Green's Functions with Applications 9781482251036, 1482251035

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GREEN’S FUNCTIONS WITH APPLICATIONS Second Edition

Advances in Applied Mathematics Series Editor: Daniel Zwillinger Published Titles Stochastic Partial Differential Equations, Second Edition Pao-Liu Chow Markov Processes James R. Kirkwood Green’s Functions with Applications, Second Edition Dean G. Duffy

Advances in Applied Mathematics

GREEN’S FUNCTIONS WITH APPLICATIONS Second Edition

DEAN G. DUFFY Former Instructor, US Naval Academy Annapolis, Maryland, USA

MATLAB® is a trademark of The MathWorks, Inc. and is used with permission. The MathWorks does not warrant the accuracy of the text or exercises in this book. This book’s use or discussion of MATLAB® software or related products does not constitute endorsement or sponsorship by The MathWorks of a particular pedagogical approach or particular use of the MATLAB® software.

CRC Press Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487-2742 © 2015 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: 20150114 International Standard Book Number-13: 978-1-4822-5103-6 (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 www.copyright.com (http://www.copyright.com/) or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. 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 http://www.taylorandfrancis.com and the CRC Press Web site at http://www.crcpress.com

Dedicated to My Dear Friends Pete, Philippe, Ed, Dave, Karl, Michael, El, John, and Ralph

v

Contents Acknowledgments

xi

Author

xiii

Preface

xv

List of Definitions

xvii

Chapter 1: Historical Development

1

1.1 Mr. Green’s Essay

1

1.2 Potential Equation

4

1.3 Heat Equation

7

1.4 Helmholtz’s Equation

14

1.5 Wave Equation

15

1.6 Ordinary Differential Equations

20 vii

g(x

) f( ) L x

Chapter 2: Background Material

27

2.1 Fourier Transform

27

2.2 Laplace Transform

35

2.3 Bessel Functions

50

2.4 Legendre Polynomials

60

2.5 The Dirac Delta Function

66

2.6 Green’s Formulas

76

2.7 What Is a Green’s Function?

79

Chapter 3: Green’s Functions for Ordinary Differential Equations

91

3.1 Initial-Value Problems

92

3.2 The Superposition Integral

97

3.3 Regular Boundary-Value Problems

99

3.4 Eigenfunction Expansion for Regular Boundary-Value Problems

110

3.5 Singular Boundary-Value Problems

117

3.6 Maxwell’s Reciprocity

127

3.7 Generalized Green’s Function

129

3.8 Integro-Differential Equations

137

Chapter 4: Green’s Functions for the Wave Equation 4.1 One-Dimensional Wave Equation in an Unlimited Domain viii

161 164

4.2 One-Dimensional Wave Equation on the Interval 0 < x < L

183

4.3 Axisymmetric Vibrations of a Circular Membrane

188

4.4 Two-Dimensional Wave Equation in an Unlimited Domain

190

4.5 Three-Dimensional Wave Equation in an Unlimited Domain

200

4.6 Asymmetric Vibrations of a Circular Membrane

212

4.7 Thermal Waves

218

4.8 Diffraction of a Cylindrical Pulse by a Half-Plane

219

4.9 Leaky Modes

221

4.10 Water Waves

240

Chapter 5: Green’s Functions for the Heat Equation

283

5.1 Heat Equation over Infinite or Semi-Infinite Domains

288

5.2 Heat Equation within a Finite Cartesian Domain

310

5.3 Heat Equation within a Cylinder

323

5.4 Heat Equation within a Sphere

340

5.5 Product Solution

344

5.6 Absolute and Convective Instability

350

Chapter 6: Green’s Functions for the Helmholtz Equation

397

6.1 Free-Space Green’s Functions for Helmholtz’s and Poisson’s Equation

401

6.2 Method of Images

422

6.3 Two-Dimensional Poisson’s Equation over Rectangular and Circular Domains

430

ix

6.4 Two-Dimensional Helmholtz Equation over Rectangular and Circular Domains

450

6.5 Poisson’s and Helmholtz’s Equations on a Rectangular Strip

467

6.6 Three-Dimensional Problems in a Half-Space

477

6.7 Three-Dimensional Poisson’s Equation in a Cylindrical Domain

483

6.8 Poisson’s Equation for a Spherical Domain

496

6.9 Improving the Convergence Rate of Green’s Functions

505

6.10 Mixed Boundary Value Problems

519

Numerical solution − Exact solution 1

0.4

0.9

0.3

0.8 0.2

0.7

t−

0.6

0.1

0.5

0

0.4 −0.1

0.3 0.2

−0.2

0.1 0

−0.3 0.2

0.4

0.6

Chapter 7: Numerical Methods

589

0.8

x

7.1 Discrete Wavenumber Representation

590

7.2 Laplace Transform Method

592

7.3 Finite Difference Method

600

7.4 Hybrid Method

604

7.5 Galerkin Method

607

7.6 Evaluation of the Superposition Integral

615

7.7 Mixed Boundary Value Problems

626

Appendix: Relationship between Solutions of Helmholtz’s and Laplace’s Equations in Cylindrical and Spherical Coordinates

635

Answers to Some of the Problems

639

Author Index

653

Subject Index

663

x

Acknowledgments Special thanks go to Prof. Michael D. Marcozzi for his many useful suggestions for improving this book. Dr. Tim DelSole provided outstanding guidance in the section on convective/absolute instability. Drs. Chris Linton and Geordie D. McBain made several useful suggestions regarding Section 6.9 and the numerical methods chapter, respectively. Many of the plots and R calculations were done using MATLAB! . MATLAB is a registered trademark of The MathWorks Inc. 24 Prime Park Way Natick, MA 01760-1500 Phone: (508) 647-7000 Email: [email protected] www.mathworks.com

xi

Author Dean G. Duffy received his bachelor of science in geophysics from Case Institute of Technology (Cleveland, Ohio) and his doctorate of science in meteorology from the Massachusetts Institute of Technology (Cambridge, Massachusetts). He served in the United States Air Force from September 1975 to December 1979 as a numerical weather prediction officer. After his military service, he began a twenty-five year (1980 to 2005) association with NASA at the Goddard Space Flight Center (Greenbelt, Maryland) where he focused on numerical weather prediction, oceanic wave modeling and dynamical meteorology. He also wrote papers in the areas of Laplace transforms, antenna theory, railroad tracks, and heat conduction. In addition to his NASA duties he taught engineering mathematics, differential equations and calculus at the United States Naval Academy (Annapolis, Maryland) and the United States Military Academy (West Point, New York). Drawing from his teaching experience, he has written several books on transform methods, engineering mathematics, Green’s functions and mixed boundary value problems.

xiii

Preface This book had its origin in some electronic mail that I received from William S. Price a number of years ago. He needed to construct a Green’s function and asked me if I knew a good book that might assist him. In suggesting several standard texts, I could not help but think that, based on my own experiences utilizing Green’s functions alone and in conjunction with numerical solvers, I had my own ideas on how to present this material. It was this thought that ultimately led to the development of this monograph. Since its original publication 14 years ago, the purpose of this book has remained the same: to provide applied scientists and engineers with a systematic presentation of the various methods available for deriving a Green’s function. However, I now understand many of the weaknesses of my original text and have sought to address these flaws in this new edition. The book opens with a new chapter on the historical development of the Green’s function. References are made to many seminal papers and books that helped establish Green’s function in its current use. The early history was slow and difficult until the mathematical concepts of transform methods and the Dirac delta function became well established. All books are written with assumptions about the background of the reader. Chapter 2 was written so that everyone would start on a flat playing field with regards to Fourier and Laplace transforms and the classical special functions of Bessel functions and Legendre polynomials. We also include a review of the Dirac delta function. We close this chapter by anticipating future results and introducing some simple applications of Green’s function. Chapters 3 through 6 are the heart of the book. For each class of differential equation (ordinary differential, wave, heat and Helmholtz equations) xv

we present Green’s functions according to the number of spatial dimensions and the geometry of the domain. In each chapter I added solutions which I missed in my first edition and new results that have appeared since 2001. Of particular interest is the use of Green’s functions in the life sciences. Each chapter closes with special sections devoted to topics where Green’s functions are particularly useful. For example, in the case of the wave equation, Green’s functions are very useful in describing diffraction and wave waves. These chapters also include problem sets. They are there so that the reader can master Green’s function as well as a source of additional results. A new aspect to this book is the use of the symbol " # to denote the end of an example, proof, or remark. The only exception occurs if an example concludes a section. In this case its use is superfluous. A few years after the publication of the original book, I received an email from a graduate student asking about numerical methods for computing Green’s functions for an aeronautical problem that he was solving. This incident motivated me to write the chapter on numerical methods that closes this book. Clearly this is an underdeveloped area of study and further work is expected.

xvi

Definitions of the Most Commonly Used Functions Function

Definition

δ(t − a)

=

H(t − a)

=

!

∞, 0,

!

1, 0,

"

t = a, t &= a, t > a, t < a,

∞ −∞

δ(t − a) dt = 1

a≥0

In (x)

modified Bessel function of the first kind and order n

Jn (x)

Bessel function of the first kind and order n

Kn (x)

modified Bessel function of the second kind and order n

Pn (x)

Legendre polynomial of order n

r


= max(r,ρ )

x


= max(x,ξ )

Yn (x)

Bessel function of the second kind and order n

y


= max(y,η )

z


= max(z,ζ )

xvii

Chapter 1 Historical Development One of the fundamental problems of field theory1 is the construction of solutions to linear differential equations when there is a specified source and the differential equation must satisfy certain boundary conditions. The purpose of this book is to show how Green’s functions provide a powerful method for obtaining these solutions. In this chapter, we present a historical overview of their evolution. 1.1 MR. GREEN’S ESSAY In 1828 George Green (1793–1841) published an Essay on the Application of Mathematical Analysis to the Theory of Electricity and Magnetism. In this seminal work of mathematical physics, Green sought to determine the electric potential within a vacuum bounded by conductors with specified potentials. In today’s notation we would say that he examined the solutions of ∇2 u = −f within a volume V that satisfy certain boundary conditions along the boundary S. 1

Any theory in which the basic quantities are fields, such as electromagnetic theory.

1

2

Green’s Functions with Applications

To solve this problem, Green first considered a problem where the source is a point charge. In modern notation, he sought to solve the partial differential equation: ∇2 g(r|r0 ) = −4πδ(r − r0 ), (1.1.1) where δ(r − r0 ) is the Dirac delta function. We now know that the solution to Equation 1.1.1 is g = 1/R, where R2 = (x − ξ)2 + (y − η)2 + (z − ζ)2 . Recognizing the singular nature of g, he proceeded as follows: First Green proved the theorem that bears his name: """ "" # 2 $ 2 ϕ∇ χ − χ∇ ϕ dV = ⊂⊃ (ϕ∇χ − χ∇ϕ) · n dS, (1.1.2) V

S

where the outwardly pointing normal is denoted by n and χ and ϕ are scalar functions that possess bounded derivatives. Then, by introducing a small ball about the singularity at r0 (because Equation 1.1.2 cannot apply there) and then excluding it from the volume V , he obtained """ "" """ "" g∇2 u dV + ⊂⊃ g∇u · n dS = u∇2 g dV + ⊂⊃ u∇g · n dS − 4πu(r0 ) V

S

V

S

(1.1.3) because the surface integral over the small ball is 4πu(r0 ) as the radius of the ball tends to zero. Next, Green required that g satisfies the homogeneous boundary condition g = 0 along the surface S. Since ∇2 u = −f and ∇2 g = 0 within V (recall that the point r0 is excluded from V ), he found that "" 1 u(r) = ⊂⊃ u ∇g · n dS, (1.1.4) 4π S when f = 0 (Laplace’s equation) for any point r within S. Here u denotes the value of u on S. This solved the boundary-value problem once g was found. Green knew that g had to exist; it physically described the electrical potential from a point charge located at r0 . Green’s essay remained relatively unknown until it was published2 at the urging of Kelvin between 1850 and 1854. Later Poincar´e3 summarized our knowledge of Green’s functions near the turn of the twentieth century. The subsequent evolution of Green’s functions can be divided into two parts: before and after the publication in 1946 of Methods of Theoretical Physics by P. M. Morse and H. Feshbach.4 In this paper-back version of classnotes that 2 Green, G., 1850, 1852, 1854: An essay on the application of mathematical analysis to the theories of electricity and magnetism. J. Reine Angew. Math., 39, 73–89; 44, 356–374; 47, 161–221. 3

Poincar´e, H., 1894: Sur les ´ equations de la physique math´ ematique. Rend. Circ. Mat. Palermo, 8, 57–156. 4 Morse, P. M., and H. Feshbach, 1946: Methods of Theoretical Physics. MIT Technology Press, 497 pp.

Historical Development

3

they developed since the late 1930s to teach mathematical methods to physics graduate students, they laid out the four properties that a Green’s function must possess. Using the Sturm-Liouville problem given by % & d dy f (x) + p(x)y = −q(x), dx dx

(1.1.5)

these four properties are: • The Green’s function satisfies the homogeneous differential equation when x &= ξ, the source point. • The Green’s function satisfies homogeneous boundary conditions. • The Green’s function is symmetric in the variables x,ξ . • The Green’s function g(x|ξ) satisfies the condition ' ' dg '' dg '' 1 − =− . ' ' dx x=ξ+ dx x=ξ− f (ξ)

(1.1.6)

Prior to the publication of Morse and Feshbach’s notes, authors used various tricks to find Green’s functions that satisfied these four properties. Morse and Feshbach’s great contribution was to show that the “Green’s function is the point source solution [to a boundary-value problem] satisfying appropriate boundary conditions.” Thus the Green’s function could be found by simply solving (in the case of Sturm-Liouville problem) % & d dg f (x) + p(x)g = −δ(x − ξ) dx dx

(1.1.7)

with homogeneous boundary conditions, where δ(x − ξ) was the recently introduced delta function by Dirac. The advantage of this formulation was that the powerful techniques of eigenvalue expansions and transform methods could be used in a straightforward manner to find Green’s functions. They will be the primary methods used in this book. By the 1960’s many textbooks began to champion the use of Green’s functions. For example, in Mackie’s 1965 book5 he sought “to give a general account of how certain mathematical techniques, notably those of Green’s functions and of integral transforms, can be used to solve important and commonly occurring boundary value problems in ordinary and partial differential equations.” In the following sections we turn to the development of Green’s functions as they evolved within each general class of differential equations.

5

Mackie, A. G., 1965: Boundary Value Problems. Oliver & Boyd, 252 pp.

4

Green’s Functions with Applications

1.2 POTENTIAL EQUATION Shortly after the publication of Green’s monograph on the European continent, the German mathematician and pedagogue Carl Gottfried Neumann (1832–1925) developed the concept of Green’s function as it applies to the two-dimensional (in contrast to three-dimensional) potential equation.6 He defined the two-dimensional Green’s function, showed that it possesses the property of reciprocity, and found that it behaves as ln(r) as r → ∞. Using elliptic coordinates he rederived Poisson’s integral formula and developed an eigenfunction expansion for the two-dimensional Green’s function. In 1875 Paul Meutzner (1849–1914) extended Neumann’s work.7 In particular, he obtained the Green’s function for the region within an ellipse (Ellipsenfl¨ ache) and a circle (Ringfl¨ ache). Finally, in his book on the logarithmic potential, A. Harnack8 (1851–1888) gave the Green’s function for a circle and rectangle. All of these authors used a technique that would become one of the fundamental techniques in constructing a Green’s function, namely eigenfunction expansions. The investigator would first find an eigenfunction expansion that satisfied both the homogeneous differential equation and boundary conditions. The geometry of the problem would determine the coordinate system that was used. Then the Fourier coefficients would be chosen so that the Green’s function exhibited the proper behavior (such as 1/r) near the source point. Later on,9 John Dougall (1867–1960) derived three-dimensional Green’s functions in cylindrical and spherical coordinates. In 1879 Alfred George Greenhill10 (1847–1927) applied the method of images to construct the Green’s function for a rectangular parallelepiped. Because his results are expressed as an infinite summation of theta functions, it was not very useful and has essentially been forgotten. Hector Munro Macdonald11 (1865–1935) took a slightly different approach in the 1890s. As before, he began with the eigenfunction expansion 6 2

∂ Φ ∂y 2

Neumann, C., 1861: Ueber die Integration der partiellen Differentialgleichung:

∂2 Φ ∂x2

+

= 0. J. Reine Angew. Math., 59, 335–366.

7 Meutzner, P., 1875: Untersuchungen im Gebiete des logarithmischen Potentiales. Math. Ann., 8, 319–338. For an alternative derivation, see Sections 15 and 17 in Neu¨ mann, C., 1906: Uber das logarithmische Potential. Ber. Verh. K. Sachs. Ges. Wiss. Leipzig, Math.-Phys. Klasse, 58, 482–559. 8 See Chapter 2 in Harnack, A., 1887: Die Grundlagen der Theorie des logarithmischen Potentiales und der eindeutigen Potentialfunktion in der Ebene. Leipzig, B. G. Teubner, 170 pp. 9 Dougall, J., 1900: The determination of Green’s function by means of cylindrical or spherical harmonics. Proc. Edinburgh Math. Soc., Ser. 1 , 18, 33–83. 10

Greenhill, A. G., 1879: On Green’s function for a rectangular parallelepiped. Proc. Cambridge Philos. Soc., 3, 289–293. 11 Macdonald, H. M., 1895: The electrical distribution on a conductor bounded by two spherical surfaces cutting at any angle. Proc. London Math. Soc., Ser. 1 , 26, 156–172;

Historical Development

5

Figure 1.2.1: Carl Gottfried Neumann (1832–1925) was a leading German mathematician and teacher. Today he is best known for his work on the Dirichlet principle and integral equations, and his co-founding with Alfred Clebsch of Mathematische Annalen. Left c c photograph !Universit¨ atarchiv Leipzig; right photograph !Photo Deutsches Museum.

that satisfied the boundary conditions. But now, the Fourier coefficients were chosen so that the expansion satisfied the general Poisson equation. Then he considered the special case of a point source. We illustrate his method in Example 6.8.2. Because you must solve the general Poisson equation first, his technique never became popular. In the late 1890’s Arnold Sommerfeld12 (1868–1951) developed a technique using integration on the complex plane to extend the method of images to several other useful geometries in three dimensions. Ernst William Hobson (1856–1933) then used this method13 to find the Green’s function for a circular disk. Later, Ludwig Waldmann (1913–1980), a young assistant to Sommerfeld, applied this technique in electrostatic calculations of an electron lens.14 Unfortunately this technique is very complicated and we will present Macdonald, H. M., 1900: Demonstration of Green’s formula for electric density near the vertex of a right cone. Trans. Cambridge Philos. Soc., 18, 292–297. 12 Sommerfeld, A., 1897: Uber ¨ verzweigte Potentiale im Raum. Proc. London Math. Soc., Ser. 1, 28, 395–429. 13 Hobson, E. W., 1900: On Green’s function for a circular disc, with applications to electrostatic problems. Trans. Cambridge Philos. Soc., 18, 277–291. 14 Waldmann, L., 1937: Zwei Anwendungen der Sommerfeld’schen Methode der verzweigten Potentiale. Phys. Z., 38, 654–663.

6

Green’s Functions with Applications

an improved version in Example 6.2.5. At the beginning of twentieth century the method of bilinear expansions was developed: g(x, y, z|ξ, η,ζ ) =

∞ ( ψn (x, y, z)ψn (ξ, η,ζ ) , λn n=1

(1.2.1)

where λn and ψn (x, y, z) are the nth eigenvalue and eigenfunction, respectively. Adolf Kneser15 (1862–1930) showed that the Green’s function was the symmetric kernel of the integral equation ψn (ξ, η,ζ ) = λn

"""

g(x, y, z|ξ, η,ζ )ψn (x, y, z) dx dy dz.

(1.2.2)

Assuming that the Green’s function can be expressed as an eigenfunction expansion, Equation 1.2.1 follows. As examples, Kneser found the bilinear expansion for rectangular and circular areas and for the surface of a sphere. In summary then, by 1950 there were essentially three methods16 for finding Green functions. The first method simply used a Green’s function developed for Helmholtz’s equation ∇2 u + k02 u = 0 and took the limit as k0 → 0. The second method wrote the Green’s function as a sum of eigenfunctions that satisfied the boundary conditions. The coefficients were then chosen so that the correct singular behavior occurred at the source point. Finally, the third method wrote the Green’s function as the sum of the free-space solution plus a harmonic function.17 The harmonic solution was chosen so that the Green’s function satisfied the boundary conditions. Later on, Kelvin’s classic inversion18 that maps the interior of a circle or sphere to the exterior and vice versa was developed to find the Green’s function for Poisson’s equation. We will illustrate this technique following Equation 6.3.37 and in Section 6.8. Finally Green’s functions have been used to solve mixed boundary-value problems involving the two-dimensional Poisson’s equation. These problems occur when the boundary condition changes along a given boundary from a 15 Kneser, A., 1911: Integralgleichungen und ihre Anwendungen in der mathematischen Physik . Braunschweig, 293 pp. 16 See, for example, Bouwkamp, C. J., and N. G. de Bruijn, 1947: The electrostatic field of a point charge inside a cylinder, in connection with wave guide theory. J. Appl. Phys., 18, 562–577. This paper is of particular note because of its use of the modern definition of the delta function. See their Equation 41. 17 Weber, E., 1939: The electrostatic potential produced by a point charge on the axis of a cylinder. J. Appl. Phys., 10, 663–666. 18 Thomson, W., 1845: Extrait d’une lettre de M. William Thomson ` a M. Liouville. J. Math. Pures Appl., 10, 364–367; Thomson, W., 1847: Extraits de deux lettres address´ees a M. Liouville. J. Math. Pures Appl., 12, 256–264. `

Historical Development

7

Dirichlet condition to a Neumann condition and vice versa. We will explore this topic in Section 6.10. 1.3 HEAT EQUATION The development of using Green’s function to solve the heat equation consists of two parts. In the nineteenth century a synthetical method was developed which replaces the actual distribution by sets of sources and sinks distributed over the boundaries and throughout the region under investigation. The twentieth century has been dominated by the use of transform methods. Our tale begins with William Thomson (Lord Kelvin)19 (1824–1907) and his solution of the one-dimensional heat equation ∂u ∂ 2u = , ∂t ∂x2 with the boundary conditions ! V, 0 < t < T, u(0, t) = 0, T < t,

0 < x < ∞,

0 < t,

lim u(x, t) → 0,

x→∞

(1.3.1)

0 < t,

(1.3.2)

and the initial conditions u(x, 0) = 0,

0 < x < ∞.

Using Fourier integrals, Kelvin obtained the solution % & " T Vx dτ x2 u(x, t) = √ exp − . 4(t − τ ) 2 π 0 (t − τ )3/2

(1.3.3)

(1.3.4)

He also showed that the solution could also be obtained “synthetically” for small T if he introduced a source at x = 0 and a sink at x = α and took the limit as α → 0. Furthermore, reporting on some correspondence with George G. Stokes (1819–1903), he gave the solution in a form that we now call the superposition integral: % & " t x x2 f (τ ) u(x, t) = √ exp − dτ, (1.3.5) 4(t − τ ) t − τ 2 π 0 where u(0, t) = f (t). In 1887 Ernst William Hobson20 (1856–1933) generalized Kelvin’s synthetical method to two and three dimensions. In general, his effort must 19

Thomson, W., 1854/55: On the theory of the electric telegraph. London, 7, 382–399.

Proc. R. Soc.

20 Hobson, E. W., 1887: Synthetical solutions in the conduction of heat. Proc. London Math. Soc., Ser. 1 , 19, 279–299.

8

Green’s Functions with Applications

Figure 1.3.1: William Thomson, Baron Kelvin of Largs, (1824–1907) was one of the leading mathematical physicists of the nineteenth century. During his work on thermodynamics he realized that there was a lower limit to temperature. He became well known to the public for his prediction on the speed of a signal in a transatlantic submarine cable that was being laid in the 1850s. (Portrait courtesy of the Royal Society.)

be considered a failure since it resulted in expressions that were difficult to interpret. For example, in another paper,21 Hobson redid Kelvin’s problem of one-dimensional heat conduction when the boundary condition at x = 0 changed to ux (0, t) = h[u(0, t) − f (t)], where h denotes the external conductivity. He found that ! % & % &) " ∞" ∞ h (x + ζ − ξ)2 (x + ζ + ξ)2 u(x, t) = √ e−hζ exp − − exp − 2 π 0 0 4t 4t × [u(ξ, 0) − ux (ξ, 0)] dξd ζ. (1.3.6) Hobson’s solution fails if the initial temperature distribution is discontinuous. In 1892, George H. Bryan22 (1864–1928) improved the synthetical method. He cleverly wrote the solution as the sum of two parts: a source term 21

Hobson, E. W., 1888: On a radiation problem. Math. Proc. Cambridge Philos. Soc., 6, 184–187. 22 Bryan, G. H., 1892: Note on a problem in the linear conduction of heat. Math. Proc. Cambridge Philos. Soc., 7, 246–248.

Historical Development

9

(located at x = ξ) plus a homogeneous solution so that the total solution satisfies the boundary condition. Using this technique to redo Hobson’s 1888 calculation, he found that % & % &) " ∞! 1 (x − ξ)2 (x + ξ)2 u(x, t) = √ exp − + exp − f (ξ) dξ 4t 4t 2 πt 0 % & " ∞" ∞ h (x + ζ + ξ)2 −√ e−hζ exp − f (ξ) dξd ζ. (1.3.7) 4t πt 0 0 Later on, Sommerfeld23 generalized Kelvin’s results by considering substances that have different thermal properties. He considered two cases: (a) two semi-infinite domains with the interface at x = 0 and (b) two semi-infinite slabs separated a finite layer lying between a < x < b. In 1939 P. V. Solovieff 24 used mirror images to construct Green’s functions for a n+1 dimensional heat equation when one or more boundaries are moving. By the turn of the twentieth century, John Dougall25 (1867–1960) introduced the concept of contour integration to find the mathematical description of how a sphere cools in a well-stirred liquid. Although Dougall’s analysis made no use of Green’s functions, it provided the necessary insight that allowed H. S. Carslaw to synthesize all of the ideas on how to apply Green’s function to heat conduction problems. Carslaw’s 1902 paper26 began by deriving how any conduction problem without sources can be expressed in terms of its Green’s function, the initial condition and the solution’s value along the boundary surrounding the domain of interest. The question then turned on the question of finding the Green’s function for various geometries and boundary conditions. In the case of unbounded domains, he used the Green’s functions given by the synthetical method. The remaining Green’s functions were obtained using Dougall’s method of contour integration. For example, to find the Green’s function for linear heat flow over the interval (0, a), when the Green’s function vanishes at both ends, Carslaw first introduced the Green’s function ! % & % &) 1 (x − ξ)2 (x + ξ)2 g1 (x, t|ξ, 0) = √ exp − − exp − . (1.3.8) 4κt 4κt 2 πκt 23 Sommerfeld, A., 1894: Zur analytischen Theorie der W¨ armeleitung. Math. Ann., 45, 263–277. 24 Solovieff, P. V., 1939: Die Greensche Funktion der W¨ armeleitungsgleichung. Dokl. Acad. Sci. USSR, 23, 132–134; Solovieff, P. V., 1939: Fonctions de Green des ´equations paraboliques. Dokl. Acad. Sci. USSR, 24, 107–109. 25 Dougall, J., 1901: Note on the application of complex integration to the equation of conduction of heat, with special reference to Dr. Peddie’s problem. Proc. Edinburgh Math. Soc., Ser. 1 , 19, 50–56. 26 Carslaw, H. S., 1902: The use of Green’s functions in the mathematical theory of the conduction of heat. Proc. Edinburgh Math. Soc., Ser. 1 , 21, 40–64.

10

Green’s Functions with Applications y

P x

Figure 1.3.2: The contour P and closed contour used in Equation 1.3.9 and Equation 1.3.10, respectively.

The first term is the free-space Green’s function given by Kelvin while the second term assures that g1 (0, t|ξ, 0) = 0. Carslaw then showed that this Green’s function could be expressed by the contour integral 1 g1 (x, t|ξ, 0) = πi

"

2

e−κz t sin(x< z)e−izx> dz,

(1.3.9)

P

where P is the contour shown in Figure 1.3.2. On the right side, the contour must lie between 0 < arg(z) < π /4 as |z| → ∞ while on the left side the contour lies between 3π/4 < arg(z) < π as |z| → ∞. Next, Carslaw introduced the new Green’s function g2 (x, t|ξ, 0) = −

"

1 πi

e−κz

P

2

t sin(x< z) sin(x> z) iaz

sin(az)

e

dz.

(1.3.10)

Why did Carslaw create this new Green’s function? Because a linear combination of g1 (x, t|ξ, 0) and g2 (x, t|ξ, 0) yields the Green’s function 1 g(x, t|ξ, 0) = πi

"

P

e−κz

2

t sin(x< z) sin[(a

sin(az)

− x> )z]

dz

(1.3.11)

which satisfies the boundary conditions g(0, t|ξ, 0) = g(a, t|ξ, 0) = 0. Furthermore, Carslaw was also able to show that this Green’s function satisfied the initial condition and had the correct behavior at the singularity x = ξ. Using Cauchy’s residue theorem and the closed contour showed in Figure 1.3.2, this Green’s function could be rewritten in the convenient form of , % & ∞ 2 ( * nπx + nπξ κn2 π 2 (t − τ ) g(x, t|ξ,τ ) = sin sin exp − . a n=1 a a a2

(1.3.12)

Historical Development

11

The difficulty of this method is quite apparent: It is not easy to choose a complex representation for the Green’s function that satisfies the boundary conditions, initial condition and the singular nature at the point of excitation. This difficulty became academic with the advent of Laplace transforms. In the mid-1920s, Gustav Doetsch (1892–1977) wrote a series of papers on heat conduction.27 In particular, he considered the heat conduction problem ∂u ∂2u = , ∂t ∂x2

0 < x < c,

0 < t,

(1.3.13)

with the boundary conditions u(0, t) = A(t),

u(c, t) = B(t),

0 < t,

(1.3.14)

and the initial condition u(x, 0) = Φ(x),

0 < x < c.

(1.3.15)

He showed that the solution u(x, t) could be expressed by the one-dimensional version of Equation 5.0.11 (see Problem 1 in Chapter 5) and the Green’s function ∞ 2 ( −n2 π2 t/c2 g(x, t|ξ, 0 ) = e sin(nπξ/c) sin(nπx/c). c n=1 +

(1.3.16)

Furthermore, he showed that the Green’s function is symmetric. Finally, the limit of c → ∞ yielded the Green’s functions on a semi-infinite domain found by Kelvin and Carslaw. The revolutionary aspect of Doetsch’s approach was his use of Laplace transforms. In 1932, Sydney Goldstein28 (1903–1989) showed how Laplace transforms could be used to solve many heat conduction problems whose derivation up to that time had been very cumbersome. Indeed he pointed out that many of the Green’s functions found by Carslaw’s contour integral method was in reality the “operational method.” 27 Bernstein, F., and G. Doetsch, 1925: Probleme aus der Theorie der W¨ armeleitung. I. Mitteilung. Eine neue Methode zur Integration partieller Differentialgleichungen. Der lineare W¨ armeleiter mit verschwindender Anfangstemperatur. Math. Z., 22, 285–292; Doetsch, G., 1925: Probleme aus der Theorie der W¨ armeleitung. II. Mitteilung. Der lineare W¨ armeleiter mit verschwindender Anfangstemperatur. Die allgemeinste L¨ osung und die Frage der Eindeutigkeit. Math. Z., 22, 293–306; Doetsch, G., 1925: Probleme aus der Theorie der W¨ armeleitung. III. Mitteilung. Der lineare W¨ armeleiter mit beliebiger Anfangstemperatur. Die zeitliche Fortsetzung des W¨ armezustandes. Math. Z., 25, 608–626; Bernstein, F., and G. Doetsch, 1927: Probleme aus der Theorie der W¨ armeleitung. IV. Mitteilung. Die r¨ aumliche Fortsetzung des Temperaturablaufs (Bolometerproblem). Math. Z., 26, 89–98. 28 Goldstein, S., 1932: Some two-dimensional diffusion problems with circular symmetry. Proc. London Math. Soc., Ser. 2 , 34, 51–88.

12

Green’s Functions with Applications Consider, for example, the one-dimensional heat conduction problem ∂u ∂ 2u = a2 2 , ∂t ∂x

0 < x < ∞,

0 < t.

(1.3.17)

Taking the Laplace transform of Equation 1.3.17, we have that d2 U = q 2 U, dx2

0 < x < ∞,

(1.3.18)

where s = a2 q 2 . The solution to Equation 1.3.18 is U (x, s) = Ae−qx ,

(1.3.19)

where we have discarded the exponentially growing solution as x → ∞. Now, " x # $ 2 U (x, s) dx = 2A 1 − e−qx /q. (1.3.20) 0

To find the Green’s function, we must choose A so that it represents an instantaneous plane source of heat from x = −∞ to x = ∞ in the limit of t → 0. In that case, the left side of Equation 1.3.20 equals one as q → ∞, or 2A = q. Therefore, G(x, s|0, 0) = 12 qe−qx . (1.3.21) Taking the inverse of G(x, s|0, 0), we have that , 1 x2 g(x, t|0, 0) = √ exp − 2 , 4a t 2 πa2 t

(1.3.22)

the same result that Kelvin found. Goldstein used similar methods to find the Green’s function for an axisymmetric problem in the plane and outside of a cylinder. During the 1930s and 1940s several authors found the Green’s functions for cylindrical and spherical geometries by using Byran’s method of writing the Green’s function as a sum of a free-space Green’s function plus a homogeneous solution which satisfies the boundary conditions. They obtained the homogeneous solution using Laplace transforms. For example,29 Lowan redid Bryan’s original problem of finding the Green’s function in a semi-infinite planar solid that is radiating at the x = 0 face. Later, Lowan applied this technique to heat conduction in cylindrical30 and spherical31 coordinates. 29 Lowan, A. N., 1937: On the operational determination of Green’s functions in the theory of heat conduction. Philos. Mag., Ser. 7 , 24, 62–70. 30

Lowan, A. N., 1938: On the operational determination of two dimensional Green’s function in the theory of heat conduction. Bull. Amer. Math. Soc., 44, 125–133. 31 Lowan, A. N., 1939: On Green’s functions in the theory of heat conduction in spherical coordinates. Bull. Amer. Math. Soc., 45, 310–315 and 951–952.

Historical Development

13

Figure 1.3.3: John Conrad Jaeger’s (1907–1979) (right portrait) association with Horatio Scott Carslaw (1870–1954) began during Jaeger’s undergraduate education at the University of Sydney. After Jaeger’s undergraduate education and subsequent studies at the University of Cambridge, the contact with Carslaw was renewed by irregular trips to Carslaw’s retirement farm when Jaeger returned to Tasmania in 1936. These meetings lead to a collaboration on the application of Laplace transforms to find the Green’s function for the heat equation. (Carslaw’s portrait courtesy of the University of Sydney Archives, Image G3/224/0695; Jaeger’s portrait courtesy of the Royal Society.)

During this same period Carslaw and Jaeger also found Green’s functions using Laplace transforms. In their earliest paper32 they found the Green’s function for the region outside of a cylinder using both the Laplace transforms and the contour method. In a subsequent paper Carslaw33 found the Green’s function for heat conduction in two semi-infinite solids of different materials with a common boundary at x = 0 as well as the case when the two semiinfinite solids are separated by a third solid of thickness 2a. In the case of three dimension problems, they34 wrote the Green’s function as a sum of the free-space Green’s function plus a homogeneous solution of the heat equation. They then used Laplace transforms to find the homogeneous solution. Finally Carslaw and Jaeger35 applied Bryan’s technique to cylindrical problems where 32 Carslaw, H. S., and J. C. Jaeger, 1939: On Green’s functions in the theory of heat conduction. Bull. Amer. Math. Soc., 45, 407–413. 33 Carslaw, H. S., 1940: A simple application of the Laplace transformation. Philos. Mag., Ser. 7 , 30, 414–417. 34 Carslaw, H. S., and J. C. Jaeger, 1940: The determination of Green’s function for the equation of conduction of heat in cylindrical coordinates by the Laplace transformation. J. London Math. Soc., 15, 273–281. 35

Carslaw, H. S., and J. C. Jaeger, 1941: The determination of Green’s function for

14

Green’s Functions with Applications

they used line sources for the free-space Green’s functions. 1.4 HELMHOLTZ’S EQUATION A partial differential equation which is quite similar to Laplace’s equation is Helmholtz’s equation. It arises in the study of forced (steady-state) vibrations governed by the wave equation; the most famous application is the diffraction of acoustic and visible light waves. The history of Green’s function involving Helmholtz’s equation begins with the theoretical work of Hermann von Helmholtz (1821–1894) during his study of acoustics.36 He showed that the free-space Green’s function is g(x, y, z|ξ, η,ζ ) = cos(k0 r)/r, where r2 = (x − ξ)2 + (y − η)2 + (z − ζ)2 . Helmholtz used this Green’s function to express the solution of ∇2 u + k02 u = 0 in a region R with the boundary S which gives the solution and its derivative on the boundary. F. Pockels (1865–1913) epic book37 on Helmholtz’s equation summarized our knowledge of Green’s function at the end of the nineteenth century. In Part IV (Bestimmung der Functionen u aus gegebenen Randwerthen und verwandten Bedingungen) he reviewed the Dirichlet principle and showed how Green’s functions could be used to solve this problem. Next Pockels discussed the expansion of Green’s function in term of eigenfunctions. For example, he gave the Green’s function within a circle in terms of Fourier series. For us, Section 4 (L¨ osung der Randwerthaufgaben f¨ ur die Functionen u mit H¨ ulfe verallgemeinerter Green’scher Functionen) of Part IV is of particular interest. In this section Pockels gave the free-space Green’s function in three-dimensional space and on xy-plane, discussed reciprocity and presented the boundary-value solution in terms of the Green’s function and the type of boundary condition. In summary, the nineteenth century saw the full development of the concept of the Green’s functions but presented few actual functions. This began to change in the twentieth century. In the early twentieth century several major studies appeared on the Green’s functions for Helmholtz’s equation. The first paper38 was by A. Sommerfeld (1868–1951). It consisted of two parts: Green’s function for a bounded and unbounded region. For a finite domain he showed that the Green’s funcline sources for the equation of conduction of heat in cylindrical coordinates by the Laplace transformation. Philos. Mag., Ser. 7 , 31, 204–208. 36 Helmholtz, H., 1860: Theorie der Luftschwingungen in R¨ ohren mit offenen Enden. J. Reine Angew. Math., 57, 1–72. 37 Pockels, F., 1891: Uber ¨ die partielle Differentialgleichung ∆u + k2 u = 0 und deren Auftreten in der mathematischer Physik . Leipzig, Teubner, 339 pp. 38 Sommerfeld, A., 1912: Die Greensche Funktion der Schwingungsgleichung. Jahresber. Deutsch. Math.-Verein., 21, 309–353.

Historical Development tion can be expressed by the eigenfunction expansion: ( um (O)um (P ) g= , 2 k02 − km m

15

(1.4.1)

where um (O) and um (P ) are the mth eigenfunction to the eigenvalue problem 2 ∇2 um + km um = 0 with um = 0 along the boundary and point O denotes a general point within the domain while the point P is the location of the singularity. Sommerfeld proved that this expansion satisfies the partial differential equation, g = 0 at the boundary, is indefinite when point P and point O are collocated, and satisfies reciprocity. As examples, he gave freespace Green’s functions in two and three dimensions as well as for a circular membrane with a fixed boundary and a three-dimensional parallelepiped with Neumann boundary conditions. In two subsequent papers H. S. Carslaw39 extended Sommerfeld’s results for a wide variety of three dimensional spaces involving cylindrical and spherical coordinates. Turning to unlimited domains, Sommerfeld gave the free-space Green’s function in two and three dimensions. More importantly, he derived his famous “radiation condition” that required outwardly propagating waves from physical considerations.40 By 1950 Green’s functions for Helmholtz’s equation were used to find the wave motions due to flow over a mountain41 and in acoustics.42 1.5 WAVE EQUATION Soon after the publication of Green’s essay, Green’s functions were used to solve the wave equation. In 1860 Bernhard Riemann43 (1826–1866) applied the method of Green’s functions to integrate the hyperbolic equation that describes the propagation of sound waves.44 39 Carslaw, H. S., 1912: Integral equations and the determination of Green’s functions in the theory of potential. Proc. Edinburgh Math. Soc., Ser. 1 , 31, 71–89; Carslaw, H. S., 1914: The Green’s function for the equation ∇2 u + k 2 u = 0. Proc. London Math. Soc., Ser. 2 , 15, 236–257. 40 See Schot, S. M., 1992: Eighty years of Sommerfeld’s radiation condition. Hist. Math., 19, 385–401. 41 Lyra, G., 1943: Theorie der station¨ aren Leewellenstr¨ omung in freier Atmosph¨ are. Zeit. Angew. Math. Mech., 23, 1–28. 42 Foldy, L. L., and H. Primakoff, 1945: A general theory of passive linear electroacoustic transducers and the electroacoustic reciprocity theorem. I. J. Acoust. Soc. Am., 17, 109– 120. 43

Riemann, B., 1860: Ueber die Fortpflanzung ebener Luftwellen von endlicher Schwingungsweite. Abh. d. K¨ on. Ges. der Wiss. zu G¨ ottingen, 8, 43–65. An English translation appears in Johnson, J. N., and R. Ch´ eret, 1998: Classic Papers in Shock Compression Science. Springer-Verlag, 524 pp. 44 Mackie, A. G., 1964/65: Green’s function and Riemann’s method. Proc. Edinburgh Math. Soc., Ser. 2 , 14, 293–302.

16

Green’s Functions with Applications

Figure 1.5.1: Originally drawn to mathematics, Arnold Johannes Wilhelm Sommerfeld (1868–1951) migrated into physics due to Klein’s interest in applying the theory of complex variables and other pure mathematics to a range of physical topics from astronomy to dynamics. Later on, Sommerfeld contributed to quantum mechanics and statistical mechanics. (AIP Emilio Segr` e Visual Archives, Margrethe Bohr Collection.)

For a linear hyperbolic equation of second order in two independent variables ∂ 2u ∂ 2u ∂u ∂u − 2 + 2a − 2b + cu = 0, (1.5.1) ∂x2 ∂y ∂x ∂y where x and y are chosen so that the two families of characteristic are x ± y = constant and a, b and c are functions only of x and y, the solution45 at the point P with coordinates (ξ,η ) is (uG)|A + 12 (uG)|B (1.5.2) " + 12 (Guy − uGy + 2b uG) dx + (Gux − uvx + 2a uG) dy.

u(ξ,η ) =

1 2

AB

Here G denotes the Riemann-Green function and is given by the adjoint equation ∂2G ∂ 2G ∂(aG) ∂(bG) − −2 +2 + cG = 0 (1.5.3) ∂x2 ∂y 2 ∂x ∂y 45 For the derivation, see Section 73 in Webster, A. G., 1966: Partial Differential Equations of Mathematical Physics. Dover, 446 pp.

Historical Development such that

17

∂G ∂G + = (a + b)G ∂x ∂y

on y − x = η − ξ,

(1.5.4)

∂G ∂G − = (a − b)G ∂x ∂y

on y + x = η + ξ,

(1.5.5)

and G(ξ,η ) = 1. The values of u and its first derivative are specified along the arc AB which is chosen so that no characteristic cuts it more than at one point; the arcs P A and P B are characteristics. Although there are several methods for finding Riemmann-Green functions,46 actually finding one is very difficult. The greatest success with this technique involved the equation of telegraphy.47 The next important development of Green’s functions for the wave equation lies with Gustav Robert Kirchhoff 48 (1824–1887), who used it during his study of the three-dimensional wave equation. Starting with Green’s second formula, he was able to show that the three-dimensional Green’s function is g(x, y, z, t|ξ, η, ζ,τ ) =

δ(t − τ − R/c) , 4πR

(1.5.6)

. where R = (x − ξ)2 + (y − η)2 + (z − ζ)2 (modern terminology). Although he did not call his solution a Green’s function,49 he clearly grasped the concept that this solution involved a function that we now call the Dirac delta function (see pg. 667 of his Annalen d. Physik ’s paper). He used this solution to derive his famous Kirchhoff ’s theorem, which is the mathematical expression for Huygen’s principle: energy always propagates out to infinity. The early twentieth century saw the development of Laplace transforms to solve the wave equation in addition to the previous known method of Fourier 46 See Copson, E. T., 1958: On the Riemann-Green function. Arch. Rat. Mech. Anal., 1, 324–348.

´ 1894: Sur l’´ Picard, E., equation aux d´eriv´ ees partielles qui se recontre dans la th´eorie de la propagation de l’´ electricit´ e. Acad. Sci., Compt. Rend., 118, 16–17; Bois-Reymond, ¨ P. du, 1889: Uber lineare partielle Differentialgleichungen zweiter Ordnung. J. Reine Angew. Math., 104, 241–301; Voigt, W., 1899: Ueber die Aenderung der Schwingungsform des Lichtes beim Fortschreiten in einem dispergirenden oder absorbirenden Mittel. Ann. Phys., 304, 598–603; Gray, M. C., 1923: The equation of telegraphy. Proc. Edinburgh Math. Soc., Ser. 2 , 42, 14–28; Rademacker, H., and R. Iglisch, 1961: Randwertprobleme der partiellen Differentialgleichungen zweiter Ordnung, 779–828 in Frank, Ph., and R. von Mises, 1961: Die Differential- und Integralgleichungen der Mechanik und Physik. I. Mathematischer Teil . Dover, 916 pp.; Section 74 in Webster, A. G., 1966: Partial Differential Equation of Mathematical Physics. Dover, 446 pp.; Wahlberg, C., 1977: Riemann’s function for a Klein-Gordon equation with a non-constant coefficient. J. Phys., Ser. A, 10, 867–878; Asfar, O. R., 1990: Riemann-Green function solution of transient electromagnetic plane waves in lossy media. IEEE Trans. Electromagn. Compat., EMC-32, 228–231. 47

48

Kirchhoff, G., 1882: Zur Theorie der Lichtstrahlen. Sitzber. K. Preuss. Akad. Wiss. Berlin, 641–669; reprinted a year later in Ann. Phys. Chem., Neue Folge, 18, 663–695. 49 This appears to have been done by Gutzmer, A., 1895: Uber ¨ den analytischen Ausdruck des Huygens’schen Princips. J. Reine Angew. Math., 114, 333–337.

18

Green’s Functions with Applications

Figure 1.5.2: Gustav Robert Kirchhoff’s (1824–1887) most celebrated contributions to physics are the joint founding with Robert Bunsen of the science of spectroscopy, and the discovery of the fundamental law of electromagnetic radiation. Kirchhoff’s work on light coincides with his final years as a professor of theoretical physics at Berlin. (Portrait taken from frontispiece of Kirchhoff, G., 1882: Gesammelte Abhandlungen. J. A. Barth, 641 pp.)

transforms.50 In 1914 T. J. I’A. Bromwich (1875–1929) showed how Laplace transforms51 can be used to solve the wave equation by eliminating the temporal dependence, leaving a boundary-value problem. Interestly he then solved this boundary-value problem using Green’s functions. Then, unknowingly he found as an example the Green’s function for the one-dimensional wave equation with fixed ends (see his Example 5 on page 438). A. N. Lowan (1898–1962) applied Bromwich’s idea to finding the wave motions within a wedge52 of infinite radius and an infinite solid53 which is exterior to a cylinder or sphere. 50

Poincar´e, H., 1893: Sur la propagation de l’´ electricit´ e. Acad. Sci., Compt. Rend., 117, 1027–1032; Webster, A. G., 1966: Partial Differential Equation of Mathematical Physics. Dover, Section 46. 51 Bromwich, T. J. I’A., 1914: Normal coordinates in dynamical systems. Proc. London Math. Soc., Ser. 2 , 15, 401–448. 52

Lowan, A. N., 1941: On the problem of wave-motion for the wedge of an angle. Philos. Mag., Ser.7 , 31, 373–381. 53 Lowan, A. N., 1939: On wave motion in an infinite solid bounded internally by a cylinder or a sphere. Bull. Amer. Math. Soc., 45, 316–325.

Historical Development

19

One difficulty of finding the Green’s function for the wave equation lies in its definition. Around 1950 A. G. Walters wrote a series of papers54 (that have been essentially forgotten) on finding the Green’s function of transient partial differential equations. In the case of the wave equation, his Green’s vibrational function g(P, P1 , t − τ ) satisfied the wave equation ∂ 2g = c2 D(g), ∂t2

(1.5.7)

where D(·) is a differential operator in one or more dimensions, and satisfies the integral conditions " lim g(P, P1 , t − τ ) dV = 0, (1.5.8) τ →t

and lim

τ →t

"

V

V

gt (P, P1 , t − τ ) dV = 1,

(1.5.9)

where V is any region enclosed by the boundaries and contains the point P1 , which is defined below. He then showed how the solution to the wave equation can be expressed in terms of volume integrals involving the initial conditions, boundary conditions, and any source terms. To compute g(P, P1 , t − τ ), Walters proved that the Green’s function is the inverse Laplace transform of Φ(P, P1 , s2 /c2 ), where Φ satisfies the boundary-value problem D(Φ) =

s2 Φ. c2

(1.5.10)

Note that Φ satisfies Equation 1.5.10 except for a singular point located at P1 . Our modern definition of the Green’s function for the wave equation as the response of this equation to impulse forcing appears to originate with H. J. Bhabha55 (1909–1966) in his study of the meson field of neutrons. As we will show in Example 4.1.1, he used transform methods to find the Green’s function for the Klein-Gordon equation. Van der Pol and Bremmer were the first to introduce the general community to the concept that the Green’s function56 of the wave equation is a 54 Walters, A. G., 1949: The solution of some transient differential equations by means of Green’s functions. Proc. Cambridge Philos. Soc., 45, 69–80; Walters, A. G., 1951: On the propagation of disturbances from moving sources. Proc. Cambridge Philos. Soc., 47, 109–126. 55 Bhabha, H. J., 1939: Classical theory of mesons. Proc. R. Soc. London, Ser. A, 172, 384–409. 56 Van der Pol, B., and H. Bremmer, 1964: Operational Calculus Based on the Two-Sided Laplace Transform. Cambridge, 415 pp.

20

Green’s Functions with Applications

particular solution to the wave equation when it is forced by a point source both in space and time. They then derived the Green’s function for the ndimensional wave equation as well as the three-dimensional wave equation with dispersion. Shortly after Van der Pol’s book, P. M. Morse and H. Feshbach further developed the theory of Green’s functions as it applies to the wave equation. From Van der Pol’s definition, they obtained the reciprocity relation and the free-space Green’s function in one, two, and three dimensions.57 Significantly Morse and Feshbach did not use transform methods but derived their results from heuristic arguments and repeated integration. The use of transform methods to find Green’s function for the wave equation rapidly occurred after the publication of Van der Pol’s book. For example, in Friedlander’s examination58 of pulses by a circular cylinder, he found the approximate Green’s function for the two-dimensional wave equation exterior to a cylinder of radius 1 where gr (1, θ , t|ρ,θ $ , τ ) = 0. He used many of the techniques in this book. In particular, Laplace transforms were used to eliminate time and Fourier series were employed to give the θ dependence. An important class of wave propagation involves the diffraction of a direct wave by an infinitesimally thin barrier along the x-axis. In 1935 L. Cagniard used Laplace transforms to find the diffraction of a step function direction by a half-plane.59 The Green’s function follows by simply taking the time derivative of Cagnaird’s solution.60 The Green’s function for the corresponding two-dimensional problem is more difficult. R. D. Turner found the earliest representation using Laplace transforms.61 G. Schouten62 has given a closed form solution for the Green’s function. See Section 4.8. 1.6 ORDINARY DIFFERENTIAL EQUATIONS The application of Green’s functions to ordinary differential equations began in 1894. Noting the use of Green’s functions in solving the two and three 57

Morse, P. M., and H. Feshbach, 1953: Methods of Theoretical Physics. Part I: Chapters 1 to 8. McGraw-Hill, 997 pp. 58 Friedlander, F. G., 1954: Diffraction of pulses by a circular cylinder. Commun. Pure Appl. Math., 7, 705–732. 59 Cagniard, L., 1935: Diffraction d’une onde progressive par un ´ ecran en forme de demiplan. J. Phys. Radium, Ser. 7 , 6, 310–318; Cagniard, L., 1935: Diffraction d’une onde harmonique par un ´ ecran en forme de demi-plan. J. Phys. Radium, Ser. 7 , 6, 369–372. 60 Schouten, G., 1999: Two-dimensional effects in the edge sound of vortices and dipoles. J. Acoust. Soc. Am., 106, 3167–3177. 61

Turner, R. D., 1956: The diffraction of a cylindrical pulse by a half-plane. Q. Appl. Math., 14, 63–73. 62

Schouten, op. cit., p. 3170.

Historical Development

21

dimensional Poisson equation, H. Burkhardt63 (1861–1914) asked whether they could be used to solve d2 y = f (x), dx2

a < x < b.

(1.6.1)

He showed that the solution to this problem can be written y(x) = −

"

a

x

(b − x)(ξ − a) f (ξ) dξ − b−a

which he wrote y(x) = − where g(x|ξ) =

"

"

b x

(b − ξ)(x − a) f (ξ) dξ, b−a

(1.6.2)

b

g(x|ξ)f (ξ) dξ,

(1.6.3)

(b − x> )(x< − a) . b−a

(1.6.4)

a

Burkhardt’s Green’s function g(x|ξ) enjoyed the classic properties: • The Green’s function satisfies the differential equation g $$ = 0. • The Green’s function is finite and continuous on the interval (a, b) except for x = ξ. • The first derivative of g(x|ξ) is continuous except at x = ξ where it possesses two distinct values that differ by 1. • g(a|ξ) = g(b|ξ) = 0. • The Green’s function has the symmetry property: g(x|ξ) = g(ξ|x).

In 1903 M. Bˆocher64 (1867–1918) gave the properties of the Green’s function for homogeneous, linear, n-th order ordinary differential equation dn y dn−1 y dy + a1 n−1 + · · · + an−1 + an y = 0, n dx dx dx

a ≤ x ≤ b,

(1.6.5)

where a1 , a2 , . . . , an are real functions of the real variable x and the coefficients of the n linearly independent boundary conditions are constants. William M. Whyburn (1901–1972) would list65 the properties of the Green’s function for the general self-adjoint linear, second-order homogeneous differential equation % & d dy p(x,λ ) − q(x,λ )y = 0, a < x < b, (1.6.6) dx dx 63 Burkhardt, M. H., 1894: Sur les fonctions de Green relatives a un domaine d’une dimension. Bull. Soc. Math., 22, 71–75. 64

Bˆ ocher, M., 1901: Green’s function in space of one dimension. Bull. Amer. Math. Soc., Ser. 2, 7, 297–299. 65 Whyburn, W. M., 1924: An extension of the definition of the Green’s function in one dimension. Ann. Math., Ser. 2 , 26, 125–130.

22

Green’s Functions with Applications

Figure 1.6.1: Educated at Harvard and G¨ ottingen, Maxime Bˆ ocher (1867–1918) spent his entire career teaching at Harvard. He published approximately 100 papers on differential equations, series and algebra. Photograph courtesy of the American Mathematical Society (www.ams.org).

with self-adjoint boundary conditions. The jump in the first derivative at the point of excitation x = ξ now becomes 1/p(x,λ ). Burkhardt also found the Green’s function for Equation 1.6.1 when the boundary condition reads y(a) + χy $ (a) = 0,

y(b) + χy $ (b) = 0.

(1.6.7)

In that case, (b + χ − x> )(x< − a + χ) . (1.6.8) b − a + 2χ He noted special difficulties when χ = ∞ and χ = (a − b)/2. In the first case, the boundary condition becomes g $ (a|ξ) = g $ (b|ξ) = 0. For the solution to /b exist, a f (x) dx = 0. In that case, " x y(x) = f (ξ)(x − ξ) dξ + λ, (1.6.9) g(x|ξ) =

a

where λ is arbitrary. In the second case, the solution exists if " b, a+b x− f (x) dx = 0. 2 a

(1.6.10)

Historical Development

23

If this condition holds, then y(x) =

"

x a

, a+b f (ξ)(x − ξ) dξ + λ x − . 2

(1.6.11)

All of these results were merely stated, not proved. In 1911 M. Bˆocher66 (1867–1918) provided the mathematical justification for these results. Ince,67 in his great treatise on ordinary differential equations, introduced these results to the general community. In his Section 11.1 he proved that the Green’s function for Equation 1.6.5 exists and is unique. Furthermore he showed that the Green function is symmetrical g(x|ξ) = g(ξ|x) if the ordinary differential equation is self-adjoint. In Section 11.11 Ince further proved that y(x) =

"

b

g(x|ξ)f (ξ) dξ

(1.6.12)

a

is the solution to the non-homogeneous equation dn y dn−1 y dy + a + · · · + an−1 + an y = f (x), 1 n n−1 dx dx dx

a ≤ x ≤ b.

(1.6.13)

As we will see in Section 3.4 an important concept in the theoretical development of Green’s functions as they apply to ordinary differential equations involves the adjoint . In 1908 G. D. Birkhoff 68 formulated the Green’s function in terms of the adjoint differential equations and the eigenvalues and eigenfunctions of the differential equations. He expressed the Green’s function as a ratio of determinants in terms of linearly independent solutions to the differential equation and the boundary conditions. Then, in 1909 E. Bounitzky69 extended Birkhoff results (although he did not reference his paper) to systems of first-order differential equations. Of all of the possible ordinary differential equations that possess a Green’s function, several are of fundamental importance because these arise in important physical problems. One of these is d2 u + [0(x) + λ2 ]u = 0, dx2

0 < x < 1,

(1.6.14)

66 Bˆ ocher, M., 1911/12: Boundary problems and Green’s functions for linear differential and difference equations. Ann. Math., Ser. 2 , 13, 71–88. 67

Ince, E. L., 1956: Ordinary Differential Equations. Dover, 558 pp.

68 Birkhoff, G. D., 1908: Boundary value and expansion problems of ordinary linear differential equations. Trans. Amer. Math. Soc., 9, 373–395. 69 Bounitzky, E., 1909: Sur la fonction de Green des ´ equations diff´ erentielles lin´ eaires ordinaires. J. Math. Pures Appl., Ser. 6 , 5, 65–126.

24

Green’s Functions with Applications

with u(0) = u(1) = 0. In 1911 Hilb70 (1882–1929) wrote down the Green’s functions in terms of linearly homogeneous solutions to Equation 1.6.14. Furthermore, he found the bilinear representation of the Green’s function, g(x|ξ) =

( ϕj (ξ)ϕj (x) j

λ2j − λ2

,

(1.6.15)

where λj and ϕj (x) are the eigenvalues and eigenfunctions, respectively, of the boundary-value problem. Another differential equation that arises from physics is the simple harmonic oscillator. In this case, the Green’s function is governed by g $$ + k 2 g = δ(x − ξ).

(1.6.16)

In 1910 A. Sommerfeld (1868–1951) examined the solution71 to Equation 1.6.16 over a finite interval 0 < x < L when the boundary conditions are g(0|ξ) = g(L|ξ) = 0. He expressed the Green’s function by the expansion g(x|ξ) =

∞ ( um (x)um (ξ) , 2 k 2 − km m=1

(1.6.17)

where km and um (x) are the eigenvalues and eigenfunctions of the SturmLiouville problem: u$$m + k 2 um = 0 with um (0) = um (L) = 0. The bilinear expansion found by Sommerfeld for the harmonic oscillator is a simple example of the general Sturm-Liouville problem: % & d du − p(x) + q(x)u = λu, 0 < x < 1, (1.6.18) dx dx with u(0) = u(1) = 0. In 1911, Adolf Kneser72 (1862–1930) showed how the symmetric Green’s function for this problem could be formulated as the homogeneous integral equation " 1 u(ξ) = λ g(x|ξ)u(x) dx. (1.6.19) 0

He showed that this integral equation lead directly to the general bilinear formula ∞ ( ψm (x)ψm (ξ) g(x|ξ) = , (1.6.20) λm m=1 ¨ Hilb, E., 1911: Uber Reihenentwicklungen nach den Eigenfunktionen linearer Diffeter rentialgleichungen 2 Ordnung. Math. Ann., 71, 76–87. 70

71 Sommerfeld, A., 1910: Die Greensche Funktion der Schwingungsgleichungen f¨ ur ein beliebiges Gebiet. Phys. Z., 11, 1057–1066. 72

Kneser, op. cit.

Historical Development

25

where λm and ψm (x) are the mth eigenvalue and eigenfunction, respectively, of Equation 1.6.18. In the following year Sommerfeld considered the case when the interval is infinite −∞ < x < ∞. He showed73 that the Green’s function in this case is 1 ±ik(x−ξ) g(x|ξ) = e , (1.6.21) 2ik where the positive sign holds if x > ξ and the negative sign holds when x < ξ. To find this solution, Sommerfeld74 introduced his famous “radiation condition” that requires that energy always propagates out to infinity. Another important ordinary differential equation whose Green’s function was studied in the early twentieth century is the one that governs the deflection of a beam75 with clamped ends: g iv = −δ(x − ξ),

0 < x < L,

(1.6.22)

with the boundary conditions g(0|ξ) = g(L|ξ) = g $$ (0|ξ) = g $$ (L|ξ) = 0.

(1.6.23)

In Chapter 3 we will show that g(x|ξ) = −

$ x< (x> − L) # 2 x< + x2> − 2Lx> . 6L

(1.6.24)

A quick check shows that Equation 1.6.24 satisfies the ordinary differential equation and boundary conditions. Furthermore, g $ (ξ|ξ) and g $$ (ξ|ξ) is continuous while 'ξ + d3 g '' = −1. (1.6.25) dx3 ' − ξ

76

In 1914 A. Kneser published a paper on the integral equation that expresses the vibrations of a string where the right boundary is attached to a mass and a spring. His analysis involved solving the Green’s function problem (in modern notation) of g $$ + λ2 g = −δ(x − ξ),

0 < x, ξ < L,

(1.6.26)

73 Sommerfeld, A., 1912: Die Greensche Funktion der Schwingungsgleichung. Jahrber. Deutsch. Math.-Verein., 21, 309–353. 74 Schot, S. M., 1992: Eighty years of Sommerfeld’s radiation condition. Hist. Math., 19, 385–401. 75

See Section 1.22 and 1.23 in Bateman, H., 1959: Partial Differential Equations of Mathematical Physics. Cambridge, 522 pp. See also Von Mises, R., Ph. Frank, H. Weber, and B. Riemann, 1925: Die Differential- und Integralgleichungen der Mechanik und Physik. Vol I. Braunschweig, F. Vieweg, 687 pp. 76 Kneser, A., 1914: Belastete Integralgleichungen. Rend. Circ. Matem. Palermo, 37, 169–197.

26 with

Green’s Functions with Applications

g(0|ξ) = g $ (L|ξ) + (p − qλ2 )g(L|ξ) = 0,

(1.6.27)

where λ &= λn , the eigenvalue of the system, and q > 0. This problem is of particular interest because λ appears in the differential equation and the boundary condition. He found that the corresponding Green’s function is g(x|ξ) =

{(qλ2 − p) sin[λ(L − x> )] − λ cos[λ(L − x> )]} sin(λx< ) . (1.6.28) λ(qλ2 − p) sin(λL) − λ2 cos(λL)

He also found the bilinear expansion for this Green’s function. In the papers cited above, the differential equations were incompatible - there are no nonvanishing solutions which satisfy both the homogeneous differential equation and the boundary conditions. However, as early as 1904, Hilbert77 gave simple examples of compatible differential equations for linear second-order, boundary-value problems of the form L[u] = [p(t)u$ ]$ − q(t)u = 0.

(1.6.29)

He then constructed a generalized Green’s function (Greensche Funktionen im erweiterten Sinne) to treat such cases. In 1909, Westfall78 proved Hilbert’s results. Almost two decades later, Elliott79 generalized Hilbert’s results by finding the generalized Green’s function for a nth order differential system. His results are not limited to self-adjoint systems. Since these early studies, the study of generalized Green’s functions has languished. Loud80 used regular differential operator theory to explain generalized Green’s functions and developed techniques to find them. In 1977 Locker81 explored the properties of generalized Green’s functions.

77 See pp. 44−45 in Hilbert, D., 1912: Grundz¨ uge einer allgemeiner Theorie der linearen Integralgleichungen. Leipzig, B. G. Teubner, 312 pp. 78 Westfall, W. D. A., 1909: Existence of the generalized Green’s function. Ann. Math., Ser. 2 , 10, 177–180. 79 Elliott, W. W., 1928: Generalized Green’s function for compatible differential systems. Amer. J. Math., 50, 243–258; Elliott, W. W., 1929: Green’s function for differential systems containing a parameter. Amer. J. Math., 51, 397–416. 80

Loud, W. S., 1970: Some examples of generalized Green’s functions and generalized Green’s matrices. SIAM Rev., 12, 194–210. 81 Locker, J., 1977: The generalized Green’s function for the nth order linear differential operator. Trans. Amer. Math. Soc., 228, 243–268.

g(x

) f( ) L x

Chapter 2 Background Material One of the fundamental problems of field theory1 is the construction of solutions to linear differential equations when there is a specified source and the differential equation must satisfy certain boundary conditions. The purpose of this book is to show how Green’s functions provide a method for obtaining these solutions. In this chapter, some of the mathematical aspects necessary for developing this technique are presented. 2.1 FOURIER TRANSFORM The Fourier transform is the natural extension of Fourier series to a function f (t) of infinite period and is defined by the pair of integrals: f (t) =

1 2π

and F (ω) =

"

"

∞

F (ω)eiωt dω,

(2.1.1)

−∞ ∞

f (t)e−iωt dt.

(2.1.2)

−∞

Equation 2.1.2 is the Fourier transform of f (t) while Equation 2.1.1 is the inverse Fourier transform, which converts a Fourier transform back to f (t). 1

Any theory in which the basic quantities are fields, such as electromagnetic theory.

27

28

Green’s Functions with Applications

Table 2.1.1: Some General Properties of Fourier Transforms

function, f (t)

Fourier transform, F(ω)

c1 f (t) + c2 g(t)

c1 F (ω) + c2 G(ω)

2. Complex conjugate

f ∗ (t)

F ∗ (−ω)

3. Scaling

f (αt)

F (ω/α)/|α|

4. Delay

f (t − τ )

e−iωτ F (ω)

5. Frequency translation

eiω0 t f (t)

F (ω − ω0 )

6. Duality in time and frequency

F (t)

2πf (−ω)

7. Time differentiation

f $ (t)

iωF (ω)

1. Linearity

Hamming2 suggested the following analog in understanding the Fourier transform. Let us imagine that f (t) is a light beam. Then the Fourier transform, like a glass prism, breaks up the function into its component frequencies ω, each of intensity F (ω). In optics, the various frequencies are called colors; by analogy the Fourier transform gives us the color spectrum of a function. On the other hand, the inverse Fourier transform blends a function’s spectrum to give back the original function. • Example 2.1.1 Repeatedly in this book, we must find the Fourier transform of the derivative of a function f (t) that is differentiable for all t and vanishes as t → ± ∞. From the definition of the Fourier transform, " ∞ $ F[f (t)] = f $ (t)e−iωt dt (2.1.3) −∞ " ∞ '∞ = f (t)e−iωt '−∞ + iω f (t)e−iωt dt (2.1.4) = iωF (ω),

2

−∞

Hamming, R. W., 1977: Digital Filters. Prentice-Hall, p. 136.

(2.1.5)

Background Material

29

where F (ω) is the Fourier transform of f (t). Similarly, F[f $$ (t)] = −ω 2 F (ω).

(2.1.6) # "

In principle, we can compute any Fourier transform from the definition. However, it is far more efficient to derive some simple relationships that relate transforms to each other. Some of the most useful properties are given in Table 2.1.1. Although we can find the inverse by direct integration or partial fractions, in many instances, the Fourier transform does not lend itself to these techniques. On the other hand, if we view the inverse Fourier transform as a line integral along the real axis in the complex ω-plane, then we can use complex variable theory to evaluate the integral. To this end, we rewrite the inversion integral, Equation 2.1.1, as " ∞ 0 " 1 1 1 f (t) = F (ω)eitω dω = F (z)eitz dz − F (z)eitz dz, 2π −∞ 2π C 2π CR (2.1.7) where C denotes a closed contour consisting of the entire real axis plus a new contour CR that joins the point (∞, 0) to (−∞, 0). There are countless possibilities for CR . For example, it could be the loop (∞, 0) to (∞, R) to (−∞, R) to (−∞,/0) with R > 0. However, any choice of CR must be such that we can compute CR F (z)eitz dz. When we take that constraint into account, the number of acceptable contours decreases to just a few. The best is given by Jordan’s lemma: Jordan’s lemma: Suppose that, on a circular arc CR with radius R and center at the origin, f (z) → 0 uniformly as R → ∞. Then " (1) lim f (z)eimz dz = 0, (m > 0) (2.1.8) R→∞

CR

if CR lies in the first and/or second quadrant; " (2) lim f (z)e−imz dz = 0, R→∞

if CR lies in the third and/or fourth quadrant; " (3) lim f (z)emz dz = 0, R→∞

(m > 0)

(m > 0)

(2.1.10)

CR

if CR lies in the second and/or third quadrant; and " (4) lim f (z)e−mz dz = 0, (m > 0) R→∞

(2.1.9)

CR

CR

(2.1.11)

30

Green’s Functions with Applications

if CR lies in the first and/or fourth quadrant. The proof is given elsewhere. Consider now the following inversions of Fourier transforms: • Example 2.1.2 Let us find the inverse for F (ω) =

1 , ω 2 − 2ibω − a2 − b2

a, b > 0.

From the inversion integral, " ∞ 1 eitω f (t) = dω, 2π −∞ ω 2 − 2ibω − a2 − b2

(2.1.12)

(2.1.13)

or 0

1 2π

eitz 1 dz − 2 z − 2ibz − a2 − b2 2π

"

eitz dz, − 2ibz − a2 − b2 C CR (2.1.14) where C denotes a closed contour consisting of the entire real axis plus CR . Because f (z) = 1/(z 2 − 2ibz − a2 − b2 ) tends to zero uniformly as |z| → ∞ and m = t, the second integral in Equation 2.1.14 vanishes by Jordan’s lemma if CR is a semicircle of infinite radius in the upper half of the z-plane when t > 0 and a semicircle in the lower half of the z-plane when t < 0. Next, we must find the location and nature of the singularities. They are located at f (t) =

z 2 − 2ibz − a2 − b2 = 0,

or

z2

z = ±a + bi.

