxEα,1  −κ k2 t α dk, (12.17.32) and G2 (x, t) = 1 2π
∞ −∞ e ikxEα,α  −κ k2 t α dk. (12.17.33) In particular, when α = 1, the classical solution of the nonhomogeneous diffusion equation (12.17.18) is obtained in the form u (x, t) =
∞ −∞ G1 (x − ξ, t) f (ξ) dξ +
t 0 dτ
∞ −∞ G2 (x − ξ, t − τ ) q (ξ, τ ) dξ, (12.17.34) where G1 (x, t) = G2 (x, t) = 1 √ 4πκt exp  − x 2 4κt . (12.17.35) In the case of classical homogeneous diffusion equation (12.17.18), solutions (12.17.30) and (12.17.34) are in perfect agreement with those of Mainardi (1996), who obtained the solution by using the Laplace transform method together with complicated evaluation of the Laplace inversion integral and the auxiliary function M (z,α). He obtained the solution in terms of M  z, α 2 and discussed the nature of the solution for different values of α. He made some comparisons between ordinary diffusion (α = 1) and fractional diffusion  α = 1 2 and α = 2 3 . For cases α = 4 3 and α = 3 2 , the solution exhibits a striking difference from ordinary diffusion with a transition from the Gaussian function centered at z = 0 (ordinary diffusion) to the Dirac delta function centered at z = 1 (wave propagation). This indicates a possibility of an intermediate process between diffusion and wave propagation. A special difference is observed between the solutions of the fractional diffusion equation (0 < α ≤ 1) and the fractional wave equation (1 < α ≤ 2). In addition, the solution exhibits a slow process for the case with 0 < α ≤ 1 and an intermediate process for 1 < α ≤ 2. (c) Fractional-Order Diffusion Equation in Semi-Infinite Medium We consider the fractional-order diffusion equation in a semi-infinite medium x > 0, when the boundary is kept at a temperature u0f (t) and the initial temperature is zero in the whole medium. Thus, the initial boundaryvalue problem is governed by the equation ∂ αu ∂tα = κ ∂ 2u ∂x2 , 0 <x< ∞,="" t=""> 0, (12.17.36) 12.17 Solution of Fractional Partial Differential Equations 515 with u (x, 0) = 0, x > 0, (12.17.37) u (0, t) = u0f (t), t> 0, and u (x, t) → 0 as x → ∞. (12.17.38) Application of the Laplace transform with respect to t gives d 2u dx2 −  s α κ  u (x, s)=0, x > 0, (12.17.39) u (0, s) = u0f (s), u (x, s) → 0 as x → ∞.(12.17.40) Evidently, the solution of this transformed boundary-value problem is u (x, s) = u0 f (s) exp (−ax), (12.17.41) where a = (s α/κ) 1 2 . Thus, the solution (12.17.41) is given by u (x, t) = u0
t 0 f (t − τ ) g (x, τ ) dτ = u0f (t) ∗ g (x, t),(12.17.42) where g (x, t) = L −1 {exp (−ax)} . When α = 1 and f (t) = 1, solution (12.17.41) becomes u (x, s) = u0 s exp  −x 2 s κ  , (12.17.43) which yields the classical solution in terms of the complementary error function (see Debnath 1995) u (x, t) = u0 erfc  x 2 √ κt . (12.17.44) In the classical case (α = 1), the more general solution is given by u (x, t) = u0
t 0 f (t − τ ) g (x, τ ) dτ = u0f (t) ∗ g (x, t),(12.17.45) where g (x, t) = L −1  exp  −x 2 s κ 0 = x 2 √ πκt3 exp  − x 2 4κt . (12.17.46) (d) The Fractional Stokes and Rayleigh Problems in Fluid Dynamics The classical Stokes problem (see Debnath 1995) deals with the unsteady boundary layer flows induced in a semi-infinite viscous fluid bounded 516 12 Integral Transform Methods with Applications by an infinite horizontal disk at z = 0 due to nontorsional oscillations of the disk in its own plane with a given frequency ω. When ω = 0, the Stokes problem reduces to the classical Rayleigh problem where the unsteady boundary layer flow is generated in the fluid from rest by moving the disk impulsively in its own plane with constant velocity U. We consider the unsteady fractional boundary layer equation for the fluid velocity u (z, t) that satisfies the equation ∂ αu ∂tα = ν ∂ 2u ∂z2 , 0 <z< ∞,="" t=""> 0, (12.17.47) with the given boundary and initial conditions u (0, t) = Uf (t), u (z, t) → 0 as z → ∞, t> 0, (12.17.48) u (z, 0) = 0 for all z > 0, (12.17.49) where ν is the kinematic viscosity, U is a constant velocity, and f (t) is an arbitrary function of time t. Application of the Laplace transform with respect to t gives s α u (z, s) = ν d 2u dz2 , 0 <z< ∞,="" (12.17.50)="" u="" (0,="" s)="U" f="" (s),="" (z,="" →="" 0="" as="" z="" ∞.="" (12.17.51)="" use="" of="" the="" fourier="" sine="" transform="" (see="" debnath="" 1995)="" with="" respect="" to="" yields="" us="" (k,="" 2="" π="" ν="" u5="" k="" (s)="" (s="" α="" +="" νk2)="" .="" (12.17.52)="" inverse="" leads="" solution="" u=""
="" ∞="" sin="" kz="" dk,="" (12.17.53)="" and="" laplace="" gives="" for="" velocity="" t)="" dk="" t="" (t="" −="" τ="" )="" α−1eα,α="" ="" −νk2="" dτ.="" (12.17.54)="" when="" (t)="exp" (iωt),="" fractional="" stokes="" problem="" is="" 2νu="" ="" e="" iωt="" −iωτ="" (12.17.55)="" 12.17="" partial="" differential="" equations="" 517="" reduces="" classical=""
4="" 1="" −νtk2=""
5="" (iω="" dk.="" (12.17.56)="" rayleigh="" problem,="" follows="" from="" in="" form="" (12.17.57)="" this="" e1,1="" −ντk2="" dτ="" exp="" 2u="" which="" (by="" (2.10.10)="" 1995),="" π="" erf="" ="" √="" νt="Uerfc" (12.17.58)="" above="" analysis="" full="" agreement="" solutions="" problems="" 1995).="" (e)="" unsteady="" couette="" flow="" we="" consider="" viscous="" fluid="" between="" plate="" at="" rest="" motion="" parallel="" itself="" a="" variable="" x-direction.="" satisfies="" equation="" ∂="" αu="" ∂tα="P" ∂z2="" ,="" ≤="" h,=""> 0, (12.17.59) with the boundary and initial conditions u (0, t) = 0 and u (h, t) = U (t), t > 0, (12.17.60) u (z, t) = 0 at t ≤ 0 for 0 ≤ z ≤ h, (12.17.61) where − 1 ρ px = P (t) and ν is the kinematic viscosity of the fluid. We apply the joint Laplace transform with respect to t and the finite Fourier sine transform with respect to z defined by u¯˜s (n, s) =
∞ 0 e −stdt
h 0 u (z, t) sin
4nπz h
5 dz (12.17.62) 518 12 Integral Transform Methods with Applications to the system (12.17.59)–(12.17.61) so that the transform solution is u¯˜s (n, s) = P (s) 1 a [1 − (−1)n ] (s α + νa2) + νa (−1)n+1 U (s) (s α + νa2) , (12.17.63) where a =  nπ h and n is the finite Fourier sine transform variable. Thus, the inverse Laplace transform yields u˜s (n, t) = 1 a [1 − (−1)n ]
t 0 P (t − τ ) τ α−1Eα,α  −νa2 τ α dτ +νa (−1)n+1
t 0 U (t − τ ) τ α−1Eα,α  −νa2 τ α dτ. (12.17.64) Finally, the inverse finite Fourier sine transform leads to the solution u (z, t) = 2 h ∞ n=1 u˜s (n, t) sin
4nπz h
5 . (12.17.65) In particular, when α = 1, P (t) = constant, and U (t) = constant, then solution (12.17.65) reduces to the solution of the generalized Couette flow (see p. 277 Debnath 1995). (f) Fractional Axisymmetric Wave-Diffusion Equation The fractional axisymmetric wave-diffusion equation in an infinite domain ∂ αu ∂tα = a  ∂ 2u ∂r2 + 1 r ∂u ∂r  , 0 <r< ∞,="" t=""> 0, (12.17.66) is called the diffusion or wave equation accordingly as a = κ or a = c 2 . For the fractional diffusion equation, we prescribe the initial condition u (r, 0) = f (r), 0 <r< ∞.="" (12.17.67)="" application="" of="" the="" joint="" laplace="" transform="" with="" respect="" to="" t="" and="" hankel="" zero="" order="" (see="" section="" 12.12)="" r="" (12.17.66)="" gives="" solution="" u¯˜="" (k,="" s)="s" α−1="" ˜f="" (k)="" (s="" α="" +="" κk2)="" ,="" (12.17.68)="" where="" k,="" s="" are="" variables="" respectively.="" inverse="" leads="" u="" (r,="" t)="
" ∞="" 0="" kj0="" (kr)="" eα,1="" ="" −κk2="" dk,="" (12.17.69)="" 12.17="" fractional="" partial="" differential="" equations="" 519="" j0="" is="" bessel="" function="" first="" kind="" zero-order="" f="" (r).="" when="" reduces="" classical="" that="" was="" obtained="" by="" debnath="" p="" 66,="" 2005).="" on="" other="" hand,="" we="" can="" solve="" wave="" equation="" a="c" 2="" initial="" conditions="" 0)="f" (r),="" ut="" (r)="" for="" <r<="" ∞,="" (12.17.70)="" provided="" transforms="" g="" exist.="" c="" 2k="" 2)="" α−2="" g˜="" .="" (12.17.71)="" transformation="" r)="" −c="" k="" dk=""
="" ˜g="" (k)teα,2="" dk.="" (12.17.72)="" (12.13.6).="" in="" finite="" domain="" ≤="" a,="" diffusion="" has="" boundary="" data="" (t)=""> 0, (12.17.73) u (r, 0) = 0 for all r in (0, a). (12.17.74) Application of the joint Laplace and finite Hankel transform of zero order (see pp. 317, 318, Debnath 1995) yields the solution u (r, t) = 2 a 2 ∞ i=1 u˜ (ki , t) J0 (rki) J 2 1 (aki) , (12.17.75) where u˜ (ki , t)=(aκ ki) J1 (aki)
t 0 f (t − τ ) τ α−1Eα,α  −κ k2 i τ α dτ. (12.17.76) When α = 1, (12.17.75) reduces to (12.16.7). Similarly, the fractional wave equation (12.17.66) with a = c 2 in a finite domain 0 ≤ r ≤ a with the boundary and initial conditions u (r, t) = 0 on r = a, t > 0, (12.17.77) u (r, 0) = f (r) and ut (r, 0) = g (r) for 0 < r < a, (12.17.78) 520 12 Integral Transform Methods with Applications can be solved by means of the joint Laplace and finite Hankel transforms. The solution of this problem is u (r, t) = 2 a 2 ∞ i=1 u˜ (ki , t) J0 (rki) J 2 1 (aki) , (12.17.79) where u˜ (ki , t) = ˜f (ki) Eα,1  −c 2 k 2 i t α + ˜g (ki)tEα,2  −c 2 k 2 i t α . (12.17.80) When α = 2, solution (12.17.79) reduces to the solution (11.4.26) obtained by Debnath (1995). (g) The Fractional Schr¨odinger Equation in Quantum Mechanics The one-dimensional fractional Schr¨odinger equation for a free particle of mass m is i
∂ αψ ∂tα = −
2 2m ∂ 2ψ ∂x2 , −∞ <x< ∞,="" t=""> 0, (12.17.81) ψ (x, 0) = ψ0 (x), −∞ <x< ∞,="" (12.17.82)="" ψ="" (x,="" t)="" →="" 0="" as="" |x|→∞,="" (12.17.83)="" where="" is="" the="" wave="" function,="" h="2π
" =="" 6.625="" ×="" 10−27erg="" sec="4.14" 10−21mev="" planck="" constant,="" and="" ψ0="" (x)="" an="" arbitrary="" function.="" application="" of="" joint="" laplace="" fourier="" transform="" to="" (12.17.81)–="" gives="" solution="" in="" space="" form="" Ψ="" (k,="" s)="s" α−1Ψ0="" (k)="" s="" α="" +="" ak2="" ,="" a="i
" 2m="" (12.17.84)="" k,="" represent="" transforms="" variables.="" use="" inverse="" yields="" √="" 2π=""
="" ∞="" −∞="" e="" ikxψ˜="" eα,1="" ="" −ak2="" t="" dk.="" (12.17.85)="F" −1="" (="" ψ˜="" )="" (12.17.86)="" which="" is,="" by="" theorem="" 12.4.1,="" convolution="" g="" (x="" −="" ξ,="" (ξ)="" dξ,="" (12.17.87)="" f="" &="" '="1" ikxeα,1="" (12.17.88)="" 12.18="" exercises="" 521="" when="" becomes="" (12.17.89)="" green’s="" function="" given="" ikxe1,1="" dk="1" exp="" ikx="" atk2="" 4πat="" ="" x="" 2="" 4at="" .="" (12.17.90)="" this="" perfect="" agreement="" with="" classical="" obtained="" debnath="" (1995).="" 1.="" find="" (a)="" −ax2="" (b)="" (−a="" |x|),="" constant.="" 2.="" gate="" fa="" ⎨="" ⎩="" 1,="" |x|="" <="" a,="" positive="" 0,="" ≥="" a.="" 3.="" −a<x<a="" otherwise,="" (c)="" 1="" ≤="" (d)="" (x2+a2)="" 4.="" 522="" 12="" integral="" methods="" applications="" 5.="" show="" that="" i="
" −a="" 2x="" dx="√" π="" 2a,=""> 0, by noting that I 2 =
∞ 0
∞ 0 e −a 2 (x 2+y 2 )dx dy =
π/2 0
∞ 0 e −a 2 r 2 r dr dθ. 6. Show that
∞ 0 e −a 2x 2 cos bx dx = √ π/2a e −b 2/4a 2 , a > 0. 7. Prove that (a) f (x) = √ 1 2π
∞ −∞ e ikxF (k) dk = F −1 {F (k)}, (b) F [f (ax − b)] = 1 |a| e ikb/aF (k/a). 8. Prove the following properties of the Fourier convolution: (a) f (x) ∗ g (x) = g (x) ∗ f (x), (b) f ∗ (g ∗ h)=(f ∗ g) ∗ h, (c) f ∗ (ag + bh) = a (f ∗ g) + b (f ∗ h), where a and b are constants, (d) f ∗ 0=0 ∗ f = 0, (e) f ∗ 1 = f, (f) f ∗ √ 2π δ = f = √ 2π δ ∗ f, (g) F {f (x) g (x)} = (F ∗ G) (k) = √ 1 2π
∞ −∞ F (k − ξ) G (ξ) dξ, 9. Prove the following properties of the Fourier convolution: (a) d dx {f (x) ∗ g (x)} = f ′ (x) ∗ g (x) = f (x) ∗ g ′ (x), (b) d 2 dx2 [(f ∗ g) (x)] = (f ′ ∗ g ′ ) (x)=(f ′′ ∗ g) (x), (c) (f ∗ g) (m+n) (x) = f (m) ∗ g (n) ! (x), (d)
∞ −∞ (f ∗ g) (x) dx =
∞ −∞ f (u) du
∞ −∞ g (v) dv. (e) If g (x) = 1 2a H (a − x), then (f ∗ g) (x) is the average value of f (x) in [x − a, x + a]. 12.18 Exercises 523 (f) If Gt (x) = √ 1 4πkt exp
4 − x 2 4κt
5 , then (Gt ∗ Gs) (x) = Gt+s (x). 10. Prove the following results: (a) √ 1 2π
∞ −∞ e −k 2 t−ikx dk = √ 1 2t e −x 2/4t , (b)
∞ −∞ F (k) g (k) e ikx dk =
∞ −∞ f (y) G (y − x) dy, (c)
∞ −∞ F (k) g (k) dk =
∞ −∞ f (y) G (y) dy, (d) sin x ∗ e −a|x| = % 2 π a sin x (1+a2) , (e) e ax ∗ χ[0,∞) (x) = 1 a e ax √ 2π , a> 0 , (f) √ 1 2a exp
4 − x 2 4a
5 ∗ √ 1 2b exp
4 − x 2 4b
5 = √ 1 2(a+b) exp
4 − x 2 4(a+b)
5 . 11. Determine the solution of the initial-value problem utt = c 2uxx, −∞ <x< ∞,="" t=""> 0, u (x, 0) = f (x), ut (x, 0) = g (x), −∞ <x< ∞.="" 12.="" solve="" ut="uxx," x=""> 0, t > 0, u (x, 0) = f (x), u (0, t)=0. 13. Solve utt = c 2uxxxx = 0, −∞ <x< ∞,="" t=""> 0, u (x, 0) = f (x), ut (x, 0) = 0, −∞ <x< ∞.="" 14.="" solve="" utt="" +="" c="" 2uxxxx="0," x=""> 0, t > 0, u (x, 0) = 0, ut (x, 0) = 0, x > 0, u (0, t) = g (t), uxx (0, t)=0, t > 0. 15. Solve φxx + φyy = 0, −a < x < a, 0 <y< ∞,="" φy="" (x,="" 0)="⎧" ⎨="" ⎩="" δ0,="" 0="" <="" |x|="" a,="" 0,=""> a. φ (x, y) → 0 uniformly in x as y → ∞. 524 12 Integral Transform Methods with Applications 16. Solve ut = uxx + t u, −∞ <x< ∞,="" t=""> 0, u (x, 0) = f (x), u (x, t) is bounded, −∞ <x< ∞.="" 17.="" solve="" ut="" −="" uxx="" +="" hu="δ" (x)="" δ="" (t),="" −∞="" <x<="" ∞,="" t=""> 0, u (x, 0) = 0, u (x, t) → 0 uniformly in t as |x|→∞. 18. Solve ut − uxx + h (t) ux = δ (x) δ (t), 0 <x< ∞,="" t=""> 0, u (x, 0) = 0, ux (0, t)=0, u (x, t) → 0 uniformly in t as x → ∞. 19. Solve uxx + uyy = 0, 0 <x< ∞,="" 0="" <y<="" u="" (x,="" 0)="f" (x),="" ≤="" x="" <="" ux="" (0,="" y)="g" (y),="" y="" →="" uniformly="" in="" as="" ∞="" and="" ∞.="" 20.="" solve="" uxx="" +="" uyy="0," −∞="" <x<="" a,="" a)="0," |x|→∞.="" 21.="" ut="uxx,"> 0, t> 0, u (x, 0) = 0, x > 0, u (0, t) = f (t), t> 0, u (x, t) is bounded for all x and t. 22. Solve uxx + uyy = 0, x > 0, 0 <y< 1,="" u="" (x,="" 0)="f" (x),="" 1)="0," x=""> 0, u (0, y)=0, u (x, y) → 0 uniformly in y as x → ∞. 12.18 Exercises 525 23. Find the Laplace transform of each of the following functions: (a) t n, (b) cos ωt, (c) sinh kt, (d) cosh kt, (e) teat, (f) e at sin ωt, (g) e at cos ωt, (h) tsinh kt, (i) t cosh kt, (j) % 1 t , (k) √ t, (l) sin at t . 24. Find the inverse transform of each of the following functions: (a) s (s 2+a2)(s 2+b 2) , (b) 1 (s 2+a2)(s 2+b 2) , (c) 1 (s−a)(s−b) , (d) 1 s(s+a) 2 , (e) 1 s(s+a) , (f) s 2−a 2 (s 2+a2) 2 . 25. The velocity potential φ (x, z, t) and the free-surface evaluation η (x, t) for surface waves in water of infinite depth satisfy the Laplace equation φxx + φzz = 0, −∞ <x< ∞,="" −∞="" <="" z="" ≤="" 0,="" t=""> 0, with the free-surface, boundary, and initial conditions φz = ηt on z = 0, t > 0, φt + gη = 0 on z = 0, t > 0, φz → 0 as z → −∞, φ (x, 0, 0) = 0 and η (x, 0) = f (x), −∞ <x< ∞,="" where="" g="" is="" the="" constant="" acceleration="" due="" to="" gravity.="" show="" that="" φ="" (x,="" z,="" t)="−" √g="" √="" 2π=""
="" ∞="" −∞="" k="" −="" 1="" 2="" f="" (k)="" e="" |k|z−ikx="" sin=""
4="" |k|t=""
5="" dk,="" η="" −ikx="" cos="" represents="" fourier="" transform="" variable.="" find="" asymptotic="" solution="" for="" as="" t="" →="" ∞.="" 26.="" use="" method="" of="" one-dimensional="" schr¨odinger="" equation="" a="" free="" particle="" mass="" m,="" i
ψt="−" 2m="" ψxx,="" <x<=""> 0, ψ (x, 0) = f (x), −∞ <x< ∞,="" 526="" 12="" integral="" transform="" methods="" with="" applications="" where="" ψ="" and="" ψx="" tend="" to="" zero="" as="" |x|→∞,="" h="2π
" is="" the="" planck="" constant,="" given="" by="" (x,="" t)="1" √="" 2π=""
="" ∞="" −∞="" f="" (ξ)="" g="" (x="" −="" ξ)="" dξ,="" 2="" γt="" exp="" "="" x="" 4iγt="" #="" green’s="" function="" γ="
" 2m="" .="" 27.="" prove="" following="" properties="" of="" laplace="" convolution:="" (a)="" ∗="" f,="" (b)="" (g="" h)="(f" g)="" h,="" (c)="" (αg="" +="" βh)="α" (f="" β="" h),="" α="" are="" constants,="" (d)="" 0="0" (e)="" d="" dt="" [(f="" (t)]="f" ′="" (t)="" (0)="" (t),="" (f)="" dt2="" ′′="" (g)="" n="" dtn="" (n)="" n−1="" k="0" (k)="" (n−k−1)="" (t).="" 28.="" obtain="" solution="" problem="" utt="c" 2uxx,="" <x<="" t=""> 0, u (x, 0) = f (x), ut (x, 0) = 0, u (0, t)=0, u (x, t) → 0 uniformly in t as x → ∞. 29. Solve utt = c 2uxx, 0 < x < l, t > 0, u (x, 0) = 0, ut (x, 0) = 0, u (0, t) = f (t), u (l, t)=0, t ≥ 0. 30. Solve ut = κuxx, 0 <x< ∞,="" t=""> 0, u (x, 0) = f0, 0 <x< ∞,="" u="" (0,="" t)="f1," (x,="" →="" f0="" uniformly="" in="" t="" as="" x=""> 0. 31. Solve ut = κuxx, 0 <x< ∞,="" t=""> 0, u (x, 0) = x, x > 0, u (0, t)=0, u (x, t) → x uniformly in t as x → ∞, t> 0. 12.18 Exercises 527 32. Solve ut = κuxx, 0 <x< ∞,="" t=""> 0, u (x, 0) = 0, 0 <x< ∞,="" u="" (0,="" t)="t" 2="" ,="" (x,="" →="" 0="" uniformly="" in="" t="" as="" x="" ≥="" 0.="" 33.="" solve="" ut="κuxx" −="" hu,="" <x<=""> 0, h = constant, u (x, 0) = f0, x > 0, u (0, t)=0, ux (0, t) → 0 uniformly in t as x → ∞, t> 0. 34. Solve ut = κuxx, 0 <x< ∞,="" t=""> 0, u (x, 0) = 0, 0 <x< ∞,="" u="" (0,="" t)="f0," (x,="" →="" 0="" uniformly="" in="" t="" as="" x=""> 0. 35. Solve utt = c 2uxx, 0 <x< ∞,="" t=""> 0, u (x, 0) = 0, ut (x, 0) = f0, 0 <x< ∞,="" u="" (0,="" t)="0," ux="" (x,="" →="" 0="" uniformly="" in="" t="" as="" x=""> 0. 36. Solve utt = c 2uxx, 0 <x< ∞,="" t=""> 0, u (x, 0) = f (x), ut (x, 0) = 0, 0 <x< ∞,="" u="" (0,="" t)="0," ux="" (x,="" →="" 0="" uniformly="" in="" t="" as="" x=""> 0. 37. A semi-infinite lossless transmission line has no initial current or potential. A time dependent EMF, V0 (t) H (t) is applied at the end x = 0. Find the potential V (x, t). Then determine the potential for cases: (i) V0 (t) = V0 = constant, and (ii) V0 (t) = V0 cos ωt. 38. Solve the Blasius problem of an unsteady boundary layer flow in a semiinfinite body of viscous fluid enclosed by an infinite horizontal disk at z = 0. The governing equation, boundary, and initial conditions are ∂u ∂t = ν ∂ 2u ∂z2 , z> 0, t > 0, u (z, t) = U t on z = 0, t > 0, u (z, t) → 0 as z → ∞, t > 0, u (z, t) = 0 at t ≤ 0, z > 0. Explain the implication of the solution. 528 12 Integral Transform Methods with Applications 39. The stress-strain relation and equation of motion for a viscoelastic rod in the absence of external force are ∂e ∂t = 1 E ∂σ ∂t + σ η , ∂σ ∂x = ρ ∂ 2u ∂t2 , where e is the strain, η is the coefficient of viscosity, and the displacement u (x, t) is related to the strain by e = ∂u/∂x. Prove that the stress σ (x, t) satisfies the modified wave equation ∂ 2σ ∂x2 − ρ η ∂σ ∂t = 1 c 2 ∂ 2σ ∂t2 , c2 = E/ρ. Show that the stress distribution in a semi-infinite viscoelastic rod subject to the boundary and initial conditions, u˙ (0, t) = U H (t), σ (x, t) → 0 as x → ∞, t > 0, σ (x, 0) = 0, u˙ (x, 0) = 0, is given by σ (x, t) = −U ρc exp  − Et 2η  I0 1 Et 2η  t 2 − x 2 c 2 1 2 3 H
4 t − x c
5 . 40. An elastic string is stretched between x = 0 and x = l and is initially at rest in the equilibrium position. Show that the Laplace transform solution for the displacement field subject to the boundary conditions y (0, t) = f (t) and y (l, t) = 0, t > 0 is y (x, s) = f (s) sinh &s c (l − x) ' sinh sl c . 41. The end x = 0 of a semi-infinite submarine cable is maintained at a potential V0H (t). If the cable has no initial current and potential, determine the potential V (x, t) at point x and at time t. 42. Obtain the solution of the Stokes–Ekman problem (see Debnath, 1995) of an unsteady boundary layer flow in a semi-infinite body of viscous fluid bounded by an infinite horizontal disk at z = 0, when both the fluid and the disk rotate with a uniform angular velocity Ω about the zaxis. The governing boundary layer equation, the boundary conditions, and the initial conditions are ∂q ∂t + 2Ωiq = ν ∂ 2 q ∂z2 , z > 0, t > 0, q (z, t) = aeiωt + be−iωt on z = 0, t > 0, q (z, t) → 0 as z → ∞, t > 0, q (z, t) = 0 at t ≤ 0, for all z > 0, 12.18 Exercises 529 where q = u + iv, is the complex velocity field, ω is the frequency of oscillations of the disk, and a and b are complex constants. Hence, deduce the steady-state solution, and determine the structure of the associated boundary layers. 43. Show that, when ω = 0 in the Stokes–Ekman problem 42, the steady flow field is given by q (z, t) ∼ (a + b) exp / −  2iΩ ν 1 2 z 0 . Hence determine the thickness of the Ekman layer. 44. For problem 14 (e) (iii) in 3.9 Exercises, show that the potential V (x, t) and the current I (x, t) satisfy the partial differential equation  ∂ 2 ∂t2 + 2k ∂ ∂t + k 2  (V, I) = c 2 ∂ 2 ∂x2 (V, I). Find the solution for V (x, t) with the boundary and initial data V (x, t) = V0 (t) at x = 0, t > 0, V (x, t) → 0 as x → ∞, t > 0, V (x, 0) = Vt (x, 0) = 0 for 0 ≤ x < ∞. 45. Use the Laplace transform to solve the Abel integral equation g (t) =
t 0 f ′ (τ ) (t − τ ) −α dτ, 0 <α< 1. 46. Solve Abel’s problem of tautochronous motion described in problem 17 of 14.11 Exercises. 47. The velocity potential φ (r, z, t) and the free-surface elevation η (r, t) for axisymmetric surface waves in water of infinite depth satisfy the equation φrr + 1 r φr + φzz = 0, 0 ≤ r < ∞, −∞ < z ≤ 0, t > 0, with the free-surface, boundary, and initial conditions φz = ηz on z = 0, t > 0, φt + gη = 0, on z = 0, t > 0, φz → 0, z → −∞, φ (r, 0, 0) = 0, and η (r, 0) = f (r), 0 ≤ r < ∞, where g is the acceleration due to gravity and f (r) represents the initial elevation. Show that 530 12 Integral Transform Methods with Applications φ (r, z, t) = − √ g
∞ 0 √ k f5(k) J0 (kr) e kz sin
4 gk t
5 dk, η (r, t) =
∞ 0 k f5(k) J0 (kr) cos
4 gk t
5 dk, where f5(k) is the zero-order Hankel transform of f (r). Derive the asymptotic solution η (r, t) ∼ gt2 2 3 2 r 3 f5  gt2 4r 2  cos  gt2 4r  as t → ∞. 48. Write the solution for the Cauchy–Poisson problem where the initial elevation is concentrated in the neighborhood of the origin, that is, f (r)=(a/2πr) δ (r), where a is the total volume of the fluid displaced. 49. The steady temperature distribution u (r, z) in a semi-infinite solid with z ≥ 0 is governed by the system urr + 1 r ur + uzz = −Aq (r), 0 <r< ∞,="" z=""> 0, u (r, 0) = 0, where A is a constant and q (r) represents the steady heat source. Show that the solution is given by u (r, z) = A
∞ 0 q5(k) J0 (kr) k −1  1 − e −kz dk, where q5(k) is the zero-order Hankel transform of q (r). 50. Find the solution for the small deflection u (r) of an elastic membrane subjected to a concentrated loading distribution which is governed by urr + 1 r ur − κ 2u = 1 2π δ (r) r , 0 ≤ r < ∞, where u and its derivatives vanish as r → ∞. 51. Obtain the solution for the potential v (r, z) due to a flat electrified disk of radius unity with the center of the disk at the origin and the axis along the z-axis. The function v (r, z) satisfies the Laplace equation vrr + 1 r vr − vzz = 0, 0 <r< ∞,="" z=""> 0, with the boundary conditions v (r, 0) = v0, 0 ≤ r < 1, vz (r, 0) = 0, r > 1. 12.18 Exercises 531 52. Prove that the Fourier sine and cosine transforms are linear. 53. If Fs (n) is the Fourier sine transform of f (x) on 0 ≤ x ≤ l, show that Fs [f ′′ (x)] = 2nπ l 2 [f (0) − (−1)n f (l)] −
4nπ l
52 Fs (n). 54. If Fc (n) is the Fourier cosine transform of f (x) on 0 ≤ x ≤ l, show that Fc [f ′′ (x)] = 2 l [(−1)n f ′ (l) − f ′ (0)] −
4nπ l
52 Fc (n). When l = π, show that Fc [f ′′ (x)] = 2 π [(−1)n f ′ (π) − f ′ (0)] − n 2Fc (n). 55. By the transform method, solve ut = uxx + g (x, t), 0 < x < π, t > 0, u (x, 0) = f (x), 0 ≤ x ≤ π, u (0, t)=0, u (π, t) → 0 t > 0. 56. By the transform method, solve ut = uxx + g (x, t), 0 < x < π, t > 0, u (x, 0) = 0, 0 < x < π, u (0, t)=0, ux (π, t) + hu (π, t)=0, t > 0. 57. By the transform method, solve ut = uxx + g (x, t), 0 < x < π, t > 0, u (x, 0) = 0, 0 < x < π, u (0, t)=0, ux (π, t)=0, t > 0. 58. By the transform method, solve ut = uxx − hu, 0 < x < π, t > 0, u (x, 0) = sin x, 0 ≤ x ≤ π, u (0, t)=0, u (π, t)=0, t > 0. 59. By the transform method, solve utt = uxx + h, 0 < x < π, t > 0, h = constant, u (x, 0) = 0, ut (x, 0) = 0, 0 < x < π, ux (0, t)=0, ux (π, t)=0, t> 0. 532 12 Integral Transform Methods with Applications 60. By the transform method, solve utt = uxx + g (x), 0 < x < π, t > 0, u (x, 0) = 0, ut (x, 0) = 0, 0 < x < π, u (0, t)=0, u (π, t)=0, t > 0. 61. By the transform method, solve utt + c 2uxxxx = 0, 0 < x < π, t > 0, u (x, 0) = 0, ut (x, 0) = 0, 0 < x < π, u (0, t)=0, u (π, t)=0, t > 0. uxx (0, t)=0, uxx (π, t) = sin t, t ≥ 0. 62. Find the temperature distribution u (r, t) in a long cylinder of radius a when the initial temperature is constant, u0, and radiation occurs at the surface into a medium with zero temperature. Here u (r, t) satisfies the initial boundary-problem ut = κ  urr + 1 r ur  , 0 ≤ r < a, t > 0, ur + α u = 0 at r = a, t > 0, u (r, 0) = u0 for 0 ≤ r < a, where κ and α are constants. 63. Apply the finite Fourier sine transform to solve the longitudinal displacement field in a uniform bar of length l and cross section A subjected to an external force F A applied at the end x = l. The governing equation and boundary and initial conditions are c 2uxx = utt,  c 2 = E ρ  , 0 < x < l, t > 0, u (0, t)=0 E u (l, t) = F, t > 0, u (x, 0) = ut (x, 0) = 0, 0 < x < l, where E is the constant Young’s modulus, ρ is the density, and F is constant. 64. Use the finite Fourier cosine transform to solve the heat conduction problem ut = κuxx, 0 < x < l, t > 0, ux (x, t) = 0 at x = 0 and x = l, t > 0, u (x, 0) = u0 for 0 < x < l, where u0 and κ are constant. 12.18 Exercises 533 65. Use the Mellin transform to find the solution of the integral equation
∞ −∞ f (x) k (xt) dx = g (t), t > 0. 66. Use the Mellin transform to show the following results: (a) ∞ n=1 f (n) = 1 2πi
c+i∞ c−i∞ ζ (p) F (p) dp, (b) ∞ n=1 f (nx) = M−1 [ζ (p) F (p)], where ζ (s) is the Riemann zeta function defined by (6.7.13). 67. Show that the solution of the boundary-value problem urr + 1 r ur + uzz = 0, r ≥ 0, z> 0, u (r, 0) = u0 for 0 ≤ r ≤ a, u (r, z) → 0 as z → ∞, is u (r, z) = au0
∞ 0 J1 (ak) J0 (kr) e −kzdk. 68. Show that the asymptotic representation of the Bessel function Jn (kr) for large kr is Jn (kr) = 1 π
π 0 cos (nθ − kr sin θ) dθ ∼  2 πkr1 2 cos
4 kr − nπ 2 − π 4
5 . 69. (a) Use the Laplace transform to solve the heat conduction problem ut = κuxx, 0 <x< ∞,="" t=""> 0, u (x, 0) = 0, x> 0, u (0, t) = f (t), u (x, t) → 0 as x → ∞, t> 0. (b) Derive Duhamel’s formula u (x, t) =
t 0 f (t − τ )  ∂u0 ∂τ  dτ, where  ∂u0 ∂t  = x √ 4πκ t −3/2 exp  − x 2 4κt . 13 Nonlinear Partial Differential Equations with Applications “True Laws of Nature cannot be linear.” Albert Einstein “... the progress of physics will to a large extent depend on the progress of nonlinear mathematics, of methods to solve nonlinear equations ... and therefore we can learn by comparing different nonlinear problems.” Werner Heisenberg 13.1 Introduction The three-dimensional linear wave equation utt = c 2 ∇2u, (13.1.1) arises in the areas of elasticity, fluid dynamics, acoustics, magnetohydrodynamics, and electromagnetism. The general solution of the one-dimensional equation (13.1.1) is u (x, t) = φ (x − ct) + ψ (x + ct), (13.1.2) where φ and ψ are determined by the initial or boundary conditions. Physically, φ and ψ represent waves moving with constant speed c and without change of shape, along the positive and the negative directions of x respectively. The solutions φ and ψ correspond to the two factors when the onedimensional equation (13.1.1) is written in the form  ∂ ∂t + c ∂ ∂x ∂ ∂t − c ∂ ∂x u = 0. (13.1.3) 536 13 Nonlinear Partial Differential Equations with Applications Obviously, the simplest linear wave equation is ut + c ux = 0, (13.1.4) and its solution u = φ (x − ct) represents a wave moving with a constant velocity c in the positive x-direction without change of shape. 13.2 One-Dimensional Wave Equation and Method of Characteristics The simplest first-order nonlinear wave equation is given by ut + c (u) ux = 0, −∞ <x< ∞,="" t=""> 0, (13.2.1) where c (u) is a given function of u. We solve this nonlinear equation subject to the initial condition u (x, 0) = f (x), −∞ <x< ∞.="" (13.2.2)="" before="" we="" discuss="" the="" method="" of="" solution,="" following="" comments="" are="" in="" order.="" first,="" unlike="" linear="" differential="" equations,="" principle="" superposition="" cannot="" be="" applied="" to="" find="" general="" solution="" nonlinear="" partial="" equations.="" second,="" effect="" nonlinearity="" can="" change="" entire="" nature="" solution.="" third,="" a="" study="" above="" initial-value="" problem="" reveals="" most="" important="" ideas="" for="" hyperbolic="" waves.="" finally,="" large="" number="" physical="" and="" engineering="" problems="" governed="" by="" system="" or="" an="" extension="" it.="" although="" (13.2.1)–(13.2.2)="" looks="" simple,="" it="" poses="" nontrivial="" mathematics,="" leads="" surprisingly="" new="" phenomena.="" solve="" characteristics.="" order="" construct="" continuous="" solutions,="" consider="" total="" du="" given="" ∂t="" dt="" +="" ∂u="" ∂xdx,="" (13.2.3)="" so="" that="" points="" (x,="" t)="" assumed="" lie="" on="" curve="" Γ.="" then,="" dx="" represents="" slope="" Γ="" at="" any="" point="" p="" thus,="" equation="" becomes="" ="" ="" ux.="" (13.2.4)="" follows="" from="" this="" result="" (13.2.1)="" regarded="" as="" ordinary="" (13.2.5)="" 13.2="" one-dimensional="" wave="" characteristics="" 537="" along="" member="" family="" curves="" which="" (u).="" (13.2.6)="" these="" called="" characteristic="" main="" (13.2.1).="" has="" been="" reduced="" pair="" simultaneous="" equations="" (13.2.6).="" clearly,="" both="" speed="" depend="" u.="" implies="" u="constant" each="" Γ,="" c="" (u)="" remains="" constant="" therefore,="" shows="" form="" straight="" lines="" t)-plane="" with="" indicates="" depends="" finding="" lines.="" also,="" line="" corresponds="" value="" if="" initial="" is="" denoted="" ξ="" one="" intersects="" t="0" x="ξ," then="" (ξ,="" 0)="f" (ξ)="" whole="" shown="" figure="" 13.2.1.="" have="" Γ:="" (u),="" (0)="ξ," (13.2.7)="" (ξ).="" (13.2.8)="" constitute="" coupled="" solved="" independently="" because="" function="" however,="" readily="" obtain="" 13.2.1="" curve.="" 538="" 13="" applications="" hence,="" (ξ),="" (13.2.9)="" where="" f="" (f="" (ξ)).="" (13.2.10)="" integrating="" gives="" ξ.="" (13.2.11)="" whose="" not="" constant,="" but="" combining="" results,="" parametric="" 0="" ,="" (13.2.12)="" next="" verify="" final="" analytic="" expression="" differentiating="" respect="" t,="" ux="f" ′="" ξx,="" ut="f" ξt,="" 1="{1" tf′="" (ξ)}="" 0="F" {1="" ξt.="" elimination="" ξx="" ξt="" .="" (13.2.13)="" since="" (ξ)),="" satisfied="" provided="" 1+tf′="" also="" satisfies="" condition="" unique.="" suppose="" v="" two="" solutions.="" tf="" t).="" proved="" following:="" 539="" theorem="" −∞="" <x<="" ∞,=""> 0, u (x, t) = f (x), at t = 0, −∞ <x< ∞,="" has="" a="" unique="" solution="" provided="" 1+tf′="" (ξ)="0," f="" and="" c="" are="" 1="" (r)="" functions="" where="" (f="" (ξ)).="" the="" is="" given="" in="" parametric="" form:="" u="" (x,="" t)="f" (ξ),="" x="ξ" +="" tf="" (ξ).="" remark:="" when="" (u)="constant" ==""> 0, equation (13.2.1) becomes the linear wave equation (13.1.4). The characteristic curves are x = ct + ξ and the solution u is given by u (x, t) = f (ξ) = f (x − ct). Physical Significance of (13.2.12). We assume c (u) > 0. The graph of u at t = 0 is the graph of f. In view of the fact u (x, t) = u (ξ + tF (ξ), t) = f (ξ) the point (ξ,f (ξ)) moves parallel to the x-axis in the positive direction through a distance tF (ξ) = ct, and the distance moved (x = ξ + ct) depends on ξ. This is a typical nonlinear phenomenon. In the linear case, the curve moves parallel to the x-axis with constant velocity c, and the solution represents waves travelling without change of shape. Thus, there is a striking difference between the linear and the nonlinear solution. Theorem 13.2.1 asserts that the solution of the nonlinear initial-value problem exists provided 1 + tF′ (ξ) = 0, x = ξ + tF′ (ξ). (13.2.14) However, the former condition is always satisfied for sufficiently small time t. By a solution of the problem, we mean a differentiable function u (x, t). It follows from results (13.2.13) that both ux and ut tend to infinity as 1 + tF′ (ξ) → 0. This means that the solution develops a singularity (discontinuity) when 1 + tF′ (ξ) = 0. We consider a point (x, t)=(ξ, 0) so that this condition is satisfied on the characteristics through the point (ξ, 0) at a time t such that t = − 1 F′ (ξ) (13.2.15) 540 13 Nonlinear Partial Differential Equations with Applications which is positive provided F ′ (ξ) = c ′ (f) f ′ (ξ) < 0. If we assume c ′ (f) > 0, the above inequality implies that f ′ (ξ) < 0. Hence, the solution (13.2.12) ceases to exist for all time if the initial data is such that f ′ (ξ) < 0 for some value of ξ. Suppose t = τ is the time when the solution first develops a singularity (discontinuity) for some value of ξ. Then τ = − 1 min−∞<ξ<∞ {c ′ (f) f ′ (ξ)} > 0. We draw the graphs of the nonlinear solution u (x, t) = f (ξ) below for different values of t = 0, τ , 2τ , .... The shape of the initial curve for u (x, t) changes with increasing values of t, and the solution becomes multiplevalued for t ≥ τ . Therefore, the solution breaks down when F ′ (ξ) < 0 for some ξ, and such breaking is a typical nonlinear phenomenon. In linear theory, such breaking will never occur. More precisely, the development of a singularity in the solution for t ≥ τ can be seen by the following consideration. If f ′ (ξ) < 0, we can find two values of ξ = ξ1, ξ2 (ξ1 < ξ2) on the initial line such that the characteristics through them have different slopes 1/c (u1) and 1/c (u2) where u1 = f (ξ1) and u2 = f (ξ2) and c (u2) < c (u1). Thus, these two characteristics will intersect at a point in the (x, t)-plane for some t > 0. Since the characteristics carry constant values of u, the solution ceases to be single-valued at their point of intersection. Figure 13.2.2 shows that the wave profile progressively distorts itself, and at any instant of time there exists an interval on the x-axis, where u assumes three values for a given x. The end result is the development of a nonunique solution, and this leads to breaking. Therefore, when conditions (13.2.14) are violated the solution develops a discontinuity known as a shock. The analysis of shock involves extension of a solution to allow for discontinuities. Also, it is necessary to impose on the solution certain restrictions to be satisfied across its discontinuity. This point will be discussed further in a subsequent section. 13.3 Linear Dispersive Waves We consider a single linear partial differential equation with constant coefficients in the form P  ∂ ∂t, ∂ ∂x, ∂ ∂y , ∂ ∂z  u (x, t)=0, (13.3.1) where P is a polynomial in partial derivatives and x = (x, y, z). We seek an elementary plane wave solution of (13.3.1) in the form u (x, t) = a ei(κ·x−ωt) , (13.3.2) where a is the amplitude, κ = (k,l, m) is the wavenumber vector, ω is the frequency and a, κ, ω are constants. When this plane wave solution is 13.3 Linear Dispersive Waves 541 Figure 13.2.2 The solution u (x, t) for different times t = 0, τ and 2τ ; the characteristics are shown by the dotted lines; two of them from x = ξ1 and x = ξ2 intersect at t>τ . substituted in the equation, ∂/∂t, ∂/∂x, ∂/∂y, and ∂/∂z produce factors −iω, ik, il, and im respectively, and the solution exists provided ω and κ are related by an equation P (−iω, ik, il, im)=0. (13.3.3) This equation is known as the dispersion relation. Evidently, we have a direct correspondence between equation (13.3.1) and the dispersion relation (13.3.3) through the correspondence ∂ ∂t ↔ −iω,  ∂ ∂x, ∂ ∂y , ∂ ∂z  ↔ i(k,l, m). (13.3.4) Equation (13.3.1) and the corresponding dispersion relation (13.3.3) indicate that the former can be derived from the latter and vice-versa by using (13.3.4). The dispersion relation characterizes the plane wave motion. In many problems, the dispersion relation can be written in the explicit form ω = W (k,l, m). (13.3.5) The phase and the group velocities of the waves are defined by Cp (κ) = ω κ κ, 6 (13.3.6) Cg (κ) = ∇κω, (13.3.7) 542 13 Nonlinear Partial Differential Equations with Applications where κ6 is the unit vector in the direction of wave vector κ. In the one-dimensional case, (13.3.5)–(13.3.7) become ω = W (k), Cp = ω k , Cg = dω dk . (13.3.8) The one-dimensional waves given by (13.3.2) are called dispersive if the group velocity Cg ≡ ω ′ (k) is not constant, that is, ω ′′ (k) = 0. Physically, as time progresses, the different waves disperse in the medium with the result that a single hump breaks into wavetrains. Example 13.3.1. (i) Linearized one-dimensional wave equation utt − c 2uxx = 0, ω = + ck. (13.3.9) (ii) Linearized Korteweg and de Vries (KdV) equation for long water waves ut + αux + βuxxx = 0, ω = αk − βk3 . (13.3.10) (iii) Klein–Gordon equation utt − c 2uxx + α 2u = 0, ω = +  c 2 k 2 + α 2 1 2 . (13.3.11) (iv) Schr¨odinger equation in quantum mechanics and de Broglie waves i
ψt −  V −
2 2m ∇2  ψ = 0,
ω =
2κ 2 2m + V, (13.3.12) where V is a constant potential energy, and h = 2π
is the Planck constant. The group velocity of de Broglie wave is (
κ/m), and through the correspondence principle,
ω is to be interpreted as the total energy, 
2κ 2/2m as the kinetic energy, and
κ as the particle momentum. Hence, the group velocity is the classical particle velocity. (v) Equation for vibration of a beam utt + α 2uxxxx = 0, ω = + αk2 . (13.3.13) (vi) The dispersion relation for water waves in an ocean of depth h ω 2 = gk tanh kh, (13.3.14) where g is the acceleration due to gravity. 13.3 Linear Dispersive Waves 543 (vii) The Boussinesq equation utt − α 2∇2u − β 2∇2utt = 0, ω = + ακ  1 + β 2κ 2 . (13.3.15) This equation arises in elasticity for longitudinal waves in bars, long water waves, and plasma waves. (viii) Electromagnetic waves in dielectrics  utt + ω 2 0u utt − c 2 0uxx − ω 2 putt = 0,  ω 2 − ω 2 0 ω 2 − c 2 0k 2 − ω 2 pω 2 = 0, (13.3.16) where ω0 is the natural frequency of the oscillator, c0 is the speed of light, and ωp is the plasma frequency. In view of the superposition principle, the general solution can be obtained from (13.3.2) with the dispersion solution (13.3.3). For the one-dimensional case, the general solution has the Fourier integral representation u (x, t) =
∞ −∞ F (k) e i[kx−tW(k)]dk, (13.3.17) where F (k) is chosen to satisfy the initial or boundary data provided the data are physically realistic enough to have Fourier transforms. In many cases, as cited in Example 13.3.1, there are two modes ω = + W (k) so that the solution (13.3.17) has the form u (x, t) =
∞ −∞ F1 (k) e i[kx−tW(k)]dk +
∞ −∞ F2 (k) e i[kx−tW(k)]dk, (13.3.18) with the initial data at t = 0 u (x, t) = φ (x), ut (x, t) = ψ (x). (13.3.19) The initial conditions give φ (x) =
∞ −∞ [F1 (k) + F2 (k)] e ikxdk, ψ (x) = −i
∞ −∞ [F1 (k) + F2 (k)] W (k) e ikxdk. Applying the Fourier inverse transformations, we have F1 (k) + F2 (k) = Φ(k) = 1 √ 2π
∞ −∞ φ (x) e −ikxdx, −iW (k) [F1 (k) − F2 (k)] = Ψ (k) = 1 √ 2π
∞ −∞ ψ (x) e −ikxdx, 544 13 Nonlinear Partial Differential Equations with Applications so that [F1 (k) + F2 (k)] = 1 2 Φ(k) + i Ψ (k) W (k) . (13.3.20) The asymptotic behavior of u (x, t) for large t with fixed (x/t) can be obtained by the Kelvin stationary phase approximation. For real φ (x), ψ (x), Φ(−k) = Φ ∗ (k) and Ψ (−k) = Ψ ∗ (k), where the asterisk denotes a complex conjugate. It follows from (13.3.20) that, for W (k) even [F1 (−k), F2 (−k)] = [F ∗ 2 (k), F∗ 1 (k)] , (13.3.21) and for W (k) odd, [F1 (−k), F2 (−k)] = [F ∗ 1 (k), F∗ 2 (k)] . (13.3.22) In particular, when φ (x) = δ (x) and ψ (x) ≡ 0, then F1 (k) = F2 (k) = 1/ √ 8π, and the solution (13.3.18) reduces to the form u (x, t) = 2 2 π
∞ 0 cos kx cos {tW (k)} dk. (13.3.23) In order to obtain the asymptotic approximation by the Kelvin stationary phase method (see Section 12.7) for t → ∞, we consider both cases when W (k) is even (W′ (k) is odd) and when W (k) is odd (W′ (k) is even) and make an extra reasonable assumption that W′ (k) is monotonic and positive for k > 0. It turns out that the asymptotic solution for t → ∞ is u (x, t) ∼ 2 Re / F1 (k)  2π t|W′′ (k)| 01 2 exp " i ( θ (x, t) − π 4 sgn W′′ (k) )#0 + O  1 t  , = Re " a (x, t) e iθ(x,t) # , (13.3.24) where k (x, t) is the positive root of the equation W′ (k) = x t , ω = W (k), x t > 0, (13.3.25ab) θ (x, t) = x k (x, t) − t ω (x, t), (13.3.26) and a (x, t)=2F1 (k)  2π t|W′′ (k)| 01 2 exp  − iπ 4 sgn W′′ (k) 0 . (13.3.27) It is important to point out that solution (13.3.24) has a form similar to that of the elementary plane wave solution, but k, ω, and a are no 13.4 Nonlinear Dispersive Waves and Whitham’s Equations 545 longer constants; they are functions of space variable x and time t. The solution still represents an oscillatory wavetrain with the phase function θ (x, t) describing the variations between local maxima and minima. Unlike the elementary plane wavetrain, the present asymptotic result (13.3.24) represents a nonuniform wavetrain in the sense that the amplitude, the distance, and the time between successive maxima are not constant. It also follows from (13.3.25a) that kt k = − W′ (k) kW′′ (k)t 1 t ∼ O  1 t  , (13.3.28) kx k = − 1 kW′′ (k) 1 t ∼ O  1 t  . (13.3.29) These results indicate the k (x, t) is a slowly varying function of x and t as t → ∞. Applying a similar argument to ω and a, we conclude that k (x, t), ω (x, t), and a (x, t) are slowly varying functions of x and t as t → ∞. Finally, all these results seem to provide an important clue for natural generalization of the concept of nonlinear and nonuniform wavetrains. 13.4 Nonlinear Dispersive Waves and Whitham’s Equations To describe a slowly varying nonlinear and nonuniform oscillatory wavetrain in a medium (see Whitham, 1974), we assume the existence of a solution in the form (13.3.24) so that u (x, t) = a (x, t) e iθ(x,t) + c.c., (13.4.1) where c.c. stands for the complex conjugate, a (x, t) is the complex amplitude given by (13.3.27), and the phase function θ (x, t) is θ (x, t) = x k (x, t) − t ω (x, t), (13.4.2) and k, ω, and a are slowly varying function of x and t. Due to slow variations of k and ω, it is reasonable to assume that these quantities still satisfy the dispersion relation ω = W (k). (13.4.3) Differentiating (13.4.2) with respect to x and t respectively, we obtain θx = k + {x − t W′ (k)} kx, (13.4.4) θt = −W (k) + {x − t W′ (k)} kt. (13.4.5) In the neighborhood of stationary points defined by (13.3.25a), these results become 546 13 Nonlinear Partial Differential Equations with Applications θx = k (x, t), θt = −ω (x, t). (13.4.6) These results can be used as a definition of local wavenumber and local frequency of the slowly varying nonlinear wavetrain. In view of (13.4.6), relation (13.4.3) gives a nonlinear partial differential equation for the phase θ in the form ∂θ ∂t + W  ∂θ ∂x = 0. (13.4.7) The solution of this equation determines the geometry of the wave pattern. However, it is convenient to eliminate θ from (13.4.6) to obtain ∂k ∂t + ∂ω ∂t = 0. (13.4.8) This is known as the Whitham equation for the conservation of waves, where k represents the density of waves and ω is the flux of waves. Using the dispersion relation (13.4.3), we obtain ∂k ∂t + Cg (k) ∂k ∂x = 0, (13.4.9) where Cg (k) = W′ (k) is the group velocity. This represents the simplest nonlinear wave (hyperbolic) equation for the propagation of k with the group velocity Cg (k). Since equation (13.4.9) is similar to (13.2.1), we can use the analysis of Section 13.2 to find the general solution of (13.4.9) with the initial condition k (x, 0) = f (x) at t = 0. In this case, the solution has the form k (x, t) = f (ξ), x = ξ + tF (ξ), (13.4.10) where F (ξ) = Cg (f (ξ)). This further confirms the propagation of k with the velocity Cg. Some physical interpretations of this kind of solution have already been discussed in Section 13.2. Equations (13.4.9) and (13.4.3) reveal that ω also satisfies the nonlinear wave (hyperbolic) equation ∂ω ∂t + W′ (k) ∂ω ∂x = 0. (13.4.11) It follows from equations (13.4.9) and (13.4.11) that both k and ω remain constant on the characteristic curves defined by dx dt = W′ (k) = Cg (k), (13.4.12) in the (x, t) plane. Since k and ω is constant on each curve, the characteristic curves are straight lines with slope Cg (k). The solution for k is given by (13.4.10). 13.4 Nonlinear Dispersive Waves and Whitham’s Equations 547 Finally, it follows from the above analysis that any constant value of the phase θ propagates according to θ (x, t) = constant, and hence, θt +  dx dt  θx = 0, (13.4.13) which gives, by (13.4.6), dx dt = − θt θx = ω k = Cp. (13.4.14) Thus, the phase of the waves propagates with the phase speed Cp. On the other hand, (13.4.9) ensures that the wavenumber k propagates with the group velocity Cg (k)=(dω/dk) = W′ (k). We next investigate how the wave energy propagates in the dispersive medium. We consider the following integral involving the square of the wave amplitude (energy) given by (13.3.24) between any two points x = x1 and x = x2 (0 < x1 < x2) Q (t) =
x2 x1 |a| 2 dx =
x2 x1 aa∗ dx, (13.4.15) = 8π
x2 x1 F1 (k) F ∗ 1 (k) t|W′′ (k)| dx, (13.4.16) which is, due to a change of variable x = t W′ (k), = 8π
k2 k1 F1 (k) F ∗ 1 (k) dk, (13.4.17) where kr = t W′ (kr), r = 1, 2. When kr is kept fixed as t varies, Q (t) remains constant so that 0 = dQ dt = d dt
x2 x1 |a| 2 dx, =
x2 x1 ∂ ∂t |a| 2 dx + |a| 2 2 W′ (k2) − |a| 2 1 W′ (k1). (13.4.18) In the limit x2 − x1 → 0, this result reduces to the partial differential equation ∂ ∂t |a| 2 + ∂ ∂x " W′ (k)|a| 2 # = 0. (13.4.19) This represents the equation for the conservation of wave energy, where |a| 2 and |a| 2 W′ (k) are the energy density and energy flux respectively. It also follows that the energy propagates with the group velocity W′ (k). It has been shown that the wavenumber k also propagates with the group velocity. Evidently, the group velocity plays a double role. 548 13 Nonlinear Partial Differential Equations with Applications The above analysis reveals another important fact; equations (13.4.3), (13.4.8), and (13.4.19) constitute a closed set of equations for the three quantities k, ω, and a. Indeed, these are the fundamental equations for nonlinear dispersive waves and are known as Whitham’s equations. 13.5 Nonlinear Instability For infinitesimal waves, the wave amplitude (ak ≪ 1) is very small, so that nonlinear effects can be neglected altogether. However, for finite amplitude waves the terms involving a 2 cannot be neglected, and the effects of nonlinearity become important. In the theory of water waves, Stokes first obtained the connection due to inherent nonlinearity between the waveprofile and the frequency of a steady periodic wave system. According to the Stokes theory, the remarkable fact is the dependence of ω on a which couples (13.4.8) to (13.4.19). This leads to a new nonlinear phenomenon. For finite amplitude waves, the frequency ω has the Stokes expansion ω = ω0 (k) + a 2ω2 (k) + ... = ω  k, a2 . (13.5.1) This can be regarded as the nonlinear dispersion relation which depends on both k and a 2 . In the linear case, the amplitude a → 0, (13.5.1) gives the linear dispersion relation (13.4.3), that is, ω = ω0 (k). In order to discuss nonlinear instability, we substitute (13.5.1) into (13.4.8) and retain (13.4.19) in the linear approximation to obtain the following coupled system: ∂k ∂t + ∂ ∂x + & ω0 (k) + ω2 (k) a 2 ' = 0, (13.5.2) ∂a2 ∂t + ∂ ∂x & ω ′ 0 (k) a 2 ' = 0, (13.5.3) where W (k) ≡ ω0 (k). These equations can be further approximated to obtain ∂k ∂t + ω ′ 0 ∂k ∂x + ω2 ∂a2 ∂x = O  a 2 , (13.5.4) ∂a2 ∂t + ω ′ 0 ∂a2 ∂x + ω ′′ 0 a 2 ∂k ∂x = 0. (13.5.5) In matrix form, these equations read ⎛ ⎝ ω ′ 0 ω2 ω ′′ 0 a 2 ω ′ 0 ⎞ ⎠ ⎛ ⎝ ∂k ∂x ∂a2 ∂x ⎞ ⎠ + ⎛ ⎝ 1 0 0 1 ⎞ ⎠ ⎛ ⎝ ∂k ∂t ∂a2 ∂t ⎞ ⎠ = 0. (13.5.6) Hence, the eigenvalues λ are the roots of the determinant equation 13.6 The Traffic Flow Model 549 |aij − λ bij | =       ω ′ 0 − λ ω2 ω ′′ 0 a 2 ω ′ 0 − λ       = 0, (13.5.7) where aij and bij are the coefficient matrices of (13.5.6). This equation gives the characteristic velocities λ = dx dt = ω ′ 0 + (ω2ω ′′ 0 ) 1 2 a + O  a 2 . (13.5.8) If ω2ω ′′ 0 > 0, the characteristics are real and the system is hyperbolic. The double characteristic velocity splits into two separate real velocities. This provides a new extension of the group velocity to nonlinear problems. If the disturbance is initially finite in extent, it would eventually split into two disturbances. In general, any initial disturbance or modulating source would introduce disturbances in both families of characteristics. In the hyperbolic case, compressive modulation will progressively distort and steepen so that the question of breaking will arise. These results are remarkably different from those found in linear theory, where there is only one characteristic velocity and any hump may distort, due to the dependence of ω ′ 0 (k) on k, but would never split. On the other hand, if ω2ω ′′ 0 < 0, the characteristics are complex and the system is elliptic. This leads to ill-posed problems. Any small perturbations in k and a will be given by the solutions of the form exp [iα (x − λt)] where λ is calculated from (13.5.8) for unperturbed values of k and a. In this elliptic case, λ is complex, and the perturbation will grow as t → ∞. Hence, the original wavetrain will become unstable. In the linear theory, the elliptic case does not arise at all. Example 13.5.1. For Stokes waves in deep water, the dispersion relation is ω = (gk) 1 2  1 + 1 2 k 2 a 2  , (13.5.9) so that ω0 (k)=(gk) 1 2 and ω2 (k) = 1 2 √g k 3 2 . In this case, ω ′′ 0 (k) = − 1 4 √g k− 3 2 so that ω ′′ 0ω2 = − g 8 k < 0. The conclusion is that Stokes waves in deep water are definitely unstable. This is one of the most remarkable results in the theory of nonlinear water waves discovered during the 1960’s. 13.6 The Traffic Flow Model We consider the flow of cars on a long highway under the assumptions that cars do not enter or leave the highway at any one of its points. We take the x-axis along the highway and assume that the traffic flows in the positive 550 13 Nonlinear Partial Differential Equations with Applications direction. Suppose ρ (x, t) is the density representing the number of cars per unit length at the point x of the highway at time t, and q (x, t) is the flow of cars per unit time. We assume a conservation law which states that the change in the total amount of a physical quantity contained in any region of space must be equal to the flux of that quantity across the boundary of that region. In this case, the time rate of change of the total number of cars in any segment x1 ≤ x ≤ x2 of the highway is given by d dt
x2 x1 ρ (x, t) dx =
x2 x1 ∂ρ ∂t dx. (13.6.1) This rate of change must be equal to the net flux across x1 and x2 given by q (x1, t) − q (x2, t) (13.6.2) which measures the flow of cars entering the segment at x1 minus the flow of cars leaving the segment at x2. Thus, we have the conservation equation d dt
x2 x1 ρ (x, t) dx = q (x1, t) − q (x2, t), (13.6.3) or
x2 x1 ∂ρ ∂t dx = −
x2 x1 ∂q ∂x dx, or
x2 x1  ∂ρ ∂t + ∂q ∂x dx = 0. (13.6.4) Since the integrand in (13.6.4) is continuous, and (13.6.4) holds for every segment [x1, x2], it follows that the integrand must vanish so that we have the partial differential equation ∂ρ ∂t + ∂q ∂x = 0. (13.6.5) We now introduce an additional assumption which is supported by both theoretical and experimental findings. According to this assumption, the flow rate q depends on x and t only through the density, that is, q = Q (ρ) for some function Q. This assumption seems to be reasonable in the sense that the density of cars surrounding a given car indeed controls the speed of that car. The functional relation between q and ρ depends on many factors, including speed limits, weather conditions, and road characteristics. Several specific relations are suggested by Haight (1963). We consider here a particular relation q = ρ v where v is the average local velocity of cars. We assume that v is a function of ρ to a first approximation. In view of this relation, (13.6.5) reduces to the nonlinear hyperbolic equation 13.6 The Traffic Flow Model 551 ∂ρ ∂t + c (ρ) ∂ρ ∂x = 0, (13.6.6) where c (ρ) = q ′ (ρ) = v + ρ v′ (ρ). (13.6.7) In general, the local velocity v (ρ) is a decreasing function of ρ so that v (ρ) has a finite maximum value vmax at ρ = 0 and decreases to zero at ρ = ρmax = ρm. For the value of ρ = ρm, the cars are bumper to bumper. Since q = ρ v, q (ρ) = 0 when ρ = 0 and ρ = ρm. This means that q is an increasing function of ρ until it attains a maximum value qmax = qM for some ρ = ρM and then decreases to zero at ρ = ρm. Both q (ρ) and v (ρ) are shown in Figure 13.6.1. Equation (13.6.6) is similar to (13.2.1) with the wave propagation velocity c (ρ) = v (ρ) + ρ v′ (ρ). Since v ′ (ρ) < 0, c (ρ) < v (ρ), that is, the propagation velocity is less than the car velocity. In other words, waves propagate backwards through the stream of cars, and drivers are warned of disturbances ahead. It follows from Figure 13.6.1a that q (ρ) is an increasing function in [0, ρM], a decreasing function in [ρM, ρm], and attains a maximum at ρM. Hence, c (ρ) = q ′ (ρ) is positive in [0, ρM], zero at ρM and negative in [ρM, ρm]. All these mean that waves propagate forward relative to the highway in [0, ρM], are stationary at ρM, and then travel backwards in [ρM, ρm]. We use Section 13.1 to solve the initial-value problem for the nonlinear equation (13.6.6) with the initial condition ρ (x, 0) = f (x). The solution is ρ (x, t) = f (ξ), x = ξ + tF (ξ), (13.6.8) where F (ξ) = c (f (ξ)). Figure 13.6.1 Graphs of q (ρ) and v (ρ). 552 13 Nonlinear Partial Differential Equations with Applications Since c ′ (ρ) = q ′′ (ρ) < 0, q (ρ) is convex, and c (ρ) is a decreasing function of ρ. This means that breaking occurs at the left due to formation of shock at the back. Waves propagate slower than the cars, so drivers enter such a local density increase from behind; they must decelerate rapidly through the shock but speed up slowly as they get out from the crowded area. These conclusions are in accord with observational results. Actual observational data of traffic flow indicate that a typical result on a single lane highway is ρm ≈ 225 cars per mile, ρM ≈ 80 cars per mile, and qM ≈ 1590 cars per hour. Thus, the maximum flow rate qM occurs at a low velocity v = qM/ρM ≈ 20 miles per hour. 13.7 Flood Waves in Rivers We consider flood waves in a long rectangular river of constant breadth. We take the x-axis along the river which flows in the positive x-direction and assume that the disturbance is approximately the same across the breadth. In this problem, the depth h (x, t) of the river plays the role of density in the traffic flow model discussed in Section 13.6. Let q (x, t) be the flow per unit breadth and per unit time. According to the Conservation Law, the rate of change of the mass of the fluid in any section x1 ≤ x ≤ x2 must be balanced by the net flux across x2 and x1 so that the conservation equation becomes d dt
