Solution Techniques for Elementary Partial Differential Equations. Third Edition

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2 Solution Techniques for Elementary Partial Differential Equations Third Edition

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4 Solution Techniques for Elementary Partial Differential Equations Third Edition Christian Constanda The Charles W. Oliphant Professor of Mathematical Sciences The University of Tulsa Tulsa, Oklahoma, USA

5 CRC Press Taylor & Francis Group 6 Broken Sound Parkway NW, Suite 3 Boca Raton, FL by Taylor & Francis Group, LLC CRC Press is an imprint of Taylor & Francis Group, an Informa business No claim to original U.S. Government works Version Date: International Standard Book Number-13: (ebook - PDF) This book contains information obtained from authentic and highly regarded sources. Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint. Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, please access com ( or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 1923, CCC is a not-for-profit organization that provides licenses and registration for a variety of users. For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged. Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. Visit the Taylor & Francis Web site at and the CRC Press Web site at

6 For Lia and Jack

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8 Contents Foreword Preface to the Third Edition Preface to the Second Edition Preface to the First Edition xiii xv xvii xix Chapter 1 ORDINARY DIFFERENTIAL EQUATIONS: BRIEF REVIEW First-Order Equations Homogeneous Linear Equations with Constant Coefficients Nonhomogeneous Linear Equations with Constant Coefficients Cauchy Euler Equations Functions and Operators 9 Chapter 2 FOURIER SERIES The Full Fourier Series Fourier Sine and Cosine Series Fourier Sine Series Fourier Cosine Series Convergence and Differentiation Series Expansion of More General Functions 28 Chapter 3 STURM LIOUVILLE PROBLEMS Regular Sturm Liouville Problems Eigenvalues and Eigenfunctions of Some Basic Problems 33 vii

9 viii Contents Generalized Fourier Series Other Problems Bessel Functions Legendre Polynomials Spherical Harmonics 6 Chapter 4 SOME FUNDAMENTAL EQUATIONS OF MATHEMATICAL PHYSICS The Heat Equation The Laplace Equation The Wave Equation Other Equations 84 Chapter 5 THE METHOD OF SEPARATION OF VARIABLES The Heat Equation Rod with Zero Temperature at the Endpoints Rod with Insulated Endpoints Rod with Mixed Boundary Conditions Rod with an Endpoint in a Zero-Temperature Medium Thin Uniform Circular Ring The Wave Equation Vibrating String with Fixed Endpoints Vibrating String with Free Endpoints Vibrating String with Other Types of Boundary Conditions The Laplace Equation Laplace Equation in a Rectangle Laplace Equation in a Circular Disc Other Equations Equations with More than Two Variables Vibrating Rectangular Membrane Vibrating Circular Membrane Equilibrium Temperature in a Solid Sphere 144

10 Contents ix Chapter 6 LINEAR NONHOMOGENEOUS PROBLEMS Equilibrium Solutions Temperature Prescribed at the Endpoints Flux Prescribed at the Endpoints Mixed Boundary Conditions Equilibrium Temperature in a Circular Annulus Nonhomogeneous Problems Time-Independent Sources and Boundary Conditions The General Case 158 Chapter 7 THE METHOD OF EIGENFUNCTION EXPANSION The Nonhomogeneous Heat Equation Rod with Zero Temperature at the Endpoints Rod with Insulated Endpoints Rod with Mixed Boundary Conditions The Nonhomogeneous Wave Equation Vibrating String with Fixed Endpoints Vibrating String with Free Endpoints The Nonhomogeneous Laplace Equation Equilibrium Temperature in a Rectangle Equilibrium Temperature in a Circular Disc Other Nonhomogeneous Equations 186 Chapter 8 THE FOURIER TRANSFORMATIONS The Full Fourier Transformation Cauchy Problem for an Infinite Rod Vibrations of an Infinite String Equilibrium Temperature in an Infinite Strip The Fourier Sine and Cosine Transformations Heat Conduction in a Semi-Infinite Rod Vibrations of a Semi-Infinite String Equilibrium Temperature in a Semi-Infinite Strip 216

11 x Contents 8.3 Other Applications 22 Chapter 9 THE LAPLACE TRANSFORMATION Definition and Properties Applications The Signal Problem for the Wave Equation Heat Conduction in a Semi-Infinite Rod Finite Rod with Temperature Prescribed on the Boundary Diffusion-Convection Problems Loss Transmission Line 244 Chapter 1 THE METHOD OF GREEN S FUNCTIONS The Heat Equation The Time-Independent Problem The Time-Dependent Problem The Laplace Equation The Wave Equation 263 Chapter 11 GENERAL SECOND-ORDER LINEAR EQUATIONS The Canonical Form Classification Reduction to Canonical Form Hyperbolic Equations Parabolic Equations Elliptic Equations Other Problems 285 Chapter 12 THE METHOD OF CHARACTERISTICS First-Order Linear Equations First-Order Quasilinear Equations The One-Dimensional Wave Equation The d Alembert Solution The Semi-Infinite Vibrating String The Finite String 37

12 Contents xi 12.4 Other Hyperbolic Equations One-Dimensional Waves Spherical Waves 312 Chapter 13 PERTURBATION AND ASYMPTOTIC METHODS Asymptotic Series Regular Perturbation Problems Singular Perturbation Problems 326 Chapter 14 COMPLEX VARIABLE METHODS Elliptic Equations Systems of Equations 342 Appendix 347 A.1 Useful Integrals 347 A.2 Table of Fourier Transforms 348 A.3 Table of Fourier Sine Transforms 349 A.4 Table of Fourier Cosine Transforms 35 A.5 Table of Laplace Transforms 351 A.6 Second-Order Linear Equations 352 Further Reading 353 Index 355

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14 Foreword It is often difficult to persuade undergraduate students of the importance of mathematics. Engineering students in particular, geared toward the practical side of learning, often have little time for theoretical arguments and abstract thinking. Mathematics is the language of engineering and applied science, the vehicle by which ideas are analyzed, developed, and communicated. It is no accident, therefore, that any undergraduate engineering curriculum requires several mathematics courses, each one designed to provide the necessary analytic tools to deal with questions raised by engineering problems of increasing complexity, for example, in the modeling of physical processes and phenomena. The most effective way to teach students how to use these mathematical tools is by example. The more worked examples and practice exercises a textbook contains, the more effective it will be in the classroom. Such is the case with Solution Techniques for Elementary Partial Differential Equations by Christian Constanda. The author, a skilled classroom performer with considerable experience, understands exactly what students want and has given them just that: a textbook that explains the essence of the method briefly and then proceeds to show it in action. The book contains a wealth of worked examples and exercises (half of them with answers) and is accompanied by an Instructor s Manual with full solutions to each problem, as well as a PDF file for use on a computer-linked projector. A new and very helpful element added to the third edition is the insertion of Mathematica R software code at the end of each worked example, which shows how the computed solution to a problem can be verified by computer. The distribution of the exercises to each relevant section and subsection instead of having them bunched up together at the end of the chapter is also a welcome development. In my opinion, this is quite simply the best book of its kind that I have seen thus far. The book not only contains solution methods for some very important classes of PDEs, in easy-to-read format, but is also student-friendly and teacher-friendly at the same time. It is definitely a textbook for adoption. Professor Peter Schiavone Department of Mechanical Engineering University of Alberta Edmonton, AB, Canada xiii

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16 Preface to the Third Edition Apart from some cosmetic adjustments, the changes in the third edition are aimed at making the book more user-friendly. Sections 2.4 and 11.5, and Subsections and 8.1.3, are new. Many other sections (2.2, 3.1, 5.1, 5.2, 5.3, 5.5, 6.1, 6.2, 8.1, 8.2, 9.2, 1.1, 11.1, 12.3, and 12.4) have been split into several subsections, to make it easier for the reader to navigate the contents. The exercises have been distributed to each section/subsection, as appropriate, instead of being listed at the end of the chapters. The lists of exercises have been revamped, with some replacements and some new ones added in to enhance variety. The same treatment has been applied to the worked examples. The number of both has increased slightly (exercises from 64 to 615 and worked examples from 143 to 159). Almost all of the worked examples end with a brief Mathematica R program that shows how the results can be verified by computer. The last item is meant to offer the students a way of checking the accuracy of their work without interfering with the solution procedure, which in this book, with a few exceptions, still has to be carried out by hand. Students have to come to terms with the fact that a computer is only a tool doing what its human handler instructs it to do. To write a program that solves a given analytic problem requires solid theoretical expertise and practical experience with the solution method, both acquired only by first energizing the timetested combination of brain, hand, pencil, and paper. The exceptions referred to in the preceding paragraph are some of the worked examples and the exercises with an italicized number. These need computer help for their solution, and in all of them, the adopted convention is that numerical results are rounded to the fourth decimal place. Also, to keep the notation simple, and without danger of ambiguity, the equality sign is used where such numbers occur, instead of the approximate equality symbol. I would like to express my thanks to Sunil Nair, Alex Edwards, and Charlotte Byrnes at Taylor & Francis for their assistance with various issues during xv

17 xvi Preface to the Third Edition the completion of this project, and to Shashi Kumar and Marcus Fontaine for helping me tweak the L A TEX class file until it bowed to my aesthetic demands. Classroom users of the book also deserve acknowledgement for their comments and constructive suggestions. I would be very pleased to continue receiving feedback from them and, in particular, to hear if anyone has discovered how to make students stop misusing the word like. I don t have enough words of praise and gratitude for my wife, who stoically endured my numerous episodes of computer time and showed no ill feelings toward my laptop. Finally, special thanks are due to my grandson Jack, who, with his cheerful nature and nascent curiosity, provided many a welcome moment of distraction and entertainment during the tedious proofreading process. I hope that, in the fullness of time, he will be able to understand the content of this book and make good use of it. Christian Constanda The University of Tulsa October 215

18 Preface to the Second Edition Following constructive suggestions received from some of the users of the book, I have made a number of changes in the second edition. Section 1.4 (Cauchy Euler Equations) has been added to Chapter 1. Chapter 3 contains three new sections: 3.3 (Bessel Functions), 3.4 (Legendre Polynomials), and 3.5 (Spherical Harmonics). The new Section 4.4 in Chapter 4 lists additional mathematical models based on partial differential equations. Sections 5.4 and 7.4 have been added to Chapters 5 and 7, respectively, to show by means of examples how the methods of separation of variables and eigenfunction expansion work on equations other than heat, wave, and Laplace. Supplementary applications of the Fourier transformations are now shown in Section 8.3. The method of characteristics is applied to more general hyperbolic equations in the additional Section Chapter 14 (Complex Variable Methods) is entirely new. The number of worked examples has increased from 11 to 143, and that of the exercises has almost quadrupled from 165 to 64. The tables of Fourier and Laplace transforms in the Appendix have been considerably enhanced. The first coefficient of the Fourier series is now 1 2 a instead of the previous a. Similarly, the direct and inverse full Fourier transformations are now defined with the normalizing factor 1/ 2π in front of the integral; the Fourier sine and cosine transformations are defined with the factor 2/π. While I still believe that students should be encouraged not to use electronic computing devices in their learning of the fundamentals of partial differential equations, I have made a concession when it comes to examples and exercises that involve special functions or exceedingly lengthy integration. The xvii

19 xviii Preface to the Second Edition (new) exercises that require computational help because they are not solvable by elementary means have been given italicized numerical labels. I wish to thank all the readers who sent me their comments and urge them to continue to do so in the future. It is only with their help that this book may undergo further improvement. I would also like to thank Sunil Nair and Sarah Morris at Taylor & Francis for their professional and expeditious handling of this project. Christian Constanda The University of Tulsa January 21

20 Preface to the First Edition There are many textbooks on partial differential equations on the market. The great majority of them are well written and very rigorous, with full background explanations, detailed proofs, and lots of comments. But they also tend to be rather voluminous and daunting for the average student. When I ask the undergraduates what they want from a book, their most common answers are (i) to understand without excessive effort most of what is being said; (ii) to be given full yet concise explanations of the essence of the topics discussed, in simple words; (iii) to have many worked examples, preferably of the type found in test papers, so they could learn the various techniques by seeing them in action and thus improve their chances of passing examinations; and (iv) to pay as little as possible for it in the bookstore. I do not wish to comment on the validity of these answers, but I am prepared to accept that even in higher education the customer may sometimes be right. This book is an attempt to meet all the above requirements. It is designed as a no-frills text that explains a number of major methods completely but succinctly, in everyday classroom language. It does not indulge in multi-page, multicolored spiels. It includes many practical applications with solutions, and exercises with selected answers. It has a reasonable number of pages and is produced in a format that facilitates digital reproduction, thus helping keep costs down. Teachers have their own individual notions regarding what makes a book ideal for use in coursework. They say with good reason that the perfect text is the one they themselves sketched in their classroom notes but never had the time or inclination to polish up and publish. We each choose our own material, the order in which the topics are presented, and how long we spend on them. This book is no exception. It is based on my experience of the subject for many years and the feedback received from my teaching s beneficiaries. The use in combat of an earlier version seems to indicate that average students can work from it independently, with some occasional instructor guidance, while the high flyers get a basic and rapid grounding in the fundamentals of the subject before progressing to more advanced texts (if they are interested in further details and want to get a truly sophisticated picture of the field). A list, by no means exhaustive, of such texts can be found in the Bibliography. xix

21 xx Preface to the First Edition This book contains no example or exercise that needs a calculating device in its solution. Computing machines are now part of everyday life and we all use them routinely and extensively. However, I believe that if you really want to learn what mathematical analysis is all about, then you should exercise your mind and hand the long way, without any electronic help. (In fact, it seems that quite a few of my students are convinced that computers are better used for surfing the Internet than for solving homework problems.) The only prerequisites for reading this book are a first course in calculus and some basic knowledge of certain types of ordinary differential equations. The topics are arranged in the order I have found to be the most convenient. After some essential but elementary ordinary differential equations, Fourier series, and Sturm Liouville problems are discussed briefly, the heat, Laplace, and wave equations are introduced in quick succession as mathematical models of physical phenomena, and then a number of methods (separation of variables, eigenfunction expansion, Fourier and Laplace transformations, and Green s functions) are applied in turn to specific initial/boundary value problems for each of these equations. There follows a brief discussion of the general second-order linear equation with two independent variables. Finally, the method of characteristics and perturbation (asymptotic expansion) methods are presented. A number of useful tables and formulas are listed in the Appendix. The style of the text is terse and utilitarian. In my experience, the teacher s classroom performance does more to generate undergraduate enthusiasm and excitement for a topic than the cold words in a book, however skillfully crafted. Since the aim here is to get the students well drilled in the main solution techniques and not in the physical interpretation of the results, the latter hardly gets a mention. The examples and exercises are formal, and in many of them the chosen data may not reflect plausible real-life situations. Due to space pressure, some intermediate steps particularly the solutions of simple ODEs are given without full working. It is assumed that the readers know how to derive them, or that they can refer without difficulty to the summary provided in Chapter 1. Personally, in class I always go through the full solution regardless, which appears to meet with the approval of the audience. Details of a highly mathematical nature, including formal proofs, are kept to a minimum, and when they are given, an assumption is made that any conditions required by the context (for example, the smoothness and behavior of functions) are satisfied. The textbook will be accompanied by an Instructor s Manual containing the complete solutions of all the exercises. Also, on adoption of the book, a PDF file of the text (in suitably large sans-serif fonts) will be made available to instructors for use on classroom projectors. My own lecturing routine consists in (i) using a projector to present a skeleton of the theory, so the students do not need to take notes and can follow the live explanations, and (ii) doing a selection of examples on the board with full details, which the students take down by hand. I found that

22 Preface to the First Edition xxi this sequence of talking periods and writing periods helps the audience to maintain concentration and makes the lecture more enjoyable (if what the end-of-semester evaluations say is true). Wanting to offer students complete, rigorous, and erudite expositions is highly laudable, but the market priorities appear to have shifted of late. With the current standards of secondary education manifestly lower than in the past, students come to us less and less equipped to tackle the learning of mathematics from a fundamental point of view. When this becomes unavoidable, they seem to prefer a concise text that shows them the method and then, without fuss and niceties of form, goes into as many worked examples as possible. Whether we like it or not, it seems that we have entered the era of the digest. It is to this uncomfortable reality that the present book seeks to offer a solution. The last stages of preparation of this book were completed while I was a Visiting Professor in the Department of Mathematical and Computer Sciences at the University of Tulsa. I wish to thank the authorities of this institution and the faculty in the department for providing me with the atmosphere, conditions, and necessary facilities to finish the work on time. Particular thanks go to the following: Bill Coberly, the head of the department, who helped me engineer several summer visits and a couple of successful sabbatical years in Tulsa; Pete Cook, who heard my daily moans and groans from across the corridor and did not complain about it; Dale Doty, the resident Mathematica wizard who drew some of the figures and showed me how to do the others; and the sui generis company at the lunch table in the Faculty Club for whom, in time-honored academic fashion, no discussion topic was too trivial or taboo and no explanation too implausible. I also wish to thank Sunil Nair, Helena Redshaw, Andrea Demby, and Jasmin Naim from Chapman & Hall/CRC for their help with technical advice and flexibility over deadlines. Finally, I would like to state for the record that this book project would not have come to fruition had I not had the full support of my wife, who, not for the first time, showed a degree of patience and understanding far beyond the most reasonable expectations. Christian Constanda

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24 C H A P T E R 1 Ordinary Differential Equations: Brief Review In the process of solving partial differential equations (PDEs), we usually reduce the problem to the solution of certain classes of ordinary differential equations (ODEs). Here we mention without proof some basic methods for integrating simple ODEs of the types encountered later in the text. We restrict our attention to real solutions of ODEs with real coefficients. In what follows, the set of real numbers is denoted by R. 1.1 First-Order Equations Variables separable equations. The general form of this type of ODE is y = dy dx = f(x)g(y). Taking standard precautions, we can rewrite the equation as dy g(y) = f(x)dx and then integrate each side with respect to its corresponding variable. 1.1 Example. For the equation the above procedure leads to y 2 y 2x =, y 2 dy = 2 x dx, 1

25 2 1 Ordinary Differential Equations: Brief Review which yields or 1 3 y3 = x 2 + c, c = const, y(x) = (3x 2 + C) 1/3, C = const. Linear equations. Their general (normal) form is y + p(x)y = q(x), where p and q are given functions. Computing an integrating factor µ by means of the formula } µ(x) = exp p(x)dx, we obtain the general solution y(x) = 1 µ(x) µ(x)q(x) dx. An equivalent formula for the general solution is y(x) = 1 [ x ] µ(t)q(t) dt + C, C = const, µ(x) a where a is any point in the interval where the ODE is satisfied. 1.2 Example. The normal form of the equation xy + 2y x 2 =, x =, is Here, y + 2 x y = x. p(x) = 2, q(x) = x, x so an integrating factor is µ(x) = exp } dx 2 = e ln x2 = x 2. x Then the general solution of the equation is y(x) = 1 x 2 x 3 dx = 1 4 x2 + C x 2, C = const.

26 Exercises 1.2 Homogeneous Linear Equations with Constant Coefficients 3 Find the general solution of the given equation. 1 (x 2 + 1)y = 2xy. 2 y 3x 2 (y + 1) =. 3 (x 1)y + 2y = x, x 1. 4 x 2 y 2xy = x 5 e x. Answers to Odd-Numbered Exercises 1 y(x) = C(x 2 + 1). 3 y(x) = (x 1) 2 (x 3 /3 x 2 /2 + C). 1.2 Homogeneous Linear Equations with Constant Coefficients First-order equations. These are equations of the form y + ay =, a = const. Such equations can be solved by means of an integrating factor or separation of variables, or by means of the characteristic equation s + a =, whose root s = a yields the general solution y(x) = Ce ax, C = const. 1.3 Example. The characteristic equation for the ODE y 3y = is s 3 = ; hence, the general solution of the equation is y(x) = Ce 3x, C = const.

27 4 1 Ordinary Differential Equations: Brief Review Second-order equations. Their general form is y + ay + by =, a, b = const. If the characteristic equation s 2 + as + b = has two distinct real roots s 1 and s 2, then the general solution of the given ODE is y(x) = C 1 e s1x + C 2 e s2x, C 1, C 2 = const. If s 1 = s 2 = s, then y(x) = (C 1 + C 2 x)e sx, C 1, C 2 = const. Finally, if s 1 and s 2 are complex conjugate that is, s 1 = α + iβ, s 2 = α iβ, where α and β are real numbers (see Section 14.1) then the general solution is y(x) = e αx [C 1 cos(βx) + C 2 sin(βx)], C 1, C 2 = const. 1.4 Remarks. (i) When s 1 = s 2 = s, s real, the general solution of the equation can also be written as y(x) = C 1 y 1 (x) + C 2 y 2 (x), C 1, C 2 = const, where y 1 (x) and y 2 (x) are any two of the functions cosh ( s (x α) ), sinh ( s (x α) ), cosh ( s (x β) ), sinh ( s (x β) ) and α and β are any distinct real numbers. Normally, α and β are chosen as the points where boundary conditions are given. (ii) This can be generalized to the case of any two distinct real roots s 1 and s 2, where direct computation shows that y 1 (x) and y 2 (x) can be any two of the functions e (s1+s2)(x α)/2 cosh[ 1 2 (s 1 s 2 )(x α)], e (s1+s2)(x α)/2 sinh[ 1 2 (s 1 s 2 )(x α)], e (s1+s2)(x β)/2 cosh[ 1 2 (s 1 s 2 )(x β)], e (s1+s2)(x β)/2 sinh[ 1 2 (s 1 s 2 )(x β)]. 1.5 Example. The characteristic equation for the ODE is y 3y + 2y = s 2 3s + 2 =, with roots s 1 = 1 and s 2 = 2, so the general solution of the ODE is y(x) = C 1 e x + C 2 e 2x, C 1, C 2 = const.

28 1.2 Homogeneous Linear Equations with Constant Coefficients Example. The general solution of the equation y 4y = is y(x) = C 1 e 2x + C 2 e 2x, C 1, C 2 = const, since the roots of the characteristic equation are s 1 = s 2 = 2. According to Remark 1.4(i), we have alternative expressions in terms of hyperbolic functions. Thus, if the ODE is accompanied by, say, the boundary conditions y() = 1, y (1) = 6, then, with α = and β = 1, a useful way to write the general solution is y(x) = C 1 sinh(2x) + C 2 cosh ( 2(x 1) ), C 1, C 2 = const, because sinh(2x) is equal to zero at x = and the derivative of cosh ( 2(x 1) ) is equal to zero at x = 1, which helps us determine C 1 and C 2 very easily as C 1 = 3 sech 2, C 2 = sech Example. The roots of the characteristic equation for the ODE y 4y + y = are s 1 = and s 2 = 2 3. Then s 1 + s 2 = 4, s 1 s 2 = 2 3, so, if y() and y(1) are prescribed, we may prefer to use the statement in Remark 1.4(ii) with α = and β = 1 and write the general solution as y(x) = C 1 e 2x sinh ( 3 x ) + C 2 e 2(x 1) sinh ( 3 (x 1) ), C 1, C 2 = const. 1.8 Example. Since the roots of the characteristic equation for the ODE y + 4y + 4y = are s 1 = s 2 = 2, the general solution is y(x) = (C 1 + C 2 x)e 2x, C 1, C 2 = const. 1.9 Example. The characteristic equation for the ODE y + 2y + 1y = has the roots s 1 = 1 + 2i and s 2 = 1 2i, so the general solution of the ODE is y(x) = e x[ C 1 cos(2x) + C 2 sin(2x) ], C 1, C 2 = const.

29 6 1 Ordinary Differential Equations: Brief Review 1.1 Remark. The characteristic equation method can also be applied to find the general solution of homogeneous linear ODEs of higher order. Exercises Find the general solution of the given equation. 1 2y + 5y =. 2 3y 2y =. 3 y 4y + 3y =. 4 2y 5y + 2y =. 5 4y + 4y + y =. 6 y 6y + 9y =. 7 y + 2y + 5y =. 8 y 6y + 13y =. Answers to Odd-Numbered Exercises 1 y(x) = Ce 5x/2. 3 y(x) = C 1 e x + C 2 e 3x. 5 y(x) = (C 1 + C 2 x)e x/2. 7 y = e x [C 1 cos(2x) + C 2 sin(2x)]. 1.3 Nonhomogeneous Linear Equations with Constant Coefficients The first-order equations in this category are of the form y + ay = f, a = const, and the second-order equations can be written as y + ay + by = f, a, b = const, where f is a given function. The general solution of such equations is the sum of the complementary function (the general solution of the corresponding homogeneous equation) and a particular integral (a particular solution of the nonhomogeneous equation). The latter is usually guessed from the structure of the function f, or may be found by some other method, such as variation of parameters.

30 1.3 Nonhomogeneous Linear Equations with Constant Coefficients Example. The complementary function for the ODE y 3y = e x is y CF (x) = Ce 3x, We seek a particular integral of the form y P I (x) = ae x, C = const. a = const. Replacing it in the equation and collecting the like terms, we have 4ae x = e x, so a = 1/4. Consequently, the general solution of the given ODE is y(x) = Ce 3x 1 4 e x, C = const Example. If the function on the right-hand side in Example 1.11 is replaced by e 3x, then we cannot find a particular integral of the form ae 3x, a = const, since this is a solution of the corresponding homogeneous equation. Instead, we try y P I (x) = axe 3x, a = const, and deduce, by replacing in the ODE, that a = 1; consequently, the general solution is y(x) = Ce 3x + xe 3x = (C + x)e 3x, C = const Example. For the equation y + 4y = 8x 2, we seek a particular integral of the form y P I (x) = ax 2 + bx + c, a, b, c = const. Proceeding as in Example 1.11, we arrive at the equality 4ax 2 + 4bx + 2a + 4c = 8x 2, which yields a = 2, b =, and c = 1. Since the complementary function is y CF (x) = C 1 cos(2x) + C 2 sin(2x), it follows that the general solution of the given ODE is y(x) = C 1 cos(2x) + C 2 sin(2x) + 2x 2 1, C 1, C 2 = const.

31 8 1 Ordinary Differential Equations: Brief Review Exercises Find the general solution of the given equation. 1 y + 2y = 2x + e 4x. 2 2y 3y = 3x 4 + e x. 3 2y y = e x/2. 4 y + y = x + 2e x. 5 y y = x 2 x y 2y 8y = 4 + 4x 8x 2. 7 y 25y = 3e 5x. 8 4y + y = 8 cos(x/2). Answers to Odd-Numbered Exercises 1 y(x) = Ce 2x + x 1/2 + e 4x /6. 3 y(x) = (C + x/2)e x/2. 5 y(x) = C 1 cosh x + C 2 sinh x x 2 + x 4. 7 y(x) = C 1 e 5x + (C 2 3x)e 5x. 1.4 Cauchy Euler Equations The general form of a second-order Cauchy Euler equation is x 2 y + αxy + βy =, α, β = const, where, for simplicity, we assume that x >. We try a solution of the form y(x) = x r, r = const. Substituting in the equation, we arrive at r 2 + (α 1)r + β =. If the roots r 1 and r 2 of this quadratic equation are real and distinct, which is the case of interest for us, then the general solution of the given ODE is y(x) = C 1 x r1 + C 2 x r2, C 1, C 2 = const Example. Following the above procedure, we see that the differential equation 2x 2 y + xy y = yields r 1 = 1 and r 2 = 1/2, so its general solution is y(x) = C 1 x + C 2 x 1/2, C 1, C 2 = const.

32 1.5 Functions and Operators 9 Exercises Find the general solution of the given equation. 1 2x 2 y + xy 3y =. 2 x 2 y + 2xy 6y =. Answers to Odd-Numbered Exercises 1 y(x) = C 1 x 3/2 + C 2 x Functions and Operators Throughout this book, we refer to a function as either f or f(x), although, strictly speaking, the latter denotes the value of f at x. To avoid complicated notation, we also write f(x) = c to designate a function f that takes the same value c = const at all points x in its domain. When c =, we sometimes simplify this further to f =. In the preceding sections we mentioned linear equations. Here we clarify the meaning of this concept Definition. Let X be a space of functions, and let L be an operator acting on the functions in X according to some rule. The operator L is called linear if L(c 1 f 1 + c 2 f 2 ) = c 1 (Lf 1 ) + c 2 (Lf 2 ) for any functions f 1, f 2 in X and any numbers c 1, c 2. Otherwise, L is called nonlinear. The space X will not normally be specified explicitly, its nature being inferred from the context Example. The operators of differentiation and definite integration acting on suitable functions of one independent variable are linear, since L(c 1 f 1 + c 2 f 2 ) = (c 1 f 1 + c 2 f 2 ) = c 1 f 1 + c 2f 2 = c 1(Lf 1 ) + c 2 (Lf 2 ), L(c 1 f 1 + c 2 f 2 ) = b a = c 1 [ c1 f 1 (x) + c 2 f 2 (x) ] dx b a f 1 (x) dx + c 2 b a f 2 (x) dx = c 1 (Lf 1 ) + c 2 (Lf 2 ) Example. Let α, β, and γ be given functions. In view of the preceding example, it is easy to verify that the operator L defined by is linear. Lf = αf + βf + γf

33 1 1 Ordinary Differential Equations: Brief Review 1.18 Example. Let L be the operator defined by Since Lf = ff. L(c 1 f 1 + c 2 f 2 ) = (c 1 f 1 + c 2 f 2 )(c 1 f 1 + c 2 f 2 ) = c 2 1f 1 f 1 + c 1 c 2 (f 1 f 2 + f 1f 2 ) + c 2 2f 2 f 2 and c 1 (Lf 1 ) + c 2 (Lf 2 ) = c 1 f 1 f 1 + c 2f 2 f 2 are not equal for all suitable functions f 1, f 2 and all numbers c 1, c 2, the operator L is nonlinear Definition. A differential equation of the form Lu = g, where L is a linear differential operator and g is a given function, is called a linear equation. If the operator L is nonlinear, then the equation is also called nonlinear. The following almost obvious result forms the basis of what is known as the principle of superposition. 1.2 Theorem. If Lu = g is a linear equation and u 1 and u 2 are solutions of this equation with g = g 1 and g = g 2, respectively, then u 1 + u 2 is a solution of the equation with g = g 1 + g 2 ; in other words, if then Lu 1 = g 1, Lu 2 = g 2, L(u 1 + u 2 ) = g 1 + g 2. Proof. This follows immediately from the definition of a linear operator. Exercises Verify whether the given ODE is linear or nonlinear. 1 xy y sin x = xe x. 2 y + 2x sin y = 1. 3 y y xy = 2x. 4 y + x y = ln x. Answers to Odd-Numbered Exercises 1 Linear. 3 Nonlinear.

34 C H A P T E R 2 Fourier Series It is well known that an infinitely differentiable function f(x) can be expanded in a Taylor series around a point x in the interval where it is defined. This series has the form f(x) c n (x x ) n, (2.1) n= where the coefficients c n are given by c n = f (n) (x ), n = 1, 2,..., f (n) = dn f n! dx n. If certain conditions are satisfied, then the above series converges to f pointwise (that is, at every point x) in an open interval centered at x, and we can use the equality sign between the two sides in (2.1). In this chapter we discuss a different class of expansions, which are particulary useful in the study and solution of PDEs and can be constructed for functions with a much lower degree of smoothness. 2.1 The Full Fourier Series This is an expansion of the form f(x) 1 2 a + ( a n cos nπx L + b n sin nπx ), (2.2) L where L is a positive number and a, a n, and b n are constant coefficients. 2.1 Definition. A function f defined on R is called periodic if there is a number T > such that f(x + T ) = f(x) for all x in R. 11

35 12 2 Fourier Series The smallest number T with this property is called the fundamental period (or, simply, the period) of f. It is obvious that a periodic function also satisfies f(x + nt ) = f(x) for any integer n and all x in R. 2.2 Example. As is well known, the functions sin x and cos x are periodic with period 2π since, for all x R, sin(x + 2π) = sin x, cos(x + 2π) = cos x. We see that for each positive integer n = 1, 2,..., ( ) nπ(x + 2L) nπx cos = cos L L + 2nπ = cos nπx L, ( ) nπ(x + 2L) nπx sin = sin L L + 2nπ = sin nπx L, so the right-hand side in (2.2) is periodic with period 2L. This suggests the following method of construction for the full Fourier series of a given function. Let f be defined on [ L, L] (see Fig. 2.1). Figure 2.1 f(x), L x L. We construct the periodic extension of f from ( L, L] to R, of period 2L (see Fig. 2.2, which shows the three central periods). The value of f at x = L is left out so that the extension is correctly defined as a function. Figure 2.2 f(x + 2L) = f(x) for all x in R. For the extended function f it now makes sense to seek an expansion of the form (2.2). All that we need to do is compute the coefficients a, a n, and b n, and discuss the convergence of the series. It is easy to check by direct calculation that L L cos nπx L dx =, L L sin nπx L dx =, n = 1, 2,..., (2.3)

36 2.1 The Full Fourier Series 13 L L L L cos nπx L sin nπx L L L mπx cos L dx =, n m, L, n = m, mπx sin L dx =, n m, L, n = m, cos nπx L n, m = 1, 2,..., (2.4) n, m = 1, 2,..., (2.5) mπx sin dx =, n, m = 1, 2,.... (2.6) L Regarding (2.2) formally as an equality, we integrate it term by term over [ L, L] and use (2.3) to obtain a = 1 L L L f(x) dx. (2.7) If we multiply (2.2) by cos(mπx/l), integrate the new relation over [ L, L], and take (2.3), (2.4), and (2.6) into account, we see that all the integrals on the right-hand side vanish except that for which the summation index n is equal to m, when the integral is equal to La m. Replacing m by n, we find that a n = 1 L L L f(x) cos nπx L dx, n = 1, 2,.... (2.8) Clearly, (2.8) incorporates (2.7) if we allow n to take the value as well; that is, n =, 1, 2,.... Finally, we multiply (2.2) by sin(mπx/l) and repeat the above procedure, where this time we use (2.3), (2.5), and (2.6). The result is b n = 1 L L L f(x) sin nπx L dx, n = 1, 2,..., (2.9) which completes the construction of the (full) Fourier series for f. The numbers a, a n, and b n given by (2.7) (2.9) are called the Fourier coefficients of f. The series on the right-hand side in (2.2) is of interest to us only on [ L, L], where the original function f is defined, and may be divergent, or may have a different sum than f(x). 2.3 Definition. A function f is said to be piecewise continuous on an interval [a, b] if it is continuous at all but finitely many points in [a, b], where it has jump discontinuities; that is, at any discontinuity point x, the function has distinct right-hand side and left-hand side (finite) limits f(x+) and f(x ).

37 14 2 Fourier Series If both f and f are continuous on [a, b], then f is called smooth on [a, b]. If at least one of f, f is piecewise continuous on [a, b], then f is said to be piecewise smooth on [a, b]. 2.4 Remarks. (i) The function shown on the left in Fig. 2.3 is piecewise smooth; the one on the right is not, since f( ) does not exist: the graph indicates that the function increases without bound as the variable approaches the origin from the left. (ii) If f is piecewise continuous on [ L, L], then the values of f at its points of discontinuity do not affect the construction of its Fourier series. L More precisely, f(x) dx exists for such a function and is independent of the L values assigned to it at its (finitely many) discontinuity points. Figure 2.3 Left: both f( ) and f(+) exist. Right: f( ) does not exist. 2.5 Theorem. If f is piecewise smooth on [ L, L], then its (full) Fourier series converges pointwise to (i) the periodic extension of f to R at all points x where this extension is continuous; (ii) 1 2[ f(x ) + f(x+) ] at the points x where the periodic extension of f has a discontinuity jump. This means that at each x in ( L, L) where f is continuous, the sum of series (2.2) is equal to f(x). Also, if the function f is continuous on [ L, L] and such that f( L) = f(l), then the sum of (2.2) is equal to f(x) at all points x in [ L, L]. 2.6 Example. Consider the function defined by x + 2, 2 x <, f(x) =, x 2, whose graph is shown in Fig Clearly, here L = 2.

38 2.1 The Full Fourier Series 15 Figure 2.4 f(x), 2 x 2. The periodic extension of period 4 of f, defined by f(x + 4) = f(x) for all x in R, is shown in Fig Figure 2.5 f(x + 4) = f(x) for all x in R. Using (2.7) (2.9) with L = 2 and integration by parts, we find that a = 1 2 a n = 1 2 b n = f(x) dx = f(x) cos nπx 2 dx = 1 2 f(x) sin nπx 2 dx = 1 2 so (2.2) yields the Fourier series f(x) (x + 2) dx = 1, 2 2 (x + 2) cos nπx 2 dx = [ 1 ( 1) n] 2 n 2 π 2, (x + 2) sin nπx 2 dx = 2, n = 1, 2,..., nπ [1 ( 1) n ] 2 nπx n 2 cos π2 2 2 } nπx sin. nπ 2 By Theorem 2.5, on 2 x 2 x + 2, 2 x <, (series) = 1, x =,, < x 2; that is, the series converges to f(x) for 2 x < and < x 2, where the periodic extension of f (see Fig. 2.5) is continuous; at the point x =, where that extension has a jump discontinuity, the series converges to 1[ ] 2 f( ) + f(+) = 1 (2 + ) = 1. 2

39 16 2 Fourier Series The graph of the function defined by the sum of the above series on R is shown in Fig Figure 2.6 Graphic representation of the sum of the series. Verification with Mathematica R. We compare the graph of f with that of the function obtained by adding the first few terms of its series representation, to see if the latter is a reasonable approximation of the former. This procedure will be followed throughout the book without further explanation where infinite series are involved. Here, with truncation after n = 5, the input GenTerm = ((1 - ( - 1) n) 2/(n 2 Pi 2)) Cos[n Pi x/2] - (2/(n Pi)) Sin[n Pi x/2]; f = Piecewise[x + 2, - 2<x<},, <x<2}}]; Plot[1/2 + Sum[GenTerm,n,1,5}],f},x, - 2,2}, PlotRange -> - 2.2,2.2}, -.5,2.4}}, PlotStyle -> Black, ImageSize -> Scaled[.25],Ticks -> - 2,2},2}}, AspectRatio ->.5] generates the graphical output which confirms that our computation is correct. Exercises Construct the full Fourier series of the given function f. In each case, discuss the convergence of the series on the interval [ L, L] where f is defined, and sketch the function to which the series converges pointwise on the interval [ 3L, 3L]. 1 f(x) = 1, 1 x,, < x 1.

40 2.1 The Full Fourier Series 17, π x, 2 f(x) = 3, < x π. 2, 1 x, 3 f(x) = 3, < x 1. 1, π/2 x, 4 f(x) = 2, < x π/2. 5 f(x) = x + 1, 1 x 1. 6 f(x) = 1 2x, 2 x 2. 1, 2 x, 7 f(x) = 2 x, < x x, 1/2 x, 8 f(x) = 4, < x 1/2. x, 1 x, 9 f(x) = 2x 1, < x 1. 1 f(x) = x 1, π x, 2x + 1, < x π. 11 f(x) = x 2 2x + 3, 1 x 1. x 2, 1 x, 12 f(x) = 1 + 2x, < x f(x) = e x, 2 x 2. 1, π/2 x, 14 f(x) = sin x, < x π/2., 2 x 1, 15 f(x) = 1 + x, 1 < x f(x) = 3, π x π/2, 1, π/2 < x π. Answers to Odd-Numbered Exercises 1 f(x) 1/2 + [( 1) n 1](1/(nπ)) sin(nπx). 3 f(x) 1/2 + [1 ( 1) n ](5/(nπ)) sin(nπx).

41 18 2 Fourier Series 5 f(x) 1 + ( 1) n+1 (2/(nπ)) sin(nπx). 7 f(x) [1 ( 1) n ](2/(n 2 π 2 )) cos(nπx/2) + [3 ( 1) n ](1/(nπ)) sin(nπx/2)}. 9 f(x) 1/4 + [( 1) n 1](1/(n 2 π 2 )) cos(nπx) [2( 1) n + 1](1/(nπ)) sin(nπx)}. 11 f(x) 1/3 + ( 1) n (4/(n 2 π 2 ))[cos(nπx) + nπ sin(nπx)]. 13 f(x) (e 4 1)/(4e 2 ) + ( 1) n [(e 4 1)/(e 2 (4 + n 2 π 2 ))] [2 cos(nπx/2) nπ sin(nπx/2)]. 15 f(x) 9/8 + (2/(n 2 π 2 ))[( 1) n cos(nπ/2)] cos(nπx/2) + [( 1) n+1 (3/(nπ)) + (2/(n 2 π 2 )) sin(nπ/2)] sin(nπx/2)}. 2.2 Fourier Sine and Cosine Series For some classes of functions, series (2.2) has a simpler form. 2.7 Definition. A function f defined on an interval symmetric with respect to the origin is called odd if f( x) = f(x) for all x in the given interval; if, on the other hand, f( x) = f(x) for all x in that interval, then f is called an even function. 2.8 Examples. (i) The functions sin(nπx/l), n = 1, 2,..., are odd. The functions cos(nπx/l), n =, 1, 2,..., are even. (ii) The graph (symmetric with respect to the origin) shown on the left in Fig. 2.7 represents an odd function; the one (symmetric with respect to the y-axis) on the right represents an even function. The function graphed in Fig. 2.4 is neither odd nor even. Figure 2.7 Left: an odd function. Right: an even function.

42 2.2 Fourier Sine and Cosine Series Remarks. (i) It is easy to verify that the product of two odd functions is even, the product of two even functions is even, and the product of an odd function and an even function is odd. (ii) If f is odd on [ L, L], then f(x) dx = f( x) d( x) = L f(x) dx; L L hence, L L f(x) dx = f(x) dx + f(x) dx =. L If f is even on [ L, L], then L L f(x) dx = f( x) d( x) = f(x) dx; L L consequently, L L L f(x) dx = f(x) dx + f(x) dx = 2 f(x) dx. L L (iii) If f is odd, then so is f(x) cos(nπx/l) and, in view of (2.7), (2.8), and (ii) above, we have a n =, n =, 1, 2,... ; that is, the Fourier series of an odd function on [ L, L] contains only sine terms. Similarly, if f is even, then, by Remark 2.9(i), f(x) sin(nπx/l) is odd, so (2.9) implies that b n =, n = 1, 2,..., which means that the Fourier series of an even function on [ L, L] has only cosine terms, including the constant term Fourier Sine Series Remark 2.9(iii) implies that if f is defined on [, L], then it can be expanded in a Fourier sine series. Let f be the function represented by the graph in Fig Figure 2.8 f(x), x L.

43 2 2 Fourier Series We construct the odd extension of f from (, L] to [ L, L] by setting f( x) = f(x) for all x in [ L, L], x =, and f() = (see Fig. 2.9). Figure 2.9 f( x) = f(x), L x L. Next, we construct the periodic extension with period 2L of this odd function from ( L, L] to R, by requiring that f(x + 2L) = f(x) for all x in R (see Fig. 2.1). Figure 2.1 f(x + 2L) = f(x) for all x in R. Finally, we construct the Fourier (sine) series of this last function, which is of the form f(x) b n sin nπx L, where, according to Remarks 2.9(ii) and (iii) and formula (2.9), the coefficients are given by b n = 2 L L f(x) sin nπx L 2.1 Example. Consider the function graphed in Fig f(x) = 1 x, x 1, dx, n = 1, 2,.... (2.1) Figure 2.11 f(x), x 1.

44 2.2 Fourier Sine and Cosine Series 21 As explained above, the odd extension of this function to [ 1, 1] is defined by f( x) = f(x) for all x in [ 1, 1], x, and f() =, and is shown in Fig Figure 2.12 f( x) = f(x), 1 x 1. The periodic extension to R, of period 2L = 2, of the function in Fig. 2.12, defined by f(x + 2) = f(x) for all x in R, is shown in Fig hence, Figure 2.13 f(x + 2) = f(x) for all x in R. Using (2.1) with L = 1 and integration by parts, we find that b n = 2 1 (1 x) sin(nπx) dx = 2, n = 1, 2,... ; nπ f(x) By Theorem 2.5, for x 1,, x =, (series) = 1 x, < x 1; 2 sin(nπx). (2.11) nπ that is, the series converges to f(x) for < x 1, where the periodic extension of f to R is continuous, and to 1 [ ] 2 f( ) + f(+) = 1 ( 1 + 1) = 2 at x =, where the periodic extension has a jump discontinuity. The graph in Fig is also the representation of the function defined by the sum of the above series on R.

45 22 2 Fourier Series Verification with Mathematica R. The input GenTerm = (2/(n Pi)) Sin[n Pi x]; f = 1 - x; Plot[Sum[GenTerm,n,1,7}],f},x,,1}, PlotRange -> -.1,1.1}, -.2,1.1}}, PlotStyle -> Black, ImageSize -> Scaled[.2], Ticks ->1},1}}, AspectRatio ->.9] generates the graphical output Fourier Cosine Series By Remark 2.9(iii), a function defined on [, L] can also be expanded in a Fourier cosine series. The construction is similar to that of a Fourier sine series. Let f be a function defined on [, L], as shown graphically in Fig Figure 2.14 f(x), x L. We extend f to an even function on [ L, L] by setting f( x) = f(x) for all x in [ L, L] (see Fig. 2.15). Figure 2.15 f( x) = f(x), L x L. Next, we construct the periodic extension of this even function from ( L, L] to R, of period 2L, which is defined by f(x + 2L) = f(x) for all x in R (see Fig. 2.16).

46 2.2 Fourier Sine and Cosine Series 23 Figure 2.16 f(x + 2L) = f(x) for all x in R. Finally, we write the Fourier (cosine) series of this last function, which is of the form f(x) 1 2 a + a n cos nπx L, where, according to (2.7), (2.8), and Remark 2.9(ii), a = 2 L a n = 2 L L L f(x) dx, f(x) cos nπx L dx, n = 1, 2,.... (2.12) As noticed earlier, the first equality (2.12) is covered by the second one if we let n =, 1, 2, Example. Consider the function graphed in Fig f(x) = 1 x, x 1, Figure 2.17 f(x), x 1. The even extension of f to [ 1, 1], defined by f( x) = f(x) for all x in [ 1, 1], is shown in Fig Figure 2.18 f( x) = f(x), 1 x 1.

47 24 2 Fourier Series The periodic extension of period 2L = 2 of this even function to R, which is defined by f(x + 2) = f(x) for all x in R, is shown in Fig Figure 2.19 f(x + 2) = f(x) for all x in R. By (2.12) with L = 1 and integration by parts, a = 2 a n = (1 x) dx = 1, (1 x) cos(nπx) dx = [ 1 ( 1) n] 2 n 2, n = 1, 2,..., π2 so the Fourier cosine series of f is By Theorem 2.5, f(x) [ 1 ( 1) n ] 2 n 2 cos(nπx). (2.13) π2 (series) = 1 x for x 1, because the periodic extension of f to R is continuous everywhere. The graph in Fig is also the representation of the function defined by the sum of the above series on R. Verification with Mathematica R. The input GenTerm = ((1 - ( - 1) n) 2/(n 2 Pi 2)) Cos[n Pi x]; f = 1 - x; Plot[1/2 + Sum[GenTerm,n,1,1}],f},x,,1}, PlotRange -> -.2,1.1}, -.2,1.1}}, PlotStyle -> Black, ImageSize -> Scaled[.2], Ticks ->1},1}}, AspectRatio ->.9] generates the graphical output

48 2.2 Fourier Sine and Cosine Series 25 Exercises Construct the Fourier sine series and the Fourier cosine series of the given function f. In each case, discuss the convergence of the series on the interval [, L] where f is defined, and sketch the functions to which the series converge pointwise on [ 3L, 3L]. 1 f(x) = 2 f(x) = 3 f(x) = 4 f(x) =, x 1, 1, 1 < x 2. 2, x π,, π < x 2π. 1, x 1, 1, 1 < x 2. 2, x π/2, 3, π/2 < x π. 5 f(x) = 2 x, x 1. 6 f(x) = 3x + 1, x 2π. 7 f(x) = x, x 1, 2, 1 < x 2. 8 f(x) = 9 f(x) = 2x 1, x 1, 2x + 1, 1 < x x, x 1, 1 x, 1 < x 2. 1 f(x) = x 2 + x 1, x f(x) = x + 1, x 1, x 2 2x, 1 < x f(x) = 1 + e x, x f(x) = x + sin x, x π. 14 f(x) = cos x, x π/2, 1, π/2 < x π.

49 26 2 Fourier Series Answers to Odd-Numbered Exercises 1 f(x) (2/(nπ))[cos(nπ/2) ( 1) n ] sin(nπx/2); f(x) 1/2 (2/(nπ)) sin(nπ/2) cos(nπx/2). 3 f(x) (2/(nπ))[1 + ( 1) n 2 cos(nπ/2)] sin(nπx/2); f(x) (4/(nπ)) sin(nπ/2) cos(nπx/2). 5 f(x) [2 ( 1) n ](2/(nπ)) sin(nπx); f(x) 3/2 + [1 ( 1) n ](2/(n 2 π 2 )) cos(nπx). 7 f(x) (2/(nπ))[ 3 cos(nπ/2) + (2/(nπ)) sin(nπ/2) + ( 1) n 2] sin(nπx/2); f(x) 3/4 + (2/(nπ))[3 sin(nπ/2) + (2/(nπ)) cos(nπ/2) 2/(nπ)] cos(nπx/2). 9 f(x) (2/(nπ))[2 + ( 1) n 3 cos(nπ/2) + (4/(nπ)) sin(nπ/2)] f(x) 1 + (6/(nπ)) sin(nπ/2) + (8/(n 2 π 2 )) cos(nπ/2) sin(nπx/2); + [( 1) n+1 1](4/(n 2 π 2 ))} cos(nπx/2). 11 f(x) (2/(n 3 π 3 ))[8( 1) n + n 2 π 2 (8 + 3n 2 π 2 ) cos(nπ/2) + 2nπ sin(nπ/2)] sin(nπx/2); f(x) 5/12 + (2/(n 3 π 3 ))[2( 1) n 1]2nπ + 2nπ cos(nπ/2) + (8 + 3n 2 π 2 ) sin(nπ/2)} cos(nπx/2). 13 f(x) 3 sin x + ( 1) n+1 (2/n) sin(nx); n=2 f(x) 2/π + π/2 (4/π) cos x + [2(1 ( 1) n 2n 2 )/((n 2 1)n 2 π)] cos(nx). n=2

50 2.3 Convergence and Differentiation 2.3 Convergence and Differentiation 27 There are several important points that need to be made at this stage Remarks. (i) It is obvious that a function f defined on [, L] can be expanded in a full Fourier series, or in a sine series, or in a cosine series. While the last two are unique to the function, the first one is not: since the full Fourier series requires no symmetry, we may extend f from [, L] to [ L, L] in infinitely many ways. We usually choose the type of series that seems to be the most appropriate for the problem. A function defined on [ L, L] can be expanded, in general, only in a full Fourier series. (ii) Earlier, we saw that we cannot automatically use the equality sign between a function f and its Fourier series. According to Theorem 2.5, for a full series we can do so at all the points in [ L, L] where f is continuous; we can also do it at x = L and x = L provided that the periodic extension of f is continuous at these points from the right and from the left, respectively, and f( L) = f(l). For a sine series, this can be done at x = L if f is continuous there from the left and f(l) =. For a cosine series, this is always possible if f is continuous from the left at x = L. Similar arguments can be used for the sine and cosine series at x =. In what follows, we work exclusively with piecewise smooth functions; therefore, to avoid cumbersome notation and possible confusion, we will use the equality sign in all circumstances, assuming tacitly that equality is understood in the sense of Theorem 2.5, Remark 2.4(ii), and the comments above. In practical applications, we often need to differentiate Fourier series term by term. Consequently, it is important to know when this operation may be performed Theorem. (i) If f is continuous on [ L, L], f( L) = f(l), and f is piecewise smooth on [ L, L], then the full Fourier series of f can be differentiated term by term, and the result is the full Fourier series of f, which converges to f (x) at all x where f (x) exists. (ii) If f is continuous on [, L], f() = f(l) =, and f is piecewise smooth on [, L], then the Fourier sine series of f can be differentiated term by term, and the result is the Fourier cosine series of f, which converges to f (x) at all x where f (x) exists. (iii) If f is continuous on [, L] and f is piecewise smooth on [, L], then the Fourier cosine series of f can be differentiated term by term, and the result is the Fourier sine series of f, which converges to f (x) at all x where f (x) exists.

51 28 2 Fourier Series 2.14 Remark. In the PDE problems that we discuss in the rest of the book, the conditions laid down in Theorem 2.13 will always be satisfied and we will formally differentiate the corresponding Fourier series term by term without explicit mention of these conditions. 2.4 Series Expansion of More General Functions It is important to know how functions defined on various types of intervals and having the necessary degree of smoothness can be expanded in an infinite series by means of the procedures set out in Sections 2.1 and 2.2. Function defined on [ L,L]. If a function like this is neither odd nor even, then it can be expanded only in a full Fourier series. Function defined on [,L]. We can expand such a function in a Fourier sine series or a Fourier cosine series. The latter requires the computation of an additional term (a ), but, because the extension of the function to R is even and, therefore, symmetric with respect to the y-axis, the cosine series generally converges to the function on a larger interval than the sine series. We can also expand the function in a full Fourier series if we first extend it by, say, to [ L, L]. Function defined on [a,b]. A function of this type can be expanded by means of the full Fourier series technique if we make the substitution x = 1 2L (b a)x + 1 (b + a), 2 which maps the interval [a, b] for x onto the interval [ L, L] for X. We can also expand the function using a Fourier sine or cosine series construction if we substitute x = 1 (b a)x + a, L which maps the interval [a, b] for x onto the interval [, L] for X. Function defined on an infinite interval. Obviously, this type of function cannot be expanded by means of the Fourier series method. Its proper handling is explained in Chapter 8.

52 C H A P T E R 3 Sturm Liouville Problems A particular class of problems involving second-order ODEs plays an essential role in the solution of partial differential equations. Below we present the main results characterizing such problems, which will be used extensively in the methods of separation of variables and eigenfunction expansion developed in Chapters 5 and Regular Sturm Liouville Problems To avoid complicated notation, in what follows we denote intervals generically by I, regardless of whether they are closed, open, or half-open, finite or infinite. Specific intervals are described either in terms of their endpoints or by means of inequalities. The functions defined on such intervals are assumed to be integrable on them. 3.1 Definition. Let X be a space of functions defined on the same interval I. A linear differential operator L acting on X is called symmetric on X if [ f1 (x)(lf 2 )(x) f 2 (x)(lf 1 )(x) ] dx = for any functions f 1 and f 2 in X. I 3.2 Example. Consider the space X of functions that are twice continuously differentiable on [, 1] and equal to zero at x = and x = 1. If L is the linear operator defined by L = d2 dx 2, then, using integration by parts, we find that for any f 1 and f 2 in X, 29

53 3 3 Sturm Liouville Problems 1 [ f1 (Lf 2 ) f 2 (Lf 1 ) ] dx = Hence, L is symmetric on X. 1 (f 1 f 2 f 2f 1 ) dx = [ 1 f 1 f 2 ] 1 f 1f 2 dx [ 1 f 2 f 1 ] 1 + f 1f 2 dx =. 3.3 Remark. An operator may be symmetric on a space of functions but not on another. If no restriction is imposed on the values at x = and x = 1 of the functions in Example 3.2, then L is not symmetric on the new space X. 3.4 Definition. Let σ be a function defined on I, with the property that σ(x) > for all x in I. Two functions f 1 and f 2, also defined on I, are called orthogonal with weight σ on I if f 1 (x)f 2 (x)σ(x) dx =. I If σ(x) = 1, then f 1 and f 2 are simply called orthogonal. A set of functions that are pairwise orthogonal on I is called an orthogonal set. 3.5 Example. The functions f 1 (x) = 1 and f 2 (x) = 9x 5 are orthogonal with weight σ(x) = x + 1 on [, 1] since 1 (9x 5)(x + 1) dx = 1 (9x 2 + 4x 5) dx =. Alternatively, we can say that the functions f 1 (x) = x + 1 and f 2 (x) = 9x 5 are orthogonal on [, 1]. 3.6 Example. The functions sin(3x) and cos(2x) are orthogonal on the interval [ π, π] because π π sin(3x) cos(2x) dx = π π 1[ ] 2 sin(5x) + sin x dx =. 3.7 Definition. Let L be a linear differential operator acting on a space X of functions defined on (a, b). An equation of the form (Lf)(x) + λσ(x)f(x) =, a < x < b, (3.1) where λ is a parameter and σ is a given function such that σ(x) > for all x in (a, b), is called an eigenvalue problem. The numbers λ for which (3.1) has nonzero solutions in X are called eigenvalues; the corresponding nonzero solutions are called eigenfunctions.

54 3.1 Regular Sturm Liouville Problems Theorem. If the operator L of the eigenvalue problem (3.1) is symmetric, then (i) all the eigenvalues λ are real; (ii) the eigenvalues form an infinite sequence λ 1 < λ 2 < < λ n < such that λ n as n ; (iii) eigenfunctions associated with distinct eigenvalues are orthogonal with weight σ on (a, b). 3.9 Definition. Let [a, b] be a finite interval, let p, q, and σ be real-valued functions, and let κ 1, κ 2, κ 3, and κ 4 be real numbers such that (i) p is continuously differentiable on [a, b] and p(x) > for all x in [a, b]; (ii) q and σ are continuous on [a, b] and σ(x) > for all x in [a, b]; (iii) κ 1, κ 2 are not both zero and κ 3, κ 4 are not both zero. An eigenvalue problem of the form [ p(x)f (x) ] + q(x)f(x) + λσ(x)f(x) =, a < x < b, (3.2) with the boundary conditions (BCs) κ 1 f(a) + κ 2 f (a) =, (3.3) κ 3 f(b) + κ 4 f (b) =, (3.4) is called a regular Sturm Liouville (S L) eigenvalue problem. 3.1 Example. The choice p(x) = 1, q(x) =, σ(x) = 1, κ 1 = 1, κ 2 =, κ 3 = 1, κ 4 =, a =, b = L generates the regular S L problem f (x) + λf(x) =, < x < L, f() =, f(l) = Example. If p, q, σ, a, and b are as in Example 3.1 but κ 1 =, κ 2 = 1, κ 3 =, and κ 4 = 1, then we have the regular S L problem f (x) + λf(x) =, < x < L, f () =, f (L) = Example. Taking p, q, σ, a, and b as in Example 3.1 and the coefficients κ 1 = 1, κ 2 =, κ 3 = h, κ 4 = 1, we obtain the regular S L problem f (x) + λf(x) =, < x < L, f() =, f (L) + hf(l) =.

55 32 3 Sturm Liouville Problems 3.13 Example. The eigenvalue problem f (x) + 2f (x) + λf(x) =, < x < 1, f() =, f(1) = may, at first glance, not seem to be a regular S L problem. However, if we use the coefficient of f to construct the function e 2x and multiply every term in the equation by this function, we bring the equation to the form ( e 2x f (x) ) + λe 2x f(x) =, < x < 1, which indicates that we have, in fact, a regular S L problem with coefficients } p(x) = σ(x) = exp 2 dx = e 2x, q(x) = satisfying the requirements in Definition Theorem. The operator Lf = (pf ) + qf (3.5) defined by the left-hand side in (3.2) and acting on a space X of functions satisfying (3.3) and (3.4) is symmetric on X. Proof. Suppose, for simplicity, that κ 1 = and κ 4. Then, by (3.3) and (3.4), any function f in X satisfies f(a) = κ 2 κ 1 f (a), f (b) = κ 3 κ 4 f(b). Consequently, using (3.5) and integration by parts, we find that for any f 1 and f 2 in X, b [ f1 (Lf 2 ) f 2 (Lf 1 ) ] dx a = b a f1 [ (pf 2 ) + qf 2 ] f2 [ (pf 1 ) + qf 1 ]} dx = [ f 1 (pf 2) ] b b a pf 2f 1 dx [ f 2 (pf 1) ] b b a + pf 1f 2 dx a = p(b) [ f 1 (b)f 2 (b) f 2(b)f 1 (b)] p(a) [ f 1 (a)f 2 (a) f 2(a)f 1 (a)] [ = p(b) κ 3 f 1 (b)f 2 (b) + κ ] 3 f 2 (b)f 1 (b) κ 4 κ 4 [ p(a) κ 2 f 1 κ (a)f 2 (a) + κ ] 2 f 2 1 κ (a)f 1 (a) =, 1 a as required. All other combinations of nonzero constants κ 1, κ 2, κ 3, and κ 4 are treated similarly.

56 3.1 Regular Sturm Liouville Problems Corollary. The eigenvalues and eigenfunctions of a regular Sturm Liouville problem have all the properties listed in Theorem 3.8. Exercises Verify that the given problem is a S L eigenvalue problem and specify whether it is regular or singular. 1 f (x) + λf(x) =, < x < 1, f() + 2f () =, f (1) =. 2 f (x) xf(x) + λ(x 2 + 1)f(x) =, < x < 1, f() =, f (1) =. 3 (x 2)f (x) + f (x) + (1 + λ)f(x) =, < x < 2, f () =, f(x), f (x) bounded as x 2. 4 f (x) f (x) + λf(x) =, < x < 2, f () =, f(2) f (2) =. 5 2f (x) + f (x) + (λ + x)f(x) =, < x <, f() =, f(x), f (x) bounded as x. 6 x 2 f (x) + xf (x) + (2λx 1)f(x) =, < x < 1, f(x), f (x) bounded as x +, f(1) f (1) =. Answers to Odd-Numbered Exercises 1 Regular. 3 Singular. 5 Singular Eigenvalues and Eigenfunctions of Some Basic Problems Before actually computing the eigenvalues and eigenfunctions of specific Sturm Liouville problems, we derive a very useful formula relating these quantities. Multiplying equation (3.2) by f and integrating over (a, b), we easily see that

57 34 3 Sturm Liouville Problems = = b a b a [ f(pf ) + qf 2] b dx + λ a σf 2 dx [ (pff ) p(f ) 2 + qf 2] dx + λ = [ pff ] b b a a [ p(f ) 2 qf 2] dx + λ b a b a σf 2 dx σf 2 dx. Since f is an eigenfunction (nonzero solution of (3.2)) and σ(x) > for all x in (a, b), the coefficient of λ in the last term is strictly positive. Hence, we can solve for λ and obtain b [ p(f ) 2 qf 2] dx [ pff ] b a a λ =. (3.6) b a σf 2 dx This expression is known as the Rayleigh quotient Example. For the regular S L problem considered in Example 3.1, we have b a so (3.6) yields [ p(x)(f ) 2 (x) q(x)f 2 (x) ] dx = L (f ) 2 (x) dx, [ p(x)f(x)f (x) ] b a = [ f(x)f (x) ] L =, b a σ(x)f 2 (x) dx = L f 2 (x) dx >, L (f ) 2 (x) dx λ = L. (3.7) f 2 (x) dx It is clear that λ = if and only if f (x) = on [, L], which means that f(x) = const. But this is unacceptable because, according to the BCs, the only constant solution is the zero solution. Consequently, λ >, and the general solution of the equation in Example 3.1 is f(x) = C 1 cos ( λ x ) + C 2 sin ( λ x ), C 1, C 2 = const. (3.8) Using the condition f() =, we find that C 1 f(l) = leads to C 2 sin ( λ L ) =. =. Then the condition

58 3.1 Regular Sturm Liouville Problems 35 We cannot have C 2 = because this would imply that f =, and we are seeking nonzero solutions; therefore, we must conclude that sin ( λ L ) =, from which we obtain λ L = nπ, n = 1, 2,..., or λ n = ( ) 2 nπ, n = 1, 2,.... L These are the eigenvalues of the problem. The corresponding eigenfunctions (nonzero solutions) are f n (x) = sin nπx, n = 1, 2,.... L For convenience, we have taken C 2 = 1; any other nonzero value of C 2 would simply produce a multiple of sin(nπx/l). In other words, the space of eigenfunctions corresponding to each eigenvalue is one-dimensional. It is easy to see that the properties in Theorem 3.8 are satisfied. The orthogonality of the eigenfunctions on [, L] follows immediately from (2.5) since the integrand is an even function Example. The same technique is used to compute the eigenvalues and eigenfunctions of the S L problem in Example Inequality (3.7) remains valid, but now we cannot reject the case λ = since the functions f(x) = const corresponding to it satisfy both BCs. Taking, say, c = 1/2, we conclude that the problem has the eigenvalue-eigenfunction pair so λ =, f (x) = 1 2. As in Example 3.16, for λ > the general solution of the equation is (3.8), f (x) = C 1 λ sin ( λ x ) + C2 λ cos ( λ x ). The condition f () = immediately yields C 2 = ; therefore, f (L) = implies that C 1 λ sin ( λ L ) =. Since we want nonzero solutions, it follows that sin ( λ L ) =. This means that λl = nπ, n = 1, 2,..., so, by (3.8) with C 2 = and C 1 = 1, the eigenvalue-eigenfunction pairs are λ n = ( ) 2 nπ, f n (x) = cos nπx, n = 1, 2,.... L L We remark that the full set of eigenvalues and eigenfunctions (with f (x) = 1 instead of 1/2) of the problem can be written as above, but with n =, 1, 2,.... Once again, we note that the eigenvalues and eigenfunctions satisfy the properties in Theorem 3.8. The orthogonality of the latter on [, L] follows from the first formula (2.3) and (2.4).

59 36 3 Sturm Liouville Problems 3.18 Example. Suppose that h > in the S L problem considered in Example Then, writing the second BC of that problem in the form f (L) = hf(l), we see that [ p(x)f(x)f (x) ] b a = [ f(x)f (x) ] L Consequently, (3.6) yields = f(l)f (L) f()f () = hf 2 (L). with λ = if and only if λ = L (f ) 2 (x) dx + hf 2 (L), L f 2 (x) dx f (x) = on [, L], f(l) =. But this implies that f =, which is not acceptable. Hence, the eigenvalues of this S L problem are positive. To find them, we note that the general solution of the equation is again given by (3.8) and that the condition f() = leads to C 1 =. Using the second BC, namely, we arrive at f (L) + hf(l) =, C 2 [ λ cos ( λ L ) + h sin ( λ L )] =. For nonzero solutions, we must have C 2 = ; in other words, λ cos ( λ L ) + h sin ( λ L ) =. Since cos ( λ L ) (otherwise sin ( λ L ) would be zero as well, which is impossible), we can divide the above equality by cos ( λ L ) and conclude that λ must be a solution of the transcendental equation tan ( λ L ) λ = h = 1 ( ) λ L. hl Setting λ L = ζ, we see that the values of ζ can computed from the points of intersection of the graphs of the functions y = tan ζ, y = ζ hl.

60 3.1 Regular Sturm Liouville Problems 37 Figure 3.1 line). The graphs of y = tan ζ (heavy line) and y = ζ/(hl) (light As Fig. 3.1 shows, there are countably many such points ζ n, so this S L problem has an infinite sequence of positive eigenvalues ( ) 2 ζn λ n =, n = 1, 2,..., λ n as n, L with corresponding eigenfunctions (given by (3.8) with C 1 = and C 2 = 1) f n (x) = sin ( λn x ) = sin ζ nx, n = 1, 2,.... L By Theorem 3.8(iii), the set } f n is orthogonal on [, L] Example. For the S L problem introduced in Example 3.13, we use a slightly modified approach. Here, the roots of the characteristic equation s 2 + 2s + λ = are s 1 = λ, s 2 = 1 1 λ. Since the nature of the roots changes according as λ < 1, λ = 1, or λ > 1, we consider these cases separately. (i) If λ < 1, then 1 λ > ; hence, s 1 and s 2 are real and distinct, and the general solution of the equation is f(x) = C 1 e s1x + C 2 e s2x, C 1, C 2 = const. Using the conditions f() = and f(1) =, we arrive at C 1 + C 2 =, C 1 e s1 + C 2 e s2 =. Since s 1 = s 2, this system has only the solution C 1 = C 2 = ; consequently, f =, and we conclude that there are no eigenvalues λ < 1.

61 38 3 Sturm Liouville Problems (ii) If λ = 1, then f(x) = (C 1 + C 2 x)e x, C 1, C 2 = const. The condition f() = yields C 1 =, while f(1) = gives C 2 e 1 =, which means that C 2 =. This generates f =, so λ = 1 is not an eigenvalue. (iii) If λ > 1, then the general solution of the equation is f(x) = e x[ C 1 cos ( λ 1 x ) + C 2 sin ( λ 1 x )], C 1, C 2 = const. Since f() =, we find that C 1 =, and from the condition f(1) = it follows that C 2 e 1 sin λ 1 =. For nonzero solutions we must have sin λ 1 = ; in other words, λ 1 = nπ, n = 1, 2,.... This implies that the eigenvalues of the problem are with corresponding eigenfunctions λ n = n 2 π 2 + 1, n = 1, 2,..., f n (x) = e x sin ( λn 1 x ) = e x sin(nπx), n = 1, 2,.... Verification with Mathematica R. The input λn = n 2 Pi 2 + 1; fn = E ( - x) Sin[n Pi x]; D[fn,x,x] + 2 D[fn,x] + λn fn,fn /. x ->, fn /. x -> 1} // FullSimplify generates the output,, }. 3.2 Remark. In Example 3.19, we may have been tempted to discard case (i) without detailed investigation because the restriction λ < 1 seems to contradict Corollary 3.15, which states that the eigenvalues of a regular S L problem form a sequence whose terms increase without bound. In fact, for certain types of BCs and values of the constant coefficients a, b, and c, an equation of the form f (x) + af (x) + bf(x) + cλf(x) = may produce an eigenvalue-eigenfunction pair even when the roots of the characteristic equation are real and distinct.

62 3.1 Regular Sturm Liouville Problems Example. The characteristic equation for the regular S L problem f (x) + 2f (x) f(x) + λf(x) =, < x < 1, f() =, 2f(1) 3f (1) = is s 2 + 2s (1 λ) =, with roots s 1 = λ, s 2 = 1 2 λ. Suppose that λ < 2, which means that 2 λ >. Then s 1 and s 2 are real and distinct; therefore, the general solution of the equation can be written as f(x) = C 1 e s1x + C 2 e s2x. However, this leads to rather untidy computation, so instead we use Remark 1.4(ii) with the values α = β =, s 1 + s 2 = 2, and s 1 s 2 = 2 2 λ, and work with the alternative form f(x) = e x[ C 1 cosh ( 2 λ x ) + C 2 sinh ( 2 λ x )]. The BC at x = yields C 1 = ; hence, applying the BC at x = 1 (with, as before, a nonzero value for C 2 ), we arrive at the equality 5 sinh 2 λ 3 2 λ cosh 2 λ =, or, since the hyperbolic cosine does not vanish for any value of its argument, tanh 2 λ = λ. As the graphs of the functions defined by the two sides clearly show (see Fig. 3.2), this equation has only one root satisfying λ < 2, which is approximately Consequently, we have the eigenvalue-eigenfunction pair λ =.2868, f (x) = e x sinh ( 2 λ x ) = e x sinh( x). Figure 3.2 The graphs of y = tanh 2 λ (heavy line) and y = λ (light line). The cases λ = 2 and λ > 2 are treated as in Examples 3.19 and 3.18.

63 4 3 Sturm Liouville Problems Verification with Mathematica R. The input λ = ; f = E ( - x) Sinh[ x]; D[f,x,x] + 2 D[f,x] - f + λ f,f /. x ->, (2 f - 3 D[f,x])/. x -> 1} // FullSimplify generates, to four decimal places, the output,, } Remark. Using the same techniques as in Examples 3.13 and 3.19, we find that the eigenvalue problem f (x) + af (x) + bf(x) + cλf(x) =, < x < L, f() =, f(l) =, where a, b, c = const, c >, is a regular S L problem with eigenvalues and eigenfunctions p(x) = e ax, q(x) = be ax, σ(x) = ce ax, λ n = 1 4c ( 4n 2 π 2 L 2 ) + a 2 4b, n = 1, 2,..., f n (x) = e (a/2)x sin nπx, n = 1, 2,.... L Exercises In 1 1, compute the eigenvalues and eigenfunctions of the given regular S L problem. 1 f (x) + λf(x) =, < x < π, f() =, f (π) =. 2 f (x) + λf(x) =, < x < 1, f () =, f(1) =. 3 f (x) + λf(x) =, < x < 1, f () f() =, f(1) =. 4 f (x) + λf(x) =, < x < 1, f () =, f (1) + f(1) =.