Therefore we can rewrite Equation 2.1.14 as 0 1 eitz f (t) = dz. 2π C (z − a − bi)(z + a − bi)

(2.1.15)

(2.1.16)

Thus, all of the singularities are simple poles. Consider now t > 0. As stated earlier, we close the line integral with an infinite semicircle in the upper half-plane. See Figure 2.1.1. Inside this closed contour there are two singularities: z = ±a + bi. For these poles, , eitz Res 2 ; a + bi z − 2ibz − a2 − b2 eitz = lim (z − a − bi) (2.1.17) z→a+bi (z − a − bi)(z + a − bi) =

eiat e−bt e−bt = [cos(at) + i sin(at)], 2a 2a

(2.1.18)

Background Material

31

y

-a+bi

CR for t > 0

a+bi

x

original contour

CR for t < 0

Figure 2.1.1: Contour used to find the inverse of the Fourier transform, Equation 2.1.12. The contour C consists of the line integral along the real axis plus CR .

where we used Euler’s formula to eliminate eiat . Similarly, , Res

eitz e−bt ; −a + bi = − [cos(at) − i sin(at)]. z 2 − 2ibz − a2 − b2 2a

(2.1.19)

Consequently, the inverse Fourier transform can be found from Equation 2.1.16 after applying the residue theorem and equals f (t) = −

e−bt sin(at) 2a

(2.1.20)

for t > 0. For t < 0 the semicircle is in the lower half-plane because the contribution from the semicircle vanishes as R → ∞. Because there are no singularities within the closed contour, f (t) = 0. Therefore, we can write in general that f (t) = −

e−bt sin(at)H(t). 2a

(2.1.21) # "

• Example 2.1.3 Let us find the inverse of the Fourier transform F (ω) = where a is real and positive.

e−ωi , ω 2 + a2

(2.1.22)

32

Green’s Functions with Applications From the inversion integral, "

∞

1 2π

ei(t−1)ω 1 dω = ω 2 + a2 2π

0

ei(t−1)z 1 dz − z 2 + a2 2π

"

ei(t−1)z dz, 2 2 −∞ C CR z + a (2.1.23) where C denotes a closed contour consisting of the entire real axis plus CR . The contour CR is determined by Jordan’s lemma because 1/(z 2 + a2 ) → 0 uniformly as |z| → ∞. Since m = t − 1, the semicircle CR of infinite radius lies in the upper half-plane if t > 1 and in the lower half-plane if t < 1. Thus, if t > 1, f (t) =

% i(t−1)z & 1 e e−a(t−1) f (t) = (2πi)Res 2 ; ai = , 2π z + a2 2a

(2.1.24)

whereas for t < 1, % i(t−1)z & 1 e ea(t−1) f (t) = (−2πi)Res 2 ; −ai = . 2π z + a2 2a

(2.1.25)

The minus sign in front of the 2πi arises from the clockwise direction or negative sense of the contour. We can write the inverse as the single expression f (t) =

e−a|t−1| . 2a

(2.1.26) # "

As with Laplace transforms, we can use Fourier transforms to solve differential equations. However, this method gives only the particular solution and we must find the complementary solution separately. Consider the differential equation y $ + y = 12 e−|t| ,

−∞ < t < ∞.

(2.1.27)

Taking the Fourier transform of both sides of Equation 2.1.27, iωY (ω) + Y (ω) =

1 2

"

0

et−iωt dt +

−∞

1 2

"

∞

e−t−iωt dt =

0

1 , ω2 + 1

(2.1.28)

where we used the time differentiation rule from Table 2.1.1 to obtain the transform of y $ and Y (ω) = F [y(t)]. Therefore, Y (ω) =

1 . (ω 2 + 1)(1 + ωi)

(2.1.29)

Background Material

33

Applying the inversion integral to Equation 2.1.29, y(t) =

1 2π

"

∞ −∞

(ω 2

eitω dω. + 1)(1 + ωi)

(2.1.30)

We evaluate Equation 2.1.30 by contour integration. For t > 0 we close the line integral with an infinite semicircle in the upper half of the ω-plane. The integration along this arc equals zero by Jordan’s lemma. Within this closed contour we have a second-order pole at z = i. Therefore, %

& % & eitz d eitz 2 Res ; i = lim (z − i) z→i dz (z 2 + 1)(1 + zi) i(z − i)2 (z + i) −t −t te e = + , 2i 4i and

% −t & 1 te e−t e−t y(t) = (2πi) + = (2t + 1). 2π 2i 4i 4

(2.1.31) (2.1.32)

(2.1.33)

For t < 0, we again close the line integral with an infinite semicircle, but this time it is in the lower half of the ω-plane. The contribution from the line integral along the arc vanishes by Jordan’s lemma. Within the contour, we have a simple pole at z = −i. Therefore, % Res

& eitz eitz et ; −i = lim (z + i) = − , z→−i (z 2 + 1)(1 + zi) i(z + i)(z − i)2 4i

and y(t) =

, t1 e et (−2πi) − = . 2π 4i 4

(2.1.34)

(2.1.35)

The minus sign in front of the 2πi results from the clockwise direction or negative sense in which the closed contour is taken. Using the Heaviside step function, we can combine Equation 2.1.33 and Equation 2.1.35 into the single expression y(t) = 14 e−|t| + 12 te−t H(t). (2.1.36) Note that we found only the particular or forced solution to Equation 2.1.27. The most general solution therefore requires that we add the complementary solution Ae−t , yielding y(t) = Ae−t + 14 e−|t| + 12 te−t H(t).

(2.1.37)

The arbitrary constant A would be determined by the initial condition, which we have not specified. Just as we can solve ordinary differential equations by Fourier transforms, similar considerations hold for partial differential equations. To illustrate this

34

Green’s Functions with Applications

technique, let us calculate the sound waves3 radiated by a sphere of radius a whose surface expands radially with an impulsive acceleration v0 δ(t). If we assume radial symmetry, the wave equation is , ∂u 1 ∂ 2u r2 = 2 2, ∂r c ∂t

1 ∂ r2 ∂r

(2.1.38)

subject to the boundary condition ∂u = −ρv0 δ(t) ∂r

(2.1.39)

at r = a, where u(r, t) is the pressure field, c is the speed of sound, and ρ is the average density of the fluid. Assuming that the pressure field possesses a Fourier transform, we may re-express it by the Fourier integral u(r, t) =

1 2π

"

∞

U (r,ω )eiωt dω.

(2.1.40)

−∞

Substituting into Equation 2.1.38 and Equation 2.1.39, they become 1 2π

"

∞

−∞

and

%

1 d r2 dr

1 2π

"

∞

−∞

,

r2

%

& dU (a,ω ) + ρv0 eiωt dω = 0, dr

dU dr

-

& + k02 U eiωt dω = 0,

(2.1.41)

(2.1.42)

with k0 = ω/c. Because Equation 2.1.41 and Equation 2.1.42 must be true for any time t, the bracketed quantities must vanish and we have , 1 d 2 dU r + k02 U = 0, r2 dr dr

with

dU (a,ω ) = −ρv0 . dr

(2.1.43)

The most general solution to the ordinary differential equation is eik0 r e−ik0 r + B(ω) , r r

(2.1.44)

& eiωt+ik0 r eiωt−ik0 r + B(ω) dω. r r

(2.1.45)

U (r,ω ) = A(ω) and u(r, t) =

1 2π

"

∞ −∞

%

A(ω)

3 See Hodgson, D. C., and J. E. Bowcock, 1975: Billet expansion as a mechanism for noise production in impact forming machines. J. Sound Vib., 42, 325–335.

Background Material

35

At this point, we note that the first and second terms on the right side of Equation 2.1.44 represent inwardly and outwardly propagating waves, respectively. Because there is no source of energy at infinity, the inwardly propagating wave is nonphysical and we discard it. Upon substituting the solution into the boundary condition with A(ω) = 0, U (r,ω ) =

ρa2 v0 e−iω(r−a)/c , 2πr 1 + iωa/c

(2.1.46)

"

(2.1.47)

or 1 u(r, t) = 2π

"

∞

iωt

U (r,ω )e

−∞

ρa2 v0 dω = 2πr

∞

−∞

eiω[t−(r−a)/c] dω. 1 + iωa/c

To evaluate Equation 2.1.47, we employ the residue theorem. For t < (r−a)/c Jordan’s lemma dictates that we close the line integral by an infinite semicircle in the lower half-plane. For t > (r − a)/c, we close the contour with an semicircle in the upper half-plane. The final result is u(r, t) =

% , -& , ρacv0 c r−a r−a exp − t− H t− . r a c c

(2.1.48)

In this example, we eliminated the temporal dependence by using Fourier transforms. The ordinary differential equation was then solved using homogeneous solutions. In those cases when the differential equation is nonhomogeneous, an alternative would be to solve the ordinary differential equation by Fourier or Hankel transforms. This is particularly true in the case of Green’s functions because the forcing function is a delta function. In the two-dimensional case, we obtain an algebraic equation that we solve to find the joint transform. We then compute the inverses successively. The order in which the transforms are inverted depends upon the problem. This repeated application of transforms or Fourier series to a linear partial differential equation to reduce it to an algebraic or ordinary differential equation can be extended to higher spatial dimensions. 2.2 LAPLACE TRANSFORM If a function is nonzero only when t > 0, we can replace the Fourier transform with the Laplace transform. It is particularly useful in solving initial-value problems involving linear, constant coefficient, ordinary and partial differential equations. We summarize the main points here. Consider a function f (t) such that f (t) = 0 for t < 0. Then the Laplace integral " L[f (t)] = F (s) =

∞

0

f (t)e−st dt

(2.2.1)

36

Green’s Functions with Applications

defines the Laplace transform of f (t), which we write L[f (t)] or F (s). The Laplace transform of f (t) exists, for sufficiently large s, provided f (t) satisfies the following conditions: • f (t) = 0 for t < 0,

• f (t) is continuous or piece-wise continuous in every interval,

• tn |f (t)| < ∞ as t → 0 for some number n, where n < 1,

• e−s0 t |f (t)| < ∞ as t → ∞, for some number s0 , the abscissa of convergence. • Example 2.2.1 Let us find the Laplace transform for the Heaviside step function: H(t − a) =

!

1, 0,

t > a, t < a,

(2.2.2)

where a ≥ 0. The Heaviside step function is essentially a bookkeeping device that gives us the ability to “switch on” and “switch off” a given function. For example, if we want a function f (t) to become nonzero at time t = a, we represent this process by the product f (t)H(t − a). From the definition of the Laplace transform, L[H(t − a)] =

"

∞

e−st dt =

a

e−as , s

s > 0.

(2.2.3) # "

• Example 2.2.2 Let us find the Laplace transform of the Dirac delta function or impulse function. From Equation 2.2.2, the Laplace transform of the delta function is L[δ(t − a)] =

"

0

∞

n n→∞ 2

δ(t − a)e−st dt = lim

"

a+1/n

e−st dt

(2.2.4)

a−1/n

, n −as+s/n −as−s/n = lim e −e (2.2.5) n→∞ 2s , n e−as s s2 s s2 = lim 1 + + 2 + · · · − 1 + − 2 + · · · (2.2.6) n→∞ 2s n 2n n 2n = e−as .

(2.2.7) # "

Background Material

37

Table 2.2.1: Some General Properties of Laplace Transforms with a > 0

1. Linearity

function, f (t)

Laplace transform, F(s)

c1 f (t) + c2 g(t)

c1 F (s) + c2 G(s)

f (t/a)/a

F (as)

ebt f (t)

F (s − b)

f (t − a)H(t − a)

e−as F (s)

f (n) (t)

sn F (s) − sn−1 f (0) −sn−2 f $ (0) − · · · −f (n−1) (0)

2. Scaling 3. Multiplication by ebt 4. Translation 5. Differentiation

/t

6. Integration 7. Convolution

f (τ ) dτF

(s)/s

f (t − τ )g(τ ) dτF

(s)G(s)

0

/t 0

• Example 2.2.3 Although we could compute Equation 2.2.1 for every function that has a Laplace transform, these results have already been tabulated and are given in many excellent tables.4 However, there are four basic transforms that the reader should memorize. They are " ∞ eat e−st dt = e−(s−a)t dt 0 0 '∞ e−(s−a)t '' 1 =− = , s > a, s − a '0 s−a

L(eat ) =

L[sin(at)] = =

"

0

∞

"

∞

sin(at)e−st dt = −

a , s2 + a2

s > 0,

(2.2.8) (2.2.9)

'∞ ' e−st ' (2.2.10) [s sin(at) + a cos(at)] ' 2 2 s +a 0 (2.2.11)

4 The most complete set is given by Erd´ elyi, A., W. Magnus, F. Oberhettinger, and F. G. Tricomi, 1954: Tables of Integral Transforms, Vol I . McGraw-Hill Co., 391 pp.

38

Green’s Functions with Applications

L[cos(at)] = =

"

∞

cos(at)e

−st

0

s , s2 + a2

'∞ ' e−st dt = 2 [−s cos(at) + a sin(at)]'' (2.2.12) 2 s +a 0

s > 0,

(2.2.13)

and n

L(t ) =

"

∞

n −st

t e

dt = n!e

0

where n is a positive integer.

−st

'∞ ' tn−m ' = n! , m+1 ' (n − m)!s sn+1 0 m=0 n (

s > 0, (2.2.14) # "

The Laplace transform inherits two important properties from its integral definition. First, the transform of a sum equals the sum of the transforms: L[c1 f (t) + c2 g(t)] = c1 L[f (t)] + c2 L[g(t)].

(2.2.15)

This linearity property holds with complex numbers and functions as well. The second important property deals with derivatives. Suppose f (t) is continuous and has a piece-wise continuous derivative f $ (t). Then L[f $ (t)] =

"

0

∞

'∞ f $ (t)e−st dt = e−st f (t)'0 + s

"

∞

f (t)e−st dt

(2.2.16)

0

by integration by parts. If f (t) is of exponential order,5 e−st f (t) tends to zero as t → ∞, for large enough s, so that L[f $ (t)] = sF (s) − f (0).

(2.2.17)

Similarly, if f (t) and f $ (t) are continuous, f $$ (t) is piece-wise continuous, and all three functions are of exponential order, then L[f $$ (t)] = sL[f $ (t)] − f $ (0) = s2 F (s) − sf (0) − f $ (0).

(2.2.18)

In general, L[f (n) (t)] = sn F (s) − sn−1 f (0) − · · · − sf (n−2) (0) − f (n−1) (0)

(2.2.19)

on the assumption that f (t) and its first n − 1 derivatives are continuous, f (n) (t) is piece-wise continuous, and all are of exponential order so that the Laplace transform exists. 5

By exponential order we mean that there exist some constants, M and k, for which

|f (t)| ≤ M ekt for all t > 0.

Background Material

39

Consider the transform of the function e−at f (t), where a is any real number. Then, by definition,

or

1 2 L e−at f (t) =

"

∞

e−st e−at f (t) dt =

0

"

∞

e−(s+a)t f (t) dt,

(2.2.20)

0

1 2 L e−at f (t) = F (s + a).

(2.2.21)

Equation 2.2.21 is known as the first shifting theorem and states that if F (s) is the transform of f (t) and a is a constant, then F (s + a) is the transform of e−at f (t). • Example 2.2.4 Let us find the Laplace transform of f (t) = e−at sin(bt). Because the Laplace transform of sin(bt) is b/(s2 + b2 ), 1 2 L e−at sin(bt) =

b , (s + a)2 + b2

(2.2.22)

where we have simply replaced s by s + a in the transform for sin(bt). # " • Example 2.2.5 Let us find the inverse of the Laplace transform F (s) =

s2

s+2 . + 6s + 1

(2.2.23)

Rearranging terms, √ s+2 s+2 s+3 1 2 2 F (s) = 2 = = − √ . s + 6s + 1 (s + 3)2 − 8 (s + 3)2 − 8 2 2 (s + 3)2 − 8 (2.2.24) Immediately, from the first shifting theorem, √ √ e−3t f (t) = e−3t cosh(2 2t) − √ sinh(2 2t). 2 2

(2.2.25) # "

The second shifting theorem states that if F (s) is the transform of f (t), then e−bs F (s) is the transform of f (t −b)H(t − b), where b is real and positive.

40

Green’s Functions with Applications

To show this, consider the Laplace transform of f (t − b)H(t − b). Then, from the definition, "

∞

f (t − b)H(t − b)e−st dt " ∞ " ∞ = f (t − b)e−st dt = e−bs e−sx f (x) dx b 0 " ∞ −bs −sx =e e f (x) dx,

L[f (t − b)H(t − b)] =

(2.2.26)

0

(2.2.27) (2.2.28)

0

or L[f (t − b)H(t − b)] = e−bs F (s),

(2.2.29)

where we have set x = t − b. This theorem is of fundamental importance because it allows us to write down the transforms for “delayed” time functions. These functions “turn on” b units after the initial time. • Example 2.2.6 Let us find the inverse of the transform e−πs /[s2 (s2 + 1)]. Because e−πs e−πs e−πs = − , s2 (s2 + 1) s2 s2 + 1

−1

L

%

& , −πs , −πs e−πs e −1 e −1 =L −L s2 (s2 + 1) s2 s2 + 1 = (t − π)H(t − π) − sin(t − π)H(t − π) = (t − π)H(t − π) + sin(t)H(t − π),

(2.2.30)

(2.2.31) (2.2.32) (2.2.33)

since L−1 (1/s2 ) = t, and L−1 [1/(s2 + 1)] = sin(t). # " In this section, we turn to a fundamental concept in Laplace transforms: convolution. We begin by formally introducing the mathematical operation of the convolution product : f (t) ∗ g(t) =

"

t 0

f (t − x)g(x) dx =

"

0

t

f (x)g(t − x) dx.

(2.2.34)

Background Material

41

• Example 2.2.7 Let us find the convolution between cos(t) and sin(t). " t cos(t) ∗ sin(t) = sin(t − x) cos(x) dx 0 " t = 12 [sin(t) + sin(t − 2x)] dx 0

= =

1 2 1 2

"

t

sin(t) dx +

0

't sin(t) x'0 +

1 4

1 2

"

0

(2.2.35) (2.2.36)

t

sin(t − 2x) dx

't cos(t − 2x)'0 = 12 t sin(t).

(2.2.37) (2.2.38) # "

The reason we have introduced convolution derives from the following fundamental theorem (often called Borel’s theorem6 ). If w(t) = u(t) ∗ v(t),

(2.2.39)

W (s) = U (s)V (s).

(2.2.40)

then In other words, we can invert a complicated transform by convoluting the inverses to two simpler functions. • Example 2.2.8 Let us find the inverse of the transform , 1 1 a a = 2 × 2 (s2 + a2 )2 a s2 + a2 s + a2 1 = 2 L[sin(at)]L[sin(at)]. a Therefore, % & " t 1 1 −1 L = 2 sin[a(t − x)] sin(ax) dx (s2 + a2 )2 a 0 " t " t 1 1 = 2 cos[a(t − 2x)] dx − 2 cos(at) dx 2a 0 2a 0 't 't ' ' 1 1 = − 3 sin[a(t − 2x)]'' − 2 cos(at) x'' 4a 0 2a 0 1 = 3 [sin(at) − at cos(at)]. 2a 6

(2.2.41) (2.2.42)

(2.2.43) (2.2.44) (2.2.45) (2.2.46)

´ 1901: Le¸cons sur les s´ Borel, E., eries divergentes. Gauthier-Villars, p. 104.

42

Green’s Functions with Applications # "

The primary use of Laplace transforms is the solution of ordinary, constant coefficient, linear differential equations. For example, let us solve the initial-value problem dn y dn−1 y dy + a + · · · + an−1 + an y = f (t), 1 dtn dtn−1 dt

t > 0,

(2.2.47)

by Laplace transforms, where a1 , a2 , . . . are constants and we know the value of y, y $ , . . . , y (n−1) at t = 0. The procedure is as follows. Applying the derivative rule Equation 2.2.19 to Equation 2.2.47, we reduce the differential equation to an algebraic one involving the constants a1 , a2 , . . . , an , the parameter s, the Laplace transform of f (t), and the values of the initial conditions. We then solve for the Laplace transform of y(t), Y (s). Finally, we apply one of the many techniques of inverting a Laplace transform to find y(t). Similar considerations hold with systems of ordinary differential equations. The Laplace transform of the system of ordinary differential equations results in an algebraic set of equations containing Y1 (s), Y2 (s), . . . , Yn (s). By some method, we solve this set of equations and invert each transform Y1 (s), Y2 (s), . . . , Yn (s) in turn, to give y1 (t), y2 (t), . . . , yn (t). The following example illustrates the details of the process. • Example 2.2.9 Let us solve the ordinary differential equation y $$ + 2y $ = 8t

(2.2.48)

subject to the initial conditions that y $ (0) = y(0) = 0. Taking the Laplace transform of both sides of Equation 2.2.48, L(y $$ ) + 2L(y $ ) = 8L(t), or s2 Y (s) − sy(0) − y $ (0) + 2sY (s) − 2y(0) =

(2.2.49) 8 , s2

(2.2.50)

where Y (s) = L[y(t)]. Substituting the initial conditions into Equation 2.2.50 and solving for Y (s), 8 A B C D = 3+ 2+ + s3 (s + 2) s s s s+2 8 (s + 2)A + s(s + 2)B + s2 (s + 2)C + s3 D = 3 = . s (s + 2) s3 (s + 2)

Y (s) =

(2.2.51) (2.2.52)

Background Material

43

Matching powers of s in the numerators of Equation 2.2.52, C + D = 0, B + 2C = 0, A + 2B = 0, and 2A = 8 or A = 4, B = −2, C = 1, and D = −1. Therefore, 4 2 1 1 Y (s) = 3 − 2 + − . (2.2.53) s s s s+2 Finally, performing term-by-term inversion of Equation 2.2.53, the final solution equals y(t) = 2t2 − 2t + 1 − e−2t . (2.2.54) # " Usually, we can find the inverse of the Laplace transform F (s) by looking it up in a table. In this section, we show an alternative method that inverts Laplace transforms through the powerful method of contour integration. Consider the piece-wise differentiable function f (x) that vanishes for x < 0. We can express the function e−cx f (x) by the complex Fourier representation of %" ∞ & " ∞ 1 −cx iωx −ct −iωt f (x)e = e e f (t)e dt dω, (2.2.55) 2π −∞ 0 for any value of the real constant c, where the integral I=

"

∞

0

e−ct |f (t)| dt

(2.2.56)

exists. By multiplying both sides of Equation 2.2.55 by ecx and bringing it inside the first integral, f (x) =

1 2π

"

∞

e(c+ωi)x

−∞

%"

∞

0

& f (t)e−(c+ωi)t dt dω.

(2.2.57)

With the substitution z = c + ωi, where z is a new, complex variable of integration, %" ∞ & " c+∞i 1 zx −zt f (x) = e f (t)e dt dz. (2.2.58) 2πi c−∞i 0 The quantity inside the square brackets is the Laplace transform F (z). Therefore, we can express f (t) in terms of its transform by the complex contour integral:

f (t) =

1 2πi

"

c+∞i

c−∞i

F (z)etz dz.

(2.2.59)

44

Green’s Functions with Applications

This line integral, Bromwich’s integral ,7 runs along the line x = c parallel to the imaginary axis and c units to the right of it, the so-called Bromwich contour. We select the value of c sufficiently large so that Equation 2.2.59 exists; subsequent analysis shows that this occurs when c is larger than the real part of any of the singularities of F (z). We must now evaluate the contour integral. Because of the power of the residue theorem in complex variables, the contour integral is usually transformed into a closed contour through the use of Jordan’s lemma. The following examples illustrate the proper use of Equation 2.2.59. • Example 2.2.10 Let us invert F (s) =

e−3s . s2 (s − 1)

(2.2.60)

From Bromwich’s integral, f (t) = =

1 2πi 1 2πi

"

c+∞i

e(t−3)z dz z 2 (z − 1)

c−∞i 0 (t−3)z C

e 1 dz − z 2 (z − 1) 2πi

(2.2.61) "

CR

e(t−3)z dz, − 1)

z 2 (z

(2.2.62)

where CR is a semicircle of infinite radius in either the right or left half of the z-plane and C is the closed contour that includes CR and Bromwich’s contour. See Figure 2.2.1. Our first task is to choose an appropriate contour so that the integral along CR vanishes. By Jordan’s lemma, this requires a semicircle in the right half-plane if t − 3 < 0 and a semicircle in the left half-plane if t − 3 > 0. Consequently, by considering these two separate cases, we force the second integral in Equation 2.2.62 to zero and the inversion simply equals the closed contour. Consider the case t < 3 first. Because Bromwich’s contour lies to the right of any singularities, there are no singularities within the closed contour and f (t) = 0. Consider now the case t > 3. Within the closed contour in the left halfplane, there is a second-order pole at z = 0 and a simple pole at z = 1. Therefore, f (t) = Res

%

& % (t−3)z & e(t−3)z e ; 0 + Res ; 1 , z 2 (z − 1) z 2 (z − 1)

(2.2.63)

7 Bromwich, T. J. I’A., 1916: Normal coordinates in dynamical systems. Proc. London Math. Soc., Ser. 2, 15, 401–448.

Background Material

45

t>3

t 0. Consequently, the general solution to Equation 2.3.1 is y(x) = AJn (µx) + BYn (µx).

(2.3.3)

Figure 2.3.1 illustrates the functions J0 (x), J1 (x), J2 (x), and J3 (x) while Figure 2.3.2 gives Y0 (x), Y1 (x), Y2 (x), and Y3 (x). 9 The classic reference on Bessel functions is Watson, G. N., 1966: A Treatise on the Theory of Bessel Functions. Cambridge University Press, 804 pp. 10 See Press, W. H., B. P. Flannery, S. A. Teukolsky, and W. T. Vetterling, 1986: Numerical Recipes: The Art of Scientific Computing. Cambridge University Press, Section 6.4.

Background Material

51

1.0

J0 (x) J1(x) 0.5

J2(x)

J3(x)

0.0

-0.5 0.0

2.0

4.0

x

6.0

8.0

Figure 2.3.1: The first four Bessel functions of the first kind over 0 ≤ x ≤ 8.

The Bessel functions Jn and Yn enjoy certain recurrence relations where functions of order n are related to functions of order n + 1 and n − 1. If we denote Jn or Yn by Cn , the most useful relationships are Cn−1 (z) + Cn+1 (z) = 2nCn (z)/z, Cn−1 (z) − Cn+1 (z) = 2Cn$ (z), d n [z Cn (z)] = z n Cn−1 (z), dz

n = 1, 2, 3, . . . , n = 1, 2, 3, . . . , n = 1, 2, 3, . . . ,

(2.3.4) (2.3.5) (2.3.6)

and

2 d 1 −n z Cn (z) = −z −n Cn+1 (z), n = 1, 2, 3, . . . , dz with the special case of J0$ (z) = −J1 (z), and Y0$ (z) = −Y1 (z). An equation that is very similar to Equation 2.3.1 is

(2.3.7)

d2 y dy +x − (n2 + x2 )y = 0. (2.3.8) dx2 dx √ If we substitute ix = t (where i = −1 ) into Equation 2.3.8, it becomes Bessel’s equation: x2

t2

d2 y dy +t + (t2 − n2 )y = 0. 2 dt dt

(2.3.9)

52

Green’s Functions with Applications

1.0

0.5

Y0 (x)

Y1 (x)

Y2 (x)

Y3(x)

0.0

-0.5

-1.0 0.0

2.0

4.0

x

6.0

8.0

Figure 2.3.2: The first four Bessel functions of the second kind over 0 ≤ x ≤ 8.

Consequently, we can immediately write the solution to Equation 2.3.8 as y(x) = c1 Jn (ix) + c2 Yn (ix),

(2.3.10)

if n is an integer. Traditionally the solution to Equation 2.3.10 has been written y(x) = c1 In (x) + c2 Kn (x) (2.3.11) rather than in terms of Jn (ix) and Yn (ix). The function In (x) is the modified Bessel function of the first kind, of order n, while Kn (x) is the modified Bessel function of the second kind, of order n. Figure 2.3.3 illustrates I0 (x), I1 (x), I2 (x), and I3 (x), while in Figure 2.3.4 K0 (x), K1 (x), K2 (x), and K3 (x) have been graphed. Again, both In and Kn can be treated as tabulated functions.11 Note that Kn (x) has no real zeros while In (x) equals zero only at x = 0 for n > 0. As our derivation suggests, modified Bessel functions are related to ordinary Bessel functions via complex variables. In particular, Jn (iz) = in In (z) and In (iz) = in Jn (z) for z complex. As in the case of Jn and Yn , there are recurrence relations where functions of order n are related to functions of order n + 1 and n − 1. If we denote by 11

Ibid., Section 6.5.

Background Material

53

5.0

4.0

3.0

I0

2.0

I1 I2

1.0

I3

0.0 0.0

1.0

x

2.0

3.0

Figure 2.3.3: The first four modified Bessel functions of the first kind over 0 ≤ x ≤ 3.

Cn either In or enπi Kn , we have Cn−1 (z) − Cn+1 (z) = 2nCn(z)/z, Cn−1 (z) + Cn+1 (z) = 2Cn$ (z), d n [z Cn (z)] = z n Cn−1 (z), dz and

2 d 1 −n z Cn (z) = z −n Cn+1 (z), dz

n = 1, 2, 3, . . . , n = 1, 2, 3, . . . , n = 1, 2, 3, . . . ,

n = 1, 2, 3, . . . ,

(2.3.12) (2.3.13) (2.3.14)

(2.3.15)

with the special case of I0$ (z) = I1 (z), and K0 (z) = −K1 (z). Table 2.3.1 lists many of the most useful relationships that involve Bessel functions of integer order. Hankel functions are an alternative to Jn and Yn . To understand why they are useful, one of the fundamental equations studied in any differential equations course is y $$ + k 2 y = 0. (2.3.16) Its solution is generally written y(x) = A cos(kx) + B sin(kx),

(2.3.17)

54

Green’s Functions with Applications

4.0

3.0

2.0

K1

K2

K3

1.0

K0 0.0 0.0

1.0

x

2.0

3.0

Figure 2.3.4: The first four modified Bessel functions of the second kind over 0 ≤ x ≤ 3.

where A and B are arbitrary constants. Another, less often quoted, solution is y(x) = Ceikx + De−ikx , (2.3.18) where C and D are generally complex constants. The advantage of Equation 2.3.18 over Equation 2.3.17 is the convenience of dealing with exponentials rather than sines and cosines. Let us return to Equation 2.3.1: # $ x2 y $$ + xy $ + x2 − n2 y = 0,

(2.3.19)

where n is a positive (including zero) integer. Its general solution is most often written y(x) = A Jn (x) + B Yn (x). (2.3.20) Again A and B are arbitrary constants. However, another, equivalent solution is y(x) = CHn(1) (x) + DHn(2) (x), (2.3.21) where Hn(1) (x) = Jn (x) + iYn (x),

(2.3.22)

Hn(2) (x) = Jn (x) − iYn (x).

(2.3.23)

and

Background Material

55

Table 2.3.1: Some Useful Relationships Involving Bessel Functions of Integer Order

Jn−1 (z) + Jn+1 (z) =

2n Jn (z), z

n = 1, 2, 3, . . .

Jn−1 (z) − Jn+1 (z) = 2Jn$ (z), n = 1, 2, 3, . . . ; J0$ (z) = −J1 (z) % & d n z Jn (z) = z n Jn−1 (z), n = 1, 2, 3, . . . dz % & d −n z Jn (z) = −z −nJn+1 (z), n = 0, 1, 2, 3, . . . dz In−1 (z) − In+1 (z) =

2n In (z), z

n = 1, 2, 3, . . .

In−1 (z) + In+1 (z) = 2In$ (z), n = 1, 2, 3, . . . ; Kn−1 (z) − Kn+1 (z) = −

2n Kn (z), z

I0$ (z) = I1 (z) n = 1, 2, 3, . . .

Kn−1 (z) + Kn+1 (z) = −2Kn$ (z), n = 1, 2, 3, . . . ;

(2) (3) (4) (5) (6) (7)

Jn (zemπi ) = enmπi Jn (z)

K0$ (z) = −K1 (z) (8) (9)

In (zemπi ) = enmπi In (z)

(10)

cos(mnπ) In (z) cos(nπ)

(11)

−π < arg(z) ≤ π/2

(12)

π 2 < /arg(z) ≤ π

(13)

Kn (zemπi ) = e−mnπi Kn (z) − mπi In (z) = e−nπi/2 Jn (zeπi/2 ), In (z) = e3nπi/2 Jn (ze−3πi/2 ),

(1)

(1)

(2)

These functions Hn (x), Hn (x) are referred to as Bessel functions of the third kind or Hankel functions, after the German mathematician Hermann Hankel (1839–1873). Table 2.3.2 lists some of their properties. The similarity between Equation 2.3.18 and Equation 2.3.21 is clarified by the asymptotic expansions for the Hankel functions: 3 2 i(z−nπ/2−π/4) (1) Hn (z) ∼ e , (2.3.24) πz and

3

2 −i(z−nπ/2−π/4) e . (2.3.25) πz A Fourier-Bessel series is a Fourier series where Jn (·) replaces the sine and cosine. The exact form of the expansion depends upon the boundary Hn(2) (z) ∼

56

Green’s Functions with Applications

Table 2.3.2: Some Useful Recurrence Relations for Hankel Functions. d 4 n (p) 5 d 4 (p) 5 (p) (p) x Hn (x) = xn Hn−1 (x), n = 1, 2, . . . ; H0 (x) = −H1 (x) dx dx (1) d 4 −n (p) 5 (p) x Hn (x) = −x−n Hn+1 (x), dx (p)

(p)

(p)

(p)

Hn−1 (x) + Hn+1 (x) =

2n (p) H (x), x n

n = 0, 1, 2, 3, . . .

(2)

n = 1, 2, 3, . . .

(3)

n = 1, 2, 3, . . .

(4)

(p)

Hn−1 (x) − Hn+1 (x) = 2

dHn (x) , dx

condition at x = L. There are three possible cases. One of them is that y(L) = 0 and results in the condition that Jn (µk L) = 0. Another condition is y $ (L) = 0 and gives Jn$ (µk L) = 0. Finally, if hy(L) + y $ (L) = 0, then hJn (µk L) + µk Jn$ (µk L) = 0. In all of these cases, the eigenfunction expansion is the same, namely

f (x) =

∞ (

Ak Jn (µk x),

(2.3.26)

k=1

where µk is the kth positive solution of either Jn (µk L) = 0, Jn$ (µk L) = 0, or hJn (µk L) + µk Jn$ (µk L) = 0. We now need a mechanism for computing Ak . We begin by multiplying Equation 2.3.26 by xJn (µm x) dx and integrate from 0 to L. This yields ∞ (

k=1

Ak

"

L

xJn (µk x)J(µm x) dx = 0

"

L

xf (x)Jn (µm x) dx.

(2.3.27)

0

From the general orthogonality condition "

L

xJn (µk x)Jn (µm x) dx = 0 0

(2.3.28)

Background Material

57

if k &= m, Equation 2.3.27 simplifies to Am

"

0

L

xJn2 (µm x) dx =

"

L

xf (x)Jn (µm x) dx,

(2.3.29)

0

or

Ak =

1 Ck

"

L

xf (x)Jn (µk x) dx,

(2.3.30)

0

where Ck =

"

L

xJn2 (µk x) dx

0

(2.3.31)

and we replaced m by k in Equation 2.3.31. The factor Ck depends upon the boundary conditions at x = L. It can be shown that 2 Ck = 12 L2 Jn+1 (µk L), (2.3.32) if Jn (µk L) = 0. If Jn$ (µk L) = 0, then µ2k L2 − n2 2 Jn (µk L). 2µ2k

(2.3.33)

µ2k L2 − n2 + h2 L2 2 Jn (µk L), 2µ2k

(2.3.34)

Ck = Finally, Ck =

if µk Jn$ (µk L) = −hJn (µk L). All of the preceding results must be slightly modified when n = 0 and the boundary condition is J0$ (µk L) = 0 or µk J1 (µk L) = 0. This modification arises from the additional eigenvalue µ0 = 0 being present and we must add the extra term A0 to the expansion. For this case, Equation 2.3.26 becomes f (x) = A0 +

∞ (

Ak J0 (µk x),

(2.3.35)

k=1

where the equation for finding A0 is 2 A0 = 2 L

"

L

f (x) x dx.

(2.3.36)

0

Equations 2.3.30 and 2.3.32 with n = 0 give the remaining coefficients.

58

Green’s Functions with Applications

• Example 2.3.1 Let us expand f (x) = x, 0 < x < 1, in the series f (x) =

∞ (

Ak J1 (µk x),

(2.3.37)

k=1

where µk denotes the kth zero of J1 (µ). From Equation 2.3.30 and Equation 2.3.32, " 1 2 Ak = 2 x2 J1 (µk x) dx. (2.3.38) J2 (µk ) 0 However, from Equation 2.3.14, 2 d 1 2 x J2 (x) = x2 J1 (x), dx

(2.3.39)

if n = 2. Therefore, Equation 2.3.38 becomes

'µ 2x2 J2 (x) '' k 2 Ak = 3 2 = , ' µk J2 (µk ) 0 µk J2 (µk )

(2.3.40)

and the resulting expansion is x=2

∞ ( J1 (µk x) , µk J2 (µk ) k=1

0 ≤ x < 1.

(2.3.41)

Figure 2.3.5 shows the Fourier-Bessel expansion of f (x) = x in truncated form when we only include one, two, three, and four terms. # " • Example 2.3.2 Let us expand the function f (x) = x2 , 0 < x < 1, in the series f (x) =

∞ (

Ak J0 (µk x),

(2.3.42)

k=1

where µk denotes the kth positive zero of J0 (µ). From Equation 2.3.30 and Equation 2.3.32, Ak =

2 J12 (µk )

"

1

x3 J0 (µk x) dx = 0

2 µ4k J12 (µk )

"

0

µk

t3 J0 (t) dt,

(2.3.43)

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59

1.25 1.00 0.75 0.50 0.25 0.00

one term

two terms

three terms

four terms

-0.25 1.25 1.00 0.75 0.50 0.25 0.00 -0.25 0.0

0.2

0.4

x

0.6

0.8

1.0 0.0

0.2

0.4

x

0.6

0.8

1.0

Figure 2.3.5: The Fourier-Bessel series representation, Equation 2.3.41, for f (x) = x, 0 ≤ x < 1, when we truncate the series so that it includes only the first, first two, first three, and first four terms.

where we set t = µk x to obtain the second integral. We now let u = t2 and dv = tJ0 (t) dt so that integration by parts gives % & " µk 'µk 2 3 2 ' Ak = 4 2 t J1 (t) 0 − 2 t J1 (t) dt µk J1 (µk ) 0 % & " µk 2 3 2 = 4 2 µ J1 (µk ) − 2 t J1 (t) dt , µk J1 (µk ) k 0

(2.3.44) (2.3.45)

because v = tJ1 (t) from Equation 2.3.14. If we integrate by parts once more, we find that % & % & 2 2 J1 (µk ) 2J2 (µk ) 3 2 Ak = 4 2 µ J1 (µk ) − 2µk J2 (µk ) = 2 − . µk J1 (µk ) k J1 (µk ) µk µ2k (2.3.46) However, Equation 2.3.12 gives J1 (µk ) = 12 µk [J2 (µk ) + J0 (µk )] if n = 1, or J2 (µk ) = 2 J1 (µk )/µk because J0 (µk ) = 0. Therefore, Ak =

2(µ2k − 4)J1 (µk ) , µ3k J12 (µk )

and

x2 = 2

∞ ( (µ2k − 4)J0 (µk x) , µ3k J1 (µk )

0 < x < 1.

k=1

(2.3.47)

60

Green’s Functions with Applications 1.00 0.75 0.50 0.25 0.00 -0.25

one term

two terms

three terms

four terms

-0.50 1.00 0.75 0.50 0.25 0.00 -0.25 -0.50 0.0

0.2

0.4

x

0.6

0.8

1.0 0.0

0.2

0.4

x

0.6

0.8

1.0

Figure 2.3.6: The Fourier-Bessel series representation, Equation 2.3.47, for f (x) = x2 , 0 < x < 1, when we truncate the series so that it includes only the first, first two, first three, and first four terms.

Figure 2.3.6 shows the representation of x2 by the Fourier-Bessel series, Equation 2.3.47, when we truncate it so that it includes only one, two, three, or four terms. As we add each additional term in the orthogonal expansion, the expansion fits f (x) better in the “least squares” sense. 2.4 LEGENDRE POLYNOMIALS In the solution of problems involving spherical geometries, we will repeatedly encounter Legendre polynomials. In this appendix, we highlight their most useful properties. Legendre’s differential equation is (1 − x2 ) or

d2 y dy − 2x + n(n + 1)y = 0, dx2 dx

% & d 2 dy (1 − x ) + n(n + 1)y = 0, dx dx

−1 ≤ x ≤ 1, −1 ≤ x ≤ 1.

(2.4.1)

(2.4.2)

Equation 2.4.1 does not have a simple general solution. [If n = 0, then y(x) = 1 is a solution.]

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61

Table 2.4.1: The First Ten Legendre Polynomials P0 (x) = 1 P1 (x) = x P2 (x) = 12 (3x2 − 1) P3 (x) = 12 (5x3 − 3x) P4 (x) = 18 (35x4 − 30x2 + 3) P5 (x) = 18 (63x5 − 70x3 + 15x) P6 (x) = P7 (x) = P8 (x) = P9 (x) = P10 (x) =

1 (231x6 16

1 7 16 (429x

1 8 128 (6435x

1 (12155x9 128

1 10 256 (46189x

− 315x4 + 105x2 − 5)

− 693x5 + 315x3 − 35x)

− 12012x6 + 6930x4 − 1260x2 + 35)

− 25740x7 + 18018x5 − 4620x3 + 315x)

− 109395x8 + 90090x6 − 30030x4 + 3465x2 − 63)

Because we require that the solution remains finite at both x = −1 and x = 1 so that we can use them in an eigenfunction expansion, the only solutions are polynomials, called Legendre polynomials,12 and we may compute them by the power series:

Pn (x) =

m (

k=0

(−1)k

(2n − 2k)! xn−2k , 2n k!(n − k)!(n − 2k)!

(2.4.3)

where m = n/2, or m = (n − 1)/2, depending upon which is an integer. Table 2.4.1 gives the first ten Legendre polynomials. Although we have Equation 2.4.3 to compute Pn (x), there are several

12 Legendre, A. M., 1785: Sur l’attraction des sph´ ero¨ıdes homog´ enes. M´ em. math. phys. pr´ esent´ es ` a l’Acad. sci. pars divers savants, 10, 411–434. The best reference on Legendre polynomials is Hobson, E. W., 1965:The Theory of Spherical and Ellipsoidal Harmonics. Chelsea Publishing Co., 500 pp.

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Green’s Functions with Applications 1.5

P0 (x)

1.0

0.5

P3 (x)

P1 (x)

0.0

-0.5

P2 (x) -1.0 -1.0

-0.5

0.0

x

0.5

1.0

Figure 2.4.1: The first four Legendre polynomials.

alternative methods. The first method, known as Rodrigues’s formula,13 is

Pn (x) =

1 2n n!

dn 2 (x − 1)n . dxn

(2.4.4)

The second method for computing Pn (x) is given by the recurrence formulas

(n + 1)Pn+1 (x) − (2n + 1)xPn (x) + nPn−1 (x) = 0

(2.4.5)

with n = 1, 2, 3, . . ..

$ $ Pn+1 (x) − Pn−1 (x) = (2n + 1)Pn (x), n = 1, 2, 3, . . . .

(2.4.6)

Given any two of the polynomials Pn+1 (x), Pn (x), and Pn−1 (x), Equation 2.4.5 or Equation 2.4.9 yields the third. 13 Rodrigues, O., 1816: M´ ´ emoire sur l’attraction des sph´ero¨ıdes. Correspond. l’Ecole Polytech., 3, 361–385.

Background Material

63

Table 2.4.2: Some Useful Relationships Involving Legendre Polynomials Rodrigues’s formula Pn (x) =

1 2n n!

dn 2 (x − 1)n dxn

(1)

Recurrence formulas (n+1)Pn+1 (x)−(2n+1)xPn (x)+nPn−1 (x) = 0, $ $ Pn+1 (x) − Pn−1 (x) = (2n + 1)Pn (x),

n = 1, 2, 3, . . . (2) n = 1, 2, 3, . . .

(3)

Orthogonality condition "

1

Pn (x)Pm (x) dx = −1

    

0,

m &= n,

2 , m = n. 2n + 1

(4)

In addition to Rodrigues’s formula and the recurrence formulas, we have an orthogonality condition between two different Legendre polynomials, namely "

1

−1

Pn (x)Pm (x) dx =

!

0, 2 2n+1 ,

m &= n, m = n.

(2.4.10)

All of these results have been summarized in Table 2.4.2. • Example 2.4.1 Let us use Rodrigues’s formula to compute P2 (x). From Equation 2.4.4 with n = 2, P2 (x) =

1 22 2!

d2 1 d2 4 1 [(x2 − 1)2 ] = (x − 2x2 − 1) = (3x2 − 1). (2.4.11) 2 dx 8 dx2 2 # "

• Example 2.4.2 Let us compute P3 (x) from a recurrence relation. From Equation 2.4.5 with n = 2, 3P3 (x) − 5xP2 (x) + 2P1 (x) = 0. (2.4.12)

64

Green’s Functions with Applications

But P2 (x) = (3x2 − 1)/2, and P1 (x) = x, so that 3P3 (x) = 5xP2 (x) − 2P1 (x) = 5x[(3x2 − 1)/2] − 2x =

15 3 2 x

− 92 x, (2.4.13)

or P3 (x) = (5x3 − 3x)/2.

(2.4.14) # "

With the orthogonality condition, Equation 2.4.7, we are ready to show that we can represent a function f (x), which is piece-wise differentiable in the interval (−1, 1), by the series:

f (x) =

∞ (

−1 ≤ x ≤ 1.

Am Pm (x),

m=0

(2.4.15)

To find Am we multiply both sides of Equation 2.4.12 by Pn (x) and integrate from −1 to 1: "

1

f (x)Pn (x) dx =

−1

∞ (

Am

m=0

"

1

Pn (x)Pm (x) dx.

(2.4.16)

−1

All of the terms on the right side vanish except for n = m because of the orthogonality condition, Equation 2.4.7. Consequently, the coefficient An is An

"

1

−1

Pn2 (x) dx =

"

1

f (x)Pn (x) dx,

(2.4.17)

−1

or

An =

2n + 1 2

"

1

f (x)Pn (x) dx.

−1

(2.4.18)

In the special case when f (x) and its first n derivatives are continuous throughout the interval (−1, 1), we may use Rodrigues’s formula to evaluate "

1

f (x)Pn (x) dx = −1

=

1 n 2 n!

"

(−1)n 2n n!

1

f (x) −1

"

dn (x2 − 1)n dx dxn

(2.4.19)

1

−1

(x2 − 1)n f (n)(x) dx

(2.4.20)

Background Material

65

by integrating by parts n times. Consequently, 2n + 1 An = n+1 2 n!

"

1

−1

(1 − x2 )n f (n) (x) dx.

(2.4.21)

A particularly useful result follows from Equation 2.4.18 if f (x) is a polynomial of degree k. Because all derivatives of f (x) of order n vanish identically when n > k, An = 0 if n > k. Consequently, any polynomial of degree k can be expressed as a linear combination of the first k + 1 Legendre polynomials [P0 (x), . . . , Pk (x)]. Another way of viewing this result is to recognize that any polynomial of degree k is an expansion in powers of x. When we expand in Legendre polynomials we are merely regrouping these powers of x into new groups that can be identified as P0 (x), P1 (x), P2 (x), . . . , Pk (x). • Example 2.4.3 Let us express f (x) = x2 in terms of Legendre polynomials. The results from Equation 2.4.18 mean that we need only worry about P0 (x), P1 (x), and P2 (x): x2 = A0 P0 (x) + A1 P1 (x) + A2 P2 (x). (2.4.22) Substituting for the Legendre polynomials, x2 = A0 + A1 x + 12 A2 (3x2 − 1),

(2.4.23)

and A0 = 13 ,

and A2 = 23 .

A1 = 0,

(2.4.24) # "

• Example 2.4.4 Let us find the expansion in Legendre polynomials of the function: f (x) =

!

0, 1,

−1 < x < 0, 0 < x < 1.

(2.4.25)

We could have done this expansion as a Fourier series but in the solution of partial differential equations on a sphere we must make the expansion in Legendre polynomials. In this problem, we find that 2n + 1 An = 2

"

0

1

Pn (x) dx.

(2.4.26)

66

Green’s Functions with Applications 1.5

1.5

one term f(x)

1 0.5

0.5

0

0

−0.5 −1

0

1.5

1

−0.5 −1

0.5

0

0 0 x

1

1

six terms f(x)

1

0.5

−0.5 −1

0

1.5

four terms f(x)

1

two terms f(x)

1

−0.5 −1

0 x

1

Figure 2.4.2: Representation of the function f (x) = 1 for 0 < x < 1 and 0 for −1 < x < 0 by various partial summations of its Legendre polynomial expansion. The dashed lines denote the exact function.

Therefore, A0 =

A2 =

5 2

"

1 2

"

1 0

1 dx = 12 ,

A1 =

3 2

"

1

x dx = 34 ,

0

1 0

1 2 2 (3x

− 1) dx = 0,

and A3 =

7 2

"

(2.4.27)

1 0

1 3 2 (5x

7 − 3x) dx = − 16 ,

(2.4.28)

so that f (x) = 12 P0 (x) + 34 P1 (x) −

7 P (x) 16 3

+

11 P (x) 32 5

+ ···.

(2.4.29)

Figure 2.4.2 illustrates the expansion, Equation 2.4.29, where we used only the first four terms. 2.5 THE DIRAC DELTA FUNCTION Since the 1950s, when Schwartz14 (1915–2002) published his theory of distributions, the concept of generalized functions has had an enormous impact on many areas of mathematics, particularly on partial differential equations. In this section, we introduce probably the most important generalized function, the Dirac delta function. As we shall shortly see, the entire concept of Green’s functions is intimately tied to this most “unusual” function. 14

Schwartz, L., 1973: Th´ eorie des distributions. Hermann, 418 pp.

Background Material

67

´ Figure 2.5.1: Educated at the Ecole Normale Sup´ erieure and University of Strasbourg, Laurent Schwartz (1915–2002) developed his theory of distributions during the late 1940s. For this work Harvard awarded him the 1950 Fields medal. In 1952 Schwartz returned to ´ Paris from Nancy to teach at the Sorbonne (1952–1959), Ecole Polytechnique (1959–1980), c and University of Paris VII (1980–1983). (!Collections Ecole polytechnique - Palaiseau, France - Philippe Lavialle, photogr.)

For many, the Dirac delta function had its birth with the quantum mechanics of Dirac15 (1902–1984). Modern scholarship16 showed, however, that this is simply not true. During the nineteenth century, both physicists and mathematicians used the delta function, although physicists viewed it as a purely mathematical idealization that did not exist in nature, while mathematicians used it as an intuitive physical notion without any mathematical reality. It was the work of Oliver Heaviside (1850–1925) and the birth of electrical engineering that brought the delta function to the attention of the broader scientific and engineering community. In his treatment of a cable that is grounded at both ends, Heaviside17 introduced the delta function via its sifting property, Equation 2.5.9. Consequently, as Laplace transforms became 15

Dirac, P., 1926-7: The physical interpretation of the quantum dynamics. Proc. R. Soc. London, Ser. A, 113, 621–641. 16 L¨ utzen, J., 1982: The Prehistory of the Theory of Distributions. Springer-Verlag, 232 pp. See Chapter 4, Part 2. 17 Heaviside, O., 1950: Electromagnetic Theory. Dover Publications, Inc., Section 267. See Equation 24.

68

Green’s Functions with Applications

Figure 2.5.2: Paul Adrien Maurice Dirac (1902–1984) ranks among the giants of twentiethcentury physics. Awarded the 1933 Nobel Prize in physics for his relativistic quantum mechanics, Dirac employed the delta function during his work on quantum mechanics. In later years, Dirac also helped to formulate Fermi-Dirac statistics and contributed to the quantum theory of electromagnetic radiation. (AIP Emilio Segr` e Visual Archives, Margrethe Bohr Collection.)

a fundamental tool of electrical engineers, so too did the use of the delta function. Despite the delta function’s fundamental role in electrical engineering and quantum mechanics, by 1945 there existed several schools of thought concerning its exact nature because Dirac’s definition: δ(t) = such that

"

!

∞

∞, 0,

t = 0, t &= 0,

δ(t) dt = 1,

(2.5.1)

(2.5.2)

−∞

was unsatisfactory; no conventional function could be found that satisfied Equations 2.5.1 and 2.5.2.

Background Material

69

60 50 40

(a)

(b)

(c)

(d)

30 20 10 0 −10 35 30 25 20 15 10 5 0 −5 −0.1

−0.05

0

x

0.05

0.1 −0.1

−0.05

0

x

0.05

0.1

Figure 2.5.3: Frames (a), (b) and (d) illustrate the first, second, and third sequence in Equation 2.5.4 as a function of x, Frame (c) is the sequence [1 − cos(nx)]/(nπx 2 ). The solid, dotted and dashed lines correspond to n = 1, n = 10 and n = 100, respectively.

One approach, especially popular with physicists because it agreed with their physical intuition of a point mass or charge, sought to view the delta function as the limit of the sequence of strongly peaked functions δn (t): δ(t) = lim δn (t),

(2.5.3)

n→∞

Candidates18 included 1 sin2 (nt) . nπ t2 (2.5.4) The difficulty with this approach was that the limits of these sequences may not exist. δn (t) =

n 1 , π 1 + n 2 t2

2 2 n δn (t) = √ e−n t , π

and

δn (t) =

18 Kirchhoff [Kirchhoff, G., 1882: Zur Theorie der Lichtstrahlen. Sitzber. K. Preuss. Akad. Wiss. Berlin, 641–669; reprinted a year later in Ann. Phys. Chem., Neue Folge, 18, 663–695.] gave the middle expression in Equation 2.5.4 in the limit of n → ∞ as an example of a delta function.

70

Green’s Functions with Applications

Another approach, favored by electrical engineers, involved the Heaviside step function: ! 1, t > 0, H(t) = (2.5.5) 0, t < 0. The delta function was now merely the derivative of H(t), δ(t) =

dH(t) . dt

(2.5.6)

The difficulty here was that the derivative does not exist at t = 0. Finally, some defined the delta function on the basis of its sifting property "

∞

−∞

δ(t − a)f (t) dt = f (a).

(2.5.7)

This property is given its name because δ(t − a) acts as a sieve, selecting from all possible values of f (t) its value at the point t = a. Unfortunately, there is no conventional function δ(t − a) with this property. • Example 2.5.1 Although viewing the delta function as the limit of a sequence of conventional functions lacks mathematical rigor, useful results can still be obtained by this method.19 For example, using the periodic set of pulses with period 2L and defined over the interval [−L, L] by δn (x) =

!

n/2, 0,

|x| < 1/n, |x| > 1/n,

(2.5.8)

the Fourier series representation of δn (x) is δn (x) =

∞ * mπ + * mπx + ( 1 n + sin cos . 2L m=1 mπ nL L

(2.5.9)

Consequently, the Fourier series representation for the delta function follows by letting n → ∞ and we find that δ(x) =

19

∞ * nπx + 1 1 ( + cos . 2L L n=1 L

(2.5.10)

T. B. Boykin [Boykin, T. B., 2003: Derivatives of the Dirac delta function by explicit construction of sequences. Am. J. Phys., 71, 462–468.] used a finite difference formula to construct a sequence that approaches the derivative of the Dirac delta function in one dimension.

Background Material

71

A quick check shows that this series is divergent, as it should be. If it were convergent, then δ(x) would be a conventional function. # " • Example 2.5.2 Let us show that our Fourier series representation, Equation 2.5.10, of δ(x) possesses the sifting property, Equation 2.5.7. /L Consider the integral −L f (x)δ(x) dx where f (x) is a conventional function. Thus, "

L

f (x)δ(x) dx =

−L

=

1 2L

"

L

f (x) dx + −L

∞

∞ " * nπx + 1 ( L f (x) cos dx (2.5.11) L n=1 −L L

A0 ( + An = f (0), 2 n=1

(2.5.12)

where A0 and An are the Fourier coefficients of f (x), and our demonstration is complete. # " • Example 2.5.3 Let us use the delta sequence technique to compute the Fourier transform of the delta function. As in the case of Fourier series, we first find the Fourier transform of δn (x) with L → ∞ and then take the limit as n → ∞. Therefore, ∆n (ω) =

"

∞

−∞

δn (t)e−iωt dt =

"

1/n

−1/n

n −iωt sin(ω/n) e dt = . 2 ω/n

(2.5.13)

Finally, the Fourier transform of the delta function follows by taking n → ∞, or F[δ(t)] = lim ∆n (ω) = 1. (2.5.14) n→∞

# " Although our use of delta sequences netted several useful results, they are suspect because the limits of these sequences do not exist according to common definitions of convergence. It is the purpose of what is known as the theory of distributions or generalized functions to put δ(x) on a firm mathematical basis and to unify the many ad hoc mathematical approaches used by engineers and

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Green’s Functions with Applications

Table 2.5.1: Some Useful Relationships Involving the Delta Function

δ(t − a) = "

!

∞ −∞

∞, 0,

"

t=a , t &= a

∞ −∞

δ(t − a) dt = 1

δ (n) (t − a)f (t) dt = (−1)n f (n) (a) δ[f (t)] =

n ( δ(t − tn ) , |f $ (tn )| m=1

(1)

(2)

(3)

where t1 , t2 , . . . , tn are the distinct zeros of f (t). δ (n) (t) = (−1)n δ (n) (−t)

(4)

f (t)δ(t − a) = f (a)δ(t − a)

(5)

δ(t2 − a2 ) =

xn δ (m) (x) =

!

δ(t + a) + δ(t − a) 2|a|

0, m! (−1)n (m−n)! δ (m−n) (x),

δ (n) (x)f (t) =

n (

m=0

(−1)m

,

n m

-

if if

(6)

0≤m τ + ∆τ , the circuit’s performance obeys the homogeneous differential equation L whose solution is

di + Ri = 0, dt

t>

i(t) = I0 e−Rt/L ,

+ ∆τ, τ

t>

(2.7.2)

+ ∆τ, τ

(2.7.3)

where I0 is a constant and L/R is the time constant of the circuit. Because the voltage v(t) during τ < t < τ + ∆τ is V0 /∆τ , then "

τ +∆τ

v(t) dt = V0 .

(2.7.4)

τ

Therefore, over the interval τ < t < τ + ∆τ , Equation 2.7.1 can be integrated to yield L or

"

τ +∆τ

di + R τ

"

τ +∆τ

i(t) dt =

τ

"

τ +∆τ

v(t) dt,

(2.7.5)

τ

L [i(τ + ∆τ ) − i(τ )] + R

"

τ +∆τ

i(t) dt = V0 .

(2.7.6)

τ

If i(t) remain continuous as∆ τ becomes small, then R

"

τ +∆τ τ

i(t) dt ≈ 0.

(2.7.7)

Finally, because i(τ ) = 0, and i(τ + ∆τ ) = I0 e−R(τ +∆τ )/L ≈ I0 e−Rτ/L for small∆ τ , Equation 2.7.6 reduces to LI0 e−Rτ/L = V0 ,

or

I0 =

V0 Rτ/L e . L

(2.7.8)

Background Material

81

i(t) V0 /L

+

t

Figure 2.7.2: The current i(t) within a RL circuit when the voltage V0 /∆τ is introduced between τ < t < τ + ∆τ .

Therefore, Equation 2.7.3 can be written as i(t) =


, x< ) , λ2 + ω 2

(3.3.29)

where F (x,ξ ) = A[ω 2 cosh(λx) + λ2 cos(ωx)] + B[λ sinh(λx) + ω sin(ωx)] + C[cosh(λx) − cos(ωx)] + D[sinh(λx)/λ − sin(ωx)/ω] + sinh[λ(x − ξ)]/λ − sin[ω(x − ξ)]/ω. (3.3.30) # "

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Green’s Functions with Applications

Table 3.3.1: Some Common One-Dimensional Green’s Functions

1. Harmonic oscillator: d2 g + k 2 g = −δ(x − ξ), dx2

0 < x, ξ < L,

g(0|ξ) = g(L|ξ) = 0 g(x|ξ) = 2.

sin(kx< ) sin[k(L − x> )] k sin(kL)

d2 g − k 2 g = −δ(x − ξ), dx2

0 < x, ξ < L,

g(0|ξ) = g(L|ξ) = 0 g(x|ξ) =

sinh(kx< ) sinh[k(L − x> )] k sinh(kL)

3. Damped harmonic oscillator: d2 g dg + 2γ + (k 2 − γ 2 )g = −δ(x − ξ), 2 dx dx

0 < x, ξ < L,

g(0|ξ) = g(L|ξ) = 0 g(x|ξ) = e−γ(x−ξ)

sin(kx< ) sin[k(L − x> )] k sin(kL)

• Example 3.3.5: Airy functions Consider the Green’s function problem10 g $$ − xg = δ(x − ξ),

0 < a < x, ξ < ∞,

(3.3.31)

subject to the boundary conditions g $ (a|ξ) = 0,

lim g(x|ξ) → 0.

x→∞

(3.3.32)

Let us apply Equation 3.3.17 to find the Green’s function. Here, g(x|ξ) = c1 y1 (x) for a ≤ x ≤ ξ while g(x|ξ) = c2 y2 (x) for ξ ≤ x ≤ ∞, where 10 Taken from Cole, K. D., 2008: Flush-mounted steady-periodic heated film with application to shear-stress measurement. J. Heat Transfer , 130, Art. No. 111601.

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107

y1 (x) = Ai$ (a)Bi(x) − Bi$ (a)Ai(x) and y2 (x) = Ai(x). Because p(x) = 1, direct substitution yields g(x|ξ) =

[Ai$ (a)Bi(x< ) − Bi$ (a)Ai(x< )]Ai(x> ) . [Bi$ (ξ)Ai(ξ) − Ai$ (ξ)Bi(ξ)]Ai$ (a)

(3.3.33) # "

So far, we showed that the Green’s function for Equation 3.3.6 exists, is symmetric and enjoys certain properties. See the material in the boxes after Equation 3.3.6 and Equation 3.3.10. But how does this help us solve Equation 3.3.1? We now prove that y(x) =

"

b

g(x|ξ)f (ξ) dξ

(3.3.34)

a

is the solution to the nonhomogeneous differential equation, Equation 3.3.1, and the homogeneous boundary conditions, Equation 3.3.7. We begin by noting that in Equation 3.3.34 x is a parameter while ξ is the dummy variable. As we perform the integration, we must switch from the form for g(x|ξ) for ξ ≤ x to the second form for ξ ≥ x when ξ equals x; thus, y(x) =

"

x

g(x|ξ)f (ξ) dξ + a

"

b

g(x|ξ)f (ξ) dξ.

(3.3.35)

x

Differentiation yields " x " x d dg(x|ξ) g(x|ξ)f (ξ) dξ = f (ξ) dξ + g(x|x− )f (x), dx a dx a

(3.3.36)

and d dx

"

b

g(x|ξ)f (ξ) dξ =

x

"

b

x

dg(x|ξ) f (ξ) dξ − g(x|x+ )f (x). dx

(3.3.37)

Because g(x|ξ) is continuous everywhere, we have that g(x|x+ ) = g(x|x− ) so that " x " b dy dg(x|ξ) dg(x|ξ) = f (ξ) dξ + f (ξ) dξ. (3.3.38) dx dx dx a x Differentiating once more gives d2 y = dx2 +

"

"

x a b

x

d2 g(x|ξ) dg(x|x− ) f (ξ) dξ + f (x) 2 dx dx d2 g(x|ξ) dg(x|x+ ) f (ξ) dξ − f (x). dx2 dx

(3.3.39)

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Green’s Functions with Applications

The second and fourth terms on the right side will not cancel in this case; to the contrary, dg(x|x− ) dg(x|x+ ) 1 − =− . (3.3.40) dx dx p(x) To show this, we note that the term dg(x|x− )/dx denotes a differentiation of g(x|ξ) with respect to x using the x > ξ form and then letting ξ → x. Thus, dg(x|x− ) y $ (x)y1 (ξ) y $ (x)y1 (x) = − lim 2 =− 2 , ξ→x p(ξ)W (ξ) dx p(x)W (x) ξx

(3.3.42)

Upon introducing these results into the differential equation p(x)

d2 y dy + p$ (x) + s(x)y = −f (x), dx2 dx

(3.3.43)

we have " x [p(x)g $$ (x|ξ) + p$ (x)g $ (x|ξ) + s(x)g(x|ξ)]f (ξ) dξ a

+

"

b

[p(x)g $$ (x|ξ) + p$ (x)g $ (x|ξ) + s(x)g(x|ξ)]f (ξ) dξ

x

− p(x)

f (x) = −f (x). p(x)

(3.3.44)

Because p(x)g $$ (x|ξ) + p$ (x)g $ (x|ξ) + s(x)g(x|ξ) = 0,

(3.3.45)

except for x = ξ, Equation 3.3.44, and thus Equation 3.3.1, is satisfied. Although Equation 3.3.45 does not hold at the point x = ξ, the results are still valid because that one point does not affect the values of the integrals. As for the boundary conditions, y(a) =

"

a

b

g(a|ξ)f (ξ) dξ,

y $ (a) =

"

a

b

dg(a|ξ) f (ξ) dξ, dx

(3.3.46)

and α1 y(a)+α2 y $ (a) = 0 from Equation 3.3.2. A similar proof holds for x = b. Finally, let us consider the solution for the nonhomogeneous boundary conditions α1 y(a) + α2 y $ (a) = α, and β1 y(b) + β2 y $ (b) = β. The solution in

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Table 3.3.2: Additional One-Dimensional Green’s Functions

4. Bessel’s equation: d2 g 1 dg + + (1 − n2 /x2 )g = −δ(x − ξ)/x, dx2 x dx lim |g(x|ξ)| < ∞,

x→0

g(x|ξ) =

0 < x, ξ < L,

g(L|ξ) = 0,

2 [Yn (L)Jn (x> ) − Jn (L)Yn (x> )] Jn (x< ) π Jn (L)

5. Confluent hypergeometric equation: x

d2 g dg + (b − x) − ag = −δ(x − ξ), 2 dx dx

0 < x, ξ < L,

g(0|ξ) = g(L|ξ) = 0, Γ(a) 1 Γ(b) M (a, b; L) × [M (a, b; L)U (a, b; x> ) − U (a, b; L)M (a, b; x> )] M (a, b; x< ),

g(x|ξ) =

where M (a, b; x) and U (a, b; x) are confluent hypergeometric functions of the first and second kinds, respectively.

this case is

αy2 (x) βy1 (x) y(x) = + + α1 y2 (a) + α2 y2$ (a) β1 y1 (b) + β2 y1$ (b)

"

a

b

g(x|ξ)f (ξ) dξ. (3.3.47)

A quick check shows that Equation 3.3.47 satisfies the differential equation and both nonhomogeneous boundary conditions. • Example 3.3.6 Consider the differential equation y $$ − (1 + 2λ)y = f (x),

−∞ < x < ∞,

(3.3.48)

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Green’s Functions with Applications

or

y $$ − y = 2λy + f (x),

−∞ < x < ∞,

(3.3.49)

with lim|x|→∞ y(x) → 0. In Problem 16, we will find that the Green’s function to g $$ − g = −δ(x − ξ) with lim|x|→∞ g(x|ξ) → 0 is g(x|ξ) = 12 exp(−|x − ξ|). Consequently, " ∞ " ∞ −|x−ξ| 1 y(x) = λ e y(ξ) dξ + 2 e−|x−ξ| f (ξ) dξ; (3.3.50) −∞

or

y(x) − λ

"

∞

−∞

e−|x−ξ| y(ξ) dξ =

−∞

1 2

"

∞

e−|x−ξ| f (ξ) dξ.

(3.3.51)

−∞

Therefore, the solution to Equation 3.3.48 is also the solution to the integral equation, Equation 3.3.51. For example, if f (x) = − cos(nx), y(x) = cos(nx)/(1 + n2 − 2λ) satisfies y $$ − (1 + 2λ)y = − cos(nx), and y(x) − λ

"

∞

e−|x−ξ| y(ξ) dξ =

−∞

(3.3.52)

cos(nx) . n2 + 1

(3.3.53)

Heywood et al.11 found these results using an entirely different method. 3.4 EIGENFUNCTION EXPANSION FOR REGULAR BOUNDARY-VALUE PROBLEMS In the previous section we showed how Green’s functions can be used to solve the nonhomogeneous linear differential equation. The next question is how do you find the Green’s function? In this section we present the most common method: series expansion. Consider the nonhomogeneous problem y $$ = −f (x),

with

y(0) = y(L) = 0.

(3.4.1)

The Green’s function g(x|ξ) must therefore satisfy g $$ = −δ(x − ξ),

with

g(0|ξ) = g(L|ξ) = 0.

(3.4.2)

Because g(x|ξ) vanishes at the ends of the interval (0, L), this suggests that it can be expanded in a series of suitably chosen orthogonal functions such as, for instance, the Fourier sine series g(x|ξ) =

∞ (

n=1

Gn (ξ) sin

* nπx + L

,

(3.4.3)

11 See Section 28 in Heywood, H. B., M. Fr´ echet, and J. Hadamand, 1912: L’´ equation de Fredholm et ses applications a ` la physique math´ ematique. Paris, A. Hermann, 165 pp.

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111

where the expansion coefficients Gn are dependent on the parameter ξ. Although we chose the orthogonal set of functions sin(nπx/L), we could have used other orthogonal functions as long as they vanish at the endpoints. Because ∞ , * nπx + ( n2 π 2 g $$ (x|ξ) = − 2 Gn (ξ) sin , (3.4.4) L L n=1 and

δ(x − ξ) = where An (ξ) =

2 L

"

0

L

∞ (

n=1

An (ξ) sin

* nπx + L

,

, * nπx + 2 nπξ δ(x − ξ) sin dx = sin , L L L

we have that , - * ∞ , 2 2∞ * nπx + ( n π 2 ( nπξ nπx + − G (ξ) sin = − sin sin , n 2 L L L n=1 L L n=1

(3.4.5)

(3.4.6)

(3.4.7)

after substituting Equation 3.4.4 through Equation 3.4.6 into the differential equation, Equation 3.4.2. Since Equation 3.4.7 must hold for any arbitrary x, , 2 2, n π 2 nπξ Gn (ξ) = sin . (3.4.8) L2 L L Thus, the Green’s function is g(x|ξ) =

, - * ∞ 2L ( 1 nπξ nπx + sin sin . 2 2 π n=1 n L L

(3.4.9)

How might we use Equation 3.4.9? We can use this series to construct the solution of the nonhomogeneous equation, Equation 3.4.1, via the formula y(x) =

"

L

g(x|ξ) f (ξ) dξ.