x2 x1 h (x, t) dx + q (x2, t) − q (x1, t)=0. (13.7.1) An argument similar to the previous section gives ∂h ∂t + ∂q ∂x = 0. (13.7.2) Although the fluid flow is extremely complicated, we assume a simple function relation q = Q (h) as a first approximation to express the increase in flow as the water level arises. Thus, equation (13.7.2) becomes ht + c (h) hx = 0, (13.7.3) where c (h) = Q′ (h) and Q (h) is determined from the balance between the gravitational force and the frictional force of the river bed. This equation is similar to (13.2.1) and the method of solution has already been obtained in Section 13.2. Here we discuss the velocity of wave propagation for some particular values of Q (h). One such result is given by the Chezy result as Q (h) = hv, (13.7.4) 13.8 Riemann’s Simple Waves of Finite Amplitude 553 where v = α √ h is the velocity of fluid flow and α is a constant, so that the propagation velocity of flood waves is given by c (h) = Q ′ (h) = 3 2 α √ h = 3 2 v. (13.7.5) Thus, the flood waves propagate one and a half times faster than the stream velocity. For a general case where v = αhn, Q (h) = hv = αhn+1 , (13.7.6) so the propagation velocity of flood waves is c (h) = Q ′ (h)=(n + 1) v. (13.7.7) This result also indicates that flood waves propagate faster than the fluid. 13.8 Riemann’s Simple Waves of Finite Amplitude We consider a one-dimensional unsteady isentropic flow of gas of density ρ and pressure p with the direction of motion along the x-axis. Suppose u (x, t) is the x-component of the velocity at time t and A is an areaelement of the (y, z)-plane. The volume of the rectangular cylinder of height dx standing on the element A is then A dx and its mass A ρt dx dt is determined by the mass entering it, which is equal to −A (∂/∂x) (ρu) dx dt. Its acceleration is (ut + uux) and the force impelling it in the positive xdirection is −pxA dx = −c 2ρxA dx, where p = f (ρ) and c 2 = f ′ (ρ). These results lead to two coupled nonlinear partial differential equations ∂ρ ∂t + ∂ ∂x (ρu)=0, (13.8.1) (ut + uux) + c 2 ρ ρx = 0. (13.8.2) In matrix form, this system is A ∂U ∂x + I ∂U ∂t = 0, (13.8.3) where U, A and I are matrices given by U = ⎛ ⎝ ρ u ⎞ ⎠ , A = ⎛ ⎝ u ρ c 2/ρ u ⎞ ⎠ and I = ⎛ ⎝ 1 0 0 1 ⎞ ⎠ . (13.8.4) The concept of characteristic curves introduced briefly in Section 13.2 requires generalization if it is to be applied to quasi-linear systems of firstorder partial differential equations (13.8.1)–(13.8.2). 554 13 Nonlinear Partial Differential Equations with Applications It is of interest to determine how a solution evolves with time t. Hence, we leave the time variable unchanged and replace the space variable x by some arbitrary curvilinear coordinate ξ so that the semi-curvilinear coordinate transformation from (x, t) to (ξ, t′ ) can be introduced by ξ = ξ (x, t), t′ = t. (13.8.5) If the Jacobian of this transformation is nonzero, we can transform (13.8.3) by the following correspondence rule: ∂ ∂t ≡ ∂ξ ∂t ∂ ∂ξ + ∂t′ ∂t · ∂ ∂t′ = ∂ξ ∂t ∂ ∂ξ + ∂ ∂t′ , ∂ ∂x ≡ ∂ξ ∂x ∂ ∂ξ + ∂t′ ∂x ∂ ∂t′ = ∂ξ ∂x ∂ ∂ξ . This rule transforms (13.8.3) into the form I ∂U ∂t′ +  ∂ξ ∂t I + ∂ξ ∂xA  ∂U ∂ξ = 0. (13.8.6) This equation can be used to determine ∂U/∂ξ provided that the determinant of its coefficient matrix is non-zero. Obviously, this condition depends on the nature of the curvilinear coordinate curves ξ (x, t) = constant, which has been kept arbitrary. We assume now that the determinant vanishes for the particular choice ξ = η so that     ∂η ∂t I + ∂η ∂x A     = 0. (13.8.7) In view of this, ∂U/∂η will become indeterminate on the family of curves η = constant, and hence, ∂U/∂η may be discontinuous across the curves η = constant. This implies that each element of ∂U/∂η will be discontinuous across any of the curves η = constant. It is then necessary to find out how these discontinuities in the elements of ∂U/∂η are related across the curve η = constant. We next consider the solutions U which are everywhere continuous with discontinuous derivatives ∂U/∂η across the particular curve η = constant = η0. Since U is continuous, elements of the matrix A are not discontinuous across η = η0 so that A can be determined in the neighborhood of a point P on η = η0. And since ∂U/∂t′ is continuous everywhere, it is continuous across the curve η = η0 at P. In view of all of the above facts, it follows that differential equation (13.8.6) across the curve ξ = η = η0 at P becomes  ∂η ∂t I + ∂η ∂x A  P ∂U ∂η P = 0, (13.8.8) where [f]P = f (P+)−f (P−) denotes the discontinuous jump in the quantity f across the curve η = η0, and f (P−) and f (P+) represent the values 13.8 Riemann’s Simple Waves of Finite Amplitude 555 to the immediate left and immediate right of the curve at P. Since P is any arbitrary point on the curve, ∂/∂η denotes the differentiation normal to the curves η = constant so that equation (13.8.8) can be regarded as the compatibility condition satisfied by ∂U/∂η on either side of and normal to these curves in the (x, t)-plane. Obviously, equation (13.8.8) is a homogeneous system of equations for the two jump quantities [∂U/∂η]. Therefore, for the existence of a nontrivial solution, the coefficient determinant must vanish, that is,     ∂η ∂t I + ∂η ∂xA     = 0. (13.8.9) However, along the curves η = constant, we have 0 = dη = ηt +  dx dt  ηx, (13.8.10) so that these curves have the constant slope, λ dx dt = − ηt ηx = λ. (13.8.11) Consequently, equations (13.8.9) and (13.8.8) can be expressed in terms of λ in the form |A − λI| = 0, (13.8.12) (A − λI) ∂U ∂η = 0, (13.8.13) where λ represents the eigenvalues of the matrix A, and [∂U/∂η] is proportional to the corresponding right eigenvector of A. Since A is a 2 × 2 matrix, it must have two eigenvalues. If these are real and distinct, integration of (13.8.11) leads to two distinct families of real curves Γ1 and Γ2 in the (x, t)-plane: Γr : dx dt = λr, r = 1, 2. (13.8.14) The families of curves Γr are called the characteristic curves of the system (13.8.3). Any one of these families of curves Γr may be chosen for the curvilinear coordinate curves η = constant. The eigenvalues λr have the dimensions of velocity, and the λr associated with each family will then be the velocity of propagation of the matrix column vector [∂U/∂η] along the curves Γr belonging to that family. In this particular case, the eigenvalues λ of the matrix A are determined by (13.8.12), that is,       u − λ ρ c 2/ρ u − λ       = 0, (13.8.15) 556 13 Nonlinear Partial Differential Equations with Applications so that λ = λr = u+c, r = 1, 2. (13.8.16) Consequently, the families of the characteristic curves Γr (r = 1, 2) defined by (13.8.14) become Γ1 : dx dt = u + c, and Γ2 : dx dt = u − c. (13.8.17) In physical terms, these results indicate that disturbances propagate with the sum of the velocities of the fluid and sound along the family of curves Γ1. In the second family Γ2, they propagate with the difference of the fluid velocity u and the sound velocity c. The right eigenvectors µr ≡ ⎛ ⎜⎝ µ (1) r µ (2) r ⎞ ⎟⎠ are solutions of the equations (A − λrI) µr = 0, r = 1, 2, (13.8.18) or, ⎛ ⎝ u − λr ρ c 2/ρ u − λr ⎞ ⎠ ⎛ ⎜⎝ µ (1) r µ (2) r ⎞ ⎟⎠ = 0, r = 1, 2. (13.8.19) This result combined with (13.8.13) gives ⎛ ⎝ [ρη] [uη] ⎞ ⎠ = ⎛ ⎜⎝ µ (1) r µ (2) r ⎞ ⎟⎠ = α ⎛ ⎝ 1 + c/ρ ⎞ ⎠ , r = 1, 2, (13.8.20) where α is a constant. In other words, across a wavefront in the Γ1 family of characteristic curves, [∂ρ/∂η] 1 = [∂u/∂η] c/ρ , (13.8.21) and across a wavefront in the Γ2 family of characteristic curves, [∂ρ/∂η] 1 = [∂u/∂η] −c/ρ , (13.8.22) where c and ρ have values appropriate to the wavefront. The above method of characteristics can be applied to a more general system 13.8 Riemann’s Simple Waves of Finite Amplitude 557 ∂U ∂t + A ∂U ∂x = 0, (13.8.23) where U is an n × 1 matrix with elements u1, u2, ..., un and A is an n × n matrix with elements aij . An argument similar to that given above leads to n eigenvalues of (13.8.13). If these eigenvalues are real and distinct, integration of equations (13.8.14) with r = 1, 2, ..., n gives n distinct families of real curves Γr in the (x, t)-plane so that Γr : dx dt = λr, r = 1, 2, . . . n. (13.8.24) When the eigenvalues λr of A are all real and distinct, there are n distinct linearly independent right eigenvectors µr of A satisfying the equation Aµr = λrµr, where µr is an n × 1 matrix with elements µ (1) r , µ (2) r , ..., µ (n) r . Then across a wavefront belonging to the Γr family of characteristics, it turns out that [∂u1/∂η] µ (1) r = [∂u2/∂η] µ (2) r = ... = [∂un/∂η] µ (n) r , (13.8.25) where the elements of µr are known on the wavefront. In order to introduce the Riemann invariants, we form the linear combination of the eigenvectors (+c/ρ, 1) with equations (13.8.1)–(13.8.2) to obtain + c ρ (ρt + ρux + uρx) +  ut + uux + c 2 ρ ρx  = 0. (13.8.26) We use ∂u/∂ρ = + c/ρ from (13.8.21)–(13.8.22) and rewrite (13.8.26) as + c ρ [ρt + (u + c) ρx]+[ut + (u + c) ux]=0. (13.8.27) In view of (13.8.17), equation (13.8.27) becomes du + c ρ dρ = 0 on Γr, r = 1, 2, (13.8.28) or, d [F (ρ) + u] = 0 on Γr, (13.8.29) where F (ρ) =
ρ ρ0 c (ρ) ρ dρ. (13.8.30) Integration of (13.8.29) gives 558 13 Nonlinear Partial Differential Equations with Applications F (ρ) + u = 2r on Γ1 and F (ρ) − u = 2s on Γ2, (13.8.31) where 2r and 2s are constants of integration on Γ1 and Γ2, respectively. The quantities r and s are called the Riemann invariants. As stated above, r is an arbitrary constant on characteristics Γ1, and hence, in general, r will vary on each Γ2. Similarly, s is constant on each Γ2 but will vary on Γ1. It is natural to introduce r and s as new curvilinear coordinates. Since r is constant on Γ1, s can be treated as the parameter on Γ1. Similarly, r can be regarded as the parameter on Γ2. Then, dx = (u + c) dt on Γr implies that dx ds = (u + c) dt ds on Γ1, (13.8.32) dx dr = (u − c) dt dr on Γ2. (13.8.33) In fact, r is a constant on Γ1, and s is a constant on Γ2. Therefore, the derivatives in the two equations are really partial derivations with respect to s and r so that we can rewrite them as ∂x ∂s = (u + c) ∂t ∂s , (13.8.34) ∂x ∂r = (u − c) ∂t ∂r . (13.8.35) These two first-order PDE’s can, in general, be solved for x = x (r, s), t = t(r, s), and then, by inversion, r and s as functions x and t can be obtained. Once this is done, we use (13.8.31) to determine u (x, t) and ρ (x, t) in terms of r and s as u (x, t) = r − s, F (ρ) = r + s. (13.8.36) When one of the Riemann invariants r and s is constant throughout the flow, the corresponding solution is tremendously simplified. The solutions are known as simple wave motions representing simple waves in one direction only. The generating mechanisms of simple waves with their propagation laws can be illustrated by the piston problem in gas dynamics. Example 13.8.1. Determine the Riemann invariants for a polytropic gas characterized by the law p = kργ , where k and γ are constants. In this case c 2 = dp dρ = kγργ−1 , F (ρ) =
ρ 0 c (ρ) ρ = 2c γ − 1 . Hence, the Riemann invariants are given by  2c γ − 1  c + u = (2r, 2s) on Γr. (13.8.37) 13.8 Riemann’s Simple Waves of Finite Amplitude 559 It is also possible to express the dependent variables u and c in terms of the Riemann invariants. It turns out that u = r − s, c = γ − 1 2 (r + s). (13.8.38) Example 13.8.2. (The Piston Problem in a Polytropic Gas). The problem is to determine how a simple wave is produced by the prescribed motion of a piston in the closed end of a semi-infinite tube filled with gas. This is a one-dimensional unsteady problem in gas dynamics. We assume that the gas is initially at rest with a uniform state u = 0, ρ = ρ0, and c = c0. The piston starts from rest at the origin and is allowed to withdraw from the tube with a variable velocity for a time t1, after which the velocity of withdrawal remains constant. The piston path is shown by a dotted curve in Figure 13.8.1. In the (x, t)-plane, the path of the piston is given by x = X (t) with X (0) = 0. The fluid velocity u is equal to the piston velocity X˙ (t) on the piston x = X (t), which will be used as the boundary condition for the piston. The initial state of the gas is given by u = u0, ρ = ρ0, and c = c0 at t = 0, in x ≥ 0. The characteristic line Γ0 that bounds it and passes through the origin is determined by the equation dx dt = (u + c) t=0 = c0 so that the equation of the characteristic line Γ0 is x = c0t. Figure 13.8.1 Simple waves generated by the motion of a piston. 560 13 Nonlinear Partial Differential Equations with Applications In view of the uniform initial state, all the Γ2 characteristics start on the x-axis so that the Riemann invariants s in (12.8.37b) must be constant and of the form 2c γ − 1 − u = 2c0 γ − 1 , (13.8.39) or, u = 2 (c − c0) γ − 1 , c = c0 + (γ − 1) 2 u. (13.8.40ab) The characteristics Γ1 meeting the piston are given by 2c γ − 1 + u = 2r on each Γ1 and Γ1 : dx dt = u + c, (13.8.41) which is, since (13.8.40ab) holds everywhere, u = constant on Γ1 and Γ1 : dx dt = c0 + 1 2 (γ + 1) u. (13.8.42) Since the flow is continuous with no shocks, u = 0 and c = c0 ahead of and on Γ0, which separates those Γ1 meeting the x-axis from those meeting the piston. The family of lines Γ1 through the origin has the equation (dx/dt) = ξ, where ξ is a parameter with ξ = c0 on Γ0. The Γ1 characteristics are also defined by (dx/dt) = u+c so that ξ = u+c. Hence, elimination of c from (13.8.40b) gives u =  2 γ + 1 (ξ − c0). (13.8.43) Substituting this value of u in (13.8.40b), we obtain c =  γ − 1 γ + 1 ξ + 2c0 γ + 1 . (13.8.44) It follows from c 2 = γkργ−1 and (13.8.40b) with the initial data, ρ = ρ0, c = c0 that ρ = ρ0 1 + γ − 1 2c0 u 2/(γ−1) . (13.8.45) With ξ = (x/t), results (13.8.43) through (13.8.45) give the complete solution of the piston problem in terms of x and t. Finally, the equation of the characteristic line Γ1 is found by integrating the second equation of (13.8.42) and using the boundary condition on the piston. When a line Γ1 intersects the piston path at time t = τ , then u = X˙ (τ ) along it, and the equation becomes 13.9 Discontinuous Solutions and Shock Waves 561 x = X (τ ) +  c0 + γ + 1 2 X˙ (τ ) 0 (t − τ ). (13.8.46) It is noted that the family Γ1 represents straight lines with slope dx/dt increasing with velocity u. Consequently, the characteristics are likely to overlap on the piston, that is, X˙ (τ ) > 0 for any τ . If u increases, so do c, ρ, and p so that instability develops. It shows that shocks will be formed in the compressive part of the disturbance. 13.9 Discontinuous Solutions and Shock Waves The development of a nonunique solution of a nonlinear hyperbolic equation has already been discussed in connection with several different problems. In real physical situations, this nonuniqueness usually manifests itself in the formation of discontinuous solutions which propagate in the medium. Such discontinuous solutions across some surface are called shock waves. These waves are found to occur widely in high speed flows in gas dynamics. In order to investigate the nature of discontinuous solutions, we reconsider the nonlinear conservation equation (13.6.5) that is, ∂ρ ∂t + ∂q ∂x = 0. (13.9.1) This equation has been solved under two basic assumptions: (i) There exists a functional relation between q and ρ, that is, q = Q (ρ); (ii) ρ and q are continuously differentiable. In some physical situations, the solution of (13.9.1) leads to breaking phenomenon. When breaking occurs, questions arise about the validity of these assumptions. To examine the formation of discontinuities, we consider the following: (a) we assume the relation q = Q (ρ) but allow jump discontinuity for ρ and q; (b) in addition to the fact that ρ and q are continuously differentiable, we assume that q is a function of ρ and ρx. One of the simplest forms is q = Q (ρ) − νρx, ν > 0. (13.9.2) In case (a), we assume the conservation equation (13.6.1) still holds and has the form d dt
x2 x1 ρ (x, t) dx + q (x2, t) − q (x1, t)=0. (13.9.3) We now assume that there is a discontinuity at x = s (t) where s is a continuously differentiable function of t, and x1 and x2 are chosen so that x2 > s (t) > x1, and U (t)= ˙s (t). Equation (13.9.3) can be written as d dt 1
s − x1 ρ dx +
x2 s+ ρ dx3 + q (x2, t) − q (x1, t)=0, 562 13 Nonlinear Partial Differential Equations with Applications which implies that
s − x1 ρt dx + ˙sρ  s −, t +
x2 s+ ρt dx − sρ˙  s +, t + q (x2, t) − q (x1, t)=0, (13.9.4) where ρ (s −, t), ρ (s +, t) are the values of ρ (x, t) as x → s from below and above respectively. Since ρt is bounded in each of the intervals separately, the integrals tend to zero as x1 → s − and x2 → s +. Thus, in the limit, q  s +, t − q  s −, t = U & ρ  s +, t − ρ  s −, t ' . (13.9.5) In the conventional notation of shock dynamics, this can be written as q2 − q1 = U (ρ2 − ρ1), (13.9.6) or −U [ρ]+[q]=0, (13.9.7) where subscripts 1 and 2 are used to denote the values behind and ahead of the shock respectively, and [ ] denotes the discontinuous jump in the quantity involved. Equation (13.9.7) is called the shock condition. Thus, the basic problem can be written as ∂ρ ∂t + ∂q ∂x = 0 at points of continuity, (13.9.8) −U [ρ]+[q] = 0 at points of discontinuity. (13.9.9) Therefore, we have a nice correspondence ∂ρ ∂t ↔ −U [ ] , ∂ ∂x ↔ [ ] , (13.9.10) between the differential equation and the shock condition. It is now possible to find discontinuous solutions of (13.9.3). In any continuous part of the solution, equation (13.9.1) is still satisfied and the assumption q = Q (ρ) remains valid. But we have q1 = Q (ρ1) and q2 = Q (ρ2) on the two sides of any shock, and the shock condition (13.9.6) has the form U (ρ2 − ρ1) = Q (ρ2) − Q (ρ1). (13.9.11) Example 13.9.1. The simplest example in which breaking occurs is ρt + c (ρ) ρx = 0, with discontinuous initial data at t = 0 13.10 Structure of Shock Waves and Burgers’ Equation 563 ρ = ⎧ ⎨ ⎩ ρ2, x< 0 ρ1, x> 0 , (13.9.12) and F (x) = ⎧ ⎨ ⎩ c2 = c2 (ρ), x< 0 c1 = c1 (ρ), x> 0 , (13.9.13) where ρ1 > ρ2 and c2 > c1. In this case, breaking will occur immediately and this can be seen from Figure 13.9.1ab with c ′ (ρ) > 0. The multivalued region begins at the origin ξ = 0 and is bounded by the characteristics x = c1t and x = c2t with c1 < c2. This corresponds to a centered compression wave with overlapping characteristics in the (x, t)-plane. On the other hand, if the initial condition is expansive with c2 < c1, there is a continuous solution obtained from (13.2.12) in which all values of F (x) in [c2, c1] are taken on characteristics through the origin ξ = 0. This corresponds to a centered fan of characteristics x = ct, c2 ≤ c ≤ c1 in the (x, t)-plane so that the solution has the explicit form c = (x/t), c2 < (x/t) < c1. The density distribution and the expansion wave are shown in Figure 13.9.2ab. In this case, the complete solution is given by c = ⎧ ⎪⎪⎪⎪⎪⎨ ⎪⎪⎪⎪⎪⎩ c2, x ≤ c2t x t , c2t<x 0. (13.10.1) Note that near breaking where ρx is large, (13.10.1) gives a better approximation. With (13.10.1), the basic equation (13.9.1) becomes ρt + c (ρ) ρx = νρxx, (13.10.2) 564 13 Nonlinear Partial Differential Equations with Applications Figure 13.9.1ab Density distribution and centered compression wave with overlapping characteristics. 13.10 Structure of Shock Waves and Burgers’ Equation 565 Figure 13.9.2ab Density distribution and centered expansion wave. 566 13 Nonlinear Partial Differential Equations with Applications where c (ρ) = Q′ (ρ), the second and the third terms represent the effects on nonlinearity and diffusion. We first solve (13.10.2) for two simple cases: (i) c (ρ) = constant = c, and (ii) c (ρ) ≡ 0. In the first case, equation (13.10.1) becomes linear and we seek a plane wave solution ρ (x, t) = a exp {i(kx − ωt)} . (13.10.3) Substituting this solution into the linear equation (13.10.2), we have the dispersion relation ω = ck − iνk2 , (13.10.4) where Im ω = −νk2 < 0, since ν > 0. Thus, the wave profile has the form ρ (x, t) = a e−νk2 t exp [ik (x − ct)] (13.10.5) which represents a diffusive wave (Im ω < 0) with wavenumbers k and phase velocity c whose amplitude decays exponentially with time t. The decay time is given by t0 =  νk2 −1 which becomes smaller and smaller as k increases with fixed ν. Thus, the waves of smaller wavelengths decay faster than waves of longer wavelengths. On the other hand, for a fixed wavenumber k, t0 decreases as ν increases so that waves of a given wavelength attenuate faster in a medium with larger ν. The quantity ν may be regarded as a measure of diffusion. Finally, after a sufficiently long time (t ≫ t0) only disturbances of long wavelength will survive, while all short wavelength disturbances will decay rapidly. In the second case, (13.10.2) reduces to the linear diffusion equation ρt = νρxx. (13.10.6) This equation with the initial data at t = 0 ρ = ⎧ ⎨ ⎩ ρ1, x< 0 ρ1, x> 0, ρ1 > ρ2, (13.10.7) can readily be solved, and the solution for t > 0 is ρ (x, t) = ρ1 2 √ πνt
0 −∞ e −(x−ξ) 2/4νtdξ + ρ2 2 √ πνt
∞ 0 e −(x−ξ) 2/4νtdξ. (13.10.8) After some manipulation involving changes of variables of integration (x − ξ) /2 √ νt = η, the solution is simplified to the form 13.10 Structure of Shock Waves and Burgers’ Equation 567 u (x, t) = 1 2 (ρ1 + ρ2)+(ρ2 − ρ1) 1 √ π
x/2 √ νt 0 e −η 2 dη, (13.10.9) = 1 2 (ρ1 + ρ2) + 1 2 (ρ2 − ρ1) erf  x 2 √ νt . (13.10.10) This shows that the effect of the term νρxx is to smooth out the initial distribution (νt) − 1 2 . The solution tends to values ρ1, ρ2 as x → +∞. The absence of the term νρxx in (13.10.2) leads to nonlinear steepening and breaking. Indeed, equation (13.10.2) combines the two opposite effects of breaking and diffusion. The sign of ν is important; indeed, solutions are stable or unstable according as ν > 0 or ν < 0. In order to investigate solutions that balance between steepening and diffusion, we seek solutions of (13.10.2) in the form ρ = ρ (X), X = x − U t, (13.10.11) where U is a constant to be determined. It follows from (13.10.2) that [c (ρ) − U] ρX = νρXX. (13.10.12) Integrating this equation gives Q (ρ) − U ρ + A = νρX, (13.10.13) where A is a constant of integration. Integrating (13.10.13) with respect to X gives an implicit relation for ρ (X) in the form X ν =
dρ Q (ρ) − U ρ + A . (13.10.14) We would like to have a solution which tends to ρ1, ρ2 as X → +∞. If such a solution exists with ρX → 0 as |X|→∞, the quantities U and A must satisfy Q (ρ1) − U ρ1 + A = Q (ρ2) − U ρ2 + A = 0, (13.10.15) which implies that U = Q (ρ1) − Q (ρ2) ρ1 − ρ2 . (13.10.16) This is exactly the same as the shock velocity obtained before. Result (13.10.15) shows that ρ1, ρ2 are zeros of Q (ρ) − U ρ + A. In the limit ρ → ρ1 or ρ2, the integral (13.10.14) diverges and X → + ∞. If c ′ (ρ) > 0, then Q (ρ)−U ρ+A ≤ 0 in ρ2 ≤ ρ ≤ ρ1 and then ρX ≤ 0 because 568 13 Nonlinear Partial Differential Equations with Applications Figure 13.10.1 Shock structure and shock thickness. of (13.10.11). Thus, ρ decreases monotonically from ρ1 at X = −∞ to ρ2 at X = ∞ as shown in Figure 13.10.1. Physically, a continuous waveform carrying an increase in ρ will progressively distort itself and eventually break forward and require a shock with ρ1 < ρ2 provided c ′ (ρ) > 0. It will break backward and require a shock with ρ1 > ρ2 and c ′ (ρ) < 0. Example 13.10.1. Obtain the solution of (13.10.2) with the initial data (13.10.7) and Q (ρ) = αρ2 + βρ + γ, α > 0. We write Q (ρ) − U ρ + A = −α (ρ1 − ρ) (ρ − ρ2), where U = β + α (ρ1 + ρ2) and A = αρ1ρ2 − γ. Integral (13.10.14) becomes X ν = − 1 α
dρ (ρ − ρ2) (ρ1 − ρ) = 1 α (ρ1 − ρ2) log  ρ1 − ρ ρ − ρ2  , which gives the solution ρ (X) = ρ2 + (ρ1 − ρ2) exp αX ν (ρ2 − ρ1) ! 1 + exp αX ν (ρ2 − ρ1) ! . (13.10.17) 13.10 Structure of Shock Waves and Burgers’ Equation 569 The exponential factor in the solution indicates the existence of a transition layer of thickness δ of the order of ν/ [α (ρ1 − ρ2)]. This can also be referred to as the shock thickness. The thickness δ increases as ρ1 → ρ2 for a fixed ν. It tends to zero as ν → 0 for a fixed ρ1 and ρ2. In this case, the shock velocity (13.10.16) becomes U = α (ρ1 − ρ2) + β = 1 2 (c1 + c2), (13.10.18) where c (ρ) = Q′ (ρ), c1 = c (ρ1), and c2 = c (ρ2). We multiply (13.10.2) by c ′ (ρ) and simplify to obtain ct + ccx = νcxx − ν c′′ (ρ) ρ 2 x . (13.10.19) Since Q (ρ) is a quadratic expression in ρ, then c (ρ) = Q′ (ρ) becomes linear in ρ and c ′′ (ρ) = 0. Thus, (13.10.19) leads to Burgers’ equation replacing c with u ut + uux = ν uxx. (13.10.20) This equation incorporates the combined opposite effects of nonlinearity and diffusion. It is the simplest nonlinear model equation for diffusive waves in fluid dynamics. Using the Cole–Hopf transformation u = −2ν φx φ . (13.10.21) Burgers’ equation can be solved exactly, and the opposite effects of nonlinearity and diffusion can be investigated in some detail. We introduce the transformation in two steps. First, we write u = ψx so that (13.10.20) can readily be integrated to obtain ψt + 1 2 ψ 2 x = ν ψxx. (13.10.22) The next step is to introduce ψ = −2ν log φ and to transform this equation into the so called diffusion equation φt = ν φxx. (13.10.23) This equation was solved in earlier chapters. We simply quote the solution of the initial-value problem of (13.10.23) with the initial data φ (x, 0) = Φ(x), −∞ <x< ∞.="" (13.10.24)="" the="" solution="" for="" φ="" is="" (x,="" t)="1" 2="" √="" πνt=""
="" ∞="" −∞="" Φ(ζ)="" exp="" 1="" −="" (x="" ζ)="" 4νt="" 3="" dζ,="" (13.10.25)="" 570="" 13="" nonlinear="" partial="" differential="" equations="" with="" applications="" where="" can="" be="" written="" in="" terms="" of="" initial="" value="" u="" 0)="F" (x)="" by="" using="" (13.10.21).="" it="" turns="" out="" that,="" at="" t="0," =="" ="" 1="" 2ν="" x="" 0="" f="" (α)="" dα0="" .="" (13.10.26)="" then="" convenient="" to="" write="" down="" form="" ="" ="" (13.10.27)="" (ζ,="" x,="" ζ="" dα="" +="" 2t="" (13.10.28)="" consequently,="" φx="" 4ν="" dζ.="" (13.10.29)="" therefore,="" follows="" from="" (13.10.21)="" and="" has=""
4="" x−ζ=""
5="" dζ="" *="" (13.10.30)="" although="" this="" exact="" burgers’="" equation,="" physical="" interpretation="" hardly="" given="" unless="" a="" suitably="" simple="" specified.="" even="" then,="" finding="" an="" evaluation="" integrals="" formidable="" task.="" necessary="" resort="" asymptotic="" methods.="" before="" we="" deal="" analysis,="" following="" example="" may="" considered="" investigation="" shock="" formation.="" 13.10.2.="" find="" equation="" significance="" case="" ⎨="" ⎩="" δ="" (x),="" x<="" 0,=""> 0. We first find f (ζ, x, t) = A
ζ 0+ δ (α) dα + (x − ζ) 2 t = ⎧ ⎪⎨ ⎪⎩ (x−ζ) 2 2t − A, ζ < 0 (x−ζ) 2 2t , ζ > 0. 13.10 Structure of Shock Waves and Burgers’ Equation 571 Thus,
∞ −∞ x − ζ t exp  − f 2ν  dζ =
0 −∞  x − ζ t  exp 1 A 2ν − (x − ζ) 2 4νt 3 dζ +
∞ 0  x − ζ t  exp 1 − (x − ζ) 2 4νt 3 dζ = 2ν
4 e A/2ν − 1
5 exp  − x 2 4νt , which is obtained by substitution, x − ζ 2 √ νt = α. Similarly,
∞ −∞ exp  − f 2ν  dζ = 2√ νt √ π +
4 e A/2ν − 1
5 erfc  x 2 √ νt , where erfc (x) is the complementary error function defined by erfc (x) = 2 √ π
∞ x e −η 2 dη. (13.10.31) Therefore, the solution for u (x, t) is u (x, t) = 2 ν t  e A/2ν − 1 exp
4 − x 2 4νt
5 √ π +  eA/2ν − 1 ( √ π/2) erfc
4 x 2 √ νt
5. (13.10.32) In the limit as ν → ∞, the effect of diffusion would be more significant than that of nonlinearity. Since erfc  x 2 √ νt → 0, eA/2ν ∼ 1 + A 2ν as ν → ∞, the solution (13.10.32) tends to the limiting value u (x, t) ∼ A 2 √ πνt exp  − x 2 4νt . (13.10.33) This represents the well-known source solution of the classical linear heat equation ut = ν uxx. On the other hand, when ν → 0 nonlinearity would dominate over diffusion. It is expected that solution (13.10.32) tends to that of Burgers’ equation as ν → 0. 572 13 Nonlinear Partial Differential Equations with Applications We next introduce the similarity variable η = x/√ 2At to rewrite (13.10.32) in the form u (x, t) =
4ν t
5 1 2  e A/2ν − 1 exp
4 − Aη2 2ν
5 √ π +  eA/2ν − 1 ( √ π/2) erfc
4% A 2ν η
5, (13.10.34) ∼
4ν t
5 1 2 exp & A 2ν  1 − η 2 ' √ π + (√ π/2) exp  A 2ν erfc
4% A 2ν η
5 as ν → 0 for all η, (13.10.35) ∼ 0 as ν → 0, for η < 0 and η > 1. (13.10.36) Invoking the asymptotic result, erfc (x) ∼  2/ √ π e −x 2 2x as x → ∞, (13.10.37) the solution (13.10.34) for 0 <η< 1 has the form, u (x, t) ∼
4ν t
5 1 2 2η  A 2ν 1 2 exp & A 2ν  1 − η 2 ' 2η  Aπ 2ν 1 2 + exp & A 2ν (1 − η 2) ' . =  2A t 1 2 η 1+2η  Aπ 2ν 1 2 exp & A 2ν (η 2 − 1)' ∼  2A t 1 2 as ν → 0. The final asymptotic solution as ν → 0 is u (x, t) ∼ ⎧ ⎨ ⎩ x t , 0 <x< (2at)="" 1="" 2="" 0,="" otherwise.="" (13.10.38)="" this="" result="" represents="" a="" shock="" at="" x="(2At)" with="" the="" velocity="" u="(A/2t)" .="" solution="" has="" jump="" from="" 0="" to="" t="(2A/t)" so="" that="" condition="" is="" fulfilled.="" asymptotic="" behavior="" of="" burgers’="" as="" ν="" →="" 0.="" we="" use="" kelvin="" stationary="" phase="" approximation="" method="" discussed="" in="" section="" 12.7="" examine="" (13.10.30).="" according="" method,="" significant="" contribution="" integrals="" involved="" (13.10.30)="" comes="" points="" for="" fixed="" and="" t,="" is,="" roots="" equation="" 13.11="" korteweg–de="" vries="" solitons="" 573="" ∂f="" ∂ζ="F" (ζ)="" −="" (x="" ζ)="" (13.10.39)="" suppose="" ζ="ξ" (x,="" t)="" root.="" (12.7.8),="" yield=""
="" ∞="" −∞="" ="" ξ="" ="" exp="" f="" 2ν="" dζ="" ∼="" ="" 4πν="" |f="" ′′="" (ξ)|="" 01="" (ξ)="" 0="" therefore,="" final="" ,="" (13.10.40)="" where="" satisfies="" (13.10.39).="" other="" words,="" can="" be="" rewritten="" form="" +="" tf="" (13.10.41)="" identical="" (13.2.12)="" which="" was="" obtained="" 13.2.="" case,="" point="" characteristic="" variable.="" although="" exact="" single-valued="" continuous="" function="" all="" time="" exhibits="" instability.="" it="" already="" been="" shown="" progressively="" distorts="" itself="" becomes="" multiple-valued="" after="" sufficiently="" long="" time.="" eventually,="" breaking="" will="" definitely="" occur.="" follows="" analysis="" nonlinear="" diffusion="" terms="" show="" opposite="" effects.="" former="" introduces="" steepening="" profile,="" whereas="" latter="" tends="" diffuse="" (spread)="" sharp="" discontinuities="" into="" smooth="" profile.="" view="" property,="" diffusive="" wave.="" context="" fluid="" flows,="" denotes="" kinematic="" viscosity="" measures="" viscous="" dissipation.="" finally,="" arises="" many="" physical="" problems,="" including="" one-dimensional="" turbulence="" (where="" had="" its="" origin),="" sound="" waves="" media,="" fluid-filled="" elastic="" pipes,="" magnetohydrodynamic="" media="" finite="" conductivity.="" celebrated="" dispersion="" relation="" (13.3.14)="" dispersive="" surface="" on="" water="" constant="" depth="" h0="" 574="" 13="" partial="" differential="" equations="" applications="" ω="(gk" tanh="" kh0)="" 3="" k="" 2h="" 1="" ≈="" c0k="" 6="" (13.11.1)="" c0="(gh0)" shallow="" wave="" speed.="" motions="" small="" exhibit="" such="" term="" contrast="" linearized="" theory="" value="" c0k.="" an="" free="" elevation="" η="" given="" by="" ηt="" c0ηx="" σηxxx="0," (13.11.2)="" σ="1" c0h="" fairly="" waves.="" called="" (kdv)="" moving="" positive="" direction="" only.="" group="" velocities="" are="" found="" they="" cp="ω" σk2="" (13.11.3)="" cg="dω" dk="c0" 3σk2="" (13.11.4)="" noted=""> Cg, and the dispersion comes from the term involving k 3 in the dispersion relation (13.11.1) and hence, from the term σηxxx. For sufficiently long waves (k → 0), Cp = Cg = c0, and hence, these waves are nondispersive. In 1895, Korteweg–de Vries derived the nonlinear equation for long water waves in a channel of depth h0 which has the remarkable form ηt + c0  1 + 3 2 η h0  ηx + σηxxx = 0. (13.11.5) This is the simplest nonlinear model equation for dispersive waves, and combines nonlinearity and dispersion. The KdV equation arises in many physical problems, which include water waves of long wavelengths, plasma waves, and magnetohydynamics waves. Like Burgers’ equation, the nonlinearity and dispersion have opposite effects on the KdV equation. The former introduces steepening of the wave profile while the latter counteracts waveform steepening. The most remarkable features is that the dispersive term in the KdV equation does allow the solitary and periodic waves which are not found in shallow water wave theory. In Burgers’ equation the nonlinear term leads to steepening which produces a shock wave; on the other hand, in the KdV equation the steepening process is balanced by dispersion to give a rise to a steady solitary wave. We now seek the traveling wave solution of the KdV equation (13.11.5) in the form 13.11 The Korteweg–de Vries Equation and Solitons 575 η (x, t) = h0f (X), X = x − U t, (13.11.6) for some function f and constant wave velocity U. We determine f and U by substitution of the form (13.11.6) into (13.11.5). This gives, with σ = 1 6 c0h 2 0 , 1 6 h 2 0 f ′′′ + 3 2 ff′ +  1 − U c0  f ′ = 0, (13.11.7) and then integration leads to 1 6 h 2 0 f ′′ + 3 4 f 2 +  1 − U c0  f + A = 0, where A is an integrating constant. We next multiply this equation by f ′ and integrate again to obtain 1 3 h 2 0 f ′2 + f 3 + 2  1 − U c0  f 2 + 4 Af + B = 0, (13.11.8) where B is a constant of integration. We now seek a solitary wave solution under the boundary conditions f, f ′ , f ′′ → 0 as |X|→∞. Therefore, A = B = 0 and (13.11.8) assumes the form 1 3 h 2 0 f ′2 + f 2 (f − α)=0, (13.11.9) where α = 2  U c0 − 1  . (13.11.10) Finally, we obtain X =
f 0 df f ′ =  h 2 0 3 1 2
f 0 df f  (α − f) , which is, by the substitution f = α sech2 θ, X − X0 =  4h 2 0 3α 1 2 θ, (13.11.11) for some integrating constant X0. Therefore, the solution for f (X) is f (X) = α sech2 1 3α 4h 2 0 1 2 (X − X0) 3 . (13.11.12) 576 13 Nonlinear Partial Differential Equations with Applications Figure 13.11.1 A soliton. The solution f (X) increases from f = 0 as X → −∞ so that it attains a maximum value f = fmax = α at X = 0, and then decreases symmetrically to f = 0 as X → ∞ as shown in Figure 13.11.1. These features also imply that X0 = 0, so that (13.11.12) becomes f (X) = α sech2 1 3α 4h 2 0 1 2 X 3 . (13.11.13) Therefore, the final solution is η (x, t) = η0 sech2 1 3η0 4h 3 0 1 2 (x − U t) 3 , (13.11.14) where η0 = (αh0). This is called the solitary wave solution of the KdV equation for any positive constant η0. However, it has come to be known as soliton since Zabusky and Kruskal coined the term in 1965. Since η > 0 for all X, the soliton is a wave of elevation which is symmetrical about X = 0. It propagates in the medium without change of shape with velocity U = c0
4 1 + α 2
5 = c0  1 + 1 2 η0 h0  , (13.11.15) which is directly proportional to the amplitude η0. The width,  3η0/4h 3 0 − 1 2 is inversely proportional to √η0. In other words, the solitary wave propagates to the right with a velocity U which is directly proportional to the amplitude, and has a width that is inversely proportional to the square root of the amplitude. Therefore, taller solitons travel faster and are narrower than the shorter (or slower) ones. They can overtake the shorter ones, and surprisingly, they emerge from the interaction without change of shape as shown in Figure 13.11.2. Indeed the discovery of soliton interactions confirms that solitons behave like elementary particles. 13.11 The Korteweg–de Vries Equation and Solitons 577 Figure 13.11.2 Interaction of two solitons (U1 > U2, t2 > t1,). General Waves of Permanent Form. We now consider the general case given by (13.11.8) which can be written  h 2 0 3  f ′2 = −f 3 + αf 2 − 4Af − B ≡ F (f). We seek real bounded solutions for f (X). Therefore, f ′2 ≥ 0 and varies monotonically until f ′ is zero. Hence, the zeros of the cubic F (f) are crucial. For bounded solutions, all the three zeros f1, f2, f3 must be real. Without loss of generality, we choose f1 = 0 and f2 = α. The third zero must be negative so we set f3 = α − β with 0 <α<β. Therefore, the equation for f (X) is 1 3 h 2 0  df dX 2 = f (α − f) (f − α + β), (13.11.16) or $ 3 h 2 0 dX = − df [f (α − f) (f − α + β)] 1 2 , (13.11.17) where U = c0  1 + 2α − β 2  . (13.11.18) 578 13 Nonlinear Partial Differential Equations with Applications We put α − f = p 2 in (13.11.17) to obtain  3 4h 2 0 1 2 dX = dp [(α − p 2) (β − p 2)] 1 2 . (13.11.19) We next substitute p = √ α q into (13.11.19) to transform it into the standard form  3β 4h 2 0 1 2 X =
q 0 dq [(1 − q 2) (1 − m2q 2)] 1 2 (13.11.20) where m = (α/β) 1 2 . The right hand side is an integral of the first kind, and hence, q can be expressed in terms of the Jacobian sn function (see Dutta and Debnath (1965)) q = sn 1 3β 4h 2 0 1 2 X, m3 , (13.11.21) where m is the modulus of the Jacobian elliptic function sn (z,m). Therefore, f (X) = α 1 1 − sn2 / 3β 4h 2 0 1 2 X 03 = α cn2 1 3β 4h 2 0 1 2 X 3 , (13.11.22) where cn (z,m) is also the Jacobian elliptic function of modulus m and cn2 (z)=1 − sn2 (z). From (13.11.20), the period P is given by P = 2  4h 2 0 3β 1 2
1 0 dq [(1 − q 2) (1 − m2q 2)] 1 2 (13.11.23) = 4h0 √ 3β K (m) ≡ λ, (13.11.24) where K (m) is the complete elliptic integral of the first kind defined by K (m) =
π/2 0  1 − m sin2 θ − 1 2 dθ (13.11.25) and λ denotes the wavelength of the cnoidal wave. It is important to note that cn (z,m) is periodic, and hence, η (X) represents a train of periodic waves in shallow water. Thus, these waves are called cnoidal waves with wavelength 13.11 The Korteweg–de Vries Equation and Solitons 579 Figure 13.11.3 A cnoidal wave. λ = 2  4h 3 3b 1/2 K (m). (13.11.26) The outcome of this analysis is that solution (13.11.22) represents a nonlinear wave whose shape and wavelength (or period) all depend on the amplitude of the wave. A typical cnoidal wave is shown in Figure 13.11.3. Sometimes, the cnoidal waves with slowly varying amplitude are observed in rivers. More often, wavetrains behind a weak bore (called an undular bore) can be regarded as cnoidal waves. Two limiting cases are of special interest: (i) m → 1 and (ii) m → 0. When m → 1 (α → β), it is easy to show that cn (z) → sech z. Hence, the cnoidal wave solution (13.11.22) tends to the solitary wave with the wavelength λ, given by (13.11.24) which approaches infinity because K (1) = ∞, K (0) = π/2. The solution identically reduces to (13.11.14) with (13.11.15). In the other limit m → 0 (α → 0), sn z → sin z and cn z → cos z so that solution (13.11.22) becomes f (X) = α cos2 1 3β 4h 2 0 1 2 X 3 , (13.11.27) where U = c0  1 − β 2  . (13.11.28) Using cos 2θ = 2 cos2 θ − 1, we can rewrite (13.11.27) in the form f (X) = α 2 1 + cos √ 3β h0  X . (13.11.29) We next introduce k = √ 3β/h0 (or β = 1 3 k 2h 2 0 ) to simplify (13.11.29) as f (X) = α 2 [1 + cos (kx − ωt)] , (13.11.30) where ω = U k = c0k  1 − 1 6 k 2h 2 0  . (13.11.31) This corresponds to the first two terms of the series of (gk tanh kh0) 1/2 . Thus, these results are in perfect agreement with the linearized theory. 580 13 Nonlinear Partial Differential Equations with Applications Remark: It is important to point out that the phase velocity (13.11.3) becomes negative for k 2 > (c0/σ) which indicates that waves propagate in the negative x direction. This contradicts the original assumption of forward travelling waves. Moreover, the group velocity given by (13.11.4) assumes large negative values for large k so that the fine-scale features of the solution are propagated in the negative x direction. The solution of (13.11.2) involves the Airy function which shows fiercely oscillatory character for large negative arguments. This leads to a lack of continuity and a tendency to emphasize short wave components which contradicts the KdV model representing fairly long waves. In order to eliminate these physically undesirable features of the KdV equation, Benjamin, Bona, and Mahony (1972) proposed a new nonlinear model equation in the form ηt + ηx + ηηx − ηxxt = 0. (13.11.32) This is known as the Benjamin, Bona and Mahony (BBM) equation. The advantage of this model over the KdV equation becomes apparent when we examine their linearized forms and the corresponding solutions. The linearized form (13.11.32) gives the dispersion relation ω = k 1 + k 2 , (13.11.33) which shows that both the phase velocity and the group velocity are bounded for all k, and both velocities tend to zero for large k. In other words, the model has the desirable feature of responding very insignifi- cantly to short wave components that may be introduced into the initial wave form. Thus, the BBM model seems to be a preferable long wave model of physical interest. However, whether the BBM equation is a better model than the KdV equation has not yet been established. Another important property of the KdV equation is that it satisfies the conservation law of the form Tt + Xx = 0, (13.11.34) where T is called the density and the X is called the flux. If T and X are integrable in −∞ <x< ∞,="" and="" x="" →="" 0="" as="" |x|→∞,="" then="" d="" dt=""
="" ∞="" −∞="" t="" dx="−" |x|="" therefore,="" so="" that="" the="" density="" is="" conserved.="" 13.12="" nonlinear="" schr¨odinger="" equation="" solitary="" waves="" 581="" canonical="" form="" of="" kdv="" ut="" −="" 6uux="" +="" uxxx="0," (13.11.35)="" can="" be="" written="" (u)="" ="" −3u="" 2="" uxx="" uxx.="" (13.11.36)="" if="" we="" assume="" u="" periodic="" or="" its="" derivatives="" decay="" very="" rapidly="" this="" leads="" to="" conservation="" mass,="" is,="" (13.11.37)="" often="" called="" time="" invariant="" function="" solutions="" equation.="" second="" law="" for="" (13.11.34)="" obtained="" by="" multiplying="" it="" ="" 1="" ="" −2u="" 3="" uux="" (13.11.38)="" gives="" (13.11.39)="" principle="" energy.="" well="" known="" has="" an="" infinite="" number="" polynomial="" laws.="" generally="" believed="" existence="" a="" soliton="" solution="" closely="" related="" first="" derive="" one-dimensional="" linear="" from="" fourier="" integral="" representation="" plane="" wave="" 582="" 13="" partial="" differential="" equations="" with="" applications="" φ="" (x,="" t)="
" f="" (k)="" exp="" [i(kx="" ωt)]="" dk,="" (13.12.1)="" where="" spectrum="" determined="" given="" initial="" boundary="" conditions.="" slowly="" modulated="" propagates="" in="" dispersive="" medium.="" such="" wave,="" most="" energy="" confined="" neighborhood="" k="k0" dispersion="" relation="" ω="ω" expanded="" about="" point="" (k="" k0)="" ′="" ′′="" ...,="" (13.12.2)="" ω0="" ≡="" (k0),="" (k0).="" view="" (13.12.2),="" rewrite="" [i(k0x="" ω0t)]="" ,="" (13.12.3)="" amplitude="" ψ="" i(k="" i="" ="" 0="" dk.="" (13.12.4)="" evidently,="" represents="" varying="" (or="" modulated)="" part="" basic="" wave.="" simple="" computation="" ψt,="" ψx,="" ψxx="" ψt="−i" ψx="i(k" i(ψt="" ψx)="" (13.12.5)="" associated="" ω′="" 2ω="" .="" (13.12.6)="" choose="" frame="" reference="" moving="" group="" velocity,="" ∗="x" t,="" term="" involving="" dropped="" then,="" satisfies="" equation,="" dropping="" asterisks,="" 583="" (13.12.7)="" next="" both="" frequency="" general="" k,="" a2="" (13.12.8)="" expand="" taylor="" series="" |a|="" ∂ω="" ∂k="" ∂="" ∂k2="" 4="" 5="" (13.12.9)="" now="" replace="" i(∂="" ∂t)="" k0="" −i(∂="" ∂x),="" resulting="" operators="" act="" on="" a,="" obtain="" i(at="" ax)="" axx="" γ="" (13.12.10)="" ω′′="" constant.="" (nls)="" velocity="" ax="" will="" drop="" out="" (13.12.10),="" normalized="" nls="" at="" (13.12.11)="" corresponding="" (13.12.12)="" according="" stability="" criterion="" established="" section="" 13.5,="" modulation="" stable="" <="" unstable=""> 0. To study the solitary wave solution, it is convenient to use the NLS equation in the standard form i ψt + ψxx + γ |ψ| 2 ψ = 0, −∞ <x< ∞,="" t="" ≥="" 0.="" (13.12.13)="" 584="" 13="" nonlinear="" partial="" differential="" equations="" with="" applications="" we="" seek="" waves="" of="" permanent="" form="" by="" assuming="" the="" solution="" ψ="f" (x)="" e="" i(mx−nt)="" ,="" x="x" −="" u="" (13.12.14)="" for="" some="" functions="" f="" and="" constant="" wave="" speed="" to="" be="" determined,="" m,="" n="" are="" constants.="" substitution="" into="" gives="" ′′="" +="" i(2m="" u)="" ′="" ="" m2="" γ="" |f|="" 2="" (13.12.15)="" eliminate="" setting="" 2m="" then,="" write="" α="" so="" that="" can="" assumed="" real.="" thus,="" equation="" becomes="" αf="" 3="0." (13.12.16)="" multiplying="" this="" 2f="" integrating,="" find="" ′2="A" 4="" ≡="" (f),="" (13.12.17)="" where="" (f)="" α1="" α2f="" β1="" β2f="" (α1β2="" α2β1),="" a="α1β1," (α2β2),="" s="" β="" real="" distinct.="" evidently,="" 0="" df="" ="" (α1="" 2)="" (β1="" .="" (13.12.18)="" putting="" (α2="" α1)="" 1="" in="" integral,="" deduce="" following="" elliptic="" integral="" first="" kind="" (see="" dutta="" debnath="" (1965)):="" σx="
" du="" (1="" κ="" 2u="" (13.12.19)="" σ="(α2β1)" (β1α2).="" final="" expressed="" terms="" jacobian="" sn="" function="" (σx,="" κ),="" or,="" α2="" 1="" κ).="" (13.12.20)="" particular,="" when=""> 0, and γ > 0, we obtain a solitary wave solution. In this case, equation (13.12.17) can be rewritten as 13.12 The Nonlinear Schr¨odinger Equation and Solitary Waves 585 √ α X =
f 0 df f  1 − γ 2α f 2 1 2 . (13.12.21) Substitution of (γ/2α) 1 2 f = sech θ in this integral gives the exact solution f (X) =  2α γ 1 2 sech √ α (x − U t) ! . (13.12.22) This represents a solitary wave which propagates without change of shape with constant velocity U. Unlike the solution of the KdV equation, the amplitude and the velocity of the wave are independent parameters. It is noted that the solitary wave exists only for the unstable case (γ > 0). This means that small modulations of the unstable wavetrain lead to a series of solitary waves. The nonlinear dispersion relation for deep water waves is ω =  gk  1 + 1 2 a 2 k 2  . (13.12.23) Therefore, ω ′ 0 = ω0 2k0 , ω′′ 0 = − ω0 4k 2 0 , and γ = − 1 2 ω0k 2 0 , (13.12.24) and the NLS equation for deep water waves is obtained from (13.12.10) in the form i  at + ω0 2k0 ax  −  ω0 8k 2 0  axx − 1 2 ω0 k 2 0 |a| 2 a = 0. (13.12.25) The normalized form of this equation in a frame of reference moving with the linear group velocity ω ′ 0 is i at −  ω0 8k 2 0  axx = 1 2 ω0 k 2 0 |a| 2 a. (13.12.26) Since γ ω′′ 0 =  ω 2 0/8 > 0, this equation confirms the instability of deep water waves. This is one of the most remarkable recent results in the theory of water waves. We next discuss the uniform solution and the solitary wave solution of (13.12.26). We look for solutions in the form a (x, t) = A (X) exp  i γ2 t , X = x − ω ′ 0 t (13.12.27) and substitute this into equation (13.12.26) to obtain AXX = −  8k 2 0 ω0 γ 2A + 1 2 ω0 k 2 0 A 3  . (13.12.28) 586 13 Nonlinear Partial Differential Equations with Applications We multiply this equation by 2AX and then integrate to find A 2 X = −  A 4 0 m′2 + 8 ω0 γ 2 k 2 0A 2 + 2k 4 0A 4  =  A 2 0 − A 2 A 2 − m′2A 2 0 , (13.12.29) where  A4 0m′2 is an integrating constant and 2k 4 0 = 1, m′2 = 1 − m2 , and A2 0 = 4γ 2/ω0k 2 0  m2 − 2 which relates A0, γ, and m. Finally, we rewrite equation (13.12.29) in the form A 2 0 dX = dA "
41 − A2 A2 0
5
4 A2 A2 0 − m′2
5# 1 2 , (13.12.30) or, A0 (X − X0) =
′ ds [(1 − s 2) (s 2 − m′2)] 1 2 , s = (A/A0). This can readily be expressed in terms of the Jacobian dn function (see Dutta and Debnath (1965)) A = A0 dn [A0 (X − X0), m] , (13.12.31) where m is the modulus of the dn function. In the limit m → 0, dn z → 1 and γ 2 → −1 2 ω0k 2 0A2 0 . Hence, the solution is a (x, t) = A (t) = A0 exp  − 1 2 i ω0 k 2 0 A 2 0 t  . (13.12.32) On the other hand, when m → 1, dn z → sech z and γ 2 → −1 4 ω0k 2 0A2 0 . Therefore, the solitary wave solution is a (x, t) = A0 sech [A0 (x − ω ′ 0 t − X0)] exp  − 1 4 ω0 k 2 0 A 2 0 t  .(13.12.33) We next use the NLS equation (13.12.26) to discuss the instability of deep water waves, which is known as the Benjamin and Feir instability. We consider a perturbation of (13.12.32) and write a (X, t) = A (t) [1 + B (X, t)] , (13.12.34) where A (t) is the uniform solution given by (13.12.32). Substituting equation (13.12.34) into (13.12.26) gives iAt (1 + B) + iA (t) Bt −  ω0 8k 2 0  A (t) BXX = 1 2 ω0k 2 0A 2 0 [(1 + B) + BB∗ (1 + B)+(B + B ∗ ) B + (B + B ∗ )] A, 13.12 The Nonlinear Schr¨odinger Equation and Solitary Waves 587 where B∗ is the complex conjugate of B. Neglecting squares of B, it follows that i Bt −  ω0 8k 2 0  BXX = 1 2 ω0k 2 0A 2 0 (B + B ∗ ). (13.12.35) We now seek a solution of the form B (X, t) = B1 e Ωt+iκX + B2 e Ωt−iκX, (13.12.36) where B1, B2 are complex constants, κ is a real wavenumber, and Ω is a growth rate (possibly complex) to be determined. Substitution of B into (13.12.35) leads to the pair of coupled equations  iΩ + ω0κ 2 8k 2 0  B1 − 1 2 ω0k 2 0A 2 0 (B1 + B ∗ 2 )=0, (13.12.37)  iΩ + ω0κ 2 8k 2 0  B2 − 1 2 ω0k 2 0A 2 0 (B ∗ 1 + B2)=0. (13.12.38) It is convenient to take the complex conjugate of (13.12.38) so that it assumes the form  −iΩ + ω0κ 2 8k 2 0  B ∗ 2 − 1 2 ω0k 2 0A 2 0 (B1 + B ∗ 2 )=0. (13.12.39) The pair of linear homogeneous equations (13.12.37) and (13.12.39) for B1 and B∗ 2 admits a nontrivial solution provided         iΩ +
4 ω0κ 2 8k 2 0
5 − 1 2 ω0k 2 0A2 0 − 1 2 ω0k 2 0A2 0 − 1 2 ω0k 2 0A2 0 iΩ +
4 ω0κ 2 8k 2 0
5 − 1 2 ω0k 2 0A2 0         = 0, or Ω 2 =  ω 2 0κ 2 8k 2 0 k 2 0A 2 0 − κ 2 8k 2 0  . (13.12.40) The growth rate Ω is purely imaginary or real and positive depending on whether  κ 2/k2 0 > 8k 2 0A2 0 or  κ 2/k2 0 < 8k 2 0A2 0 . The former case corresponds to a wave (an oscillatory solution) for B, and the latter case represents the Benjamin and Feir instability criterion with κ5 = (κ/k0) as the non-dimensional wavenumber so that κ5 2 < 8k 2 0A 2 0 . (13.12.41) The range of instability is given by 0 < κ <5 κ5c = 2√ 2 (k0A0). (13.12.42) 588 13 Nonlinear Partial Differential Equations with Applications Since Ω is a function of κ5, maximum instability occurs at κ5 = κ5max = 2k0A0 with a maximum growth rate given by (Re Ω)max = 1 2 ω0k 2 0A 2 0 . (13.12.43) To establish the connection with the Benjamin–Feir instability, we have to find the velocity potential for the fundamental wave mode multiplied by exp (kz). It turns out that the term proportional to B1 is the upper sideband, whereas that proportional to B2 is the lower sideband. The main conclusion of the preceding analysis is that Stokes water waves are definitely unstable. In 1967, Benjamin and Feir (see Whitham (1976) or Debnath (2005)) confirmed these remarkable results both theoretically and experimentally. Conservation Laws for the NLS Equation Zakharov and Shabat (1972) proved that equation (13.12.13) has an infinite number of polynomial conservation laws. Each has the form of an integral, with respect to x, of a polynomial expression in terms of the function ψ (x, t) and its derivatives with respect to x. These laws are somewhat similar to those already proved for the KdV equation. Therefore, the proofs of the conservation laws are based on similar assumptions used in the context of the KdV equation. We prove here three conservation laws for the nonlinear Schr¨odinger equation (13.12.13):
∞ −∞ |ψ| 2 dx = constant = C1, (13.12.44)
∞ −∞ i  ψ ψx − ψ ψx dx = constant = C2, (13.12.45)
∞ −∞  |ψx| 2 − 1 2 γ |ψ| 4  dx = constant = C3, (13.12.46) where the bar denotes the complex conjugate. We multiply (13.12.13) by ψ and its complex conjugate by ψ and subtract the latter from the former to obtain i d dt  ψ ψ + d dx  ψx ψ − ψx ψ = 0. (13.12.47) Integration with respect to x in −∞ <x< ∞="" gives="" i="" d="" dt=""
="" −∞="" |ψ|="" 2="" dx="0." this="" proves="" result="" (13.12.44).="" 13.12="" the="" nonlinear="" schr¨odinger="" equation="" and="" solitary="" waves="" 589="" we="" multiply="" (13.12.13)="" by="" ψx="" its="" complex="" conjugate="" then,="" add="" them="" to="" obtain="" ="" ψt="" −="" +="" ψxx="" γ="" ψ="" (13.12.48)="" differentiate="" with="" respect="" x,="" former="" latter="" then="" together.="" leads="" ψxψxt="" ψxt="" ψxxx="" +γ="" "=""
4=""