64 3.1 Regular Sturm Liouville Problems 41 5 f (x) + λf(x) =, < x < 1, f () f() =, f (1) =. 6 2f (x) + (λ 1)f(x) =, < x < 2, f() =, f(2) =. 7 f (x) + 4f (x) + 3λf(x) =, < x < 1, f() =, f(1) =. 8 f (x) f (x) + λf(x) =, < x < 1, f () =, f (1) =. 9 2f (x) + 3f (x) + λf(x) =, < x < 1, f() =, f (1) =. 1 f (x) 5f (x) + 2λf(x) =, < x < π/2, f () =, f(π/2) =. In 11 16, show that the regular S L problem f (x) + af (x) + bf(x) + cλf(x) =, < x < 1, κ 1 f() + κ 2 f () =, κ 3 f(1) + κ 4 f (1) =, with the constants a, b, c and κ 1, κ 2, κ 3, κ 4 as indicated, has an exponentialtype eigenfunction, and compute the corresponding eigenvalue-eigenfunction pair, rounding all numbers to the fourth decimal place. 11 a = 3, b = 1, c = 2, κ 1 = 1, κ 2 =, κ 3 = 1, κ 4 = a = 4, b = 2, c = 1, κ 1 = 1, κ 2 =, κ 3 = 2, κ 4 = a = 2, b = 2, c = 5, κ 1 = 1, κ 2 =, κ 3 = 4, κ 4 = a = 1, b = 2, c = 2, κ 1 = 1, κ 2 =, κ 3 = 5, κ 4 = a = 1, b = 2, c = 5, κ 1 = 2, κ 2 = 1, κ 3 = 1, κ 4 =. 16 a = 3, b = 1, c = 1, κ 1 = 3, κ 2 = 2, κ 3 = 1, κ 4 =. Answers to Odd-Numbered Exercises 1 λ n = (2n 1) 2 /4, f n (x) = sin ( (2n 1)x/2 ), n = 1, 2, λ n = ζn 2, where ζ n are the roots of the equation tan ζ = ζ, f n (x) = ζ n cos(ζ n x) + sin(ζ n x), n = 1, 2, λ n = ζn, 2 where ζ n are the roots of the equation cot ζ = ζ, f n (x) = ζ n cos(ζ n x) + sin(ζ n x), n = 1, 2,....

65 42 3 Sturm Liouville Problems 7 λ n = (4 + n 2 π 2 )/3, f n (x) = e 2x sin(nπx), n = 1, 2, λ n = 2ζn 2 + 9/8, where ζ n are the roots of the equation tan ζ = 4ζ/3, f n (x) = e 3x/4 sin(ζ n x), n = 1, 2, λ =.286, f (x) = e 3x/2 sinh(1.915 x). 13 λ = , f (x) = e x sinh( x). 15 λ =.6817, f (x) = e x/2 sinh ( (x 1) ) Generalized Fourier Series Regular Sturm Liouville problems have two important additional properties Theorem. (i) Only one linearly independent eigenfunction f n exists for each eigenvalue λ n, n = 1, 2,.... (ii) The set } f n of eigenfunctions is complete; that is, any piecewise smooth function u on [a, b] has a unique generalized Fourier series (or eigenfunction) expansion of the form u(x) c n f n (x), a x b, (3.9) which converges pointwise to 1 2 [u(x ) + u(x+)] for a x b (in particular, to u(x) if u is continuous at x) Remarks. (i) The coefficients c n are computed by means of the properties mentioned in Theorem 3.23(i) and Theorem 3.8(iii). Thus, treating (3.9) formally as an equality, multiplying it by f m (x)σ(x), and integrating over [a, b], we obtain b a u(x)f m (x)σ(x) dx = = c m Replacing m by n, we can now write c n = b a c n b a b a f n (x)f m (x)σ(x) dx f 2 m(x)σ(x) dx. u(x)f n (x)σ(x) dx, n = 1, 2,.... (3.1) b a f 2 n(x)σ(x) dx

66 3.1 Regular Sturm Liouville Problems 43 The denominator in (3.1) cannot be zero for any n since the f n are nonzero solutions of (3.2) (3.4) and σ(x) >. (ii) In the case of the S L problems discussed in Examples 3.1 and 3.16, and in Examples 3.11 and 3.17, expansion (3.9) coincides, respectively, with the Fourier sine and cosine series of u, and (3.1) yields formulas (2.1) and (2.12). (For the problem discussed in Examples 3.11 and 3.17, we have n =, 1, 2,....) 3.25 Example. Suppose that we want to expand the function u(x) = x + 1, x 1, in the eigenfunctions of the regular S L problem discussed in Examples 3.12 and 3.18 with L = h = 1. The first five positive roots of the equation tan ζ = ζ mentioned in the latter, rounded to the fourth decimal place, are ζ 1 = 2.288, ζ 2 = , ζ 3 = , ζ 4 = , ζ 5 = Then, using (3.1) with σ(x) = 1 and f n (x) = sin(ζ n x), we find the approximate coefficients so (3.9) takes the form c 1 = , c 2 =.1572, c 3 =.399, c 4 =.137, c 5 =.1696, u(x) sin(2.288 x) sin( x) sin( x) sin( x) sin( x) +. (3.11) Verification with Mathematica R. The input (based on the computed expansion (3.11)) Approx = Sin[2.288 x] Sin[ x] Sin[ x] Sin[ x] Sin[ x]; Plot[Approx, x+1},x,,1}, PlotRange -> -.1,1.1}, -.2,2.2}}, PlotStyle -> Black, ImageSize -> Scaled[.22], Ticks ->1},1,2}}, AspectRatio ->.9]

67 44 3 Sturm Liouville Problems generates the graphical output 3.26 Example. Consider the function u(x) = e x, x 1. If f n are the eigenfunctions of the regular S L problem discussed in Examples 3.13 and 3.19, that is, then (3.1) with yields the coefficients f n (x) = e x sin(nπx), n = 1, 2,..., σ(x) = e 2x c n = [ 1 ( 1) n] 2, n = 1, 2,..., nπ so expansion (3.9) is [ u(x) 1 ( 1) n ] 2 nπ e x sin(nπx). (3.12) Verification with Mathematica R. The input (based on the computed expansion (3.12)) GenTerm = (1 - ( - 1) n) (2/(n Pi)) E ( - x) Sin[n Pi x]; Plot[GenTerm,n,1,7}],E ( - x)},x,,1}, PlotRange -> -.1,1.1}, -.2,1.2}}, PlotStyle -> Black, ImageSize -> Scaled[.22], Ticks ->1},1}}, AspectRatio ->.9] generates the graphical output

68 3.1 Regular Sturm Liouville Problems 45 Exercises In 1 8, construct the generalized Fourier series expansion for the given function u in the eigenfunctions sin ( (2n 1)x/2 )} of the S L problem in Exercise 1, Subsection u(x) = 1, x π. 2 u(x) = 2x 1, x π. 3 u(x) = 3x + 2, x π. 1, x π/2, 4 u(x) = 2, π/2 < x π. 2, x π/2, 5 u(x) = 3, π/2 < x π. 2x 1, x π/2, 6 u(x) = 1, π/2 < x π. 2, x π/2, 7 u(x) = x + 1, π/2 < x π. 8 u(x) = 2x + 1, x π/2, 3 2x, π/2 < x π. In 9 16, construct the generalized Fourier series expansion for the given function u in the eigenfunctions cos ( (2n 1)πx/2 )} of the S L problem in Exercise 2, Subsection u(x) = 1/2, x 1. 1 u(x) = x + 1, x u(x) = 2 x, x 1. 3, x 1/2, 12 u(x) = 4, 1/2 < x 1. 2, x 1/2, 13 u(x) = 1, 1/2 < x x, x 1/2, 14 u(x) = 2, 1/2 < x 1. 1, x 1/2, 15 u(x) = 2x 3, 1/2 < x 1.

69 46 3 Sturm Liouville Problems 16 u(x) = 1 x, x 1/2, 2x + 1, 1/2 < x 1. In 17 24, compute the first five terms of the generalized Fourier series expansion for the given function u in the eigenfunctions of the S L problem discussed in Example 3.18, with h = 1 and the interval for x as indicated, rounding all numbers to the fourth decimal place. 17 u(x) = 1, x u(x) = 1 2x, x u(x) = 2 x, x 2. 1, x 1/2, 2 u(x) = 3, 1/2 < x 1. 2, x 1/2, 21 u(x) = 4, 1/2 < x 1. 1, x 1, 22 u(x) = x + 2, 1 < x 2. 2x, x 1/2, 23 u(x) = 1, 1/2 < x 1. 1 x, x 1/2, 24 u(x) = 1 + x, 1/2 < x 1. In 25 32, compute the first five terms of the generalized Fourier series expansion for the given function u in the eigenfunctions e 2x sin(nπx) } of the S L problem discussed in Exercise 7, Subsection 3.1.1, rounding all numbers to the fourth decimal place. 25 u(x) = 1, x u(x) = x, x u(x) = 2x + 1, x 1. 2, x 1/2, 28 u(x) = 1, 1/2 < x 1. 1, x 1/2, 29 u(x) = 3, 1/2 < x 1. 2, x 1/2, 3 u(x) = 2x + 1, 1/2 < x 1.

70 31 u(x) = x 1, x 1/2, 1, 1/2 < x u(x) = x + 1, x 1/2, 3 2x, 1/2 < x Regular Sturm Liouville Problems 47 Answers to Odd-Numbered Exercises 1 u(x) [4/((2n 1)π)] sin((2n 1)x/2). 3 u(x) [8(2n 1 3( 1) n )/((2n 1) 2 π)] sin((2n 1)x/2). 5 u(x) [8 2 sin((3 2n)π/4)]/((2n 1)π)} sin((2n 1)x/2). 7 u(x) [2/((2n 1) 2 π)](2n 1)(π 2) sin((3 2n)π/4) 4[1 + ( 1) n 2n + sin((2n 1)π/4)]} sin((2n 1)x/2). 9 u(x) [( 1) n+1 2/((2n 1)π)] cos((2n 1)πx/2). 11 u(x) [(8 4( 1) n (2n 1)π)/((2n 1) 2 π 2 )] cos((2n 1)πx/2). 13 u(x) 4[( 1) n+1 3 sin((2n 1)π/4)]/((2n 1)π)} cos((2n 1)πx/2). 15 u(x) [4/((2n 1) 2 π 2 )]4(2n 1)π sin((2n 1)π/4) 4 sin((3 2n)π/4) + (2n 1)π[( 1) n sin((2n 1)π/4)]} cos((2n 1)πx/2). 17 u(x) sin(2.288 x) sin( x) sin( x) sin( x) sin( x) u(x).9639 sin( x) sin( x) sin(4.481 x) sin( x) sin( x) u(x) sin(2.288 x) sin( x) sin( x) sin( x) sin( x) u(x) sin(2.288 x) sin( x).157 sin( x).3785 sin( x) sin( x) u(x) e 2x [3.84 sin(πx) sin(2πx) sin(3πx).9917 sin(4πx) sin(5πx) + ].

71 48 3 Sturm Liouville Problems 27 u(x) e 2x [8.331 sin(πx) sin(2πx) sin(3πx) sin(4πx) sin(5πx) + ]. 29 u(x) e 2x [ sin(πx) sin(2πx) sin(3πx) sin(4πx) sin(5πx) + ] 31 u(x) e 2x [ sin(πx) sin(2πx) sin(3πx) sin(4πx) 1.14 sin(5πx) + ]. 3.2 Other Problems Many important Sturm Liouville problems are not regular Definition. With the notation in Definition 3.9, an eigenvalue problem that consists in solving equation (3.2) with the conditions is called a periodic Sturm Liouville problem. f(a) = f(b), f (a) = f (b) (3.13) 3.28 Remark. It can be shown that the differential operator L introduced in (3.5) is also symmetric on the new space of functions defined by conditions (3.13), so Theorem 3.8 is again applicable. However, in contrast to regular S L problems, here we may have two linearly independent eigenfunctions for the same eigenvalue. Nevertheless, we can always choose a pair of eigenfunctions that are orthogonal with weight σ on [a, b]. For such a choice, Theorem 3.23(ii) remains valid Example. Consider the periodic S L problem where p(x) = 1, q(x) =, σ(x) = 1, a = L, and b = L, L > ; that is, f (x) + λf(x) =, L < x < L, f( L) = f(l), f ( L) = f (L). Using the Rayleigh quotient argument and the above BCs, we arrive again at inequality (3.7). We accept the case λ = since, as in Example 3.17, the corresponding constant solutions satisfy both BCs; therefore, the problem has the eigenvalue-eigenfunction pair λ =, f (x) = 1 2. For λ >, the general solution of the equation is f(x) = C 1 cos ( λ x ) + C 2 sin ( λ x ), C 1, C 2 = const,

72 3.2 Other Problems 49 with derivative f (x) = C 1 λ sin ( λ x ) + C2 λ cos ( λ x ). From the BCs and the fact that cos( α) = cos α and sin( α) = sin α, it follows that C 1 and C 2 satisfy the system C 1 cos ( λ L ) C 2 sin ( λ L ) = C 1 cos ( λ L ) + C 2 sin ( λ L ), C 1 sin ( λ L ) + C 2 cos ( λ L ) = C 1 sin ( λ L ) + C 2 cos ( λ L ), which reduces to C 1 sin ( λ L ) =, C 2 sin ( λ L ) =. Since we want nonzero solutions, we cannot have sin ( λ L ) = because this would imply C 1 = C 2 = that is, f = ; hence, sin ( λ L ) =. As we have seen in previous examples, this yields the eigenvalues ( ) 2 nπ λ n =, n = 1, 2,.... L Given that both C 1 and C 2 remain arbitrary, from the general solution we can extract (by taking C 1 = 1, C 2 = and then C 1 =, C 2 = 1) for every λ n the two linearly independent eigenfunctions f 1n (x) = cos nπx L, f 2n(x) = sin nπx, n = 1, 2,..., L which, as noted in Chapter 2, are orthogonal on [ L, L]. Expansion (3.9) in this case becomes [ u(x) c f (x) + c1n f 1n (x) + c 2n f 2n (x) ] = 1 2 c + ( c 1n cos nπx L + c 2n sin nπx ), L which we recognize as the full Fourier series of u. The coefficients c, c 1n, and c 2n, computed by means of (3.1) with σ(x) = 1, a = L, b = L, and n =, 1, 2,... (to allow for the additional eigenfunction f ), coincide with those given by (2.7) (2.9). 3.3 Remark. It is easily shown that the eigenvalues and eigenfunctions computed in Example 3.29 remain the same if the interval L < x < L is replaced by < x < 2L or any other interval of length 2L Definition. An eigenvalue problem involving equation (3.2) is called a singular Sturm Liouville problem if it exhibits one or more of the following features:

73 5 3 Sturm Liouville Problems (i) p(a) = or p(b) = or both; (ii) any of p(x), q(x), and σ(x) becomes infinite as x a+ or as x b or both; (iii) the interval (a, b) is infinite at a or at b or at both Remark. In the case of a singular Sturm Liouville problem, some new boundary conditions need to be chosen to ensure that the operator L defined by (3.5) remains symmetric. For example, if p(a) = but p(b), then we may require f(x) and f (x) to remain bounded as x a+ and to satisfy (3.4). If the interval where the equation holds is of the form (a, ), then we may require f(x) and f (x) to be bounded as x and to satisfy (3.3). Other types of boundary conditions, dictated by analytic and/or physical necessity, may also be considered Example. The eigenvalue problem (x 2 + 1)f (x) + 2xf (x) + (λ x)f(x) =, x > 1, f (1) =, f(x), f (x) bounded as x, is a singular S L problem: the ODE can be written in the form (3.2) with p(x) = x 2 + 1, q(x) = x, σ(x) = 1, and (a, b) = (1, ), and the boundary conditions are of the type indicated in the latter part of Remark A number of singular S L problems occurring in the study of mathematical models give rise to important classes of functions. We discuss some of them explicitly. 3.3 Bessel Functions Consider the singular S L problem x 2 f (x) + xf (x) + (λx 2 m 2 )f(x) =, < x < a, (3.14) f(x), f (x) bounded as x +, f(a) =, (3.15) where a is a given positive number and m is a nonnegative integer. Writing (3.14) in the form [ xf (x) ] m 2 f(x) + λxf(x) =, x we verify that this is (3.2) with p(x) = x, q(x) = m 2 /x, and σ(x) = x. First, we consider the case m =, when q = and (3.14) reduces to xf (x) + f (x) + λxf(x) =, < x < a.

74 3.3 Bessel Functions 51 Taking (3.15) into account, we see that p(x)(f ) 2 (x) q(x)f 2 (x) = x(f ) 2 (x), < x < a, [ p(x)f(x)f (x) ] a = [ xf(x)f (x) ] a = af(a)f (a) lim xf(x)f (x) = ; x + consequently, according to the Rayleigh quotient (3.6), a x(f ) 2 (x) dx λ = a. xf 2 (x) dx In fact, as the formula above shows, we have λ = if and only if f (x) =, < x a; that is, if and only if f(x) = const. But this is impossible, since the boundary condition f(a) = would imply that f is the zero solution, which contradicts the definition of an eigenfunction. We must therefore conclude that λ >. Now let m 1. Multiplying (3.14) by f(x) and integrating from to a, we arrive at a [ x 2 f (x)f(x) + xf (x)f(x) + λx 2 f 2 (x) m 2 f 2 (x) ] dx =. (3.16) Integration by parts, the interpretation of f(x) and f (x) at x = in the limiting sense as above, and (3.15) yield the equalities a a xf (x)f(x) dx = a = x(f 2 (x)) dx = 1 [ 2 xf 2 (x) ] a 1 2 a f 2 (x) dx, a f 2 (x) dx x 2 f (x)f(x) dx = [ x 2 f (x)f(x) ] a a [ x 2 (f (x)) 2 + 2xf (x)f(x) ] dx = a which, replaced in (3.16), lead to [ x 2 (f (x)) 2 f 2 (x) ] dx, a [ x 2 (f (x)) 2 + ( ) m f 2 (x) ] dx λ = a. x 2 f 2 (x) dx The same argument used for m = rules out the value λ = ; hence, λ >.

75 52 3 Sturm Liouville Problems For m, let ξ = λ x and f(x) = f ( ξ/ λ ) = g(ξ). By the chain rule, and (3.14) becomes f (x) = λ g (ξ), f (x) = λg (ξ), ξ 2 g (ξ) + ξg (ξ) + (ξ 2 m 2 )g(ξ) =, (3.17) which is Bessel s equation of order m. Two linearly independent solutions of this equation are the Bessel functions of the first and second kind and order m, denoted by J m and Y m Remarks. (i) For m 1, the J m satisfy the recurrence relations J m 1 (x) + J m+1 (x) = 2m x J m(x), J m 1 (x) J m+1 (x) = 2J m (x), from which we deduce that [ x m J m (x) ] = x m J m 1 (x). Similar equalities are satisfied by the Y m. (ii) The J m and Y m can also be computed by means of the Bessel integrals J m (x) = 1 π Y m (x) = 1 π π π (iii) As x +, cos(mτ x sin τ) dτ = 1 2π sin(x sin τ mτ) dτ 1 π π π e i(mτ x sin τ) dτ, J (x) 1, J m (x), m = 1, 2,..., Y m (x), m =, 1, 2,.... (iv) J m has the Taylor series expansion ( 1) k ( ) 2k+m x J m (x) =. k!(k + m)! 2 k= [ e mτ + ( 1) m e mτ] e x sinh τ dτ. (v) The Bessel functions can be defined for any real, and even complex, order m. For example, when the order is a negative integer m, we set J m (x) = ( 1) m J m (x). (vi) When the order is an integer, the J m are generated by the function e (x/2)(t 1/t) = J m (x)t m. m=

76 3.3 Bessel Functions 53 The graphs of J, J 1 and Y, Y 1 are shown in Fig Figure 3.3 Left: J (light line) and J 1 (heavy line). Right: Y (light line) and Y 1 (heavy line). Going back to (3.17), we can write its general solution as g(ξ) = C 1 J m (ξ) + C 2 Y m (ξ), C 1, C 2 = const, from which f(x) = C 1 J m ( λ x ) + C2 Y m ( λ x ). In view of Remark 3.34(iii), the first condition (3.15) implies that C 2 =. The second one then yields C 1 J m ( λ a ) =. Since we want nonzero solutions, we must have J m ( λ a ) =. The function J m has infinitely many positive zeros, which form a sequence ξ mn, n = 1, 2,..., such that ξ mn as n for each m =, 1, 2,... ; therefore, the singular S L problem (3.14), (3.15) has the eigenvalue-eigenfunction pairs λ mn = ( ξmn a ) 2, f mn(x) = J m ( ξmn x a ), n = 1, 2,.... It turns out that } f mn is a complete system for each m =, 1, 2,..., and that a ( ) ( ξmn x ξmp x J m J m a a = ) x dx, n p, 1 2 a2 Jm+1 2 (ξ mn), n = p, m =, 1, 2,.... (3.18) These orthogonality (with weight σ(x) = x) relations allow us to expand any suitable function u in a generalized Fourier series of the form ( ) ξmn x u(x) c mn J m (3.19) a

77 54 3 Sturm Liouville Problems for each m =, 1, 2,.... To find the coefficients c mn, we follow the procedure described in Remark 3.24(i) with (a, b) replaced by (, a), c n by c mn, f n (x) by J m (ξ mn x/a), and σ(x) by x. Making use of (3.18), in the end we find that c mn = a u(x)j m (ξ mn x/a)x dx 1 2 a2 Jm+1 2 (ξ, n = 1, 2,.... (3.2) mn) 3.35 Remark. Series (3.19) converges to u(x) at all points x where u is continuous in the interval (, a). If m =, the point x = is also included. If m >, then x = is included if u() =. The point x = a is included for any m if u(a) = Example. Let u(x) = 2x 1, and let a = 1 and m =. The values, rounded to the fourth decimal place, of the first five zeros of J (ξ) are ξ 1 = 2.448, ξ 2 = 5.521, ξ 3 = , ξ 4 = , ξ 5 = , which, when used in (3.2), generate the coefficients c 1 =.329, c 2 = , c 3 =.7452, c 4 =.7644, c 5 = Therefore, by (3.19), we have the expansion u(x).329 J (2.448 x) J (5.521 x) J ( x).7644 J ( x) J ( x) +. (3.21) We now construct a second expansion for u by taking m = 1. The approximate values of the first five zeros of J 1 are ξ 11 = , ξ 12 = 7.156, ξ 13 = , ξ 14 = , ξ 15 = , so, by (3.2), we arrive at the coefficients c 11 =.3788, c 12 = , c 13 =.47, and the corresponding expansion c 14 =.9199, c 15 =.4248 u(x).3788 J 1 ( x) J 1 (7.156 x) +.47 J 1 ( x).9199 J 1 ( x) J 1 ( x) +. (3.22)

78 3.3 Bessel Functions 55 Verification with Mathematica R. The input (based on (3.21) and (3.22)) ApproxJ =.329 BesselJ[,2.448 x] BesselJ[,5.521 x] BesselJ[, x] BesselJ[, x] BesselJ[, x]; p = Plot[ApproxJ,2 x - 1},x,,1},PlotRange ->-.1,1.1}, - 1.2,1.2}}, PlotStyle -> Black, ImageSize -> Scaled[.25], Ticks ->1}, - 1,1}},AspectRatio ->.9]; ApproxJ1 =.3788 BesselJ[1, x] BesselJ[1,7.156 x] +.47 BesselJ[1, x] BesselJ[1, x] BesselJ[1, x]; p1 = Plot[ApproxJ1,2 x - 1},x,,1},PlotRange -> -.1,1.1}, - 1.2,1.2}}, PlotStyle -> Black, ImageSize -> Scaled[.25], Ticks ->1}, - 1,1}},AspectRatio ->.9]; Show[GraphicsRow[p,p1}]] generates the graphical output Exercises Compute the first five terms of the generalized Fourier series expansion for the given function u in the eigenfunctions constructed with the Bessel functions (i) J and (ii) J 1. 1 u(x) = 1, x 1. 2 u(x) = 2x + 1, x 1.

79 56 3 Sturm Liouville Problems 3 u(x) = 1 3x, x 1. 4 u(x) = 3, x 1/2, 1, 1/2 < x 1. 5 u(x) = 1, x 1/2, 2, 1/2 < x 1. 6 u(x) = x + 3, x 1/2, 2, 1/2 < x 1. 7 u(x) = 1, x 1/2, 2 2x, 1/2 < x 1. 8 u(x) = x 2, x 1/2, x + 1, 1/2 < x 1. Answers to Odd-Numbered Exercises 1 (i) u(x) 1.62 J (2.448 x) J (5.521 x) J ( x).7296 J ( x) J ( x) +. (ii) u(x) J 1 ( x).5171 J 1 (7.156 x) J 1 ( x).455 J 1 ( x) J 1 ( x) +. 3 (i) u(x).853 J (2.448 x) J (5.521 x) J ( x) J ( x) J ( x) +. (ii) u(x) J 1 ( x) J 1 (7.156 x) J 1 ( x) J 1 ( x) J 1 ( x) +. 5 (i) u(x).8947 J (2.448 x) 4.54 J (5.521 x) J ( x).663 J ( x) +.79 J ( x) +. (ii) u(x) J 1 ( x) J 1 (7.156 x) J 1 ( x).1883 J 1 ( x) J 1 ( x) +. 7 (i) u(x) J (2.448 x).2378 J (5.521 x).983 J ( x) J ( x) +.62 J ( x) +. (ii) u(x) J 1 ( x)+.436 J 1 (7.156 x)+.291 J 1 ( x) J 1 ( x) J 1 ( x) +.

80 3.4 Legendre Polynomials Legendre Polynomials Consider the singular S L problem (1 x 2 )f (x) 2xf (x) + λf(x) =, 1 < x < 1, (3.23) f(x), f (x) bounded as x 1+ and as x 1. (3.24) Since (3.23) can be written as [ (1 x 2 )f (x) ] + λf(x) =, it is clear that p(x) = 1 x 2, q(x) =, and σ(x) = 1. A detailed analysis of problem (3.23), (3.24), which is beyond the scope of this book, shows that its eigenvalues are λ n = n(n + 1), n =, 1, 2,..., and that the general solution of (3.23) with λ = λ n is of the form f n (x) = C 1 P n (x) + C 2 Q n (x), C 1, C 2 = const, (3.25) where P n are polynomials of degree n, called the Legendre polynomials. They are computed by means of the formula where P n (x) = 1 2 n s ( 1) k k= k! s = (2n 2k)! (n 2k)!(n k)! xn 2k, n =, 1, 2,..., n/2, n even, (n 1)/2, n odd. The Q n, called the Legendre functions of the second kind, have the representation dx Q n (x) = P n (x) (1 x 2 )Pn(x) 2, n =, 1, 2, Remarks. (i) The P n are given by the Rodrigues formula P n (x) = 1 2 n n! thus, for n =, 1,... 5, d n dx n (x2 1) n, n =, 1, 2,... ; P (x) = 1, P 1 (x) = x, P 2 (x) = 1 2 ( 1 + 3x2 ), P 3 (x) = 1 2 ( 3x + 5x3 ), P 4 (x) = 1 8 (3 3x2 + 35x 4 ), P 5 (x) = 1 8 (15x 7x3 + 63x 5 ).

81 58 3 Sturm Liouville Problems (ii) The P n may also be computed by means of the generating function (1 2tx + t 2 ) 1/2 = P n (x)t n, x < 1, t < 1. n= (iii) The P n satisfy the recursive relation (n + 1)P n+1 (x) (2n + 1)xP n (x) + np n 1 (x) =, n = 1, 2,.... (iv) The set P n } n= is orthogonal over [ 1, 1]; specifically, 1, n = m, P n (x)p m (x) dx = 2 2n + 1, n = m. (3.26) 1 (v) } P n is a complete system. n= (vi) Q n (x) becomes unbounded as x 1+ and as x 1. The graphs of P 4, P 5 and Q 4, Q 5 are shown in Fig Figure 3.4 Left: P 4 (light line) and P 5 (heavy line). Right: Q 4 (light line) and Q 5 (heavy line). In view of Remark 3.37(vi), f n given by (3.25) satisfies the BCs (3.24) only if C 2 = ; therefore, the eigenfunctions of the S L problem (3.23), (3.24) are f n (x) = P n (x), n =, 1, 2,.... As before, Remark 3.37(v) allows us to expand any suitable function u in a series of the form u(x) c n P n (x), (3.27) n= where, applying the technique described in Remark 3.24(i) and taking (3.26) into account, we see that the coefficients c n are computed by means of the formula c n = 2n u(x)p n (x) dx, n =, 1, 2,.... (3.28)

82 3.4 Legendre Polynomials Remark. If u is piecewise smooth on ( 1, 1), series [ (3.27) converges ] to u(x) at all points x where u is continuous, and to 1 2 u(x ) + u(x+) at the points x where u has a jump discontinuity Example. Consider the function 2x + 1, 1 x, u(x) = 3, < x 1. By (3.28) and the expressions of the P n in Remark 3.37(i), we find the coefficients c = 3 2, c 1 = 5 2, c 2 = 5 8, c 3 = 7 8, c 4 = 3 16, c 5 = 11 16, so (3.27) yields the expansion u(x) 3 2 P (x) P 1(x) 5 8 P 2(x) 7 8 P 3(x) P 4(x) P 5(x) +. (3.29) Verification with Mathematica R. The input (based on (3.29)) u = Piecewise[2*x + 1, - 1<x<},3, <x<1}}]; Approx = (3/2) LegendreP[,x] + (5/2) LegendreP[1,x] - (5/8) LegendreP[2,x] - (7/8) LegendreP[3,x] + (3/16) LegendreP[4,x] + (11/16) LegendreP[5,x]; Plot[Approx,u},x, - 1,1},PlotRange -> - 1.1,1.1}, - 1.2,3.4}}, PlotStyle -> Black, ImageSize -> Scaled[.25], Ticks -> - 1,1}, - 1,1,3}},AspectRatio ->.5] generates the graphical output Exercises Compute the first six terms of the generalized Fourier series expansion for the given function u, in the Legendre polynomials P n.

83 6 3 Sturm Liouville Problems 1 u(x) = x 2 3x + 4, 1 x 1. 2 u(x) = x 3 + 2x 2 3, 1 x 1. 3 u(x) = 3, 1 x, 1, < x 1. 4 u(x) = 2, 1 x, 5, < x 1. 5 u(x) = 2, 1 x, x 1, < x 1. 6 u(x) = 2x + 3, 1 x, 1, < x 1. 7 u(x) = 2x + 3, 1 x, 1 2x, < x 1. 8 u(x) = 3x + 1, 1 x, x + 4, < x 1. Answers to Odd-Numbered Exercises 1 u(x) = (13/3)P (x) 3P 1 (x) + (2/3)P 2 (x). 3 u(x) P (x) + 3P 1 (x) (7/4)P 3 (x) + (11/8)P 5 (x) +. 5 u(x) (5/4)P (x) + (5/4)P 1 (x) + (5/16)P 2 (x) (7/16)P 3 (x) (3/32)P 4 (x) + (11/32)P 5 (x) +. 7 u(x) P (x) (3/2)P 1 (x) (5/4)P 2 (x) + (7/8)P 3 (x) + (3/8)P 4 (x) (11/16)P 5 (x) Spherical Harmonics A more general form of the singular S L problem (3.23), (3.24) is obtained if (3.23) is replaced by the associated Legendre equation [ (1 x 2 )f (x) ] ) + (λ m2 1 x 2 f(x) =, 1 < x < 1, (3.3) where m is a nonnegative integer. Obviously, here we have p(x) = 1 x 2, q(x) = m 2 /(1 x 2 ), and σ(x) = 1.

84 3.5 Spherical Harmonics 61 It can be shown that the eigenvalue-eigenfunction pairs of problem (3.3), (3.24) are λ n = n(n + 1), n = m, m + 1,..., f n (x) = Pn m (x) = (1 x2 m/2 dm ) dx m P n(x) = ( 1)m 2 n n! (1 x 2 m/2 dn+m [ ) (x 2 dx n+m 1) n]. (3.31) For each m =, 1,..., n, the Pn m, called the associated Legendre functions, satisfy the orthogonality formula 1, n k, Pn m (x)p k m (x) dx = 2(n + m)! (2n + 1)(n m)!, n = k. (3.32) 1 The definition of the Pn m can be extended to negative-integer superscripts by means of the last expression on the right-hand side in (3.31). A straightforward computation shows that Pn m (x) = ( 1) m 2 n (1 x 2 m/2 dn m [ ) (x 2 n! dx n m 1) n] m (n m)! = ( 1) (n + m)! P n m (x). (3.33) The spherical harmonics are complex-valued functions (see Section 14.1) that occur in problems formulated in terms of spherical coordinates, which, as is well known, are linked to the Cartesian coordinates x, y, z by the expressions x = r cos θ sin ϕ, y = r sin θ sin ϕ, z = r cos ϕ. Here, r, r, is the radius, θ, θ < 2π, is the azimuthal angle, and ϕ, ϕ π, is the polar angle. The spherical harmonics are defined by Y m n (θ, ϕ) = P m n (cos ϕ)e imθ, n =, 1, 2,..., where m =, 1,..., n and, by Euler s formula, e imθ = cos(mθ) + i sin(mθ). In view of (3.32), it is not difficult to verify that 2π π 4π(n + m)! Yn m l, m = l, n = k, (θ, ϕ)ȳk (θ, ϕ) sin ϕ dϕ dθ = (2n + 1)(n m)! otherwise, where the superposed bar denotes complex conjugation. The above equality allows us to normalize the Yn m by writing [ ] 1/2 (2n + 1)(n m)! Y n,m (θ, ϕ) = Pn m 4π(n + m)! (cos ϕ)eimθ. (3.34)

85 62 3 Sturm Liouville Problems This is the form by which the spherical harmonics are usually known. The Y n,m satisfy the orthonormality formula 2π π 1, m = l, n = k, Y n,m (θ, ϕ)ȳk,l(θ, ϕ) sin ϕ dϕ dθ = (3.35) otherwise. In view of (3.33), for nonnegative integers m and n we can also define Y n, m (θ, ϕ) = ( 1) m Ȳ n,m (θ, ϕ). (3.36) 3.4 Example. By the second formula (3.31), (3.34), and (3.36), Y, (θ, ϕ) = π, Y 1, 1 (θ, ϕ) = π e iθ sin ϕ = 1 3 x iy, 2 2π r Y 1, (θ, ϕ) = π cos ϕ = 1 3 z 2 π r, Y 1,1 (θ, ϕ) = π eiθ sin ϕ = 1 3 x + iy, 2 2π r Y 2, 2 (θ, ϕ) = π e 2iθ sin 2 ϕ = 1 15 (x iy) 2 4 2π r 2, Y 2, 1 (θ, ϕ) = π e iθ sin ϕ cos ϕ = 1 15 (x iy)z 2 2π r 2, Y 2, (θ, ϕ) = π (3 cos2 ϕ 1) = 1 5 2z 2 x 2 y 2 4 π r 2, Y 2,1 (θ, ϕ) = π eiθ sin ϕ cos ϕ = 1 15 (x + iy)z 2 2π r 2, Y 2,2 (θ, ϕ) = π e2iθ sin 2 ϕ = (x + iy) 2 2π r Remark. Any suitable function u defined on a sphere can be expanded in a spherical harmonics series of the form n u(θ, ϕ) c n,m Y n,m (θ, ϕ), (3.37) n= m= n which converges to u(θ, ϕ) at all points where u is continuous. To find the coefficients c k,l, we multiply every term in (3.37) by Ȳk,l(θ, ϕ) sin ϕ, integrate over [, 2π] [, π], and take (3.35) into account. This (after k, l are replaced by n, m) leads to c n,m = 2π π u(θ, ϕ)ȳn,m(θ, ϕ) sin ϕ dϕ dθ. (3.38)

86 3.5 Spherical Harmonics Example. Let v(x, y, z) = x 2 + 2y be defined on the sphere with center at the origin and radius 1. In terms of spherical coordinates, this becomes u(θ, ϕ) = cos 2 θ sin 2 ϕ + 2 sin θ sin ϕ. Then, using (3.38) and the explicit expressions of the Y n,m given in Example 3.4, we find that c, = 2 π 2π 3, c 1, =, c 1,1 = c 1, 1 = 2 3 i, c 2, 1 = c 2,1 =, c 2, = 2 π 2π 3 5, c 2,2 = c 2, 2 = 15, so, by (3.37), u(θ, ϕ) = 2 π 3 2π Y, (θ, ϕ) i Y 1, 1(θ, ϕ) 2π 2π i Y 1,1(θ, ϕ) + 15 Y 2, 2(θ, ϕ) 2 π 3 5 Y 2,(θ, ϕ) + 2π 15 Y 2,2(θ, ϕ). (3.39) This is a terminating series because f is a linear combination of the spherical harmonics listed in Example 3.4. Verification with Mathematica R. The input (based on (3.39)) (2 Pi (1/2)/3) SphericalHarmonicY[,,ϕ,θ]] + 2 ((2 Pi/3) (1/2)) I SphericalHarmonicY[1, - 1,ϕ,θ] + 2 ((2 Pi/3) (1/2))as I SphericalHarmonicY[1,1,ϕ,θ] + ((2 Pi/15) (1/2)) SphericalHarmonicY[2, - 2,ϕ,θ] - ((2/3) (Pi/5) (1/2)) SphericalHarmonicY[2,,ϕ,θ] + ((2 Pi/15) (1/2)) SphericalHarmonicY[2,2,ϕ,θ] // FullSimplify generates the output cos 2 θ sin 2 ϕ + 2 sin θ sin ϕ, which confirms the statement above Example. If the function in Example 3.42 is replaced by x, z, v(x, y, z) = yz, z <, then its equivalent form in spherical coordinates is cos θ sin ϕ, ϕ π/2, u(θ, ϕ) = sin θ sin ϕ cos ϕ, π/2 ϕ π.

87 64 3 Sturm Liouville Problems In this case, (3.38) yields the coefficients π c, = c 1, =, c 1,1 = c 1, 1 = ( i ), π c 2, 2 = c 2, = c 2,2 =, c 2,1 = c 2, 1 = 3 (15 8 i ), so expansion (3.37) becomes π u(θ, ϕ) ( 1 3 π 6 8 i ) Y 1, 1 (θ, ϕ) ( i ) Y 1,1 (θ, ϕ) π + 3 ( i ) Y 2, 1 (θ, ϕ) π 3 (15 8 i ) Y 2,1 (θ, ϕ) +. (3.4) Verification with Mathematica R. The input (based on (3.4)) u = Piecewise[Cos[θ] Sin[ϕ],<θ<2 Pi && <ϕ<pi/2}, Sin[θ] Sin[ϕ] Cos[ϕ],<θ<2 Pi,&& Pi/2<ϕ<Pi}}]; approx = Plot3D[ComplexExpand[(Pi/6) (1/2) (1 - (3/8) I) SphericalHarmonicY[1, - 1,ϕ,θ] - (Pi/6) (1/2) (1 + (3/8) I) SphericalHarmonicY[1,1,ϕ,θ] + (Pi/3) (1/2) (15/8 + I) SphericalHarmonicY[2, - 1,ϕ,θ] - (Pi/3) (1/2) (15/8 - I) SphericalHarmonicY[2,1,ϕ,θ],ϕ,,Pi},θ,,2 Pi}, ImageSize -> Scaled[.3], Ticks ->,2,4,6},,1,2,3}, -.8,,.8}}]; Graphu = Plot3D[u,ϕ,,Pi},θ,,2 Pi}, ImageSize -> Scaled[.3], Ticks ->,2,4,6},,1,2,3},-1,,1}}]; GraphicsRow[Graphu, approx}] generates the graphical output As can be seen, although the approximation is the sum of only the four terms in (3.4), its graph shows reasonable agreement with that of u away from the discontinuity line ϕ = π/2 of the latter. The approximating function is, of course, continuous across that line.

88 3.5 Spherical Harmonics Remark. A very general class of what we have called suitable functions in this chapter consists of square-integrable functions, namely, functions u such that u(x) 2 dx <, D where D is the domain where the functions are defined and x and dx are, respectively, a generic point in D and the element of length/area/volume, as appropriate. If series (3.9), (3.19), (3.27), and (3.37) are written collectively in the form u(x) c n f n (x), then they are convergent in the sense that lim N D N u(x) 2 c n f n (x) dx =. As already mentioned, when u is sufficiently smooth, the series also converge to u(x) in the classical pointwise sense at every x in D where u is continuous. To keep the notation simple, since the functions of interest to us meet the smoothness requirement, from now on we use the equality sign between them and their generalized Fourier series representations, although, strictly speaking, equality holds only at their points of continuity. Exercises The given function v is defined on the sphere with center at the origin and radius 1. Writing it as u(θ, ϕ) in terms of spherical coordinates, compute the first nine terms of its expansion in the spherical harmonics. (Use the spherical harmonics listed in Example 3.4.) 1 v(x, y, z) = y 2 2z 2. 2 v(x, y, z) = 2 + xz. 3x 3 v(x, y, z) = 1, z,, z <. 4 v(x, y, z) = 2, z, 1, z <. 5 v(x, y, z) = 2, z, x y, z <.

89 66 3 Sturm Liouville Problems 6 v(x, y, z) = 2x + z, z, 1, z <. 7 v(x, y, z) = 2y + yz, z, x 2 3y, z <. 8 v(x, y, z) = xz + z 2, z, xy, z <. Answers to Odd-Numbered Exercises 1 u(θ, ϕ) = (2 π/3)y, (θ, ϕ) 2π/15 Y 2, 2 (θ, ϕ) (2 5π/3)Y 2, (θ, ϕ) 2π/15 Y 2,2 (θ, ϕ). 3 u(θ, ϕ) π Y, (θ, ϕ) + ( 3π/2)Y 1, (θ, ϕ) +. 5 u(θ, ϕ) 2 π Y, (θ, ϕ) + π/6 (1 i)y 1, 1 (θ, ϕ) + 3π Y 1, (θ, ϕ) π/6 (1 + i)y 1,1 (θ, ϕ) (1/8) 15π/2 (1 i)y 2, 1 (θ, ϕ) + (1/8) 15π/2 (1 + i)y 2,1 (θ, ϕ) +. 7 u(θ, ϕ) ( π/3)y, (θ, ϕ) (5/8) π/6 iy 1, 1 (θ, ϕ) ( 3π/8)Y 1, (θ, ϕ) (5/8) π/6 iy 1,1 (θ, ϕ) + π/3 Y 2, 2 (θ, ϕ) + (83/8) π/3 iy 2, 1 (θ, ϕ) (1/3) π/5 Y 2, (θ, ϕ) + (83/8) π/3 iy 2,1 (θ, ϕ) + π/3 Y 2,2 (θ, ϕ) +.

90 C H A P T E R 4 Some Fundamental Equations of Mathematical Physics The great majority of processes and phenomena in the real world are studied by means of mathematical models. An investigation of this kind generally consists of three stages. (i) The model is set up in terms of mathematical expressions describing the quantitative relationships between the physical quantities involved. (ii) The equations of the model are solved by means of various mathematical methods. (iii) The mathematical results are interpreted from a physical point of view in relation to the original process. In this chapter, we show how three simple but fundamental mathematical models are derived, which involve partial differential equations. These models are important because each of them is representative of an entire class of linear second-order PDEs. 4.1 Definition. A partial differential equation is an equation that contains an unknown function of several variables and one or more of its partial derivatives. To simplify the notation, we will denote the partial derivatives of functions almost exclusively by means of subscripts; thus, for u = u(x, t), and so on. u tt 2 u t 2, u t u t, u x u x, u xt 2 u x t, u xx 2 u x 2, 67

91 68 4 Some Fundamental Equations of Mathematical Physics 4.1 The Heat Equation Heat conduction in a one-dimensional rod. Consider a heatconducting rod in the shape of a thin cylinder described by the following physical parameters: L : length; A : cross-sectional area, assumed constant; E(x, t) : heat energy density (heat energy per unit volume); ϕ(x, t) : heat flux (heat energy per unit area, flowing to the right per unit time); q(x, t) : heat sources or sinks (heat energy per unit volume, generated or lost inside the rod per unit time). Throughout what follows, we assume that the lateral (cylindrical) surface of the rod is insulated; that is, no heat exchange takes place across it. We also use the generic term sources to describe both sources and sinks. The physical law we are using to set up the mathematical model in this case is the law of conservation of heat energy, which states that rate of change of heat energy in body = heat flow across boundary per unit time + heat generated by sources per unit time. Figure 4.1 An arbitrary segment of the rod. For an arbitrary length of the rod between x = a and x = b on the x-axis (see Fig. 4.1), this law translates as d b E(x, t)a dx = [ ϕ(a, t) ϕ(b, t) ] b A + q(x, t)a dx dt a a b = a b ϕ x (x, t) A dx + a q(x, t)a dx, t >.

92 4.1 The Heat Equation 69 Dividing through by A and moving all the terms to the left-hand side, we find that for t > and any a, b, < a < b < L, b [ Et (x, t) + ϕ x (x, t) q(x, t) ] dx =. a In view of the arbitrariness of a and b, this means that E t (x, t) = ϕ x (x, t) + q(x, t), < x < L, t >. (4.1) We need to introduce further physical parameters for the rod; thus, u(x, t) : temperature; c(x) : specific heat (the heat energy that raises the temperature of one unit of mass by one unit), assumed, for simplicity, to be independent of temperature; K (x) : thermal conductivity, also assumed to be independent of temperature; ρ(x) : mass density. Then the total heat energy in the section of rod between x = a and x = b is b a E(x, t)a dx = as before, this means that b a c(x)ρ(x)u(x, t)a dx, t > ; E(x, t) = c(x)ρ(x)u(x, t), < x < L, t >. (4.2) At the same time, according to Fourier s law of heat conduction, ϕ(x, t) = K (x)u x (x, t), < x < L, t >. (4.3) Using (4.2) and (4.3), we can now write the conservation of heat energy expressed by (4.1) as c(x)ρ(x)u t (x, t) = ( K (x)u x (x, t) ) + q(x, t), < x < L, t >. (4.4) x Assuming that the rod is uniform (c, ρ, and K are constants) and that there are no internal heat sources (q = ), we see that (4.4) reduces to u t (x, t) = ku xx (x, t), < x < L, t >, (4.5) where k = K /(cρ) = const is the thermal diffusivity of the rod. Equation (4.5) is called the heat (diffusion) equation.

93 7 4 Some Fundamental Equations of Mathematical Physics If sources are present in the rod, then this equation is replaced by its nonhomogeneous version u t (x, t) = ku xx (x, t) + q(x, t), < x < L, t >, where q now incorporates the constant 1/(cρ). Initial condition. Since (4.5) is a first-order equation with respect to time, it needs only one initial condition, which is normally taken to be u(x, ) = f(x), < x < L. This means prescribing the initial distribution of temperature in the rod. Boundary conditions. The equation is of second order with respect to the space variable, so we need two boundary conditions. There are three main types of physically meaningful conditions that are usually prescribed at the near endpoint x = and far endpoint x = L. (i) The temperature may be given at one endpoint; for example, u(, t) = α(t), t >. (ii) If the rod is insulated at an endpoint, then the condition must mean that the heat flux there is zero. In view of (4.3), this is equivalent to the derivative u x being equal to zero; for example, u x (L, t) =, t >. More generally, if a (nonzero) heat flux through the endpoint x = L is prescribed, then the above condition is replaced by u x (L, t) = β(t), t >. (iii) When one of the endpoints is in contact with another medium, we use Newton s law of cooling, which states that the heat flux at that endpoint is proportional to the difference between the temperature of the rod and the temperature of the external medium; for example, K u x (, t) = H [ u(, t) U(t) ], t >, where U(t) is the (known) temperature of the external medium and H > is the heat transfer coefficient. Owing to the convention concerning the direction of the heat flux, at the far endpoint this type of condition becomes K u x (L, t) = H [ u(l, t) U(t) ], t >. 4.2 Remarks. (i) Only one boundary condition is prescribed at each endpoint. (ii) The boundary condition at x = may differ from that at x = L. (iii) It is easily verified that the heat equation is linear. (iv) The one-dimensional heat equation is the simplest example of a socalled parabolic equation.

94 4.1 The Heat Equation 71 The model. As we have seen, the mathematical model for heat conduction in a rod consists of several elements. 4.3 Definition. A partial differential equation (PDE) and the initial conditions (ICs) and boundary conditions (BCs) associated with it form an initial boundary value problem (IBVP). If only initial conditions or boundary conditions are present, then we have an initial value problem (IVP) or a boundary value problem (BVP), respectively. 4.4 Example. The IBVP modeling heat conduction in a one-dimensional uniform rod with sources, insulated lateral surface, and temperature prescribed at both endpoints is of the form u t (x, t) = ku xx (x, t) + q(x, t), < x < L, t >, (PDE) u(, t) = α(t), u(l, t) = β(t), t >, (BCs) u(x, ) = f(x), < x < L, (IC) where q, α, β, and f are given functions. 4.5 Example. If the near endpoint is insulated and the far one is kept in a medium of constant zero temperature, and if the rod contains no sources, then the corresponding IBVP is u t (x, t) = ku xx (x, t), < x < L, t >, (PDE) u x (, t) =, u x (L, t) + hu(l, t) =, t >, (BCs) u(x, ) = f(x), < x < L, (IC) where f is a given function and h is a known positive constant. 4.6 Definition. A classical solution of an IBVP is a function u that satisfies the PDE, BCs, and ICs pointwise everywhere in the region where the problem is formulated. For brevity, in what follows we refer to such functions simply as solutions. It is obvious that an IBVP has solutions in this sense only if the data functions have a certain degree of smoothness. 4.7 Example. Suppose that q, α, β, and f in Example 4.4 are continuous functions. Then a solution of that IBVP has the following properties: (i) It is continuously differentiable with respect to t and twice continuously differentiable with respect to x at all points in the domain (semi-infinite strip) in the (x, t)-plane defined by G = (x, t) : < x < L, t > }, and satisfies the PDE at all points in G. (ii) Its continuity extends to the spatial boundary lines of G, that is, to the two half-lines G x = (x, t) : x =, t > or x = L, t > }, and it satisfies the appropriate BC at every point on G x.

95 72 4 Some Fundamental Equations of Mathematical Physics (iii) Its continuity also extends to the temporal boundary line of G, namely G t = (x, t) : < x < L, t = }, and it satisfies the IC at all points on G t. 4.8 Remarks. (i) We note that nothing is said in Example 4.7 about the behavior of u at the corner points (, ) and (L, ) of G. If, for example, lim u(, t) = lim u(x, ), t x that is, if lim α(t) = lim f(x), t x then u can be defined at (, ) by the common value of the above limits. If, on the other hand, the two limits are distinct, then u cannot be defined at (, ) in the classical sense. In the applications that follow we do not concern ourselves with these two corner points. (ii) Suppose that f has a jump discontinuity at a point (x, ) on G t. In this case, it is impossible to find a solution u that satisfies the IC at this point in the classical sense; however, we might find a type of solution that does so in some average way. We call this a weak solution. (iii) The definition of a solution can be generalized in the obvious manner to IBVPs with more than two independent variables. 4.9 Theorem (Uniqueness). If the functions α, β, and f are sufficiently smooth to ensure that u, u t, u x, and u xx are continuous in G and up to the boundary of G, including the two corner points, then the IBVP in Example 4.4 has at most one solution. Proof. Suppose that there are two solutions u 1 and u 2 of the problem in question. Then, because of linearity, it is obvious that u = u 1 u 2 is a solution of the fully homogeneous IBVP; that is, for any arbitrarily fixed number T >, the function u satisfies u t (x, t) = ku xx (x, t), < x < L, < t < T, u(, t) =, u(l, t) =, < t < T, u(x, ) =, < x < L. Multiplying the PDE by u, integrating over [, L], and using the BCs, the smoothness properties of the solutions for x L and t T, and integration by parts, we find that

96 4.1 The Heat Equation 73 therefore, 1 2 d dt = = = L L L L (uu t kuu xx ) dx [ 1 2 (u2 ) t + ku 2 x] dx k [ uu x ] x=l x= [ 1 2 (u2 ) t + ku 2 x] dx; u 2 dx = k which means that the function W (t) = L L u 2 x dx, t T, u 2 (x, t) dx, t T, is nonincreasing. Since W (t) and W () = (because of the IC satisfied by u), this is possible only if W (t) =, t T. In view of the nonnegative integrand in the definition of W above and the arbitrariness of T >, we then conclude that u = ; in other words, the solutions u 1 and u 2 coincide. 4.1 Remarks. (i) Theorem 4.9 states that if the IBVP in Example 4.4 has a solution, then that solution is unique. The existence issue is resolved in Chapter 5, where we actually construct the solution. (ii) The properties required of the solution in Theorem 4.9 are stronger than those mentioned in Example 4.7. (iii) Uniqueness can also be proved for the solutions of IBVPs with other types of BCs. (iv) We can easily convince ourselves that the change of variables reduces the heat equation to the form ξ = x L, τ = k L 2 t, ( ) u(x, t) = u Lξ, L2 τ = v(ξ, τ) k v τ (ξ, τ) = v ξξ (ξ, τ), < ξ < 1, τ >. Hence, without loss of generality, in most of our applications we consider this simpler version, with x, t, and u in place of ξ, τ, and v, respectively.

97 74 4 Some Fundamental Equations of Mathematical Physics Exercises 1 Show that Theorem 4.9 also holds for the IBVP in Example 4.4 with the BCs replaced by (i) u x (, t) hu(, t) = α(t), u(l, t) = β(t), t > ; (ii) u x (, t) = α(t), u x (L, t) + hu(l, t) = β(t), t >, where α and β are given functions and h = const >. 2 Verify the statement in Remark 4.1(iv). 4.2 The Laplace Equation The higher-dimensional heat equation. For simplicity, we study the case of two space dimensions. Consider a heat-conducting body (thin plate) occupying a finite region D in the (x, y)-plane, bounded by a smooth, simple, closed curve D, and let R be an arbitrary element of D with a smooth boundary R (see Fig. 4.2). We adopt the same notation for the physical parameters of this body as in Section 4.1. Heat flow depends on direction, so here the heat flux is a vector ϕ, which we can write in the form ϕ = normal component + tangential component = ( ϕ n) n + ( ϕ τ) τ, n = τ = 1, where n and τ are the unit normal and tangent vectors to R, directed as shown in Fig Clearly, the tangential component makes no contribution to the heat exchange between R and the rest of the body. Figure 4.2 A two-dimensional body configuration.

98 4.2 The Laplace Equation 75 As in the case of a rod, the conservation of heat energy states that rate of change of heat energy R = heat flow across boundary per unit time + heat generated by sources per unit time. This translates mathematically as d E(x, y, t) da = ϕ(x, y, t) n(x, y) ds + q(x, y, t) da, dt R where da and ds are, respectively, the elements of area and arc length. Taking (4.2) into account and applying the divergence theorem, that is, (div ϕ)(x, y, t) da = ϕ(x, y, t) n(x, y) ds, R we can rewrite the above equality as [ c(x, y)ρ(x, y)ut (x, y, t) + (div ϕ)(x, y, t) q(x, y, t) ] da =. R R R In view of the arbitrariness of R, we conclude that for (x, y) in D and t >, c(x, y)ρ(x, y)u t (x, y, t) = (div ϕ)(x, y, t) + q(x, y, t). (4.6) Once again, we now use Fourier s law of heat conduction, which in two dimensions is ϕ(x, y, t) = K (x, y)(grad u)(x, y, t); consequently, (4.6) becomes c(x, y)ρ(x, y)u t (x, y, t) = div ( K (x, y)(grad u)(x, y, t) ) + q(x, y, t). For a uniform body (c, ρ, and K are constants) with no internal sources (q = ), the last formula reduces to or u t (x, y, t) = k(div gradu)(x, y, t), k = K cρ = const, u t (x, y, t) = k( u)(x, y, t), (x, y) in D, t >, (4.7) where the differential operator, defined for smooth functions v(x, y) by ( v)(x, y) = v xx (x, y) + v yy (x, y), is called the Laplacian. Equality (4.7) is the two-dimensional heat (diffusion) equation.

99 76 4 Some Fundamental Equations of Mathematical Physics If the body contains sources, then (4.7) is replaced by its nonhomogeneous counterpart u t (x, y, t) = k( u)(x, y, t) + q(x, y, t), (x, y) in D, t >, where, as before, q now incorporates the factor 1/(cρ). Initial condition. Here, this takes the form u(x, y, ) = f(x, y), (x, y) in D, and represents the initial distribution of temperature in the body. Boundary conditions. The main types of BCs are similar in nature to those for a rod. (i) When the temperature is prescribed on the boundary, u(x, y, t) = α(x, y, t), (x, y) on D, t >. This is called a Dirichlet boundary condition. (ii) If the flux through the boundary is prescribed, then or, equivalently, (grad u)(x, y, t) n(x, y) = β(x, y, t), (x, y) on D, t >, u n (x, y, t) = β(x, y, t), (x, y) on D, t >, where n is the unit outward normal to D and u n = u/ n. This is called a Neumann boundary condition. In particular, when the boundary is insulated, we have u n (x, y, t) =, (x, y) on D, t >. (iii) Newton s law of cooling takes the form K u n (x, y, t) = H [ u(x, y, t) U(x, y, t) ], (x, y) on D, t >. This is referred to as a Robin boundary condition. Sometimes we may have one type of condition prescribed on some part of the boundary, and another type on the remaining part. Equilibrium temperature. An equilibrium (steady-state) temperature is a time-independent solution u = u(x, y) of (4.7); in other words, it is a function satisfying the Laplace equation ( u)(x, y) =, (x, y) in D, and a time-independent boundary condition, for example, u(x, y) = α(x, y), (x, y) on D.

100 4.2 The Laplace Equation 77 If steady-state sources q(x, y) are present in the body, then the equilibrium temperature satisfies the Poisson equation ( u)(x, y) = 1 q(x, y), (x, y) in D. k In what follows, when we discuss the Poisson equation that is, the nonhomogeneous Laplace equation we omit the factor 1/k on the right-hand side, regarding it as incorporated in the source term q. An equilibrium temperature may or may not exist Remark. The Poisson equation also arises in electrostatics. According to Maxwell s equations, a time-independent, two-dimensional electric field (u, v) satisfies v x u y =, u x + v y = ρ, where ρ is the charge density. The first equation shows that udx vdy is the differential of a function ϕ, called the electric potential; that is, dϕ = ϕ x dx + ϕ y dy = udx vdy, which means that ϕ x = u and ϕ y = v. If we replace this in the second equation, we arrive at ϕ = ϕ xx + ϕ yy = ρ Remarks. (i) Problems in three space variables are formulated similarly, with u = u(x, y, z, t). (ii) In polar coordinates x = r cos θ, y = r sin θ, for u = u(r, θ) we have u = r 1 (ru r ) r + r 2 u θθ = u rr + r 1 u r + r 2 u θθ. (iii) In cylindrical coordinates x = r cos θ, y = r sin θ, z, the Laplacian of u = u(r, θ, z) takes the form u = r 1 (ru r ) r + r 2 u θθ + u zz. (iv) In circularly (axially) symmetric problems, the function u is independent of θ, so u = r 1 (ru r ) r + u zz = u rr + r 1 u r + u zz. (v) In spherical coordinates x = r cos θ sin ϕ, y = r sin θ sin ϕ, z = r cos ϕ, for u = u(r, θ, ϕ) we have u = r 2 (r 2 u r ) r + (r 2 sin 2 ϕ) 1 u θθ + (r 2 sin ϕ) 1 ((sin ϕ)u ϕ ) ϕ. (vi) It is obvious that the Laplace and Poisson equations are linear. (vii) The Laplace equation is the simplest example of a so-called elliptic equation.

101 78 4 Some Fundamental Equations of Mathematical Physics 4.13 Example. The equilibrium temperature distribution in a thin, uniform finite plate with sources, insulated upper and lower faces, and prescribed temperature on the boundary is modeled by the BVP ( u)(x, y) = q(x, y), (x, y) in D, (PDE) u(x, y) = α(x, y), (x, y) on D, (BC) where D is a simple closed curve. The Laplacian is written either in Cartesian or in polar coordinates, depending on the geometry of the plate. When polar coordinates are used, the nature of the ensuing PDE may require us to consider additional boundary conditions, suggested by the physics of the process Remark. A (classical) solution of the BVP in Example 4.13 has the following properties: (i) It is twice continuously differentiable in D and satisfies the PDE at every point in D. (ii) It is continuous up to the boundary D of D and satisfies the BC at every point of D. As mentioned in Remark 4.8(ii), a boundary data function with discontinuities may generate a less smooth solution Theorem. If u is a solution of the BVP in Example 4.13, then u attains its maximum and minimum values on the boundary D of D. Proof. We anticipate a result established in Section 5.3, according to which (see Remark 5.18) the temperature at the center of a circular disc is equal to the average of the temperature on its boundary circle. Suppose that the maximum of the solution u occurs at a point P inside D. Regarding P as the center of a small disk lying within D, we deduce that the value of u at P is the average of all the values of u on the boundary circle of that disc, and so it cannot be greater than all of those values. We have thus arrived at a contradiction, which implies that u must attain its maximum on the boundary D. Considering v = u, we also conclude that u attains its minimum on D. This assertion is known in the literature as the maximum principle for the Laplace equation Corollary. If a solution u of the BVP in Example 4.13 is identically zero on D, then u is also identically zero in D. Proof. By Theorem 4.15, the maximum and minimum of u are zero, since they occur at points on D. Hence, u is zero in D Theorem (Uniqueness). The BVP in Example 4.13 has at most one solution.

102 4.2 The Laplace Equation 79 Proof. Suppose that there are two solutions u 1 and u 2. Because of the linearity of the PDE and BC, the difference u = u 1 u 2 is a solution of the fully homogenous BVP ( u)(x, y) =, (x, y) in D, u(x, y) =, (x, y) on D; therefore, by Corollary 4.16, u =, which means that u 1 and u 2 are the same function Definition. A solution of a BVP (IVP, IBVP) is said to depend continuously on the data (or to be stable) if a small variation in the data (BCs, ICs, nonhomogeneous term in the PDE) induces only a small variation in the solution Theorem. The solution of the BVP in Example 4.13 depends continuously on the boundary data. Proof. Consider the BVPs and ( u)(x, y) = q(x, y), (x, y) in D, u(x, y) = α(x, y), (x, y) on D, ( v)(x, y) = q(x, y), (x, y) in D, v(x, y) = α(x, y) + ε(x, y), (x, y) on D, where ε is a small perturbation of the boundary function α. Then, taking the linearity of the PDE and BC into account, we see that w = u v satisfies ( w)(x, y) =, (x, y) in D, w(x, y) = ε(x, y), By Theorem 4.15, for all (x, y) in D, (x, y) on D. min ε(x, y) w(x, y) = u(x, y) v(x, y) max ε(x, y); (x,y) on D (x,y) on D that is, the perturbation of the solution u is small. 4.2 Remark. A BVP (IVP, IBVP) is said to be well posed if it has a unique solution that depends continuously on the data. In the following chapters, we construct a solution for the BVP in Example 4.13, which, according to Theorem 4.17, is the only solution of that problem. Furthermore, by Theorem 4.19, this solution depends continuously on the boundary data, so the Dirichlet problem for the Laplace equation is well posed.

103 8 4 Some Fundamental Equations of Mathematical Physics Exercises 1 Show that any (classical) solution of the Poisson equation u = q in D, which is continuously differentiable (rather than merely continuous) up to the boundary D, satisfies (u 2 x + u2 y ) da + uq da = uu n ds. D Using this formula, show that D (i) each of the Dirichlet and Robin BVPs for the Poisson equation in D has at most one solution of this type; (ii) any two such solutions of the Neumann BVP for the Poisson equation in D differ by a constant. D 4.3 The Wave Equation Vibrating string. Consider the motion of a tightly stretched elastic string, in which the horizontal displacement of the points of the string is negligible (see Fig. 4.3). Figure 4.3 An arbitrary small segment of an elastic string. We define the following physical parameters for the string: L : length; u(x, t) : vertical displacement;

104 4.3 The Wave Equation 81 ρ (x) : mass density (mass per unit length); F(x, t): tension; q(x, t) : vertical component of the body force per unit mass. Assuming that the vertical displacement from the equilibrium position (the segment [, L] along the x-axis) is small and writing Newton s second law force = mass acceleration in vertical projection for the string segment between two close points corresponding to x and x + x, we obtain the approximate equality ρ (x)( x)u tt (x, t) = F(x + x, t) sin ( θ(x + x, t) ) F(x, t) sin ( θ(x, t) ) + ρ (x)( x)q(x, t). If we divide by x and then let x, we arrive at the equality ρ (x)u tt (x, t) = [ F(x, t) sin ( θ(x, t) )] x + ρ (x)q(x, t). (4.8) For a small angle θ, the slope of the string satisfies u x = tan θ = sin θ cos θ = sin θ = 1 1 sin θ, 2 θ2 + so we can rewrite (4.8) in the form ρ (x)u tt (x, t) = ( F(x, t)u x (x, t) ) x + ρ (x)q(x, t). (4.9) Assuming that the string is perfectly elastic (that is, when θ is small we may take F(x, t) = F = const) and homogeneous (ρ = const), and that the body force is negligible compared to tension (q = ), we see that (4.9) becomes u tt (x, t) = c 2 u xx (x, t), < x < L, t >, (4.1) where c 2 = F /ρ. This is the one-dimensional wave equation, in which c has the dimensions of velocity. If the body force is not negligible, then (4.1) is replaced by u tt (x, t) = c 2 u xx (x, t) + q(x, t), < x < L, t >. Initial conditions. Since (4.1) is of second order with respect to time, we need two initial conditions. These usually are u(x, ) = f(x), u t (x, ) = g(x), < x < L; that is, we prescribe the initial position and velocity of the points in the string.

105 82 4 Some Fundamental Equations of Mathematical Physics Boundary conditions. As in the case of the heat equation, there are three main types of BCs for (4.1) that are physically meaningful. (i) The displacement may be prescribed at an endpoint; for example, u(l, t) = β(t). (ii) An endpoint may be free (no vertical tension); that is, F sin θ = F tan θ = F u x =, so we may have a condition of the form u x (, t) =, t >. (iii) An endpoint may have an elastic attachment, described by F u x (, t) = ku(, t) or F u x (L, t) = ku(l, t), t >, where k = const > Remarks. (i) There is an obvious analogy between the BCs associated with the wave equation and those associated with the heat equation. (ii) The condition prescribed at one endpoint may be different from that prescribed at the other. (iii) It is clear that the wave equation is linear. (iv) The one-dimensional wave equation is the simplest example of a linear second-order hyperbolic equation. The model. As we have seen, the vibrations of an elastic string are governed by the wave equation and appropriate BCs and ICs Example. The IBVP for a vibrating string with fixed endpoints when the effect of the body force is taken into account is of the form u tt (x, t) = c 2 u xx (x, t) + q(x, t), < x < L, t >, (PDE) u(, t) =, u(l, t) =, t >, (BCs) u(x, ) = f(x), u t (x, ) = g(x), < x < L, (ICs) where q, f, and g are given functions Example. If the string in Example 4.22 has an elastic attachment at its near endpoint, a free far endpoint, and a negligible body force, then the corresponding IBVP is u tt (x, t) = c 2 u xx (x, t), < x < L, t >, (PDE) u x (, t) ku(, t) =, u x (L, t) =, t >, (BCs) u(x, ) = f(x), u t (x, ) = g(x), < x < L, (ICs) where k is a known positive constant.

106 4.3 The Wave Equation Remark. Let the domain G and its boundary lines G x and G t be as defined in Example 4.7. A (classical) solution u of the IBVP in Example 4.22 has the following properties: (i) It is twice continuously differentiable with respect to x and t in G and satisfies the PDE at every point in G. (ii) It is continuous up to G x and satisfies the BCs at every point on G x. (iii) It is continuous together with its first-order time derivative up to G t and satisfies the ICs at every point on G t. As in the case of the heat equation and the Laplace equation, the presence of jump discontinuities in the data functions leads to a solution with reduced smoothness Theorem (Uniqueness). If the functions q, f, and g are such that u and all its first-order and second-order derivatives are continuous up to the boundary of G, including the corner points (, ) and (L, ), then the IBVP in Example 4.22 has at most one solution. Proof. Suppose that the given IBVP has two solutions u 1 and u 2. Then their difference u = u 1 u 2 satisfies the fully homogeneous IBVP u tt (x, t) = c 2 u xx (x, t), < x < L, t >, u(, t) =, u(l, t) =, t >, u(x, ) =, u t (x, ) =, < x < L. Multiplying the PDE by u t, integrating the result with respect to x over [, L] and with respect to t over [, T ], where T > is an arbitrarily fixed number, using integration by parts, and taking the BCs and the smoothness properties of u into account, we see that = = = 1 2 T L T L T L hence, for any T >, L (u tt u t c 2 u xx u t ) dx dt (u tt u t + c 2 u x u xt ) dx dt c 2 (u 2 t + c 2 u 2 x) t dx dt = 1 2 [ u 2 t (x, T ) + c 2 u 2 x (x, T )] dx = L L T [ ux u t ] x=l x= dt [ u 2 t + c 2 u 2 x] t=t t= dx; [ u 2 t (x, ) + c 2 u 2 x (x, )] dx,

107 84 4 Some Fundamental Equations of Mathematical Physics which implies that V (t) = L [ u 2 t (x, t) + c 2 u 2 x (x, t)] dx = κ = const, t. From the ICs and the smoothness properties of u it follows that V () =, so κ = ; in other words, V =. Since the integrand in V is nonnegative, we see that this is possible if and only if u t (x, t) =, u x (x, t) =, x L, t. Consequently, u is constant in G and on its boundary lines. Since u is zero on these lines, we conclude that u = ; thus, u 1 and u 2 coincide Remarks. (i) As for the heat and Laplace equations, in what follows we construct a solution for the IBVP mentioned in Example By Theorem 4.25, this will be the only solution of that problem. (ii) Uniqueness can also be proved for the solutions of IBVPs with other types of BCs. (iii) The substitution ξ = 1 L x, τ = c t, u(x, t) = u(lξ, Lτ/c) = v(ξ, τ) L reduces the wave equation to v ττ (ξ, τ) = v ξξ (ξ, τ), < ξ < 1, τ >. It is mostly this simpler form that we use in applications. Exercises 1 Show that Theorem 4.25 also holds for the IBVP in Example 4.22 with h = const > and the BCs replaced by (i) u x (, t) hu(, t) =, u(l, t) =, t > ; (ii) u x (, t) =, u x (L, t) + hu(l, t) =, t >. 2 Verify the statement in Remark 4.26(iii). 4.4 Other Equations Below is a list of other linear partial differential equations that occur in important mathematical models. Their classification as parabolic, hyperbolic, or elliptic is explained in Chapter 11.

108 4.4 Other Equations 85 Brownian motion. The function u(x, t) that specifies the probability that a particle undergoing one-dimensional motion in a fluid is located at point x at time t satisfies the second-order parabolic PDE u t (x, t) = au xx (x, t) bu x (x, t), where the coefficients a, b > are related to the average displacement of the particle per unit time and the variance of the observed displacement around the average. Diffusion-convection problems. The one-dimensional case is described by the second-order parabolic PDE u t (x, t) = ku xx (x, t) au x (x, t) + bu(x, t), where u(x, t) is the temperature and the coefficients k, a >, and b are expressed in terms of the physical properties of the medium, the heat flow rate, and the strength of the source. When a = and b >, the equation above describes a diffusion process with a chain reaction. Stock market prices. The price V (S, t) of a derivative on the stock market is the solution of the second-order parabolic PDE V t (S, t) σ2 S 2 V SS (S, t) + rsv S (S, t) rv (S, t) =, where S is the price of the stock, σ is the volatility of the stock, r is the continuously compounded risk-free interest per annum, and time t is measured in years. This is known as the Black Scholes equation. Evolution of a quantum state. In quantum mechanics, the wave function ψ(x, t) that characterizes the one-dimensional motion of a particle under the influence of a potential V (x) satisfies the Schrödinger equation i ψ t (x, t) = 2 2m ψ xx(x, t) + V (x)ψ(x, t). This is a second-order parabolic PDE where is the reduced Planck constant, m is the mass of the particle, and i 2 = 1. Motion of a quantum scalar field. The wave function ψ(x, y, z, t) of a free moving particle in relativistic quantum mechanics is the solution of the (second-order hyperbolic) Klein Gordon equation ψ tt (x, y, z, t) = c 2 ψ(x, y, z, t) m2 c 4 ψ(x, y, z, t), where c is the speed of light. The one-dimensional version of this equation is of the form where a = const >. 2 u tt (x, t) = c 2 u xx (x, t) au(x, t),

109 86 4 Some Fundamental Equations of Mathematical Physics Loss transmission line. The current and voltage in the propagation of signals on a telegraph line satisfy the second-order hyperbolic PDE au tt (x, t) + bu t (x, t) + cu(x, t) = u xx (x, t), where the coefficients a, b >, and c are expressed in terms of the resistance, inductance, capacitance, and conductance characterizing the line. This is known as the telegraph (or telegrapher s) equation. Dissipative waves. In the one-dimensional case, the propagation of such waves is governed by a second-order hyperbolic PDE of the form u tt (x, t) + au t (x, t) + bu(x, t) = c 2 u xx (x, t) du x (x, t), where the coefficients a, b, d are not all zero and c >. Transverse vibrations of a rod. The deflection u(x, t) of a generic point in the rod satisfies the fourth-order PDE u tt (x, t) + c 2 u xxxx (x, t) =, where c is a physical constant related to the rigidity of the rod material. Scattered waves. The Helmholtz equation u(x, y, z) + k 2 u(x, y, z) =, where k = const, plays an important role in the study of scattering of acoustic, electromagnetic, and elastic waves. It is a second-order elliptic PDE. The equation u(x, y, z) k 2 u(x, y, z) = is called the modified Helmholtz equation. Steady-state convective heat. In the two-dimensional case, the temperature distribution u(x, y) governing this process satisfies a second-order elliptic PDE of the form u(x, y) au x (x, y) bu y (x, y) + cu(x, y) =, where the coefficients a, b (not both zero), and c are related to the thermal properties of the medium, the heat flow rates in the x and y directions, and the strength of the heat source. Plane problems in continuum mechanics. The Airy stress function in plane elasticity and the stream function in the slow flow of a viscous incompressible fluid are solutions of the elliptic fourth-order biharmonic equation u(x, y) = u xxxx (x, y) + 2u xxyy (x, y) + u yyyy (x, y) =.