(3.4.10)

0

This leads to , ∞ * nπx + " L nπξ 2L ( 1 y(x) = 2 sin f (ξ) sin dξ, π n=1 n2 L L 0 or y(x) =

∞ * nπx + L2 ( a n sin , π 2 n=1 n2 L

(3.4.11)

(3.4.12)

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Green’s Functions with Applications

where an are the Fourier sine coefficients of f (x). • Example 3.4.1 Consider now the more complicated boundary-value problem y $$ + k 2 y = −f (x),

with

y(0) = y(L) = 0.

(3.4.13)

The Green’s function g(x|ξ) must now satisfy g $$ + k 2 g = −δ(x − ξ),

and

g(0|ξ) = g(L|ξ) = 0.

(3.4.14)

Once again, we use the Fourier sine expansion g(x|ξ) =

∞ (

n=1

Gn (ξ) sin

* nπx + L

.

(3.4.15)

Direct substitution of Equation 3.4.15 and Equation 3.4.5 into Equation 3.4.14 and grouping by corresponding harmonics yields , n2 π 2 2 nπξ 2 − 2 Gn (ξ) + k Gn (ξ) = − sin , L L L or Gn (ξ) =

2 sin(nπξ/L) . L n2 π 2 /L2 − k 2

(3.4.16)

(3.4.17)

Thus, the Green’s function12 is g(x|ξ) =

∞ 2 ( sin(nπξ/L) sin(nπx/L) . L n=1 n2 π 2 /L2 − k 2

(3.4.18)

Examining Equation 3.4.18 more closely, we note that it enjoys the symmetry property that g(x|ξ) = g(ξ|x). # " • Example 3.4.2 Let us find the series expansion for the Green’s function for , m2 xg $$ + g $ + k 2 x − g = −δ(x − ξ), x

0 < x < L,

(3.4.19)

12 Sommerfeld, A., 1912: Die Greensche Funktion der Schwingungsgleichung. Jahresber. Deutsch. Math.-Verein., 21, 309–353.

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113

Figure 3.4.1: Equation 3.4.23 as functions of x/L and ξ/L when kL = 10 and m = 1.

where m ≥ 0 and is an integer. The boundary conditions are lim |g(x|ξ)| < ∞,

x→0

and g(L|ξ) = 0.

(3.4.20)

To find this series, consider the Fourier-Bessel series g(x|ξ) =

∞ (

Gn (ξ)Jm (knm x),

(3.4.21)

n=1

where knm is the nth root of Jm (knm L) = 0. This series enjoys the advantage that it satisfies the boundary conditions and we will not have to introduce any homogeneous solutions so that g(x|ξ) satisfies the boundary conditions. Substituting Equation 3.4.21 into Equation 3.4.19 after we divide by x and using the Fourier-Bessel expansion for the delta function, we have that 2 (k 2 − knm )Gn (ξ) = −

2 2knm Jm (knm ξ) 2Jm (knm ξ) =− 2 $ , 2 2 L [Jm+1 (knm L)] L [Jm (knm L)]2

so that g(x|ξ) =

∞ 2 ( Jm (knm ξ)Jm (knm x) . 2 − k 2 )[J $ (k 2 L2 n=1 (knm m nm L)]

(3.4.22)

(3.4.23)

Figure 3.4.1 illustrates Equation 3.4.23. # "

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Green’s Functions with Applications

We summarize the expansion technique as follows: Suppose that we want to solve the differential equation Ly(x) = −f (x),

(3.4.24)

with some condition By(x) = 0 along the boundary, where L now denotes the Sturm-Liouville differential operator L=

% & d d p(x) + [q(x) + λr(x)], dx dx

(3.4.25)

and B is the boundary condition operator  d    α1 + α2 , dx B=  d   β1 + β2 , dx

at x = a, (3.4.26) at x = b.

We begin by seeking a Green’s function g(x|ξ), which satisfies Lg = −δ(x − ξ),

Bg = 0.

(3.4.27)

To find the Green’s function, we utilize the set of eigenfunctions ϕn (x) associated with the regular Sturm-Liouville problem % & d dϕn p(x) + [q(x) + λn r(x)]ϕn = 0, dx dx

(3.4.28)

where ϕn (x) satisfies the same boundary conditions as y(x). If g exists and if the set {ϕn } is complete, then g(x|ξ) can be represented by the series g(x|ξ) =

∞ (

Gn (ξ)ϕn (x).

(3.4.29)

n=1

Applying L to Equation 3.4.29, Lg(x|ξ) =

∞ (

Gn (ξ)Lϕn (x) =

n=1

∞ (

n=1

Gn (ξ)(λ − λn )r(x)ϕn (x) = −δ(x − ξ),

(3.4.30) provided that λ does not equal any of the eigenvalues λn . Multiplying both sides of Equation 3.4.30 by ϕm (x) and integrating over x, ∞ (

n=1

Gn (ξ)(λ − λn )

"

a

b

r(x)ϕn (x)ϕm (x) dx = −ϕm (ξ).

(3.4.31)

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115

If the eigenfunctions are orthonormal , "

b

r(x)ϕn (x)ϕm (x) dx = a

!

1, 0,

n = m, n &= m,

and

Gn (ξ) =

ϕn (ξ) . λn − λ (3.4.32)

This leads directly to the bilinear formula:

g(x|ξ) =

∞ ( ϕn (ξ)ϕn (x) , λn − λ n=1

(3.4.33)

which permits us to write the Green’s function at once if the eigenvalues and eigenfunctions of L are known. One of the intriguing aspects of the bilinear formula is that the Green’s function possesses poles, called the point spectrum of the Green’s function, at λn . For regular second-order differential equations, there are an infinite number of simple poles that all lie along the real λ-axis according to Sturm’s oscillation theorem.13 This fact leads to the following important theorem: Theorem 3.4.1:

1 2πi

0

C

g(x|ξ) dλ = −

δ(x − ξ) , r(ξ)

(3.4.34)

where C is a closed contour in the complex λ-plane that encloses all of the singularities of g(x|ξ) and g(x|ξ) is the Green’s function governed by Equation 3.4.27. Proof : Starting with the bilinear formula, Equation 3.4.33, we apply Cauchy’s integral formula and find that 1 2πi

0

0 ∞ 1 ( dλ g(x|ξ) dλ = − ϕn (ξ)ϕn (x) 2πi λ − λn C C n=1 =−

13

10.6.

∞ (

ϕn (ξ)ϕn (x).

(3.4.35) (3.4.36)

n=1

Ince, E. L., 1956: Ordinary Differential Equations. Dover Publications, Inc., Section

116

Green’s Functions with Applications

-plane n2 2 poles a2

C Figure 3.4.2: Closed contour C enclosing the poles of the Green’s function, Equation 3.4.41.

Next, let us expand δ(x − ξ) in terms of ϕn (x). From Sturm-Liouville theory, δ(x − ξ) = where cn =

/b a

∞ (

cn ϕn (x),

(3.4.37)

n=1

r(x)δ(x − ξ)ϕn (x) dx = r(ξ)ϕn (ξ), /b r(x)ϕ2n (x) dx a

(3.4.38)

because ϕn (x) is an orthonormal eigenfunction. Thus, ∞ δ(x − ξ) ( = ϕn (ξ)ϕn (x), r(ξ) n=1

(3.4.39) # "

which completes the proof. • Example 3.4.3

In Example 3.3.3, we showed that the Green’s function for the problem g $$ + λg = −δ(x − ξ),

0 < x < L,

(3.4.40)

with g(0|ξ) = g(L|ξ) = 0, is √ √ sin( λ x< ) sin[ λ (L − x> )] √ √ g(x|ξ) = . λ sin( λ L)

(3.4.41)

Let us verify our theorem using the solution over the interval [0, ξ]. For the verification we must evaluate √ √ 0 sin( λ x) sin[ λ (L − ξ)] √ √ dλ, λ sin( λ L) C where the closed contour C is shown in Figure 3.4.2.

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117

Our first concern is the presence of square roots in the integrand; we might have a multivalued function. However, when the Taylor expansion for sine is substituted into the integral, the square roots disappear and we have a single-valued integrand. Next, we √ must find the location and nature of singularities. Clearly λ = 0 is one, while λn L = nπ, or λn = n2 π 2 /L2 , n = 1, 2, 3, . . ., is another. To discover the nature of the singularities, we use the infinite product formula for sine and find that √ √ sin( λ x) sin[ λ (L − ξ)] √ √ (3.4.42) λ sin( λ L) +* + 4 54 5 √ * 2 2 2 2 λ x 1 − λx 1 − λx · · · 1 − λ(L−ξ) 1 − λ(L−ξ) ··· π2 9π 2 π2 9π 2 √ # = . $# $ λL2 λL2 λ 1 − π2 1 − 9π2 · · · Consequently, λ = 0 is a removable singularity; λn = n2 π 2 /L2 are simple poles. Therefore, 1 2πi

> √ √ sin( λ x) sin[ λ (L − ξ)] n2 π 2 √ √ g(x|ξ) dλ = Res ; 2 (3.4.43) L λ sin( λ L) C n=1 √ √ ∞ ( λ − λn sin( λ x) sin[ λ (L − ξ)] √ √ = lim lim λ→λn sin( λ L) λ→λn λ n=1

0

∞ (


) π = Pλ (−x< )Pλ (x> ), p(ξ)W [Pλ (−ξ), Pλ (ξ)] 2 sin(πλ)

(3.5.28)

since p(ξ) = 1 − ξ 2 and W [Pλ (−ξ), Pλ (ξ)] = −W [Pλ (ξ), Pλ (−ξ)]. # " • Example 3.5.4: Confluent hypergeometric equation The confluent hypergeometric equation arises in problems such as the three-dimensional harmonic oscillator.20 Let us find the fundamental or freespace Green’s function21 governed by the confluent hypergeometric equation22 zg $$ + (c − z)g $ − ag = δ(z − ζ),

0 < z, ζ < ∞.

(3.5.29)

Motivated by Equation 3.3.17, we guess that the Green’s function is of the form g(z|ζ) = CΦ(a, c; z< )Ψ(a, c; z> ), (3.5.30) 19 See Equation 7.7.3 and Equation 7.7.6 in Lebedev, N. N., 1972: Special Functions and Their Applications. Dover Publ., 308 pp. 20

See Negro, J., L. M. Nieto, and O. Rosas-Ortiz, 2000: Confluent hypergeometric equations and related solvable potentials in quantum mechanics. J. Math. Phys., 41, 7964–7996. 21 In electromagnetic theory, a free-space Green’s function is the particular solution of the differential equation valid over a domain of infinite extent, where the Green’s function remains bounded as we approach infinity, or satisfies a radiation condition there. 22 See Granovskii, I. Ya., and V. I. Nechet, 1974: Non-Hermitian nature of the degenerate hypergeometric equation. Sov. Phys. J., 17, 1085–1087.

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Green’s Functions with Applications

whereΦ( a, c; z) and Ψ a, ( c; z) are the homogeneous solutions to Equation 3.5.29 and are known as confluent hypergeometric functions23 of the first and second kind, respectively. To compute C, we integrate Equation 3.5.29 over the small interval [ζ − , ζ + ], where ζ + and ζ − denote points just above and below z = ζ, respectively. Thus, ' ' dg '' dg '' z −z = Cζ [Φ(a, c; ζ)Ψ$ (a, c; ζ) − Φ$ (a, c; ζ)Ψ(a, c; ζ)] = 1. dz 'z=ζ + dz 'z=ζ − (3.5.31) Because Φ(a, c; z)Ψ$ (a, c; z) − Φ$ (a, c; z)Ψ(a, c; z) = −

Γ(c) −c z z e , Γ(a)

(3.5.32)

where Γ(·) is the gamma function, C=− and g(z|ζ) = −

Γ(a) c−1 −ζ ζ e , Γ(c)

Γ(a) c−1 −ζ ζ e Φ(a, c; z< )Ψ(a, c; z> ). Γ(c)

(3.5.33)

(3.5.34)

The Green’s function may be re-expressed in terms of Whittaker functions since Φ(a, c; z) = z −c/2 ez/2 M 2c −a, 2c − 12 (z), (3.5.35) and

Ψ(a, c; z) = z −c/2 ez/2 W 2c −a, 2c − 12 (z),

(3.5.36)

so that , Γ(a) −c/2 c −1 z−ζ z ζ 2 exp M 2c −a, 2c − 12 (z< )W c2 −a, c2 − 12 (z> ). Γ(c) 2 (3.5.37) Finally, we use the integral formulas for the product of two Whittaker functions to obtain the following integral representation: g(z|ζ) = −

, -c−1 , -" ∞, -c −a , ζ 2 z−ζ ξ+1 2 z+ζ g(z|ζ) = − exp exp − ξ (3.5.38) z 2 ξ−1 2 1 4. 5 dξ × Ic−1 zζ(ξ 2 − 1) . , ξ2 − 1 23 See Lebedev, N. N., 1972: Special Functions and Their Applications. Dover Publications, Inc., Sections 9.9 and 9.10.

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125

Figure 3.5.2: Equation 3.5.37 as functions of z and ζ when a =

3 , 2

and c = 3.

when 3(c) > 3(a) > 0. Equation 3.5.38 is asymmetric in z and ζ; this asymmetry results from the non-self-adjoint nature of the confluent hypergeometric equation. # " • Example 3.5.5: Whittaker equation

In this example, we find the free-space Green’s function22 governed by Whittaker’s equation , 2 1 − µ2 d 1 κ 4 − + + g = δ(z − ζ), 0 < z, ζ < ∞. (3.5.39) dz 2 4 z z2 Again, motivated by Equation 3.3.17, we assume that the Green’s function has the form g(z|ζ) = CMκ,µ (z< )Wκ,µ (z> ), (3.5.40) where Mκ,µ (z) and Wκ,µ (z) are the homogeneous solutions and are known as Whittaker’s functions.24 To compute C, we integrate Equation 3.5.40 over the small interval [ζ − , ζ + ], where ζ + and ζ − denote points just above and below z = ζ, respectively. Thus, ' ' dg '' dg '' − = 1. (3.5.41) dz ' + dz ' − z=ζ

z=ζ

24 See Whittaker, E. T., and G. N. Watson, 1963: A Course of Modern Analysis. Cambridge University Press, Chapter 16.

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Green’s Functions with Applications

Figure 3.5.3: Equation 3.5.43 as functions of z and ζ when κ = 0, and µ = 1.

Carrying out the calculation, C=− and g(z|ζ) = −

Γ(µ − κ + 12 ) , Γ(2µ + 1)

Γ(µ − κ + 12 ) Mκ,µ (z< )Wκ,µ (z> ), Γ(2µ + 1)

(3.5.42)

(3.5.43)

where Γ(·) is the gamma function. Finally, we use the integral formulas for the product of two Whittaker functions to obtain the integral representation , , 4. 5 . " ∞ ξ+1 κ z+ζ dξ g(z|ζ) = − zζ exp − ξ I2µ zζ(ξ 2 − 1) . , ξ−1 2 ξ2 − 1 1 (3.5.44) when 3(κ −µ + 12 ) > 0. This Green’s function is symmetric in z and ζ because Whittaker’s equation is self-adjoint. In his study of a charged particle in a magnetic field, Ueta25 found the Green’s function governed by the ordinary differential equation , - * d dg z+ z + a2 − g = −δ(z − 0+ ), 0 < z < ∞. (3.5.45) dz dz 4 25 Ueta, T., 1992: Green’s function of a charged particle in magnetic fields. J. Phys. Soc. Japan, 61, 4314–4324.

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127

He showed that the Green’s function can be expressed in terms of the Whittaker Wκ,µ (z) by g(z|0+ ) =

W 2 (z) π 1 # $ a√,0 . cos(πa2 ) Γ a2 + 12 z

(3.5.46)

3.6 MAXWELL’S RECIPROCITY From the theory of ordinary differential equations, we know that every linear differential equation of the form L(u) = a0 (x)

dn u dn−1 u du + a (x) + · · · + an−1 (x) + an (x)u = 0 1 n n−1 dx dx dx

(3.6.1)

has associated with it an adjoint equation L(v) = (−1)n

dn dn−1 [a0 (x)v] + (−1)n−1 n−1 [a1 (x)v] + · · · n dx dx d − [an−1 (x)v] + an (x)v = 0. dx

(3.6.2)

The concept of adjoints is useful because vL(u) − uL(v) =

d P (u, v), dx

(3.6.3)

where P (u, v) is known as the bilinear concomitant.26 If we apply these theoretical considerations to Green’s functions, we discover the important property of reciprocity. We prove it here for second-order differential equations; the results can be generalized to higher orders. Consider the Green’s function to the second-order differential equation a0 (ξ)

d2 g(ξ|x) dg(ξ|x) + [2a$0 (ξ) − a1 (ξ)] 2 dξ dξ + [a$$0 (ξ) − a$1 (ξ) + a2 (ξ)]g(ξ|x) = δ(ξ − x).

(3.6.4)

We assume that g(ξ|x) satisfies homogeneous boundary conditions at ξ = a and ξ = b. The associated adjoint problem of Equation 3.6.4 is a0 (ξ)

d2 g ∗ (ξ|x$ ) dg ∗ (ξ|x$ ) + a (ξ) + a2 (ξ)g ∗ (ξ|x$ ) = δ(ξ − x$ ) 1 dξ 2 dξ

(3.6.5)

26 See Ince, E. L., 1956: Ordinary Differential Equations. Dover Publications, Inc., Section 5.3.

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Green’s Functions with Applications

g(

x ) = g( x

)

0

x

L

Figure 3.6.1: Maxwell’s reciprocity.

plus suitable homogeneous boundary conditions. Here we add a prime to x because x and x$ are not necessarily the same. We now compute Equation 3.6.3 and find % & d2 g ∗ (ξ|x$ ) dg ∗ (ξ|x$ ) ∗ $ g(ξ|x) a0 (ξ) + a1 (ξ) + a2 (ξ)g (ξ|x ) dξ 2 dξ ! d2 g(ξ|x) dg(ξ|x) − g ∗ (ξ|x$ ) a0 (ξ) + [2a$0 (ξ) − a1 (ξ)] 2 dξ dξ ) $$ $ + [a0 (ξ) − a1 (ξ) + a2 (ξ)]g(ξ|x) (3.6.6) ! d dg ∗ (ξ|x$ ) = a0 (ξ)g(ξ|x) + [a1 (ξ) − a$0 (ξ)] g(ξ|x)g ∗ (ξ|x$ ) dξ dξ ) ∗ $ dg(ξ|x) − a0 (ξ)g (ξ, x ) . dξ Upon integrating from a to b, we obtain % & " b d2 g ∗ (ξ|x$ ) dg ∗ (ξ|x$ ) ∗ $ g(ξ|x) a0 (ξ) + a1 (ξ) + a2 (ξ)g (ξ|x ) dξ dξ 2 dξ a ! " b d2 g(ξ|x) dg(ξ|x) = g ∗ (ξ|x$ ) a0 (ξ) + [2a$0 (ξ) − a1 (ξ)] 2 dξ dξ a ) + [a$$0 (ξ) − a$1 (ξ) + a2 (ξ)]g(ξ|x) dξ, (3.6.7) because both g(ξ|x) and g ∗ (ξ|x$ ) satisfy homogeneous boundary conditions. Upon substituting Equation 3.6.4 and Equation 3.6.5, we have that " b " b $ δ(ξ − x )g(ξ|x) dξ = g ∗ (ξ|x$ )δ(ξ − x) dξ, (3.6.8) a

or

a

g(x$ |x) = g ∗ (x|x$ ).

(3.6.9)

Ordinary Differential Equations

129

Figure 3.6.2: Equation 3.6.11 as functions of x and ξ. The symmetry of the solution in x and ξ is quite apparent.

If the differential operators in Equation 3.6.4 and Equation 3.6.5 are selfadjoint, i.e., L(·) = L(·), then g(x|ξ) = g(ξ|x). • Example 3.6.1 The Green’s function governed by the ordinary differential equation g $$ (x|ξ) = δ(x − ξ),

g(0|ξ) = g(L|ξ) = 0,

(3.6.10)

is g(x|ξ) = (L − x> )x< .

(3.6.11)

A quick check shows that g(x|ξ) = g(ξ|x) is symmetric because the differential operator L = d2 /dx2 is self-adjoint. Figures 3.6.1 and 3.6.2 illustrate this result graphically. 3.7 GENERALIZED GREEN’S FUNCTION So far in this chapter we assumed that the Green’s function to an ordinary differential equation exists. This is not true if the adjoint equation and the adjoint boundary conditions possesses a nontrivial solution (a compatible differential system). The purpose of this section is to modify the traditional formulation for finding the Green’s function so that it includes compatible differential systems. To fix ideas, consider the self-adjoint differential equation u$$ = −f (x),

a < x < b,

(3.7.1)

with the boundary conditions u$ (a) = u$ (b) = 0. If we attempt to construct the Green’s function for this problem, we would fail because there is a nontrivial homogeneous solution u0 (x) = 1. What are we to do in these special cases? Must we abandon the concept of Green’s function here?

130

Green’s Functions with Applications Consider the Sturm-Liouville problem % & d du p(x) + q(x)u = 0, dx dx

(3.7.2)

where p(x) and q(x) are continuous on the interval (a, b) and p(x) > 0. Let the boundary conditions consist of U (u) = u(a) − γu(b) = 0, V (u) = p(a)u$ (a) −

p(b) $ u (b) = 0; γ

(3.7.3) (3.7.4)

or U (u) = u(a) − γp(b)u$ (b) = 0,

(3.7.5)

u(b) = 0, γ

(3.7.6)

V (u) = p(a)u$ (a) −

where γ is a nonzero real number. In the case when there is no solution u0 (x) &= 0 to Equation 3.7.2 satisfying the boundary conditions, Equations 3.7.3 and 3.7.4 or Equations 3.7.5 and 3.7.6, we have already shown in the previous sections that the Green’s function exists for this incompatible system. On the other hand, if there exists a solution u0 (x) &= 0 for Equation 3.7.2 through Equation 3.7.6, then Harada27 proved that there is a function g(x|ξ), called the generalized Green’s function that has the following properties: /b For any function ϕ(x) such that a ϕ(x)u0 (x) dx = 0, the solution to % & d du p(x) + q(x)u = −ϕ(x), dx dx

(3.7.7)

with the boundary conditions given by Equations 3.7.3 and 3.7.4 or Equations 3.7.5 and 3.7.6 is given by u(x) =

"

b

g(x|ξ)ϕ(ξ) dξ.

(3.7.8)

a

The generalized Green’s function in the present case satisfies the following conditions: 1. g(x|ξ) is continuous in x and ξ and its first partial derivative relative to x is bounded for x &= ξ. 27 Harada, S., 1953: An existence proof of the generalized Green’s function. Osaka Math. J., 5, 59–63.

Ordinary Differential Equations

131

2. For any fixed ξ, g(x|ξ) satisfies the boundary conditions as a function of x. 3. For any x &= ξ, g(x|ξ) satisfies % & d dg p(x) + q(x)g = αu0 (ξ)u0 (x) dx dx

(3.7.9)

as the function of x. 4. gx (ξ + |ξ) − gx (ξ − |ξ) = − 5.

"

1 . p(ξ)

(3.7.10)

b

g(x|ξ)u0 (x) dx = 0.

(3.7.11)

a

Given the generalized Green’s function, Harada showed that the solution to Equation 3.7.7 is u(x) = Au0 (x) +

"

b

g(x|ξ)ϕ(ξ) dξ.

(3.7.12)

a

It is important to recall that ϕ(x) must satisfy the condition "

b

u0 (x)ϕ(x) dx = 0.

(3.7.13)

a

How do we find the generalized Green’s function that satisfies these conditions? Here we assume that the number of independent boundary conditions equals the order of the system.28 Second we assume that there is a single homogeneous solution to the adjoint problem that we shall denote by v0 ; the generalization to the case of multiple homogeneous solutions is easily done. The procedure consists of two steps: (1) Solving the boundary-value problem % & d dg p(x) + q(x)g = −δ(x − ξ) + v0 (ξ)v0 (x), (3.7.14) dx dx with the appropriate boundary conditions, where v0 (x) denotes the nonzero normalized homogeneous solution. (2) Choose the arbitrary constant so that "

b

g(x|ξ)v0 (x) dx = 0.

(3.7.15)

a

28 When this is not true, see Greub, W., and W. C. Reinbolut, 1960: Non-self-adjoint boundary value problems in ordinary differential equations. J. Res. Nat. Bur. Standards, Sect. B, 64, 83–90; Wyler, O., 1965: On two-point boundary-value problems. Ann. Mat. Pura Appl., 67, 127–142.

132

Green’s Functions with Applications

• Example 3.7.1 Let us return to our original, self-adjoint boundary-value problem u$$ = −f (x),

0 < x < 1,

(3.7.16)

with u$ (0) = u$ (1) = 0. The homogeneous adjoint problem is v $$ = 0 with v$ (0) = v $ (1) = 0. The normalized homogeneous solution is v0 (x) = 1. Consequently the generalized Green’s function is given by g $$ = −δ(x − ξ) + 1,

g $ (0|ξ) = g $ (1|ξ) = 0.

(3.7.17)

For x < ξ, g(x|ξ) = x2 /2 + A while g(x|ξ) = x2 /2 − x + B for x > ξ. From the continuity condition at x = ξ, B − A = ξ. From the orthogonality condition, ξA + (1 − ξ)B = 5/6 − ξ 2 /2. The generalized Green’s function is then 5 x2 ξ2 + + − x> . 6 2 2

g(x|ξ) =

(3.7.18)

The generalized Green’s function given by Equation 3.7.18 can be used to write down the solution to Equation 3.7.16 as u(x) = A +

"

b

g(x|ξ)f (ξ) dξ

(3.7.19)

a

provided that

"

b

f (x) dx = 0.

(3.7.20)

a

For example, the function f (x) = 12 − x satisfies this condition. Upon substituting Equation 3.7.18 and this f (x) into Equation 3.7.19 and carrying out the integration, we find that u(x) = x3 /6 − x4 /4 + K, where K is an arbitrary constant. The same result follows integrating Equation 3.7.16 twice. # " • Example 3.7.2 Consider the self-adjoint boundary-value problem:29 u$$ + n2 u = −f (x),

u(0) = u(π) = 0.

(3.7.21)

The homogeneous adjoint problem is v $$ + n2 v = 0,

v(0) = v(π) = 0.

(3.7.22)

29 Loud, W. S., 1970: Some examples of generalized Green’s functions and generalized Green’s matrices. SIAM Review, 12, 194–210.

Ordinary Differential Equations

133

The normalized homogeneous solution is 3 2 v0 (x) = sin(nx). π

(3.7.23)

Consequently, the governing equation for the generalized Green’s function is g $$ + n2 g = −δ(x − ξ) +

2 sin(nξ) sin(nx) π

(3.7.24)

with g(0|ξ) = g(π|ξ) = 0. The generalized Green’s function is g(x|ξ) = C1 cos(nx) + C2 sin(nx) +

x cos(nx) sin(nξ), nπ

x < ; (3.7.25) ξ

x cos(nx) sin(nξ), nπ

x > ξ (3.7.26) .

and g(x|ξ) = C3 cos(nx) + C4 sin(nx) +

Applying continuity at x = ξ, Equation 3.7.26 becomes g(x|ξ) = C1 cos(nx) + C2 sin(nx) + +

x cos(nx) sin(nξ), nπ

sin[n(x − ξ)] n

x>ξ

.

(3.7.27)

Because g(0|ξ) = g(π|ξ) = 0, C1 = 0. Upon applying the orthogonality condition, ξ 1 C2 = cos(nξ) − cos(nξ). (3.7.28) nπ n Therefore, the generalized Green’s function is 1 [x< cos(nx< ) sin(nx> ) − (π − x> ) cos(nx> ) sin(nx< )] . nπ (3.7.29) This generalized Green’s function can be used to solve Equation 3.7.21 provided " π f (x) sin(nx) dx = 0. (3.7.30) g(x|ξ) =

0

An alternative form of the generalized Green’s function follows from the bilinear representation of the delta function: δ(x − ξ) =

∞ 2 ( sin(mξ) sin(mx), π 2 m=1

0 < x < 1.

(3.7.31)

Consequently, Equation 3.7.24 can be written g $$ + n2 g = −

∞ 2 ( sin(mξ) sin(mx). π 2 m=1 m&=n

(3.7.32)

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Green’s Functions with Applications

Solving Equation 3.7.32, we find that g(x|ξ) =

∞ 2 ( sin(mξ) sin(mx) . π 2 m=1 m2 − n2

(3.7.33)

m&=n

Finally, we can find the generalized Green’s function using Equation 3.3.18 when L = π: g1 (x|ξ) =

sin(kx< ) sin[k(π − x> )] k sin(kπ)

(3.7.34)

by taking the limit as k → n. Hence, % & ∂ 2 2 g(x|ξ) = lim (k − n )g1 (x|ξ) (3.7.35) k2 →n2 ∂(k 2 ) % & 1 ∂ (k − n)(k + n) sin(kx< ) sin[k(π − x> )] = lim (3.7.36) 2n k→n ∂k k sin(kπ) % & 1 ∂ (k − n)(k + n)] = lim sin(nx< ) sin[n(π − x> )] (3.7.37) 2n k→n ∂k k sin(kπ) % & 1 (k − n)(k + n) + lim x< cos(kx< ) sin[k(π − x> )] 2n k→n k sin(kπ) % & 1 (k − n)(k + n) + lim sin(kx< )(π − x> ) cos[k(1 − x> )]. 2n k→n k sin(kπ) Upon taking the derivatives and limits in Equation 3.7.37, we obtain sin(nξ) sin(nx) x< cos(nx< ) sin(nx> ) − 2n2 π nπ (π − x> ) sin(nx< ) cos(nx> ) + . nπ

g(x|ξ) =

(3.7.38)

Equation 3.7.38 agrees with Equation 3.7.29 except for the first term in Equation 3.7.38. This term is a homogeneous solution to Equation 3.7.24. This illustrates the fact30 that the generalized Green’s function for a compatible system such as Equation 3.7.24 is not unique. # "

30 See Theorem 2 in Elliott, W. W., 1928: Generalized Green’s functions for compatible differential systems. Amer. J. Math., 50 243–258.

Ordinary Differential Equations

135

• Example 3.7.3 Consider the self-adjoint boundary-value problem:31 u$$ + π 2 u = −f (x),

u(−1) = u(1),

u$ (−1) = u$ (1).

(3.7.39)

The homogeneous adjoint problem is v$$ + π 2 v = 0,

v(−1) = v(1),

v $ (−1) = v(1).

(3.7.40)

The interesting aspect of this problem is that it has two homogeneous solutions: v0 (x) = sin(πx), v1 (x) = cos(πx). (3.7.41) Consequently, the governing equation for the generalized Green’s function is g $$ + π 2 g = −δ(x − ξ) + sin(πξ) sin(πx) + cos(πξ) cos(πx) = cos[π(x − ξ)], (3.7.42) with g(−1|ξ) = g(1|ξ) and g $ (−1|ξ) = g $ (1|ξ). The extended Green’s function is g(x|ξ) =

sin[π|ξ − x|] x − ξ cos[π(x − ξ)] + sin[π(x − ξ)] + . 2π 2π 4π 2

(3.7.43)

A quick check shows that this Green’s function satisfies the boundary conditions and is orthogonal to sin(πx) and cos(πx) over the interval [−1, 1]. # " • Example 3.7.4: Legendre operator In Example 3.5.3 we found the Green’s function to the problem % & d 2 dg (1 − x ) + λ(λ + 1)g = δ(x − ξ), −1 < x, ξ < 1, (3.7.44) dx dx with the boundary conditions |g(−1|ξ)| < ∞,

|g(1|ξ)| < ∞.

(3.7.45)

There we assumed that λ &= N , a nonnegative integer. In the case when λ = N , we must modify Equation 3.7.44 to read % & d dg 2N + 1 (1 − x2 ) +N (N +1)g = δ(x−ξ)− PN (ξ)PN (x), −1 < x, ξ < 1, dx dx 2 (3.7.46) 31 Bounitzky, E., 1909: Sur la fonction de Green des ´ equations diff´ erentielles lin´ eaires ordinaires. J. Math. Pures Appl., Ser. 6 , 5, 65–126.

136

Green’s Functions with Applications

subject to the boundary conditions, Equation 3.7.45, and the orthogonality condition " 1 PN (x)g(x|ξ) dx = 0. (3.7.47) −1

The reason for this orthogonality condition and the extra term in Equation 3.7.46 follows from the fact that we have a null eigenvalue in the spectrum of the Legendre operator. There are two ways of finding the extended Green’s function. The first method uses Equation 3.4.46 and taking g(x|ξ) =

∞ (

An Pn (ξ)Pn (x).

(3.7.48)

n=0 n&=N

Upon substituting Equation 3.5.26 and Equation 3.7.48 into Equation 3.7.46, we obtain the extended Green’s function g(x|ξ) =

1 2

∞ (

n=0 n&=N

2n + 1 Pn (ξ)Pn (x). N (N + 1) − n(n + 1)

(3.7.49)

The second approach begins by using Equation 3.5.27 in the limit as λ → N . Because the denominator of the N th term vanishes in this limit, we find that ∂ {[λ(λ + 1) − N (N + 1)]g(x|ξ)} . ∂[λ(λ + 1)] (3.7.50) Using Equation 3.5.28, we have that % & π ∂ (λ − N )(λ + N + 1) gN (x|ξ) = lim Pλ (−x< )Pλ (x> ) . 4N + 2 λ→N ∂λ sin(πλ) (3.7.51) Taking the limit via l’Hospital rule, we find that ' ' (−1)N ∂Pλ (−x< ) '' (−1)N ∂Pλ (−x< ) '' gN (x|ξ) = PN (x> ) + PN (−x< ) ' ' 2 ∂λ 2 ∂λ λ=N λ=N Pλ (−x< )Pλ (x> ) + . (3.7.52) 4N + 2 gN (x|ξ) =

lim

λ(λ+1)→N (N +1)

Szmytkowski32 showed how to evaluate the partial derivatives in Equation 3.7.52. This yields gN (x|ξ) = 12 PN (x< )PN (x> ) ln[(1 − x< )(1 + x> )/4] + 12 WN (−x< )PN (−x> ) PN (x< )PN (x> ) + 12 PN (x< )WN (x> ) + , (3.7.53) 4N + 2 32

Szmytkowski, op. cit.

Ordinary Differential Equations

137

where WN (x) = 2

N −1 (

(−1)N+1

n=0

2n + 1 [Pn (x) − PN (x)]. (N − n)(N + n + 1)

(3.7.54)

If the upper limit of the summation is less than the lower one, then the sum equals zero. In Problem 47, you will derive the result in the classic case when N = 0. 3.8 INTEGRO-DIFFERENTIAL EQUATIONS If we simply view Green’s functions as the response of a linear system to an impulse forcing, then integro-differential equations can have Green’s functions. The following examples illustrate the method of finding these Green’s functions. • Example 3.8.1: Transport equation In the case of neutron diffusion, the Green’s function33 for this process is given by µ

∂g c +g = ∂x 2

"

1 −1

g(x, µ$ |0, µ0 ) dµ$ +

δ(x) δ(µ − µ0 ), 4π

(3.8.1)

where −∞ < x < ∞, −1 < µ, µ0 < 1, and c is a constant and is taken here to be less than 1. Before tackling Equation 3.8.1, consider the simpler equation ∂u c µ +u= ∂x 2

"

1

u(x, µ$ ) dµ$ .

(3.8.2)

−1

Guessing the solution u(x, µ) = e−x/ν ϕν (µ), direct substitution yields (ν − µ)ϕν (µ) = 12 cν

(3.8.3)

provided we choose ϕν such that "

1

ϕν (µ$ ) dµ$ = 1.

(3.8.4)

−1

Substituting Equation 3.8.3 into Equation 3.8.4, we find cν tanh−1 (1/ν) = 1.

(3.8.5)

33 See Section III of Case, K. M., 1960: Elementary solutions of the transport equation. Ann. Phys., 9, 1–23.

138

Green’s Functions with Applications

There are two roots to Equation 3.8.5 which we will denote by ν = ±ν0 . The most general solution to Equation 3.8.3 is % & c ν ϕν = PV + λ(ν)δ(µ − ν), (3.8.6) 2 ν−µ where PV denotes that we take the principal value when integrating an expression involving ϕν (µ). Therefore, "

1

cν ϕν (µ) dµ = PV 2 −1

"

1

−1

dµ + λ(ν) ν−µ

"

1 −1

δ(µ − ν) dµ = 1.

(3.8.7)

Consider now two possibilities: • Case 1 : If ν is not real and between −1 and 1. The second term of Equation 3.8.7 vanishes and we have that " cν 1 dµ = 1. (3.8.8) 2 −1 ν − µ Thus, there are two roots ±ν0 and u± (x, µ) = ϕ± (µ)e∓x/ν0 , where ϕ± (µ) =

c ν0 . 2 ν0 ∓ µ

• Case 2 : ν is real and lies −1 and 1. Then " 1 cν dµ PV + λ(ν) = 1. 2 −1 ν − µ

(3.8.9) (3.8.10)

(3.8.11)

The solutions are uν (x, µ) = ϕν (µ)e−x/ν , where

and

−1 ≤ ν ≤ 1,

(3.8.12)

, cν 1 ϕν (µ) = PV + λ(ν)δ(µ − ν), 2 ν −µ

(3.8.13)

λ(ν) = 1 − cν tanh−1 (ν).

(3.8.14)

In summary: There are two discrete solutions given by Equation 3.8.9 and a continuum of solutions given by Equation 3.8.12. Case then proved the orthogonality condition: , " 1 c c ϕ2± (µ) µ dµ = N± = ± ν02 − 1 , (3.8.15) 2 ν02 − 1 −1

Ordinary Differential Equations and

"

1

−1

139

ϕν $ (µ)ϕν (µ) µ dµ = Nν δ(ν − ν $ ),

(3.8.16)

where the normalization constant is % & π 2 c2 ν 2 2 Nν = ν λ (ν) + . 4

(3.8.17)

Next, Case proved that ϕ± and ϕν (−1 ≤ ν ≤ 1) are complete for functions ψ(µ) defined in the interval −1 ≤ µ ≤ 1. Consequently, the function ψ(µ) is ψ(µ) = a+ ϕ+ (µ) + a− ϕ− (µ) + where

1 a± = N±

"

"

1

A(ν)ϕν(µ) dν,

(3.8.18)

−1

1

ϕ± (µ)ψ(µ) µ dµ.

−1

(3.8.19)

With these results, we are ready to solve Equation 3.8.1. We begin by integrating Equation 3.8.1 across the plane x = 0. This gives the jump condition 1 2 δ(µ − µ0 ) µ g(0+ , µ|0, µ$ ) − g(0− , µ|0, µ$ ) = . 4π

(3.8.20)

If we require that our Green’s function vanishes as |x| → ∞, the Green’s function is given by $

g(x, µ|0, µ ) = C+ u+ (x, µ) +

"

1

A(ν)uν (x, µ) dν,

x > 0,

(3.8.21)

0

and $

g(x, µ|0, µ ) = −C− u− (x, µ) −

"

0

A(ν)uν (x, µ) dν,

x < 0,

(3.8.22)

−1

Substituting Equation 3.8.21 and Equation 3.8.22 into the jump condition, we find that " 1 δ(µ − µ0 ) = C+ ϕ+ (µ) + C− ϕ− (µ) + A(ν)ϕµ (µ) dν. (3.8.23) 4πµ −1 Therefore, C± =

1 N±

"

1 −1

µ ϕ± (µ)δ(µ − µ0 ) 1 ϕ± (µ0 ) dµ = , 4πµ 4π N±

(3.8.24)

140

Green’s Functions with Applications

and A(ν) =

1 ϕν (µ0 ) . 4π Nν

(3.8.25)

In summary, the Green’s function for the integro-differential equation, Equation 3.8.1, is g(x, µ|0, µ$ ) =

% & " 1 1 ϕ+ (µ0 )ϕ+ (µ) −x/ν0 ϕν (µ0 )ϕν (µ) −x/ν e + e dν 4π N+ Nν 0 (3.8.26)

if x > 0, and % & " 0 1 ϕ− (µ0 )ϕ− (µ) x/ν0 ϕν (µ0 )ϕν (µ) x/ν g(x, µ|0, µ ) = − e + e dν 4π N− Nν −1 (3.8.27) if x < 0. Several authors have extended Case’s work. Erdmann and Siewert34 used Case’s singular eigenfunction expansion technique to find the Green’s function for the transport equation in a spherical geometry. Smith and Siewert35 found the half-space Green’s function for the matrix transport equation that describes the flow of polarized radiation in a free-electron atmosphere. Qin et al.36 computed the Green’s function for a model atmosphere consisting of a single layer and having sources located at the middle of the atmosphere. $

# " • Example 3.8.2: Nonlocal wave interaction Integro-differential equations of the form d2 u + α2 u + dx2

"

∞ 0

K(|x − x$ |)u(x$ ) dx$ = 0,

0 < x < ∞,

(3.8.28)

subject to the boundary conditions u(0) = 1 and limx→∞ u(x) → 0, occur in electromagnetic wave problems37 involving a bounded medium where the relationship between the electric current and the field driving it is nonlocal. 34 Erdmann, R. C., 1968: Green’s functions for the one-speed transport equation in spherical geometry. J. Math. Phys., 9, 81–89. 35 Smith, O. J., and C. E. Siewert, 1967: The half-space Green’s function for an atmosphere with a polarized radiation field. J. Math. Phys., 8, 2467–2474. 36 Qin, Y., M. A. Box, and P. Douriaguine, 2004: Computation of Green’s function for radiative transfer. J. Quant. Spectr. Rad. Transf., 84, 159–168. 37 Reuter, G. E. H., and E. H. Sondheimer, 1948: The theory of the anomalous skin effect in metals. Proc. R. Soc. London, Ser. A, 195, 336–364.

Ordinary Differential Equations

141

0.2

0.1

0

−0.2

0

g (x| )

−0.1

−0.3

−0.4

−0.5

−0.6 −30

−20

−10

0

x

10

20

30

Figure 3.8.1: The Green’s function g0 (x|ξ) for Equation 3.8.29 when a = 0.01, α = 1, ξ = 10, ∆x = 0.02, and the number of grid points N = 3001.

In this example we will find the Green’s function for this nonlocal wave interaction. We begin by finding the Green’s function for an infinite medium given by the integro-differential equation: " ∞ d2 g 0 2 + α g0 + K(|x − x$ |)g0 (x$ |ξ) dx$ = δ(x − ξ), −∞ < x, ξ < ∞, dx2 −∞ (3.8.29) with the boundary conditions: lim g0 (x|ξ) → 0.

(3.8.30)

|x|→∞

We can solve Equation 3.8.29 numerically by introducing centered finite differences and Simpson’s rule and modeling the delta function by 1/(∆x) at x = ξ. We approximated the end conditions by g(−L|ξ) = g(L|ξ) = 0. Figure 3.8.1 illustrates the numerical solution when the kernel is K(|x|) = −

1 −a|x| e , 2a

a > 0.

(3.8.31)

As a becomes larger, the magnitude of the peak grows as a. Turning to a semi-infinite medium 0 < x, ξ < ∞, the governing equation is now " ∞ d2 g 2 +α g+ K(|x−x$ |)g(x$ |ξ) dx$ = δ(x−ξ), 0 < x, ξ < ∞, (3.8.32) dx2 0

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Green’s Functions with Applications

with the boundary condition g(0|ξ) = 0, and limx→∞ g(x|ξ) → 0. We can solve this equation numerically by introducing centered finite differences, Simpson’s rule to evaluate the integral, and approximating δ(x − ξ) = 1/(∆x) at x = ξ. We approximated the right boundary condition with g(L|ξ) = 0. The MATLAB code is: alpha sq = alpha*alpha; n source = xi/h; AA = zeros(N,N); for i = 1:N AA(i,i) = -2/(h*h) + alpha sq; bb(i) = 0; end bb(n source) = 1/h; for i = 1:N-1 AA(i,i+1) = 1/(h*h); AA(i+1,i) = 1/(h*h); end for x for x

i = 1:N = i*h; j = 1:N prime = j*h; K = - exp(-a*abs(x-x prime))/(2*a); if (mod(j,2) == 0) AA(i,j) = AA(i,j) + 2*h*K/3; else AA(i,j) = AA(i,j) + 4*h*K/3; end end; end green = AA\bb’; Here h = ∆x and N = L/h. Figure 3.8.2 illustrates the Green’s function. Baraff38 used the Wiener-Hopf technique to compute this Green’s function and found that g(x|ξ) = −

"

g(x|ξ) = −

"

and

ξ

0

u(η)u(η + x − ξ) dη,

x > ξ,

(3.8.33)

u(η)u(η + ξ − x) dη,

x < ξ,

(3.8.34)

x 0

where u(x) is the solution to Equation 3.8.28. Here we use the boundary condition u(0) = 1 and u(L) = 0. The MATLAB for computing u(x) is: 38 Baraff, G. A., 1970: Green’s function for the nonlocal wave equation. J. Math. Phys., 11, 1538–1541; 12, 927.

Ordinary Differential Equations

143

0.2 0.1 0

g(x| )

−0.1 −0.2 −0.3 −0.4 −0.5 −0.6 0

5

10

15

20

25

30

x Figure 3.8.2: The Green’s function for Equation 3.8.32 when a = 0.01, α = 1, ξ = 10, h = 0.01, and N = 3000. The solid line is given by Baraff’s approach while the crosses denote the numerical solution of Equation 3.8.32.

N s = xi/(2*h); M = N/2; AA = zeros(N,N); for i = 1:N AA(i,i) = -2/(h*h) + alpha sq; bb(i) = 0; end bb(1) = -1/(h*h); %

u(0) = 1

for i = 1:N-1 AA(i,i+1) = 1/(h*h); AA(i+1,i) = 1/(h*h); end for x for x

i = 1:N = i*h; j = 1:N prime = j*h; K = - exp(-a*abs(x-x prime)) / (2*a); if (mod(j,2) == 0) AA(i,j) = AA(i,j) + 2*h*K/3; else

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Green’s Functions with Applications

AA(i,j) = AA(i,j) + 4*h*K/3; end end; end u = AA\b’; Next, we compute the integrals given by Baraff via the MATLAB: %%%%% Equation 3.8.34 %%%%% for m = 1:N s n = 2*m-1; mid = 2*N s-n; integrand = h*u(mid)/3 + 4*h*u(1)*u(1+mid)/3 ... + h*u(2)*u(2+mid)/3; % N.B. u(0) = 1 green(m) = - integrand; if (m > 1) for j = 2:m integrand = h*u(2*j-2)*u(2*j-2+mid)/3 ... + 4*h*u(2*j-1)*u(2*j-1+mid)/3 ... + h*u(2*j)*u(2*j+mid)/3; green(m) = green(m) - integrand; end; end end %%%%% Equation 3.8.33 %%%%% for m = N s+1:M n = 2*m-1; mid = n-2*N s; integrand = h*u(mid)/3 + 4*h*u(1)*u(1+mid)/3 ... + h*u(2)*u(2+mid)/3; % u(0) = 1 green(m) = - integrand; for j = 2:N s integrand = h*u(2*j-2)*u(2*j-2+mid)/3 ... + 4*h*u(2*j-1)*u(2*j-1+mid)/3 ... + h*u(2*j)*u(2*j+mid)/3; green(m) = green(m) - integrand; end; end

Ordinary Differential Equations

145 Problems

For the following initial-value problems, find the Green’s function. Assume that the initial conditions equal zero and τ > 0. 1. g $ + kg = δ(t − τ )

2. g $$ − 2g $ − 3g = δ(t − τ )

3. g $$ + 4g $ + 3g = δ(t − τ )

4. g $$ − 2g $ + 5g = δ(t − τ )

5. g $$ − 3g $ + 2g = δ(t − τ )

6. g $$ + 4g $ + 4g = δ(t − τ )

7. g $$ − 9g = δ(t − τ )

8. g $$ + g = δ(t − τ )

9. g $$ − g $ = δ(t − τ ) Hilbert presented a number of Green’s function in his great tome on integral equations.39 In Problem 10 to Problem 16, verify his solutions: 10. Show that the Green’s function governed by g $$ = −δ(x − ξ),

0 < x, ξ < 1,

subject to the boundary conditions g(0|ξ) = g(1|ξ) = 0, is g(x|ξ) = (1 − x> )x< . 11. Show that the Green’s function governed by g $$ = −δ(x − ξ),

0 < x, ξ < 1,

subject to the boundary conditions g(0|ξ) = g $ (1|ξ) = 0, is g(x|ξ) = x< . 12. Show that the Green’s function governed by g $$ = −δ(x − ξ),

−1 < x, ξ < 1,

subject to the boundary conditions g(−1|ξ) = g(1|ξ) = 0, is g(x|ξ) = − 12 (|x − ξ| + xξ − 1) . 13. Show that the Green’s function governed by g $$ = −δ(x − ξ),

0 < x, ξ < 1,

39 Hilbert, D., 1912: Grundz¨ uge einer allgemeinen Theorie der linearen Integralgleichungen. Leipzig, B. G. Teubner, 282 pp. See pp. 43 and 44.

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Green’s Functions with Applications

subject to the boundary conditions g(0|ξ) = −g(1|ξ) and g $ (0|ξ) = −g $ (1|ξ), is g(x|ξ) = − 12 |x − ξ| + 14 . 14. Show that the Green’s function governed by xg $$ + g $ = −δ(x − ξ),

0 < x, ξ < 1,

subject to the boundary conditions limx→0 |g(x|ξ)| < ∞, and g(1|ξ) = 0, is g(x|ξ) = ln(x> ). 15. Show that the Green’s function governed by % & d 4α2 2 dg (1 − x ) − g = −δ(x − ξ), dx dx 1 − x2

−1 < x, ξ < 1,

where α is a positive constant, subject to the boundary conditions lim |g(x|ξ)| < ∞,

lim |g(x|ξ)| < ∞,

x→−1

is g(x|ξ) =

1 4α

,

x→1

1 + x< 1 − x> 1 − x< 1 + x>

-α

.

16. Show that the Green’s function governed by g $$ − g = −δ(x − ξ),

−∞ < x, ξ < ∞,

subject to the boundary conditions lim|x|→∞ g(x|ξ) → 0, is g(x|ξ) = 12 e−|x−ξ| . 17. Show that the Green’s function for the periodic boundary-value problem40 g $$ + m2 g = δ(x − ξ),

m &= 0, ±1, ±2, . . . ,

0 ≤ x,ξ ≤ 2π,

subject to the periodic boundary conditions g(0|ξ) = g(2π|ξ) and g $ (0|ξ) = g $ (2π|ξ), is sin(m|x − ξ|) + sin[m(2π − |x − ξ|)] g(x|ξ) = . 2m[1 − cos(2mπ)] Find the Green’s function and the corresponding bilinear expansion using eigenfunctions from the regular Sturm-Liouville problem yn$$ + kn2 yn = 0 for g $$ = −δ(x − ξ),

0 < x, ξ < L,

40 Graef, J. R., L. Kong, and H. Wang, 2008: A periodic boundary value problem with vanishing Green’s function. Appl. Math. Lett., 21, 176–180.

Ordinary Differential Equations

147

which satisfy the following boundary conditions: 18. g(0|ξ) − αg $ (0|ξ) = 0, α &= 0, −L,

g(L|ξ) = 0,

19. g(0|ξ) − g $ (0|ξ) = 0,

g(L|ξ) − g $ (L|ξ) = 0,

20. g(0|ξ) − g $ (0|ξ) = 0,

g(L|ξ) + g $ (L|ξ) = 0.

21. Find the Green’s function for g $$ −

1 = −δ(x − ξ), a

0 < x, ξ < a,

with the boundary conditions g $ (0|ξ) = g $ (a|ξ) = 0. Is there a bilinear expansion? Why or why not? Find the Green’s function41 and the corresponding bilinear expansion [using eigenfunctions from the regular Sturm-Liouville problem yn$$ + kn2 yn = 0] for g $$ − k 2 g = −δ(x − ξ),

0 < x, ξ < L,

which satisfy the following boundary conditions: 22. g(0|ξ) = 0,

g(L|ξ) = 0,

23. g $ (0|ξ) = 0,

g $ (L|ξ) = 0,

24. g(0|ξ) = 0,

g(L|ξ) + g $ (L|ξ) = 0,

25. g(0|ξ) = 0,

g(L|ξ) − g $ (L|ξ) = 0,

26. ag(0|ξ) + g $ (0|ξ) = 0,

g $ (L|ξ) = 0,

27. g(0|ξ) + g $ (0|ξ) = 0,

g(L|ξ) − g $ (L|ξ) = 0.

28. Show that the Green’s function42 for g $$ + k 2 g = −δ(θ − θ$ ),

−π < θ, θ$ < π ,

subject to 2π periodicity is g(θ|θ $ ) = −

cos[k(π − |θ − θ$ |)] . 2k sin(kπ)

41

For an application of Problem 26, see Chakrabarti, A., and T. Sahoo, 1996: Reflection of water waves by a nearly vertical porous wall. J. Austral. Math. Soc., Ser. B , 37, 417– 429. 42 See Felsen, F. B., and N. Marcuvitz, 1973: Radiation and Scattering of Waves. Prentice-Hall, Inc., p. 310.

148

Green’s Functions with Applications

Problem 28

Then use Equation 3.4.34 to prove that δ(θ − θ$ ) =

∞ 1 ( in(θ−θ $ ) e . 2π n=−∞

The figure captioned Problem 28 illustrates this Green’s function. The Green’s function does not exist when k is an integer and there are corresponding gaps in the figure. 29. Show that the Green’s function43 governed by g $$ − ag $ = −δ(x − ξ),

0 < x, ξ < L,

subject to the boundary conditions a g(0|ξ) − g $ (0|ξ) = 0, and g $ (L|ξ) = 0, is g(x|ξ) = eax< −aξ /a. 30. Show that the Green’s function44 governed by g $$ + k 2 g $ = −δ(x − ξ),

0 < x, ξ < L,

subject to the boundary conditions g $ (0|ξ) = g(L|ξ) = 0 is g(x|ξ) =

sin[k(L − x> )] cos(kx< ) . k cos(kL)

43 See Kafarov, V. V., V. V. Shestopalov, I. N. Dorokhov, and G. L. Zheleznova, 1967: Mathematical model of fluid flow in packed beds and a method for determining its parameters. Theoret. Found. Chem. Engng., 1, 240–251. 44 See Harker, K. J., G. S. Kino, and D. L. Eitelbach, 1968: Longitudinal oscillations in bounded one-dimensional nonuniform plasmas. Phys. Fluids, 11, 425–431.

Ordinary Differential Equations

149

Problem 31

31. Find the Green’s function and corresponding bilinear expansion for g $$ + ag $ − k 2 g = −δ(x − ξ),

0 < x, ξ < π ,

subject to the boundary conditions g(0|ξ) = g(π|ξ) = 0. The figure captioned Problem 31 illustrates this Green’s function as x and ξ vary when a = 3 and k = 1. 32. Find the free-space Green’s function for νg $$ − g $ − sg = −δ(x − ξ),

−∞ < x, ξ < ∞,

where s, ν> 0. The figure captioned Problem 32 illustrates this Green’s function as x and ξ vary when ν = 1 and νs = 2.

Problem 32

33. Consider the boundary-value problem45 y $$ − y = 2,

0 < x < 1,

45 Lonseth, A. T., 1954: Approximate solutions to Fredholm-type integral equations. Bull. Amer. Math. Soc., 60, 415–430.

150

Green’s Functions with Applications

with y(0) = 0 and y(1) = 1, Step 1 : Use standard methods and show that y(x) = 2 cosh(x) − 2 + [3 − 2 cosh(1)] sinh(x)/ sinh(1). Step 2 : Rewriting the problem as y $$ = −f (x, y),

y(0) = 0,

y(1) = 1,

where f (x, y) = −2 − y(x), show that y(x) satisfies the integral equation " 1 y(x) + g(x|ξ)y(ξ) dξ = x2 , 0

where g(x|ξ) = x< (1 − x> ). Hint: Use Equation 3.3.47. Step 3 : Verify that the y(x) found in Step 1 does indeed satisfy the integral equation in Step 2 by direct substitution.

Problem 34

34. Show that the free-space Green’s function for , 1 d dg δ(r − ρ) r − k2 g = − , 0 < r, ρ < ∞, r dr dr r is g(r|ρ) = I0 (kr< )K0 (kr> ). The figure captioned Problem 34 illustrates this Green’s function as r and ρ vary when k = 1. 35. Show46 that the Green’s function for , d dg r2 = δ(r − ρ), dr dr

a < r, ρ

1,
) g(r|ρ) = , γ + a + (β − 1)a2 where γ and β are constants. 36. Show47 that the Green’s function for , d 1 dg = δ(r − ρ), dr r dr

0 ≤ r, ρ < a,

with the boundary conditions g(0|ρ) = g(a|ρ) = 0, is r2 g(r|ρ) = < 2

,

2 r> −1 . a2

Problem 37

37. Show48 that the Green’s function for , 1 d dg m2 δ(r − ρ) r − 2 g=− , r dr dr r r

0 < r, ρ < a,

with the boundary conditions limr→0 |g(r|ρ)| < ∞, and g(a|ρ) = 0, is 1 g(r|ρ) = 2m

,

r< r>

-m %

1−

* r +2m & >

a

.

47 See Rostoker, N., and A. Qerushi, 2002: Equilibrium of field reversed configurations with rotation. I. One space dimension and one type of ion. Phys. Plasmas, 9, 3057–3067. 48 See Schecter, D. A., D. H. E. Dubin, A. C. Cass, C. F. Driscoll, I. M. Lansky, and T. M. O’Neil, 2000: Inviscid damping of asymmetries on a two-dimensional vortex. Phys. Fluids, 12, 2397–2412.

152

Green’s Functions with Applications

Problem 38

The figure captioned Problem 37 illustrates this Green’s function as r/a and ρ/a vary while m = 3. 38. Show that the Green’s function for d2 g 1 dg δ(r − ρ) + + α2 g = , dr2 r dr r

0 < r, ρ < 1,

with the boundary conditions g $ (0|ρ) = g $ (1|ρ) = 0, is g(r|ρ) = [aJ0 (αr> ) + bY0 (αr> )] J0 (αr< ) , where aJ1 (α) + bY1 (α) = 0, and b = π/2. The figure captioned Problem 38 illustrates this Green’s function as r and ρ vary while α = 20.

Problem 39

39. Find the Green’s function49 and its corresponding bilinear expansion for 49 See Cockran, J. A., 1988: Unusual identities for special functions from waveguide propagation analyses. IEEE Trans. Microwave Theory Tech., 36, 611–614.

Ordinary Differential Equations

153

the singular differential equation , m2 xg $$ + g $ + k 2 x − g = −δ(x − ξ), x

0 < x < L,

where m > −1. The boundary conditions are limx→0 |g $ (x|ξ)| < ∞, and g $ (L|ξ) = 0. The figure captioned Problem 39 illustrates this Green’s function (divided by k 2 L2 ) as x/L and ξ/L vary while kL = 10.

Problem 40

40. Find the eigenfunction expansion for the Green’s function within a sphere of radius a: , 1 d 1 d2 δ(r − ρ) 2 dg r = (rg) = − , 0 < r, ρ < a, 2 2 r dr dr r dr 4πρr with the boundary conditions limr→0 |g(r|ρ)| < ∞, and g(a|ρ) = 0. The figure captioned Problem 40 illustrates this Green’s function (multiplied by 2π 3 a) for 0.02 ≤ r/a, ρ/a ≤ 1. We have not plotted values that exceed 50. 41. Consider the nonhomogeneous boundary-value problem , d 1 dy 2 + 4 y = −f (x), 1 < x < 2, 2 dx x dx x with the boundary conditions y(1) = y(2) = 0. Step 1 : Show that g(x|ξ) = c1 (x − x2 ) for 1 ≤ x ≤ ξ and g(x|ξ) = c2 (2x − x2 ) for ξ ≤ x ≤ 2. Step 2 : Using the conditions on the Green’s function at x = ξ, show that g(x|ξ) = xξ(2 − x> )(x< − 1).

154

Green’s Functions with Applications

Step 3 : Using Equation 3.3.34, show that " x " y(x) = x(2 − x) ξ(ξ − 1)f (ξ) dξ + x(x − 1) 1

2

x

ξ(2 − ξ)f (ξ) dξ.

Step 4 : If f (x) = 2/x, show that y(x) = x(x − 1)(2 − x). Step 5 : Using Equation 3.3.47, show that the solution becomes y(x) = x(1 + x − x2 ) when the boundary conditions change to y(1) = 1 and y(2) = −2. 42. Consider the singular boundary-value problem50 , d α dy x = f (x, y), 0 < α, x < 1, dx dx

with the boundary conditions y(0) = A, and y(1) = B. Using the previous problem as a guide, show that the solution is given by the integral equation " 1 y(x) = A + (B − A)x1−α + g(x|ξ)f [ξ, y(ξ)] dξ, 0

where

# 1−α $ g(x|ξ) = x1−α x> − 1 /(1 − α).
0, is * + . (ξ/x)p± √ g(x|ξ) = , with p± = 12 −1 ± 1 + 4/a2 . ax a2 + 4 The negative sign applies for 0 < x ≤ ξ while the positive sign holds for ξ ≤ x < ∞. The figure captioned Problem 45 illustrates this Green’s function when 0.1 ≤ x,ξ ≤ 2.5 and a = 1. 46. Find the free-space Green’s function54 governed by g $$ − [s + ρH(−x)]g = −δ(x − ξ), 52 See Chakrabarti, A., and T. Sahoo, 1998: Reflection of water waves in the presence of surface tension by a nearly vertical porous wall. J. Austral. Math. Soc., Ser. B , 39, 308–317. 53

See Kamb, B., and K. A. Echelmeyer, 1986: Stress-gradient coupling in glacier flow: I. Longitudinal averaging of the influence of ice thickness and surface slope. J. Glaciol., 32, 267–284. 54

Linetsky, V., 1999: Step options. Math. Finance, 9, 55–96.

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Green’s Functions with Applications

where s, ρ> 0. Hint: Consider the four separate cases: (1) x ≥ 0, ξ ≥ 0; (2) x ≤ 0, ξ ≥ 0; (3) x ≥ 0, ξ ≤ 0; and (4) x ≤ 0, ξ ≤ 0. 47. Find the free-space Green’s function55 governed by g $$ − [κ2 + CV (x)]g = −δ(x − ξ),

0,< −∞ < x < ∞,

ξ

subject to the boundary condition lim|x|→∞ g(x|ξ) → 0, when (a) V (x) = H(x), (b) V (x) = δ(x), and (c) V (x) = e−|x| when 2κ is not an integer. Hint: For (c), the equation y $$ − (a2 − bex )y = 0 becomes

z 2 u$$ + zu$ + (z 2 − 4a2 )u = 0 √ if y(x) = u(z) and z = w b ex/2 . 48. Show that the generalized Green’s function56 governed by % & d dg (1 − x2 ) = −δ(x − ξ) + 12 , −1 < x, ξ < 1, dx dx subject to the conditions that |g(−1|ξ)| < ∞,

|g(1|ξ)| < ∞,

"

1

g(x|ξ) dx = 0, −1

is g(x|ξ) = − 21 ln[(1 − x< )(1 + x> )] −

1 2

+ ln(2).

49. Find the Green’s function57 for g iv = δ(x − ξ),

0 < x, ξ < b,

subject to the boundary conditions g(0|ξ) = g $ (0|ξ) = g(b|ξ) = g $ (b|ξ) = 0. 55 Koppe, H., 1951: Die Streuung eines Teilchens an einer Potentialschwelle. Naturforschg, Ser. A, 6, 229–233.

Zeit.

56

Kneser, A., 1911: Integralgleichungen und ihre Anwendungen in der mathematischen Physik . Braunschweig, p. 114; Courant, R., and D. Hilbert, 1953: Methods of Mathematical Physics, Vol. I. Interscience Publ., p. 372. 57 See Singh, R., J. Kumar, and G. Nelakanti, 2014: Approximate series solution of fourth-order boundary value problems using decomposition method with Green’s function. J. Math. Chem., 52, 1099–1118.

Ordinary Differential Equations

157

Problem 51

50. Find the Green’s function58 for g iv = δ(x − ξ),

0 < x, ξ < 1,

subject to the boundary conditions g(0|ξ) = g $ (0|ξ) = g $$ (1|ξ) = g $$$ (1|ξ) = 0. 51. Find the Green’s function59 for g iv = δ(x − ξ),

0 < x, ξ < L,

subject to the boundary conditions g(0|ξ) = g $$ (0|ξ) = g(L|ξ) = g $$ (L|ξ) = 0. The figure captioned Problem 51 illustrates this Green’s function (multiplied by 6/L3 ) as x/L and ξ/L vary. 52. Use Fourier transforms to find the Green’s function for g iv − k 4 g = −δ(x − ξ),

−∞ < x, ξ < ∞.

53. Find the Green’s function60 and its corresponding bilinear expansion for g iv − k 4 g = δ(x − ξ),

0 < x, ξ < L,

with the boundary conditions g(0|ξ) = g(L|ξ) = g $$ (0|ξ) = g $$ (L|ξ) = 0. The figure captioned Problem 53 illustrates this Green’s function (multiplied by 100/L3) as x/L and ξ/L vary when kL = 4. 58 See Xu, X., and A. Nadim, 1994: Deformation and orientation of an elastic slender body sedimenting in a viscous liquid. Phys. Fluids, 6, 2889–2893. 59 For an application, see Gao, J., A. S. Selvarathinam, and Y. J. Weitsman, 1999: Analysis of adhesively joined composite beams. J. Sandwich Struct. Mater., 1, 323–339. 60 See Nicholson, J. W., and L. A. Bergman, 1985: On the efficacy of the modal series representation for Green functions of vibrating continuous structures. J. Sound Vibr., 98, 299–304.

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Green’s Functions with Applications

Problem 53

54. Show that the Green’s function61 for g iv − 2α2 g $$ + α4 g = δ(x − ξ),

−1 < x, ξ < 1,

with the boundary conditions g(−1|ξ) = g(1|ξ) = g $ (−1|ξ) = g $ (1|ξ) = 0, is g(x|ξ) = ΦT (x> − 1)GΦ(x< + 1), where



 cosh(αx)  sinh(αx)  Φ(x) =  , x cosh(αx) x sinh(αx)



0 0 1 0 µ11 G=  ∆ 0 −αµ11 0 µ21

0 −αµ11 α2 µ11 −αµ21

 0 µ12  , −αµ12 µ22

∆ = 2α3 [sinh2 (2α) − 4α2 ], µ11 = −2α cosh(2α) − sinh(2α), µ12 = −µ21 = −2α2 sinh(2α), and µ22 = −α2 [2α cosh(2α) − sinh(2α)]. The figure captioned √ Problem 54 illustrates this Green’s function as x and α vary when ξ = 1/ 7.

0 −0.1

g(x| )

−0.2 −0.3 −0.4 −0.5 −0.6 −0.7 10 8

1 6

0.5 4

0 2

−0.5 0

−1

x

Problem 54 61 Li, J.-C., 1982: Solving the Orr-Sommerfeld equation with integral equation method in terms of the Green function. Appl. Math. Mech.-Engl. Ed., 3, 647–655.

Ordinary Differential Equations

159

55. Find the Green’s function62 for g iv − a4 g + Rδ(x − x0 )g = δ(x − ξ),

−∞ < x, x0 , ξ < ∞,

where a > 0, and x0 &= ξ. Step 1 : Taking the Fourier transform of the ordinary differential equation, show that " R ∞ −ix0 (k−τ ) 4 4 (k − a )G(k|ξ) + e G(τ |ξ) dτ = e−ikξ , 2π −∞ where k is the transform variable, or G(k|ξ) =

e−ikξ Rf (ξ)e−ikx0 − , 4 4 k −a k 4 − a4

where f (ξ) =

1 2π

"

∞

−∞

G(τ |ξ)eix0 τ dτ.

Step 2 : Show that f (ξ) = where 1 g0 (x|y) = 2π

"

∞

−∞

g0 (x0 |ξ) , 1 + Rg0 (x0 |x0 )

+ ei(x−y)τ 1 * ia|x−y| −a|x−y| dτ = ie − e . τ 4 − a4 4a3

Hint: Set k = τ in the second equation in Step 1. Then multiply the equation by eix0 τ dτ and integrate from −∞ to ∞. Step 3 : Show that g(x|ξ) = g0 (x|ξ) −

Rg0 (x0 |ξ) g0 (x|x0 ). 1 + Rg0 (x0 |x0 )

56. Find the Green’s function for the singular differential equation , 1 d dg m2 δ(r − ρ) r − 2 g − ωδ(r − 1)g = − , 0 < r, ρ < ∞, r dr dr r r with the boundary conditions limr→0 |g(r|ρ)| < ∞, and limr→∞ g(r|ρ) → 0. Step 1 : If g(r|ρ) = g1 (r|ρ) + g2 (r|ρ), show that g1 and g2 are governed by , 1 d dg1 m2 δ(r − ρ) r − 2 g1 = − , r dr dr r r 62 See DiPerna, D. T., and D. Feit, 2000: A perturbation technique for the prediction of the displacement of a line-driven plate with discontinuities. J. Acoust. Soc. Am., 107, 2004–2010.

160 and

Green’s Functions with Applications , 1 d dg2 m2 r − 2 g2 − ω δ(r − 1)g2 = ω δ(r − 1)g1 . r dr dr r

Step 2 : Show that 1 2m

,

r< r>

ω f (ρ) g2 (r|ρ) = − 2m(2m + ω)

!

rm , r−m ,

g1 (r|ρ) =

-m .

Step 3 : Show that

where f (ρ) = Hint: Show that

!

ρ−m , ρm ,

ρ ρ

r < 1, r > 1,

1,< 1. >

'r=1+ dg2 '' ω − ω g2 (1|ρ) = f (ρ). ' dr r=1− 2m

57. Show that any nonhomogeneous, linear ordinary differential equation A(x)

d2 y dy + B(x) + C(x)y = f (x) dx2 dx

can 1/ be written as Equation 3.3.1 by first multiplying it by the exponential 2 x exp B(ξ) dξ/A(ξ) /A(x) and then factoring.