5="" x="" #="0." (13.12.49)="" if="" subtract="" from="" simplify,="" have="" 4="" .="" integrating="" second="" result.="" resulting="" equations="" derive="" or,="" |ψx|="" (13.12.50)="" (13.12.46).="" above="" three="" conservation="" integrals="" a="" simple="" physical="" meaning.="" in="" fact,="" constants="" of="" motion="" c1,="" c2="" c3="" are="" related="" number="" particles,="" momentum,="" energy="" system="" governed="" equation.="" 590="" 13="" partial="" differential="" applications="" an="" analysis="" section="" reveals="" several="" remarkable="" features="" can="" also="" be="" used="" investigate="" instability="" phenomena="" many="" other="" systems.="" like="" various="" forms="" kdv="" equation,="" nls="" arises="" problems,="" including="" water="" ocean="" waves,="" plasma,="" propagation="" heat="" pulses="" solid,="" self-trapping="" optics,="" fluid="" filled="" viscoelastic="" tube,="" fluids="" plasmas="" (see="" debnath="" (2005)).="" 13.13="" lax="" pair="" zakharov="" shabat="" scheme="" his="" 1968="" seminal="" paper,="" developed="" elegant="" formalism="" for="" finding="" isospectral="" potentials="" as="" solutions="" evolution="" all="" integrals.="" work="" deals="" some="" new="" fundamental="" ideas,="" deeper="" results,="" their="" application="" model.="" subsequently="" paved="" way="" generalizations="" technique="" method="" solving="" equations.="" introducing="" heisenberg="" picture,="" inverse="" scattering="" based="" upon="" abstract="" formulation="" certain="" properties="" operators="" on="" hilbert="" space,="" which="" familiar="" context="" quantum="" mechanics.="" has="" feature="" associating="" linear="" that="" analogs="" formulate="" lax’s="" (1968),="" consider="" two="" l="" m.="" eigenvalue="" operator="" corresponds="" general="" form="" is="" lψ="λψ," (13.13.1)="" where="" eigenfunction="" λ="" corresponding="" eigenvalue.="" m="" describes="" change="" eigenvalues="" parameter="" t,="" usually="" represents="" time="" (13.13.2)="" differentiating="" t="" ltψ="" lψt="λtψ" λψt.="" (13.13.3)="" next="" eliminate="" using="" lmψ="λtψ" λmψ="λtψ" mλψ="λtψ" mlψ,="" (13.13.4)="" 591="" equivalently,="" ∂l="" ∂t="" (ml="" lm)="" ψ.="" (13.13.5)="" thus,="" constant="" nonzero="" eigenfunctions="" only="" (lm="" ml)="" [l,="" m]="" ψ,="" (13.13.6)="" called="" commutator="" m,="" derivative="" left-hand="" side="" interpreted="" alone.="" it="" picture="" problem,="" course,="" how="" determine="" these="" given="" there="" no="" systematic="" solution="" problem.="" negative="" integrable="" hierarchy,="" qiao="" (1995)="" strampp="" (2002)="" suggest="" approach="" generate="" equations;="" they="" devise="" strategies="" spectral="" initial-value="" problem="" u="" (x,="" t)="" satisfies="" ut="N" (u)="" (13.13.7)="" 0)="f" (x),="" (13.13.8)="" ∈="" y="" suitable="" function="" n="" :="" →="" independent="" but="" may="" involve="" or="" derivatives="" x.="" must="" assume="" expressed="" lt="" (13.13.9)="" space="" h="" depend="" scalar="" operator.="" self-adjoint="" so="" (lφ,="" ψ)="(φ," lψ)="" φ="" (·,="" ·)="" inner="" product.="" now="" h:="" (t)="" ≥="" 0,="" r.="" (13.13.10)="" making="" use="" (13.13.9),="" λtψ="(L" λ)="" (ψt="" mψ).="" (13.13.11)="" product="" yields="" (ψ,="" λt="((L" mψ),="" (13.13.12)="" 592="" which,="" since="" self-adjoint,="" mψ)="0." hence,="" confirming="" each="" constant.="" consequently,="" becomes="" l(ψt="" (13.13.13)="" shows="" λ.="" always="" possible="" redefine="" adding="" identity="" original="" remains="" unchanged.="" 0.="" (13.13.14)="" following.="" theorem="" 13.13.1.="" (13.13.15)="" holds,="" (13.13.14).="" not="" yet="" clear="" find="" satisfy="" preceding="" conditions.="" illustrate="" method,="" choose="" ≡="" ∂="" ∂x2="" u,="" (13.13.16)="" sturm–liouville="" l.="" l,="" theory="" unitary="" h,="" chosen="" antisymmetric,="" (mφ,="" (φ,="" h.="" so,="" combination="" odd="" natural="" choice="" follows="" nφ="" ∂xn="" ψdx="−" nψ="" mψ),(13.13.17)="" provided="" n,="" φ,="" tend="" zero,="" |x|→∞.="" moreover,="" require="" sufficient="" freedom="" any="" unknown="" functions="" make="" multiplicative="" operator,="" is,="" degree="" zero.="" simplest="" (∂="" ∂x),="" c="" automatically="" 593="" cux="0," (13.13.18)="" one-dimensional="" wave="" (13.13.19)="" associated="" motion.="" 3="" ∂x3="" ∂x="" ∂xa="" b,="" (13.13.20)="" constant,="" t),="" b="B" third="" term="" right-hand="" dropped,="" retain="" convenience.="" algebraic="" calculation="" uxxx="" axxx="" bxx="" uxa="" (3auxx="" 4axx="" 2bx)="" (3aux="" 4ax)="" would="" au="" (t).="" auux="" (13.13.21)="" standard="" defined="" reduces="" ="" ∂xu="" ="" (13.13.22)="" simplified="" (u="" uψ)x="" 3ψx="" +3(uψ)x="" bψ="2(u" 2λ)="" uxψ="" bψ.="" (13.13.23)="" close="" comments.="" first,="" solvable="" transform="" (ist),="" form.="" however,="" main="" difficulty="" completely="" determining="" whether="" produces="" and,="" indeed,="" proved="" infinite="" operators,="" one="" order="" ∂x,="" family="" flows="" under="" spectrum="" 594="" preserved.="" second,="" study="" choosing="" alternative="" third,="" restriction="" should="" limited="" class="" could="" removed.="" matrix="" operators.="" already="" been="" extended="" such="" fourth,="" (1972,="" 1974)="" published="" series="" notable="" papers="" field="" extending="" (nls)="" first="" time,="" generalized="" more="" than="" spatial="" variable.="" extension="" known="" (zs)="" scheme,="" essentially,="" recasts="" form,="" leading="" marchenko="" finally,="" briefly="" discuss="" zs="" nonself-adjoint="" n-soliton="" introduced="" ingenious="" (13.13.24)="" ml),="" (13.13.25)="" include="" coefficients,="" refers="" t.="" lφ="λφ." (13.13.26)="" differentiation="" dλ="" (iφt="" mφ).="" (13.13.27)="" initially="" changes="" manner="" iφt="Mφ," (13.13.28)="" (13.13.26).="" coupling="" coefficients="" nature="" determines="" potential="" (13.13.26),="" (13.13.28).="" although="" quite="" general,="" crucial="" step="" factor="" according="" (13.13.25).="" (1972)="" introduce="" ×="" matrices="" follows:="" ⎡="" ⎣="" 1="" α="" 0="" ⎤="" ⎦="" ∗="" ⎦,="" (13.13.29)="" ⎢="" |u|="" 1+α="" iu∗="" −iux="" −|u|="" 1−α="" ⎥="" (13.13.30)="" 13.14="" exercises="" 595="" iut="" uxx="" (13.13.31)="" complete="" solved="" initial="" condition="" 0).="" seems="" significant="" contribution="" come="" point="" large="" times="" (t="" ∞).="" physically,="" disturbance="" tends="" disintegrate="" into="" waves.="" mathematical="" asymptotic="" |x|→∞,="" expected="" end="" wavetrains="" modulations.="" 1.="" flow="" density="" relation="" q="vρ" (1="" ρ="" ρ1),="" traffic="" (x)="" let="" f="" ⎪⎪⎪⎪⎨="" ⎪⎪⎪⎪⎩="" ,="" ≤="" 5="" 12x,="" show="" (i)="" ρ0="" along="" characteristic="" lines="" ct="3(x" x0),="" x0="" (ii)="" x),="" 1,="" (iii)="" 12x0ρ0="" (x="" what="" happens="" at="" intersection="" x).="" draw="" versus="" 2.="" mountain="" height="" vulnerable="" erosion="" slope="" hx="" very="" large.="" ht="" functional="" (hx),="" ux="0," ′="" (u).="" 3.="" river="" carrying="" particles="" through="" solid="" bed.="" during="" sedimentation="" process,="" will="" 596="" deposited="" assuming="" v="" velocity="" ρb="" density,="" ρf="" carried="" material="" bed,="" law="" ∂ρ="" ∂q="" (a)="" ∂ρf="" (ρf="" )="" q′="" (b)="" chemical="" engineering="" between="" ∂ρb="" (α="" ρb)="" k2="" (β="" ρb,="" k1,="" represent="" reaction="" rates="" α,="" β="" values="" saturation="" levels="" bed="" respectively.="" speed="" (k1α="" k2β)="" densities="" small.="" 4.="" steady="" (ζ),="" ζ="x" burgers’="" boundary="" conditions="" +∞="" tanh="" ="" ∞)="" 4ν="" ct)="" 0="" 2a="" roots="" 2cf="" integration.="" 5.="" transformations="" −6γ="" reduce="" uux="" γuxxx="0" canonical="" 6uux="" hence="" otherwise,="" prove="" zero="" |x|→∞="" sech2="" (a="" 597="" 6.="" verify="" riccati="" transformation="" +vx="" transforms="" vt="" 6v="" vx="" vxxx="0." (without="" energy-level="" term)="" 7.="" apply="" characteristics="" solve="" ∂u="" ∂v="" data="" −x="" riemann="" invariants="" r="" (α)="2" cosh="" s="" (β)="2" sinh="" β,="" also,="" t).="" 8.="" isentropic="" flow,="" euler="" ρt="" (ρu)x="0," px="0," st="" usx="0," direction,="" p="" pressure,="" entropy.="" families="" characteristics.="" Γ0="" Γ+="" c,="" ∂p="" ∂ρ
5="" following="" full="" set="" ds="" Γ0,="" dp="" ρc="" du="" Γ+.="" particular,="" when="" (s="constant" everywhere),="" (ρ)="" dρ="" c.="" 598="" 9.="" (13.8.32)–(13.8.36),="" second-order="" trs="" (r="" s)="" (tr="" ts)="0," 2c="" dc="" dρ="" polytropic="" gas,="" 4c="α" (c)="" γ−1="r" s,="" 9(b)="" euler–poisson–darboux="" tuu="" 2="" tcc="" 2α="" tc="" variables.="" 10.="" uxx,="" 11.="" uuxx="" 3u="" 2uxx="" xx="" uxut.="" 12.="" laws="" (v)="" vxx="" vvxx="" 599="" 13.="" (13.12.13),="" 14.="" (13.10.20)="" ν="" 15.="" uxxt="0" uxt="" uuxt="" 16.="" <x<="" ∞,=""> 0, ψ → 0, |x|→∞, ψ (x, 0) = ψ (x) with
∞ −∞ |ψ| 2 dx = 1. has the conservation law
4 i|ψ| 2
5 t + (ψ ∗ψx − ψψ∗ x )x = 0 and the energy integral
∞ −∞ |ψ| 2 dx = 1. 17. Seek a dispersive wave solution of the telegraph equation (see problem 14, 3.9 Exercises) utt − c 2uxx + ac2ut + bc2u = 0 in the form u (x, t) = A exp [i(kx − ωt)] . (a) Show that ω = − 1 2  iac2 + 1 2 4c 2k 2 +  4b − c 2 c 2 ! 1 2 . (b) If 4b = a 2 c 2 , show that the solution u (x, t) = A exp  − 1 2 ac2 t  exp [ik (x + ct)] represents nondispersive waves with attenuation. 14 Numerical and Approximation Methods “The strides that have been made recently, in the theory of nonlinear partial differential equations, are as great as in the linear theory. Unlike the linear case, no wholesale liquidation of broad classes of problems has taken place; rather, it is steady progress on old fronts and on some new ones, the complete solution of some special problems, and the discovery of some brand new phenomena. The old tools – variational methods, fixed point theorems, mapping degree, and other topological tools have been augmented by some new ones. Pre-eminent for discovering new phenomena is numerical experimentation; but it is likely that in the future numerical calculations will be parts of proofs.” Peter Lax “Almost everyone using computers has experienced instances where computational results have sparked new insights.” Norman J. Zabusky 14.1 Introduction The preceding chapters have been devoted to the analytical treatment of linear and nonlinear partial differential equations. Several analytical methods to find the exact analytical solution of these equations within simple domains have been discussed. The boundary and initial conditions in these problems were also relatively simple, and were expressible in simple mathematical form. In dealing with many equations arising from the modelling of physical problems, the determination of such exact solutions in a simple domain is a formidable task even when the boundary and/or initial data are simple. It is then necessary to resort to numerical or approximation methods in order to deal with the problems that cannot be solved analytically. In 602 14 Numerical and Approximation Methods view of the widespread accessibility of today’s high speed electronic computers, numerical and approximation methods are becoming increasingly important and useful in applications. In this chapter some of the major numerical and approximation approaches to the solution of partial differential equations are discussed in some detail. These include numerical methods based on finite difference approximations, variational methods, and the Rayleigh–Ritz, Galerkin, and Kantorovich methods of approximation. The chapter also contains a large section on analytical treatment of variational methods and the Euler– Lagrange equations and their applications. A short section on the finite element method is also included. 14.2 Finite Difference Approximations, Convergence, and Stability The Taylor series expansion of a function u (x, y) of two independent variables x and y is u (xi + h, yj ) = ui + 1,j = ui,j + h (ux) i,j + h 2 2! (uxx) i,j + h 3 3! (uxxx) i,j + ..., (14.2.1ab) u (xi , yj + k) = ui,j + 1 = ui,j + k (uy) i,j + k 2 2! (uyy) i,j + k 3 3! (uyyy) i,j + ..., (14.2.2ab) where ui,j = u (x, y), ui + 1,j = u (x + h, y), and ui,j + 1 = u (x, y + k). We choose a set of uniformly spaced rectangles with vertices at Pi,j with coordinates (ih, jk), where i, j, are positive or negative integers or zero, as shown in Figure 14.2.1. We denote u (ih, jk) by ui,j . Using the above Taylor series expansion, we write approximate expressions for ux at the vertex Pi,j in terms of ui,j , ui + 1,j : ux = 1 h [u (x + h, y) − u (x, y)] ∼ 1 h (ui+1,j − ui,j ) + O (h), (14.2.3) ux = 1 h [u (x, y) − u (x − h, y)] ∼ 1 h (ui,j − ui−1,j ) + O (h), (14.2.4) ux = 1 2h [u (x + h, y) − u (x − h, y)] ∼ 1 2h (ui+1,j − ui−1,j ) + O  h 2 . (14.2.5) These expressions are called the forward first difference, backward first difference, and central first difference of ux, respectively. The quantity O (h) or O  h 2 is known as the truncation error in this discretization process. 14.2 Finite Difference Approximations, Convergence, and Stability 603 Figure 14.2.1 Uniformly spaced rectangles. A similar approximate result for uxx at the vertex Pi,j is uxx = 1 h 2 [u (x + h, y) − 2 u (x, y) + u (x − h, y)] ∼ 1 h 2 [ui+1,j − 2 ui,j + ui−1,j ] + O  h 2 . (14.2.6) Similarly, the approximate formulas for uy and uyy at Pi,j are uy = 1 k [u (x, y + k) − u (x, y)] ∼ 1 k (ui,j+1 − ui,j ) + O (k), (14.2.7) uy = 1 k [u (x, y) − u (x, y − k)] ∼ 1 k (ui,j − ui,j−1) + O (k), (14.2.8) uy = 1 2k [u (x, y + k) − u (x, y − k)] ∼ 1 2k (ui,j+1 − ui,j−1) + O  k 2 , (14.2.9) uyy = 1 k 2 [u (x, y + k) − 2 u (x, y) + u (x, y − k)] ∼ 1 k 2 [ui,j+1 − 2ui,j + ui,j−1] + O  k 2 . (14.2.10) All these difference formulas are extremely useful in finding numerical solutions of first or second order partial differential equations. Suppose U (x, y) represents the exact solution of a partial differential equation L(U) = 0 with independent variables x and y, and ui,j is the exact solution of the corresponding finite difference equation F (ui,j ) = 0. Then, the finite difference scheme is said to be convergent if ui,j tends to U as h and k tend to zero. The difference, di,j ≡ (Ui,j − ui,j ) is called the cummulative truncation (or discretization) error. This error can generally be minimized by decreasing the grid sizes h and k. However, this error depends not only on h and k, but also on the 604 14 Numerical and Approximation Methods number of terms in the truncated series which is used to approximate each partial derivative. Another kind of error is introduced when a partial differential equation is approximated by a finite difference equation. If the exact finite difference solution ui,j is replaced by the exact solution Ui,j of the partial differential equation at the grid points Pi,j , then the value F (Ui,j ) is called the local truncation error at Pi,j . The finite difference scheme and the partial differential equation are said to be consistent if F (Ui,j ) tends to zero as h and k tend to zero. In general, finite difference equations cannot be solved exactly because the numerical computation is carried out only up to a finite number of decimal places. Consequently, another kind of error is introduced in the finite difference solution during the actual process of computation. This kind of error is called the round-off error, and it also depends upon the type of computer used. In practice, the actual computational solution is u ∗ i,j , but not ui,j , so that the difference ri,j =  ui,j − u ∗ i,j is the roundoff error at the grid point Pi,j . In fact, this error is introduced into the solution of the finite difference equation by round-off errors. In reality, the round-off error depends mainly on the actual computational process and the finite difference itself. In contrast to the cummulative truncation error, the round-off error cannot be made small by allowing h and k to tend to zero. Thus, the total error involved in the finite difference analysis at the point Pi,j is given by  Ui,j − u ∗ i,j = (Ui,j − ui,j ) +  ui,j − u ∗ i,j = di,j − ri,j . (14.2.11) Usually the discretization error di,j is bounded when ui,j is bounded because the value of Ui,j is fixed for a given partial differential equation with the prescribed boundary and initial data. This fact is used or assumed in order to introduce the concept of stability. The finite difference algorithm is said to be stable if the round-off errors are sufficiently small for all i as j → ∞, that is, the growth of ri,j can be controlled. It should be pointed out again that the round-off error depends not only on the actual computational process and the type of computer used, but also on the finite difference equation itself. Lax (1954) proved a remarkable theorem which establishes the relationship between consistency, stability, and convergence for the finite difference algorithm. Theorem 14.2.1. (Lax’s Equivalence Theorem). Given a properly posed linear initial-value problem and a finite difference approximation to it that satisfies the consistency criterion, stability is a necessary and sufficient condition for convergence. 14.3 Lax–Wendroff Explicit Method 605 Von Neumann’s Stability Method This method is essentially based upon a finite Fourier series. It expresses the initial errors on the line t = 0 in terms of a finite Fourier series and then examines the propagation of errors as t → ∞. It is convenient to denote the error function by er,s instead of ei,j so that er,s gives the initial values er,0 = e (rh) = er on the line t = 0 between x = 0 and x = l, where r = 0, 1, 2, ..., N and Nh = l. The finite Fourier series expansion of er is er =  N n=0 An exp (inπx/l) =  N n=0 An exp (iαnrh), (14.2.12) where αn = (nπ/l), x = rh, and An are the Fourier coefficients which are determined from the (N + 1) equations (14.2.12). Since we are concerned with the linear finite difference scheme, errors form an additive system so that the total error can be found by the superposition principle. Thus, it is sufficient to consider a single term exp (iαrh) in the Fourier series (14.2.12). Following the method of separation of variables commonly used for finding the analytical solution of a partial differential equation, we seek a separable solution of the finite difference equation for er,s in the form er,s = exp (iαrh + βsk) = exp (iαrh) p s (14.2.13) which reduces to exp (iαrh) at s = 0(t = sk = 0), where p = exp (βk), and β is a complex constant. This shows that the error is bounded as (t → ∞) provided that |p| ≤ 1 (14.2.14) is satisfied. This condition is found to be necessary and sufficient for the stability of the finite difference algorithm. 14.3 Lax–Wendroff Explicit Method To describe this method, we consider the first-order conservation equation ∂u ∂t + c ∂u ∂x = 0 (14.3.1) where u ≡ u (x, t) is some physical function of space variable x and time t. This equation occurs frequently in applied mathematics. Lax and Wendroff use the Taylor series expansion in t in the form ui,j+1 = ui,j + k (ut) i,j + k 2 2! (utt) i,j + k 3 3! (uttt) i,j + ..., (14.3.2) 606 14 Numerical and Approximation Methods where k ≡ δt. The partial derivatives in t in (14.3.2) can easily be eliminated by using ut = −c ux so that (14.3.2) becomes ui,j+1 = ui,j − c k (ux) i,j + c 2k 2 2 (uxx) i,j − .... (14.3.3) Replacing ux, uxx by the central difference formulas, (14.3.3) becomes ui,j+1 = ui,j −  ck 2h  (ui+1,j − ui−1,j ) + 1 2  ck h 2 (ui+1,j − 2 ui,j + ui−1,j ), or ui,j+1 =  1 − ε 2 ui,j + ε 2 (1 + ε) ui−1,j − ε 2 (1 + ε) ui+1,j + O  ε 3 ,(14.3.4) where ε = (ck/h). This is called the Lax–Wendroff second-order finite difference scheme; it has been widely used to solve first-order hyperbolic equations. Von Neumann criterion (14.2.14) can be applied to investigate the stability of the Lax–Wendroff scheme. It is noted that the error function er,s given by (14.2.13) satisfies the finite difference equation (14.3.4). We then substitute (14.2.13) into (14.3.4) and cancel common factors to obtain p =  1 − ε 2 + ε 2 (1 + ε) e −iαh − (1 − ε) e iαh! = 1 − 2 ε 2 sin2  αh 2  − 2iε sin  αh 2  cos  αh 2  , so that |p| 2 = 1 − 4ε 2  1 − ε 2 sin4  αh 2  . (14.3.5) According to the von Neumann criterion, the Lax–Wendroff scheme (14.3.4) is stable as t → ∞ if |p| ≤ 1, which gives 4ε 2  1 − ε 2 ≥ 0, that is, 0 < ε ≤ 1. The local truncation error of the Lax–Wendroff equation (14.3.4) at Pi,j is Ti,j = 1 k (ui,j+1 − ui−1,j ) which is, by (14.2.2a) and (14.2.1b) with ck = h (ε = 1), = (ut + cux) i,j + k 2  utt − c 2uxxx i,j + k 2 6  uttt + c 3uxxx i,j + O
4 (ck) 3
5 . (14.3.6) 14.3 Lax–Wendroff Explicit Method 607 The first two terms on the right side of (14.3.6) vanish by equation (14.3.1) so that the local truncation error becomes Ti,j = 1 6  k 2uttt + c h2uxxx i,j . (14.3.7) Another approximation to (14.3.1) with first-order accuracy is 1 k (ui,j+1 − ui,j ) + c h (ui,j − ui−1,j )=0. (14.3.8) A final explicit scheme for (14.3.1) is based on the central difference approximation. This scheme is called the leap frog algorithm. In this method, the finite difference approximation to (14.3.1) is 1 2k (ui,j+1 − ui,j−1) + c 2h (ui+1,j − ui−1,j )=0, or, ui,j+1 = ui,j−1 − ε (ui+1,j − ui−1,j ). (14.3.9) As shown in Figure 14.3.1, this equation shows that the value of u at Pi,j+1 is computed from the previously computed values at three grid points at two previous time steps. Figure 14.3.1 Grid system for the leap frog algorithm. 608 14 Numerical and Approximation Methods 14.4 Explicit Finite Difference Methods (A) Wave Equation and the Courant–Friedrichs–Lewy Convergence Criterion The method of characteristics provides the most convenient and accurate procedure for solving Cauchy problems involving hyperbolic equations. One of the main advantages of this method is that discontinuities of the initial data propagate into the solution domain along the characteristics. However, when the initial data are discontinuous, the finite difference algorithm for the hyperbolic systems is not very convenient. Problems concerning hyperbolic equations with continuous initial data can be solved successfully by finite difference methods with rectangular grid systems. A commonly cited problem is the propagation of a one-dimensional wave governed by the system utt = c 2 uxx, −∞ <x< ∞,="" t=""> 0, (14.4.1) u (x, 0) = f (x), ut (x, 0) = g (x) for all x ∈ R. (14.4.2) Using a rectangular grid system with h = δx, k = δt, ui,j = u (ih, jk), −∞ <x< ∞,="" and="" 0="" ≤="" j="" <="" the="" central="" difference="" approximation="" to="" equation="" (14.4.1)="" is="" 1="" k="" 2="" (ui,j+1="" −="" ui,j="" +="" ui,j−1)="c" h="" (ui+1,j="" ui−1,j="" ),="" or,="" ui,j+1="ε" )+2="" ="" ε="" ui,j−1,="" (14.4.3)="" where="" ≡="" (ck="" h),="" often="" called="" courant="" parameter.="" this="" explicit="" formula="" allows="" us="" determine="" approximate="" values="" at="" grid="" points="" on="" lines="" t="2k," 3k,="" 4k,="" ...,="" when="" have="" been="" obtained.="" of="" initial="" data="" line="" are="" ui,0="fi" ,="" 2k="" (ui,1="" ui,−1)="gi,0" (14.4.4)="" so="" that="" second="" result="" gives="" ui,−1="ui,1" gi,0.="" (14.4.5)="" in="" used,="" we="" obtain="" ui,1="1" (fi−1="" fi+1)="" fi="" (14.4.6)="" determines="" 14.4="" finite="" methods="" 609="" figure="" 14.4.1="" computational="" systems="" characteristics.="" value="" u="" pi,j+1="" obtained="" terms="" its="" previously="" calculated="" pi="" 1,j="" pi,j="" pi,j−1,="" which="" determined="" from="" computed="" 1)="" k,="" (j="" 2)="" 3)="" k.="" thus,="" computation="" suggests="" will="" represent="" a="" function="" within="" domain="" bounded="" by="" drawn="" back="" toward="" p="" whose="" gradients="" (+="" ε)="" as="" shown="" 14.4.1.="" thus="" triangular="" regions="" ab,="" pcd="" domains="" dependence="" solutions="" differential="" (14.4.1).="" analogy="" with="" real="" characteristic="" c="" d="" equation,="" straight="" b="" numerical="" it="" follows="" ∆p="" ab="" lies="" inside="" cd,="" means="" solution="" system="" would="" remain="" unchanged="" even="" along="" changed.="" courant,="" friedrichs="" lewy="" (cfl,="" 1928)="" proved="" converges="" tend="" zero="" provided="" partial="" equation.="" condition="" for="" convergence="" known="" cfl="" condition,="" ≥="" h,="" is,="" 1.="" if="" parameter="" reduces="" simple="" form="" 610="" 14="" ui,j−1.="" (14.4.7)="" 14.3.1,="" shows="" three="" two="" previous="" time="" steps.="" leap="" frog="" algorithm.="" (5.3.4)="" chapter="" 5,="" know="" cauchy="" problem="" wave="" has="" (x,="" t)="φ" (x="" ct)="" ψ="" ct),="" functions="" φ="" waves="" propagating="" without="" changing="" shape="" negative="" positive="" x="" directions="" constant="" speed="" c.="" slope="" (dt="" dx)="+" (1="" c)="" plane,="" trace="" progress="" waves,="" characteristics="" point="" (xi="" tj="" above="" )="" ).="" (14.4.8)="" 14.3.1="" xi="x1" (i="" takes="" (α="" ih="" jck)="" (β="" jck),="" (14.4.9)="" α="(x1" h)="" (t1="" k),="" β="(x1" k).="" since="" ck="h," becomes="" j)="" h).="" satisfies="" (14.4.7).="" method="" exact="" apply="" von="" neumann="" stability="" analysis="" investigate="" seek="" separable="" error="" er,s="" (iαrh)="" s="" (14.4.10)="" (βk).="" (14.4.3).="" substituting="" into="" cancelling="" common="" factors,="" quadratic="" +1="0," (14.4.11)="" 2ε="" sin2="" (αh="" 2),="" all="" α.="" complex="" roots="" p1="" p2="" ·="" one="" always="" modulus="" greater="" 611="" than="" unless="" |p1|="|p2|" =="" scheme="" unstable="" →="" ∞="" exceeds="" unity.="" other="" hand,="" −1="" 1,="" then="" stable="" leads="" useful="" cosec2="" ="" αh="" ="" .="" (14.4.12)="" limit="" space-grid="" size="" h.="" however,="" true="" example="" find="" utt="" uxx="0," <x<=""> 0, with the boundary conditions u (0, t) = u (1, t)=0, t ≥ 0, and the initial conditions u (x, 0) = sin πx, ut (x, 0) = 0, 0 ≤ x ≤ 1. Compare the numerical solution with the analytical solution u (x, t) = cos πtsin πx at several points. The explicit finite difference approximation to the wave equation with ε = (k/h) = 1 is found from (14.4.3) in the form ui,j+1 = ui−1,j + ui+1,j − ui,j−1, j ≥ 1. The problem is symmetric with respect to x = 1 2 , so we need to calculate the solution only for 0 ≤ x ≤ 1 2 . We take h = k = 1 10 = 0.1. The boundary conditions give u0,j = 0 for j = 0, 1, 2, 3, 4, 5. The initial condition ut (x, 0) = 0 yields ut (x, 0) = 1 2 (ui,1 − ui,−1)=0, or, ui,1 = ui,−1. The explicit formula with j = 0 gives ui,1 = 1 2 (ui−1,0 + ui+1,0), i = 1, 2, 3, 4, 5. Thus, 612 14 Numerical and Approximation Methods u1,1 = 1 2 (u0,0 + u2,0) = 1 2 u2,0 = 1 2 sin (0.2π)=0.2939. Similarly, u2,1 = 0.5590, u3,1 = 0.7695, u4,1 = 0.9045, u5,1 = 0.9511. We next use the basic explicit formula to compute u1,2 = u0,1 + u2,1 − u1,0 =0+0.5590 − 0.3090 = 0.2500, u2,2 = u1,1 + u3,1 − u2,0 = 0.2939 + 0.7695 − 0.5878 = 0.4756. Similarly, we compute other values for ui,j which are shown in Table 14.4.1. The analytical solutions at (x, t) = (0.1, 0.1) and (0.2, 0.3) are given by u (0.1, 0.1) = cos (0.1π) sin (0.1π) = (0.9511) (0.3090) = 0.2939, u (0.2, 0.2) = cos (0.2π) sin (0.2π) = (0.8090) (0.5878) = 0.4577, u (0.2, 0.3) = cos (0.3π) sin (0.2π) = (0.5878) (0.5878) = 0.3455. Comparison of the analytical solutions with the above finite difference solutions shows that the latter results are very accurate. (B) Parabolic Equations As a prototype diffusion problem, we consider ut = κ uxx, 0 <x< 1,="" t=""> 0, (14.4.13) u (0, t) = u (1, t)=0, for all t, (14.4.14) u (x, 0) = f (x), for all x in (0, 1), (14.4.15) where f (x) is a given function. Table 14.4.1. i 0 1 2 3 4 5 x 0.0 0.1 0.2 0.3 0.4 0.5 j t 1 0.1 0 0.2939 0.5590 0.7695 0.9045 0.9511 2 0.2 0 0.2500 0.4577 0.6545 0.7695 0.8090 3 0.3 0 0.1817 0.3455 0.4756 0.5590 0.5878 4 0.4 0 0.9045 0.7695 0.2500 0.2939 0.3090 5 0.5 0 0 0 0 0 0 14.4 Explicit Finite Difference Methods 613 The explicit finite difference approximation to (14.4.13) is 1 k (ui,j+1 − ui,j ) = κ h 2 (ui+1,j − 2 ui,j + ui−1,j ), (14.4.16) or, ui,j+1 = ε (ui+1,j + ui−1,j ) + (1 − 2 ε) ui,j , (14.4.17) where ε =  κk/h2 . This explicit finite difference formula gives approximate values of u on t = (j + 1) k in terms of values on t = jk with given ui,0 = fi . Thus, ui,j can be obtained for all j by successive use of (14.4.17). The problems of stability and convergence of the parabolic equation are similar to those of the wave equation. It can be shown that the solution of the finite difference equation converges to that of the differential equation system (14.4.13)–(14.4.15) as h and k tend to zero provided ε ≤ 1 2 . In particular, when ε = 1 2 , equation (14.4.17) takes a simple form ui,j+1 = 1 2 (ui+1,j + ui−1,j ). (14.4.18) This is called the Bender–Schmidt explicit formula which determines the solution at (xi , tj+1) as the mean of the values at the grid points (i + 1, j). However, more accurate results can be found from (14.4.17) for ε < 1 2 . To investigate the stability of the numerical scheme, we assume that the error function is er,s = exp (iαrh) p s , (14.4.19) where p = e βk. The error function and urs satisfy the same difference equation. Hence, we substitute (14.4.19) into (14.4.17) to obtain p = 1 − 4 ε sin2  αh 2  . (14.4.20) Clearly, p is always less than 1 because ε > 0. If p ≥ 0, the function given by (14.4.19) will decay steadily as s = j → ∞. If −1 <p< 0,="" then="" the="" solution="" will="" have="" a="" decaying="" amplitude="" as="" s="" →="" ∞.="" therefore,="" finite="" difference="" scheme="" be="" stable="" if="" p=""> −1, that is, if 0 < ε ≤ 1 2 cosec2  αh 2  . (14.4.21) This shows that the stability limit depends on h. However, in view of the inequality ε ≤ 1 2 ≤ 1 2 cosec2  αh 2  , we conclude that the stability condition is ε ≤ 1 2 . Finally, if p < −1, the solution oscillates with increasing amplitude as s → ∞, and hence, the scheme will be unstable for ε > 1 2 . 614 14 Numerical and Approximation Methods Example 14.4.2. Show that the Richardson explicit finite difference scheme for (14.4.13) is unconditionally unstable. The Richardson finite difference approximation to (14.4.13) is 1 2k (ui,j+1 − ui,j−1) = κ h 2 (ui+1,j − 2 ui,j + ui−1,j ). (14.4.22) To establish the instability of this equation, we use the Fourier method and assume that er,s = exp (iαrh) p s , p = e βk . This function satisfies the Richardson difference equation as does ur,s. Consequently, p − 1 p = −8 ε sin2  αh 2  , or p 2 + 8 p ε sin2  αh 2  − 1=0. This quadratic equation has two roots p1, p2 = −4 ε sin2  αh 2  +  1 + 16 ε 2 sin4 αh 2 1 2 , (14.4.23) or p1, p2 = + 1 − 4 ε sin2  αh 2 1 + 2 ε sin2 αh 2  + O  ε 4 . This gives |p1| ≤ 1 and |p2| > 1+4 ε sin2  αh 2  > 1 for all positive ε and, consequently, the Richardson scheme is always unstable. The unstable feature of the Richardson scheme can be eliminated by replacing ui,j with 1 2 (ui,j+1 + ui,j−1) in (14.4.22), which now becomes (1 + 2 ε) ui,j+1 = 2 ε (ui+1,j + ui−1,j ) + (1 − 2 ε) ui,j−1. (14.4.24) This is called the Du Fort–Frankel explicit algorithm, and it can be shown to be stable for all ε. 14.4 Explicit Finite Difference Methods 615 Example 14.4.3. Prove that the solution of the finite difference equation for the diffusion equation (14.4.13) in −∞ <x< ∞="" with="" the="" initial="" condition="" u="" (x,="" 0)="e" iαx="" converges="" to="" exact="" solution="" of="" (14.4.13)="" as="" h="" and="" k="" tend="" zero.="" we="" obtain="" by="" seeking="" a="" separable="" form="" t)="e" v="" (t),="" where="" (t)="" is="" function="" t="" alone="" which="" be="" determined.="" substituting="" this="" into="" gives="" dv="" dt="" +="" κ="" α2="" admits="" solutions="" e−κα2="" ,="" an="" integrating="" constant.="" (0)="1" hence,="" ="" −="" α="" 2κt="" .="" (14.4.25)="" now="" solve="" corresponding="" finite="" difference="" equation="" (14.4.17)="" replacing="" i,="" j="" r,="" s.="" seek="" ur,s="e" iαrh="" vs="" ur,0="e" v0,="" so="" that="" v0="1." (14.4.7)="" yields="" vs+1="" 1="" 4="" ε="" sin2="" αh="" 2="" ="" vs,="" can="" obtained="" simple="" inspection="" s="" (14.4.26)="" ="" 4ε="" (14.4.27)="" κk="" h2="" for="" small="" h,="" (αh="" 2)="" ∼="" α2h="" ≈="" exp="" −ε="" α2k.="" consequently,="" final="" becomes="" 616="" 14="" numerical="" approximation="" methods="" e="" iαrh−κα2ks="" →="" 0.="" (14.4.28)="" identical="" differential="" rh="x" sk="t." example="" shows="" reasonably="" good.="" 14.4.4.="" calculate="" boundaryvalue="" problem="" ut="uxx," 0="" <x<="" 1,=""> 0, with the boundary conditions u (0, t) = u (1, t)=0, t ≥ 0, and the initial condition u (x, 0) = x (1 − x), 0 ≤ x ≤ 1. Compare the numerical solution with the exact analytical solution at x = 0.04 and t = 0.02. The explicit finite difference approximation to the parabolic equation is ui,j+1 = ε ui−1,j + (1 − 2 ε) ui,j + ε ui+1,j , where ε =  k/h2 . This gives the unknown value ui,j+1 at the (i, j + 1) th grid point in terms of given values of u along the jth time row. We set h = 1 5 and k = 1 100 so that ε =  k/h2 = 1 4 and the above formula becomes ui,j+1 = 1 4 (ui−1,j + 2 ui,j + ui+1,j ). With the notation ui,0 = u (ih, 0), the initial condition gives u4,0 = 0.16, and u5,0 = 0. The boundary conditions yield u0,j = u (0, jk) = 0 and u5,j = u (5h, jk) = u (1, jk) = 0, for all j = 0, 1, 2, .... Using these initial and boundary data, we calculate ui,j as follows: u1,1 = 1 4 (u0,0 + 2u1,0 + u2,0)=0.14, u1,2 = 1 4 (u0,1 + 2u1,1 + u2,1) = 0.125, u2,1 = 1 4 (u1,0 + 2u2,0 + u3,0)=0.22, u2,2 = 1 4 (u1,1 + 2u2,1 + u3,1) = 0.200, u3,1 = 1 4 (u2,0 + 2u3,0 + u4,0)=0.22, u3,2 = 1 4 (u2,1 + 2u3,1 + u4,1) = 0.200, u4,1 = 1 4 (u3,0 + 2u4,0 + u5,0)=0.14, u4,2 = 1 4 (u3,1 + 2u4,1 + u5,1) = 0.125, 14.4 Explicit Finite Difference Methods 617 u1,3 = 1 4 (u0,2 + 2u1,2 + u2,2)=0.1125, u1,4 = 1 4 (u0,3 + 2u1,3 + u2,3) = 0.1016, u2,3 = 1 4 (u1,2 + 2u2,2 + u3,2)=0.1813, u2,4 = 1 4 (u1,3 + 2u2,3 + u3,3) = 0.1641, u3,3 = 1 4 (u2,2 + 2u3,2 + u4,2)=0.1813, u3,4 = 1 4 (u2,3 + 2u3,3 + u4,3) = 0.1641, u4,3 = 1 4 (u3,2 + 2u4,2 + u5,2)=0.1125, u4,4 = 1 4 (u3,3 + 2u4,3 + u5,3) = 0.1016. The method of separation of variables gives the analytical solution of the problem as u (x, t) = 8 π 3 ∞ n=0 1 (2n + 1)3 exp " − (2n + 1)2 π 2 t # sin (2n + 1) πx. This exact solution u (x, t) at x = 0.4 (i = 2) and t = 0.02 (j = 2) gives u ∼ 8 3 1 1 3 exp  −0.02 π 2 sin (0.4) π + 1 3 3 exp  −0.18 π 2 sin (1.2) π = 0.2000. The analytical solution is seen to be identical with the numerical value. Example 14.4.5. Obtain the numerical solution of the initial boundaryvalue problem ut = κ uxx, 0 ≤ x ≤ 1, t > 0, u (0, t)=1, u (1, t)=0, t ≥ 0, u (x, 0) = 0, 0 ≤ x ≤ 1. We use the explicit finite-difference formula (14.4.17) ui.j+1 = ε (ui+1,j + ui−1,j ) + (1 − 2 ε) ui,j , where ε =  κk/h2 . We set h = 0.25 = 1 4 and ε = 2 5 = 0.4 to compute ui.j for i, j = 0, 1, 2, 3, 4 as follows: 618 14 Numerical and Approximation Methods i 0 1 2 3 4 j 0 1.000 0.000 0.000 0.000 0.000 1 1.000 0.400 0.000 0.000 0.000 2 1.000 0.480 0.160 0.000 0.000 3 1.000 0.560 0.224 0.064 0.000 4 1.000 0.602 0.295 0.103 0.000 (C) Elliptic Equations As a prototype boundary-value problem, we consider the Dirichlet problem for the Laplace equation ∇2u ≡ uxx + uyy = 0, 0 ≤ x ≤ a, 0 ≤ y ≤ b, (14.4.29) where the value of u (x, y) is prescribed everywhere on the boundary of the rectangular domain. The rectangular grid system is the most common and convenient system for this problem. We choose the vertices of the rectangular domain as the nodal points and set h = a/m and k = b/n where m and n are positive integers so that the domain is divided into mn subrectangles. The finite difference approximation to the Laplace equation (14.4.29) is 1 h 2 (ui+1,j − 2ui,j + ui−1,j ) + 1 k 2 (ui,j+1 − 2ui,j + ui,j−1)=0,(14.4.30) or, 2  h 2 + k 2 ui,j = k 2 (ui+1,j + ui−1,j ) + h 2 (ui,j+1 + ui,j−1), (14.4.31) where 1 ≤ i ≤ m − 1 and 1 ≤ j ≤ n − 1. The prescribed conditions on the boundary of the rectangular domain determine the values u0,j , um,j , ui,0, and ui,n. For a square grid system (k = h), equation (14.4.30) becomes ui,j = 1 4 (ui+1,j + ui−1,j + ui,j+1 + ui,j−1). (14.4.32) This means that the value of u at an interior point is equal to the average of the value of u at four adjacent points. This is the well known mean value theorem for harmonic functions that satisfy the Laplace equation. As i and j vary, the present scheme reduces to a set of (m − 1) (n − 1) linear non-homogeneous algebraic equations for (m − 1) (n − 1) unknown values of u at interior grid points. It can be shown the solution of the finite 14.4 Explicit Finite Difference Methods 619 difference equation (14.4.31) converges to the exact solution of the problem as h, k → 0. The proof of the existence of a solution and its convergence to the exact solution as h and k tend to zero is essentially based on the Maximum Modulus Principle. It follows from the finite difference equation (14.4.30) or (14.4.31) that the value of |u| at any interior grid point does not exceed its value at any of the four adjoining nodal points. In other words, the value of u at Pi,j cannot exceed its values at the four adjoining points Pi + 1,j and Pi,j + 1. The successive application of this argument at all interior grid points leads to the conclusion that |u| at the interior grid points cannot be greater than the maximum value of |u| on the boundary. This may be recognized as the finite difference analogue of the Maximum Modulus Principle discussed in Section 9.2. Thus, the success of the numerical method is directly associated with the existence of the Maximum Modulus Principle. Clearly, the present numerical algorithm deals with a large number of algebraic equations. Even though numerical accuracy can be improved by making h and k sufficiently small, there is a major computational difficulty involved in the numerical solution of a large number of equations. It is possible to handle such a large number of algebraic equations by direct methods or by iterative methods, but it would be very difficult to obtain a numerical solution with sufficient accuracy. It is therefore necessary to develop some alternative methods of solution that can be conveniently and efficiently carried out on a computer. In order to eliminate some of the drawbacks stated above, one of the numerical schemes, the Liebmann’s iterative method, is useful. In this method values of u are first guessed for all interior grid points in addition to those given as the boundary points on the edges of the given domain. These values are denoted by u (0) i,j where the superscript 0 indicates the zeroth iteration. It is convenient to choose a square grid so that the simplified finite difference equation (14.4.32) can be used. The values of u are calculated for the next iteration by using (14.4.32) at every interior point based on the values of u at the present iteration. The sequence of computation starts from the interior grid point located at the lowest left corner, proceeds upward until reaching the top, and then goes to the bottom of the next vertical line on the right. This process is repeated until the new value of u at the last interior grid point at the upper right corner has been obtained. At the starting point, formula (14.4.32) gives u (1) 2,2 = 1 4
4 u (0) 3,2 + u (0) 1,2 + u (0) 2,3 + u (0) 2,1
5 , (14.4.33) where u (0) 1,2 and u (0) 2,1 are boundary values which remain constant during the iteration process. They may be replaced, respectively, with u (1) 1,2 , u (1) 2,1 in (14.4.33). The computation at the next step involves u (0) 2,2 . Since an improved value u (1) 2,2 is available at this time, it will be utilized instead. Hence, 620 14 Numerical and Approximation Methods u (1) 2,3 = 1 4
4 u (0) 3,3 + u (1) 1,3 + u (0) 2,4 + u (1) 2,2
5 , (14.4.34) where u (1) 1,3 is used to replace the constant boundary value u (0) 1,3 . We repeat this argument to obtain a general iteration formula for computation of u at step (n + 1) u (n+1) i,j = 1 4
4 u (n) i+1,j + u (n+1) i−1,j + u (n) i,j+1 + u (n+1) i,j−1
5 . (14.4.35) This result is valid for any interior point, whether it is next to some boundary point or not. If Pi,j is a true point, the second and fourth terms on the right side of (14.4.35) represent, respectively, the values of u at the grid points to the left of and below that point. These values have already been recomputed according to our scheme, and therefore, carry the superscript (n + 1). Result (14.4.35) is known as the Liebmann (n + 1) th iteration formula. It can be proved that u (n) i,j converges to ui,j as n → ∞. Another iteration scheme similar to (14.4.35) is given by u (n+1) i,j = 1 4
4 u (n) i+1,j + u (n) i−1,j + u (n) i,j+1 + u (n) i,j−1
5 . (14.4.36) This is called the Richardson iteration formula, and it is also useful. However, this scheme converges more slowly than that based on (14.4.35). One of the major difficulties of the above methods is the slow rate of convergence. An improved numerical method, the Successive Over-Relaxation (SOR) scheme gives a faster convergence than the Liebmann or Richardson method in solving the Laplace (or the Poisson) equation. For a rectangular domain of square grids, the successive iteration scheme is given by u (n+1) i,j = u (n) i,j + ω 4
4 u (n+1) i−1,j + u (n) i+1,j + u (n+1) i,j−1 + u (n) i,j+1 − 4 u (n) i,j
5 , (14.4.37) where ω is called the acceleration parameter (or relaxation factor ) to be determined. In general, ω lies in the range 1 ≤ ω < 2. The successive iterations converge fairly rapidly to the desired solution for 1 ≤ ω < 2. The most rapid rate of convergence is achieved for the optimum value of ω. Example 14.4.6. Obtain the standard five-point formula for the Poisson equation uxx + uyy = −f (x, y) in D ⊂ R 2 with the prescribed value of u (x, y) on the boundary ∂D. We assume that the domain D is covered by a system of squares with sides of length h parallel to the x and y axes. Using the central difference approximation to the Laplace operator, we obtain 1 h 2 (ui+1,j − 2 ui,j + ui−1,j ) + 1 h 2 (ui,j+1 − 2 ui,j + ui,j−1) = −fi,j , 14.4 Explicit Finite Difference Methods 621 or, ui,j = 1 4 (ui+1,j + ui−1,j + ui,j+1 + ui,j−1) + 1 4 h 2 fi,j where fi,j = f (ih, jh). This is known as the five-point formula. Example 14.4.7. Find the numerical solution of the torsion problem in a square beam governed by ∇2u = −2 in D = {(x, y):0 ≤ x ≤ 1, 0 ≤ y ≤ 1} with u (x, y) = 0 on ∂D. From the above five-point formula, we obtain ui,j = 1 4 (ui+1,j + ui−1,j + ui,j+1 + ui,j−1) − 1 2 h 2 where h is the side-length of the unit square net. We choose h = 1 2 , 1/2 2 , 1/2 3 , 1/2 4 to calculate the corresponding numerical values ui,j = 0.1250, 0.1401, 0.1456, 0.1469. Note that the known exact analytical solution is 0.1474. Example 14.4.8. Using the explicit finite difference method, find the solution of the Dirichlet problem uxx + uyy = 0, in 0 <x< 1,="" 0="" <y<="" u="" (x,="" 0)="x," 1)="0," on="" ≤="" x="" y)="0," for="" and="" y="" 1.="" we="" use="" four="" interior="" grid="" points="" (that="" is,="" i,="" j="1," 2,="" 3,="" 4)="" as="" shown="" in="" figure="" 14.4.2="" the="" y)-plane.="" apply="" explicit="" finite="" difference="" formula="" (14.4.32)="" to="" obtain="" algebraic="" equations="" −4u2,2="" +="" u3,2="" u1,2="" u2,3="" u2,1="0," −4u2,3="" u3,3="" u1,3="" u2,4="" u2,2="0," −4u3,2="" u4,2="" u3,1="0," −4u3,3="" u4,3="" u3,4="" given="" boundary="" conditions="" imply="" that="" =="" 0,="" 3="" so="" above="" system="" of="" becomes="" 1="" 2="" 622="" 14="" numerical="" approximation="" methods="" square="" system.="" matrix="" notation,="" this="" reads="" ⎡="" ⎢="" ⎣="" −41="" −40="" −4="" 011="" ⎤="" ⎥="" ⎦="" −="" .="" solutions="" are="" 72="" ,="" (d)="" simultaneous="" first-order="" recall="" wave="" equation="" (14.4.1)="" <x<="" t=""> 0. Introducing two auxiliary variables v and w by v = ut and w = c 2ux, the wave equation gives two simultaneous first-order equations vt = wx, wt = c 2 vx. (14.4.38) 14.4 Explicit Finite Difference Methods 623 The initial values of v and w are given at t = 0 for all x in 0 <x< 1.="" the="" boundary="" condition="" on="" v="" and="" w="" is="" also="" prescribed="" lines="" x="0" for="" t=""> 0. The explicit finite difference method can be used to determine v and w in the triangular domain of dependence bounded by the characteristics x − ct = 0 and x + ct = 1. The finite difference approximations to the differential equations (14.4.38) are 1 k (vi,j+1 − vi,j ) = 1 2h (wi+1,j − wi−1,j ), (14.4.39) 1 k (wi,j+1 − wi,j ) = c 2 2h (vi+1,j − vi−1,j ), (14.4.40) where the forward difference for vt or wt and the central difference for vx or wx are used. However, the central difference approximations to (14.4.38) can also be utilized to obtain 1 2k (vi,j+1 − vi,j−1) = 1 2h (wi+1,j − wi−1,j ), (14.4.41) 1 2k (wi,j+1 − wi,j−1) = c 2 2h (vi+1,j − vi−1,j ). (14.4.42) We examine the stability of the above two sets of finite difference formulas with c = 1. The von Neumann stability method is applied by replacing i and j by r and s respectively. The error function er,s given by (14.4.10) is substituted in (14.4.39)–(14.4.40) to obtain the stability relations A (p − 1) = εiB sin αh, (14.4.43) B (p − 1) = εiA sin αh, (14.4.44) where the initial perturbations in v and w along t = 0 are A exp (iαrh) and B exp (iαrh) respectively with two different constants A and B. Elimination of A and B from the above relations gives (p − 1)2 + ε 2 sin2 αh = 0 or p = 1+ iε sin αh, and |p| =  1 + ε 2 sin2 αh 1 2 ∼ 1 + 1 2 ε 2 sin2 αh =1+ O  ε 2 . (14.4.45) Since |p| > 1 + O (ε), the finite difference scheme for the finite time-step t = sk would be unstable as the grid sizes tend to zero. A similar stability analysis for (14.4.41)–(14.4.42) leads to the condition 624 14 Numerical and Approximation Methods  p − 1 p 2 + 4 ε 2 sin2 αh = 0. (14.4.46) This scheme is stable for ε ≤ 1. Another finite difference approximation to the coupled system (14.4.38) is 1 2h (vr+1,s − vr−1,s) = 1 k wr,s+1 − 1 2 (wr+1,s − wr−1,s) , (14.4.47) 1 2h (wr+1,s − wr−1,s) = 1 k vr,s+1 − 1 2 (vr+1,s − vr−1,s) . (14.4.48) A similar stability analysis can be carried out for these systems by substituting vr,s = Aps e iαrh and wr,s = Bps e iαrh into the equations. Elimination of A/B yields the stability equation p = cos αh + i ε sin αh, or |p| 2 = cos2 αh + 1 ε 2 sin2 αh ≤ 1. (14.4.49) Hence, the scheme is stable provided that ε ≥ 1, that is, k ≤ h. (14.4.50) 14.5 Implicit Finite Difference Methods From a computational point of view, the explicit finite difference algorithm is simple and convenient. However, as shown in Section 14.4(B), the major difficulty in the method for solving parabolic partial differential equations is the severe restriction on the time-step imposed by the stability condition ε ≤ 1 2 or k ≤ h 2/2κ. This difficulty is also present in the explicit finite difference method for the solution of hyperbolic equations. In order to overcome the above difficulty, we develop implicit finite difference schemes for solving partial differential equations. (A) Parabolic Equations One of the successful implicit finite difference schemes is the Crank and Nicolson Method (1947), which is based on six grid points. This method eliminates the major difficulty involved in the explicit scheme. When the Crank–Nicolson implicit scheme is applied to the parabolic equation (14.4.13), uxx is replaced by the mean value of the finite difference values 14.5 Implicit Finite Difference Methods 625 in the jth and the (j + 1) th row so that the finite difference approximation (14.4.13) becomes 1 k (ui,j+1 − ui,j ) = κ 2h 2 [(ui+1,j+1 − 2ui,j+1 + ui−1,j+1) + (ui+1,j − 2ui,j + ui−1,j )] , (14.5.1) or 2 (1 + ε) ui,j+1 − ε (ui−1,j+1 + ui+1,j+1) = 2 (1 − ε) ui,j + ε (ui−1,j + ui+1,j ), (14.5.2) where ε =  kκ/h2 is a parameter. The left side of (14.5.2) is a linear combination of three unknowns in the (j + 1) th row, and the right side involves three known values of u in the jth row of the grid system in the (x, t)-plane. Equation (14.5.2) is called the Crank–Nicolson implicit formula. This formula (or its suitable modification) is widely used for solving parabolic equations. If there are n internal grid points along each jth row, then, for j = 0 and i = 1, 2, 3, ..., n, the implicit formula (14.5.2) gives n simultaneous algebraic equations for n unknown values of u along the first jth row (j = 0) in terms of given boundary and initial data. Similarly, if j = 1 and i = 1, 2, 3, ..., n, equation (14.5.2) represents n unknown values of u along the second jth row (j = 1) and so on. This means that the method involves the solution of a system of simultaneous algebraic equations. In practice, the Crank–Nicolson scheme is convergent and unconditionally stable for all finite values of ε, and has the advantage of reducing the amount of numerical computation. This implicit scheme can be further generalized by introducing a numerical weight factor λ in the modified version of the explicit equation (14.4.16) which is written below by approximating uxx in (14.4.13) in the (j + 1) th row instead of the jth row. 1 k (ui,j+1 − ui,j ) = κ h 2 (ui+1,j+1 − 2ui,j+1 + ui−1,j+1), (14.5.3) or ui,j+1 − ui,j = ε (ui+1,j+1 − 2ui,j+1 + ui−1,j+1). (14.5.4) Introducing the numerical factor λ, this can be replaced by a more general difference equation in the form ui,j+1 − ui,j = ε λ δ2 x ui,j+1 + (1 − λ) δ 2 x ui,j ! , (14.5.5) where 0 ≤ λ ≤ 1 and δ 2 x is the difference operator defined by δ 2 xui,j = ui+1,j − 2ui,j + ui−1,j . (14.5.6) Another equivalent form of (14.5.5) is 626 14 Numerical and Approximation Methods (1 + 2ελ) ui,j+1 − ελ (ui+1,j+1 + ui−1,j+1) = {1 − 2ε (1 − λ)} ui,j + ε (1 − λ) (ui+1,j − ui−1,j ). (14.5.7) This is a fairly general implicit formula which reduces to (14.5.4) when λ = 1. When λ = 1 2 , (14.5.7) becomes the Crank–Nicolson formula (14.5.2). Finally, if λ = 0, this implicit difference equation reduces to the explicit equation (14.4.17). The Richardson explicit scheme was found to be unconditionally unstable in Section 14.4. This undesirable feature of the scheme can be eliminated by considering the corresponding implicit scheme. In terms of δ 2 x , the Richardson equation (14.4.22) can be expressed as ui,j+1 = 2 ε δ2 x ui,j + ui,j−1. (14.5.8) To obtain the implicit Richardson formula, we replace δ 2 x ui,j by 1 3 δ 2 x (ui,j+1 + ui,j + ui,j−1) in (14.5.8) and we obtain  1 − 2 ε 3 δ 2 x  ui,j+1 = 2ε 3 δ 2 xui,j +  1 + 2ε 3  ui,j−1. (14.5.9) This implicit scheme can be shown to be unconditionally stable. To prove this result, we apply the von Neumann stability method with the error function (14.5.9) to obtain the equation for p as (1 + a) p 2 + ap + (a − 1) = 0, (14.5.10) where a ≡  8ε 3  sin2  αh 2  . (14.5.11) The roots of the quadratic equation are p = −a +  4 − 3a 2 1 2 2 (1 + a) . (14.5.12) This gives |p| ≤ 1 for all values of a. Hence, the result is proved. Example 14.5.1. Obtain the numerical solution of the following parabolic system by using the Crank–Nicolson method ut = uxx, 0 <x< 1,="" t=""> 0, u (0, t) = u (1, t)=0, t ≥ 0, u (x, 0) = x (1 − x), 0 ≤ x ≤ 1. We recall the Crank–Nicolson equation (14.5.2) and then set h = 0.2 and k = 0.01 so that ε = 1 4 . The boundary and initial conditions give 14.5 Implicit Finite Difference Methods 627 u0,0 = u5,0 = u0,1 = u5,1 = 0 and ui,0 = u (ih, 0) = ih (1 − ih), i = 1, 2, 3, 4. Consequently, formula (14.5.2) leads to the following system of four equations: −u0,1 − u2,1 + 10u1,1 = u0,0 + u2,0 + 6u1,0 −u1,1 − u3,1 + 10u2,1 = u1,0 + u3,0 + 6u2,0 −u2,1 − u4,1 + 10u3,1 = u2,0 + u4,0 + 6u3,0 −u3,1 − u5,1 + 10u4,1 = u3,0 + u5,0 + 6u4,0. Using the boundary and initial conditions, the above system becomes −u2,1 + 10u1,1 = 1.20 −u1,1 + 10u2,1 − u3,1 = 1.84 −u2,1 − u4,1 + 10u3,1 = 1.84 −u3,1 + 10u4,1 = 1.20. These equations can be solved by direct elimination to obtain the solutions as u1,1 = 0.1418, u2,1 = 0.2202, u3,1 = 0.2202, u4,1 = 0.1420. (B) Hyperbolic Equations We consider an implicit finite difference scheme to solve the initial boundaryvalue problem consisting of the first-order hyperbolic equation ∂u ∂t + c ∂u ∂x = 0, (c > 0), (14.5.13) with the initial data u (x, 0) = U (x) and the boundary condition u (0, t) = V (t) where 0 ≤ x, t < ∞. The implicit finite difference approximation to (14.5.13) is 1 k (ui,j+1 − ui,j ) + c h (ui,j+1 − ui−1,j+1)=0, or ui,j = (1 + ε) ui,j+1 − ε ui−1,j+1, (14.5.14) where ε = (ck/h). The stability of the scheme can be examined by using the von Neumann method with the error function (14.4.10). It turns out that p = [1 − ε + ε exp (−iαh)]−1 , (14.5.15) from which it follows that |p| ≤ 1 for all h. Hence, the implicit scheme is unconditionally stable. 628 14 Numerical and Approximation Methods We next solve the wave equation utt = c 2uxx by an implicit finite difference scheme. In this case, utt is replaced by the central difference formula, and uxx by the mean value of the central difference values in the (j − 1) th and (j + 1) th rows. Consequently, the implicit difference approximation to the wave equation is ui,j+1 − 2 ui,j + ui,j−1 = ε 2 2 [(ui+1,j+1 − 2ui,j+1 + ui−1,j+1) + (ui+1,j−1 − 2 ui,j−1 + ui−1,j−1)] , (14.5.16) where ε = (ck/h). Expressing the solution for the (j + 1) th step in terms of the two preceding steps gives 2  1 + ε 2 ui,j+1 − ε 2 (ui−1,j+1 + ui+1,j+1) = 4ui,j + ε 2 (ui−1,j−1 + ui+1,j−1) − 2  1 + ε 2 ui,j−1. (14.5.17) The N grid points along the time step, j = 0, i = 1, 2, 3, ..., N, (14.5.17) along with the finite difference approximation to the boundary condition give N simultaneous equations for the N unknown values of u along the first time step. This constitutes a tridiagonal system of equations that can be solved by direct or iterative numerical methods. To investigate the stability of the implicit scheme, we apply the von Neumann stability method with the error function (14.4.10). This leads to the equation p + 1 p = 2  1+2 ε 2 sin2 αh 2 −1 , or p 2 − 2bp +1=0, (14.5.18) where b =  1+2ε 2 sin2 αh/2 −1 so that 0 < b ≤ 1. Hence, the stability condition is |p| ≤ 1 (14.5.19) which is always satisfied provided 0 < b ≤ 1, that is, ε < 1 for all positive h. This confirms the unconditional stability of the scheme. A more general implicit scheme can be introduced by replacing uxx in the wave equation (14.4.1) with uxx ∼ 1 h 2 λ  δ 2 x ui,j+1 + δ 2 x ui,j−1 + (1 + 2λ) δ 2 x ui,j ! , (14.5.20) 14.6 Variational Methods and the Euler–Lagrange Equations 629 where λ is a numerical weight (relaxation) factor and the central difference operator δ 2 x is given by (14.5.6). This general scheme allows us to approximate the wave equation with c = 1 by the form δ 2 t ui,j = ε 2 λ  δ 2 x ui,j+1 + δ 2 x ui,j−1 + (1 − 2λ) δ 2 x ui,j ! , (14.5.21) where ε = k/h. This equation reduces to (14.5.16) when λ = 1 2 , and to the explicit finite difference result when λ = 0. It follows from von Neumann stability analysis that the implicit scheme is unconditionally stable for λ ≥ 1 4 . Von Neumann introduced another fairly general finite difference algorithm for the wave equation (14.4.1) in the form δ 2 t ui,j = ε 2 δ 2 xui,j + ω h 2 δ 2 t δ 2 xui,j . (14.5.22) This equation with appropriate boundary conditions can be solved by the tridiagonal method. Von Neumann discussed the question of stability of this implicit scheme and proved that the scheme is conditionally stable if ω ≤ 1 4 and unconditionally stable if ω > 1 4 . 14.6 Variational Methods and the Euler–Lagrange Equations To describe the variational methods and Rayleigh–Ritz approximate method, it is convenient to introduce the concepts of the inner product (pre-Hilbert) and Hilbert spaces. An inner product space X consisting of elements u, v, w, ... over the complex number field C is a complex linear space with an inner product u, v : X × X → C such that (i) u, v = v, u, where the bar denotes the complex conjugate of v, u, (ii) αu + βv, w = α u, w + β v, w for any scalars α, β ∈ C, (iii) u, u ≥ 0; equality holds if and only if u = 0. By (i) u, u = u, u, and so u, u is real. We denote u, u 1 2 = u, which is called the norm of u. Thus, the norm is induced by the inner product. Thus, every inner product space is a normed linear space under the norm u =  u, u. Let X be an inner product space. A sequence {un} where un ∈ X for every n is called a Cauchy sequence in X if and only if for every given ε > 0 (no matter how small) we can find an N (ε) such that un − um < ε for all n, m > N (ε). The space X is called complete if every Cauchy sequence converges to a point in X. A complete normed linear space is called a Banach Space. A complete linear inner product space is called a Hilbert Space and is usually denoted by H. 630 14 Numerical and Approximation Methods Example 14.6.1. Let C n be the set of all n-tuples of complex numbers. Thus, C n is an n-dimensional Hilbert space with the inner product x, y = n k=1 xk yk. Obviously, the set of all n-tuples of real numbers Rn is an n-dimensional Hilbert space. Example 14.6.2. Let l2 be the set of all sequences with entries from C such that 2∞ k=1 |xk| 2 < ∞. This forms a Hilbert space with the inner product x, y = ∞ k=1 xk yk. Example 14.6.3. Let L2 ([a, b]) be the set of all square integrable functions in the Lebesgue sense in an interval [a, b]. L2 ([a, b]) is a Hilbert space with the inner product u, v =
b a u (x) v (x) dx. We next introduce the notion of an operator in a Hilbert space H. An operator A is a mapping from H to H (that is, A : H → H). It assigns to an element u in H a new element Au in H. An operator A is called linear if it satisfies the property A (αu + βv) = αAu + βAv for every α, β ∈ C. An operator is said to be bounded if there exists a constant k such that Au ≤ k u for all u ∈ H. We consider a bounded operator A on a Hilbert space H. For a fixed element v in H, the inner product Au, v in H can be regarded as a number I (u) which varies with u. Thus, Au, v = I (u) is a linear functional on H. If there exists an operator A∗ on a Hilbert space (A∗ : H → H) such that Au, v = u, A∗ v for all u, v ∈ H, then A∗ is called the adjoint of A. In general, A = A∗ . If A = A∗ , that is, Au, v = u, Av for all u, v in H, then A is called self-adjoint. It is important to note that any bounded operator T on a real Hilbert space (T : H → H) of the form T = A∗A is self-adjoint. This follows from the fact that T u, v = A ∗Au, v = Au, Av = u, A∗Au = u, T v. 14.6 Variational Methods and the Euler–Lagrange Equations 631 A self-adjoint operator A on a Hilbert space H is said to be positive if Au, u ≥ 0 for all u in H, where equality implies that u = 0 in H. Further, if there exists a positive constant k such that Au, u ≥ k u, u for all u in H, then A is called positive definite in H. The rest of this section is essentially concerned with linear operators in a real Hilbert space, which means that the associated scalar field involved is real. Some specific inner products which will be used in the subsequent sections include u, v =
b a u (x) v (x) dx, u, v =