110 4.4 Other Equations 87 Plane transonic flow. The transonic flow of a compressible gas is described by the Euler Tricomi equation u xx (x, y) = xu yy (x, y), where u(x, y) is a function of speed. This is a second-order PDE of mixed type: it is hyperbolic for x >, elliptic for x <, and parabolic for x =. Mathematical biology. The Fisher equation u t (x, t) = Du xx (x, t) + ru(x, t)(1 u(x, t)), where r, D = const >, models the advance of advantageous genes in a population. This is a nonlinear equation of parabolic type. Fluid dynamics. The propagation of waves on a free-moving fluid surface is governed by the nonlinear Boussinesq equation ( ) 3 η tt (x, t) ghη xx (x, t) gh 2h η2 (x, t) + h2 3 η xx(x, t) =, where η = η(x, t) is the free surface elevation, h = const is the fluid depth, and g is the acceleration of gravity Remark. In the rest of the book we assume, without explicit mention, that all functions representing nonhomogeneous terms in PDEs, ICs, and BCs are prescribed and reasonably behaved, in the sense that they possess the degree of smoothness required by the implementation of the solution procedure. xx Exercises In 1 4, find real numbers α and β such that the function substitution u(x, t) = e αx+βt v(x, t) reduces the given diffusion-convection equation to the heat equation for v. 1 u t (x, t) = 2u xx (x, t) 3u x (x, t) u(x, t). 2 u t (x, t) = u xx (x, t) 6u x (x, t) 2u(x, t). 3 u t (x, t) = 1 2 u xx(x, t) 4u x (x, t) 2u(x, t). 4 u t (x, t) = 3u xx (x, t) 2u x (x, t) 3u(x, t). In 5 8, determine if there are real numbers α and β such that the function substitution u(x, t) = e αx+βt v(x, t) in the given dissipative wave equation eliminates (i) both first-order derivatives; (ii) the unknown function and its firstorder t-derivative; (iii) the unknown function and its first-order x-derivative.

111 88 4 Some Fundamental Equations of Mathematical Physics (In all cases, the coefficients of the reduced equation must satisfy the restrictions mentioned in Section 4.4.) When such a substitution exists, write out the reduced equation. 5 u tt (x, t) + u t (x, t) + u(x, t) = u xx (x, t) 2u x (x, t). 6 u tt (x, t) + 3u t (x, t) + u(x, t) = 2u xx (x, t) u x (x, t). 7 u tt (x, t) + 2u t (x, t) u(x, t) = 2u xx(x, t) 2 u x (x, t). 8 u tt (x, t) + 2u t (x, t) + 2u(x, t) = u xx (x, t) u x (x, t). In 9 12, find real numbers α and β such that the function substitution u(x, y) = e αx+βy v(x, y) reduces the given steady-state convective heat equation to the Helmholtz (modified Helmholtz) equation. In each case, write out the reduced equation. 9 u xx (x, y) + u yy (x, y) 2u x (x, y) 4u y (x, y) + 7u(x, y) =. 1 u xx (x, y) + u yy (x, y) u x (x, y) 2u y (x, y) u(x, y) =. 11 u xx (x, y) + u yy (x, y) 3u x (x, y) u y (x, y) + 2u(x, y) =. 12 u xx (x, y) + u yy (x, y) 2u x (x, y) 3u y (x, y) + 4u(x, y) =. Answers to Odd-Numbered Exercises 1 α = 3/4, β = 17/8. 3 α = 4, β = 1. 5 (i) α = 1, β = 1/2, v tt (x, t) + (7/4)v(x, t) = v xx (x, t); (ii) α = 1 7/2, β = 1/2, v tt (x, t) = v xx (x, t) 7 v x (x, t). 7 (iii) α = 2/4, β = 1/2, v tt (x, t) + v t (x, t) = 2v xx (x, t). 9 α = 1, β = 2, v xx (x, y) + v yy (x, y) + 2v(x, y) =. 11 α = 3/2, β = 1/2, v xx (x, y) + v yy (x, y) (1/2)v(x, y) =.

112 C H A P T E R 5 The Method of Separation of Variables Separation of variables is one of the oldest and most efficient solution techniques for a certain class of PDE problems. Below, we show it in application to initial boundary value problems for the heat and wave equations, and to boundary value problems for the Laplace equation. 5.1 The Heat Equation We investigate several cases, identified in terms of the BCs prescribed in the mathematical model Rod with Zero Temperature at the Endpoints Consider the IBVP u t (x, t) = ku xx (x, t), < x < L, t >, (PDE) u(, t) =, u(l, t) =, t >, (BCs) u(x, ) = f(x), < x < L, (IC) where f =. We remark that the PDE and BCs are linear and homogeneous, and seek a solution of the form u(x, t) = X(x)T (t). Substituting into the PDE, we obtain the equality X(x)T (t) = kx (x)t (t). It is clear that neither X nor T can be the zero function: if either of them were, then so would u, which is impossible, because the zero solution does not 89

113 9 5 The Method of Separation of Variables satisfy the nonhomogeneous IC. Hence, we may divide the preceding equality by kx(x)t (t) to arrive at 1 T (t) k T (t) = X (x) X(x). Since the left-hand side above is a function of t and the right-hand side a function of x alone, this equality is possible if and only if both sides are equal to one and the same constant, say, λ. In other words, we must have 1 T (t) k T (t) = X (x) X(x) = λ, where λ is called the separation constant. This leads to separate equations for the functions X and T : From the first BC we see that X (x) + λx(x) =, < x < L, (5.1) T (t) + λkt (t) =, t >. (5.2) u(, t) = X()T (t) = for all t >. As T =, it follows that X() =. (5.3) Similarly, the second BC yields X(L) =. (5.4) We are now ready to find X and T. We want nonzero functions X that satisfy the regular Sturm Liouville problem (5.1), (5.3), and (5.4), in other words, the eigenfunctions X n corresponding to the eigenvalues λ n, both computed in Example 3.16: λ n = ( ) 2 nπ, X n (x) = sin nπx, n = 1, 2,.... (5.5) L L For each λ n, from (5.2) we find the corresponding time component T n (t) = e k(nπ/l)2t, n = 1, 2,.... (5.6) We have taken the arbitrary constant of integration to be 1 since these functions are used in the next stage with arbitrary numerical coefficients. Combining (5.5) and (5.6), we conclude that all functions of the form u n (x, t) = X n (x)t n (t) = sin nπx L e k(nπ/l)2t, n = 1, 2,...,

114 5.1 The Heat Equation 91 satisfy both the PDE and the BCs. By the principle of superposition, so does any finite linear combination N b n u n (x, t) = N b n sin nπx 2t L e k(nπ/l), (5.7) where b n are arbitrary numbers. The IC is satisfied by such an expression if f(x) = N b n sin nπx L. But this is impossible unless f is a finite linear combination of the eigenfunctions, which, in general, is not the case. This implies that (5.7) is not a good representation for the solution u(x, t) of the IBVP. However, we recall (see Section 2.2) that if f is piecewise smooth, then it can be written as an infinite linear combination of the eigenfunctions (its Fourier sine series): where, by (2.1), f(x) = b n = 2 L L b n sin nπx, < x < L, (5.8) L f(x) sin nπx L Therefore, the solution is given by the infinite series dx, n = 1, 2,.... (5.9) u(x, t) = b n sin nπx L e k(nπ/l)2 t (5.1) with the coefficients b n computed from (5.9). 5.1 Example. Consider the IBVP u t (x, t) = u xx (x, t), < x < 1, t >, u(, t) =, u(1, t) =, t >, u(x, ) = sin(3πx) 2 sin(5πx), < x < 1. Here k = 1, L = 1, and the function on the right-hand side in the IC is a linear combination of the eigenfunctions. Using (5.8) and Theorem 3.23(ii), we find that b 3 = 1, b 5 = 2, b n = (n 3, 5); so, by (5.1), the solution of the problem is u(x, t) = sin(3πx)e 9π2t 2 sin(5πx)e 25π2t.

115 92 5 The Method of Separation of Variables Equally, this result can be obtained from (5.9) and (5.1). Verification with Mathematica R. The input u = Sin[3 Pi x] E ( - 9 Pi 2 t) - 2 Sin[5 Pi x] E ( - 25 Pi 2 t); D[u,t] - D[u,x,x],u /.x ->,u /.x -> 1,(u /.t -> ) - (Sin[3 Pi x] - 2 Sin[5 Pi x])} // FullSimplify generates the output,,, }, which confirms that the PDE, BCs, and IC are satisfied. 5.2 Example. To find the solution of the IBVP u t (x, t) = u xx (x, t), < x < 1, t >, u(, t) =, u(1, t) =, t >, u(x, ) = x, < x < 1, we first use (5.9) with f(x) = x and L = 1, and integration by parts, to compute the coefficients b n : 1 b n = 2 x sin(nπx) dx = ( 1) n+1 2, n = 1, 2,... ; nπ then, by (5.1), u(x, t) = ( 1) n+1 2 nπ sin(nπx) π 2 t e n2 is the solution of the IBVP. Verification with Mathematica R. The input un = ( - 1) (n + 1) (2/(n Pi)) Sin[n Pi x] E ( - n 2 Pi 2 t); Simplify[D[un,t] - D[un,x,x],un /.x ->,un /.x -> 1},n Integers] Plot[Sum[un /.t ->,n,1,7}],x},x,,1}, PlotRange -> -.1,1.1}, -.2,1.1}}, PlotStyle -> Black, ImageSize -> Scaled[.2], Ticks ->1},1}}, AspectRatio ->.9] generates the output,, } and the graph

116 5.1 The Heat Equation 93 which confirm that the PDE and BCs are satisfied and that the truncation (after 9 terms) of the series solution at t = is a good approximation of the IC function. Exercises Use separation of variables to solve the IBVP with function f as indicated. u t (x, t) = u xx (x, t), < x < 1, t >, u(, t) =, u(1, t) =, t >, u(x, ) = f(x), < x < 1, 1 f(x) = sin(2πx) 3 sin(6πx). 2 f(x) = 3 sin(πx) sin(4πx). 3 f(x) = 2. 4 f(x) = 1 2x. 5 f(x) = 2x + 1., < x 1/2, 6 f(x) = 2, 1/2 < x < 1. x, < x 1/2, 7 f(x) =, 1/2 < x < 1. 2, < x 1/2, 8 f(x) = x 1, 1/2 < x < 1. Answers to Odd-Numbered Exercises 1 u(x, t) = sin(2πx) e 4π2t 3 sin(6πx) e 36π2t. 3 u(x, t) = [( 1) n 1](4/(nπ)) sin(nπx) e n2 π 2t. 5 u(x, t) = [1 ( 1) n 3](2/(nπ)) sin(nπx) e n2 π 2t. 7 u(x, t) = [(2/(n 2 π 2 )) sin(nπ/2) (1/(nπ)) cos(nπ/2)] sin(nπx) e n2 π 2t.

117 94 5 The Method of Separation of Variables Rod with Insulated Endpoints The heat conduction problem for a uniform rod with insulated endpoints is described by the IBVP u t (x, t) = ku xx (x, t), < x < L, t >, (PDE) u x (, t) =, u x (L, t) =, t >, (BCs) u(x, ) = f(x), < x < L. (IC) Since the PDE and BCs are linear and homogeneous, we again seek a solution of the form u(x, t) = X(x)T (t) with X, T. Just as in Subsection 5.1.1, from the PDE and BCs we now find that X and T satisfy, respectively, and X (x) + λx(x) =, < x < L, X () =, X (L) =, T (t) + kλt (t) =, t >, where λ is the separation constant. The nonzero solutions X are the eigenfunctions of the above Sturm Liouville problem, which were computed in Example 3.17: λ =, X (x) = 1 2, ( ) 2 nπ λ n =, X n (x) = cos nπx L L, n = 1, 2,.... (5.11) Integrating the equation satisfied by T with λ = λ n, n =, 1, 2,..., we obtain the associated time components T n (t) = e k(nπ/l)2t, n =, 1, 2,.... (5.12) In view of (5.11), (5.12), and the argument used in the preceding case, we now expect the solution of the IBVP to have the series representation u(x, t) = 1 2 a + a n cos nπx L e k(nπ/l)2t, (5.13) where each term satisfies the PDE and the BCs. The IC is satisfied if u(x, ) = f(x) = 1 2 a + a n cos nπx L, < x < L. This shows that a /2 and a n are the Fourier cosine series coefficients of f, given by (2.12): a n = 2 L f(x) cos nπx dx, n =, 1, 2,.... (5.14) L L Therefore, (5.13) with the a n computed by means of (5.14) is the solution of the IBVP.

118 5.1 The Heat Equation Example. The solution of the IBVP u t (x, t) = u xx (x, t), < x < 1, t >, u x (, t) =, u x (1, t) =, t >, u(x, ) = x, < x < 1, is obtained from (5.13) and (5.14) with Specifically, a = 2 a n = so, as in Example 5.2, x dx = 1, L = 1, f(x) = x. x cos(nπx) dx = [ ( 1) n 1 ] 2 n 2, n = 1, 2,... ; π2 u(x, t) = [ ( 1) n 1 ] 2 n 2 π 2 cos(nπx) e n2 π 2t. Verification with Mathematica R. The input un = (( - 1) n - 1) (2/(n 2 Pi 2)) Cos[n Pi x] E ( - n 2 Pi 2 t); Simplify[D[un,t] - D[un,x,x], D[un,x] /.x ->, D[un,x] /.x -> 1}, n Integers] Plot[(1/2)+Sum[un /.t ->,n,1,2}],x},x,,1}, PlotRange -> -.1,1.1}, -.2,1.1}}, PlotStyle -> Black, ImageSize -> Scaled[.2], Ticks ->1},1}}, AspectRatio ->.9] generates the output,, } and the graph

119 96 5 The Method of Separation of Variables Exercises Use separation of variables to solve the IBVP with function f as indicated. 1 f(x) = 3 2 cos(4πx). u t (x, t) = u xx (x, t), < x < 1, t >, u x (, t) =, u x (1, t) =, t >, u(x, ) = f(x), < x < 1, 2 f(x) = cos(2πx) 3 cos(3πx). 3 f(x) = 2 3x. 4 f(x) = 3 2x. 2, < x 1/2, 5 f(x) =, 1/2 < x < 1. 3, < x 1/2, 6 f(x) = 1, 1/2 < x < 1., < x 1/2, 7 f(x) = 2x, 1/2 < x < 1. 1, < x 1/2, 8 f(x) = x + 1, 1/2 < x < 1. Answers to Odd-Numbered Exercises 1 u(x, t) = 3 2 cos(4πx) e 16π2t. 3 u(x, t) = 1/2 + [1 ( 1) n ](6/(n 2 π 2 )) cos(nπx) e n2 π 2t ; 5 u(x, t) = 1 (4/(nπ)) sin(nπ/2) cos(nπx) e n2 π 2t. 7 u(x, t) = 3/4+ (2/(nπ)) sin(nπ/2)+(4/(n 2 π 2 ))[( 1) n cos(nπ/2)]} cos(nπx) e n2 π 2t Rod with Mixed Boundary Conditions The method of separation of variables can also be used in the case where one endpoint is held at zero temperature while the other one is insulated.

120 5.1 The Heat Equation Example. Consider the IBVP u t (x, t) = u xx (x, t), < x < 1, t >, u(, t) =, u x (1, t) =, t >, u(x, ) = 1, < x < 1. Since the PDE and BCs are linear and homogeneous, we seek a solution of the form u(x, t) = X(x)T (t), X, T =. Proceeding as in the other cases, from the PDE and BCs we find that X and T satisfy, respectively, X (x) + λx(x) =, < x < 1, X() =, X (1) =, (5.15) and T (t) + λt (t) =, t >, where λ is the separation constant. To find X, we go back to the Rayleigh quotient (3.6) for the regular Sturm Liouville problem (5.15). As in Example 3.16, [ p(x)f(x)f (x) ] b a = [ X(x)X (x) ] 1 =, and we arrive at inequality (3.7), from which, in view of the BCs in (5.15), we deduce that λ >. Hence, the general solution of the differential equation in (5.15) is X(x) = C 1 cos( λ x) + C 2 sin( λ x), C 1, C 2 = const. Since X() =, it follows that C 1 =. Using the condition X (1) =, we now find that cos λ = ; therefore, (2n 1)π λ =, n = 1, 2, This means that the eigenvalue-eigenfunction pairs of (5.15) are λ n = (2n 1)2 π 2, X n (x) = sin 4 and that the corresponding time components are Consequently, the functions u n (x, t) = X n (x)t n (t) = sin (2n 1)πx, n = 1, 2,..., 2 T n (t) = e (2n 1)2 π 2 t/4, n = 1, 2,.... (2n 1)πx 2 e (2n 1)2 π 2 t/4, n = 1, 2,...,

121 98 5 The Method of Separation of Variables satisfy the PDE and the BCs. Considering the usual arbitrary linear combination of all the (countably many) functions u n, that is, u(x, t) = c n u n (x, t) = c n sin we see that the IC is satisfied if u(x, ) = 1 = c n sin (2n 1)πx 2 e (2n 1)2 π 2 t/4, (5.16) (2n 1)πx. (5.17) 2 In other words, (5.16) is the solution of the given IBVP if c n are the coefficients of the generalized Fourier series for the function f(x) = 1, x 1. These coefficients are computed by means of (3.1) with, according to (5.17), u(x) = 1, σ(x) = 1, f n (x) = sin Direct calculation shows that so 1 1 c n = 2 (2n 1)πx. 2 2 (2n 1)πx sin dx = 1, n = 1, 2,..., 2 2 sin (2n 1)πx 2 dx = 4, n = 1, 2,.... (2n 1)π Thus, the solution of the IBVP has the series representation u(x, t) = 4 (2n 1)πx sin e (2n 1)2 π 2 t/4. (2n 1)π 2 Verification with Mathematica R. The input un = (4/((2 n - 1) Pi)) Sin[(2 n - 1) Pi x/2] E ( - (2 n - 1) 2 Pi 2 t/4); Simplify[D[un,t] - D[un,x,x], un /.x ->, D[un,x] /.x -> 1}, n Integers] Plot[Sum[un /.t ->,n,1,7}],1},x,,1}, PlotRange -> -.1,1.1}, -.2,1.3}}, PlotStyle -> Black, ImageSize -> Scaled[.2], Ticks ->1},1}}, AspectRatio ->.9]

122 5.1 The Heat Equation 99 generates the output,, } and the graph 5.5 Example. The solution of the IBVP u t (x, t) = u xx (x, t), < x < 1, t >, u x (, t) =, u(1, t) =, t >, u(x, ) = x, < x < 1, is constructed just as in Example 5.4, except here the BCs and the Rayleigh quotient yield the eigenvalue-eigenfunction pairs Since λ n = (2n 1)2 π 2, X n (x) = cos 4 1 (2n 1)πx, n = 1, 2, (2n 1)πx cos dx = 1, n = 1, 2,..., 2 2 the coefficients of the generalized Fourier series for the IC function, given by (3.1) with u(x) = x, σ(x) = 1, k = 1, and L = 1, are c n = 2 1 x cos (2n 1)πx 2 dx = ( 1) n+1 4 (2n 1)π 8 (2n 1) 2, n = 1, 2,.... π2 Consequently, the solution of the IBVP is [ ] u(x, t) = ( 1) n+1 4 (2n 1)π 8 (2n 1) 2 π 2 cos Verification with Mathematica R. The input (2n 1)πx 2 e (2n 1)2 π 2 t/4. un = ((( - 1) (n + 1) 4/((2*n - 1) Pi)) - (8/((2*n - 1) 2 Pi 2))) Cos[(2 n - 1) Pi x/2] E ( - (2 n - 1) 2 Pi 2 t/4);

123 1 5 The Method of Separation of Variables Simplify[D[un,t] - D[un,x,x], D[un,x] /.x ->, un /.x -> 1}, n Integers] Plot[Sum[un /.t ->,n,1,7}],1},x,,1}, PlotRange -> -.1,1.1}, -.2,1.1}}, PlotStyle -> Black, ImageSize -> Scaled[.2], Ticks ->1},1}}, AspectRatio ->.9] generates the output,, } and the graph Exercises Use separation of variables to solve the IBVP consisting of the PDE and IC u t (x, t) = u xx (x, t), < x < 1, t >, u(x, ) = f(x), < x < 1, with BCs and function f as indicated. 1 u(, t) =, u x (1, t) =, f(x) = 3 sin(πx/2) sin(5πx/2). 2 u(, t) =, u x (1, t) =, f(x) = 3. 3 u(, t) =, u x (1, t) =, f(x) = 2 + x., < x 1/2, 4 u(, t) =, u x (, t) =, f(x) = 1, 1/2 < x < 1. 5 u x (, t) =, u(1, t) =, f(x) = 2 cos(5πx/2). 6 u x (, t) =, u(1, t) =, f(x) = 4. 7 u x (, t) =, u(1, t) =, f(x) = 2x 3. 2 x, < x 1/2, 8 u x (, t) =, u(1, t) =, f(x) = 1, 1/2 < x < 1.

124 5.1 The Heat Equation 11 Answers to Odd-Numbered Exercises 1 u(x, t) = 3 sin(πx/2) e π2t/4 sin(5πx/2) e 25π2t/4. 3 u(x, t) = [8/((2n 1)π) + ( 1) n+1 8/((2n 1) 2 π 2 )] sin((2n 1)πx/2) e (2n 1)2 π 2 t/4. 5 u(x, t) = 2 cos(5πx/2) e 25π2t/4. 7 u(x, t) = [( 1) n 4/((2n 1)π) 16/((2n 1) 2 π 2 )] cos((2n 1)πx/2) e (2n 1)2 π 2 t/ Rod with an Endpoint in a Zero-Temperature Medium Consider the problem of heat flow in a uniform rod without internal sources, when the near endpoint is kept at zero temperature and the far endpoint is kept in open air of zero temperature. The corresponding IBVP (see Section 4.1) is u t (x, t) = ku xx (x, t), < x < L, t >, (PDE) u(, t) =, u x (L, t) + hu(l, t) =, t >, (BCs) u(x, ) = f(x), < x < L, (IC) where h = const >. Since the PDE and BCs are linear and homogeneous, we use separation of variables and seek a solution of the form u(x, t) = X(x)T (t), X, T. In the usual way, from the PDE and BCs we find that X satisfies the regular S L problem X (x) + λx(x) =, < x < L, X() =, X (L) + hx(l) =, (5.18) and that T is the solution of the equation T (t) + kλt (t) =, t >. The eigenvalue-eigenfunction pairs of (5.18) were computed in Example 3.18: ( ) 2 ζn λ n =, X n (x) = sin ζ nx, n = 1, 2,..., L L where ζ n are the positive roots of the equation tan ζ = ζ hl.

125 12 5 The Method of Separation of Variables Hence, T n (t) = e k(ζn/l)2t, n = 1, 2,..., so we expect the solution of the IBVP to have a series representation of the form u(x, t) = c n X n (x)t n (t) = c n sin ζ nx L e k(ζn/l)2t. (5.19) The coefficients c n are found from the IC, according to which u(x, ) = f(x) = c n sin ζ nx L. Since the eigenfunctions of (5.18) are orthogonal on [, L], the c n are computed by means of (3.1): L f(x) sin(ζ n x/l) dx c n =, n = 1, 2,.... (5.2) L sin 2 (ζ n x/l) dx Therefore, the solution of the IBVP is (5.19) with the c n given by (5.2). 5.6 Example. Consider the IBVP u t (x, t) = u xx (x, t), < x < 1, t >, u(, t) =, u x (1, t) + u(1, t) =, t >, u(x, ) =, < x 1/2, 1, 1/2 < x < 1. From Examples 3.18 and (3.24) with L = h = 1, we know that the eigenvalueeigenfunction pairs associated with this problem are λ n = ζ 2 n, X n(x) = sin(ζ n x), n = 1, 2,..., where, rounded to the fourth decimal place, ζ 1 = 2.288, ζ 2 = , ζ 3 = , ζ 4 = , ζ 5 = Using (5.2) and f(x) = u(x, ), we now compute the approximate coefficients c 1 =.81, c 2 =.3813, c 3 =.1326, c 4 =.116, c 5 =.153. Consequently, by (5.19), we obtain the expansion

126 5.1 The Heat Equation 13 u(x, t) =.81 sin(2.288 x)e t.3813 sin( x)e t.1326 sin( x)e t sin( x)e t sin( x)e t +. (5.21) Verification with Mathematica R. Each term in (5.21) is an approximation of the corresponding term in (5.19), so it is not expected to satisfy the PDE and BCs exactly. However, computation shows that the error for the truncation consisting of the terms shown in (5.21) is less than 1 3, and that the graph of the function represented by this truncation at t = is close to that of u(x, ). Specifically, the input Approx =.81 Sin[2.288 x] E ( t) Sin[ x] E ( t) Sin[ x] E ( t) Sin[ x] E ( t) Sin[ x] E ( t); Chop[Simplify[D[Approx,t] - D[Approx,x,x], Approx /. x ->, (D[Approx,x]/. x -> 1) + Approx/. x->1}, n Integers],.1] Plot[Approx/. t ->, Piecewise[,<x<1/2},1,1/2<x<1}}]}, x,,1},plotrange -> -.1,1.1}, -.2,1.1}}, PlotStyle -> Black, ImageSize -> Scaled[.2], Ticks ->.5,1},1}}, AspectRatio ->.9] generates the output,, } and the graph Exercises Use separation of variables to solve the IBVP u t (x, t) = u xx (x, t), < x < 1, t >, u x (, t) =, u x (1, t) + hu(1, t) =, t >, u(x, ) = f(x), < x < 1,

127 14 5 The Method of Separation of Variables with coefficient h and function f as indicated. Compute only the first five terms of the series solution, rounding all numbers to the fourth decimal place. 1 h = 1, f(x) = x h = 2, f(x) = 1 2x. 1, < x 1/2, 3 h = 1/2, f(x) = 2, 1/2 < x < 1.. x < x 1/2, 4 h = 1, f(x) = 1 x, 1/2 < x < 1. Answers to Odd-Numbered Exercises 1 u(x, t) = sin(2.288 x)e t sin( x)e t sin( x)e t sin( x)e t sin( x)e t +. 3 u(x, t) = sin( x)e t sin( x)e t sin( x)e t.1945 sin(11.48 x)e t.1625 sin( x)e t Thin Uniform Circular Ring Physically, the ring of circumference 2L shown in Fig. 5.1 can be regarded as a rod of length 2L that, for continuity reasons, has the same temperature and heat flux at the two endpoints x = L and x = L. Figure 5.1 Representation of a circular ring as a rod. Thus, the corresponding IBVP is u t (x, t) = ku xx (x, t), L < x < L, t >, (PDE) u( L, t) = u(l, t), u x ( L, t) = u x (L, t), t >, (BCs) u(x, ) = f(x), L < x < L. (IC)

128 5.1 The Heat Equation 15 The PDE and BCs are homogeneous, so, as before, we seek a solution of the form u(x, t) = X(x)T (t) with X, T =. Then from the PDE and BCs we find by the standard argument that and X (x) + λx(x) =, L < x < L, X( L) = X(L), X ( L) = X (L), (5.22) T (t) + λkt (t) =, t >. (5.23) The nonzero function X must be a solution of the periodic Sturm Liouville problem (5.22), which was discussed in Example The eigenvalues of this S L problem are ( ) 2 nπ λ =, λ n =, n = 1, 2,..., L with corresponding eigenfunctions X (x) = 1 2, X 1n(x) = cos nπx L, X 2n(x) = sin nπx, n = 1, 2,.... L Therefore, the time components given by (5.23) with λ = λ n are T n (t) = e k(nπ/l)2t, n =, 1, 2,..., so we consider a series solution of the form u(x, t) = 1 2 a [ + an X 1n (x) + b n X 2n (x) ] T n (t) = 1 2 a + The IC requires that u(x, ) = f(x) = 1 2 a + ( a n cos nπx L + b n sin nπx ) e k(nπ/l)2t. (5.24) L ( a n cos nπx L + b n sin nπx ), L < x < L. L As mentioned in Example 3.29, the coefficients of this full Fourier series of f are computed by means of formulas (2.7) (2.9): a n = 1 L b n = 1 L L L L L f(x) cos nπx L f(x) sin nπx L dx, n =, 1, 2,..., dx, n = 1, 2,.... Consequently, the solution of the IBVP is given by (5.24) and (5.25). (5.25)

129 16 5 The Method of Separation of Variables 5.7 Example. Consider the IBVP u t (x, t) = u xx (x, t), 1 < x < 1, t >, u( 1, t) = u(1, t), u x ( 1, t) = u x (1, t), t >, u(x, ) = x + 1, 1 < x < 1. Here L = 1 and f(x) = x + 1, so, by (5.25), a = a n = b n = (x + 1) dx = 2, (x + 1) cos(nπx) dx =, (x + 1) sin(nπx) dx = ( 1) n+1 2 nπ. Hence, by (5.24), the solution of the IBVP is u(x, t) = 1 + ( 1) n+1 2 nπ sin(nπx) π 2t. e n2 Verification with Mathematica R. The input un = ( - 1) (n + 1) (2/(n Pi)) Sin[n Pi x] E ( - n 2 Pi 2 t); Simplify[D[un,t] - D[un,x,x],(un /. x -> - 1) - (un/. x -> 1), (D[un,x]/. x -> - 1) - (D[un,x]/. x -> 1)},n Integers] Plot[1 + Sum[un/. t ->,n,1,5}],x + 1},x, - 1,1}, PlotRange -> - 1.1,1.1}, -.2,2.1}}, PlotStyle -> Black, ImageSize -> Scaled[.3],Ticks -> - 1,1},2}}, AspectRatio ->.5] generates the output,, } and the graph

130 5.2 The Wave Equation 17 Exercises Use separation of variables to solve the IBVP u t (x, t) = u xx (x, t), 1 < x < 1, t >, u( 1, t) = u(1, t), u x ( 1, t) = u x (1, t), t >, u(x, ) = f(x), 1 < x < 1, with function f as indicated. 1 f(x) = 2 sin(2πx) cos(5πx). 2 f(x) = 3x 2. 3, 1 < x, 3 f(x) =, x < 1. 2, 1 < x, 4 f(x) = x 1, < x < 1. Answers to Odd-Numbered Exercises 1 u(x, t) = 2 sin(2πx) e 4π2t cos(5πx) e 25π2t. 3 u(x, t) = 3/2 + [( 1) n 1](3/(nπ)) sin(nπx) e n2 π 2t. 5.2 The Wave Equation Generally speaking, the method of separation of variables is applied to IBVPs for the wave equation in much the same way as for the heat equation. Here, however, the time component satisfies a second-order ODE, which introduces a second IC Vibrating String with Fixed Endpoints We know from Section 4.3 that the corresponding IBVP is u tt (x, t) = c 2 u xx (x, t), < x < L, t >, (PDE) u(, t) =, u(l, t) =, t >, (BCs) u(x, ) = f(x), u t (x, ) = g(x), < x < L. (ICs) Since, as in the earlier problems, the PDE and BCs are linear and homogeneous, we seek a solution of the form u(x, t) = X(x)T (t),

131 18 5 The Method of Separation of Variables where, clearly, neither X nor T can be the zero function. Replacing this in the PDE, we obtain X(x)T (t) = c 2 X (x)t (t), which, divided by c 2 X(x)T (t), yields 1 T (t) c 2 T (t) = X (x) = λ, (5.26) X(x) where λ = const is the separation constant. At the same time, it can be seen that the BCs lead to X() = and X(L) = ; hence, X satisfies X (x) + λx(x) =, < x < L, X() =, X(L) =. This regular S L problem was solved in Example 3.16, where it was shown that its eigenvalue-eigenfunction pairs are ( ) 2 nπ λ n =, X n (x) = sin nπx, n = 1, 2,.... (5.27) L L From (5.26) we deduce that for each eigenvalue λ n, we obtain a function T n that satisfies the equation T n (t) + c 2 λ n T n (t) =, n = 1, 2,..., whose general solution (since λ n > ) is T n (t) = b 1n cos nπct L + b 2n sin nπct L, b 1n, b 2n = const. (5.28) In view of (5.27) and (5.28), we assume that the solution of the IBVP has the series representation u(x, t) = = X n (x)t n (t) sin nπx ( b 1n cos nπct L L + b 2n sin nπct ). (5.29) L Each term of this series satisfies the PDE and the BCs. To satisfy the ICs, we set t = in (5.29) and in the expression of u t (x, t) obtained by differentiating (5.29) term by term: u(x, ) = f(x) = u t (x, ) = g(x) = b 1n sin nπx L, b 2n nπc L sin nπx L.

132 5.2 The Wave Equation 19 We see that the numbers b 1n and b 2n (nπc)/l are the Fourier sine series coefficients of f and g, respectively; so, by (2.1), b 1n = 2 L L b 2n = 2 nπc f(x) sin nπx L L g(x) sin nπx L dx, n = 1, 2,..., dx, n = 1, 2,.... (5.3) Therefore, the solution of the IBVP is given by (5.29) with the b 1n and b 2n computed from (5.3). 5.8 Remark. The terms in the expansion of u are called normal modes of vibration. Solutions of this type are called standing waves. 5.9 Example. We notice that in the IBVP u tt (x, t) = u xx (x, t), < x < 1, t >, u(, t) =, u(1, t) =, t >, u(x, ) = 3 sin(πx) 2 sin(3πx), u t (x, ) = 4π sin(2πx), < x < 1, the functions f and g are linear combinations of the eigenfunctions. Setting c = 1 and L = 1 in their eigenfunction expansions above, from Theorem 3.23(ii) we deduce that b 11 = 3, b 13 = 2, b 1n = (n = 1, 3), b 22 = 2, b 2n = (n 2). Hence, by (5.29), the solution of the IBVP is u(x, t) = 3 sin(πx) cos(πt) 2 sin(3πx) cos(3πt) + 2 sin(2πx) sin(2πt). 5.1 Example. If the initial conditions in the IBVP in Example 5.9 are replaced by x, < x 1/2, u(x, ) = u t (x, ) =, < x < 1, 1 x, 1/2 < x < 1, then, using (5.3) (with c = 1 and L = 1) and integration by parts, we find that 1/2 b 1n = 2 f(x) sin(nπx) dx + 1/2 = 2 x sin(nπx) dx + 1 1/2 1 1/2 } f(x) sin(nπx) dx (1 x) sin(nπx) dx } = 4 n 2 π 2 sin nπ 2,

133 11 5 The Method of Separation of Variables b 2n = 2 1 g(x) sin(nπx) dx =, n = 1, 2,... ; nπ so, by (5.29), the solution of the IBVP is u(x, t) = 4 nπ n 2 sin sin(nπx) cos(nπt). π2 2 Verification with Mathematica R. The input un = (4/(n 2APi 2)) Sin[n Pi/2] Sin[n Pi x] Cos[n Pi t]; Simplify[D[un,t,t] - D[un,x,x],un /. x ->, un /. x -> 1, D[un,t] /. t -> },n Integers] f = Piecewise[x,<x<1/2},1 - x, 1/2<x<1}}]; Plot[Sum[un/. t ->,n,1,3}],f},x,,1}, PlotRange -> -.1,1.1}, -.1,.6}}, PlotStyle -> Black, ImageSize -> Scaled[.3], Ticks ->.5,1},.5}}, AspectRatio ->.9] generates the output,,, } and the graph Exercises Use separation of variables to solve the IBVP u tt (x, t) = u xx (x, t), < x < 1, t >, u(, t) =, u(1, t) =, t >, u(x, ) = f(x), u t (x, ) = g(t), < x < 1, with functions f and g as indicated.

134 5.2 The Wave Equation f(x) = 3 sin(2πx) + 4 sin(7πx), g(x) = sin(3πx). 2 f(x) = 1, g(x) = 3 sin(πx). 3 f(x) = 2 sin(3πx), g(x) = 2. 4 f(x) = 1, g(x) = x. 5 f(x) = 1, < x 1/2, 2, 1/2 < x < 1, g(x) = 3 sin(2πx). 6 f(x) = 1, g(x) = 2, < x 1/2, 1, 1/2 < x < 1. 7 f(x) = x, < x 1/2, 1, 1/2 < x < 1, g(x) = 1. 8 f(x) = x + 1, g(x) = 1, < x 1/2, 2x, 1/2 < x < 1. Answers to Odd-Numbered Exercises 1 u(x, t) = 3 sin(2πx) cos(2πt) + 4 sin(7πx) cos(7πt) + (1/(3π)) sin(3πx) sin(3πt). 3 u(x, t) = 2 sin(3πx) cos(3πt) + [1 ( 1) n ](4/(n 2 π 2 )) sin(nπx) sin(nπt). 5 u(x, t) = (2/(nπ))[1 ( 1) n 2 + cos(nπ/2)] sin(nπx) cos(nπt) + (3/(2π)) sin(2πx) sin(2πt). 7 u(x, t) = sin(nπx)(1/(n 2 π 2 ))[( 1) n+1 2nπ + nπ cos(nπ/2) + 2 sin(nπ/2)] cos(nπt) + [( 1) n 1](2/(n 2 π 2 )) sin(nπt)} Vibrating String with Free Endpoints The IBVP corresponding to this case is (see Section 4.3) u tt (x, t) = c 2 u xx (x, t), < x < L, t >, (PDE) u x (, t) =, u x (L, t) =, t >, (BCs) u(x, ) = f(x), u t (x, ) = g(x), < x < L. (ICs)

135 112 5 The Method of Separation of Variables Separating the variables, we are led to the eigenvalue-eigenfunction pairs λ n = ( ) 2 nπ, X n (x) = cos nπx, n =, 1, 2,.... L L For λ =, the general solution of the equation satisfied by T is T (t) = 1 2 a a 2t, a 1, a 2 = const, while for λ n, n 1, the functions T n are given by (5.28). Hence, we expect the solution of the IBVP to be of the form u(x, t) = 1 2 a a 2t + cos nπx L ( a 1n cos nπct L + a 2n sin nπct ), (5.31) L where the a 1n and a 2n, n =, 1, 2,..., are constants. Every term in (5.31) satisfies the PDE and the BCs. To satisfy the ICs, we require that consequently, by (2.12), u(x, ) = f(x) = 1 2 a 1 + u t (x, ) = g(x) = 1 2 a 2 + a 1n cos nπx L, a 2n nπc L cos nπx L ; a 1 = 2 L a 1n = 2 L a 2 = 2 L L L L a 2n = 2 nπc f(x) dx, f(x) cos nπx L g(x) dx, L g(x) cos nπx L dx, n = 1, 2,..., dx, n = 1, 2,..., (5.32) and we conclude that the solution of the IBVP is given by (5.31) with the a 1n and a 2n computed by means of (5.32).

136 5.2 The Wave Equation Example. To find the solution of the IBVP u tt (x, t) = u xx (x, t), < x < 1, t >, u x (, t) =, u x (1, t) =, t >, u(x, ) = cos(2πx), u t (x, ) = 2π cos(πx), < x < 1, we remark that c = 1 and L = 1, and that the initial data functions f and g are linear combinations of the eigenfunctions; thus, a 12 = 1, a 1n = (n 2), a 21 = 2, a 2n = (n = 1), so, by (5.31), the solution of the IBVP is u(x, t) = 2 cos(πx) sin(πt) + cos(2πx) cos(2πt) Example. Consider the IBVP u tt (x, t) = u xx (x, t), < x < 1, t >, u x (, t) =, u x (1, t) =, t >, u(x, ) =, < x < 1, u t (x, ) = Again, here we have c = 1 and L = 1; so, by (5.32), a 1n =, n =, 1, 2,..., 1/4 a 2 = 2 g(x) dx + 3/4 1/4 g(x) dx + a 2n = 2 1/4 g(x) cos(nπx) dx + nπ = 2 nπ 3/4 1/4 1 3/4 3/4 1/4 1, 1/4 x 3/4, otherwise. } 3/4 g(x) dx = 2 ( 1) dx = 1, 1/4 g(x) cos(nπx) dx + 1 3/4 ( 1) cos(nπx) dx = 4 n 2 π 2 sin nπ 4 cos nπ 2 Hence, by (5.31), the solution of the IBVP is u(x, t) = 1 2 t 4 n 2 π 2 sin nπ 4 cos nπ 2 } g(x) cos(nπx) dx, n = 1, 2,.... cos(nπx) sin(nπt).

137 114 5 The Method of Separation of Variables Verification with Mathematica R. The input un = - (4/(n 2 Pi 2)) Sin[n Pi/4] Cos[n Pi/2] Cos[n Pi x] Sin[n Pi t]; Simplify[D[un,t,t] - D[un,x,x],D[un,x]/. x ->, D[un,x] /. x -> 1, un /. t -> }, n Integers] f = Piecewise[,<x<1/4}, - 1, 1/4<x<3/4},, 3/4<x<1}}]; Plot[D[ - (1/2) t + Sum[un,n,1,11}],t] /. t ->, f},x,,1}, PlotRange -> -.1,1.1}, - 1.1,.1}}, PlotStyle -> Black, ImageSize -> Scaled[.25],Ticks ->1}, - 1}}, AspectRatio ->.9] generates the output,,, } and the graph 5.13 Remark. IBVPs with mixed BCs are handled similarly. Exercises Use separation of variables to solve the IBVP u tt (x, t) = u xx (x, t), < x < 1, t >, u x (, t) =, u x (1, t) =, t >, u(x, ) = f(x), u t (x, ) = g(t), < x < 1, with functions f and g as indicated. 1 f(x) = 2 3 cos(4πx), g(x) = 2 cos(3πx). 2 f(x) = x 1, g(x) = 2 cos(πx). 3 f(x) = 3 cos(2πx), g(x) = 2x 1. 4 f(x) = x, g(x) = 2x 1.

138 5 f(x) = 2, < x 1/2, 3, 1/2 < x < 1, g(x) = cos(3πx). 6 f(x) = x, g(x) = 2, < x 1/2, 1, 1/2 < x < 1. 7 f(x) = x + 1, < x 1/2, g(x) = 1 x. 8 f(x) = 2x, g(x) = 2, 1/2 < x < 1, 2x + 1, < x 1/2, x + 2, 1/2 < x < The Wave Equation 115 Answers to Odd-Numbered Exercises 1 u(x, t) = 2 3 cos(4πx) cos(4πt) + (2/(3π)) cos(3πx) sin(3πt). 3 u(x, t) = 3 cos(2πx) cos(2πt) + [( 1) n 1](4/(n 3 π 3 )) cos(nπx) sin(nπt). 5 u(x, t) = 5/2 + (2/(nπ)) sin(nπ/2) cos(nπx) cos(nπt) (1/(3π)) cos(3πx) sin(3πt). 7 u(x, t) = 13/8 + t/2 + cos(nπx)(1/(n 2 π 2 ))[2 cos(nπ/2) nπ sin(nπ/2) 2] cos(nπt) + [1 ( 1) n ](2/(n 3 π 3 )) sin(nπt)} Vibrating String with Other Types of Boundary Conditions By way of illustration, we consider a string with one endpoint free and the other fixed, and then a string where one endpoint is fixed and the other has an elastic attachment Example. To apply the method of separation of variables to the IBVP u tt (x, t) = u xx (x, t), < x < 1, t >, u x (, t) =, u(1, t) =, t >, 2, < x 1/2, u(x, ) = u t (x, ) = cos 3πx 1, 1/2 < x < 1, 2, < x < 1, we start, as usual, by seeking the solution in the form u(x, t) = X(x)T (t) with X, T. Replaced in the PDE and BCs, this yields the equation T (t) + λt (t) =, t >,

139 116 5 The Method of Separation of Variables and the regular S L problem X (x) + λx(x) =, < x < 1, X () =, X(1) =. The eigenvalue-eigenfunction pairs of the latter, computed as in Example 5.4, are λ n = (2n 1)2 π 2 (2n 1)πx, X n (x) = cos, n = 1, 2,..., 4 2 so the equation for T generates the solutions T n (t) = c 1n cos (2n 1)πt 2 + c 2n sin Following the general procedure, we now write u(x, t) = cos (2n 1)πt, n = 1, 2, (2n 1)πx (2n 1)πt c 1n cos + c 2n sin 2 2 and use the ICs to determine the coefficients c 1n and c 2n. The first IC leads to u(x, ) = c 1n cos (2n 1)πx. 2 } (2n 1)πt 2 Given that, according to Theorem 3.23(ii), the eigenfunctions X n are orthogonal on the interval x 1, we multiply the above equation by cos((2m 1)πx/2), integrate the new equality term by term from to 1, and arrive at [ ] 4 c 1n = ( 1) n+1 (2n 1)π + sin. (2n 1)π 4 The second IC is (2n 1)π 2 c 2n cos (2n 1)πx 2 so c 22 = 2 3π, c 2n = for n = 2. Hence, the solution of the IBVP is u(x, t) = = cos 3πx 2, [ ] 4 ( 1) n+1 (2n 1)π + sin (2n 1)π 4 cos (2n 1)πx 2 cos (2n 1)πt π cos 3πx 2 sin 3πt 2.

140 5.2 The Wave Equation 117 Verification with Mathematica R. The input un = (4/((2 n - 1) Pi)) (( - 1) (n + 1) + Sin[(2 n - 1) Pi/4]) Cos[(2 n - 1) Pi x/2] Cos[(2 n - 1) Pi t/2]; AddTerm = (2/(3 Pi)) Cos[3 Pi x/2] Sin[3 Pi t/2]; Simplify[D[un + AddTerm,t,t] - D[un + AddTerm,x,x], D[un + AddTerm,x]/. x ->, (un + AddTerm)/. x -> 1, (D[(un + AddTerm),t]/. t -> ) - Cos[3 Pi x/2]},n Integers] f = Piecewise[2,<x<1/2},1, 1/2<x<1}}]; Plot[AddTerm + Sum[un,n,1,11}]/. t ->, f},x,,1}, PlotRange -> -.1,1.1}, -.1,2.1}}, PlotStyle -> Black, ImageSize -> Scaled[.25], Ticks ->.5,1},1,2}}, AspectRatio ->.9] generates the output,,, } and the graph 5.15 Example. In the IBVP u tt (x, t) = u xx (x, t), < x < 1, t >, u(, t) =, u x (1, t) + u(1, t) =, t >, 1, < x 1/2, u(x, ) = u t (x, ) = 1, < x < 1,, 1/2 < x < 1, the substitution u(x, t) = X(x)T (t), X, T, leads to the equation and the regular S L problem T (t) + λt (t) =, t >, X (x) + λx(x) =, < x < 1, X() =, X (1) + X(1) =. As shown in Example 3.18, the eigenvalues and eigenfunctions of the latter,

141 118 5 The Method of Separation of Variables for h = L = 1, are, respectively, λ n = ζ 2 n and X n (x) = sin(ζ n x), n = 1, 2,..., where the first five numbers ζ n, computed in Example 3.25 with rounding at the fourth decimal place, are ζ 1 = 2.288, ζ 2 = , ζ 3 = , ζ 4 = , ζ 5 = Then T n (t) = c 1n cos(ζ n t) + c 2n sin(ζ n t), n = 1, 2,..., so the series form of the solution is u(x, t) = sin(ζ n x)[c 1n cos(ζ n t) + c 2n sin(ζ n t)]. Using this series and the ICs, and constructing the generalized Fourier series expansions u(x, ) = a n sin(ζ n x), u t (x, ) = b n sin(ζ n x), we find that c 1n = a n and c 2n = b n /ζ n. Completing the calculation, in the end we arrive at the approximate solution u(x, t) = sin(2.288 x)[.3891 cos(2.288 t) sin(2.288 t)] + sin( x)[.6947 cos( t) sin( t)] + sin( x)[.412 cos( t) sin( t)] + sin( x)[.469 cos( t) sin( t)] + sin( x)[.446 cos( t) +.16 sin( t)] +. Verification with Mathematica R. The input Approx = Sin[2.288 x] (.3891 Cos[2.288 t] Sin[2.288 t]) + Sin[ x] (.6947 Cos[ t] Sin[ t]) + Sin[ x] (.412 Cos[ t] Sin[ t]) + Sin[ x] (.46 Cos[ t] Sin[ t]) + Sin[ x] (.446 Cos[ t] +.16 Sin[ t]); f = Piecewise[1,<x<1/2},, 1/2<x<1}}]; Chop[FullSimplify[D[Approx,t,t] - D[Approx,x,x], Approx /. x ->, (D[Approx,x]/. x -> 1) + Approx/. x -> 1}, n Integers],.1] p1 = Plot[Approx/. t ->, f},x,,1},plotrange -> -.1,1.1},

142 5.2 The Wave Equation ,1.3}}, PlotStyle -> Black, ImageSize -> Scaled[.2], Ticks ->.5,1},1}}, AspectRatio ->.9]; p2 = Plot[D[Approx,t]/. t ->, 1},x,,1}, PlotRange -> -.1,1.1}, -.1,1.3}}, PlotStyle -> Black, ImageSize -> Scaled[.2], Ticks ->1},1}}, AspectRatio ->.9]; Show[GraphicsRow[p1,p2}]] generates the output,, } and the graphs Exercises Use separation of variables to solve the IBVP consisting of the PDE and ICs u tt (x, t) = u xx (x, t), < x < 1, t >, u(x, ) = f(x), u t (x, ) = g(x), < x < 1, with BCs and functions f and g as indicated. Where appropriate, round the numbers to the fourth decimal place. 1 u x (, t) =, u(1, t) =, f(x) = 3 cos(πx/2) cos(3πx/2), g(x) = 1. 2 u x (, t) =, u(, t) =, f(x) = 3 u(, t) =, u x (1, t) =, f(x) = x 1, g(x) = 2 sin(πx/2) 4 sin(5πx/2). 1, < x 1/2,, 1/2 < x < 1, 4 u(, t) =, u x (1, t) =, f(x) = 1, g(x) = x. 5 u(, t) =, u x (1, t) + u(1, t) =, f(x) = 1, g(x) = x. 6 u(, t) =, u x (1, t) + u(1, t) =, f(x) = x 1,, < x 1/2, g(x) = 2, 1/2 < x < 1. g(x) = x.

143 12 5 The Method of Separation of Variables Answers to Odd-Numbered Exercises 1 u(x, t) = [( 1) n 8/((2n 1) 2 π 2 )] cos((2n 1)πx/2) sin((2n 1)πt/2) + 3 cos(πx/2) cos(πt/2) cos(3πx/2) cos(3πt/2). 3 u(x, t) = [( 1) n+1 8 4(2n 1)π]/((2n 1) 2 π 2 ) sin((2n 1)πx/2) cos((2n 1)πt/2) + (4/π) sin(πx/2) sin(πt/2) (8/(5π)) sin(5πx/2) sin(5πt/2). 5 u(x, y) = sin(2.288 x)[ cos(2.288 t) sin(2.288 t)] + sin( x)[.3134 cos( t).318 sin( t)] + sin( x)[.2775 cos( t) +.77 sin( t)] + sin( x)[.1629 cos( t).29 sin( t)] + sin( x)[.1499 cos( t) +.14 sin( t)] The Laplace Equation In this section, we discuss the solution of two types of problems: one formulated in terms of Cartesian coordinates, and one where the use of polar coordinates is more appropriate Laplace Equation in a Rectangle Consider the equilibrium temperature in a uniform rectangle D = (x, y) : x L, y K}, L, K = const, with no sources and with temperature prescribed on the boundary. The corresponding problem is (see Section 4.2) u xx (x, y) + u yy (x, y) =, < x < L, < y < K, u(, y) = f 1 (y), u(l, y) = f 2 (y), < y < K, u(x, ) = g 1 (x), u(x, K) = g 2 (x), < x < L. Although this is a boundary value problem, it turns out that the method of separation of variables can still be applied. However, since here the BCs are nonhomogeneous, we need to do some additional preliminary work. By the principle of superposition (see Theorem 1.2), we can seek the solution of the above BVP as u(x, y) = u 1 (x, y) + u 2 (x, y), where both u 1 and u 2 satisfy the PDE, u 1 satisfies the BCs with f 1 (y) =,

144 5.3 The Laplace Equation 121 f 2 (y) =, and u 2 satisfies the BCs with g 1 (x) =, g 2 (x) = ; thus, the BVP for u 1 is (u 1 ) xx (x, y) + (u 1 ) yy (x, y) =, < x < L, < y < K, (PDE) u 1 (, y) =, u 1 (L, y) =, < y < K, u 1 (x, ) = g 1 (x), u 1 (x, K) = g 2 (x). We seek a solution of this BVP of the form u 1 (x, y) = X(x)Y (y). (BCs) Clearly, for the same reason as in the case of the heat and wave equations, neither X nor Y can be the zero function. Replacing in the PDE, we obtain X (x)y (y) + X(x)Y (y) =, from which, on division by X(x)Y (y), we find that X (x) X(x) = Y (y) Y (y) = λ, where λ = const is the separation constant. The two homogeneous BCs yield X()Y (y) =, X(L)Y (y) =, < y < K. Since, as already mentioned, Y (y) cannot be identically equal to zero, we conclude that X() =, X(L) =. Hence, X is the solution of the regular Sturm Liouville problem X (x) + λx(x) =, < x < L, X() =, X(L) =. From Example 3.16 we know that the eigenvalue-eigenfunction pairs of this problem are ( ) 2 nπ λ n =, X n (x) = sin nπx, n = 1, 2,.... (5.33) L L Then for each n = 1, 2,..., we seek functions Y n that satisfy ( ) 2 nπ Y n (y) Y n (y) =, < y < K. L By Remark 1.4(i), the general solution of the equation for Y n can be written as Y n (y) = C 1n sinh nπ(y K) L + C 2n sinh nπy L, C 1n, C 2n = const. (5.34)

145 122 5 The Method of Separation of Variables We have chosen this combination of hyperbolic functions because when the nonhomogeneous BCs are applied, one term vanishes at y = and the other at y = K, which is very convenient for the computation of the coefficients C 1n and C 2n. In view of (5.33) and (5.34), it seems reasonable to expect u 1 to be of the form u 1 (x, y) = X n (x)y n (y) = sin nπx ( nπ(y K) C 1n sinh + C 2n sinh nπy ), (5.35) L L L where each term satisfies the PDE and the two homogeneous BCs. To satisfy the two remaining BCs, we must have u 1 (x, ) = g 1 (x) = u 1 (x, K) = g 2 (x) = ( C 1n sinh nπk L ( C 2n sinh nπk L ) sin nπx L ) sin nπx L = = b 1n sin nπx L, b 2n sin nπx L ; in other words, we need the Fourier sine series of g 1 and g 2. By (2.1), for n = 1, 2,..., therefore, b 1n = C 1n sinh nπk L b 2n = C 2n sinh nπk L C 1n = sinh(nπk/l) = 2 L C 2n = b 1n = 2 L = 2 L L L csch nπk L b 2n sinh(nπk/l) = 2 nπk csch L L g 1 (x) sin nπx L dx, g 2 (x) sin nπx L dx; L L g 1 (x) sin nπx L g 2 (x) sin nπx L dx, (5.36) dx. (5.37) This means that the function u 1 (x, y) is given by (5.35) with the coefficients C 1n and C 2n computed from (5.36) and (5.37). The procedure for finding u 2 is similar, with the obvious modifications.

146 5.3 The Laplace Equation Example. For the BVP we have u xx (x, y) + u yy (x, y) =, < x < 1, < y < 2, u(, y) =, u(1, y) =, < y < 2, u(x, ) = x, u(x, 2) = 3 sin(πx), < x < 1, L = 1, K = 2, f 1 (y) =, f 2 (y) =, g 1 (x) = x, g 2 (x) = 3 sin(πx). First, integrating by parts, we see that so, by (5.36), 1 x sin(nπx) dx = ( 1) n+1 1 nπ ; C 1n = ( 1) n 2 csch(2nπ), n = 1, 2,.... nπ Second, using (5.37), we easily convince ourselves that C 21 = 3 csch(2π), C 2n =, n = 2, 3,.... Consequently, by (5.35), the solution of the given BVP is u(x, y) = ( 1) n 2 nπ csch(2nπ) sin(nπx) sinh ( nπ(y 2) ) Verification with Mathematica R. The input + 3 csch(2π) sin(πx) sinh(πy). un = ( - 1) n (2/(n Pi)) Csch[2 n Pi] Sin[n Pi x] Sinh[n Pi (y - 2)]; AddTerm = 3 Csch[2 Pi] Sin[Pi x] Sinh[Pi y]; Simplify[D[un + AddTerm,x,x] + D[un + AddTerm,y,y], (un + AddTerm)/. x ->, (un + AddTerm)/. x -> 1, ((un + AddTerm)/. y -> 2) - 3 Sin[Pi x]},n Integers] Plot[AddTerm + Sum[un,n,1,9}] /. y ->, x},x,,1}, PlotRange -.1,1.1}, -.1,1.2}},PlotStyle -> Black, ImageSize -> Scaled[.25], Ticks ->1},1}}, AspectRatio ->.9]

147 124 5 The Method of Separation of Variables generates the output,,, } and the graph 5.17 Remark. As mentioned in Remark 1.4(i) and Example 1.6, the general solution of Y n should be expressed in the most convenient form for the given nonhomogeneous BCs. We have already indicated the best choice when u 1 (x, ) and u 1 (x, K) are prescribed. The other combinations are as follows: (i) If (u 1 ) y (x, ) = g 1 (x), (u 1 ) y (x, K) = g 2 (x), then Y n (y) = C 1n cosh nπ(y K) L + C 2n cosh nπy L. (ii) If u 1 (x, ) = g 1 (x), (u 1 ) y (x, K) = g 2 (x), then Y n (y) = C 1n cosh nπ(y K) L + C 2n sinh nπy L. (iii) If (u 1 ) y (x, ) = g 1 (x), u 1 (x, K) = g 2 (x), then Y n (y) = C 1n cosh nπy L + C 2n sinh nπ(y K). L Exercises Use separation of variables to solve the BVP consisting of the PDE u xx (x, y) + u yy (x, y) =, < x < 1, < y < 2, and BCs as indicated. 1 u(, y) =, u(1, y) =, u(x, ) = 3 sin(2πx), u(x, 2) = sin(3πx). 2 u x (, y) = 2 sin(πy), u x (1, y) = sin(2πy), u(x, ) =, u(x, 2) =. 3 u x (, y) = y 2, u(1, y) = 3 cos(2πy), u y (x, ) =, u y (x, 2) =.

148 5.3 The Laplace Equation u x (, y) =, u x (1, y) =, u(x, ) = 2 cos(πx), u y (x, 2) = x. 5 u x (, y) =, u(1, y) =, 1, < x 1/2, u(x, ) = 2, 1/2 < x < 1, 6 u x (, y) = sin(3πy/4), u x (1, y) = u(x, ) =, u y (x, 2) =. u y (x, 2) = 3 cos(πx/2)., < y 1, 1, 1 < y < 2, Answers to Odd-Numbered Exercises 1 u(x, y) = 3 csch(4π) sin(2πx) sinh(2π(y 2)) + csch(6π) sin(3πx) sinh(3πy). 3 u(x, y) = 3 sech(2π) cosh(2πx) cos(2πy) + 1 x + [( 1) n 1](8/(n 3 π 3 )) sech(nπ/2) sinh(nπ(x 1)/2) cos(nπy/2). 5 u(x, y) = (6/π) sech π cos(πx/2) sinh(πy/2) + (4/(2n 1)π)[( 1) n+1 2 sin((2n 1)π/4)] sech((2n 1)π) cos((2n 1)πx/2) cosh((2n 1)π(y 2)/2) Laplace Equation in a Circular Disc Consider the equilibrium temperature in a uniform circular disc D = (r, θ) : r < α, π < θ π} with no sources, under the condition of prescribed temperature on the boundary r = α = const. Because of the geometry of the body, it is advisable that we write the modeling BVP in terms of polar coordinates r and θ with the pole at the center of the disc. Thus, recalling the form of the Laplacian in polar coordinates given in Remark 4.12(ii), we have u rr (r, θ) + r 1 u r (r, θ) + r 2 u θθ (r, θ) =, < r < α, π < θ < π, (PDE) u(α, θ) = f(θ), π < θ π. (BC) In Subsection 5.3.1, we solved the BVP by fully using the boundary data prescribed at the endpoints of the intervals on which the variables x and y

149 126 5 The Method of Separation of Variables were defined. This time, we have a BC given only at r = 1, so we need to consider additional boundary conditions, given at the remaining endpoints for r and θ, namely, r =, θ = π, and θ = π. The form of these conditions is suggested by analytic requirements based on physical considerations: u (the temperature) and u r, u θ (the heat flux) in a BVP in a disc with a reasonably behaved function f (see Remark 4.27) are expected to be continuous (therefore bounded) everywhere in the disc. Thus, our additional conditions can be formulated as u(r, θ), u r (r, θ) bounded as r +, π < θ < π, u(r, π) = u(r, π), u θ (r, π) = u θ (r, π), < r < α. The PDE and all the BCs except u(α, θ) = f(θ) are homogeneous; consequently, it seems advisable to seek a solution of the form Then the last two (periodic) BCs become u(r, θ) = R(r)Θ(θ), R, Θ. (5.38) Θ( π)r(r) = Θ(π)R(r), Θ ( π)r(r) = Θ (π)r(r), < r < α, from which, since R =, we deduce that Θ( π) = Θ(π), Θ ( π) = Θ (π). In view of (5.38), the PDE can be written as [ R (r) + r 1 R (r) ] Θ(θ) + r 2 R(r)Θ (θ) =, which, on division by r 2 R(r)Θ(θ) and in accordance with the usual argument, yields Θ (θ) Θ(θ) = R (r) + rr (r) r2 = λ, (5.39) R(r) where λ = const is the separation constant. Then Θ is a solution of the periodic Sturm Liouville problem Θ (θ) + λθ(θ) =, π < θ < π, Θ( π) = Θ(π), Θ ( π) = Θ (π). This problem was discussed in Example Hence, for L = π, the eigenvalues and their corresponding eigenfunctions are, respectively, λ =, λ n = n 2, n = 1, 2,..., Θ (θ) = 1, Θ 1n (θ) = cos(nθ), Θ 2n (θ) = sin(nθ), n = 1, 2,.... From the second equality in (5.39) we see that for each n =, 1, 2,..., the corresponding radial component R n satisfies r 2 R n(r) + rr n(r) n 2 R n =, < r < α. (5.4)

150 5.3 The Laplace Equation 127 This is a Cauchy Euler equation, for which, if n, we seek solutions of the form (see Section 1.4) R n (r) = r p, p = const. Replacing in (5.4), we find that [p(p 1) + p n 2 ]r p =, < r < α, or p 2 n 2 =, with roots p 1 = n and p 2 = n. Therefore, the general solution of (5.4) in this case is R n (r) = C 1 r n + C 2 r n, C 1, C 2 = const. (5.41) If n =, then (5.4) reduces to (rr (r)) =, so rr (r) = C 2 = const; in turn, this leads to R (r) = r 1 C2, and then to the general solution R (r) = C 1 + C 2 ln r, C1, C2 = const. (5.42) From (5.38) and the condition of boundedness at r =, we deduce that all the R n () must be finite. As (5.41) and (5.42) indicate, this happens only if C 2 = and C 2 = ; in other words, R (r) = C 1, R n (r) = C 1 r n, n = 1, 2,.... Taking C 1 = 1/2 and C 1 = 1, we write R (r) = 1 2, R n(r) = r n, n = 1, 2,.... In accordance with the separation of variables procedure, we expect the solution of the IBVP to have a series representation of the form u(r, θ) = a R (r)θ (θ) + = 1 2 a + R n (r) [ a n Θ 1n (θ) + b n Θ 2n (θ) ] r n[ a n cos(nθ) + b n sin(nθ) ], (5.43) where each term satisfies the PDE and the three homogeneous BCs. The unknown numerical coefficients a, a n, and b n are determined from the remaining (nonhomogeneous) condition: u(α, θ) = f(θ) = 1 2 a + α n[ a n cos(nθ) + b n sin(nθ) ], π < θ π. (5.44)

151 128 5 The Method of Separation of Variables This is the full Fourier series for f with L = π and coefficients a /2, α n a n, and α n b n ; so, by (2.7) (2.9), a = 1 π π π a n = 1 πα n b n = 1 πα n f(θ) dθ, π π π π f(θ) cos(nθ) dθ, n = 1, 2,..., f(θ) sin(nθ) dθ, n = 1, 2,.... (5.45) Consequently, the solution of the BVP is given by (5.43) with the a, a n, and b n computed by means of (5.45) Remark. From (5.43) and (5.45) it follows that u(, θ) = 1 2 a = 1 2π π π f(θ) dθ; in other words, the temperature at the center of the disk is equal to the average of the temperature on the boundary circle. This result, already used in the proof of Theorem 4.15, is called the mean value property for the Laplace equation Example. In the BVP we have ( u)(r, θ) =, < r < 2, π < θ < π, u(2, θ) = sin θ 32 cos(4θ), π < θ < π, α = 2, f(θ) = sin θ 32 cos(4θ). Since f is a linear combination of the eigenfunctions associated with this problem, we use (5.44) and Theorem 3.23(ii) to deduce that 1 2 a = 1, 2 4 a 4 = 32, 2b 1 = 8, while the remaining coefficients are zero; therefore, a = 2, a 4 = 2, a n = (n =, 4), b 1 = 4, b n = (n = 1), which means that, by (5.43), the solution of the given BVP is u(r, θ) = 1 + 4r sin θ 2r 4 cos(4θ).

152 5.4 Other Equations 129 Exercises Use separation of variables to solve the BVP u rr (r, θ) + r 1 u r (r, θ) + r 2 u θθ (r, θ) =, < r < α, π < θ < π, u(r, θ), u r (r, θ) bounded as r +, u(α, θ) = f(θ), π < θ < π, u(r, π) = u(r, π), u θ (r, π) = u θ (r, π), < r < α, with constant α and function f as indicated. 1 α = 1/2, f(θ) = 3 4 cos(3θ). 2 α = 3, f(θ) = 2 cos(4θ) + sin(2θ). 1, π < θ, 3 α = 2, f(θ) = 3, < θ < π. 1, π < θ, 4 α = 1, f(θ) = θ, < θ < π. Answers to Odd-Numbered Exercises 1 u(r, θ) = 3 32r 3 cos(3θ). 3 u(r, θ) = 1 + [1 ( 1) n ](1/(2 n 2 nπ))r n sin(nθ). 5.4 Other Equations The method of separation of variables can also be applied to problems involving some of the equations mentioned in Section Example. Consider the diffusion-convection IBVP u t (x, t) = 2u xx (x, t) u x (x, t), < x < 1, t >, (PDE) u(, t) =, u(1, t) =, t >, (BCs) u(x, ) = 2e x/4 sin(3πx), < x < 1. (IC) Applying the arguments developed earlier in this chapter, we seek the solution in the form u(x, t) = X(x)T (t) with X, T and, from the PDE and BCs, deduce that the components X and T satisfy 2X (x) X (x) + λx(x) =, < x < 1, X() =, X(1) =, (5.46)

153 13 5 The Method of Separation of Variables and T (t) + λt (t) =, t >. (5.47) The regular Sturm Liouville problem (5.46) is solved by the method used in Example 3.19, and its eigenvalues and eigenfunctions are found to be λ n = 1 8 (16n2 π 2 + 1), X n (x) = e x/4 sin(nπx), n = 1, 2,.... Then (5.47) becomes with general solution T n(t) (16n2 π 2 + 1)T n (t) =, T n (t) = c n e (16n2 π 2 +1)t/8, c n = const. Combining the X n and T n in the usual way, we find that the function u(x, t) = c n e x/4 (16n2 π 2 +1)t/8 sin(nπx) (5.48) satisfies both the PDE and the BCs. The coefficients c n are found from the IC; thus, for t = we must have from which u(x, ) = c n e x/4 sin(nπx) = 2e x/4 sin(3πx), c 3 = 2, c n = (n = 3), so the solution, given by (5.48), of the IBVP is u(x, t) = 2e x/4 (144π2 +1)t/8 sin(3πx). Verification with Mathematica R. The input u = - 2 E (x/4 - (144 Pi 2 + 1) t/8) Sin[3 Pi x]; FullSimplify[D[u,t] - 2 D[u,x,x] + D[u,x],u /. x ->, u /. x -> 1, (u /. t -> ) + 2 E (x/4) Sin[3 Pi x]}] generates the output,,, } Example. Consider the dissipative wave propagation problem u tt (x, t) + 2u t (x, t) = u xx (x, t) 3u x (x, t), < x < 1, t >, u(, t) =, u(1, t) =, t >, u(x, ) = e 3x/2 sin(2πx), u t (x, ) = 2e 3x/2 sin(πx), < x < 1.

154 5.4 Other Equations 131 Writing u(x, t) = X(x)T (t), X, T, and proceeding as in Example 5.2, we find that X (x) 3X (x) + λx(x) =, < x < 1, X() =, X(1) =, (5.49) and T (t) + 2T + λt (t) =, t >. (5.5) The regular S L problem (5.49), solved as in Example 3.19 (see also Remark 3.22), yields the eigenvalues and eigenfunctions λ n = 1 4 (4n2 π 2 + 9), X n (x) = e 3x/2 sin(nπx), n = 1, 2,.... Then, from (5.5) with λ replaced by λ n = 1 4 (4n2 π 2 + 9), we find that T n (t) = e t[ c 1n cos(α n t) + c 2n sin(α n t) ], c 1n, c 2n = const, where α n = 1 2 4n2 π 2 + 5, n = 1, 2,.... Putting together the X n and T n, we now see that the function u(x, t) = e 3x/2 t sin(nπx) [ c 1n cos(α n t) + c 2n sin(α n t) ] (5.51) satisfies both the PDE and the BCs. We now set t = in (5.51) and in its t-derivative, and compute the coefficients c 1n and c 2n from the ICs: hence, u(x, ) = c 1n e 3x/2 sin(nπx) = e 3x/2 sin(2πx), u t (x, ) = ( ) c1n + α n c 2n e 3x/2 sin(nπx) = 2e 3x/2 sin(πx); c 12 = 1, c 1n = (n 2), c 11 + α 1 c 21 = 2, c 1n + α n c 2n = (n 1). From these equalities we find that c 21 = 2 α 1, c 22 = 1 α 2, c 2n = (n 1, 2), so the solution of the IBVP, obtained from (5.51), is [ 2 u(x, t) = e 3x/2 t sin(α 1 t) sin(πx) cos(α 2 t) sin(2πx) α 1 1 ] sin(α 2 t) sin(2πx), α 2

155 132 5 The Method of Separation of Variables where α 1 = 1 2 4π2 + 5, α 2 = π Verification with Mathematica R. The input α1,α2} =(1/2) (4 Pi 2 + 5) (1/2), (1/2) (16 Pi 2 + 5) (1/2)}; u = FullSimplify[E (3 x/2 - t) ((2/α1) Sin[α1 t] Sin[Pi x] - Cos[α2 t] Sin[2 Pi x] - (1/α2) Sin[α2 t] Sin[2 Pi x])]; FullSimplify[D[u,t,t] + 2 D[u,t] - D[u,x,x] + 3 D[u,x], u /. x ->, u /. x -> 1, (u /. t -> ) + E (3 x/2) Sin[2 Pi x], (D[u,t] /. t -> ) - 2 E (3 x/2) Sin[Pi x]}] generates the output,,,, } Example. Consider the two-dimensional steady-state diffusion-convection problem u xx (x, y) + u yy (x, y) 2u x (x, y) =, < x < 1, < y < 2, (PDE) u(, y) =, u(1, y) = 3 sin(πy), < y < 2, u(x, ) =, u(x, 2) =, < x < 1. (BCs) Seeking the solution in the form u(x, y) = X(x)Y (y) with X, Y =, from the PDE and the last two (homogeneous) BCs we see that X (x) 2X (x) λx(x) =, < x < 1, (5.52) and Y (y) + λy (y) =, < y < 2, Y () =, Y (2) =. The regular S L problem for Y has been solved in Example 3.16 and has the eigenvalues and eigenfunctions so (5.52) becomes with general solution λ n = n2 π 2 4, Y n(y) = sin nπy, n = 1, 2,..., 2 X n (x) 2X n (x) n2 π 2 X n (x) =, 4 X n (x) = c 1n e s1nx + c 2n e s2nx,

156 5.4 Other Equations 133 where s 1n = n2 π 2 + 4, s 2n = n2 π 2 + 4, c 1n, c 2n = const. Consequently, the function u(x, y) = sin nπy 2 ( c1n e s1nx + c 2n e s2nx) (5.53) satisfies the PDE and the last two BCs. The coefficients c 1n and c 2n are determined from the first two BCs; thus, setting x = and then x = 1 in (5.53), we find that so Then (c 1n + c 2n ) sin nπy =, 2 ( c1n e s1n + c 2n e s2n) sin nπy = 3 sin(πy), 2 c 1n + c 2n =, n = 1, 2,..., c 12 e s12 + c 22 e s22 = 3, c 1n e s1n + c 2n e s2n = (n 2). c 12 = c 22 = 3 e s22 e s12, c 1n = c 2n = (n = 2), which, substituted in (5.53), yield the solution of the given BVP: where u(x, y) = 3 e s22 e s12 ( e s 12x e s22x) sin(πy), s 12 = 1 + π 2 + 1, s 22 = 1 π Verification with Mathematica R. The input s12,s22} =1 + (Pi 2 + 1) (1/2), 1 - (Pi 2 + 1) (1/2)}; u = FullSimplify[(3/(E s22 - E s12)) (E (s12 x) - E (s22 x)) Sin[Pi y]]; FullSimplify[D[u,x,x] + D[u,y,y] - 2 D[u,x],u /. x ->, (u /. x -> 1) + 3 Sin[Pi y],(u /. y -> ),u /. y -> 2}] generates the output,,,, }.