Chapter 4 Green’s Functions for the Wave Equation In Chapter 3, we showed how Green’s functions could be used to solve initial- and boundary-value problems involving ordinary differential equations. When we approach partial differential equations, similar considerations hold, although the complexity increases. In the next three chapters, we work through the classic groupings of the wave, heat and Helmholtz’s equations. Of these three groups, we start with the wave equation ∇2 u −

1 ∂ 2u = −q(r, t), c2 ∂t2

(4.0.1)

where ∇ is the three-dimensional gradient operator, t denotes time, r is the position vector, c is the phase velocity of the wave, and q(r, t) is the source density. In addition to Equation 4.0.1 it is necessary to state boundary and initial conditions to obtain a unique solution. The condition on the boundary can be either Dirichlet or Neumann or a linear combination of both (Robin condition). The conditions in time must be Cauchy—that is, we must specify the value of u and its time derivative at t = t0 for each point of the region under consideration. 161

162

Green’s Functions with Applications

The purpose of this introductory section is to prove that we can express the solution to Equation 4.0.1 in terms of boundary conditions, initial conditions and the Green’s function, which is found by solving 1 ∂ 2g = −δ(r − r0 )δ(t − τ ), (4.0.2) c2 ∂t2 where r0 denotes the position of the source. Equation 4.0.2 expresses the effect of an impulse as it propagates from r = r0 as time increases from t = τ . For t < τ, causality requires that g(r, t|r0 , τ ) = gt (r, t|r0 , τ ) = 0 if the impulse is the sole source of the disturbance. We also require that g satisfies the homogeneous form of the boundary condition satisfied by u. ∇2 g −

Remark. Although we will use Equation 4.0.2 as our fundamental definition of the Green’s function as it applies to the wave equation, we can also find it by solving the initial-value problem: 1 ∂ 2u = 0, t > τ , (r, u τ ) = 0, ut (r, τ ) = δ(r − r0 ). (4.0.3) c2 ∂t2 Then g(r, t|r0 , τ ) = u(r, t−τ)H(t−τ ). This is most easily seen by introducing a new time variable t$ = t − τ into Equation 4.0.2 and Equation 4.0.3 and noting that the Laplace transform of Equation 4.0.2 and Equation 4.0.3 are identical. # " ∇2 u −

Before we turn to the solution of Equation 4.0.2, we first deal with reciprocity, namely that g(r, t|r0 , τ ) = g(r0 , −τ |r, −t). Our analysis starts with the two equations: ∇2 g(r, t|r0 , τ0 ) −

1 ∂ 2 g(r, t|r0 , τ0 ) = −δ(r − r0 )δ(t − τ0 ), c2 ∂t2

(4.0.4)

and 1 ∂ 2 g(r, −t|r1 , −τ1 ) = −δ(r − r1 )δ(t − τ1 ). (4.0.5) c2 ∂t2 Equation 4.0.5 holds because the delta function is an even function. Multiplying Equation 4.0.4 by g(r, −t|r1 , −τ1 ) and Equation 4.0.5 by g(r, t|r0 , τ0 ), subtracting, and integrating over the volume V and over the time t from −∞ to t$ , where t$ is greater than either τ0 or τ1 , then " t$ """ % g(r, t|r0 , τ0 )∇2 g(r, −t|r1 , −τ1 ) ∇2 g(r, −t|r1 , −τ1 ) −

−∞

V

− g(r, −t|r1 , −τ1 )∇2 g(r, t|r0 , τ0 ) 1 ∂ 2 g(r, −t|r1 , −τ1 ) + 2 g(r, t|r0 , τ0 ) c ∂t2 & 1 ∂ 2 g(r, t|r0 , τ0 ) − 2 g(r, −t|r1 , −τ1 ) dV dt c ∂t2 = g(r0 , −τ0 |r1 , −τ1 ) − g(r1 , τ1 |r0 , τ0 ).

(4.0.6)

The Wave Equation

163

The left side of Equation 4.0.6 can be transformed by Green’s second formula and the identity % & ∂ ∂g(r, −t|r1 , −τ1 ) ∂g(r, t|r0 , τ0 ) g(r, t|r0 , τ0 ) − g(r, −t|r1 , −τ1 ) (4.0.7) ∂t ∂t ∂t ∂ 2 g(r, −t|r1 , −τ1 ) ∂ 2 g(r, t|r0 , τ0 ) = g(r, t|r0 , τ0 ) − g(r, −t|r1 , −τ1 ) 2 ∂t ∂t2 to become " t$ "" % ⊂⊃ g(r, t|r0 , τ0 ) ∇g(r, −t|r1 , −τ1 ) −∞ S & − g(r, −t|r1 , −τ1 ) ∇g(r, t|r0 , τ0 ) · n dS dt """ % 1 ∂g(r, −t|r1 , −τ1 ) + 2 g(r, t|r0 , τ0 ) c ∂t V &t=t$ ∂g(r, t|r0 , τ0 ) − g(r, −t|r1 , −τ1 ) dV. (4.0.8) ∂t t=−∞ The surface integral in Equation 4.0.8 vanishes because both Green’s functions satisfy the same homogeneous boundary conditions on S. The volume integral in Equation 4.0.8 also vanishes for the following reasons: (1) At the lower limit both g(r, −∞|r0 , τ0 ) and its time derivative vanish from the causality condition. (2) At the time t = t$ , g(r, −t$ |r1 , τ1 ) and its time derivative vanish because −t$ is earlier than −τ1 . Thus, the left side of Equation 4.0.6 equals zero and g(r, t|r0 , τ ) = g(r0 , −τ |r, −t). We now establish that the solution to the nonhomogeneous wave equation can be expressed in terms of the Green’s function, boundary conditions and initial conditions. We begin with the equations ∇20 u(r0 , t0 ) −

1 ∂ 2 u(r0 , t0 ) = −q(r0 , t0 ), c2 ∂t20

(4.0.9)

and ∇20 g(r, t|r0 , t0 ) −

1 ∂ 2 g(r, t|r0 , t0 ) = −δ(r − r0 )δ(t − t0 ), c2 ∂t20

(4.0.10)

where we obtain Equation 4.0.9 from a combination of Equation 4.0.2 plus reciprocity. As we did above, we multiply Equation 4.0.9 by g(r, t|r0 , t0 ) and Equation 4.0.10 by u(r0 , t0 ) and subtract. Integrating over the volume V0 and over t0 from 0 to t+ , where t+ denotes a time slightly later than t so that we avoid ending the integration exactly at the peak of the delta function, we obtain , -& " t+ """ % 1 ∂2g ∂2u g ∇20 u − u ∇20 g + 2 u 2 − g dV0 dt0 c ∂t0 ∂t20 0 V0 " t+ """ = u(r, t) − q(r0 , t0 ) g(r, t|r0 , t0 ) dV0 dt0 . (4.0.11) 0

V0

164

Green’s Functions with Applications

Again employing Green’s second formula, Gauss’s divergence theorem and rewriting some terms, we find that "

0

t+

&t+ "" """ % 1 ∂g ∂u ⊂⊃ (g ∇0 u − u ∇0 g) · n dS0 dt0 + 2 u −g dV0 c ∂t0 ∂t0 0 S0 V0 " t+ """ + q(r0 , t0 ) g(r, t|r0 , t0 ) dV0 dt0 = u(r, t). (4.0.12) 0

V0

The integrand in the first integral is specified by the boundary conditions. In the second integral, the integrand vanishes at t = t+ from the initial conditions on g. The limit at t = 0 is determined by the initial conditions. Hence, u(r, t) = +

"

"

0

0

−

t+

1 c2

"""

q(r0 , t0 ) g(r, t|r0 , t0 ) dV0 dt0

V0

t+

"" % ⊂⊃ g(r, t|r0 , t0 ) ∇0 u(r0 , t0 ) S0

""" ! V0

&

− u(r0 , t0 ) ∇0 g(r, t|r0 , t0 ) · n dS0 dt0 % & ∂g(r, t|r0 , 0) u(r0 , 0) ∂t0 % &) ∂u(r0 , 0) − g(r, t|r0 , 0) dV0 . (4.0.13) ∂t0

Equation 4.0.13 gives the complete solution of the nonhomogeneous problem including initial conditions. In the case of the surface integrals, the surface value equals the limit of the value of the function as the surface is approached from the interior. The first two integrals on the right side of this equation represent the effect of the source and the boundary conditions, respectively. The last term involves the initial conditions; it can be interpreted as asking what sort of source is needed so that the function u starts in the desired manner. 4.1 ONE-DIMENSIONAL WAVE EQUATION IN AN UNLIMITED DOMAIN The simplest possible example of Green’s functions for the wave equation is the one-dimensional vibrating string problem.1 In this problem the Green’s function is given by the equation 2 ∂2g 2∂ g − c = c2 δ(x − ξ)δ(t − τ ), ∂t2 ∂x2

(4.1.1)

1 See also Graff, K. F., 1991: Wave Motion in Elastic Solids. Dover Publications, Inc., Section 1.1.8.

The Wave Equation

165

where −∞ < x, ξ < ∞, and 0 < t, τ. If the initial conditions equal zero, the Laplace transform of Equation 4.1.1 is d2 G s2 − 2 G = −δ(x − ξ)e−sτ , dx2 c

(4.1.2)

where G(x, s|ξ,τ ) is the Laplace transform of g(x, t|ξ,τ ). To solve Equation 4.1.2 we take its Fourier transform and obtain the algebraic equation G(k, s|ξ,τ ) =

exp(−ikξ − sτ ) . k 2 + s2 /c2

(4.1.3)

Having found the joint Laplace-Fourier transform of g(x, t|ξ,τ ), we must work our way back to the Green’s function. From the definition of the Fourier transform, we have that e−sτ G(x, s|ξ,τ ) = 2π

"

∞

−∞

eik(x−ξ) dk. k 2 + s2 /c2

(4.1.4)

To evaluate the Fourier-type integral, Equation 4.1.4, we apply the residue theorem. Performing the calculation, G(x, s|ξ,τ ) =

c exp(−sτ − s|x − ξ|/c) . 2s

(4.1.5)

Finally, applying the second shifting theorem, g(x, t|ξ,τ ) =

c H (t − τ − |x − ξ|/c) , 2

(4.1.6)

or

g(x, t|ξ,τ ) =

c H [c(t − τ ) + (x − ξ)] H [c(t − τ ) − (x − ξ)] , 2

(4.1.7)

where H(·) is the Heaviside step function, Equation 2.2.2. Let us show that Equation 4.1.6 satisfies the initial conditions and the differential equation. We begin by noting that g(x, 0|ξ,τ ) =

c H(−τ − |x − ξ|/c) = 0, 2

(4.1.8)

since the argument of the Heaviside step function is always negative. Similarly, gt (x, 0|ξ,τ ) =

c δ(−τ − |x − ξ|/c) = 0, 2

(4.1.9)

166

Green’s Functions with Applications

0.5

g(x,t| , )/c

0.4 0.3 0.2 0.1 0 10 8

10 6

8 6

4

c(t− )/

4

2 0

2

x/

0

Figure 4.1.1: The Green’s function g(x, t|ξ,τ )/c given by Equation 4.1.16 for the one-dimensional wave equation for x > 0 at different distances x/ξ and times c(t − τ ) subject to the boundary condition g(0, t|ξ,τ ) = 0.

as the argument of the delta function is nonzero. To show that the solution satisfies the differential equation, we use Equation 4.1.7 and find that ∂ 2g 1 ∂ 2g − 2 2 = −2c δ[c(t − τ ) + (x − ξ)] δ[c(t − τ ) − (x − ξ)] 2 ∂x c ∂t = −2c δ[c(t − τ ) + (x − ξ)] δ[2(x − ξ)] = −c δ[c(t − τ ) + (x − ξ)] δ(x − ξ) = −δ(t − τ ) δ(x − ξ) .

(4.1.10) (4.1.11) (4.1.12) (4.1.13)

We can use Equation 4.1.6 and the method of images to obtain the Green’s function for ∂ 2g 1 ∂ 2g − 2 2 = δ(x − ξ)δ(t − τ ), 2 ∂x c ∂t

0 < x, t, ξ, τ ,

(4.1.14)

subject to the boundary condition g(0, t|ξ,τ ) = 0. We begin by noting that the free-space Green’s function,2 Equation 4.1.6, is the particular solution to Equation 4.1.14. Therefore, we need only find a 2 In electromagnetic theory, a free-space Green’s function is the particular solution of the differential equation valid over a domain of infinite extent, where the Green’s function remains bounded as we approach infinity, or satisfies a radiation condition there.

The Wave Equation

167

homogeneous solution f (x, t|ξ,τ ) so that g(x, t|ξ,τ ) =

c H (t − τ − |x − ξ|/c) + f (x, t|ξ,τ ) 2

(4.1.15)

satisfies the boundary condition at x = 0. To find f (x, t|ξ,τ ), let us introduce a source at x = −ξ at t = τ . The corresponding free-space Green’s function is H(t − τ − |x + ξ|/c). If, along the boundary x = 0 for any time t, this Green’s function destructively interferes with the free-space Green’s function associated with the source at x = ξ, then we have our solution. This will occur if our new source has a negative sign, resulting in the combined Green’s function g(x, t|ξ,τ ) =

c [H(t − τ − |x − ξ|/c) − H(t − τ − |x + ξ|/c)] . 2

(4.1.16)

See Figure 4.1.1. Because Equation 4.1.16 satisfies the boundary condition, we need no further sources. Let us check and see that Equation 4.1.16 is the solution to our problem. First, direct substitution into Equation 4.1.14 yields δ(x − ξ)δ(t − τ ) − δ(x + ξ)δ(t − τ ), which equals δ(x − ξ)δ(t − τ ) if x > 0. Second, an evaluation of Equation 4.1.16 at x = 0 gives g(0, t|ξ,τ ) = 0. Thus, Equation 4.1.16 is the Green’s function for the problem, Equation 4.1.14, in the half-plane x > 0 because it (1) satisfies the wave equation in the domain x > 0, (2) remains finite as x → ∞, and (3) satisfies the boundary condition g(0, t|ξ,τ ) = 0. In a similar manner, we can use Equation 4.1.6 and the method of images to find the Green’s function for ∂ 2g 1 ∂ 2g − 2 2 = δ(x − ξ)δ(t − τ ), 2 ∂x c ∂t

0 < x, t, ξ, τ ,

(4.1.17)

subject to the boundary condition gx (0, t|ξ,τ ) = 0. We begin by examining the related problem ∂2g 1 ∂ 2g − = δ(x − ξ)δ(t − τ ) + δ(x + ξ)δ(t − τ ), ∂x2 c2 ∂t2

(4.1.18)

where −∞ < x, ξ < ∞, and 0 < t, τ. In this particular case, we have chosen an image that is the mirror reflection of δ(x − ξ). This was dictated by the fact that the Green’s function must be an even function of x along x = 0 for any time t. In line with this argument, g(x, t|ξ,τ ) =

c [H (t − τ − |x − ξ|/c) + H (t − τ − |x + ξ|/c)] . 2

(4.1.19)

To show that this is the correct solution, we compute gx (x, t|ξ,τ ) and find that gx (x, t|ξ,τ ) = − 12 sgn(x − ξ)δ (t − τ − |x − ξ|/c)

− 12 sgn(x + ξ)δ (t − τ − |x + ξ|/c) ,

(4.1.20)

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Green’s Functions with Applications

where the signum function sgn(t) is ! 1, sgn(t) = −1,

t > 0, t < 0.

(4.1.21)

Consequently, gx (0, t|ξ,τ ) = 0. Thus, Equation 4.1.19 is the correct Green’s function because it satisfies Equation 4.1.17 if x, ξ> 0 and the boundary condition gx (0, t|ξ,τ ) = 0. • Example 4.1.1: One-dimensional Klein-Gordon equation The Klein-Gordon equation is a form of the wave equation that arose in particle physics as the relativistic scalar wave equation describing particles with nonzero rest mass. In this example, we find its Green’s function when there is only one spatial dimension: , ∂ 2g 1 ∂ 2g 2 − + a g = −δ(x − ξ)δ(t − τ ), (4.1.22) ∂x2 c2 ∂t2 where −∞ < x, ξ < ∞, 0 < t, τ, c is a real, positive constant (the wave speed) and a is a real, nonnegative constant. The corresponding boundary conditions are lim g(x, t|ξ,τ ) → 0, (4.1.23) |x|→∞

and the initial conditions are g(x, 0|ξ,τ ) = gt (x, 0|ξ,τ ) = 0. that

(4.1.24)

We begin by taking the Laplace transform of Equation 4.1.22 and find , 2 d2 G s + a2 − G = −δ(x − ξ)e−sτ . (4.1.25) dx2 c2

Applying Fourier transforms to Equation 4.1.25, we obtain G(x, s|ξ,τ ) = =

c2 −sτ e 2π c2 −sτ e π

"

"

∞

eik(x−ξ) dk s2 + a2 + k 2 c 2

(4.1.26)

cos[k(x − ξ)] dk. s2 + a2 + k 2 c2

(4.1.27)

−∞ ∞

0

Inverting the Laplace transform and employing the second shifting theorem, c2 g(x, t|ξ,τ ) = H(t − τ ) π

"

∞ 0

√ 1 2 sin (t − τ ) a2 + k 2 c2 cos[k(x − ξ)] √ dk a2 + k 2 c 2 (4.1.28)

The Wave Equation

169

Figure 4.1.2: The free-space Green’s function g(x, t|ξ,τ )/c for the one-dimensional KleinGordon equation at different distances a(x − ξ)/c and times a(t − τ ).

√ 1 2 sin (t − τ ) a2 + k 2 c2 − k|x − ξ| √ dk a2 + k 2 c2 0 √ 1 2 " ∞ sin (t − τ ) a2 + k 2 c2 + k|x − ξ| c2 √ + H(t − τ ) dk 2π a2 + k 2 c2 0 (4.1.29) √ 1 2 " ∞ 2 − a2 |x − ξ|/c sin ω(t − τ ) − ω c √ = H(t − τ ) dω 2π ω 2 − a2 a √ 1 2 " ∞ sin ω(t − τ ) + ω 2 − a2 |x − ξ|/c c √ + H(t − τ ) dω. 2π ω 2 − a2 a (4.1.30) c2 = H(t − τ ) 2π

"

∞

Equation 4.1.28 represents a superposition of homogeneous solutions (normal modes) to Equation 4.1.22. An intriguing aspect of Equation 4.1.29 and Equation 4.1.30 is that this solution occurs everywhere after t > τ . If |x − ξ| > c(t − τ ), these wave solutions destructively interfere so that we have zero there while they constructively interfere at those times and places where the physical waves are present. Applying integral tables to Equation 4.1.28, the final result is

g(x, t|ξ,τ ) =

5 c 4 . J0 a (t − τ )2 − (x − ξ)2 /c2 2 × H[c(t − τ ) − |x − ξ|].

(4.1.31)

170

Green’s Functions with Applications

Figure 4.1.2 illustrates this Green’s function. Let us return to Equation 4.1.26 and invert the Fourier transform via the residue theorem. We obtain . exp[−|x − ξ| (s2 + a2 )/c2 − sτ ] . G(x, s|ξ,τ ) = . (4.1.32) 2 (s2 + a2 )/c2 Instead of using tables to invert the Laplace transform, we employ Bromwich’s integral with the contour running just to the right of the imaginary axis. The square root is computed by taking the branch cuts to run from s = ai out to +∞i and from s = −ai out to −∞i; the cuts are along the imaginary axis. Setting s = ωi, √ 1 2 " ∞ sin ω(t − τ ) − ω 2 − a2 |x − ξ|/c c √ g(x, t|ξ,τ ) = H(t − τ ) dω 2π ω 2 − a2 a √ 1 2 " a exp −|x − ξ| a2 − ω 2 /c cos[ω(t − τ )] c √ + H(t − τ ) dω. 2π a2 − ω 2 0 (4.1.33) Comparing Equation 4.1.30 and Equation 4.1.33, we see that the backward propagating waves in Equation 4.1.30 have become evanescent waves in Equation 4.1.33. This suggests that in certain transient problems there exists a correspondence between the evanescent modes and backward propagating solutions. # " • Example 4.1.2: Equation of telegraphy When the vibrating string problem includes the effect of air resistance,3 Equation 4.1.1 becomes ∂2g ∂g ∂ 2g + 2γ − c2 2 = c2 δ(x − ξ)δ(t − τ ), 2 ∂t ∂t ∂x

(4.1.34)

where −∞ < x, ξ < ∞, and 0 < t, τ, with the boundary conditions, Equation 4.1.23, and the initial conditions, Equation 4.1.24. Let us find the Green’s function. Our analysis begins by introducing an intermediate dependent variable w(x, t|ξ,τ ), where g(x, t|ξ,τ ) = e−γtw(x, t|ξ,τ ). Substituting for g(x, t|ξ,τ ), we now have ∂2w ∂2w − γ 2 w − c2 2 = c2 δ(x − ξ)δ(t − τ )eγτ . 2 ∂t ∂x

(4.1.35)

3 For an alternative derivation, see Walters, A. G., 1951: On the propagation of disturbances from moving sources. Proc. Cambridge Philos. Soc., 47, 109–126.

The Wave Equation

171

Figure 4.1.3: The free-space Green’s function g(x, t|ξ,τ )/c for the one-dimensional equation of telegraphy with γ = 1 at different distances (x − ξ)/c and times t − τ .

Taking the Laplace transform of Equation 4.1.35, we obtain d2 W − dx2

,

s2 − γ 2 c2

-

W = −δ(x − ξ)eγτ −sτ .

(4.1.36)

Using Fourier transforms as in Example 4.1.1, the solution to Equation 4.1.36 is . exp[−|x − ξ| (s2 − γ 2 )/c2 + γτ − sτ ] . W (x, s|ξ,τ ) = . (4.1.37) 2 (s2 − γ 2 )/c2 Employing tables to invert the Laplace transform and the second shifting theorem, we have that w(x, t|ξ,τ ) = or

g(x, t|ξ,τ ) =

5 c γτ 4 . e I0 γ (t − τ )2 − (x − ξ)2 /c2 H[c(t − τ ) − |x − ξ|], 2 (4.1.38)

5 c −γ(t−τ ) 4 . e I0 γ (t − τ )2 − (x − ξ)2 /c2 H[c(t − τ ) − |x − ξ|]. 2 (4.1.39) # "

172

Green’s Functions with Applications

• Example 4.1.3: D’Alembert’s solution Consider the case of an unbounded space that is free of sources. If the initial values are u(x0 , 0) = F (x0 ),

and

∂u(x0 , 0) = G(x0 ), ∂t0

(4.1.40)

then by Equation 4.0.13, 1 2c

"

∞

G(x0 )H(t − |x − x0 |/c) dx0 " ∞ 1 + F (x0 ) δ(t − |x − x0 |/c) dx0 (4.1.41) 2c −∞ " x+ct 1 = G(x0 ) dx0 + [cF (x + ct) − (−c)F (x − ct)] /(2c) (4.1.42) 2c x−ct " 1 x+ct 1 = 2 [F (x + ct) + F (x − ct)] + G(x0 ) dx0 . (4.1.43) 2c x−ct

u(x, t) =

−∞

Equation 4.1.43 is known as d’Alembert’s solution. Note that u(x, t) depends on all of the values of ut(x, 0) within a distance ct of x but depends only on the value of u(x, 0) at the two points exactly at x0 = x ± ct. # " • Example 4.1.4: Wave propagation in a visco-elastic medium Visco-elastic media exhibit the important property of memory that is expressed mathematically in the dynamical equations by time convolution. In this example, we examine the effect of visco-elasticity by finding the freespace Green’s function governed by the protypical equation4 ∂ 2g ∂2g − 2 = [1 + K(t)∗]δ(t − τ )δ(x − ξ), (4.1.44) 2 ∂t ∂x √ with −∞ < x, ξ < ∞, and 0 < t, τ, where K(t) = (a2 + 2a/ πt )H(t). We begin by taking the Laplace transform of Equation 4.1.44 or [1 + K(t)∗]

√ d2 G (s + a s )2 G − = dx2

, a2 2a 1+ + √ e−sτ δ(x − ξ). s s

(4.1.45)

4 See Hanyga, A., and M. Seredy´ nska, 1999: Some effects of the memory kernel singularity on wave propagation and inversion in poroelastic media–I. Forward problems. Geophys. J. Int., 137, 319–335.

The Wave Equation

173

Figure 4.1.4: The free-space Green’s function g(x, t|ξ,τ ) for the one-dimensional wave equation in a visco-elastic medium at different distances x − ξ and times t− τ when a = 0.05

Solving Equation 4.1.45, we have that , √ 1 a G(x, s|ξ,τ ) = + 3/2 e−(s+a s )|x−ξ|−sτ . 2s 2s

(4.1.46)

Taking the inverse Laplace transform and using the second shifting theorem, we obtain the Green’s function g(x, t|ξ,τ ) 1 2

#

2

$

F

a|x − ξ|

G

. H(t − τ − |x − ξ|) 2 t − τ − |x − ξ| 3 % & a t − τ − |x − ξ| a2 (x − ξ)2 + exp − H(t − τ − |x − ξ|). 2 π 4(t − τ − |x − ξ|) (4.1.47) =

1 − a |x − ξ| erfc

Let us compare our solution, Equation 4.1.47, with the solution when the visco-elastic effect is absent, namely Equation 4.1.6. As Figure 4.1.4 shows, the sharp wave front at t − τ = |x − ξ| becomes progressively smoother as time increases. # " • Example 4.1.5: Wave propagation into a plasma So far, we treated the case where the same wave equation holds over the entire domain. Here, we consider the problem of the wave equation ∂ 2 g1 ∂ 2 g1 − = δ(t − τ )δ(x + ξ), ∂t2 ∂x2

ξ

0,>

(4.1.48)

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Green’s Functions with Applications

governing the semi-infinite domain x < 0 while the Klein-Gordon equation ∂ 2 g2 ∂ 2 g2 − + g2 = 0 ∂t2 ∂x2

(4.1.49)

holds for x > 0. At the interface, we have the conditions that g1 (0, t|ξ,τ ) = g2 (0, t|ξ,τ ),

and g1 x (0, t|ξ,τ ) = g2 x (0, t|ξ,τ ).

(4.1.50)

Of course, we also have the boundary conditions at infinity that lim g1 (x, t|ξ,τ ) → 0,

x→−∞

and

lim g2 (x, t|ξ,τ ) → 0,

x→∞

(4.1.51)

and all of the initial conditions equal zero. This problem5 models the propagation of progressive waves into a region where a plasma is present. As usual, we begin by taking the Laplace transform of the governing equations and find that

and

d2 G1 − s2 G1 = −δ(x + ξ)e−sτ , dx2

(4.1.52)

d2 G2 − (s2 + 1)G2 = 0, dx2

(4.1.53)

with G1 (0, s|ξ,τ ) = G2 (0, s|ξ,τ ),

and

G$1 (0, s|ξ,τ ) = G$2 (0, s|ξ,τ ).

(4.1.54)

The solution to Equation 4.1.52 and Equation 4.1.53 is G1 (x, s|ξ,τ ) =

e−s|x+ξ|−sτ + Aesx , 2s

and G2 (x, s|ξ,τ ) = Be−x

√ s2 +1

.

(4.1.55)

(4.1.56)

The first term in Equation 4.1.55 represents the particular solution to Equation 4.1.52 while the remaining terms in Equation 4.1.55 and Equation 4.1.56 5 See Nodland, B., and C. J. McKinstrie, 1997: Propagation of a short laser pulse in a plasma. Phys. Review E , 56, 7174–7178; Khrushchinskii, A. A., and D. Yu. Churmakov, 2000: Passage of an electromagnetic pulse through a layer of homogeneous plasma (exact solution). J. Engng. Phys. Thermophys., 73, 554–562.

The Wave Equation

175

Figure 4.1.5: The free-space Green’s function g(x, t|ξ,τ ) governed by the one-dimensional wave equation for x < 0 and the one-dimensional Klein-Gordon equation for x > 0. The source of excitation at time t = τ occurs at x = −ξ.

are necessary to satisfy the interfacial conditions. The solution e−sx in Equa√ x s2 +1 tion 4.1.55 and e in Equation 4.1.56 are discarded because they violate the boundary conditions at infinity. After satisfying the interfacial condition at x = 0, Equation 4.1.55 and Equation 4.1.56 become √ e−s|x+ξ|−sτ s − s2 + 1 √ $ e−s(ξ−x+τ ), G1 (x, s|ξ,τ ) = + # (4.1.57) 2s 2s s + s2 + 1

and

√

2

e−x s +1−s(ξ+τ ) √ G2 (x, s|ξ,τ ) = . (4.1.58) s + s2 + 1 The physical interpretation of Equation 4.1.57 and Equation 4.1.58 is as follows: The leading term in Equation 4.1.57 represents the direct wave radiating out from the source point x = −ξ. At the interface, some of the energy passes into the region x > 0; this is given by Equation 4.1.58. Some of the energy is reflected and this reflected wave is given by the second term in Equation 4.1.57. Taking the inverse of Equation 4.1.57 and Equation 4.1.58, the corresponding Green’s functions are " t−τ −(ξ−x) J2 (z) g1 (x, t|ξ,τ ) = 12 H(t − τ − |x + ξ|) − H[t − τ − (ξ − x)] dz, z 0 (4.1.59)

176

Green’s Functions with Applications

and 4. 5 (t − τ − ξ)2 − x2 . g2 (x, t|ξ,τ ) = H(t − τ − x − ξ) (t − τ + x − ξ) (t − τ − ξ)2 − x2 4. 5 xJ1$ (t − τ − ξ)2 − x2 + H(t − τ − x − ξ). t−τ +x−ξ (4.1.60) (t − τ − ξ) J1

Figure 4.1.5 illustrates Equations 4.1.59 and 4.1.60 when ξ = 1. The flip side of this problem, the source located x = ξ in the plasma, is left as Problem 7. # " • Example 4.1.6 In the study of the propagation of sound in fluid-saturated porous media with rigid solid frames, the wave equation must be modified to include the attenuation of the wave without dispersion and the dispersion due to the thermal interactions between fluid and structure. In their recent study of this phenomena, Fellah et al.6 solved the Green’s function problem: c2

∂ 2g ∂2g ∂g ∂ 3 g − 2 −a + 3 = 0, 2 ∂x ∂t ∂t ∂t

0 < x < ∞,

0 < t,

(4.1.61)

subject to the boundary conditions g(0, t|0, 0+) = δ(t − 0+ ),

lim g(x, t|0, 0+ ) → 0,

x→∞

(4.1.62)

and the initial conditions g(x, 0|0, 0+) = gt (x, 0|0, 0+ ) = gtt (x, 0|0, 0+ ) = 0.

(4.1.63)

Our analysis begins by taking the Laplace transform of Equation 4.1.61 and Equation 4.1.62

with

$ d2 G 1 # − 2 s2 + as − bs3 G = 0, 2 dx c G(0, s|0, 0+ ) = 1,

0 < x < ∞,

lim G(x, s|0, 0+ ) → 0.

x→∞

(4.1.64)

(4.1.65)

6 Fellah, Z. E. A., M. Fellah, F. G. Mitri, N. Sebaa, W. Lauriks, and C. Depollier, 2007: Transient acoustic wave propagation in air-saturated porous media at low frequencies. J. Appl. Phys., 102, Art. No. 084906.

The Wave Equation

177

The solution to Equation 4.1.64 and Equation 4.1.65 is % . H & + 2 3 G(x, s|0, 0 ) = exp −x s + as − bs c (4.1.66)   LF G √ M √ -2 M √∆ √ 2 , √ s   x bN = exp− s − s s−  , (4.1.67) c 2b 2b

where ∆ = 1 + 4ab > 0. Next, consider the integral "

y

∞

−aζ

e

# √ $ * . + 2 − b2 exp −y a √ I0 b ζ 2 − y 2 dζ = . a 2 − b2

(4.1.68)

Taking the derivative of Equation 4.1.68, " * . + exp −y a2 − b2 = e−ay + by

∞ y

* . + I1 b ζ 2 − y 2 e−aζ . dζ. ζ 2 − y2

(4.1.69)

Consequently, % . H & exp −x s2 + as − bs3 c F √ G √ , √ x ∆s x b √ s = exp − √ + s s− (4.1.70) c 2b 2c b S . √ F √ G R √ " ∞ 2 − bx2 /c2 I [s s − s/(2b)] ζ 1 ζ ∆s . × √ exp − dζ. 2b ζ 2 − bx2 /c2 x b/c For the first term in Equation 4.1.70, : F √ G; 3 , x ∆s x ∆ x2 ∆ −1 L exp − √ = exp − . 16c2bt 4ct3/2 bπ 2c b

(4.1.71)

To invert the second term, we first consider ! %, √ - &) √ √ s L−1 e−ϑ s I0 s s − α (4.1.72) 2b ! % , √ -&) " √ √ 1 1 −1 s dτ √ = L exp −ϑ s − ατ s s − , π −1 2b 1 − τ2 where

√ ζ ∆ ϑ= 2b

and

α=

3

ζ2 −

bx2 . c2

(4.1.73)

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Green’s Functions with Applications

The inversion of the Laplace transform on the right side of Equation √ 4.1.72 is computed via Bromwich’s integral. Due to the presence of s we introduce a branch cut along the negative s-axis. When the complex integration is done, the only nonzero contributions appear from the integrations along the branch cut. Carrying out the calculations, we find that L

−1

!

% , √ -&) √ √ s exp −ϑ s − ατ s s − 2b % , √ -& " √ √ 1 ∞ −zt z = e sin ϑ z − ατ z z + dz. π 0 2b

(4.1.74)

Consequently, −1

L

!

%, √ - &) √ s e I0 s s − α (4.1.75) 2b !" 1 % , ) √ -& " ∞ √ √ 1 z τdτ √ = 2 e−zt sin ϑ z − ατ z z + dz. π 0 2b 1 − τ2 −1 √ −ϑ s

Taking the derivative of Equation 4.1.75, we obtain L−1

!, %, √ √ - &) √ √ s −ϑ√s s s s− e I1 s s − α 2b 2b , √ " ∞ √ 1 z =− 2 e−zt z z + (4.1.76) π 0 2b !" 1 % , ) √ -& √ √ z τdτ √ × cos ϑ z − ατ z z + dz. 2b 1 − τ2 −1

Substituting for ϑ and α in Equation 4.1.76 and utilizing Equation 4.1.70 and Equation 4.1.71, the Green’s function is √

% √ , bx ∆ ∆ x2 g(x, t|0, 0 ) = √ exp − π c 4bt3/2 16bc2 t & " ∞ 2 F (ζ, t) √ . − dζ , π π √bx/c ζ 2 − bx2 /c2 +

where

F (ζ, t) =

"

∞ 0

√

, 1 s s+ N (ζ, s)e−st ds, 2b

(4.1.77)

(4.1.78)

and N (ζ, s) =

"

1

−1

:√

cos

; , -3 2 √ ∆ √ 1 bx y dy . ζ s−y s s+ ζ2 − 2 . 2b 2b c 1 − y2 (4.1.79)

The Wave Equation

179

A problem similar to the previous one involves the propagation of ultrasound in a porous media having a rigid frame. Here the wave equation must be modified to include viscous and thermal interactions between the frame and fluid. Mathematically this problem is of interest due to the presence of the convolution operator: " t 2 2 ∂2g ∂g b ∂ g dτ 2∂ g √ c − 2 −a −√ = 0, 0 < x < ∞, 0 < t, 2 ∂x ∂t ∂t π 0 ∂τ 2 t − τ (4.1.80) subject to the boundary conditions g(0, t|0, 0+) = δ(t − 0+ ),

lim g(x, t|0, 0+ ) → 0,

x→∞

(4.1.81)

and the initial conditions g(x, 0|0, 0+ ) = gt (x, 0|0, 0+ ) = 0.

(4.1.82)

Fellah et al.7 solved this problem. Their method was similar to that shown above; the interested reader can find the details in their paper. Setting ∆2 = b2 − 4a, they showed that % & bx b 2 x2 + g(x, t|0, 0 ) = √ exp − H(t − x/c) 16c2 (t − x/c) 4 π c(t − x/c)3/2 " t−x/c ∆x − H(t − x/c) h(ξ) dξ, (4.1.83) 4cπ 3/2 0

where

h(ξ) =

ξ 3/2

and

.

1 (t −

f (µ) =

• Example 4.1.7

ξ)2

−

x2 /c2

"

1

µ dµ e−f (µ)/2 [f (µ) − 1] . , 1 − µ2 −1

4 . 52 µ∆ (t − ξ)2 − x2 /c2 + b(t − ξ) 8ξ

.

(4.1.84)

(4.1.85) # "

Let us find8 the Green’s function for the cylindrical wave equation , -2 ∂ 2 g 1 ∂g λ ∂ 2g δ(r − 1) + − g − =− δ(t − 0+ ), 0 ≤ r < ∞, 0 < t, ∂r2 r ∂r r ∂t2 r (4.1.86) 7

See Fellah, Z. E. A., M. Fellah, W. Lauriks, C. Depollier, J.-Y. Chapelon and Y. C. Angel, 2003: Solution in time domain of ultrasonic propagation equation in a porous material. Wave Motion, 38, 151–161. 8 See Watanabe, K., 2001: Some exact closed form solutions for transient response of a cylindrically anisotropic elastic solid subjected to three types of a ring source. Zeit. Angew. Math. Mech., 81, 788–792.

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Green’s Functions with Applications

subject to the boundary conditions lim |g(r, t|1, 0+ )| < ∞,

r→0

lim g(r, t|1, 0+ ) → 0,

r→∞

0 < t,

(4.1.87)

and initial conditions g(r, 0|1, 0+) = gt (r, 0|1, 0+ ) = 0,

0 < r < ∞.

(4.1.88)

Here λ is a nonnegative constant. We begin by taking the Laplace transform of Equation 4.1.86 and Equation 4.1.87 and discover that : , -2 ; d2 G 1 dG λ δ(r − 1) + − s2 + G=− , (4.1.89) dr2 r dr r r with the transformed boundary conditions lim |G(r, s|1, 0+ )| < ∞,

lim G(r, s|1, 0+ ) → 0,

r→∞

r→0

where G(r, s|1, 0+ ) =

"

∞

g(r, t|1, 0+)e−st dt.

(4.1.90)

(4.1.91)

0

The solution to Equation 4.1.89 is ! AIλ (sr), + G(r, s|1, 0 ) = BKλ (sr),

0 ≤ r < 1, 1 < r < ∞.

(4.1.92)

Because the Green’s function must be continuous at r = 1, we have that AIλ (s) = BKλ (s).

(4.1.93)

Upon multiplying Equation 4.1.89 by r and integrating from 1− to 1+ , we also have that '1+ dG '' = sBKλ$ (s) − sAIλ$ (s) = −1. (4.1.94) dr ' − 1

Solving for A and B and substituting into Equation 4.1.92, the Laplace transform G(r, s|1, 0+ ) becomes G(r, s|1, 0+ ) = Iλ (sr< )Kλ (sr> ).

(4.1.95)

Upon substituting the integral representation9 " 4. 5 1 π Iλ (a)Kλ (b) = K0 a2 + b2 − 2ab cos(ϕ) cos(λϕ) dϕ (4.1.96) π 0 " ∞ 4. 5 1 − sin(λπ) K0 a2 + b2 + 2ab cosh(y) e−λy dy, π 0 9 Oberhettinger, F., 1958: On the diffraction and reflection of waves and pulses by wedges and corners. J. Res. NBS, Sec. D, 61, 343–365.

The Wave Equation

181

into Equation 4.1.95, we obtain " 4 . 5 1 π + G(r, s|1, 0 ) = K0 s r2 + 1 − 2r cos(ϕ) cos(λϕ) dϕ π 0 " 4 . 5 sin(λπ) ∞ − K0 s r2 + 1 + 2r cosh(y) e−λy dy. π 0

(4.1.97)

Taking the inverse Laplace transform, we find that 4 5 . " π H t − r2 + 1 − 2r cos(ϕ) 1 . g(r, t|1, 0+) = cos(λϕ) dϕ (4.1.98) π 0 t2 − [r2 + 1 − 2r cos(ϕ)] 4 5 . " 2 sin(λπ) ∞ H t − r + 1 + 2r cosh(y) −λy . − e dy. π t2 − [r2 + 1 + 2r cosh(y)] 0 Further simplification yields

" 1 π/2cos{2λ arcsin[p sin(ϕ)]} = π g(r, t|1, 0+ ) = H(r + 1 − t)H(t − |r − 1|) √ dϕ r 0 1 − p2 sin2 (ϕ) 1 + H(t − r − 1) √ r

"

π

cos(2λϕ) dϕ (4.1.99) 2 + cos2 (ϕ) q 0 %= &2λ " π/2 1 + q 2 sin2 (ϕ) − q sin(ϕ) sin(λπ) = − H(t − r − 1) √ dϕ, r 0 1 + q 2 sin2 (ϕ)

where p=

.

t2 − (r − 1)2 4r

and

.

q=

.

t2 − (r + 1)2 . 4r

(4.1.100)

Figure 4.1.6 illustrates this Green’s function when λ = 1.5. # " • Example 4.1.8 Let us solve10 the one-dimensional wave equation on an infinite domain: ∂ 2u ∂ 2u − = cos(ωt)δ[x − X(t)], ∂t2 ∂x2

(4.1.101)

10 See Knowles, J. K., 1968: Propagation of one-dimensional waves from a source in random motion. J. Acoust. Soc. Am., 43, 948–957.

182

Green’s Functions with Applications

2.5

+

g(r,t|1,0 )

5

0

−2.5

−5 3

3 2

2 1

t

1 0

0

r

Figure 4.1.6: The Green’s function g(r, t|1, 0+ ) governed by Equation 4.1.86 through Equation 4.1.88 when λ = 1.5.

subject to the boundary conditions lim|x|→∞ u(x, t) → 0, 0 < t, and initial conditions u(x, 0) = ut (x, 0) = 0, −∞ < x < ∞. Here ω is a constant and X(t) is some function of time. In Problem 1 you will show that the solution to the one-dimensional wave equation ∂ 2u 1 ∂2u − 2 2 = −q(x, t) (4.1.102) 2 ∂x c ∂t is u(x, t) =

"

t+

"

b

q(ξ,τ )g(x, t|ξ,τ ) dξdτ

0 a " t+ %

&ξ=b ∂u(ξ,τ ) ∂g(x, t|ξ,τ ) + g(x, t|ξ,τ ) − u(ξ,τ ) dτ ∂ξ ∂ξ 0 ξ=a & " % 1 b ∂g(x, t|ξ, 0) ∂u(ξ, 0) − 2 u(ξ, 0) − g(x, t|ξ, 0) dξ, c a ∂τ ∂τ

(4.1.103)

where g(x, t|ξ,τ ) is the Green’s function given by ∂ 2g 1 ∂ 2g − = −δ(x − ξ)δ(t − τ ), ∂x2 c2 ∂t2

(4.1.104)

The Wave Equation

183

and a and b are the endpoints of the domain. In the present case, only the first integral in Equation 4.1.103 does not vanish. Substituting the source term q(x, t) = cos(ωt)δ[x − X(t)] and the Green’s function given by Equation 4.1.6, we have that u(x, t) =

" t" 0

= =

1 2 1 2

"

0

"

0

∞

q(ξ,τ )g(x, t|ξ,τ ) dξdτ

−∞ t" ∞ t

−∞

(4.1.105)

cos(ωτ )δ[ξ − X(τ )]H(t − τ ) − |x − ξ|) dξdτ

H[t − τ − |X(τ ) − x|] cos(ωτ ) dτ,

(4.1.106) (4.1.107)

since c = 1. 4.2 ONE-DIMENSIONAL WAVE EQUATION ON THE INTERVAL 0 < x < L One of the classic problems of mathematical physics involves finding the displacement of a taut string between two supports when an external force is applied. The governing equation is ∂2u ∂ 2u − c2 2 = f (x, t), 2 ∂t ∂x

0 < x < L,

0 < t,

(4.2.1)

where c is the constant phase speed. In this section, we find the Green’s function for this problem by considering the following problem: ∂ 2g ∂2g − c2 2 = δ(x − ξ)δ(t − τ ), 2 ∂t ∂x

0 < x, ξ < L,

0 < t, τ ,

(4.2.2)

with the boundary conditions α1 g(0, t|ξ,τ ) + β1 gx (0, t|ξ,τ ) = 0,

0 < t,

(4.2.3)

α2 g(L, t|ξ,τ ) + β2 gx (L, t|ξ,τ ) = 0,

0 < t,

(4.2.4)

0 < x < L.

(4.2.5)

and and the initial conditions g(x, 0|ξ,τ ) = gt (x, 0|ξ,τ ) = 0,

We start by taking the Laplace transform of Equation 4.2.2 and find that d2 G s2 δ(x − ξ) −sτ − 2G = − e , 2 dx c c2

0 < x < L,

(4.2.6)

184

Green’s Functions with Applications

with and

α1 G(0, s|ξ,τ ) + β1 G$ (0, s|ξ,τ ) = 0,

(4.2.7)

α2 G(L, s|ξ,τ ) + β2 G$ (L, s|ξ,τ ) = 0.

(4.2.8)

Problems similar to Equation 4.2.6 through Equation 4.2.8 were considered in the previous chapter. There, solutions were developed in terms of an eigenfunction expansion. Applying the same technique here, G(x, s|ξ,τ ) = e−sτ

∞ ( ϕn (ξ)ϕn (x) , s2 + c2 kn2 n=1

(4.2.9)

where ϕn (x) is the nth orthonormal eigenfunction to the regular SturmLiouville problem $$

ϕ (x) + k 2 ϕ(x) = 0,

0 < x < L,

(4.2.10)

subject to the boundary conditions

and

α1 ϕ(0) + β1 ϕ$ (0) = 0,

(4.2.11)

α2 ϕ(L) + β2 ϕ$ (L) = 0.

(4.2.12)

Taking the inverse of Equation 4.2.9, we have that the Green’s function is

g(x, t|ξ,τ ) =




H(t − τ ).

(4.2.13)

Let us verify that Equation 4.2.13 is indeed the solution to Equation 4.2.2. We begin by computing

( ∂g = ϕn (ξ)ϕn (x) cos[kn c(t − τ )] H(t − τ ) ∂t n=1

( sin[kn c(t − τ )] + ϕn (ξ)ϕn (x) δ(t − τ ) (4.2.14) kn c n=1

( = ϕn (ξ)ϕn (x) cos[kn c(t − τ )] H(t − τ ), (4.2.15) n=1

The Wave Equation

185

because the bracketed term multiplying the delta function equals zero since f (t)δ(t − a) = f (a)δ(t − a). Next,

( ∂2g =− ϕn (ξ)ϕn (x)kn c sin[kn c(t − τ )] H(t − τ ) ∂t2 n=1

( + ϕn (ξ)ϕn (x) cos[kn c(t − τ )] δ(t − τ ) (4.2.16) =− + =−




ϕn (ξ)ϕn (x)kn c sin[kn c(t − τ )] H(t − τ )

+ δ(x − ξ)δ(t − τ ).

(4.2.18)

Therefore, ∂2g ∂2g − c2 2 2 ∂t < ∂x > ∞ 4 5 sin[k c(t − τ )] ( n 2 2 2 $$ = ϕn (ξ) −kn c ϕn (x) − c ϕn (x) H(t − τ ) kn c n=1 + δ(x − ξ)δ(t − τ ) = δ(x − ξ)δ(t − τ )

(4.2.19)

and Equation 4.2.13 satisfies the differential equation, Equation 4.2.2. • Example 4.2.1 Let us use our results to find the Green’s function11 for ∂ 2g ∂ 2g − c2 2 = δ(x − ξ)δ(t − τ ), 2 ∂t ∂x

(4.2.20)

with the boundary conditions g(0, t|ξ,τ ) = g(L, t|ξ,τ ) = 0,

0 < t,

(4.2.21)

and the initial conditions g(x, 0|ξ,τ ) = gt (x, 0|ξ,τ ) = 0,

0 < x < L.

(4.2.22)

11 For an alternative derivation, see Walters, A. G., 1949: The solution of some transient differential equations by means of Green’s functions. Proc. Cambridge Philos. Soc., 45, 69–80.

186

Green’s Functions with Applications

Figure 4.2.1: The Green’s function cg(x, t|ξ,τ ) given by Equation 4.2.25 for the one-dimensional wave equation over the interval 0 < x < L as a function of location x/L and time c(t − τ )/L with ξ/L = 0.2. The boundary conditions are g(0, t|ξ,τ ) = g(L, t|ξ,τ ) = 0.

For this example, the Sturm-Liouville problem is ϕ$$ (x) + k 2 ϕ(x) = 0,

0 < x < L,

(4.2.23)

with the boundary conditions ϕ(0) = ϕ(L) = 0. The nth orthonormal eigenfunction for this problem is ϕn (x) =

3

* nπx + 2 sin . L L

(4.2.24)

Consequently, from Equation 4.2.13, the Green’s function is 2 g(x, t|ξ,τ ) = πc


∞ ( 1 nπξ nπx + nπc(t − τ ) sin sin sin H(t − τ ). n L L L n=1 (4.2.25) # "

• Example 4.2.2 Let us find the Green’s function for 2 ∂ 2g 2∂ g − c = δ(x − ξ)δ(t − τ ), ∂t2 ∂x2

(4.2.26)

with the boundary conditions gx (0, t|ξ,τ ) = gx (L, t|ξ,τ ) = 0,

0 < t,

(4.2.27)

The Wave Equation

187

and the initial conditions g(x, 0|ξ,τ ) = gt (x, 0|ξ,τ ) = 0,

0 < x < L.

(4.2.28)

We begin by taking the Laplace transform of Equation 4.2.26 and Equation 4.2.27. This yields d2 G s2 δ(x − ξ) −sτ − 2G = − e , 2 dx c c2

(4.2.29)

with the boundary conditions G$ (0, s|ξ,τ ) = G$ (L, s|ξ,τ ) = 0.

(4.2.30)

There are two ways that we can express the solution to Equation 4.2.29 and Equation 4.2.30. Applying the technique given in Section 3.3, we have G(x, s|ξ,τ ) =

cosh[s(L − |x − ξ|)/c] + cosh[s(L − x − ξ)/c] −sτ e . (4.2.31) 2sc sinh(sL/c)

At this point, we note that cosh[s(L − χ)/c] es(L−χ)/c + e−s(L−χ)/c = sinh(sL/c) esL/c − e−sL/c 4 5 = e−sχ/c 1 + e−2sL/c + e−4sL/c + · · · 4 5 + e−s(2L−χ)/c 1 + e−2sL/c + e−4sL/c + · · · .

(4.2.32)

Applying this result to each of the hyperbolic terms in Equation 4.2.31 and then inverting each term separately, we obtain g(x, t|ξ,τ ) =

∞ ! 1 ( H(t − τ − |x − ξ|/c − 2nL/c) 2c n=0

+ H(t − τ − (x + ξ)/c − 2nL/c)

+ H[t − τ + |x − ξ|/c − 2(n + 1)L/c]

) + H[t − τ + (x + ξ)/c − 2(n + 1)L/c] .

(4.2.33)

Equation 4.2.33 can be interpreted in terms of multiple reflections of the initial wave at the boundaries. We can also express the solution to Equation 4.2.29 and Equation 4.2.30 as the eigenfunction expansion G(x, s|ξ,τ ) =

∞ 1 −sτ ( cos(nπξ/L) cos(nπx/L) e 2n , L s2 + n2 π 2 c2 /L2 n=0

(4.2.34)

188

Green’s Functions with Applications

Figure 4.2.2: The Green’s function cg(x, t|ξ,τ ) given by Equation 4.2.33 or Equation 4.2.35 for the one-dimensional wave equation over the interval 0 < x < L as functions of location x/L and time c(t − τ )/L with ξ/L = 0.2. The boundary conditions are gx (0, t|ξ,τ ) = gx (L, t|ξ,τ ) = 0.

where 20 = 1 and 2n = 2 if n > 0. Equation 4.2.34 follows from Equation 4.2.31 by employing the Mittag-Leffler expansion theorem from complex variable theory. A term-by-term inversion of the Laplace transform yields the Green’s function (t − τ )H(t − τ ) (4.2.35) L , ∞ * nπx + 4 nπc 5 2H(t − τ ) ( 1 nπξ + cos cos sin (t − τ ) . cπ n L L L n=1

g(x, t|ξ,τ ) =

4.3 AXISYMMETRIC VIBRATIONS OF A CIRCULAR MEMBRANE An important one-dimensional problem involves the oscillations of a circular membrane that is clamped along the boundary r = a. To find its Green’s function, we must solve the partial differential equation , 1 ∂ 2g ∂g ∂ 2g 1 ∂g δ(r − ρ)δ(t − τ ) 2 + 2b + α g − − = , (4.3.1) c2 ∂t2 ∂t ∂r2 r ∂r 2πr with 0 < r, ρ < a, and 0 < t, τ, subject to the boundary conditions lim |g(r, t|ρ,τ )| < ∞,

r→0

g(a, t|ρ,τ ) = 0,

0 < t,

(4.3.2)

and the initial conditions g(r, 0|ρ,τ ) = gt (r, 0|ρ,τ ) = 0,

0 < r < a.

(4.3.3)

The Wave Equation

189

Figure 4.3.1: The Green’s function ag(x, t|ξ,τ )/c for a circular membrane as functions of location r/a and time c(t − τ )/a when ρ/a = 0.3, b = 1, and a2 (α2 − b2 )/c2 = 1.

We begin by taking the Laplace transform of Equation 4.3.1 and find that d2 G 1 dG s2 + 2bs + α2 δ(r − ρ) −sτ + − G=− e . (4.3.4) 2 2 dr r dr c 2πr To solve Equation 4.3.4, we first expand the right side of this equation as a Fourier-Bessel series or ∞ δ(r − ρ) 1 ( J0 (kn ρ/a)J0 (kn r/a) = , (4.3.5) 2πr πa2 n=1 J12 (kn )

where kn is the nth root of J0 (k) = 0. This series has the advantage that it vanishes along r = a. Because the delta function, Equation 4.3.5, is a Fourier-Bessel series, we anticipate that G(r, s|ρ,τ ) will be of the form G(r, s|ρ,τ ) =

∞ (

An J0 (kn r/a).

(4.3.6)

n=1

Substitution of Equation 4.3.5 and Equation 4.3.6 into Equation 4.3.4 yields ∞ c2 e−sτ ( J0 (kn ρ/a)J0 (kn r/a) G(r, s|ρ,τ ) = . (4.3.7) 2 2 πa n=1 [(s + b) + α2 − b2 + c2 kn2 /a2 ]J12 (kn )

A straightforward application of tables and the second shifting theorem gives g(r, t|ρ,τ ) =

c2 −b(t−τ ) e H(t − τ ) πa2 . ∞ ( J0 (kn ρ/a)J0 (kn r/a) sin[(t − τ ) c2 kn2 /a2 + α2 − b2 ] . × , J12 (kn ) c2 kn2 /a2 + α2 − b2 n=1

(4.3.8)

190

Green’s Functions with Applications

where we assumed that the damping is sufficiently weak so that c2 kn2 /a2 +α2 > b2 for all n. 4.4 TWO-DIMENSIONAL WAVE EQUATION IN AN UNLIMITED DOMAIN Consider the two-dimensional wave equation12 ∂ 2g ∂ 2g 1 ∂2g + − = −δ(x − ξ)δ(y − η)δ(t − τ ), ∂x2 ∂y 2 c2 ∂t2

(4.4.1)

where −∞ < x, y, ξ, η < ∞, and 0 < t, τ. If all of the initial conditions are zero, the transformed form of Equation 4.4.1 is , d2 G 1 2 2 − s α + 2 G = −δ(y − η)e−ikξ e−sτ , dy 2 c

(4.4.2)

where we took the Laplace transform with respect to time and the Fourier transform with respect to x and replaced the Fourier transform parameter k with αs. The solution of Equation 4.4.2 that tends to zero as |y − η| → ∞ is G(α, y, s|ξ, η,τ ) = where

1 −iαsξ−sβ|y−η|−sτ e , 2sβ

, -1/2 1 α2 + 2 , c

β(α) =

(4.4.3)

(4.4.4)

with 3(β) ≥ 0. Inverting the Fourier transform yields e−sτ G(x, y, s|ξ, η,τ ) = 4π

"

∞

−∞

e−s[β|y−η|−iα(x−ξ)] dα. β

(4.4.5)

With the substitution α = iw, Equation 4.4.5 becomes G(x, y, s|ξ, η,τ ) =

e−sτ 4πi

with β=

"

∞i

−∞i

,

e−s[w(x−ξ)+β|y−η|) dw, β

1 − w2 c2

-1/2

.

(4.4.6)

(4.4.7)

12 Patterned after Gopalsamy, K., and B. D. Aggarwala, 1972: Propagation of disturbances from randomly moving sources. Zeit. Angew. Math. Mech., 52, 31–35. See also Walters, A. G., 1951: On the propagation of disturbances from moving sources. Proc. Cambridge Philos. Soc., 47, 109–126; Graff, K. F., 1991: Wave Motion in Elastic Solids. Dover Publications, Inc., pp. 285–288.

The Wave Equation

191

To evaluate Equation 4.4.6, we deform the line integration from the imaginary axis in the w-plane to one where w(x − ξ) + β|y − η| = t

(4.4.8)

is real and positive. This change of variable and deformation of the original contour results in Equation 4.4.6 becoming a forward Laplace transform so that we can determine the inversion by inspection—the so-called Cagniard-de Hoop technique.13 In the present case, we must deform our integral to the new Cagniard contour with great care. There are branch points at w = 1/c and w = −1/c. To ensure that 3(β) ≥ 0, we take the branch cuts to lie along the real axis from 1/c to ∞ and from −1/c to −∞. From Equation 4.4.8, a little algebra gives (x − ξ)t i|y − η| w= ± r2 r2

3

t2 −

r2 , c2

r < t < ∞, c

(4.4.9)

where r2 = (x − ξ)2 + (y − η)2 . Furthermore, along this hyperbola |y − η|t (x − ξ)i β= ∓ r2 r2 and

3

t2 −

r2 , c2

∂w iβ = ±. . ∂t t2 − r2 /c2

(4.4.10)

(4.4.11)

Upon using the symmetry of the path of integration, Equation 4.4.6 becomes the t-integral G(x, y, s|ξ, η,τ ) =

e−sτ 2π

"

∞

r/c

.

* rs + e−st e−sτ dt = K0 . 2π c t2 − r2 /c2

(4.4.12)

Consequently, by inspection and the second shifting theorem,

g(x, y, t|ξ, η,τ ) =

1 H(t − τ − r/c) . , 2π (t − τ )2 − r2 /c2

(4.4.13)

the Green’s function of the two-dimensional wave equation. 13 De Hoop, A. T., 1960: A modification of Cagniard’s method for solving seismic pulse problems. Appl. Sci. Res., B8, 349–356.

192

Green’s Functions with Applications

An interesting aspect of Equation 4.4.13 follows from a comparison of it with the Green’s function that we found in one dimension, Equation 4.1.6. Here our Green’s function decays as (t−τ )−1 . Consequently, in addition to the abrupt change at t − τ = r/c, we also have a tail that theoretically continues forever. In the next section, when we continue to the three-dimensional case, we again have a sharp propagation boundary. Let us now modify this problem so that we have dissipation.14 The governing equation becomes ∂ 2g ∂2g 1 ∂ 2g ∂g + − −b = −δ(x − ξ)δ(y − η)δ(t − t$ ), ∂x2 ∂y 2 c2 ∂t2 ∂t

(4.4.14)

with −∞ < x, y, ξ, η < ∞, and 0 < t, t$ . Because we are finding the free-space Green’s function, we can translate the origin of the spatial and temporal coordinates to the point of excitation without loss of generality. Using polar coordinates and invoking symmetry, Equation 4.4.14 becomes ∂ 2g 1 ∂g 1 ∂ 2g ∂g 1 + − −b = − δ(r)δ(τ ), ∂r2 r ∂r c2 ∂τ 2 ∂τ r

(4.4.15)

where x − ξ = r cos(θ), y − η = r sin(θ), and τ = t − t$ . Our boundary conditions are lim |gr (r,τ |0+ , 0+ )| < ∞,

lim g(r,τ |0+ , 0+ ) → 0,

r→∞

r→0

0 < τ , (4.4.16)

and the initial conditions g(r, 0|0+ , 0+ ) = gτ (r, 0|0+ , 0+ ) = 0 for 0 ≤ r < ∞. We begin our analysis by taking the Laplace transform of Equation 4.4.15 and find that , 2 d2 G 1 dG s s 1 + − + 2 G = − δ(r), (4.4.17) 2 2 dr r dr c c ζ r where ζ = (bc2 )−1 , the relaxation time scale. Next, we introduce the Hankel transform " ∞ + + G(k, s|0 , 0 ) = G(r, s|0+ , 0+ )J0 (kr) r dr. (4.4.18) 0

Taking the Hankel transform of Equation 4.4.17, we obtain the algebraic equation: 1 G(k, s|0+ , 0+ ) = 2 , (4.4.19) $ k + s (s$ + β) 14 See Sezginer, A., and W. C. Chew, 1984: Closed form expression of the Green’s function for the time-domain wave equation for a lossy two-dimensional medium. IEEE Trans. Antennas Propagat., AP-32, 527–528; Aleixo, R., and E. Capelas de Oliveira, 2008: Green’s function for the lossy wave equation. Rev. Bras. Ens. Fis., 30, Art. No. 1302. For an alternative derivation, see Walters, A. G., 1951: On the propagation of disturbances from moving sources. Proc. Cambridge Philos. Soc., 47, 109–126.

The Wave Equation

193

where s$ = s/c and β = (cζ)−1 . Therefore, " ∞ $ + + G(r, s |0 , 0 ) = G(r, s$ |0+ , 0+ )J0 (kr) k dk 0 " ∞ k J0 (kr) = dk 2 + s$ (s$ + β) k 0 * . + = K0 r s$2 + s$ β .

(4.4.20) (4.4.21) (4.4.22)

Using the scaling property, first shifting theorem and tables for Laplace transforms, we find that # √ $ 2 2 2 + + −'cτ cosh 2 c τ − r √ g(r,τ |0 , 0 ) = e H(cτ − r) (4.4.23) c2 τ 2 − r 2 with 2 = β/2. The final result is 4 . 5 cosh 2 c2 (t − t$ )2 − r2 . g(x, y, t|ξ, η, t$ ) = e−'c(t−t ) H[c(t−t$ )−r], (4.4.24) c2 (t − t$ )2 − r2 $

where r2 = (x − ξ)2 + (y − η)2 . In the limit of b → 0, ζ → ∞, β → 0 and we recover Equation 4.4.13. • Example 4.4.1

Let us find15 the Green’s function for the cylindrical wave equation , -2 2 ∂ 2g 1 ∂g 2ν ∂ 2 g λ ∂ g 1 ∂ 2g δ(r − ρ) + + + − 2 2 =− δ(θ)δ(t − 0+ ) 2 2 ∂r r ∂r r ∂r∂θ r ∂θ c ∂t r (4.4.25) for 0 ≤ r, ρ< ∞, 0 ≤ θ < 2π, and 0 < t, subject to the boundary conditions lim |g(r, θ, t|ρ, 0, 0+ )| < ∞,

r→0

lim g(r, θ, t|ρ, 0, 0+ ) → 0,

r→∞

0 ≤ θ < 2π, 0 < t,

(4.4.26)

g(r, θ, t|ρ, 0, 0+ ) is periodic. The initial conditions are g(r, θ, 0|ρ, 0, 0+) = gt (r, θ, 0|ρ, 0, 0+ ) = 0,

0 ≤ r < ∞,

Here it is assumed that λ > ν. Introducing the Laplace transform " ∞ G(r, θ, s|ρ, 0, 0+ ) = g(r, θ, t|ρ, 0, 0+ )e−st dt,

0 ≤ θ < 2π. (4.4.27)

(4.4.28)

0

15 See Watanabe, K., and R. G. Payton, 2002: Green’s function for SH-waves in a cylindrically monoclinic material. J. Mech. Phys. Solids, 50, 2425–2439.

194

Green’s Functions with Applications

and finite Fourier transform G(r, n, s|ρ, 0, 0+ ) =

"

π

G(r, θ, s|ρ, 0, 0+ )einθ dθ,

(4.4.29)

−π

Equation 4.4.25 becomes the ordinary differential equation d2 G 1 dG 2iνn dG + − − dr2 r dr r dr

,

-2

s2 δ(r − ρ) G=− , c2 r

(4.4.30)

lim G(r, n, s|ρ, 0, 0+ ) → 0.

(4.4.31)

λn r

G−

with the transformed boundary conditions lim |G(r, n, s|ρ, 0, 0+ )| < ∞,

r→0

r→∞

The solution to Equation 4.4.30 is G(r, n, s|ρ, 0, 0+) = (r/ρ)iνn Ip|n| (r< s/c)Kp|n| (r> s/c),

(4.4.32)

√ where p = λ2 − ν 2 . See Equation 4.1.89 through Equation 4.1.92. Taking the inverse of the finite Fourier transform of Equation 4.4.32, we find that 2π 2 G(r, θ, s|ρ, 0, 0+ ) %" π ∞ ( = K0 (sr1 /c) dϕ exp {−in [θ − ν ln(r/ρ)]} cos(pnϕ) 0

−

"

∞

n=−∞ ∞ (

K0 (sr2 /c) dy

0

n=−∞

! exp −in [θ − ν ln(r/ρ)] )

& − p|n|y sin(p|n|ϕ) ,

(4.4.33)

. ρ2 + r2 + 2ρr cosh(y).

(4.4.34)

where r1 =

.

ρ2 + r2 − 2ρr cos(ϕ),

r2 =

To obtain Equation 4.4.33 we used the integral representation16

" 4. 5 1 π K0 a2 + b2 − 2ab cos(ϕ) cos(λϕ) dϕ (4.4.35) π 0 " ∞ 4. 5 1 − sin(λπ) K0 a2 + b2 + 2ab cosh(y) e−λy dy, π 0

Iλ (a)Kλ (b) =

if b > a. 16

Oberhettinger, op. cit.

The Wave Equation

195

To eliminate the summations in Equation 4.4.33, we employ 1 π

∞ (

∞ (

exp(−ina) cos(bn) =

n=−∞

δ(a + b + 2mπ) +

m=−∞

∞ (

m=−∞

δ(a − b + 2mπ), (4.4.36)

and 2

∞ (

n=−∞

exp(−ina − b|n|) sin(c|n|) =

sin(a + c) cosh(b) − cos(a + c)

−

sin(a − c) . cosh(b) − cos(a − c)

and find that +

2πG(r, θ, s|ρ, 0, 0 ) =

" ∞ (

(4.4.37)

π

m=−∞ −π " ∞

K0 (sr1 /c)δ[θ − ν ln(r/ρ) + pϕ + 2mπ] dϕ

! K0 (sr2 /c)

sin[θ − ν ln(r/ρ) + πp] cosh(py) − cos[θ − ν ln(r/ρ) + πp] 0 ) sin[θ − ν ln(r/ρ) − πp] − dy. (4.4.38) cosh(py) − cos[θ − ν ln(r/ρ) − πp]

1 − 2π

Upon taking the inverse Laplace transform, " π ∞ ( H(ct − r1 ) + 2πg(r, θ, t|ρ, 0, 0 ) = c δ[θ − ν ln(r/ρ) + pϕ + 2mπ] . dϕ c2 t2 − r12 m=−∞ −π ! " ∞ 1 H(ct − r2 ) sin[θ − ν ln(r/ρ) + πp] . − 2 cosh(py) − cos[θ − ν ln(r/ρ) + πp] 2 2 2π 0 c t − r2 ) sin[θ − ν ln(r/ρ) − πp] − dy. cosh(py) − cos[θ − ν ln(r/ρ) − πp] (4.4.39) Simplifying the integrals, we obtain the final answer 2πg(r, θ, t|ρ, 0, 0+ ) =

H(ct − r − ρ) − 2π

"

∞ c ( H[pπ − |θ − ν ln(r/ρ) + 2mπ|] p m=−∞ * + . H ct − ρ2 + r2 − 2ρr cos{[θ − ν ln(r/ρ) + 2mπ]/p} × . c2 t2 − ρ2 − r2 + 2ρr cos{[θ − ν ln(r/ρ) + 2mπ]/p} cosh−1 [(c2 t2 −r 2 −ρ2 )/(2ρr)]

dy . 2 2 2 c t − ρ − r2 − 2ρr cosh(y) 0 ! sin[θ − ν ln(r/ρ) + πp] × (4.4.40) cosh(py) − cos[θ − ν ln(r/ρ) + πp] ) sin[θ − ν ln(r/ρ) − πp] − . cosh(py) − cos[θ − ν ln(r/ρ) − πp]

196

Green’s Functions with Applications # "

• Example 4.4.2 Let us find17 the Green’s function for the cylindrical wave equation , ∂ 2g 1 + p ∂g 1 ∂ 2g 1 ∂ 2g δ(r − 1) + + 2 2 − 2ν 2 = − p+1 δ(θ)δ(t − 0+ ) (4.4.41) 2 ∂r r ∂r r ∂θ r ∂t r for the domain 0 ≤ r < ∞, 0 ≤ θ < 2π and 0 < t, subject to the boundary conditions lim |g(r, θ, t|1, 0, 0+)| < ∞, lim g(r, θ, t|1, 0, 0+ ) → 0, r→∞

r→0

0 ≤ θ < 2π, 0 < t,

(4.4.42)

and initial conditions g(r, θ, 0|1, 0, 0+) = gt (r, θ, 0|1, 0, 0+ ) = 0,

0 ≤ θ < 2π. (4.4.43) Here it is assumed that ν &= 1; Watanabe18 treated this as a special case. We begin by noting that the Green’s function must be periodic. Therefore, we introduce the finite Fourier transform: 1 fn = 2π

"

π

f (θ)e

−inθ

dθ,

f (θ) =

−π

0 ≤ r < ∞,

∞ (

fn einθ .

(4.4.44)

n=−∞

Taking the Laplace transform and finite Fourier transform of Equation 4.4.41, we obtain , %* + * + & d2 G 1 + p dG n 2 s 2 δ(r − 1) + − + ν G=− , (4.4.45) 2 dr r dr r r 2πrp+1 with the transformed boundary conditions lim |G(r, n, s|1, 0, 0+ )| < ∞,

r→0

lim G(r, n, s|1, 0, 0+ ) → 0.

r→∞

The solution to Equation 4.4.45 is # $ ! r−p/2 Iνn βsr1−ν Kνn (βs), # $ G(r, n, s|1, 0, 0+) = Iνn (βs)Kνn βsr1−ν , 2πβ where νn = β

.