D u (x, y) v (x, y) dx dy, where D ⊂ R2 . Example 14.6.4. Determine whether the differentiable operators (i) A = d/dx, (ii) A = d 2/dx2 , and (iii) A = ∇2 =  ∂ 2/∂x2 +  ∂ 2/∂y2 are selfadjoint for functions that are differentiable in a ≤ x ≤ b or in D ⊂ R2 and vanish on the boundary. (i) Au, v =
b a  du dx v dx =
b a u  − dv dx dx + [u, v] b a = u, A∗ v where A ∗ = − d dx = A. Hence, A is not self-adjoint. (ii) Au, v =
b a  d 2u dx2  v dx =
b a  du dx− dv dx dx + v du dxb a =
b a u  d 2v dx2  dx + v du dx − u dv dxb a =
b a u  d 2v dx2  dx = u, Av. Thus, A is self-adjoint. (iii) Au, v =

D  ∇2u v dx dy =

D [∇ · (∇u) v − ∇u · ∇v] dx dy =
∂D (n6 · ∇u) v dS −

D ∇u · ∇v dx dy = −

D (∇u · ∇v) dx dy, where the divergence theorem is used with the unit outward normal vector n6. Noting the symmetry of the right hand side in u and v, it follows that Au, v = Av, u = u, Av. This means that the operator A = ∇2 is self-adjoint. 632 14 Numerical and Approximation Methods Example 14.6.5. Use the inner product u, v = *** D (u · v) dV and the operator A = grad to show that A∗ = −div provided the functions vanish on the boundary surface ∂D of D. We use the divergence theorem to obtain Aφ, v =


D (grad φ · v) dV =


D [div (φv) − φ div v] dV, =


D φ (−div v) dV +

∂D (n6 · φv) dS, =


D φ (−div v) dV = φ, −div v = φ, A∗ v. In the theory of calculus of variations it is a common practice to use δu, δ 2u, etc. to denote the first and second variations of a function u. Thus, δ can be regarded as an operator that changes u into δu, ux into δ (ux), and uxx into δ (uxx) with the meaning, δu = εv, δ (ux) = εvx, δ (uxx) = εvxx, where ε is a small arbitrary real parameter. The operators δ, δ 2 are called the first and second variational operators respectively. Some simple properties of the operator δ are given by ∂ ∂x (δu) = ∂ ∂x (εv) = ε ∂v ∂x = δ  ∂u ∂x , (14.6.1) δ 1
b a u dx3 = ε
b a v dx =
b a ε v dx =
b a (δu) dx. (14.6.2) The variational operator can be interchanged with the differential and integral operators, and proves to be very useful in the calculation of the variation of a functional. The main task of the calculus of variations is concerned with the problem of minimizing or maximizing functionals involved in mathematical, physical and engineering problems. The variational principles have their origins in the simplest kind of variational problem, which was first considered by Euler in 1744 and Lagrange in 1760-61. The classical Euler–Lagrange variational problem is to determine the extremum value of the functional I (u) =
b a F (x, u, u′ ) dx, u′ = du dx, (14.6.3) with the boundary conditions u (a) = α and u (b) = β, (14.6.4) where u belongs to the class C 2 ([a, b]) of functions which have continuous derivatives up to second-order in a ≤ x ≤ b, and F has continuous secondorder derivatives with respect to all of its arguments. 14.6 Variational Methods and the Euler–Lagrange Equations 633 We assume that I (u) has an extremum at some u ∈ C (2) ([a, b]). Then we consider the set of all variations u + ǫv for fixed u where v is an arbitrary function belonging to C 2 ([a, b]) such that v (a) = v (b) = 0. We next consider the increment of the functional δI = I (u + εv) − I (u) =
b a [F (x, u + ε v, u′ + ε v′ ) − F (x, u, u′ )] dx. (14.6.5) From the Taylor series expansion F (x, u + ε v, u′ + ε v′ ) = F (x, u, u′ ) + ε  v ∂F ∂u + v ′ ∂F ∂u′  + ε 2 2!  v ∂F ∂u + v ′ ∂F ∂u′ 2 + ··· , it follows from (14.6.5) that I (u + εv) = I (u) + ε δI + ε 2 2! δ 2 I + ··· , (14.6.6) where the first and second variations of I are given by δI =
b a  v ∂F ∂u + v ′ ∂F ∂u′  dx, (14.6.7) δ 2 I =
b a  v ∂F ∂u + v ′ ∂F ∂u′ 2 dx. (14.6.8) The necessary condition for the functional I (u) to have an extremum (that is, I (u) is stationary at u) is that the first variation becomes zero at u so 0 = δI =
b a  v ∂F ∂u + v ′ ∂F ∂u′  dx (14.6.9) which is, by partial integration of the second integral, =
b a ∂F ∂u − d dx  ∂F ∂u′  v dx + v ∂F ∂u′ b a . Because v (a) = v (b) = 0, this means that
b a ∂F ∂u − d dx  ∂F ∂u′  v dx = 0. (14.6.10) Since v is arbitrary in a ≤ x ≤ b, it follows from (14.6.10) that ∂F ∂u − d dx  ∂F ∂u′  = 0. (14.6.11) This is the famous Euler–Lagrange equation. We therefore can state: 634 14 Numerical and Approximation Methods Theorem 14.6.1. A necessary condition for the functional I (u) to be stationary at u is that u is a solution of the Euler–Lagrange equation ∂F ∂u − d dx  ∂F ∂u′  = 0, a ≤ x ≤ b (14.6.12) with u (a) = α, u (b) = β. (14.6.13) This is called the Euler–Lagrange variational principle. Note that, in general, equation (14.6.12) is a nonlinear second-order ordinary differential equations, and, although such an equation is very difficult to solve, still it seems to be more accessible analytically than the functional (14.6.3) from which it is derived. The derivative d dx in (14.6.12) can be computed by recalling u = u (x) and u ′ = du dx , and equation (14.6.12) becomes ∂F ∂u − ∂ 2F ∂x∂u′ − ∂ 2F ∂u∂u′  du dx − ∂ 2F ∂u′2 d 2u ∂x2 = 0. It is left to the reader to verify that the functional with one independent variable and nth-order derivatives in the form I (u) =
b a F (x, u, ux, uxx,...,uxn ,...,) dx admits the Euler–Lagrange equation ∂F ∂u − d dx  ∂F ∂ux  + d 2 dx2  ∂F ∂uxx  − ...(−1)n d n dxn  ∂ 2F ∂uxn  = 0. After we have determined the function u which makes I (u) stationary, the question of the nature of the extremum arises, that is, its minimum, maximum, or saddle point properties. To answer this question, we look at the second variation defined in (14.6.8). If terms of O  ε 3 can be neglected in (14.6.6), or if they vanish for the case of quadratic F, it follows from (14.6.6) that a necessary condition for the functional I (u) to have a minimum I (u) ≥ I (u0) at u = u0 is that δ 2 I ≥ 0, for I (u) to have a maximum I (u) ≤ I (u0) at u = u0 is that δ 2 I ≤ 0 at u = u0 respectively for all admissible values of v. These results enable us to determine the upper or lower bounds for the stationary value I (u0) of the functional. Example 14.6.6. Find out the shortest distance between given points A and B in the (x, y)-plane. Suppose AP B is any curve in the plane through A and B, and s = arcAP. The problem is to determine the curve for which the functional 14.6 Variational Methods and the Euler–Lagrange Equations 635 I (y) =
B A ds, (14.6.14) is a minimum. Since ds/dx =  1 + y ′2 1 2 , functional (14.6.14) becomes I (y) =
x2 x1  1 + y ′2 1 2 dx. (14.6.15) In this case, F =  1 + y ′2 1 2 which depends on y ′ only, so ∂F/∂y = 0. Hence, the Euler–Lagrange equation (14.6.12) becomes d dx  ∂F ∂y′  = 0. This gives the differential equation y ′′ = 0. (14.6.16) This means that the curvature for all points on the curve AB is zero. Hence, the path AB is a straight line. It follows from the integration of (14.6.16) that y = mx + c is a two-parameter family of straight lines. Example 14.6.7. (Fermat principle in optics). In an optically homogeneous isotro-pic medium, light travels from one point A to another point B along the path for which the travel time is minimum. The velocity of light v is the same at all points of the medium; hence, the minimum time is equivalent to the minimum path length. For simplicity, consider a path joining the two points A and B in the (x, y)-plane. The time to travel an elementary arc length ds is ds/v. Thus, the variational problem is to find the path for which
B A ds v =
x2 x1  1 + y ′2 1 2 dx v =
x2 x1 F (y, y′ ) dx (14.6.17) is a minimum, where y ′ = dy/dx, and v = v (y). When F is a function of y and y ′ , the Euler–Lagrange equation (14.6.12) becomes d dx (F − y ′Fy′ )=0. (14.6.18) This follows from the result d dx (F − y ′Fy′ ) = d dxF (y, y′ ) − y ′′Fy′ − y ′ d dx (Fy′ ) = y ′Fy + y ′′Fy − y ′′Fy − y ′ d dx (Fy′ ) = y ′ Fy − d dx (Fy′ ) = 0, by (14.6.12). 636 14 Numerical and Approximation Methods Hence, F − y ′Fy′ = constant, (14.6.19) or  1 + y ′2 1 2 v − y ′2 v (1 + y ′2) 1 2 = constant, or v −1  1 + y ′2 − 1 2 = constant. (14.6.20) In order to give a simple physical interpretation, we rewrite (14.6.20) in terms of the angle φ made by the tangent to the minimum path with the vertical y-axis so that sin φ =  1 + y ′2 − 1 2 . Hence, 1 v sin φ = constant = K (14.6.21) for all points on the minimum curve. For a ray of light, (1/v) must be directly proportional to the refractive index n of the medium through which light is travelling. Equation (14.6.21) is called the Snell law of refraction of light. Often this law is stated as n sin φ = constant. (14.6.22) (A) Hamilton Principle The difference between the kinetic energy T and the potential energy V of a dynamical system is denoted by L = T − V . The quantity L is called the Lagrangian of the system. The Hamilton principle states that the first variation of the time integral of L is zero, that is, δ
t2 t1 L dt = δ
t2 t1 (T − V ) dt = 0. (14.6.23) This result is supposed to be valid for all dynamical systems whether they are conservative or nonconservative. For a conservative system the force field F = −∇V and T + V = C, where C is a constant, and so (14.6.23) gives the principle of least action δA = 0, A =
t2 t1 L dt, (14.6.24) where A is called the action integral or simply the action of the system. 14.6 Variational Methods and the Euler–Lagrange Equations 637 Example 14.6.8. Derive the Newton second law of motion from the Hamilton principle. Consider a particle of mass m at the position r = (x, y, z) which is moving under the action of a field of force F. The kinetic energy of the particle is T = 1 2mr˙ 2 , and the variation of work done is δW = F · δr and δV = −δW. Thus, the Hamilton principle for the system is 0 = δ
t2 t1 (T − V ) dt =
t2 t1 (δT − δV ) dt =
t2 t1 (mr˙ · δr˙ + F · δr) dt. Integrating this result by parts and noting that δr vanishes at t = t1 and t = t2, we obtain
t2 t1 (m¨r − F) · δr dt = 0. This is true for every virtual displacement δr, and hence, the integrand must vanish, that is, m¨r = F. (14.6.25) This is the celebrated Newton second law of motion. Example 14.6.9. Derive the equation for a simple harmonic oscillator in a non-resisting medium from the Hamilton principle. For a simple harmonic oscillator, T = 1 2mx˙ 2 and V = 1 2mω2x 2 . According to the Hamilton principle δ
t2 t1  1 2 mx˙ 2 − 1 2 mω2x 2  dt = δ
t2 t1 F (x, x˙) dt = 0. This leads to the Euler–Lagrange equation ∂F ∂x − d dt (mx˙)=0, or x¨ + ω 2x = 0. (14.6.26) This is the equation for the simple harmonic oscillator. Example 14.6.10. A straight uniform elastic beam of length l, line density ρ, cross-sectional moment of inertia I, and modulus of elasticity E is fixed at each end. The beam performs small transverse oscillations in the horizontal (x, y)-plane. Derive the equation of motion of the beam. The potential energy of the elastic beam is V = 1 2
l 0 M2 EI dx = 1 2
l 0 EIy′′2 dx, 638 14 Numerical and Approximation Methods where the bending moment M is proportional to the curvature so that M = EI y ′′ (1 + y ′2) 1 2 ∼ EIy′′ for small y ′ . The variational principle gives δ
t2 t1 (T − V ) dt = δ
t2 t1 F (y ′′ , y˙) dt = 0, where F (y ′′ , y˙) = 1 2
l 0  ρy˙ 2 − EIy′′2 dx. This principle leads to the Euler–Lagrange equation −
l 0
4 ρy¨ + EIy(iv)
5 dx = 0, or ρy¨ + EIy(iv) = 0. (14.6.27) This represents the partial differential equation of the transverse vibration of the beam. (B) The Generalized Coordinates, Lagrange Equation, and Hamilton Equation The Euler–Lagrange analysis of a dynamical system can be extended to more complex cases where the configuration of the system is described by generalized coordinates q1, q2, ..., qn. Without loss of generality, we consider a system of three variables where the familiar Cartesian coordinates x, y, z can be expressed in terms of the generalized coordinates q1, q2, q3 as x = x (q1, q2, q3), y = y (q1, q2, q3), z = z (q1, q2, q3). (14.6.28) For example, if (q1, q2, q3) represents the cylindrical polar coordinates (r, θ, z), the above result becomes x = r cos θ, y = r sin θ, z = z. Since the coordinates are functions of time t, we obtain the following result by differentiation x˙ = ∂x ∂q1 q˙1 + ∂x ∂q2 q˙2 + ∂x ∂q3 q˙3 (14.6.29) 14.6 Variational Methods and the Euler–Lagrange Equations 639 with similar expressions for ˙y and ˙z. If these results are substituted into T = 1 2m  x˙ 2 + ˙y 2 + ˙z 2 and V = V (x, y, z), then both T and V can be written in terms of the generalized coordinates qi and the generalized velocities ˙qi , as T = T (q1, q2, q3; ˙q1, q˙2, q˙3), V = V (q1, q2, q3), (14.6.30) so that the Lagrangian has the form L = T − V = L(qi , q˙i). (14.6.31) The Hamilton principle gives δ
t2 t1 L(qi , q˙i) dt = 0. (14.6.32) The simple variation of this integral with fixed end points, the interchange of the variation operations and time derivatives for the variation of the generalized velocities, and then integration by parts yield
t2 t1 1 3 i=1  ∂L ∂qi − d dt  ∂L ∂q˙i 0 δqi 3 dt = 0, (14.6.33) where the integrated components vanish because of the conditions δqi = 0 (i = 1, 2, 3) at t = t1 and t = t2. When the generalized coordinates are independent and the variations δqi are independent for all t in (t1, t2), the coefficients of the variations δqi vanish independently for arbitrary values of t1 and t2. This means that the integrand in (14.6.33) vanishes, that is, d dt  ∂L ∂q˙i  − ∂L ∂qi = 0, i = 1, 2, 3. (14.6.34) These are called the Lagrange equations of motion. If a particle of mass m at position r = (x1, x2, x3) moves under the action of a conservative force field Fi = −∂V /∂xi , the Lagrangian function is L = T − V = 1 2 m  x˙ 2 1 + ˙x 2 2 + ˙x 2 3 − V (x1, x2, x3). (14.6.35) Consequently, ∂L ∂x˙ i = mx˙ i , ∂L ∂xi = − ∂V ∂xi = Fi . (14.6.36) The former represents the momentum of the particle and the latter is the force acting on the particle. In view of (14.6.36), the Lagrange equation (14.6.34) gives the Newton second law of motion in the form d dt (mx˙ i) = Fi . (14.6.37) 640 14 Numerical and Approximation Methods Example 14.6.11. Apply the Lagrange equations of motion to derive the equations of motion of a particle under the action of a central force, −mF (r) where r is the distance of the particle of mass m from the center of force. It is convenient to use the polar coordinates r and θ. In terms of the generalized coordinates q1 = r and q2 = θ, we write x = r cos θ = q1 cos q2, y = r sin θ = q1 sin q2. The kinetic energy T is T = 1 2 m  x˙ 2 + ˙y 2 = 1 2 m
4 r˙ 2 + r 2 ˙θ 2
5 = 1 2 m  q˙ 2 1 + q 2 1 q˙ 2 2 . (14.6.38) Since F = ∇V , the potential is V (r) =
r F (r) dr =
q1 F (q1) dq1. (14.6.39) Then, the Lagrangian L is L = T − V = 1 2 m  q˙ 2 1 + q 2 1 q˙ 2 2 − 2
q1 F (q1) dq1 . (14.6.40) Thus, the Lagrange equations (14.6.34) with i = 1, 2, 3 give the equations of motion q¨1 − q1q˙ 2 2 + F (q1)=0, d dt  q 2 1 q˙2 = 0. (14.6.41) In term of the polar coordinates, these equations become r¨ − r ˙θ 2 = −F (r), d dt
4 r 2 ˙θ
5 = 0. (14.6.42ab) Equation (14.6.42b) gives immediately r 2 ˙θ = h, (14.6.43) where h is a constant. In this case, r ˙θ represents the transverse velocity component, and mr2 ˙θ = mh is the constant angular momentum of the particle about the center of force. Introducing r = 1/u, we obtain r˙ = dr dt = − 1 u 2 du dt = − 1 u 2 du dθ · dθ dt = −h du dθ , r¨ = d 2 r dt2 = −h d dt  du dθ  = −h d 2u dθ2 dθ dt = −h 2u 2 d 2u dθ2 . Substituting these into (14.6.42a) gives 14.6 Variational Methods and the Euler–Lagrange Equations 641 −h 2u 2 d 2u dθ2 − h 2u 3 = −F  1 u  , or d 2u dθ2 + u = 1 h 2u 2 F  1 u  . (14.6.44) This is the differential equation of the central orbit and can be solved by standard methods. In particular, if the law of force is the attractive inverse square, F (r) = µ/r2 so that the potential V (r) = −µ/r, the differential equation (14.6.44) becomes d 2u dθ2 + u = µ h 2 , (14.6.45) if the particle is projected initially from distance a with velocity V at an angle β that the direction of motion makes with the outward radius vector. Thus, the constant h in (14.6.43) is h = V a sin β. The angle φ between the tangent and radius vector of the orbit at any point is given by cot φ = 1 r dr dθ = u d dθ  1 u  = − 1 u du dθ . (14.6.46) At t = 0, the initial conditions are u = 1 a , du dθ = − 1 a cot β when θ = 0. (14.6.47) The general solution of equation (14.6.45) is u = µ h 2 [1 + e cos (θ + α)] , (14.6.48) where e and α are constants to be determined from the initial data. Finally, the solution can be written as l r =1+ e cos (θ + α), (14.6.49) where l = h 2 µ = (V a sin β) 2 /µ. (14.6.50) This represents a conic section of semi-latus rectum l and eccentricity e with its axis inclined at an angle α to the radius vector at the point of projection. The initial conditions (14.6.47) lead to 642 14 Numerical and Approximation Methods l a =1+ e cos α, − l a cot β = −e sin α, (14.6.51) which give tan α = l cot β l − a , e 2 =  l a − 1 2 + l 2 a 2 cot2 β = l 2 a 2 cosec2β − 2l a + 1, = 1 − 2aV 2 sin2 β µ + a 2V 4 sin2 β µ2 . (14.6.52) Thus, the conic is an ellipse, parabola, or hyperbola accordingly as e <=> 1 that is, V 2 < = > 2µ/a. To derive the Hamilton equations, we introduce the concept of generalized momentum, pi and generalized force, Fi as pi = ∂L ∂q˙i , Fi = ∂L ∂qi . (14.6.53ab) Consequently, the Lagrange equations (14.6.34) become ∂L ∂qi = d dt pi = ˙pi . (14.6.54) The Hamiltonian function H is defined by H = n i=1 pi q˙i − L. (14.6.55) In general, L = L(qi , q˙i , t) is a function of qi , ˙qi and t, where ˙qi enters through the kinetic energy as a quadratic term. Hence, equation (14.6.53a) will give pi as a linear function of ˙qi . This system of linear equations involving pi and ˙qi can be solved to determine ˙qi in terms of pi , and then, the q˙i can, in principle, be eliminated from (14.6.55). This means that H can always be expressed as a function of pi , qi and t so that H = H (pi , qi , t). Thus, dH =  ∂H ∂pi dpi +  ∂H ∂qi dqi + ∂H ∂t dt. (14.6.56) On the other hand, differentiating H in (14.6.55) with respect to t gives dH dt = pi d dt q˙i + q˙i d dt pi −  ∂L ∂qi d dt qi −  ∂L ∂q˙i d dt q˙i − ∂L ∂t , (14.6.57) or 14.6 Variational Methods and the Euler–Lagrange Equations 643 dH = pi dq˙i + q˙i dpi −  ∂L ∂qi dqi −  ∂L ∂q˙i dq˙i − ∂L ∂t dt, (14.6.58) which becomes, in view of (14.6.53a), dH = q˙i dpi −  ∂L ∂qi dqi − ∂L ∂t dt. (14.6.59) Evidently, two expressions of dH in (14.6.56) and (14.6.59) must be equal so that the coefficients of the corresponding differentials can be equated to obtain q˙i = ∂H ∂pi , − ∂L ∂qi = ∂H ∂qi , − ∂L ∂t = ∂H ∂t . (14.6.60abc) Using the Lagrange equation (14.6.54), the first two of the above equations become q˙i = ∂H ∂pi , p˙i = − ∂H ∂qi . (14.6.61ab) These are commonly known as the Hamilton canonical equations of motion. They play a fundamental role in advanced analytical dynamics. Finally, the Lagrange–Hamilton theory can be used to derive the law of conservation of energy. In general, the Lagrangian L is independent of time t and hence, (14.6.60c) implies that H = constant. Again, T involved in L = T − V is given by T = 1 2 n i=1 n j=1 aij q˙i q˙j , (14.6.62) where the coefficients aij are symmetric functions of the generalized coordinates qij , that is, aij = aji. On the other hand, V is, in general, independent of qi and hence, pi = ∂L ∂q˙i = ∂T ∂q˙i = n j=1 aij q˙j . (14.6.63) Thus, the Hamiltonian H becomes H = n i=1 piq˙i − L = n i=1 ⎛ ⎝ n j=1 aij q˙j ⎞ ⎠ q˙i − L = 2T − L = T + V. (14.6.64) Thus, H is equal to the total energy. It has already been observed that, if L does not contain t explicitly, H is a constant. This means that the sum of the potential and kinetic energies is constant. This is the law of the conservation of energy. 644 14 Numerical and Approximation Methods Example 14.6.12. Use the Hamiltonian equations to derive the equations of motion for the problem stated in Example 14.6.11. The Lagrangian L for this problem is given by (14.6.40) with q1 = r and q2 = θ. It follows from the definition (14.6.53a) of the generalized momentum that p1 = mq˙1 = mr, p ˙ 2 = mq2 1 q˙2 = mr2 ˙θ. (14.6.65) Expressing the results of the kinetic energy (14.6.38) and the potential energy (14.6.39) in terms of p1 and p2 the Hamiltonian H = T + V can be written as H = 1 2m  p 2 1 + p 2 2 q 2 1  + m
q1 F (q1) dq1. (14.6.66) Then, equations (14.6.65) and the Hamilton equation (14.6.61b) give p1 = mr, p ˙ 2 = mr2 ˙θ, (14.6.67) p˙1 = 1 m p 2 2 q 3 1 + mF (q1), p˙2 = 0. (14.6.68) Clearly, these equations are identical with the equations of motion (14.6.42ab). Example 14.6.13. Derive the equation of a simple pendulum by using (i) the Lagrange equations and (ii) the Hamilton equations. We consider the motion of simple pendulum of mass m attached at the end of a rigid massless string of length l that pivots about a fixed point. We suppose that the pendulum makes an angle θ with its vertical position. The force F acting on the mass m is F = −mg sin θ, so that the potential V is obtained from F = −∇V as V = mgl(1 − cos θ). The kinetic energy T = 1 2ml2 ˙θ 2 . Thus the Lagrangian L is L = T − V = 1 2 ml2 ˙θ 2 − mgl(1 − cos θ) = L
4 θ, ˙θ
5 . (14.6.69) The Lagrange equation is ∂L ∂θ − d dt  ∂L ∂ ˙θ  = 0, (14.6.70) or −mglsin θ − d dt
4 ml2 ˙θ
5 = 0, or ¨θ + ω 2 sin θ = 0, ω2 = g/l. (14.6.71ab) 14.6 Variational Methods and the Euler–Lagrange Equations 645 This is the equation of the simple pendulum. To derive the same equation from the Hamilton equations, we choose q1 = l( ˙q1 = 0) and q2 = θ as the generalized (polar) coordinates. The kinetic and potential energies are T = 1 2 ml2 q˙ 2 2 , V = mgl(1 − cos q2). (14.6.72ab) Thus, H = T + V and L = T − V are given by (H, L) = 1 2 ml2 q˙ 2 2 + mgl(1 − cos q2). (14.6.73ab) From the definition of the generalized momentum, we find that p2 = ∂L ∂q˙2 = ml2 q˙2 so that the Hamiltonian H in terms of p2 and q2 is H = 1 2 p 2 2 ml2 + mgl(1 − cos q2). Thus, the Hamilton equation (14.6.61ab) gives ¨θ + ω 2 sin θ = 0, ω2 = g l . (14.6.74) The variational methods can be further extended for functionals depending on functions or more independent variables in the form I [u (x, y)] =

D F (x, y, u, ux, uy) dx dy (14.6.75) where the values of the function u (x, y) are prescribed on the boundary ∂D of a finite domain D in the (x, y)-plane. We assume that F is differentiable and the surface u = u (x, y) giving an extremum is also continuously differentiable twice. The first variation δI of I is defined by δI [u, ε] = I (u + ε) − I (u) (14.6.76) which is, by Taylor’s expansion theorem =

D [εFu + εxFp + εyFq] dx dy (14.6.77) where ε ≡ ε (x, y) is small and p = ux and q = uy. According to the variational principle, δI = 0 for all admissible values of ε. The partial integration of (14.6.77) combined with ε = 0 on ∂D gives 646 14 Numerical and Approximation Methods 0 = δI =

D Fu − ∂ ∂xFp − ∂ ∂yFq ε (x, y) dx dy. (14.6.78) This is true for all arbitrary ε, and hence, the integrand must vanish, that is ∂ ∂xFp + ∂ ∂yFq − Fu = 0. (14.6.79) This is the Euler–Lagrange equation which is the second-order partial differential equation to be satisfied by the extremizing function u (x, y). Example 14.6.14. Derive the equation of motion for the free vibration of an elastic string of length l. The potential energy V of the string is V = 1 2 T ∗
l 0 u 2 xdx (14.6.80) where u = u (x, y) is the displacement of the string from its equilibrium position and T ∗ is the constant tension of the string. The kinetic energy T is T = 1 2
l 0 ρu2 t dx (14.6.81) where ρ is the constant line-density of the string. According to the Hamilton principle δI = δ
t2 t1 (T − V ) dt = δ
t2 t1
l 0 1 2  ρu2 t − T ∗u 2 x dx dt = 0 (14.6.82) which has the form δ
t2 t1
l 0 L(ut, ux)=0, (14.6.83) where L = 1 2  ρ u2 t − T ∗u 2 x . Then the Euler–Lagrange equation is given by ∂ ∂t (ρut) − ∂ ∂x (T ∗ux)=0, (14.6.84) or utt − c 2uxx = 0, c2 = T ∗ /ρ. (14.6.85) This is the wave equation of motion of the string. 14.7 The Rayleigh–Ritz Approximation Method 647 Example 14.6.15. Derive the Laplace equation from the functional I (u) =

D  u 2 x + u 2 y dx dy with a boundary condition u = f (x, y) on ∂D. The variational principle gives δI = δ

D  u 2 x + u 2 y dx dy = 0. This leads to the Euler–Lagrange equation uxx + uyy = 0 in D. Similarly, the functional I [u (x, y, z)] =


D  u 2 x + u 2 y + u 2 z dx dy dz will lead to the three-dimensional Laplace equation ∇2u = uxx + uyy + uzz = 0. 14.7 The Rayleigh–Ritz Approximation Method We consider the boundary-value problem governed by the differential equation Au = f in D (14.7.1) with the boundary condition B (u) = 0 on ∂D (14.7.2) where A is a self-adjoint differential operator in a Hilbert space H and f ∈ H. In general, the determination of the exact solution of the problem is often a difficult task. However, it can be shown that the solution of (14.7.1)– (14.7.2) is equivalent to finding the minimum of a functional I (u) associated with the differential system. In other words, the solution can be characterized as the function which minimizes (or maximizes) the functional I (u). A simple and efficient method for an approximate solution of the extremum problem was independently formulated by Lord Rayleigh and W. Ritz. We next prove a fundamental result which states that the solution of the equation (14.7.1) is equivalent to finding the minimum of the quadratic functional 648 14 Numerical and Approximation Methods I (u) ≡ A u, u − 2 f, u. (14.7.3) Suppose that u = u0 is the solution of (14.7.1) so that Au0 = f. Consequently, I (u) ≡ A u, u − 2 Au0, u = A (u − u0), u−Au0, u. Since the inner product is symmetrical and Au0, u = u0, Au = Au, u0, I (u) can be written as I (u) = A (u − u0), u−Au, u0 + Au0, u−Au0, u0, = A (u − u0), u − u0−Au0, u0, = A (u − u0), u − u0 + I (u0). (14.7.4) Since A is a positive operator, A (u − u0), u − u0 ≥ 0 where equality holds if and only if u − u0 = 0. It follows that I (u) ≥ I (u0), (14.7.5) where equality holds if and only if u = u0. We conclude from this inequality that I (u) assumes its minimum at the solution u = u0 of equation (14.7.1). Conversely, the function u = u0 that minimizes I (u) is a solution of equation (14.7.1). Clearly, I (u) ≥ I (u0), that is, in particular, I (u0 + αv) ≥ I (u0) for any real α and any function v. Explicitly, I (u0 + αv) = A (u0 + αv), u0 + αv − 2 f, u0 + αv, = Au0, u0 + 2α Au0, v + α 2 Av, v − 2 f, u0 − 2α f,v. This means that I (u0 + αv) is a quadratic expression in α. Since I (u) is minimum at u = u0, then δI (u0, v) = 0, that is, 0 = d dαI (u0 + αv) α=0 = 2 Au0, v − 2 f,v = 2 Au0 − f,v. This is true for any arbitrary but fixed v. Hence, Au0 − f = 0. This proves the assertion. In the Rayleigh–Ritz method an approximate solution of (14.7.1)– (14.7.2) is sought in the form un (x) = n i=1 aiφi (x), (14.7.6) where a1, a2, ..., an are n unknown coefficients to be determined so that I (un) is minimum, and φ1, φ2, ..., φn represent a linearly independent and 14.7 The Rayleigh–Ritz Approximation Method 649 complete set of arbitrarily chosen functions that satisfy (14.7.2). This set of functions is often called a trial set. We substitute (14.7.6) into (14.7.3) to obtain I (un) = ;n i=1 aiA (φi), n j=1 ajφj < − 2 ; f, n i=1 aiφi < . Then the necessary condition for I to obtain a minimum (or maximum) is that ∂I ∂aj (a1, a2,...,an)=0, j = 1, 2, . . . , n, (14.7.7) or ∂I ∂aj ⎡ ⎣ ;n i=1 aiA (φi), n j=1 ajφj < − 2 ; f, n i=1 aiφi <⎤ ⎦ = 0, or n i=1 A (φi), φj  ai + n i=1 A (φj ), φi aj − 2 f,φj  = 0, or 2 n i=1 A (φi), φj  ai = 2 f,φj . Therefore, n i=1 A (φi), φj  ai = f,φj , j = 1, 2, . . . , n. (14.7.8) This is a linear system of n equations for the n unknown coefficients aj . Once a1, a2, ..., an are determined, the approximate solution is given by (14.7.6). In particular, when A (φi), φj  = ⎧ ⎨ ⎩ 0, i = j 1, i = j, (14.7.9) equation (14.7.8) gives aj as aj = φj , f, (14.7.10) so that the Rayleigh–Ritz approximate series (14.7.6) becomes 650 14 Numerical and Approximation Methods un (x) = n i=1 φi , f φi (x). (14.7.11) This is similar to the Fourier series solution with known Fourier coefficients ai . In the limit n → ∞, a limit function can be obtained from (14.7.6) as u (x) = limn→∞ n i=1 ai φi (x) = ∞ i=1 ai φi (x), (14.7.12) provided that the series converges. Under certain assumptions imposed on the functional I (u) and the trial functions φ1, φ2, ..., φn, the limit function u (x) represents an exact solution of the problem. In any event, (14.7.6) or (14.7.11) gives a reasonable approximate solution. In the simplest case corresponding to n = 1, the Rayleigh–Ritz method gives a simple form of the functional I (u1) = I (a1φ1) = a 2 1 Aφ1, φ1 − 2a1 f,φ1, where a1 is readily determined from the necessary condition for extremum 0 = ∂ ∂a1 I (a1φ1)=2a1 Aφ1, φ1 − 2 f,φ1, or a1 = f,φ1 Aφ1, φ1 . (14.7.13) The corresponding minimum value of the functional is given by I (a1φ1) = − f,φ1 2 Aφ1, φ1 . (14.7.14) Thus, the essence of the Rayleigh–Ritz method is as follows. For a given boundary-value problem, an approximate series solution is sought so that the trial functions φi satisfy the boundary conditions. We solve the system of algebraic equations (14.7.7) to determine the coefficients ai . We now illustrate the method by several examples. Example 14.7.1. Find an approximate solution of the Dirichlet problem ∇2u ≡ uxx + uyy = 0 in D u = f on ∂D, where D ⊂ R2 , and f is a given function. This problem is equivalent to finding the minimum of the associated functional 14.7 The Rayleigh–Ritz Approximation Method 651 I (u) =

D  u 2 x + u 2 y dx dy. We seek an approximate series solution in the form u2 (x, y) = a1φ1 + a2φ2 with a1 = 1 so that u2 satisfies the given boundary conditions, that is, φ1 = f and φ2 = 0 on ∂D. Substituting u2 into the functional gives I (u2) =

D 1 ∂u2 ∂x 2 +  ∂u2 ∂y 2 3 dx dy =

D (∇φ1) 2 dx dy + 2a2

D (∇φ1 · ∇φ2) dx dy + a 2 2

D |∇φ2| 2 dx dy. The necessary condition for an extremum of I (u2) is ∂I ∂a2 = 0, or 2