157 134 5 The Method of Separation of Variables Exercises In 1 4, use separation of variables to solve the IBVP u t (x, t) = ku xx (x, t) au x (x, t) + bu(x, t), < x < 1, t >, u(, t) =, u(1, t) =, t >, u(x, ) = f(x), < x < 1, with numbers k, a, b and function f as indicated. 1 k = 1, a = 1, b =, f(x) = 3e x/2 sin(2πx). 2 k = 1, a = 2, b =, f(x) = 1., < x 1/2, 3 k = 1, a = 1, b = 1, f(x) = 1, 1/2 < x < 1. 1, < x 1/2, 4 k = 2, a = 2, b = 1, f(x) = 1, 1/2 < x < 1. In 5 8, use separation of variables to solve the IBVP u tt (x, t) + au t (x, t) + bu(x, t) = c 2 u xx (x, t) du x (x, t), u(, t) =, u(1, t) =, t >, < x < 1, t >, u(x, ) = f(x), u t (x, ) = g(x), < x < 1, with numbers a, b, c, d and functions f, g as indicated. 5 a = 1, b = 1, c = 2, d = 2, f(x) = 2e x/4 sin(3πx), g(x) = e x/4 sin(2πx). 6 a = 1, b = 1, c = 1, d = 1, f(x) = 3, g(x) = 1. 1, < x 1/2, 7 a = 1, b = 2, c = 1, d = 2, f(x) =, 1/2 < x < 1, g(x) = e x sin(2πx). 8 a = 2, b = 1, c = 1/2, d = 1, f(x) = 3e 2x sin(4πx),, < x 1/2, g(x) = 1, 1/2 < x < 1. In 9 12, use separation of variables to solve the BVP consisting of the PDE u xx (x, y) + u yy (x, y) au x (x, y) bu y (x, y) + cu(x, y) =, numbers L, K, a, b, c, and BCs as indicated. < x < L, < y < K,

158 5.5 Equations with More than Two Variables L = 2, K = 1, a =, b = 2, c =, u(, y) =, u(2, y) =, u(x, ) = sin(πx/2), u(x, 1) = 2 sin(πx). 1 L = 1, K = 2, a = 1, b = 1, c =, u(, y) =, u(2, y) =, u(x, ) = 1, u(x, 2) = e x/2 sin(πx). 11 L = 2, K = 1, a =, b = 2, c =, u(, y) = 3e y sin(2πy), u(2, y) = 2e y sin(πy), u(x, ) =, u(x, 1) =. 12 L = 1, K = 2, a = 1, b = 1, c = 1, u(, y) = 2e y/2 sin(3πy/2), 1, < y 1, u(1, y) =, u(x, ) =, u(x, 2) =., 1 < y < 2, Answers to Odd-Numbered Exercises 1 u(x, t) = 3e (1+16π2 )t/4+x/2 sin(2πx). 3 u(x, t) = (1 + 4n 2 π 2 ) 1 ( 1) n+1 8nπ + 4e 1/4 [2nπ cos(nπ/2) + sin(nπ/2)]} e 1/2+(3/4 n2 π 2 )t+x/2 sin(nπx). 5 u(x, t) = (1 + 16π 2 ) 1/2 e (x 2t)/4 sin(2πx) sin((1 + 16π 2 ) 1/2 t) e (x 2t)/4 sin(3πx)[2 cos((1 + 36π 2 ) 1/2 t) + (1 + 36π 2 ) 1/2 sin((1 + 36π 2 ) 1/2 t)]. 7 u(x, t) = 2( π 2 ) 1/2 e (2x t)/2 sin(2πx) sin((1/2)( π 2 ) 1/2 t) + 2(1 + n 2 π 2 ) 1 (11 + 4n 2 π 2 ) 1/2 [ e 1/2 nπ + nπ cos(nπ/2) + sin(nπ/2)]e (2x t 1)/2 sin(nπx) [(11 + 4n 2 π 2 ) 1/2 cos((1/2)(11 + 4n 2 π 2 ) 1/2 t) + sin((1/2)(11 + 4n 2 π 2 ) 1/2 t)]. 9 u(x, y) = e y [csch((4 + π 2 ) 1/2 /2) sin(πx/2) sinh(((4 + π 2 ) 1/2 /2)(y 1)) + 2e 1 csch((1 + π 2 ) 1/2 ) sin(πx) sinh((1 + π 2 ) 1/2 y)]. 11 u(x, y) = e y [3 csch(2(1 + 4π 2 ) 1/2 ) sinh((1 + 4π 2 ) 1/2 (x 2)) sin(2πy) + 2 csch(2(1 + π 2 ) 1/2 ) sinh((1 + π 2 ) 1/2 x) sin(πy)]. 5.5 Equations with More than Two Variables In Section 4.2, we derived a model for heat conduction in a uniform plate, where the unknown temperature function depends on the time variable and two space variables. Here we consider the two-dimensional analog of the wave equation and solve it by the method of separation of variables in terms of both Cartesian and polar coordinates.

159 136 5 The Method of Separation of Variables Vibrating Rectangular Membrane The vibrations of a rectangular membrane with negligible body force and clamped edge are modeled by the IBVP u tt (x, t) = c 2 ( u)(x, y) = c 2[ u xx (x, y) + u yy (x, y) ], < x < L, < y < K, t >, u(, y, t) =, u(l, y, t) =, < y < K, t >, u(x,, t) =, u(x, K, t) =, < x < L, t >, u(x, y, ) = f(x, y), u t (x, y, ) = g(x, y), < x < L, < y < K. (PDE) (BCs) (ICs) Since the PDE and BCs are linear and homogeneous, we seek the solution as a product of a function of the space variables and a function of the time variable: u(x, y, t) = S(x, y)t (t), S, T. (5.54) Reasoning as in previous similar situations, from the PDE and BCs we find that T satisfies the ODE and that S is a solution of the BVP T (t) + λc 2 T (t) =, t >, S xx (x, y) + S yy (x, y) + λs(x, y) =, < x < L, < y < K, S(, y) =, S(L, y) =, < y < K, S(x, ) =, S(x, K) =, < x < L, (5.55) where λ = const is the separation constant. We notice that the PDE and BCs for S are themselves linear and homogeneous, so we use separation of variables once more and seek a solution for (5.55) of the form The PDE in (5.55) then becomes S(x, y) = X(x)Y (y), X, Y =. (5.56) X (x)y (y) + X(x)Y (y) = λx(x)y (y), which, on division by X(x)Y (y), yields X (x) X(x) = λ Y (y) Y (y) = µ, where µ = const is a second separation constant.

160 5.5 Equations with More than Two Variables 137 From the above chain of equalities and the BCs in (5.55) we find in the usual way that X and Y are, respectively, solutions of the regular Sturm Liouville problems and X (x) + µx(x) =, < x < L, X() =, X(L) =, Y (y) + (λ µ)y (y) =, < y < K, Y () =, Y (K) =. (5.57) (5.58) The nonzero solutions that we require for (5.57) are the eigenfunctions of this S L problem, which were computed in Example 3.16 together with the corresponding eigenvalues: µ n = ( nπ L ) 2, X n (x) = sin nπx, n = 1, 2,.... L For each fixed value of n, the function Y satisfies the S L problem (5.58) with µ = µ n. Since this problem is similar to (5.57), its eigenvalues and eigenfunctions (nonzero solutions) are λ nm µ n = ( mπ K ) 2, Y nm(y) = sin mπy, m = 1, 2,.... K We remark that (5.55) may be regarded as a two-dimensional eigenvalue problem with eigenvalues ( ) 2 ( ) 2 ( ) 2 mπ nπ mπ λ nm = µ n + = +, n, m = 1, 2,..., K L K and corresponding eigenfunctions S nm (x, y) = X n (x)y nm (y) = sin nπx L mπy sin, n, m = 1, 2,.... K Going back to the equation satisfied by T, for λ = λ nm we obtain the functions T nm (t) = a nm cos ( λnm ct ) + b nm sin ( λnm ct ), a nm, b nm = const. From this, (5.54), and (5.56), we now conclude that the solution of the original IBVP should have a representation of the form u(x, y, t) = S nm (x, y)t nm (t) = m=1 m=1 sin nπx L sin mπy K [ a nm cos ( λnm ct ) + b nm sin ( λnm ct )]. (5.59)

161 138 5 The Method of Separation of Variables Each term in (5.59) satisfies the PDE and the BCs. To satisfy the first IC, we must have u(x, y, ) = f(x, y) ( = a nm sin nπx ) sin mπy L K. m=1 Multiplying the second equality by sin(pπy/k), p = 1, 2,..., integrating the result with respect to y over [, K], and taking into account formula (2.5) with L replaced by K, we find that K f(x, y) sin pπy K dy = K 2 a np sin nπx L. Multiplying this new equality by sin(qπx/l), q = 1, 2,..., and integrating with respect to x over [, L], we arrive at L K f(x, y) sin pπy K sin qπx L dy dx = LK 4 a qp, from which, replacing q by n and p by m, we obtain a nm = 4 LK L K The second IC is satisfied if which yields b nm = f(x, y) sin nπx L m=1 sin mπy K dy dx, n, m = 1, 2,.... (5.6) u t (x, y, ) = g(x, y) ( = λnm c b nm sin nπx ) sin mπy L K, 4 LK λ nm c L K g(x, y) sin nπx L mπy sin dy dx, K n, m = 1, 2,.... (5.61) In conclusion, the solution of the IBVP is given by series (5.59) with the coefficients a nm and b nm computed by means of (5.6) and (5.61). Other homogeneous BCs are treated similarly.

162 5.5 Equations with More than Two Variables Example. In the IBVP u tt (x, y, t) = u xx (x, y, t) + u yy (x, y, t), < x < 1, < y < 2, t >, u(, y, t) =, u(1, y, t) =, < y < 2, t >, u(x,, t) =, u(x, 2, t) =, < x < 1, t >, [ u(x, y, ) = sin(2πx) sin πy ] 2 2 sin(πy), < x < 1, < y < 2, u t (x, y, ) = 2 sin(2πx) sin πy, < x < 1, < y < 2, 2 we have c = 1, L = 1, and K = 2; thus, comparing the initial data functions with the double Fourier series expansions of f and g, we see that a 21 = 1, a 22 = 2, a 2m = (m = 1, 2), λ21 b 21 = 2, λ2m b 2m = (m 1). Since λ 21 = (2π) 2 + (π/2) 2 = 17π 2 /4, formula (5.59) yields the solution u(x, y, t) = sin(2πx) sin πy 17 πt 2 cos 2 2 sin(2πx) sin(πy) cos ( 5 πt ) π sin(2πx) sin πy 2 sin Verification with Mathematica R. The input 17 πt. 2 u = Sin[2 Pi x] Sin[Pi y/2] Cos[(17) (1/2) Pi t/2] - 2 Sin[2 Pi x] Sin[Pi y] Cos[5 (1/2) Pi t] + (4/(17 (1/2) Pi)) Sin[2 Pi x] Sin[Pi y/2] Sin[17 (1/2) Pi t/2]; FullSimplify[D[u,t,t] - D[u,x,x] - D[u,y,y],u /. x ->, u /. x -> 1, u /. y ->, u /. y -> 2, (u /. t -> ) - Sin[2 Pi x] (Sin[Pi y/2] - 2 Sin[Pi y]),(d[u,t]/. t -> ) - 2 Sin[2 Pi x] Sin[Pi y/2]}] generates the output,,,, }. Exercises Use separation of variables to solve the IBVP consisting of the PDE u tt (x, y, t) = u xx (x, y, t) + u yy (x, y, t), < x < 1, < y < 1, t >, and BCs and ICs as indicated.

163 14 5 The Method of Separation of Variables 1 u(, y, t) =, u(1, y, t) =, u(x,, t) =, u(x, 1, t) =, u(x, y, ) = sin(πx) sin(2πy), u t (x, y, ) = 2 sin(2πx) sin(πy). 2 u(, y, t) =, u(1, y, t) =, u(x,, t) =, u(x, 1, t) =, u(x, y, ) = 3 sin(2πx) sin(πy), u t (x, y, ) = sin(3πx) [ sin(2πy) 2 sin(3πy) ]. 3 u(, y, t) =, u(1, y, t) =, u y (x,, t) =, u y (x, 1, t) =, u(x, y, ) = sin(πx) cos(2πy), u t (x, y, ) = 3 sin(2πx). 4 u x (, y, t) =, u x (1, y, t) =, u(x,, t) =, u(x, 1, t) =, u(x, y, ) = cos(2πx), u t (x, y, ) = 2 cos(πx) sin(3πy). 5 u(, y, t) =, u(1, y, t) =, u(x,, t) =, u(x, 1, t) =, u(x, y, ) = 1, u t (x, y, ) = xy. 6 u(, y, t) =, u(1, y, t) =, u y (x,, t) =, u y (x, 1, t) =, u(x, y, ) = (x 1)y, u t (x, y, ) = 2xy. Answers to Odd-Numbered Exercises 1 u(x, y, t) = sin(πx) sin(2πy) cos( 5 πt) (2/( 5 π)) sin(2πx) sin(πy) sin( 5 πt). 3 u(x, y, t) = sin(πx) cos(2πy) cos( 5 πt) + (3/(2π)) sin(2πx) sin(2πt). 5 u(x, y, t) = (4/π 2 ) (1/(mn))[1 ( 1) m ][1 ( 1) n ] m=1 cos((m 2 + n 2 ) 1/2 πt) + ( 1) m+n (π(m 2 + n 2 )) 1/2 sin((m 2 + n 2 ) 1/2 πt)} sin(nπx) sin(mπy) Vibrating Circular Membrane A two-dimensional body of this type is defined in terms of polar coordinates by r α, π < θ π. If the body force is negligible and the boundary r = α is clamped, then the vertical vibrations of the membrane are described by the IBVP u tt (r, θ, t) = c 2 ( u)(r, θ, t), < r < α, π < θ < π, t >, (PDE) u(α, θ, t) =, u(r, θ, t), u r (r, θ, t) bounded as r +, u(r, π, t) = u(r, π, t), u(r, θ, ) = f(r, θ), π < θ < π, t >, u θ (r, π, t) = u θ (r, π, t), u t (r, θ, ) = g(r, θ), < r < α, t >, < r < α, π < θ < π. (BCs) (ICs)

164 5.5 Equations with More than Two Variables 141 As in Subsection 5.3.2, the BCs at r = and θ = π, π express the continuity (and therefore boundedness) of the displacement (u) and tension (u r, u θ ) in the membrane for reasonably behaved functions f and g. Since the PDE and BCs are linear and homogeneous, we try the method of separation of variables and seek a solution of the form u(r, θ, t) = S(r, θ)t (t), S, T. (5.62) Then, following the standard procedure, from the PDE and BCs we find that S is a solution of the problem ( S)(r, θ) + λs(r, θ) =, < r < α, π < θ < π, S(α, θ) =, S(r, θ), S r (r, θ) bounded as r +, π < θ < π, S(r, π) = S(r, π), S θ (r, π) = S θ (r, π), < r < α, (5.63) where λ = const is the separation constant, while T satisfies T (t) + λc 2 T (t) =, t >. Since the PDE and BCs for S are linear and homogeneous, we seek S of the form S(r, θ) = R(r)Θ(θ), R, Θ. (5.64) Next, recalling the expression of the Laplacian in polar coordinates (see Remark 4.12(ii)), we write the PDE in (5.63) as [ R (r) + r 1 R (r) ] Θ(θ) + r 2 R(r)Θ (θ) + λr(r)θ(θ) =, which, on division by r 2 R(r)Θ(θ), yields Θ (θ) Θ(θ) = R (r) + rr (r) r2 λr 2 = µ, R(r) where µ = const is a second separation constant. Applying the usual argument to the BCs, we now conclude that R and Θ are solutions of Sturm Liouville problems. First, Θ satisfies the periodic S L problem Θ (θ) + µθ(θ) =, π < θ < π, Θ( π) = Θ(π), Θ ( π) = Θ (π), whose eigenvalues and eigenfunctions, computed in Example 3.29, are µ =, Θ (θ) = 1 2, µ n = n 2, Θ 1n (θ) = cos(nθ), Θ 2n (θ) = sin(nθ), n = 1, 2,.... For each n = 1, 2,..., we compute a solution R n of the singular S L problem r 2 R n(r) + rr n(r) + (λr 2 n 2 )R n (r) =, < r < α, R n (α) =, R n (r), R n (r) bounded as r +.

165 142 5 The Method of Separation of Variables This is the singular Sturm Liouville problem (3.14), (3.15) (with m replaced by n) discussed in Section 3.3, which has countably many eigenvalues and a complete set of corresponding eigenfunctions λ nm = ( ξnm α ) 2, R nm(r) = J n ( ξnm r α ), n =, 1, 2,..., m = 1, 2,..., where J n is the Bessel function of the first kind and order n and ξ nm are the zeros of J n. Returning to the ODE satisfied by T, for λ = λ nm = (ξ nm /α) 2 we now obtain the solutions T nm (t) = A nm cos ξ nmct α + B nm sin ξ nmct α, A nm, B nm = const. Suppose, for simplicity, that g = ; then it is clear that the sine term in T nm must vanish. Hence, if we piece together the various components of u according to (5.62) and (5.64), we conclude that the solution of the IBVP should be of the form [ u(r, θ, t) = anm Θ 1n (θ) + b nm Θ 2n (θ) ] R nm (r)t nm (t) = n= m=1 n= m=1 [ anm cos(nθ) + b nm sin(nθ) ] ( ) ξnm r J n cos ξ nm ct α α. Each term in this sum satisfies the PDE, the BCs, and the second IC. To satisfy the first IC, we need to have u(r, θ, ) = f(r, θ) [ = anm cos(nθ) + b nm sin(nθ) ]} ( ) ξnm r J n. α m=1 n= The a nm and b nm are now computed as in other problems of this type, by means of the orthogonality properties of cos(nθ), sin(nθ), and J n (ξ nm r/α) expressed by (2.4) (2.6) with L = π, and by (3.18) Remark. In the circularly symmetric case (that is, when the data functions are independent of θ) we use the solution split u(r, t) = R(r)T (t), R, T. Here, the problem reduces to solving Bessel s equation of order zero. In the end, we obtain [ u(r, t) = am cos ( λm ct ) + b m sin ( λm ct )] ( J λm r ), m=1 where λ m is the same as λ m = (ξ m /α) 2 in the general problem and the a m and b m are determined from the property of orthogonality with weight r on [, α] of the functions J ( λm r ).

166 5.5 Equations with More than Two Variables Example. With α = 1, c = 1, f(r, θ) = 1, and g(r, θ) =, the formula in Remark 5.24 yields ( u(r, ) = a m J λm r ) = 1, u t (r, ) = m=1 ( b m λm J λm r ) =, m=1 so the a m are the coefficients of the expansion of the constant function 1 in the functions J ( λm r ) on (, 1), and b m =. Since, with the numbers rounded to the fourth decimal place, 1 = 1.62 J (2.448 r) J (5.521 r) J ( r) +, it follows that a 1 = 1.62, a 2 = 1.463, a 3 =.8514,..., so the solution of the problem is u(r, t) = 1.62 cos(2.448 t)j (2.448 r) cos(5.521 t)j (5.521 r) cos( t)j ( r) +. Exercises Use separation of variables to compute the first five terms of the series solution of the IBVP u tt (r, θ, t) = u rr (r, θ, t) + r 1 u r (r, θ, t) + r 2 u θθ (r, θ, t), < r < 1, π < θ < π, t >, u(1, θ, t) =, u(r, θ, t), u r (r, θ, t) bounded as r +, u(r, π, t) = u(r, π, t), u(r, θ, ) = f(r, θ), with functions f and g as indicated. 1 f(r, θ) = 2 sin(2θ), g(r, θ) =. 2 f(r, θ) =, g(r, θ) = cos θ. π < θ < π, t >, u θ (r, π, t) = u θ (r, π, t), u t (r, θ, ) = g(r, θ), < r < 1, t >, < r < 1, π < θ < π,

167 144 5 The Method of Separation of Variables 3 f(r, θ) = r sin θ, g(r, θ) =. 4 f(r, θ) =, g(r, θ) = (r 1) cos(2θ). Answers to Odd-Numbered Exercises 1 u(r, θ, t) = [ J 2 ( r) cos( t).3168 J 2 ( r) cos( t) J 2 ( r) cos( t).4762 J 2 ( r) cos( t) J 2 ( r) cos( t) + ] sin(2θ). 3 u(r, θ, t) = [.3382 J 1 ( r) cos( t) J 1 (7.156 r) cos(7.156 t).774 J 1 ( r) cos( t) J 1 ( r) cos( t).375 J 1 ( r) cos( t) + ] sin θ Equilibrium Temperature in a Solid Sphere The steady-state distribution of heat inside a homogeneous sphere of radius α when the (time-independent) temperature is prescribed on the surface and no sources are present is the solution of the BVP u(r, θ, ϕ) =, < r < α, < θ < 2π, < ϕ < π, (PDE) u(α, θ, ϕ) = f(θ, ϕ), u(r, θ, ϕ), u r (r, θ, ϕ) bounded as r +, u(r,, ϕ) = u(r, 2π, ϕ), u θ (r,, ϕ) = u θ (r, 2π, ϕ), < θ < 2π, < ϕ < π, < r < α, < ϕ < π, u(r, θ, ϕ), u ϕ (r, θ, ϕ) bounded as ϕ + and as ϕ π, < r < α, < θ < 2π, (BCs) where f is a known function. The first boundary condition represents the given surface temperature; the remaining ones have been adjoined on the basis of arguments similar to those used in the case of a uniform circular disc. In terms of the spherical coordinates r, θ, ϕ, the PDE above is written as (see Remark 4.12(v)) 1 r 2 (r2 u r ) r + 1 r 2 sin 2 ϕ u θθ + 1 r 2 sin ϕ ((sin ϕ)u ϕ) ϕ =. Since the PDE and all but one of the BCs are homogeneous, we try a solution of the form u(r, θ, ϕ) = R(r)Θ(θ)Φ(ϕ), R, Θ, Φ.

168 5.5 Equations with More than Two Variables 145 As before, the nonhomogeneous BC implies that none of R, Θ, Φ can be the zero function. Replacing in the PDE and multiplying every term by (r 2 sin 2 ϕ)/(rθφ), we find that Θ Θ = (sin2 ϕ) (r2 R ) (sin ϕ) R ( ) (sin ϕ)φ, Φ where the left-hand side is a function of θ and the right-hand side is a function of r and ϕ. Consequently, both sides must be equal to one and the same constant, which, for convenience, we denote by µ. This leads to the pair of equations Θ (θ) + µθ(θ) =, < θ < 2π, (5.65) (r 2 R ) R + 1 ( ) (sin ϕ)φ µ sin ϕ Φ sin 2 ϕ =, < r < α, < ϕ < π. (5.66) (In (5.66), we have also divided by sin 2 ϕ.) Since R, Φ, from the two periodicity conditions in the BVP we deduce in the usual way that Θ() = Θ(2π), Θ () = Θ (2π). (5.67) The periodic S L problem (5.65), (5.67) for Θ is similar to that discussed in Example According to Remark 3.3, it has the eigenvalues µ = m 2, m =, 1, 2,..., and corresponding eigenfunctions (expressed in their most general form) Θ m (θ) = C 1m cos(mθ) + C 2m sin(mθ), which, using Euler s formula, we can rewrite as Θ m (θ) = C 1m eimθ + C 2m e imθ. (5.68) We now operate a second separation of variables, this time in (5.66): ( 1 ) (sin ϕ)φ m2 sin ϕ Φ sin 2 ϕ = (r2 R ) = λ = const; R this yields the equations 1 ( (sin ϕ)φ (ϕ) ) ) + (λ m2 sin ϕ sin 2 Φ(ϕ) =, ϕ < ϕ < π, (5.69) (r 2 R ) λr =, < r < α. (5.7) In (5.69), we substitute ξ = cos ϕ, dξ = sin ϕ dϕ, sin 2 ϕ = 1 cos 2 ϕ = 1 ξ 2, d dξ = 1 sin ϕ d dϕ, Φ(ϕ) = Ψ(ξ)

169 146 5 The Method of Separation of Variables and bring the equation to the form ( (1 ξ 2 )Ψ (ξ) ) ) + (λ m2 1 ξ 2 Ψ(ξ) =, 1 < ξ < 1. (5.71) The boundary conditions for Ψ are obtained from those at ϕ = and ϕ = π in the original problem by recalling that Φ cannot be the zero function: Ψ(ξ), Ψ (ξ) bounded as ξ 1+ and as ξ 1. (5.72) We remark that (5.71), (5.72) is the singular S L problem for the associated Legendre equation mentioned in Section 3.5, with eigenvalues and eigenfunctions λ n = n(n + 1), Ψ(ξ) = P m n (ξ), n = m, m + 1,..., where P m n are the associated Legendre functions; consequently, Φ n (ϕ) = P m n (cos ϕ). (5.73) Turning to the Cauchy Euler equation (5.7) with λ = n(n + 1), we find that its general solution (see Section 1.4) is R n (r) = C 1 r n + C 2 r (n+1), C 1, C 2 = const. When the condition that R(r) and R (r) be bounded as r +, which follows from the second BC in the given BVP, is applied, we find that C 2 =, so R n (r) = r n, n = m, m + 1,.... (5.74) Here, as usual, we have taken C 1 = 1. Finally, we combine (5.68), (5.73), (5.74), (3.33), and (3.34) and write the solution of the original problem in the form u(r, θ, ϕ) = R n (r)θ m (θ)φ nm (ϕ) or, equivalently, = m=,1,2,... n=m,m+1,... m=,1,2,... n=m,m+1,... u(r, θ, ϕ) = = r n( C 1me imθ + C 2me imθ) P m n (cos ϕ), m=n n= m= n m=n n= m= n C nm r n e imθ P m n (cos ϕ) c n,m r n Y n,m (θ, ϕ), (5.75) where Y n,m are the orthonormalized spherical harmonics defined by (3.34).

170 5.5 Equations with More than Two Variables 147 According to the nonhomogeneous BC in the given BVP, we must have u(α, θ, ϕ) = m=n n= m= n c n,m α n Y n,m (θ, ϕ) = f(θ, ϕ). The coefficients c n,m are now determined in the standard way, by means of the orthonormality relations (3.35): c n,m = 1 α n 2π π f(θ, ϕ)ȳn,m(θ, ϕ) sin ϕ dϕ dθ. (5.76) 5.26 Example. If in the preceding problem the upper and lower hemispheres of the sphere with the center at the origin and radius α = 1 are kept at constant temperatures of 2 and 1, respectively, then 2, < θ < 2π, < ϕ < π/2, f(θ, ϕ) = 1, < θ < 2π, π/2 < ϕ < π; so, using (5.76), we find that the first two nonzero coefficients c n,m are c, = π, c 1, = 3 3π. 2 By (5.75) and the expressions of the spherical harmonics given in Example 3.4, the solution of the problem is u(r, θ, ϕ) = π Y, (θ, ϕ) π ry1, (θ, ϕ) + = r cos ϕ +. Exercises Use separation of variables to compute, up to r 2 -terms, the series solution of the BVP 1 1 r 2 (r2 u r ) r + r 2 sin 2 ϕ u 1 ( ) θθ + (sin ϕ)uϕ r 2 sin ϕ =, ϕ < r < 1, < θ < 2π, < ϕ < π, u(1, θ, ϕ) = f(θ, ϕ), u(r, θ, ϕ), u r (r, θ, ϕ) bounded as r +, u(r,, ϕ) = u(r, 2π, ϕ), < θ < 2π, < ϕ < π, u θ (r,, ϕ) = u θ (r, 2π, ϕ), < r < 1, < ϕ < π, u(r, θ, ϕ), u ϕ (r, θ, ϕ) bounded as ϕ + and as ϕ π, with function f as indicated. < r < 1, < θ < 2π,

171 148 5 The Method of Separation of Variables 1 f(θ, ϕ) = 2 cos 2 θ sin 2 ϕ sin θ sin ϕ cos ϕ cos 2 ϕ. 1, < ϕ π/2, 2 f(θ, ϕ) = sin θ sin ϕ cos ϕ, π/2 < ϕ < π. Answers to Odd-Numbered Exercises 1 u(r, θ, ϕ) = 1/3+r 2 (2/3 2 cos 2 ϕ sin θ sin ϕ cos ϕ+cos(2θ) sin 2 ϕ)+.

172 C H A P T E R 6 Linear Nonhomogeneous Problems The method of separation of variables can be applied only if the PDE and BCs are homogeneous. However, a mathematical model may include nonhomogeneous terms (that is, terms not containing the unknown function or its derivatives) in the equation, and/or may have nonhomogeneous (nonzero) data prescribed at its boundary points. In this chapter we show how, in certain cases, such problems can be reduced at least in part to their corresponding homogeneous versions. 6.1 Equilibrium Solutions We have already discussed equilibrium (or steady-state, or time-independent) solutions for the higher-dimensional heat equation (see Section 4.2). Below, we consider equilibrium temperature distributions for a uniform rod Temperature Prescribed at the Endpoints The general IBVP for a rod with internal sources in this case is of the form u t (x, t) = ku xx (x, t) + q(x, t), < x < L, t >, u(, t) = α(t), u(l, t) = β(t), t >, u(x, ) = f(x), < x < L, where α, β, and f are given functions. If α and β are constant and q = q(x), then we may have an equilibrium solution u = u (x), which satisfies the 149

173 15 6 Linear Nonhomogeneous Problems boundary value problem u (x) + q(x) =, < x < L, u () = α, u (L) = β, with q incorporating the factor 1/k. As can be seen, here the IC plays no role in the computation of u. For a time-dependent problem, it seems physically reasonable to expect that if an equilibrium solution exists, then lim u(x, t) = u (x). t This justifies the use of the subscript in the symbol of the equilibrium solution. 6.1 Example. The equilibrium temperature for the IBVP u t (x, t) = 4u xx (x, t) + 2x 1, < x < 2, t >, u(, t) = 1, u(2, t) = 2, t >, u(x, ) = sin πx 2, < x < 2, is obtained by solving the BVP 4u (x) + 2x 1 =, < x < 2, u () = 1, u (2) = 2. Integrating the ODE twice, we find that u (x) = 1 12 x x2 + C 1 x + C 2, C 1, C 2 = const. The constants C 1 and C 2 are now easily found from the BCs. The equilibrium solution is u (x) = 1 12 x x x Flux Prescribed at the Endpoints The general IBVP is u t (x, t) = ku xx (x, t) + q(x, t), < x < L, t >, u x (, t) = α(t), u x (L, t) = β(t), t >, u(x, ) = f(x), < x < L. As above, an equilibrium solution may exist if α and β are constant and q is independent of t. Such a solution satisfies the BVP u (x) + q(x) =, < x < L, u () = α, u (L) = β,

174 6.1 Equilibrium Solutions 151 with q incorporating the factor 1/k, and, if it exists, may once again be regarded as the limit, as t, of the solution u(x, t) of the IBVP. It turns out that in this case the IC cannot be set aside. In fact, this condition now plays an essential role in the computation of the equilibrium temperature. Its input comes through the law of conservation of heat energy in the full timedependent problem, which is equivalent to the PDE itself. Thus, under the assumption made above on q, α, and β, integrating the PDE term by term from to L and using the BCs, we obtain d dt L u(x, t) dx = k L u xx (x, t) dx + L = k [ u x (L, t) u x (, t) ] + = k [ β(t) α(t) ] + L q(x) dx L q(x) dx. q(x) dx Given that the lateral (cylindrical) surface of the rod is insulated, an equilibrium temperature cannot physically exist unless the total contribution of the sources in the rod and of the heat flux through its endpoints is zero; that is, k [ β(t) α(t) ] + L q(x, t) dx =. If this happens, then, as the above equality shows, L u(x, t) dx is constant for all t >, and so, in other words, L u(x, ) dx = lim L u (x) dx = L t L u(x, t) dx; 6.2 Example. The equilibrium temperature for the IBVP f(x) dx. (6.1) u t (x, t) = 4u xx (x, t) + γx + 24, < x < 2, t >, u x (, t) = 2, u x (2, t) = 14, t >, u(x, ) = π sin πx 2 + 1, < x < 2,

175 152 6 Linear Nonhomogeneous Problems where γ = const, satisfies the BVP The ODE yields 4u (x) + γx + 24 =, < x < 2, u () = 2, u (2) = 14. u (x) = 1 8 γx2 6x + C 1, and the BCs imply that C 1 = 2 and γ = 48. This is the only value of γ for which an equilibrium temperature exists. Integrating again, we get u (x) = 2x 3 3x 2 + 2x + C 2. Since L L f(x) dx = u (x) dx = 2 2 ( π sin πx ) dx = 6, ( 2x 3 3x 2 + 2x + C 2 ) dx = 2C2 + 4, it follows that, by (6.1), C 2 = 1; hence, u (x) = 2x 3 3x 2 + 2x + 1. It is easily verified that, as expected, the total contribution of the source term γx + 24 = 48x + 24 over the length of the rod, and of the heat flux at the endpoints, is zero Mixed Boundary Conditions The same method can be applied to compute the equilibrium solution in the case of mixed BCs. 6.3 Example. To find the equilibrium temperature for the IBVP we need to solve the BVP u t (x, t) = 4u xx (x, t) 16, < x < 2, t >, u(, t) = 5, u x (2, t) = 4, t >, u(x, ) = sin πx 2, < x < 2, u (x) 4 =, < x < L, u () = 5, u (2) = 4. Direct integration shows that u (x) = 2x 2 4x 5.

176 6.1 Equilibrium Solutions Example. The equilibrium temperature for the IBVP satisfies u t (x, t) = 4u xx (x, t) + 32, < x < 2, t >, u x (, t) = 2[u(, t) ], u(2, t) = 1, t >, u(x, ) = sin πx 2, < x < 2, 4u (x) + 32 =, < x < 2, u () = 2[u () ], u (2) = 1, so u (x) = 4x 2 + 7x + 3. Equilibrium temperatures can sometimes also be computed explicitly for heat IBVPs in higher dimensions Equilibrium Temperature in a Circular Annulus In view of the geometry of the domain, in this case it is advisable to use the Laplacian in polar coordinates (see Remark 4.12(ii)). 6.5 Example. The IBVP u t (r, t) = k( u)(r, t) = kr 1 (ru r (r, t)) r, 1 < r < 2, t >, u(1, t) = 3, u(2, t) = 1, t >, u(r, ) = f(r), 1 < r < 2, models heat conduction in a uniform circular annulus with no sources when the inner and outer boundary circles r = 1 and r = 2 are held at two different constant temperatures, and the IC function depends only on r. Since the body and the IC have circular symmetry, we may assume that the solution u is independent of the polar angle θ. Then the equilibrium temperature u (r) satisfies the BVP (ru (r)) =, 1 < r < 2, u (1) = 3, u (2) = 1. Integrating the ODE once, we find that ru (r) = C 1 = const, while a second integration yields the general solution u (r) = C 1 ln r + C 2, C 2 = const. Using the BC at r = 1, we see that C 2 = 3; then the condition at r = 2 leads to C 1 = 4/(ln 2), so u (r) = 4 ln r + 3. ln 2

177 154 6 Linear Nonhomogeneous Problems Verification with Mathematica R. The input u = - (4/Log[2]) Log[r] + 3; FullSimplify[r ( - 1) (D[r D[u,r],r]), (u /. r->1) - 3, (u /. r->2) + 1}] generates the output,, }. Exercises In 1 6, find the equilibrium solution for the IBVP consisting of the PDE u t (x, t) = u xx (x, t) + q(x), < x < 1, t >, and function q, BCs, and IC as indicated. (In 5 and 6, also find the value of γ for which an equilibrium solution exists.) 1 q(x) = 6 12x, u(, t) = 1, u(1, t) = 2, u(x, ) = f(x). 2 q(x) = 4 6x, u(, t) = 3, u x (1, t) = 1, u(x, ) = f(x). 3 q(x) = 4 12x 2, u x (, t) = 3, u x (1, t) = 3[u(1, t) 23/3], u(x, ) = f(x). 4 q(x) = 24x 12x 2, u x (, t) = 3[u(, t) + 1/3], u(1, t) = 2, u(x, ) = f(x). 5 q(x) = γx 4, u x (, t) = 3, u x (1, t) = 4, u(x, ) = 3x 1/12. 6 q(x) = γ + 6x 12x 2, u x (, t) = 2, u x (1, t) = 3, u(x, ) = (157π/12) sin(πx). In 7 12, find the equilibrium solution of the IBVP consisting of the PDE and IC u t (x, t) = ku xx (x, t) au x (x, t) + bu(x, t) + q(x), < x < 1, t >, u(x, ) = f(x), < x < 1, with numbers k, a, b, function q, and BCs as indicated. 7 k = 1, a = 3, b = 2, q(x) = 2x 2 + 2x + 6, u(, t) = 1, u(1, t) = 2. 8 k = 1, a = 1, b = 2, q(x) = 1x+5, u(, t) = 2, u x (1, t) = 5 2/e. 9 k = 4, a = 4, b = 1, q(x) = x x 2, u x (, t) = 1, u x (1, t) = 1 + (7/2) e. 1 k = 4, a = 4, b = 5, q(x) = 9 5x, u x (, t) = 3/2, u(1, t) = e cos 1.

178 6.2 Nonhomogeneous Problems k = 1, a = 2, b = 1, q(x) = 2π cos(πx) + (1 π 2 ) sin(πx), u(, t) = 2, u x (1, t) = 2 [ u(1, t) e π/2 ]. 12 k = 1, a = 1, b = 6, q(x) = 6x 17, u x (, t) = u(, t) + 8, u x (1, t) = 1 + 6e 3. Answers to Odd-Numbered Exercises 1 u (x) = 2x 3 3x 2 + 4x 1. 3 u (x) = x 4 + 2x 2 + 3x 2. 5 γ = 6, u (x) = x 3 + 2x 2 3x 1. 7 u (x) = x 2 + 2x 1. 9 u (x) = (3x 2)e x/2 + x 2 3x. 11 u (x) = (2 x)e x sin(πx). 6.2 Nonhomogeneous Problems Fully nonhomogeneous linear IBVPs are much easier to solve if we can bring them to a form where the PDE and BCs, or at least the latter, are homogeneous. Sometimes, this can be accomplished by means of a substitution involving the equilibrium solution, if one exists, or some other suitably chosen function Time-Independent Sources and Boundary Conditions Consider the IBVP u t (x, t) = ku xx (x, t), < x < L, t >, u(, t) = α, u(l, t) = β, t >, u(x, ) = f(x), < x < L, (6.2) where k, α, β = const and f is a given function. To be able to use the method of separation of variables, we need to make the BCs homogeneous, while keeping the PDE also homogeneous. Proceeding as in Section 6.1, it is easily seen that the equilibrium solution for this problem is u (x) = α + β α L x, < x < L. (6.3)

179 156 6 Linear Nonhomogeneous Problems Then the function v = u u is a solution of the IBVP v t (x, t) = kv xx (x, t), < x < L, t >, v(, t) =, v(l, t) =, t >, v(x, ) = f(x) u (x), < x < L. This new IBVP was solved in Section 5.1 and, by (5.1), its solution is v(x, t) = where, by (5.9) with f replaced by f u, b n = 2 L L b n sin nπx 2 L e k(n π 2 /L)t, [f(x) u (x)] sin nπx L Consequently, the solution of (6.2) is u(x, t) = u (x) + v(x, t) = u (x) + with u and the b n given by (6.3) and (6.4), respectively. The above method also works for an IBVP of the form dx, n = 1, 2,.... (6.4) b n sin nπx L e k(nπ/l)2t, u t (x, t) = ku xx (x, t) + q(x), < x < L, t >, u(, t) = α, u(l, t) = β, t >, u(x, ) = f(x), < x < L, since, as seen in Section 6.1, the computation of the equilibrium temperature distribution takes the source term into account. Other types of BCs are treated similarly if a steady-state solution exists. 6.6 Example. The equilibrium solution u of the IBVP satisfies u t (x, t) = u xx (x, t) + π 2 sin(πx), < x < 1, t >, u(, t) = 1, u(1, t) = 3, t >, u(x, ) = x 2, < x < 1, u (x) + π 2 sin(πx) =, < x < 1, u () = 1, u (1) = 3.

180 6.2 Nonhomogeneous Problems 157 The solution of this BVP is Then the substitution reduces the given IBVP to u (x) = sin(πx) 4x + 1. v(x) = u(x) sin(πx) + 4x 1 v t (x, t) = v xx (x, t), < x < 1, t >, v(, t) =, v(1, t) =, t >, v(x, ) = x 2 + 4x 1 sin(πx), < x < 1, which can be solved by the method of separation of variables. 6.7 Example. If the IBVP u t (x, t) = u xx (x, t) + x γ, < x < 1, t >, u x (, t) =, u x (1, t) =, t >, u(x, ) = x x2 1 6 x3, < x < 1, where γ = const, has a steady-state solution u, then this solution must satisfy u (x) + x γ =, < x < 1, u () =, u (1) =. Integrating the equation once, we find that u (x) = 1 2 x2 + γx + C 1, C 1 = const, which satisfies both boundary conditions only if γ = 1/2 and C 1 =. Another integration yields u (x) = 1 6 x x2 + C 2, C 2 = const. Following the procedure in Example 6.2, we deduce that C 2 =. Then the substitution v(x) = u(x) x3 1 4 x2 reduces the original problem to the IBVP v t (x, t) = v xx (x, t), < x < 1, t >, v x (, t) =, v x (1, t) =, t >, v(x, ) = x 1 2, < x < 1, to which we can apply the method of separation of variables.

181 158 6 Linear Nonhomogeneous Problems The General Case Consider the IBVP u t (x, t) = ku xx (x, t) + q(x, t), < x < L, t >, u(, t) = α(t), u(l, t) = β(t), t >, u(x, ) = f(x), < x < L, where α, β, f, and q are given functions. Also, let p(x, t) = C 1 (t) + C 2 (t)x be a linear polynomial in x satisfying the BCs; that is, p(, t) = α(t) and p(l, t) = β(t). Then C 1 = α and C 2 = (β α)/l, and the substitution u = v + p reduces the given problem to the new IBVP v t (x, t) = kv xx (x, t) + [ q(x, t) p t (x, t) ], < x < L, t >, v(, t) =, v(l, t) =, t >, v(x, ) = f(x) p(x, ), < x < L, which has homogeneous BCs but a nonhomogeneous PDE. Such problems, and similar ones with other types of BCs, are solved by the method of eigenfunction expansion, discussed in the next chapter. 6.8 Example. The IBVP u t (x, t) = u xx (x, t) + xt, < x < 1, t >, u(, t) = 1 t, u(1, t) = t 2, t >, u(x, ) = x, < x < 1, is of the above form, with L = 1, α(t) = 1 t, β(t) = t 2, and q(x, t) = xt; therefore, p(x, t) = 1 t + (t 2 + t 1)x. The substitution u(x, t) = v(x, t) + 1 t + (t 2 + t 1)x now reduces the given IBVP to the problem with homogeneous BCs. v t (x, t) = v xx (x, t) + 1 xt x, < x < 1, t >, v(, t) =, v(1, t) =, t >, v(x, ) = 2x 1, < x < 1,

182 6.2 Nonhomogeneous Problems Example. The IBVP u t (x, t) = u xx (x, t) + xt, < x < 1, t >, u x (, t) = t, u x (1, t) = t 2, t >, u(x, ) = x + 1, < x < 1, needs slightly different handling. Here, we take p x instead of p to be a linear polynomial in x satisfying the BCs; that is, p x (x, t) = C 1 (t) + C 2 (t)x, p x (, t) = t, p x (1, t) = t 2. Then p x (x, t) = t + (t 2 t)x, from which, by integration, we obtain p(x, t) = xt x2 (t 2 t). (Since we need only one such function p, we have taken the constant of integration to be zero.) The substitution u(x, t) = v(x, t) + xt x2 (t 2 t) now reduces the original problem to the IBVP v t (x, t) = v xx (x, t) + t 2 t + xt 1 2 x2 (2t 1) x, v x (, t) =, v x (1, t) =, t >, v(x, ) = x + 1, < x < 1. < x < 1, t >, 6.1 Remark. This technique can also be applied to reduce other types of IBVPs to simpler versions where the PDE is nonhomogeneous but the BCs are homogeneous. In the case of BVPs, we can similarly make two of the BCs homogeneous by means of a suitably chosen linear polynomial. Exercises In 1 6, find a substitution that reduces the IBVP consisting of the PDE u t (x, t) = u xx (x, t) + q(x, t), < x < 1, t >, function q, BCs, and IC as indicated, to an equivalent IBVP with homogeneous BCs. In each case, compute the nonhomogeneous term q in the transformed PDE, and the transformed IC. 1 q(x, t) = x + t 2, u(, t) = t + 1, u(1, t) = 2t t 2, u(x, ) = 2x. 2 q(x, t) = 2x t, u x (, t) = 2 t 2, u x (1, t) = t 3, u(x, ) = x 1.

183 16 6 Linear Nonhomogeneous Problems 3 q(x, t) = 3 2xt, u x (, t) = 3t, u x (1, t) = t 2 + t, u(x, ) = 2 x. 4 q(x, t) = x 2 t + 2, u(, t) = t + 2, u x (1, t) = 1 2t, u(x, ) = x 2 x. 5 q(x, t) = 3x + 2t 4, u x (, t) = 2[u(, t) t], u x (1, t) = 2t + 3, u(x, ) = 2x 3. 6 q(x, t) = 2xt 3t, u(, t) = t 2, u x (1, t) = 4[u(1, t) 3], u(x, ) = x+3. In 7 1, find a substitution that reduces the IBVP consisting of the PDE u tt (x, t) = u xx (x, t) + q(x, t), < x < 1, t >, function q, BCs, and ICs as indicated, to an equivalent IBVP with homogeneous BCs. In each case, compute the nonhomogeneous term q in the transformed PDE, and the transformed ICs. 7 q(x, t) = 2x xt, u(, t) = 2t + 3, u(1, t) = t 2 4, u(x, ) = x + 1, u t (x, ) = x. 8 q(x, t) = x + 2t 1, u(, t) = 3t 1, u(1, t) = t t 2, u(x, ) = x, u t (x, ) = 2x q(x, t) = 2 t + xt, u x (, t) = t 1, u x (1, t) = t 2 + 2, u(x, ) = 4x, u t (x, ) = 2x 1. 1 q(x, t) = t 2 + 2x, u x (, t) = t 2 t, u x (1, t) = 2t + 3, u(x, ) = 1 x, u t (x, ) = x + 2. In 11 14, find a substitution that reduces the BVP consisting of the PDE u xx (x, y) + u yy (x, y) = q(x, y), < x < L, < y < K, numbers L, K, function q, and BCs as indicated, to an equivalent BVP with homogeneous BCs at x = and x = L. In each case, compute the nonhomogeneous term q in the transformed PDE, and the transformed BCs at y = and y = K. 11 L = 1, K = 1, q(x, y) = 2x + y, u(, y) = 2y 2 1, u(1, y) = y y 2, u(x, ) = 3x, u(x, 1) = 1 2x. 12 L = 1, K = 2, q(x, y) = x x 2 y 1, u(, y) = y 2 +2, u(1, y) = 3y 1, u(x, ) = 1 x, u(x, 2) = x L = 2, K = 1, q(x, y) = 2x+3y, u x (, y) = y y 2, u x (2, y) = y 2 +1, u(x, ) = 2x, u(x, 1) = x L = 2, K = 1, q(x, y) = x 4y 1, u x (, y) = 2y 2, u x (2, y) = 1 y 2, u(x, ) = 2x + 1, u(x, 1) = 3 2x.

184 6.2 Nonhomogeneous Problems 161 In 15 18, find a substitution that reduces the IBVP consisting of the PDE u t (x, t) = ku xx (x, t) au x (x, t) + bu(x, t) + q(x, t), < x < 1, t >, numbers k, a, b, function q, BCs, and IC as indicated, to an equivalent IBVP with homogeneous BCs. In each case, compute the nonhomogeneous term q in the transformed PDE, and the transformed IC. 15 k = 1, a = 2, b =, q(x, t) = xt x + 1, u(, t) = 2t + 3, u(1, t) = 4 3t, u(x, ) = x k = 2, a = 1, b = 1, q(x, t) = 3x 2t, u(, t) = t + 4, u(1, t) = 2t 1, u(x, ) = 1 2x. 17 k = 2, a = 1, b = 2, q(x, t) = x 2 + 2xt, u x (, t) = 1 2t, u x (1, t) = t + 3, u(x, ) = x k = 1, a = 2, b = 3, q(x, t) = xt + 3t 2, u x (, t) = t 2, u x (1, t) = 3t + 2, u(x, ) = 2x. Answers to Odd-Numbered Exercises 1 v(x, t) = u(x, t) t 1+(t 2 t+1)x, q(x, t) = 2xt+t 3, v(x, ) = 3x 1. 3 v(x, t) = u(x, t) 3xt (t 2 /2 t)x 2, q(x, t) = x 2 t + x 2 + t 2 2xt 3x 2t + 3, v(x, ) = 2 x. 5 v(x, t) = u(x, t) 2t 3/2 (2t + 3)x, q(x, t) = x + 2t 6, v(x, ) = x 9/2. 7 v(x, t) = u(x, t) 2t 3 (t 2 2t 7)x, q(x, t) = xt, v(x, ) = 8x 2, v t (x, ) = 3x 2. 9 v(x, t) = u(x, t) (t 1)x (1/2)(t 2 t + 3)x 2, q(x, t) = x 2 + xt + t 2 2t + 5, v(x, ) = 5x 3x 2 /2, v t (x, ) = x 2 /2 + x v(x, y) = u(x, y) 2y x(1 + y 3y 2 ), q(x, y) = 4x + y 4, v(x, ) = 2x + 1, v t (x, ) = x. 13 v(x, y) = u(x, y) x(y y 2 ) (x 2 /4)(2y 2 y + 1), q(x, y) = x 2 y 2 + 7y/2 1/2, v(x, ) = 2x x 2 /4, v t (x, ) = 3 + x x 2 /2. 15 v(x, t) = u(x, t) 2t 3 (1 5t)x, q(x, t) = xt + 4x + 1t 3, v(x, ) = v(x, t) = u(x, t) + (2t 1)x (3t/2 + 1)x 2, q(x, t) = 3x 2 t + 3x 2 /2 5xt + 2x + 8t + 3, v(x, ) = x.

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186 C H A P T E R 7 The Method of Eigenfunction Expansion Separation of variables cannot be performed if the PDE and/or BCs are not homogeneous. As we saw in Chapter 6, there are particular situations where we can reduce the problem to an equivalent one with homogeneous PDE and BCs, but this is not always possible. In the general case, the best we can do is make the boundary conditions homogeneous. The eigenfunction expansion technique is designed for IBVPs with a nonhomogeneous equation and homogeneous BCs. 7.1 The Nonhomogeneous Heat Equation Below, we consider problems with several types of BCs Rod with Zero Temperature at the Endpoints We start with the IBVP u t (x, t) = ku xx (x, t) + q(x, t), < x < L, t >, (PDE) u(, t) =, u(l, t) =, t >, (BCs) u(x, ) = f(x), < x < L. (IC) The eigenvalues and eigenfunctions of the corresponding homogeneous problem (q = ) are, respectively (see Section 5.1), λ n = ( ) 2 nπ, X n (x) = sin nπx, n = 1, 2,.... L L 163

187 164 7 The Method of Eigenfunction Expansion Since } X n is a complete set, we may consider for the solution an expansion of the form u(x, t) = c n (t)x n (x). (7.1) Differentiating series (7.1) term by term and recalling that X n + λ n X n =, from the PDE we obtain or c n(t)x n (x) = k c n (t)x n(x) + q(x, t) = k c n (t)λ n X n (x) + q(x, t), [ c n (t) + kλ n c n (t) ] X n (x) = q(x, t). Multiplying this equality by X m (x), integrating from to L, and taking the orthogonality of the X n on [, L] (see Theorem 3.8(iii)) into account, we find that [ c m (t) + kλ m c m (t) ] L L Xm(x) 2 dx = q(x, t)x m (x) dx, which (with m replaced by n) yields the equations c n (t) + kλ nc n (t) = L q(x, t)x n (x) dx, t >, n = 1, 2,.... (7.2) L Xn(x) 2 dx The BCs are automatically satisfied since each of the X n in (7.1) satisfies them. From (7.1) and the IC we see that u(x, ) = f(x) = c n ()X n (x), so, proceeding as above, we arrive at the initial conditions c n () = L f(x)x n (x) dx, n = 1, 2,.... (7.3) L Xn 2(x) dx

188 7.1 The Nonhomogeneous Heat Equation 165 Clearly, this is the same as finding formal expansions q(x, t) = q n (t)x n (x), f(x) = f n X n (x), where the q n (t) and f n are given by the right-hand sides in (7.2) and (7.3), respectively. Then the solution of the IBVP is of the form (7.1) with the coefficients c n (t) computed from (7.2) and (7.3), which, for our problem, reduce to c n (t) + k ( nπ L c n () = f n = 2 L ) 2 c n (t) = q n (t) = 2 L L L f(x) sin nπx L dx. IBVPs with other types of BCs are treated similarly. q(x, t) sin nπx L dx, 7.1 Example. The eigenvalues and eigenfunctions for the IBVP u t (x, t) = u xx (x, t) + π 2 e 24π2t sin(5πx), < x < 1, t >, u(, t) =, u(1, t) =, t >, u(x, ) = 3 sin(4πx), < x < 1, are (here k = 1 and L = 1) Since λ n = n 2 π 2, X n (x) = sin(nπx), n = 1, 2,.... q(x, t) = π 2 e 24π2t sin(5πx), f(x) = 3 sin(4πx) coincide with their eigenfunction expansions, the formulas above yield the IVPs c n (t) + n2 π 2 c n (t) = π 2 e 24π2t, n = 5,, n = 5, t >, c n () = 3, n = 4,, n = 4. Hence, for n = 4 we have c 4 (t) + 16π2 c 4 (t) =, t >, c 4 () = 3,

189 166 7 The Method of Eigenfunction Expansion with solution c 4 (t) = 3e 16π2t ; for n = 5, c 5 (t) + 25π2 c 5 (t) = π 2 e 24π2t, t >, c 5 () =, with solution c 5 (t) = e 24π2t e 25π2t ; and for n 4, 5, c n (t) + n2 π 2 c n (t) =, t >, c n () =, with solution c n (t) =. Consequently, by (7.1), the solution of the IBVP is u(x, t) = c 4 (t)x 4 (x) + c 5 (t)x 5 (x) = 3e 16π2t sin(4πx) + (e 24π2t e 25π2t ) sin(5πx). Verification with Mathematica R. The input u = 3 E ( - 16 Pi 2 t) Sin[4 Pi x] + (E ( - 24 Pi 2 t) - E ( - 25 Pi 2 t)) Sin[5 Pi x]; q, f} =Pi 2 E ( - 24 Pi 2 t) Sin[5 Pi x],3 Sin[4 Pi x]}; Simplify[D[u,t] - D[u,x,x] - q,u /. x ->, u /. x -> 1, (u /. t -> ) - f}] generates the output,,, }. 7.2 Example. The IBVP u t (x, t) = u xx (x, t) + xe t, < x < 1, t >, u(, t) =, u(1, t) =, t >, u(x, ) = x 1, < x < 1, has the same eigenvalues and eigenfunctions as that in Example 7.1. Here, q(x, t) = xe t and f(x) = x 1. Hence, by (7.2), (7.3), and (2.5), and using integration by parts, we find that and c n(t) + n 2 π 2 c n (t) = 2e t c n () = x sin(nπx) dx = ( 1) n+1 2 nπ e t, t >, (7.4) (x 1) sin(nπx) dx = 2 nπ. (7.5)

190 7.1 The Nonhomogeneous Heat Equation 167 The solution of the IVP (7.4), (7.5) is c n (t) = 2 ] } ( 1) n+1 nπ n 2 π 2 1 e t [1 + ( 1)n+1 n 2 π 2 e n2 π 2 t, 1 which means that, by (7.1), the solution of the given IBVP is u(x, t) = 2 ( 1) n+1 nπ n 2 π 2 1 e t [1 + ( 1)n+1 n 2 π 2 1 ] } e n2 π 2 t sin(nπx). (7.6) Verification with Mathematica R. As remarked in Chapter 5, it is impractical to use the infinite series (7.6) in the PDE, BCs, and ICs to show that the function u defined by its sum is the solution of our IBVP. Instead, we cut off the series after, say, seven terms and show graphically that this truncation is a good approximation of the exact solution. First, we write the PDE in the operator form Lu = q; then the input un = (2/(n Pi)) ((( - 1) (n + 1)/(n 2 Pi 2-1)) E ( - t) - (1 + ( - 1) (n + 1)/(n 2 Pi 2-1)) E ( - n 2 Pi 2 t)) Sin[n Pi x]; EqApprox = D[Approx,t] - D[Approx,x,x]; p1 = Plot3D[EqApprox,t,,1},x,,1}, PlotRange ->,1}, Ticks ->1},1},1}}]; p2 = Plot3D[x E ( - t),t,,1},x,,1},plotrange ->,1}, Ticks ->1},1},1}}]; GraphicsRow[p1, p2}] generates the 3D graphs (for < x < 1 and < t < 2) of the function obtained by applying the partial differential operator L to the truncation, and of the nonhomogeneous term q(x, t) = xe t : As can be seen, the two graphs show good agreement, the observable discrepancy between them occurring near t =. Next, the input

191 168 7 The Method of Eigenfunction Expansion Approx /. x ->, Approx /. x -> 1} generates the output, }, confirming that the truncation satisfies the BCs. Finally, the input Plot[Approx/. t ->, x - 1},x,,1},PlotRange -> -.18,1.1}, - 1.1,.1}}, PlotStyle -> Black, ImageSize -> Scaled[.2], Ticks ->1},-1}},AspectRatio ->.9] outputs the graphs of the chosen truncation at t = and of the IC function f(x) = x 1: These show good agreement away from the point x =, where the infinite series (7.6) with t = converges to f() = 1. Exercises Use the method of eigenfunction expansion to find the solution of the IBVP consisting of the PDE and BCs u t (x, t) = u xx (x, t) + q(x, t), < x < 1, t >, u(, t) =, u(1, t) =, t >, and IC and function q as indicated. 1 u(x, ) = sin(2πx) 5 sin(4πx), q(x, t) = 2t sin(2πx). 2 u(x, ) = 2 sin(πx), q(x, t) = [ 3 + π 2 (3t 2) ] sin(πx) + (9π 2 t 2 + 2t) sin(3πx). 3 u(x, ) = sin(πx) + 2 sin(3πx), q(x, t) = e t sin(3πx) sin(5πx). 4 u(x, ) = sin(2πx) + 3 sin(4πx), q(x, t) = [ (π 2 1)e t π 2] sin(πx) + (4π 2 t + 4π 2 + 1) sin(2πx) 5 u(x, ) = sin(πx) + 2 sin(2πx), q(x, t) = (t 1) sin(πx). + 48π 2 sin(4πx).

192 7.1 The Nonhomogeneous Heat Equation u(x, ) = 3 sin(πx) 2 sin(2πx), q(x, t) = π 2 e 4π2t sin(2πx) + (2t + 3) sin(3πx). 1, < x 1/2, 7 u(x, ) = q(x, t) = xt., 1/2 < x < 1, 8 u(x, ) = x, q(x, t) = (x 1)t/2. Answers to Odd-Numbered Exercises 1 u(x, t) = [((8π 4 + 1)/(8π 4 ))e 4π2t + (1/(2π 2 ))t 1/(8π 4 )] sin(2πx) 5e 16π2t sin(4πx). 3 u(x, t) = e π2t sin(πx) + [((18π 2 3)/(9π 2 1))e 9π2t + (1/(9π 2 1))e t ] sin(3πx) + (1/(25π 2 ))(e 25π2t 1) sin(5πx). 5 u(x, t) = [(1/π 2 )(t 1) 1/π 4 + (1 + 1/π 2 + 1/π 4 )e π2t ] sin(πx) + 2e 4π2t sin(2πx). 7 u(x, t) = (2/(n 5 π 5 ))[( 1) n+1 + n 4 π 4 (1 cos(nπ/2))]e n2 π 2 t + ( 1) n (1 n 2 π 2 t)} sin(nπx) Rod with Insulated Endpoints Since the general discussion here is similar to that in Subsection 7.1.1, we proceed directly to an illustration. 7.3 Example. The eigenvalues and eigenfunctions associated with the IBVP are (see Section 5.1) u t (x, t) = u xx (x, t) + 2t + cos(2πx), < x < 1, t >, u x (, t) =, u x (1, t) =, t >, u(x, ) = 1 cos(2πx), < x < 1, 2π2 λ n = n 2 π 2, X n (x) = cos(nπx), n =, 1, 2,... (for convenience, this time we have taken X (x) = 1 instead of 1/2). Reasoning just as in the case of the sine eigenfunctions, we deduce that (7.1) (7.3) still hold except that now n =, 1, 2,... and the X n are cosines. Since q(x, t) = 2t + cos(2πx), f(x) = 1 2π 2 cos(2πx)

193 17 7 The Method of Eigenfunction Expansion are linear combinations of the eigenfunctions, (7.2) (with k = 1) and (7.3) lead to the IVPs 2t, n =, c n(t) + n 2 π 2 c n (t) = 1, n = 2, t >,, n =, 2, 1 c n () = 2π 2, n = 2,, n 2. Thus, for n = we have c (t) = 2t, t >, c () =, with solution c (t) = t 2 ; for n = 2, c 2(t) + 4π 2 c 2 (t) = 1, t >, c 2 () = 1 2π 2, with solution c 2 (t) = (4π 2 ) 1 (1 + e 4π2t ); and for n =, 2, c n(t) + n 2 π 2 c n (t) =, t >, c n () =, with solution c n (t) =. Hence, the solution of the given IBVP is u(x, t) = c (t)x (x) + c 2 (t)x 2 (x) = t (1 + 2t e 4π ) cos(2πx). 4π2 Verification with Mathematica R. The input u = t 2 + (1/(4 Pi 2)) (1 + E ( - 4 Pi 2 t)) Cos[2 Pi x]; q, f} = 2 t + Cos[2 Pi x],(1/(2 Pi 2)) Cos[2 Pi x]; Simplify[D[u,t] - D[u,x,x] - q,d[u,x] /. x ->, D[u,x] /. x -> 1, (u /. t -> ) - f}] generates the output,,, }. Exercises Use the method of eigenfunction expansion to find the solution of the IBVP consisting of the PDE and BCs u t (x, t) = u xx (x, t) + q(x, t), < x < 1, t >, u x (, t) =, u x (1, t) =, t >, and IC and function q as indicated.

194 7.1 The Nonhomogeneous Heat Equation u(x, ) = 2 cos(πx) cos(2πx), q(x, t) = 2 + cos(2πx). 2 u(x, ) = 4 cos(2πx) + 2 cos(3πx), q(x, t) = π 2 cos(2πx) + [ (9π 2 1)e t + 9π 2] cos(3πx). 3 u(x, ) = cos(4πx), q(x, t) = t t cos(πx). 4 u(x, ) = 2 cos(2πx), q(x, t) = π 2 e π2t cos(πx) + (1 2t) cos(2πx). 5 u(x, ) = 2 cos(3πx), q(x, t) = e t cos(πx). 6 u(x, ) = 1, < x 1/2, 1, 1/2 < x < 1, q(x, t) = 3t 4. 7 u(x, ) = 1 3 cos(2x), q(x, t) = 2xt. 8 u(x, ) = x, q(x, t) = (1 x)t. Answers to Odd-Numbered Exercises 1 u(x, t) = 2t + 2e π2t cos(πx) + [1/(4π 2 ) ((4π 2 + 1)/(4π 2 ))e 4π2t ] 3 u(x, t) = (1/2)t [1/π 4 (1/π 2 )t (1/π 4 )e π2t ] cos(πx) + 3e 16π2t cos(4πx). cos(2πx). 5 u(x, t) = 2 + (1/(π 2 1))(e t e π2t ) cos(πx) e 9π2t cos(3πx). 7 u(x, t) = 1 t 2 /2 3e 4π2t cos(2πx) + [1 ( 1) n ](4/(n 6 π 6 ))(e n2 π 2t 1 + n 2 π 2 t) cos(nπx) Rod with Mixed Boundary Conditions The method of eigenfunction expansion works equally well for IBVPs with other combinations of BCs. 7.4 Example. The eigenvalues and eigenfunctions for the IBVP u t (x, t) = u xx (x, t) sin 3πx 5πx + t sin, < x < 1, t >, 2 2 u(, t) =, u x (1, t) =, t >, u(x, ) = sin πx 3πx + 2 sin 2 2, < x < 1, are (see Section 5.1) λ n = (2n 1)2 π 2, X n (x) = sin 4 (2n 1)πx, n = 1, 2,.... 2

195 172 7 The Method of Eigenfunction Expansion Following the general scheme (with k = 1 and L = 1), and noticing that q(x, t) = sin(3πx/2) + t sin(5πx/2), f(x) = sin(πx/2) + 2 sin(3πx/2) are already linear combinations of the X n, we deduce that c n(t) + (2n 1)2 π 2 1, n = 2, c n (t) = t, n = 3, t >, 4, n = 2, 3, 1, n = 1, c n () = 2, n = 2,, n 1, 2. Thus, for n = 1 we have c π2 1 (t) + 4 c 1(t) =, t >, c 1 () = 1, with solution c 1 (t) = e π2 t/4 ; for n = 2, c 2(t) + 9π2 4 c 2(t) = 1, t >, c 2 () = 2, with solution c 2 (t) = [2 + (4/(9π 2 ))]e 9π2 t/4 (4/(9π 2 )); for n = 3, c 3(t) + 25π2 4 c 3 () =, c 3 (t) = t, t >, with solution c 3 (t) = (16/(625π 4 )) e 25π2 t/4 +(4/(25π 2 ))t (16/(625π 4 )); and finally, for n 1, 2, 3, c n (t) + (2n 1)2 π 2 4 c n () =, c n (t) =, t >, with solution c n (t) =. Hence, by (7.1), the solution of the given IBVP is u(x, t) = e π2 t/4 sin πx ( 18π π 2 + e 9π 2 t/4 4 ) 9π 2 sin 3πx 2 ) ( 4 25π 2 t π π 4 e 25π2 t/4 sin 5πx 2.

196 7.1 The Nonhomogeneous Heat Equation 173 Verification with Mathematica R. The input u = E ( - Pi 2 t/4) Sin[Pi x/2] + (((18 Pi 2 + 4)/(9 Pi 2)) E ( - 9 Pi 2 t/4) - 4/(9 Pi 2)) Sin[3 Pi x/2] + ((4/(25 Pi 2)) t - (16/(625 Pi 4)) + (16/(625 Pi 4)) E ( - 25 Pi 2 t/4)) Sin[5 Pi x/2]; q, f} =-Sin[3 Pi x/2] + t Sin[5 Pi x/2],sin[pi x/2] + 2 Sin[3 Pi x/2]}; Simplify[D[u,t] - D[u,x,x] - q,u /. x ->, D[u,x] /. x -> 1, (u /. t -> ) - f}] generates the output,,, }. Exercises Use the method of eigenfunction expansion to find the solution of the IBVP consisting of the PDE u t (x, t) = u xx (x, t) + q(x, t), < x < 1, t >, BCs, IC, and function q as indicated. 1 u(, t) =, u x (1, t) =, u(x, ) = sin(3πx/2), q(x, t) = sin(3πx/2) 2 sin(5πx/2). 2 u(, t) =, u x (1, t) =, u(x, ) = 2 sin(5πx/2), q(x, t) = (1/4)[9π 2 t 2 + 2(4 9π 2 )t 8] sin(3πx/2) + (1/2)(25π 2 4)e t sin(5πx/2). 3 u x (, t) =, u(1, t) =, u(x, ) = cos(πx/2) + 2 sin(5πx/2), q(x, t) = t cos(πx/2). 4 u x (, t) =, u(1, t) =, u(x, ) = cos(πx/2), q(x, t) = (1/4)(2π 2 t + 8 π 2 ) cos(πx/2) + (1/4)(9π 2 t 2 + 8t) cos(3πx/2). Answers to Odd-Numbered Exercises 1 u(x, t) = [((9π 2 4)/(9π 2 ))e 9π2 t/4 + 4/(9π 2 )] sin(3πx/2) + (8/(25π 2 ))(e 25π2 t/4 1) sin(5πx/2). 3 u(x, t) = [(1 + 16/π 4 )e π2 t/4 + 4t/π 2 16/π 4 ] cos(πx/2) + 2e 25π2 t/4 cos(5πx/2).