(4.4.46)

0 ≤ r < 1, 1 < r < ∞, (4.4.47)

n2 + (p/2)2 and β = |1 − ν|−1 .

17 See Watanabe, K., and R. G. Payton, 2004: Green’s function and its non-wave nature for SH-wave in inhomogeneous elastic solid. Int. J. Engng. Sci., 42, 2087–2106. 18

Watanabe, K., 1982: Transient response of an inhomogeneous elastic solid to an impulsive SH-source (variable SH-wave velocity). Bull. JSME , 25, 315–320.

The Wave Equation

197

Upon substituting the integral representation for the product of Bessel functions, Equation 4.1.96, into Equation 4.4.47, we obtain ! " π 1 2πG(r, n, s|1, 0, 0+) = βr−p/2 K0 [βsr1 (ϕ)] cos(νn ϕ) dϕ (4.4.48) π 0 ) " sin(νn π) ∞ − K0 [βsr2 (η)] e−νn η dη , π 0 where = = r1 (ϕ) = 1 + r2(1−ν) − 2r1−ν cos(ϕ), r2 (η) = 1 + r2(1−ν) + 2r1−ν cosh(η). (4.4.49) Taking the inverse Laplace transform, we find that ! " π H[t − βr1 (ϕ)] + −p/2 1 . 2πg(r, n, t|1, 0, 0 ) = βr cos(νn ϕ) dϕ (4.4.50) π 0 t2 − β 2 r12 (ϕ) ) " sin(νn π) ∞ H[t − βr2 (η)] −νn η . − e dη . π t2 − β 2 r22 (η) 0 Taking the inverse of the finite Fourier transform and interchanging the order of summation and integration, ! " π ∞ ( H[t − βr1 (ϕ)] + −p/2 1 . 2πg(r, θ, t|1, 0, 0 ) = βr dϕ cos(νn ϕ)einθ π 0 t2 − β 2 r12 (ϕ) n=−∞ ) " ∞ ∞ ( 1 H[t − βr2 (η)] inθ−νn η . − dη e . π 0 t2 − β 2 r22 (η) n=−∞ (4.4.51)

Finally, evaluating the two summations, we obtain the Green’s function: ∞ ( 2πrp/2 g(r, θ, t|1, 0, 0+) = H(βπ − |θ − 2nπ|) n=−∞

%

! × H(t − βR1 ) . ×

"

(β/2)

0

√

1

t2 − β 2 R12

β 2 ϕ2∗ −(θ−2nπ)2

1 # $ 2 + H β 1 + r1−ν − t

J1 (x) dx . t2 − β 2 R22 $2 1−ν

)

(4.4.52)

1 # +H t−β 1+r & " (β/2)√β 2 π2 −(θ−2nπ)2 J1 (x) dx . × t2 − β 2 R22 0 " ∞ η ∗ # $2 ( β2p 1 dη . − H t − β 1 + r1−ν 2 − β 2 R2 π t 0 3 n=−∞ 4 5 ! (η − πi)K (p/2).β 2 (η − πi)2 + (θ − 2nπ)2 ) 1 . ×5 , β 2 (η − πi)2 + (θ − 2nπ)2

198

Green’s Functions with Applications

where

= R1 = 1 + r2(1−ν) − 2r1−ν cos[(θ − 2nπ)/β], L  T  M , -2 M 1 2x M R2 = N1 + r2(1−ν) − 2r1−ν cos + (θ − 2nπ)2 , β p = 1 + r2(1−ν) + 2r1−ν cosh(η), T  2 − (1 − r1−ν )2 (t/β) , φ∗ = 2 arcsin 4r1−ν R3 =

and

−1

η∗ = cosh

%

(t/β)2 − 1 − r2(1−ν) 2r1−ν

&

(4.4.53)

(4.4.54)

(4.4.55)

(4.4.56)

.

(4.4.57) # "

• Example 4.4.3 Let us find19 the free-space Green’s function for the axisymmetric wave equation , ∂ 2g 1 ∂ ∂g ∂ 2g δ(r − ρ) − r − 2 = δ(z)δ(t − 0+ ) 2 ∂t r ∂r ∂r ∂z 2πr

(4.4.58)

for 0 ≤ r, ρ< ∞, −∞ < z < ∞, and 0 < t, subject to the boundary conditions lim |g(r, z, t|ρ, 0, 0+)| < ∞,

lim g(r, z, t|ρ, 0, 0+) → 0,

r→∞

r→0

|z| < ∞, 0 < t,

(4.4.59)

and lim g(r, z, t|ρ, 0, 0+) → 0,

|z|→∞

0 ≤ r < ∞,

0 < t.

(4.4.60)

The initial condition is g(r, z, 0|ρ, 0, 0+) = gt (r, z, 0|ρ, 0, 0+) = 0,

0 ≤ r < ∞,

|z| < ∞. (4.4.61)

Introducing the Hankel transform G(k, z, t|ρ, 0, 0+) =

"

∞

g(r, z, t|ρ, 0, 0+)J0 (kr) r dr,

(4.4.62)

0

19 Borisov, V. V., and I. I. Simonenko, 1994: Transient waves generated by a source on a circle. J. Phys., Ser. A, 27, 6243–6252.

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199

or +

g(r, z, t|ρ, 0, 0 ) =

"

∞

G(k, z, t|ρ, 0, 0+)J0 (kr) k dk,

(4.4.63)

0

Equation 4.4.58 becomes the partial differential equation ∂ 2G ∂ 2G J0 (kρ) − + k2 G = δ(z)δ(t − 0+ ) 2 2 ∂t ∂z 2π for −∞ < z < ∞, and 0 < t, subject to the boundary condition lim G(k, z, t|ρ, 0, 0+) → 0,

|z| < ∞,

|z|→∞

(4.4.64)

(4.4.65)

and the initial condition G(k, z, 0|ρ, 0, 0+) = Gt (k, z, 0|ρ, 0, 0+) = 0,

|z| < ∞.

(4.4.66)

Using the Riemann method,20 we find that " t " −τ +z+t 1 g(r, z, t|ρ, 0, 0+) = δ(τ ) dτ δ(ζ) dζ (4.4.67) 4π 0 τ +z−t " ∞ 4 . 5 × J0 k (t − τ )2 − (z − ζ)2 J0 (kρ)J0 (kr) k dk =

=

= where t1 = t −

.

1 4π

"

0 t−z

δ(τ ) dτ

0

1 4π 2 rρ

"

0

"

4 . 5 J0 k (t − τ )2 − z 2 J0 (kρ)J0 (kr) k dk

∞

(4.4.68)

t2

max(0,t1 )

.

1−

[r2

+

ρ2

δ(τ ) dτ + z 2 − (t − τ )2 ]2 /(4ρ2 r2 ) (4.4.69)

1 H(t2 ) − H(t1 ) . , 2π 2 4ρ2 r2 − (r2 + ρ2 + z 2 − t2 )2

(r + ρ)2 + z 2

and t2 = t −

.

(r − ρ)2 + z 2 .

(4.4.70)

(4.4.71)

To obtain Equation 4.4.69 from Equation 4.4.68 we applied results from Gradshteyn and Ryzhik.21 We can now generalize results to any arbitrary source. Our Green’s function becomes g(r, z, t|ρ, ζ,τ ) =

1 H(t2 ) − H(t1 ) . , 2 2 2 2 2π 4ρ r − [r + ρ2 + (z − ζ)2 − (t − τ )2 ]2

(4.4.72)

20 Martin, M. H., 1951: Riemann’s method and the problem of Cauchy. Bull. Amer. Math. Soc., Ser. 2 , 57, 238–249; E. C. Young, 1968: Riemann’s method and the characteristic value and Cauchy problems for the damped wave equation. Glasgow Math. J., 10, 147–152. 21 Gradshteyn, I. S., and I. M. Ryzhik, 1965: Tables of Integrals, Series, and Products. Academic Press, Formula 6.578, Number 8.

200

Green’s Functions with Applications

where t1 and t2 now equal . t1 = t − τ − (r − ρ)2 + (z − ζ)2

and t2 = t − τ −

. (r + ρ)2 + (z − ζ)2 . (4.4.73)

• Example 4.4.4 Watanabe and Payton22 found the Green’s function for the cylindrical wave equation ∂ 2 g 1 ∂g g ∂2g β ∂g ∂2g 1 ∂2g δ(r − ρ) + − 2 +2β + +α2 2 − 2 2 = − δ(z)δ(t−0+ ) 2 ∂r r ∂r r ∂r∂z r ∂z ∂z c ∂t r (4.4.74) for 0 ≤ r, ρ< ∞, |z| < ∞, and 0 < t, subject to the boundary conditions lim |g(r, z, t|ρ, 0, 0+)| < ∞, lim g(r, z, t|ρ, 0, 0+) → 0, r→∞

r→0

|z| < ∞,

0 < t, (4.4.75)

and lim g(r, z, t|ρ, 0, 0+) → 0,

|z|→∞

0 ≤ r < ∞,

0 < t,

(4.4.76)

The initial conditions are g(r, θ, 0|ρ, 0, 0+) = gt (r, θ, 0|ρ, 0, 0+ ) = 0,

0 ≤ r < ∞,

|z| < ∞. (4.4.77) Here it is assumed that α > β. Using Laplace and Fourier transforms, they showed that T :T ; 2 2 t2 2 t2 − R 2 cH(R − ct)H(ct − r ) R − c c 2 1 2 1 g(r, z, t|ρ, 0, 0+) = − , 4πγρr c2 t2 − R12 R22 − c2 t2 (4.4.78) 2 where R1,2 = (r ∓ ρ)2 + [z − β(r − ρ)]2 /γ 2 and γ 2 = α2 − β 2 . 4.5 THREE-DIMENSIONAL WAVE EQUATION IN AN UNLIMITED DOMAIN In this section, we find the Green’s function for the three-dimensional wave equation resulting from a point source in a domain of infinite extent. Mathematically, the problem is ∂ 2g ∂2g ∂ 2g 1 ∂2g + + − = −δ(x − ξ)δ(y − η)δ(z − ζ)δ(t − τ ), ∂x2 ∂y 2 ∂z 2 c2 ∂t2

(4.5.1)

where −∞ < x, y, z, ξ, η, ζ < ∞, 0 < t, τ, and c is the phase speed, with g(x, y, z, 0|ξ, η, ζ,τ ) = gt (x, y, z, 0|ξ, η, ζ,τ ) = 0.

(4.5.2)

22 Watanabe, K., and R. G. Payton, 2005: Green’s function for torsional waves in a cylindrically monoclinic material. Int. J. Engng. Sci., 43, 1283–1291.

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201

It arises in such diverse fields as acoustics and seismology. We begin by taking the Laplace transform of Equation 4.5.1. This yields ∂ 2 G ∂ 2 G ∂ 2 G s2 + + − 2 G = −δ(x − ξ)δ(y − η)δ(z − ζ)e−sτ . ∂x2 ∂y 2 ∂z 2 c

(4.5.3)

Next, we introduce the three-dimensional Fourier transform G(x,y, z, s|ξ, η, ζ,τ ) " ∞" ∞" ∞ 1 = G(k, 0, m, s|ξ, η, ζ,τ )ei(kx+4y+mz) dk d0dm. (2π)3 −∞ −∞ −∞ (4.5.4) Direct substitution of Equation 4.5.4 into Equation 4.5.3 gives G(k, 0, m, s|ξ, η, ζ,τ ) =

s2

c2 e−ikξ−i4η−imζ−sτ , + c2 κ2

(4.5.5)

so that G(x, y, z, s|ξ, η, ζ,τ ) =

e−sτ (2π)3

"

∞

"

∞

−∞ −∞

"

∞

ei[k(x−ξ)+4(y−η)+m(z−ζ)] dk d0dm, κ2 + s2 /c2 −∞ (4.5.6)

where κ2 = k 2 + 02 + m2 . To evaluate Equation 4.5.6 we introduce spherical coordinates in such a manner that the polar axis (θ = 0) lies along the half line from the origin to (x − ξ, y − η, z − ζ) so that x − ξ = 0, y − η = 0, z − ζ = R, k = κ cos(ϕ) sin(θ), 0 = κ sin(ϕ) sin(θ), and m = κ cos(θ). Then, Equation 4.5.6 becomes G(x, y, z, s|ξ, η, ζ,τ ) =

e−sτ (2π)3

"

∞ 0

"

0

π

"

2π

0

κ2 eiκR cos(θ) sin(θ) dϕd θdκ κ2 + s2 /c2 (4.5.7)

" ∞ " π 2 iκR cos(θ) e−sτ κ e sin(θ) dθdκ 2 (2π) 0 κ2 + s2 /c2 0 " ∞ e−sτ κ eiκR = dκ, 2 2 (2π) iR −∞ κ + s2 /c2 =

(4.5.8) (4.5.9)

. where R = (x − ξ)2 + (y − η)2 + (z − ζ)2 . Finally, we use contour integration to evaluate Equation 4.5.9 and find that G(x, y, z, s|ξ, η, ζ,τ ) =

e−sR/c−sτ 4πR

(4.5.10)

202

Green’s Functions with Applications

for R &= 0. Taking the inverse of Equation 4.5.10 and applying the second shifting theorem, the Green’s function arising from a point source is

g(x, y, z, t|ξ, η, ζ,τ ) =

δ(t − τ − R/c) . 4πR

(4.5.11)

• Example 4.5.1 Our first application of Equation 4.5.11 is to verify the Green’s function that we found for the two-dimensional wave equation, Equation 4.4.1. If we integrate Equation 4.5.11 over ζ, the two-dimensional Green’s function is " ∞ 1 δ[c(t − τ ) − R] g(x, y, t|ξ, η,τ ) = dζ. (4.5.12) 4π −∞ R Introducing the change of variable s = z − ζ, we have that " ∞ 1 δ[c(t − τ ) − R] g(x, y, t|ξ, η,τ ) = ds (4.5.13) 4π −∞ R " ∞ 1 δ[c(t − τ ) − R] = ds, (4.5.14) 2π 0 R . where R = (x − ξ)2 + (y − η)2 + s2 . Equation 4.5.14 follows from the fact that the integrand is an even function of s. If we now change to R2 = ρ2 + s2 , where ρ2 = (x − ξ)2 + (y − η)2 , so that

then

ds dR dR = = . , R s R2 − ρ2 g(x, y, t|ξ, η,τ ) =

1 2π

"

ρ

∞

δ[c(t − τ ) − R] . dR. R2 − ρ2

(4.5.15)

(4.5.16)

If c(t−τ ) < ρ, the integral vanishes because the argument of the delta function is always negative. If c(t − τ ) > ρ, the sifting property of the delta function yields 1 H[c(t − τ ) − ρ] . g(x, y, t|ξ, η,τ ) = . (4.5.17) 2π c2 (t − τ )2 − ρ2

The Heaviside function allows us to combine everything into a single expression. This agrees with the result, Equation 4.4.13, that we found earlier. # "

The Wave Equation

203

• Example 4.5.2: Kirchhoff’s formula Consider the wave equation, Equation 4.0.1, where there are no source terms and the initial conditions are zero. Then, by Equation 4.0.13, u(r, t) =

"

t+

0

"" % ⊂⊃ g(r, t|r0 , t0 ) ∇0 u(r0 , t0 ) S0

&

− u(r0 , t0 ) ∇0 g(r, t|r0 , t0 ) · n dS0 dt0 .

(4.5.18)

If we now use Equation 4.5.11 in Equation 4.5.18, u(r, t) =

1 4π

"

t+ 0

& "" !% δ(t0 − t + R/c) ⊂⊃ ∇0 u(r0 , t0 ) (4.5.19) R S0 % &) δ(t0 − t + R/c) − u(r0 , t0 ) ∇0 · n dS0 dt0 , R

where R = |r − r0 |. Carrying out the t0 integration, we find that "

t+ 0

δ(t0 − t + R/c) ∇0 u(r0 , t − R/c) ∇0 u(r0 , t0 ) dt0 = . R R

(4.5.20)

On the other hand, "

0

t+

& δ(t0 − t + R/c) dt0 R % & " t+ ∂ δ(t0 − t + R/c) = u(r0 , t0 ) ∇0 R dt0 (4.5.21) ∂R R 0 , -% & " t+ R R $ = u(r0 , t0 ) δ (t0 − t + R/c) − δ(t0 − t + R/c) dt0 R3 c 0 (4.5.22) < > % & R R ∂u(r0 , t0 ) = − 3 u(r0 , t − R/c) + , (4.5.23) R c ∂t0 t0 =t−R/c

u(r0 , t0 ) ∇0

%

where R = r − r0 . Substituting Equation 4.5.20 and Equation 4.5.23 into Equation 4.5.19, the final result is ' ' "" % ' 1 1 R ∂u(r0 , t0 ) '' u(r, t) = ⊂⊃ ∇u(r0 , t0 )'' − ' 4π S0 R cR2 ∂t0 t0 =t−R/c t0 =t−R/c ' , & ' R + u(r0 , t0 )'' · n dS0 , (4.5.24) R3 t0 =t−R/c

204

Green’s Functions with Applications

where the normal vector points outward. Equation 4.5.24 is the general form of Kirchhoff ’s theorem;23 it states that if u and ut are known at all points on a closed surface at any time, then the value of u can be calculated for all points at all times. Consider now the simple case when S0 is a sphere of radius R = ct centered at r. Because dS0 = R2 sin(θ) dθdϕ , Equation 4.5.24 becomes 1 u(r, t) = 4π

!

% " π " 2π & ∂ ct u(r0 , t0 )|t0 =0 sin(θ) dϕdθ ∂(ct) 0 0 ' ) " π " 2π ∂u(r0 , t0 ) '' +t sin(θ) dϕdθ . (4.5.25) ' ∂t0 0 0 t0 =0

This result, Equation 4.5.25, is known as Poisson’s formula.24 An alternative derivation25 obtains this result by solving the radially symmetric wave equation in an unlimited domain using Laplace transforms. Equation 4.5.25 says that the value of u(r, t) depends only upon the values of u(r0 , 0) and ut0 (r0 , 0) on the spherical surface R = ct and not on these quantities inside or outside of this sphere. This statement can be inverted to say that the values of u and ut0 at a spatial point r influence the solution only on the surface R = ct of the light or sound cone that emanates from (r, 0). Consequently, light or sound are carried through the medium at exactly a fixed speed c without “echoes.” # " • Example 4.5.3: Weyl and Sommerfeld integrals For our third application, let us now invert the Fourier transform, Equation 4.5.6, without using spherical coordinates. We perform the m integration by closing the line integral along the real axis as dictated by Jordan’s lemma. For z > 0, this requires an infinite semicircle in the upper half-plane; for z < 0, an infinite semicircle in the lower half-plane. One source of difficulty . is the presence of singularities along the real axis at m = ± k 2 + l2 + s2 /c2 . If we set s = −iω so that the temporal behavior is e−iωt , we must pass un. 2 2 2 2 derneath . the singularity at m = k + l + s /c and over the singularity at 2 2 2 2 m = − k + l + s /c so that we have radiating waves. This is clearly seen 23 Kirchhoff, G., 1882: Zur Theorie der Lichtstrahlen. Sitzber. K. Preuss. Akad. Wiss. Berlin, 641–669; reprinted a year later in Ann. Phys. Chem., Neue Folge, 18, 663–695. 24 Poisson, S. D., 1818: Sur l’int´ egration de quelques ´equations lin´eares aux diff´ erences partielles, et particuli` erement de l’´equation g´en´ erale du mouvement des fluides ´ elastiques. M´ em. Acad. Sci. Paris, 3, 121–176. 25 Bromwich, T. J. I’A., 1927: Some solutions of the electromagnetic equations, and of the elastic equations, with applications to the problem of secondary waves. Proc. London Math. Soc., Ser. 2 , 28, 438-475. See Section 5.

The Wave Equation

205

after applying the residue theorem; we find that the Laplace transform of the Green’s function is G(x, y, z, s|ξ, η, ζ,τ ) √ 2 2 2 2 " " e−sτ ∞ ∞ eik(x−ξ)+il(y−η)−|z−ζ| k +l +s /c . = dk dl 8π 2 −∞ −∞ k 2 + l2 + s2 /c2

(4.5.26)

with the condition that the square root has a real part ≥ 0. Because the G(x, y, z, s|ξ, η, ζ,τ )’s given by Equation 4.5.10 and Equation 4.5.26 must be equivalent, we immediately obtain the Weyl integral :26 " ∞ " ∞ ik(x−ξ)+il(y−η)−|z−ζ|√k2 +l2 −ω2 /c2 eiωR/c 1 e . = dk dl, (4.5.27) R 2π −∞ −∞ k 2 + l2 − ω 2 /c2

if we substitute s = −iω. We can further simplify Equation 4.5.27 by introducing k = ρ cos(ϕ), l = ρ sin(ϕ), x − ξ = r cos(θ), and y − η = r sin(θ) in which case " ∞ " 2π iρr cos(θ−ϕ)+i|z−ζ|√ω2 /c2 −ρ2 eiωR/c 1 e . = dϕ ρ dρ. (4.5.28) R 2π 0 −i ω 2 /c2 − ρ2 0 If we now carry out the ϕ integration, we obtain the Sommerfeld integral :27 √ 2 2 2 " ∞ eiωR/c J0 (ρr)ei|z−ζ| ω /c −ρ . =i ρ dρ , (4.5.29) R ω 2 /c2 − ρ2 0 where the imaginary part of the square root must be positive. Both the Weyl and Sommerfeld integrals are used extensively in electromagnetism and elasticity as an integral representation of spherical waves propagating from a point source. # " • Example 4.5.4: Dispersive wave equation

Let us find the Green’s function28 for the three-dimensional wave equation , ∂2g ∂ 2g ∂ 2g 1 ∂2g 2 + + − + a g = −δ(x − ξ)δ(y − η)δ(z − ζ)δ(t − τ ), ∂x2 ∂y 2 ∂z 2 c2 ∂t2 (4.5.30) 26

Weyl, H., 1919: Ausbreitung elektromagnetischer Wellen u ¨ber einem ebenen Leiter. Ann. Phys., 4te Folge, 60, 481–500. ¨ Sommerfeld, A., 1909: Uber die Ausbreitung der Wellen in der draftlosen Telegraphie. Ann. Phys., 4te Folge, 28, 665–736. 27

28 Bhabha, H. J., 1939: Classical theory of mesons. Proc. R. Soc. London, Ser. A, 172, 384–409. An alternative derivation is given by Chambers, Ll. G., 1966: Derivation of solutions of the Klein-Gordon equation from solutions of the wave equation. Proc. Edinburgh Math. Soc., Ser. 2 , 15, 125–129.

206

Green’s Functions with Applications

where −∞ < x, y, z, ξ, η, ζ < ∞, 0 < t, τ, c is a real, positive constant (the wave speed), and a is a real, nonnegative constant. This equation arises in plasma physics. We begin by taking the Laplace transform of Equation 4.5.30 and find that , 2 ∂ 2G ∂ 2G ∂ 2G s + a2 + + − G = −δ(x−ξ)δ(y −η)δ(z −ζ)e−sτ . (4.5.31) ∂x2 ∂y 2 ∂z 2 c2 Using the techniques that gave Equation 4.5.10, the solution to Equation 4.5.31 is G(x, y, z, s|ξ, η, ζ,τ ) =

e−R

√ s2 +a2 /c−sτ

4πR

,

R &= 0,

. where R = (x − ξ)2 + (y − η)2 + (z − ζ)2 . Consider now the transform √ exp(−R s2 + a2 /c) √ U (x, y, z, s|ξ, η, ζ,τ ) = . s2 + a2

(4.5.32)

(4.5.33)

which has the inverse * . + u(x, y, z, t|ξ, η, ζ,τ ) = J0 a t2 − R2 /c2 H(t − R/c).

(4.5.34)

Consequently, the transform

√ ∂U exp(−R s2 + a2 /c) =− ∂R c

(4.5.35)

has the inverse

so that

. aR J1 (a t2 − R2 /c2 ) ∂u δ(t − R/c) . = H(t − R/c) − , 2 2 2 2 ∂R c c t − R /c

δ(t − τ − R/c) 4πR . a J1 [a (t − τ )2 − R2 /c2 ] . − H(t − τ − R/c). 4πc (t − τ )2 − R2 /c2

(4.5.36)

g(x, y, z, t|ξ, η, ζ,τ ) =

(4.5.37)

The Wave Equation

207

Thus, after the initial passage of the wave front at c(t − τ ) = R, there remains a contribution from the tail due to the dispersive effects introduced by the a2 term in Equation 4.5.30. # " • Example 4.5.5: Dissipative wave equation As sound waves propagate through a medium they may suffer frictional and viscous losses. A prototypical equation describing this phenomenon is ∂ 2u ∂2u ∂ 2u 1 + 2+ 2 − 2 ∂x2 ∂y ∂z c

,

∂ +α ∂t

-,

∂ + β u = −f (x, y, z, t), ∂t

(4.5.38)

where −∞ < x, y, z, ξ, η, ζ < ∞, and 0 < t, τ, and α and β are real, nonnegative constants. Thus, the Green’s function29 for this dissipative wave equation is ∂ 2g ∂2g ∂ 2g 1 + + − 2 2 2 2 ∂x ∂y ∂z c

,

-, ∂ ∂ +α +β g ∂t ∂t = −δ(x − ξ)δ(y − η)δ(z − ζ)δ(t − τ ),

(4.5.39)

with the initial condition g(x, y, z, 0|ξ, η, ζ,τ ) = gt (x, y, z, 0|ξ, η, ζ,τ ) = 0. We begin by taking the Laplace transform of Equation 4.5.39 or ∂2G ∂ 2G ∂2G + + − γ 2 G = −δ(x − ξ)δ(y − η)δ(z − ζ)e−sτ , ∂x2 ∂y 2 ∂z 2

(4.5.40)

. where γ = (s + α)(s + β)/c. Applying the same technique used in the derivation of Equation 4.5.10, the solution to Equation 4.5.40 is G(x, y, z, s|ξ, η, ζ,τ ) =

e−γR−sτ , 4πR

R &= 0,

(4.5.41)

. where R = (x − ξ)2 + (y − η)2 + (z − ζ)2 . If we introduce a ≡ (α + β)/2 and b ≡ |β − α|/2, we may write Equation 4.5.41 as G(x, y, z, s|ξ, η, ζ,τ ) =

exp[−sτ − R

.

(s + a)2 − b2 /c] . 4πR

(4.5.42)

Consider now the transform √ exp(−R s2 − b2 /c) √ U (x, y, z, s|ξ, η, ζ,τ ) = . s 2 − b2

(4.5.43)

29 For an alternative derivation, see Walters, A. G., 1951: On the propagation of disturbances from moving sources. Proc. Cambridge Philos. Soc., 47, 109–126.

208

Green’s Functions with Applications

Its inverse is

* . + u(x, y, z, t|ξ, η, ζ,τ ) = I0 b t2 − R2 /c2 H(t − R/c).

(4.5.44)

Consequently, the transform

√ ∂U exp(−R s2 − b2 /c) =− ∂R c

(4.5.45)

has the inverse

. bR I1 (b t2 − R2 /c2 ) ∂u δ(t − R/c) . =− H(t − R/c) − . 2 2 2 2 ∂R c c t − R /c

(4.5.46)

Therefore, L and

−1

*

e

+ √ −R s2 −b2 /c

. bR I1 (b t2 − R2 /c2 ) . = δ(t − R/c) + H(t − R/c), c t2 − R2 /c2 (4.5.47)

e−a(t−τ ) δ(t − τ − R/c) 4πR . 2 2 2 −a(t−τ ) I1 [b (t − τ ) − R /c ] . + be H(t − τ − R/c), 4πc (t − τ )2 − R2 /c2

g(x, y, z, t|ξ, η, ζ,τ ) =

(4.5.48) after applying the second shifting theorem. In the lossless case30 when a = b = 0, we recover Equation 4.5.11. The Green’s function for the general case of a, b &= 0 and a &= b consists of two parts. The first term in Equation 4.5.48 is an attenuated delta function that is propagating outward at the speed c. The second term represents a tail or residue that will remain after the pulse and wave front pass the point R = c(t − τ ). In a similar vein, Hanyga31 found the Green’s function for , 2 ∂2g ∂g ∂ g ∂ 2g ∂ 2g 2 +a + bg − c + 2 + 2 = δ(x − ξ)δ(y − η)δ(z − ζ)δ(t − τ ) ∂t2 ∂t ∂x2 ∂y ∂z (4.5.49) 30

In telegraphy, this is the case when the telegraphic line has no resistance or leakage.

31 Hanyga, A., 2002: Propagation of pulses in viscoelastic media. Pure Appl. Geophys., 159, 1749–1769.

The Wave Equation

209

with the initial condition g(x, y, z, 0|ξ, η, ζ,τ ) = gt (x, y, z, 0|ξ, η, ζ,τ ) = 0. He found that g(x, y, z, t|ξ, η, ζ,τ ) = 0 if t − τ < R/c. On the other hand, if t − τ > R/c and ∆= b − a2 /4 > 0, % e−a(t−τ )/2 δ(t − τ − R/c) 4πRc2 R. S √ ∆(R/c)J1 ∆[(t − τ )2 − R2 /c2 ] & . − . (t − τ )2 − R2 /c2

g(x, y, z, t|ξ, η, ζ,τ ) =

(4.5.50)

If t − τ > R/c and∆ < 0,

% e−a(t−τ )/2 δ(t − τ − R/c) 4πRc2 R. S . |∆|(R/c)I1 |∆|[(t − τ )2 − R2 /c2 ] & . + . (t − τ )2 − R2 /c2

g(x, y, z, t|ξ, η, ζ,τ ) =

(4.5.51) # "

• Example 4.5.6 Let us find32 the free-space Green’s function for the wave equation in cylindrical coordinates , ∂ 2g 1 ∂ ∂g 1 ∂ 2g ∂ 2g δ(r − ρ) − r − 2 2 − 2 = δ(θ)δ(z)δ(t − 0+ ) (4.5.52) 2 ∂t r ∂r ∂r r ∂θ ∂z r for 0 ≤ r, ρ< ∞, 0 ≤ θ < 2π, −∞ < z < ∞, and 0 < t, subject to the boundary conditions lim |g(r, θ, z, t|ρ, 0, 0, 0+)| < ∞,

r→0

lim g(r, θ, z, t|ρ, 0, 0, 0+) → 0,

r→∞

(4.5.53)

for 0 ≤ θ < 2π, |z| < ∞, and 0 < t; and lim g(r, θ, z, t|ρ, 0, 0, 0+) → 0,

|z|→∞

(4.5.54)

for 0 ≤ r < ∞, 0 ≤ θ < 2π, and 0 < t; and a periodicity condition in θ. The initial condition is g(r, θ, z, 0|ρ, 0, 0, 0+) = gt (r, θ, z, 0|ρ, 0, 0, 0+) = 0, for 0 ≤ r < ∞, 0 ≤ θ < 2π, and |z| < ∞. 32

Borisov and Simonenko, op. cit.

(4.5.55)

210

Green’s Functions with Applications

Using the periodicity condition, we begin by writing the delta function and Green’s function as δ(θ) = and

∞ 1 ( imθ e , 2π m=−∞

∞ (

g(r, θ, z, t|ρ, 0, 0, 0+) =

(4.5.56)

gm (r, z, t|ρ, 0, 0+ )eimθ .

(4.5.57)

m=−∞

The equation governing for gm (r, z, t|ρ, 0, 0+) is , ∂ 2 gm 1 ∂ ∂gm ∂ 2 gm m2 δ(r − ρ) − r − + 2 gm = δ(z)δ(t − 0+ ) (4.5.58) 2 2 ∂t r ∂r ∂r ∂z r 2πr for 0 ≤ r, ρ< ∞, −∞ < z < ∞, and 0 < t, subject to the boundary conditions lim |gm (r, z, t|ρ, 0, 0+)| < ∞,

lim gm (r, z, t|ρ, 0, 0+ ) → 0,

r→∞

r→0

|z| < ∞, 0 < t,

(4.5.59)

and lim gm (r, z, t|ρ, 0, 0+ ) → 0,

|z|→∞

0 ≤ r < ∞,

0 < t.

(4.5.60)

The initial condition is gm (r, z, 0|ρ, 0, 0+) = ∂gm (r, z, 0|ρ, 0, 0+)/∂t = 0,

0 ≤ r < ∞,

|z| < ∞. (4.5.61)

Introducing the Hankel transform Gm (k, z, t|ρ, 0, 0+) =

"

∞

gm (r, z, t|ρ, 0, 0+)Jm (kr) r dr,

(4.5.62)

Gm (k, z, t|ρ, 0, 0+)Jm (kr) k dk,

(4.5.63)

0

or gm (r, z, t|ρ, 0, 0+) =

"

∞

0

Equation 4.5.58 becomes the partial differential equation ∂ 2 Gm ∂ 2 Gm Jm (kρ) − + k 2 Gm = δ(z)δ(t − 0+) 2 2 ∂t ∂z 2π

(4.5.64)

for −∞ < z < ∞, and 0 < t, subject to the boundary condition lim Gm (k, z, t|ρ, 0, 0+) → 0,

|z|→∞

|z| < ∞,

(4.5.65)

The Wave Equation

211

and the initial condition Gm (k, z, 0|ρ, 0, 0+) = ∂Gm (k, z, 0|ρ, 0, 0+)/∂t = 0,

|z| < ∞.

(4.5.66)

Using Riemann’s method,33 we find that gm (r, z, t|ρ, 0, 0+) =

1 4π

"

t

δ(τ ) dτ

0

"

−τ +z+t

δ(ζ) dζ

(4.5.67)

τ +z−t

" ∞ 4 . 5 × J0 k (t − τ )2 − (z − ζ)2 Jm (kρ)Jm (kr) k dk 0 " t−z " ∞4 . 5 1 = δ(τ ) dτ J0 k (t − τ )2 − z 2 Jm (kρ)Jm (kr) k dk 4π 0 0 (4.5.68) " t2 1 cos(mϑ) = δ(τ ) dτ (4.5.69) 4π 2 rρ max(0,t1 ) sin(ϑ) = where

1 H(t2 ) − H(t1 ) . Tm [cos(ϑ)], 2π 2 4ρ2 r2 − (r2 + ρ2 + z 2 − t2 )2 cos(ϑ) =

sin(ϑ) = t1 = t −

.

T

r2 + ρ2 + z 2 − (t − τ )2 , 2rρ

1−

(r + ρ)2 + z 2 ,

(4.5.70)

(4.5.71)

[r2 + ρ2 + z 2 − (t − τ )2 ]2 , 4r2 ρ2 . and t2 = t − (r − ρ)2 + z 2 .

(4.5.72) (4.5.73)

The function Tm (·) denotes the mth Chebyshev polynomials of the first kind. To obtain Equation 4.5.69 from Equation 4.5.68, we applied results from Gradshteyn and Ryzhik.34 We can generalize results to any arbitrary source. Our Green’s function becomes g(r, θ, z, t|ρ,θ $ , ζ , τ) =

∞ (

$

gm (r, z, t|ρ, ζ,τ )eim(θ−θ ) ,

(4.5.74)

m=−∞

where gm (r, z, t|ρ, ζ,τ ) =

1 H(t2 ) − H(t1 ) . Tm [cos(ϑ)], 2π 2 4ρ2 r2 − (r2 + ρ2 + (z − ζ)2 − (t − τ )2 )2 (4.5.75)

33

See Martin, op. cit.; Young, op. cit.

34

Gradshteyn and Ryzhik, op. cit.

212

Green’s Functions with Applications

r2 + ρ2 + (z − ζ)2 − (t − τ )2 , (4.5.76) 2rρ . . t1 = t − τ − (r + ρ)2 + (z − ζ)2 , and t2 = t − τ − (r − ρ)2 + (z − ζ)2 . (4.5.77) # " cos(ϑ) =

• Example 4.5.7 M. Ochmann35 found the Green’s function for the three-dimensional wave equation, Equation 4.5.1 in the half-space z > 0. For a impulse forcing at (0, 0, ζ) with ζ > 0, the Green’s function is δ[t − r(ζ)/c] δ[t − r(−ζ)/c] + (4.5.78) 4π r(ζ) 4π r(−ζ) . cγ γ(z+ζ) exp(−γ c2 t2 − ρ2 ) . − e H[t − r(−ζ)/c], 2π c2 t2 − ρ2

g(x, y, z, t|0, 0, ζ ,0+) =

if the boundary condtion on the xy-plane is

gz (x, y, 0, t|0, 0, ζ ,0+ ) = γ g(x, y, 0, t|0, 0, ζ ,0+).

(4.5.79)

Here r2 (h) = ρ2 + (z − h)2 and ρ2 = x2 + y 2 . On the other hand, if the boundary condition now reads gz (x, y, 0, t|0, 0, ζ ,0+) = gt (x, y, 0, t|0, 0, ζ ,0+ )/(cZ0 ),

(4.5.80)

then the Green’s function now becomes δ[t − r(ζ)/c] δ[t − r(−ζ)/c] + (4.5.81) 4π r(ζ) 4π r(−ζ) < > 1 ∂ H[t − r(−ζ)/c] . − . 2π ∂t (Z0 ct + z + ζ)2 + (1 − Z02 )ρ2

g(x, y, z, t|0, 0, ζ ,0+) =

4.6 ASYMMETRIC VIBRATIONS OF A CIRCULAR MEMBRANE A problem of considerable practical importance involves the oscillations of a circular membrane that is clamped along the boundary r = a. To find its Green’s function, we must solve the partial differential equation , 1 ∂ 2g ∂g ∂ 2 g 1 ∂g 1 ∂2g δ(r − ρ)δ(θ − θ$ )δ(t − τ ) 2 + 2b + α g − 2− − 2 2 = , 2 2 c ∂t ∂t ∂r r ∂r r ∂θ r (4.6.1) 35 Ochmann, M., 2011: Closed form solutions for the acoustical impulse response over a masslike or an absorbing plane. J. Acoust. Soc. Am., 129, 3502–3512.

The Wave Equation

213

with 0 < r, ρ < a, 0 ≤ θ − θ$ ≤ 2π, and 0 < t, τ, subject to the boundary conditions lim |g(r, θ, t|ρ,θ $ , τ )| < ∞,

r→0

g(a, θ, t|ρ,θ $ , τ ) = 0,

0 < t,

(4.6.2)

and the initial conditions g(r, θ, 0|ρ,θ $ , τ ) = gt (r, θ, 0|ρ,θ $ , τ ) = 0,

0 ≤ r < a.

(4.6.3)

We begin by taking the Laplace transform of Equation 4.6.1 and find that ∂ 2 G 1 ∂G 1 ∂ 2 G s2 + 2bs + α2 δ(r − ρ)δ(θ − θ$ ) −sτ + + 2 − G=− e . (4.6.4) 2 2 2 ∂r r ∂r r ∂θ c r To solve Equation 4.6.4, we first expand its right side as a double FourierBessel series or ∞ ∞ δ(r − ρ)δ(θ − θ$ ) 1 ( ( Jm (knm ρ/a)Jm (knm r/a) = 2m cos[m(θ − θ $ )], 2 r πa2 m=0 n=1 Jm+1 (knm ) (4.6.5) where 20 = 1, 2m = 2 for m > 0, and knm is the nth root of Jm (k) = 0. This series has the advantage that it vanishes along r = a. With this Fourier-Bessel series representation of the delta functions, we anticipate that

G(r, θ, s|ρ,θ $ , τ ) =

∞ ∞ 1 (( Jm (knm ρ/a)Jm (knm r/a) 2m Gnm (s|τ ) 2 πa2 m=0 n=1 Jm+1 (knm )

× cos[m(θ − θ $ )].

(4.6.6)

Substitution of Equation 4.6.5 and Equation 4.6.6 into Equation 4.6.4 yields Gnm (s|τ ) =

c2 e−sτ . (s + b)2 + α2 − b2 + c2 kn2 /a2

(4.6.7)

A straightforward application of tables and the shifting theorems gives g(r, θ, t|ρ,θ $ , τ ) =

c2 −b(t−τ ) e H(t − τ ) πa2 ∞ ( ∞ ( Jm (knm ρ/a)Jm (knm r/a) × 2m cos[m(θ − θ$ )] 2 J (k ) nm m+1 m=0 n=1 . 2 /a2 + α2 − b2 ] sin[(t − τ ) c2 knm . × , (4.6.8) 2 2 c knm /a2 + α2 − b2

214

Green’s Functions with Applications 3 2 1 0 −1 −2

(a)

(b)

(c)

(d)

−3 3 2 1 0 −1 −2 −3

0

1

2 TIME

3

4 0

1

2 TIME

3

4

Figure 4.6.1: The Green’s function ag(x, t|ξ,τ )/(πc) for a circular membrane with asymmetric vibrations as a function of time c(t − τ )/a when θ − θ ! has the values of (a) 0, (b) π/3, (c) 2π/3, and (d) π when b = 0, a2 β 2 /c2 = 1, r = 0.8, and ρ = 0.5. The dashed line gives the free-space Green’s function [the first term in Equation 4.6.12] for this particular problem. 2 where we have assumed that the damping is sufficiently weak so that c2 knm /a2 2 2 +α > b for all n and m. Although Equation 4.6.8 solves the problem, what is occurring physically? To address this question, let us retrace our steps and approach the problem in a different manner. Once again, we take the Laplace transform of Equation 4.6.1 and use the Fourier cosine expansion for the delta function. This suggests that our Green’s function has the form ∞ ( G(r, θ, s|ρ,θ $ , τ ) = e−sτ Gn (r|ρ)2n cos[n(θ − θ$ )], (4.6.9) n=0

where Gn (r|ρ) is given by

d2 Gn 1 dGn n2 s$2 + β 2 δ(r − ρ) + − 2 Gn − Gn = − , (4.6.10) 2 dr r dr r c2 2πr where s$ = s +b, and β 2 = α2 −b2 . We now solve Equation 4.6.10 in a manner similar to that shown in Example 3.5.2 and find that + *r . + 1 * r< . $2 > Gn (r|ρ) = In s + β 2 Kn s$2 + β 2 (4.6.11) 2π c c

The Wave Equation −

Kn (a

215 .

. . s$2 + β 2 /c)In (ρ s$2 + β 2 /c)In (r s$2 + β 2 /c) . . 2πIn (a s$2 + β 2 /c)

The interpretation of Equation 4.6.11 is straightforward. The first term is the free-space Green’s function and the particular solution to Equation 4.6.10 while the second term is a homogeneous solution of Equation 4.6.10 required so that the Green’s function satisfies the boundary condition, Equation 4.6.2. Upon substituting Equation 4.6.11 into Equation 4.6.9, it remains for us to invert the Laplace transform and find g(r, θ, t|ρ,θ $ , τ ). The combination of Equation 4.6.9 and Equation 4.6.11 consists of two terms. The first term is ∞ *r . + *r . + e−sτ ( < > 2n In s$2 + β 2 Kn s$2 + β 2 cos[n(θ − θ$ )]. 2π n=0 c c

Using the result from Equation 5.3.45, this summation equals * . + e−sτ K0 R s$2 + β 2 , 2π

where R2 = r2 + ρ2 − 2rρ cos(θ − θ$ ). Finally, using both shifting theorems and tables, we find that the free-space portion of g(r, θ, t|ρ,θ $ , τ ) is 4 . 5 cos β (t − τ )2 − R2 /c2 . e−b(t−τ ) H(t − τ − R/c). 2π (t − τ )2 − R2 /c2

We invert the second term using residues. From the asymptotic representation for modified Bessel functions, this term will “turn on” for c(t − τ ) > r + ρ − 2a and corresponds to the minimum time needed for a wave emanating from the source to reflect from the boundary at r = a and the reflected wave to travel back to the observational point. Carrying out the inversion, we find that an alternative to Equation 4.6.8 is 4 . 5 cos β (t − τ )2 − R2 /c2 . g(r, θ, t|ρ,θ $ , τ ) = e−b(t−τ ) H(t − τ − R/c) 2π (t − τ )2 − R2 /c2 c + e−b(t−τ ) H[c(t − τ ) + r + ρ − 2a] 2a ∞ ( × 2m cos[m(θ − θ$ )] m=0

×

!( ∞

knm Ym (knm )Jm (knm ρ/a)Jm (knm r/a) Jm+1 (knm ) n=1 . ) 2 + a2 β 2 /c2 /a] sin[c(t − τ ) knm . × , (4.6.12) 2 + a2 β 2 /c2 knm

216

Green’s Functions with Applications

where knm is the nth root of Jm (knm ) = 0. • Example 4.6.1: Vibration of a circular plate Another problem of keen practical interest is the vibration of a circular plate with a clamped edge. To find the Green’s function for this problem, we must solve ,

-2 ∂2 1 ∂ 1 ∂2 1 ∂ 2g δ(r − ρ)δ(θ − θ$ )δ(t − τ ) + + g+ 2 2 = , 2 2 2 ∂r r ∂r r ∂θ c ∂t r

(4.6.13)

with α < r, ρ < 1, 0 ≤ θ − θ$ ≤ 2π, and 0 < t, τ, subject to boundary conditions at r = α and r = 1, where 0 < α < 1. The exact form of the boundary conditions varies considerably. If the plate is clamped along the boundary at r = α, then g(α, θ, t|ρ,θ $ , τ ) = gr (α, θ, t|ρ,θ $ , τ ) = 0.

(4.6.14)

On the other hand, if the plate is allowed to vibrate freely there, then the boundary conditions become , , ∂ ∂ 2g 1 ∂g 1 ∂2g 1 − ν ∂ 2 ∂g g + + + − = 0, ∂r ∂r2 r ∂r r2 ∂θ2 r2 ∂θ2 ∂r r and

∂ 2g +ν ∂r2

,

1 ∂g 1 ∂2g + 2 2 r ∂r r ∂θ

-

=0

(4.6.15)

(4.6.16)

at r = α, where ν denotes Poisson’s ratio. The initial conditions are the familiar g(r, θ, 0|ρ,θ $ , τ ) = gt (r, θ, 0|ρ,θ $ , τ ) = 0,

α < r < 1.

(4.6.17)

gnm (t)Rnm (r) cos[n(θ − θn )].

(4.6.18)

We begin by writing the solution as g(r, θ, t|ρ,θ $ , τ ) =

∞ ( ∞ (

n=0 m=1

Here Rnm (r) = c1 Jn (knm r) + c2 Yn (knm r) + c3 In (knm r) + c4 Kn (knm r). (4.6.19) The coefficients ci and knm are computed from substituting Rnm (r) into the boundary conditions. This results in four homogeneous linear equations whose

The Wave Equation

217

determinant must equal zero.36 Because these computations must be done numerically for even the simplest boundary conditions, we presently assume that they have been computed by some numerical algorithm. Upon substituting Equation 4.6.18 into Equation 4.6.13, we have that ∞ ( ∞ , $$ ( gnm δ(r − ρ)δ(θ − θ$ )δ(t − τ ) 4 + k g . nm nm Rnm (r) cos[n(θ − θn )] = 2 c r n=0 m=1 (4.6.20) We now multiply Equation 4.6.20 by r RN M (r) cos[N (θ − θN )] and integrate over r and θ. If we choose the normalization constant of Rnm (r) so that "

1

Rnm (r)Rn4 (r) r dr =

α

δm4 , π

(4.6.21)

where δm4 is the Kronecker delta function, Equation 4.6.21 simplifies to $$ 2 2 $ gN M + ωN M gN M = c RN M (ρ) cos[N (θ − θN )]δ(t − τ ),

(4.6.22)

4 2 where c2 kN M = ωN M , and N, M are arbitrary integers such that 0 ≤ N < ∞ and 1 ≤ M < ∞. The solution to Equation 4.6.22 is

gnm (t) = c2 RN M (ρ) cos[N (θ $ − θN )]

sin[ωnm (t − τ )] H(t − τ ). ωnm

(4.6.23)

Therefore, g(r, θ, t|ρ,θ $ , τ ) = c2 H(t − τ )

∞ ( ∞ (

n=0 m=1

Rnm (ρ)Rnm (r) cos[n(θ$ − θn )]

× cos[n(θ − θn )]

sin[ωnm (t − τ )] . ωnm

(4.6.24)

Quite often θn is set to zero for all n. For the special case when α = 0, so that there is no longer an annulus, the problem simplifies because c2 = c4 = 0, and Rnm (r) = c1 Jn (knm r) + c3 In (knm r).

(4.6.25)

If the edge at r = 1 is clamped, knm is given by solution to $ $ Jm (knm )Im (knm ) − Im (knm )Jm (knm ) = 0.

(4.6.26)

36 See, for example, Vogel, S. M., and D. W. Skinner, 1965: Natural frequencies of transversely vibrating uniform annular plates. J. Appl. Mech., 32, 926–931 and Anderson, G., 1969: On the determination of finite integral transforms for forced vibrations of circular plates. J. Sound Vib., 9, 126–144.

218

Green’s Functions with Applications

4.7 THERMAL WAVES In the classical theory of heat conduction, Fourier’s law implies that knowledge about changes in a temperature field propagates infinitely fast. Although this is clearly unrealistic, classical theory works well in most situations because the thermal diffusivity is usually 10 orders of magnitude slower than the speed at which this information travels.37 One case where diffusivity and the speed of these thermal waves becomes comparable is in superfluids such as liquid helium. Another example is shortpulse laser technology used in modern microfabrication technologies. In these cases the heat equation becomes 1 ∂2u 1 ∂u + − ∇2 u = f (r, t), c2 ∂t2 α ∂t

(4.7.1)

where α is the thermal conductivity and c is the velocity of the thermal wave. The function f (r, t) represents an internal heat source. In this section we find38 the Green’s function for Equation 4.7.1. The mathematical problem is 1 ∂2g 1 ∂g + − ∇2 g = δ(r − r0 )δ(t − τ ). c2 ∂t2 α ∂t

(4.7.2)

We begin by assuming that we can express the Green’s function in terms of the set of eigenfunction g(r, t|r0 , τ ) =

∞ (

Cn (t)Gn (r),

(4.7.3)

n=1

where Gn (r) satisfies ∇2 Gn = −λ2n Gn as well as time-independent homogeneous boundary conditions, and Cn (t) are the presently unknown, timedependent Fourier coefficients. Although Gn (r) depends upon the domain’s geometry and boundary conditions, there is always an orthogonality condition that we can write as ! """ Nn (λn ), n = m, Gn (r) Gm (r) dV = (4.7.4) 0, otherwise. V Upon substituting Equation 4.7.3 into Equation 4.7.2, we obtain ∞ , ( 1 d2 Cn 1 dCn + Gn (r) − Cn (t)∇2 Gn (r) = δ(r − r0 )δ(t − τ ), (4.7.5) 2 dt2 c α dt n=1 37

Nettleton, R. E., 1960: Relaxation theory of thermal conduction in liquids. Phys. Fluids, 3, 216–225; Francis, P. H., 1972: Thermo-mechanical effects in elastic wave propagation: A survey. J. Sound Vibr., 21, 181–192. 38 For a slightly different derivation, see Haji-Sheikh, A., and J. V. Beck, 1994: Green’s function solution for thermal wave equation in finite bodies. Int. J. Heat Mass Transfer , 37, 2615–2626.

The Wave Equation

219

or ∞ , ( 1 d2 Cn 1 dCn 2 + + λn Gn (r) = δ(r − r0 )δ(t − τ ), c2 dt2 α dt n=1

(4.7.6)

since ∇2 Gn = −λ2n Gn . At this point, we multiply each side of Equation 4.7.6 by Gm (r) and integrate over the volume. Applying Equation 4.7.4, Equation 4.7.6 simplifies to 1 d2 Cn 1 dCn Gn (r0 )δ(t − τ ) + + λ2n Cn = , 2 2 c dt α dt Nn (λn )

(4.7.7)

where we have substituted n for m. Solving Equation 4.7.7 by Laplace transforms, we find that . 2 4 2 2 −c2 (t−τ )/(2α) Gn (r0 ) sin[(t − τ ) λn − c /(4α ) ] . Cn (t) = c e H(t−τ ). (4.7.8) Nn (λn ) λ2n − c4 /(4α2 )

Consequently, the Green’s function for Equation 4.7.2 is 2

g(r, t|r0 , τ ) = c2 e−c (t−τ )/2α H(t − τ ) . ∞ ( Gn (r0 )Gn (r) sin[(t − τ ) λ2n − c4 /(4α2 ) ] . × . Nn (λn ) λ2n − c4 /(4α2 ) n=1

(4.7.9)

A simplified version of this problem is left as Problem 27.

4.8 DIFFRACTION OF A CYLINDRICAL PULSE BY A HALF-PLANE One of the classic problems involving Green’s function is the diffraction of waves by a wedge. In 1956 Turner39 attacked this problem by examining an impulse forcing to the wave equation when an infinitely thin wedge lies along the x-axis. It is particularly noteworthy because of its use of transform methods to obtain the solution. Consider the wave equation in the xy-plane which is impulsively forced at (ρ,θ $ ). The governing equation is , 1 ∂ 2g 1 ∂ ∂g 1 ∂ 2g δ(r − ρ)δ(θ − θ$ )δ(t − 0+ ) + r + = , (4.8.1) c2 ∂t2 r ∂r ∂r r2 ∂θ2 r subject to the boundary conditions lim g(r, θ, t|ρ,θ $ , 0+) → 0,

r→∞

(4.8.2)

39 Turner, R. D., 1956: The diffraction of a cylindrical pulse by a half-plane. Quart. Appl. Math., 14, 63–73.

220

Green’s Functions with Applications

and

g(r, 0, t|ρ,θ $ , 0+) = g(r, 2π, t|ρ,θ $ , 0+) = 0.

(4.8.3)

The initial condition is g(r, θ, 0|ρ,θ $ , 0+ ) = gt (r, θ, 0|ρ,θ $ , 0+ ) = 0.

(4.8.4)

Setting R2 = r2 + ρ2 − 2rρ cos(θ − θ$ ) and R02 = x2 + ρ2 − 2xρ cos(θ − θ $ ), (4.8.5) we write the Green’s function as the sum g(r, θ, t|ρ,θ $ , 0+ ) = g0 (r, θ, t|ρ,θ $ , 0+ ) + u(r, θ, t),

(4.8.6)

where g0 is the free-space Green’s function g0 (r, θ, t|ρ,θ $ , 0+ ) = and u(r, θ, t) is given by

H(t − R/c) . , 2π t2 − R2 /c2

, 1 ∂2u 1 ∂ ∂u 1 ∂2u = r + c2 ∂t2 r ∂r ∂r r2 ∂θ2 with u(r, 0, t) = u(r, 2π, t) = −

H(t − R0 /c) . . t2 − R02 /c2

2π

Defining the Laplace transform of u(r, θ, t) by " ∞ U (r, θ, s) = u(r, θ, t)e−st dt,

(4.8.7)

(4.8.8)

(4.8.9)

(4.8.10)

0

the Laplace transform of Equation 4.8.8 and Equation 4.8.9 is , 1 ∂ ∂U 1 ∂ 2U s2 r + 2 − U = 0, r ∂r ∂r r ∂θ2 c2

(4.8.11)

subject to the boundary condition U (r, θ, s) = −

K0 (sR0 /c) . 2π

(4.8.12)

Turner showed (see his Equation 20) that U (r, θ, s) equals U (r, θ, s) = −

∞ 1# $ 2 1# $ 2 1( Kn+ 12 (sr> /c)In+ 12 (sr< /c) sin n + 12 θ sin n + 12 θ$ π n=0

K0 (sR/c) + K0 (sR1 /c) , 4π

(4.8.13)

The Wave Equation

221

where R12 = r2 + ρ2 − 2rρ cos(θ + θ$ ). Finally, Turner inverted Equation 4.8.13 to yield u(r, θ, t) = where v(r, θ, t) =

H(t − R/c) H(t − R1 /c) . − . + v(r, θ, t), 4π t2 − R2 /c2 4π t2 − R12 /c2 c √

∞ (

2π rρ n=0

,

Pn

r2 + ρ2 − c2 t2 2rρ

-

(4.8.14)

1# $ 2 1# $ 2 sin n + 12 θ sin n + 12 θ $

(4.8.15) if |r − ρ| < ct < r + ρ; v(r, θ, t) = 0, otherwise. Figure 4.8.1 illustrates Equation 4.8.14 for various θ’s when ρ = π and √ θ $ = π. For 0 < θ < π we clear see the direct wave and the reflected wave off the wedge. For π < θ < 2π we lose the reflected wave and the direct wave is much smaller as θ increases because of the diffraction of the waves around the wedge. K. Watanabe40 redid this problem for an elastic solid and computed the reflected SH waves arising from an edge that moves as x = V t + L. Using the standard Cagniard-de Hoop technique, he found the regular reflected wave solution plus a pulse-like wave when a ray at a critical incident angle hits the moving edge. 4.9 LEAKY MODES One of the most common structures in nature is a stratified medium which consists of homogeneous layers. Disturbances originating in one of these layers will reflect off interfaces as well as leak some of their energy into adjacent layers. A particularly interesting case occurs when one of the adjacent layers has a phase speed larger than that of the layer of the incident wave. As a wave enters the layer with higher phase speed, it can race ahead of its counterpart in the source layer and reappear as a precursor to the direct wave. These “leaky modes”41 are essential in such disciplines as acoustics and electromagnetics. Green’s functions play a fundamental role42 in understanding leaky modes. In this section we highlight their use in explaining the reflection and transmission of electromagnetic waves in various stratified media. 40 Watanabe, K., 2010: Reflection of an impulsive SH wave at a moving edge. Quart. J. Mech. Appl. Mech., 63, 335–348. 41 Tamir, T., and A. A. Oliner, 1963: Guided complex waves. Part 1. Fields at an interface. Proc. IEE, 110, 310–324. 42 Marcuvitz, N., 1956: On field representations in terms of leaky modes or eigenmodes. IRE Trans. Antennas Propag., 4, 192–194.

222

Green’s Functions with Applications

0.3

0.3 = 2 /9

= 4 /9

0.2

0.2

0.1

0.1

0

0

−0.1 0

2

0.3

4

6

8

−0.1 0 0.3

= 2 /3

0.2

0.2

0.1

0.1

0

0

−0.1 0

2

4

0.3

6

8

−0.1 0

0.2

0.1

0.1

0

0 2

4

6

4

8

0.3

−0.1 0

2

4

8

6

8

= 4 /3

2

4

6

8

0.3 = 14 /9

= 16 /9

0.2

0.2

0.1

0.1

0

0

−0.1 0

6

= 8 /9

0.3

= 10 /9

0.2

−0.1 0

2

2

4

r

6

8

−0.1 0

2

4

6

8

r

Figure 4.8.1: The Green’s function for the diffraction of waves due to wedge located along the positive x-axis. Each frame gives the Green’s function along a specific θ when the source √ √ is located at ρ = π and θ ! = π. No value for the Green’s function greater than 0.3 was plotted. One hundred terms was used to evaluate v(r,θ ).

The Wave Equation

223 z

Region 2

z=

z=a

Region 1

x Figure 4.9.1: Schematic of a dielectric of thickness a lying above a perfect conductor at z = 0 and below some external medium a < z < ∞.

A simple example involves electromagnetic wave propagation within an one-dimensional, open waveguide. Our model consists of a one-dimensional slab43 , as shown in Figure 4.9.1. Within the slab 0 < z < a, waves travel at the speed of c/n, where c is the speed of light in region 2. The governing equations for each layer are ∂ 2 g 1 n2 ∂ 2 g 1 − 2 = 0, ∂z 2 c ∂t2

0 < z < a,

0 < t,

(4.9.1)

and ∂ 2 g2 1 ∂ 2 g2 − 2 = −δ(z − ζ)δ(t − τ ), 2 ∂z c ∂t2

a < z < ∞,

0 < t,

(4.9.2)

where ζ > a, z is the vertical distance and t is time. At z = 0, we assume a perfect conductor and g1 (0, t|ζ,τ ) = 0,

0 < t.

(4.9.3)

At the interface z = a, we require continuity of tangential components or g1 (a, t|ζ,τ ) = g2 (a, t|ζ,τ ), and

∂g1 (a, t|ζ,τ ) ∂g2 (a, t|ζ,τ ) = , ∂z ∂z

0 < t,

0 < t.

(4.9.4)

(4.9.5)

43 See Section 2 of Veselov, G. I., A. I. Kirpa and N. I. Platonov, 1986: Transient-field representation in terms of steady-improper solutions. IEE Proc., Part H, 133, 21–25.

224

Green’s Functions with Applications

Initially, the system is undisturbed so that g1 (z, 0|ζ,τ ) =

∂g1 (z, 0|ζ,τ ) = 0, ∂t

0 < z < a,

(4.9.6)

and

∂g2 (z, 0|ζ,τ ) = 0, a < z < ∞. (4.9.7) ∂t At infinity, limz→∞ g2 (z, t|ζ,τ ) → 0. Taking the Laplace transform of Equation 4.9.1 through Equation 4.9.5, g2 (z, 0|ζ,τ ) =

d2 G1 n2 s 2 − G1 = 0, dz 2 c2

(4.9.8)

d2 G2 s2 − G2 = −δ(z − ζ)e−sτ , dz 2 c2

(4.9.9)

G1 (0, s|ζ,τ ) = 0,

(4.9.10)

G1 (a, s|ζ,τ ) = G2 (a, s|ζ,τ ),

(4.9.11)

G$1 (a, s|ζ,τ ) = G$2 (a, s|ζ,τ ).

(4.9.12)

with

and

The difficulty in solving this set of differential equations is the presence of the delta function in Equation 4.9.9. We solve this problem by further breaking down the region a < z < ∞ into two separate regions a < z < ζ and ζ < z < ∞. At z = ζ, we integrate Equation 4.9.9 across a very narrow strip from ζ − to ζ + . This gives the additional conditions that

and

G2 (ζ − , s|ζ,τ ) = G2 (ζ + , s|ζ,τ ),

(4.9.13)

G$2 (ζ + , s|ζ,τ ) − G$2 (ζ − , s|ζ,τ ) = −e−sτ .

(4.9.14)

Solutions that satisfy the differential equations, Equation 4.9.8 and Equation 4.9.9, plus the boundary conditions, Equation 4.9.10 and Equation 4.9.14, are G1 (z, s|ζ,τ ) =

sinh(nsz/c) −s(ζ−a)/c−sτ e , ∆(s)

0 < z < a,

(4.9.15)

and * s + c exp − |z − ζ| − sτ 2s % c & c s(ζ − a) s(z − a) − exp − − − sτ (4.9.16) 2s c c % & sinh(nsa/c) s(ζ − a) s(z − a) + exp − − − sτ , ∆(s) c c

G2 (z, s|ζ,τ ) =

The Wave Equation

225

Figure 4.9.2: Plot of the Green’s function (divided by c) at various distances z/a and times c(t − τ )/a for ζ = 2a, and n = 1.5811.

where a < z < ∞, and ∆(s) =

that

* nsa + ns * nsa + s sinh + cosh . c c c c

(4.9.17)

We invert the first term of Equation 4.9.16 by inspection and discover , c |z − ζ| (d) g2 (z, t|ζ,τ ) = H t − τ − . (4.9.18) 2 c

This portion of the solution represents the direct wave emitted from z = ζ. The second term in G2 (z, s|ζ,τ ) contains a simple pole at s = 0. Performing the inversion by inspection, , c z−a ζ −a (s) g2 (z, t|ζ,τ ) = − H t − τ − − . (4.9.19) 2 c c This is a steady-state field set up in region 2 due to the reflection of the direct wave from the interface at z = a. The inversion of the last term in Equation 4.9.16 depends upon the value of n. We will do the case for n > 1 while the case n < 1 is reserved as Problem 17. In either case, this term contains simple poles that are the zeros of s−1 ∆(s). For n > 1, a little algebra gives , asm 1 n−1 mπ = ln + i, c 2n n+1 2n

(4.9.20)

226

Green’s Functions with Applications

where m = ±1, ±3, . . .. These poles at s = sm give the transient solutions to the problem. Applying the residue theorem in conjunction with Bromwich’s integral, , c n(z − a) ζ − a (t) g1 (z, t|ζ,τ ) = 2 H t−τ + − (4.9.21) n −1 c c !( ) sinh(nsm z/c) exp[sm (t − τ ) − sm (ζ − a)/c] × , (asm /c) sinh(nsm a/c) m and (t) g2 (z, t|ζ,τ )

, c z−a ζ −a = 2 H t−τ − − (4.9.22) n −1 c c !( ) exp[sm (t − τ ) − sm (ζ − a)/c − sm (z − a)/c] × . asm /c m

(t)

The solution g1 (z, t|ζ,τ ) gives the total field in region 1 while the total field (d) (s) (t) within region 2 equals the sum of g2 (z, t|ζ,τ ), g2 (z, t|ζ,τ ) and g2 (z, t|ζ,τ ). (t) Let us examine the transient solution g2 (z, t|ζ,τ ) more closely. Its origins are in Equation 4.9.16, which contain terms that behave as e−sm z/c . Because 3(sm ) < 0 for n > 1, this solution grows exponentially as z → ∞. Have we made a mistake in deriving Equation 4.9.22? No, because there is an exponential decay est in time for each fixed space point at which the wave front passed. This exponential time decay is sufficiently large to counteract the exponential growth in z so that the net effect yields solutions that are always finite. We see this in a plot of the solution given in Figure 4.9.2. Figure 4.9.3 illustrates individual episodes. Initially, a square pulse emanates from z = 2a, propagating both inward and outward from the source. When the inwardly propagating wave strikes the interface, a portion enters the dielectric while the remaining energy is reflected, as shown in frame c(t − τ )/a = 2. After reflecting from the conductor at z = 0, a portion of the energy escapes the dielectric while the remaining portion is reflected back into the dielectric, as frame c(t − τ )/a = 5 shows. This repeated process of reflection and transmission continues until all of the energy escapes the dielectric. Although our one-dimensional problem illustrates internal reflections, the addition of another spatial dimension introduces us to a new phenomenon: head waves. Consider44 two dielectric half-spaces where the speed of light is c/n in region 1 and c in region 2. (See Figure 4.9.4.) We locate the impulse forcing 44

See Section 2.1 of Shirai, H., 1995: Transient scattering responses from a plane interface between dielectric half-spaces. Electron. Comm. Japan, Part 2 , 78(9), 52–62. See also De Hoop, A. T., 1979: Pulsed electromagnetic radiation from a line source in a two-media configuration. Radio Sci., 14, 253–268.

The Wave Equation

227

time = 10.

0.60 0.20 −0.20

time = 7.0

0.60 0.20 −0.20

time = 5.0

Green’s function / c

0.60 0.20 −0.20

time = 3.5

0.60 0.20 −0.20

time = 2.0

0.60 0.20 −0.20

time = 0.5

0.60 0.20 −0.20 0.0

2.0

4.0

6.0

8.0

distance

10.0

Figure 4.9.3: Snapshots of the Green’s function (divided by c) at various distances z/a and times c(t − τ )/a for ζ = 2a and n = 1.5811.

at x = 0 and z = ζ. The Green’s functions then satisfy the equations ∂ 2 g1 ∂ 2 g 1 n2 ∂ 2 g 1 + − 2 = 0, 2 ∂x ∂z 2 c ∂t2

z < 0,

(4.9.23)

and ∂ 2 g2 ∂ 2 g2 1 ∂ 2 g2 + − 2 = −δ(z − ζ)δ(x)δ(t), z > 0, (4.9.24) 2 2 ∂x ∂z c ∂t2 where x is the horizontal distance. Continuity in the electric and magnetic fields at the interface z = 0 implies g1 (x, 0, t|0, ζ ,0) = g2 (x, 0, t|0, ζ ,0),

(4.9.25)

and

∂g1 (x, 0, t|0, ζ ,0) ∂g2 (x, 0, t|0, ζ ,0) = . (4.9.26) ∂z ∂z At infinity, lim|z|→∞ gj (x, z, t|0, ζ ,0) → 0. We solve Equation 4.9.23 through Equation 4.9.26 by the joint application of Laplace and Fourier transforms. We denote this joint transform of gj (x, z, t|0, ζ ,0) by Gj (k, z, s|0, ζ ,0) with j = 1, 2, where s and k are the transform variables of the Laplace and Fourier transforms, respectively. Assuming that the system is initially at rest, d2 G1 − ν12 G1 = 0, dz 2

z < 0,

(4.9.27)

228

Green’s Functions with Applications

Region 2

z (0, )

x Region 1

Figure 4.9.4: A line impulse lying above a dielectric half-space.

and

d2 G2 − ν22 G2 = −δ(z − ζ), dz 2 with the boundary conditions

z > 0,

G1 (k, 0, s|0, ζ ,0) = G2 (k, 0, s|0, ζ ,0),

(4.9.28)

(4.9.29)

and

G1$ (k, 0, s|0, ζ ,0) = G2$ (k, 0, s|0, ζ ,0), (4.9.30) . 2 2 2 2 2 2 2 where ν1 = k + n s /c , and ν2 = k + s /c . Solutions that satisfy Equation 4.9.27 through Equation 4.9.30 are .