D (∇φ1 · ∇φ2) dx dy + 2a2

D |∇φ2| 2 dx dy = 0. Therefore, a2 = − ** (∇φ1 · ∇φ2) dx dy ** |∇φ2| 2 dx dy . This a2 minimizes the functional and the approximate solution is obtained. However, this procedure can be generalized by seeking an approximate solution in the form un = n i=1 aiφi (a1 = 1) so that φ1 = f and φi = 0(i = 2, 3,...,n) on ∂D. The coefficients ai can be obtained by solving the system (14.7.7) with j = 2, 3, ..., n. Example 14.7.2. A uniform elastic beam of length l carrying a uniform load W per unit length is freely hinged at x = 0 and x = l. Find the approximate solution of the boundary-value problem EI (iν) y (x) = W, y = y ′′ = 0 at x = 0 and x = l, 652 14 Numerical and Approximation Methods where y = y (x) is the displacement function. This problem is equivalent to finding a function y (x) that minimizes the energy functional I (y) =
l 0  W y − EI 2 y ′′2  dx. We seek an approximate solution yn (x) = n r=1 ar sin
4rπx l
5 which satisfies the boundary conditions. Substitution of this solution into the energy functional gives I (yn) = n r=1 1
l 0 W ar sin
4rπx l
5 dx − EI 2
l 0 r 4π 4 l 4 a 2 r sin2
4rπx l
5 dx3 = 2Wl π n r=1 ar r − EIπ4 4l 3 n r=1 r 4 a 2 r . The necessary conditions for extremum are 0 = ∂I ∂ar = 2Wl rπ − EIπ4 4l 3 2arr 4 , r = 1, 2,...,n which give ar as ar = 4Wl4 π 5r 5EI , r = 1, 2, . . . , n. Thus, the approximate function y (x) is yn (x) = 4Wl4 π 5EI n r=1 1 r 5 sin
4rπx l
5 . The maximum deflection at x = l/2 is ymax = 4Wl4 π 5EI  1 − 1 3 5 + 1 5 5 − ... . In this case, the first term of the series solution gives a reasonably good approximate solution as y1 (x) ∼ 4Wl4 EIπ5 sin
4πx l
5 . 14.7 The Rayleigh–Ritz Approximation Method 653 Example 14.7.3. Apply the Rayleigh–Ritz method to investigate the free vibration of a fixed elastic wedge of constant thickness governed by the energy functional I (y) =
1 0  αx3 y ′′2 − ωxy2 dx, y (1) = y ′′ (1) = 0, where the free vibration is described by the function u (x, t) = e iωty (x), ω is the frequency. We seek an approximate solution in the form yn (x) = n r=1 aryr (x) = n r=1 ar (x − 1)2 x r−1 which satisfies the given boundary conditions. We take only the first two terms so that y2 (x) = a1y1 +a2y2 = (x − 1)2 (a1 + a2x). Substituting y2 into the functional we obtain I2 = I (y2) =
1 0 " αx3 (6a2x + 2a1 − 4a2) 2 − ωx (x − 1)4 (a1 + a2x) 2 # dx = α (a1 − 2a2) 2 + 24 5 (a1 − 2a2) a2 + 6a 2 2 − ω 5 a 2 1 6 + 2a1a2 21 + a 2 2 56 . The necessary conditions for an extremum are ∂I2 ∂a1 = 2a1
4 α − ω 30
5 + 2 5 a2
4 2α − ω 21
5 = 0, ∂I2 ∂a2 = 2a1 5
4 2α − ω 21
5 + 2a2 5
4 2α − ω 56
5 = 0. For nontrivial solutions, the determinant of this algebraic system must be zero, that is,       α − ω 30 1 5  2α − ω 21 2α − ω 21 2α − ω 56       = 0, or 5
4 α − ω 30
5
42α − ω 56
5 −
4 2α − ω 21
52 = 0. This represents the frequency equation of the vibration which has two roots ω1 and ω2. The smaller of these two frequencies gives an approximate value of the fundamental frequency of the vibration of the wedge. Example 14.7.4. An elastic beam of length l, density ρ, cross-sectional area A, and modulus of elasticity E has its end x = 0 fixed and the other end 654 14 Numerical and Approximation Methods connected to a rigid support through a linear elastic spring with spring constant k. Apply the Rayleigh–Ritz method to investigate the harmonic axial motion of the beam. The kinetic energy and the potential energy associated with the axial motion of the beam are T =
l 0 ρA 2 U 2 t dx, V =
l 0 EA 2 U 2 x dx + k 2 U 2 (l, t), where U (x, t) is the displacement function. Since the axial motion is simple harmonic, U (x, t) = u (x) e iωt, where ω is the frequency of vibration. Consequently, the expressions for T and V can be written in terms of u (x). We then apply the Hamilton variational principle δI (u) = δ 1
t2 t1
l 0 1 2  ρAω2 u 2 − EAu2 x dx − k 2 u 2 (l) 3 dt = 0. The Euler–Lagrange equation for the variational principle is d dx  EA du dx + ρAω2 u = 0, 0 < x < l, EA du dx + ku = 0, at x = l. In terms of nondimensional variables (x ∗ , u∗ ) = (1/l) (x, u) and parameters λ =  ω 2ρl2/E and α = (kl/EA), this system becomes, dropping the asterisks, uxx + λ u = 0, 0 <x< 1,="" ux="" +="" α="" u="0," at="" x="1." the="" associated="" functional="" for="" system="" is="" i="" (u)="1" 2=""
="" 1="" 0="" ="" λ="" u2="" −="" dx="" (1).="" according="" to="" rayleigh–ritz="" method,="" we="" seek="" approximate="" solution="" with="" in="" form="" (x)="a1x" a2x="" so="" that="" (u2)="" minimum.="" substitute="" into="" obtain="" i2="I" l="" "="" a1x="" (a1="" 2a2x)="" #="" a2)="" .="" 14.8="" galerkin="" approximation="" method="" 655="" necessary="" conditions="" extremum="" of="" are="" ∂a1="a1" ="" 3="" ="" a2="" 4="" ,="" ∂a2="a1" 5="" 7="" nontrivial="" solutions,="" determinant="" must="" be="" zero,="" is,="" ="" or="" 3λ="" 128λ="" 480="0." this="" quadratic="" equation="" gives="" two="" solutions:="" λ1="4.155," λ2="38.512." corresponding="" values="" frequency="" given="" by="" ω1="2.038" e="" ρ="" l2="" 1="" ω2="6.206" exact="" determined="" transcendental="" √="" tan="" first="" roots="" can="" obtained="" graphically="" as="" ω01="" ∼="" 2.0288="" ω02="" 4.9132="" an="" extension="" formulated="" ingenious="" which="" may="" applied="" a="" problem="" no="" simple="" variational="" principle="" exists.="" differential="" operator="" (14.7.1)="" need="" not="" linear="" equation.="" order="" solve="" boundary-value="" (14.7.1)–(14.7.2),="" construct="" un="" +n="" aiφi="" (x),="" (14.8.1)="" 656="" 14="" numerical="" and="" methods="" where="" φi="" known="" functions,="" u0="" introduced="" satisfy="" boundary="" conditions,="" coefficients="" ai="" determined.="" substituting="" non-zero="" residual="" rn="" (a1,="" a2,...,an,="" x,="" y)="A" (un)="A" (u0)="" aia="" (φi).="" (14.8.2)="" unknown="" solving="" following="" equations="" rn,="" φj="" ="0," j="1," 2,...,n.="" (14.8.3)="" since="" linear,="" written="" n="" a="" (φi),="" au0,="" ,="" (14.8.4)="" determines="" ′="" s.="" substitution="" s="" from="" required="" un.="" find="" interesting="" connection="" between="" fourier="" representation="" function="" u.="" (14.8.5)="" special="" restriction="" on="" satisfies="" condition="" aφi="" ⎨="" ⎩="" 0,="" (14.8.6)="" thus,="" application="" (14.8.7)="" (14.8.6),="" aj="f,φj" (14.8.8)="" becomes="" f,φi="" (x).="" (14.8.9)="" evidently,="" just="" finite="" series="" solution.="" 657="" finally,="" shall="" cite="" example="" show="" equivalence="" methods.="" consider="" poisson="" uxx="" uyy="f" (x,="" d="" ⊂="" r="" (14.8.10)="" homogeneous="" ∂d.="" equivalent="" finding="" minimum="" y="" 2fu="" dy.="" (14.8.11)="" trial="" y),="" (14.8.12)="" functions="" chosen="" they="" then="" use="" ∂i="" ∂ak="0," k="1," 2,="" ...,="" n=""

="" ∂un="" ∂x="" ∂φk="" ∂y="" fφk="" dy="0." (14.8.13)="" greens="" theorem="" leads="" ∇2un="" f="" φk="" (14.8.14)="" φk="0," ≡="" f.="" (14.8.15)="" undetermined="" ak.="" establishes="" 14.8.1.="" ∇2u="" :="" |x|="" <="" a,="" |y|="" b}="" ∂d="{(x," 658="" m,n="1" ="" 3,5,...="" amn="" φmn=""
4mπx="" 2a=""
5="" cos=""
4nπy="" 2b="" case="" +1="∇2uN" 1="" m="1" m2π="" 4a="" 2π="" 4b="" φmn3="" 1.="" φkl="
" −a="" b="" −b="" kπx="" lπy="" akl="" 16ab="" π="" 2kl="" (−1){(k+l)="" 2}−1="" 8ab="" 2="" (−1)="" (k+l)−1="" (b="" 2k="" 2l="" 2)="" (m+n)−1="" 2m2="" 2n2)="" particular,="" derived="" square="" domain="" 1}.="" limit="" →="" ∞,="" these="" solutions="" perfect="" agreement="" those="" double="" series.="" 14.8.2.="" 14.8.1="" using="" algebraic="" polynomials="" functions.="" appropriate="" y="" a1="" a3y="" a4x="" ...="" obviously,="" conditions.="" approximation,="" assumes="" u1="" a1φ1="a1" 14.9="" kantorovich="" 659="" coefficient="" a1="" integral="" ∇2u1="" φ1dx="" 2a1="" 1!="" x="" evaluation="" −1="" hence,="" 1932,="" gave="" generalization="" rayleigh–="" ritz="" partial="" terms="" coefficients.="" essence="" reduce="" ordinary="" governed="" (14.7.1)–(14.7.2).="" it="" has="" been="" shown="" section="" 14.7="" (14.7.3).="" when="" problem,="" (14.7.6)="" ak="" constants.="" determine="" minimize="" (un).="" assume="" longer="" constants="" but="" one="" independent="" variables="" (14.9.1)="" products="" same="" minimizing="" (un="" (x))="I" 4n="" 5="" (14.9.2)="" 660="" perform="" integration="" respect="" all="" except="" ¯i="" (x),...,="" depending="" variable="" x.="" a2,...,an).="" under="" certain="" converges="" ∞.="" describe="" more="" precisely,="" dimensions:="" d,="" (14.9.3)="" ∂d,="" (14.9.4)="" closed="" bounded="" curves="" vertical="" lines="" (14.9.5)="" (14.9.6)="" condition,="" 1="" 2fun="" 3="" β(x)="" α(x)="" ⎧="" 1n="" 32="" 32="" 2f="" akφk="" ⎫="" ⎬="" ⎭="" (14.9.7)="
" ak,="" a′="" )="" dx,="" (14.9.8)="" integrand="" assumed="" have="" performed="" result="" denoted="" ).="" reduced="" determining="" found="" euler="" equations:="" 661="" ∂f="" ∂a′="" 3,="" n.="" (14.9.9)="" solved="" (a)="ak" (b)="0," consequently,="" 14.9.1.="" torsion="" rectangle="" −a<x<a,="" −b<y<b}="" =="" +a,="" b.="" next="" (−a)="0." 4u="" yields="" (u1)="
" "="" 4y="" 16="" 15="" 8="" dx.="" ′′="" constant="" coeffi-="" cients,="" general="" cosh="" kx="" sinh="" 2="" ka.="" ka="" 662="" torsional="" moment="" 2µα="" u1dx="" µαb3="" tanh="" (ak)="" 14.9.2.="" triangular="" 2u="" −x="" 1y="" 2xu1="" 02="" 135√=""
4="" 2x="" 5u="" 10x="" 4u1u="" 30x="" 3u="" 15x="" 3u1="" 5xu′="" 5u1="15." nonhomogeneous="" two.="" (r="" 1)="" 5)="0." particular="" bx−5="" solution,="" 0.="" implies="" therefore,="" final=""
4x="" 14.10="" element="" 663="" many="" problems="" mathematics,="" science="" engineering="" cannot="" closed-form="" analytical="" formulas.="" often="" asymptotic="" rather="" than="" solutions.="" evolved="" over="" years="" discrete="" easily="" computer.="" however,="" if="" carefully="" chosen,="" numerically="" computed="" anywhere="" close="" true="" another="" computation="" difficult="" take="" long="" impractical="" computer="" carry="" out.="" most="" commonly="" used="" differences="" give="" pointwise="" approximations="" governing="" equations.="" successfully="" fairly="" problems,="" their="" major="" weakness="" suitable="" irregular="" geometries,="" curved="" boundaries="" unusual="" example,="" difference="" particularly="" effective="" circular="" because="" circle="" accurately="" partitioned="" rectangles.="" there="" other="" including="" method.="" unlike="" methods,="" effectively="" accurate="" wide="" variety="" defined="" regions.="" entire="" modeled="" analytically="" approximated="" replacing="" small,="" interconnected="" elements="" (hence="" name="" element).="" extremely="" (linear="" functions)="" small="" such="" triangles.="" collected="" together="" requirements="" continuity="" equilibrium="" satisfied="" neighboring="" elements.="" nutshell,="" basic="" idea="" (fem)="" consists="" decomposing="" set="" arbitrary="" shape="" size.="" decomposition="" usually="" called="" mesh="" grid="" overlap="" nor="" leave="" any="" part="" uncovered.="" each="" element,="" number="" points="" located="" edges="" inside.="" nodes="" vertices="" triangles="" figure="" 14.10.1.="" consideration="" whole="" interpolation="" historically,="" was="" developed="" originally="" study="" stress="" fields="" complicated="" aircraft="" structures="" early="" 1960s.="" subsequently,="" extended="" widely="" science,="" engineering.="" richard="" courant="" (1888–1972)="" who="" piecewise="" continuous="" domains="" 1943;="" he="" combined="" potential="" en-="" 664="" ergy="" st.="" venant="" continuum="" mechanics.="" also="" described="" properties="" based="" principle.="" 1965,="" received="" even="" broader="" interpretation="" zienkiewicz="" cheung="" (1965)="" suggested="" applicable="" field="" cast="" form.="" during="" late="" 1960s="" 1970s,="" considerable="" attention="" errors,="" bounds="" convergence="" criteria="" various="" develop="" recall="" celebrated="" euler–lagrange="" (14.6.12)="" divide="" interval="" ≤="" parts="" rn+1="" set:="" x1="" x2="" xn="b." subinterval="" element.="" general,="" length="" equal,="" though="" simplicity,="" equal="" h="1" a).="" uk="u" (xk),="" 2,...,n="" (x0)="α" (xn)="β," while="" u1,="" u2,="" un−1="" quantities.="" rewrite="" (14.6.3)="" x0="" u,="" u′="" xn−1="" (14.10.1)="" define="" interpolating="" l(x)="" ui="" [a,="" b]="" whose="" graph="" straight="" line="" segments="" joining="" consecutive="" pairs="" (xk,="" uk),="" (xk+1,="" uk+1)="" (n="" 1),="" (uk+1="" uk)="" (x="" xk),="" xk="" xk+1,="" (14.10.2)="" 1).="" assuming="" integrals="" exactly="" in−1="In−1" (u1,="" u2,...,un−1).="" (14.10.3)="" ∂in−1="" ∂uk="0," 2,...,(n="" (14.10.4)="" substituted="" continuous,="" (the="" dirichlet="" plane).="" 665="" △="" (14.10.5)="" (14.10.6)="" region="" triangulated="" dn="" union="" figures="" 14.10.1="" (b).="" denote="" interior="" v1,="" v2,="" vn.="" choose="" v1="" v2="" vn="" vertex.="" vm="" its="" zero="" (c).="" c,="" b,="" c="" different="" triangle.="" requirement="" uniquely.="" indeed,="" simply="" pyramid="" unit="" height="" peak="" do="" touch="" vm.="" combination="" a2v2="" anvn="" amvm="" (14.10.7)="" a1,="" a2,="" multiply="" v="" green’s="" identity="" ∇u="" ·="" ∇v="" (14.10.8)="" valid="" only="" (a),="" (b),="" 666="" am="" (∇um="" ∇vk)="" vk="" (14.10.9)="" equations,="" am,="" rewritten="" αmk="" n,="" (14.10.10)="" dy,="" fk="

" (14.10.11)="" (14.10.10).="" value="" (14.10.7).="" several="" comments="" order.="" first,="" depend="" geometry="" completely="" known.="" second,="" vanishes="" ∂dn.="" third,="" vertex="" vi="(xi" yi),="" (xi="" yi)="aivi" arvr="" vr="" yk)="⎧" fourth,="" yi).="" 14.10.2.="" extremes="" 6="" ′2="" 2xu="" (14.10.12)="" (0)="1" (6)="7." three="" x3="6." (xk)="" u3="u" (14.10.13)="" 667="" (14.10.14)="" ⎪⎪⎪⎪⎨="" ⎪⎪⎪⎪⎩="" (u1="" u0)="" (u2="" u1)="" 2),="" (u3="" 4),="" 6,="" (14.10.15)="" derivative="" u0),="" u1),="" 6.="" (14.10.16)="" (14.10.15)–(14.10.16)="" get="" ="" 0="" −2x="" 2)02="" −2="" 2)0="" 2)0dx="" 4)02="" 4)0="" 4)0dx.="" integrating="" u1u2="" 37="" 67="" 101="" ∂i2="" ∂u1="14" ∂u2="−" 668="" putting="" (14.10.17)="" (1="" x).="" (14.10.18)="" identical="" due="" simplicity="" problem.="" will="" different.="" adding="" some="" technique="" (boundary="" method).="" research="" solid="" mechanics,="" fluid="" theory="" electromagnetic="" theory.="" breakthrough="" came="" 1963="" classic="" papers="" were="" published="" jaswon="" (1963)="" symm="" (1963).="" mathematical="" aspect="" prescribed="" uses="" volume="" surface="" very="" useful="" especially="" dimensional="" rapidly="" changing="" fracture="" contact="" computationally="" less="" efficient="" industry.="" popular="" acoustic="" problems.="" 1970s="" continued="" fast="" pace="" include="" nonlinear="" 14.11="" exercises="" explicit="" utt="" 4uxx="0," <x<="" t=""> 0, u (0, t) = u (1, t)=0, t ≥ 0, u (x, 0) = sin 2πx, ut (x, 0) = 0, 0 ≤ x ≤ 1. Compare the numerical solution with the analytical solution u (x, t) = cos 4πtsin 2πx at several points. 14.11 Exercises 669 2. (a) Calculate an explicit finite difference solution of the wave equation uxx − utt = 0, 0 <x< 1,="" t=""> 0, satisfying the boundary conditions u (0, t) = u (1, t)=0, t ≥ 0, and the initial conditions u (x, 0) = 1 8 sin πx, ut (x, 0) = 0, 0 ≤ x ≤ 1. Show that the exact solution of the problem is u (x, t) = 1 8 cos πtsin πx. Compare the two solutions at several points. (b) Solve the wave equation in (a) with the same boundary data and the initial data u (x, 0) = sin πx, ut (x, 0) = 0, 0 ≤ x ≤ 1. 3. Use the Lax–Wendroff method to find a numerical solution of the problem ux + ut = 0, x > 0, t > 0, u (x, 0) = 2 + x, x > 0, u (0, t)=2 − t, t > 0. Show that the exact solution of the problem is u (x, t)=2+(x − t). Compare the two solutions at various points. 4. Show that the finite difference approximation to the equation aut + bux = f (x, t) is ui,j+1 − 1 2 (ui+1,j + ui−1,j ) +  εb 2a  (ui+1,j − ui−1,j ) − fi,j = 0, where a, b are constants and ε = k/h. 670 14 Numerical and Approximation Methods 5. Obtain a finite difference solution of the heat conduction problem ut = κ uxx, 0 < x < l, t > 0, with the boundary conditions u (0, t) = u (l, t)=0, t> 0, and the initial condition u (x, 0) = 4x l (l − x), 0 ≤ x ≤ l. 6. (a) Find an explicit finite difference solution of the parabolic system ut = uxx, 0 <x< 1,="" t=""> 0, u (0, t) = u (1, t)=0, t > 0, u (0, t) = sin xπ on 0 ≤ x ≤ 1. Compare the numerical results with the analytical solution u (x, t) = e −π 2 t sin πx, at t = 0.5 and t = 0.05. (b) Prove that the Richardson finite difference scheme for problem 6(a) is ui,j+1 = ui,j−1 + 2 ε δ2 x ui,j . Hence, show that the exact solution of this equation is ui,j =
4 A1α j 1 + A2α j 2
5 sin πhi, where α1 and α2 are the roots of the quadratic equation x 2 + 8εx sin2 (πh/2) − 1=0. 7. Using four internal grid points, find the explicit finite difference solution of the Dirichlet problem ∇2u ≡ uxx + uyy = 0, 0 <x< 1,="" 0="" <y<="" u="" (x,="" 0)="x" (1="" −="" x),="" 1)="0" on="" ≤="" x="" (0,="" y)="u" (1,="" y="" 1.="" compare="" the="" numerical="" solution="" with="" exact="" analytical="" π="" 3="" ∞="" n="0" 2="" (2n="" +="" 1)3="" sin="" nπx="" sinh="" nπ="" at="" point="" 1="" ,="" .="" 14.11="" exercises="" 671="" 8.="" solve="" dirichlet="" problem="" by="" explicit="" finite="" difference="" method="" uxx="" uyy="0," <x<="" πx,="" for="" and="" 9.="" using="" a="" square="" grid="" system="" h="1" find="" of="" laplace="" equation="" quarter-disk="" given="" x2="" <=""> 0, u (x, 0) = 0, −1 <x< 1,="" u="" (x,="" y)="102" ,="" x2="" +="" y="" 2="1,"> 0. 10. Find a finite difference solution of the wave problem utt − uxx = 0, 0 <x< 1,="" t=""> 0, u (0, t) = u (1, t)=0, t ≥ 0, u (x, 0) = 1 2 x (1 − x), ut (x, 0) = 0, 0 ≤ x ≤ 1. Compare the numerical results with the exact analytical solution u (x, t) = 2 π 3 ∞ r=1 1 r 3 {1 − (−1)r } cos πrtsin πrx, at various points. 11. Obtain a finite difference solution of the problem utt = c 2uxx, 0 <x< 1,="" t=""> 0, u (0, t) = sin πct, u (1, t)=0, t ≥ 0, u (x, 0) = ut (x, 0) = 0, 0 ≤ x ≤ 1. 12. Show that the transformation v = log u transforms the nonlinear system vt = vxx + v 2 x , 0 <x< 1,="" t=""> 0, vx (0, t)=1, v (1, t)=0, t ≥ 0, v (x, 0) = 0, 0 ≤ x ≤ 1 into the linear system ut = uxx, 0 <x< 1,="" t=""> 0, ux (0, t) = u (0, t), u (x, 1) = 1, t ≥ 0, u (x, 0) = 1, 0 ≤ x ≤ 1. Solve the linear system by the explicit finite difference method with the derivative boundary condition approximated by the central difference formula. 672 14 Numerical and Approximation Methods 13. Solve the following parabolic system by the Crank–Nicolson method ut = uxx, 0 <x< 1,="" t=""> 0, u (0, t) = u (1, t)=0, t ≥ 0, with the initial condition (a) u (x, 0) = 1, 0 ≤ x ≤ 1, (b) u (x, 0) = sin πx, 0 ≤ x ≤ 1. (c) u (x, 0) = sin πx, 0 ≤ x ≤ 1 with 0 ≤ t ≤ 0.2 and in formula (14.5.2) κ = 1, k = h 2 . 14. Use the Crank–Nicolson implicit method with the central difference formula for the boundary conditions to find a numerical solution of the differential system ut = uxx, 0 <x< 1,="" t=""> 0, ux (0, t) = ux (1, t) = −u, t ≥ 0, u (x, 0) = 1, 0 ≤ x ≤ 1. 15. Find a numerical solution of the wave equation utt = c 2uxx, 0 < x < l, t > 0, with the boundary and initial conditions u = 1 20 ux at x = 0 and x = l, t > 0, u (x, 0) = 0, ut (x, 0) = a sin
4πx l
5 0 ≤ x ≤ l. 16. Determine the function representing a curve which makes the following functional extremum: (a) I (y (x)) =
1 0  y ′2 + 12xy dx, y (0) = 0, y (1) = 1, (b) I (y (x)) =
π/2 0  y ′2 − y 2 dx, y (0) = 0, y
4π 2
5 = 1, (c) I (y (x)) =
x1 x0 1 x  1 + y ′2 1 2 dx. 17. In the problem of tautochroneous motion, find the equation of the curve joining the origin O and a point A in the vertical (x, y)-plane so that a particle sliding freely from A to O under the action of gravity reaches the origin O in the shortest time, friction and resistance of the medium being neglected. 14.11 Exercises 673 18. In the problem of minimum surface of revolution, determine a curve with given boundary points (x0, y0) and (x1, y1) such that rotation of the curve about the x-axis generates a surface of revolution of minimum area. 19. Show that the Euler equation of the variational principle δI [u (x, y)] = δ

D F (x, y, u, p, q, l, m, n) dx dy = 0 is Fu − ∂ ∂xFp − ∂ ∂yFq + ∂ 2 ∂x2 Fl + ∂ 2 ∂x∂yFm + ∂ 2 ∂y2 Fn = 0, where p = ux, q = uy, l = uxx, m = uxy, n = uyy. 20. Prove that the Euler–Lagrange equation for the functional I =


R F (x, y, z, u, p, q, r, l, m, n, a, b, c) dx dy dz is Fu − ∂ ∂xFp − ∂ ∂yFq + ∂ 2 ∂z Fr + ∂ 2 ∂x2 Fl + ∂ 2 ∂y2 Fm + ∂ 2 ∂z2 Fn + ∂ 2 ∂x∂yFa + ∂ 2 ∂y∂z Fb + ∂ 2 ∂z∂xFc = 0, where (p, q, r)=(ux, uy, uz), (l, m, n)=(uxx, uyy, uzz), and (a, b, c) = (uxy, uyz, uzx). 21. In each of the following cases apply the variational principle or its simple extension with appropriate boundary conditions to derive the corresponding equations: (a) F = u 2 x + u 2 y + 2u 2 xy. (b) F = 1 2 u 2 t − α  u 2 x + u 2 y − β 2u 2 ! , (c) F = 1 2  utux + αu2 x − βu2 xx , (d) F = 1 2  u 2 t − α 2u 2 xx , (e) F = p (x) u ′2 + d dx  q (x) u 2 − [r (x) + λs (x)] u 2 , where p, q, r, and s are given functions of x, and α, β are constants. 674 14 Numerical and Approximation Methods 22. Derive the Schr¨odinger equation from the variational principle δ


R
2 2m  ψ 2 x + ψ 2 y + ψ 2 z + (V − E) ψ 2 dx dy dz = 0, where h = 2π
is the Planck constant, m is the mass of a particle moving under the action of a force field described by the potential V (x, y, z) and E is the total energy of the particle. 23. Derive the Poisson equation ∇2u = F (x, y) from the variational principle with the functional I (u) =

D u 2 x + u 2 y + 2uF (x, y) ! dx dy, where u = u (x, y) is given on the boundary ∂D of D. 24. Derive the equation of motion of a vibrating string of length l under the action of an external force F (x, t) from the variational principle δ
t2 t1
l 0 1 2 ρ u2 t − T ∗u 2 x  + ρ uF (x, t) dx dt = 0, where ρ is the line density and T ∗ is the constant tension of the string. 25. The kinetic and potential energies associated with the transverse vibration of a thin elastic plate of constant thickness h are T = 1 2 ρ

D u˙ 2 dx dy, V = 1 2 µ0

D " (∇u) 2 − 2 (1 − σ)  uxxuyy − u 2 xy # dx dy, where ρ is the surface density and µ0 = 2h 3E/3  1 − σ 2 . Use the variational principle δ
t2 t1

D [(T − V ) + fu] dx dy dt = 0 to derive the equation of motion of the plate ρu¨ + µ0∇4u = f (x, y, t), where f is the transverse force per unit area acting on the plate. 26. The kinetic and potential energies associated with the wave motion in elastic solids are 14.11 Exercises 675 T = 1 2


D ρ  u 2 t + v 2 t + w 2 t dx dy dz V = 1 2


D " λ (ux + vy + wz) 2 + 2µ  u 2 x + v 2 y + w 2 z +µ ( (vx + uy) 2 + (wy + vz) 2 + (uz + wx) 2 )# dx dy dz. Use the variational principle δ
t2 t1


D (T − V ) dx dy dz = 0 to derive the equation of wave motion in an elastic medium (λ + µ) grad divu + µ ∇2u = ρ utt, where u = (u, v, w) is the displacement vector. 27. From the variational principle δ

D L dx dt = 0 with L = −ρ
η −h  φt + 1 2 (∇φ) 2 + gz0 dz derive the basic equations of water waves ∇2φ = 0, −h (x, y) <z<η (x,="" y,="" t),="" t=""> 0, ηt + ∇φ · ∇η − φz = 0, on z = η, φz + 1 2 (∇φ) 2 + gz = 0, on z = η, φz = 0, on z = −h, where φ (x, y, z, t) is the velocity potential, and η (x, y, t) is the free surface displacement function in a fluid of depth h. 28. Derive the Boussinesq equation for water waves utt − c 2uxx − µ uxxtt = 1 2  u 2 xx from the variational principle δ

L dx dt = 0, where L ≡ 1 2 φ 2 t − 1 2 c 2 φ 2 x + 1 2 µ φ2 xt − 1 6 φ 3 x and φ is the potential for u (u = φx). 29. Determine an approximate solution of the problem of finding an extremum of the functional I (y (x)) =
1 0  y ′2 − y 2 − 2xy dx, y (0) = y (1) = 0. 676 14 Numerical and Approximation Methods 30. Find an approximate solution of the torsion problem of a cylinder with an elliptic base; the domain of integration D is the interior of the ellipse with the major and minor axes 2a and 2b respectively. The associated functional is I (u (x, y)) =