197 174 7 The Method of Eigenfunction Expansion 7.2 The Nonhomogeneous Wave Equation As an illustration, we examine two different sets of BCs Vibrating String with Fixed Endpoints Consider the IBVP u tt (x, t) = c 2 u xx (x, t) + q(x, t), < x < L, t >, (PDE) u(, t) =, u(l, t) =, t >, (BCs) u(x, ) = f(x), u t (x, ) = g(x), < x < L. (ICs) The eigenvalues and eigenfunctions associated with this problem are (see Section 5.2) ( ) 2 nπ λ n =, X n (x) = sin nπx, n = 1, 2,.... L L Starting with an expansion of the form (7.1) and following the same general procedure as for the heat equation, this time also including the second IC, we find that (7.2) is replaced by L q(x, t)x n (x) dx c n (t) + c2 λ n c n (t) =, t >, n = 1, 2,..., (7.7) L Xn(x) 2 dx that (7.3) remains unchanged, and that c n() = L g(x)x n (x) dx, n = 1, 2,.... (7.8) L Xn 2(x) dx The solution of the IBVP is then given by (7.1) with the c n computed by means of (7.7), (7.3), and (7.8), which in our case take the form c n(t) + c 2 ( nπ L c n () = 2 L c n() = 2 L L L ) 2 c n (t) = 2 L L f(x) sin nπx L dx, g(x) sin nπx L dx. q(x, t) sin nπx L dx, t >,

198 7.2 The Nonhomogeneous Wave Equation Example. For the IBVP u tt (x, t) = u xx (x, t) + π 2 sin(πx), < x < 1, t >, u(, t) =, u(1, t) =, t >, u(x, ) = π, u t (x, ) = 2π sin(2πx), < x < 1, the formulas above with c 2 = 1, L = 1, and functions q(x, t) = π 2 sin(πx), f(x) = π, g(x) = 2π sin(2πx) yield c n (t) + n2 π 2 c n (t) = c n () = 2 c n() = 1 π 2, n = 1,, n = 1, t >, π sin(nπx) dx = [ 1 ( 1) n] 2, n = 1, 2,..., n 2π, n = 2,, n = 2. Hence, for n = 1, c 1(t) + π 2 c 1 (t) = π 2, t >, c 1 () = 4, c 1() =, with solution c 1 (t) = 3 cos(πt) + 1; for n = 2, c 2(t) + 4π 2 c 2 (t) =, t >, c 2 () =, c 2() = 2π, with solution c 2 (t) = sin(2πt); and for n 1, 2, c n(t) + n 2 π 2 c n (t) =, t >, c n () = [ 1 ( 1) n] 2 n, c n () =, with solution c n (t) = [ 1 ( 1) n] 2 n cos(nπt). Consequently, by (7.1), u(x, t) = [3 cos(πt) + 1] sin(πx) + sin(2πt) sin(2πx) [ + 1 ( 1) n ] 2 cos(nπt) sin(nπx). (7.9) n n=3

199 176 7 The Method of Eigenfunction Expansion Verification with Mathematica R. The input un = (1 - ( - 1) n) (2/n) Cos[n Pi t] Sin[n Pi x]; Approx = (3 Cos[Pi t] + 1) Sin[Pi x] + Sin[2 Pi t] Sin[2 Pi x] + Sum[un,n,3,7}]; D[Approx,t,t] - D[Approx,x,x] - Pi 2 Sin[Pi x] // Simplify, Approx /. x->, Approx /. x->1}, (D[Approx,t]/. t->) - 2 Pi Sin[2 Pi x]} Plot[Approx/. t->, Pi},x,,1}, PlotRange->-.2,1.1}, -.7,3.8}}, PlotStyle->Black, ImageSize->Scaled[.2], Ticks->1},Pi}}, AspectRatio->[.7] generates the numerical output,, }, }, which shows that the truncation after 7 terms of u given by (7.9) satisfies the PDE, the BCs, and the second IC exactly, and the graph which indicates that, at t =, this truncation is also a good approximation of the first IC function. Exercises Use the method of eigenfunction expansion to find the solution of the IBVP consisting of the PDE u tt (x, t) = u xx (x, t) + q(x, t), < x < 1, t >, BCs, ICs, and function q as indicated. 1 u(, t) =, u(1, t) =, u(x, ) = sin(πx), u t (x, ) = 3 sin(2πx), q(x, t) = 2 sin(2πx). 2 u(, t) =, u(1, t) =, u(x, ) = 3 sin(πx) + sin(3πx), u t (x, ) = sin(3πx), q(x, t) = (3π 2 4 2π 2 t 2 ) sin(πx) + 9π 2 (1 + t) sin(3πx). 3 u(, t) =, u(1, t) =, u(x, ) = sin(πx), u t (x, ) = 2 sin(πx) + 4 sin(3πx), q(x, t) = t sin(πx).

200 7.2 The Nonhomogeneous Wave Equation u(, t) =, u(1, t) =, u(x, ) =, u t (x, ) = 2 sin(2πx) + 2 sin(4πx), q(x, t) = 4 [ π 2 (π 2 + 1)e 2t] sin(2πx) + 2(16π 2 1) sin t sin(4πx). 5 u(, t) =, u(1, t) =, u(x, ) =, u t (x, ) = 2x 1, q(x, t) = (t + 1) sin(2πx). 6 u(, t) =, u(1, t) =, u(x, ) = u t (x, ) = 1, q(x, t) = 2xt. Answers to Odd-Numbered Exercises 1 u(x, t) = cos(πt) sin(πx), < x 1/2, 2 1/2 < x < 1, +[1/(2π 2 ) (1/(2π 2 )) cos(2πt) (3/(2π)) sin(2πt)] sin(2πx). 3 u(x, t) = [cos(πt) + ((2π 2 1)/π 3 ) sin(πt) + (1/π 2 )t] sin(πx) + (4/(3π)) sin(3πt) sin(3πx). 5 u(x, t) = (1/(8π 3 ))[2π(1 + t) 2π cos(2πt) sin(2πt)] sin(2πx) + [( 1) n+1 1](2/(n 2 π 2 )) sin(nπt) sin(nπx) Vibrating String with Free Endpoints The solution procedure here is similar to that in Subsection 7.2.1, with the obvious changes. 7.6 Example. The IBVP u tt (x, t) = u xx (x, t) t cos(πx), < x < 1, t >, u x (, t) =, u x (1, t) =, t >, u(x, ) = 2, u t (x, ) = 2 cos(2πx), < x < 1, gives rise to the eigenvalues and eigenfunctions (see Section 5.3) λ n = n 2 π 2, X n (x) = cos(nπx), n =, 1, 2,.... Taking c 2 = 1 and L = 1 and seeing that q(x, t) = 1 + t cos(πx), f(x) = 2, g(x) = 2 cos(2πx) coincide with their eigenfunction expansions, we find that (7.7), (7.3), and (7.8) lead to the IVPs 1, n =, c n(t) + n 2 π 2 c n (t) = t, n = 1, t >,, n, 1,

201 178 7 The Method of Eigenfunction Expansion c n () = 2, n =,, n, c n () = 2, n = 2,, n = 2. So for n = we have c (t) = 1, t >, c () = 2, c () =, with solution c (t) = t 2 /2 + 2; for n = 1, c 1 (t) + π2 c 1 (t) = t, t >, c 1 () =, c 1 () =, with solution c 1 (t) = π 3 sin(πt) + π 2 t; for n = 2, c 2(t) + 4π 2 c 2 (t) =, t >, c 2 () =, c 2 () = 2, with solution c 2 (t) = π 1 sin(2πt); and for n, 1, 2, c n(t) + n 2 π 2 c n (t) =, t >, c n () =, c n () =, with solution c n (t) =. Consequently, by (7.1), the solution of the IBVP is [ u(x, t) = t π 2 t 1 ] π 3 sin(πt) cos(πx) 1 sin(2πt) cos(2πx). π Verification with Mathematica R. The input u = (1/2) t ((1/(Pi 2)) t - (1/(Pi 3)) Sin[Pi t]) Cos[Pi x] - (1/Pi) Sin[2 Pi t] Cos[2 Pi x]; q, f, g} =1 + t Cos[Pi x],2, - 2 Cos[2 Pi x]}; Simplify[D[u,t,t] - D[u,x,x] - q,d[u,x] /. x ->, D[u,x] /. x -> 1, (u /. t -> ) - f,(d[u,t] /. t -> ) - g}] generates the output,,,, }. Exercises Use the method of eigenfunction expansion to find the solution of the IBVP consisting of the PDE u tt (x, t) = u xx (x, t) + q(x, t), < x < 1, t >, BCs, ICs, and function q as indicated.

202 7.3 The Nonhomogeneous Laplace Equation u x (, t) =, u x (1, t) =, u(x, ) = 1+2 cos(2πx), u t (x, ) = cos(3πx), q(x, t) = 3. 2 u x (, t) =, u x (1, t) =, u(x, ) = 3 cos(πx), u t (x, ) = cos(πx) + 6π cos(3πx), 3 u x (, t) =, u x (1, t) =, u(x, ) = 2 cos(2πx), u t (x, ) = 1 + cos(πx) cos(2πx), q(x, t) = π 2 (3 t) cos(πx). q(x, t) = 1 + t cos(2πx). 4 u x (, t) =, u x (1, t) =, u(x, ) = cos(3πx), u t (x, ) = 1 + cos(2πx), q(x, t) = (t 2)e t + 4π 2 t cos(2πx). 5 u x (, t) =, u x (1, t) =, u(x, ) = x 1, u t (x,, q(x, t) = x. ) = 6 u x (, t) =, u x (1, t) =, u(x, ) = x, u t (x, ) =, < x 1/2, 1, 1/2 < x < 1, q(x, t) = 2xt. Answers to Odd-Numbered Exercises 1 u(x, t) = 3t 2 / cos(2πt) cos(2πx) + (1/(3π)) sin(3πt) cos(3πx). 3 u(x, t) = t 2 /2 + t + (1/π) sin(πt) cos(πx) + [2 cos(2πt) ((4π 2 + 1)/(8π 3 )) sin(2πt) + (1/(4π 2 ))t] cos(2πx). 5 u(x, t) = 1/2 t 2 /4 + [( 1) n 1](2/(n 4 π 4 ))[(n 2 π 2 +1) cos(nπt) 1] cos(nπx). 7.3 The Nonhomogeneous Laplace Equation We solve some typical boundary value problems for this type of equation in both Cartesian and polar coordinates Equilibrium Temperature in a Rectangle Consider the BVP u xx (x, y) + u yy (x, y) = q(x, y), < x < L, < y < K, u(, y) =, u(l, y) =, < y < K, u(x, ) = f 1 (x), u(x, K) = f 2 (x), < x < L, where, for convenience, the source term has been shifted to the right-hand side of the PDE. The eigenvalues and eigenfunctions associated with this problem, found in Section 5.3, are

203 18 7 The Method of Eigenfunction Expansion λ n = Seeking a solution of the form ( ) 2 nπ, X n (x) = sin nπx, n = 1, 2,.... L L u(x, y) = c n (y) sin nπx L, we deduce, as in Sections 7.1 and 7.2, that the coefficients c n, n = 1, 2,..., satisfy the ODE boundary value problems c n(y) n 2 π 2 c n (y) = 2 L c n () = 2 L c n (K) = 2 L L L L f 1 (x) sin nπx L dx, f 2 (x) sin nπx L dx. q(x, y) sin nπx L dx, < y < K, (7.1) 7.7 Example. In the BVP u xx (x, y) + u yy (x, y) = π 2 sin(πx), < x < 1, < y < 2, u(, y) =, u(1, y) =, < y < 2, u(x, ) = 2 sin(3πx), u(x, 2) = sin(πx), < x < 1, we have L = 1, K = 2, and q(x, y) = π 2 sin(πx), f 1 (x) = 2 sin(3πx), f 2 (x) = sin(πx). Since q, f 1, and f 2 are linear combinations of the eigenfunctions, we can bypass (7.1) and see directly that c n satisfies c n(y) n 2 π 2 π 2, n = 1, c n (y) = < y < 2,, n 1, 2, n = 3, c n () =, n 3, 1, n = 1, c n (2) =, n = 1. Then for n = 1, c 1 (y) π2 c 1 (y) = π 2, < y < 2, c 1 () =, c 1 (2) = 1,

204 7.3 The Nonhomogeneous Laplace Equation 181 with solution c 1 (y) = csch(2π) sinh ( π(y 2) ) 1; for n = 3, c 3(y) 9π 2 c 3 (y) =, < y < 2, c 3 () = 2, c 3 (2) =, with solution c 3 (y) = 2 csch(6π) sinh ( 3π(y 2) ) ; and for n = 1, 3, c n (y) n2 π 2 c n (y) =, < y < 2, c n () =, c n (2) =, with solution c n (y) =. Hence, the solution of the BVP is u(x, y) = [ csch(2π) sinh ( π(y 2) ) + 1 ] sin(πx) Verification with Mathematica R. The input 2 csch(6π) sinh ( 3π(y 2) ) sin(3πx). u = - (Csch[2 Pi] Sinh[Pi (y - 2)] + 1) Sin[Pi x] - 2 Csch[6 Pi] Sinh[3 Pi (y - 2)] Sin[3 Pi x]; q, f, g} =Pi 2 Sin[Pi x],2 Sin[3 Pi x], - Sin[Pi x]}; Simplify[D[u,x,x] + D[u,y,y] - q,u /. x ->, u /. x -> 1, (u /. y -> ) - f,(u /. y -> 2) - g}] generates the output,,,, }. Exercises Use the method of eigenfunction expansion to find the solution of the BVP consisting of the PDE u xx (x, y) + u yy (x, y) = q(x, y), < x < 1, < y < 2, function q, and BCs as indicated. 1 q(x, y) = sin(2πx), u(, y) =, u(1, y) =, u(x, ) = sin(πx) 2 sin(3πx), u(x, 2) = sin(2πx). 2 q(x, y) = π 2 (2y + 3) sin(πx) + (1 4π 2 )e y sin(2πx), u(, y) =, u(1, y) =, u(x, ) = 3 sin(πx) + sin(2πx), 3 q(x, y) = 2y sin(πx), u(, y) =, u(1, y) =, u y (x, ) = x 1, u y (x, 2) =. u y (x, 2) = 2 sin(πx) e 2 sin(2πx).

205 182 7 The Method of Eigenfunction Expansion 4 q(x, y) = π 2 /2, u(, y) =, u(1, y) =, u(x, ) =, u(x, 2) = x/2. 5 q(x, y) = π 2 sin(πy), u(, y) = 2 sin(πy), u(1, y) = sin(πy/2), u(x, ) =, u(x, 2) =. 6 q(x, y) = [ 2(1 π 2 )e x + π 2] sin(πy) (9π 2 /4)(x + 3) sin(3πy/2), u(, y) = sin(πy) + 3 sin(3πy/2), u x (1, y) = 2e sin(πy) + sin(3πy/2), u(x, ) =, u(x, 2) =. 7 q(x, y) = (x 1) sin(πy), u x (, y) =, u(1, y) = y 2, u(x, ) =, u(x, 2) =. 8 q(x, y) = 3x sin(πy/2), u(, y) = y, u(1, y) =, u(x, ) =, u(x, 2) =. 9 q(x, y) = 2 4π 2 (2y 1) cos(2πx), u x (, y) =, u x (1, y) =, u(x, ) = 1 cos(2πx), u(x, 2) = cos(2πx). 1 q(x, y) = 2 + π 2 (2 y) cos(πx) + (4 9π 2 )e 2y cos(3πx), u x (, y) =, u x (1, y) =, u y (x, ) = cos(πx) + 2 cos(3πx), u(x, 2) = 3 + e 4 cos(3πx). 11 q(x, y) = y 2 cos(2πx), u x (, y) =, u x (1, y) =, u(x, ) = 2 cos(πx), u y (x, 2) = cos(2πx). 12 q(x, y) = sin y cos(πx), u x (, y) =, u x (1, y) =, u(x, ) = cos(πx), u(x, 2) = q(x, y) = (1 x 2 ) cos(πy), u(, y) = cos(πy), u(1, y) = 2 cos(πy/2), u y (x, ) =, u y (x, 2) =. 14 q(x, y) = 2 (3π 2 /4) cos(πy/2) + 2(π 2 1)e x cos(πy), u(, y) = 3 cos(πy/2) 2 cos(πy), u x (1, y) = 2e 1 cos(πy), u y (x, ) =, u y (x, 2) =. 15 q(x, y) = x cos(3πy/2), u x (, y) =, u(1, y) = 2 y, u y (x, ) =, u y (x, 2) =. 16 q(x, y) = 4y 2, u(, y) = cos(πy), u(1, y) =, u y (x, ) =, u y (x, 2) =.

206 Answers to Odd-Numbered Exercises 7.3 The Nonhomogeneous Laplace Equation u(x, y) = cosech(2π) sinh(π(y 2)) sin(πx) + [(1/(4π 2 ) 1) cosech(4π) sinh(2πy) (1/(4π 2 )) cosech(4π) sinh(2π(y 2)) 1/(4π 2 )] sin(2πx) + 2 cosech(6π) sinh(3π(y 2)) sin(3πx). 3 u(x, y) = (2/π 3 )csch(2π)[cosh(πy) cosh(π(y 2))] πy} sin(πx) + (2/(n 2 π 2 )) csch(2nπ) sin(nπx) cosh(nπ(y 2)). 5 u(x, y) = cosech(π/2) sinh(πx/2) sin(πy/2) + 1 (cosech π)[sinh(πx) + sinh(π(x 1))]} sin(πy). 7 u(x, y) = (1/π 3 )[sech π sinh(π(x 1)) + π(1 x)] sin(πy) (4/(nπ)) sech(nπ/2) cosh(nπx/2) sin(nπy/2). 9 u(x, y) = 1 y 2 + (2y 1) cos(2πx). 11 u(x, y) = 2 sech(2π) cos(πx) cosh(π(y 2)) + (1/(8π 4 )) cos(2πx) [4π(1 π 2 ) sech(4π) sinh(2πy) + sech(4π) cosh(2π(y 2)) 2π 2 y 2 1]. 13 u(x, y) = 2 csch(π/2) sinh(πx/2) cos(πy/2) (1/π 4 )[π 2 2 π 2 x csch π sinh(πx) (π 4 π 2 + 2) csch π sinh(π(x 1))] cos(πy). 15 u(x, y) = (4/(27π 3 )) sech(3π/2)[3π cosh(3πx/2) + 2 sinh(3π(x 1)/2) 3πx cosh(3π/2)] cos(3πy/2) [1 ( 1) n ](4/(n 2 π 2 )) sech(nπ/2) cosh(nπx/2) cos(nπy/2) Equilibrium Temperature in a Circular Disc The same method can be applied to the nonhomogeneous Laplace equation in polar coordinates. 7.8 Example. In Section 5.3, we established that the eigenvalues and eigenfunctions associated with a BVP such as u rr (r, θ) + r 1 u r (r, θ) + r 2 u θθ (r, t) = 4, u(1, θ) = 2 cos θ sin(2θ), π < θ < π, < r < 1, π < θ < π,

207 184 7 The Method of Eigenfunction Expansion are λ =, Θ (θ) = 1, λ n = n 2, Θ 1n (θ) = cos(nθ), Θ 2n (θ) = sin(nθ), n = 1, 2,.... Seeking a solution of the form u(r, θ) = c (r) + [ c1n (r)θ 1n (θ) + c 2n (r)θ 2n (θ) ] and reasoning exactly as in Example 7.7, we deduce that the coefficients c, c 1n, and c 2n, n = 1, 2,..., satisfy, respectively, c (r) + r 1 c (r) = 4, < r < 1, c (1) =, c 1n (r) + r 1 c 1n (r) n2 r 2 c 1n (r) =, < r < 1, 2, n = 1, c 1n (1) =, n 1, c 2n (r) + r 1 c 2n (r) n2 r 2 c 2n (r) =, < r < 1, 1, n = 2, c 2n (1) =, n 2. Since the BC function is smooth, it follows that, according to the explanation given in Subsection 5.3.2, u(r, θ), u r (r, θ), and u θ (r, θ) must be continuous (hence, bounded) for r 1, π < θ π; consequently, we look for solutions c (r), c 1n (r), and c 2n (r) that remain bounded together with their derivatives as r +. The ODE for n = is integrated by first noting that, after multiplication by r, the left-hand side can be written as (rc ) ; in the end, we find that the desired bounded solution is c (r) = r 2 1. Multiplying the ODEs for n 1 by r 2, we arrive at BVPs for Cauchy Euler equations. Thus, for n = 1, and r 2 c 11(r) + rc 11(r) c 11 (r) =, < r < 1, c 11 (1) = 2, r 2 c 21(r) + rc 21(r) c 21 (r) =, < r < 1, c 21 (1) =,

208 7.3 The Nonhomogeneous Laplace Equation 185 with bounded solutions c 11 (r) = 2r and c 21 (r) = ; for n = 2, and r 2 c 12 (r) + rc 12 (r) 4c 12(r) =, < r < 1, c 12 (1) =, r 2 c 22(r) + rc 22(r) 4c 22 (r) =, < r < 1, c 22 (1) = 1, with bounded solutions c 12 (r) = and c 22 (r) = r 2 ; and for n = 3, 4,..., and r 2 c 1n(r) + rc 1n(r) n 2 c 1n (r) =, < r < 1, c 1n (1) =, r 2 c 2n(r) + rc 2n(r) n 2 c 2n (r) =, < r < 1, c 2n (1) =, with bounded solutions c 1n (r) = and c 2n (r) =. Therefore, the solution of the BVP is u(r, θ) = c (r) + c 1n (r)θ 1n (θ) + c 2n (r)θ 2n (θ) = r r cos θ r 2 sin(2θ). Verification with Mathematica R. The input u = r r Cos[θ] - r 2 Sin[2 θ]; q, f} =4, 2 Cos[θ] - Sin[2 θ]}; Simplify[D[u,r,r] + r ( - 1) D[u,r] + r ( - 2) D[u,θ, θ] - q, (u /. r -> 1) - f}] generates the output, }. Exercises Use the method of eigenfunction expansion to find the solution of the BVP consisting of the PDE u rr (r, θ) + r 1 u r (r, θ) + r 2 u θθ (r, θ) = q(r, θ), < x < 1, π < θ < π, u(r, θ), u r (r, θ) bounded as r +, u(1, θ) = f(θ), π < θ < π, u(r, π) = u(r, π), u θ (r, π) = u θ (r, π), < r < 1, and functions q and f as indicated.

209 186 7 The Method of Eigenfunction Expansion 1 q(r, θ) = 8, f(θ) = sin θ + 2 cos(3θ). 2 q(r, θ) = 9r + 6 cos θ + 3 sin θ, f(θ) = 1 3 sin θ. 3 q(r, θ) = 18r 4 + 8r cos θ + (7r 2 5) sin(3θ), f(θ) = 1 + cos θ + 2 sin(3θ). 4 q(r, θ) = 3 cos θ + 1 cos(3θ), f(θ) = cos θ cos(3θ) sin(4θ). 5 q(r, θ) = (1 r) sin(2θ), f(θ) =, π < θ, 1, < θ < π. 6 q(r, θ) = r cos θ 2 sin(3θ), f(θ) = θ, π < θ, 1, < θ < π. Answers to Odd-Numbered Exercises 1 u(r, θ) = 1 2r 2 + 2r sin θ + 2r 3 cos(3θ). 3 u(r, θ) = 2r 3 r 2 + r 3 cos θ + (r 4 + r 2 ) sin(3θ). 5 u(r, θ) = (1/2)( 4r 3 + 4r 2 + 5r 2 ln r) sin(2θ) + 1/2 + [1 ( 1) n ](1/(nπ))r n sin(nθ). 7.4 Other Nonhomogeneous Equations The method of eigenfunction expansion can be extended to some IBVPs or BVPs for more general partial differential equations. Below, we make a selection from among the linear ones mentioned in Section Example. Consider the IBVP u t (x, t) = u xx (x, t) 2u x (x, t) + u(x, t) + 2te x sin(2πx), u(, t) =, u(1, t) =, t >, u(x, ) =, < x < 1. < x < 1, t >, Separating the variables in the associated homogenous PDE, we arrive at the regular S L problem X (x) 2X (x) + (1 + λ)x(x) =, < x < 1, X() =, X(1) =,

210 7.4 Other Nonhomogeneous Equations 187 so, by the formulas in Remark 3.22 with a = 2, b = 1, c = 1, and L = 1, λ n = n 2 π 2, X n (x) = e x sin(nπx), n = 1, 2,.... This suggests that we should seek the solution of our IBVP in the form u(x, t) = c n (t)e x sin(nπx). The standard procedure now leads to the IVPs c n(t) + n 2 π 2 2t, n = 2, c n (t) =, n = 2, c n () =. Specifically, for n = 2 we have c 2(t) + 4π 2 c 2 (t) = 2t, t >, c 2 () =, t >, from which c 2 (t) = (1/(8π 4 )) ( e 4π2t + 4π 2 t 1 ). For n 2, we find that c n (t) =. Replaced in the expansion above, these coefficients give rise to the solution u(x, t) = 1 8π 4 ( e 4π 2t + 4π 2 t 1 ) e x sin(2πx). Verification with Mathematica R. The input u = (1/(8 Pi 4)) (E ( - 4 Pi 2 t) + 4 Pi 2 t - 1) E x Sin[2 Pi x]; q, f} =2 t E x Sin[2 Pi x],}; Simplify[D[u,t] - D[u,x,x] + 2 D[u,x] - u - q,u /. x ->, u /. x -> 1, (u /. t -> ) - f}] generates the output,,, }. 7.1 Example. The same technique applied to the IBVP u tt (x, t) + 2u t (x, t) = u xx (x, t) u x (x, t) + [2 + (9π )t]ex/2 sin(3πx), < x < 1, t >, u(, t) =, u(1, t) =, t >, u(x, ) =, u t (x, ) = e x/2 sin(3πx), < x < 1, yields the associated S L problem X (x) X (x) + λx(x) =, < x < 1, X() =, X(1) =,

211 188 7 The Method of Eigenfunction Expansion which leads to λ n = n 2 π , X n(x) = e x/2 sin(nπx), n = 1, 2,... ; therefore, we seek the solution of the IBVP in the form u(x, t) = c n (t)e x/2 sin(nπx), where the coefficients c n (t) satisfy c n(t) + 2c n(t) + (n 2 π )c n(t) = c n () =, c n () = 1, n = 3,, n (9π )t, n = 3,, n = 3, t >, The IVP for n = 3 is c 3 (t) + 2c 3 (t) + (9π )c 3(t) = 2 + (9π )t, t >, c 3 () =, c 3() = 1, with solution c 3 (t) = t; the fully homogeneous IVP for n 3 produces the coefficients c n (t) =. Hence, the solution of our problem is u(x, t) = te x/2 sin(3πx). Verification with Mathematica R. The input u = t E (x/2) Sin[3 Pi x]; q, f, g} =(2 + (9 Pi 2 + 1/4) t) E (x/2) Sin[3 Pi x],, E (x/2) Sin[3 Pi x]}; Simplify[D[u,t,t] + 2 D[u,t] - D[u,x,x] + D[u,x] - q,u /. x ->, u /. x -> 1, (u /. t -> ) - f,(d[u,t] /. t -> ) - g}] generates the output,,,, } Example. In the application of the method of eigenfunction expansion to the BVP u xx (x, y) + u yy (x, y) 4u x (x, y) = (y 2 y)e 2x sin(πx), u(, y) =, u(1, y) =, < y < 2, < x < 1, < y < 2, u(x, ) = 2e 2x sin(2πx), u(x, 2) =, < x < 1,

212 7.4 Other Nonhomogeneous Equations 189 the associated S L problem is X (x) 4X (x) + λx(x) =, < x < 1, X() =, X(1) =, from which we find that λ n = n 2 π 2 + 4, X n (x) = e 2x sin(nπx), n = 1, 2,... ; thus, we seek the solution in the form u(x, y) = c n (y)e 2x sin(nπx), where the coefficients c n (y) are shown to satisfy c n(y) (n 2 π 2 + 4)c n (y) = c n () = 2, n = 2,, n = 2, y 2 y, n = 1,, n 1, c n (2) =. < y < 2, The individual coefficients are now easily computed. For n = 1, with solution c 1 (y) = c 1(y) (π 2 + 4)c 1 (y) = y 2 y, < y < 2, c 1 () =, c 1 (2) =, 1 ( 2 csch 2 π2 (π 2 + 4) )[ (π 2 + 5) sinh ( π2 + 4 y ) sinh ( π2 + 4 (y 2) )] + (π 2 + 4)(y y 2 ) 2 } ; for n = 2, c 2(y) 4(π 2 + 1)c 2 (y) =, < y < 2, c 2 () = 2, c 2 (2) =, with solution c 2 (y) = 2 csch ( 4 π ) sinh ( 2 π (y 2) ) ; and for n = 1, 2, the homogeneous data imply, as usual, that c n (y) =. Consequently, the solution of the given BVP is u(x, y) = c 1 (y)e 2x sin(πx) + c 2 (y)e 2x sin(2πx), with c 1 (y) and c 2 (y) as determined above.

213 19 7 The Method of Eigenfunction Expansion Verification with Mathematica R. The input c1 = (1/(Pi 2 + 4) 2) (2 Csch[2 (Pi 2 + 4) (1/2)] ((Pi 2 + 5) Sinh[((Pi 2 + 4) (1/2)) y] - Sinh[((Pi 2 + 4) (1/2)) (y - 2)]) + (Pi 2 + 4) (y - y 2) - 2); c2 = - 2 Csch[4 (Pi 2 + 1) (1/2)] Sinh[2 ((Pi 2 + 1) (1/2)) (y - 2)]; u = c1 E (2 x) Sin[Pi x] + c2 E (2 x) Sin[2 Pi x]; q, f, g} =(y 2 - y) E (2 x) Sin[Pi x],2 E (2 x) Sin[2 Pi x],}; Simplify[D[u,x,x] + D[u,y,y] - 4 D[u,x] - q,u /. x ->, u /. x -> 1, (u /. y -> ) - f,(u /. y -> 2) - g}] generates the output,,,, }. Exercises In 1 4, use the method of eigenfunction expansion to find the solution of the IBVP u t (x, t) = ku xx (x, t) au x (x, t) + bu(x, t) + q(x, t), u(, t) =, u(1, t) =, t >, u(x, ) = f(x), < x < 1, with numbers k, a, b and functions q, f as indicated. < x < 1, t >, 1 k = 1, a = 1, b = 2, q(x, t) = (1/4)(16π 2 11)e t+x/2 sin(2πx), f(x) = e x/2 sin(2πx). 2 k = 2, a = 1, b = 1, q(x, t) = [2 + (18π 2 7/8)(2t 3)]e x/4 sin(3πx), f(x) = 3e x/4 sin(3πx). 3 k = 1, a = 2, b = 2, q(x, t) = 2[1 + (π 2 1)t]e x sin(πx), f(x) = e x sin(πx). 4 k = 1, a = 1, b =, q(x, t) = [4 + (1 + 16π 2 )t]e x/2 sin(2πx), f(x) = 1.

214 7.4 Other Nonhomogeneous Equations 191 In 5 8, use the method of eigenfunction expansion to find the solution of the IBVP u tt (x, t) + au t (x, t) + bu(x, t) = c 2 u xx (x, t) du x (x, t) + q(x, t), u(, t) =, u(1, t) =, t >, u(x, ) = f(x), u t (x, ) = g(x), < x < 1, < x < 1, t >, with numbers a, b, c, d and functions q, f, g as indicated. 5 a = 1, b = 1, c = 1, d = 2, q(x, t) = [ (π 2 + 2)t + 1 ] e x sin(πx), f(x) =, g(x) = e x sin(πx). 6 a = 1, b = 2, c = 1, d = 1, q(x, t) = 4(4π 2 + 3)e t/2+x/2 sin(2πx), f(x) = 4e x/2 sin(2πx), g(x) = 2e x/2 sin(2πx). 7 a = 2, b = 1, c = 1/2, d = 1, q(x, t) = (π 2 + 2)e 2x sin(2πx), f(x) =, g(x) = e 2x sin(2πx). 8 a = 2, b =, c = 1, d = 4, q(x, t) = [(9π 2 + 4)t 2 + 4t + 2]e 2x sin(3πx), f(x) =, g(x) = 1. In 9 12, use the method of eigenfunction expansion to find the solution of the BVP u xx (x, y) + u yy (x, y) au x (x, y) bu y (x, y) + cu(x, y) = q(x, y), u(, y) =, u(l, y) =, < y < K, u(x, ) = f(x), u(x, K) = g(x), < x < L, with numbers L, K and functions q, f, g as indicated. < x < L, < y < K, 9 L = 1, K = 2, a = 2, b = 1, c = 1, q(x, y) = (6 π 2 )e x 2y sin(πx), f(x) = e x sin(πx), g(x) = e x 4 sin(πx). 1 L = 2, K = 1, a = 1, b = 1, c = 2, q(x, y) = [9/2 8π 2 + (4π 2 7/4)y ]e x/2 sin(2πx), f(x) = 2e x/2 sin(2πx), g(x) = e x/2 sin(2πx). 11 L = 2, K = 1, a = 2, b = 1, c = 1, q(x, y) = (5/4)(4π 2 + 1)e x+y/2 sin(πx) sin ( (4π 2 + 1) 1/2 y ), f(x) =, g(x) = e x+1/2 sin ( (4π 2 + 1) 1/2) sin(πx).

215 192 7 The Method of Eigenfunction Expansion 12 L = 1, K = 2, a =, b =, c = 1, q(x, y) = (π 2 + 1) sinh ( 2(π 2 + 1) 1/2) y sin(πx), f(x) = 1, g(x) =. Answers to Odd-Numbered Exercises 1 u(x, t) = e t+x/2 sin(2πx). 3 u(x, t) = [e (1 π2)t 2t]e x sin(πx). 5 u(x, t) = te x sin(πx). 7 u(x, t) = [ e t cos ( (π 2 + 1) 1/2 t ) 1 ] e 2x sin(2πx). 9 u(x, y) = e x 2y sin(πx). 11 u(x, y) = e x+y/2 sin(πx) sin ( (4π 2 + 1) 1/2 y ).

216 C H A P T E R 8 The Fourier Transformations Some problems of practical importance are beyond the reach of the method of eigenfunction expansion. This is the case, for example, when the space variable is defined on the entire real line and where, as a consequence, there are no boundary points. Such problems may have a continuum of eigenvalues instead of a countable set, which means that they require other techniques of solution. The Fourier transformations developed, in fact, from the Fourier series representations of functions are particularly useful tools when dealing with infinite or semi-infinite spatial regions because they are designed for exactly this type of setup and have the added advantage that they reduce the number of active variables in the given PDE problem. 8.1 The Full Fourier Transformation Consider, for simplicity, a function f that is continuous and periodic with period 2L on R. Then we have the representation (see Chapter 2) where f(x) = 1 2 a + ( a n cos nπx L + b n sin nπx ), (8.1) L a n = 1 L b n = 1 L L L L L f(x) cos nπx L f(x) sin nπx L dx, n =, 1, 2,..., dx, n = 1, 2,.... (8.2) 193

217 194 8 The Fourier Transformations Using Euler s formula (see Section 14.1) e iθ = cos θ + i sin θ, i 2 = 1, and its version with θ replaced by θ, we get cos θ = 1 2 (eiθ + e iθ ), sin θ = 1 2 i(eiθ e iθ ); consequently, (8.1) with θ = nπx/l becomes f(x) = 1 2 a (a n ib n )e inπx/l (a n + ib n )e inπx/l. Replacing n by n in the first sum above and noticing from (8.2) that a n = a n and b n = b n, from the last equality we obtain or f(x) = 1 2 a + where, by (8.2), hence, n= (a n + ib n )e inπx/l + f(x) = c n = 1 2 (a n + ib n ) = 1 2L f(x) = n= = 1 2L [ 1 2L n= L L L L L L c n e inπx/l, ( f(x) cos nπx L 1 2 (a n + ib n )e inπx/l, + i sin nπx ) dx L f(x)e inπx/l dx (with b = ); ] f(ξ)e inπξ/l dξ e inπx/l. (8.3) If f is not periodic in the proper sense of the word, then we may regard it as periodic with an infinite period. Using some advanced calculus arguments, we find that, as L, representation (8.3) for such a function takes the form [ ] 1 f(x) = f(ξ) e iωξ dξ e iωx dω. (8.4) 2π We define the (full) Fourier transform of f by F[f](ω) = F(ω) = 1 2π f(x)e iωx dx. (8.5)

218 8.1 The Full Fourier Transformation 195 From (8.4) it is clear that the inverse Fourier transform of F is F 1 [F](x) = f(x) = 1 2π F(ω)e iωx dω. (8.6) The integral operators F and F 1 are called the Fourier transformation and inverse Fourier transformation, respectively. The variable ω is called the transformation parameter. 8.1 Remarks. (i) Formulas (8.5) and (8.6) are valid, respectively, if f(x) dx <, F(ω) dω < ; that is, if f (in (8.5)) and F (in (8.6)) are absolutely integrable onr. However, the Fourier transform (inverse Fourier transform) may also be defined for some functions that do not have the above property. This is done in a generalized sense through a limiting process involving transforms (inverse transforms) of absolutely integrable functions. (ii) The construction of the full Fourier transform can be extended to piecewise [ continuous ] functions f. In this case, f(x) must be replaced by f(x ) + f(x+) in (8.4) and (8.6). 1 2 (iii) The pair F, F 1 is defined slightly differently in some branches of science and engineering. Thus, in classical physics, systems engineering, and signal processing the preferred definitions are, respectively, F[f](ω) = 1 2π F[f](ω) = F[f](ω) = f(x)e iωx dx, F 1 [F](x) = f(x)e iωx dx, F 1 [F](x) = 1 2π f(x)e 2πiωx dx, F 1 [F](x) = F(ω)e iωx dω, F(ω)e iωx dω, F(ω)e 2πiωx dω. Mathematicians can work with any of these alternatives. Here, we selected formulas (8.5) and (8.6) since they are the default in the Mathematica R package used in this book. 8.2 Example. The Fourier transform of the function 1, a x a, f(x) = otherwise,

219 196 8 The Fourier Transformations where a > is a constant, is F(ω) = 1 2π f(x)e iωx dx = 1 2π = 1 2π 1 iω (eiωa e iωa ) = a a 2 π e iωx dx sin(aω). ω 8.3 Definition. Let f and g be absolutely integrable on R. By the convolution of f and g we understand the function f g defined by (f g)(x) = 1 2π f(x ξ)g(ξ) dξ, < x <. 8.4 Remark. Making the change of variable x ξ = η and then replacing η by ξ, we easily convince ourselves that (f g)(x) = 1 2π f(ξ)g(x ξ) dξ = (g f)(x). This means that the operation of convolution is commutative. 8.5 Theorem. (i) F is linear; that is, F[c 1 f 1 + c 2 f 2 ] = c 1 F[f 1 ] + c 2 F[f 2 ] for any functions f 1 and f 2 (to which F can be applied) and any numbers c 1 and c 2. (ii) If u = u(x, t), u(x, t) as x ±, and F[u](ω, t) = U(ω, t), then F[u x ](ω, t) = iωu(ω, t). (iii) If, in addition, u x (x, t) as x ±, then F[u xx ](ω, t) = ω 2 U(ω, t). (iv) Time differentiation and the Fourier transformation with respect to x commute: F[u t ](ω, t) = ( F[u] ) t (ω, t) = U (ω, t). (v) The Fourier transform of a convolution is F[f g] = F[f]F[g]. 8.6 Remarks. (i) In general, F[f g] = F[f]F[g]. (ii) It is obvious from its definition that F 1 is also linear. (iii) The Fourier transforms of a few elementary functions are given in Table A.2 in the Appendix. The properties of the Fourier transformation listed in Theorem 8.5 play an essential role in the solution of certain types of PDE problems. The solution strategy in such situations is best illustrated by examples.

220 8.1.1 Cauchy Problem for an Infinite Rod 8.1 The Full Fourier Transformation 197 Heat conduction in a very long, uniform rod where the diffusion activity diminishes toward the endpoints is modeled mathematically by the IVP (called a Cauchy problem) Adopting the notation u t (x, t) = ku xx (x, t), < x <, t >, u(x, t), u x (x, t) as x ±, t >, u(x, ) = f(x), < x <. F[u](ω, t) = U(ω, t), F[f](ω) = F(ω), we apply F to the PDE and IC and use the properties of F in Theorem 8.5 to reduce the given IVP to an initial value problem for an ordinary differential equation in the so-called transform domain. This new IVP (in which ω is an inert parameter) is with solution U (ω, t) + kω 2 U(ω, t) =, t >, U(ω, ) = F(ω), U(ω, t) = F(ω)e kω2t. (8.7) To find the solution of the original IVP, we now need to compute the inverse Fourier transform of U. Let P (ω, t) = e kω2t. Using formula 7 (with a = (4kt) 1/2 ) in Table A.2, we easily see that the inverse transform of P is F 1 [P ](x, t) = p(x, t) = 1 2kt e x2 /(4kt). By Theorem 8.5(v), we can now write (8.7) as F[u](ω, t) = F[f](ω)F[p](ω, t) = F[f p](ω, t), so u = f p, or, by Definition 8.3 and Remark 8.4, u(x, t) = 1 2π An alternative form for this is u(x, t) = 1 f(ξ) e (x ξ)2 /(4kt) dξ. 2kt G(x, t; ξ, )u(ξ, ) dξ, (8.8) where G(x, t; ξ, ) = πkt e (x ξ) /(4kt)

221 198 8 The Fourier Transformations is called the Gauss Weierstrass kernel, or influence function. Formula (8.8) shows how the initial temperature distribution u(x, ) influences the subsequent evolution of the temperature in the rod. This type of representation formula will be discussed in more detail in Chapter Example. In the IVP we have k = 2, so which, replaced in (8.8), yields u t (x, t) = 2u xx (x, t), < x <, t >, u(x, t), u x (x, t) as x ±, t >, 3, x 1, u(x, ) = f(x) =, x > 1, G(x, t; ξ, ) = πt e (x ξ) /(8t), u(x, t) = 3 2 2πt 1 1 e (x ξ)2 /(8t) dξ. If instead of (8.8) we use formula 12 (with a = 1) in Table A.2 to compute F(ω) = 3 2/π (sin ω)/ω and then solve the transformed problem we find that U (ω, t) + 2ω 2 U(ω, t) =, t >, 2 sin ω U(ω, ) = 3 π ω, 2 U(ω, t) = 3 π sin ω ω e 2ω2t. From Euler s formula for e iω and e iω, it follows that so sin ω = 1 2 i(eiω e iω ), U(ω, t) = 3 2 i 2 π Using formula 16 with a = 1/(2 2t) applied to 1 ω (eiω e iω )e 2ω 2t. we deduce that 2 V (ω) = i π 1 2t ω e 2ω, v(x) = F 1 [V ](x) = erf x 2 2t ;

222 8.1 The Full Fourier Transformation 199 hence, by formula 3 with a = 1, b = 1 and then a = 1, b = 1, which yields F 1 [e iω V (ω)] = v(x 1) = erf x 1 2 2t, F 1 [e iω V (ω)] = v(x + 1) = erf x t, u(x, t) = F 1 [U](x, t) = 3 2 ( erf x 1 2 2t erf x + 1 ) 2. 2t The function y = erf(x) is called the error function. The graphs of the error function and of the complementary error function defined on (, ), respectively, by y = erf(x) = 2 x e ξ2 dξ, π are shown in Fig y = erfc(x) = 2 e ξ2 dξ = 1 erf(x) π x Figure 8.1 The graphs of erf(x) and erfc(x). Verification with Mathematica R. Because of the piecewise definition of the IC function, we use the input u = (3/2) (Erf[(x - 1)/(2 (2 t) (1/2))] - Erf[(x + 1)/(2 (2 t) (1/2))]); f1, f2, f3} =, - 3,}; Simplify[D[u,t] - 2 D[u,x,x], Assuming[x< - 1, Limit[u,t -> ] - f1], Assuming[ - 1<x<1, Limit[u,t -> ] - f2], Assuming[x>1, Limit[u,t -> ] - f3]}] which generates the output,,, }.

223 2 8 The Fourier Transformations 8.8 Example. To solve the IBVP u t (x, t) = 2u xx (x, t) + (9 32x 2 )e t 2x2, < x <, t >, u(x, t), u x (x, t) as x ±, t >, u(x, ) = e 2x2, < x <, we set F[u](x, t) = U(ω, t) and transform the nonhomogeneous terms by means of formulas 7 and 9 (with a = 2) in Table A.2: F [ (9 32x 2 )e t 2x2] = 9F[e 2x2 ] 32F[x 2 e 2x2 ] } e t = e ω2 / (4 ω2 )e ω2 /8 } e t = 1 2 (1 + 2ω2 )e ω2 /8 e t, F[e 2x2 ] = 1 2 e ω2 /8. Then, using formula 2, we convert the IBVP to the ODE problem U (ω, t) + 2ω 2 U(ω, t) = 1 2 (1 + 2ω2 )e ω2 /8 e t, t >, U(ω, ) = 1 2 e ω2 /8, where the general solution of the equation is of the form U(ω, t) = C(ω)e 2ω2t + A(ω)e t. After we find that A(ω) = 1 2 e ω2 /8, we apply the IC and see that C(ω) = ; consequently, U(ω, t) = 1 2 e ω2 /8 e t, so, by formula 7, u(x, t) = F 1 [U](x, t) = e 2x2 e t = e t 2x2. Verification with Mathematica R. The input u = E (t - 2 x 2); q, f} =(9-32 x 2) E (t - 2 x 2), E ( - 2 x 2)}; Simplify[D[u,t] - 2 D[u,x,x] - q,(u /. t -> ) - f}] generates the output, }. Exercises In 1 6, use the full Fourier transformation to find the solution of the IVP consisting of the PDE u t (x, t) = ku xx (x, t) + q(x, t), < x <, t >,

224 8.1 The Full Fourier Transformation 21 constant k, function q, and IC as indicated. In the exercises with underlined numerical labels, express the answer as an integral. 1 k = 2, q(x, t) =, u(x, t), u x (x, t) as x ±, u(x, ) = 1, x 1, 2 k = 2, q(x, t) =, x > 1, u(x, t), u x (x, t) as x ±, u(x, ) =. 3 k = 1, q(x, t) =, 1, x 1,, x > 1. u(x, t), u x (x, t) as x ±, u(x, ) = 3e x2. 4 k = 1, q(x, t) = (4x 2 5)e 3t x2, u(x, t), u x (x, t) as x ±, u(x, ) = e x2. 5 k = 1, q(x, t) = (1 + 4t 16x 2 t)e 2x2, u(x, t), u x (x, t) as x ±, u(x, ) =. 6 k = 2, q(x, t) = 2(8x 2 t 4x 2 4t + 1)e x2, u(x, t), u x (x, t) as x ±, u(x, ) = e x2. Answers to Odd-Numbered Exercises 1 u(x, t) = (1/π) ((sin ω)/ω)e 2ω2t e iωx dω = (1/2) [ erf((x + 1)/(2 2t)) erf((x 1)/(2 2t)) ]. 3 u(x, t) = 3(1 + 4t) 1/2 e x2 /(1+4t). 5 u(x, t) = te 2x Vibrations of an Infinite String The vibrations of a very long string initially at rest, with negligible body force and where the effects of mechanical activity at the endpoints are insignificant, are modeled by the IVP u tt (x, t) = c 2 u xx (x, t), < x <, t >, u(x, t), u x (x, t) as x ±, t >, u(x, ) = f(x), u t (x, ) =, < x <.

225 22 8 The Fourier Transformations As before, let F[u](ω, t) = U(ω, t) and F[f](ω) = F(ω). Applying F to the PDE and ICs, we arrive at the ODE problem with general solution U (ω, t) + c 2 ω 2 U(ω, t) =, t >, U(ω, ) = F(ω), U (ω, ) =, U(ω, t) = C 1 (ω) cos(cωt) + C 2 (ω) sin(cωt), where C 1 (ω) and C 2 (ω) are arbitrary functions of the transformation parameter. Hence, using the ICs, we find that U(ω, t) = F(ω) cos(cωt). By Euler s formula, we have cos(cωt) = 1 2 (eicωt + e icωt ), so from (8.6) it follows that if f is continuous, then u(x, t) = F 1 [U](x, t) = 1 2π = 1 2 2π F(ω) cos(cωt) e iωx dω F(ω)[e iω(x ct) + e iω(x+ct) ] dω = 1 [f(x ct) + f(x + ct)]. (8.9) 2 This problem will be approached from a different angle in Chapter Example. In the IVP u tt (x, t) = u xx (x, t), < x <, t >, u(x, t), u x (x, t) as x ±, t >, u(x, ) = e x2, u t (x, ) =, < x <, we have c = 1 and f(x) = e x2, so, by (8.9), u(x, t) = 1 2 [ e (x t) 2 + e (x+t)2]. Verification with Mathematica R. The input u = (1/2) ( ( - (x - t) 2) + E ( - (x + t) 2)); q, f, g} =,E ( - x 2), }; Simplify[D[u,t,t] - D[u,x,x]-q,(u /. t -> ) - f, (D[u,t] /. t -> ) - g}] generates the output,, }.

226 8.1 The Full Fourier Transformation Example. Consider the IVP u tt (x, t) = 4u xx (x, t) + 2(2 + 2t 2 x 2 t 2 )e x2 /4, u(x, t), u x (x, t) as x ±, t >, u(x, ) =, u t (x, ) =, < x <. < x <, t >, Setting F[u](ω, t) = U(ω, t) and using formulas 2, 7, and 9 (with a = 1/2) in Table A.2, we change the IVP into the ODE problem U (ω, t) + 4ω 2 U(ω, t) = 4 2(1 + 2ω 2 t 2 )e ω2, t >, U(ω, ) =, U (ω, ) =. Here, the general solution of the equation has the form U(ω, t) = C 1 (ω) cos(2ωt) + C 2 (ω) sin(2ωt) + A(ω)t 2 + B(ω)t + C(ω), and the usual technique yields A(ω) = 2 2 e ω2 and B(ω) = C(ω) =. Applying the ICs now leads to C 1 (ω) = C 2 (ω) =, so U(ω, t) = 2 2 t 2 e ω2. Hence, by formula (7), the solution of the IVP is u(x, t) = F 1 [U](x, t) = 2t 2 e x2 /4. Verification with Mathematica R. The input u = 2 t 2 E ( - x 2/4); q, f, g} =2 (2 + 2 t 2 - x 2 t 2) E ( - x 2/4),, }; Simplify[D[u,t,t] - 4 D[u,x,x] - q,(u /. t -> ) - f, (D[u,t] /. t -> ) - g}] generates the output,, }. Exercises Use the full Fourier transformation to find the solution of the IVP consisting of the PDE u tt (x, t) = c 2 u xx (x, t) + q(x, t), < x <, t >, constant c, function q, and ICs as indicated. In the exercises with underlined numerical labels, express the answer as an integral. 1 c = 1, q(x, t) =, u(x, t), u x (x, t) as x ±, u(x, ) = (2x 1)e x2, u t (x, ) =.

227 24 8 The Fourier Transformations 2 c = 1, q(x, t) =, u(x, t), u x (x, t) as x ±, u(x, ) =, u t (x, ) = x 2 e x2. 1, x 1, 3 c = 2, q(x, t) =, x > 1, u(x, t), u x (x, t) as x ±, u(x, ) =, u t (x, ) =. 4 c = 2, q(x, t) =, u(x, t), u x (x, t) as x ±, 1, x 2, u(x, ) = u t (x, ) =., x > 2, 5 c = 2, q(x, t) = 2(9 16x 2 )e t x2, u(x, t), u x (x, t) as x ±, u(x, ) = 2e x2, u t (x, ) = 2e x2. 6 c = 1, q(x, t) = 2(2x 2 2x 2 t + t 1)e x2, u(x, t), u x (x, t) as x ±, u(x, ) = e x2, u t (x, ) = e x2. Answers to Odd-Numbered Exercises 1 u(x, t) = (1/2)[(2x + 2t 1)e (x+t)2 + (2x 2t 1)e (x t)2 ]. 3 u(x, t) = (1/(4π)) ω 3 sin ω[1 cos(2ωt)]e iωx dω. 5 u(x, t) = 2e t x Equilibrium Temperature in an Infinite Strip To solve the BVP u xx (x, y) + u yy (x, y) =, < x < 1, < y <, u(, y) = f 1 (y), u(1, y) = f 2 (y), < y <, u(x, y), u x (x, y), f 1 (y), f 2 (y) as y ±, < x < 1, we set F[u](x, ω) = U(x, ω), F[f 1 ](ω) = F 1 (ω), and F[f 2 ](ω) = F 2 (ω), and reduce the BVP to the ODE problem U (x, ω) ω 2 U(x, ω) =, < x < 1, U(, ω) = F 1 (ω), U(1, ω) = F 2 (ω).

228 8.1 The Full Fourier Transformation 25 Writing the general solution of the equation in the form (see Remark 1.4(i)) U(x, ω) = C 1 (ω) sinh(ωx) + C 2 (ω) sinh ( ω(x 1) ) and applying the BCs, we find that from which U(x, ω) = (csch ω) [ F 2 (ω) sinh(ωx) F 1 (ω) sinh ( ω(x 1) ) ], u(x, y) = F 1 [U](x, y) = 1 (csch ω) [ F 2 (ω) sinh(ωx) 2π F 1 (ω) sinh ( ω(x 1) ) ]e iωy dω. (8.1) 8.11 Remarks. (i) When f 1 and f 2 have certain specific forms, the inversion formula (8.1) may yield a result in terms of simple, familiar functions. (ii) The same procedure also works when the PDE is nonhomogeneous, as well as for other types of BCs Example. Consider the BVP u xx (x, y) + u yy (x, y) =, < x < 1, < y <, u(, y) =, u(1, y) = e y2, < y <, u(x, y), u x (x, y) as y ±, < x < 1. Here, f 1 (y) = and f 2 (y) = e y2, so therefore, by (8.1), u(x, y) = 1 2 π F 1 (ω) =, F 2 (ω) = 1 2 e ω2 /4 ; e ω2 /4 csch ω sinh(ωx)e iωy dω Example. Applying the same method to the BVP u xx (x, y) + u yy (x, y) = 4(x 2xy 2 )e y2, < x < 1, < y <, u(, y) =, u x (1, y) = 2e y2, < y <, u(x, y), u y (x, y) as y ±, < x < 1, we convert it into the ODE problem U (x, ω) ω 2 U(x, ω) = 2 ω 2 xe ω2 /4, < x < 1, U(, ω) =, U (1, ω) = 2 e ω2 /4,

229 26 8 The Fourier Transformations where the nonzero right-hand sides have been computed with formula 7 in Table A.2. The general solution of the equation is (see Remark 1.4(i)) U(x, ω) = C 1 (ω) sinh(ωx) + C 2 (ω) cosh ( ω(x 1) ) 2 xe ω2 /4, from which, using the BCs, we now obtain C 1 (ω) = C 2 (ω) =. Hence, so, again by formula 7, U(x, ω) = 2 xe ω2 /4, u(x, y) = F 1 [U](x, y) = 2xe y2. Verification with Mathematica R. The input u = - 2 x E ( - y 2); q, f1, f2} = 4 (x - 2 x y 2) E ( - y 2),, - 2 E ( - y 2); Simplify[D[u,x,x] + D[u,y,y] - q,(u /. x -> ) - f1, (D[u,x] /. x -> 1) - f2}] generates the output,, }. Exercises Use the full Fourier transformation to find the solution of the BVP consisting of the PDE u xx (x, y) + u yy (x, y) = q(x, y), < x < 1, < y <, function q, and BCs as indicated. In the exercises with underlined numerical labels, express the answer as an integral. 1 q(x, y) =, u(, y) =, u(1, y) = 2e 3y2, u(x, y), u y (x, y) as y ±. 2 q(x, y) =, u(, y) = y 2 e y2, u(1, y) =, u(x, y), u y (x, y) as y ±. x, y 1, 3 q(x, y) = u x (, y) =, u(1, y) =, y > 1, u(x, y), u y (x, y) as y ±. 2, y 1, 4 q(x, y) =, u(, y) =, y > 1, u(x, y), u y (x, y) as y ±. u x (1, y) =,

230 8.2 The Fourier Sine and Cosine Transformations 27 5 q(x, y) = 2(8x 2 y 2 2x y 2 3)e 2y2, u(, y) = 2e 2y2, u(1, y) = 3e 2y2, u(x, y), u y (x, y) as y ±. 6 q(x, y) = 2(2y 2 + 1)e 2x y2, u(, y) = e y2, u x (1, y) = 2e 2 y2, u(x, y), u y (x, y) as y ±. Answers to Odd-Numbered Exercises 1 u(x, y) = (1/ 3π) csch ω sinh(ωx) e ω2 /12 iωy dω. 3 u(x, y) = (1/π) 5 u(x, y) = (x 2 + 2)e 2y2. ω 4 sin ω sech ω [ ω cosh(ωx) + sinh ( ω(x 1) )] ωx } e iωy dω. 8.2 The Fourier Sine and Cosine Transformations The Fourier sine and cosine transforms, constructed as generalizations of the corresponding sine and cosine series, are defined for functions f that are piecewise continuous on < x < by F S [f](ω) = F(ω) = F C [f](ω) = F(ω) = 2 π 2 π f(x) sin(ωx) dx, f(x) cos(ωx) dx. It can be shown that these transforms exist if f is absolutely integrable on (, ); that is, f(x) dx <. The corresponding inverse transforms are 2 F 1 S [F](x) = f(x) = π 2 F 1 C [F](x) = f(x) = π F(ω) sin(ωx) dω, F(ω) cos(ωx) dω.

231 28 8 The Fourier Transformations The operators F S and F C are called the Fourier sine transformation and Fourier cosine transformation, respectively, while F 1 S and F 1 C are their inverse transformations. The comment made at the end of Remark 8.1(ii) applies here as well Example. The Fourier sine and cosine transforms of the function 1, x a, f(x) = a = const,, x > a, are 2 F S [f](ω) = f(x) sin(ωx) dx = π 2 1 [ ] = 1 cos(aω), π ω 2 F C [f](ω) = f(x) cos(ωx) dx = π 2 1 = π ω sin(aω). 2 π 2 π a a sin(ωx) dx cos(ωx) dx 8.15 Remark. F S (F C ) can also be defined for functions on R if these functions are odd (even) Theorem. (i) F S and F C are linear operators. (ii) If u = u(x, t) and u(x, t) as x, then F S [u x ](ω, t) = ωf C [u](ω, t), 2 F C [u x ](ω, t) = π u(, t) + ωf S[u](ω, t). (iii) If, in addition, u x (x, t) as x, then 2 F S [u xx ](ω, t) = π ωu(, t) ω2 F S [u](ω, t), 2 F C [u xx ](ω, t) = π u x(, t) ω 2 F C [u](ω, t). (iv) Time differentiation commutes with both the Fourier sine and cosine transformations: F S [u t ](ω, t) = ( F S [u] ) (ω, t), t F C [u t ](ω, t) = ( F C [u] ) (ω, t). t

232 8.2 The Fourier Sine and Cosine Transformations 29 As Theorem 8.16(iii) indicates, the choice between using the sine transformation or the cosine transformation in the solution of a given IBVP depends on the type of BC prescribed at x =. Brief lists of Fourier sine and cosine transforms are given in Tables A.3 and A.4 in the Appendix Heat Conduction in a Semi-Infinite Rod The process of heat conduction in a long rod for which the temperature at the near endpoint is prescribed while the effects of the conditions at the far endpoint are negligible is modeled by the IBVP u t (x, t) = ku xx (x, t), x >, t >, u(, t) = g(t), u(x, t), u x (x, t) as x, t >, u(x, ) = f(x), x >. In view of the given BC, we use the Fourier sine transformation; thus, let F S [u](ω, t) = U(ω, t), F S [f](ω) = F(ω). Applying F S to the PDE and IC and using the properties in Theorem 8.16, we arrive at the ODE problem 2 U (ω, t) + kω 2 U(ω, t) = kωg(t), t >, π U(ω, ) = F(ω). After finding U in the transform domain, we obtain the solution of the original IBVP as u(x, t) = F 1 [U](x, t). S 8.17 Example. The IBVP u t (x, t) = u xx (x, t), x >, t >, u(, t) = 1, u(x, t), u x (x, t) as x, t >, u(x, ) =, x >, requires the use of the Fourier sine transformation because it is the temperature that is prescribed in the BC. By Theorem 8.16, the transform U(ω, t) = F S [u](ω, t) of u satisfies 2 U (ω, t) + ω 2 U(ω, t) = ω, t >, π U(ω, ) =. The solution of the transformed problem is 2 1 ( U(ω, t) = 1 e ω 2 t ) ; π ω

233 21 8 The Fourier Transformations so, by formula 12 (with a = 1/(2 t )) in Table A.3, u(x, t) = F 1 x S [U](x, t) = erfc 2 t. Verification with Mathematica R. The input u = Erfc[x/(2 t (1/2))]; f, g} =, 1}; Simplify[D[u,t] - D[u,x,x], Assuming[x>, Limit[u, t ->, Direction -> - 1]] - f,(u /. x -> ) - g}] generates the output,, } Example. The IBVP u t (x, t) = 2u xx (x, t) + 2(x 2t 1)e x, x >, t >, u x (, t) = 2t 1, u(x, t), u x (x, t) as x, t >, u(x, ) = xe x, x >, can be solved by means of the Fourier cosine transformation since the BC at x = prescribes the value of u x and not that of u. Thus, with the notation F C [u](ω, t) = U(ω, t), using formulas 2, 5, and 6 in Table A.4, and performing a simple algebraic manipulation of the coefficients, we convert the IBVP into the ODE problem 2 2ω U (ω, t) + 2ω 2 2 U(ω, t) = 2 π 2 1 ω 2 U(ω, ) = π (1 + ω 2 ) ω 2 t ω4 (1 + ω 2 ) 2 The general solution of the equation is of the form U(ω, t) = C(ω)e 2ω2t + A(ω)t + B(ω). }, t >, Determining first the coefficients A(ω) and B(ω) and then applying the IC, we find that C(ω) = and that U(ω, t) = which, by formulas 5 and 6, yields 2 2 π 1 + ω 2 t 1 } ω2 (1 + ω 2 ) 2, u(x, t) = F 1 C [U](x, t) = (2t x)e x.

234 8.2 The Fourier Sine and Cosine Transformations 211 Verification with Mathematica R. The input u = (2 t - x) E ( - x); q, f, g} =2 (x - 2 t - 1) E ( - x), - x E ( - x), - 2 t - 1}; Simplify[D[u,t] - 2 D[u,x,x] - q,(u /. t -> ) - f, (D[u,x] /. x -> ) - g}] generates the output,, }. Exercises Use a suitable Fourier transformation to find the solution of the IBVP consisting of the PDE u t (x, t) = ku xx (x, t) + q(x, t), x >, t >, constant k, function q, BC, and IC as indicated. In the exercises with underlined numerical labels, express the answer as an integral. 1 k = 1, q(x, t) =, u(, t) =, u(x, t), u x (x, t) as x, u(x, ) = xe x. 2 k = 1, q(x, t) =, u(, t) = 1, u(x, t), u x (x, t) as x, u(x, ) = e x/2. 1, < x 1, 3 k = 2, q(x, t) = u(, t) = t,, x > 1, u(x, t), u x (x, t) as x, u(x, ) =. 4 k = 2, q(x, t) =, u(, t) = 1 t, u(x, t), u x (x, t) as x, u(x, ) = 5 k = 2, q(x, t) = 2(x 2t 1)e x, u(, t) = 2t, u(x, t), u x (x, t) as x, u(x, ) = xe x. 1, < x 1,, x > 1. 6 k = 1, q(x, t) = 2(4xt x 4t + 2)e 2x, u(, t) = 1, u(x, t), u x (x, t) as x, u(x, ) = e 2x. 7 k = 1, q(x, t) =, u x (, t) = t, u(x, t), u x (x, t) as x, u(x, ) = 2e x. 8 k = 1, q(x, t) =, u x (, t) = 1, u(x, t), u x (x, t) as x, u(x, ) = xe 2x.

235 212 8 The Fourier Transformations 9 k = 2, q(x, t) = 1, < x 1,, x > 1, u x (, t) = t 1, u(x, t), u x (x, t) as x, u(x, ) =. 1 k = 2, q(x, t) =, u x (, t) = t, u(x, t), u x (x, t) as x, u(x, ) = 1, < x 2,, x > k = 2, q(x, t) = 2(8xt x 8t 4)e 2x, u x (, t) = 2t 2, u(x, t), u x (x, t) as x, u(x, ) = e 2x. 12 k = 1, q(x, t) = e 2t x+1, u x (, t) = e 2t+1, u(x, t), u x (x, t) as x, u(x, ) = e 1 x. Answers to Odd-Numbered Exercises 1 u(x, t) = (4/π) ω(ω 2 + 1) 2 e ω2t sin(ωx) dω. 3 u(x, t) = (1/π) ω 3 [(e 2ω2t 1) cos ω + 2ω 2 t] sin(ωx) dω. 5 u(x, t) = (2t x)e x. 7 u(x, t) = (2/π) (ω 6 + ω 4 ) 1 [(ω 4 + ω 2 )t ω (1 + ω 2 2ω 4 )e ω2t ] cos(ωx) dω. 9 u(x, t) = (1/π) ω 4 [ (ω sin ω + 2ω 2 + 1)e 2ω2 t 11 u(x, t) = (1 2xt)e 2x. + ω sin ω + 2ω 2 (1 t) + 1] cos(ωx) dω Vibrations of a Semi-Infinite String The method applied in this case is similar to that used above Example. The solution of the IBVP u tt (x, t) = 4u xx (x, t), x >, t >, u x (, t) =, u(x, t), u x (x, t) as x, t >, u(x, ) = e x, u t (x, ) =, x >, is obtained by means of the Fourier cosine transformation because the BC prescribes the x-derivative of u at x =. According to formula 5 (with

236 8.2 The Fourier Sine and Cosine Transformations 213 a = 1) in Table A.4, the Fourier cosine transform of the function e x is 2/π (1 + ω 2 ) 1 ; hence, by Theorem 8.16, U(ω, t) = F C [u](ω, t) satisfies with solution U (ω, t) + 4ω 2 U(ω, t) =, t >, 2 1 U(ω, ) = π 1 + ω 2, U(ω, t) = 2 cos(2ωt) π 1 + ω 2. Then, by formula 14 (with a = 2t and b = 1), we find that the solution of the IBVP is u(x, t) = F 1 C [U](x, t) = e x cosh(2t), t x/2, e 2t cosh x, t > x/2. Verification with Mathematica R. Because u is defined piecewise, we use the input u1, u2} =E ( - x) Cosh[2 t],e ( - 2 t) Cosh[x]}; q, f, g, h} =,E ( - x),, }; Simplify[D[u1,u2},t,t] - 4 D[u1,u2},x,x] - q, (D[u2,x]/. x -> ) - h,(u1 /. t -> ) - f,(d[u1,t]/. t -> ) - g}] to generate the output, },,, }. 8.2 Example. In view of the BC at x =, we solve the IBVP u tt (x, t) = u xx (x, t) + (x + t 3)e x, x >, t >, u(, t) = 1 t, u(x, t), u x (x, t) as x, t >, u(x, ) = (1 x)e x, u t (x, ) = e x, x >, by means of the Fourier sine transformation. Therefore, making the notation F S [u](ω, t) = U(ω, t), using formulas 2, 6, and 7 in Table A.3, and reorganizing the coefficients, we reduce the IBVP to the ODE problem 2 ω U (ω, t) + ω 2 3 U(ω, t) = π 1 + ω 2 t + ω3 (1 ω 2 } ) (1 + ω 2 ) 2, t >, 2 ω(1 ω 2 ) 2 U(ω, ) = π (1 + ω 2 ) 2, U ω (ω, ) = π 1 + ω 2. The general solution of the equation is U(ω, t) = C 1 (ω) cos(ωt) + C 2 (ω) sin(ωt) + A(ω)t + B(ω), where, in fact, A(ω)t + B(ω) is equal to the right-hand side divided by ω 2.

237 214 8 The Fourier Transformations Applying the ICs, we find that C 1 (ω) = C 2 (ω) = ; hence, using partial fraction decomposition, we have so 2 U(ω, t) = π ω(1 ω 2 ) (1 + ω 2 ) 2 = ω 1 + ω 2 + 2ω (1 + ω 2 ) 2, ω 1 + ω 2 t+ω(1 } } ω2 ) 2 ω (1 + ω 2 ) 2 = π 1 + ω 2 (1 t) 2ω (1 + ω 2 ) 2, from which, by formulas 6 and 7, u(x, t) = F 1 S [U](x, t) = (1 t x)e x. Verification with Mathematica R. The input u = (1 - t - x) E ( - x); q, f, g, h} =(x + t - 3) E ( - x), (1 - x) E ( - x), - E ( - x), 1 - t}; Simplify[D[u,t,t] - D[u,x,x] - q,(u /. x -> ) - h,(u /. t -> ) - f, (D[u,t] /. t -> ) - g}] generates the output,,, }. Exercises Use a suitable Fourier transformation to find the solution of the IBVP consisting of the PDE u tt (x, t) = c 2 u xx (x, t) + q(x, t), x >, t >, constant c, function q, BC, and ICs as indicated. In the exercises with underlined numerical labels, express the answer as an integral. 1 c = 1, q(x, t) =, u(, t) = 1, u(x, t), u x (x, t) as x, u(x, ) = 3e 2x, u t (x, ) =. 2 c = 1, q(x, t) =, u(, t) = t, u(x, t), u x (x, t) as x, u(x, ) =, u t (x, ) = xe x. 1, < x 1, 3 c = 2, q(x, t) = u(, t) =,, x > 1, u(x, t), u x (x, t) as x, 1, < x 1, u(x, ) =, u t (x, ) =, x > 1.

238 8.2 The Fourier Sine and Cosine Transformations c = 2, q(x, t) =, u(, t) = t 1, u(x, t), u x (x, t) as x, 1, < x 2, u(x, ) = u t (x, ) =., x > 2, 5 c = 1, q(x, t) = 3e 2t x 1, u(, t) = e 2t 1, u(x, t), u x (x, t) as x, u(x, ) = e x 1, u t (x, ) = 2e x 1. 6 c = 1, q(x, t) = 4(2 + t xt)e 2x, u(, t) = 2, u(x, t), u x (x, t) as x, u(x, ) = 2e 2x, u t (x, ) = xe 2x. 7 c = 1, q(x, t) =, u x (, t) = 1, u(x, t), u x (x, t) as x, u(x, ) = 3e x, u t (x, ) =. 8 c = 1, q(x, t) =, u x (, t) = t, u(x, t), u x (x, t) as x, u(x, ) =, u t (x, ) = xe 2x. t, < x 2, 9 c = 2, q(x, t) = u x (, t) =,, x > 2, u(x, t), u x (x, t) as x, 1, < x 2, u(x, ) =, u t (x, ) =, x > 2. 1 c = 2, q(x, t) =, u x (, t) = 2t, u(x, t), u x (x, t) as x, 1, < x 1, u(x, ) = u t (x, ) =., x > 1, 11 c = 1, q(x, t) = (xt 2t 2)e x, u x (, t) = t 2, u(x, t), u x (x, t) as x, u(x, ) = 2e x, u t (x, ) = xe x. 12 c = 1, q(x, t) = 4(3x xt + t 4)e 2x, u x (, t) = t 5, u(x, t), u x (x, t) as x, u(x, ) = (1 3x)e 2x, u t (x, ) = xe 2x. Answers to Odd-Numbered Exercises 1 u(x, t) = (2/π) (ω 3 + 4ω) 1 [ω (ω 2 2) cos(ωt)] sin(ωx) dω. 3 u(x, t) = (1/π) ω 3 sin 2 (ω/2)[1 2ω sin(2ωt) cos(2ωt)] sin(ωx) dω.

239 216 8 The Fourier Transformations 5 u(x, t) = e 2t x 1. 7 u(x, t) = (2/π) (ω 4 + ω 2 ) 1 [1 + ω 2 + (2ω 2 1) cos(ωt)] cos(ωx) dω. 9 u(x, t) = (1/(4π)) ω 4 sin(2ω)[2ωt (4ω 2 + 1) sin(2ωt)] cos(ωx) dω. 11 u(x, t) = (2 xt)e x Equilibrium Temperature in a Semi-Infinite Strip The steady-state temperature distribution in a semi-infinite strip x L, y is modeled by the BVP u xx (x, y) + u yy (x, y) =, < x < L, y >, u(, y) = g 1 (y), u(l, y) = g 2 (y), y >, g 1 (y), g 2 (y) as y, u(x, ) = f(x), u(x, y), u y (x, y) as y, < x < L. Using the principle of superposition, we write the solution in the form u(x, y) = u 1 (x, y) + u 2 (x, y), where u 1 satisfies the given BVP with g 1 = and g 2 =, and u 2 satisfies it with f =. Thus, the first BVP is (u 1 ) xx (x, y) + (u 1 ) yy (x, y) =, < x < L, y >, u 1 (, y) =, u 1 (L, y) =, y >, u 1 (x, ) = f(x), u 1 (x, y), (u 1 ) y (x, y) as y, < x < L, which is solved by the method of separation of variables. Proceeding as in Section 5.3, from the PDE and BCs at x = and x = L we obtain u 1 (x, y) = = sin nπx ( A n cosh nπy L L sin nπx L + B n sinh nπy ) L (a n e nπy/l + b n e nπy/l ). Since u 1 (x, y) as y, we must have a n =, n = 1, 2,... ; hence, u 1 (x, y) = b n sin nπx L e nπy/l. (8.11)

240 8.2 The Fourier Sine and Cosine Transformations 217 The BC at y = now yields u 1 (x, ) = f(x) = b n sin nπx L, where, by (2.1), the Fourier sine series coefficients b n are b n = 2 L L f(x) sin nπx L dx, n = 1, 2,.... Consequently, the solution of the first BVP is given by (8.11) with the b n computed by means of the formula above. The second BVP is (u 2 ) xx (x, y) + (u 2 ) yy (x, y) =, < x < L, y >, u 2 (, y) = g 1 (y), u 2 (L, y) = g 2 (y), y >, g 1 (y), g 2 (y) as y, u 2 (x, ) =, u 2 (x, y), (u 2 ) y (x, y) as y, < x < L. In view of the BC at y =, we use F S with respect to y and make the notation F S [u 2 ](x, ω) = U(x, ω), F S [g 1 ](ω) = G 1 (ω), F S [g 2 ](ω) = G 2 (ω). Then the above BVP reduces to U (x, ω) ω 2 U(x, ω) =, < x < L, U(, ω) = G 1 (ω), U(L, ω) = G 2 (ω). The general solution of the equation can be written as (see Remark 1.4(i)) U(x, ω) = C 1 (ω) sinh(ωx) + C 2 (ω) sinh ( ω(l x) ), where C 1 and C 2 are arbitrary functions of ω. Using the BCs at x = and x = L, we find that U(x, ω) = csch(ωl) [ G 2 (ω) sinh(ωx) + G 1 (ω) sinh ( ω(l x) )]. (8.12) The solution u 2 is now obtained by applying F 1 S 8.21 Example. For the BVP u xx (x, y) + u yy (x, y) =, < x < 1, y >, u(, y) =, u(1, y) = g(y), y >, to the equality above. u(x, ) =, u(x, y), u y (x, y) as y, < x < 1,

241 218 8 The Fourier Transformations where g(y) = 1, < y 2,, y > 2, we have L = 1, g 1 (y) =, and g 2 (y) = g(y), so G 1 (ω) = and, by formula 1 (with a = 2) in Table A.3, 2 1 [ ] 2 2 G 2 (ω) = 1 cos(2ω) = π ω π ω sin2 ω; consequently, by (8.12), which yields U(x, ω) = 2 π 2 ω sin2 ω csch ω sinh(ωx), u(x, y) = F 1 S = 4 π [U](x, y) 1 ω sin2 ω csch ω sinh(ωx) sin(ωy) dω. Exercises Use a suitable Fourier transformation to find the solution of the BVP consisting of the PDE u xx (x, y) + u yy (x, y) = q(x, y), < x < 1, y >, function q, and BCs as indicated. In the exercises with underlined numerical labels, express the answer as an integral. 1 q(x, y) =, u(, y) = e 2y, u(1, y) =, u(x, ) = x, u(x, y), u y (x, y) as y. 2 q(x, y) =, u(, y) =, u(1, y) = 2e y, u(x, ) = x 1, u(x, y), u y (x, y) as y. 3 q(x, y) = 1, < y 1,, y > 1, u(, y) = 2, < y 1,, y > 1, u(1, y) =, u(x, ) =, u(x, y), u y (x, y) as y.