G1 (k, z, s|0, ζ ,0) =

2ν2 e−ν2 ζ+ν1 z , ν1 + ν2 2ν2

(4.9.31)

and

e−ν2 |z−ζ| ν2 − ν1 e−ν2 (z+ζ) + . (4.9.32) 2ν2 ν1 + ν2 2ν2 The physical interpretation of these equations is straightforward. Equation 4.9.31 gives the wave that has been transmitted into region 1 from region 2. On the other hand, Equation 4.9.32 gives the direct wave emanating from the source (the first term) and a reflected wave (the second term). G2 (k, z, s|0, ζ ,0) =

(a) Direct wave Let us now invert Equation 4.9.31 and Equation 4.9.32. We begin with (d) the first term in Equation 4.9.32. If we denote this term by G2 (k, z, s|0, ζ ,0), then the inverse Fourier transform gives " ∞ ikx−ν2 |z−ζ| 1 e (d) G2 (x, z, s|0, ζ ,0) = dk. (4.9.33) 4π −∞ ν2

The Wave Equation

229

Setting x = r sin(ϕ), |z − ζ| = r cos(ϕ), where r2 = x2 + (z − ζ)2 and k = s sinh(w)/c, then Equation 4.9.33 becomes (d)

G2 (x, z, s|0, ζ ,0) =

1 4π

"

∞

e−sr cosh(w−ϕi)/c dw,

(4.9.34)

−∞

with the requirement that 3[cosh(w − ϕi)] > 0. Introducing w − ϕi = β, (d)

G2 (x, z, s|0, ζ ,0) =

1 4π

"

∞−ϕi

e−sr cosh(β)/c dβ

−∞−ϕi " ∞ −sr cosh(β)/c

1 e dβ 4π −∞ " ∞ 1 = e−sr cosh(β)/c dβ 2π 0 " ∞ 1 e−st . = dt, 2π r/c t2 − r2 /c2 =

(4.9.35) (4.9.36) (4.9.37) (4.9.38)

where t = r cosh(β)/c. We can deform the contour in Equation 4.9.35 to the one in Equation 4.9.36 because the integrand has no singularities. Examining the integrand in Equation 4.9.38 . closely, we note that it is the Laplace transform of the function H(t − r/c)/ t2 − r2 /c2 . Consequently, we immediately have that (d)

g2 (x, z, t|0, ζ ,0) =

H(t − r/c) . . 2π t2 − r2 /c2

(4.9.39)

Equation 4.9.39 agrees with the two-dimensional, free-space Green’s function, Equation 4.4.13, that we found for a line source. This technique of deforming the inversion contour of the Fourier transform into the definition of the Laplace transform so that we can invert the Laplace transform by inspection was first developed by Cagniard45 and then subsequently improved by De Hoop.13 (b) Reflected wave We now turn our attention to the waves that are reflected as the result of the interface at z = 0. If we denote the Laplace transform of that portion of the Green’s function by G(r) (x, z, s|0, ζ ,0), it is governed by the equation (r) G2 (x, z, s|0, ζ ,0)

1 = 4π

"

∞

−∞

,

ν2 − ν1 ν2 + ν1

-

eikx−ν2 (z+ζ) dk. ν2

(4.9.40)

45 The most accessible version is Cagniard, L., E. A. Flinn and C. H. Dix, 1962: Reflection and Refraction of Progressive Seismic Waves. McGraw-Hill, 282 pp.

230

Green’s Functions with Applications

Using the same substitution that we used to evaluate the direct wave, we have that 1 (r) G2 (x, z, s|0, ζ ,0) = 4π

"

∞

−∞

 

cosh(w) − cosh(w) +

= =

n2 + sinh2 (w) n2

2

+ sinh (w)

× e−sρ cosh(w−ϕr i)/c dw,

 

(4.9.41)

where ρ2 = x2 + (z + ζ)2 , x = ρ sin(ϕr ) and z + ζ = ρ cos(ϕr ). Because the integrand of Equation 4.9.41 has branch points at ± sin−1 (n)i, two branch cuts must be introduced. We take one to run from sin−1 (n)i out to ∞i while the other runs from − sin−1 (n)i out to −∞i. We will do the analysis for n < 1; the case n > 1 is left as Problem 18. When n < 1, we introduce w = β + ϕr i so that (r)

G2 (x, z, s|0, ζ ,0) =

1 4π

"

∞−ϕr i

e−sρ cosh(β)/c dβ

−∞−ϕr i



×

cosh(β + ϕr i) − cosh(β + ϕr i) +

= =

(4.9.42)

n2 + sinh2 (β + ϕr i) n2 + sinh2 (β + ϕr i)



.

Turning to the case sin−1 (n) > |ϕr | first, we can still deform the contour β − ϕr i to the real axis because we will not cross any branch points. Then

" ∞ 1 F (β)e−sρ cosh(β)/c dβ (4.9.43) 4π −∞ " ∞ 1 = [F (β) + F (−β)]e−sρ cosh(β)/c dβ (4.9.44) 4π 0 " ∞ U V 1 = F [cosh−1 (ct/ρ)] + F [− cosh−1 (ct/ρ)] 4π ρ/c

(r)

G2 (x, z, s|0, ζ ,0) =

where F (β) =

cosh(β + ϕr i) − cosh(β + ϕr i) +

ρ2 = x2 + (z + ζ)2 ,

= =

e−st ×. dt, t2 − ρ2 /c2

n2 + sinh2 (β + ϕr i) n2 + sinh2 (β + ϕr i)

ϕr = tan−1

,

x z+ζ

-

,

,

t = ρ cosh(β)/c, and the real part of the radical must be positive.

(4.9.45)

(4.9.46)

(4.9.47)

The Wave Equation

231

iy (| r |-w b )i

x

-(| r |-w b )i

original contour Figure 4.9.5: Schematic of the deformation of the contour used in Equation 4.9.42 to the one used in Equation 4.9.50.

Because we recognize Equation 4.9.45 as the definition of the Laplace transform, we can immediately write down U V (r) g2 (x, z, t|0, ζ ,0) = F [cosh−1 (ct/ρ)] + F [− cosh−1 (ct/ρ)] H(t − ρ/c) × . (4.9.48) 4π t2 − ρ2 /c2 ! % , -&) ct H(t − ρ/c) . = 3 F cosh−1 , (4.9.49) ρ 2π t2 − ρ2 /c2

since F (−β) = F ∗ (β). Consider now the case when |ϕr | > wb = sin−1 (n). Let us take the branch cut associated with the branch points ±(|ϕr | − wb )i to run along the imaginary axis between the two branch points. Consequently, when we deform the original contour from β − ϕr i to the real axis, we have additional contour integrals arising from integrations along the branch cut. For example, if ϕr > 0, then the branch cut integrals would be: (1) a line integral running from the real axis (and just to the left of the branch cut) down to the branch point, (2) an integration around the branch point, and (3) a line integral from the branch point (and just to the right of the branch cut) running up to the real axis. See Figure 4.9.5. Since the integration around the branch point equals zero, we have that " 0 1 (r) G2 (x, z, s|0, ζ ,0) = F (β)e−sρ cosh(β)/c dβ 4π −∞ = " |ϕr |−wb cos(|ϕ | − y) − n2 − sin2 (|ϕ | − y) r r 1 = − 4π 0 cos(|ϕr | − y) + n2 − sin2 (|ϕr | − y) × e−sρ cos(y)/c dy

232

Green’s Functions with Applications 1 − 4π

"

cos(|ϕr | − y) +

0

|ϕr |−wb

cos(|ϕr | − y) −

× e−sρ cos(y)/c dy

+

"

1 = 4π

"

= =

n2 − sin2 (|ϕr | − y) n2 − sin2 (|ϕr | − y)

∞ 1 F (β)e−sρ cosh(β)/c dβ (4.9.50) 4π 0 " ∞ 1 = [F (β) + F (−β)]e−sρ cosh(β)/c dβ 4π 0 = " |ϕn |−wb cos(|ϕ | − y) n2 − sin2 (|ϕ | − y) r r 1 + 2 4π 0 1−n

+

1 π

× e−sρ cos(y)/c dy

ρ/c

"

(4.9.51) V F [cosh−1 (ct/ρ)] + F [− cosh−1 (ct/ρ)]

∞U

e−st ×. dt t2 − ρ2 /c2

ρ cos(|ϕr |−wb )/c

ρ/c

where G(β) =

G[|ϕr | − cos−1 (ct/ρ)]

e−st ×. dt, ρ2 /c2 − t2 cos(β)|n2 − sin2 (β)|1/2 . 1 − n2

(4.9.52)

(4.9.53)

In particular, we recognize Equation 4.9.52 as the definition of the Laplace transform, in which case we can immediately write down ! % , -&) ct H(t − ρ/c) (r) −1 . g2 (x, z, t|0, ζ ,0) = 3 F cosh ρ 2π t2 − ρ2 /c2 +

G[|ϕr | − cos−1 (ct/ρ)]H(|ϕr | − wb ) . (4.9.54) π ρ2 /c2 − t2 × {H[t − ρ cos(wb − |ϕr |)/c] − H(t − ρ/c)} .

Let us examine Equation 4.9.54 more closely. The first term is merely the reflected wave from the interface that is felt at any point in region 2 after the time t = ρ/c. The second term is new and radically different. First, it exists before and disappears when the conventional reflected wave arrives at those points where |ϕr | > sin−1 (n); it is a precursor to the arrival of the reflected wave. Second, this wave occurs only for sufficiently large |ϕr |; this corresponds to large |x|. Consequently, these waves occur near the lateral sides of the expanding reflected wave field. For this reason, they are called lateral waves.

The Wave Equation

233

Figure 4.9.6: The minimum arrival times ctmin /ζ for the laterally reflected [z > 0 and |ϕr | > sin−1 (n)] and the transmitted (z < 0) waves as a function of position when n = 0.5.

In geophysics, similar waves occur due to different elastic properties within the solid earth. Because these lateral waves are recorded at the beginning or head of a seismographic record, they are often called head waves. Why do these lateral or head waves exist? Recall that the case n < 1 corresponds to light traveling faster in region 1 than in region 2. Consequently, some of the wave energy that enters region 1 near x = 0 can outrun its counterpart in region 2 and then reemerge into region 2 where it is observed as head waves. Thus lateral waves can reach an observer sooner than the direct wave much as a car may reach a destination sooner by taking an indirect route involving an expressway rather than a direct route that consists of residential streets: the higher speeds of the highway more than compensate for the increased distance traveled. (c) Transmitted wave We complete our analysis by finding the Green’s function for the waves that pass through the interface into region 1. If G(t) (x, y, s|0, ζ ,0) denotes the Laplace transform of the Green’s function, it is found by evaluating - ikx−ν2 ζ+ν1 z " ∞, 1 ν2 e (t) G1 (x, z, s|0, ζ ,0) = dk. (4.9.55) 2π −∞ ν2 + ν1 ν2 At this point, we introduce the new variable p such that k = isp and " ∞i −s(px+ζ √1/c2 −p2 −z √n2 /c2 −p2 ) 1 e (t) . . G1 (x, z, s|0, ζ ,0) = dp. (4.9.56) 2πi −∞i n2 /c2 − p2 + 1/c2 − p2

234

Green’s Functions with Applications

Figure 4.9.7: The Green’s function ζgj (x, z, t|0, ζ ,0)/c for various values of x/ζ and z/ζ when ct/ζ = 4. The shaded region denotes that region where head waves are present.

We next change to the variable 3 t = px + ζ

1 − p2 − z c2

3

n2 − p2 . c2

(4.9.57)

Just as we have done twice before, we would like to deform the original contour (along the imaginary axis) into a contour C along which t is real so that " 1 e−st (t) . . G1 (x, z, s|0, ζ ,0) = dp (4.9.58) 2πi C n2 /c2 − p2 + 1/c2 − p2 F G " 1 ∞ 1 dp . = 5 . e−st dt, π tmin n2 /c2 − p2 + 1/c2 − p2 dt (4.9.59)

and

dp 1 . . = . dt x − ζpc/ 1 − p2 c2 + zpc/ n2 − p2 c2

(4.9.60)

What is the nature of C? From a detailed analysis of Equation 4.9.57, we find the following properties: • C must remain on the same Riemann surface as the original contour. We take the branch cuts along the real axis from the branch points (±1/c, 0) and (±n/c, 0) out to infinity. • For x > 0, the contour C lies in the first and fourth quadrants of the complex p-plane.

The Wave Equation

235

Figure 4.9.8: The reflected wave and ct/ζ when z/ζ = 0.05.

.

(r)

|t2 − ρ2 /c2 | g2 (x, z, t|0, ζ ,0) for various values of x/ζ

• For x > 0, the contour C crosses the real axis with 0 < p < 1/c or n/c, depending upon which is smaller. At this crossing, t assumes its smallest value, tmin , which corresponds to the minimum travel time for a transmitted wave to arrive at a point in region 1. At that time, dp/dt is of infinite magnitude and ζpc zpc x− . +. = 0. (4.9.61) 2 2 2 1−p c n − p2 c2 Combining this result with Equation 4.9.57, we have

ctmin 1 n2 (z/ζ) =. −. . ζ 1 − p2 c2 n2 − p2 c2

(4.9.62)

Figure 4.9.6 gives this minimum arrival time for the reflected lateral and transmitted waves when n = 0.5. Because the head waves have their origin in the leaking of energy from region 1 into region 2, the arrival times must match at the interface. Having found tmin , we now find the transmitted wave solution for any time t > tmin , given x, z, ζ, c, and n. To do this, we must find the corresponding value of p that is the (complex) zero of the analytic function, Equation 4.9.57. Once the value of p is found, then F G 1 c dp . = 5 . H (t − tmin ) . π n2 − p2 c2 + 1 − p2 c2 dt (4.9.63) Figure 4.9.7 illustrates the Green’s function ζgj (x, z, t|0, ζ ,0)/c for various values of x and z when ct/ζ = 4. The shaded region of the wave field denotes (t) g1 (x, z, t|0, ζ ,0)

236

Green’s Functions with Applications z Region 2 (0, )

A

B z=a

Region 1

C

D x

Figure 4.9.9: A dielectric slab on a perfectly conducting half-space with a line source of current above it. The four labels A, B, C, D give the locations at which the Green’s function will be subsequently displayed.

that region where head waves are present. We clearly see how these wave are at the lateral sides of the reflected wave field. A similar, but much more difficult, problem arises if we wish to find the Green’s function for the wave equation due to a point source within one of two semi-infinite media. Towne46 and Volodicheva and Lopukhov47 presented solutions to this problem using Cagniard’s method for inverting the joint transform. The reflection and transmission of electromagnetic waves within two semiinfinite dielectrics differs from our first problem because of an additional spatial dimension and the replacement of the slab in Region 1 with a semi-infinite region. We now combine the two problems together by replacing one of the semi-infinite regions (region 1) with one of finite depth a. See Figure 4.9.9. Because the dielectric in region 1 now lies on a conducting half-plane, we have the new boundary condition g1 (x, 0, t|0, ζ ,0) = 0 at z = 0. We anticipate not only leaky modes but also internally reflecting waves. This problem is nearly identical to the one that we just solved. The governing equations are the same, Equation 4.9.23 through Equation 4.9.26, except that the interface occurs at z = a rather than z = 0. The mathematical analysis begins as before. After a joint application of Laplace and Fourier

46 Towne, D. H., 1968: Pulse shapes of spherical waves reflected and refracted at a plane interface separating two homogeneous fluids. J. Acoust. Soc. Am., 44, 65–76. 47 Volodicheva, M. I., and K. V. Lopukhov, 1994: Effect of the sphericity of an acoustic wave on its reflection coefficient from a plane interface of two liquid media. Acoust. Phys., 40, 681–684.

The Wave Equation

237

2.5

GREEN’S FUNCTION

2.0

1.5

1.0

0.5

0.0

−0.5

0

1

2 3 4 NONDIMENSIONAL TIME

5

6

Figure 4.9.10: The temporal evolution of the Green’s function at the point (a, 3a ). The 2 dashed line corresponds to the Green’s functions given by Equation 4.9.39 and Equation 4.9.54 for the problem of two half-spaces.

transforms, we have that d2 G1 − ν12 G1 = 0, dz 2 and

0 < z < a,

d2 G2 − ν22 G2 = −δ(z − ζ), dz 2

a < z,

(4.9.64)

(4.9.65)

with the boundary conditions G1 (k, 0, s|0, ζ ,0) = 0, G1 (k, a, s|0, ζ ,0) = G2 (k, a, s|0, ζ ,0),

(4.9.66)

and G1$ (k, a, s|0, ζ ,0) = G2$ (k, a, s|0, ζ ,0), (4.9.67) . . where ν1 = k 2 + n2 s2 /c2 , and ν2 = k 2 + s2 /c2 . Solutions that satisfy Equation 4.9.64 through Equation 4.9.67 are G1 =

sinh(ν1 z)e−ν2 (ζ−a) , ν1 cosh(ν1 a) + ν2 sinh(ν1 a)

0 < z < a,

(4.9.68)

and G2 =

% & e−ν2 |z−ζ| e−ν2 (z+ζ−2a) ν1 cosh(ν1 a) − ν2 sinh(ν1 a) − , 2ν2 2ν2 ν1 cosh(ν1 a) + ν2 sinh(ν1 a)

a < z. (4.9.69)

238

Green’s Functions with Applications 1.5

GREEN’S FUNCTION

1.0

0.5

0.0

−0.5

−1.0

−1.5

0

1

2 3 4 NONDIMENSIONAL TIME

5

6

Figure 4.9.11: The temporal evolution of the Green’s function at the point (3a, 3a ). The 2 dashed line corresponds to the Green’s functions given by Equation 4.9.39 and Equation 4.9.54 for the problem of two half-spaces. The shaded area corresponds to the period of time when leaky modes are present.

Because Equation 4.9.68 and Equation 4.9.69 are even functions of ν1 , no branch points are associated with ν1 , but there are branch points and cuts with ν2 . To satisfy the radiation condition as z → ∞, 3(ν2 ) ≥ 0. Some physical insight concerning these equations can be achieved by rewriting the hyperbolic functions in terms of exponentials and then expanding the resulting expressions as a geometric series in e−2ν1 a : $ eν1 (z−a)−ν2 (ζ−a) # 1 − Re−2ν1 a + R2 e−4ν1 a − R3 e−6ν1 a + . . . ν1 + ν2 −ν1 (z+a)−ν2 (ζ−a) # $ e − 1 − Re−2ν1 a + R2 e−4ν1 a − R3 e−6ν1 a + . . . , ν1 + ν2 (4.9.70)

G1 =

and G2 =

e−ν2 |z−ζ| e−ν2 (z+ζ−2a) 1 − R + (1 − R2 )e−2ν1 a − R(1 − R2 )e−4ν1 a 2ν2 2ν2 2 + R2 (1 − R2 )e−6ν1 a − · · · , (4.9.71)

where R = (ν1 − ν2 )/(ν1 + ν2 ). The first term in Equation 4.9.70 and the first two terms in Equation 4.9.71 equal Equation 4.9.31 and Equation 4.9.32, respectively, when the displacement of the interface is taken into account, and correspond to the solutions when there are simply two semi-infinite dielectrics

The Wave Equation

239

1.5

GREEN’S FUNCTION

1.0

0.5

0.0

−0.5

−1.0

0

1

2 3 4 NONDIMENSIONAL TIME

5

6

Figure 4.9.12: The temporal evolution of the Green’s function at the point (a, a2 ). The dashed line corresponds to the Green’s function, Equation 4.9.63, for the problem of two half-spaces.

present. Consequently, the remaining terms give the contribution from internal reflections within region 1. This illustrates how Green’s functions within a finite domain can be constructed from an infinite series of free-space Green’s functions. Let us now find gj (x, z, t|0, ζ ,0). Equations 4.9.68 and 4.9.69 cannot be inverted in closed form except for the leading term in Equation 4.9.69. Therefore, we must invert these two transforms numerically via the inversion integrals ) " γ+∞i !" ∞ 1 st ikx gj (x, z, t|0, ζ ,0) = 2 e Gj (k, z, s|0, ζ ,0)e dk ds, 4π i γ−∞i −∞ (4.9.72) where γ > 0 so that the Laplace transform converges. If we introduce s = ωi, we may rewrite Equation 4.9.72 as !" ∞−γi %" ∞ & ) 1 gj (x, z, t|0, ζ ,0) = 2 3 Gj (k, z,ω |0, ζ ,0)ei(kx+ωt) dk dω . 2π 0−γi −∞ (4.9.73) The numerical inversion was performed in a two-step procedure. First, for fixed ω, the inner integral was computed using Simpson’s rule with∆ k = 0.01 and k running from 0 to 20,000. Then the frequency integral was done with γ = 0.05,∆ ω = 0.1 and ω running from 0 to 3840 using Weddle’s rule. To illustrate the inversion of Equation 4.9.68 and Equation 4.9.69, the temporal behavior of the Green’s function is given at four points A–D labeled on Figure 4.9.9. Point A is the closest to the source and its behavior with time is shown in Figure 4.9.10. The first peak corresponds to the direct wave and the second, negative peak is the reflected wave off the interface. Not

240

Green’s Functions with Applications 1.0

GREEN’S FUNCTION

0.5

0.0

−0.5

−1.0

−1.5

0

1

2 3 4 NONDIMENSIONAL TIME

5

6

Figure 4.9.13: Same as Figure 4.9.12 except for the point (3a, a2 ).

surprisingly, the solution is in complete agreement with our earlier solution for the two dielectric half spaces (included as a dotted line) because the effect of the boundary at z = 0 has not yet been felt. After ct/a = 2.8, we see a series of gradually diminishing pulses. These are the portion of the energy that has passed through the interface from internal reflections off the boundaries at z = 0 and z = a. Point B is still outside of the slab but located farther away from the source than point A. The temporal evolution of the Green’s function there is shown in Figure 4.9.11. This point was chosen because leaky waves are present. Indeed, the leaky waves arrive at ct/a = 3, while the direct wave does not arrive until ct/a = 3.04. After that time, they coexist (highlighted with shading) until ct/a = 3.35 when the reflected wave off the interface arrives. Again, we have our train of pulses from the internal reflections. However, the amplitudes are smaller than those shown in Figure 4.9.10 because some of the energy has already left the slab during earlier internal reflections. Both points C and D are located within the slab and directly below points A and B, respectively. The temporal evolution of the Green’s function is given in Figures 4.9.12 and 4.9.13, respectively. At both points, we first note the passage of the transmitted wave that we found in the case of two dielectric half spaces. After that, we have a train of pulses as internal reflections propagate through the slab. 4.10 WATER WAVES If classical electrostatics gave birth to the concept of Green’s function, it has enjoyed great success in the equally classic field of irrotational water waves. This section illustrates its use.

The Wave Equation

241

The theory of surface water waves stands on two assumptions: First, the fluid is irrotational. This allows us to introduce a velocity potential ϕ such that the Eulerian three-dimensional velocity v is given by v = ∇ϕ. Second, the fluid is incompressible so that ∇ · v = 0, or ∂ 2ϕ ∂ 2 ϕ ∂ 2 ϕ + + = 0. ∂x2 ∂y 2 ∂z 2

(4.10.1)

In addition to the field equation, Equation 4.10.1, we must include boundary conditions. Assuming that the fluid lies on an infinitely flat surface, we have that ∂ϕ = 0, z = −h. (4.10.2) ∂z Finally, from the linearization of Bernoilli’s equation along the top surface, we find that ∂ 2ϕ ∂ϕ 1 ∂p +g =− = f (x, y, t), 2 ∂t ∂z ρ ∂t

z = 0,

(4.10.3)

where g temporarily denotes the acceleration due to gravity. Here p(x, y, 0, t) is a prescribed surface pressure. From Equation 4.10.3, we immediately see that ϕ and ∂ϕ/∂t must be known at t = 0. The method that we will use to solve this initial-boundary-value problem involves the time-dependent Green’s function. This Green’s function is the velocity potential that gives the wave motions arising from an impulse forcing at (ξ, η,ζ ) in a fluid that is initially at rest. After the Green’s function is constructed, then we can write the solution ϕ(x, y, z, t) in terms of any arbitrary initial conditions.48 We now consider several classic problems involving Equation 4.10.1 through Equation 4.10.3. • Example 4.10.1: The Cauchy-Poisson problem One of the simplest problems of surface waves involves finding the waves that result from the impulse forcing of the surface. Solved at the beginning of the nineteenth century by Cauchy49 and Poisson,50 the Green’s function method provides an insightful means of presenting their results. Although the original problem was for two spatial dimensions, we will solve it here in three dimensions, reserving the two-dimensional problem as Problem 20. 48 See Wehausen, J. V., and E. V. Laitone, 1960: Surface waves in Handbuch der Physik. Springer-Verlag, Section 22α. 49 Cauchy, A.-L., 1827: Th´ eorie de la propagation des ondes ` a la surface d’un fluide pesant d’une profondeur ind´ efinie. M´ em. pr´ esent´ es par divers savans ` a l’Acad. Roy. Sci. Inst. France, 1, 3–123. 50 Poisson, S. D., 1818: M´ emoire sur la th´ eorie des ondes. M´ em. Acad. Roy. Sci. Inst. France, Ser. 2 , 1, 71–186.

242

Green’s Functions with Applications

We now determine the water waves that arise from an impulse forcing51 at the point x = ξ, y = η and at the depth z = ζ. The water is at rest at t = τ in its equilibrium position and the surface pressure equals zero. To construct our Green’s function we must find the solution to the following problem: The Green’s function satisfies the partial differential equations ∂ 2g ∂2g ∂2g + + = −δ(x − ξ)δ(y − η)δ(z − ζ), ∂x2 ∂y 2 ∂z 2

(4.10.4)

where −∞ < x, y, ξ, η < ∞, −h < z, ζ < 0, and 0 ≤ τ < t, while at the free surface ∂ 2g ∂g + = 0, z = 0, (4.10.5) ∂t2 ∂z and at the bottom ∂g = 0, z = −h. (4.10.6) ∂z Here, we have nondimensionalized the equations so that g again denotes the Green’s function. At infinity, we require that g and ∂g/∂t tend to zero. As initial conditions we take g(x, y, 0, τ |ξ, η,ζ ) = gt (x, y, 0, τ |ξ, η,ζ ) = 0. Introducing the Fourier transform G of g, " ∞" ∞ G(k, 0, z, t|ξ, η,ζ ) = g(x, y, z, t|ξ, η,ζ ) e−i(kx+4y) dx dy, −∞

(4.10.7)

(4.10.8)

−∞

or g(x, y, z, t|ξ, η,ζ ) =

1 4π 2

"

∞ −∞

"

∞

G(k, 0, z, t|ξ, η,ζ ) ei(kx+4y) dk d0, (4.10.9)

−∞

Equation 4.10.4 through Equation 4.10.7 become ∂ 2G − κ2 G = −e−i(kξ+4η) δ(z − ζ), ∂z 2 ∂ 2 G ∂G + = 0, z = 0, ∂t2 ∂z ∂G = 0, z = −h, ∂z and G=

∂G = 0, ∂t

t = τ,

z = 0,

(4.10.10) (4.10.11) (4.10.12)

(4.10.13)

51 See Finkelstein, A. B., 1957: The initial value problem for transient water waves. Commun. Pure Appl. Math., 10, 511–522.

The Wave Equation

243

with κ2 = k 2 + 02 . To solve Equation 4.10.10 through Equation 4.10.13, we apply techniques from Section 3.3. The general solution to Equation 4.10.10 can be written  A(t) cosh[κ(z + h)], z< ,  ? = ei(kξ+4η) G = G  B(t) cosh(κz) − B $$ (t) sinh(kz)/κ, z >ζ . (4.10.14) Requiring that the Green’s function is continuous at z = ζ, ? = cosh[κ(z< + h)] [C(t) cosh(κz> ) − C $$ (t) sinh(κz> )/κ] . G

Because

'z=ζ + ? '' ∂G = −1, ' ∂z ' −

(4.10.15)

(4.10.16)

z=ζ

C(t) is governed by

cosh(κh)C $$ (t) + κ sinh(κh)C(t) = 1.

(4.10.17)

Solving Equation 4.10.17 with C(τ ) = C $ (τ ) = 0,

C(t) =

4 5 . 1 − cos (t − τ ) κ tanh(κh) κ sinh(κh)

.

Substituting Equation 4.10.14 into Equation 4.10.9, " ∞" ∞ 1 ? 0, z, t|ξ, η,ζ ) ei[k(x−ξ)+4(y−η)] dk d0. g= G(k, 4π 2 −∞ −∞

(4.10.18)

(4.10.19)

Upon introducing k = κ cos(α), 0 = κ sin(α), x − ξ = r cos(θ), y − η = r sin(θ), with r2 = (x − ξ)2 + (y − η)2 , we find that 1 g= 4π 2

"

∞ 0

? 0, z, t|ξ, η,ζ ) G(k,

:"

π+θ

e

iκr cos(α−θ)

;

dα κ dκ .

−π+θ

(4.10.20)

Using the integral definition for J0 (·), J0 (x) =

1 2π

"

π

eix cos(α) dα,

(4.10.21)

−π

Equation 4.10.20 simplifies to " ∞ 1 ? 0, z, t|ξ, η,ζ )J0 (κr) κ dκ , g= G(k, 2π 0

(4.10.22)

ζ

244

Green’s Functions with Applications

or 1 2π "

"

∞

sinh(κz> ) cosh[κ(z< + h)] J0 (κr) dκ cosh(κh) 0 ∞ 1 cosh[κ(z + h)] cosh[κ(ζ + h)] + (4.10.23) 2π 0 cosh2 (κh) 4 5  .  1 − cos (t − τ ) κ tanh(κh)  × J0 (κr) κ dκ .   κ tanh(κh)

g=−

Since cosh[κ(ζ + h)] = eκζ cosh(κh) − e−κh sinh(κζ), " ∞ 1 g=− sinh(κz< )eκz> J0 (κr) dκ 2π 0 " ∞ 1 sinh(κz) sinh(κζ) + e−κh J0 (κr) dκ 2π 0 cosh(κh) " ∞ 1 cosh[κ(z + h)] cosh[κ(ζ + h)] + (4.10.24) 2π 0 cosh2 (κh) 4 5  .  1 − cos (t − τ ) κ tanh(κh)  × J0 (κr) κ dκ .   κ tanh(κh) Finally, we have that

"

0

∞

# $−1/2 eκa J0 (κb) dκ = a2 + b2 ,

(4.10.25)

% & " ∞ −κh 1 1 1 e sinh(κz) sinh(κζ) − $ +2 J0 (κr) dκ 4π R R cosh(κh) 0 " ∞ 1 cosh[κ(z + h)] cosh[κ(ζ + h)] + (4.10.26) 2π 0 cosh2 (κh) 4 5   .  1 − cos (t − τ ) κ tanh(κh)  × J0 (κr) κ dκ ,   κ tanh(κh)

g=

where

and R$2 = (x − ξ)2 + (y − η)2 + (z + ζ)2 . (4.10.27) This is not the Green’s function in the conventional sense because the source, although localized in position, is left “turned on.” Consequently, another version of the problem is R2 = (x − ξ)2 + (y − η)2 + (z − ζ)2 ,

∂ 2g ∂2g ∂2g + + = −δ(x − ξ)δ(y − η)δ(z − ζ)δ(t − τ ), ∂x2 ∂y 2 ∂z 2

(4.10.28)

The Wave Equation

245

with g(x, y, z, 0|ξ, η, ζ,τ ) = gt (x, y, z, 0|ξ, η, ζ,τ ) = 0,

(4.10.29)

Equation 4.10.5, and Equation 4.10.6. Upon taking the Fourier transform in both the x and y directions and the Laplace transform in time, we now have d2 G − κ2 G = −e−i(kξ+4η)−sτ δ(z − ζ), dz 2 s2 G(k, 0, 0, s|ξ, η, ζ,τ ) + G$ (k, 0, 0, s|ξ, η, ζ,τ ) = 0, and 2

2

2

G$ (k, 0, −h, s|ξ, η, ζ,τ ) = 0,

(4.10.30) (4.10.31) (4.10.32)

with κ = k + 0 . We now solve Equation 4.10.30 through Equation 4.10.32 as we did earlier by piecing together solutions valid for z < ζ and ζ < z and choosing the ? is continuous and possesses a jump in the first derivative constants so that G equal to −1 at z = ζ. This yields 1 2 cosh[κ(z< + h)] cosh(κz> ) − s2 sinh(κz> )/κ i(kξ+4η)+sτ ? G=e G= . s2 cosh(κh) + κ sinh(κh) (4.10.33) Taking the inverse Fourier transform in k and 0, the Laplace transform of g(x, y, z, t|ξ, η, ζ,τ ) is 1 2 " " e−sτ ∞ ∞ cosh[κ(z< + h)] cosh(κz> ) − s2 sinh(κz> )/κ L(g) = 4π 2 −∞ −∞ cosh(κh)[s2 + κ tanh(κh)] × ei[k(x−ξ)+4(y−η)] dk d0.

Upon introducing polar coordinates, we find that " e−sτ ∞ sinh(κz> ) cosh[κ(z< + h)] L(g) = − J0 (κr) dκ 2π 0 cosh(κh) " e−sτ ∞ cosh[κ(z + h)] cosh[κ(ζ + h)] + J0 (κr) κ dκ . 2π 0 cosh2 (κh)[s2 + κ tanh(κh)]

(4.10.34)

(4.10.35)

Finally, taking the inverse Laplace transform and simplifying the first integral in Equation 4.10.35, we obtain % & " ∞ −κh δ(t − τ ) 1 1 e sinh(κz) sinh(κζ) g= − +2 J0 (κr) dκ 4π R R$ cosh(κh) 0 " H(t − τ ) ∞ cosh[κ(z + h)] cosh[κ(ζ + h)] + J0 (κr) κ d κ (4.10.36) 2π cosh2 (κh) 0 4 5 . sin (t − τ ) κ tanh(κh) . × , κ tanh(κh)

246

Green’s Functions with Applications

where R and R$ are defined by Equation 4.10.27. # " • Example 4.10.2: Time harmonic water waves Having just found the transient water waves that arise from an impulse source that “turns on” at t = τ or acts for only an instant, we now seek the “steady-state” solution when harmonic water waves have established themselves.U In the case of threeVspatial dimensions, the Green’s function is given by 3 g(x, y, z|ξ, η,ζ )e−iωt , where g is governed by ∂ 2g ∂2g ∂2g + + = −δ(x − ξ)δ(y − η)δ(z − ζ), ∂x2 ∂y 2 ∂z 2

(4.10.37)

with −∞ < x, y, ξ, η < ∞, and −h < z, ζ < 0, while the free surface condition becomes ∂g = ω 2 g, z = 0, (4.10.38) ∂z and at the bottom

∂g = 0, ∂z

z = −h.

(4.10.39)

Finally, we require that ' ' ' ' 'g(x, y, z|ξ, η,ζ ) − 1 ' < ∞, ' R' (x,y,z)→(ξ,η,ζ) lim

(4.10.40)

where R is defined by Equation 4.10.27. In the previous example, we applied Fourier transforms in both the x and y directions. Due to axial symmetry these Fourier transforms became a Hankel transform in r. Because we have similar axial symmetry here, we anticipate that a similar process will occur and apply Hankel transforms from the outset. This leads to d2 G δ(z − ζ) − k2 G = − , 2 dz 2π

(4.10.41)

G$ (k, 0|0, ζ) = ω 2 G(k, 0|0, ζ),

(4.10.42)

G$ (k, −h|0, ζ) = 0,

(4.10.43)

with and where G(k, z|0, ζ) =

"

0

∞

g(r, z|0, ζ) J0 (kr) r dr,

(4.10.44)

The Wave Equation

247 ki k= kr

Figure 4.10.1: The line integral used in the evaluation of Equation 4.10.46.

and r2 = (x − ξ)2 + (y − η)2 . The solution to Equation 4.10.41 through Equation 4.10.43 is G(k, z|0, ζ) =

ω 2 sinh(kz> ) + k cosh(kz> ) cosh[k(z< + h)] . k sinh(kh) − ω 2 cosh(kh) 2πk

(4.10.45)

Taking the inverse Hankel transform, the Green’s function is " 1 ∞ ω 2 sinh(kz> ) + k cosh(kz> ) g(r, z|0, ζ) = ∪ cosh[k(z< + h)] J0 (kr) dk, 2π 0 k sinh(kh) − ω 2 cosh(kh) (4.10.46) where the special integral sign denotes an integration along the real axis except for a small semicircle below the singularity at k = κ, where κ tanh(κh) = ω 2 . See Figure 4.10.1. Equation 4.10.46 does not lend itself to numerical evaluation. An alternative form is as follows: Because the integrand is an even function of k, we convert it to an integral along the entire real k axis except for passing over the singularity at k = −κ and below the singularity at k = κ. We then close the line integration with a semicircular path of infinite radius joining k = +∞ to k = −∞. Because the contribution from the arc at infinity vanishes, the Green’s function is given by a sum of the residues and g(r, z|0, ζ) = where

∞ 1 ( K0 (κn r) cos[κn (ζ + h)] cos[κn (z + h)], 2π n=0 Nn

Nn =

% & h sin(2κnh) 1+ , 2 2κn h

(4.10.47)

(4.10.48)

and κn is the nth positive root of κ tan(κh) = −ω 2 . Another alternative to Equation 4.10.46 rewrites the integrand so that we separate out explicitly the singular nature of the Green’s function. When this is done, we obtain , 1 1 1 g(r, z|0, ζ) = + $$ 4π R R " 1 ∞ (ω 2 + k)e−kh − ∪ 2 (4.10.49) 2π 0 ω cosh(kh) − k sinh(kh) × cosh[k(z + h)] cosh[k(ζ + h)]J0 (kr) dk,

248

Green’s Functions with Applications

where R$$2 = (x − ξ)2 + (y − η)2 + (z + ζ + 2h)2 . In the limit of infinitely deep water, Equation 4.10.46 and Equation 4.10.49 become " 1 1 ∞ ω 2 + k k(z+ζ) g(r, z|0, ζ) = − ∪ e J0 (kr) dk, (4.10.50) 4πR 4π 0 ω 2 − k and g(r, z|0, ζ) =

" 1 1 1 ∞ k − + ∪ ek(z+ζ) J0 (kr) dk, 4πR 4πR$ 2π 0 k − ω 2

(4.10.51)

respectively. # " • Example 4.10.3: Rapidly converging Green’s function representations In the previous examples, we found several Green’s function representations without considering how difficult they would be to evaluate. For example, if we were to compute the Green’s function given by Equation 4.10.46 and Equation 4.10.47, we would find that direct integration is difficult when |z| and |ζ| are both small and that the eigenfunction expansion converges slowly when r is small. Linton and McIver52 developed a method for finding efficient representations of the Green’s functions for water waves. To understand their technique, consider the initial-value problem ut = ∇2 u,

0 < t,

(4.10.52)

where ∇ is the three-dimensional gradient operator. We take as our initial condition u(x, 0) = δ(x − x0 ), (4.10.53) where x is the position vector and x0 is the point of excitation. The solution satisfies time-independent boundary conditions. Linton and McIver’s first insight was to observe that the Green’s function governed by ∇2 g = −δ(x − x0 ) (4.10.54) for the same domain and boundary conditions that govern Equation 4.10.52 equals " ∞ g(x|x0 ) = u(x, t) dt. (4.10.55) 0

52 Linton, C. M., and P. McIver, 2000: Green’s functions for water waves in porous structures. Appl. Ocean Res., 22, 1–12.

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249

This technique depends, of course, upon the integral, Equation 4.10.55, existing. In point of /fact, it does not in our particular problem and we must ∞ choose a u ˜ so that 0 (u + u ˜) dt does exist. With this correction, " ∞ g(x|x0 ) = [u(x, t) + u ˜(x, t)] dt − g˜(x|x0 ), (4.10.56) 0

where ∇2 g˜(x) = −˜ u(x, 0). Why have we introduced Equation 4.10.52 and Equation 4.10.53? This new initial-boundary-value problem is generally more difficult to solve than Equation 4.10.54. This leads us to Linton and McIver’s second insight. Let us split Equation 4.10.55 into two parts: " a " ∞ g(x|x0 ) = u(x, t) dt + u(x, t) dt, (4.10.57) 0

a

where a is a free parameter. If we could find two representations of u such that first one, u1 , converges rapidly for small values of t and the second one, u2 , converges rapidly for large t, then the resulting expansion would converge more rapidly than using a single u for all t. As Linton and McIver showed, there is an optimal value for a that minimizes the total number of terms needed for a desired accuracy. Let us now apply these theoretical considerations to find the Green’s function for water waves. We start by solving the initial-boundary-value problem ∂u ∂2u ∂2u ∂2u = + 2 + 2, ∂t ∂x2 ∂y ∂z

(4.10.58)

where −∞ < x, y < ∞, and −h < z < 0, with the boundary conditions uz (x, y, 0, t) = ω 2 u(x, y, 0, t),

(4.10.59)

uz (x, y, −h, t) = 0,

(4.10.60)

and and the initial condition u(x, y, z, 0) = δ(x − ξ)δ(y − η)δ(z − ζ).

(4.10.61)

As we will show in Chapter 5, Equation 4.10.58 through Equation 4.10.61 correspond to finding the Green’s function for a three-dimensional heat equation. One of the techniques for solving this problem is product solutions of the form u(x, y, z, t) = u1 (x, y, t)w(z, t)H(t), (4.10.62) where u1 (x, y, t) is the free-space Green’s function for the two-dimensional problem ∂u1 ∂ 2 u1 ∂ 2 u1 = + , (4.10.63) ∂t ∂x2 ∂y 2

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Green’s Functions with Applications

with u1 (x, y, 0) = δ(x − ξ)δ(y − η), and w(z, t) is the Green’s function for the one-dimensional problem ∂w ∂2w = , (4.10.64) ∂t ∂z 2 with the boundary conditions wz (0, t) = ω 2 w(0, t),

and wz (−h, t) = 0,

and the initial condition w(z, 0) = δ(z − ζ). The free-space Green’s function is found in Section 5.1 and % & H(t) (x − ξ)2 + (y − η)2 u1 (x, y, t) = exp − . 4πt 4t

(4.10.65)

(4.10.66)

To find w(z, t), we take the Laplace transform of Equation 4.10.64 and Equation 4.10.65 and find that d2 W − sW = −δ(z − ζ), dz 2

(4.10.67)

with W $ (0, s) = ω 2 W (0, s), and W $ (−h, s) = 0. The solution to Equation 4.10.67 is [ω 2 sinh(qz> ) + q cosh(qz> )] cosh[q(z< + h)] (4.10.68) q[ω 2 cosh(qh) − q sinh(qh)] -n ( ∞ , 4 e−q|z−ζ| e−qχ04 1 ( q + ω2 = + + e−qχnm , (4.10.69) 2q 2q 2q n=1 q − ω 2 m=1

W (z, s) = −

where χn1 = 2(n − 1)h − ζ − z,χ χn3 = 2nh + ζ − z,χ

n4

= 2nh − ζ + z,

(4.10.70)

= 2(n + 1)h + ζ + z,

(4.10.71)

n2

and q 2 = s. Our next step is to invert the Laplace transforms. From Equation 4.10.68 the inversion yields w(z, t) = where

2 ∞ ( e−µn t cos[µn (z + h)] cos[µn (ζ + h)], Nn n=0

% & h sin(2µn h) Nn = 1+ , 2 2µnh

(4.10.72)

(4.10.73)

and µn is the nth root of µ tan(µh) = −ω 2 . Equation 4.10.72 was derived assuming that all of the poles are simple; Linton and McIver discuss the case

The Wave Equation

251

when there are higher-order poles. Note that this equation converges rapidly if t is large. Turning to Equation 4.10.69, we invert it term by term. This yields 2

w(z, t) =

2

e−(z−ζ) /(4t) e−(2h+z+ζ) √ √ + 4πt 4πt

where In (χ) =

n ( j=0

/(4t)

+

∞ (

(−1)n

n=1

4 (

In (χnm ),

m=1

n! (−1)j (2ω 2 )n−j I n−j (χ), j!(n − j)!

(4.10.74) (4.10.75)

and I n (χ) is given by the recurrence relationship (n − 1)I n (χ) = 2tI n−2 (χ) + (χ − 2ω 2 t)I n−1 (χ),

(4.10.76)

with the initial values √ χ √ − ω2 t . 2 t (4.10.77) Equation 4.10.74 converges rapidly for small t. It also enjoys the property that it is independent of any dispersion relationship. If we now substitute Equation 4.10.62 in Equation 4.10.55, we find that it does not converge because of the n = 0 term in Equation 4.10.72. Consequently, by choosing 2

I 0 (χ) =

e−χ /(4t) √ , 4πt

I 1 (χ) = − 12 eω

and

4

t−ω 2 χ

erfc

,

2

u ˜(x, t) = −u1 (x, y, t) the integral

/∞ 0

e−µ0 t cos[µ0 (z + h)] cos[µ0 (ζ + h)], N0

(4.10.78)

(u + u ˜) dt now converges. To find g˜(x), we must solve

δ(x − ξ)δ(y − η) cos[µ0 (z + h)] cos[µ0 (ζ + h)], N0 (4.10.79) because u1 (x, y, 0) = δ(x − ξ)δ(y − η). Equation 4.10.79 is a two-dimensional Poisson equation where z is a parameter. Using techniques that we will develop in Chapter 6, the solution is ∇2 g˜(x) = −˜ u(x, 0) =

K0 (µ0 r) cos[µ0 (z + h)] cos[µ0 (ζ + h)]. (4.10.80) 2πN0 /∞ Our final task remains to evaluate 0 (u + u ˜) dt. We have two choices for w(z, t): there is Equation 4.10.72 or Equation 4.10.74. As pointed out earlier, Linton and McIver introduced the technique of breaking up the integral into two parts with the break point at t = a. In the first integral, we apply the g˜(x) =

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Green’s Functions with Applications

representation that converges rapidly for small t, namely Equation 4.10.74, while, in the second integral, we use the representation that converges rapidly for large t, Equation 4.10.72. Carrying out the integrations, we find that M ( Λm cos[µm (z + h)] cos[µm (ζ + h)] N m m=0 . , $- ( 4 * r + 1 1 r erfc[ R2 + χ21i /(ah)] . + erfc + erfc + 4πr ah 4πr$ ah 4π R2 + χ21i i=1 " 4 *χ + du ω 2 ah/2 ω4 u2 −R2 /(4u2 ) ( −ω2 χ1 i 1i + e e erfc − ω2u 2π 0 2u u i=1

g(x|x0 ) =

+ ···, where Λ0 =

(4.10.81)

" a2 h2 /4 −R2 /(4t) 2 K0 (µ0 R) e − e−µ0 t dt, 2π 4πt 0 " ∞ 2 e−R /(4t) −µ2m t Λm = e dt, 4πt a2 h2 /4

(4.10.82)

(4.10.83)

and M is a truncation parameter. Linton53 used the same technique to find rapidly convergent representations for the free-space Green’s function describing water waves in channels of constant depth and width. Problems 1. Following the derivation presented in the introduction to this chapter, show that the solution to the one-dimensional wave equation ∂2u 1 ∂ 2u − 2 2 = −q(x, t), 2 ∂x c ∂t is u(x, t) = +

"

"

b

q(ξ,τ )g(x, t|ξ,τ ) dξdτ

0 a " t+ %

g(x, t|ξ,τ )

0

−

t+

1 c2

" b% a

a < x < b,

u(ξ, 0)

∂u(ξ,τ ) ∂g(x, t|ξ,τ ) − u(ξ,τ ) ∂ξ ∂ξ

&ξ=b

dτ

ξ=a

& ∂g(x, t|ξ, 0) ∂u(ξ, 0) − g(x, t|ξ, 0) dξ, ∂τ ∂τ

53 Linton, C. M., 1999: A new representation for the free-surface channel Green’s function. Appl. Ocean Res., 21, 17–25.

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253

where g(x, t|ξ,τ ) is the one-dimensional Green’s function given by ∂ 2g 1 ∂ 2g − = −δ(x − ξ)δ(t − τ ). ∂x2 c2 ∂t2 2. In their study of photoacoustic spectroscopy, Cywiak et al.54 solved the nonhomogeneous wave equation ∂ 2u 1 ∂ 2u − = q(x, t), ∂x2 c2 ∂t2

−∞ < x < ∞,

0 < t,

with p(x, t) = −ut (x, t), subject to the boundary conditions lim|x|→∞ u(x, t) → 0, 0 < t, and initial conditions u(x, 0) = ut (x, 0) = 0, −∞ < x < ∞. Step 1 : Using the previous problem, show that the solution to the wave equation is " ∞" ∞ u(x, t) = g(x, t|ξ,τ )q(ξ,τ ) dτd ξ, −∞

0

where g(x, t|ξ,τ ) is the one-dimensional Green’s function given by ∂ 2g 1 ∂2g − 2 2 = δ(x − ξ)δ(t − τ ), 2 ∂x c ∂t or

c g(x, t|ξ,τ ) = − H (t − τ − |x − ξ|/c) . 2 Step 2 : Show that p(x, t) is given by " " c x c ∞ p(x, t) = q[ξ, t − (x − ξ)/c] dξ + q[ξ, t − (x − ξ)/c] dξ. 2 −∞ 2 x Unlimited Domains 3. By direct substitution, show55 that √ g(x, t|0, 0) = J0 ( xt )H(x)H(t) is the free-space Green’s function governed by ∂ 2g + 1 g = δ(x)δ(t), ∂x∂t 4

−∞ < x, t < ∞.

54 Cywiak, D., M. D. Barreiro-Arg¨ uelles, M. Cywiak, A. Landa-Curiel, C. Garcia-Segundo, and G. Gutierrez-Ju´ arez, 2013: A one-dimensional solution of the photoacoustic wave equation and its relationship with optical absorption. Int. J. Thermophys., 34, 1473–1480. 55 First proven by Picard, E., ´ 1894: Sur une ´ equation aux d´ eriv´ ees partielles de la th´eorie de la propagation de l’´ electricit´e. Bull. Soc. Math., 22, 2–8.

254

Green’s Functions with Applications

Problem 3

The figure captioned Problem 3 illustrates this Green’s function. 4. Find the Green’s function for the wave equation 2 ∂ 2g 2∂ g − c = −δ(x − ξ)δ(t − τ ), ∂t2 ∂x2

0 < x, ξ < ∞,

0 < t, τ ,

subject to the boundary conditions k g(0, t|ξ,τ ) − T gx (0, t|ξ,τ ) = 0,

lim g(x, t|ξ,τ ) → 0,

x→∞

0 < t, τ ,

and the initial conditions g(x, 0|ξ,τ ) = gt (x, 0|ξ,τ ) = 0, 0 < x < ∞, where k and T are nonnegative constants. Step 1 : Taking the Laplace transform of the partial differential equation and the boundary conditions, show that d2 G s2 − 2 G = −δ(x − ξ)e−sτ , dx2 c with

k G(0, s|ξ,τ ) − T G$ (0, s|ξ,τ ) = 0,

0 < x, ξ < ∞, lim G(x, s|ξ,τ ) → 0.

x→∞

Step 2 : Show that G(x, s|ξ,η ) =

1 −s(x> −x< )/c−sτ 1 −s(x> +x< )/c−sτ 1 e−s(x> +x< )/c−sτ e − e + . 2sc 2sc c s + kc/T

The Wave Equation

255

Step 3 : Taking the inverse of G(x, s|ξ,τ ) from Step 2, show that g(x, t|ξ,τ ) = H[t − τ − (x> − x< )/c ] /(2c) − H[t − τ − (x> + x< )/c ] /(2c) % & k kc + exp (x> + x< ) − (t − τ ) H[t − τ − (x> + x< )/c ] /c. T T

Yen56 solved a generalized version of this problem. First, he examined the damped wave equation 2 ∂ 2g ∂g 2∂ g − 2 − c = −δ(x − ξ)δ(t − 0+ ), 0 < x, ξ < L, 0 < t. ∂t2 ∂t ∂x2 He considered both the cases when L is finite and infinite. Second, for finite L, he changed the boundary conditions so that they read

and

k1 g(0, t|ξ, 0+ ) − T gx (0, t|ξ, 0+) = 0,

0 < t,

k2 g(L, t|ξ, 0+) − T gx (L, t|ξ, 0+) = 0,

0 < t.

57

5. Find the free-space Green’s function equation

for the hyperbolic partial differential

∂2g ∂ 2g − + g = δ(x − ξ)δ(y − η), −∞ < x, y, ξ, η < ∞, ∂y 2 ∂x2 subject to the boundary conditions lim|x|→∞ g(x, y|ξ,η ) → 0, −∞ < y < ∞, and lim|y|→∞ g(x, y|ξ,η ) → 0, −∞ < x < ∞, Step 1 : Taking the Fourier transform of the partial differential equation, show that the joint transform G(k,0 |ξ,η ) is

e−ikξ−i4η , k 2 − 02 + 1 where k and 0 are the transform variables in the x and y directions, respectively. G(k,0 |ξ,η ) =

Step 2 : Show that the Green’s function can be written as " ∞ " ∞ ik(x−ξ)+i4(y−η) 1 e g(x, y|ξ,η ) = d0dk 2 4π −∞ −∞ k 2 − 02 + 1 " ∞" ∞ 1 cos[k(x − ξ)] cos[0(y − η)] = 2 d0dk π 0 (k 2 + 1) − 02 0 √ " ∞ 1 sin[(y − η) k 2 + 1 ] cos[k(x − ξ)] √ = dk 2π 0 k2 + 1 4. 5 = 12 J0 (y − η)2 − (x − ξ)2 H (|y − η| − |x − ξ|) .

56 Yen, D. H. Y, 2003: Green’s function for the damped wave equation on a finite interval subject to two Robin boundary conditions. Chinese J. Mech., Ser. A, 19, 257–262. 57 See Shay, L. K., R. L. Elsberry and P. G. Black, 1989: Vertical structure of the ocean current response to a hurricane. J. Phys. Oceanogr., 19, 649–669.

256

Green’s Functions with Applications

Problem 6

Figure 4.1.2 illustrates this Green’s function with a = c = 1 where y and η replace t and τ , respectively. 6. Find the free-space Green’s function58 governed by the partial differential equation ∂ 2g ∂ 2g + + g = δ(x − ξ)δ(t − τ ), −∞ < x, ξ < ∞, 0 < t, τ , ∂t2 ∂x∂t subject to the boundary conditions lim|x|→∞ g(x, t|ξ,τ ) → 0, 0 < t, and the initial conditions g(x, 0|ξ,τ ) = gt (x, 0|ξ,τ ) = 0, −∞ < x < ∞, Step 1 : Taking the Laplace transform and Fourier transform of the partial differential equation, show that the joint transform G(k, s|ξ,τ ) is e−ikξ−sτ . iks + s2 + 1 Step 2 : By inverting the Fourier transform, show that G(k, s|ξ,τ ) =

e−(x−ξ)/s−s[τ +(x−ξ)] H(x − ξ). s Step 3 : Using tables and the second shifting theorem, show that the Green’s function equals 4 . 5 . g(x, t|ξ,τ ) = J0 2 x − ξ t − τ − (x − ξ) H(x − ξ)H[t − τ − (x − ξ)]. G(x, s|ξ,τ ) =

The figure captioned Problem 6 illustrates this Green’s function.

58 See Moore, D. W., R. C. Kloosterziel and W. S. Kessler, 1998: Evolution of mixed Rossby gravity waves. J. Geophys. Res., 103, 5331–5346.

The Wave Equation

257

7. Find the free-space Green’s function when the domain x < 0 is governed by ∂ 2 g1 ∂ 2 g1 − + g1 = δ(t − τ )δ(x + ξ), ξ 0,> ∂t2 ∂x2 while the domain x > 0 obeys ∂ 2 g2 ∂ 2 g2 − = 0. ∂t2 ∂x2 At the interface, we have the conditions that g1 (0, t|ξ,τ ) = g2 (0, t|ξ,τ ),

and g1 x (0, t|ξ,τ ) = g2 x (0, t|ξ,τ ).

At infinity, the Green’s functions must be finite or lim g1 (x, t|ξ,τ ) → 0,

x→−∞

and

lim g2 (x, t|ξ,τ ) → 0.

x→∞

All of the initial conditions equal zero. Step 1 : Taking the Laplace transform of the partial differential equations and boundary conditions, show that they reduce to the ordinary differential equations d2 G1 − (s2 + 1)G1 = −δ(x + ξ)e−sτ , dx2

and

d2 G2 − s2 G2 = 0, dx2

with the boundary conditions G1 (0, s|ξ,τ ) = G2 (0, s|ξ,τ ),

and

G$1 (0, s|ξ,τ ) = G$2 (0, s|ξ,τ ).

Step 2 : Show that the Laplace transform of g1 (x, t|ξ,τ ) and g2 (x, t|ξ,τ ) are √ exp(−|x + ξ| s2 + 1 − sτ ) √ G1 (x, s|ξ,τ ) = 2 s2 + 1 √ exp[−(ξ − x) s2 + 1 − sτ ] −s(ξ−x+τ ) + # e , √ $2 √ 2 s + s2 + 1 s2 + 1 and

√ exp[−ξ s2 + 1 − s(x + τ )] √ G2 (x, s|ξ,τ ) = . s + s2 + 1

Step 3 : Show that g1 (x, t|ξ,τ ) and g2 (x, t|ξ,τ ) are 4. 5 g1 (x, t|ξ,τ ) = 12 H(t − τ − |x + ξ|)J0 (t − τ )2 − (x + ξ)2 % & 4 5 . 1 t−τ −ξ+x + J2 (t − τ )2 − (x − ξ)2 H(t − τ + x − ξ), 2 t−τ +ξ−x

258

Green’s Functions with Applications

Problem 7

and

5 (t − τ − x)2 − ξ 2 . g2 (x, t|ξ,τ ) = H(t − τ − x − ξ) (t − τ + ξ − x) (t − τ − x)2 − ξ 2 4. 5 ξJ1$ (t − τ − x)2 − ξ 2 + H(t − τ − x − ξ). t−τ +ξ−x (t − τ − x) J1

4.

The figure captioned Problem 7 illustrates this Green’s function with ξ = 1. 8. Find the Green’s function59 governed by the partial differential equation , -, 2 ∂ ∂ ∂ 2 + + ω g + 2iωγ 2 g = δ(x − 0+ )δ(t − 0+ ), ∂t ∂x ∂t2 where 0 < x < ∞, and 0 < t, subject to the boundary conditions g(0, t|0+, 0+ ) = 0, limx→∞ g(x, t|0+ , 0+) → 0, 0 < t, and the initial conditions g(x, 0|0+ , 0+ ) = gt (x, 0|0+, 0+ ) = 0, 0 < x < ∞. Step 1 : Take the Laplace transform of the partial differential equation and show that it becomes the ordinary differential equation dG 2iωγ 2 δ(x − 0+ ) + sG + 2 G = , dx s + ω2 s2 + ω 2

0 < x < ∞,

59 Suggested by Kanaev, A. V., and C. J. McKinstrie, 1998: Exact Green’s function for a class of parametric instabilities. Phys. Plasmas, 5, 4511–4514.

The Wave Equation

259

subject to the boundary conditions G(0, s|0+ , 0+) = 0 and limx→∞ G(x, s| 0+ , 0+ ) → 0. Step 2 : Show that , e−sx 2iωγ 2 x exp − s2 + ω 2 s2 + ω 2 % , 2 -& % , -& 1 γ x 1 γ 2x = e−sx exp exp − . s + ωi s + ωi s − ωi s − ωi

G(x, s|0+ , 0+ ) =

Step 3 : Invert the Laplace transform using the second shifting theorem, tables and convolution and show that the Green’s function is g(x, t|0+ , 0+ ) =eiω(t−x) H(t − x) " t−x 5 # √ $ 4 . × e−2iωτ I0 2γ xτ J0 2γ x(t − x − τ ) dτ. 0

9. Find the free-space Green’s function60 for the system ∂g1 ∂g1 +a − bg2 = δ(x − ξ)δ(t − τ ), ∂t ∂x

and

∂g2 + cg2 − bg1 = 0. ∂t

Step 1 : Taking the Laplace transform in time and the Fourier transform in space, show that G1 (k, s|ξ,τ ) =

(s + c)e−ikξ−sτ , (s + ika)(s + c) − b2

G2 (k, s|ξ,τ ) =

b e−ikξ−sτ . (s + ika)(s + c) − b2

Step 2 : Using the residue theorem, take the inverse Fourier transform and show that % & 1 c(x − ξ) G1 (x, s|ξ,τ ) = exp + cτ H(x − ξ) a a % & (x − ξ + aτ )(s + c) b2 (x − ξ) × exp − + , a a(s + c) and G2 (x, s|ξ,τ ) =

% & b c(x − ξ) H(x − ξ) exp + cτ a a s+c % & (x − ξ + aτ )(s + c) b2 (x − ξ) × exp − + . a a(s + c)

60 See Mounaix, Ph., D. Pesme, W. Rozmus, and M. Casanova, 1993: Space and time behavior of parametric instabilities for a finite pump wave duration in a bounded plasma. Phys. Fluids B , 5, 3304–3318.

260

Green’s Functions with Applications

Problem 9

Step 3 : Using tables and the second shifting theorem, take the inverse Laplace transform and show that % , -& 1 x−ξ g1 (x, t|ξ,τ ) = exp −c t − τ − H(x − ξ) a a , ! ) d 2b . × I0 (x − ξ)[a(t − τ ) − (x − ξ)] dt a × H[t − τ − (x − ξ)/a] g1 (x, t|ξ,τ ) = δ[a(t − τ ) − (x − ξ)]H(t − τ ) T % , -& b x−ξ x−ξ + exp −c t − τ − H(x − ξ) a a(t − τ ) − (x − ξ) a ! ) , 2b . x−ξ × I1 (x − ξ)[a(t − τ ) − (x − ξ)] H t − τ − , a a

and

% , -& b x−ξ exp −c t − τ − H(x − ξ) a a ! ) , 2b . x−ξ × I0 (x − ξ)[a(t − τ ) − (x − ξ)] H t − τ − . a a

g2 (x, t|ξ,τ ) =

The figure captioned Problem 9 illustrates ag2 (x, t|ξ,τ )/b as functions of distance (x − ξ)/a and time t − τ when c = 0.1, and b = 1.

The Wave Equation

261

10. Find the free-space Green’s function61 governed by the equation ,

∂2 1 ∂ + ∂r2 r ∂r

-2

g+

1 ∂ 2g δ(r)δ(t − τ ) = , L4 ∂t2 2πr

0 < r < ∞,

0 < t, τ ,

subject to the boundary conditions limr→0 g(r, t|0, τ ) → 0, limr→∞ g(r, t|0, τ ) → 0, 0 < t, and the initial conditions g(r, 0|0, τ ) = gt (r, 0|0, τ ) = 0, 0 < r < ∞. Step 1 : Let us introduce the joint Laplace-Hankel transform: " ∞ G(k, s|0, τ ) = G(r, s|0, τ ) J0 (kr) r dr. 0

Show that G(k, s|0, τ ) =

, e−sτ L2 e−sτ e−sτ = − . 2π(k 4 + s2 /L4 ) 4πis k 2 + se−πi/2 /L2 k 2 + seπi/2 /L2

Step 2 : Using the integral representation " ∞ J0 (ak) k dk = K0 (az), k2 + z 2 0

a > 0,

3(z) > 0,

show that G(r, s|0, τ ) =

% , −πi/4 √ , πi/4 √ -& L2 e−sτ e r s e r s K0 − K0 . 4πis L L

Step 3 : Using the second shifting theorem and the relationships that F (s) = s

"

t

f (τ ) dτ,

0

1 2 e−α/(4t) √ and L−1 K0 ( αs ) = , 2t

if α &= 0, and 3(α) ≥ 0, show that H(t − τ )L2 g(r, t|0, τ ) = 4π

"

t−τ

0

, 2 % & r du H(t − τ )L2 (r/L)2 sin = si , 4L2 u u 4π 4(t − τ )

where si(·) is the sine integral si(u) =

"

∞

u

sin(u) du. u

61 See Wen, P. H., M. H. Aliabadi and A. Young, 2000: A boundary element method for dynamic plate bending problems. Int. J. Solids Struct., 37, 5177–5188. See also Graff, K. F., 1991: Wave Motion in Elastic Solids. Dover Publications, Inc., Section 4.2.5.

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Green’s Functions with Applications

Problem 10

The figure captioned Problem 10 illustrates 4πg(r, t|0, τ )/L2 as functions of distance r/L and time t − τ . 11. Find the Green’s function62 governed by the equation D11

∂4g ∂ 4g ∂4g + 4D + 2(D + 2D ) 16 12 66 ∂x4 ∂x3 ∂y ∂x2 ∂y 2 4 4 ∂ g ∂ g ∂ 2g + 4D26 + D + ρh = δ(x − ξ)δ(y − η)δ(t − τ ), 22 ∂x∂y3 ∂y 4 ∂t2

where −∞ < x, y, ξ, η < ∞, and 0 < t, τ. The boundary conditions are lim g(x, y, t|ξ, η,τ ) → 0,

|x|→∞

and

lim g(x, y, t|ξ, η,τ ) → 0.

|y|→∞

The initial conditions are g(x, y, 0|ξ, η,τ ) = gt (x, y, 0|ξ, η,τ ) = 0. Step 1 : If G(k, 0, t|ξ, η,τ ) denotes the double Fourier transform of the Green’s function g(x, y, t|ξ, η,τ ), show that ρh

d2 G + ρhγ 2 G = e−ikξ−i4η δ(t − τ ), dt2

62 See F¨ allstr¨ om, K.-E., and O. Lindblom, 1998: Transient bending wave propagation in anisotropic plates. J. Appl. Mech., 65, 930–938; Sun, L., 2003: Dynamic response of Kirchhoff plate on a viscoelastic foundation to harmonic circular loads. J. Appl. Mech., 70, 595–600.

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where ρhγ 2 = D11 k 4 + 4D16 k 3 0 + 2(D12 + 2D66 )k 2 02 + 4D26 k03 + D22 04 . Step 2 : Using Laplace transforms, show that the solution to the ordinary differential equation in Step 1 is G(k, 0, t|ξ, η,τ ) =

e−ikξ−i4η sin[γ(t − τ )] H(t − τ ). ρh γ

Step 3 : Show that H(t − τ ) g(x, y, t|ξ, η,τ ) = 4π 2 ρh

"

∞

−∞

"

∞

sin[γ(t − τ )] γ −∞

× cos[k(x − ξ) + 0(y − η)] dk d0,

or g(r, θ, t|0, 0, τ ) =

H(t − τ ) 4π 2 ρh

"

0

2π

"

∞

0

sin[γ(t − τ )] cos[σr cos(ϕ − θ)] σ d σdϕ , γ

where x − ξ = r cos(θ), y − η = r sin(θ), k = σ cos(ϕ), 0 = σ sin(ϕ), and ρhγ 2 = D11 cos4 (ϕ) + 4D16 cos3 (ϕ) sin(ϕ) + 2(D12 + 2D66 ) cos2 (ϕ) sin2 (ϕ) σ4 + 4D26 cos(ϕ) sin3 (ϕ) + D22 sin4 (ϕ). Step 4 : If D11 = D12 = D22 = D, and D16 = D66 = D26 = 0, show that we recover g(r, t|0, τ ) from the previous problem if D/(ρh) = L4 . 12. For the two-dimensional Klein-Gordon equation, , ∂2g ∂ 2g 1 ∂ 2g 2 + − + a g = −δ(x − ξ)δ(y − η)δ(t − τ ), ∂x2 ∂y 2 c2 ∂t2 find its free-space Green’s function where −∞ < x, y, ξ, η < ∞, and 0 < t, τ. Step 1 : Using Laplace transforms, show that our partial differential equation reduces to , 2 ∂ 2G ∂2G s + a2 + − G = −δ(x − ξ)δ(y − η)e−sτ . ∂x2 ∂y 2 c2 Step 2 : Using polar coordinates, show that G(x, y, s|ξ, η,τ ) =

* . + e−sτ K0 r s2 + a2 /c . 2π

264

Green’s Functions with Applications

Problem 12

Here r =

.

(x − ξ)2 + (y − η)2 .

Step 3 : Invert the Laplace transform and show that 4 . 5 cos a (t − τ )2 − r2 /c2 . g(x, y, t|ξ, η,τ ) = H(t − τ − r/c). 2π (t − τ )2 − r2 /c2

The figure captioned Problem 12 illustrates this Green’s function (times 2π) as functions of position r/c and time t − τ . The value of a is 1.5; no value was allowed to be greater than two. 13. Find the Green’s function63 governed by the equation , 2 ∂2 ∂2g ∂ 2g 2∂ g + + N = δ(x − ξ)δ(z − ζ)δ(t − τ ), ∂t2 ∂x2 ∂z 2 ∂x2 where −∞ < x, ξ < ∞, 0 < z, ζ < L, and 0 < t, τ. The boundary conditions are g(x, 0, t|ξ, ζ,τ ) = g(x, L, t|ξ, ζ,τ ) = 0, and lim|x|→∞ g(x, z, t|ξ, ζ,τ ) → 0. The initial conditions are g(x, z, 0|ξ, ζ,τ ) = gt (x, z, 0|ξ, ζ,τ ) = 0. Step 1 : If the Green’s function can be written as the Fourier half-range sine expansion ∞ * nπz + ( g(x, z, t|ξ, ζ,τ ) = Gn (x, t|ξ,τ ) sin , L n=1 63 See Grigor’ev, P. L., and V. A. Yakovlev, 1986: Solution of the Cauchy problem for internal waves in a medium with rigid boundaries. Sov. Phys. Tech. Phys., 31, 1254–1256.

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Problem 13

which fulfills the boundary conditions in the z direction, show that we can find Gn (x, t|ξ,τ ) from the partial differential equation , , 2 ∂ 2 ∂ 2 Gn n2 π 2 2 nπζ 2 ∂ Gn − Gn + N = sin δ(x − ξ)δ(t − τ ). ∂t2 ∂x2 L2 ∂x2 L L Step 2 : Taking the Fourier transform of the partial differential equation in Step 1, show that the Fourier transform of Gn (x, t|ξ,τ ), Gn (k, t|ξ,τ ), is governed by , , n2 π 2 2 nπζ $$ 2 2 2 k + Gn + N k Gn = − sin e−ikξ δ(t − τ ). L2 L L Step 3 : Show that , 2 nπζ Gn (k, t|ξ,τ ) = − sin e−ikξ H(t − τ ) NL L . sin[N k(t − τ )/ k 2 + n2 π 2 /L2 ] . × . k k 2 + n2 π 2 /L2

Step 4 : Taking the inverse Fourier transform of Gn (k, t|ξ,τ ), show that , 2 nπζ Gn (x, t|ξ,τ ) = − sin H(t − τ ) NLπ L . " ∞ sin[N k(t − τ )/ k 2 + n2 π 2 /L2 ] . × cos[k(x − ξ)] dk k k 2 + n2 π 2 /L2 0

266

Green’s Functions with Applications , 2 nπζ |x − ξ| Gn (x, t|ξ,τ ) = − sin H(t − τ ) NLπ L n . " ∞ sin[N (t − τ )η/ η 2 + π 2 (x − ξ)2 /L2 ] . × cos(nη) dη. η η2 + π 2 (x − ξ)2 /L2 0

Hint: Let |x − ξ|k = nη.