D 1 ∂u ∂x − y 2 u +  ∂u ∂y + x 2 u 3 dx dy. 31. Use the Rayleigh–Ritz method to find an approximate solution of the problem ∇2u = 0, 0 <x< 1,="" 0="" <y<="" u="" (0,="" y)="0=" (1,="" y),="" (x,="" 0)="x" (1="" −="" x).="" 32.="" find="" an="" approximate="" solution="" of="" the="" boundary-value="" problem="" ∇2u="0," x=""> 0, y> 0, x + 2y < 2, u (0, y)=0, u (x, 0) = x (2 − x), u (2 − 2y, y)=0. 33. In the torsion problem in elasticity, the Prandtl stress function Ψ (x, y) = ψ (x, y) − 1 2  x 2 + y 2 satisfies the boundary value problem ∇2Ψ = −2 in D Ψ = 0 on ∂D. Use the Galerkin method to find an approximate solution of the problem in a rectangular domain D = {(x, y) : −a ≤ x ≤ a, −b ≤ y ≤ b}. 34. Apply the Galerkin approximation method to find the first eigenvalue of the problem of a circular membrane of radius a governed by the equation ∇2u ≡ d 2u dr2 + 1 r du dr = λu in 0 <r 0, u (0) = u (l)=0, with given initial conditions. (a) Show that an appropriate requirement is that n i=1 A ′′ i (t)
l 0 vi (x) vj (x) dx + n i=1 Ai (t)
l 0 ∂vi ∂x · ∂vj ∂x dx = 0, where j = 1, 2, ..., n and that the approximate solution is given by un (x) = A1 (t) v1 (x) + ... + An (t) vn (t) = n i=1 Ai (t) vi (x). (b) Show that the finite element method leads to a system of ordinary differential equations B d 2A dt2 + C A (t)=0, A (0) = D where B and C are n × n matrices, A (t) is a n-vector function and D is an n-vector. 15 Tables of Integral Transforms In this chapter we provide a set of short tables of integral transforms of the functions that are either cited in the text or are in most common use in mathematical, physical, and engineering applications. For exhaustive lists of integral transforms, the reader is referred to Erd´elyi et al. (1954), Campbell and Foster (1948), Ditkin and Prudnikov (1965), Doetsch (1970), Marichev (1983), Debnath (1995), and Oberhettinger (1972). 15.1 Fourier Transforms f (x) F (k) = √ 1 2π
∞ −∞ exp (−ikx) f (x) dx 1 exp (−a |x|), a > 0
4% 2 π
5 a  a 2 + k 2 −1 2 x exp (−a |x|), a > 0
4% 2 π
5 (−2aik)  a 2 + k 2 −2 3 exp  −ax2 , a > 0 √ 1 2a exp
4 − k 2 4a
5 4  x 2 + a 2 −1 , a > 0 π 2 exp(−a|k|) a 5 x  x 2 + a 2 −1 π 2  ik 2a exp (−a |k|) 682 15 Tables of Integral Transforms f (x) F (k) = √ 1 2π
∞ −∞ exp (−ikx) f (x) dx 6  c, a ≤ x ≤ b 0, outside. √ ic 2π 1 k  e −ibk − e −iak 7 |x| exp (−a |x|), a > 0 % 2 π  a 2 − k 2 a 2 + k 2 −2 8 sin ax x π 2 H (a − |k|) 9 exp {−x (a − iω)} H (x) √ 1 2π i (ω−k+ia) 10  a 2 − x 2 − 1 2 H (a − |x|) π 2 J0 (ak) 11 sin
b(x 2+a 2 ) 1 2 (x2+a2) 1 2 π 2 J0  a √ b 2 − k 2 H (b − |k|) 12 cos(b √ a2−x2) (a2−x2) 1 2 H (a − |x|) π 2 J0  a √ b 2 + k 2 13 e −ax H (x), a > 0 √ 1 2π (a − ik)  a 2 + k 2 −1 14 √ 1 |x| exp (−a |x|), a > 0  a 2 + k 2 − 1 2 " a +  a 2 + k 2 1 2 # 1 2 15 δ (n) (x − a), n = 0, 1, 2,... √ 1 2π (ik) n exp (−iak) 16 exp (iax) √ 2π δ (k − a) 15.2 Fourier Sine Transforms 683 15.2 Fourier Sine Transforms f (x) Fs (k) = % 2 π
∞ 0 sin (kx) f (x) dx 1 exp (−ax), a> 0 % 2 π k  a 2 + k 2 −1 2 x exp (−ax), a> 0 % 2 π (2ak)  a 2 + k 2 −2 3 x α−1 , 0 <α< 1 % 2 π k −αΓ (α) sin  πα 2 4 √ 1 x √ 1 k , k> 0 5 x α−1 e −ax , α > −1, a > 0 % 2 π Γ (α) r −α sin (αθ), where r =  a 2 + k 2 1 2 , θ = tan−1  k a 6 x −1 e −ax, a> 0 % 2 π tan−1  k a , k> 0 7 x exp  −a 2x 2 2 −3/2  k a3 exp
4 − k 2 4a2
5 8 erfc (ax) % 2 π 1 k " 1 − exp
4 − k 2 4a2
5# 9 x  a 2 + x 2 −1 π 2 exp (−ak), a> 0 10 x  a 2 + x 2 −2 √ 1 2π  k a exp (−ak), (a > 0) 684 15 Tables of Integral Transforms f (x) Fs (k) = % 2 π
∞ 0 sin (kx) f (x) dx 11 H (a − x), a> 0 % 2 π 1 k (1 − cos ak) 12 x −1J0 (ax) ⎧ ⎪⎨ ⎪⎩ % 2 π sin−1  k a , 0 <k 0, b> 0 π 2 e −akI0 (ab), a<k< ∞="" 14="" j0="" (a="" √="" x),="" a=""> 0 % 2 π 1 k cos
4 a 2 4k
5 15  x 2 − a 2 ν− 1 2 H (x − a), |ν| < 1 2 2 ν− 1 2  a k ν Γ  ν + 1 2 J−ν (ak) 16 x 1−ν  x 2 + a 2 −1 Jν (ax), ν > − 3 2 , a, b > 0 π 2 a −ν exp (−ak) Iν (ab), a<k< ∞="" 17="" x="" −νjν+1="" (ax),="" ν=""> − 1 2 k(a 2−k 2 ) ν− 1 2 2 ν− 1 2 aν+1Γ(ν+ 1 2 ) H (a − k) 18 erfc (ax) % 2 π 1 k " 1 − exp
4 − k 2 4a2
5# 19 x −α, 0 < Re α < 2 % 2 π Γ (1 − α) k α−1 cos  απ 2 20  ax − x 2 α− 1 2 H (a − x), α > − 1 2 √ 2 Γ  α + 1 2  a k α sin  ak 2 Jα  ak 2 15.3 Fourier Cosine Transforms 685 15.3 Fourier Cosine Transforms f (x) Fc (k) = % 2 π
∞ 0 cos (kx) f (x) dx 1 exp (−ax), a> 0
4% 2 π
5 a  a 2 + k 2 −1 2 x exp (−ax), a > 0
4% 2 π
5  a 2 − k 2 a 2 + k 2 −2 3 exp  −a 2x 2 1 a √ 2 exp
4 − k 2 4a2
5 4 H (a − x) % 2 π  sin ak k 5 x a−1 , 0 <a< 1="" %="" 2="" π="" Γ="" (a)="" k="" −a="" cos="" ="" aπ="" 6="" ax2="" ,="" a=""> 0 1 2 √ a " cos
4 k 2 4a
5 + sin
4 k 2 4a
5# 7 sin  ax2 , a> 0 1 2 √ a " cos
4 k 2 4a
5 − sin
4 k 2 4a
5# 8  a 2 − x 2 ν− 1 2 H (a − x),ν> − 1 2 2 ν− 1 2 Γ  ν + 1 2  a k ν Jν (ak) 9  a 2 + x 2 −1 J0 (bx), a, b > 0 π 2 a −1 exp (−ak) I0 (ab), b<k< ∞="" 10="" x="" −νjν="" (ax),ν=""> − 1 2 (a 2−k 2 ) ν− 1 2 H(a−k) 2 ν− 1 2 aν Γ(ν+ 1 2 ) 686 15 Tables of Integral Transforms f (x) Fc (k) = % 2 π
∞ 0 cos (kx) f (x) dx 11  x 2 + a 2 − 1 2 exp " −b  x 2 + a 2 1 2 # K0 " a  k 2 + b 2 1 2 # , a> 0, b> 0 12 x ν−1 e −ax, ν> 0, a> 0 % 2 π Γ (ν) r −ν cos nθ, where r =  a 2 + k 2 1 2 , θ = tan−1  k a 13 2 x e −x sin x % 2 π tan−1  2 k2 14 sin " a  b 2 − x 2 1 2 H (b − x) # π 2 (ab)  a 2 + k 2 − 1 2 × J1 " b  a 2 + k 2 1 2 # 15 (1−x 2 ) (1+x2) 2 π 2 k exp (−k) 16 x −α, 0 <α< 1 π 2 k α−1 Γ(α) sec  πα 2 17  1 a + x e −ax, a> 0 π 2 2a 2 (a2+k2) 2 18 log
4 1 + a 2 x2
5 , a> 0 √ 2π (1−e −ak) k 19 log
4 a 2+x 2 b 2+x2
5 , a, b > 0 √ 2π (e −bk−e −ak) k 20 a  x 2 + a 2 −1 , a> 0 π 2 exp (−ak), k> 0 15.4 Laplace Transforms 687 15.4 Laplace Transforms f (t) f (s) =
∞ 0 exp (−st) f (t) dt 1 f (n) (t) s nf (s) − n−1 r=0 s n−r−1f (r) (0) 2
t 0 f (t − τ ) g (τ ) dτ f (s) g (s) 3 t nf (t) (−1)n d n dsn f (s) 4 f (t − a) H (t − a) exp (−as) f (s) 5 t n (n = 0, 1, 2, 3,...) n! sn+1 6 e at 1 s−a 7 t ne −at Γ(n+1) (s+a) n+1 8 t a (a > −1) Γ(a+1) s a+1 9 e at cos bt s−a (s−a) 2+b 2 10 e at sin bt b (s−a) 2+b 2 11 √ 1 t π s 12 2 √ t 1 s π s 688 15 Tables of Integral Transforms f (t) f (s) =
∞ 0 exp (−st) f (t) dt 13 t −1/2 exp  − a t π s exp (−2 √ as) 14 t −3/2 exp  − a t π a exp (−2 √ as) 15 √ 1 πt (1 + 2at) e at s (s−a) √ s−a 16 1 2 √ πt3  e bt − e at √ s − a − √ s − b 17 exp  a 2 t erf  a √ t √ a s(s−a2) 18 exp  a 2 t erfc  a √ t 1 √ s( √ s+a) 19 √ 1 πt + a exp  a 2 t erf  a √ t √ s (s−a2) 20 √ 1 πt − a exp  a 2 t erfc  a √ t √ 1 s+a 21 exp(−at) √ b−a erf
4 (b − a)t
5 1 (s+a) √ s+b 22 1 2 e iωt " exp (−λz) erfc
4 ζ − √ iωt
5 + exp (λz) erfc  ζ + √ iωt ! , where ζ = z/2 √ νt, λ = % iω ν (s − iω) −1 exp  −z s ν 23 1 2 " exp (−ab) erfc
4 b−2at 2 √ t
5 + exp (ab) erfc
4 b+2at 2 √ t
5# exp " −b  s + a 2 1 2 # 15.4 Laplace Transforms 689 f (t) f (s) =
∞ 0 exp (−st) f (t) dt 24 J0 (at)  s 2 + a 2 − 1 2 25 I0 (at)  s 2 − a 2 − 1 2 26 t α−1 exp (−at), α > 0 Γ (α) (s + a) −α 27 t −1Jν (at) ν −1a ν √ s 2 + a 2 + s −ν , Re ν > − 1 2 28 J0  a √ t 1 s exp
4 − a 2 4s
5 29  2 a ν t ν/2Jν  a √ t s −(ν+1) exp
4 − a 2 4s
5 , Re ν > − 1 2 30 a 2t √ πt exp
4 − a 2 4t
5 exp (−a √ s), a> 0 31 √ 1 πt exp
4 − a 2 4t
5 √ 1 s exp (−a √ s), a ≥ 0 32 exp
4 − a 2 t 2 4
5 √ π a exp
4 s 2 a2
5 erfc  s a , a ≥ 0 33  t 2 − a 2 − 1 2 H (t − a) K0 (as), a> 0 34 δ (n) (t − a), n = 0, 1,... s n exp (−as) 690 15 Tables of Integral Transforms f (t) f (s) =
∞ 0 exp (−st) f (t) dt 35 t mα+β−1E (m) α,β (+ at), m = 0, 1, 2,... m! s α−β (sα+ a) m+1 36 √ π Γ(ν+ 1 2 )  t 2a ν Jν (at)  s 2 + a 2 −(ν+ 1 2 ) , Re ν > − 1 2 37 1 2 e −ct" exp  −a √ b − c ×erfc( √a 4t −  (b − c)t ) − exp  a √ b − c ×erfc( √a 4t +  (b − c)t )# exp(−a √ s+b) (s+c) √ (s+b) 38 1 2 e −ct" exp  −a √ b − c ×erfc( √a 4t − t √ b − c ) − exp  a √ b − c ×erfc( √a 4t + t √ b − c )# exp(−a √ s+b) (s+c) 39 e −bt "% 4t π exp
4 − a 2 4t
5 −a erfc
4 √a 4t
5# exp(−a √ s+b) (s+b) 3/2 40 e −bt " t + 1 2 a 2 erfc
4 √a 4t
5 − % ta2 π exp
4 − a 2 4t
5 exp(−a √ s+b) (s+b) 2 15.5 Hankel Transforms 691 15.5 Hankel Transforms f (r) order n ˜fn (k) =
∞ 0 r Jn (kr) f (r) dr 1 H (a − r) 0 a k J1 (ak) 2 exp (−ar) 0 a  a 2 + k 2 − 3 2 3 1 r exp (−ar) 0  a 2 + k 2 − 1 2 4  a 2 − r 2 H (a − r) 0 4a k3 J1 (ak) − 2a 2 k2 J0 (ak) 5 a  a 2 + r 2 − 3 2 0 exp (−ak) 6 1 r cos (ar) 0  k 2 − a 2 − 1 2 H (k − a) 7 1 r sin (ar) 0  a 2 − k 2 − 1 2 H (a − k) 8 1 r 2 (1 − cos ar) 0 cosh−1  a k H (a − k) 9 1 r J1 (ar) 0 1 a H (a − k), a> 0 10 Y0 (ar) 0  2 π a 2 − k 2 −1 11 K0 (ar) 0  a 2 + k 2 −1 692 15 Tables of Integral Transforms f (r) order n ˜fn (k) =
∞ 0 r Jn (kr) f (r) dr 12 δ(r) r 0 1 13  r 2 + b 2 − 1 2 × exp ( −a  r 2 + b 2 1 2 ) 0  k 2 + a 2 − 1 2 exp ( −b  k 2 + a 2 1 2 ) 14  r 2 + a 2 − 1 2 0 1 k exp (−ak) 15 exp (−ar) 1 k  a 2 + k 2 −3/2 16 sin ar r 1 a H(k−a) k(k2−a2) 1 2 17 1 r exp (−ar) 1 1 k 1 − a (k2+a2) 1 2 18 1 r 2 exp (−ar) 1 1 k " k 2 + a 2 1 2 − a # 19 r n H (a − r) > −1 1 k a n+1 Jn+1 (ak) 20 r n exp (−ar), (Re a > 0) > −1 √ 1 π 2 n+1 Γ(n+ 3 2 ) a kn (a2+k2) n+ 3 2 21 r n exp  −ar2 > −1 k n (2a) n+1 exp
4 − k 2 4a
5 15.5 Hankel Transforms 693 f (r) order n ˜fn (k) =
∞ 0 r Jn (kr) f (r) dr 22 r a−1 > −1 2a Γ[ 1 2 (a+n+1)] ka+1 Γ[ 1 2 (1−a+n)] 23 r n  a 2 − r 2 m−n−1 × H (a − r) > −1 2 m−n−1 Γ (m − n) a mk n−mJm (ak) 24 r m exp  −r 2/a2 > −1 k n am+n+2 2n+1 Γ(n+1) Γ  1 + m 2 + n 2 × 1F1  1 + m 2 + n 2 ; n + 1; − 1 4 a 2k 2 25 1 r Jn+1 (ar) > −1 k na −(n+1)H (a − k), a > 0 26 r n  a 2 − r 2 m H (a − r), m > −1 > −1 2 ma nΓ (m + 1)  a k m+1 × Jn+m+1 (ak) 27 1 r 2 Jn (ar) > 1 2 ⎧ ⎨ ⎩ 1 2n  k a n , 0 < k ≤ a 1 2n  a k n , a<k< ∞="" 28="" r="" n="" (a2+r="" 2)m+1="" ,="" a=""> 0 > −1  k 2 m a n−m Γ(m+1) Kn−m (ak) 29 exp  −p 2 r 2 Jn (ar) > −1  2p 2 −1 exp
4 − a 2+k 2 4p2
5 In
4 ak 2p2
5 30 1 r exp (−ar) > −1
(k 2+a 2 ) 1 2 −a n kn(k2+a2) 1 2 31 r n (r 2+a2) n+1 > −1  k 2 n K0(ak) Γ(n+1) 694 15 Tables of Integral Transforms f (r) order n ˜fn (k) =
∞ 0 r Jn (kr) f (r) dr 32 r n (a2−r 2) n+ 1 2 H (a − r) < 1 √ 1 π  k 2 n Γ  1 2 − n  sin ak k 33 f (ar) n 1 a2 ˜fn  k a 34 r −1 exp  −ar2 1 1 k " 1 − exp
4 − k 2 4a
5# 35 r −1 sin  ar2 , a > 0 1 1 k sin
4 k 2 4a
5 36 r −1 cos  ar2 , a > 0 1 1 − cos
4 k 2 4a
5 37 exp (−ar), a > 0 > −1 (a+n √ k2+a2) (k2+a2) 3/2
4 k a+ √ a2+k2
5n 38 exp  −ar2 J0 (br) 0 a 2 exp
4 − k 2−b 2 4a
5 I0  bk 2a 39 H(a−r) √ a2−r 2 0 aπ 2k J 1 2 (ak), a > 0 40 r nH(a−r) √ a2−r 2 > −1  π 2k a n+ 1 2 Jn+1 (ak), a > 0 41 r −2 sin r 0 sin−1  1 k , (k > 1) 15.6 Finite Hankel Transforms 695 15.6 Finite Hankel Transforms f (r) order n ˜fn (ki) =
a 0 r Jn (r ki) f (r) dr 1 c, where c is a constant 0
4 ac ki
5 J1 (aki) 2  a 2 − r 2 0 4a k 3 i J1 (aki) 3  a 2 − r 2 − 1 2 0 k −1 i sin (aki) 4 J0(αr) J0(αa) 0 − aki (α2−k 2 i ) J1 (aki) 5 1 r 1 k −1 i {1 − J0 (aki)} 6 r −1  a 2 − r 2 − 1 2 1 (1−cos aki) (aki) 7 r n > −1 a n+1 ki Jn+1 (aki) 8 Jν (αr) Jν (αa) > −1 aki (α2−k 2 i ) J ′ ν (aki) 9 r −n  a 2 − r 2 − 1 2 > −1 π 2 & J n 2  aki 2 '2 10 r n  a 2 − r 2 −(n+ 1 2 ) < 1 2 Γ( 1 2 −n) √ π 2n k n−1 i sin (aki) 11 r n−1  a 2 − r 2 n− 1 2 > − 1 2 √ π 2 Γ  n + 1 2
4 2 ki
5n a 2nJ 2 n  aki 2 Answers and Hints to Selected Exercises 1.6 Exercises 1. (a) Linear, nonhomogeneous, second-order; (b) quasi-linear, first-order; (c) nonlinear, first-order; (d) linear, homogeneous, fourth-order; (e) linear, nonhomogeneous, second-order; (f) quasi-linear, third-order; (g) nonlinear, second-order; and (h) nonlinear, homogeneous. 5. u (x, y) = f (x) cos y + g (x) sin y. 6. u (x, y) = f (x) e −y + g (y). 7. u (x, y) = f (x + y) + g (3x + y). 8. u (x, y) = f (y + x) + g (y − x). 11. ux = vy ⇒ uxx = vxy, vx = −uy ⇒ vyx = −uyy. Thus, uxx + uyy = 0. Similarly, vxx + vyy = 0. 12. Since u (x, y) is a homogeneous function of degree n, u = x nf  y x . ux = n xn−1f  y x − x n−2y f′  y x , and uy = x n−1f ′  y x . Thus, x ux + y uy = n xnf  y x = n u. 23. ux = − 1 b exp  − x b f (ax − by) + exp  − x b d d(ax−by) f (ax − by) · d(ax−by) dx = − 1 b exp  − x b f + a exp  − x b f ′ (ax − by) uy = (−b) exp  − x b f ′ (ax − by). Thus, b ux + a uy + u = 0. 698 Answers and Hints to Selected Exercises 24. V ′′ (t)+2b V ′ (t) + k 2 c 2V (t)=0. 25. Differentiating with respect to r and t partially gives V ′′ (r) + n 2V (r) = 0. 2.8 Exercises 2. (a) xp − yq = x − y, (d) yp − xq = y 2 − x 2 . 3. (a) u = f (y), (b) u = f (bx − ay), (c) u = f (y e−x ), (d) u = f  y − tan−1 x , (e) u = f
4 x 2−y 2 x
5 , (f) Hint: dx y+u = dy y = du x−y = d(x+u) x+u = d(u+y) x , x dx = (u + y) d (u + y) ⇒ (u + y) 2 − x 2 = c1. d(u+x) u+x = dy y ⇒ u+x y = c2, f
4 u+x y ,(u + y) 2 − x 2
5 = 0. (g) dx y2 = dy −xy = du xu−2xy = d(u−y) x(u−y) . From the second and the fourth, (u − y) y = c1 and x 2 + y 2 = c2. Hence, (u − y) y = f  x 2 + y 2 . Thus, u = y + y −1f  x 2 + y 2 . (h) u + log x = f (xy), (i) f  x 2 + u 2 , y3 + u 3 = 0. 4. u (x, y) = f  x 2 + y −1 . Verify by differentiation that u satisfies the original equation. 5. (a) u = sin  x − 3 2 y , (b) u = exp  x 2 − y 2 , (c) u = xy + f  y x , u = xy + 2 −  y x 3 , (d) u = sin  y − 1 2 x 2 , (e) u = ⎧ ⎪⎨ ⎪⎩ 1 2 y 2 + exp −  x 2 − y 2 ! for x > y, 1 2 x 2 + exp −  y 2 − x 2 ! for x < y. (f) Hint: y = 1 2 x 2 + C1, u = C 2 1 x + C2, u = x  y − 1 2 x 2 2 +f  y − 1 2 x 2 , u = x  y − 1 2 x 2 2 + exp  y − 1 2 x 2 . (g) y x = C1 and u+1 y = C2, C2 = 1+ 1 C2 1 . Thus, u = y + x 2 y − 1, y = 0. (h) Hint: x + y = C1, dy −u = du u2+C2 1 , u 2 + C 2 1 = C2 exp (−2y). 2.8 Exercises 699 From the Cauchy data, it follows that 1 + C 2 1 = C2, and hence, u = "(1+(x + y) 2 ) e −2y − (x + y) 2 # 1 2 . (i) dy dx − y x = 1, d dx  y x = 1 x which implies that x = C1 exp  y x . u+1 x = C2. Hence, f  u+1 x , x exp  − y x = 0. Initial data imply x = C1 and x 2+1 x = C2. Hence C2 = C1 + 1 C1 . u+1 x = x exp  − y x + 1 x exp  y x . Thus, u = x 2 exp  − y x + exp  y x − 1. (j) √ dx x = dy u = du −u2 . The second and the third give y = − log (Au) and hence, A = 1 and u = exp (−y). The first and the third yield u −1 = 2√ x − B. At (x0, 0), x0 > 0, B = 2√ x0 − 1. Hence, u −1 = 2 √ x − √ x0 +1= 1 y . The solution along the characteristic is u = exp (−y) or u −1 = 2 √ x − √ x0 + 1. (k) dx ux2 = dy exp(−y) = du −u2 . The first and the third give x −1 = log u + A and hence, A = 1 x0 , x0 > 0. The second and third yield u = exp (−y). Or, eliminating u gives y =  x −1 0 − x −1 . 6. u 2 − 2ut + 2x = 0, and hence, u = t + √ t 2 − 2x. 7. u (x, y) = exp
4 x x2−y2
5 . 8. (a) u = f  y x , z x (b) Hint: u1 = x−y xy = C1, d(x−y) x2−y2 = dz z(x+y) gives u = x−z z = C2. Hence, u = f
4 x−y xy , x−y z
5 . (c) φ = (x + y + z) = C1. Hint: ( dx x ) y−z = ( dy y ) z−x = ( dz z ) x−y = dx x + dy y + dz z 0 = d log(xyz) 0 , ψ = xyz = C2, and hence, u = f (x + y + z, xyz) is the general solution. (d) Hint: x dx + y dy = 0, x 2 + y 2 = C1 z dz = −  x 2 + y 2 y dy = −C1 y dy, z 2 +  x 2 + y 2 y 2 = C2, u = f  x 2 + y 2 , z2 +  x 2 + y 2 y 2 . (e) x −1dx y2−z 2 = y −1dy z 2−x2 = z −1dz y2−x2 = d(log xyz) 0 . u = f  x 2 + y 2 + z 2 , xyz . 700 Answers and Hints to Selected Exercises 9. (a) Hint: y − x 2 2 = C1, u = xy − x 3 3 +C2, φ
4 u − xy + x 3 3 , y − x 2 2
5 = 0. u = xy − x 3 3 + f
4 y − x 2 2
5 , u = xy − x 3 3 +
4 y − x 2 2
52 . (b) u = xy − 1 3 x 3 + y − x 2 2 + 5 6 . 11. x+u y = C1, u 2 − (x − y) 2 = C2, u 2 − 2u y − (x − y) 2 − 2 y (x − y) = 0. u = 2 y + (x − y), y > 0. 12. (a) x = τ 2 2 + τs + s, y = τ + 2s, u = τ + s = (2x−2y+y 2 ) 2(y−1) (b) x = τ 2 2 + τs + s 2 , y = τ + 2s, u = τ + s, (y − s) 2 = 2x − s 2 , which is a set of parabolas. (c) x = 1 2 (τ + s) 2 , y = u = τ + s. 13. Hint: The initial curve is a characteristic, and hence, no solution exists. 14. (a) u = exp
4 xy x+y
5 , (b) u = sin
4 x 2−y 2+1 2
5 1 2 , (c) u = 2  xy 3 1 2 + 1 2 log  y 3x , (e) u = 1 2 x 2 − 1 4 y 2 + 1 2 x 2y + 1 4 . (f) Hint: dx 1 = dy 2 = du 1+u , y − 2x = c1 and (1 + u) e −x = c2, (1 + u) e −x = f (y − 2x), 1+u = exp (3x − y + 1) [1 + sin (y − 2x − 1)]. (g) Hint: dx 1 = dy 2 = du u , y−2x = c1, and u e−x = c2, u e−x = f (y − 2x), u = exp  y−x 2 cos  y−3x 2 . (h) dx 1 = dy 2x = du 2x u , (y − x) 2 = c1, and u e−x 2 = c2, u e−x 2 = f  y − x 2 , u (x, y) =  x 2 − y e y . (i) dx u = dy 1 = du u , u − x = c1, and u e−y = c2, f (u e−y , u − x) = 0, u ey = g (u − x), u = 2x ey 2e y−1 , dx dy = u, x = A (2e y − 1) is the family of characteristics. (j) dx 1 = dy 1 = du u2 , y − x = c1, and 1 u + x = c2, 1 u + x = f (y − x), f (x) = −  1−tanh x tanh x , u (x, y) = tanh(x−y) 1−y tanh(x−y) . 15. 3uy = u 2 + x 2 + y 2 . Hint: x dx+y dy+u du 0 , x 2 + y 2 + u 2 = c1, dy y = − du u gives uy = c2. x 2 + y 2 + u 2 = f (uy), and hence, 3u 2 = f  u 2 . 2.8 Exercises 701 16. (a) x (s, τ ) = τ , y (s, τ ) = τ 2 2 + aτs + s, u (s, τ ) = τ + as. τ = x, s = (1 + ax) −1  y − 1 2 x 2 a, and hence, u (x, y) = x + as = (1 + ax) −1 & x + a  y + 1 2 x 2 ', singular at x = − 1 a . (b) y = u 2 2 + f (u − x), 2y = u 2 + (u − x) 2 , u (0, y) = √y. 17. (a) Hint: d(x+y+u) 2(x+y+u) = d(y−u) −(y−u) = d(u−x) −(u−x) (x + y + u) (y − u) 2 = c1 and (x + y + u) (u − x) 2 = c2. (b) Hint: dx x = dy −y . Hence, xy = a. dx xu(u2+a) = du x4 . So, dx du = u(u 2+a) x3 giving x 4 = u 4 + 2au2 + b and, thus, x 4 − u 4 − 2u 2xy = b. (c) dx x+y = dy x−y = dy 0 (exact equation). u = f  x 2 − 2xy − y 2 . (d) f  x 2 − y 2 , u − 1 2 y 2  x 2 − y 2 = 0. (e) f  x 2 + y 2 + z 2 , ax + by + cz = 0. 18. Hint: dx x = dy y = dz z , and hence, x z = c, y z = d. x 2 + y 2 = a 2 and z = tan−1  y x give  c 2 + d 2 z 2 = a 2 and z = b tan−1  d c . c =  a z cos θ, d =  a z sin θ, and z = b tan−1 (tan θ) = bθ. Thus, the curves are xbθ = az cos θ and ybθ = az sin θ. 19. F ( x + y + u,(x − 2y) 2 + 3u 2 ) = 0. Hint: (dx−2dy) 9u = du −3(x−2y) . (x − 2y) 2 + 3u 2 = (x + y + u) 2 . 20. F  x 2 + y, yu = 0,  x 2 + y 4 = yu. 21. Hint: x − y + z = c1, dz −(x+y+z) = (dx+dy+dz) 8z , and hence, 8z 2 + (x + y + z) 2 = c2. F ( (x − y + z), 8z 2 + (x + y + z) 2 ) = 0. c 2 1 + c2 = 2a 2 , or (x − y + z) 2 + (x + y + z) 2 + 8z 2 = 2a 2 . 22. F  x 2 + y 2 + z 2 , y2 − 2yz − z 2 = 0. (a) y 2 − 2yz − z 2 = 0, two planes y =  1+√ 2 z. (b) x 2 + 2yz + 2z 2 = 0, a quadric cone with vertex at the origin. (c) x 2 − 2yz + 2y 2 = 0, a quadric cone with vertex at the origin. 702 Answers and Hints to Selected Exercises 23. Use the Hint of 17(c). dx dt = x + y, dy dt = x − y, d 2x dt2 = 2x.  dx dt 2 = 2x 2 + c. When x =0= y, dx dt = √ 2 x. √ 2 u = ln x + x 2 − 2xy + 2y. 24. (a) a = f  x + 3 2 y . (b) x = at + c1, y = bt, u = c2 e ct , c2 = f (c1), u (x, y) = f  x − a b y exp  cy b . (c) u = f
4 x 1−y
5 (1 − y) c . (d) x = 1 2 t 2 + αst + s, y = t; u = y + 1 2 α (αy + 1)−1  2x − y 2 . 26. (a) Hint: (f ′ ) 2 = 1 − (g ′ ) 2 = λ 2 ; f ′ (x) = λ and g ′ (y) = √ 1 − λ2. f (x) = λx + c1 and g (y) = y √ 1 − λ2 + c2. Hence, u (x, y) = λx + y √ 1 − λ2 + c. (b) Hint: (f ′ ) 2 + (g ′ ) 2 = f (x)+g (y) or (f ′ ) 2 −f (x) = g (y)−(g ′ ) 2 = λ. Hence, (f ′ ) 2 = f (x) + λ and g ′ =  g (y) − λ. Or, √ df f+λ = dx and √ dg g−λ = dy. f (x) + λ =  x+c1 2 2 and g (y) − λ =  y+c2 2 2 . u (x, y) =  x+c1 2 2 +  y+c2 2 2 . (c) Hint: (f ′ ) 2 + x 2 = −g ′ (y) = λ 2 . Or f ′ (x) = √ λ2 − x 2, and g (y) = −λ 2y + c2. Putting x = λ sin θ, we obtain f (x) = 1 2 λ 2 sin−1  x λ + x 2 √ λ2 − x 2 + c1, u (x, y) = 1 2 λ 2 sin−1  x λ + x 2 √ λ2 − x 2 − λ 2y + (c1 + c2). (d) Hint: x 2 (f ′ ) 2 = λ 2 and 1 − y 2 (g ′ ) 2 = λ 2 . Or, f (x) = λ ln x + c1 and g (y) = √ 1 − λ2 ln y + c2. 27. (a) Hint: v = ln u gives vx = 1 u · ux, and vy = 1 u · uy. x 2  ux u 2 + y 2  uy u 2 = 1. Or, x 2v 2 x + y 2v 2 y = 1 gives x 2 (f ′ ) 2 + y 2 (g ′ ) 2 = 1. 2.8 Exercises 703 x 2 {f ′ (x)} 2 = 1 − y 2 (g ′ ) 2 = λ 2 . Or, f (x) = λ ln x + c1 and g (y) = √ 1 − λ2 (ln y) + c2. Thus, v (x, y) = λ ln x + √ 1 − λ2 (ln y) + ln c, (c1 + c2 = ln c). u (x, y) = c xλ y √ 1−λ2 . (b) Hint: v = u 2 and v (x, y) = f (x) + g (y) may not work. Try u = u (s), s = λxy, so that ux = u ′ (y)·(λy) and uy = u ′ (s)·(λx). Consequently, 2λ 2  1 u du ds 2 = 1. Or, 1 u du ds = √ 1 2 1 λ . Hence, u (s) = c1 exp
4 s λ √ 2
5 . u (x, y) = c1 exp
4 √xy 2
5 . 28. Hint: vx = 1 2 √ux u , vy = 1 2 √ uy u . This gives x 4 (f ′ ) 2 + y 2 (g ′ ) 2 = 1. Or, x 4 (f ′ ) 2 = 1 − y 2 (g ′ ) 2 = λ 2 . Or, x 4 (f ′ ) 2 = λ 2 and y 2 (g ′ ) 2 = 1 − λ 2 . Hence, f (x) = − λ x + c1 and g (y) = √ 1 − λ2 ln y + c2 u (x, y) =  − λ x + √ 1 − λ2 ln y + c 2 . 29. Hint: vx = ux u , vy = uy u . v 2 x x2 + v 2 y y2 = 1, and v = f (x) + g (y). Or, (f ′ ) 2 x2 = 1 − 1 y2 (g ′ ) 2 = λ 2 . f ′ (x) = λx, and g ′ (y) = √ 1 − λ2 y. Or, f (x) = λ 2 x 2 + c1, and g (y) = 1 2 y 2 √ 1 − λ2 + c2. v (x, y) = λ 2 x 2 + y 2 2 √ 1 − λ2 + c = ln u. u (x, y) = c exp
4 λ 2 x 2 + y 2 2 √ 1 − λ2
5 , c1 + c2 = ln c. e x 2 = u (x, 0) = c e λ 2 x 2 , which gives c = 1 and λ = 2. 30. (a) Hint: ξ = x − y, η = y; u (x, y) = e yf (x − y), (b) ξ = x, η = y − x 2 2 , uξ = η + 1 2 ξ 2 , u = ξη + 1 6 ξ 3 + f (η). u (x, y) = xy − 1 3 x 3 + f
4 y − x 2 2
5 . (c) ξ = y exp  −x 2 , η = y, and e 2uf (ξ) = η, e 2uf
4 y e−x 2
5 = y, (d) dx 1 = dy −y = du 1+u , ξ = y ex , η = y. Thus, (1 + u) f (ξ) = 1 η . Or, (1 + u) f (y ex ) = y −1 . 31. (c) u (x, y) = α exp  βx − a b βy . 704 Answers and Hints to Selected Exercises 32. (a) v (x, t) = x + ct, u (x, t) = (6x+3ct2+5ct3 ) 6(1+2t) . (b) v (x, t) = x + ct, u (x, t) = (6x+3ct2+4ct3 ) 6(1+2t) . 33. (a) v (x, t) = e x+at , u − 1 a e at = c1, and u − 1 a e at = f  x − ut + t a e at − 1 a2 e at . u (x, t) = (1 + t) −1 & (x − ut) +  1 a + t a − 1 a2 e at +  1 a2 − 1 a '. (b) v = x − ct, u (x, t) = (6x−3ct2+4ct3 ) 6(1−2t) . 34. dt 1 = dy −x = du u , t + ln x = c1, and xu = c2. g (xu, t + ln x) = 0. Or, u = 1 x h (t + ln x). u (x, t) = e t ln (xet ), where g and h are arbitrary functions. 3.9 Exercises 11. Hint: Differentiate the first equation with respect to t to obtain ρtt + ρ0divut = 0. Take gradient of the last equation to get ∇ρ = −  ρ0/c2 0 ut. We next combine these two equations to obtain ρtt = c 2 0 ∇2ρ. Application of ∇2 to p − p0 = c 2 0 (ρ − ρ0) leads to ∇2p = c 2 0 ∇2ρ. Also ptt = c 2 0 ρtt = c 4 0 ∇2ρ = c 2 0 ∇2p. Using u = ∇φ in the first equation gives ρt + ρ0∇2φ = 0, and differenting the last equation with respect to t yields ρt = −  ρ0/c2 0 φtt. Combining these two equations produces the wave equation for φ. Finally, we take gradient of the first and the last equations to obtain ∇ρt + ρ0∇2u = 0 and ∇ρ = −  ρ0/c2 0 ut that leads to the wave equation for ut. 14. (a) Differentiate the first equation with respect to t and the second equation with respect to x. Then eliminate Vxt and Vx to obtain the desired telegraph equation. (e) (i) ∂ 2 ∂x2 (I,V ) = 1 c 2 ∂ 2 ∂t2 (I,V ), c2 = 1 LC . 3.9 Exercises 705 (ii) ∂ ∂t (I,V ) = κ ∂ 2 ∂x2 (I,V ), κ = 1 RC . (iii)
4 ∂ 2 ∂t2 + 2k ∂ ∂t + k 2
5 (I,V ) = c 2 ∂ 2 ∂x2 (I,V ). 17. (a) The two-dimensional unsteady Euler equations are du dt = − 1 ρ ∂p ∂x , dv dt = − 1 ρ ∂p ∂y , where d dt = ∂ ∂t + u · ∇ = ∂ ∂t + u ∂ ∂x + v ∂ ∂y , and u = (u, v). (b) For two-dimensional steady flow, the Euler equations are u ux + v uy = − 1 ρ px, u vx + v vy = − 1 ρ py. Using dp dρ = c 2 , these equations become u ux + v uy = −c 2 (ρx/ρ), u vx + v vy = −c 2 (ρy/ρ). Multiply the first equation by u and the second by v and add to obtain u 2 ux + uv (uy + vx) + v 2vy = −
4 c 2 ρ
5 (uρx + vρy). Using the continuity equation (ρu)x + (ρv)y = 0, the right hand of this equation becomes c 2 (ux + vy). Hence is the desired equation. (c) Using u = ∇φ = (φx, φy), the result follows. (d) Substitute ρx and ρy from 17(b) into the continuity equation uρx + vρy + ρ (ux + vy) = 0 to obtain  c 2 − u 2 φxx − 2uvφxy +  c 2 − v 2 φyy = 0. Also du = ux dx + uy dy = −φxxdx − φxydy, dv = vx dx + vy dy = −φxydx − φyydy. Denoting D for the coefficient determinant of the above equations for φxx, φxy and φyy gives the solutions φxx = − D1 D , φxy = D2 D , φyy = −D3 D . D = 0 gives a quadratic equation for the slope of the characteristic C, that is,  c 2 − u 2
4 dy dx
52 + 2uv
4 dy dx
5 +  c 2 − v 2 = 0. Thus, directions are real and distinct provided 706 Answers and Hints to Selected Exercises 4u 2v 2 − 4  c 2 − u 2 c 2 − v 2 > 0, or  u 2 + v 2 > c2 . D2 = 0 gives − dy dx = (c 2−v 2 ) (c 2−u2)  dv du . Substitute into the quadratic equation to obtain  c 2 − v 2  dv du 2 − 2uv  dv du +  c 2 − u 2 = 0. Note that when D1 = D2 = D3 = D = 0, any one of the second order φ derivatives can be discontinuous. 18. (a) Hint: Use ∇ × ∂u ∂t = ∂ω ∂t , ∇ × (u · ∇u) u · ∇ω − ω∇u, where we have used ∇ · u = 0 and ∇ · ω = 0. Since ∇×∇f = 0 for any scalar function f, these lead to the vorticity equation in this simplified model. (b) The rate of change of vorticity as we follow the fluid is given by the term ω · ∇u. (c) u = i u (x, y) + j v (x, y) and ω = ω (x, y) k and hence, ω · ∇u = ω (x, y) ∂ ∂z [i u (x, y) + j v (x, y)] = 0. This gives the result. 20. We differentiate the first equation partially with respect to t to find Ett = c curl Ht. We then substitute Ht from the second equation to obtain Ett = −c 2 curl (curl E). Using the vector identity curl (curl E) = grad (div E)−∇2E with div E = 0 gives the desired equation. A similar procedure shows that H satisfies the same equation. 21. When Hooke’s law is used to the rod of variable cross section, the tension at point P is given by TP = λ A (x) ux, where λ is a constant. A longitudinal vibration would displace each cross sectional element of the rod along the x-axis of the rod. An element P Q of length δx and mass m = ρ A (x) δx will be displaced to P ′Q′ with length (δx + δu) with the same mass m. The acceleration of the element P ′Q′ is utt so that the difference of the tensions at P ′ and Q′ must be equal to the product m utt. Hence, m utt = TQ′ − TP ′ =  ∂ ∂t TP ′ δx = ∂ ∂x (λ A (x) ux) δx. This gives the equation. 4.6 Exercises 707 4.6 Exercises 1. (a) x < 0, hyperbolic; uξη = 1 4
4 ξ−η 4
54 − 1 2
4 1 ξ−η
5 (uξ − uη), x = 0, parabolic, the given equation is then in canonical form; x > 0, elliptic and the canonical form is uαα + uββ = 1 β uβ + β 4 16 . (b) y = 0, parabolic; y = 0, elliptic, and hence, uαα + uββ = uα + e α. (d) Parabolic everywhere and hence, uηη = 2ξ η2 uξ + 1 η2 e ξ/η . (f) Elliptic everywhere for finite values of x and y, then uαα + uββ = u − 1 α uα − 1 β uβ. (g) Parabolic everywhere uηη = 1 1−e 2(η−ξ) sin−1  e η−ξ − uξ ! . (h) B2 − 4AC = y − 4x. Equation is hyperbolic if y > 4x, parabolic if y = 4x and elliptic if y < 4x. (i) y = 0, parabolic; and y = 0, hyperbolic, uξη = (1+ξ−ln η) η uξ + uη + 1 η u. 2. (i) u (x, y) = f (y/x) + g (y/x) e −y , (ii) Hint: ϕ = ru and check the solution by substitution. u (r, t) = (1/r) f (r + ct) + (1/r) g (r − ct); (iii) A = 4, B = 12, C = 9. Hence, B2 − 4AC = 0. Parabolic at every point in (x, y)-plane. dy dx = 3 2 or y = 3 2 x+c ⇒ 2y−3x = c1, ξ = 2y−3x, η = y. The canonical form is uηη − u = 1 ⇒ u (ξ,η) = f (ξ) cosh η + g (ξ) sinh η−1. Or, u (x, y) = f (2y − 3x) cosh y+g (2y − 3x) sinh y−1. (iv) Hyperbolic at all points in the (x, y)-plane. ξ = y − 2x, η = y + x. Thus, uξη + uη = ξ, u (ξ,η) = η (ξ − 1) + f (ξ) e −ξ + g (ξ). 708 Answers and Hints to Selected Exercises u (x, y)=(x + y) (y − 2x − 1) + f (x + y) exp (2x − y) + g (y − 2x). (v) Hyperbolic. ξ = y, η = y−3x. ξuξη+uη = 0, u (ξ,η) = 1 ξ f (η)+g (ξ). u (x, y) = 1 y fu (y − 3x) + g (y). (vi) A = 1, B = 0, C = 1, B2 − 4AC = −4 < 0. So, this equation is elliptic. dy dx = + i or dx dy = + i or ξ = x + i y = c1 and η = x − i y = c2. The general solution is u = φ (ξ) + ψ (η) = φ (x + iy) + ψ (x − iy). (vii) u = φ (x + 2iy) + ψ (x − 2iy). (viii) B2 − 4AC = 0. Equation is parabolic. The general solution is given by (4.3.16), where λ =  B 2A = −1 and hence, the general solution becomes u = φ (y + x) + y ψ (y + x). (xi) B2 − 4AC = 25 > 0. Hyperbolic. The general solution is u = φ  y − 3 2 x + ψ  y − 1 6 x . 3. (a) ξ = (y − x) + i √ 2 x, η = (y − x) − i √ 2 x, α = y − x, β = √ 2 x, uαα + uββ = − 1 2 uα − 2 √ 2 uβ − 1 2 u + 1 2 exp  β/√ 2 . (b) ξ = y + x, η = y; uηη = − 3 2 u. (c) ξ = y − x, η = y − 4x; uξη = 7 9 (uξ + uη) − 1 9 sin [(ξ − η) /3]. (d) ξ = y + ix, η = y − ix. Thus, α = y, β = x. The given equation is already in canonical form. (e) ξ = x, η = x − (y/2); uξη = 18uξ + 17uη − 4. (f) ξ = y + (x/6), η = y; uξη = 6u − 6η 2 . (g) ξ = x, η = y; the given equation is already in canonical form. (h) ξ = x, η = y; the given equation is already in canonical form. (i) Hyperbolic in the (x, y) plane except the axes x = y = 0. ξ = xy, η = (y/x); y = √ ξη, x =  ξ/η; uξη = 1 2
4 1 + 1 2 %η ξ
5 uη − 1 4 √ 1 ξη − 1 4ξη − 1 2 . (j) Elliptic when y > 0; dy dx = + i √y, α = 2√y and β = −x; uαα + uββ = α 2uβ. Parabolic when y = 0; uxx + 1 2 uy = 0. 4.6 Exercises 709 Hyperbolic when y < 0; ξ = x − 2 √ −y, η = −x − 2 √ −y. The canonical form is uξη = 1 16 (ξ + η) 2 (uη − uξ). (k) Parabolic, dy dx = (xy) −1 . Integrating gives 1 2 y 2 = ln x+ln ξ, where ξ is an integrating constant. Hence, ξ = 1 x exp  1 2 y 2 , η = x. uxx = x −4 e y 2 uξξ − 2x −2 exp  1 2 y 2 uξη + uηη + 2x −2 exp  1 2 y 2 uξ, uxy = −yx−3 exp  y 2 uξξ + yx−1 exp  1 2 y 2 uξη − yx−2 exp  1 2 y 2 uξ, uyy =  y 2x −2 e y 2 uξξ +  y 2x −2 exp  1 2 y 2 uξ. uηη +  ξ/η2 uξ = 0. (l) Elliptic if y > 0, ξ = x + 2i √y, η = x − 2i √y, α = 1 2 (ξ + η) = x, β = 1 2i (ξ − η)=2√y; uαα + uββ = 1 β uβ. Hyperbolic y < 0, ξ = x + 2i √y, η = x − 2i √y, ξ − η = 4i √y; uξη + 1 2
4 uξ−uη ξ−η
5 = 0. The equations of the characteristic curves are dy dx = + i √y that gives 2√y = + i(x − c), or y = + 1 4 (x − c) 2 , where c is an integrating constant. Two branches of parabolas with positive or negative slopes. 4. (i) u (x, y) = f (x + cy)+g (x − cy); (ii) u (x, y) = f (x + iy)+f (x − iy); (iii) Use z = x + iy. Hence, u (x, y)=(x − iy) f1 (x + iy) + f2 (x + iy) + (x + iy) + f3 (x − iy) + f4 (x − iy) (iv) u (x, y) = f (y + x)+g (y + 2x); (v) u (x, y) = f (y)+g (y − x); (vi) u (x, y)=(−y/128) (y − x) (y − 9x) + f (y − 9x) + g (y − x). 5. (i) vξη = − (1/16) v, (ii) vξη = (84/625) v. 7. (ii) Use α = 3y 2 , β = −x 3/2. 8. x = r cos θ, y = r sin θ; r =  x 2 + y 2, θ = tan−1  y x . ∂u ∂x = ∂u ∂r · ∂r ∂x + ∂u ∂θ · ∂θ ∂x = ur · x r + uθ ·  − y r 2 . uxx = (ux)x = (ux) r · ∂r ∂x + (ux) θ ∂θ ∂x =  x r ur − y r 2 uθ r =  x r +  x r ur − y r 2 uθ − y r 2 =  x r urr − x r 2 ur + 1 r ur ∂x ∂r x r −
4 y r 2 urθ − 2y r 3 uθ + 1 r 2 ∂y ∂r uθ
5 x r +  x r urθ + 1 r ur · ∂x ∂θ − y r 2 +
4 y r 2 uθθ + 1 r 2 uθ · ∂y ∂θ
5 y r 2 710 Answers and Hints to Selected Exercises = x 2 r 2 urr − 2xy r 3 urθ + y 2 r 4 uθθ + y 2 r 3 ur + 2xy r 4 uθ. Similarly, uyy = y 2 r 2 urr +  2xy r 3 urθ + x 2 r 4 uθθ + x 2 r 3 ur − 2xy r 4 uθ. Adding gives the result: ∇2u = uxx + uyy = urr + 1 r ur + 1 r 2 uθθ = 0. 9. (c) Use Exercise 8. 10. (a) ux = uξξx + uηηx = a uξ + c uη =
4 a ∂ ∂ξ + c ∂ ∂η
5 u, uy = uξξy + uηηy = b uξ + d uη =
4 b ∂ ∂ξ + d ∂ ∂η
5 u. uxx = (ux)x =
4 a ∂ ∂ξ + c ∂ ∂η
5
4a ∂ ∂ξ + c ∂ ∂η
5 u =  a 2uξξ + 2ac uξη + c 2uηη . uyy = (uy) y =
4 b ∂ ∂ξ + d ∂ ∂η
5
4b ∂ ∂ξ + d ∂ ∂η
5 u = b 2uξξ + 2bd uξη + d 2uηη. uxy = (uy)x =
4 a ∂ ∂ξ + c ∂ ∂η
5
4b ∂ ∂ξ + d ∂ ∂η