242 8.2 The Fourier Sine and Cosine Transformations q(x, y) =, u(, y) =, u(1, y) = 1, < y 2,, y > 2, u(x, ) = 1 2x, u(x, y), u y (x, y) as y. 5 q(x, y) = (x 2 y + 4)e y, u(, y) = ye y, u(1, y) = (1 y)e y, u(x, ) = x 2, u(x, y), u y (x, y) as y. 6 q(x, y) = 4(x 2 + y 2 2y + 1)e 2y, u(, y) = y 2 e 2y, u(1, y) = (y 2 + 1)e 2y, u(x, ) = x 2, u(x, y), u y (x, y) as y. 7 q(x, y) = (5y + 1)e x 2y, u(, y) = (y + 1)e 2y, u x (1, y) = (y + 1)e 1 2y, u(x, ) = e x, u(x, y), u y (x, y) as y. 8 q(x, y) = 4(1 + x xy)e 2y, u x (, y) = ye 2y, u(1, y) = (1 y)e 2y, u(x, ) = 1, u(x, y), u y (x, y) as y. 9 q(x, y) =, u(, y) = 2e y, u(1, y) =, u y (x, ) = x 1, u(x, y), u y (x, y) as y. 1 q(x, y) =, u(, y) =, u(1, y) = e 2y, u y (x, ) = 2x, u(x, y), u y (x, y) as y. 1, < y 1, 11 q(x, y) = u(, y) =, u(1, y) =,, y > 1, u y (x, ) = 1, u(x, y), u y (x, y) as y. 1, < y 2, 12 q(x, y) =, u(, y) = u(1, y) =,, y > 2, u y (x, ) = x 2 1, u(x, y), u y (x, y) as y. 13 q(x, y) = 2(4xy + 2y 2 4x 4y + 1)e 2y, u(, y) = y 2 e 2y, u(1, y) = (y 2 + 2y)e 2y, u y (x, ) = 2x, u(x, y), u y (x, y) as y. 14 q(x, y) = (5y 2)e 2x y, u(, y) = ye y, u(1, y) = ye 2 y, u y (x, ) = e 2x, u(x, y), u y (x, y) as y.

243 22 8 The Fourier Transformations 15 q(x, y) = (x 2 + x + 1)e y, u x (, y) = e y, u(1, y) = e y, u y (x, ) = 1 x x 2, u(x, y), u y (x, y) as y. 16 q(x, y) = 4(3xy 3x 2)e 2y, u(, y) = 2e 2y, u x (1, y) = 3ye 2y, u y (x, ) = 3x + 4, u(x, y), u y (x, y) as y. Answers to Odd-Numbered Exercises 1 u(x, y) = (1/π) (ω 3 + 4ω) 1 csch ω[ 2(ω 2 + 4) sinh(ωx) + 2ω 2 sinh(ω(x 1))] + 2(ω 2 + 4)x} sin(ωy) dω. 3 u(x, y) = (2/π) ω 3 (1 cos ω) 1 + csch ω[sinh(ωx) + (2ω 2 1) sinh(ω(x 1))]} sin(ωy) dω. 5 u(x, y) = (x 2 y)e y. 7 u(x, y) = (y + 1)e x 2y. 9 u(x, y) = (2/π) (ω 4 + ω 2 ) 1 [(1 ω 2 ) csch ω sinh(ω(x 1)) + (ω 2 + 1)(1 x)] cos(ωy) dω. 11 u(x, y) = (2/π) ω 3 (ω sin ω) 13 u(x, y) = (2xy + y 2 )e 2y. 15 u(x, y) = (x 2 + x 1)e y. csch ω[sinh(ωx) sinh(ω(x 1))] 1} cos(ωy) dω. 8.3 Other Applications The Fourier transformation method may also be applied to other suitable problems, such as those mentioned in Section 4.4, including some with nonhomogeneous PDEs Example. The IVP (Cauchy problem) u t (x, t) = u xx (x, t) + 2u(x, t) + (1 4x 2 t)e x2, u(x, t), u x (x, t) as x ±, t >, u(x, ) =, < x <, < x <, t >,

244 8.3 Other Applications 221 models a diffusion process with a chain reaction and a source in a onedimensional medium. By formulas 7 and 9 (with a = 1) in Table A.2, F[(1 4x 2 t)e x2 ] = 1 2 [ (ω 2 2)t + 1 ] e ω2 /4 ; hence, setting F[u](ω, t) = U(ω, t) and using formula 2, we see that U is the solution of the transformed problem U (ω, t) + (ω 2 2)U(ω, t) = 1 2 [ (ω 2 2)t + 1 ] e ω2 /4, t >, U(ω, ) =. Direct calculation shows that U(ω, t) = 1 2 te ω2 /4, so, by formula 8, the solution of the given IVP is u(x, t) = F 1 [U](x, t) = te x2. Verification with Mathematica R. The input u = t E ( - x 2); q, f} =(1-4 x 2 t) E ( - x 2), }; Simplify[D[u,t] - D[u,x,x] - 2 u - q,(u /. t -> ) - f}] generates the output, } Example. The IBVP u tt (x, t) + 2u t (x, t) = u xx (x, t) + (2 + 4t 4t 2 )e 2x, u(, t) = t 2, t >, x >, t >, u(x, t), u x (x, t) as x, t >, u(x, ) =, u t (x, ) =, x >, describes the propagation of a dissipative wave along a semi-infinite string initially at rest, under the action of a prescribed displacement of the near endpoint and an external force. In view of the BC, we use the Fourier sine transformation to find the solution. First, by formula 6 (with a = 2) in Table A.3, 2 F S [(2 + 4t 4t 2 )e 2x ] = 2 π ω 4 + ω 2 (1 + 2t 2t2 );

245 222 8 The Fourier Transformations hence, by Theorem 8.16 and the ICs, the transform U(ω, t) = F S [u](ω, t) of u satisfies U (ω, t) + 2U (ω, t) + ω 2 U(ω, t) 2 2 ω = π ωt2 + 2 π 4 + ω 2 (1 + 2t 2t2 ), t >, U(ω, ) =, U (ω, ) =. Solving this ODE problem in the usual way, we find that so, again by formula 6, U(ω, t) = 2 π ω 4 + ω 2 t2, u(x, t) = F 1 S [U](x, t) = t2 e 2x. Verification with Mathematica R. The input u = t 2 E ( - 2 x); q, f, g, h} =(2 + 4 t - 4 t 2) E ( - 2 x),,, t 2}; Simplify[D[u,t,t] + 2 D[u,t] - D[u,x,x] - q,(u /. t -> ) - f, (D[u,t] /. t -> ) - g,(u /. x -> ) - h}] generates the output,,, } Example. The BVP u xx (x, y) + u yy (x, y) + u(x, y) = x(y4 + 8y 2 1) (1 + y 2 ) 3, u(, y) =, u(1, y) = y 2, y >, u y (x, ) =, < x < 1, u(x, y), u y (x, y) as y, < x < 1, < x < 1, y >, models the steady-state distribution of heat in a semi-infinite strip with a time-dependent source and insulated base. Since the BC at y = prescribes the derivative u y, we use the Fourier cosine transformation with respect to y to find the solution. By formulas 9 12 (with a = 1) in Table A.4, [ x(y 4 + 8y 2 ] 1) π F C (1 + y 2 ) 3 = 2 (1 ω2 )e ω x, [ ] 1 π F C 1 + y 2 = 2 e ω ;

246 8.3 Other Applications 223 consequently, by Theorem 8.16, the transform U(x, ω) = F C [u](x, ω) of u satisfies π U (x, ω) (ω 2 1)U(x, ω) = 2 (1 ω2 )e ω x, < x < 1, π U(, ω) =, U(1, ω) = 2 e ω, with solution so, by formula 9, U(x, ω) = π 2 xe ω ; u(x, y) = F 1 x C [U](x, y) = 1 + y 2. Verification with Mathematica R. The input u = x/(1 + y 2); q, f1, f2, g} =(x (y y 2 + 1))/((1 + y 2) 3),, 1/(1 + y 2), }; Simplify[D[u,x,x] + D[u,y,y] + u - q,(u /. x -> ) - f1, (u /. x -> 1) - f2,(d[u,y]/. y -> ) - g}] generates the output,,, }. Exercises In 1 4, solve the IVP (IBVP) consisting of the PDE u t (x, t) = ku xx (x, t) + bu(x, t) + q(x, t), domain in the (x, t)-plane, numbers k, b, function q, and IC (IC and BC) as specified. In the exercises with underlined numerical labels, express the answer as an integral. 1 < x <, t >, k = 1, b = 2, q(x, t) = (9 + 4t 8x 2 4x 2 t)e x2, u(x, t), u x (x, t) as x ±, u(x, ) = 2e x2. 2 < x <, t >, k = 1, b = 1, q(x, t) = e x2, u(x, t), u x (x, t) as x ±, u(x, ) =. 3 x >, t >, k = 2, b = 2, q(x, t) =, u x (, t) =, u(x, t), u x (x, t) as x, u(x, ) = e x.

247 224 8 The Fourier Transformations 4 x >, t >, k = 2, b = 1, q(x, t) = (x + 4t 3xt)e x, u(, t) =, u(x, t), u x (x, t) as x, u(x, ) =. In 5 8, solve the IVP (IBVP) consisting of the PDE u tt (x, t) + au t (x, t) + bu(x, t) = c 2 u xx (x, t) + q(x, t), domain in the (x, t)-plane, numbers a, b, c, function q, and ICs (ICs and BC) as specified. In the exercises with underlined numerical labels, express the answer as an integral. 5 < x <, t >, a =, b = 1, c = 1, q(x, t) =, u(x, t), u x (x, t) as x ±, u(x, ) = 2e 2x2, u t (x, ) =. 6 < x <, t >, a =, b = 1, c = 1, q(x, t) = (1 x 2 )e t x2 /4, u(x, t), u x (x, t) as x ±, u(x, ) = 4e x2 /4, u t (x, ) = 4e x2 /4. 7 x >, t >, a = 1, b =, c = 1, q(x, t) = (3 2t)e x, u(, t) = 2t 1, u(x, t), u x (x, t) as x, u(x, ) = e x, u t (x, ) = 2e x. 8 x >, t >, a = 1, b =, c = 2, q(x, t) = (9 2x t)e x/2, u x (, t) = 2 t/2, u(x, t), u x (x, t) as x, u(x, ) = 2xe x/2, u t (x, ) = e x/2. In 9 12, solve the BVP consisting of the PDE u xx (x, y) + u yy (x, y) au x (x, y) + cu(x, y) = q(x, y), domain in the (x, y)-plane, numbers a, c, function q, and BCs as specified. In the exercises with underlined numerical labels, express the answer as an integral. 9 < x < 1, < y <, a = 1, c =, q(x, y) = (3 + 2x 4y 2 4xy 2 )e y2, u(, y) = e y2, u(1, y) = 2e y2, u(x, y), u y (x, y) as y ±, < x < 1. 1 < x < 1, < y <, a =, c = 2, q(x, y) =, 1, 2 y 2, u(, y) = u(1, y) =, otherwise, u(x, y), u y (x, y) as y ±, < x < 1.

248 8.3 Other Applications < x < 1, y >, a =, c = 1, q(x, y) = xe y/2, u(, y) =, u(1, y) =, u y (x, ) =, u(x, y), u y (x, y) as y, < x < < x < 1, y >, a = 2, c = 1, q(x, y) = 2(x + 2y x 2 2xy)e y, u(, y) =, u(1, y) =, u(x, ) =, u(x, y), u y (x, y) as y, < x < 1. Answers to Odd-Numbered Exercises 1 u(x, t) = (t + 2)e x2. 3 u(x, t) = (2/π) (1 + ω 2 ) 1 e 2(1+ω2 )t cos(ωx) dω. 5 u(x, t) = (2π) 1/2 e ω2 /8 cos((1 + ω 2 ) 1/2 t) e iωx dω. 7 u(x, t) = (2t 1)e x. 9 u(x, y) = (x + 1)e y2. 11 u(x, y) = (4/π) (1 + 5ω 2 + 4ω 4 ) 1 [x csch((1+ω 2 ) 1/2 ) sinh((1+ω 2 ) 1/2 x)] cos(ωy) dω.

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250 C H A P T E R 9 The Laplace Transformation The Fourier transformations are used mainly with respect to the space variables. In certain circumstances, however, for reasons of expedience or necessity, it is desirable to eliminate time as an active variable. This is achieved by means of the Laplace transformation. Problems where the spatial part of the domain is unbounded but the solution is not expected to decay fast enough away from the origin are particularly suited to this method. 9.1 Definition and Properties First, we introduce a couple of useful mathematical entities. 9.1 Definition. The function H defined by, t <, H(t) = 1, t, is called the Heaviside (unit step) function. It is clear that, more generally, for any real number t,, t < t, H(t t ) = 1, t t. 9.2 Remark. H is piecewise continuous. As mentioned in Remark 2.4(ii), the type of analysis we are performing is not affected by the value of a piecewise continuous function at its points of discontinuity. For our purposes, therefore, the value H() = 1 is chosen purely for convenience, to have H correctly defined as a function on the entire real line, but it is otherwise unimportant. 227

251 228 9 The Laplace Transformation In mathematical modeling it is often necessary to handle a special type of physical data such as unit impulses and point sources. Let us assume, for example, that a unit impulse is produced by a constant force of magnitude 1/ε acting over a very short time interval (t ε/2, t + ε/2), ε >. We may express this mathematically by taking the force to be 1/ε, t ε/2 < t < t + ε/2, g t,ε(t) = otherwise, and computing the total impulse as g t,ε(t) dt = t +ε/2 t ε/2 1 ε dt = 1 ε ε = 1. We note that the integral above is equal to 1 irrespective of the value of ε. If we now want to regard the impulse as being produced at the single moment t = t, we need to consider a limiting process and introduce some sort of limit of g t,ε as ε, which we denote by δ(t t ). 9.3 Definition. The mathematical object δ defined by (i) δ(t t ) = for all t = t, (ii) δ(t t ) dt = 1 is called the Dirac delta. 9.4 Remarks. (i) From Definition 9.3 it is obvious that δ cannot be ascribed a finite value at t = t, because then its integral over R would be, not 1. Consequently, δ is not a function. Strictly speaking, δ is a so-called distribution (generalized function) and its proper handling requires a special formalism that goes beyond the scope of this book. (ii) If t < t, we have t δ(τ t ) dτ = t dτ = ; if t > t, we can find ε > sufficiently small so that t + ε/2 < t; hence, using the function g t,ε introduced above, we see that t δ(τ t ) dτ = lim t ε g t,ε(τ) dτ = lim Therefore, combining these results, we may write t t +ε/2 ε t ε/2 δ(τ t ) dτ = H(t t ), 1 dτ = 1. ε

252 9.1 Definition and Properties 229 which means that, in a certain generalized (distributional) sense, H (t t ) = δ(t t ). (iii) If f is continuous, then, according to the mean value theorem, there is t, t ε/2 < t < t + ε/2, such that t+ε/2 f(τ)δ(t τ) dτ = lim f(τ)g t,ε (τ) dτ = lim f(τ) 1 ε dτ ε 1 = lim ε ε [( t + ε 2 ) ( t ε 2 ε t ε/2 )] f(t ) = f(t). (9.1) In distribution theory, δ is, in fact, defined rigorously by a formula of this type and not as in Definition 9.3. Like differentiation, integration on the left-hand side above is understood in a generalized, distributional sense. (iv) The Dirac delta may also be used in problems formulated on semiinfinite or finite intervals. In such cases, its symbol stands for the restriction of this distribution to the corresponding interval. 9.5 Definition. The Laplace transform of a function f(t), < t <, is defined by L[f](s) = F(s) = f(t)e st dt. Here, s is the transformation parameter. The corresponding inverse Laplace transform, computed by means of complex variable techniques (see Section 14.1), is L 1 [F](t) = f(t) = 1 c+i F(s)e st ds. 2πi c i The operators L and L 1 are called the Laplace transformation and inverse Laplace transformation, respectively. 9.6 Remark. L is applicable to a wider class of functions than F. The following assertion gives sufficient conditions for the existence of the Laplace transform of a function. 9.7 Theorem. If (i) f is piecewise continuous on [, ); (ii) there are constants C and α such that f(t) Ce αt, < t <, then L[f](s) = F(s) exists for all s > α, and F(s) as s.

253 23 9 The Laplace Transformation 9.8 Examples. (i) The function f(t) = 1, t >, satisfies the conditions in Theorem 9.7 with C = 1 and α =, and we have F(s) = e st dt = 1 s [ e st ] = 1 s, s >. (ii) For f(t) = e 2t, t >, Theorem 9.7 holds with C = 1 and α = 2. In this case, we have F(s) = e 2t e st dt = e (2 s)t dt = 1 s 2, s > 2. (iii) The rate of growth of the function f(t) = e t2, t >, as t exceeds the exponential growth prescribed in Theorem 9.7. It turns out that this function does not have a Laplace transform. 9.9 Theorem. (i) L is linear; that is, L[c 1 f 1 + c 2 f 2 ] = c 1 L[f 1 ] + c 2 L[f 2 ] for any functions f 1, f 2 (to which L can be applied) and any numbers c 1, c 2. (ii) If u = u(x, t) and L[u](x, s) = U(x, s), then L[u t ](x, s) = su(x, s) u(x, ), L[u tt ](x, s) = s 2 U(x, s) su(x, ) u t (x, ). (iii) For the same type of function u, differentiation with respect to x and the Laplace transformation commute: L[u x ](x, s) = (L[u]) x (x, s) = U (x, s). (iv) If we adopt a definition of the convolution f g of two functions f and g, which is slightly different from Definition 8.3, namely, (f g)(t) = t f(τ)g(t τ) dτ = t f(t τ)g(τ) dτ = (g f)(t), (9.2) then L[f g] = L[f]L[g]. 9.1 Remarks. (i) As in the case of the Fourier transformations, in general L[fg] L[f]L[g]. (ii) Clearly, L 1 is also linear. The Laplace transforms of some frequently used functions are listed in Table A.5 in the Appendix.

254 9.1 Definition and Properties Example. To find the inverse Laplace transform of the function 1/ ( s(s 2 + 1) ), we first see from Table A.5 that 1/s is the transform of the constant function 1 and 1/(s 2 + 1) is the transform of the function sin t. Therefore, by Theorem 9.9(iv), we can write symbolically so 1 s(s 2 + 1) = 1 s 1 s = L[1]L[sin t] = L[ 1 (sin t) ], [ ] L 1 1 s(s 2 = 1 (sin t) = + 1) t sin τ dτ = 1 cos t. Alternatively, we can split the given function into partial fractions as 1 s(s 2 + 1) = 1 s s s and then use the linearity of L 1 to obtain the result above Example. The simplest way to find the inverse Laplace transform of (3s 2 + 2s + 12)/ ( s(s 2 + 4) ) by direct calculation is to establish the partial fraction decomposition 3s 2 + 2s + 12 s(s 2 + 4) = 3 s + 2 s and then apply formulas 5 and 8 in Table A.5 to arrive at [ 3s L 1 2 ] + 2s + 12 s(s 2 = 3 + sin(2t). + 4) The next assertion lists two other helpful properties of the Laplace transformation Theorem. If L[f](s) = F(s), then (i) L[e at f](s) = F(s a), s > a = const; (ii) L[H(t b)f(t b)](s) = e bs F(s), b = const > Example. Since L[sin(2t)] = 2 s 2 + 4, L[cos(3t)] = s s 2 + 9, L[t2 ] = 2 s 3, from Theorem 9.13 it follows that L [ e t sin(2t) + e 5t cos(3t) 2(t 2) 2 H(t 2) ] = 2 (s + 1) s 5 (s 5) s 3 e 2s.

255 232 9 The Laplace Transformation 9.15 Example. Similarly, we have [ L 1 s 1 s 2 2s ] s e s [ ] = L 1 s 1 (s 1) s e s = e t cos(3t) 1 2 H(t 1) sin ( 2(t 1) ). Exercises In 1 4, use Table A.5 to compute the Laplace transform of the given function f. 1 f(t) = e t sin(3t) 3t 4. 2 f(t) = e 4t cos(2t) + 4(t 3) 3 H(t 3). 3 f(t) = e 2t (cos t 3 sin t) 2t 2 H(t 1). 4 f(t) = t 3 e t 2 3 sin 2 (t/2). In 5 8, use Table A.5 to compute the inverse Laplace transform of the given function F. 5 F(s) = 6 F(s) = 7 F(s) = 3s + 1 s 8 F(s) = 2 s s 2 2s + 1 s 2 2s s + 2 s 2 + 6s s (s 1) 2 e s. e s/2. e 3 s. Answers to Odd-Numbered Exercises 1 3/(s 2 + 2s + 1) 72/s 5. 3 (s 1)/(s 2 + 4s + 5) 2((s 2 + 2s + 2)/s 3 )e s. 5 e t [2 cos(5t) + (3/5) sin(5t)]. 7 (3/(2 2π))t 3/2 e 1/(8t) + erfc(1/(2 2t )).

256 9.2 Applications 9.2 Applications 233 In this section, we examine the use of the Laplace transformation in the study of some typical linear mathematical models The Signal Problem for the Wave Equation Consider a very long elastic string of negligible weight, initially at rest, where the vertical displacement (signal) is given at the near endpoint. This is described by an IBVP of the form Making the notation u tt (x, t) = c 2 u xx (x, t), x >, t >, u(, t) = f(t), t >, u(x, ) =, u t (x, ) =, x >. L[u](x, s) = U(x, s), L[f](s) = F(s), applying L to the PDE and BC, and using the properties of L listed in Theorem 9.9, we arrive at the transformed problem s 2 U(x, s) = c 2 U (x, s), x >, U(, s) = F(s). The ODE above can be rewritten in the form and its general solution is U (x, s) (s/c) 2 U(x, s) =, U(x, s) = C 1 (s)e (s/c)x + C 2 (s)e (s/c)x, where C 1 (s) and C 2 (s) are arbitrary functions of the transformation parameter. Since x, c > and, according to Theorem 9.7, U(x, s) as s, we must have C 1 (s) =. Then the BC yields C 2 (s) = F(s), so U(x, s) = F(s)e (s/c)x = F(s)e (x/c)s. Consequently, by Theorem 9.13(ii), the solution of the original IBVP is u(x, t) = L 1[ F(s)e (x/c)s] = H(t x/c)f(t x/c), < t < x/c, = f(t x/c), t x/c. This solution can also be expressed as f(t x/c), x ct, u(x, t) =, x > ct. (9.3)

257 234 9 The Laplace Transformation We see that u(x, t) is constant when x ct = const. Physically, this means that the solution is a wave of fixed shape (determined by the BC function f) with velocity dx/dt = c. Formula (9.3) indicates that at time t the signal originating from x = has not reached the points x > ct, which are still in the initial state of rest Remark. The same inversion result can also be obtained by means of convolution. Since, by formula 12 in Table A.5, e (x/c)s = L [ δ(t x/c) ], we can write U(x, s) = F(s)e (x/c)s = L[f]L [ δ(t x/c) ] = L [ f δ(t x/c) ], from which we conclude that u(x, t) = L 1 [U](x, t) = f δ(t x/c) = t f(τ)δ(t x/c τ) dτ. If t < x/c, then t x/c τ < for τ t, so δ(t x/c τ) = ; hence, u(x, t) = for t < x/c, or, equivalently, for x > ct. If t > x/c, then t x/c τ = at < τ = t x/c < t; therefore, by (9.1), for x < ct, u(x, t) = t f(τ)δ(t x/c τ) dτ = This result is the same as (9.3). f(τ)δ(t x/c τ) dτ = f(t x/c). Exercises Use the Laplace transformation to find the solution of the IBVP for the PDE u tt (x, t) = c 2 u xx (x, t) + q(x, t), x >, t >, with coefficient c, function q, BC, and ICs as indicated. 1 c = 1, q(x, t) = 1, u(, t) = t, u(x, ) =, u t (x, ) = 1. 2 c = 1, q(x, t) = 1, u(, t) = t + 1, u(x, ) = 2, u t (x, ) =. 3 c = 2, q(x, t) = e 3t, u x (, t) = 2t, u(x, ) =, u t (x, ) = 1. 4 c = 2, q(x, t) = t, u x (, t) = 2e t, u(x, ) = 1, u t (x, ) =. 5 c = 1, q(x, t) = e 2t, u(, t) = 1, u(x, ) = 1, u t (x, ) = 2. 6 c = 1, q(x, t) = e 3t, u(, t) = 1, u(x, ) = 2, u t (x, ) = 2. 7 c = 1, q(x, t) = (t + 1)e t, u(, t) = 4, u(x, ) = 2, u t (x, ) = 1.

258 9.2 Applications c = 1, q(x, t) = (2t+1)e 2t, u(, t) = 1, u(x, ) = 1, u t (x, ) = 2. 9 c = 1, q(x, t) = (2 + 3e 2t )e x, u(, t) = 2 e 2t, u(x, ) = e x, u t (x, ) = 2e x. 1 c = 2, q(x, t) = 2(2t + 5e t ) sin x, u x (, t) = 2e t + t, u(x, ) = 2 sin x, u t (x, ) = sin x. 11 c = 1, q(x, t) = e t/2 4 sin(2x), u x (, t) = 2, u(x, ) = 4 sin(2x), u t (x, ) = c = 2, q(x, t) = 8e x sin(2t), u(, t) = sin(2t), u(x, ) =, u t (x, ) = 2e x. 13 c = 2, q(x, t) = 2, u x (, t) = cos t, u(x, ) =, u t (x, ) = c = 1, q(x, t) = e x, u(, t) = e t, u(x, ) = 2, u t (x, ) =. Answers to Odd-Numbered Exercises 1 u(x, t) = (1/2)[t 2 2t (t x)(t x 4)H(t x)]. 3 u(x, t) = (1/9)(e 3t 6t 1) (1/2)(x 2t) 2 H(t x/2). 5 u(x, t) = 5/4 + 3t/2 e 2t /4 + [ 1/4 3(t x)/2 + e 2(t x) /4]H(t x). 7 u(x, t) = 5 t (t + 3)e t + [t x 1 + (3 + t x)e (t x) ]H(t x). 9 u(x, t) = (2 e 2t )e x. 11 u(x, t) = 4e t/2 sin(2x). 13 u(x, t) = t 2 + t 2H(t x/2) sin(t x/2) Heat Conduction in a Semi-Infinite Rod A very long rod without sources, with the near endpoint kept in open air of zero temperature, and with a constant initial temperature distribution, is modeled by the IBVP u t (x, t) = u xx (x, t), x >, t >, u x (, t) u(, t) =, t >, u(x, ) = u = const, x >. Let L[u](x, s) = U(x, s). Applying L to the PDE and BCs, we arrive at the transformed problem U (x, s) su(x, s) + u =, x >, U (, s) U(, s) =.

259 236 9 The Laplace Transformation The general solution of the equation is U(x, s) = C 1 (s)e s x + C 2 (s)e s x + 1 s u, where C 1 (s) and C 2 (s) are arbitrary functions of the transformation parameter. Since U(x, s) as s, we must have C 1 (s) =. Then, differentiating U and using the BC at x =, we see that C 2 (s) s C 2 (s) 1 s u =, from which C 2 (s) = u / ( s ( s + 1 )) ; hence, [ 1 U(x, s) = u s ( s + 1 ) s x e + 1 ]. s After some manipulation (coupled with the use of a more comprehensive Laplace transform table than A.5), it can be shown that the solution of the original IBVP is ( ) ( ] x t u(x, t) = L 1 [U](x, t) = u [1 erfc 2 x erfc + )e t 2 x+t. t Other IBVPs for a semi-infinite rod can be solved by the same method Example. Consider the IBVP u t (x, t) = u xx (x, t) + sin x, x >, t >, u(, t) = 2t 1, t >, u(x, ) = 1, x >. Setting L[u](x, s) = U(x, s), as before, and applying the Laplace transformation to the PDE and BC, we arrived at the ODE problem U (x, s) su(x, s) = 1 1 s sin x, x >, U(, s) = 2 s 2 1 s. The general solution of the equation, written as the sum of the complementary function and a particular integral, is U(x, s) = C 1 (s)e s x + C 2 (s)e s x + 1 s + 1 s(s + 1) sin x. Since U(x, s) as s, it follows that C 1 (s) =. Then, applying the BC, we find that C 2 (s) = 2/s 2 2/s; hence, given the partial fraction decomposition 1 s(s + 1) = 1 s 1 s + 1,

260 9.2 Applications 237 we arrive at U(x, s) = ( 2 s 2 2 ) e s x + 1 ( 1 s s + s 1 ) sin x. s + 1 By formulas 5 7, 15, and 16 in Table A.5, the solution of the original IBVP is u(x, t) = L 1 [U](x, t) ( ) t = 2x 2 x π e x /(4t) + (x 2 + 2t 2) erfc 2 t (1 e t ) sin x. Verification with Mathematica R. The input u = - 2 x ((t/pi) (1/2)) E ( - x 2/(4 t)) + (x t - 2) Erfc[x/(2 t (1/2))] (1 - E ( - t)) Sin[x]; q, f, g} =Sin[x],2 t - 1, 1}; Simplify[D[u,t] - D[u,x,x] - q,(u /. x -> ) - f,assuming[x>, Limit[u,t ->, Direction -> - 1]] - g}] generates the output,, }. Exercises Use the Laplace transformation to find the solution of the IBVP consisting of the PDE u t (x, t) = ku xx (x, t) + q(x, t), x >, t >, coefficient k, function q, BC, and IC as indicated. 1 k = 1, q(x, t) = 1, u(, t) = t + 1, u(x, ) = sin(2x). 2 k = 4, q(x, t) = 1, u(, t) = 2 t, u(x, ) = 2. 3e 3t + 2(x 1) 2 4t, < x 1, 3 k = 1, q(x, t) = 3e 3t, x > 1, u x (, t) = 4t, ) = 1, u, u x continuous at x = 1. u(x, 4 k = 1, q(x, t) = (1 t)e t + 6, < x 1, (1 t)e t, x > 1, u x (, t) = 6, u(x, ) = 3(2x x 2 ), < x 1, 3, x > 1, u, u x continuous at x = 1. 5 k = 1, q(x, t) = (2t + 1) cos x, u x (, t) =, u(x, ) = cos x.

261 238 9 The Laplace Transformation 6 k = 2, q(x, t) = (2t + 3)e x, u(, t) = t + 2, u(x, ) = 2e x. 7 k = 1, q(x, t) = 5e x[ cos(2t) + sin(2t) ], u(, t) = 3 cos(2t) + sin(2t), u(x, ) = 3e x. 8 k = 1, q(x, t) = e 2t (4 cos t + 3 sin t) sin(2x), u x (, t) = 2e 2t (cos t + 2 sin t), u(x, ) = sin(2x). 9 k = 4, q(x, t) = 3e 3t, u(, t) = 3 e 3t, u(x, ) = 1. 1 k = 4, q(x, t) = 2e 2t x, u(, t) = e 2t 2, u(x, ) = e x. Answers to Odd-Numbered Exercises 1 u(x, t) = t + erfc(x/(2 t )) + e 4t sin(2x). 3 u(x, t) = e 3t + 2t(x 1) 2 H(1 x). 5 u(x, t) = (2t 1) cos x. 7 u(x, t) = e x [sin(2t) 3 cos(2t)]. 9 u(x, t) = 2 e 3t + erfc(x/(4 t )) Finite Rod with Temperature Prescribed on the Boundary The IBVP w t (x, t) = w xx (x, t), < x < 1, t >, w(, t) =, w(1, t) = 1, t >, w(x, ) =, < x < 1, is of a type that we have already encountered. The equilibrium solution in this case, computed as in Section 6.1, is w (x) = x. Using this solution, we reduce the problem to a similar one where both BCs are homogeneous and which can thus be solved by the method of separation of variables. Putting all the results together, we obtain w(x, t) = x + ( 1) n 2 nπ sin(nπx)e n2 π 2t. (9.4) The same IBVP can also be solved by using the Laplace transformation with respect to t. If we write L[w](x, s) = W (x, s), then from the PDE and BCs we find that W is the solution of the BVP W (x, s) sw(x, s) =, < x < 1, W (, s) =, W(1, s) = 1 s.

262 9.2 Applications 239 The general solution of the transformed equation can be written in the form (see Remark 1.4(i)) W (x, s) = C 1 (s) cosh( s x) + C 2 (s) sinh( s x), with C 1 (s) and C 2 (s) determined from the BCs. Applying these conditions leads to C 1 (s) =, C 2 (s) sinh s = 1 s, from which hence, C 2 (s) = 1 s sinh s ; W (x, s) = sinh ( s x ) s sinh s. Then, using the inverse transformation and comparing with (9.4), we see that [ 1 w(x, t) = L 1 sinh ( s x ) ] s sinh = x + ( 1) n 2 π 2t. s nπ sin(nπx)e n2 Now consider the more general IBVP u t (x, t) = u xx (x, t), < x < 1, t >, u(, t) =, u(1, t) = f(t), t >, u(x, ) =, < x < 1, and let L[u](x, s) = U(x, s) and L[f](s) = F(s). Applying L to the PDE and BCs, we arrive at the BVP U (x, s) su(x, s) =, < x < 1, U(, s) =, U(1, s) = F(s). Proceeding exactly as before, we find that U(x, s) = F(s) sinh ( s x ) [ 1 sinh sinh ( s x ) ]} = F(s) s s s sinh s Since w(x, ) = in the IBVP for w, it follows that = F(s) [ sw(x, s) ]. (9.5) L[w t ](x, s) = sw(x, s) w(x, ) = sw(x, s); therefore, by (9.5) and Theorem 9.9(iv), L[u] = U = F(sW) = L[f]L[w t ] = L[f w t ].

263 24 9 The Laplace Transformation Using (9.2) and integration by parts, we now obtain u(x, t) = (f w t )(x, t) = t f(t τ)w τ (x, τ) dτ = f(t τ)w(x, τ) t τ=t τ= + w(x, τ)f (t τ) dτ = f()w(x, t) f(t)w(x, ) + = t t w(x, t τ)f (τ) dτ + f()w(x, t), w(x, t τ)f (τ) dτ where we have used the condition w(x, ) = and the commutativity of the convolution operation. This result shows how the solution of a problem with more general BCs can sometimes be obtained from that of a problem with simpler ones Remark. If we replace the BC w(1, t) = 1 by then the formula above becomes u(x, t) = w(1, t) = δ(t), t w(x, t τ)f(τ) dτ Diffusion-Convection Problems Suppose that a chemical substance is being poured at a constant rate into a straight, narrow, clean river that flows with a constant velocity. The concentration u(x, t) of the substance at a distance x downstream at time t is the solution of the IBVP u t (x, t) = σu xx (x, t) vu x (x, t), x >, t >, u(, t) = α = const, t >, u(x, ) =, x >, where σ is the diffusion coefficient, v = const > is the velocity of the river, α = const > is related to the substance discharge rate, and the second term on the right-hand side in the PDE accounts for the convection effect of the water flow on the substance.

264 9.2 Applications 241 If the river is slow, then the convection term is much smaller than the diffusion term and the PDE assumes the approximate form u t (x, t) = σu xx (x, t), x >, t >, which is the diffusion equation. If the river is fast, then the approximation is given by the convection equation u t (x, t) = vu x (x, t), x >, t >. Since we have already studied the diffusion (heat) equation, we now turn our attention to the convection and combined cases. (i) The IBVP for pure convection is u t (x, t) = vu x (x, t), x >, t >, u(, t) = α, t >, u(x, ) =, x >. Let L[u](x, s) = U(x, s). Applying the Laplace transformation to the PDE and BC, we arrive at the problem vu (x, s) + su(x, s) =, x >, U(, s) = 1 s α, with solution U(x, s) = 1 s αe (s/v)x = 1 s αe (x/v)s. Since L 1[ e as /s ] = H(t a), we set a = x/v to find that the solution of the original IBVP is u(x, t) = L 1 [U](x, t) = L 1 [ 1 s αe (x/v)s ] = αh(t x/v) =, < t < x/v, α, t x/v. Thus, the substance reaches a fixed position x at time t = x/v; after that, the concentration of the substance at x remains constant (equal to the concentration of the substance at the point where it is poured into the river). The line t = x/v in the (x, t)-plane is the advancing wave front of the substance (see Fig. 9.1).

265 242 9 The Laplace Transformation Figure 9.1 The front wave and the regions behind it and ahead of it. (ii) We now consider a very long river with the substance already uniformly distributed in it from the source of the river up to the observation point x =, and assume that both diffusion and convection effects are significant. This mixed diffusion-convection problem in an infinite one-dimensional medium is modeled by the IVP u t (x, t) = σu xx (x, t) vu x (x, t), < x <, t >, u(x, t), u x (x, t) as x ±, t >, u(x, ) = 1 H(x), < x <. We already know two possible methods for solving this problem: we can apply the Laplace transformation with respect to t or the (full) Fourier transformation with respect to x. In light of the discussion in (i), however, we indicate a third one, which consists in changing the x-coordinate by connecting it to the wave front through the combination ξ = x vt. (9.6) Clearly, ξ = means that the point (x, t) is on the wave front, ξ > means that (x, t) is ahead of the wave front, and ξ < means that (x, t) is behind the wave front. We also write Hence, by the chain rule, u(x, t) = u(ξ + vt, t) = w(ξ, t). u t = w ξ ξ t + w t = vw ξ + w t, u x = w ξ ξ x = w ξ, u xx = (w ξ ) ξ ξ x = w ξξ.

266 9.2 Applications 243 Since t = yields x = ξ, the IVP becomes w t (ξ, t) = σw ξξ (ξ, t), < ξ <, t >, w(ξ, t), w ξ (ξ, t) as ξ ±, t >, w(ξ, ) = 1 H(ξ), < ξ <. This problem was solved earlier by means of the Fourier transformation (see Section 8.1), and its solution is 1 w(ξ, t) = 2 πσt 1 = 2 πσt [ 1 H(y) ] e (ξ y) 2 /(4σt) dy e (ξ y)2 /(4σt) dy. Then, by (9.6), we find that the solution of our IVP in terms of the original variables x and t is 1 u(x, t) = 2 πσt e (x vt y)2 /(4σt) dy. Exercises Use the Laplace transformation to find the solution of the IBVP for the PDE u t (x, t) = ku xx (x, t) au x (x, t) + bu(x, t) + q(x, t), x >, t >, with coefficients k, a, b, function q, BC, and IC as indicated. 1 k = 1, a = 2, b = 1, q(x, t) = (6t + 1) cos x + (2t + 3) sin x, u(, t) = t, u(x, ) = sin x. 2 k = 1, a = 4, b = 4, q(x, t) = (3 2t)e 2t + (1 9t)e x, u x (, t) = t, u(x, ) = 1. 3 k = 2, a = 1, b = 1/8, q(x, t) =, u(, t) = 1, u(x, ) =. 4 k = 2, a = 4, b = 2, q(x, t) = δ(t), u(, t) = δ(t), u(x, ) = 1. Answers to Odd-Numbered Exercises 1 u(x, t) = t cos x + (3t + 1) sin x. 3 u(x, t) = e x/4 erfc ( x/ 8t ).

267 244 9 The Laplace Transformation Loss Transmission Line Linear problems for equations that contain additional, lower-order terms can also be solved by the Laplace transformation method Example. Consider the IBVP u tt (x, t) + 4u t (x, t) + 4u(x, t) = u xx (x, t) 1, x >, t >, u(, t) =, u(x, ) = 1, u t (x, ) =, x >. If we write, as usual, L[u](x, s) = U(x, s) and apply the Laplace transformation to the PDE and BC, we arrive at the ODE problem U (x, s) (s + 2) 2 U(x, s) = U(, s) =, with general solution 1 4s s2, x >, s U(x, s) = C 1 (s)e (s+2)x + C 2 (s)e (s+2)x + s2 + 4s 1 s(s + 2) 2. The property that U(x, s) as s implies that C 1 (s) =, and the BC yields U(x, s) = s2 + 4s 1 s(s + 2) 2 [ 1 e (s+2)x ]. This can also be written in the form where, using partial fractions, we have U(x, s) = F(s) F(s)e sx e 2x, (9.7) F(s) = s2 + 4s 1 s(s + 2) 2 = 1 4s + 5 4(s + 2) + 5 2(s + 2) 2. By formulas 3 and 5 7 in Table A.5, f(t) = L 1 [F](t) = e 2t te 2t = 1 4 Applying formula 2 in (9.7), we now find that [ 5(2t + 1)e 2t 1 ]. u(x, t) = L 1 [U](x, t) = f(t) f(t x)h(t x)e 2x = 1 [ 4 5(2t + 1)e 2t 1 ] 1 [ 4 5(2t 2x + 1)e 2t e 2x] H(t x). Verification with Mathematica R. Because of the split form of the solution, we use the input

268 9.2 Applications 245 u1, u2} =(1/4) (5 (2 t + 1) E ( - 2 t) - 1), (1/4) (5 (2 t + 1) E ( - 2 t) - 1) - (1/4) ((5 (2 t - 2 x + 1)) E ( - 2 t) - E (2 x))}; q, f, g, h} =-1,1,, }; Simplify[D[u1,u2},t,t] + 4 D[u1,u2},t] + 4 u1,u2} - D[u1,u2},x,x] - q,(u1 /. t -> ) - f,(d[u1,t]/. t -> ) - g, (u2 /. x -> ) - h}] which generates the output, },,, }. Exercises Use the Laplace transformation to find the solution of the IBVP consisting of the PDE u tt (x, t) + au t (x, t) + bu(x, t) = c 2 u xx (x, t) du x (x, t) + q(x, t), coefficients a, b, c, d, function q, BC, and ICs as indicated. 1 a = 1, b =, c = 2, d = 1, q(x, t) = 2x 2 + 4xt 16t, u x (, t) =, u(x, ) = 1, u t (x, ) = 2x a = 1, b = 2, c = 1, d = 1, q(x, t) = (8t + 5)e x+2t, u x (, t) = te 2t, u(x, ) =, u t (x, ) = e x. 3 a = 4, b = 4, c = 1, d =, q(x, t) = 2e 2t, u(, t) = 4t, u(x, ) =, u t (x, ) = 1. 4 a = 2, b = 1, c = 1, d =, q(x, t) = 1, u(, t) = t + 2, u(x, ) = 1, u t (x, ) =. 5 a = 3, b = 2, c = 1, d = 1, q(x, t) = 4t, u(, t) = 2t 2, u(x, ) = 2, u t (x, ) = 4. 6 a = 2, b = 3/4, c = 1/2, d = 1/2, q(x, t) = 3, u(, t) = t + 2, u(x, ) = 1, u t (x, ) = 1. x >, t >, Answers to Odd-Numbered Exercises 1 u(x, t) = e t + 2x 2 t. 3 u(x, t) = (t 2 + t)e 2t + e 2x [1 2(t x) e 2(t x) ]H(t x). 5 u(x, t) = 2t 3 + e 2t + 4e t + e x [1 4e (t x) e 2(t x) ]H(t x).

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270 C H A P T E R 1 The Method of Green s Functions The types of problems we have considered for the heat, wave, and Laplace equations have solutions that are determined uniquely by the prescribed data (boundary conditions, initial conditions, and any nonhomogeneous term in the equation). It is natural, therefore, to seek a formula that gives the solution directly in terms of the data. Such closed-form solutions are constructed by means of the so-called Green s function of the given problem, and are of great importance in practical applications. 1.1 The Heat Equation We begin by examining the simplest case The Time-Independent Problem The equilibrium temperature distribution in a finite rod with internal sources and zero temperature at the endpoints is modeled by a BVP of the form (see Section 6.1) u (x) = 1 q(x), < x < L, k u() =, u(l) =. (1.1) For convenience, we have omitted the subscript from the symbol of the steady-state solution, but have kept the factor 1/k since we will later make a comparison between the solutions of the equilibrium and time-dependent problems. If we have just one unit source concentrated at a point ξ, < ξ < L, then q(x) = δ(x ξ), 247

271 248 1 The Method of Green s Functions and the corresponding two-point solution G(x, ξ) of the above BVP satisfies G xx (x, ξ) = 1 δ(x ξ), < x < L, k G(, ξ) =, G(L, ξ) =. (1.2) The function G(x, ξ) can be computed explicitly. Since (see Remark 9.4(ii)) H x (x ξ) = δ(x ξ), from (1.2) it follows that G x (x, ξ) = 1 C 1 (ξ), x < ξ, k H(x ξ) + C 1(ξ) = 1 k + C 1(ξ), x > ξ, from which xc 1 (ξ) + C 2 (ξ), x < ξ, [ G(x, ξ) = x C 1 (ξ) 1 ] + C 3 (ξ), x > ξ, k (1.3) where C 1, C 2, and C 3 are arbitrary functions of ξ. Using the BCs in (1.2), we find that [ C 2 (ξ) =, L C 1 (ξ) 1 ] + C 3 (ξ) =. k Hence, C 3 (ξ) = L [ C 1 (ξ) 1/k ], and (1.3) becomes xc 1 (ξ), x < ξ, [ G(x, ξ) = (x L) C 1 (ξ) 1 ], x > ξ. k (1.4) If G(x, ξ) had a jump (H-type) discontinuity at x = ξ, then G x would have a δ-type singularity at x = ξ. Since this is not the case, we must conclude that G(x, ξ) is continuous at x = ξ; in other words, G(ξ, ξ) = G(ξ+, ξ), which, in view of (1.4), leads to [ ξc 1 (ξ) = (ξ L) C 1 (ξ) 1 ]. k Thus, C 1 (ξ) = (L ξ)/(kl), and (1.4) yields x (L ξ), x ξ, kl G(x, ξ) = ξ (1.5) kl (L x), x > ξ. Clearly, G(x, ξ) = G(ξ, x).

272 1.1 The Heat Equation 249 Using integration by parts, we find that for two smooth functions u and v on [, L], L (u v v u) dx = u v L L u v dx v u L L + v u dx = [ u (L)v(L) v (L)u(L) ] [ u ()v() v ()u() ]. (1.6) This is known as Green s formula. If u is now the solution of (1.1) and v = G is the solution of (1.2), then the right-hand side in (1.6) vanishes and we arrive at L [ u(x)δ(x ξ) G(x, ξ)q(x)] dx =. By (9.1), interchanging x and ξ and recalling that G(x, ξ) = G(ξ, x), we obtain u(x) = L G(x, ξ)q(ξ) dξ. (1.7) G(x, ξ), called the Green s function of the BVP (1.1), is the temperature at x due to a concentrated unit heat source at ξ. Formula (1.7) shows the aggregate influence of all the sources q(ξ) in the rod on the temperature at x. A representation formula similar to (1.7) can also be derived for nonhomogeneous BCs. Suppose that the BCs in (1.1) are replaced by u() = a and u(l) = b. Then (1.6) with the same choice of u and v as above becomes so L [ u(x)δ(x ξ) G(x, ξ)q(x) ] dx = k [ u(x)gx (x, ξ) ] x=l x=, u(x) = L G(x, ξ)q(ξ) dξ k [ bg ξ (x, L) ag ξ (x, ) ]. By (1.5), x kl, x < ξ, G ξ (x, ξ) = x L kl, x > ξ; consequently, the desired representation formula is L u(x) = G(x, ξ)q(ξ) dξ + b x ( L + a 1 x ). (1.8) L

273 25 1 The Method of Green s Functions 1.1 Example. To compute the steady-state solution for the IBVP u t (x, t) = u xx (x, t) + x 1, < x < 1, t >, u(, t) = 2, u(1, t) = 1, t >, u(x, ) = f(x), < x < 1, we use (1.8) with k = 1, L = 1, a = 2, b = 1, and q(x) = x 1. First, by (1.5), x(1 ξ), x ξ, G(x, ξ) = ξ(1 x), x > ξ, x, x < ξ, G ξ (x, ξ) = 1 x, x > ξ; hence, 1 G(x, ξ)q(ξ) dξ = x = (1 x) ξ(1 x)(ξ 1) dξ + x 1 (ξ 2 ξ) dξ x = 1 6 x x2 1 3 x. x 1 x x(1 ξ)(ξ 1) dξ (ξ 1) 2 dξ Since bx/l + a(1 x/l) = x + 2(1 x) = 2 3x, the equilibrium solution (1.8) of the given IBVP is u(x) = 1 6 x x2 1 3 x + 2. Verification with Mathematica R. The input u = - (1/6) x 3 + (1/2) x 2 - (1/3) x + 2; q, f1, f2} =x - 1,2, - 1}; Simplify[D[u,x,x] + q,(u /. x -> ) - f1,(u /. x -> 1) - f2}] generates the output,, }. 1.2 Example. In the case of the IBVP u t (x, t) = u xx (x, t) + q(x), < x < 1, t >, u(, t) = 1, u(1, t) = 3, t >, u(x, ) = f(x), < x < 1, with q(x) = 2, < x 1/2, 1, 1/2 < x < 1,

274 1.1 The Heat Equation 251 we notice that the function G is the same as in the preceding example, whereas a = 1 and b = 3. Given that the expressions of q and G change at x = 1/2 and x = ξ, respectively, we split the computation of the equilibrium solution (1.8) into two parts. (i) If < x < 1/2, the first term on the right-hand side in (1.8) is written as a sum of three integrals, one for each of the intervals < ξ x, x < ξ 1/2, and 1/2 < ξ < 1. Thus, (1.8) yields u(x) = x 2ξ(1 x) dξ + = x x /2 x 2x(1 ξ) dξ + 1 1/2 x(1 ξ) dξ + 2x + 1 (ii) If 1/2 < x < 1, the three integrals are over the intervals < ξ 1/2, 1/2 < ξ x, and x < ξ < 1, so u(x) = 1/2 2ξ(1 x) dξ + = 1 2 x x It is easy to verify that but that u( 1 2 ) = u( 1 2 x 1/2 ξ(1 x) dξ + 1 x x(1 ξ) dξ + 2x ) = 16, u ( 1 ) 2 = u ( ) = 2 8, u ( 1 2 ) = 2, u ( 1 2 +) = 1, which confirms that, as expected, the discontinuity of q at x = 1/2 has reduced the smoothness of the solution. Verification with Mathematica R. In view of the split form of the source term and solution, we use the input u1, u2 =-x 2 + (21/8) x + 1,(1/2) x 2 + (9/8) x + (11/8)}; q1, q2},f1, f2}} =2,-1},1,3}}; Simplify[D[u1,u2},x,x] +q1,q2}, (u1 /. x -> ) - f1, (u2 /. x -> 1) - f2}] which generates the output, },, }. 1.3 Remark. The Green s function can be expanded in a double Fourier series. In view of the eigenfunctions of the Sturm Liouville problem associated with (1.1) (see Section 5.1), the continuity of G, and the symmetry relation G(x, ξ) = G(ξ, x), it seems reasonable to seek a series representation of the form ( G(x, ξ) = b mn sin nπx ) sin mπξ L L. (1.9) m=1

275 252 1 The Method of Green s Functions Differentiating (1.9) term by term twice with respect to x and substituting in the ODE in (1.2), we obtain [ m=1 ( nπ L ) 2 b mn sin nπx ] sin mπξ L L = 1 δ(x ξ). k Next, we multiply this equality by sin(pπx/l), p = 1, 2,..., integrate over [, L], and use (2.5) and (9.1) to find that ( ) 2 pπ kl L 2 m=1 b mp sin mπξ L pπξ = sin, p = 1, 2,.... L Theorem 3.23(ii) now implies that the only nonzero coefficients above are b pp = 2L/(kp 2 π 2 ), so series (1.9) is G(x, ξ) = 2L nπx nπξ kn 2 sin sin π2 L L. Exercises In 1 1, construct the appropriate Green s function and solve the BVP for the one-dimensional steady-state heat equation ku (x) + q(x) =, < x < L, with constants k, L, function q, and BCs as indicated. 1 k = 2, L = 1, q(x) = 2x 1, u() = 1, u(1) = 2. 1, < x 1, 2 k = 1, L = 2, q(x) = u() = 3, u(2) = 1. x 1 < x < 2, 3 k = 1, L = 2, q(x) = 4, u() = 2, u (2) = 3. 4 k = 2, L = 1, q(x) = 1 x, u() = 1, u (1) = 1. 1, < x 1, 5 k = 2, L = 2, q(x) = u() = 2, u (2) = 3. 2, 1 < x < 2, 1, < x 1/2, 6 k = 1, L = 1, q(x) = u() = 4, u (1) = 2. x + 1, 1/2 < x < 1, 7 k = 3, L = 1, q(x) = 2, u () = 3, u(1) = 4. 8 k = 2, L = 2, q(x) = x 2, u () = 1, u(2) = 3. 2, < x 1/2, 9 k = 1, L = 1, q(x) = u () = 2, u(1) = 5. 3, 1/2 < x < 1,

276 1.1 The Heat Equation k = 1, L = 2, q(x) = x 1, < x 1, 1, 1 < x < 2, u () = 3, u(2) = 3. Answers to Odd-Numbered Exercises x(1 ξ)/2, x ξ, 1 G(x, ξ) = u(x) = (1/12)(12 37x + 3x 2 2x 3 ). ξ(1 x)/2, x > ξ, x, x ξ, 3 G(x, ξ) = u(x) = 2 + 5x 2x 2. ξ, x > ξ, x/2, x ξ, x 2 /4 + 7x/2 2, x 1, x 2 /2 + 5x 11/4, 1 < x 2. 5 G(x, ξ) = u(x) = ξ/2, x > ξ, (1 ξ)/3, x ξ, 7 G(x, ξ) = u(x) = (1/3)(2 9x + x 2 ). (1 x)/3, x > ξ, 1 ξ, x ξ, 9 G(x, ξ) = u(x) = 1 x, x > ξ, 59/8 2x x 2, x 1/2, 8 9x/2 + 3x 2 /2, 1/2 < x The Time-Dependent Problem A finite rod with internal sources and zero temperature at the endpoints is modeled by the IBVP u t (x, t) = ku xx (x, t) + q(x, t), < x < L, t >, u(, t) =, u(l, t) =, t >, u(x, ) = f(x), < x < L. (According to the arguments presented in Chapter 6, we may consider homogeneous BCs without loss of generality.) This problem can be solved by the method of eigenfunction expansion (see Section 7.1), so let where q n (t) = 2 L L u(x, t) = q(x, t) = f(x) = q(x, t) sin nπx L dx, u n (t) sin nπx L, q n (t) sin nπx L, f n sin nπx L, f n = 2 L L f(x) sin nπx L dx. (1.1)

277 254 1 The Method of Green s Functions Replacing these series in the PDE, we find in the usual way that the u n, n = 1, 2,..., satisfy ( ) 2 nπ u n (t) + k u n (t) = q n (t), t >, L u n () = f n. This problem is solved, for example, by means of the integrating factor exp ( ) 2 nπ k dt} = e k(nπ/l)2t. L Thus, taking the IC for u n into account, we obtain so, by (1.1), u(x, t) = = = [ t u n (t) = e k(nπ/l)2 t L = f n e k(nπ/l)2t + e k(nπ/l)2 t [ t f n e k(nπ/l)2t + e k(nπ/l)2 t [ 2 L L ] q n (τ)e k(nπ/l)2τ dτ + C t f(ξ) sin nπξ ] L dξ e k(nπ/l)2 t + e k(nπ/l)2 t t [ 2 L L q n (τ)e k(nπ/l)2τ dτ; ] q n (τ)e k(nπ/l)2τ dτ sin nπx L q(ξ, τ) sin nπξ ] } L dξ e k(nπ/l)2τ dτ [ ] 2 nπx nπξ f(ξ) sin sin 2 L L L e k(nπ/l) t dξ + L t [ q(ξ, τ) sin nπx L ] 2 nπx nπξ sin sin 2 L L L e k(nπ/l) (t τ) dτ dξ. If we now define the Green s function of this problem by G(x, t; ξ, τ) = 2 nπx nπξ sin sin 2 L L L e k(nπ/l) (t τ), τ < t, (1.11)

278 1.1 The Heat Equation 255 then the solution of the IBVP can be written in the form u(x, t) = L G(x, t; ξ, )f(ξ) dξ + L t G(x, t; ξ, τ)q(ξ, τ) dτ dξ. (1.12) The first term in this formula represents the influence of the initial temperature in the rod on the subsequent temperature at any point x and any time t. The second term represents the influence of all the sources in the rod at all times < τ < t on the temperature at x and t. This expresses what is known as the causality principle. 1.4 Example. We use (1.11) and (1.12) to compute the solution of the IBVP u t (x, t) = u xx (x, t) + t(x 1), < x < 1, t >, u(, t) =, u(1, t) =, t >, u(x, ) = x, < x < 1. Here, k = 1, L = 1, q(x, t) = t(x 1), and f(x) = x, so which leads to u(x, t) = = 1 G(x, t; ξ, τ) = + 2 sin(nπx) sin(nπξ)e n2 π 2 (t τ), G(x, t; ξ, )ξ dξ + [ 1 2 sin(nπx) 2 sin(nπx) Integrating by parts, we have 1 1 t 1 t [ 1 ] ξ sin(nπξ) dξ ξ sin(nπξ) dξ = ( 1) n+1 1 nπ, (ξ 1) sin(nπξ) dξ = 1 nπ, G(x, t; ξ, τ)τ(ξ 1) dτ dξ e n2 π 2 t ][ t (ξ 1) sin(nπξ) dξ τe n2 π 2 (t τ) dτ = 1 n 2 π 2 t 1 n 4 π 4 ( 1 e n 2 π 2 t ). ] τe n2 π 2 (t τ) dτ.

279 256 1 The Method of Green s Functions We now combine these results and obtain the solution of the IBVP in the form u(x, t) = [ 2 ( 1) n+1 e n2 π 2 t nπ 1 n 2 π 2 t + 1 ( 1 e n 2 π 2 t ) ] n 4 π 4 sin(nπx). (1.13) Verification with Mathematica R. Since (1.13) is the Fourier sine series representation of u, we check that the nth term satisfies the PDE and the IC with t(x 1) and x replaced, respectively, by the corresponding terms in their own expansions. The BCs are also tested. Specifically, the input un = ((2/(n Pi)) (( - 1) (n + 1) E ( - n 2 Pi 2 t) - (1/(n 2 Pi 2)) t + (1/(n 4 Pi 4)) (1 - E ( - n 2 Pi 2 t)))) Sin[n Pi x]; q, f1, f2, g} =t (x - 1),,, x}; qn = Simplify[2 Integrate[q Sin[n Pi x],x,,1}],n Integers] Sin[n Pi x]; gn = Simplify[2 Integrate[g Sin[n Pi x],x,,1}],n Integers] Sin[n Pi x]; Simplify[D[un,t] - D[un,x,x] - qn,(un /. x ) - f1,(un /. x -> 1) - f2, (un /. t -> ) - gn},n Integers] generates the output,,, }. 1.5 Remark. Green s functions and representation formulas in terms of such functions can also be constructed for IBVPs with other types of BCs, and for IBVPs where the space variable takes values in a semi-infinite or infinite interval (see, for example, (8.8)). Exercises Construct a series representation of the appropriate Green s function and solve the IBVP consisting of the PDE u t (x, t) = ku xx (x, t) + q(x, t), < x < L, t >, constants k, L, function q, BCs, and IC as indicated. 1 k = 2, L = 1, q(x, t) = t sin(πx), u(, t) =, u(1, t) =, u(x, ) = sin(2πx). 2 k = 1, L = 2, q(x, t) = x(t 2), u(, t) =, u(2, t) =, u(x, ) = 1.

280 3 k = 1, L = 2, q(x, t) = 1, < x 1, 1, 1 < x < 2, u(, t) =, u(2, t) =, u(x, ) = 1 x. 4 k = 3, L = 1, q(x, t) = x, < x 1/2,, 1/2 < x < 1, u(, t) =, u(1, t) =, u(x, ) =, < x 1/2, 2, 1/2 < x < 1. 5 k = 1, L = 1, q(x, t) = 2xt, u x (, t) =, u x (1, t) =, u(x, ) = x. 1, < x 1/2, 6 k = 2, L = 1, q(x, t) = 2, 1/2 < x < 1, u x (, t) =, u x (1, t) =, u(x, ) = x 1. 7 k = 2, L = 2, q(x, t) = t, u(, t) =, u x (2, t) =, u(x, ) = 1., < x 1, 8 k = 1, L = 2, q(x, t) = 1, 1 < x < 2, 1.1 The Heat Equation 257 u(, t) =, u x (2, t) =, u(x, ) = x. 9 k = 1, L = 1, q(x, t) = 1, < x 1/2,, 1/2 < x < 1, u x (, t) =, u(1, t) =, u(x, ) =, < x 1/2, 2, 1/2 < x < 1. 1 k = 1, L = 2, q(x, t) = 1, u x (, t) =, u(2, t) =, u(x, ) = x 2. Answers to Odd-Numbered Exercises 1 G(x, t; ξ, τ) = 2e 2n2 π 2 (t τ) sin(nπx) sin(nπξ), u(x, t) = (1/(4π 4 ))(e 2π2t + 2π 2 t 1) sin(πx) e 8π2t sin(2πx). 3 G(x, t; ξ, τ) = e n2 π 2 (t τ)/4 sin(nπx/2) sin(nπξ/2), u(x, t) = (2/(n 3 π 3 ))[1 + ( 1) n ]n 2 π 2 + 4(e n2 π 2 t/4 1) [1 + ( 1) n 2 cos(nπ/2)]}e n2 π 2 t/4 sin(nπx/2).

281 258 1 The Method of Green s Functions 5 G(x, t; ξ, τ) = 1 + 2e n2 π 2 (t τ) cos(nπx) cos(nπξ), u(x, t) = 1/2 + t 2 /2 + [( 1) n 1](2/(n 6 π 6 )) [2 + n 4 π 4 + 2(n 2 π 2 t 1)e n2 π 2t ]e n2 π 2t cos(nπx). 7 G(x, t; ξ, τ) = e (2n 1)2 π 2 (t τ)/8 sin((2n 1)πx/4) sin((2n 1)πξ/4), u(x, t) = (4/((2n 1) 5 π 5 )) (2n 1) 4 π 4 8[8+((2n 1) 2 π 2 t 8)e (2n 1)2 π 2 t/8 ]} e (2n 1)2 π 2 t/8 sin((2n 1)πx/4). 9 G(x, t; ξ, τ) = 2e (2n 1)2 π 2 (t τ)/4 cos((2n 1)πx/2) cos((2n 1)πξ/2), u(x, t) = [8/((2n 1) 3 π 3 )]( 1) n (2n 1) 2 π The Laplace Equation [2 (2n 1) 2 π 2 2e (2n 1)2 π 2 t/4 ] sin((2n 1)π/4)} e (2n 1)2 π 2 t/4 cos((2n 1)πx/2). The equilibrium temperature in a thin, uniform rectangular plate with timeindependent sources and zero temperature on the boundary is the solution of the BVP ( u)(x, y) = q(x, y), < x < L, < y < K, u(x, ) =, u(x, K) =, < x < L, u(, y) =, u(l, y) =, < y < K. As in Section 1.1, let G(x, y; ξ, η) be the effect at (x, y) produced by just one unit source located at a point (ξ, η), < ξ < L, < η < K. Then G is the solution of the BVP (x, y)g(x, y; ξ, η) = δ(x ξ, y η), < x < L, < y < K, G(x, ; ξ, η) =, G(x, K; ξ, η) =, < x < L, G(, y; ξ, η) =, G(L, y; ξ, η) =, < y < K, where δ(x ξ, y η) = δ(x ξ)δ(y η) and (x, y) indicates that the Laplacian is applied with respect to the variables x, y. Also, let D be the rectangle where the problem is formulated, that is, D = (x, y) : < x < L, < y < K }, and let D be the four-sided boundary of D.

282 1.2 The Laplace Equation 259 Since D is piecewise smooth, we use the divergence theorem to find that for a pair of smooth functions u and v, (u v v u) da = (u div grad v v div grad u) da D D [ = div(u grad v) (grad u) (grad v) D = D D div(v grad u) + (grad v) (grad u) ] da [ ] (u grad v) n (v grad u) n ds = (uv n vu n ) ds, (1.14) where da and ds are the elements of area and arc, respectively, and the subscript n denotes the derivative in the direction of the unit outward normal to the boundary. (The normal is not defined at the four corner points, but this does not influence the outcome.) Equality (1.14) is Green s formula for functions of two space variables. If u is the solution of the given BVP and v is replaced by G, then the homogeneous BCs satisfied by both u and G make the right-hand side in (1.14) vanish, and the formula reduces to [ ] u(x, y)δ(x ξ)δ(y η) q(x, y)g(x, y; ξ, η) da(x, y) =, D where da(x, y) indicates that integration is performed with respect to x, y. By (9.1), this yields u(ξ, η) = G(x, y; ξ, η)q(x, y) da(x, y). (1.15) D At the same time, applying (1.14) with u(x, y) replaced by G(x, y; ξ, η) and v(x, y) replaced by G(x, y; ρ, σ) and making use once more of (9.1), we obtain the symmetry G(ξ, η; ρ, σ) = G(ρ, σ; ξ, η). Then, as a simple interchange of variables shows, (1.15) becomes the representation formula u(x, y) = G(x, y; ξ, η)q(ξ, η) da(ξ, η). (1.16) D G(x, y; ξ, η) is called the Green s function of the given BVP. Formula (1.16) shows the effect of all the sources in D on the temperature at the point (x, y).

283 26 1 The Method of Green s Functions 1.6 Remark. To find a Fourier series representation for G, we recall that the two-dimensional eigenvalue problem (5.55) associated with our BVP has the eigenvalue-eigenfunction pairs ( ) 2 nπ λ nm = + L S nm = sin nπx L ( ) 2 mπ, K mπy sin, n, m = 1, 2,.... K Consequently, it seems reasonable to seek an expansion of the form G(x, y; ξ, η) = = m=1 m=1 c nm (ξ, η)s nm (x, y) c nm (ξ, η) sin nπx L sin mπy K. If we replace this series in the equation satisfied by G, then (x, y)g(x, y; ξ, η) = m=1 = c nm (ξ, η)( S nm )(x, y) m=1 λ nm c nm (ξ, η)s nm (x, y) = δ(x ξ)δ(y η). Multiplying both sides above by S pq (x, y), integrating over D, and taking (9.1) into account, we arrive at c pq (ξ, η) = 4 LKλ pq S pq (ξ, η). Hence, the double Fourier sine series for G is G(x, y; ξ, η) = 4 LK m=1 sin(nπξ/l) sin(mπη/k) (nπ/l) 2 + (mπ/k) Example. To compute the solution of the BVP we notice that here sin nπx L u xx (x, y) + u yy (x, y) = 5π 2 sin(πx) sin(2πy), < x < 1, < y < 2, u(x, ) =, u(x, 2) =, < x < 1, u(, y) =, u(1, y) =, < y < 2, L = 1, K = 2, q(x, t) = 5π 2 sin(πx) sin(2πy). sin mπy K. (1.17)

284 1.2 The Laplace Equation 261 Consequently, by (1.17), G(x, y; ξ, η) = 2 m=1 sin(nπξ) sin(mπη/2) n 2 π 2 + m 2 π 2 /4 sin(nπx) sin mπy 2, so from (1.16)) and (2.5) it follows that u(x, y) = 2 1 = 4 ( = ( 2) m=1 m=1 4 sin(nπξ) sin(mπη/2) π 2 (4n 2 + m 2 ) ( 5π 2 ) sin(πξ) sin(2πη) dξ δη ( 1 1 4n 2 + m 2 ( ) sin(πξ) sin(nπξ) dξ Verification with Mathematica R. The input sin(nπx) sin mπy 2 sin(2πη) sin mπη ) dη sin(nπx) sin mπy 2 2 ) sin(πx) sin(2πy) = sin(πx) sin(2πy). u = Sin[Pi x] Sin[2 Pi y]; q, f1, f2, g1, g2} =-5 Pi 2 Sin[Pi x] Sin[2 Pi y],,,, }; Simplify[D[u,x,x] + D[u,y,y] - q,(u /. x -> ) - f1,(u /. x -> 1) - f2, (u /. y -> )mi g1, (u /. y -> 2) - g2}] generates the output,,,, }. Exercises Construct a series representation of the appropriate Green s function and solve the BVP for the nonhomogeneous Laplace equation u xx (x, y) + u yy (x, y) = q(x, y), < x < L, < y < K, with constants L, K, function q, and BCs as indicated. 1 L = 1, K = 2, q(x, y) = sin(πy), u(, y) =, u(1, y) =, u(x, ) =, u(x, 2) =. 2 L = 1, K = 1, q(x, y) = (y 1) cos(πx), u x (, y) =, u x (1, y) =, u(x, ) =, u(x, 1) =.

285 262 1 The Method of Green s Functions 3 L = 1, K = 1, q(x, y) = x sin(3πy/2), u(, y) =, u(1, y) =, u(x, ) =, u y (x, 1) =. 4 L = 2, K = 1, q(x, y) = cos(πx/4), u x (, y) =, u(2, y) =, u(x, ) =, u(x, 1) =. 5 L = 2, K = 1, q(x, y) = 2 cos(3πy), u(, y) =, u x (2, y) =, u y (x, ) =, u y (x, 1) =. 6 L = 1, K = 2, q(x, y) = sin(3πx/2), u(, y) =, u x (1, y) =, u y (x, ) =, u(x, 2) =. 7 L = 1, K = 1, q(x, y) = 1, u(, y) =, u(1, y) =, u(x, ) =, u(x, 1) =. 8 L = 1, K = 2, q(x, y) = 2, u(, y) =, u(1, y) =, u(x, ) =, u y (x, 2) =. 9 L = 1, K = 1, q(x, y) = x, u x (, y) =, u(1, y) =, u(x, ) =, u y (x, 1) =. 1 L = 2, K = 1, q(x, y) = y, u(, y) =, u(2, y) =, u y (x, ) =, u y (x, 1) =. Answers to Odd-Numbered Exercises 1 G(x, y; ξ, η) = [8/((m 2 + 4n 2 )π 2 )] sin(nπx) sin(mπy/2) m=1 sin(nπξ) sin(mπη/2), u(x, y) = [( 1) n 1]2[n(n 2 + 1)π 3 ] 1 sin(nπx) sin(πy). 3 G(x, y; ξ, η) = m=1 [16/((4m 2 + 4n 2 4m + 1)π 2 )] sin(nπx) sin((2m 1)πy/2) sin(nπξ) sin((2m 1)πη/2), u(x, y) = ( 1) n 8[n(4n 2 + 9)π 3 ] 1 sin(nπx) sin(3πy/2). 5 G(x, y; ξ, η) = m= [32/((16m 2 + 4n 2 4n + 1)π 2 )] sin((2n 1)πx/4) cos(mπy) sin((2n 1)πξ/4) cos(mπη), u(x, y) = 128[(2n 1)(4n 2 4n + 145)π 3 ] 1 7 G(x, y; ξ, η) = m=1 sin((2n 1)πx/4) cos(3πy). (4/((m 2 + n 2 )π 2 )) sin(nπx) sin(mπy) sin(nπξ) sin(mπη),

286 1.3 The Wave Equation 263 u(x, y) = 4[( 1) m 1][( 1) n 1][mn(m 2 + n 2 )π 4 ] 1 m=1 sin(nπx) sin(mπy). 9 G(x, y; ξ, η) = [8/((2m 2 + 2n 2 2m 2n + 1)π 2 )] m=1 cos((2n 1)πx/2) sin((2m 1)πy/2) cos((2n 1)πξ/2) sin((2m 1)πη/2), u(x, y) = 32[2 + ( 1) n (2n 1)π] m=1 [(2m 1)(2n 1) 2 (2m 2 + 2n 2 2m 2n + 1)π 5 ] 1 cos((2n 1)πx/2) sin((2m 1)πy/2). 1.3 The Wave Equation The vibrations of an infinite string are described by the IVP u tt (x, t) = c 2 u xx (x, t) + q(x, t), < x <, t >, u(x, t), u x (x, t) as x ±, t >, u(x, ) = f(x), u t (x, ) = g(x), < x <. If we have a unit force acting at a point ξ at time τ >, then its influence G(x, t; ξ, τ) on the vertical vibration of a point x at time t is the solution of the IVP where G tt (x, t; ξ, τ) = c 2 G xx (x, t; ξ, τ) + δ(x ξ, t τ), < x <, t >, G(x, t; ξ, τ), G x (x, t; ξ, τ) as x ±, t >, G(x, t; ξ, τ) =, < x <, t < τ, δ(x ξ, t τ) = δ(x ξ)δ(t τ) and the IC reflects the physical reality that the displacement of the point x is not affected by the unit force at ξ until this force has acted at time τ. As we did in Chapter 8 in the case of the Cauchy problem for the heat equation, we find the function G by means of the full Fourier transformation. First, we note that, by (9.1), F [ δ(x ξ) ] = 1 2π δ(x ξ)e iωx dx = 1 2π e iωξ.