Step 5 : Using the relationship that ') , - * !' ∞ ( ' cos(η) − cos[π(z + ζ)/L] ' cos(nη) nπζ nπz + 1 ' , sin sin = ln '' n L L 8 cos(η) − cos[π(z − ζ)/L] ' n=1 show that

or

2|x − ξ| g(x, z, t|ξ, ζ,τ ) = − H(t − τ ) NLπ . " ∞ sin[N (t − τ )η/ η 2 + π 2 (x − ξ)2 /L2 ] . × η η 2 + π 2 (x − ξ)2 /L2 0 , - * ∞ ( cos(nη) nπζ nπz + × sin sin dη, n L L n=1 H(t − τ ) g(x, z, t|ξ, ζ,τ ) = 4N π 2

"

π/2

0

sin[N (t − τ ) sin(ϕ)] dϕ sin(ϕ) ') !' ' cos(η) − cos[π(z − ζ)/L] ' ' ' , × ln ' cos(η) − cos[π(z + ζ)/L] '

where η = |x − ξ|π tan(ϕ)/L. The figure captioned Problem 13 illustrates the Green’s function N g(x, z, t|ξ, ζ,τ ) as functions of |x − ξ|/L and z/L when ζ/L = 1/π, and N (t−τ ) = 50. The integration was performed using Weddle’s rule with an increment of 0.001 after setting χ = tan(ϕ). 14. Find the free-space Green’s function64 governed by the equation , ∂ 2 g 1 + λ ∂g 1 ∂ 2g ∂ 2g δ(r − 1)δ(θ − θ$ )δ(t − τ ) + + − = − , ∂r2 r ∂r r2 ∂θ2 ∂t2 rλ over the domain 0 < r < ∞, 0 ≤ θ,θ $ ≤ 2π, and 0 < t, τ. The boundary conditions are lim g(r, θ, t|1, θ$ , τ ) → 0,

r→0

lim g(r, θ, t|1, θ $ , τ ) → 0,

r→∞

64 See Watanabe, K., 1981: Transient response of an inhomogeneous elastic solid to an impulse SH-source (variable SH-wave velocity)(in Japanese). Nihon Kikai Gakkai Rombumshu (Trans. Japan Soc. Mech. Engrs.), Ser. A, 47, 740–746.

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and g(r, θ, t|1, θ $ , τ ) = g(r,θ + 2nπ, t|1, θ $ , τ ). The initial conditions are g(r, θ, 0|1, θ$ , τ ) = gt (r, θ, 0|1, θ $ , τ ) = 0. Step 1 : Taking the Laplace transform of our partial differential equation, show that it becomes , ∂ 2 G 1 + λ ∂G 1 ∂ 2G δ(r − 1)δ(θ − θ$ ) −sτ 2 + + − s G =− e , 2 2 2 ∂r r ∂r r ∂θ rλ with the boundary conditions lim G(r, θ, s|1, θ$ , τ ) → 0,

lim G(r, θ, s|1, θ$ , τ ) → 0,

r→∞

r→0

and G(r, θ, s|1, θ$ , τ ) = G(r,θ + 2nπ, s|1, θ$ , τ ). Step 2 : Introducing the complex Fourier coefficient 1 G(r, n, s|1, θ , τ ) = 2π $

"

2π

G(r, θ, s|1, θ $ , τ ) e−inθ dθ,

0

show that the partial differential equation in Step 1 reduces to the ordinary differential equation d2 G 1 + λ dG (n2 + s2 ) δ(r − 1) −inθ $ −sτ + − G=− e , dr2 r dr r2 2πrλ with the boundary conditions lim |G(r, n, s|1, θ $ , τ )| < ∞,

r→0

lim G(r, n, s|1, θ $ , τ ) → 0.

r→∞

Step 3 : Introducing the new independent variable x = ln(r), show that the ordinary differential equation in Step 2 becomes d2 G dG δ(x) −inθ $ −sτ +λ − (n2 + s2 )G = − e , 2 dx dx 2π

−∞ < x < ∞,

with the boundary conditions lim|x|→∞ G(x, n, s|0, θ$ , τ ) → 0. Step 4 : Taking the Fourier transform of G(x, n, s|0, θ$ , τ ) with respect to x, show that $ e−inθ −sτ G(k, n, s|0, θ$ , τ ) = . 2π(k 2 − iλk + n2 + s2 ) Step 5 : Using the residue theorem, show that

. r−λ/2 exp[−| ln(r)| s2 + λ2n ] −sτ . G(r, n, s|1, θ , τ ) = e , 4π s2 + λ2n $

268

Green’s Functions with Applications 2.0 1.5 1.0 0.5 0.0 −0.5 −1.0 −1.5

(a)

(b)

(d)

(e)

(c)

−2.0 2.0 1.5 1.0 0.5 0.0 −0.5 −1.0 −1.5 −2.0

0

5

10 15 20 25 0 (a) θ − θ ! = 0,

(d) θ − θ ! = 3π/5,

5

(f)

10 15 20 25 0 TIME

(b) θ − θ ! = π/5,

(e) θ − θ ! = 4π/5,

5

10 15 20 25

(c) θ − θ ! = 2π/5,

Problem 14

(f) θ − θ ! = π

. where λ2n = (λ/2)2 + n2 , and 3( s2 + λ2n ) ≥ 0.

Step 6 : Taking the inverse Laplace transform, show that g(r, n, t|1, θ $ , τ ) =

% = & r−λ/2 H [t − τ − | ln(r)| ] J0 λn (t − τ )2 − ln2 (r) . 4π

Step 7 : Constructing the Fourier series, show that g(r, θ, t|1, θ$ , τ ) =

r−λ/2 H [t − τ − | ln(r)| ] 2π % = & ∞ ( 2 2 × 2n J0 λn (t − τ ) − ln (r) cos[n(θ − θ$ )], n=0

where 20 = 12 , and 2n = 1 if n > 0. The figure captioned Problem 14 illustrates 2πrλ/2 g(r, θ, t|1, θ$ , τ ) as a function of time t − τ for six values of θ − θ$ , r = 4, and λ = 2.

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269

15. Find the Green’s function65 governed by the equation ∂2g ∂ 2g ∂ 2g 1 ∂ 2g + + − = −δ(x − ξ)δ(y − η)δ(z − ζ)δ(t − τ ), ∂x2 ∂y 2 ∂z 2 c2 ∂t2 in the rectangular waveguide defined by 0 < x, ξ < a, 0 < y, η < b, −∞ < z, ζ< ∞, and 0 < t, τ. The boundary conditions are g(0, y, z, t|ξ, η, ζ,τ ) = g(a, y, z, t|ξ, η, ζ,τ ) = 0, g(x, 0, z, t|ξ, η, ζ,τ ) = g(x, b, z, t|ξ, η, ζ,τ ) = 0, and lim |g(x, y, z, t|ξ, η, ζ,τ ) → 0.

|z|→∞

The initial conditions are g(x, y, z, 0|ξ, η, ζ,τ ) = gt (x, y, z, 0|ξ, η, ζ,τ ) = 0. Step 1 : If we write the Green’s function as the double Fourier half-range sine expansion g(x, y, z, t|ξ, η, ζ,τ ) =

∞ ∞ 4 (( Gnm (z, t|ζ,τ ) ab m=1 n=1 * nπx + , nπξ - * mπy + * mπη + × sin sin sin sin , a a b b

show that Gnm (z, t|ζ,τ ) satisfies the equation , ∂ 2 Gnm 1 ∂ 2 Gnm 2 − 2 + ωnm Gnm = −δ(z − ζ)δ(t − τ ), ∂z 2 c ∂t2 where

3*

ωnm = c Step 2 : Show that Gnm (z, t|ζ,τ ) =

nπ +2 * mπ +2 + . a b 

c H[c(t − τ ) − |z − ζ|] J0 ωnm 2

Step 3 : Finish the problem and show that g(x, y, z, t|ξ, η, ζ,τ ) =

T

(t − τ )2 −

,

z−ζ c

-2



.

2c H[c(t − τ ) − |z − ζ|] ab   T , -2 ∞ ( ∞ ( z − ζ  × J0 ωnm (t − τ )2 − c m=1 n=1 * nπx + , nπξ - * mπy + * mπη + × sin sin sin sin . a a b b

65 See Dokuchaev, V. P., 1998: Excitation of a rectangular waveguide by pulsed electric and magnetic currents. Radiophys. Quantum Electronics, 41, 301–309.

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Green’s Functions with Applications

16. Find the Green’s function for , -2 ∂2g ∂ 2g ∂ 2g 1 ∂ ∂ + 2+ 2− +M g = δ(x − ξ)δ(y − η)δ(z − ζ)δ(t − τ ), ∂x2 ∂y ∂z c ∂t ∂x where −∞ < x, y, z, ξ, η, ζ < ∞, 0 < t, τ, c and M are real, positive constants, and 0 ≤ M < 1. The boundary conditions are lim g(x, y, z, t|ξ, η, ζ,τ ) → 0,

lim g(x, y, z, t|ξ, η, ζ,τ ) → 0,

|x|→∞

|y|→∞

and lim g(x, y, z, t|ξ, η, ζ,τ ) → 0.

|z|→∞

The initial conditions are g(x, y, z, 0|ξ, η, ζ,τ ) = gt (x, y, z, 0|ξ, η, ζ,τ ) = 0. Step 1 : Take the Laplace transform of the partial differential equation and show that it equals (1 − M 2 )

∂ 2 G 2sM ∂G ∂ 2 G ∂ 2 G s2 − + + − 2 G = δ(x − ξ)δ(y − η)δ(z − ζ)e−sτ ∂x2 c ∂x ∂y 2 ∂z 2 c

plus the boundary conditions lim G(x, y, z, s|ξ, η, ζ,τ ) → 0,

lim G(x, y, z, s|ξ, η, ζ,τ ) → 0,

|x|→∞

|y|→∞

and lim G(x, y, z, s|ξ, η, ζ,τ ) → 0.

|z|→∞

Step 2 : By introducing x = (1−M 2 )1/2 x$ , ξ = (1−M 2 )1/2 ξ $ , s = (1−M 2 )1/2 s$ $ $ and G = Ges Mx /c , show that the partial differential equation in Step 1 becomes ∂2G ∂ 2G ∂2G s$2 δ(x$ − ξ $ )δ(y − η)δ(z − ζ) −sτ −s$ M ξ$ /c + + − G = e . ∂x$2 ∂y 2 ∂z 2 c2 (1 − M 2 )1/2 Step 3 : Show that $

G(x, y, z, s|ξ, η, ζ,τ ) = where R$ =

.

$

$

$

$

e−s R /c+s (x −ξ )M/c−sτ , 4π(1 − M 2 )1/2 R$

(x$ − ξ $ )2 + (y − η)2 + (z − ζ)2 .

Step 4 : Show that g(x, y, z, t|ξ, η, ζ,τ ) =

δ(t − τ − τ $ ) , 4π(1 − M 2 )1/2 R$

The Wave Equation

271

where τ $ = [R$ − M (x$ − ξ $ )]/[c(1 − M 2 )1/2 ]. 17. Show that transient solutions for the one-dimensional dielectric problem governed by Equation 4.9.15 through Equation 4.9.17 are (t) g1 (z, t|ζ,τ )

% & c n(z − a) ζ − a = 2 H t−τ + − n −1 c c sinh(nsn z/c) exp[sn (t − τ ) − sn (ζ − a)/c] × , (asn /c) sinh(nsn a/c)

and (t) g2 (z, t|ζ,τ )

where

, c z−a ζ −a = 2 H t−τ − − n −1 c c exp[sn (t − τ ) − sn (ζ − a)/c − sn (z − a)/c] × , asn /c asn 1 = ln c 2n

,

1−n 1+n

-

,

when n < 1. Plot this solution as a function z/a for various c(t − τ )/a. Compare and contrast this solution with the solution that we found for n > 1. 18. Find the reflected and transmitted waves for the half-space problem solved in Section 4.9, Equation 4.9.31 and Equation 4.9.32, when n > 1. Do you still have head waves? Explain your solution in terms of the physical processes that are occurring. 19. Construct the Green’s function51 governed by the two-dimensional Poisson equation ∂2g ∂ 2g + 2 = −δ(x − ξ)δ(z − ζ), 2 ∂x ∂z where −∞ < x, ξ < ∞, −h < z, ζ < 0, and 0 ≤ τ < t, with the boundary conditions ∂ 2g ∂g + = 0, z = 0, τ < t, ∂t2 ∂z ∂g = 0, ∂z

z = −h,

τ < t,

and the initial conditions g(x, z,τ |ξ, ζ,τ ) = 0, gt (x, z,τ |ξ, ζ,τ ) = 0. As |x| → ∞, we require that g and its first two derivatives with respect to x should tend to zero in such a manner so that g possesses a Fourier transform with respect to x.

272

Green’s Functions with Applications

Step 1 : Applying the Fourier transform to the problem, show that d2 G − k 2 G = −e−ikξ δ(z − ζ), dz 2

−h < z, ζ < 0.

Step 2 : Show that e−ikξ G =

−

sinh(kz> ) cosh[k(z< + h)] k cosh(kh)

4 5  .   1 − cos (t − τ ) k tanh(kh) cosh[k(z + h)] cosh[k(ζ + h)] cosh2 (kh)



k tanh(kh)



.

Step 3 : By inverting the Fourier transform, show that the Green’s function is % , & " ∞ 1 R −kh sinh(kz) sinh(kζ) g(x, z, t|ξ, ζ,τ ) = ln $ + 2 e cos(kr) dk 2π R k cosh(kh) 0 1 − π

"

∞ 0

cosh[k(z + h)] cosh[k(ζ + h)] cos(kr) dk cosh2 (kh) 4 5  .  1 − cos (t − τ ) k tanh(kh)  × ,   k tanh(kh)

where R2 = r2 + (z − ζ)2 , R$2 = r2 + (z + ζ)2 , and r = |x − ξ|.

20. Find the harmonic waves in the two-dimensional case by solving ∂2g ∂ 2g + 2 = −δ(x − ξ)δ(z − ζ), 2 ∂x ∂z where −∞ < x, ξ < ∞, and −h < z, ζ < 0, while, at the free surface, ∂g = ω 2 g, ∂z

z = 0,

and

∂g = 0, ∂z

z = −h.

Finally, we require that lim|x|→∞ g(x, z|ξ,ζ ) → 0. Step 1 : Applying the Fourier transform in the x direction, show that d2 G − k 2 G = −e−ikξ δ(z − ζ), dz 2

−h < z, ζ < 0.

Step 2 : Solve the ordinary differential equation in Step 1 and show that G(k, z|ξ,ζ ) =

e−ikξ ω 2 sinh(kz> ) + k cosh(kz> ) cosh[k(z< + h)]. k k sinh(kh) − ω 2 cosh(kh)

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273

Step 3 : Taking the inverse Fourier transform, show that " 1 ∞ ω 2 sinh(kz> ) + k cosh(kz> ) cos[k(x − ξ)] g(x, z|ξ,ζ ) = ∪ cosh[k(z< +h)] dk. π 0 k sinh(kh) − ω 2 cosh(kh) k Our special integral sign denotes an integration along the real k axis except for passing below the singularity at k = κ0 , where κ0 tanh(κ0 h) = ω 2 . Mei66 showed that we can rewrite this expression as , 2! " ∞ 1 h dk cos[k(x − ξ)] g(x, z|ξ,ζ ) = ln −2∪ 2 cosh(kh) − k sinh(kh) 2π rr$ k ω 0 ) −kh × cosh[k(z + h)] cosh[k(ζ + h)] − e , where r2 = (x − ξ)2 + (z − ζ)2 , and r$2 = (x − ξ)2 + (z + ζ + 2h)2 . Step 4 : Using the residue theorem,67 show that g(x, z|ξ,ζ ) =

∞ ( cos[κn (z + h)] cos[κn (ζ + h)] −κn |x−ξ| e , 2κn Nn n=0

where Nn is given by Equation 4.10.48 and κn is the nth simple root to κ tan(κh) = −ω 2 . Note that the root κ0 lies on the real k axis. Therefore, we pass below the singularity at k = κ0 and above the singularity at k = −κ0 . Finite Domains 21. Use Equation 4.2.13 to construct the Green’s function for the one-dimensional wave equation ∂2g ∂ 2g − = δ(x − ξ)δ(t − τ ), ∂t2 ∂x2

0 < x, ξ < L,

0 < t, τ ,

subject to the boundary conditions g(0, t|ξ,τ ) = gx (L, t|ξ,τ ) = 0, 0 < t, and the initial conditions that g(x, 0|ξ,τ ) = gt(x, 0|ξ,τ ) = 0, 0 < x < L. The 66

Mei, C. C., 1989: The Applied Dynamics of Ocean Surface Waves. World Scientific, 740 pp. See pp. 379–382. 67 See Appendix B in Dalrymple, R. A., M. A. Losada, and P. A. Martin, 1991: Reflection and transmission from porous structures under oblique wave attack. J. Fluid Mech., 224, 625–644.

274

Green’s Functions with Applications

Problem 21

figure captioned Problem 20 illustrates this Green’s function as functions of position x/L and time (t − τ )/L when ξ = 0.2. 22. Use Equation 4.2.13 to construct the Green’s function for the one-dimensional wave equation ∂2g ∂ 2g − = δ(x − ξ)δ(t − τ ), 2 ∂t ∂x2

0 < x, ξ < L,

0 < t, τ ,

subject to the boundary conditions gx (0, t|ξ,τ ) = gx (L, t|ξ,τ ) = 0, 0 < t, and the initial conditions that g(x, 0|ξ,τ ) = gt (x, 0|ξ,τ ) = 0, 0 < x < L. 23. Use the Green’s function given by Equation 4.2.25 to write down the solution to the wave equation utt = uxx on the interval 0 < x < L with the boundary conditions u(0, t) = u(L, t) = 0, 0 < t, and the initial conditions u(x, 0) = cos(πx/L) and ut (x, 0) = 0, 0 < x < L. 24. Use the Green’s function given by Equation 4.2.25 and the results from Problem 1 to write down the solution to the wave equation utt = uxx on the interval 0 < x < L with the boundary conditions u(0, t) = e−t and u(L, t) = 0, 0 < t, and the initial conditions u(x, 0) = sin(πx/L) and ut (x, 0) = 1, 0 < x < L. 25. Use the Green’s function that you found in Problem 21 and the results from Problem 1 to write down the solution to the wave equation utt = uxx

The Wave Equation

275

on the interval 0 < x < L with the boundary conditions u(0, t) = 0 and ux (L, t) = 1, 0 < t, and the initial conditions u(x, 0) = x and ut (x, 0) = 1, 0 < x < L. 26. Use the Green’s function that you found in Problem 22 and the results from Problem 1 to write down the solution to the wave equation utt = uxx on the interval 0 < x < L with the boundary conditions ux (0, t) = 1 and ux (L, t) = 0, 0 < t, and the initial conditions u(x, 0) = 1 and ut (x, 0) = 0, 0 < x < L. 27. Find the Green’s function68 governed by ∂2g ∂g ∂ 2g +2 − = δ(x − ξ)δ(t − τ ), 2 ∂t ∂t ∂x2

0 < x, ξ < L,

0 < t, τ ,

subject to the boundary conditions gx (0, t|ξ,τ ) = gx (L, t|ξ,τ ) = 0, 0 < t, and the initial conditions g(x, 0|ξ,τ ) = gt(x, 0|ξ,τ ) = 0, 0 < x < L. Step 1 : If the Green’s function can be written as the Fourier half-range cosine series ∞ * nπx + 1 2 ( g(x, t|ξ,τ ) = G0 (t|τ ) + Gn (t|τ ) cos , L L n=1 L

so that it satisfies the boundary conditions, show that Gn (t|τ ) is governed by , n2 π 2 nπξ $$ $ Gn + 2Gn + Gn = cos δ(t − τ ), n≥0 L2 L Step 2 : Show that G0 (t|τ ) = e−(t−τ ) sinh(t − τ )H(t − τ ), and , nπξ sin[βn (t − τ )] Gn (t|τ ) = cos e−(t−τ ) H(t − τ ), n ≥ 1, L βn . where βn = (nπ/L)2 − 1. Step 3 : Combine the results from Steps 1 and 2 and show that

g(x, t|ξ,τ ) = e−(t−τ ) sinh(t − τ )H(t − τ )/L , ∞ * nπx + ( sin[βn (t − τ )] nπξ + 2e−(t−τ )H(t − τ )/L cos cos . βn L L n=1 68 See Ozi¸ ¨ sik, M. N., and B. Vick, 1984: Propagation and reflection of thermal waves in a finite medium. Int. J. Heat Mass Transfer , 27, 1845–1854. See also Tang, D.-W., and N. Araki, 1996: Propagation of non-Fourier temperature wave in finite medium under laser-pulse heating (in Japanese). Nihon Kikai Gakkai Rombumshu (Trans. Japan Soc. Mech. Engrs.), Ser. B , 62, 1136–1141.

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Green’s Functions with Applications

Problem 27

The figure captioned Problem 27 illustrates this Green’s function with L = 1 and ξ = 0.2. 28. Find the Green’s function governed by ∂ 2g ∂ 2g − c2 2 = c2 δ(x − ξ)δ(t − 0+ ), 2 ∂t ∂x

0 < x, ξ < L,

0 < t,

where a, b, c are positive constants, with the boundary conditions gx (0, t|ξ, 0+ ) − a/c gt (0, t|ξ, 0+ ) = 0,

0 < t,

gx (L, t|ξ, 0+ ) + b/c gt (L, t|ξ, 0+ ) = 0,

0 < t,

and +

+

and the initial conditions g(x, 0|ξ, 0 ) = gt (x, 0|ξ, 0 ) = 0 Step 1 : Taking the Laplace transform of the partial differential equation and boundary conditions, show that d2 G s2 − 2 G = −δ(x − ξ), dx2 c with

G$ (0, s|ξ, 0+ ) −

as G(0, s|ξ, 0+ ) = 0, c

The Wave Equation

277

and

bs G(L, s|ξ, 0+ ) = 0, c Step 2 : Show that the solution to the boundary-value problem in Step 1 is 4 5 G(x, s|ξ, 0+ ) = ϕ(x, s) = 12 (1 + a)esx/c + (1 + a)e−sx/c , 0 0 and the same initial and boundary conditions. 29. The Green’s function for a beam of length L with internal damping is given by ∂ 2g ∂4g ∂ 5g + + a 4 = δ(x − ξ)δ(t − τ ), 2 4 ∂t ∂x ∂x ∂t

0 < x, ξ < L,

0 < t, τ ,

where a > 0, with the boundary conditions g(0, t|ξ,τ ) = g(L, t|ξ,τ ) = gxx (0, t|ξ,τ ) = gxx (L, t|ξ,τ ) = 0, and the initial conditions g(x, 0|ξ,τ ) = gt (x, 0|ξ,τ ) = 0. Find the Green’s function. Step 1 : Assuming that the Green’s function has the form g(x, t|ξ,τ ) =

, - * ∞ 2 ( nπξ nπx + Gn (t) sin sin , L n=1 L L

show that Gn (t) is governed by * nπ +4 dG * nπ +4 d2 Gn n + a + Gn = δ(t − τ ). dt2 L dt L

Why have we chosen this particular eigenfunction expansion?

Step 2 : Solve the ordinary differential equation in Step 1 and show that .  − τ ) 1 − a2 ωn2 /4 ]   sin[ωn (t . , a nω < 2,    ωn 1 − a2 ωn2 /4    2 (t − τ ), a n = 2, Gn (t) = e−aωn (t−τ )/2   .    sinh[ωn (t − τ ) a2 ωn2 /4 − 1 ]   . , 2 < a ωn ,  ωn a2 ωn2 /4 − 1

ω

if t > τ and zero otherwise, where ωn = n2 π 2 /L2 .

69 Jovanovic, V., and S. Koshkin, 2014: D’Alembert sums for a vibrating bar with viscous ends. J. Eng. Math., 85, 99–114.

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279

Problem 29

Step 3 : Show that g(x, t|ξ,τ ) =

, - * ∞ ( 2 2 nπξ nπx + H(t − τ ) e−aωn (t−τ )/2 sin sin L L L n=1 .  sin[ωn (t − τ ) 1 − a2 ωn2 /4 ]   . , a nω < 2,    ωn 1 − a2 ωn2 /4    (t − τ ), a n = 2, ×   .    sinh[ωn (t − τ ) a2 ωn2 /4 − 1 ]   . , 2 < a ωn .  ωn a2 ωn2 /4 − 1

ω

The figure captioned Problem 29 illustrates this Green’s function as functions of position x and time t − τ when a = 0.2, L = 1, and ξ = 0.2. 30. Find the Green’s function within a spherical cavity of radius a that is governed by the partial differential equation ∂ 2 g 2 ∂g 1 ∂ 2g 1 ∂ 2 (rg) 1 ∂2g δ(r − ρ)δ(t − τ ) + − = − =− , ∂r2 r ∂r c2 ∂t2 r ∂r2 c2 ∂t2 4πr2 where 0 < r, ρ < a, and 0 < t, τ, subject to the boundary conditions lim g(r, t|ρ,τ ) → 0,

r→0

g(a, t|ρ,τ ) = 0,

0 < t,

and the initial conditions g(r, 0|ρ,τ ) = gt (r, 0|ρ,τ ) = 0, 0 < r < a.

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Step 1 : By setting ϕ(r, t|ρ,τ ) = rg(r, t|ρ,τ ), show that the problem can be rewritten as ∂ 2ϕ 1 ∂ 2ϕ δ(r − ρ)δ(t − τ ) − =− , 2 2 2 ∂r c ∂t 4πr

0 < r, ρ < a,

0 < t, τ ,

subject to the boundary conditions ϕ(0, t|ρ,τ ) = ϕ(a, t|ρ,τ ) = 0, 0 < r < a, and the initial conditions ϕ(r, 0|ρ,τ ) = ϕt (r, 0|ρ,τ ) = 0, 0 < t. Step 2 : Take the Laplace transform of the partial differential equation in Step 1 and show that it becomes the ordinary differential equation d2 Φ s2 δ(r − ρ) −sτ − 2Φ = − e , dr2 c 4πr

0 < r, ρ < a,

subject to the boundary conditions Φ(0, s|ρ,τ ) = Φ (a, s|ρ,τ ) = 0. Step 3 : Show that ∞ δ(r − ρ) 1 ( * nπρ + * nπr + = sin sin . 4πr 2πaρ n=1 a a

Step 4 : Show that the eigenfunction expansion of the ordinary differential equation in Step 2 is G(r, s|ρ,τ ) =

∞ e−sτ ( sin(nπρ/a) sin(nπr/a) . 2πaρr n=1 s2 /c2 + n2 π 2 /a2

Step 5 : Invert the Laplace transform and show that the Green’s function is ∞ * nπρ + * nπr + % nπc(t − τ ) & c H(t − τ ) ( 1 g(r, t|ρ,τ ) = sin sin sin . 2π 2 ρr n=1 n a a a

The figure captioned Problem 30 illustrates this Green’s function (divided by c/a2 ) for a/100 ≤ r ≤ a. 31. Find the Green’s function for a damped axially moving wire70 governed by the partial differential equation 2 ∂2g ∂2g ∂g ∂g 2 ∂ g + 2ν − (1 − ν ) + ην +η = δ(x − ξ)δ(t − τ ), 2 2 ∂t ∂x∂t ∂x ∂x ∂t 70 See Chung, C., and I. Kao, 2011: Green’s function and forced vibration response of damped axially moving wire. J. Vibr. Control , 18, 1798–1808.

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281

Problem 30

where 0 < x, ξ < 1, 0 < t, τ, 0 ≤ ν < 1, and 0 ≤ η, subject to the boundary conditions g(0, t|ξ,τ ) = g(1, t|ξ,τ ) = 0, 0 < t, and the initial conditions g(x, 0|ξ,τ ) = gt(x, 0|ξ,τ ) = 0, 0 < x < 1. Step 1 : Take the Laplace transform of the partial differential equation and show that it becomes the ordinary differential equation d2 G ν(2s + η) dG s2 + sη δ(x − ξ) −sτ − − G=− e , 0 < x < 1, dx2 1 − ν 2 dx 1 − ν2 1 − ν2 subject to the boundary conditions G(0, s|ξ,τ ) = G(1, s|ξ,τ ) = 0, where G(x, s|ξ,τ ) is the Laplace transform of g(x, t|ξ,τ ). Step 2 : Show that G(x, s|ξ,τ ) = where

sinh(βx< ) sinh[β(1 − x> )] α(x−ξ)−sτ e , (1 − ν 2 )β sinh(β)

. ν(2s + η) 4s2 + 4sη + ν 2 η 2 α(s) = and β(s) = . 2(1 − ν 2 ) 2(1 − ν 2 ) Step 3 : Use the residue theorem to invert G(x, s|ξ,τ ) and show that ∞ ( 1 g(x, t|ξ,τ ) = 2(1 − ν 2 )H(t − τ )e−η(t−τ )/2 sin(nπξ) sin(nπx) ω n=1 n U V × sin(xn ) cos[ωn (t − τ )] + cos(xn ) sin[ωn (t − τ )] , where . (1 − ν 2 )[4n2 π 2 (1 − ν 2 ) − η 2 ] νωn (x − ξ) xn = and ωn = . 2 1−ν 2

Chapter 5 Green’s Functions for the Heat Equation In this chapter, we present the Green’s function1 for the heat equation ∂u − a2 ∇2 u = q(r, t), ∂t

(5.0.1)

where ∇ is the three-dimensional gradient operator, t denotes time, r is the position vector, a2 is the diffusivity, and q(r, t) is the source density. In addition to Equation 5.0.1, boundary conditions must be specified to ensure the uniqueness of solution; the most common ones are Dirichlet, Neumann and Robin (a linear combination of the first two). An initial condition u(r, t = t0 ) is also needed. The heat equation differs in many ways from the wave equation and the Green’s function must, of course, manifest these differences. The most notable one is the asymmetry of the heat equation with respect to time. This merely 1 See also Carslaw, H. S., and J. C. Jaeger, 1959: Conduction of Heat in Solids. Clarendon Press, Chapter 14; Beck, J. V., K. D. Cole, A. Haji-Sheikh, and B. Litkouhi, 1992: Heat ¨ sik, M. Conduction Using Green’s Functions. Hemisphere Publishing Corp., 523 pp.; Ozi¸ N., 1993: Heat Conduction. John Wiley & Sons, Inc., Chapter 6.

283

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Green’s Functions with Applications

reflects the fact that the heat equation differentiates between past and future as entropy continually increases. The purpose of this introductory section is to prove that we can express the solution to Equation 5.0.1 in terms of boundary conditions, the initial condition and the Green’s function, which is found by solving ∂g − a2 ∇2 g = δ(r − r0 )δ(t − τ ), ∂t

(5.0.2)

where r0 denotes the position of the source. From causality2 we know that g(r, t|r0 , τ ) = 0 if t < τ . We again require that the Green’s function g satisfies the homogeneous form of the boundary condition on u. For example, if u satisfies a homogeneous or nonhomogeneous Dirichlet condition, then the Green’s function will satisfy the corresponding homogeneous Dirichlet condition. Although we will focus on the mathematical aspects of the problem, Equation 5.0.2 can be given the physical interpretation of the temperature distribution within a medium when a unit of heat is introduced at r0 at time τ. Remark. Although we will use Equation 5.0.2 as our fundamental definition of the Green’s function as it applies to the heat equation, we can also find it by solving the initial-value problem: ∂u − a2 ∇2 u = 0, ∂t

t>τ

, (r, u τ ) = δ(r − r0 ).

(5.0.3)

Then g(r, t|r0 , τ ) = u(r, t−τ)H(t−τ ). This is most easily seen by introducing a new time variable t$ = t − τ into Equation 5.0.2 and Equation 5.0.3 and noting that the Laplace transform of Equation 5.0.2 and Equation 5.0.3 are identical. Therefore, the Green’s function gives the heat flow resulting from a temperature distribution at t = τ which is everywhere equal to zero except at r = r0 , where it is infinite. # " As in the case of the wave equation, we begin by studying reciprocity. We now show that g(r, t|r0 , τ ) = g(r0 , −τ |r, −t). Physically, the function g(r0 , −τ |r, −t) gives the effect at r0 and time −τ of a heat source that is introduced into the medium at r at a time −t. Because τ < t, the time sequence is still properly ordered. Consider the adjoint function g ∗ (r, t|r0 , τ ) defined by the relationship g(r, −t|r0 , −τ ) = g ∗ (r, t|r0 , τ ). This function g ∗ satisfies the time-reversed equation ∂g ∗ + a2 ∇2 g ∗ = −δ(r − r0 )δ(t − τ ), (5.0.4) ∂t 2

The principle stating that an event cannot precede its cause.

The Heat Equation

285

with g ∗ (r, t|r0 , τ ) = 0 if t > τ . In other words, g ∗ gives the development backward in time of a source placed at r0 and at time τ . The reciprocity condition now reads g(r, t|r0 , τ ) = g ∗ (r0 , τ |r, t). The function g describes the evolution as time increases, leading from the initial source to the final distribution. The function g ∗ describes the same process in reverse time order, beginning with the final distribution and going backward in time to the initial source. The proof of reciprocity patterns itself after the proof given for the wave equation. The two equations are now a2 ∇2 g(r, t|r0 , τ0 ) −

∂g(r, t|r0 , τ0 ) = −δ(r − r0 )δ(t − τ0 ), ∂t

(5.0.5)

∂g(r, −t|r1 , −τ1 ) = −δ(r − r1 )δ(t − τ1 ). ∂t

(5.0.6)

and a2 ∇2 g(r, −t|r1 , −τ1 ) +

Multiplying Equation 5.0.5 by g(r, −t|r1 , −τ1 ) and Equation 5.0.6 by g(r, t|r0 , τ0 ), subtracting, and integrating over the volume V and over the time t from −∞ to t+ 0 , we eventually obtain a

2

"

t+ 0 −∞

"" % ⊂⊃ g(r, −t|r1 , −τ1 ) ∇g(r, t|r0 , τ0 ) S

−

"

t+ 0 −∞

""" %

&

− g(r, t|r0 , τ0 ) ∇g(r, −t|r1 , −τ1 ) · n dS

∂g(r, t|r0 , τ0 ) ∂t V & ∂g(r, −t|r1 , −τ1 ) + g(r, t|r0 , τ0 ) dV dt ∂t = g(r1 , τ1 |r0 , t0 ) − g(r0 , −τ0 |r1 , −τ1 ), (5.0.7) g(r, −t|r1 , −τ1 )

where S is the surface enclosing V . The first integral vanishes due to the homogeneous boundary conditions satisfied by g. In the second integral, we perform the time integration and obtain 't+ '0 g(r, −t|r1 , −τ1 )g(r, t|r0 , t0 )''

.

t=−∞

At the lower limit, the second of the two factors vanishes because of the initial condition. At the upper limit, the first factor vanishes also because of the initial condition. Note that we tacitly assumed that τ1 is within the region of integration. The reciprocity condition now follows directly.

286

Green’s Functions with Applications

We now establish that the solution to the nonhomogeneous heat equation can be expressed in terms of the Green’s function, boundary conditions and the initial condition. We begin with the equations a2 ∇20 u(r0 , t0 ) −

∂u(r0 , t0 ) = −q(r0 , t0 ), ∂t0

(5.0.8)

and a2 ∇20 g(r, t|r0 , t0 ) +

∂g(r, t|r0 , t0 ) = −δ(r − r0 )δ(t − t0 ). ∂t0

(5.0.9)

As we did in the previous chapter, we multiply Equation 5.0.8 by g and Equation 5.0.9 by u and subtract. Integrating over the volume V0 and over t0 from 0 to t+ , where t+ denotes a time slightly later than t so that we avoid ending the integration exactly at the peak of the delta function, we find " t+ """ 1 2 2 a u ∇20 g − g ∇20 u dV0 dt0 0

V0

% , , -& ∂g ∂u + u +g dV0 dt0 ∂t0 ∂t0 0 V0 " t+ """ = q(r0 , t0 ) g(r, t|r0 , t) dV0 dt0 − u(r, t). "

t+

"""

0

(5.0.10)

V0

Applying Green’s second formula to the first integral of Equation 5.0.10 and performing the time integration in the second integral, we finally obtain " t+ """ u(r, t) = q(r0 , t0 ) g(r, t|r0 , t0 ) dV0 dt0 0

+ a2

"

V0

t+

0

+

"""

"" % ⊂⊃ g(r, t|r0 , t0 ) ∇0 u(r0 , t0 ) S0

&

− u(r0 , t0 ) ∇0 g(r, t|r0 , t0 ) · n dS0 dt0 u(r0 , 0)g(r, t|r0 , 0) dV0 ,

(5.0.11)

V0

where we used g(r, t|r0 , t+ ) = 0. The first two terms in Equation 5.0.11 represent the familiar effects of volume sources and boundary conditions, while the third term includes the effects of the initial data. • Example 5.0.1 Some care is necessary in the construction of solutions using Equation 5.0.11. Consider the heat equation problem: ∂u ∂ 2u = , ∂t ∂x2

0 < x < 1,

0 < t,

(5.0.12)

The Heat Equation

287

subject to the boundary conditions ux (0, t) = 1, ux (1, t) = 0, 0 < t, and the initial condition u(x, 0) = 0, 0 < x < 1. Using the results from Problem 1 and Problem 29, Equation 5.0.11 simplifies to " t

u(x, t) = −

g(x, t|ξ,τ )uξ (0, τ ) dτ,

(5.0.13)

;

H(t − τ ). (5.0.14)

0

with :

g(x, t|ξ,τ ) = 1 + 2

∞ (

cos(nπξ) cos(nπx)e

−n2 π 2 (t−τ )

n=1

Substituting g(x, t|ξ,τ ) and uξ (0, τ ) = 1 in Equation 5.0.13 and carrying out the integration, ∞ + 2 ( cos(nπx) * −n2 π 2 t 1 − e π 2 n=1 n2

u(x, t) = −t −

∞ x2 1 2 ( cos(nπx) −n2 π 2 t − −t+ 2 e . 2 3 π n=1 n2

=x−

(5.0.15) (5.0.16)

Equation 5.0.16 follows from Equation 5.0.15 because x−

∞ x2 1 2 ( cos(nπx) = − 2 . 2 3 π n=1 n2

(5.0.17)

Imagine now that we wanted to use our results to compute the flux within the domain 0 ≤ x ≤ 1. Now, ux (x, t) =

∞ + 2 2 2 ( sin(nπx) * 1 − e−n π t π n=1 n

= 1−x−

∞ 2 ( sin(nπx) −n2 π2 t e . π n=1 n

(5.0.18) (5.0.19)

If we use Equation 5.0.18, we have that ux (0, t) = 0 which violates the boundary condition at x = 0. On the other hand, Equation 5.0.19 yields ux (0, t) = 1, which is the correct result. The problem here is the poor convergence3 of the Fourier series in Equation 5.0.18 at x = 0.

3 See Section 5 in Chapter 1 of Kantorovich, L. V., and V. I. Krylov, 1964: Approximate Methods of Higher Analysis, New York, Interscience Publishers, 681 pp.

288

Green’s Functions with Applications

5.1 HEAT EQUATION OVER INFINITE OR SEMI-INFINITE DOMAINS The Green’s function for the one-dimensional heat equation is governed by ∂g ∂ 2g − a2 2 = δ(x − ξ)δ(t − τ ), ∂t ∂x

−∞ < x, ξ < ∞,

0 < t, τ ,

(5.1.1)

subject to the boundary conditions lim g(x, t|ξ,τ ) → 0,

0 < t,

|x|→∞

(5.1.2)

and the initial condition that g(x, 0|ξ,τ ) = 0, −∞ < x < ∞. Let us find g(x, t|ξ,τ ). We begin by taking the Laplace transform of Equation 5.1.1 and find that d2 G s δ(x − ξ) −sτ − 2G = − e . dx2 a a2

(5.1.3)

Next, we take the Fourier transform of Equation 5.1.3 so that (k 2 + q 2 )G(k, s|ξ,τ ) =

e−ikξ e−sτ , a2

(5.1.4)

where G(k, s|ξ,τ ) is the Fourier transform of G(x, s|ξ,τ ) and q 2 = s/a2 . To find G(x, s|ξ,τ ), we use the inversion integral G(x, s|ξ,τ ) =

e−sτ 2πa2

"

∞

−∞

ei(x−ξ)k dk. k2 + q2

(5.1.5)

Transforming Equation 5.1.5 into a closed contour via Jordan’s lemma, we evaluate it by the residue theorem and find that √

e−|x−ξ| s/a−sτ √ G(x, s|ξ,τ ) = . 2a s

(5.1.6)

From a table of Laplace transforms we finally obtain

% & H(t − τ ) (x − ξ)2 g(x, t|ξ,τ ) = . exp − 2 , 4a (t − τ ) 4πa2 (t − τ )

(5.1.7)

The Heat Equation

289

Table 5.1.1: Free-Space Green’s Function in Various Coordinate Systems for the Heat Equation Two-Dimensional g(x, y, t|ξ, η,τ ) =

% & H(t − τ ) (x − ξ)2 + (y − η)2 exp − 4πa2 (t − τ ) 4a2 (t − τ )

% 2 & H(t − τ ) r + ρ2 − 2rρ cos(θ − θ $ ) exp − 4πa2 (t − τ ) 4a2 (t − τ ) % & H(t − τ ) rρ g(r, z, t|ρ, ζ,τ ) = I 3/2 0 2a2 (t − τ ) 8 [πa2 (t − τ )] % 2 & r + ρ2 + (z − ζ)2 × exp − 4a2 (t − τ )

g(r, θ, t|ρ,θ $ , τ ) =

Three-Dimensional g(x, y, z, t|ξ, η, ζ,τ ) =

g(r, θ, z, t|ρ,θ $ , ζ , τ) =

H(t − τ )

[4πa2 (t − τ )]3/2 % & (x − ξ)2 + (y − η)2 + (z − ζ)2 × exp − 4a2 (t − τ ) H(t − τ )

3/2

[4πa2 (t − τ )] % 2 & r + ρ2 − 2rρ cos(θ − θ$ ) + (z − ζ)2 × exp − 4a2 (t − τ )

after applying the second shifting theorem. In the same manner, the Green’s function for , ∂g a2 ∂ ∂g δ(r − ρ)δ(t − τ ) − r = , 0 < r, ρ < ∞, ∂t r ∂r ∂r 2πr

0 < t, τ , (5.1.8)

is

g(r, t|ρ,τ ) =

H(t − τ ) 2π

"

∞ 0

J0 (kr)J0 (kρ)e−k

2 2

a (t−τ )

k dk.

(5.1.9)

290

Green’s Functions with Applications

Figure 5.1.1: Equation 5.1.15 for the one-dimensional heat equation on the semi-infinite domain 0 < x < ∞, and 0.01 ≤ t − τ , when the left boundary condition is gx (0, t|ξ,τ ) = 0, ξ = 0.5, and a2 = 1.

The primary use of the fundamental or free-space Green’s function4 is as a particular solution to the Green’s function problem. For this reason, it is often called the fundamental heat conduction solution. Consequently, we usually must find a homogeneous solution so that the sum of the free-space Green’s function plus the homogeneous solution satisfies any boundary conditions. The following examples show some commonly employed techniques. • Example 5.1.1 Let us find the Green’s function for the following problem: ∂g ∂2g − a2 2 = δ(x − ξ)δ(t − τ ), ∂t ∂x

0 < x, ξ < ∞,

0 < t, τ ,

(5.1.10)

subject to the boundary conditions g(0, t|ξ,τ ) = 0,

lim g(x, t|ξ,τ ) → 0,

x→∞

0 < x < ∞,

(5.1.11)

and the initial condition that g(x, 0|ξ,τ ) = 0, 0 < x < ∞. From the boundary condition g(0, t|ξ,τ ) = 0, we deduce that g(x, t|ξ,τ ) must be an odd function in x over the open interval (−∞, ∞). We find this Green’s function by introducing an image source of −δ(x + ξ) and resolving Equation 5.1.1 with the source δ(x − ξ)δ(t − τ ) − δ(x + ξ)δ(t − τ ). Because this 4 In electromagnetic theory, a free-space Green’s function is the particular solution of the differential equation valid over a domain of infinite extent, where the Green’s function remains bounded as we approach infinity, or satisfies a radiation condition there.

The Heat Equation

291

equation is linear, Equation 5.1.7 gives the solution for each delta function and the Green’s function for Equation 5.1.10 is ! % & % &) H(t − τ ) (x − ξ)2 (x + ξ)2 . g(x, t|ξ,τ ) = exp − 2 − exp − 2 4a (t − τ ) 4a (t − τ ) 4πa2 (t − τ ) (5.1.12) % & % & 2 2 H(t − τ ) x +ξ xξ =. exp − 2 sinh . (5.1.13) 2 (t − τ ) 2 4a (t − τ ) 2a πa (t − τ ) In a similar manner, if the boundary condition at x = 0 changes to gx (0, t|ξ,τ ) = 0, then Equation 5.1.12 and Equation 5.1.13 become ! % & % &) H(t − τ ) (x − ξ)2 (x + ξ)2 g(x, t|ξ,τ ) = . exp − 2 + exp − 2 4a (t − τ ) 4a (t − τ ) 4πa2 (t − τ ) (5.1.14) % & % & 2 2 H(t − τ ) x +ξ xξ =. exp − 2 cosh . (5.1.15) 2 (t − τ ) 2 4a (t − τ ) 2a πa (t − τ ) Figure 5.1.1 illustrates Equation 5.1.15 for the special case when a2 = 1.

# " • Example 5.1.2 Several years ago Carslaw and Jaeger5 showed how free-space Green’s functions and Laplace transforms may be used to derive additional Green’s functions. To illustrate this technique, consider the following problem: ∂g ∂2g − a2 2 = δ(x − ξ)δ(t − τ ), ∂t ∂x

0 < x, t, ξ, τ ,

(5.1.16)

subject to the boundary conditions that gx (0, t|ξ,τ ) = h g(0, t|ξ,τ ),

lim g(x, t|ξ,τ ) → 0,

x→∞

0 < t,

(5.1.17)

and the initial condition that g(x, 0|ξ,τ ) = 0, 0 < x < ∞. Because this problem is linear, we can write the solution as 2

g(x, t|ξ,τ ) =

2

e−(x−ξ) /[4a (t−τ )] . H(t − τ ) + v(x, t), 2a π(t − τ )

(5.1.18)

5 The results in this example were first derived by Bryan, G. H., 1891: Note on a problem in the linear conduction of heat. Proc. Camb. Phil. Soc., 7, 246–248. The present derivation is due to Carslaw, H. S., and J. C. Jaeger, 1938: Some problems in the mathematical theory of the conduction of heat. Philos. Mag., Ser. 7 , 26, 473–495. See Section 4.

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Green’s Functions with Applications

where the first term on the right side of Equation 5.1.18 is the free-space Green’s function, Equation 5.1.7. Since this Green’s function is a nonhomogeneous solution of Equation 5.1.16, v(x, t) needs only satisfy the homogeneous problem ∂v ∂ 2v = a2 2 , 0 < x, t, (5.1.19) ∂t ∂x with the boundary conditions that < >' 2 2 ∂ e−(x−ξ) /[4a (t−τ )] '' . −vx (0, t) + h v(0, t) = H(t − τ ) ' ∂x 2a π(t − τ ) x=0 2

−h

2

e−ξ /[4a (t−τ )] . H(t − τ ), 2a π(t − τ )

(5.1.20)

and limx→∞ v(x, t) → 0, 0 < t, and the initial condition that v(x, 0) = 0, 0 < x. To solve Equation 5.1.19 and Equation 5.1.20, we now employ Laplace transforms and obtain the ordinary differential equation d2 V − q 2 V = 0, dx2

0 < x,

(5.1.21)

with the boundary conditions that −V $ (0, s) + h V (0, s) =

q − h −qξ−sτ √ e , 2a s

(5.1.22)

and limx→∞ V (x, s) → 0, where q 2 = s/a2 . The solution to this ordinary differential equation is V (x, s) =

q − h e−q(x+ξ)−sτ √ , q+h 2a s

(5.1.23)

so that G(x, s|ξ,τ ) =

% & e−sτ 2h −q(x+ξ) √ e−q|x−ξ| + e−q(x+ξ) − e . q+h 2a s

(5.1.24)

Our final task is to invert Equation 5.1.24 term by term. We find that ! 2 2 2 2 H(t − τ ) g(x, t|ξ,τ ) = . e−(x−ξ) /[4a (t−τ )] + e−(x+ξ) /[4a (t−τ )] (5.1.25) 2a π(t − τ ) ) " ∞ 2 2 − 2h e−hη−(x+ξ+η) /[4a (t−τ )] dη . 0

The Heat Equation

293

Figure 5.1.2: The Green’s function for the one-dimensional heat equation on the semiinfinite domain 0 < x < ∞, and 0.01 ≤ t − τ when the left boundary condition is gx (0, t|ξ,τ ) − h g(0, t|ξ,τ ) = 0, ξ = 0.5, and h = 1.

The first two terms on the right side of Equation 5.1.25 are straightforward inversions from the tables; the third term follows from L−1

%

since

& " ∞4 5 he−q(x+ξ) 1 √ = 1 − e−h(aη−x−ξ) H (aη − x − ξ) 2a s (q + h) 2a 0 2 η × √ e−η /(4t) dη (5.1.26) 2 πt3 " ∞ # $ τ + x + ξ −(τ +x+ξ)2 /(4a2 t) 1 √ e = 1 − e−hτ dτ 2a 0 2a2 t πt (5.1.27) " ∞ 1 τ + x + ξ −(τ +x+ξ)2 /(4a2 t) = √ e dτ 2a2 t 2a πt 0 " ∞ 1 τ + x + ξ −hτ −(τ +x+ξ)2 /(4a2 t) − √ e dτ (5.1.28) 2a2 t 2a πt 0 " ∞ 2 2 h = √ e−hτ −(τ +x+ξ) /(4a t) dτ, (5.1.29) 2a πt 0 1 √ 2 L−1 F ( s ) =

"

0

∞

2 η f (η) √ e−η /(4t) dη, 3 2 πt

(5.1.30)

where L [f (t)] = F (s). Figure 5.1.2 illustrates Equation 5.1.25 as functions of x and t − τ when h = 1. The effect of introducing the radiative boundary condition can be seen by comparing it with Figure 5.1.1, which gives the solution when h = 0. Equation 5.1.25 yields further insight if we rewrite it in terms of tabulated

294

Green’s Functions with Applications

functions. After some work, it transforms into S 2 2 H(t − τ ) R −(x−ξ)2 /[4a2 (t−τ )] . e + e−(x+ξ) /[4a (t−τ )] 2a π(t − τ ) % & √ h(x + ξ) h(x+ξ)+a2 h2 (t−τ ) √ − |h|e erfc a|h| t − τ + 2a|h| t − τ

g(x, t|ξ,τ ) =

2

− 2hH(−h)eh(x+ξ)+a

h2 (t−τ )

,

(5.1.31)

if h &= 0. For h = 0, we already have Equation 5.1.14. The intriguing aspect of Equation 5.1.31 is the appearance of an extra term when h < 0. In this case, we have a discrete contribution to the spectrum. Bressloff6 explained this phenomenon in terms of the time dependence of the spectral data. # " • Example 5.1.3 Let us resolve the previous example taking a different approach. In particular, if a2 = h = 1, we will find the Green’s function for the heat equation ∂g ∂ 2g − = δ(x − ξ)δ(t − τ ), ∂t ∂x2

0 < x, ξ , 0 < t, τ ,

(5.1.32)

subject to the boundary conditions that gx (0, t|ξ,τ ) = g(0, t|ξ,τ ),

lim g(x, t|ξ,τ ) → 0,

x→∞

0 < t,

(5.1.33)

and the initial condition that g(x, 0|ξ,τ ) = 0, 0 < x < ∞. We begin by considering the initial-value problem7 ∂u ∂2u = , ∂t ∂x2

0 < x, t,

(5.1.34)

subject to the boundary conditions that ux (0, t) = u(0, t), and limx→∞ u(x, t) → 0, 0 < t, and the initial condition that u(x, 0) = δ(x − ξ), 0 < x. As we showed in our introductory remarks, Equation 5.1.34 is equivalent to the Green’s function problem, Equation 5.1.32 and Equation 5.1.33 if τ = 0. To solve Equation 5.1.34, we introduce the new dependent variable h(x, t) = u(x, t) − ux (x, t), or " ∞ $ x u(x, t) = e h(x$ , t)e−x dx$ . (5.1.35) x

6

Bressloff, P. C., 1997: A new Green’s function method for solving linear PDE’s in two variables. J. Math. Anal. Appl., 210, 390–415. 7 See Nadler, W., and D. L. Stein, 1996: Reaction-diffusion description of biological transport processes in general dimension. J. Chem. Phys., 104, 1918–1936.

The Heat Equation

295

Figure 5.1.3: Same as Figure 5.1.2 except that ξ = 2.

Then Equation 5.1.34 simplifies to ∂h ∂2h = , ∂t ∂x2

0 < x, t,

(5.1.36)

subject to the boundary conditions h(0, t) = 0, and limx→∞ h(x, t) → 0, 0 < t, and the initial condition that h(x, 0) = δ(x− ξ)−δ $ (x− ξ), 0 < x. Why did we introduce h(x, t)? The main advantage is a boundary condition that results in a simpler Green’s function problem. We could solve Equation 5.1.36 by Laplace transforms. As an alternative method, let us find the Green’s function for this system. Denoting this Green’s function by y(x, t|ξ, 0), we can use the method of images and immediately write down y(x, t|ξ, 0) = √

5 2 1 4 −(x−ξ)2 /(4t) e − e−(x+ξ) /(4t) . 4πt

(5.1.37)

Why did we introduce y(x, t|ξ, 0)? From Equation 5.0.11, we have that h(x, t) =

"

∞

y(x, t|ξ, 0)h(ξ, 0) dξ % & " ∞ ∂y(x, t|ξ, 0) = δ(x − ξ) y(x, t|ξ, 0) + dξ ∂ξ 0 5 2 1 4 −(x−ξ)2 /(4t) =√ e − e−(x+ξ) /(4t) 4πt 5 2 2 1 4 +√ (x − ξ)e−(x−ξ) /(4t) + (x + ξ)e−(x+ξ) /(4t) . πt

(5.1.38)

0

(5.1.39)

(5.1.40)

296

Green’s Functions with Applications

After substituting Equation 5.1.40 into Equation 5.1.35, we obtain ! % & % &) 1 (x − ξ)2 (x + ξ)2 √ u(x, t) = exp − + exp − 4t 4t 4πt % & x + ξ + 2t √ − ex+ξ+t erfc , 2 t

(5.1.41)

where erfc(·) is the complementary error function. The point here is that the introduction of transformations such as Equation 5.1.35 allows the construction of solutions to problems with complicated boundary conditions using the solutions from problems with simpler boundary conditions. To solve our original problem, Equation 5.1.32 and Equation 5.1.33, we merely shift the time axis and obtain ! % & % &) H(t − τ ) (x − ξ)2 (x + ξ)2 g(x, t|ξ,τ ) = . exp − + exp − 4(t − τ ) 4(t − τ ) 4π(t − τ ) % & x + ξ + 2(t − τ ) √ − H(t − τ )ex+ξ+t−τ erfc . (5.1.42) 2 t−τ Figure 5.1.3 illustrates Equation 5.1.42 with ξ = 2. During their study of heat transfer in tissue that includes blood perfusion, Alkhwaji et al.8 solved the Green’s function problem: ∂g ∂2g = a2 2 − bg + δ(x − ξ)δ(t − τ ), ∂t ∂x

0 < x, t, ξ, τ ,

(5.1.43)

subject to the boundary conditions that gx (0, t|ξ,τ ) = h g(0, t|ξ,τ ),

lim gx (x, t|ξ,τ ) → 0,

x→∞

0 < t,

(5.1.44)

and the initial condition that g(x, 0|ξ,τ ) = 0, 0 < x. Here b > 0. If we introduce g(x, t|ξ,τ ) = e−b(t−τ )v(x, t|ξ,τ ), our problem becomes ∂v ∂2v = a2 2 + δ(x − ξ)δ(t − τ ), ∂t ∂x

0 < x, t, ξ, τ ,

(5.1.45)

subject to the boundary conditions that vx (0, t|ξ,τ ) = h v(0, t|ξ,τ ),

lim vx (x, t|ξ,τ ) → 0,

x→∞

0 < t,

(5.1.46)

8 Alkhwaji, A., B. Vick, and T. Diller, 2012: New mathematical model to estimate tissue blood perfusion, thermal contact resistance and care temperature. J. Biomech. Eng., 134, Art. No. 081004.

The Heat Equation

297

and the initial condition that v(x, 0|ξ,τ ) = 0, 0 < x. We then use the above analysis to find v(x, t|ξ,τ ). We find that ! % & % &) e−b(t−τ ) (x − ξ)2 (x + ξ)2 g(x, t|ξ,τ ) = . exp − + exp − H(t − τ ) 4(t − τ ) 4(t − τ ) 4π(t − τ ) % & x + ξ + 2(t − τ ) √ − H(t − τ )ex+ξ+(1−b)(t−τ ) erfc . (5.1.47) 2 t−τ # " • Example 5.1.4: Moving boundaries In this example, we find the Green’s function for a semi-infinite domain where the left boundary moves as x = vt, where v is constant. Although we will only work through a problem9 when Dirichlet conditions are present, we can perform the same calculations when Neumann or Robin conditions prevail and/or the domain extends from vt < x < L + vt. Similar considerations10 hold in the case when the boundaries move as x = αt ± βt2 . Consider the Green’s function problem ∂g ∂ 2g − a2 2 = δ(x − ξ)δ(t − τ ), ∂t ∂x

vt < x, ξ,

τ < t,

(5.1.48)

τ < t,

(5.1.49)

subject to the boundary conditions that g(x, t|ξ,τ )| x=vt = 0,

lim g(x, t|ξ,τ ) → 0,

x→∞

and the initial condition that g(x, 0|ξ,τ ) = 0, 0 < x < ∞. We begin by writing the Green’s function as a linear combination of the free-space Green’s function plus a presently unknown function u(x, t). The purpose of the free-space Green’s function is to eliminate the delta functions in Equation 5.1.48 while u(x, t) is a homogeneous solution that was introduced so that g(x, t|ξ,τ ) satisfies the boundary conditions. Therefore, 2

2

e−(x−ξ) /[4a (t−τ )] . g(x, t|ξ,τ ) = H(t − τ ) + u(x, t). 2a π(t − τ )

(5.1.50)

Substituting Equation 5.1.50 into Equation 5.1.48, we find that ∂u ∂2u = a2 2 , ∂t ∂x

vt < x,

τ < t,

(5.1.51)

9 See Kartashov, E. ´ M., and G. M. Bartenev, 1969: Integral-equation construction of the Green’s function for generalized boundary-value problems involving the heat-conduction equation. Sov. Phys. J., 12, 189–198. 10 Kartashov, E. ´ M., B. Ya. Lyubov, and G. M. Bartenev, 1970: A diffusion problem in a region with a moving boundary. Sov. Phys. J., 13, 1641–1647.

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Green’s Functions with Applications

Figure 5.1.4: Equation 5.1.48 for the one-dimensional heat equation where a Dirichlet boundary condition applies along the left boundary which moves as x = vt. The parameters are ξ − vτ = 2 and a2 /v = 0.5.

with the boundary conditions that 2

u(x, t)|x=vt = −

2

e−(vt−ξ) /[4a (t−τ )] . , 2a π(t − τ )

(5.1.52)

and limx→∞ u(x, t) → 0, τ < t, and the initial condition that u(x, 0) = 0, 0 < x. To eliminate the moving boundary, we introduce the new independent variables z = x − vt and t$ = t − τ , and the dependent variable % & vz v 2 t$ u(x, t) = exp − 2 − 2 w(z, t$ ). (5.1.53) 2a 4a Substituting Equation 5.1.53 into Equation 5.1.51 and Equation 5.1.52, we obtain 2 ∂w 2∂ w = a , 0 < z, t$ , (5.1.54) ∂t$ ∂z 2 with the boundary conditions % & 1 vξ0 ξ02 w(0, t$ ) = − √ exp − , (5.1.55) 2a2 4a2 t$ 2a πt$ and limz→∞ w(z, t$ ) → 0, 0 < t$ , and the initial condition that w(z, 0) = 0, 0 < z, where ξ0 = ξ − vτ . To solve Equation 5.1.54 and Equation 5.1.55, we take their Laplace transform and find d2 W a2 2 − sW = 0, (5.1.56) dz

The Heat Equation with

299

, √ 1 ξ0 s vξ0 W (0, s) = − √ exp − + 2 , a 2a 2a s

(5.1.57)

and limz→∞ W (z, s) → 0. The solution to Equation 5.1.56 and Equation 5.1.57 is % & √ 1 (z + ξ0 ) s vξ0 W (z, s) = − √ exp − + 2 , (5.1.58) 2a s a 2a or

% & H(t$ ) (z + ξ0 )2 vξ0 √ w(z, t ) = − exp − + 2 , 4a2 t$ 2a 2a πt$ $

(5.1.59)

so that % & H(t − τ ) (x + ξ − 2vτ )2 v(ξ − vτ ) u(x, t) = − . exp − + . 4a2 (t − τ ) a2 2a π(t − τ )

(5.1.60)

Substituting Equation 5.1.60 into Equation 5.1.50 yields the Green’s function. # " • Example 5.1.5 Let us find the Green’s function for the parabolic equation11 , ∂g ∂ ∂g − x2−a = δ(x − ξ)δ(t − τ ), 0 < x, ξ < ∞, 0 < t, τ , (5.1.61) ∂t ∂x ∂x where the solution remains finite over the entire interval and is initially given by g(x, 0|ξ,τ ) = 0. We begin by taking the Laplace transform of Equation 5.1.61 or , d dG x2−a − sG = −δ(x − ξ)e−sτ , 0 < x, ξ < ∞. (5.1.62) dx dx Assuming for the moment that a &= by introducing √0, we solve Equation 5.1.62 √ new independent variables y = 2 s xa/2 /|a|, and η = 2 s ξ a/2 /|a|, and the dependent variable G(x, s|ξ,τ ) =

2(xξ)(a−1)/2 F (y,η )e−sτ . |a|

(5.1.63)

11 A portion of this was first solved by Becker, P. A., 1992: First-order Fermi acceleration in spherically symmetric flows: Solutions including quadratic losses. Astrophys. J., 397, 88–116. The special case a = n + 2, where n is an even integer, was treated by Peyraud and Coste [Peyraud, J., and J. Coste, 1976: Some diffusion equations in plasma physics. Phys. Fluids, 19, 388–393].

300

Green’s Functions with Applications

These new variables transform Equation 5.1.62 into y2

d2 F dF +y − (µ2 + y 2 )F = −η δ(y − η), dy 2 dy

where µ = 1 − 1/a. The solution to Equation 5.1.64 is ! A I|µ| (y), F (y,η ) = B K|µ| (y),

y ≤ η, y ≥ η.

(5.1.64)

(5.1.65)

Applying the condition that the Green’s function must be continuous across y = η and the jump condition 'y=η+ dF '' η = −η, dy 'y=η− 2

we find

(5.1.66)

F (y,η ) = I|µ| (y< )K|µ| (y> ).

(5.1.67)

Of course, Equation 5.1.67 does not hold if a = 0. In that case, the solution to Equation 5.1.62 is ! √ (−1+ 1+4s )/2 x ≤ ξ, G(x, s|ξ,τ ) = A x (−1−√1+4s )/2, (5.1.68) Bx , x ≥ ξ. The values of A and B are found by again requiring that the Green’s function is continuous at x = ξ and satisfies the jump condition that 'x=ξ+ , -√s+ 12 −1/2 −sτ ' dG (xξ) e x< ' = ξ2 = −e−sτ , and G(x, s|ξ,τ ) = . ' dx x=ξ− x> 2 s+ 1 2

(5.1.69) The final step requires the inversion of Equation 5.1.63 and Equation 5.1.69. Applying tables,12 we find % & % & (xξ)(a−1)/2 xa + ξ a 2(xξ)a/2 g(x, t|ξ,τ ) = exp − 2 I|µ| 2 H(t − τ ), |a|(t − τ ) a (t − τ ) a (t − τ ) (5.1.70) if a &= 0. To invert Equation 5.1.69, we note that √ 1 (xξ)−1/2 e−sτ −ln(x> /x< ) s+ 2 = G(x, s|ξ,τ ) = . (5.1.71) 2 s + 12 12 Erd´ elyi, A., W. Magnus, F. Oberhettinger, and F. G. Tricomi, 1954: Tables of Integral Transforms, Vol. I. McGraw-Hill Co., Section 5.16, Formula 56.

The Heat Equation

301

Employing the first and second shifting theorems, g(x, t|ξ,τ ) = if a = 0.

% & (xξ)−1/2 H(t − τ ) ln2 (x> /x< ) t − τ . exp − − , 4(t − τ ) 2 2 π(t − τ )

(5.1.72)

# " • Example 5.1.6 A problem13 similar to the previous one consists of finding the Green’s function governed by % , -& ∂g 1 ∂ ∂g 4 = 2 x +g , 0 < x, ξ < ∞, 0 < t, (5.1.73) ∂t x ∂x ∂x subject to the boundary conditions lim |g(x, t|ξ, 0+ )| < ∞,

x→0

lim g(x, t|ξ, 0+ ) → 0,

x→∞

0 < t,

(5.1.74)

and the initial condition that g(x, 0|ξ, 0+ ) = δ(x − ξ)/ξ 2 . We begin by taking the Laplace transform of Equation 5.1.73 and the boundary condition, Equation 5.1.74. This yields % , -& 1 d 4 dG δ(x − ξ) x + G − sG = − , 0 < x, ξ < ∞, (5.1.75) 2 x dx dx ξ2 and lim |G(x, s|ξ, 0+ )| < ∞ and

where G(x, s|ξ, 0+ ) =

"

lim G(x, s|ξ, 0+ ) → 0,

(5.1.76)

g(x, t|ξ, 0+ )e−st dt.

(5.1.77)

x→∞

x→0

∞

0

The solution to Equation 5.1.75 and Equation 5.1.76 is ! AM2,µ (x), x ≤ ξ, G(x, s|ξ, 0+ ) = x−2 e−x/2 BW2,µ (x), x ≥ ξ,

(5.1.78)

. where M2,µ (·) and W2,µ (·) are Whittaker functions and µ(s) = s + 9/4. The values of A and B follow from continuity: AM2,µ (ξ) = BW2,µ (ξ), and an 13 Becker, P. A., 2003: Exact solution for the Green’s function describing time-dependent thermal Comptonization. Mon. Not. R. Astron. Soc., 343, 215–240.

302

Green’s Functions with Applications

integration of Equation 5.1.75 over the interval [ξ − 2,ξ + 2]. This gives the condition lim G$ (ξ + 2, s|ξ, 0+ ) − lim G$ (ξ − 2, s|ξ, 0+ ) = −

'→0

'→0

1 . ξ4

(5.1.79)

Substituting for G(x, s|ξ, 0+ ) into Equation 5.1.79 and simplifying, we find that G(x, s|ξ, 0+ ) =

Γ(µ − 3/2) e(ξ−x)/2 M2,µ (x< )W2,µ (x> ). Γ(1 + 2µ) ξ 2 x2

(5.1.80)

It remains for us to invert G(x, s|ξ, 0+ ) to obtain g(x, t|ξ, 0+ ). An important consideration is the branch cut that arises from the square root in the definition of µ(s). We take the branch cut along the negative real axis in the s-plane from −∞ to −9/4. In addition to the integral arising from the integration along the branch cut, there are simple poles at s = 0 and s = −2. These simple poles yield the contribution 1 −x 2e

+ 12 e−x−2t

(2 − x)(2 − ξ) . ξx

From Appendix A3 from Becker’s paper, the final Green’s function is (2 − x)(2 − ξ) (5.1.81) ξx " 2 32 (ξ−x)/2−9t/4 ∞ η sinh(πη)e−tη + e W2,iη (ξ)W2,iη (x) dη. π ξ 2 x2 (1 + 4η2 )(9 + 4η2 ) 0

g(x, t|ξ, 0+ ) = 12 e−x + 12 e−x−2t

# " • Example 5.1.7 In his study of diffusion-controlled reactions where there is loss of population due to desorption, Agmon14 found the Green’s function for the following problem: ∂g ∂ 2g ∂g − − 2c = δ(x − ξ)δ(t − τ ), 2 ∂t ∂x ∂x

0 < x, t, ξ, τ ,

subject to the boundary conditions " t g(0, t|ξ,τ ) = κ [gx (0, t$ |ξ,τ ) + 2c g(0, t$ |ξ,τ )] dt$ ,

0 < t,

(5.1.82)

(5.1.83)

0

14 Agmon, N., 1984: Diffusion with back reaction. J. Chem. Phys., 81, 2811–2817; see also Kim, H., and K. J. Shin, 1999: Exact solution of the reversible diffusion-influenced reaction for an isolated pair in three dimensions. Phys. Rev. Lett., 82, 1578–1581.