5 u = ab uξξ + (ad + bc) uξη + cd uηη. Consequently, 0 = A uxx + 2B uxy + C uyy =  A a2 + 2Bab + C b2 uξξ+2 [ac A + (ad + bc) B + bd C] uξη +  A c2 + 2Bcd + C d2 uηη. Choose arbitrary constants a, b, c and d such that a = c = 1 and such that b and d are the two roots of the equation Cλ2 + 2Bλ + A = 0, and λ = −B + √ D C = b, d, D = B2 − AC. Thus, the transformed equation with a = c = 1 is given by [A + (b + d) B + bd C] uξη = 0. Or,  2 C AC − B2 uξη = 0. If B2 −AC > 0, the equation is hyperbolic, and the equation uξη = 0 is in the canonical form. The general solution of this canonical equation 4.6 Exercises 711 is u = φ (ξ) + ψ (η), where φ and ψ are arbitrary functions and the transformation becomes ξ = x + by and η = x + dy, where b, d are real and distinct. If B2 −AC < 0, the equation is elliptic, and b and d are complex conjugate numbers  d = b . With a = c = 1, the transformation is given by ξ = x+by and η = x+b y. Then α = 1 2 (ξ + η) and β = 1 2i (ξ − η) can be used to transform the equation into the canonical form uαα + uββ = 0. If B2 − AC = 0, the equation is parabolic, here b = − B C , a, c and d are arbitrary, but c and d are not both zero. Choose a = c = 1, d = 0 so that ξ = x − B C y and η = y are used to transform the equation into the form uηη = 0. The general solution is u = φ (ξ)+ηψ (η), where φ and ψ are arbitrary functions, and b is the double root of Cλ2 + 2Bλ+A = 0, and ξ = x + by. 11. Seek a trial solution u (x, y) = f (x + my) so that uxx = f ′′ , uyy = m2f ′′. Substituting into the Laplace equation yields  m2 + 1 f ′′ = 0 which gives that either f ′′ = 0 or m2 + 1 = 0. Thus, m = + i. The general solution is u (x, y) = F (x + iy) + G (x − iy). Identifying c with i gives the d’Alembert solution u (x, y) = 1 2 [f (x + iy) + f (x − iy)] + 1 2i
x+iy x−iy g (α) dα. 12. (a) Hyperbolic. (ξ,η) = 2 3  y 3/2 + x 3/2 , 3  ξ 2 − η 2 uξη = ηuη − ξuη. (b) Elliptic. dy dx = + isech2 x, ξ = y + itanh x, η = y − itanh x; α = y, β = tanh x. Thus, uαα + uββ = 2β (1−β2) uβ. (d) Hyperbolic. ξ = y + tanh x, η = y − tanh x. uξη = " 4 − (ξ − η) 2 #−1 (η − ξ) (uξ − uη) in the domain (η − ξ) 2 < 4. (e) Parabolic. ξ = y − 3x, η = y; uηη = − η 3 (uξ + uη). (f) Elliptic. α = 1 2  y 2 − x 2 , β = 1 2 x 2 . The canonical form is 712 Answers and Hints to Selected Exercises uαα + uββ = [2β (α + β)]−1 [αuα − (α + 2β) uβ]. (g) Elliptic. α = sin x + y, β = x, uαα + uββ = (sin β) uα − u. (h) Parabolic. ξ = x + cos y, η = y. Thus, uηη =  sin2 η cos η uξ. 13. The general solution is u (x, y) = e x
x 0 e −α cos (α + y eα−x ) dα + e x f (y e−x ) + g (x), where f and g are arbitrary functions. 5.12 Exercises 1. (a) u (x, t) = t, (b) u (x, t) = sin x cos ct + x 2 t + 1 3 c 2 t 3 , (c) u (x, t) = x 3 + 3c 2xt2 + xt, (d) u (x, t) = cos x cos ct + (t/e), (e) u (x, t)=2t + 1 2 log  1 + x 2 + 2cxt + c 2 t 2 + log  1 + x 2 − 2cxt + c 2 t 2 !, (f) u (x, t) = x + (1/c) sin x sin ct. 2. (a) u (x, t)=3t + 1 2 xt2 . (c) u (x, t)=5+ x 2 t + 1 3 c 2 t 3 +  1/2c 2 (e x+ct + e x−ct − 2e x ), (e) u (x, t) = sin x cos ct + (e t − 1) (xt + x) − xtet , (f) u (x, t) = x 2 + t 2  1 + c 2 + (1/c) cos x sin ct. 3. s (r, t) = ⎧ ⎪⎪⎪⎪⎨ ⎪⎪⎪⎪⎩ 0, 0 ≤ t ct. 30. (a) When ω = ck, u (x, t) = 1 (k2c 2−ω2) sin (kx − ωt) − (ω−kc) 2kc(ω2−c 2k2) sin [k (x + ct)] + (ω+kc) 2kc(ω2−c 2k2) sin [k (x − ct)] . This solution represents three harmonic waves which propagate with different amplitudes and with speeds + c and the phase velocity (ω/k). (b) When ω = ck, u (x, t) = 1 4 sin (x − t) − 1 4 sin (x + t) + 1 2 t cos (x − t). This solution represents two harmonic waves with constant amplitude and another harmonic wave whose amplitude grows linearly with time. 31. (a) u (x, t) = 1 2 [cos (x − 3t) + cos (x + 3t)] + 1 6
x+3t x−3t sin (2α) dα = cos x cos (3t) + 1 6 sin (2x) sin (6t). (c) u (x, t) = cos (3x) cos (21t) + tx. (e) u (x, t) = x 3 + 27xt2 + 1 6 [cos (x + 3t) − cos (x − 3t)] + 1 6 [(x + 3t) sin (x + 3t) − (x − 3t) sin (x − 3t)]. 714 Answers and Hints to Selected Exercises (f) u (x, t) = 1 2 [cos (x − 4t) + cos (x + 4t)] + 1 8 e −x (x + 1 − 4t) e 4t − (x +1+4t) e −4t ! . 32. Verify that u (x, t) =
t 0 v (x, t; τ ) dτ satisfies the Cauchy problem. ut (x, t) = v (x, t;t) +
t 0 vt (x, t; τ ) dτ =
t 0 vt (x, t; τ ) dτ utt (x, t) = vt (x, t;t)+
t 0 vtt (x, t; τ ) dτ = p (x, t)+
t 0 vtt (x, t; τ ) dτ uxx (x, t) =
t 0 vxx (x, t; τ ) dτ. Thus, utt − c 2uxx = p (x, t) +
t 0  vtt − c 2vxx dτ = p (x, t). 33. ut = v (x, t;t) +
t 0 vt (x, t; τ ) dτ = p (x, t) +
t 0 vt (x, t; τ ) dτ uxx =
t 0 vxx (x, t; τ ) dτ . Hence, ut − κ uxx = p (x, t) +
t 0 (vt − κ vxx) dτ = p (x, t). 34. According to the Duhamel principle u (x, t) =
t 0 v (x, t; τ ) dτ is the solution of the problem where v (x, t; τ ) satisfies vt = κ vxx, 0 ≤ x < 1, t> 0, v (0, t; τ )=0= v (1, t; τ ), v (x, τ ; τ ) = e −τ sin πx, 0 ≤ x ≤ 1. Using the separation of variables, the solution is given by v (x, t) = X (x) T (t) so that X′′ + λ 2X = 0 and T ′ + κλ2T = 0. The solution is v (x, t; τ ) = ∞ n=1 an (τ ) e −λ 2 nκt sin λnx, when λn = nπ, n = 1, 2, 3,.... 6.14 Exercises 715 Since v (x, τ ; τ ) = e −τ sin πx, e −τ sin πx = ∞ n=1 an (τ ) exp  −n 2π 2κτ sin (πnx). Equating the coefficients gives e −τ = a1 (τ ) exp  −π 2κτ , an (τ )=0, n = 2, 3,.... Consequently, v (x, t; τ ) = exp π 2κ − 1 τ ! exp  −π 2κt sin πx. Thus, u (x, t) = exp  −π 2κt sin πx
t 0 exp π 2κ − 1 τ ! dτ = e −t−exp(−π 2κt) (π2κ−1) · sin πx. 36. (a) The solution is u (x, t) = 1 n e nx sin  2n 2 t + nx + e −nx sin  2n 2 t − nx !, and u (x, t) → ∞ as n → ∞ for certain values of x and t. (b) un (x, y) = 1 n exp (− √ n ) sin nx sinh ny is the solution. For y = 0, un (x, y) → ∞ as n → ∞. But (un)y (x, 0) = exp (− √ n ) sin nx → 0 as n → ∞. 6.14 Exercises 1. (a) f (x) = − π 4 + h 2 + ∞ k=1 ( 1 πk2 " 1+(−1)k+1# cos kx + 1 πk " h + (h + π) (−1)k+1# sin kx) . (c) f (x) = sin x + ∞ k=1 2(−1)k+1 k sin kx. (e) f (x) = sinh π π 1 1 +∞ k=1 2(−1)k 1+k2 (cos kx − k sin kx) 3 . 2. (a) f (x) = ∞ k=1 2 k sin kx (b) f (x) = ∞ k=1  2 πk " 1 − 2 (−1)k + cos kπ 2 # sin kx. 716 Answers and Hints to Selected Exercises (c) f (x) = ∞ k=1 " 2 (−1)k+1 π k + 4 πk3
4 (−1)k − 1
5# sin kx. (d) f (x) = ∞ k=2 2k π " 1+(−1)k k2−1 # sin kx. 3. (a) f (x) = 3 2 π + ∞ k=1 2 πk2 " (−1)k − 1 # cos kx. (b) f (x) = π 2 + ∞ k=1 2 πk2 " (−1)k − 1 # cos kx. (c) f (x) = π 2 3 + ∞ k=1 4(−1)k k2 cos kx. (d) f (x) = 2 3π + ∞ k=1,2,4,... 6 π " 1+(−1)k 9−k2 # cos kx, k = 3. 4. (b) f (x) = ∞ k=1  2 kπ sin kπ 2 cos  kπx 6 . (c) f (x) = 2 π + ∞ k=2  2 kπ " 1+(−1)k 1−k2 # cos  kπx l . (f) f (x) = ∞ k=1 kπ 1+k2π2 (−1)k+1  e − e −1 sin (kπx). 5. (a) f (x) = ∞ k=−∞ 1 π
4 2+ik 4+k2
5 (−1)k sinh 2π eikx . (b) f (x) = ∞ k=−∞ (−1)k π(1+k2) sinh π eikx . (d) f (x) = ∞ k=−∞ (−1)k  i kπ e ikπx . 6. (a) f (x) = π 8 + ∞ k=1 " 1 2πk2 ( (−1)k − 1 ) cos kx + (−1)k+1 2k sin kx# . 7. (a) f (x) = l 2 3 + ∞ k=1 4 (−1)k  1 kπ 2 cos  kπx l . 8. (a) sin2 x = ∞ k=1,3,4,... 4(1−cos kπ) kπ(4−k2) sin kx. (b) cos2 x = ∞ k=1,3,4,... 2 kπ
4 1−k 2 4−k2
5 (1 − cos kπ) sin kx. (d) sin x cos x = ∞ k=1,3,4,... 2 π
4 1−cos kπ 4−k2
5 cos kx. 6.14 Exercises 717 9. (a) x 2 4 = π 2 12 − ∞ k=1 (−1)k+1 k2 cos kx. (c)
∞ 0 ln  2 cos x 2 dx = ∞ k=1 (−1)k+1 sin kx k2 . (e) π 2 − 4 π ∞ k=1 cos(2k−1)x (2k−1)2 = ⎧ ⎪⎨ ⎪⎩ −x, −π<x< 0="" x,="" <="" x="" π.="" 10.="" (a)="" f="" (x,="" y)="16" π2="" ∞="" m="1,3,..." n="1,3,..." ="" 1="" mn="" sin="" mx="" ny="" (c)="" 4="" 9="" +="" 2="" 8="" 3="" π="" (−1)m="" m2="" cos="" n2="" 16(−1)m+n="" m2n2="" ny.="" (e)="" 2(−1)m+1="" y.="" (g)="" dmn="" mπx="" nπy="" ,="" where="" 1.2=""
="" xy="" (mπx)=""
4nπy=""
5="" dx="" dy="2" m2π="" −="" mπ="" y="" ="" −4="" (−1)n="" nπ="" ="8" (−1)m+n="" 2mn="" .="" (h)="" 16="" [(2m="" 1)="" (2n="" 1)]−1="" "="" (2m−1)πx="" a="" #="" ×="" (2n−1)πy="" b="" (double="" fourier="" sine="" series).="" (i)="" (−1)m+1="" πmx="" 718="" answers="" and="" hints="" to="" selected="" exercises="" (−1)m+n+1="" m(4−π2n2)="" πny="" (j)="" 2∞="" mx+="" 8(−1)m+n+1="" (k)=""
4="" 4π="" 1∞="" ny3="" +16∞="" 20.="" b2n="0," b2n+1="8" π(2n+1)3="" (x)="x" (π="" x)="8" 3x="" 5x="" 5="" ...="" (b)="" put="" find="" the="" sum="" of="" series.="" 21.="" bn="2" nx="" n2π2="" 1,="" 2,....="" π2(2n+1)2="" 22.="" nπx="" (1="" nπ)="2" [1="" ]="⎧" ⎪⎨="" ⎪⎩="" for="" odd="" n,="" 0,="" even="" n.="" ∼="" πx="" 3πx="" 5πx="" ...!="" a0="" an="" (sin="" 0)="0," ·="" 2nπ="" (−1)n+1="" 2a="" ((−1)n="" 4a="" 6.14="" 719="" 23.="" −a="" π2n2="" !a="" (cos="" (−nπ))="0" ="" (−nπ)="(−1)n+1" 2π="" all="" [−1+(−1)n="" is="" odd.="" k="0" sin(2k+1)x="" (2k+1)="" 24.="" 2nx="" (4n2−1)="" 26.="" hint:="" argument="" similar="" that="" used="" in="" section="" 6.5="" can="" be="" employed="" prove="" this="" general="" parseval="" relation.="" more="" precisely,="" use="" (6.5.10)="" (f="" g)="" obtain="" −π="" (a0="" α0)="" (ak="" αk)="" (bk="" βk)="" subtracting="" later="" equality="" from="" former="" gives="" g="" (akαk="" bkβk).="" 27.="" ∞="" α="" αx="" aα="" dα,="" πα="" (1−α2)="" dα.="" 720="" 33.="" ck="1" e−ikxdx="1" 1="" e−ikx="" −ik="" ik="" e="" −ikxdx3="1" e−ikπ="" eikπ="" (−ik)="" −ikπ="" 1="" ikπ="" 3="i" kπ="" i="" c0="1" 35.="" cke="" ikx="" ibk)="1" −ikxdx,="1" i(a−k)x="" −i(a+k)x="" dx,="1" 4πi(a="" k)="" #π="" (a="" real="" quantity="" hence,="" bk="0" 2,="" 3,="" ...,="" ak="2" (−1)k="" (πa)="" 2)="" thus,="" (ax)="2a" aπ="" 2x="" ...="" since="" ax="" even,="" above="" series="" continuous,="" at="" nπ.="" putting="" treating="" variable="" cot="" or,="" (n2="" convergence="" uniform="" any="" interval="" x-axis="" does="" not="" contain="" integers,="" term-by-term="" integration="" <a<x<="" 721="" πt="" πt="" dt="ln" ln="" πa="" limit="" as="" →="" we="" or="" 3∞="" product="" representation="" πx.="" (d)="" wallis="" formula="" infinite="" (2n)="" 6="" 7="" ····="" 36.="" (x="" 2π)="f" (x).="" kx="" dx.="" evaluating="" these="" integrals="" (k="" kx)="" case,="" (−x)="−f" odd,="" periodic="" period="" 2π.="" given="" by="" 6x="" 10x="" with="" 1.="" 722="" (2πkx)="" corresponding="" represent="" value="" ....="" [f="" (n+)="" (n−)]="1" 37.="" represents="" square="" wave="" function.="" l="" −l="" πkx="" πk="" πk)="⎧" ⎨="" ⎩="" 2l="" sin=""
4πx="" ∞="" (2k="" !="" nl,="" terms="" after="" first="" vanish="" (l="" 2).="" graphs="" partial="" sums="" sn="" (=""
5)="" drawn.="" show="" how="" converges="" points="" continuity="" (x),="" ∞.="" however,="" discontinuity="" l,="" converge="" mean="" value.="" just="" beyond="" discontinuities="" sums,="" overshoot="" |l|.="" behavior="" known="" gibbs="" phenomenon.="" 723="" (πk)="" 1),="" 4l="" cos="" 1)2="" )="" [ak="" (πkx)="" (πkx)]="" 2k="" π3k3="" 38.="" have="" summing="" n="" n="" dividing="" result="" adding="" both="" sides="" result.="" kx),="" (t)="" kt="" (t="" dt.="" 724="" &n="" '="" which="" is,="" t="" =="" π−x="" −π−x="" ξ)="" ξ="" dξ,="" integrand="" has="" 2π,="" replace="" interval−π−x,="" length="" (−π,="" π).="" 7.9="" u="" t)="∞" (nπ)="" (nπct)="" (nπx).="" ctsin="" x.="" 2.="" 32[(−1)n−1]="" πcn2(n2−4)="" (nct)="" (nx).="" 5.="" antn="" tn="" ⎪⎪⎪⎪⎨="" ⎪⎪⎪⎪⎩="" −at="" cosh="" αt="" 2α="" sinh=""> 0 e −at/2  1 + at 2 , for α = 0 e −at/2
4 cos βt + a 2β sin βt
5 , for α 2 < 0, in which α = 1 2 " a 2 − 4
4 b + n 2π 2 c 2 l 2
5# 1 2 , β = 1 2 " 4
4 b + n 2π 2 c 2 l 2
5 − a 2 # 1 2 . 7.9 Exercises 725 6. u (x, t) = ∞ n=1 anTn (t) sin  nπx l , an = 2 l
l 0 g (x) sin nπx l dx, and Tn (t) = ⎧ ⎪⎪⎪⎪⎪⎨ ⎪⎪⎪⎪⎪⎩ 2e −at/2 √ (a2−α) sinh √ (a2−α) 2 t  , for a 2 > α, t e−at/2 , for a 2 = α, 2e −at/2 √ (α−a2) sin √ (α−a2) 2 t  , for a 2 < α. 7. θ (x, t) = ∞ n=1 an cos (aαnt) sin (αnx + φn), where an = 2(α 2 n+h 2 ) 2h+(α2 n+h2)l
l 0 f (x) sin (αnx + φn) dx and φn = tan−1  αn h ; αn are the roots of the equation tan αl =
4 2hα α2−h2
5 . 11. u (x, t) = v (x, t) + U (x), where v (x, t) = ∞ n=1 1 −  2 l
l 0 U (τ ) sin  nπτ l dτ3 cos  nπct l sin  nπx l and U (x) = − A c 2 sinh x + A c 2 sinh (l + k − h) x l + h. 12. u (x, t) = A 6c 2 x 2 (1 − x) + ∞ n=1 12 (nπ) 3 (−1)n cos (nπct) sin (nπx). 14. (a) u (x, t) = − hx2 2k +  2u0 + h 2k x − 4h kπ e −kπ2 t sin (πx) + ∞ n=2 ane −kn2π 2 t sin (nπx), where an = 2u0 nπ [1 + (−1)n ] + 2u0n (n2−1)π [1 + (−1)n ] + 2h kπ3n3 [(−1)n − 1]. (b) Hint: v (x, t) = e −htu (x, t). u (x, t) = e −ht 1 1 2 a0 + ∞ n=1 an cos  nπx l exp  −n 2π 2kt/l2 3 , where an = 2 l
l 0 f (ξ) cos
4 nπξ l
5 dξ. 15. (a) u (x, t) = ∞ n=1 4 n3π3 " 2 (−1)n+1 − 1 # e −4n 2π 2 t sin (nπx). (b) u (x, t) = ∞ n=1,3,4,... [(−1)n − 1] " n π(4−n2) − 1 nπ # e −n 2kt sin (nx). 726 Answers and Hints to Selected Exercises 16. u (x, t) = ∞ n=1 2l 2 n3π3 [1 − (−1)n ] e −(nπ/l) 2 t cos  nπx l . 18. v (x, t) = Ct  1 − x l − Cl2 6k " x l 3 − 3  x l 2 + 2  x l # +
4 2Cl2 π3k
5∞ k=1 e −n2π2kt/l2 n3 sin  nπx l . 21. u (x, t) = v (x, t) + w (x), where v (x, t) = e −kt sin x + ∞ n=1 an e −n 2kt sin (nx), and an = −n (n2+a2) [(−1) e −n−ax − 1] + 2A a2kπ 1 n {(−1)n − 1} ! + (−1)n n [e −aπ − 1] w (x) = A a2k 1 − e −ax + x π (e −aπ − 1)! . 36. Hint: Suppose R is the rectangle 0 ≤ x ≤ a, 0 ≤ y ≤ b and ∂R is its boundary positively oriented. Suppose that u1 and u2 are solutions of the problem, and put v = (u1 − u2). Then v satisfies the Laplace equation with v = (x, 0) = 0 = v (x, b), vx (0, y)=0= vx (a, y). 8.14 Exercises 1. (a) λn = n 2 , φn (x) = sin nx for n = 1, 2, 3,... (b) λn = ((2n − 1) /2)2 , φn (x) = sin ((2n − 1) /2) πx for n = 1, 2, 3,... (c) λn = n 2 , φn (x) = cos nx for n = 1, 2, 3,.... 2. (a) λn = 0, n 2π 2 , φn (x) = 1, sin nπx, cos nπx for n = 1, 2, 3,... (b) λn = 0, n 2 , φn (x) = 1, sin nx, cos nx for n = 1, 2, 3,... (c) λn = 0, 4n 2 , φn (x) = 1, sin 2nx, cos 2nx for n = 1, 2, 3,.... 3. (a) λn = −  3/4 + n 2π 2 , φn (x) = e −x/2 sin nπx, n = 1, 2, 3,.... 4. (a) λn =1+ n 2π 2 , φn (x) = (1/x) sin (nπ ln x), n = 1, 2, 3,.... (b) λn = 1 4+(nπ/ ln 3)2 , φn (x) = " 1/ (x + 2) 1 2 # sin [(nπ/ ln 3) ln (x + 2)], n = 1, 2, 3,.... 9.10 Exercises 727 (c) λn = 1 12 " 1 + (2nπ/ ln 2)2 # , φn (x) = " 1/ (1 + x) 1 2 # sin [(nπ/ ln 2) ln (1 + x)], n = 1, 2, 3,.... 5. (a) φ (x) = sin
4√ λ ln x
5 , λ > 0. (b) φ (x) = sin
4√ λ x
5 , λ > 0. 7. f (x) ∼ ∞ n=1 2 π " (−1)n−1 n2 # cos nx. 11. (a) G (x, ξ) = ⎧ ⎪⎪⎪⎪⎨ ⎪⎪⎪⎪⎩ x, x ≤ ξ ξ, x > ξ. 12. (a) u (x) = − cos x +  cos 1−1 sin 1 sin x + 1, (b) u (x) = − 2 5 cos 2x − 1 10  1+2 sin 2 cos 2 sin 2x + 1 5 e x . 16. G (x, ξ) = ⎧ ⎪⎪⎪⎪⎨ ⎪⎪⎪⎪⎩  x 3 ξ/2 +  xξ3/2 − (9xξ/5) + x, for 0 ≤ x<ξ  x 3 ξ/2 +  xξ3/2 − (9xξ/5) + ξ, for ξ ≤ x ≤ 1. 24. (a) Hint: Differentiate cot θ = py′ y with respect to x to find −cosec2 θ dθ dx = 1 y (py′ ) ′ − 1 y 2  py′2 = −  λr + q + 1 p cot2 θ  . dθ dx = (q + λr) sin2 θ + 1 p cos2 θ, dr dx = r 2  1 p − q − λr sin 2θ. (b) At θ = nπ, dθ dx = 1 p , and at θ =  n + 1 2 π, dθ dx = (q + λp). 9.10 Exercises 8. (a) u (r, θ) = 4 3  1 r − r 4 sin θ. (c) u (r, θ) = ∞ n=1 an sinh (nπ/ ln 3)  θ − π 2 ! sin [(nπ/ ln 3) ln r], where 728 Answers and Hints to Selected Exercises an = 2 ln 3 sinh(nπ2/2 ln 3) ( nπ ln 3 n2π2+4(ln 3)2 [9 (−1)n − 1] − 4nπ ln 3 n2π2+(ln 3)2 [3 (−1)n − 1] + 3 ln 3 nπ [(−1)n − 1]) . 9. u (r, θ) = ∞ n=1 an  r −nπ/α − b −2nπ/α r nπ/α sin  nπθ α + ∞ n=1 bn sinh " nπ ln(b/a) (θ − α) # sin " nπ ln(b/a) (ln r − ln a) # , where an = 2 α  a −nπ/α − b −2nπ/αa nπ/α !−1
α 0 f (θ) sin  nπθ α dθ, bn = −2 ln  b a sinh {αnπ/ ln (b/a)} !−1 +
b a f (r) sin [(nπ/ ln [b/a]) ln (ra)] dr r . 12. u (r, θ) = ∞ n=1 2 αJν (a)
α 0 f (θ) sin  nπτ α dτ Jν (r) sin  nπθ α , ν = nπ/α. 13. u (r, θ) = 1 2  a 2 − r 2 . 14. (a) u (r, θ) = − 1 3  r + 4 r sin θ + constant. 16. u (r, θ) = ∞ n=1 1 nRn−1 r n sin nθ. 18. u (r, θ) = a0 2 + ∞ n=1 r n (an cos nθ + bn sin nθ), where an = R 1−n (n+Rh)π
2π 0 f (θ) cos nθ dθ, n = 0, 1, 2,.... bn = R 1−n (n+Rh)π
2π 0 f (θ) sin nθ dθ, n = 1, 2, 3,.... 20. u (r, θ) = c − r 4 12 sin 2θ + 1 6
4 r 6 1−r 6 2 r 4 1−r 4 2
5 r 2 sin 2θ + 1 6
4 r 2 1−r 2 2 r 4 1−r 4 2
5 r 4 1 r 4 2 r −2 sin 2θ. 21. u (r, θ) = ∞ n=1 an  r −nπ/α − b −2nπ/αr nπ/α sin  nπθ α + ∞ n=1 bn sinh " nπ ln(b/a) (θ − α) # sin " nπ ln(b/a) (ln r − ln a) # . 9.10 Exercises 729 22. (a) u (x, y) = ∞ n=1 4[1−(−1)n] (nπ) 3 sinh nπ sin nπx sinh {nπ (y − 1)} (c) u (x, y) = ∞ n=1 an (sinh nπx − tanh nπ cosh nπx) sin nπy, where an = 1 tanh nπ " 2nπ3 n2π4−4 + 1−(−1)n nπ # . 23. (a) u (x, y) = c + ∞ n=1 an (cosh nx − tanh nπ sinh nx) cos ny, where an = 2 [1 − (−1)n ] =n 3π tanh nπ . (c) u (x, y) = − 1 tanh π [cosh y − tanh π sinh y] cos x + C. 25. u (x, y) = xy (1 − x) + ∞ n=1 4(−1)n (nπ) 3 sinh nπ sin (nπx) sinh (nπy). 27. u (x, y) = c +  x 2/2
4 x 2 3 − y 2
5 + ∞ n=1 8a 4 (−1)n+1 (nπ) 3 sinh nπ cosh  nπx a cos  nπy a . 29. u (x, y) = x [(x/2) − π] + ∞ n=1 an sin & 2n−1 2 x ' cosh & 2n−1 2 y ' , where an = 2 Aπ
π 0 " f (x) − h
4 x 2 2 − πx
5# sin  2n−1 2 x ! dx with A =  2n−1 2 sinh  2n−1 2 π + h cosh  2n−1 2 π. 32. Hint: The solution is given by (9.5.3) and the boundary conditions require sin2 θ = 1 2 a0 + ∞ n=1 [(an + bn) cos nθ + (cn + dn) sin nθ] , 0 = 1 4 b0 + ∞ n=1 n an 2 n−1 − bn 2 −n−1 cos nθ +  cn 2 n−1 − dn 2 −n−1 sin nθ! . Using sin2 θ = 1 2 (1 − cos 2θ), we equate coefficients to obtain a0 = 1, b0 = 0; a2 + b2 = − 1 2 , 2a2 − 1 8 b2 = 0; an + bn = 0 2 n−1 an − 2 −n−1 bn = 0 ⎫ ⎪⎬ ⎪⎭ n = 1, 3, 4, 5,.... 730 Answers and Hints to Selected Exercises cn + dn = 0 2 n−1 cn − 2 −n−1 dn = 0 ⎫ ⎪⎬ ⎪⎭ n = 1, 2, 3,.... Thus, a0 = 1, b0 = 1, a2 = − 1 34 , b2 = − 8 17 , and the remaining coeffi- cients are zero; finally u (r, θ) = 1 2 − 1 34  r 2 + 16 r 2 cos 2θ. 33. (a) Hint: Seek a separable solution u (r, z) = R (r)Z (z) so that r 2R′′ + r R′ − λr2R = 0, and Z ′′ + λZ = 0, with Z (0) = 0 = Z (h). The solution of this eigenvalue problem is λn =  nπ h 2 , Zn (z) = sin  nπz h , n = 1, 2, 3,.... The solution of the radial equation is Rn (r) = anI0  nπr h + bnK0  nπr h , where I0 and K0 are modified Bessel functions. Since K0 is unbounded at r = 0, all bn ≡ 0. Thus, u (r, z) = ∞ n=1 an I0  nπr h sin  nπz h . f (z) = u (1, z) = ∞ n=1 an I0  nπ h sin  nπz h . This is a Fourier sine series for f (z) and hence, anI0  nπ h = 2 h
h 0 f (z) sin  nπz h dz. (d) u (r, z) = a I0  3πr h sin  3πz h =I0  3π h . 35. (a) u (r, z)=8∞ n=1 sinh z kn z kn sinh kn J0(knr) J0(kn) . (b) u (r, z) =  4a π ∞ n=1 1 (2n−1) I0[ 1 2 (2n−1)r] I0[ 1 2 (2n−1)] sin &1 2 (2n − 1) z ' . (c) u (r, z) = ∞ n=1 anI0  nπr h sin  nπz h , where anI0  nπa h = 2 h
h 0 f (z) sin  nπz h dz. 10.13 Exercises 731 10.13 Exercises 1. u (x, y, z) = sinh[(π/b) 2+(π/c) 2 ] 1 2 (a−x) sinh[(π/b) 2+(π/c) 2 ] 1 2 a sin  πy b sin  πz c . 2. u (x, y, z) = sinh( √ 2 πz) √ 2 π − cosh( √ 2 πz) √ 2 π tanh √ 2 π cos πx cos πy. 4. (a) u (r, θ, z) = ∞ m=0 ∞ n=1 (amn cos mθ + bmn sin mθ) Jmn (amnr/a) × sinh αmn(l−z)/a sinh αmnl/a , where amn = 2 a 2πεn [Jm+1 (αmn)]2
2π 0
a 0 f (r, θ) Jm (αmnr/a) cos mθ r dr dθ bmn = 2 a 2π [Jm+1 (αmn)]2
2π 0
a 0 f (r, θ) Jm (αmnr/a) sin mθ r dr dθ with εn = ⎧ ⎪⎨ ⎪⎩ 1, for m = 0 2, for m = 0 and αmn is the nth root of the equation Jm (αmn) = 0. 5. u (r, θ) = 1 3 +  2/3a 2 r 2P2 (cos θ). 7. u (r, z) = ∞ n=1 an sinh αn(l−z)/a cosh αnl/a J0 (αnr/a), where an = 2qu kα2 nJ0(αn) and αn is the root of J0 (αn) = 0 and k is the coefficient of heat conduction. 8. u (r, z) = 4u0 π ∞ n=1 [I0(2n+1)(πr/l)] [I0(2n+1)(πa/l)] sin(2n+1)πz/l (2n+1) . 9. u (r, θ) = u2 +  u1−u2 2 ∞ n=1
4 2n+1 n+1
5 Pn−1 (0)  r a n Pn (cos θ). 11. u (r, θ, φ) = C + ∞ n=1 ∞ m=0 r nP m n (cos θ) [anm cos mφ + bnm sin nφ], where 732 Answers and Hints to Selected Exercises anm = (2n + 1) (n − m)! 2nπ (n + m)!
2π 0
π 0 f (θ, ϕ) P m n (cos θ) cos mϕ sin θ dθ dϕ bnm = (2n + 1) (n − m)! 2nπ (n + m)!
2π 0
π 0 f (θ, ϕ) P m n (cos θ) sin mϕ sin θ dθ dϕ an0 = (2n + 1) 4nπ
2π 0
π 0 f (θ, ϕ) Pn (cos θ) sin θ dθ dϕ. 12. u (x, y, t) = ∞ n=1,3,4,...  − 4 π [1−(−1)n] n(n2−4) cos
4 (n2 + 1)πct
5 (sin nπx sin nπy). 13. u (r, θ, t) = ∞ n=0 ∞ m=1 Jn (αmnr/a) cos (αmnct/a) [amn cos nθ + bmn sin nθ] + ∞ n=0 ∞ m=1 Jn (αmnr/a) sin (αmnct/a) [cmn cos nθ + dmn sin nθ], where amn = 2 πa2εn [J ′ n (αmn)]2
2π 0
a 0 f (r, θ) Jn (αmnr/a) cos nθ r dr dθ bmn = 2 πa2 [J ′ n (αmn)]2
2π 0
a 0 f (r, θ) Jn (αmnr/a) sin nθ r dr dθ cmn = 2 πacαmnεn [J ′ n (αmn)]2
2π 0
a 0 g (r, θ) Jn (αmnr/a) cos nθ r dr dθ dmn = 2 πacαmn [J ′ n (αmn)]2
2π 0
a 0 g (r, θ) Jn (αmnr/a) sin nθ r dr dθ in which αmn is the root of the equation Jn (αmn) = 0 and εn = ⎧ ⎪⎨ ⎪⎩ 2 n = 0 1 n = 0 . 10.13 Exercises 733 15. u (r, θ, t) = ∞ n=0 ∞ m=1 Jn (αmnr) exp (−αmnkt) [amn cos nθ + bmn sin nθ], where anm = 2 πεn [J ′ n (αmn)]2
2π 0
1 0 f (r, θ) Jn (αmnr) cos nθ r dr dθ, bnm = 2 π [J ′ n (αmn)]2
2π 0
1 0 f (r, θ) Jn (αmnr) sin nθ r dr dθ, where αmn is the root of the equation Jn (αmn) = 0 and εn = ⎧ ⎪⎨ ⎪⎩ 1 for n = 0 2 for n = 0. 16. u (x, y, z, t) = sin πx sin πy sin πz cos √ 3 πct . 18. u (r, θ, z, t) = ∞ n=0 ∞ m=1 ∞ l=1 Jn (αmnr/a) sin (mπz/l) cos (ωct) × [anml cos nθ + bnml sin nθ] + ∞ n=0 ∞ m=1 ∞ l=1 Jn (αmnr/a) sin (mπz/l) sin (ωct) × [cnml cos nθ + dnml sin nθ], where anml = 4 πa2lεn [J ′ n (αmn)]2
a 0
2π 0
l 0 f (r, θ, z) Jn (αmnr/a) × sin (mπz/l) cos nθ r dr dθ dz, bnml = 4 πa2l [J ′ n (αmn)]2
a 0
2π 0
l 0 f (r, θ, z) Jn (αmnr/a) × sin (mπz/l) sin nθ r dr dθ dz, cnml = 4 ω −1 πa2lεn [J ′ n (αmn)]2
a 0
2π 0
l 0 g (r, θ, z) Jn (αmnr/a) × sin (mπz/l) cos nθ r dr dθ dz, 734 Answers and Hints to Selected Exercises dnml = 4 ω −1 πa2l [J ′ n (αmn)]2
a 0
2π 0
l 0 g (r, θ, z) Jn (αmnr/a) × sin (mπz/l) sin nθ r dr dθ dz, where αmn is the root of the equation Jn (αmn) = 0 and ω = " (mπ/l) 2 + (αmn/a) 2 # 1 2 , εn = ⎧ ⎪⎨ ⎪⎩ 1; for n = 0 2; for n = 0. 20. u (r, θ, z, t) = ∞ n=0 ∞ m=1 ∞ p=1 (anmp cos nθ + bnmp sin nθ) ×Jn (αmnr/a) sin (pπz/l) e −ωt , where anmp = 4 πa2lεn [J ′ n (αmn)]2
a 0
2π 0
l 0 f (r, θ, z) Jn (αmnr/a) × sin (pπz/l) cos nθ r dr dθ dz bnmp = 4 πa2l [J ′ n (αmn)]2
a 0
2π 0
l 0 f (r, θ, z) Jn (αmnr/a) × sin (pπz/l) sin nθ r dr dθ dz, in which εn = ⎧ ⎪⎨ ⎪⎩ 1 for n = 0 2 for n = 0 and ω = " (pπ/l) 2 + (αmn/a) 2 # . 23. u (x, y, t) = ∞ m=1 ∞ n=1 umn (t) sin mx sin ny, where 10.13 Exercises 735 umn (t) = 4 (−1)m+n+1 mn αmnc sin (αmnct)  cos (1 − αmnc)t − 1 2 (1 − αmnc) + cos (1 + αmnc)t − 1 2 (1 + αmnc) 0 + cos (αmnct)  sin (1 − αmnc)t 2 (1 − αmnc) + sin (1 + αmnc)t 2 (1 + αmnc) 0 , and αmn =  m2 + n 2 1 2 . 25. u (x, y, t) = ∞ n=1 ∞ m=1 4A mnπ2 [(−1)n−1][(−1)n−1] k(n2+m2) " 1 − e −k(n 2+m2 )t # × sin nx (sin my − m cos my). 27. u (x, y, t) = x (x − π)  1 − y π sin t + ∞ n=1 ∞ m=1 vmn (t) sin nx sin my, where α 2 mn =  m2 + n 2 and vmn (t) = 8 exp  −c 2α 2 tα2 mn [1 − (−1)n ] π 2mn (1 + c 4α4 mn) c 2 n2  α 2 mn − n 2 × & cost exp  −c 2α 2 tα2 mn − 1 ' +  1 n2 + c 4α 2 mn sin t exp  −c 2α 2 tα2 mn . 30. u (x, y, t) =
4 4qb4 π5D
5 ∞ n=1,3,... 1 n5 " 1 − vn(x) 1+cosh(nπa/b) # sin (nπy/b), where vn (x) = 2 cosh
4nπa 2b
5 cosh
4nπx b
5 +
4nπa 2b
5 sinh
4nπa 2b
5 cosh
4nπx b
5 −
4nπx b
5 sinh
4nπx b
5 cosh
4nπa 2b
5 . 32. Hint: In region 1, x ≤ −a, the solution of the Schr¨odinger equation d 2ψ dx2 = κ 2ψ, κ2 = 2M
2 (V0 − E), is ψ1 (x) = A eκx + B e−κx , where A and B are constants. For boundedness of the solution as 736 Answers and Hints to Selected Exercises x → −∞, B ≡ 0, and hence, ψ1 (x) = A eκx . In region 2, x ≥ a, the solution of the Schr¨odinger equation is ψ2 (x) = C eκx + D e−κx . For boundedness as x → ∞, C ≡ 0. The solution is ψ2 (x) = D e−κx . In region 3, −a ≤ x ≤ a, the potential is zero and hence, the equation takes the simple form ψxx + k 2ψ = 0, where k 2 =  2M
2 E. The solution is ψ3 (x) = E sin kx + F cos kx. For matching conditions at x = a, ψ2 (a) = ψ3 (a), or, De−aκ = E sin ka+ F cos ak. (1) Similarly, matching conditions at x = −a gives ψ1 (−a) = ψ3 (−a), or A e−aκ = −E sin ak + F cos ak. (2) Further, matching the derivatives ψ ′ (a), ψ ′ (−a) gives ψ ′ 2 (a) = ψ ′ 3 (a) and ψ ′ 1 (−a) = ψ ′ 3 (−a), or −κDe−aκ = k (E cos ak − F sin ak), (3) κAe−κa = k (E cos ak − F sin ak). (4) Adding and subtracting (1) and (2) gives 2F cos ak = (A + D) e −aκ , 2E sin ak = − (A − D) e −aκ . Adding and subtracting (3) and (4) gives 2k E cos ak = −κ (A − D) e −aκ , 2k F sin ak = κ (A + D) e −aκ . Setting A − D = −A1 and A + D = A2, the last two sets of equations can be combined and rewritten as 2E sin ak − A1e −aκ = 0 2k E cos ak + κA1e −aκ = 0 ⎫ ⎪⎬ ⎪⎭ (5) and 2F cos ak − A2e −aκ = 0 2k F sin ak − κA2e −aκ = 0 ⎫ ⎪⎬ ⎪⎭ . (6) The set (5) has nontrivial solutions for E and A1 only if 10.13 Exercises 737           2 sin ak −e −aκ 2k cos ak κe−aκ           = 0 which gives k cot ak = −κ. Similarly, the set (6) has nontrivial solutions for F and A2 only if k tan ak = κ. Note that it is impossible to satisfy both k cot ak = −κ and k tan ak = κ simultaneously. Hence, there are two classes of solutions, and solution is possible in quantum mechanics only if the energy satisfies certain conditions. 1. Odd solutions: k cot ak = −κ. In this case, F = A2 = 0. In terms of dimensionless variables, ξ = ak and η = aκ with definitions of k and κ, it follows that ξ 2 + η 2 = a 2  k 2 + κ 2 = a 2 2M
2 (V0 − E) + 2M
2 E ! = 2M V0a 2
2 . (7) This represents a circle. In terms of ξ and η, we write k cot ak = −κ as ak cot ak = −aκ, or ξ cot ξ = −η. (8) The simultaneous solutions of equations (7) and (8) can be determined from graphs of these functions at their point of intersection. It turns out that both ξ and η assume the positive values in the first quadrant only. Clearly, in the range 0 ≤ α = 2M V0a 2
2 < π 2 4 , there is no solution. For  π 2 2 ≤ α ≤  3π 2 2 , there is one solution. Thus, the existence of solutions depends on the parameters, M, V0 and the range of the potential. A simultaneous solution determines the allowed energy for which the quantum mechanical motion is described by an odd solution. 2. Even solutions: k tan ak = κ. In this case, E = A1 = 0, and (5) still holds. We can write the above condition in terms of nondimensional variables as 738 Answers and Hints to Selected Exercises ξ tan ξ = η. (9) The simultaneous solutions of (7) and (9) can be found graphically as before. It follows from the graphical representation that (7) and (9) intersect once if 0 ≤ α<π2 in the first quadrant. There are two points of intersection if π 2 ≤ α < (2π) 2 . The number of intersections (solutions) increases with the value of the parameter α. For each such allowed value of the energy determined from the points of intersection, there is an even solution in the present case. Note also that, for both even and odd solutions, ψ (x) is nonzero outside the finite square well so that there exists a nonzero probability for finding the particle there. This result is different from what is expected in classical mechanics. Finally, if V0 → ∞, it is easy to see that the intersections occur at ξ = nπ,  n + 1 2 π which are in agreement with the analysis of the infinite square well potential discussed in Example 10.10.1. 33. Hint: The boundary conditions at x = −a yields the matching conditions A e−ika + B eika = C eaκ + D e−aκ , A e−ika − B eika =  iκ k   C eaκ − D e−aκ . These results give the desired solution. Similarly, matching conditions at x = a gives the desired answer. Combining the matching relations leads to the final matrix equation. 11.11 Exercises 739 11.11 Exercises 3. u (ρ, θ) = 1 2π
2π 0 (ρ 2−1)f(β)dβ [1−ρ2−2ρ cos(β−θ)] . 7. u (x, y) = −  2 b ∞ n=1 sin(nπy/b) sinh(nπa/b) sinh & nπ b (a − x) '
x 0 f (ξ) sinh nπξ b dξ + sinh  nπx b
a 0 f (ξ) sinh & nπ b (a − ξ) ' dξ . 8. u (r, θ) = − ∞ n=0 ∞ k=1 (R/αnk) 2 Jn (αnkr/R) (Ank cos nθ + Bnk sin nθ), where A0k = 1 πR2J 2 1 (α0k)
R 0
2π 0 rf (r, θ) J0 (α0kr/R) dr dθ Ank = 2 πR2J 2 n+1(αnk)
R 0
2π 0 rf (r, θ) Jn (αnkr/R) cos nθ dr dθ Bnk = 2 πR2J 2 n+1(αnk)
R 0
2π 0 rf (r, θ) Jn (αnkr/R) sin nθ dr dθ n = 1, 2, 3,...; k = 1, 2, 3,... and αnk are the roots of J (αnk) = 0. 9. G (r, r′ ) = e ik|r−r ′ | |r−r ′ | − e ik|ρ−r ′ | |ρ−r ′ | , where r = (ξ, η, ζ), r ′ = (x, y, z), and ρ = (ξ, η, −ζ). 10. G (r, r′ ) = e ik|r−r ′ | |r−r ′ | + e ik|ρ−r ′ | |ρ−r ′ | . 14. G = − 4a π ∞ n=1
∞ 0 1 (α2a2+n2π2) sin  nπx a sin
4 nπξ a
5 sin αy sin αη dα. 16. u (r, z) = 2C π
∞ 0
∞ 0 1 (κ2−λ2−β2) J0 (βr) J1 (βa) cos λz dβ dλ. 17. u (r, θ) = A r 1 2 sin (θ/2). 18. G = − 2 a ∞ n=1 sinh σy′ sinh σ(y−b) σ sinh σb sin  nπx a sin
4 mπx′ a
5 , σ =  (κ 2 + (n2π 2) /α2), 0 < x′ < x < a, 0 < y′ <y x/c 1 2 [f (x + ct) + f (x − ct)] for t < x/c. 29. u (x, t) = ⎧ ⎪⎨ ⎪⎩ 0, for t < x/c f (t − x/c), for x/c < t ≤ (2 − x) /c. 30. u (x, t) = f0 + (f1 − f0) erfc
4 (x 2/4κt)
5 . 31. u (x, t) = x − x erfc  x/√ 4κt . 32. u (x, t)=2
t 0
η 0 erfc  x/√ 4κξ dη dξ. 33. u (x, t) = f0 e −ht 1 − erfc  x/√ 4κt !. 34. u (x, t) = f0 erfc  x/√ 4κt . 35. u (x, t) = ⎧ ⎪⎨ ⎪⎩ f0t, for t < x/c f0x/c, for t > x/c. 36. u (x, t) = ⎧ ⎪⎪⎪⎪⎨ ⎪⎪⎪⎪⎩ 1 2 [f (x + ct) − f (ct − x)] , t < x/c 1 2 [f (x + ct) + f (x − ct)] , t > x/c. 744 Answers and Hints to Selected Exercises 37. V (x, t) = V0  t − x c H  t − x c . (i) V = V0H  t − x c , (ii) V = V0 cos & ω  t − x c ' H  t − x c . 38. u (z, t) = U t " 1+2ζ 2 erfc (ζ) − √ 2ζ π e −ζ 2 # , where ζ = z 2 √ νt . 41. V (x, t) = V0erfc
4 x 2 √ κt
5 . 42. q (z, t) = a 2 e iωt " e −λ1z erfc( ζ − [it(2Ω + ω)] 1 2 ) +e λ1z erfc( ζ + [it(2Ω + ω)] 1 2 )# + b 2 e −iωt " e −λ2z erfc( ζ − [it(2Ω − ω)] 1 2 ) +e λ2z erfc( ζ + [it(2Ω − ω)] 1 2 )# , where λ1,2 =  i(2Ω + ω) ν 0 . q (z, t) ∼ a exp (iωt − λ1z) + b exp (−iωt − λ2z), δ1,2 =  ν |2Ω + ω| 01 2 . 43.  ν 2Ω 1 2 . 45. f (t) = f (0) + 1 Γ(α)Γ(1−α)
t 0 g (x) (t − x) α−1 dx. 46. x = a (θ − sin θ), y = a (1 − cos θ). 55. u (x, t) = ∞ n=1 sin nx
t 0 e −n 2 (t−τ)an (τ ) dτ + ∞ n=1 bn (0) sin nx e−n 2 t , where an (t) = 2 π
π 0 g (x, t) sin nx dx, bn (0) = 2 π
π 0 f (x) sin nx dx. 57. u (x, t) = ∞ n=1 2 π sin &n − 1 2 x '
t 0
π 0 e −(2n−1)2 (t−τ)/4 × sin &n − 1 2 ξ ' g (ξ, τ ) dξ dτ . 61. u (x, t) = c 2 sinh x √ 1/c sinh π √ 1/c − sin x √ 1/c sin π √ 1/c sin t +  2 πc ∞ n=1 (−1)n+1 n n4−(1/c) 2 sin n 2 ctsin nx, in which  1/c is not an integer. 14.11 Exercises 745 65. f (x) =
∞ 0 g (t) h (xt) dt, where h (x) = M−1 " 1 K(1−p) # . 14.11 Exercises 13. (c) With h = 0.2, the initial values are ui,0 (ih, 0) = sin π (ih). u1,0 = sin 0.2π = 0.5878, u2,0 = sin 0.4π = 0.9511. Also, u2,0 = u3,0 and u1,0 = u4,0. In each time step, there are 4 internal mesh points. We have to solve 4 equations with 4 unknowns. However, the initial temperature distribution is symmetric about x = 0.5, and u = 0 at the endpoints for all time t. We have u3,1 = u2,1 and u4,1 = u1,1 in the first time row and similarly for the other time rows. This gives two equations with two unknowns. 16. (a) y = x 3 , (b) y = sin x, (c) x 2 + (y − β) 2 = r 2 , where β and r are constants. 17. x = a (θ − sin θ), y = a (1 − cos θ). 18. Hint: I (y (x)) = 2π
x1 x0 y  1 + y ′2 1 2 dx. x = c1t + c2, y = c1 cosh t = c1 cosh
4 x−c2 c1
5 . (A surface generated by rotation of a catenary is called a catenoid). 21. (a) ∇4u = 0 (Biharmonic equation) (b) utt − α 2∇2u + β 2u = 0 (Klein–Gordon equation) (c) φt + αφx + βφxxx = 0, (φ = ux) (KdV equation) (d) utt + α 2uxxxx = 0 (Elastic beam equation) (e) d dx (pu′ )+(r + λs) u = 0 (Sturm–Liouville equation). 22.
2 2m ∇2ψ + (E − V ) ψ = 0 (Schr¨odinger equation). 24. utt − c 2uxx = F (x, t), where c 2 = T ∗/ρ. 746 Answers and Hints to Selected Exercises 29. Hint: yn = x (1 − x) n r=1 arx r−1 . Find the solution for n = 1 and n = 2. n =1: a1 = 5 18 , y1 = a1x (1 − x). n =2: a1 = 71 369 , a2 = 7 41 , y2 = x (1 − x) (a1 + a2x). 30. Hint: u1 (x, y) = a1xy. I (u1) = πab 4 " (a1 + 1)2 a 2 + (a1 − 1)2 b 2 # , a1 =
4 b 2−a 2 b 2+a2
5 . 31. Hint: u3 = x (1 − x) (1 − y) + x (1 − x) y (1 − y) (a2 + a3y). 32. Hint: u2 = x (2 − x − 2y) + a2 xy (2 − x − 2y). 33. Hint: ΨN =  N m,n=1 amn φmn =  N m,n=1 amn cos  mπx 2a cos  nπy 2b . 34. Hint: φn = cos (2n − 1) πr 2a ! . 35. Hint: I (u) =
a −a
a −a " ∇2u 2 −  4α a2 u # dx dy = min, and un =  x 2 − a 2 2  y 2 − a 2 2  a1 + a2x 2 + a3y 3 + ... . 36. Hint: Ψ1 =  b 2 − y 2 U (x). 37. (a) Introduce two functions φ and ψ and two parameters α and β such that U = u + αφ (x) and V = v + βψ (x). Then ∂I ∂α = 0 and ∂I ∂β = 0. 40. (a) uy − uxy ′ = uy′′ 1+y′2 . (b) y ′2 = 1−A 2 (y1−y) A2(y1−y) , (d) 2y ′′ − 3y + 3xy2 = 0. 42. Seek an approximate solution un (x, y) =  a 2 − x 2 b 2 − y 2 a1 + a2x 2 + a3y 2 + ... + anx 2ry 2s . For n = 1, f = 2 0 =