287 264 1 The Method of Green s Functions Therefore, if we write F[G](ω, t; ξ, τ) = G(ω, t; ξ, τ) and apply F to the PDE and IC satisfied by G, we arrive at the transformed problem G tt (ω, t; ξ, τ) + c 2 ω 2 G(ω, t; ξ, τ) = 1 2π e iωξ δ(t τ), t >, G(ω, t; ξ, τ) =, t < τ. Since δ(t τ) = for t τ, the solution of (1.18) is, t < τ, G(ω, t; ξ, τ) = C 1 cos ( cω(t τ) ) + C 2 sin ( cω(t τ) ), t > τ, (1.18) (1.19) where C 1 and C 2 are arbitrary functions of ω, ξ, and τ. Requiring G to be continuous at t = τ yields C 1 =. To find C 2, we consider an interval [τ 1, τ 2 ] such that < τ 1 < τ < τ 2, and integrate (1.18) with respect to t over this interval: τ 2 G t (ω, τ 2 ; ξ, τ) G t (ω, τ 1 ; ξ, τ) + c 2 ω 2 G(ω, t; ξ, τ) dt By (1.19), = 1 τ 2 e iωξ 2π τ 1 δ(t τ) dt = G t (ω, τ 1 ; ξ, τ) =, τ 1 1 2π e iωξ G t (ω, τ 2 ; ξ, τ) = cωc 2 cos ( cω(τ 2 τ) ). δ(t τ) dt = 1 2π e iωξ. If we now let τ 1, τ 2 τ, from the continuity of G at t = τ it follows that C 2 = e iωξ /( 2π cω); hence,, t < τ, G(ω, t; ξ, τ) = 1 e iωξ sin ( cω(t τ) ), t > τ. 2π c ω According to formulas 12 and 3 in Table A.2 in the Appendix, F 1 [ 2 π ] sin(aω) = H(a x ), ω F 1[ e iωa F[f](ω) ] = f(x a); so, setting a = c(t τ) and a = ξ, respectively, we obtain G(x, t; ξ, τ) = 1 2c H( c(t τ) x ξ ). (1.2)

288 1.3 The Wave Equation 265 The diagram in Fig. 1.1 shows the values of G in the upper half (t > ) of the (x, t)-plane, computed from (1.2). Using a similar diagram, it is easy to see that (1.2) can also be written in the form G(x, t; ξ, τ) = 1 2c[ H ( (x ξ) + c(t τ) ) H ( (x ξ) c(t τ) )]. (1.21) Figure 1.1 The values of G(x, t; ξ, τ) in the half-plane t >. A procedure analogous to, but more involved than, that followed in the case of the Laplace equation can also be devised for the wave equation to obtain a symmetry relation for G and a representation formula for a solution u in terms of G. Thus, in its most general form, the latter is u(x, t) = t b a b + a c 2 G(x, t; ξ, τ)q(ξ, τ) dξ dτ [ G(x, t; ξ, )uτ (ξ, ) G τ (x, t; ξ, )u(ξ, ) ] dξ t [ Gξ (x, t; ξ, τ)u(ξ, τ) G(x, t; ξ, τ)u ξ (ξ, τ) ] ξ=b dτ, (1.22) ξ=a where q is the forcing term and a and b are the points where the BCs are prescribed. G(x, t; ξ, τ) is called the Green s function for the wave equation. When < x <, as in our problem, the corresponding formula is obtained from the one above by letting a and b and taking into account that G(x, t; ξ, τ) = for x sufficiently large. 1.8 Example. Consider the IBVP u tt (x, t) = u xx (x, t) + q(x, t), < x <, t >, u(x, t), u x (x, t) as x ±, t >, u(x, ) =, u t (x, ) =, < x <,

289 266 1 The Method of Green s Functions where q(x, t) = t, 1 < x < 1, t >, otherwise. By formulas (1.22) and (1.21) with c = 1, the solution of this problem is computed as that is, u(x, t) = = t t = 1 2 t G(x, t; ξ, τ)q(ξ, τ) dξ dτ 1 [ ] 2 H(x ξ + t τ) H(x ξ t + τ) q(ξ, τ) dξ dτ x+t τ q(ξ, τ) dξ dτ 1 2 u(x, t) = 1 2 t t x+t τ x t+τ x t+τ q(ξ, τ) dξ dτ. The value of the solution at, say, (x, t) = (3, 2) is u(3, 2) = τ 1+τ q(ξ, τ) dξ dτ. q(ξ, τ) dξ dτ; To calculate the integral above, we sketch the lines ξ = 1 + τ and ξ = 5 τ in the (ξ, τ) system of axes and identify the domain of integration (see Fig. 1.2). Since, as the diagram shows, q is zero in this domain, it follows that u(3, 2) =. Figure 1.2 The domain of integration for (x, t) = (3, 2).

290 1.3 The Wave Equation 267 Similarly, u(2, 2) = τ τ q(ξ, τ) dξ δτ. Following the same procedure and taking into account the intersection of the domain of integration and the semi-infinite strip where q is nonzero (see Fig. 1.3), we have u(2, 2) = τ τ dξ dτ = τξ ξ=1 ξ=τ dτ = τ(1 τ) dτ = Figure 1.3 The domain of integration for (x, t) = (2, 2). Finally, to compute u(1, 3) = τ 2+τ q(ξ, τ) dξ dτ, we make use of the sketch in Fig. 1.4 and obtain u(1, 3) = τ dξ dτ τ τ dξ dτ = Figure 1.4 The domain of integration for (x, t) = (1, 3).

291 268 1 The Method of Green s Functions Alternatively, by changing the order of integration, u(1, 3) = ξ+2 τ dτ dξ = (ξ + 2) 2 dξ = Verification with Mathematica R. To confirm the values of the integrals computed above, we use the input q = Piecewise[,t< - 1},t, - 1<x<1},, t>1}}]; (1/2) Assuming[t Reals,Integrate[Integrate[q,x,1 + t, 5 - t}], t,,2}], Integrate[Integrate[q,x,t,1}],t,,1}], (Integrate[Integrate[q,x, - 1,1}],t,,1}] + Integrate [Integrate[q,x, t,1}],t,1,3}])}] which generates the output, 1 12, 13 6 }. 1.9 Remark. If there is no forcing term (q = ) in the general IVP with < x <, then the representation formula (1.22) reduces to u(x, t) = [ G(x, t; ξ, )uτ (ξ, ) G τ (x, t; ξ, )u(ξ, ) ] dξ. (1.23) This formula can be further simplified by using the explicit form of G. By (1.21) and Remark 9.4(ii), according to which we have G τ (x, t; ξ, τ) = 1 2 H (τ a) = δ(τ a), [ δ ( (x ξ) + c(t τ) ) + δ ( (x ξ) c(t τ) )] ; so, by the definition of δ and H (see Section 9.1), (1.23) becomes u(x, t) = c [ δ(x ξ ct) + δ(x ξ + ct) ] f(ξ) dξ [ H(x ξ + ct) H(x ξ ct) ] g(ξ) dξ = 1 [ ] 1 2 f(x + ct) + f(x ct) + 2c x+ct x ct g(ξ) dξ. This formula, called d Alembert s solution, is revisited in Chapter 12, where it is established by another method.

292 1.3 The Wave Equation 269 Exercises Construct an integral representation of the solution in terms of the appropriate Green s function for the IVP u tt (x, t) = c 2 u xx (x, t) + q(x, t), < x <, t >, u(x, t), u x (x, t) as x ±, u(x, ) =, u t (x, ) =, < x <, with the constant c and function q as indicated, then compute the solution at the given point x and time t. 1 c = 1, q(x, t) = (x, t) = ( 1, 2). 2 c = 1, q(x, t) = (x, t) = (2, 3). 3 c = 2, q(x, t) = (x, t) = (1, 2). 4 c = 2, q(x, t) = (x, t) = ( 2, 1). 5 c = 1, q(x, t) = (x, t) = (3, 2). 6 c = 1, q(x, t) = (x, t) = ( 4, 2). 7 c = 2, q(x, t) = (x, t) = ( 3, 4). 8 c = 2, q(x, t) = (x, t) = (5, 3). x, 1 < x < 2, t >, otherwise, xt, 1 < x < 3, t >, otherwise, x t, 2 < x < 1, t >, otherwise, 2x + t, 3 < x < 1, t >, otherwise, 1 x + 2t, < x < 4, t > 1, otherwise, xt + 2, 5 < x <, < t < 1, otherwise, 2 t, 8 < x <, t >, otherwise, x 1, 3 < x < 1, t >, otherwise,

293 27 1 The Method of Green s Functions 9 c = 1, q(x, t) = (x, t) = (4, 2). 1 c = 1, q(x, t) = (x, t) = ( 1, 4). 2x t 1, < x < 5, 1/2 < t < 3/2, otherwise, x(t 1), 3 < x < 2, t >, otherwise, Answers to Odd-Numbered Exercises 1 u( 1, 2) = 1/3. 3 u(1, 2) = 27/32. 5 u(3, 2) = 1/3. 7 u( 3, 4) = 8/3. 9 u(4, 2) = 133/24.

294 C H A P T E R 11 General Second-Order Linear Equations Having studied several solution procedures for the heat, wave, and Laplace equations, we need to explain why we have chosen these particular models in preference to others. In Chapter 4, we mentioned that these were typical examples of what we called parabolic, hyperbolic, and elliptic equations, respectively. Below, we present a systematic discussion of the general secondorder linear PDE in two independent variables and show how such an equation can be reduced to its simplest form. It will be seen that if the equation has constant coefficients, then its dominant part that is, the sum of the terms containing the highest-order derivatives with respect to each of the variables consists of the same terms as one of the three equations named above. This gives us a good indication of what solution technique we should use and what kind of behavior to expect from the solution The Canonical Form We start by making a preliminary examination of these PDEs in terms of the coefficients of their dominant terms Classification The general form of a second-order linear PDE in two independent variables is A(x, y)u xx + B(x, y)u xy + C(x, y)u yy + D(x, y)u x + E(x, y)u y + F(x, y)u = G(x, y), (11.1) where u = u(x, y) is the unknown function and A,..., G are given coefficients. (In particular, some or all of these coefficients may be constant.) 271

295 General Second-Order Linear Equations 11.1 Definition. (i) If B 2 4AC >, (11.1) is called a hyperbolic equation. (ii) If B 2 4AC =, (11.1) is called a parabolic equation. (iii) If B 2 4AC <, (11.1) is called an elliptic equation Example. For the one-dimensional wave equation we have (considering t in place of y) u tt c 2 u xx =, A = c 2, C = 1, B = D = E = F = G = ; hence, B 2 4AC = 4c 2 >, which means that the equation is hyperbolic at all points in the (x, t)-plane Example. In the case of the one-dimensional heat equation we have u t ku xx =, A = k, E = 1, B = C = D = F = G =, so B 2 4AC = : the equation is parabolic in the entire (x, t)-plane Example. The (two-dimensional) Laplace equation is obtained for u = u xx + u yy = A = 1, C = 1, B = D = E = F = G = ; therefore, B 2 4AC = 4 <, which means that this equation is elliptic throughout the (x, y)-plane Example. The equation u xx y u xy + xu yy + (2x + y)u x 3yu y + 4u = sin(x 2 2y), y >, fits the general template with A = 1, B = y, C = x, D = 2x + y, E = 3y, F = 4, G = sin(x 2 2y), so B 2 4AC = y 4x. Consequently, (i) if y > 4x, the equation is hyperbolic; (ii) if y = 4x, the equation is parabolic; (iii) if y < 4x, the equation is elliptic. In other words, the type of this equation at (x, y) depends on where the point lies in the half-plane y > (see Fig. 11.1).

296 11.1 The Canonical Form 273 Figure 11.1 The type of the equation by region Reduction to Canonical Form We introduce new coordinates r = r(x, y), s = s(x, y), or, conversely, x = x(r, s), y = y(r, s), and write u ( x(r, s), y(r, s) ) = v(r, s). By the chain rule, and u x = v r r x + v s s x, u y = v r r y + v s s y, u xx = (u x ) x = (v r r x + v s s x ) x = (v r ) x r x + v r (r x ) x + (v s ) x s x + v s (s x ) x = [ (v r ) r r x + (v r ) s s x ] rx + v r r xx + [ (v s ) r r x + (v s ) s s x ] sx + v s s xx = v rr r 2 x + 2v rs r x s x + v ss s 2 x + v r r xx + v s s xx, u yy = (u y ) y = (v r r y + v s s y ) y = (v r ) y r y + v r (r y ) y + (v s ) y s y + v s (s y ) y = [ (v r ) r r y + (v r ) s s y ] ry + v r r yy + [ (v s ) r r y + (v s ) s s y ] sy + v s s yy = v rr r 2 y + 2v rs r y s y + v ss s 2 y + v r r yy + v s s yy, u xy = (u x ) y = (v r r x + v s s x ) y = (v r ) y r x + v r (r x ) y + (v s ) y s x + v s (s x ) y = [ (v r ) r r y + (v r ) s s y ] rx + v r r xy + [ (v s ) r r y + (v s ) s s y ] sx + v s s xy = v rr r x r y + v rs (r x s y + r y s x ) + v ss s x s y + v r r xy + v s s xy. Replacing all the derivatives in (11.1) and gathering the like terms, we arrive at an equation of the form Ā(r, s)v rr + B(r, s)v rs + C(r, s)v ss + D(r, s)v r + Ē(r, s)v s + F(r, s)v = Ḡ(r, s), (11.2)

297 General Second-Order Linear Equations where the new coefficients through the equalities Ā,..., Ḡ are connected to the old ones A,..., G Ā = Ar 2 x + Br x r y + Cr 2 y, B = 2Ar x s x + B(r x s y + r y s x ) + 2Cr y s y, C = As 2 x + Bs x s y + Cs 2 y, D = Ar xx + Br xy + Cr yy + Dr x + Er y, Ē = As xx + Bs xy + Cs yy + Ds x + Es y, F = F, Ḡ = G. (11.3) We now choose r and s so that Ā = C =. Since neither r = r(x, y) nor s = s(x, y) can be a constant, it follows that at least one of r x, r y and at least one of s x, s y must be nonzero. Suppose, for definiteness, that r y and s y are nonzero. Setting the expression of Ā in (11.3) equal to zero and dividing through by r 2 y, we find that A ( rx r y ) 2 + B ( rx r y ) + C = ; (11.4) a similar procedure applied to the expression of C in (11.3) leads to ( ) 2 ( ) sx sx A + B + C =. (11.5) s y From (11.4) and (11.5) it follows that r x /r y and s x /s y are the roots of the quadratic equation with coefficients A, B, C, so s y r x = B + r y s x s y = B B2 4AC, 2A B2 4AC. 2A (11.6) Obviously, the nature of the solutions of (11.6) depends on whether the given equation is hyperbolic, parabolic, or elliptic. Exercises Discuss the type (hyperbolic, parabolic, or elliptic) of the given PDE and sketch a graph in the (x, y)-plane to illustrate your findings. 1 u xx + (x 1)u xy + u yy 2x 2 u x + 3xyu y + 2u = sin x. 2 u xx + y u xy (x 2)u yy + 2u y (x y)u = e x sin y, y >.

298 11.2 Hyperbolic Equations yu xx xu xy + yu yy 3x 2 u y = xe xy. 4 2xu xx u xy + (y + 1)u yy xu x + (x 2y)u = x + 2y 2. 5 (x + 2)u xx + 2(x + y)u xy + 2(y 1)u yy 3x 2 u x = x 3 y xu xx + 4yu xy + (4 x)u yy 2xyu y + 2u = x(2y + 1). Answers to Odd-Numbered Exercises 1 (x, y) : 1 < x < 3}: elliptic; (x, y) : x < 1 or x > 3}: hyperbolic; (x, y) : x = 1 or x = 3}: parabolic. 3 (x, y) : x < 2 y }: elliptic; (x, y) : x > 2 y }: hyperbolic; (x, y) : x = 2 y }: parabolic. 5 (x, y) : (x + 1) 2 + (y 2) 2 < 1}: elliptic; (x, y) : (x + 1) 2 + (y 2) 2 > 1}: hyperbolic; (x, y) : (x + 1) 2 + (y 2) 2 = 1}: parabolic Hyperbolic Equations Since here B 2 4AC >, there are two distinct equations (11.6). Consider the ODEs dy dx = B B2 4AC, 2A dy dx = B + B2 4AC, 2A called the characteristic equations for (11.1), and let (11.7) ϕ(x, y) = c 1, ψ(x, y) = c 2, c 1, c 2 arbitrary constants, be the families of their solution curves, called characteristics. Along these curves we have, respectively, which yields dϕ = ϕ x dx + ϕ y dy =, dψ = ψ x dx + ψ y dy =, ϕ x = dy ϕ y dx = B + B2 4AC, 2A ψ x = dy ψ y dx = B B2 4AC. 2A

299 General Second-Order Linear Equations Consequently, the functions r = ϕ(x, y), s = ψ(x, y) (11.8) are solutions of (11.6); that is, they define the change of variables that produces Ā = C =. Using ((11.3) and (11.8), we now compute the new coefficients Ā,..., Ḡ and write out the canonical form (11.2) as B(r, s)v rs + D(r, s)v r + Ē(r, s)v s + F(r, s)v = Ḡ(r, s). (11.9) 11.6 Remark. The hyperbolic equation has an alternative canonical form. If we introduce new variables α, β by means of the formulas α = 1 2 (r + s), β = 1 (r s) 2 and write v ( r(α, β), s(α, β) ) = w(α, β), then v r = w α α r + w β β r = 1 2 w α w β, v s = w α α s + w β β s = 1 2 w α 1 2 w β, v rs = (v r ) s = 1 2 [(w α + w β ) α α s + (w α + w β ) β β s ] and (11.9) becomes = 1 2 [ 1 2 (w αα + w βα ) 1 2 (w αβ + w ββ )] = 1 4 w αα 1 4 w ββ, B(w αα w ββ ) + 2( D + Ē)w α + 2( D Ē)w β + 4 Fw = 4Ḡ. The one-dimensional wave equation u tt c 2 u xx = can be brought to this form with new coefficients D = Ē = F = Ḡ = by means of the substitution τ = ct Example. The coefficients of the PDE are u xx 7u xy + 6u yy = A = 1, B = 7, C = 6, D = E = F = G =. Hence, B 2 4AC = 25 >, so the equation is hyperbolic at all points in the (x, y)-plane. The characteristic equations (11.7) are with general solutions y (x) = 6, y (x) = 1, y = 6x + c 1, y = x + c 2, c 1, c 2 = const, respectively. Consequently, we choose the coordinate transformation r = y + 6x, s = y + x,

300 11.2 Hyperbolic Equations 277 which, by (11.3), leads to B = 25 and D = Ē = F = Ḡ = (we already know that Ā = C = ). Replacing in (11.9), we obtain the canonical form v rs =. Integrating first in terms of s, we find that v r (r, s) = C(r), where C(r) is an arbitrary function. A second integration, this time with respect to r, yields the general solution v(r, s) = ϕ(r) + ψ(s), or, in terms of the original variables x and y, u(x, y) = v(r(x, y), s(x, y)) = ϕ(y + 6x) + ψ(y + x), where ϕ and ψ are arbitrary one-variable functions. Verification with Mathematica R. The input u = ϕ [y + 6 x] + ψ [y + x]; Simplify[D[u,x,x] - 7 D[u,x,y] + 6 D[u,y,y]] generates the output Example. For the PDE we have u xx + u xy 2u yy 3u x 6u y = 18x 9y A = 1, B = 1, C = 2, D = 3, E = 6, F =, G = 18x 9y. Since B 2 4AC = 9 >, this equation is hyperbolic at all points in the (x, y)-plane. By (11.7), the characteristic equations are y (x) = 2, y (x) = 1, with solutions y = 2x+c 1 and y = x+c 2, respectively, c 1, c 2 = const; hence, the coordinate transformation is From (11.3) it follows that r = y 2x, s = y + x. B = 9, D =, E = 9, F =, Ḡ = 9r; therefore, the canonical form (11.9) of the given PDE is v rs + v s = r,

301 General Second-Order Linear Equations or (v s ) r + v s = r, which, treated as a first-order equation for v s, yields v s (r, s) = r 1 + C(s)e r. Then, after integration with respect to s, so the general solution of the PDE is u(x, y) = v ( r(x, y), s(x, y) ) v(r, s) = s(r 1) + ϕ(s)e r + ψ(r), = (x + y)(y 2x 1) + ϕ(x + y)e 2x y + ψ(y 2x), where ϕ and ψ are arbitrary one-variable functions. Verification with Mathematica R. The input u = (x + y) (y - 2 x - 1) + ϕ [x + y] E (2 x - y) + ψ [y - 2 x]; Simplify[D[u,x,x] + D[u,x,y] - 2 D[u,y,y] - 3 D[u,x] - 6 D[u,y] - 18 x + 9 y] generates the output Example. For the PDE 4u xx + 14u xy + 6u yy + 5u x + 15u y = 5 5x 15y we have A = 4, B = 14, C = 6, D = 5, E = 15, F =, G = 5(1 + x + 3y). Since B 2 4AC = 1 >, the equation is hyperbolic, and the characteristic equations (11.7) are y (x) = 1 2, y (x) = 3, with solutions y = x/2+c 1 and y = 3x+c 2 ; hence, we make the transformation r = 2y x, s = y 3x, or, conversely, x = 1 (r 2s), 5 y = 1 (3r s). 5 Then, by (11.3) and (11.9), after some simplification we arrive at the canonical form 2v rs v r = 4r 2s + 2, which, handled as the one in Example 11.8, yields the general solution v(r, s) = ϕ(r)e s/2 + ψ(s) + 2rs 2r 2 + 2r.

302 11.2 Hyperbolic Equations 279 In terms of the original variables x and y, this is u(x, y) = 2(2y x)(1 2x y) + ϕ(2y x)e (y 3x)/2 + ψ(y 3x), where ϕ and ψ are arbitrary one-variable functions. Verification with Mathematica R. The input u = 2 (2 y - x) (1-2 x - y) + ϕ [2 y - x] E ((y - 3 x)/2) + ψ [y - 3 x]; Simplify[4 D[u,x,x] + 14 D[u,x,y] + 6 D[u,y,y] + 5 D[u,x] + 15 D[u,y] x + 15 y] generates the output. Exercises Verify that the given PDE is hyperbolic everywhere in the (x, y)-plane, reduce it to its canonical form, and find its general solution. 1 2u xx 7u xy + 3u yy = 15x 5y. 2 u xx + u xy 2u yy = 72(2x 2 + xy y 2 ) u xx + 2u xy u yy = 32e 2x+2y. 4 2u xx + u xy 6u yy = 98e 7x. 5 6u xx + u xy 2u yy + 42u x 21u y =. 6 u xx + 2u xy 3u yy + 4u x 4u y = 32(3x y). 7 3u xx + u xy 2u yy 3u x + 2u y = 25(2x + 3y). 8 2u xx 3u xy 2u yy + 1u x 2u y = 25(4x + 2y + 5). 9 4u xx + 14u xy + 6u yy 1u x 5u y = 25(4y 7x 2). 1 2u xx 2u xy 4u yy 9u x + 18u y = 9(24x 6y 1). Answers to Odd-Numbered Exercises 1 v rs = 2r, u(x, y) = (3x + y) 2 (x + 2y) + ϕ(3x + y) + ψ(x + 2y). 3 v rs = 2e 2s, u(x, y) = (x + 3y)e 2x+2y + ϕ(x + 3y) + ψ(y x). 5 v rs + 3v s =, u(x, y) = ϕ(3y 2x)e 3(x+2y) + ψ(x + 2y). 7 v rs 2v s = r, u(x, y) = (1/4)(y x) + (1/2)(2x + 3y)(y x) + ϕ(2x + 3y) + ψ(y x)e 4x+6y. 9 2v rs v s = 2 r 2s, u(x, y) = (4x 3y)(3x y) + ϕ(2y x) + ψ(y 3x)e y x/2.

303 28 11 General Second-Order Linear Equations 11.3 Parabolic Equations Since here B 2 4AC =, from (11.6) we see that r and s satisfy the same ODE, which means that we can make only one of Ā and C zero. Let Ā =. Then (11.6) reduces to r x r y = B 2A, from which dy dx = r x r y = B 2A. (11.1) The general solution of this equation provides us with the function r(x, y). Now B 2 4AC = implies that AC and B = 2 AC. Without loss of generality, we may also assume that A, C. Then, by (11.3), B = 2Ar x s x + B(r x s y + r y s x ) + 2Cr y s y = 2 [ Ar x s x + A C(r x s y + r y s x ) + Cr y s y ] = 2 [ A r x ( A s x + C s y ) + C r y ( A s x + C s y ) ] = 2( A r x + C r y )( A s x + C s y ). But, according to (11.1), r x = B A C C r y 2A = 2 =, 2A A so B =. Consequently, s can be chosen arbitrarily, in any manner that does not clash with r given by (11.1) (more precisely, so that the Jacobian of the transformation is nonzero). The canonical form in this case is C(r, s)v ss + D(r, s)v r + Ē(r, s)v s + F(r, s)v = Ḡ(r, s). (11.11) 11.1 Remark. The one-dimensional heat equation u t ku xx = is of the form (11.11) with C = k, D = 1, and Ē = F = Ḡ = Example. The coefficients of the PDE are u xx + 2u xy + u yy = A = 1, B = 2, C = 1, D = E = F = G =. Since B 2 4AC =, the equation is parabolic, and its characteristic equation (11.1) is y (x) = 1, with general solution y = x+c, c = const. Hence, we take r = y x; for the function s we can make any suitable choice, for example, s = y. Then C = 1, D = Ē = F = Ḡ =

304 11.3 Parabolic Equations 281 (we already know that Ā = B = ), which, by (11.11), leads to the canonical form v ss =. Integrating twice with respect to s, we obtain the general solution or, in terms of x and y, v(r, s) = sϕ(r) + ψ(r), u(x, y) = v ( r(x, y), s(x, y) ) = yϕ(y x) + ψ(y x), where ϕ and ψ are arbitrary one-variable functions. Verification with Mathematica R. The input u = y ϕ [y - x] + ψ [y - x]; Simplify[D[u,x,x] + 2 D[u,x,y] + D[u,y,y]] generates the output Example. For the equation we have 4u xx + 12u xy + 9u yy 9u = 9, A = 4, B = 12, C = 9, D = E =, F = 9, G = 9, so B 2 4AC = ; that is, the PDE is parabolic. By (11.1), the characteristic equation is y (x) = 3 2, with general solution or y = 3 2 x + c, 2y 3x = c, c = const, c = const; consequently, we can take r = 2y 3x and, say, s = y, as above. Then C = 9, F = 9, Ḡ = 9, which, replaced in (11.11), yields the canonical form v ss v = 1. The general solution of this equation is v(r, s) = ϕ(r) cosh s + ψ(r) sinh s 1,

305 General Second-Order Linear Equations or, for the given PDE, u(x, y) = v(r(x, y), s(x, y)) = ϕ(2y 3x) cosh y + ψ(2y 3x) sinh y 1, where ϕ and ψ are arbitrary one-variable functions. Verification with Mathematica R. The input u = ϕ [2 y - 3 x] Cosh[y] + ψ [2 y - 3 x] Sinh[y] - 1; Simplify[4 D[u,x,x] + 12 D[u,x,y] + 9 D[u,y,y] - 9 u - 9] generates the output Example. The coefficients of the PDE u xx + 4u xy + 4u yy + 6u x + 12u y = x + 36y show that B 2 4AC =, so the equation is parabolic. In this case, (11.1) becomes y (x) = 2, with solution y = 2x+c, c = const. Then we perform, say, the transformation r = y 2x, s = y, or x = 1 (s r), y = s, 2 and, from (11.3) and (11.11), obtain the canonical form v ss + 3v s = 4 3r + 12s. The general solution of this equation is v(r, s) = 2s 2 rs + ϕ(r)e 3s + ψ(r), which means that u(x, y) = y(2x + y) + ϕ(y 2x)e 3y + ψ(y 2x), where ϕ and ψ are arbitrary one-variable functions. Verification with Mathematica R. The input u = y (2 x + y) + ϕ [y - 2 x] E ( - 3 y) + ψ [y - 2 x]; Simplify[D[u,x,x] + 4 D[u,x,y] + 4 D[u,y,y] + 6 D[u,x] + 12 D[u,y] x - 36 y] generates the output.

306 11.4 Elliptic Equations 283 Exercises Verify that the given PDE is parabolic everywhere in the (x, y)-plane, reduce it to its canonical form, and find its general solution. 1 u xx + 8u xy + 16u yy + 64u = u xx 24u xy + 9u yy + 36u x 27u y = u xx + 3u xy + 9u yy 45u x 27u y + 18u = 18(3xy 5y 2 ). 4 9u xx 6u xy + u yy + 12u x 4u y + 3u = 9x + 21y u xx + 4u xy + u yy 2u x u y 2u = 3y x. 6 u xx + 6u xy + 9u yy + 9u x + 27u y + 18u = 27(4x 2y 1). Answers to Odd-Numbered Exercises 1 v ss + 4v = 1, u(x, y) = 1/4 + ϕ(y 4x) cos(2y) + ψ(y 4x) sin(2y). 3 v ss 3v s + 2v = 2rs, u(x, y) = (1/2)(2y + 3)(3x 5y) + ϕ(5y 3x)e y + ψ(5y 3x)e 2y. 5 v ss v s 2v = r + s, u(x, y) = (1/4)(1 + 2x 6y) + ϕ(2y x)e y + ψ(2y x)e 2y Elliptic Equations The procedure in this case is the same as for hyperbolic equations, but, since this time B 2 4AC <, the characteristic curves are complex (see Section 14.1). However, a real canonical form can still be obtained Example. The coefficients of the PDE are u xx + 2u xy + 5u yy + u x = A = 1, B = 2, C = 5, D = 1, E = F = G =. Thus, B 2 4AC = 16 <, so the equation is elliptic. By (11.7), the characteristic equations are with general solutions y (x) = 1 2i, y (x) = 1 + 2i, y = (1 2i)x + c 1, y = (1 + 2i)x + c 2,

307 General Second-Order Linear Equations respectively. Therefore, the coordinate transformation is r = y (1 2i)x, s = y (1 + 2i)x. Then B = 16, D = (1 2i), Ē = (1 + 2i), and F = Ḡ = (Ā = C = because of the transformation), which yields the complex canonical form 16v rs (1 2i)v r (1 + 2i)v s =. Performing the second transformation α = 1 (r + s), 2 β = 1 i(r s), 2 and writing w(α, β) = v(r(α, β), s(α, β)), we have v r = w α α r + w β β r = 1 2 w α 1 2 iw β, v s = w α α s + w β β s = 1 2 w α iw β, v rs = (v r ) s = ( 1 2 w α 1 2 iw β) α α s + ( 1 2 w α 1 2 iw β) β β s = 1 2 ( 1 2 w αα 1 2 iw αβ) i( 1 2 w αβ 1 2 iw ββ) = 1 4 (w αα + w ββ ), which, replaced in the PDE satisfied by v, produce the new (real) canonical form 4(w αα + w ββ ) w α + 2w β = Remark. The Laplace equation u xx + u yy = is already in this real canonical form. Exercises Verify that the given PDE is elliptic everywhere in the (x, y)-plane and reduce it to its real canonical form. 1 u xx + 4u xy + 5u yy 2u x + u y = 3. 2 u xx + 2u xy + 1u yy + 3u x 2u y + 2u = x + y. 3 u xx + 6u xy + 1u yy u x + u y 3u = 2x y. 4 u xx + 4u xy + 13u yy + 2u x u y + u = 3x + 2y. Answers to Odd-Numbered Exercises 1 w αα + w ββ + 5w α 2w β = 3. 3 w αα + w ββ + 4w α w β 3w = α β.

308 11.5 Other Problems Other Problems In this section, we consider equations with variable coefficients, as well as equations accompanied by initial or boundary conditions Example. The coefficients of the PDE are yu xx + 3yu xy + 3u x =, y, A = y, B = 3y, C =, D = 3, E = F = G =. They yield B 2 4AC = 9y 2 >, so the equation is hyperbolic at all points (x, y) with y =. Here, the characteristic equations (11.7) are with general solutions y (x) =, y (x) = 3, y = c 1, y = 3x + c 2, c 1, c 2 = const, respectively. Hence, we can take our coordinate transformation to be r = y, s = y 3x, which, by (11.3), leads to B = 9y = 9r, Ē = 9, and D = F = Ḡ = (we already know that Ā = C = ). Replacing in (11.9), we obtain the canonical form rv rs + v s =. Writing this equation as r(v s ) r + v s = and using, for example, the integrating factor method or noting that the lefthand side is (rv s ) r, we find that v s (r, s) = 1 r c(s), where c(s) is an arbitrary function. Integration with respect to s now produces the solution v(r, s) = 1 ϕ(s) + ψ(r), r or, in terms of x and y, u(x, y) = v(r(x, y), s(x, y)) = 1 ϕ(y 3x) + ψ(y), y where ϕ and ψ are arbitrary one-variable functions. Verification with Mathematica R. The input u = (1/y) ϕ [y - 3 x] + ψ [y]; Simplify[y D[u,x,x] + 3 y D[u,x,y] + 3 D[u,x]] generates the output.

309 General Second-Order Linear Equations Example. The PDE x 2 u xx + 2xyu xy + y 2 u yy + (x + y)u x = x 2 2y, considered at all points (x, y) except the origin, has coefficients A = x 2, B = 2xy, C = y 2, D = x + y, E = F =, G = x 2 2y. Since B 2 4AC = 4x 2 y 2 4x 2 y 2 =, the equation is parabolic and its characteristic equation is y (x) = y x, with solution y = cx, c = const. Hence, we may use the transformation r = y x, s = y, under which the new coefficients, given by (11.3), are C = s 2, D = r r 2, Ē = F =, Ḡ = s2 r 2 2s. In view of (11.11), we arrive at the canonical form r 2 s 2 v ss (r 4 + r 3 )v r = s 2 2r 2 s Example. Consider the IVP (with the variable t replaced by y to facilitate the use of the general formulas already derived in this chapter) Here, 2u xx 5u xy + 2u yy = 36x 18y, < x <, y >, u(x, ) = 4x 3 + 3x, u y (x, ) = 12x 2 + 4, < x <. A = 2, B = 5, C = 2, D = E = F =, G = 36x 18y, so B 2 4AC = 9 >, which means that the equation is hyperbolic. From (11.7) it follows that the characteristic equations are y (x) = 2, y (x) = 1 2, with solutions y = 2x + c 1 and y = x/2 + c 2, where c 1, c 2 Therefore, we operate the coordinate transformation = const. r = 2x + y, s = x + 2y and, using (11.3), find that B = 9, D = Ē = F =, and Ḡ = 18r. This yields the canonical form (see (11.9)) v rs = 2r,

310 11.5 Other Problems 287 with general solution v(r, s) = r 2 s + ϕ(r) + ψ(s). Hence, the general solution of the given PDE is u(x, y) = (2x + y) 2 (x + 2y) + ϕ(2x + y) + ψ(x + 2y), (11.12) where, as before, ϕ and ψ are arbitrary one-variable functions. To apply the ICs, we first differentiate u with respect to y to obtain u y (x, y) = 2(2x + y)(x + 2y) + 2(2x + y) 2 + ϕ (2x + y) + 2ψ (x + 2y) and then set y = in u and u y. Canceling out the like terms, we arrive at the system of equations ϕ(2x) + ψ(x) = 3x, ϕ (2x) + 2ψ (x) = 4. Differentiation of the first equation with respect to x yields 2ϕ (2x) + ψ (x) = 3, which, combined with the second equation, leads to ψ (x) = 5 3 ; consequently, ψ(x) = 5 3 x + c, where c is an arbitrary constant. Next, ϕ(2x) = 3x ψ(x) = 3x 5 3 x c = 2 (2x) c, 3 so ϕ(x) = 2x/3 c. Then the solution of the IVP, given by (11.12), is u(x, y) = (2x + y) 2 (x + 2y) (2x + y) + 5 (x + 2y) 3 = (2x + y) 2 (x + 2y) + 3x + 4y. Verification with Mathematica R. The input u = (2 x + y) 2 (x + 2 y) + 3 x + 4*y; f, g} =4 x x, 12 x 2 + 4}; Simplify[2 D[u,x,x] - 5 D[u,x,y] + 2 D[u,y,y] + 36 x + 18 y, (u /. y -> ) - f,(d[u,y] /. y -> ) - g}] generates the output,, }.

311 General Second-Order Linear Equations Exercises In 1 6, identify the type of the PDE and reduce the equation to its canonical form. 1 u xx + (x + 1)u xy + xu yy + (2y x 2 )u = y + x x 2, x 1. 2 u xx + 2xu xy + (x 2 + 1)u yy = x + y. 3 u xx + 2yu xy + y 2 u yy + 2u y = x + y. 4 u xx + 2u xy + (x 2 + 1)u yy = 4(y x), x. 5 u xx + xu xy + (2x 4)u yy + (4y x 2 )u = (y 2x)(2y + 4x x 2 ), x 4. 6 u xx + 2 x u xy + xu yy yu y = 2y + 1, x >. In 7 1, identify the type of the PDE, reduce the equation to its canonical form, and compute the solution of the given IVP. 7 4u xx + 12u xy + 9u yy 6u x 9u y = 27(3x 2y), < x <, y >, u(x, ) = 3x, u y (x, ) = 1, < x <. 8 9u xx + 6u xy + u yy + 3u x + u y = 2x 6y + 1, x >, < y <, u(, y) = y, u x (, y) = 2y, < y <. 9 u xx + 6u xy + 8u yy = 8(1x 3y), < x <, y >, u(x, ) = 1(8x 3 + x), u y (x, ) = 84x 2 + 3, < x <. 1 u xx u xy 6u yy 1u x + 3u y = 5, < x <, y >, u(x, ) = 4x + e 6x, u y (x, ) = 3 + 2e 6x, < x <. Answers to Odd-Numbered Exercises 1 Hyperbolic; 2(s 2r 1)v rs 2v s + sv = s r. 3 Parabolic; s 2 v ss + (2r/s r)v r + 2v s = s ln(r/s). 5 Hyperbolic; 2(s 2r 16)v rs 2v s + (2r + s)v = rs. 7 Parabolic; v ss v s = 3r, u(x, y) = 1 12x + 8y (1 9x + 6y)e y 9xy + 6y 2. 9 Hyperbolic; v rs = 2r + 4s, u(x, y) = (3y 1x)(8x 2 6xy + y 2 + 1) + 3y 1x.

312 C H A P T E R 12 The Method of Characteristics The equations in the problems we have investigated so far are all linear, and the terms containing the unknown function and its derivatives have constant coefficients. The only exception is the type of problem where we need to make use of polar coordinates, but in such problems the polar radius is present in some of the coefficients in a very specific way, which does not disturb the solution scheme. Below, we discuss a procedure for solving first-order linear PDEs with more general variable coefficients, and first-order nonlinear PDEs of a particular form. We also re-examine the one-dimensional wave equation from the perspective of this new technique First-Order Linear Equations Consider the IVP u t (x, t) + cu x (x, t) =, < x <, t >, (PDE) u(x, ) = f(x), < x <, (IC) where c = const. If we measure the rate of change of u from a moving position given by x = x(t), then, by the chain rule, d dt u(x(t), t) = u t(x(t), t) + u x (x(t), t)x (t). The first term on the right-hand side above is the change in u at a fixed point x, while the second one is the change in u resulting from the movement of the observation position. Assuming that x (t) = c, from the PDE we see that d dt u(x(t), t) = u t(x(t), t) + cu x (x(t), t) = ; 289

313 29 12 The Method of Characteristics that is, u = const as perceived from the moving observation point. The position of this point is obtained by integrating its velocity x (t) = c: x = ct + x, x = x(). (12.1) This formula defines a family of lines in the (x, t)-plane, which are called characteristics (see Fig. 12.1). As mentioned earlier, the characteristics have the property that u(x, t) takes a constant value along each one of them (but, in general, different constant values on different characteristics). Figure 12.1 Characteristic lines. Hence, to find the value of the solution u at (x, t), we consider the characteristic through (x, t), of equation x = ct + x, which intersects the x-axis at (x, ). Since u is constant on this line, its value at (x, t) is the same as at (x, ). But the latter is known from the IC, so u(x, t) = u(x, ) = f(x ). (12.2) The parameter x is now replaced from the equation (12.1) of the characteristic line: x = x ct. So, by (12.2), the solution of the given IVP is u(x, t) = f(x ct). This formula shows that at a fixed time t, the shape of the solution is the same as at t =, but is shifted by ct along the x-axis. In other words, the shape of the initial data function travels in the positive (negative) x-direction with velocity c if c > (c < ), which means that the solution is a wave Example. In the IVP u t (x, t) u x(x, t) =, < x <, t >, sin x, x π, u(x, ) = otherwise, the ODE of the characteristics is x (t) = 1/2, so the characteristic passing through x = x at t = has equation x = t/2 + x. The dotted lines in Fig are the characteristics passing through x = and x = π at t = ; that is, the lines of equations x = t/2 and x = t/2 + π, respectively.

314 12.1 First-Order Linear Equations 291 Figure 12.2 The characteristic lines through (, ) and (π, ). Since du dt = u t + u x x = u t u x =, the solution u is constant along the characteristics: sin x, x π, u(x, t) = u(x, ) = otherwise; therefore, since x = x t/2 on the characteristic through (x, t), it follows that sin(x t/2), x t/2 π, u(x, t) = otherwise. This can also be written as sin(x t/2), t/2 x t/2 + π, u(x, t) = otherwise. Verification with Mathematica R. The input u1, u2} =Sin[x - t/2],}; f1, f2} =Sin[x],}; Simplify[D[u1,u2},t] + (1/2) D[u1,u2},x],(u1, u2} /. t -> ) -f1, f2}}] generates the output, },, }} Example. The velocity of the observation point in the IVP u t (x, t) + 3tu x (x, t) = u, < x <, t >, u(x, ) = cos x, < x <, is x (t) = 3t, so the characteristic through (x, t) is x = 3t 2 /2 + x, x = x(). Along this characteristic we have du dt = u t + u x x = u t + 3tu x = u,

315 The Method of Characteristics with solution u(x, t) = Ce t, C = const. Since the characteristic through (x, t) also passes through (x, ) and ue t = C is constant on this curve, we use the IC to write C = u(x, t)e t = u(x, )e = u(x, ) = cos x. However, x = x 3t 2 /2 on the characteristic; consequently, u(x, t) = Ce t = e t cos x = e t cos(x 3 2 t2 ). Verification with Mathematica R. The input u = E t Cos[x - 3 t 2/2]; f = Cos[x]; Simplify[D[u,t] + 3 t D[u,x] - u,(u /. t -> ) - f}] generates the output, } Example. In the IVP u t (x, t) + xu x (x, t) = 1, < x <, t >, u(x, ) = x 2, < x <, the velocity of the observation point satisfies the ODE x (t) = x, with general solution x = ce t, c = const. Thus, the characteristic through (x, t) that also passes through (x, ) has equation x = x e t. On this characteristic, du dt = u t + u x x = u t + xu x = 1; that is, u(x, t) = t + C, C = const. Using the IC, we see that C = u(x, t) t = u(x, ) = u(x, ) = x 2. The solution of the IVP is now obtained by replacing x = xe t from the equation of the characteristic: u(x, t) = t + C = t + x 2 = t + x 2 e 2t. Verification with Mathematica R. The input u = t + x 2 E ( - 2 t); f = x 2; Simplify[D[u,t] + x D[u,x] - 1,(u /. t -> ) - f}] generates the output, }.

316 12.1 First-Order Linear Equations Example. The IBVP u t (x, t) + u x (x, t) = x, x >, t >, u(, t) = t, t >, u(x, ) = sin x, x >, needs slightly different handling since here x is restricted to nonnegative values and we also have a BC at x =. First, using the standard argument, we see that the velocity of the moving observation point satisfies x (t) = 1; therefore, the equation of the family of characteristics is x = t + c, c = const. Since this problem is defined in the first quadrant of the (x, t)-plane, we notice (see Fig. 12.3) that if the point (x, t) is above the line x = t, then the characteristic passing through this point never reaches the x-axis, so the IC cannot be used for it. However, this characteristic reaches the t-axis, and we can use the BC instead. We split the discussion into three parts. Figure 12.3 Characteristic lines in the first quadrant. (i) Let x > t. Since the characteristic through (x, t) also passes through (x, ), its equation is written in the form x = t + c = t + x, and the PDE shows that on this line we have hence, Consequently, du dt = u t + u x x = u t + u x = x = t + x ; u(x, t) = 1 2 t2 + x t + C, C = const. C = u(x, t) 1 2 t2 x t = u(x, ) = sin x, which, on substituting x = x t from the equation of the characteristic, yields u(x, t) = 1 2 t2 + x t + sin x = 1 2 t2 + t(x t) + sin(x t) = tx 1 2 t2 + sin(x t).

317 The Method of Characteristics (ii) Let x < t. Then, since the characteristic through (x, t) also passes through (, t ), its equation is written as x = t + c = t t, and on it, with solution du dt = x = t t, u(x, t) = 1 2 t2 t t + C, therefore, using the BC, we deduce that C = const; C = u(x, t) 1 2 t2 + t t = u(, t ) 1 2 t2 + t2 = t t2. To find the solution at (x, t), we now need to replace the parameter t = t x from the equation of the characteristic: u(x, t) = 1 2 t2 t t + t t2 = 1 2 t2 t(t x) + t x + 1 (t x)2 2 = 1 2 x2 x + t. (iii) We see that as the point (x, t) approaches the line x = t from either side, we obtain the same limiting value u(x, t) = x 2 /2. Hence, the solution is continuous across this line, and we can write 1 2 u(x, t) = x2 x + t, x t, xt 1 2 t2 + sin(x t), x > t. Verification with Mathematica R. Given the split form of the solution, we use the input u1, u2} =x 2/2 - x + t,x t - t 2/2 + Sin[x - t]}; f, g} =t, Sin[x]}; Simplify[D[u1,u2},t] + D[u1,u2},x] - x,(u1 /. x -> ) - f, (u2 /. t -> ) - g}] which generates the output, },, } Remark. The continuity of u across the line x = t is due to the continuity of the data at (, ); that is, lim sin x = lim t =. x t When this condition is not satisfied in an IBVP of this type, then the solution u is discontinuous across the corresponding dividing line in the (x, t)-plane. Discontinuities and, in general, any perturbations in the solution always propagate along the characteristic lines.

318 12.1 First-Order Linear Equations Example. The problem u x (x, y) + u y (x, y) + 2u(x, y) =, < x, y <, u(x, y) = x + 1 on the line 2x + y + 1 = is neither an IVP nor a BVP. However, the method of characteristics works in this case as well. Thus, assuming that x = x(y), we can write d dy u(x(y), y) = u y(x(y), y) + u x (x(y), y)x (y); so, if x (y) = 1 that is, x = y + c, c = const then the PDE becomes with general solution du + 2u =, dy u(x, y) = Ce 2y, C = const. Figure 12.4 Characteristics and the data line. The equation of the characteristic through (x, y) and (x 1, y 1 ) (see Fig. 12.4) is x = y + x 1 y 1, and, in view of the prescribed data, on this line we have u(x, y)e 2y = C = u(x 1, y 1 )e 2y1 = (x 1 + 1)e 2y1. Since the point (x 1, y 1 ) lies on both the characteristic and the data lines, its coordinates satisfy the system with solution x 1 y 1 = x y, 2x 1 + y 1 = 1, x 1 = 1 3 (x y 1), y 1 = 1 ( 2x + 2y 1). 3

319 The Method of Characteristics Consequently, the solution of the given problem is u(x, y) = Ce 2y = (x 1 + 1)e 2(y1 y) = 1 3 (x y + 2)e 2(2x+y+1)/3. Verification with Mathematica R. Given that y = 2x 1 on the data line, the input u = (1/3) (x - y + 2) E ( - 2 (2 x + y + 1)/3); f = x + 1; Simplify[D[u,x] + D[u,y] + 2 u,(u /. y -> ( - 2 x - 1)) - f}] generates the output, }. Exercises In 1 1, use the method of characteristics to solve the IVP u t (x, t) + c(x, t)u x (x, t) = q(x, t, u), < x <, t >, u(x, ) = f(x), < x <, with functions c, q, and f as indicated. In each case, sketch the family of characteristics in the (x, t)-plane. 1 c(x, t) = 2, q(x, t, u) = t, f(x) = 1 x. 2 c(x, t) = 2t, q(x, t, u) = 2, f(x) = x 3. 3 c(x, t) = 2t + 1, q(x, t, u) = 2u, f(x) = sin x. 4 c(x, t) = t + 2, q(x, t, u) = u + 2t, f(x) = x c(x, t) = x, q(x, t, u) = e t, f(x) = x c(x, t) = x 1, q(x, t, u) = 3t 2, f(x) = 2x 1. 7 c(x, t) = 3, q(x, t, u) = 1 x, f(x) = e 2x. 8 c(x, t) = x + 2, q(x, t, u) = 2x + t 2, f(x) = 2x c(x, t) = 4, q(x, t, u) = u + x + t, f(x) = cos(2x). 1 c(x, t) = 2x, q(x, t, u) = u 2x, f(x) = 2 x. In 11 16, use the method of characteristics to solve the IBVP u t (x, t) + cu x (x, t) = q(x, t, u), x >, t >, u(, t) = g(t), t >, u(x, ) = f(x), x >, with constant c and functions q, f, and g as indicated. In each case, sketch the family of characteristics in the (x, t)-plane.

320 12.1 First-Order Linear Equations c = 1/2, q(x, t, u) = u, g(t) = 1, f(x) = e x. 12 c = 1, q(x, t, u) = u 1, g(t) = t 1, f(x) = x c = 2, q(x, t, u) = x 2, g(t) = t 2 + 1, f(x) = x. 14 c = 3/2, q(x, t, u) = x + t, g(t) = e t, f(x) = 1 x. 15 c = 1, q(x, t, u) = u + x, g(t) = 2t + 1, f(x) = cos x. 16 c = 2, q(x, t, u) = 2u t, g(t) = t 2, f(x) = 2x 1. In 17 22, use the method of characteristics to solve the problem u y (x, y) + cu x (x, y) = q(x, y, u), < x, y <, u(x, y) = f(x, y) on the line ax + by + d =, with constant c, functions q and f, and line ax+by+d = as indicated. In each case, sketch the family of characteristics and the data line in the (x, y)-plane. 17 c = 2, q(x, y, u) = u, f(x, y) = 2x y, x + y 1 =. 18 c = 1, q(x, y, u) = 2y, f(x, y) = xy, x 2y 1 =. 19 c = 1/2, q(x, y, u) = 2u y, f(x, y) = x + y, 2x + 3y 2 =. 2 c = 1, q(x, y, u) = u + x, f(x, y) = 2x y 1, 2x + y 3 =. 21 c = 2, q(x, y, u) = u x y, f(x, y) = x 2xy, x y 2 =. 22 c = 1/2, q(x, y, u) = 2x + y, f(x, y) = e x y, 3x y 3 =. Answers to Odd-Numbered Exercises 1 x = 2t + x, u(x, t) = 1 x + 2t + t 2 /2. 3 x = t 2 + t + x, u(x, t) = e 2t sin(x t 2 t). 5 x = x e t, u(x, t) = x 2 e 2t e t x = 3t + x, u(x, t) = xt 3t 2 /2 + t + e 2(x+3t). 9 x = 4t + x, u(x, t) = [x 4t cos(2x 8t)]e t x t x = t/2 + x, x t/2, e x+t/2, x t/2, u(x, t) = (t t )/2, x < t/2, e 2x, x < t/2. 13 x = 2t + x, x 2t, 2(t t ), x < 2t, u(x, t) = x 2 t 2xt 2 + 4t 3 /3 + x 2t, x > 2t, x 3 /6 + x 2 /4 xt + t 2 + 1, x < 2t.

321 The Method of Characteristics 15 x = t + x, x t, t t, x < t, u(x, t) = [x t cos(x t)]e t x 1, x t, 2(t x + 1)e x x 1, x < t. 17 x = 2y + x, u(x, y) = (x 2y + 1)e (x+y 1)/3. 19 x = y/2 + x, u(x, y) = (1/4)[(2x y + 1)e (2x+3y 2)/2 + 2y + 1]. 21 x = 2y+x, u(x, y) = (1/9)(2x 2 +8xy+8y 2 +7x+14y 31)e (y x+2)/3 + x + y First-Order Quasilinear Equations A PDE of the form u t (x, t) + c(x, t, u)u x (x, t) = q(x, t, u) is called quasilinear. Although technically nonlinear, it is linear in the firstorder derivatives of u. Such equations arise in the modeling of a variety of phenomena (for example, traffic flow) and can be solved by the method of characteristics Example. Consider the IVP u t (x, t) + u 3 (x, t)u x (x, t) =, < x <, t >, u(x, ) = x 1/3, < x <. If x = x(t), then the usual procedure leads to the conclusion that the characteristic line through (x, t) and (x, ) satisfies x (t) = u 3 (x(t), t), x() = x. On this line, du dt = u t + u x x = u t + u 3 u x =, which, in view of the IC, yields u(x, t) = C = u(x, ) = x 1/3. Hence, the ODE problem for the characteristic line becomes x (t) = x, x() = x,

322 12.2 First-Order Quasilinear Equations 299 with solution x = x t+x = x (t+1). Since on this line we have x = x/(t+1), we can now write the solution of the given IVP as ( ) 1/3 u(x, t) = x 1/3 x =. t + 1 Verification with Mathematica R. The input u = (x/(t + 1)) (1/3); f = x (1/3); Simplify[D[u,t] + u 3 D[u,x],(u /. t -> ) - f}] generates the output, } Example. A similar procedure is used to solve the IVP If x = x(t) satisfies u t (x, t) + u(x, t)u x (x, t) = 2t, < x <, t >, u(x, ) = x, < x <. x (t) = u(x(t), t), x() = x, then on the characteristic curve through (x, t) and (x, ) we have Therefore, u(x, t) = t 2 + C, or du dt = u t + u x x = u t + uu x = 2t. u(x, t) t 2 = C = u(x, ) = x, so u(x, t) = t 2 + x. This means that the characteristic curve satisfies with solution x (t) = t 2 + x, x() = x, x = 1 3 t3 + x t + x = 1 3 t3 + x (t + 1). Since on this curve x = x 1 3 t3 3x t3 = t + 1 3(t + 1), the solution of the IBVP is u(x, t) = t 2 + 3x t3 3(t + 1). Verification with Mathematica R. The input u = t 2 + (3 x - t 3)/(3 (t + 1)); f = x; Simplify[D[u,t] + u D[u,x] - 2 t,(u /. t -> ) - f}] generates the output, }.

323 3 12 The Method of Characteristics Exercises Use the method of characteristics to solve the quasilinear IVP u t (x, t) + c(x, t, u)u x (x, t) = q(x, t, u), < x <, t >, u(x, ) = f(x), < x <, with the functions c, q, and f as indicated. 1 c(x, t, u) = 2u 1, q(x, t, u) = 3, f(x) = 1 x. 2 c(x, t, u) = u 1, q(x, t, u) = t + 1, f(x) = 2x. 3 c(x, t, u) = u + t, q(x, t, u) = 1, f(x) = x c(x, t, u) = 1 + 2t u, q(x, t, u) = 4t 1, f(x) = 2 3x. 5 c(x, t, u) = 1, q(x, t, u) = 2tu 2, f(x) = e x. 6 c(x, t, u) = (t + 2)u, q(x, t, u) = u + 1, f(x) = x. Answers to Odd-Numbered Exercises 1 x = (2t + 1)x 3t 2 t, u(x, t) = (3t 2 + 4t x + 1)/(2t + 1). 3 x = t 2 + t + (t + 1)x, u(x, t) = (x + t + 1)/(t + 1). 5 x = t + x, u(x, t) = 1/(e x t + t 2 ) The One-Dimensional Wave Equation We now reconsider the wave equation in terms of details provided by the method of characteristics The d Alembert Solution Formally, we can write the one-dimensional wave equation as ( u tt (x, t) c 2 u xx (x, t) = t + c )( x t c ) u(x, t) x ( = t + c ) w(x, t) x = w t (x, t) + cw x (x, t) =. (12.3) We know from Section 12.1 that the general solution of the equation satisfied by w in (12.3) is w(x, t) = u t (x, t) cu x (x, t) = P (x ct), (12.4)

324 12.3 The One-Dimensional Wave Equation 31 where P is an arbitrary one-variable function. On the other hand, if we write the wave equation in the alternative form ( u tt (x, t) c 2 u xx (x, t) = t c )( x t + c ) u(x, t) x ( = t c ) v(x, t) x then, as above, = v t (x, t) cv x (x, t) =, v(x, t) = u t (x, t) + cu x (x, t) = Q(x + ct), (12.5) where Q is another arbitrary one-variable function. Adding (12.4) and (12.5) side by side, we find that Direct integration now yields u t (x, t) = 1 [P (x ct) + Q(x + ct)]. 2 u(x, t) = F(x ct) + G(x + ct), (12.6) where F and G are arbitrary one-variable functions. According to the explanation given in Section 12.1, F(x ct) is a fixedshape wave traveling to the right with velocity c, and is constant on the characteristics x ct = const. Similarly, G(x+ct) is a fixed-shape wave traveling to the left with velocity c, and is constant on the characteristics x + ct = const (see Fig. 12.5). Through every point (x, t) in the t > half-plane, there pass two characteristics, one from each family. Figure 12.5 The two families of characteristics. Consider an infinite vibrating string, which is modeled by the IVP u tt (x, t) = c 2 u xx (x, t), < x <, t >, u(x, ) = f(x), u t (x, ) = g(x), < x <. Differentiating the solution (12.6) of the PDE with respect to t and recalling that F and G are one-variable functions, we obtain u t (x, t) = c F (x ct) + c G (x + ct).

325 32 12 The Method of Characteristics From the ICs it now follows that f(x) = u(x, ) = F(x) + G(x), g(x) = u t (x, ) = c F (x) + c G (x). We solve the system of these equations for F and G. Thus, F = f G, so that g/c = f + 2G, from which we easily find that G = (f 1 + 1c ) 2 g, F = (f 1 1c ) 2 g ; hence, by integration, F(x) = 1 2 f(x) 1 2c G(x) = 1 2 f(x) + 1 2c x x g(y) dy, g(y) dy. From these expressions and (12.6) we conclude that u(x, t) = 1 2 [f(x + ct) + f(x ct)] + 1 2c x+ct x ct g(y) dy. (12.7) This is d Alembert s solution, which was derived earlier in Section 1.3 by means of the Green s function for the wave equation Remark. Technically, (12.6), which is obtained by integrating u t (x, t) with respect to t, should be written as u(x, t) = F(x ct) + G(x + ct) + ϕ(x), where ϕ is an arbitrary function of x. Replaced in the equation u tt = c 2 u xx, this leads to ϕ (x) =, so ϕ(x) = ax + b, a, b = const, or, what is the same, ϕ(x) = a[ 1 2 (x ct) (x + ct)] + b. Since the terms above can be incorporated in the arbitrary functions F(x ct) and G(x + ct), it follows that there is no need for ϕ to be specifically included in (12.6) Example. Suppose that in the IVP above we have 1, x < h, u(x, ) = f(x) = u t (x, ) = g(x) =., x > h,

326 12.3 The One-Dimensional Wave Equation 33 By (12.7), the solution is u(x, t) = 1 [f(x ct) + f(x + ct)], 2 where 1 f(x ct) = 2 1 f(x + ct) = 2 1/2, x ct < h, or h + ct < x < h + ct, otherwise, 1/2, x + ct < h, or h ct < x < h ct, otherwise. Thus, the solution is the sum of two pulses of amplitude 1/2, which move away from each other with velocity 2c. Since their endpoints are initially 2h apart, they separate after t = h/c Example. In the case of the IVP u tt (x, t) = c 2 u xx (x, t), < x <, t >, u(x, ) = sin x, u t (x, ) =, < x <, d Alembert s solution (12.7) yields u(x, t) = 1 [ ] 2 sin(x + ct) + sin(x ct) = sin x cos(ct). Waves represented by solutions like this, where the variables separate, are called standing waves Example. Also by (12.7), the solution of the IVP is u tt (x, t) = c 2 u xx (x, t), < x <, t >, u(x, ) =, u t (x, ) = sin x, < x <, u(x, t) = 1 2c = 1 2c x+ct x ct sin y dy [ ] 1 cos(x ct) cos(x + ct) = sin x sin(ct). c This solution also represents standing waves Example. For the IVP u tt (x, t) = 4u xx (x, t), < x <, t >, u(x, ) = x, u t (x, ) = 2x 2, < x <,

327 34 12 The Method of Characteristics d Alembert s formula with c = 2 produces the solution u(x, t) = 1 [ ] 2 (x 2t) + (x + 2t) x+2t x 2t Verification with Mathematica R. The input u = x + 2 x 2 t + (8/3) t 3; f1, f2} =x,2 x 2}; Simplify[D[u,t,t] - 4 D[u,x,x],(u /. t -> ) - f1, (D[u,t] /. t -> ) - f2}] generates the output,, }. 2y 2 dy = x + 2x 2 t t3. Exercises Use the method of operator decomposition to solve the IVP u tt (x, t) = c 2 u xx (x, t), < x <, t >, u(x, ) = f(x), u t (x, ) = g(x), < x <, with the constant c and the functions f and g as indicated. (Do not use d Alembert s formula directly.) 1 c = 1, f(x) = 2x + 3, g(x) = x c = 3, f(x) = x 2 + x, g(x) = 2 x. 3 c = 1/2, f(x) = e 2x, g(x) = x x 2. 4 c = 2, f(x) = x 2 3x + 1, g(x) = sin x. Answers to Odd-Numbered Exercises 1 u(x, t) = x 2 t + t 3 /3 + 2x + t u(x, t) = e t 2x /2 + e t 2x /2 x 2 t t 3 /12 + xt The Semi-Infinite Vibrating String Consider the IBVP u tt (x, t) = c 2 u xx (x, t), x >, t >, u(, t) =, t >, u(x, ) = f(x), u t (x, ) = g(x), x >.

328 12.3 The One-Dimensional Wave Equation 35 As in the preceding case, from the PDE we obtain where u(x, t) = F(x ct) + G(x + ct), F(x) = 1 2 f(x) 1 2c G(x) = 1 2 f(x) + 1 2c x x g(y) dy, x >, g(y) dy, x >. Since x > in this problem, the functions F and G are determined only for positive values of their arguments. This does not affect G(x + ct), since t >. But the argument of F(x ct) is negative if < x < ct. To obtain F(x ct) for x ct <, we use the BC. Thus, u(, t) = = F( ct) + G(ct), t >, so for ξ < we have F(ξ) = G( ξ), which means that the solution for < x < ct is u(x, t) = F(x ct) + G(x + ct) = G(ct x) + G(x + ct) = 1 2 [f(x + ct) f(ct x)] + 1 [ x+ct g(y) dy 2c = 1 2 [f(ct + x) f(ct x)] + 1 2c ct+x ct x ct x ] g(y) dy g(y) dy. (12.8) The term G(ct x) is a fixed-shape wave that travels to the right and is called the reflected wave. For x > ct, the solution of the problem is given by d Alembert s formula, as before Example. The solution of the IBVP u tt (x, t) = 4u xx (x, t), x >, t >, u(, t) =, t >, u(x, ) = 4x, u t (x, ) = 2x + 6, x >, is computed in two stages. First, for < x < 2t we use (12.8) with c = 2 to find that u(x, t) = 1 [ ] 4(2t + x) 4(2t x) t+x 2t x (2y + 6) dy = 2xt + 7x ;

329 36 12 The Method of Characteristics then, by (12.7), for x > 2t we have u(x, t) = 1 [ ] 4(x + 2t) + 4(x 2t) = 2xt + 4x + 6t. x+2t x 2t (2y + 6) dy The solution is continuous across the characteristic line x = 2t since Thus, we can write lim x u(x, t) = u(x, ) = lim u(, t) =. t 2xt + 7x, x 2t, 2xt + 4x + 6t, x > 2t. Verification with Mathematica R. The input, which reflects the split definition of u, u1, u2} =2 x t + 7 x,2 x t + 4 x + 6 t}; f1, f2} =4 x,2 x + 6}; g = ; Simplify[D[u1,u2},t,t] - 4 D[u1,u2},x,x],(u1 /. x -> ) - g, (u2 /. t -> ) - f1,(d[u2,t] /. t -> ) - f2}] generates the output, },,, }. Exercises Use the method of operator decomposition to solve the IBVP u tt (x, t) = c 2 u xx (x, t), x >, t >, u(, t) =, t >, u(x, ) = f(x), u t (x, ) = g(x), x >, with the constant c and the functions f and g as indicated. (Do not use d Alembert s formula directly.) 1 c = 1, f(x) = 2x 2 x, g(x) = 4x c = 1/2, f(x) = x + 2, g(x) = 3x 2. 3 c = 3, f(x) = 1 x 2, g(x) = cos x. 4 c = 2, f(x) = e x + 2, g(x) = 3x 2 2x.

330 12.3 The One-Dimensional Wave Equation 37 Answers to Odd-Numbered Exercises 8xt, x < t, 1 u(x, t) = 2x 2 + 4xt + 2t 2 x + t, x t. 3 u(x, t) = 6xt + (1/3) sin x cos(3t), x < 3t, 1 x 2 9t 2 + (1/3) cos x sin(3t), x > 3t The Finite String Consider the IBVP u tt (x, t) = c 2 u xx (x, t), < x < L, t >, u(, t) =, u(l, t) =, t >, u(x, ) = f(x), u t (x, ) = g(x), < x < L. Using separation of variables, in Section 5.2 we obtained the solution u(x, t) = sin nπx ( b 1n cos nπct L L + b 2n sin nπct ), L where the coefficients b 1n and b 2n are determined from the expansions f(x) = g(x) = Suppose that f and g = ; then Using the formula we write the solution in the form u(x, t) = 1 2 b 1n sin nπx L, b 2n nπc L sin nπx L. b 2n =, n = 1, 2,.... sin α cos β = 1 [sin(α + β) + sin(α β)], 2 [ nπ(x + ct) b 1n sin + sin L = 1 [f(x + ct) + f(x ct)]. 2 ] nπ(x ct) L Suppose now that f = and g ; then b 1n =, n = 1, 2,.... Since sin α sin β = 1 [cos(α β) cos(α + β)], 2

331 38 12 The Method of Characteristics we rewrite the solution as u(x, t) = 1 2 On the other hand, hence, x+ct x ct g(y) dy = = c [ nπ(x ct) b 2n cos cos L cb 2n x+ct x ct nπ L sin nπx L dx [ nπ(x ct) b 2n cos cos L u(x, t) = 1 2c x+ct x ct g(y) dy. ] nπ(x + ct). L ] nπ(x + ct) ; L Combining the two separate solutions and using the superposition principle, we now regain d Alembert s formula (12.7) Other Hyperbolic Equations The method set out in the preceding section can be extended to more general hyperbolic equations. We illustrate how this is done in two particular cases One-Dimensional Waves The solution of the type of problem discussed here is based on a decomposition of the PDE operator, which is similar to that performed for the wave equation Example. The IVP u tt (x, t) + 5u xt (x, t) + 6u xx (x, t) =, < x <, t >, u(x, ) = x + 2, u t (x, ) = 2x, < x <, is hyperbolic because B 2 4AC = = 1 >. It is easily verified that the PDE can be rewritten alternatively as ( t + 2 )( x t + 3 ) ( u(x, t) = x t + 3 )( x t + 2 ) u(x, t) =. x Setting we have u t + 3u x = w, u t + 2u x = v, w t + 2w x =, v t + 3v x =.

332 12.4 Other Hyperbolic Equations 39 By the argument developed in Section 12.1, the general solutions of these equations are, respectively, w(x, t) = P (x 2t), v(x, t) = Q(x 3t), where P and Q are arbitrary one-variable functions; hence, u t + 3u x = P (x 2t), u t + 2u x = Q(x 3t). The elimination of u x between these equations now yields u t (x, t) = 3Q(x 3t) 2P (x 2t), from which, by integration, we find that u(x, t) = F(x 2t) + G(x 3t), where F and G are a new pair of arbitrary one-variable functions. We remark that, by the chain rule, the time derivative of u is u t (x, t) = 2F (x 2t) 3G (x 3t). (12.9) At this point, we apply the ICs and arrive at the equations F(x) + G(x) = x + 2, 2F (x) 3G (x) = 2x. Differentiating the first equation term by term, we obtain F (x) + G (x) = 1, which, combined with the second equation above, leads to G (x) = 2x 2. This allows us to determine G and then F: G(x) = x 2 2x + c, c = const, F(x) = x + 2 G(x) = x 2 + 3x + 2 c. Replacing F and G in (12.9), we conclude that the solution of the given IVP is u(x, t) = [ (x 2t) 2 + 3(x 2t) + 2 c ] + [ (x 3t) 2 2(x 3t) + c ] = 2xt 5t 2 + x + 2. Verification with Mathematica R. The input u = 2 x t - 5 t 2 + x + 2; f1, f2} =x + 2,2 x}; Simplify[D[u,t,t] + 5 D[u,x,t] + 6 D[u,x,x],(u /. t -> ) - f1, (D[u,t] /. t -> ) - f2}] generates the output,, }.

333 31 12 The Method of Characteristics Example. Consider the IBVP u tt (x, t) 3u xt (x, t) 4u xx (x, t) =, x >, t >, u(, t) =, t >, u(x, ) = 1 x 2, u t (x, ) = x + 2, x >. Since B 2 4AC = = 25 >, the PDE is hyperbolic. Rewriting the equation as ( t + )( x t 4 ) ( u(x, t) = x t 4 )( x t + ) u(x, t) = x and proceeding as in Example 12.15, we find that where u(x, t) = F(x t) + G(x + 4t), (12.1) F(x) = 1 1 ( 9x2 4x + 1) c, G(x) = 1 1 ( x2 + 4x) + c c = const, are defined for positive values of their arguments. If x > t, then x t > and we replace (12.11) in (12.1): u(x, t) = 1 [ 1 9(x t) 2 4(x t) + 1 (x + 4t) 2 + 4(x + 4t) ] = x 2 + xt 5 2 t2 + 2t + 1. If x < t, then x t <, and we make use of the BC: which implies that = u(, t) = F( t) + G(4t) = F( t) + G( 4( t)), so, by (12.1) and (12.11), F(z) = G( 4z) z < ; u(x, t) = G( 4x + 4t) + G(x + 4t) = 1 [ 1 ( 4x + 4t) 2 4( 4x + 4t) (x + 4t) 2 + 4(x + 4t) ] = 3 2 x2 4xt + 2x. The two cases can be written together in the form 3 u(x, t) = 2 x2 4xt + 2x, x < t, x 2 + xt 5 2 t2 + 2t + 1, x > t. We notice that the solution is discontinuous across the line x = t. (12.11)

334 12.4 Other Hyperbolic Equations 311 Verification with Mathematica R. The split form of u prompts us to use the input u1, u2} =3 x 2/2-4 x t + 2 x, - x 2 + x t - 5 t 2/2 + 2 t + 1}; f1, f2} =1 - x 2, x + 2}; g = ; Simplify[D[u1,u2},t,t] - 3 D[u1,u2},x,t] - 4 D[u1,u2},x,x], (u1 /. x -> ) - g,(u2 /. t -> ) - f1,(d[u2,t]/. t -> ) - f2}] which generates the output, },,, } Remark. A d Alembert-type formula can also be derived for PDEs of the kind studied in Examples and Exercises In 1 4, use the method of operator decomposition to solve the IVP au tt (x, t) + bu xt (x, t) + cu xx (x, t) =, < x <, t >, u(x, ) = f(x), u t (x, ) = g(x), < x <, with constants a, b, c and functions f, g as indicated. 1 a = 1, b = 1, c = 2, f(x) = x 2, g(x) = 2x a = 1, b = 1, c = 6, f(x) = 1 2x, g(x) = 3x 2. 3 a = 1, b = 4, c = 3, f(x) = x 2 x, g(x) = 4x 2. 4 a = 2, b = 7, c = 4, f(x) = 3x + 1, g(x) = 1 2x. In 5 8, use the method of operator decomposition to solve the IBVP au tt (x, t) + bu xt (x, t) + cu xx (x, t) =, x >, t >, u(, t) =, t >, u(x, ) = f(x), u t (x, ) = g(x), x >, with constants a, b, c and functions f, g as indicated. 5 a = 1, b = 2, c = 3, f(x) = x + 1, g(x) = 2x. 6 a = 1, b = 2, c = 8, f(x) = 2x 1, g(x) = 4x a = 2, b = 1, c = 1, f(x) = x 2, g(x) = 2x a = 2, b = 3, c = 2, f(x) = x 2 + 1, g(x) = x 3.

335 The Method of Characteristics Answers to Odd-Numbered Exercises 1 u(x, t) = x 2 + 2xt + t 2 + t. 3 u(x, t) = x 2 + 4xt 11t 2 x 2t. 2x 2 /9 + 2xt/3 + x, x < 3t, 5 u(x, t) = 2xt 2t 2 + x + 1, x > 3t. x 2 + 2xt + 3x, x < t, 7 u(x, t) = x 2 + 2xt + 3t, x t Spherical Waves The three-dimensional wave equation is u tt = c 2 u, (12.12) where the Laplacian is expressed in terms of a system of coordinates appropriate for the geometry of the problem. If the amplitude of the waves depends only on the distance r of the wave front from a point source, then we choose spherical coordinates, in which case u = u(r, t) and, by Remark 4.12(v), u = r 2 (r 2 u r ) r = r 2 (2ru r + r 2 u rr ) = u rr + 2r 1 u r. Therefore, the equation governing the propagation of spherical waves is u tt (r, t) = c 2[ u rr (r, t) + 2r 1 u r (r, t) ]. Multiplying this equality by r and manipulating the right-hand side, we find that ru tt = c 2[ (ru rr + u r ) + u r ] = c 2 [ (ru r ) r + u r ] = c 2 (ru r + u) r = c 2 ((ru) r ) r = c 2 (ru) rr. The substitution ru(r, t) = v(r, t) now reduces the equation above to v tt (r, t) = c 2 v rr (r, t), so, by analogy with the argument developed in Section 12.1, v(r, t) = F(r ct) + G(r + ct). Consequently, the general solution of (12.12) is u(r, t) = 1 r [ F(r ct) + G(r + ct) ], where F and G are arbitrary one-variable functions.

336 Perturbation and Asymptotic Methods C H A P T E R 13 Owing to the complexity of the PDEs involved in some mathematical models, it is not always possible to find an exact solution to an initial/boundary value problem. The next best thing in such situations is to compute an approximate solution instead. This is the idea behind the method of asymptotic expansion, which is applicable to problems that contain a small positive parameter and relies on the expansion of the solution in a series of powers of the parameter. If the series converges, then the technique is called a perturbation method; when the series diverges but is asymptotic (in the sense explained below), we have an asymptotic method. In what follows, we discuss only the formal construction of the series solution and obtain the first few terms, without considering the question of convergence Asymptotic Series In order to state what an asymptotic series is, we need a mechanism to compare the magnitude of functions Definition. Let f and g be two functions of a real variable x. We say that f is of order g near x = a and write f(x) = O ( g(x) ) as x a if We also write f(x) g(x) is bounded as x a. f(x) = o ( g(x) ) as x a 313

337 Perturbation and Asymptotic Methods if f(x) as x a. g(x) 13.2 Examples. (i) The function sin x is of order x near x = since sin x lim x x = 1; therefore, (sin x)/x is bounded for x close to. (ii) We have x 2 ln x = o(x) near x = since x 2 ln x lim =. x x (iii) Similarly, e 1/ x = o(x n ) near x = for any positive integer n because e 1/ x lim x x n = Definition. A function f(x, ε), where < ε 1 is a small parameter, is said to have the asymptotic (power) series f(x, ε) f n (x)ε n as ε + if, for any positive integer N, f(x, ε) = N 1 n= uniformly for x in some interval. n= f n (x)ε n + O(ε N ) as ε +, (13.1) 13.4 Remarks. (i) Equality (13.1) means that the remainder after N terms is of order ε N as ε +. An alternative way to write this is f(x, ε) = N 1 n= f n (x)ε n + o(ε N 1 ) as ε +. (ii) The right-hand side of (13.1) may diverge as N, but it yields a good approximation of f(x, ε) when N is fixed and ε > is very small. We do not need convergence for the result to be acceptable. Also, the approximation may not get better if we take additional terms because, as mentioned earlier, the series may be divergent. (iii) In an asymptotic series it is assumed that the terms containing higher powers of ε are much smaller than those with lower powers. (iv) In this chapter, we assume that asymptotic series may be differentiated and integrated term by term. (v) If k 1 is a large parameter, then we can reduce the problem to one with a small parameter by setting k = 1/ε.

338 13.1 Asymptotic Series Definition. Consider an initial/boundary value problem that depends (smoothly) on a small parameter ε >. The problem obtained by setting ε = in the equation and data functions is called the reduced (unperturbed) problem. If the reduced problem is of the same type and order as the given one and both have unique solutions, then the given problem is called a regular perturbation problem; otherwise, it is called a singular perturbation problem Example. The IBVP u t (x, t) = ku xx (x, t) + εu x (x, t), < x < L, t >, u(, t) =, u(l, t) =, t >, u(x, ) = f(x), < x < L, is a regular perturbation problem since both this IBVP and its reduced version (involving the heat equation) are second-order parabolic problems with unique solutions Example. The signalling (hyperbolic) problem εu tt (x, t) c 2 u xx (x, t) + u t (x, t) =, x >, < t <, u(, t) = f(t), < t <, reduces to a parabolic one when we set ε =. Therefore, although both the given and reduced problems have unique solutions, the given IVP is a singular perturbation problem Example. Let D be a finite region bounded by a smooth, closed, simple curve D in the (x, y)-plane, and let n be the unit outward normal to D. The fourth-order BVP ( u)(x, y) =, (x, y) in D, u(x, y) = f(x, y), εu n (x, y) + u(x, y) = g(x, y), (x, y) on D, where u n = u/ n, reduces, when we set ε =, to the BVP ( u)(x, y) =, (x, y) in D, u(x, y) = f(x, y), u(x, y) = g(x, y), (x, y) on D. If f = g, then the reduced problem has no solution. On the other hand, if f = g, then the solution is not unique, since we have lost one of the BCs. Hence, the given BVP is a singular perturbation problem Definition. If the solution u of a perturbation problem has the asymptotic series u u n ε n, then u u is called a perturbation of the solution n= of the reduced problem.

339 Perturbation and Asymptotic Methods 13.2 Regular Perturbation Problems The best way to see how the method works in this case is to examine a few specific models Example. Let D = (r, θ) : r < 1, π θ < π }, D = (r, θ) : r = 1, π θ < π } be the unit disc (with the center at the origin) and its circular boundary, and consider the Dirichlet problem for the two-dimensional Helmholtz equation ( u)(r, θ) + εu(r, θ) =, (r, θ) in D, r, u(r, θ) = 1, (r, θ) on D, where < ε 1, which ensures that the BVP above has a unique solution. The reduced problem (for ε = ) is the Dirichlet problem for the Laplace equation, also uniquely solvable. Both the reduced and the perturbed problems are elliptic and of second order, so the given BVP is a regular perturbation problem. Assuming a perturbation series of the form u(r, θ) u n (r, θ)ε n, from the PDE we find that u + εu ( u n )ε n + u n ε n+1 n= = u + n= n= ( u n + u n 1 )ε n = in D, r. At the same time, the BC yields u u + u n ε n = 1 on D. Equating the coefficients of every power of ε on both sides, we then obtain the BVPs and, for n 1, u = in D, r =, u = 1 on D, u n = u n 1 in D, r, u n = on D.

340 13.2 Regular Perturbation Problemss 317 Since the region where the problem is posed and the boundary data are independent of the polar angle θ, we may assume that u n = u n (r); consequently, by Remark 4.11(ii), the problem for u can be written as ( u )(r) = u (r) + r 1 u (r) =, < r < 1, u (1) = 1. Multiplying the differential equation by r and noting that the left-hand side of the new ODE is, in fact, (ru ), we find that u (r) = C 1 ln r + C 2, C 1, C 2 = const. We know (see Section 5.3) that for this type of problem we also have additional conditions, generated by physical considerations. Since we have assumed that the solution is independent of θ, the only other condition of this kind to be satisfied here is that the solution and its derivative be continuous (hence, bounded) in D. This implies that C 1 = ; the value of C 2 is then found from the BC at r = 1, which yields u (r) = 1. Next, applying the same argument to the BVP satisfied by u 1, namely, ( u 1 )(r) = u 1(r) + r 1 u 1(r) = u (r) = 1, < r < 1, u 1 (1) =, we find that u 1 (r) = 1 4 (1 r2 ); hence, u(r) = ε(1 r2 ) + O(ε 2 ). (13.2) For this problem we can actually see how good an approximation (13.2) is. The given (perturbed) equation can be written in the form u (r) + r 1 u (r) + εu(r) =, which is Bessel s equation of order zero (see (3.14) with m =, λ = ε, and x replaced by r). Its bounded solution satisfying u(1) = 1 is u(r) = J ( ε r)/j ( ε), where J is the Bessel function of the first kind and order zero. This formula makes sense because < ε 1; that is, ε is smaller than the first zero (ξ = 2.4) of J (ξ). Since, for small values of ξ, we use the formula J (ξ) = ξ2 + O(ξ 4 ), 1 + q + q q n + = 1 1 q

341 Perturbation and Asymptotic Methods for the sum of an infinite geometric progression with ratio q, < q < 1, to find that J ( ε r) J ( ε) = εr2 + O(ε 2 ) ε + O(ε2 ) = εr2 + O(ε 2 1 ( 1 4 ε + O(ε2 ) ) = [1 1 4 εr2 + O(ε 2 )][1 + ( 1 4 ε + O(ε2 )) + O(ε 2 )] = ε(1 r2 ) + O(ε 2 ). Thus, the perturbation solution (13.2) coincides with the exact solution to O(ε)-terms throughout D. Verification with Mathematica R. This is also easily seen to be the case if we replace the truncation of (13.2) after two terms in the PDE and the BC, and restrict the computation to the first two terms of the asymptotic series in the results. Specifically, the input u = 1 + (1/4) ε (1 - r 2); f = 1; Simplify[Series[D[u,r,r] + r ( - 1) D[u,r] + ε u, (u /. r -> 1) - f},ε,,1}] // Normal] generates the output, } Example. The elliptic nonlinear BVP u(r, θ) + εu 2 (r, θ) = 36r, (r, θ) in D, u(r, θ) = 4, (r, θ) on S, where < ε 1 and D and D are the same as in Example 13.1, is a regular perturbation problem. Since we again notice that the solution u is expected to depend only on r, we assume for the solution a series of the form u(r) n= asymptotic expansion u n (r)ε n. Then, in the usual way, taking into account the u 2 = [ u + εu 1 + O(ε 2 ) ] 2 = u 2 + 2εu u 1 + O(ε 2 ) and recalling the boundedness condition mentioned in Example 13.1, we see that u and u 1 are, respectively, the solutions of the linear BVPs u (r) + r 1 u (r) = 36r, < r < 1, u (r), u (r) bounded as r +, u (1) = 4, and u 1(r) + r 1 u 1(r) = u 2 (r), < r < 1, u 1 (r), u 1(r) bounded as r +, u 1 (1) = ;

342 13.2 Regular Perturbation Problemss 319 that is, u (r) = 4r 3 and u 1 (r) = 1 4 (1 r8 ). Consequently, u(r) = 4r ε(1 r8 ) + O(ε 2 ). This example shows how, in certain circumstances, the perturbation method can reduce a nonlinear problem to a sequence of linear ones. Verification with Mathematica R. The input u = 4 r 3 + (1/4) ε (1 - r 8); f = 4; Simplify[Series[D[u,r,r] + r ( - 1) D[u,r] + ε u 2-36 r, (u /. r -> 1) - f},ε,,1}] // Normal] generates the output, }. Sometimes, the small parameter may occur not only in the equation, but also in any accompanying ICs and/or BCs. The solution procedure in such cases remains the same Example. The BVP u xx (x, y) + (2 ε)u x (x, y) εu y (x, y) 3u(x, y) = 2 3x + 3y 2 + ε(2y 1 8e x cos y), < x < 1, < y <, u(, y) = y 2 + ε, u(1, y) = 1 y 2 + εe(1 2 cos y), < y <, is a regular perturbation problem since both the given BVP and the reduced one are second-order parabolic problems with unique solutions. If we want to find a formal two-term asymptotic approximation of the solution, we replace u(x, y) = u (x, y) + εu 1 (x, y) + O(ε 2 ) in the PDE and BCs and, equating the constant terms and the coefficients of ε on both sides, find that u and u 1 are, respectively, the solutions of the BVPs and (u ) xx (x, y) + 2(u ) x (x, y) 3u (x, y) = 2 3x + 3y 2, < x < 1, < y <, u (, y) = y 2, u (1, y) = 1 y 2, < y <, (u 1 ) xx (x, y) + 2(u 1 ) x (x, y) 3u 1 (x, y) = (u ) x (x, y) + (u ) y (x, y) + 2y 1 8e x cos y, < x < 1, < y <, u 1 (, y) = 1, u 1 (1, y) = e(1 2 cos y), < y <. The solution of the first BVP is u (x, y) = C 1 (y)e x + C 2 (y)e 3x + A(y)x + B(y) = x y 2.

343 32 13 Perturbation and Asymptotic Methods Then the equation in the second one becomes (u 1 ) xx (x, y) + 2(u 1 ) x (x, y) 3u 1 (x, y) = 8e x cos y, which, together with the appropriate BCs, yields u 1 (x, y) = C 1 (y)e x + C 2 (y)e 3x + A(y)xe x = (1 2x cos y)e x. Consequently, the desired approximate solution is u(x, y) = x y 2 + ε(1 2x cos y)e x + O(ε 2 ). Verification with Mathematica R. The input u = x - y 2 + ε (1-2 x Cos[y]) E x; q = 2-3 x + 3 y 2 + ε (2 y E x Cos[y]); f1, f2} =-y 2 + ε, 1 - y 2 + ε E (1-2 Cos[y])}; Simplify[Series[D[u,x,x] + (2 - ε D[u,x] - ε D[u,y] - 3 u - q, (u /. x -> ) - f1,(u /. x -> 1) - f2},ε,,1}] // Normal] generates the output,, } Remark. The solution of the BVP in Example is well ordered and, therefore, valid as an asymptotic approximation for all real values of y since, as is easily seen, the second term is always much smaller than the first when ε is very small. However, if one of the variables is defined on an infinite domain, this is not always the case Example. Consider the regular perturbation problem u t (x, t) + 3εu x (x, t) =, < x <, t >, u(x, ) = sin x, < x <, where < ε 1. Replacing u(x, t) = u (x, t) + εu 1 (x, t) + O(ε 2 ) in the PDE and IC and following normal procedure, we find that u and u 1 are the solutions of the IVPs and (u ) t (x, t) =, < x <, t >, u (x, ) = sin x, < x <, (u 1 ) t (x, t) + 3(u ) x (x, t) =, < x <, t >, u 1 (x, ) =, < x <. From the first IVP it follows that u (x, t) = sin x, which, replaced in the second, leads to u 1 (x, t) = 3t cos x, so u(x, t) = sin x 3εt cos x + O(ε 2 ).

344 13.2 Regular Perturbation Problemss 321 This series is a good asymptotic approximation for the solution of the given IVP as ε if t is restricted to any fixed finite interval [, t ]. But in the absence of such a restriction, we see that when t = O(ε 1 ), the term 3εt cos x is of the same order of magnitude as the leading O(1)-term. (The situation becomes worse as t increases further.) Such a term is called secular, and its presence means that the perturbation series is not valid for very large values of t. To extend the validity of the series for all t >, we use the power series expansions of sin α and cos α and write sin(x 3εt) = sin x cos(3εt) cos x sin(3εt) = sin x[1 + O(ε 2 )] cos x[3εt + O(ε 3 )] = sin x 3εt cos x + O(ε 2 ). Then the asymptotic solution series above can be rearranged in the form u(x, t) = sin(x 3εt) + O(ε 2 ), which does not contain the secular term identified earlier. Higher-order terms in ε may, however, contain additional secular terms. Verification with Mathematica R. The input u = Sin[x - 3 ε t]; f = Sin[x]; Simplify[Series[D[u,t] + 3 ε D[u,x],(u /. t -> ) - f},ε,,1}] // Normal] generates the output, } Remark. According to the discussion at the beginning of Section 12.1, the exact solution of the IVP in Example is, in fact, u(x, t) = sin(x 3εt). The additional O(ε 2 )-term represents noise generated by the approximating nature of the asymptotic expansion technique Example. The first-order IVP u t (x, t) + u x (x, t) εu(x, t) =, < x <, t >, u(x, ) = cos x, < x <, where < ε 1, is a regular perturbation problem because the reduced problem is also of first order and both problems have unique solutions. Writing u(x, t) = u (x, t) + εu 1 (x, t) + O(ε 2 ), we arrive at the IVPs (u ) t (x, t) + (u ) x (x, t) =, < x <, t >, u (x, ) = cos x, < x <, and (u 1 ) t (x, t) + (u 1 ) x (x, t) = u (x, t), < x <, t >, u 1 (x, ) =, < x <.

345 Perturbation and Asymptotic Methods Using the method of characteristics (see Section 12.1), we find that the solutions of these problems are, respectively, u (x, t) = cos(x t), u 1 (x, t) = t cos(x t); hence, u(x, t) = cos(x t) + εt cos(x t) + O(ε 2 ). As explained in Example 13.12, εt cos(x t) is a secular term. Since the procedure that removed such a term in the preceding example does not work here, we try another technique, called the method of multiple scales, which consists in introducing an additional variable τ = εt. Denoting by v(x, t, τ) the new unknown function and remarking that v depends on t both directly and through τ, we arrive at the problem v t (x, t, τ) + εv τ (x, t, τ) + v x (x, t, τ) εv(x, t, τ) =, v(x,, ) = cos x, < x <. < x <, t, τ >, We now set v(x, t, τ) = v (x, t, τ)+εv 1 (x, t, τ)+o(ε 2 ) and deduce in the usual way that v and v 1 are, respectively, the solutions of the IVPs (v ) t (x, t, τ) + (v ) x (x, t, τ) =, < x <, t, τ >, v (x,, ) = cos x, and (v 1 ) t (x, t, τ) + (v 1 ) x (x, t, τ) = (v ) τ (x, t, τ) + v (x, t, τ), v 1 (x,, ) =, < x <. < x <, t, τ >, We seek the solution of the former as v (x, t, τ) = f(x, t)ϕ(τ), where ϕ satisfies the condition ϕ() = 1 but is otherwise arbitrary. Then the problem for v reduces to the IVP f t (x, t) + f x (x, t) =, < x <, t >, f(x, ) = cos x, < x <,

346 13.2 Regular Perturbation Problemss 323 with solution f(x, t) = cos(x t), so v (x, t, τ) = ϕ(τ) cos(x t). Replacing this on the right-hand side in the PDE for v 1, we arrive at the equation (v 1 ) t (x, t, τ) + (v 1 ) x (x, t, τ) = [ ϕ (τ) ϕ(τ) ] cos(x t), whose solution satisfying the condition v 1 (x,, ) =, computed by the method of characteristics, is v 1 (x, t, τ) = [ ϕ (τ) ϕ(τ) ] t cos(x t). Consequently, the solution of the modified problem is v(x, t, τ) = ϕ(τ) cos(x t) ε [ ϕ (τ) ϕ(τ) ] t cos(x t) + O(ε 2 ). To eliminate the secular term on the right-hand side above, we now choose ϕ so that ϕ ϕ =. In view of the earlier condition ϕ() = 1, we find that ϕ(τ) = e τ. Hence, v(x, t, τ) = e τ cos(x t) + O(ε 2 ), which, since τ = εt, means that u(x, t) = e εt cos(x t) + O(ε 2 ). The terms of the asymptotic solution written in this form are now well ordered with respect to magnitude for all t >. As in Remark 13.15, we easily convince ourselves that the exact solution of the original IVP is u(x, t) = e εt cos(x t), which, in the above approximation, is accompanied by computational noise. Verification with Mathematica R. The input u = E (ε t) Cos[x - t]; f = Cos[x]; Simplify[Series[D[u,t] + D[u,x] - ε u,(u /. t -> ) - f},ε,,1}] // Normal] generates the output, }. Exercises In 1 4, use the method of asymptotic expansion to compute a formal approximate solution (to O(ε) terms) for the regular perturbation problem u (r) + r 1 u (r) + εf(u) = q(r), < r < 1, u(r), u (r) bounded as r +, u(1) = g, with functions f, q, and number g as indicated, where < ε 1.

347 Perturbation and Asymptotic Methods 1 f(u) = 4u, q(r) = 1, g = 1 + ε. 2 f(u) = 2u, q(r) = 4 + 2ε(r 2 + 9r), g = 1 + 3ε. 3 f(u) = u 3, q(r) = ε, g = 3. 4 f(u) = u u 2, q(r) = 6(3r 2) + ε(12 3r 2 + 2r 3 9r r 5 4r 6 ), g = 1 + 3ε. In 5 14, use the method of asymptotic expansion to compute a formal approximate solution (to O(ε) terms) for the regular perturbation problem u t (x, t) + au x (x, t) + bu(x, t) = q(x, t), < x <, t >, u(x, ) = f(x), < x <, with functions q, f and coefficients a, b as indicated, where < ε 1. 5 a = ε, b = 1 + ε, q(x, t) = 2x + 2xt + ε(2t + 2xt), f(x) = εx. 6 a = ε, b = 2 ε, q(x, t) = ε[ 1 2t 2x+(1+x)e 2t ], f(x) = x εx. 7 a = 2ε, b = 1 + 2ε, q(x, t) = 1 + 3x 2 + 3t + ε(4x x 2 t e 3t ), f(x) = x 2 + εx. 8 a = 2ε, b = 2 + ε, q(x, t) = e 2t + ε[2x + 4xt + (x + t 2)e 2t ], f(x) = x. 9 a = 3, b = 2ε, q(x, t) =, f(x) = x. 1 a = 1, b = 1 + ε, q(x, t) =, f(x) = 2x a = 2 + ε, b = 1 + ε, q(x, t) = 2 + ε[(4t 1)e t 2], f(x) = x a = 1 ε, b = 1 + 2ε, q(x, t) = x + ε(4x 1 + 5e t ), f(x) = 1 + x + 2εx. 13 a = 2 + ε, b =, q(x, t) = 2t, f(x) = e x. 14 a = 1 3ε, b =, q(x, t) = 1 + 3x + t + ε(x 8t 2), f(x) = x + εx. In 15 24, use the method of asymptotic expansion to compute a formal approximate solution (to O(ε) terms) for the regular perturbation problem u xx (x, y) + au x (x, y) + bu y (x, y) + cu(x, y) = q(x, y), u(, y) = f(y), u(1, y) = g(y), < y <, < x < 1, < y <, with functions q, f, g and coefficients a, b, c as indicated, where < ε a =, b = ε, c = 1, q(x, y) = xy 2, f(y) = εy, g(y) = y 2 + ε(e 2)y.

348 13.2 Regular Perturbation Problemss a =, b = 2ε, c = ε, q(x, y) = 2 ε(x 2 + 2y + 4), f(y) = 2y, g(y) = 1 + 2y + εy a =, b = ε, c = 1 + ε, q(x, y) = 4 2(1 + ε)x 2, f(y) = εy, g(y) = 2 + εy cos a =, b = ε, c = 4+ε, q(x, y) = 4x 2 y 2y ε [ x 2 y x 2 +3(y+1)e x], f(y) = ε(y + 1), g(y) = y + εe(y + 1). 19 a = 2 ε, b = 3ε, c = 1, q(x, y) = ε(y 2 6y)e x, f(y) = y 2 εy, g(y) = e 1 y 2 + εe 1 (1 y). 2 a = 1, b = ε, c =, q(x, y) = ε(y 2)e x, f(y) = 2y, g(y) = (2 + ε)ey. 21 a = 2 + ε, b = 2ε, c =, q(x, y) = 6 + ε(1 + 4ye 2x ), f(y) = y + ε, g(y) = 3 + y + εe 2 (1 2y). 22 a = 3, b = ε, c = 2, q(x, y) = 2+ε(2e x +ye 2x ), f(y) = 1 2y ε, g(y) = 1 2ey + εe 2 (y 1). 23 a = 1 + ε, b = ε, c = 2, q(x, y) = 6y 2 e x + ε(2y 2 4xy 2xy 2 )e x, f(y) = ε(3 y), g(y) = 2e 1 y 2 + εe 2 (3 y). 24 a = 2 ε, b = ε, c = 5 + ε, q(x, y) = 3y(5x + 2) + ε(3x 8y + 3xy), f(y) = ε(y + 1), g(y) = 3y ε(y + e 1 cos 2). Answers to Odd-Numbered Exercises 1 u(r) = (1/4)(3 + r 2 ) + (ε/16)(29 12r 2 r 4 ) + O(ε 2 ). 3 u(r) = 3 (13ε/2)(r 2 1) + O(ε 2 ). 5 u(x, t) = 2xt + εxe t + O(ε 2 ). 7 u(x, t) = x 2 + t + ε(x t)e 3t + O(ε 2 ). 9 u(x, t) = x 3t + 2εt(x 3t) + O(ε 2 ). 11 u(x, t) = (2t x)e t 2 + εxte t + O(ε 2 ). 13 u(x, t) = e x 2t + t 2 εte x 2t + O(ε 2 ). 15 u(x, y) = xy 2 + εy(e x 2x) + O(ε 2 ). 17 u(x, y) = 2x 2 + εy cos x + O(ε 2 ). 19 u(x, y) = y 2 e x + ε(x y)e x + O(ε 2 ). 21 u(x, y) = 3x + y + ε(1 2xy)e 2x + O(ε 2 ). 23 u(x, y) = 2xy 2 e x + ε(3 y)e 2x + O(ε 2 ).

349 Perturbation and Asymptotic Methods 13.3 Singular Perturbation Problems In this type of problem, we need to construct different solutions that are valid in different regions and then match them in an appropriate manner for overall uniform validity Example. Consider the IVP ε [ u t (x, t) + 2u x (x, t) ] + u(x, t) = sin t, < x <, t >, u(x, ) = x, < x <, where < ε 1. This is a singular perturbation problem since the reduced (unperturbed) problem contains no derivatives of u. Following the standard procedure, we try to seek a solution of the form u(x, t) u n (x, t)ε n. Replacing in the PDE, we obtain n= u (x, t) = sin t, u n (x, t) = (u n 1 ) t (x, t) 2(u n 1 ) x (x, t), n = 1, 2,.... From the recurrence relation above, it is easily seen that therefore, u 2n = ( 1) n sin t, u 2n+1 = ( 1) n+1 cos t, n ; [ ] [ ] u(x, t) = ( 1) n ε 2n sin t ε ( 1) n ε 2n cos t n= = 1 (sin t ε cos t), 1 + ε2 where we have used the formula for the sum of an infinite geometric progression with ratio ε 2. However, it is immediately obvious that this function does not satisfy the initial condition. Also, when t ε, using the power series for cos ε and sin ε, we have n= sin t ε cos t sin ε ε cos ε = ( ε + O(ε 3 ) ) ε ( 1 + O(ε 2 ) ) = O(ε 2 ); in other words, for t = O(ε) the two terms in u(x, t) are of the same order, which is not allowed in an asymptotic series. Consequently, the series is not well ordered in the region where t = O(ε), so it is not a valid representation of the solution in that region. This means that we must seek a different series in a boundary layer of width O(ε) near the x-axis (t = ). To construct the new series, we make the change of variables τ = t ε, u(x, t) = u(x, ετ) = ui (x, τ),

350 13.3 Singular Perturbation Problems 327 which amounts to a stretching of the time argument near t =. Under this transformation, the IVP becomes u i τ (x, τ) + 2εui x (x, τ) + ui (x, τ) = sin ετ = ετ + O(ε 3 ), u i (x, ) = x, < x <. < x <, τ >, For the inner solution u i of this boundary layer IVP, we assume an asymptotic expansion of the form u i (x, τ) u i n(x, τ)ε n. n= Then the new PDE and IC yield, in the usual way, the sequence of IVPs (u i ) τ(x, τ) + u i (x, τ) =, < x <, τ >, u i (x, ) = x, < x <, (u i 1 ) τ(x, τ) + u i 1 (x, τ) = τ 2(ui ) x(x, τ), u i 1(x, ) =, < x <, < x <, τ >, and so on. Restricting our attention to the first two terms, from the first problem we easily obtain u i (x, τ) = xe τ, and from the second problem (seeking a particular integral of the form aτe τ, a = const), u i 1 (x, τ) = τ 1 + (1 2τ)e τ ; hence, u i (x, τ) = xe τ + ε [ τ 1 + (1 2τ)e τ] + O(ε 2 ). We rename the first solution the outer solution and denote it by u o : u o (x, t) = 1 (sin t ε cos t) 1 + ε2 = sin t ε cos t + O(ε 2 ). Now we need to match u o (valid for t = O(1)) and u i (valid for t = O(ε)) up to the order of ε considered (here, it is O(ε)), in some common region of validity. To this end, first we write the outer solution in terms of the inner variable τ = t/ε and expand it for τ fixed and ε small, listing the terms up to O(ε), after which we revert to t: (u o ) i sin t ε cos t + = sin(ετ) ε cos(ετ) + = ε(τ 1) + = t ε +.

351 Perturbation and Asymptotic Methods Next, we write the inner solution in terms of the outer variable t = ετ and expand it for t fixed and ε small to the same order: [ ( t (u i ) o xe t/ε + ε ε t ] )e t/ε + = t ε + ε (the rest of the terms are o(ε n ) for any positive integer n). Finally, we impose the condition (u o ) i = (u i ) o up to O(ε) terms. In our case this is already satisfied, so the two solutions match. Since the common region of validity is not clear, it is useful to consider a composite solution of the form This is valid uniformly for t > since Here, we have u c = u o + u i (u o ) i = u o + u i (u i ) o. (u c ) o = (u o ) o + (u i ) o ((u i ) o ) o = u o + (u i ) o (u i ) o = u o, (u c ) i = (u o ) i + (u i ) i ((u o ) i ) i = (u o ) i + u i (u o ) i = u i. u(x, t) u c (x, t) = sin t ε cos t + (x 2t + ε)e t/ε Example. Consider the elliptic problem in a semi-infinite strip ε( u)(x, y) + u x (x, y) + u y (x, y) =, x >, < y < 1, u(x, ) = e x, u(x, 1) = p(x), x >, u(, y) = y, u(x, y) bounded as x, < y < 1, where < ε 1 and p(x) = 1 x, < x < 1,, x 1. This is a singular perturbation problem because the perturbed BVP is of second order whereas the reduced BVP is of first order. Let u(x, y) u n (x, y)ε n. Then it is readily seen that u is the solution of the BVP n= (u ) x (x, y) + (u ) y (x, y) =, x >, < y < 1, u (x, ) = e x, u (x, 1) = p(x), x >, u (, y) = y, u (x, y) bounded as x, < y < 1,

352 13.3 Singular Perturbation Problems 329 u 1 is the solution of the BVP (u 1 ) x (x, y) + (u 1 ) y (x, y) = ( u )(x, y), x >, < y < 1, u 1 (x, ) =, u 1 (x, 1) =, x >, u 1 (, y) =, u 1 (x, y) bounded as x, < y < 1, and so on. We find u by the method of characteristics (see Chapter 12), using y = 1 as the data line. If x = x(y) is a characteristic curve, then on it d dy u (x(y), y) = (u ) x (x(y), y)x (y) + (u ) y (x(y), y). Hence, x (y) = 1 implies that du /dy =, so x = y + c, u (x, y) = c, c, c = const. The equation of the characteristic through the points (x, y) and (x, 1) is x = y + x 1, and on this line, u (x, y) = c = u (x, 1) = p(x ) = p(x y + 1); therefore, the solution of the given BVP is u(x, y) = u (x, y) + O(ε) = p(x y + 1) + O(ε) y x + O(ε), < x < y, = O(ε), x y. (13.3) Since it is clear that this solution does not satisfy the other BCs, we need to introduce two boundary layers. In the boundary layer near y = we make the substitution η = y/δ(ε), write u(x, y) = u(x, δ(ε)η) = v(x, η), and arrive at the new BVP εv xx + ε δ 2 (ε) v ηη + v x + 1 δ(ε) v η =, v(x, ) = e x. The unspecified boundary layer width δ(ε) must be chosen so that the v ηη term is one of the dominant terms in the equation above. Comparing the coefficients of all the terms in the new PDE, we see that there are three possibilities: (i) ε δ 2 (ε) ε; (ii) ε δ 2 (ε) 1; (iii) ε δ 2 (ε) 1 δ(ε). It is not difficult to check that cases (i) and (ii) do not satisfy our magnitude requirement, whereas (iii) does. Hence, we can take δ(ε) = ε, for simplicity, and the BVP becomes ε 2 v xx + v ηη + εv x + v η =, v(x, ) = e x.

353 33 13 Perturbation and Asymptotic Methods Writing v(x, η) = v n (x, η)ε n, from the PDE and BC satisfied by v we find that n= (v ) ηη + (v ) η =, v (x, ) = e x, with solution v (x, η) = α(x) + [ e x α(x) ] e η, where α is an arbitrary function; therefore, u(x, y) = v(x, η) = v (x, η) + O(ε) = α(x) + [ e x α(x) ] e η + O(ε). (13.4) We rename (13.3) the outer solution u o and (13.4) the first inner solution u i1, and match them by the method used in Example Thus, expanding p in powers of ε, we have (u o ) i1 = p(x εη + 1) + O(ε) = p(x + 1) + O(ε) = O(ε), (u i1 ) o = α(x) + [ e x α(x) ] e y/ε + O(ε) = α(x) + O(ε), (13.5) so (u o ) i1 = (u i1 ) o to O(1) terms if α(x) =, x >. (13.6) The second boundary layer needs to be constructed near x =. Proceeding as above (this time with the transformed u xx term as one of the dominant terms in the new PDE), we again conclude that the correct layer width is δ(ε) ε, so we set ξ = x ε. Writing we arrive at the new BVP u(x, y) = u(εξ, y) = w(ξ, y), w ξξ + ε 2 w yy + w ξ + εw y =, w(, y) = y. If we assume that w(ξ, y) w n (ξ, y)ε n, then with solution n= (w ) ξξ + (w ) ξ =, w (, y) = y, w (ξ, y) = β(y) + [ y β(y) ] e ξ, where β is another arbitrary function; hence, u(x, y) = w(ξ, y) = w (ξ, y) + O(ε) = β(y) + [ y β(y) ] e ξ + O(ε). (13.7)

354 13.3 Singular Perturbation Problems 331 We call (13.7) the second inner solution u i2 O(1) terms. As before, we have and match it with u o up to so (u o ) i2 = p(εξ y + 1) + O(ε) = p( y + 1) + O(ε) = y + O(ε), (u i2 ) o = β(y) + [ y β(y) ] e x/ε + O(ε) = β(y) + O(ε), (13.8) β(y) = y. (13.9) It can be verified that the composite solution in this case is u c = u o + u i1 + u i2 (u i1 ) o (u i2 ) o = u o + u i1 + u i2 (u o ) i1 (u o ) i2, since (u i1 ) i2 = (u i1 ) o and (u i2 ) i1 = (u i2 ) o. Thus, by (13.3) (13.9), the asymptotic solution of the given BVP to O(1) terms is u(x, y) u c (x, y) = p(x y + 1) + [ e x p(x + 1) ] e y/ε + [ y p(1 y) ] e x/ε + O(ε) y x + e x y/ε + O(ε), < x < y, = e x y/ε + O(ε), x y Remarks. (i) We did not consider effects caused by the incompatibility of the boundary values at the corner points (, ) and (, 1). (ii) If we had tried to construct the first boundary layer solution near y = 1 instead of y = (and, thus, use y = as the data line for computing the first term in u o ), we would have failed. In that region, we would need to substitute η = (1 y)/ε, and setting we would obtain the BVP from which This would yield u(x, y) = u(x, 1 εη) = v(x, η), ε 2 v xx + v ηη + εv x v η =, v(x, ) = p(x), (v ) ηη (v ) η =, v (x, ) = p(x). v (x, η) = α(x) + [ p(x) α(x) ] e η, which is not good for matching since, with y fixed, e η = e y/ε as ε +.

355 Perturbation and Asymptotic Methods Exercises In 1 6, use the method of matched asymptotic expansions to compute a formal composite solution (to O(ε) terms) for the singular perturbation problem εu xx (x, y) + au x (x, y) + bu y (x, y) + cu(x, y) =, u(, y) = f(y), u(l, y) = g(y), < y <, < x < L, < y <, with functions f, g, coefficients a, b, c, and number L as indicated, where < ε 1. (The location of the boundary layer is specified in each case.) 1 f(y) = y + 1, g(y) = 2y, a = 1 + 2ε, b = 2ε, c = 1, L = 1; boundary layer at x =. 2 f(y) = y, g(y) = 2y 1, a = 1 ε, b =, c = 1, L = 2; boundary layer at x =. 3 f(y) = 2y + 3, g(y) = y, a = ε 1, b = ε, c =, L = 2; boundary layer at x = 2. 4 f(y) = 3y, g(y) = y + 2, a = 1 ε, b = 2ε, c = 1, L = 1; boundary layer at x = 1. 5 f(y) = y 2, g(y) = 1 y, a = 1 2ε, b = ε, c = 1, L = 1; boundary layer at x =. 6 f(y) = 2, g(y) = 2 3y, a = 1, b = 2ε, c = 1, L = 2; boundary layer at x =. In 7 12, use the method of matched asymptotic expansions (in conjunction with the method of characteristics) to compute a formal composite solution (to O(ε) terms) for the singular perturbation problem ε [ u xx (x, y) + au yy (x, y) ] + bu x (x, y) + cu y (x, y) =, u(x, ) = f(x), u(x, L) = g(x), x >, u(, y) = h(y), < y < L. x >, < y < L, with functions f, g, h, coefficients a, b, c, and number L as indicated, where < ε 1. (The locations of the boundary layers are specified in each case.) 7 f(x) = x 2 + 1, g(x) = 2x 3, h(y) = y 2, a = 1, b = 1, c = 2, L = 1; boundary layers at x =, y =. 8 f(x) = x 2, g(x) = 1 x, h(y) = y 2 1, a = 2, b = 1, c = 1, L = 2; boundary layers at x =, y =.

356 13.3 Singular Perturbation Problems f(x) = 2x 2 1, g(x) = x + 2, h(y) = y 2 + 1, a = 2, b = 1, c = 1, L = 1; boundary layers at x =, y = 1. 1 f(x) = 1 x 2, g(x) = 2x + 4, h(y) = 2y 2, a = 1, b = 2, c = 2, L = 2; boundary layers at x =, y = f(x) = x + 1, g(x) = x 2 2, h(y) = 1 2y, a = 4, b = 2, c = 1, L = 1; boundary layers at x =, y =. 12 f(x) = x 2 + x, g(x) = 3x, h(y) = y 2 2y, a = 3, b = 1, c = 1, L = 2; boundary layers at x =, y = 2. Answers to Odd-Numbered Exercises 1 u c (x, y) = 2ye 1 x + 2ε(xy 2x y + 2)e 1 x + [1 x + (1 2e)y (1 2e)xy + 2εe(y 2)]e x/ε + O(ε 2 ). 3 u c (x, y) = 2y εx + (3xy + 3x 9y 9 4ε)e (2 x)/ε + O(ε 2 ). 5 u c (x, y) = (1 y)e x 1 + εy(1 x)e x 1 + (exy 2 + xy + ey 2 x + y 1 εy)e 1 x/ε + O(ε 2 ). 7 u c (x, y) = 2x y 2 + (x 2 + xy 2x y + 3)e 2y/ε + (4xy + y 2 + 2x + y + 2)e x/ε + O(ε 2 ). 9 u c (x, y) = 2x 2 + 4xy + 2y εy + ( 2x 2 + 4xy 7x + 3y 2 12ε)e (1 y)/(2ε) + (2xy y εy)e x/ε + O(ε 2 ). 11 u c (x, y) = x 2 4xy + 4y 2 + 4x 8y ε(1 y) (x 2 + 4xy + 3x + 6y ε)e y/(4ε) + [ 4xy 4y 2 + 3x + 6y ε(y 1)]e 2x/ε + O(ε 2 ).

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358 C H A P T E R 14 Complex Variable Methods Certain linear two-dimensional elliptic problems turn out to be difficult to solve in a Cartesian coordinate setup. In many such cases, it is advisable to go over to equivalent formulations in terms of complex variables, which may be able to help us find the solutions much more readily and elegantly. Although complex numbers have already been mentioned in Chapters 1, 3, 8, 9, and 11, we start by giving a brief presentation of their basic rules of manipulation and a few essential details about complex functions Elliptic Equations A complex number is an expression of the form z = x + iy, x, y real, i 2 = 1, where x and y are called, respectively, the real part and the imaginary part of z. The number z = x iy is the complex conjugate of z and r = z = (z z) 1/2 = (x 2 + y 2 ) 1/2 is the modulus of z. By Euler s formula, a complex number can also be written in polar form as z = re iθ = r(cos θ + i sin θ), where θ, π < θ π, called the argument of z, is determined from the equalities cos θ = x r, sin θ = y r. 335

359 Complex Variable Methods Obviously, z = r(cos θ i sin θ) = re iθ. Addition and multiplication of complex numbers are performed according to the usual algebraic rules for real numbers. A complex function of a complex variable has the general form f(z) = (Re f)(x, y) + i(im f)(x, y), where Re f and Im f are its real and imaginary parts. Such a function f is called holomorphic if its derivative f (z) exists at all points in its domain of definition. A holomorphic function is analytic that is, it can be expanded in a convergent power series Example. If f(z) = z 2, then, since (x + iy) 2 = x 2 + 2ixy y 2, (Re f)(x, y) = x 2 y 2, (Im f)(x, y) = 2xy Theorem. A function f is holomorphic if and only if it satisfies the Cauchy Riemann relations (Re f) x = (Im f) y, (Im f) x = (Re f) y. (14.1) In this case, both Re f and Im f are solutions of the Laplace equation Remarks. (i) Suppose that a smooth function f of real variables x and y is expressed in terms of the complex variables z and z; that is, f(x, y) = g(z, z). From the chain rule of differentiation it follows that f x = g z + g z, f y = i(g z g z ), f xx = g zz + 2g z z + g z z, f yy = g zz + 2g z z g z z, f xy = f yx = i(g zz g z z ). (14.2) (ii) The Laplace equation, which is not easily integrated in terms of real variables, has a very simple solution in terms of complex variables. Thus, if u(x, y) = v(z, z), then, by (14.2), so u(x, y) = is equivalent to u = u xx + u yy = 4v z z, (14.3) v z z (z, z) =. It is now trivial to see that the latter has the general solution v(z, z) = ϕ(z) + ψ( z), where ϕ and ψ are arbitrary analytic functions of z.

360 14.1 Elliptic Equations 337 If we want the real general solution, then we must have v(z, z) = v( z, z); that is, ϕ(z) + ψ( z) = ϕ( z) + ψ(z), or ϕ(z) ϕ( z) = ψ(z) ψ( z), which means that Im ϕ = Im ψ. Using this equality and (14.1), we find that Re ϕ = Re ψ as well, so ψ = ϕ; therefore, the real general solution of the two-dimensional Laplace equation is where ϕ is an arbitrary analytic function. v(z, z) = ϕ(z) + ϕ( z), (14.4) (iii) A similar treatment can be applied to the biharmonic equation u(x, y) =. Since, as we have seen, v = 4v z z, where v(z, z) = u(x, y), we easily deduce that v(x, y) = 16v zz z z (z, z); hence, the biharmonic equation is equivalent to v zz z z (z, z) =. Then, by (ii) above, v(z, z) = 4v z z (z, z) = Φ(z) + Φ( z). Integrating and applying the argument in (ii), we arrive at the real general solution v(z, z) = zφ(z) + z Φ( z) + ϕ(z) + ϕ( z), where Φ and ϕ are arbitrary analytic functions of z Example. Consider the BVP u = 8 in D, u = 1 2 cos θ cos(2θ) + 2 sin(2θ) on D, where D and D are the unit circular disk and its boundary, respectively, and r, θ are polar coordinates with the pole at the center of the disk. On D we have z = e iθ = σ, z = e iθ = σ 1, cos(nθ) = 1 2 (einθ + e inθ ) = 1 2 (σn + σ n ), n = 1, 2,..., sin(nθ) = 1 2 i(einθ e inθ ) = 1 2 i(σn σ n ); (14.5) hence, using (14.3), we bring the given BVP to the equivalent form v z z = 2 in D, v = ( 1 2 i)σ 2 σ σ ( i)σ2 on D.

361 Complex Variable Methods It is easily seen that 2z z is a particular solution of the PDE, so, by (14.4), the general solution of the equation is v(z, z) = ϕ(z) + ϕ( z) + 2z z. (14.6) Since the arbitrary function ϕ is analytic, it admits a series expansion of the form ϕ(z) = a n z n. (14.7) n= Replacing (14.7) in (14.6), and then v with z = σ and z = σ 1 in the BC, and performing the usual comparison of coefficients, we find that a + ā + 2 = 1, a 1 = 1, a 2 = ( i), a n = (n =, 1, 2). Therefore, a + ā = 1, and (14.6) and (14.7) yield the solution v(z, z) = a + ā + a 1 z + ā 1 z + a 2 z 2 + ā 2 z 2 + 2z z = 1 z z ( i)z2 ( 1 2 i) z2 + 2z z. In Cartesian coordinates with the origin at the center of the disc, this becomes u(x, y) = 1 2x + x 2 + 4xy + 3y 2. Verification with Mathematica R. The input u = x + x x y + 3 y 2; Polaru = u/.x -> r Cos[θ],y -> r Sin[θ]}; f = 1-2 Cos[θ] - Cos[2 θ] + 2 Sin[2 θ]; Simplify[D[u,x,x] + D[u,y,y] - 8,(Polaru /. r -> 1) - f}] generates the output, } Example. To solve the elliptic BVP u xx 2u xy + 2u yy = 4 in D, u = cos(2θ) sin(2θ) on D, where the notation is the same as in Example 14.4, we use a slightly altered procedure. First, by (14.2), we easily see that v(z, z) = u(x, y) is the solution of the problem (1 + 2i)v zz 6v z z + (1 2i)v z z = 4 in D, v = 1 4 (1 3i)σ (1 + 3i)σ2 4 on D. (14.8) We now perform a simple transformation of the form ζ = z + ᾱ z, ζ = z + αz, v(z, z) = w(ζ, ζ),

362 14.1 Elliptic Equations 339 where α is a complex number to be chosen so that the left-hand side of the PDE for w consists only of the mixed second-order derivative. Thus, by the chain rule, v zz = w ζζ + 2αw ζ ζ + α 2 w ζ ζ, v z z = ᾱw ζζ + (1 + αᾱ)w ζ ζ + αw ζ ζ, v z z = ᾱ 2 w ζζ + 2ᾱw ζ ζ + w ζ ζ. When we replace this in the PDE in (14.8), we see that the coefficients of both w ζζ and w ζ ζ vanish if α is a root of the quadratic equation (1 + 2i)α 2 6α + 1 2i = ; that is, if α = 1 2i or α = 1 (1 2i). 5 Choosing, say, the first root, we have the transformation ζ = z + (1 + 2i) z, ζ = z + (1 2i)z, (14.9) which leads to the equation with general solution w ζ ζ = 1 4, w(ζ, ζ) = ϕ(ζ) + ϕ( ζ) ζ ζ, (14.1) where ϕ(ζ) is an arbitrary analytic function. As in (14.7), and taking (14.9) into account, we now write so, by (14.9) and (14.1), v(z, z) = ϕ(ζ) = a n ζ n ( ) n, = a n z + (1 + 2i) z n= n= [ ( ) n ) n ] an z + (1 + 2i) z + ān ( z + (1 2i)z n= ( z + (1 + 2i) z )( z + (1 2i)z ). (14.11) On the boundary D, (14.11) and the BC in (14.8) give rise to the equality [ ( an σ + (1 + 2i)σ 1 ) n ( + ān σ 1 + (1 2i)σ ) n] n= [ (1 + 2i)σ (1 2i)σ 2] = 1 4 (1 3i)σ (1 + 3i)σ2.

363 34 14 Complex Variable Methods Expanding the binomials on the left-hand side and equating the coefficients of each power of σ on both sides, we immediately note that and that, in this case, a n =, n = 3, 4,..., a + ā + 2(1 + 2i)a 2 + 2(1 2i)ā = 3 2, a 1 + (1 2i)ā 1 =, a 2 (3 + 4i)ā (1 2i) = 1 (1 + 3i). 4 The second and third equalities, taken together with their conjugates, yield the systems a 1 + (1 2i)ā 1 =, (1 + 2i)a 1 + ā 1 =, and a 2 (3 + 4i)ā 2 = 1 (2 + i), 4 ( 3 + 4i)a 2 + ā 2 = 1 (2 i), 4 from which a 1 =, a 2 = 1 (2 + i). 16 Replacing a 2 in the first equality, we find that a + ā =. Hence, by (14.11), ( ) ) v(z, z) = a + ā + a 1 z + (1 + 2i) z + ā1 ( z + (1 2i)z ( ) 2 ) 2 + a 2 z + (1 + 2i) z + ā2 ( z + (1 2i)z = 1 4 (1 + 3i)z z z 1 4 (1 3i) z2, or, in Cartesian coordinates, u(x, y) = x 2 + 3xy + 2y 2. Verification with Mathematica R. The input u = x x y + 2 y 2; Polaru = u/.x -> r Cos[θ],y -> r Sin[θ]}; f = 3/2 - (1/2) Cos[2 θ] + (3/2) Sin[2 θ]; Simplify[D[u,x,x] - 2 D[u,x,y] + 2 D[u,y,y] - 4, (Polaru /. r -> 1) - f}] generates the output, }.

364 14.1 Elliptic Equations 341 Exercises In 1 6, find the solution u(x, y) of the BVP u = q in D, u = f on D, where D and D are the disc with the center at the origin and radius 1 and its circular boundary, respectively, θ is the polar angle mentioned earlier in this chapter, and the number q and function f are as indicated. 1 q = 6, f(θ) = (1/2) [ cos(2θ) ]. 2 q = 8, f(θ) = cos θ 3 cos(2θ). 3 q = 2, f(θ) = (1/2) [ sin θ 3 cos(2θ) + 3 sin(2θ) ]. 4 q = 2, f(θ) = (1/2) [ sin θ 5 cos(2θ) + sin(2θ) ]. 5 q = 4, f(θ) = cos θ 3 sin θ + 2 cos(2θ) 2 sin(2θ). 6 q = 4, f(θ) = cos θ sin θ + 2 cos(2θ) 4 sin(2θ). In 7 14, find the solution u(x, y) of the BVP u xx + au xy + bu yy = q in D, u = f on D, where D and D are the disc with the center at the origin and radius 1 and its circular boundary, respectively, θ is the polar angle, and the numbers a, b, q and function f are as indicated. 7 a = 4, b = 5, q = 18, f(θ) = (1/2) [ 3 + cos(2θ) sin(2θ) ]. 8 a = 2, b = 5, q = 8, f(θ) = 2 cos θ + 2 cos(2θ) + sin(2θ). 9 a = 2, b = 2, q = 2, f(θ) = (1/2) [ 1 2 sin θ + 3 cos(2θ) + 4 sin(2θ) ]. 1 a = 4, b = 5, q = 28, f(θ) = 2 + cos θ 2 sin θ + 3 cos(2θ) + sin(2θ). 11 a = 1, b = 5/4, q = 25/2, f(θ) = (1/2) [ cos θ 2 cos(2θ) 3 sin(2θ) ]. 12 a = 4, b = 17/4, q = 67/2, f(θ) = (1/2) [ cos θ + 5 cos(2θ) + 3 sin(2θ) ]. 13 a = 2, b = 1, q = 66, f(θ) = 3 + cos θ 2 sin θ cos(2θ) sin(2θ). 14 a = 4, b = 13, q = 4, f(θ) = cos θ 3 sin θ + 2 cos(2θ) 2 sin(2θ).

365 Complex Variable Methods Answers to Odd-Numbered Exercises 1 u(x, y) = x 2 4y u(x, y) = 2x 2 + 3xy + y 2 + y 2. 5 u(x, y) = 3x 2 4xy y 2 + 2x 3y. 7 u(x, y) = 2x 2 xy + y 2. 9 u(x, y) = x 2 + 4xy 2y 2 y. 11 u(x, y) = x 2 3xy + 3y 2 + 2x u(x, y) = x 2 2xy + 3y 2 + x 2y Systems of Equations As an illustration of the efficiency of the complex variable method in solving linear two-dimensional systems of partial differential equations, we consider the mathematical model of plane deformation of an elastic body. This state is characterized by a two-component displacement vector u = (u 1, u 2 ) defined in the two-dimensional domain D occupied by the body. In the absence of body forces, u satisfies the system of PDEs (λ + µ) [ (u 1 ) xx + (u 2 ) xy ] + µ u1 =, (λ + µ) [ (u 1 ) xy + (u 2 ) yy ] + µ u2 =, (14.12) where λ and µ are physical constants and is the Laplacian. We assume that D is finite, simply connected (roughly, this means that D has no holes ), and bounded by a simple, smooth, closed contour D, and consider the Dirichlet problem for (14.12); that is, the BVP with the displacement components prescribed on the boundary: u 1 D = f 1, u 2 D = f 2, where f 1 and f 2 are known functions. Our aim is to find u at every point (x, y) in D Example. Using the same notation as in Example 14.4, consider the system 2 [ ] (u 1 ) xx + (u 2 ) xy + u1 =, 2 [ ] (14.13) (u 1 ) xy + (u 2 ) yy + u2 = in D, with the BCs u D 1 = 1 2 sin(2θ), u 2 D = cos θ. (14.14) To treat this BVP problem in terms of complex variables, we define the complex displacement U = u 1 + iu 2

366 14.2 Systems of Equations 343 and see that (u 1 ) x + (u 2 ) y = u 1,z + u 1, z + iu 2,z iu 2, z Next, we rewrite (14.13) in the form = (u 1 + iu 2 ) z + (u 1 iu 2 ) z = U z + Ū z. (14.15) 2[(u 1 ) x + (u 2 ) y ] x + u 1 =, 2[(u 1 ) x + (u 2 ) y ] y + u 2 =. Multiplying the second equation above by i, adding it to the first one, using (14.15), (14.3), and the fact that, by (14.2), x + i y = 2 z, we have = 2( x + i y ) [ (u 1 ) x + (u 2 ) y ] + (u1 + iu 2 ) = 4 z (U z + Ū z) + U = 4 [ (U z + Ū z) z + U z z ], or with general solution (2U z + Ū z) z =, 2U z + Ū z = 1 2 α (z), where α is an analytic function of z. The algebraic system formed by this equation and its conjugate now yields therefore, by integration, U z = 1 3 α (z) 1 6 ᾱ ( z); (14.16) U(z, z) = 1 3 α(z) 1 6 zᾱ ( z) + β( z), (14.17) where β is another analytic function of z. To investigate the arbitrariness of α and β, let p and q be functions of z such that α + p and β + q generate the same displacement U as α and β. Then, by (14.16), 1 3 [ α (z) + p (z) ] 1 6 [ᾱ ( z) + p ( z) ] = 1 3 α (z) 1 6 ᾱ ( z), from which 2p (z) p ( z) =. This equation together with its conjugate yield p (z) =, so p(z) = c, where c is a complex number. Using (14.17), we now see that 1 3 α(z) c 1 6 zᾱ ( z) + β( z) + q( z) = 1 3 α(z) 1 6 zᾱ ( z) + β( z), so q(z) = c/3. The arbitrariness produced by the complex constant c in the

367 Complex Variable Methods functions α and β can be eliminated by imposing an additional condition. For example, since the origin is in D, we may ask that α() =. (14.18) As in the preceding section, let σ be a generic point on the circle D. Then, by (14.5) and the fact that σ = σ 1, the BC (14.14) can be written as U D = (u 1 + iu 2 ) D = 1 4 i(σ 2 + 2σ 1 + 2σ σ 2 ); hence, in view of (14.17), we have 1 3 α(σ) 1 6 σᾱ (σ 1 ) + β(σ 1 ) = 1 4 i(σ 2 + 2σ 1 + 2σ σ 2 ). (14.19) Since D is finite and simply connected, we can consider series expansions of the analytic functions α and β, of the form α(z) = a n z n, β(z) = b n z n. n= Substituting these series in (14.19), we arrive at n= ( 1 3 a nσ n 1 6 nā nσ n+2 + b n σ n) = 1 4 i(σ 2 + 2σ 1 + 2σ σ 2 ). n= The next step consists in equating the coefficients of each power of σ on both sides, and it is clear that a n = b n = for all n = 3, 4,.... Thus, since (14.18) implies that a =, the only nonzero coefficients are given by the equalities b2 = 1 4 i, b1 = 1 2 i, b 1 3 ā2 =, 1 3 a ā1 = 1 2 i, 1 3 a 2 = 1 4 i. The coefficient a 1 is computed by combining the equation that it satisfies with its complex conjugate form. In the end, we obtain a 1 = i, a 2 = 3 4 i, b = 1 4 i, b 1 = 1 2 i, b 2 = 1 4 i, which generate the functions α(z) = iz 3 4 iz2, β(z) = 1 4 i 1 2 iz 1 4 iz2 and, by (14.17), the complex displacement U(z, z) = 1 4 i(1 + 2z + 2 z z2 z z + z 2 ). Hence, in terms of Cartesian coordinates, the solution of our BVP is u 1 (x, y) = Re ( U(z, z) ) = xy, u 2 (x, y) = Im ( U(z, z) ) = 1 4 (1 + 4x x2 y 2 ).

368 14.2 Systems of Equations 345 Verification with Mathematica R. The input u1, u2} =x y,(1/4) (1 + 4 x - x 2 - y 2)}; Polaru1,Polaru2} =u1,u2} /.x -> r Cos[θ],y -> r Sin[θ]}; f1, f2} =(1/2) Sin[2 θ],cos[θ]}; Simplify[2 (D[u1,x,x] + D[u2,x,y]) + D[u1,x,x] + D[u1,y,y], 2 (D[u1,x,y] + D[u2,y,y]) + D[u2,x,x] + D[u2,y,y]}, (Polaru1,Polaru2} /. r -> 1) -f1,f2}}] generates the output, },, }} Remark. If the functions f 1 and f 2 prescribed on D are not finite sums of integral powers of σ, they need to be expanded in full Fourier series. Using an argument similar to that in Section 8.1, we can write such a series in the form (see Chapter 8) f(θ) = n= c n e inθ = n= c n σ n, where c n = 1 2π π π f(θ)e inθ dθ. Exercises Find the solution ( u 1 (x, y), u 2 (x, y) ) of the BVP (λ + µ) [ (u 1 ) xx + (u 2 ) xy ] + µ u1 =, (λ + µ) [ (u 1 ) xy + (u 2 ) yy ] + µ u2 = u 1 = f 1, u 2 = f 2 on D, in D, where D and D are the disc with the center at the origin and radius 1 and its circular boundary, respectively, is the Laplacian, θ is the polar angle, and the numbers λ, µ and functions f 1, f 2 are as indicated. 1 λ = 1, µ = 1/2, f 1 (θ) = 2 cos θ cos(2θ), f 2 (θ) = cos θ + sin(2θ). 2 λ = 1, µ = 1, f 1 (θ) = sin θ + cos(2θ), f 2 (θ) = sin(2θ). 3 λ = 2, µ = 1, f 1 (θ) = 2 cos θ + 2 sin(2θ), f 2 (θ) = 1 cos(2θ). 4 λ = 1, µ = 2, f 1 (θ) = cos θ + 2 sin(2θ), f 2 (θ) = cos(2θ) + sin(2θ).

369 Complex Variable Methods 5 λ = 2, µ = 3, f 1 (θ) = cos(2θ) sin(2θ), f 2 (θ) = cos θ 2 sin(2θ). 6 λ = 1, µ = 2, f 1 (θ) = 1 sin θ + cos(2θ), f 2 (θ) = cos(2θ) 2 sin(2θ). 7 λ = 1/2, µ = 1, f 1 (θ) = 2 + sin θ + sin(2θ), f 2 (θ) = 3 cos(2θ) sin(2θ). 8 λ = 1/2, µ = 1, f 1 (θ) = cos θ + 2 cos(2θ), f 2 (θ) = 2 cos(2θ) sin(2θ). 9 λ = 3, µ = 1, f 1 (θ) = cos(2θ) sin(2θ), f 2 (θ) = 1 + sin θ 2 cos(2θ). 1 λ = 1/2, µ = 2, f 1 (θ) = cos θ 2 sin(2θ), f 2 (θ) = cos(2θ) + 3 sin(2θ). Answers to Odd-Numbered Exercises 1 u 1 (x, y) = x 2 + y 2 + 2x, u 2 (x, y) = 2xy + x. 3 u 1 (x, y) = 4xy x + 2, u 2 (x, y) = (1/5)( 14x 2 4y ). 5 u 1 (x, y) = 2x 2 2xy 2y 2 +1, u 2 (x, y) = (1/7)(x 2 28xy +y 2 +7x 1). 7 u 1 (x, y) = (1/7)(3x xy + 3y 2 + 7y + 11), u 2 (x, y) = (1/7)(27x 2 14xy 15y 2 6). 9 u 1 (x, y) = (1/3)(x 2 6xy 5y 2 +2), u 2 (x, y) = (1/3)( 8x 2 +4y 2 +3y+5).

370 Appendix A.1 Useful Integrals For all m, n = 1, 2,..., L L L L L L L cos nπx L sin nπx L cos nπx L sin nπx L sin cos dx = ; L L sin mπx L dx = sin nπx L dx = ;, n = m, L/2, n = m; mπx cos L dx =, n m, L/2, n = m; cos mπx L (2n 1)πx 2L (2n 1)πx 2L sin cos dx = ; (2m 1)πx 2L (2m 1)πx 2L dx = dx =, n = m, L/2, n = m;, n m, L/2, n = m. For all real numbers a, b, c, and p, (bx + c) cos(px) dx = b p 2 cos(px) + 1 (bx + c) sin(px) + const; p (bx + c) sin(px) dx = 1 p (bx + c) cos(px) + b sin(px) + const; p2 e ax cos(px) dx = e ax sin(px) dx = eax a 2 + p 2 [ a cos(px) + p sin(px) ] + const; eax a 2 + p 2 [ p cos(px) + a sin(px) ] + const. 347

371 348 Appendix A.2 Table of Fourier Transforms f(x) = F 1 [F](x) F(ω) = F[f](ω) 1 f (x) iωf(ω) 2 f (x) ω 2 F(ω) 3 f(ax + b) (a > ) 1 a e i(b/a)ω F(ω/a) 4 (f g)(x) F(ω)G(ω) 5 δ(x) 1 2π 6 e iax f(x) F(ω + a) 7 e a2 x a e ω2 /(4a 2 ) 8 xe a2 x i 2 (a > ) 2 2 a 3 ωe ω 2 /(4a 2 ) 9 x 2 e a2 x 1 2 (a > ) 4 2 a 5 (2a2 ω 2 )e ω 2 /(4a 2 ) 1 π 1 1 x 2 + a 2 (a > ) 2 a e a ω x π 1 11 x 2 + a 2 (a > ) i 2 2a ωe a ω 1, x a 2 sin(aω) 12 H(a x ) =, x > a π ω x, x a 2 1 [ ] 13 xh(a x ) = i sin(aω) aω cos(aω), x > a π ω 2 14 e a x 2 π a a 2 + ω 2 15 e (x+b)2 /(4a) + e (x b)2 /(4a) 2 2a e aω2 cos(bω) 16 erf(ax) 2 1 /(4a i ) π ω e ω2

372 Appendix 349 A.3 Table of Fourier Sine Transforms f(x) = F 1 S [F](x) F(ω) = F S[f](ω) 1 f (x) ωf C [f](ω) 2 f (x) 2 π ωf() ω2 F(ω) 3 f(ax) (a > ) 4 f(ax) cos(bx) (a, b > ) e ax (a > ) 7 xe ax (a > ) 1 a F(ω/a) [ ( 1 ω + b F 2a a 2 1 ω π 2 π 2 π ω a 2 + ω 2 2aω (a 2 + ω 2 ) 2 ) + F 2 8 x 2 e ax 3a 2 ω ω 3 (a > ) 2 π (a 2 + ω 2 ) 3 9 x x 2 + a 2 1, x a 1 H(a x) =, x > a x, x a 11 xh(a x ) =, x > a 12 erfc(ax) (a > ) π 2 π 2 π 2 π 13 xe a2 x π 14 tan 1 (x/a) (a > ) 2 2 e aω 1 [ ] 1 cos(aω) ω ( )] ω b 1 [ ] sin(aω) aω cos(aω) ω 2 1 [ ] 1 e ω 2 /(4a 2 ) ω 1 a 3 ωe ω2 /(4a 2 ) 1 ω e aω a

373 35 Appendix A.4 Table of Fourier Cosine Transforms f(x) = F 1 C [F](x) F(ω) = F C[f](ω) 2 1 f (x) π f() + ωf S[f](ω) 2 2 f (x) π f () ω 2 F(ω) 3 f(ax) (a > ) 4 f(ax) cos(bx) (a, b > ) 5 e ax (a > ) 6 xe ax (a > ) 1 a F(ω/a) [ ( 1 ω + b F 2a a 2 a a 2 + ω 2 π 2 π 2 7 x 2 e ax (a > ) 2 π a 2 ω 2 (a 2 + ω 2 ) 2 ) + F a 3 3aω 2 (a 2 + ω 2 ) 3 8 e a2 x a e ω2 /(4a 2 ) x 2 + a 2 (a > ) 1 (a 2 + x 2 ) 3 (a > ) x 2 (a 2 + x 2 ) 3 (a > ) x 4 12 (a 2 + x 2 ) 3 (a > ) 1, x a 13 H(a x) =, x > a (1/b)e bx cosh(ab), x a, 14 (1/b)e ab cosh(bx), x < a (a, b > ) π 1 2 a e aω π 2 π 2 π π ω sin(aω) 2 cos(aω) π b 2 + ω 2 ( )] ω b 1 8a 5 (a2 ω 2 + 3aω + 3)e aω 1 8a 3 ( a2 ω 2 + aω + 1)e aω 1 8a (a2 ω 2 5aω + 3)e aω a

374 Appendix 351 A.5 Table of Laplace Transforms f(t) = L 1 [F](t) F(s) = L[f](s) 1 f (n) (t) (nth derivative) s n F(s) s n 1 f() 2 H(t a)f(t a) e as F(s) f (n 1) () 3 e at f(t) F(s a) 4 (f g)(t) F(s)G(s) t n (n positive integer) 1 s (s > ) n! s n+1 (s > ) 7 e at 1 (s > a) s a a 8 sin(at) s 2 + a 2 (s > ) 9 cos(at) 1 sinh(at) 11 cosh(at) 12 δ(t a) (a ) e as s s 2 + a 2 (s > ) a s 2 a 2 (s > a ) s s 2 a 2 (s > a ) 13 e a2t erfc ( a t ) 1 (a > ) s + a s a 14 2 π t 3/2 e a2 /(4t) (a > ) e a s ( ) a 15 erfc 2 1 (a > ) s t s e a t 16 a /(4t) + ( 1 π e a2 2 a2 + t) erfc a 2 t (a > ) 1 s s 2 e a

375 352 Appendix A.6 Second-Order Linear Equations If Au xx + Bu xy + Cu yy + Du x + Eu y + Fu = G, if new variables r = r(x, y), s = s(x, y) are defined by means of the characteristic equations and if then where dy dx = B dy dx = B + B2 4AC, 2A B2 4AC, 2A u(x, y) = u ( x(r, s), y(r, s) ) = v(r, s), Āv rr + Bv rs + Cv ss + Dv r + Ēv s + Fv = Ḡ, Ā = A(r x ) 2 + Br x r y + C(r y ) 2, B = 2Ar x s x + B(r x s y + r y s x ) + 2Cr y s y, C = A(s x ) 2 + Bs x s y + C(s y ) 2, D = Ar xx + Br xy + Cr yy + Dr x + Er y, Ē = As xx + Bs xy + Cs yy + Ds x + Es y, F = F, Ḡ = G.

376 Further Reading Andrews, L.C.: Elementary Partial Differential Equations with Boundary Value Problems. Academic Press, Orlando (1986) Bleecker, D.D., Csordas, G.: Basic Partial Differential Equations. International Press of Boston, Boston, MA (1997) Brown, J.W., Churchill, R.V.: Fourier Series and Boundary Value Problems, 6th ed. McGraw-Hill, New York (2) Davis, J.M.: Introduction to Applied Partial Differential Equations. W.H. Freeman, New York, NY (212) DuChateau, P., Zachmann, D.: Applied Partial Differential Equations. Dover Publications, New York, NY (22) Duffy, D.G.: Solutions of Partial Differential Equations. TAB Books, Blue Ridger Summit, PA (1986) Evans, L.C.: Partial Differential Equations, 2nd ed. American Mathematical Society, Providence, RI (21) Farlow, S.J.: Partial Differential Equations for Scientists and Engineers. Dover Publications, New York, NY (1993) Garabedian, P.R.: Partial Differential Equations, 2nd ed. American Mathematical Society, Providence, RI (1998) Gustafson, K.E.: Introduction to Partial Differential Equations and Hilbert Space Methods, 3rd ed. Dover Publications, New York, NY (1997) Haberman, R.: Applied Partial Differential Equations with Fourier Series and Boundary Value Problems, 5th ed. Pearson (212) Keane, M.K.: A Very Applied First Course in Partial Differential Equations. Prentice Hall, Upper Saddle River, NJ (21) Kevorkian, J., Cole, J.D.: Perturbation Methods in Applied Mathematics. Springer, New York, NY (1985) 353

377 354 Further Reading Kovach, L.D.: Boundary-Value Problems. Addison Wesley, Reading, MA (1984) Leach, J.A., Needham, D.J.: Matched Asymptotic Expansions in Reaction- Diffusion Theory. Springer, London (23) Myint-U, T., Debnath, L.: Linear Partial Differential Equations for Scientists and Engineers, 4th ed. Birkhäuser, Boston, MA (26) Nayfeh, A.H.: Introduction to Perturbation Techniques. Wiley, New York, NY (1993) Shearer, M., Levy, R.: Partial Differential Equations: An Introduction to Theory and Applications. Princeton University Press, Princeton, NJ (215) Stakgold, I., Holst, M.J.: Green s Functions and Boundary Value Problems, 3rd ed. Wiley, New York, NY (211) Strauss, W.A.: Partial Differential Equations: An Introduction. Wiley, New York, NY (27) Taylor, M.E.: Partial Differential Equations, vols. I III, 2nd ed. Springer, New York, NY (213) Vvedensky, D.: Partial Differential Equations with Mathematica. Addison Wesley, Reading, MA (1993) Wan, F.Y.M.: Mathematical Models and Their Analysis. Harper & Row, New York, NY (1989) Weinberger, H.F.: A First Course in Partial Differential Equations: with Complex Variables and Transform Methods. Dover Publications, New York, NY (1995) Zachmanoglou, E.C., Thoe, D.W.: Introduction to Partial Differential Equations with Applications. Dover Publications, New York, NY(1987) Zauderer, E.: Partial Differential Equations of Applied Mathematics, 3rd ed. Wiley-Interscience, Hoboken, NJ (26)

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