R (−u1xx − u1yy − 2)  a 2 − x 2 b 2 − y 2 dx dy = 2
a −a
b −b 1 − a1  a 2 − x 2 − a1  b 2 − y 2 ! a 2 − x 2 b 2 − y 2 dx dy = 32 9 a 2 b 3 − 128 45 (ab) 3  a 2 + b 2 a1, a1 = 5 4  a 2 + b 2 −1 and u1 = 5 4 (a 2−x 2 )(b 2−y 2 ) (a2+b 2) . 14.11 Exercises 747 The torsional moment M = 2Gθ
a −a
b −b u1dx dy =  40 9 (Gθ)
4 a 3 b 3 a2+b 2
5 , where G is the shear modulus and θ is the angle of twist per unit length. When a = b, M =  20 9 Gθa4 ∼ 0.1388 (2a) 4 Gθ. The tangential stresses are τzx = Gθ
4 ∂u1 ∂y
5 , τzy = Gθ  ∂u1 ∂x . (b) The exact solution is u (x, y) = x (a − x) − 8a 2 π3 ∞ n=1 cosh{(2n−1) πy 2a } sin{(2n−1) πx a } (2n−1)3 cosh{(2n−1) πb 2a } . M = 2Gθ 1 a 3 b 6 − 32a 4 π5 ∞ n=1 1 (2n−1)5 tanh & (2n − 1) πb 2a ' 3 . For a = b, M = 0.1406 (2a) 4 Gθ. 43. This problem deals with the expansion of a rectangular plate under tensile forces. Make the boundary conditions homogeneous. Integrating the boundary conditions gives u0 = 1 2 c y2
4 1 − y 2 6b 2
5 . Set u = u0+u5 so that ∇4u5 =  2c b 2 and the boundary conditions become u5xy =0= u5yy for x = + a, u5xy =0= u5xx for y = + b. These boundary conditions hold if u5 = 0, u5x = 0 for x = + a, u5 = 0, u5y = 0 for y = + b. By the Rayleigh–Ritz method

R  ∇4un − f φkdx dy = 0, k = 1, 2, . . . , n, where the nth approximate solution un (x, y) has the form un (x, y) =  x 2 − a 2 2  y 2 − b 2 2  a1 + a2x 2 + a3y 2 + ... . For n = 1, 0 =
a −a
b −b 24a1  y 2 − b 2 2 + 16a1  3x 2 − a 2 3y 2 − b 2 +24a1  x 2 − a 2 2 −  2c b #  x 2 − a 2 2  y 2 − b 2 2 dx dy, 748 Answers and Hints to Selected Exercises or
4 54 7 + 256 49 b 2 a2 + 64 7 b 4 a4
5 a1 = c a6b 2 . When a = b, a1 = (0.04325) c a6 . u1 ∼ u0 + u51 = 1 2 c y2
4 1 − y 2 6b 2
5 + (0.04325)  c a−6 ×  x 2 − a 2 2  y 2 − b 2 2 . 44. dF dx = ∂F ∂x + ∂F ∂u du dx + ∂F ∂u′ · du′ dx = ∂F ∂x + u ′ ∂F ∂u + u ′′ ∂F ∂u′ d dx  u ′ ∂F ∂u′ = u ′ d dx  ∂F ∂u′ + ∂F ∂u′ · u ′′ . Subtracting the latter from the former with (14.6.12) we obtain d dx  F − u ′ ∂F ∂u′ = ∂F ∂x + u ′ ∂F ∂u − d dx  ∂F ∂u′ ! = ∂F ∂x . 45. H = I − λJ =
b a F (x, y, y′ ) dx =
b 0 p (x) y ′2 − q (x) y 2 − λ r (x) y 2 ! dx. The extremum of H leads to the Euler–Lagrange equation d dx
4 ∂F ∂y′
5 − ∂F ∂y = 0. This leads to the answer. 46. (a) For simplicity, we assume that l is an integer and partition the interval into l equal subintervals. Each of the l − 1 = n interior vertices has the trial function vj (x) defined by vj (x) = ⎧ ⎪⎪⎪⎪⎨ ⎪⎪⎪⎪⎩ 1 − j + x for j − 1 ≤ x ≤ j, 1 + j − x for j ≤ x ≤ j + 1, 0 otherwise. vj (x) is continuous and piecewise linear with vj (j) = 1 and vj (k)=0 for all integers k = j. Appendix: Some Special Functions and Their Properties “One of the properties inherent in mathematics is that any real progress is accompanied by the discovery and development of new methods and simplifications of previous procedures ... The unified character of mathematics lies in its very nature; indeed, mathematics is the foundation of all exact natural sciences.” David Hilbert This appendix is a short introduction to some special functions used in the book. These functions include gamma, beta, error, and Airy functions and their main properties. Also included are Hermite and Webber–Hermite functions and their properties. Our discussion is brief since we assume that the reader is already familiar with this material. For more details, the reader is referred to appropriate books listed in the bibliography. A-1 Gamma, Beta, Error, and Airy Functions The Gamma function (also called the factorial function) is defined by a definite integral in which a variable appears as a parameter Γ (x) =
∞ 0 e −t t x−1 dt, x > 0. (A-1.1) In view of the fact that the integral (A-1.1) is uniformly convergent for all x in [a, b] where 0 < a ≤ b < ∞, Γ (x) is a continuous function for all x > 0. Integrating (A-1.1) by parts, we obtain the fundamental property of Γ (x) 750 Appendix: Some Special Functions and Their Properties Γ (x) = −e −t t x−1 !∞ 0 + (x − 1)
∞ 0 e −t t x−2 dt = (x − 1) Γ (x − 1), for x − 1 > 0. Then we replace x by x + 1 to obtain the fundamental result Γ (x + 1) = x Γ (x). (A-1.2) In particular, when x = n is a positive integer, we make repeated use of (A-1.2) to obtain Γ (n + 1) = n Γ (n) = n (n − 1) Γ (n − 1) = ··· = n (n − 1) (n − 2)··· 3 · 2 · 1 Γ (1) = n!, (A-1.3) where Γ (1) = 1. We put t = u 2 in (A-1.1) to obtain Γ (x)=2
∞ 0 exp  −u 2 u 2x−1 du, x > 0. (A-1.4) Letting x = 1 2 , we find Γ  1 2  = 2
∞ 0 exp  −u 2 du = 2 √ π 2 = √ π. (A-1.5) Using (A-1.2), we deduce Γ  3 2  = 1 2 Γ  1 2  = √ π 2 . (A-1.6) Similarly, we can obtain the values of Γ  5 2 , Γ  7 2 ,...,Γ  2n+1 2 . The gamma function can also be defined for negative values of x by rewriting (A-1.2) as Γ (x) = Γ (x + 1) x , x = 0, −1, −2,... (A-1.7) For example Γ  − 1 2  = Γ  1 2 − 1 2 = −2 Γ  1 2  = −2 √ π, (A-1.8) Γ  − 3 2  = Γ  − 1 2 − 3 2 = 4 3 √ π. (A-1.9) We differentiate (A-1.1) with respect to x to obtain A-1 Gamma, Beta, Error, and Airy Functions 751 Figure A-1.1 The gamma function. d dxΓ (x) = Γ ′ (x) =
∞ 0 d dx (t x ) e −t t dt =
∞ 0 d dx [exp (x log t)] e −t t dt =
∞ 0 t x−1 (log t) e −t dt. (A-1.10) At x = 1, this gives Γ ′ (1) =
∞ 0 e −t log t dt = −γ, (A-1.11) where γ is called the Euler constant and has the value 0.5772. The graph of the gamma function is shown in Figure A-1.1. Several useful properties of the gamma function are recorded below without proof for reference. Legendre Duplication Formula 2 2x−1 Γ (x) Γ  x + 1 2  = √ π Γ (2x), (A-1.12) 752 Appendix: Some Special Functions and Their Properties In particular, when x = n (n = 0, 1, 2,...) Γ  n + 1 2  = √ π (2n)! 2 2n n! . (A-1.13) The following properties also hold for Γ (x): Γ (x) Γ (1 − x) = π cosec πx, x is a noninteger, (A-1.14) Γ (x) = p x
∞ 0 exp (−pt)t x−1 dt, (A-1.15) Γ (x) =
∞ −∞ exp  xt − e t dt. (A-1.16) Γ (x + 1) ∼ √ 2π exp (−x) x x+ 1 2 for large x, (A-1.17) n! ∼ √ 2π exp (−n) x n+ 1 2 for large n. (A-1.18) The incomplete gamma function, γ (x, a), is defined by the integral γ (a, x) =
x 0 e −t t a−1 dt, a > 0. (A-1.19) The complementary incomplete gamma function, Γ (a, x), is defined by the integral Γ (a, x) =
∞ x e −t t a−1 dt, a > 0. (A-1.20) Thus, it follows that γ (a, x) + Γ (a, x) = Γ (a). (A-1.21) The beta function, denoted by B (x, y) is defined by the integral B (x, y) =
t 0 t x−1 (1 − t) y−1 dt, x > 0, y> 0. (A-1.22) The beta function B (x, y) is symmetric with respect to its arguments x and y, that is, B (x, y) = B (y, x). (A-1.23) This follows from (A-1.22) by the change of variable 1 − t = u, that is, B (x, y) =
1 0 u y−1 (1 − u) x−1 du = B (y, x). If we make the change of variable t = u /(1 + u) in (A-1.22), we obtain another integral representation of the beta function A-1 Gamma, Beta, Error, and Airy Functions 753 B (x, y) =
∞ 0 u x−1 (1 + u) −(x+y) du =
∞ 0 u y−1 (1 + u) −(x+y) du, (A-1.24) Putting t = cos2 θ in (A-1.22), we derive B (x, y)=2
π/2 0 cos2x−1 θ sin2y−1 θ dθ. (A-1.25) Several important results are recorded below without proof for ready reference. B (1, 1) = 1, B  1 2 , 1 2  = π, (A-1.26) B (x, y) =  x − 1 x + y − 1  B (x − 1, y), (A-1.27) B (x, y) = Γ (x) Γ (y) Γ (x + y) , (A-1.28) B  1 + x 2 , 1 − x 2  = π sec
4πx 2
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