The Heat Equation

303

Figure 5.1.5: Equation 5.1.92 for the one-dimensional heat equation on the semi-infinite domain 0 < x < ∞, and 0.01 ≤ t − τ when the left boundary is the integral condition, Equation 5.1.83. The parameters are c = κ = 1, and ξ = 2.

lim g(x, t|ξ,τ ) → 0,

x→∞

0 < t,

(5.1.84)

and the initial condition that g(x, 0|ξ,τ ) = 0, 0 < x < ∞. The interesting aspect of this problem is the integral condition that appears in the boundary condition at x = 0. We begin by introducing the intermediate dependent variable ϕ(x, t| ξ,τ ) such that g(x, t|ξ,τ ) = e−c(x−ξ+ct−cτ )ϕ(x, t|ξ,τ ). Substitution into Equation 5.1.82 through Equation 5.1.84 results in the heat equation ∂ϕ ∂ 2 ϕ − = δ(x − ξ)δ(t − τ ), ∂t ∂x2

0 < x, t, ξ, τ ,

(5.1.85)

subject to the boundary conditions " t 2 $ −c2 (t−τ ) e ϕ(0, t|ξ,τ ) = κ [ϕx (0, t$ |ξ,τ ) + c ϕ(0, t$ |ξ,τ )] e−c (t −τ ) dt$ , 0

(5.1.86) and limx→∞ ϕ(x, t|ξ,τ ) → 0, τ < t, with the initial condition ϕ(x, 0|ξ,τ ) = 0, 0 < x < ∞. We rewrite Equation 5.1.86 as " t 2 $ ϕ(0, t|ξ,τ ) = κ [ϕx (0, t$ |ξ,τ ) + c ϕ(0, t$ |ξ,τ )] ec (t−t ) dt$ . (5.1.87) 0

To find ϕ(x, t|ξ,τ ), we take the Laplace transform of Equation 5.1.85 and Equation 5.1.87 and find that d2 Φ − sΦ = −δ(x − ξ)e−sτ , dx2

0 < x, ξ < ∞,

(5.1.88)

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Green’s Functions with Applications

with (s − c2 )Φ(0, s|ξ,τ ) = κ [Φ$ (0, s|ξ,τ ) + c Φ(0, s|ξ,τ )] ,

(5.1.89)

and limx→∞ Φ(x, s|ξ,τ ) → 0. The general solution that satisfies Equation 5.1.88 is √ √ √ 1 1 Φ(x, s|ξ,τ ) = √ e−|x−ξ| s − √ e−(x+ξ) s + Ae−(x+ξ) s . 2 s 2 s

(5.1.90)

Substituting Equation 5.1.90 into Equation 5.1.89 yields an equation for A. Substituting this value into Equation 5.1.90, we obtain √ √ 1 1 Φ(x, s|ξ,τ ) = √ e−|x−ξ| s − √ e−(x+ξ) s 2 s 2 s √

√

κe−(x+ξ) s κe−(x+ξ) s √ √ + − . (κ + 2c)( s − c) (κ + 2c)( s + c + κ)

(5.1.91)

Taking the inverse Laplace transform of Equation 5.1.91, we find that H(t − τ ) −c(x−ξ+ct−cτ ) g(x, t|ξ,τ ) = . e 4π(t − τ ) ! % & % &) (x − ξ)2 (x + ξ)2 × exp − − exp − 4(t − τ ) 4(t − τ ) κ(κ + c)H(t − τ ) [2cξ+κ(x+ξ)+(κ2 +2cκ)(t−τ )] + e κ + 2c % & x + ξ + 2(κ + c)(t − τ ) √ × erfc , 2 t−τ % & cκH(t − τ ) −2cx x + ξ − 2c(t − τ ) √ + e erfc , (5.1.92) κ + 2c 2 t−τ where erfc(·) is the complementary error function. Figure 5.1.5 illustrates this Green’s function. In a similar vein, T. Pr¨ ustel and M. Meier-Schellersheim15 found the Green’s function for the radially symmetric heat equation governed by , ∂g a2 ∂ ∂g = r , b < r, ρ 0 ,< t, (5.1.93) ∂t r ∂r ∂r subject to the boundary conditions +

+

gr (b, t|ρ, 0 ) = h g(b, t|ρ, 0 ) − κd

"

t 0

gr (b,τ |ρ, 0+ ) dτ,

(5.1.94)

15 Pr¨ ustel, T., and M. Meier-Schellersheim, 2012: Exact Green’s function of the reversible diffusion-influenced reaction for an isolated pair in two dimensions. J. Chem. Phys., 137, Art. No. 054104.

The Heat Equation

305

and limr→∞ g(r, t|ρ, 0+) → 0, 0 < t, and the initial condition that g(r, 0+ |ρ, 0+ ) = δ(r − ρ)/(2πρ), b < r < ∞. Using Laplace transforms and an inversion via contour integration, they found that " ∞ 2 2 1 + g(r, t|ρ, 0 ) = e−a η t Θ(η, r)Θ(η,ρ ) η dη , (5.1.95) 2π 0 where Θ(θ, r) =

J0 (rη)β(η) − Y0 (rη)α(η) . , α2 (η) + β 2 (η)

(5.1.96)

α(η) = (η 2 − κd )J1 (aη) + hηJ 0 (aη),

(5.1.97)

β(η) = (η 2 − κd )Y1 (aη) + hηY 0 (aη).

(5.1.98)

and # " • Example 5.1.8 Let us find the Green’s function16 governed by the heat-like partial differential equation , ∂g 1 ∂ ∂g g ∂2g − r + 2 − 2 = δ(r − ρ)δ(z − ζ)δ(t − τ ), (5.1.99) ∂t r ∂r ∂r r ∂z with 0 ≤ r, ρ< ∞, −∞ < z, ζ < ∞, and 0 < t, τ, subject to the boundary conditions lim |g(r, z, t|ρ, ζ,τ )| < ∞, lim g(r, z, t|ρ, ζ,τ ) → 0, r→∞

r→0

−∞ < z < ∞, 0 < t,

(5.1.100)

and lim g(r, z, t|ρ, ζ,τ ) → 0,

|z|→∞

0 ≤ r < ∞,

0 < t,

(5.1.101)

and the initial condition that g(r, z, 0|ρ, ζ,τ ) = 0, 0 ≤ r < ∞, −∞ < z < ∞. We begin by taking the Laplace transform of Equation 5.1.99. Denoting this Laplace transform by G(r, z, s|ρ, ζ,τ ), we find that , 1 ∂ ∂G G ∂2G r − 2+ − sG = −δ(r − ρ)δ(z − ζ)e−sτ , (5.1.102) r ∂r ∂r r ∂z 2 with the boundary conditions lim |G(r, z, s|ρ, ζ,τ )| < ∞,

r→0

lim G(r, z, s|ρ, ζ,τ ) → 0,

r→∞

|z| < ∞,

(5.1.103)

16 See Tidman, D. A., 1975: Thermally generated magnetic fields in laser-driven compressions and explosions. Phys. Fluids, 18, 1454–1459.

306

Green’s Functions with Applications

and lim|z|→∞ G(r, z, s|ρ, ζ,τ ) → 0, 0 ≤ r < ∞. Next, we take the Fourier transform with respect to z. Denoting the joint Laplace-Fourier transform by G(r, k, s|ρ, ζ,τ ), we obtain , 1 d dG G r − 2 − (s + k 2 )G = −δ(r − ρ)e−ikζ e−sτ , r dr dr r

(5.1.104)

with the boundary conditions lim |G(r, k, s|ρ, ζ,τ )| < ∞,

r→0

lim G(r, k, s|ρ, ζ,τ ) → 0.

r→∞

(5.1.105)

√ √ Introducing η = ir s + k 2 and ρ$ = iρ s + k 2 , Equation 5.1.104 simplifies to , 1 d dG G ie−ikζ−sτ η +G− 2 = √ δ(η − ρ$ ), (5.1.106) η dη dη η s + k2 with the boundary conditions lim |G(η, k, s|ρ$ , ζ , τ)| < ∞,

η→0

lim G(η, k, s|ρ$ , ζ , τ) → 0.

η→∞

(5.1.107)

Note that we used the property that aδ(ax) = δ(x) in obtaining Equation 5.1.106. Solving the ordinary differential equation, Equation 5.1.106, and taking the conditions as η → 0 and ∞ into account, we find that ! AJ1 (η), η < $, ρ G(η, k, s|ρ$ , ζ , τ) = (5.1.108) (1) BH1 (η), η > $. ρ To compute A and B, we use the continuity of the Green’s function at η = (1) ρ$ to obtain the condition that A J1 (ρ$ ) = B H1 (ρ$ ). On the other hand, integrating Equation 5.1.106 over the infinitismally small interval from ρ$ − 2 to ρ$ + 2 as 2 → 0, we obtain the second condition: (1)

B

dH1 (ρ$ ) i − AJ1 (ρ$ ) = √ e−ikζ−sτ . dη s + k2

(5.1.109)

Solving for A and B and using the fact that the Wronskian of J1 (z) and (1) H1 (z) equals 2i/(πz), ! " ∞ 2iρ 1 g(r, z, t|ρ, ζ,τ ) = eik(z−ζ) dk (5.1.110) π 2π −∞ % &) " c+∞i * . + * . + 1 (1) (t−τ )s 2 2 × H ir< s + k J1 ir> s + k e ds . 2πi c−∞i 1 In deriving Equation 5.1.110, we have taken the inverse Fourier transform and inverse Laplace transform where c is sufficiently large so that all of the

The Heat Equation

307

singularities associated with the Laplace transform lie to right of the Bromwich contour. To evaluate the inverse Laplace transform, we introduce a branch cut along the negative axis from s = −∞ to s = −k 2 in the s−plane. Introducing s + k 2 = ze±πi along the top and bottom of the branch cut, Equation 5.1.110 becomes 3 % &" ∞ # √ $ # √ $ ρ π (z − ζ)2 g(r, z, t|ρ, ζ,τ ) = exp − J1 ρ z J1 r z e−z(t−τ ) dz, 4π t − τ 4(t − τ ) 0 (5.1.111) where we have used the fact that (1)

(1)

J1 (−z)H1 (−ρ) − J1 (z)H1 (ρ) = −2J1 (z)J1 (ρ).

(5.1.112)

Finally, we can evaluate17 the integral involving the Bessel functions and obtain the final solution of % & % & √ ρ π (z − ζ)2 + r2 + ρ2 ρr g(r, z, t|ρ, ζ,τ ) = I1 . 3 exp − 4(t − τ ) 2(t − τ ) 4π(t − τ ) 2 (5.1.113) # " • Example 5.1.9 In her study of diffuse light in two adjoining turbid media, where the interface is located at x = 0, Shendeleva18 solved a two-layered Green’s function problem governed by ∂g1 ∂ 2 g1 − a21 2 + β1 g1 = δ(t − 0+ )δ(x − ξ), ∂t ∂x

0 < x, ξ , 0 < t, (5.1.114)

and

∂g2 ∂ 2 g2 − a22 2 + β2 g2 = 0, ∂t ∂x subject to the boundary conditions: lim g2 (x, t|ξ, 0+ ) → 0,

x→−∞

N 2 g1 (0, t|ξ, 0+) = g2 (0, t|ξ, 0+ ),

x < 0,

0 < t,

lim g1 (x, t|ξ, 0+ ) → 0,

x→∞

γ2

(5.1.115)

(5.1.116)

∂g1 ∂g2 (0, t|ξ, 0+ ) = (0, t|ξ, 0+ ), ∂x ∂x (5.1.117)

17

Watson, G. N., 1966: A Treatise on the Theory of Bessel Functions. Cambridge University Press, Section 13.31. 18 Shendeleva, M. L., 2004: One-dimensional time-domain Green functions for diffuse light in two adjoining turbid media. Opt. Commun., 235, 233–245.

308

Green’s Functions with Applications

and the initial conditions: g1 (x, 0|ξ, 0+ ) = g2 (x, 0|ξ, 0+ ) = 0. We begin by writing down the solution as if there were no interface. This solution is % & 1 (x − ξ)2 + √ exp − gS (x, t|ξ, 0 ) = − β1 t (5.1.118) 4a21 t 2a1 πt which has the Laplace transform GS (x, s|ξ, 0+ ) =

* + . 1 √ exp −|x − ξ| s + β1 /a1 . 2a1 s + β1

(5.1.119)

Using Equation 5.1.119 and the method of images, G1 (x, s|ξ, 0+ ) =

exp(−η1 |x − ξ|) exp[−η1 (x + ξ)] − + Ψ (x, s|ξ, 0+ ), 2a21 η1 2a21 η1 (5.1.120)

and G2 (x, s|ξ, 0+ ) =

N 2γ2 exp(−η1 ξ + η2 x) , γ 2 η1 + N 2 η2 a21

where Ψ(x, s|ξ, 0+ ) =

γ 2 η1

γ2 exp[−η1 (x + ξ)] , + N 2 η2 a1

(5.1.121)

(5.1.122)

√ and ηi = s + βi /ai . Turning to medium 1, the inverse of Equation 5.1.120 is % & 1 (x − ξ)2 + √ exp − g1 (x, t|ξ, 0 ) = − β1 t (5.1.123) 4a21 t 2a1 πt % & 1 (x + ξ)2 √ exp − − − β1 t + ψ(x, t|ξ, 0+ ), 4a21 t 2a1 πt where ψ(x, t|ξ, 0+ ) is the inverse Laplace transform of Ψ(x, s|ξ, 0+ ). To compute the inverse of Ψ(x, s|ξ, 0+ ), we note that % , -& " ∞ 1 N 2 η2 = exp −ζ η + dζ, (5.1.124) 1 η1 + N 2 η2 /γ 2 γ2 0 so that Equation 5.1.122 becomes , " ∞ 1 η2 ζN 2 Ψ(x, s|ξ, 0+ ) = 2 e−η1 (ζ+x+ξ) exp − dζ. a1 0 γ2

(5.1.125)

Therefore, N 5/2 ψ(x, t|ξ, 0 ) = 4πγa41 t2 +

"

0

% & (ζ + x + ξ)2 ζ 2N 5 dζ ζ(ζ + x + ξ) exp − 2 − 2 2 4a1 t(1 − ϕ) 4γ a1 tϕ 0 dϕ × e−β1 t(1−ϕ)−β2 tϕ 3/2 . (5.1.126) ϕ (1 − ϕ)3/2

∞

"

1

The Heat Equation

309

Setting 1 G1 = √ 2 2 π [a1 tϕ(1 − ϕ)]3/2

"

0

%

& (ζ + x + ξ)2 ζ2N 5 ζ(ζ+x+ξ) exp − 2 − 2 2 dζ, 4a1 t(1 − ϕ) 4γ a1 tϕ (5.1.127)

∞

and using the formula " ∞ 2 2 (ζ + a)(ζ + b)e−k(ζ+a) −q(ζ+b) dζ 0 % & , -% & √ π (a − b)2 kq ak + bq 2kq(a − b)2 √ = exp − erfc 1 − k+q k+q 4(k + q)3/2 k+q bk + aq −a2 k−b2 q + e , (5.1.128) 2(k + q)2 we find that

! ) γ3 N 5 (x + ξ)2 G1 = 2 exp − 2 2 4a1 t[γ ϕ + N 5 (1 − ϕ)] [γ ϕ + N 5 (1 − ϕ)]3/2 : ;! ) √ γ(x + ξ) ϕ N 5 (x + ξ)2 . . × erfc 1− 2 2 2a1 t[γ ϕ + N 5 (1 − ϕ)] 2a1 t(1 − ϕ) γ 2 ϕ + N 5 (1 − ϕ) √ % & γ 2 N 5 (x + ξ) 1 − ϕ (x + ξ)2 + √ exp − . (5.1.129) a1 πtϕ[γ 2 ϕ + N 5 (1 − ϕ)]2 4a21 t(1 − ϕ)

Therefore, ψ(x, t|ξ, 0+ ) =

N 5/2 √ 2a1 γ πt

"

0

1

e−β1 t(1−ϕ)−β2 tϕ G1 dϕ.

(5.1.130)

Consequently, the Green’s function for layer 1 is ! % & % &) 1 (x − ξ)2 (x + ξ)2 + √ g1 (x, t|ξ, 0 ) = exp − − β1 t − exp − − β1 t 4a21 t 4a21 t 2a1 πt " 1 N 5/2 √ + e−β1 t(1−ϕ)−β2 tϕ G1 dϕ. (5.1.131) 2γa1 πt 0 Turning to layer 2, the Laplace transform of g2 (x, t|ξ, 0+ ) is % , 2 -& " N 2 ∞ −η1 (ζ+ξ) N G2 (x, s|ξ, 0+ ) = 2 e exp −η2 ζ − x dζ. a1 0 γ2

(5.1.132)

Therefore, N 9/2 g2 (x, t|ξ, 0 ) = 4πγa41 t2 +

"

0

, γ 2 x −β1 t(1−ϕ)−β2 tϕ dζ (ζ + ξ) ζ − 2 e N 0 % & (ζ + ξ)2 N 5 (ζ − γ 2 x/N 2 )2 × exp − 2 − 4a1 t(1 − ϕ) 4γ 2 a21 tϕ dϕ × 3/2 . (5.1.133) ϕ (1 − ϕ)3/2

∞

"

1

310

Green’s Functions with Applications

To simplify Equation 5.1.133, we introduce , " ∞ 1 γ2x G2 = √ 2 (ζ + ξ) ζ − 2 dζ (5.1.134) N 2 π[a1 tϕ(1 − ϕ)]3/2 0 : , -2 ; (ζ + ξ)2 N5 γ 2x × exp − 2 − ζ− 2 , 4a1 t(1 − ϕ) 4γ 2 a21 tϕ N or ! ) γ3 N (N 2 ξ + γ 2 x)2 G2 = 2 exp − 2 2 (5.1.135) 4a1 t[γ ϕ + N 5 (1 − ϕ)] [γ ϕ + N 5 (1 − ϕ)]3/2 < >! ) γ[ξϕ − N 3 x(1 − ϕ)] N (N 2 ξ + γ 2 x)2 . . × erfc 1− 2 2 2a1 t[γ ϕ + N 5 (1 − ϕ)] 2a1 tϕ(1 − ϕ) γ 2 ϕ + N 5 (1 − ϕ) % 2 & γ 2 [N 7 ξ(1 − ϕ) − γ 4 xϕ] ξ ϕ + γ 2 N x2 (1 − ϕ) . + exp − . 4a21 tϕ(1 − ϕ) a1 N 2 πtϕ(1 − ϕ)[γ 2 ϕ + N 5 (1 − ϕ)]2 Consequently,

g2 (x, t|ξ, 0+ ) =

N 9/2 √ 2a1 γ πt

"

1 0

e−β1 t(1−ϕ)−β2 tϕ G2 dϕ.

(5.1.136)

In 2007 Shendelev19 solved a two-dimensional version of this problem using the same mathematical techniques. She had earlier solved this twodimensional problem using the Cagniard-de Hoop method.20 The advantage of her new solution lies in the ease of the numerical computations compared to her original solution. 5.2 HEAT EQUATION WITHIN A FINITE CARTESIAN DOMAIN In this section, we find Green’s functions for the heat equation within finite Cartesian domains. These solutions can be written as series involving orthonormal eigenfunctions from regular Sturm-Liouville problems. • Example 5.2.1 Here we find the Green’s function for the one-dimensional heat equation over the interval 0 < x < L associated with the problem ∂u ∂2u − a2 2 = f (x, t), ∂t ∂x

0 < x < L,

0 < t,

(5.2.1)

19

Shendeleva, M. L., 2007: Time-domain Green functions for diffuse light in two adjoining turbid half-spaces. Appl. Opt., 46, 1641–1649. 20 Shendeleva, M. L., 2004: Green functions for diffuse photon-density waves generated by a line source in two nonabsorbing turbid media in contact. Appl. Opt., 43, 1638–1642.

The Heat Equation

311

where a2 is the diffusivity constant. To find the Green’s function for this problem, consider the following problem: ∂g ∂ 2g − a2 2 = δ(x − ξ)δ(t − τ ), ∂t ∂x with the boundary conditions

0 < x, ξ < L,

0 < t, τ ,

(5.2.2)

α1 g(0, t|ξ,τ ) + β1 gx (0, t|ξ,τ ) = 0,

0 < t,

(5.2.3)

α2 g(L, t|ξ,τ ) + β2 gx (L, t|ξ,τ ) = 0,

0 < t,

(5.2.4)

and and the initial condition that g(x, 0|ξ,τ ) = 0, 0 < x < L. We begin by taking the Laplace transform of Equation 5.2.2 and find that d2 G s δ(x − ξ) −sτ − 2G = − e , dx2 a a2 with and

0 < x < L,

(5.2.5)

α1 G(0, s|ξ,τ ) + β1 G$ (0, s|ξ,τ ) = 0,

(5.2.6)

α2 G(L, s|ξ,τ ) + β2 G$ (L, s|ξ,τ ) = 0.

(5.2.7)

Problems similar to Equation 5.2.5 through Equation 5.2.7 were considered in Chapter 3. Applying this technique of eigenfunction expansions, we have that ∞ ( ϕn (ξ)ϕn (x) G(x, s|ξ,τ ) = e−sτ , (5.2.8) s + a2 kn2 n=1 where ϕn (x) is the nth orthonormal eigenfunction to the regular SturmLiouville problem $$

ϕ (x) + k 2 ϕ(x) = 0,

0 < x < L,

(5.2.9)

subject to the boundary conditions

and

α1 ϕ(0) + β1 ϕ$ (0) = 0,

(5.2.10)

α2 ϕ(L) + β2 ϕ$ (L) = 0.

(5.2.11)

Taking the inverse of Equation 5.2.8, we have that

g(x, t|ξ,τ ) =

:

∞ (

n=1

ϕn (ξ)ϕn (x)e

2 2 −kn a (t−τ )

;

H(t − τ ).

(5.2.12)

312

Green’s Functions with Applications

Let us now verify that Equation 5.2.12 is indeed the solution to Equation 5.2.2. We begin by computing :∞ ; ( 2 2 ∂g 2 2 −kn a (t−τ ) =− kn a ϕn (ξ)ϕn (x)e H(t − τ ) ∂t n=1 :∞ ; ( 2 2 −kn a (t−τ ) + ϕn (ξ)ϕn (x)e δ(t − τ ) (5.2.13) =− + =−

:

n=1 ∞ (

n=1 :∞ ( n=1 :∞ (

kn2 a2 ϕn (ξ)ϕn (x)e

2 2 −kn a (t−τ )

;

ϕn (ξ)ϕn (x) δ(t − τ ) kn2 a2 ϕn (ξ)ϕn (x)e

H(t − τ ) (5.2.14)

2 2 −kn a (t−τ )

n=1

;

;

H(t − τ )

+ δ(x − ξ)δ(t − τ ),

(5.2.15)

because the bracketed term multiplying the delta function δ(t − τ ) equals δ(x − ξ) from Equation 3.4.39. Therefore,

4 5 2 ( 2 2 ∂g 2∂ g 2 2 2 $$ −kn a (t−τ ) −a = ϕn (ξ) −kn a ϕn (x) − a ϕn (x) e H(t − τ ) ∂t ∂x2 n=1 + δ(x − ξ)δ(t − τ ) = δ(x − ξ)δ(t − τ ),

(5.2.16)

and Equation 5.2.12 satisfies the differential equation, Equation 5.2.2. # " • Example 5.2.2 Let us find the Green’s function for the heat equation on a finite domain ∂g ∂2g − a2 2 = δ(x − ξ)δ(t − τ ), ∂t ∂x

0 < x, ξ < L,

0 < t, τ ,

(5.2.17)

with the boundary conditions g(0, t|ξ,τ ) = g(L, t|ξ,τ ) = 0,

0 < t,

(5.2.18)

and the initial condition that g(x, 0|ξ,τ ) = 0, 0 < x < L. The Sturm-Liouville problem is ϕ$$ (x) + k 2 ϕ(x) = 0,

0 < x < L,

(5.2.19)

The Heat Equation

313

with the boundary conditions ϕ(0) = ϕ(L) = 0. The nth orthonormal eigenfunction to Equation 5.2.19 is 3 * nπx + 2 ϕn (x) = sin . (5.2.20) L L Substituting Equation 5.2.20 into Equation 5.2.12, we find that

, - * 2 ( nπξ nπx + −a2 n2 π2 (t−τ )/L2 g(x, t|ξ,τ ) = sin sin e H(t − τ ). L n=1 L L (5.2.21) Let us now develop an alternative to Equation 5.2.12 using the method of images. The free-space Green’s function g(x, t|ξ,τ ) for the one-dimensional heat equation was found in the previous section and equals % & H(t − τ ) (x − ξ)2 g(x, t|ξ,τ ) = . exp − 2 . (5.2.22) 4a (t − τ ) 4πa2 (t − τ )

We will now express the Green’s function to Equation 5.2.2 in terms of a sum of Equation 5.2.22 plus other Green’s functions that we now find by the method of images. We begin by noting that we must introduce a negative image at x = −ξ so that the sum of the Green’s function resulting from the source point x = ξ plus the Green’s function resulting from the image point x = −ξ will satisfy the boundary condition at x = 0. These two source points will, in turn, require additional images so that the boundary condition will be satisfied at x = L. Eventually, we are led to an infinite sequence of source points at the points ξn = ±ξ ± 2nL, where n = 0, 1, 2, . . .. For this set of source points, the Green’s function g(x, t|ξ,τ ) will have the form % & ∞ ! ( H(t − τ ) (x − ξ − 2nL)2 g(x, t|ξ,τ ) = . exp − 4a2 (t − τ ) 4πa2 (t − τ ) n=−∞ % &) (x + ξ − 2nL)2 − exp − . (5.2.23) 4a2 (t − τ ) The term with n = 0 and a positive coefficient corresponds to our free-space Green’s function. Clearly Equation 5.2.23 equals zero for t < τ and the initial condition is satisfied. A quick check shows that the series can be differentiated term by term. Because each of the terms in the series except the one corresponding to the free-space Green’s function has its source point ξn outside the interval 0 < x < L, Equation 5.2.23 satisfies the differential equation. Does our Green’s function satisfy the boundary conditions? If we replace n by −n in the second series in Equation 5.2.23, it becomes identical to the

314

Green’s Functions with Applications

Figure 5.2.1: Equation 5.2.28 (times L/2) for the one-dimensional heat equation where the left boundary is insulated and the right boundary radiates to space when ξ/L = 0.3, and hL = 1.

first series when x = 0 and consequently the boundary condition is satisfied there. Similarly, if we replace n by −n + 1 in the second series, the two series become identical if we set x = L, and again, the boundary condition is satisfied. In summary, we showed that Equation 5.2.23 satisfies the differential equation, the initial condition and the boundary conditions. Therefore, it is a solution to Equation 5.2.2. # " • Example 5.2.3 Let us find the Green’s function21 for the heat equation on a finite domain ∂g ∂2g − a2 2 = δ(x − ξ)δ(t − τ ), ∂t ∂x

0 < x, ξ < L,

0 < t, τ ,

(5.2.24)

0 < t,

(5.2.25)

with the boundary conditions gx (0, t|ξ,τ ) = 0,

gx (L, t|ξ,τ ) + h g(L, t|ξ,τ ) = 0,

and the initial condition that g(x, 0|ξ,τ ) = 0, 0 < x < L. The Sturm-Liouville problem that we must now solve is ϕ$$ (x) + λ ϕ(x) = 0,

ϕ$ (0) = 0,

ϕ$ (L) + h ϕ(L) = 0.

(5.2.26)

21 For an application, see Deng, Z.-S., and J. Liu, 2001: Blood perfusion-based model for characterizing the temperature fluctuation in living tissues. Physica, Ser. A, 300, 521–530.

The Heat Equation

315

The nth orthonormal eigenfunction for Equation 5.2.26 is T 2(kn2 + h2 ) ϕn (x) = cos(kn x), L(kn2 + h2 ) + h

(5.2.27)

where kn is the nth root of k tan(kL) = h. We also used the identity that (kn2 + h2 ) sin2 (kn h) = h2 . Substituting Equation 5.2.27 into Equation 5.2.12, we finally obtain

2 ( [(kn L)2 + (hL)2 ] cos(kn ξ) cos(kn x) −a2 kn2 (t−τ ) g(x, t|ξ,τ ) = e H(t−τ ). L n=1 (kn L)2 + (hL)2 + hL (5.2.28) Liu and Zhou22 used this solution to solve for phase change problems involved in the freezing and thawing processes of biological skins with finite thicknesses. # " • Example 5.2.4 Consider the Green’s function problem23 for the heat equation on a finite domain ∂g ∂ 2g = + δ(x − ξ)δ(t − τ ), ∂t ∂x2

0 < x, ξ < L,

0 < t, τ ,

(5.2.29)

with the boundary conditions gt (0, t|ξ,τ ) − γ gx (0, t|ξ,τ ) = 0,

gx (L, t|ξ,τ ) = 0,

0 < t,

(5.2.30)

and the initial condition that g(x, 0|ξ,τ ) = 0, 0 < x < L. Here γ > 0. We begin by taking the Laplace transform of Equation 5.2.29 and Equation 5.2.30 and find that d2 G − sG = −δ(x − ξ)e−sτ , dx2

0 < x, ξ < L,

(5.2.31)

with the boundary conditions s G(0, s|ξ,τ ) = γ G$ (0, s|ξ,τ ), and G$ (L, s|ξ,τ ) = 0. To solve Equation 5.2.31, we use the expansion G(x, s|ξ,τ ) = e−sτ

∞ ( An (ξ)ϕn (x) . s + λ2n n=0

(5.2.32)

22

Liu, J., and Y. X. Zhou, 2002: Analytical study on the freezing and thawing processes of biological skin with finite thickness. Heat Mass Transfer , 38, 319–326. 23 See Bluman, G., and H. Tuckwell, 1987: Technique for obtaining solutions for Rall’s model neuron. J. Neurosci. Methods, 20, 151–166.

316

Green’s Functions with Applications

L g(x,t| , )

0.8 0.6 0.4 0.2 0 0 0.1

0.2

0.08

0.4

0.06

0.6

x/L

0.04 0.8

0.02 1

0

a2(t− )/L2

Figure 5.2.2: Equation 5.2.37 (times L) for the one-dimensional heat equation where the boundary conditions are given by Equation 5.2.30 when ξ/L = 0.2 and γL = 1.

This expansion satisfies the differential equation and the boundary condition at x = L if cos[λn (L − x)] ϕn (x) = , (5.2.33) cos(λn L) where λ0 = 0 and λn is the nth root of γL tan(λL) + λL = 0. Because √ γ cosh[(L − ξ) s ]e−sτ √ √ √ , G(0, s|ξ,τ ) = √ (5.2.34) s[γ sinh(L s ) + s cosh(L s )] An = lim 2 (s + λ2n )G(0, s|ξ,τ )esτ , s→−λn

(5.2.35)

or A0 =

γ , 1 + γL

and An =

2γ cos[λn (L − ξ)] . (1 + γL) cos(λn L) − λL sin(λn L)

(5.2.36)

Taking the inverse Laplace transform of Equation 5.2.32, the Green’s function is γ g(x, t|ξ,τ ) = H(t − τ ) (5.2.37) 1 + γL 2 ∞ ( cos[λn (L − ξ)] cos[λn (L − x)]e−λn (t−τ ) + 2γH(t − τ ) . [(1 + γL) cos(λn L) − γL sin(λn L)] cos(λn L) n=1 Figure 5.2.2 illustrates Equation 5.2.37.

The Heat Equation

317

Jack and Redman24 solved this problem but they inverted the transform as a series of parabolic cylinder functions. In a similar vein, Evans et al.25 modeled passive neurones as a cable which consists of N segments of length Ln in which the electrical properties change in each segment. The mathematical model consists of series of N heat equations. They found the Green’s function for each heat equation. # " • Example 5.2.5 Let us find the Green’s function26 governed by the parabolic equation ∂g ∂2g − x−q 2 = ξ −q δ(x − ξ)δ(t − τ ), ∂t ∂x

0 < x, ξ < 1,

0 < t, τ , (5.2.38)

with the boundary conditions g(0, t|ξ,τ ) = g(1, t|ξ,τ ) = 0,

0 < t,

(5.2.39)

and the initial condition that g(x, 0|ξ,τ ) = 0, 0 < x < 1. We begin by introducing the eigenfunction expansion g(x, t|ξ,τ ) =

∞ (

an (t)ϕn (x).

(5.2.40)

n=1

Substituting Equation 5.2.40 into Equation 5.2.38, we find formally that ∞ (

n=1

a$n (t)ϕn (x) − x−q an (t)ϕ$$n (x) = ξ −q δ(x − ξ)δ(t − τ ).

(5.2.41)

Consider now the Sturm-Liouville problem ϕ$$ + λ xq ϕ = 0,

ϕ(0) = ϕ(1) = 0,

(5.2.42)

where λ is the eigenvalue and xq is the weight function. If we set u = x(q+2)/2 , then , -2 d2 ϕ q dϕ 2 + +λ ϕ = 0, ϕ(0) = ϕ(1) = 0. (5.2.43) du2 (2 + q)u du 2+q 24 Jack, J. J. B., and S. J. Redman, 1971: An electrical description of the mononeurone, and its application to the analysis of synaptic potentials. J. Physiol., 215, 321–352. 25 Evans, J. D., G. C. Kember, and G. Major, 1992: Techniques for obtaining analytical solutions to the multicylinder somatic shunt cable model for passive neurones. Biophys. J., 63, 350–365. 26 See Chan, C. Y., and W. Y. Chan, 1999: Existence of classical solutions for degenerate semilinear parabolic problems. Appl. Math. Comp., 101, 125–149.

318

Green’s Functions with Applications

The orthonormal eigenfunction solutions to Equation 5.2.43 are % √ &H ' , √ -' . ' 2 λn (2+q)/2 2 λn '' ' ϕn (x) = (2 + q)x J1/(2+q) x J|1/(2+q)|+1 , ' 2+q 2+q ' 4 √ 5 (5.2.44) where λn is the nth root of the equation J1/(2+q) 2 λ/(2 + q) = 0. Note that the use of this particular eigenfunction in Equation 5.2.40 ensures that g(x, t|ξ,τ ) will satisfy the boundary conditions, Equation 5.2.39. Returning to the original problem, we now use the relationship that ϕ$$n + λn xq ϕn = 0

(5.2.45)

to transform Equation 5.2.41 into ∞ (

n=1

[a$n (t) + λn an (t)] ϕn (x) = ξ −q δ(x − ξ)δ(t − τ ),

or

a$n (t) + λn an (t) = ϕn (ξ)δ(t − τ )

(5.2.46)

(5.2.47)

for each n. In the derivation of Equation 5.2.47, we expanded δ(x − ξ) in terms of ϕn (x). Integrating this Equation 5.2.47, we have that an (t) = ϕn (ξ)e−λn (t−τ ) H(t − τ ), because an (0) = 0. The amplitude an (t) leads directly to the Green’s function g(x, t|ξ,τ ) = H(t − τ )

∞ (

ϕn (ξ)ϕn (x)e−λn (t−τ ) .

(5.2.48)

n=1

# "

• Example 5.2.6

Let us compute the Green’s function27 governing the parabolic problem , ∂g ∂ ∂g axα − bxβ = δ(x − ξ)δ(t − τ ), 0 < x, ξ < L, 0 < t, τ , ∂t ∂x ∂x (5.2.49) where a, b > 0, with the Neumann boundary conditions gx (0, t|ξ,τ ) = gx (L, t|ξ,τ ) = 0,

0 < t,

(5.2.50)

and the initial condition that g(x, 0|ξ,τ ) = 0, 0 < x < L. 27 See Lin, J.-S., and L. M. Hildemann, 1996: Analytical solutions of the atmospheric diffusion equation with multiple sources and height-dependent wind speed and eddy diffusivities. Atmos. Environ., 30, 239–254.

The Heat Equation

319

We begin by taking the Laplace transform of Equation 5.2.49 and Equation 5.2.50 or , 1 d δ(x − ξ) −sτ β dG sG − α bx = e (5.2.51) ax dx dx aξ α with the transformed boundary conditions G$ (0, s|ξ,τ ) = G$ (L, s|ξ,τ ) = 0. Consider now the Sturm-Liouville problem d2 ϕ β dϕ ak 2 xα−β + + ϕ = 0, dx2 x dx b

ϕ$ (0) = ϕ$ (L) = 0.

(5.2.52)

If we set z = x(α−β+2)/2 , and ϕ(z) = z (1−β)/(α−β+2) ϕ(z), ? Equation 5.2.52 transforms into z2

d2 ϕ ? dϕ ? 4ak 2 z 2 /b − (1 − β)2 + z + ϕ ? = 0. dz 2 dz (α − β + 2)2

(5.2.53)

The general solution to Equation 5.2.53 is

ϕ(z) ? = AJµ (ωz) + BJ−µ (ωz),

where µ=

1−β , α−β+2

and ω =

. 2k a/b . α−β+2

assuming that µ > 0. Therefore, R 4 5 4 5S ϕ(x) = x(1−β)/2 AJµ ωx(α−β+2)/2 + BJ−µ ωx(α−β+2)/2 .

(5.2.54)

(5.2.55)

(5.2.56)

Turning to the boundary conditions and using the asymptotic formula for Bessel functions for small argument, the Neumann boundary condition at x = 0 leads to A = 0. On the other hand, the Neumann boundary condition at x = L gives 4 5 4 5 $ ωL(α−β+2)/2 J−µ ωL(α−β+2)/2 + µJ−µ ωL(α−β+2)/2 = 0, (5.2.57) or

4 5 J1−µ ωL(α−β+2)/2 = 0.

Consequently, the appropriate eigenfunctions for this problem are 4 5 ϕn (x) = x(1−β)/2 J−µ ωn x(α−β+2)/2 ,

(5.2.58)

(5.2.59)

where ωn is the nth root of Equation 5.2.58. We now use these eigenfunctions to express G(x, s|ξ,τ ) and δ(x − ξ). Beginning with the delta function, we assume that δ(x − ξ) = c0 + x(1−β)/2

∞ (

n=1

4 5 cn J−µ ωn x(α−β+2)/2 .

(5.2.60)

320

Green’s Functions with Applications

This is a Dini series for δ(x − ξ) and the coefficients equal 1 2 1+α α − β + 2 (1−β)/2 J−µ ωn ξ (α−β+2)/2 1 2 . (5.2.61) c0 = , and cn = ξ 2 aL1+α aLα−β+2 J−µ ωn L(α−β+2)/2

Substituting Equation 5.2.60 and a similar expansion for G(x, s|ξ,τ ) into Equation 5.2.51, we find that 1 2 ∞ ( J−µ ωn x(α−β+2)/2 e−sτ (1−β)/2 −sτ G(x, s|ξ,τ ) = c0 +x e cn . s s + b(α − β + 2)2 ωn2 /(4a) n=1 (5.2.62) Taking the inverse Laplace transform and applying the second shifting theorem, g(x, t|ξ,τ ) =

1+α α−β+2 (1−β)/2 H(t − τ ) + (xξ) H(t − τ ) aL1+α aLα−β+2 ∞ ( J−µ [λn (x/L)(α−β+2)/2 ]J−µ [λn (ξ/L)(α−β+2)/2 ] × 2 J−µ (λn ) n=1 % & b(α − β + 2)2 λ2n (t − τ ) × exp − , (5.2.63) 4aLα−β+2

where λn is the nth root of J1−µ (λ) = 0. In a similar manner, when we have the Dirichlet boundary conditions g(0, t|ξ,τ ) = g(L, t|ξ,τ ) = 0, the Green’s function becomes g(x, t|ξ,τ ) =

α−β+2 (1−β)/2 (xξ) H(t − τ ) aLα−β+2 ∞ ( Jµ [λn (x/L)(α−β+2)/2 ]Jµ [λn (ξ/L)(α−β+2)/2 ] × 2 Jµ+1 (λn ) n=1 % & b(α − β + 2)2 λ2n (t − τ ) × exp − , 4aLα−β+2

(5.2.64)

where λn is the nth root of Jµ (λ) = 0. Finally, for the boundary conditions gx (0, t|ξ,τ ) = g(L, t|ξ,τ ) = 0,

0 < t,

(5.2.65)

g(0, t|ξ,τ ) = gx (L, t|ξ,τ ) = 0,

0 < t,

(5.2.66)

and the Green’s functions are α−β+2 (1−β)/2 g(x, t|ξ,τ ) = α−β+2 (xξ) H(t − τ ) aL ∞ ( J−µ [λn (x/L)(α−β+2)/2 ]J−µ [λn (ξ/L)(α−β+2)/2 ] × (5.2.67) 2 J1−µ (λn ) n=1 % & b(α − β + 2)2 λ2n (t − τ ) × exp − , 4aLα−β+2

The Heat Equation

321

where λn is the nth root of J−µ (λ) = 0, and g(x, t|ξ,τ ) =

α−β+2 (1−β)/2 (xξ) H(t − τ ) aLα−β+2 ∞ ( Jµ [λn (x/L)(α−β+2)/2 ]Jµ [λn (ξ/L)(α−β+2)/2 ] × Jµ2 (λn ) n=1 % & b(α − β + 2)2 λ2n (t − τ ) × exp − , 4aLα−β+2

(5.2.68)

where λn is the nth root of Jµ−1 (λ) = 0, respectively. # " • Example 5.2.7 Let us find the Green’s function28 governing the parabolic problem ∂g ∂ 2 g ∂ 2g ∂2g − 2 − KB (t) 2 − KH (t) 2 + ρ(t)g = δ(x − ξ)δ(y − η)δ(z − ζ)δ(t − τ ), ∂t ∂x ∂y ∂z (5.2.69) where −∞ < x, y, ξ, η < ∞, z1 < z, ζ < z 2 , and 0 < t, τ, with the boundary conditions g(x, y, z1 , t|ξ, η, ζ,τ ) = g(x, y, z2 , t|ξ, η, ζ,τ ) = 0,

(5.2.70)

and the initial condition that g(x, y, z, 0|ξ, η, ζ,τ ) = 0. From the form of the boundary conditions, the Green’s function can be written as the eigenfunction expansion & nπ(z − z1 ) g(x, y, z, t|ξ, η, ζ,τ ) = Gn (x, y, t|ξ, η,τ ) sin , D n=1 ∞ (

%

(5.2.71)

where D = z2 − z1 . Substituting Equation 5.2.71 into Equation 5.2.70, the equation that governs Gn (x, y, t|ξ, η,τ ) is % 2 2 & ∂Gn ∂ 2 Gn ∂ 2 Gn n π − − K (t) + K (t) + ρ(t) Gn B H ∂t ∂x2 ∂y 2 D2 % & 2 nπ(ζ − z1 ) = δ(x − ξ)δ(y − η) sin δ(t − τ ). D D

(5.2.72)

28 See Marinoschi, G., U. Jaekel, and H. Vereecken, 1999: Analytical solutions of threedimensional convection-dispersion problems with time dependent coefficients. Zeit. Angew. Math. Mech., 79, 411–421.

322

Green’s Functions with Applications

To solve Equation 5.2.72, we introduce Fn (x, y, t|ξ, η,τ ), which is defined by Gn (x, y, t|ξ, η,τ ) = Fn (x, y, t|ξ, η,τ ) ! " t% 2 2 & ) n π × exp − KH (µ) + ρ(µ) dµ , D2 τ

(5.2.73)

so that the problem now reads % & ∂Fn ∂ 2 Fn ∂ 2 Fn 2 nπ(ζ − z1 ) − − K (t) = δ(x − ξ)δ(y − η) sin δ(t − τ ). B ∂t ∂x2 ∂y 2 D D (5.2.74) Finally, we must solve Equation 5.2.74. Taking the Fourier transform of this equation in the x and y directions, we obtain the ordinary differential equation % & 2 dF n 1 2 2 −ikξ−i4η nπ(ζ − z1 ) 2 + k + KB (t)0 F n = e sin δ(t − τ ). (5.2.75) dt D D The solution to Equation 5.2.75 is % & 2 nπ(ζ − z1 ) −ikξ−i4η F n (k, 0, t|ξ, η,τ ) = sin e H(t − τ ) D D % & " t × exp −k 2 (t − τ ) − 02 KB (µ) dµ .

(5.2.76)

τ

Applying the inverse Fourier transform, " ∞" ∞ 1 Fn (x, y, t|ξ, η,τ ) = 2 F n (k, 0, t|ξ, η,τ )eikx+i4y dk d0, 4π −∞ −∞

(5.2.77)

we obtain % &" ∞ " ∞ 1 nπ(ζ − z1 ) Fn (x, y, t|ξ, η,τ ) = sin e−s1 (k) e−s2 (4) dk d0, 2Dπ 2 D −∞ −∞ (5.2.78) where % &2 √ i(x − ξ) (x − ξ)2 s1 (k) = k 2 (t − τ ) − ik(x − ξ) = k t − τ − √ + , (5.2.79) 4(t − τ ) 2 t−τ and s2 (0) = −i0(y − η) + 0

2

"

τ

t

KB (µ) dµ

(5.2.80)

 T 2 " t 2 i(y − η)  + / (y − η) = 0 KB (µ) dµ − =/ . (5.2.81) t t 4 τ KB (µ) dµ τ 2 τ KB (µ) dµ

The Heat Equation

323

Performing the integrals with respect to k and 0, " ∞ " exp[−(x − ξ)2 /4(t − τ )] ∞ −µ2 −s1 (k) √ e dk = e dµ t−τ −∞ −∞ % & √ π (x − ξ)2 =√ exp − , 4(t − τ ) t−τ

(5.2.82) (5.2.83)

and "

∞

e

−s2 (4)

−∞

we finally obtain

d0 = =/ t τ

√ π KB (µ) dµ

:

(y − η)2

;

exp − / t , 4 τ KB (µ) dµ

(5.2.84)

% & (x − ξ)2 % & exp − 2 nπ(z − z1 ) 4(t − τ ) . Gn (x, y, t|ξ, η,τ ) = sin D D 4π(t − τ ) : ; (y − η)2 exp − / t 4 KB (µ) dµ × = /τ H(t − τ ) (5.2.85) t 4π τ KB (µ) dµ ! " t% 2 2 & ) n π × exp − KH (µ) + ρ(µ) dµ . D2 τ The complete solution is given by substituting Equation 5.2.85 into Equation 5.2.71. 5.3 HEAT EQUATION WITHIN A CYLINDER In this section, we turn our attention to cylindrical domains. The techniques used here have much in common with those used in the previous section. However, in place of sines and cosines from a Sturm-Liouville problem, we will encounter Bessel functions. • Example 5.3.1 In this example, we find the Green’s function for the heat equation in cylindrical coordinates , ∂g a2 ∂ ∂g δ(r − ρ)δ(t − τ ) − r = , 0 < r, ρ < b, 0 < t, τ , (5.3.1) ∂t r ∂r ∂r 2πr subject to the boundary conditions lim |g(r, t|ρ,τ )| < ∞,

r→0

g(b, t|ρ,τ ) = 0,

0 < t,

(5.3.2)

324

Green’s Functions with Applications

Figure 5.3.1: Equation 5.3.8 for the axisymmetric heat equation, Equation 5.3.1, with a Dirichlet boundary condition at r = b. Here ρ/b = 0.3 and the graph starts at a2 (t−τ )/b2 = 0.001 to avoid the delta function at t − τ = 0.

and the initial condition that g(r, 0|ρ,τ ) = 0. As usual, we begin by taking the Laplace transform of Equation 5.3.1 or , 1 d dG s e−sτ r − 2G = − δ(r − ρ). r dr dr a 2πa2 r

(5.3.3)

Next we re-express δ(r − ρ)/r as the Fourier-Bessel expansion ∞ δ(r − ρ) ( = An J0 (kn r/b), 2πr n=1

(5.3.4)

where kn is the nth root of J0 (k) = 0, and "

b

δ(r − ρ) J0 (kn ρ/b) J0 (kn r/b) r dr = 2 2 2πr πb J1 (kn )

(5.3.5)

, ∞ 1 d dG s e−sτ ( J0 (kn ρ/b)J0 (kn r/b) r − 2G = − 2 2 . r dr dr a πa b n=1 J12 (kn )

(5.3.6)

An =

2 b2 J12 (kn )

0

so that

The solution to Equation 5.3.6 is G(r, s|ρ,τ ) =

∞ e−sτ ( J0 (kn ρ/b)J0 (kn r/b) . π n=1 (sb2 + a2 kn2 )J12 (kn )

(5.3.7)

The Heat Equation

325

Taking the inverse of Equation 5.3.7 and applying the second shifting theorem, g(r, t|ρ,τ ) =

∞ H(t − τ ) ( J0 (kn ρ/b)J0 (kn r/b) −a2 kn2 (t−τ )/b2 e . πb2 J12 (kn ) n=1

(5.3.8)

Moghadam29 used this Green’s function to find the electro-osmotic fullydeveloped flow in a circular micro-channel under an alternating electrical field. If we modify the boundary condition at r = b so that it now reads gr (b, t|ρ,τ ) + h g(b, t|ρ,τ ) = 0,

0 < t,

(5.3.9)

where h ≥ 0, our analysis now leads to g(r, t|ρ,τ ) =

∞ H(t − τ ) ( J0 (kn ρ/b)J0 (kn r/b) −a2 kn2 (t−τ )/b2 e , πb2 n=1 J02 (kn ) + J12 (kn )

(5.3.10)

where kn are the positive roots of k J1 (k) − hb J0 (k) = 0. If h = 0, we must add 1/(πb2 ) to Equation 5.3.10. # " • Example 5.3.2 Let us find the Green’s function for the heat equation within a cylindrical shell. Mathematically, this problem consists of the partial differential equation , ∂g a2 ∂ ∂g δ(r − ρ)δ(t − τ ) − r = , α < r, ρ < β ,0 < t, τ , (5.3.11) ∂t r ∂r ∂r 2πr subject to the boundary conditions that g(α, t|ρ,τ ) = g(β, t|ρ,τ ) = 0,

0 < t,

(5.3.12)

and the initial condition that g(r, 0|ρ,τ ) = 0, α < r < β. Once again, we begin by taking the Laplace transform of Equation 5.3.11 and obtaining , 1 d dG s e−sτ r − 2G = − δ(r − ρ). (5.3.13) r dr dr a 2πa2 r Next, we express δ(r − ρ)/r as a Fourier-Bessel expansion by considering the regular Sturm-Liouville problem , 1 d dϕ r + k 2 ϕ = 0, ϕ(α) = ϕ(β) = 0. (5.3.14) r dr dr 29 Moghadam, A. J., 2012: An exact solution of AC electro-kimetic-driven flow in a circular micro-channel. Eur. J. Mech., Ser. B , 34, 91–96; Moghadam, A. J., 2013: Exact solution of AC electro-osmotic flow in a microannulus. J. Fluids Eng., 135, Art. No. 091201.

326

Green’s Functions with Applications

Figure 5.3.2: Equation 5.3.22 for the axisymmetric heat equation, Equation 5.3.11, in an annulus with Dirichlet boundary conditions at r = α and r = β when α = 1, β = 2, and ρ = 1.3. The plot begins at a2 (t − τ ) = 10−5 to avoid the delta function at t − τ = 0. Values greater than one are not plotted.

The eigenfunctions that satisfy Equation 5.3.14 are ϕn (r) = Y0 (kn α)J0 (kn r) − J0 (kn α)Y0 (kn r),

(5.3.15)

provided kn is the nth zero of J0 (kα)Y0 (kβ) − J0 (kβ)Y0 (kα) = 0. Therefore, the expansion for the delta function in terms of ϕn (r) is /β ∞ δ(r − ρ)ϕn (r) dr δ(r − ρ) ( = An ϕn (r), where An = α / β . (5.3.16) r ϕ2n (r) r dr n=1 α

Using the orthogonality conditions30 that " β 1 2'β J02 (kn r) r dr = 12 r2 J02 (kn r) + J12 (kn r) 'α ,

(5.3.17)

α

"

β

1 2 r 2

J0 (kn r)Y0 (kn r) r dr =

α

and

'β [J0 (kn r)Y0 (kn r) + J1 (kn r)Y1 (kn r)]'α ,

(5.3.18)

"

β α

Y02 (kn r) r dr =

1 2 r 2

1 2 2'β Y0 (kn r) + Y12 (kn r) 'α ,

(5.3.19)

and the Wronskian relationship31 J0 (z)Y1 (z) − J1 (z)Y0 (z) = −2/(πz), we find that 30 Watson, G. N., 1966: A Treatise on the Theory of Bessel Functions. Cambridge University Press, Section 5.12, Equation 2. 31

Ibid., Section 3.6, Equation 12.

The Heat Equation

327 "

β α

ϕ2n (r) r

% 2 & 2 J0 (kn α) dr = 2 2 −1 . π kn J02 (kn β)

(5.3.20)

Therefore, G(r, s|ρ,τ ) is G(r, s|ρ,τ ) =

∞ π −sτ ( kn2 J02 (kn β)ϕn (ρ)ϕn (r) e . 2 4 [J0 (kn α) − J02 (kn β)] (s + a2 kn2 ) n=1

(5.3.21)

Inverting the Laplace transform, we obtain the final answer g(r, t|ρ,τ ) =

∞ ( π kn2 J02 (kn β)ϕn (ρ)ϕn (r) −a2 kn2 (t−τ ) H(t − τ ) e . 4 J02 (kn α) − J02 (kn β) n=1

(5.3.22)

In a similar manner, we can solve the general problem32 of Equation 5.3.1 with Robin boundary conditions a1 gr (α, t|ρ,τ ) − a2 g(α, t|ρ,τ ) = 0,

(5.3.23)

b1 gr (β, t|ρ,τ ) + b2 g(β, t|ρ,τ ) = 0,

(5.3.24)

and where a1 , a2 , b1 , b2 ≥ 0. The solution for this problem is g(r, t|ρ,τ ) =

∞ ( π kn2 2 H(t − τ ) [b1 kn J1 (kn β) − b2 J0 (kn β)] 4 F (k ) n n=1 2

× ϕn (ρ)ϕn (r)e−a

2 kn (t−τ )

,

(5.3.25)

− Y0 (kn r) [a1 kn J1 (kn α) + a2 J0 (kn α)] ,

(5.3.26)

where ϕn (r) = J0 (kn r) [a1 kn Y1 (kn α) + a2 Y0 (kn α)]

and F (kn ) = (b21 kn2 + b22 ) [a1 kn J1 (kn α) + a2 J0 (kn α)]

2

2

− (a21 kn2 + a22 ) [b1 kn J1 (kn β) − b2 J0 (kn β)] .

(5.3.27)

The positive values of kn are given by [b2 J0 (kn β) − b1 kn J1 (kn β)][a1 kn Y1 (kn α) + a2 Y0 (kn α)] (5.3.28) + [b1 kn Y1 (kn β) − b2 Y0 (kn β)][a1 kn J1 (kn α) + a2 J0 (kn α)] = 0. 32 Naji et al. [Naji, M., M. Al-Nimr, and S. Masoud, 2000: Transient thermal behavior of a cylindrical brake system. Heat Mass Transfer, 36, 45–49. ] found the Green’s function for a two-layer ring with radiative boundary conditions.

328

Green’s Functions with Applications

If a2 = b2 = 0, we must add the extra term 1/[π(β 2 − α2 )] to Equation 5.3.25. Consider the case, for example, when a1 = b1 = 1, a2 = 0, and b2 = h. This problem arose in a paper by Ma et al.33 who were studying computer chip coolant. Taking into account a factor of 2π because of differences in their definition of the forcing, the Naji et al. solution would read g(r, t|ρ,τ ) =

∞ ( π2 [kn J1 (kn β) − hJ0 (kn β)]2 ϕn (r)ϕn (ρ) H(t − τ ) 2 2 (k + h2 )J12 (kn α) − [kn J1 (kn β) − hJ0 (kn β)]2 n=1 n 2 2 kn (t−τ )

× e−a

,

(5.3.29)

where ϕn (r) = kn [Y1 (kn α)J0 (kn r) − J1 (kn α)Y0 (kn r)],

(5.3.30)

and kn is the nth positive root of [hJ0 (kβ) − kJ1 (kβ)]Y1 (kα) + [kY1 (kβ) − hY0 (kβ)]J1 (kα) = 0.

(5.3.31)

Using Equation 5.3.31, we can rewrite g(r, t|ρ,τ ) as ∞ ( π2 kn2 J12 (kn α)ϕn (r)ϕn (ρ) exp[−a2 kn2 (t − τ )] H(t − τ ) . 2 2 (kn + h2 )J12 (kn α) − [kn J1 (kn β) − hJ0 (kn β)]2 n=1 (5.3.32) where ϕn (r) now reads

g(r, t|ρ,τ ) =

ϕn (r) = [hY0 (kn β) − kn Y1 (kn β)]J0 (kn r) − [hJ0 (kn β) − kn J1 (kn β)]Y0 (kn r). (5.3.33) # " • Example 5.3.3 In Example 5.1.2, we utilized the one-dimensional, free-space Green’s function to develop additional Green’s functions by finding the homogeneous solution via Laplace transforms. Similar considerations hold in the twodimensional case. For example, let us find the Green’s function for ∂g − a2 ∂t

,

∂2g 1 ∂g 1 ∂ 2g + + 2 2 2 ∂r r ∂r r ∂θ

-

=

δ(r − ρ)δ(θ − θ$ )δ(t − τ ) , 2πr

(5.3.34)

where α < r, ρ < β, 0 ≤ θ,θ $ ≤ 2π, and 0 < t, τ. The boundary conditions are g(α, θ, t|ρ,θ $ , τ ) = g(β, θ, t|ρ,θ $ , τ ) = 0,

0 ≤ θ ≤ 2π,

0 < t,

(5.3.35)

33 Ma, K. Q., J. Liu, S. H. Xiang, K. W. Xie, Y. X. Zhou, 2009: Study of thawing behavior of liquid metal used as computer chip coolant. Int. J. Therm. Sci., 48, 964–974.,

The Heat Equation

329

Table 5.3.2: Green’s Functions for the Two-Dimensional Heat Conduction Equation ∂g −κ ∂t

,

∂ 2 g 1 ∂g 1 ∂2g + + 2 2 2 ∂r r ∂r r ∂θ

-

=

δ(r − ρ)δ(θ − θ $ )δ(t) 2πr

for Various Annuli with Dirichlet Boundary Conditions 1. 0 ≤ r < a g=

∞ ∞ 2 1 ( ( Jm (αn r)Jm (αn ρ) cos[m(θ − θ$ )]e−καn t , $ (α a)]2 πa2 m=−∞ n=1 [Jm n

where αn is the nth positive root of Jm (αa) = 0. 2. a < r < ∞

" ∞ ∞ 2 1 ( Un (αr)Un (αρ) $ g= cos[n(θ − θ )] e−κα t 2 α dα , 2π n=−∞ Jn (αa) + Yn2 (αa) 0

where Un (αr) = Jn (αr)Yn (αa) − Jn (αa)Yn (αr). 3. a < r < b g=

∞ ∞ 2 2 π ( ( 2 Jm (αn b)Um (αn r)Um (αn ρ) αn cos[m(θ − θ$ )]e−καn t , 2 2 4 m=−∞ n=1 Jm (αn a) − Jm (αn b)

where Um (αr) = Jm (αr)Ym (αa) −Jm (αa)Ym (αr) and αn is the nth positive root of Um (αb) = 0.

with the initial condition that g(r, θ, 0|ρ,θ $ , τ ) = 0,

α ) − Kn (qβ)In (qr> )]. Upon applying the residue theorem to Equation 5.3.47, we obtain ∞ ( π g(r, θ, t|ξ,θ $ , τ ) = H(t − τ ) cos[m(θ − θ$ )] (5.3.48) 4 m=−∞ !( ) ∞ J 2 (kn β)Um (kn ρ)Um (kn r) −a2 kn2 (t−τ ) × kn2 m 2 e , 2 (k β) Jm (kn α) − Jm n n=1

where Um (kr) = Jm (kr)Ym (kα) − Jm (kα)Ym (kr),

(5.3.49)

and kn is the nth positive root of Um (kβ) = 0. Carslaw and Jaeger35 used the same technique to find the Green’s functions for a sector 0 < θ, θ$ < θ0 of the annulus α < r, ρ < β. Tables 5.3.1 and 5.3.2 present their results. # "

35 Carslaw, H. S., and J. C. Jaeger, 1941: The determination of Green’s function for line sources for the equation of conduction of heat in cylindrical coordinates by the Laplace transformation. Philos. Mag., Ser. 7 , 31, 204–208.

The Heat Equation

333

Table 5.3.4: Green’s Functions for the Three-Dimensional Heat Conduction Equation ∂g −κ ∂t

,

∂ 2g 1 ∂g 1 ∂ 2g ∂ 2g + + 2 2+ 2 2 ∂r r ∂r r ∂θ ∂z

-

=

δ(r − ρ)δ(θ − θ$ )δ(z)δ(t) 2πr

for Infinitely Long Cylinders with Dirichlet Boundary Conditions 1. 0 ≤ r < a 2 ∞ ∞ 2 e−z /4κt ( ( Jm (αn r)Jm (αn ρ) √ g= cos[m(θ − θ $ )]e−καn t , $ (α a)]2 [Jm 2πa2 πκt m=−∞ n=1 n

where αn is the nth positive root of Jm (αa) = 0. 2. a < r < ∞

" ∞ 2 ∞ 2 e−z /4κt ( Un (αr)Un (αρ) $ √ g= cos[n(θ − θ )] e−κα t 2 α dα , Jn (αa) + Yn2 (αa) 4π πκt n=−∞ 0

where Un (αr) = Jn (αr)Yn (αa) − Jn (αa)Yn (αr). 3. a < r < b √ ∞ ∞ ( ( 2 π J 2 (αn b)Um (αn r)Um (αn ρ) g = √ e−z /4κt α2n m 2 2 (α b) Jm (αn a) − Jm 8 κt n m=−∞ n=1

2

× cos[m(θ − θ$ )]e−καn t ,

where Um (αr) = Jm (αr)Ym (αa) −Jm (αa)Ym (αr) and αn is the nth positive root of Um (αb) = 0.

• Example 5.3.4 In their study of laser resonators, Peng et al.36 found the Green’s function for heat conduction within an axisymmetric cylinder of radius b and height d.

36 Peng, Y., Z. Cheng, Y. Zhang and J. Qiu, 2001: Temperature distributions and thermal deformations of mirror substrates in laser resonators. Appl. Opt., 40, 4824–4830; Peng, Y., Z. Cheng, Y. Zhang, and J. Qiu, 2001: Laser-induced temperature distributions and thermal deformations in sapphire, silicon, and calcium fluoride substrates at 1.315 µm. Opt. Eng., 40, 2822–2829.

334

Green’s Functions with Applications

The governing equation is , 2 ∂g 1 ∂g ∂ 2 g δ(r − ρ)δ(z − ζ)δ(t − τ ) 2 ∂ g −a + + 2 = , ∂t ∂r2 r ∂r ∂z 2πr

(5.3.50)

where 0 ≤ r, ρ < b, 0 < z, ζ < d, and 0 < t, τ, with the boundary conditions lim |g(r, z, t|ρ, ζ,τ )| < ∞,

0 < z < d,

r→0

gr (b, z, t|ρ, ζ,τ ) + Hg(b, z, t|ρ, ζ,τ ) = 0, gz (r, 0, t|ρ, ζ,τ ) = 0,

0 < t,

0 < z < d,

0 ≤ r < b,

gz (r, d, t|ρ, ζ,τ ) + Hg(r, d, t|ρ, ζ,τ ) = 0,

(5.3.51) 0 < t,

0 < t,

0 ≤ r < b,

0 < t,

(5.3.52) (5.3.53) (5.3.54)

and the initial condition that g(r, z, 0|ρ, ζ,τ ) = 0, 0 ≤ r < b, 0 < z < d. We begin by constructing Fourier representations of δ(z − ζ) and δ(r − ρ)/r. For δ(z − ζ) we use the eigenfunction Zn (z) = cos(αn z), where αn is the nth solution of α tan(αd) = H. This eigenfunction enjoys the properties that Zn$ (0) = 0 and Zn$ (d) + HZn (d) = 0. Therefore, we set δ(z − ζ) =

∞ (

An cos(αn z),

(5.3.55)

n=1

where An =

/d 0

δ(z − ζ) cos(αn z) dz 2(α2n + H 2 ) cos(αn ζ) = . /d (α2n + H 2 )d + H cos2 (αn z) dz 0

(5.3.56)

Turning to δ(r − ρ), the eigenfunction is Rm (r) = J0 (βm r), where βm is the mth root of β J1 (βb) = HJ0 (βb). This eigenfunction satisfies the condition $ that Rm (b) + HRm (b) = 0. Therefore, the expansion for δ(r − ρ) is ∞ ( δ(r − ρ) = Bm J0 (βm r), r m=1

(5.3.57)

where Bm

2 2βm = 2 2 b (βm + H 2 )J02 (βm b)

"

0

b

δ(r − ρ)J0 (βm r) dr =

2 2βm J0 (βm ρ) . 2 + H 2 )J 2 (β b) b2 (βm m 0 (5.3.58)

Here we used Equation 2.3.30 and Equation 2.3.34. The nature of the expansions for δ(z − ζ) and δ(r − ρ) suggests that the eigenfunction expansion for g(r, z, t|ρ, ζ,τ ) is g(r, z, t|ρ, ζ,τ ) =

∞ ∞ 1 (( An Bm gnm (t|τ ) cos(αn z)J0 (βm r). 2π n=1 m=1

(5.3.59)

The Heat Equation

335

This expansion has the advantage that it satisfies the boundary conditions given by Equation 5.3.52 through Equation 5.3.54. Substituting Equation 5.3.59 into Equation 5.3.50 yields the ordinary differential equation dgnm 2 + a2 (α2n + βm )gnm = δ(t − τ ) dt

(5.3.60)

with gnm (0|τ ) = 0. The solution to this equation is 2 gnm (t|τ ) = exp[−a2 (α2n + βm )(t − τ )]H(t − τ ).

(5.3.61)

Therefore, the Green’s function is ∞ ∞ 2 2 2 H(t − τ ) ( ( An Bm cos(αn z)J0 (βm r)e−a (αn +βm )(t−τ ) . 2π n=1 m=1 (5.3.62) In a similar vein, Elliott et al.37 solved Equation 5.3.50 through Equation 5.3.54 with the modifications that Equation 5.3.53 and Equation 5.3.54 now become

g(r, z, t|ρ, ζ,τ ) =

gz (r, 0, t|ρ, ζ,τ ) + Hg(r, 0, t|ρ, ζ,τ ) = 0,

0 ≤ r < b,

0 < t,

(5.3.63)

and g(r, d, t|ρ, ζ,τ ) = 0,

0 ≤ r < b,

0 < t.

(5.3.64)

Because the boundary conditions changed only along the top and bottom, we must only modify the eigenfunction Zn (z) so that it now reads sin[αn (d − z)], where αn is the nth solution of α = H tan(αd). Therefore, δ(z − ζ) =

∞ (

n=1

An sin[αn (d − z)],

(5.3.65)

where An =

/d 0

δ(z − ζ) sin[αn (d − z)] dz 2(α2n + H 2 ) sin[αn (d − ζ)] = . (5.3.66) /d 2 (α2n + H 2 )d + H sin [αn (d − z)] dz 0

Turning to the radial dependence, the coefficient is still given by Equation 5.3.58. However, because βm J1 (βm b) = H J0 (βm b), we can rewrite Bm as Bm =

2H 2 J0 (βm ρ) . 2 + H 2 )J 2 (β b) b2 (βm m 1

(5.3.67)

37 Elliott, A., J. Schwartz, J. Wang, A. Shetty, J. Hazle, and J. R. Stafford, 2008: Analytical solution to heat equation with magnetic resonance experimental verification for nanoshell enhanced thermal therapy. Lasers Surg. Med., 40, 660–665. I have corrected some typos; there is a difference of 2π/ρ because they used a different source term.

336

Green’s Functions with Applications

Therefore, the Green’s function for this new problem is g(r, z, t|ρ, ζ,τ ) =

∞ ∞ H(t − τ ) ( ( An Bm sin[αn (d − z)]J0 (βm r) 2π n=1 m=1 2

× e−a

2 (α2n +βm )(t−τ )

,

(5.3.68)

where An and Bm are given by Equation 5.3.66 and Equation 5.3.67. # " • Example 5.3.5 In this example, we find the Green’s function for the three-dimensional equation , 2 ∂g 1 ∂g 1 ∂2g ∂2g δ(r − ρ)δ(θ − θ $ )δ(z − ζ)δ(t − τ ) 2 ∂ g −a + + + = , ∂t ∂r2 r ∂r r2 ∂θ2 ∂z 2 2πr (5.3.69) where 0 < r, ρ < ∞, 0 ≤ θ,θ $ ≤ 2π, 0 < z, ζ < L, and 0 < t, τ. The boundary conditions are lim |g(r, θ, z, t|ρ,θ $ , ζ , τ)| < ∞,

lim g(r, θ, z, t|ρ,θ $ , ζ , τ) → 0, (5.3.70)

r→∞

r→0

if 0 ≤ θ ≤ 2π, 0 < z < L, and 0 < t, and g(r, θ, 0, t|ρ,θ $ , ζ , τ) = g(r, θ, L, t|ρ,θ $ , ζ , τ) = 0,

(5.3.71)

if 0 < r < ∞, 0 ≤ θ ≤ 2π, and 0 < t. The initial condition is g(r, θ, z, 0|ρ,θ $ , ζ , τ) = 0,

0 < r < ∞,

0 ≤ θ ≤ 2π,

0 < z < L. (5.3.72)

Obviously, the solution must be periodic in θ. Because the differential equation is linear, we can write 2

g(r, θ, z, t|ρ,θ $ , ζ , τ) =

2

2

e−[R +(z−ζ) ]/[4a (t−τ )] H(t− τ )+ v(r, θ, z, t), (5.3.73) 8a3 [π(t − τ )]3/2

where R2 = r2 + ρ2 − 2ρr cos(θ − θ$ ). The first term in Equation 5.3.73 is the three-dimensional, free-space Green’s function in cylindrical coordinates; it is the particular solution. On the other hand, v(r, θ, z, t) is the homogeneous solution that satisfies , 2 ∂v ∂ v 1 ∂v 1 ∂ 2v ∂ 2v 2 =a + + 2 2 + 2 , (5.3.74) ∂t ∂r2 r ∂r r ∂θ ∂z subject to the boundary conditions

The Heat Equation

337

Table 5.3.5: Green’s Functions for the Three-Dimensional Heat Conduction Equation ∂g −κ ∂t

,

∂ 2 g 1 ∂g 1 ∂2g ∂2g + + 2 2+ 2 2 ∂r r ∂r r ∂θ ∂z

-

=

δ(r − ρ)δ(θ − θ $ )δ(z − ζ)δ(t) 2πr

for Right Cylinders between 0 < z, ζ < L with Dirichlet Boundary Conditions

1. 0 ≤ r < a

, - , ∞ 2 ( −κk2 π 2 t/L2 kπζ kπz g= e sin sin πa2 L L L k=1 < ∞ ∞ > ( ( Jm (αn r)Jm (αn ρ) $ −κα2n t × cos[m(θ − θ )]e , $ (α a)]2 [Jm n m=−∞ n=1

where αn is the nth positive root of Jm (αa) = 0. 2. a < r < ∞

, - , ∞ 1 ( −κk2 π2 t/L2 kπζ kπz g= e sin sin πL L L k=1 < ∞ > " ∞ ( 2 U (αr)U (αρ) n n × cos[n(θ − θ$ )] e−κα t 2 αdα , Jn (αa) + Yn2 (αa) 0 n=−∞

where Un (αr) = Jn (αr)Yn (αa) − Jn (αa)Yn (αr). 3. a < r < b , - , ∞ π ( −κk2 π2 t/L2 kπζ kπz e sin sin 2L L L k=1 < ∞ > ∞ 2 ( ( 2 Jm (αn b)Um (αn r)Um (αn ρ) $ −κα2n t × αn cos[m(θ − θ )]e , 2 (α a) − J 2 (α b) Jm n n m m=−∞ n=1

g=

where Um (αr) = Jm (αr)Ym (αa) −Jm (αa)Ym (αr) and αn is the nth positive root of Um (αb) = 0.

338

Green’s Functions with Applications

Table 5.3.6: Green’s Functions for the Three-Dimensional Heat Conduction Equation ∂g −κ ∂t

,

∂ 2g 1 ∂g 1 ∂ 2g ∂ 2g + + 2 2+ 2 2 ∂r r ∂r r ∂θ ∂z

-

=

δ(r − ρ)δ(θ − θ$ )δ(z)δ(t) 2πr

for the Sector 0 < θ, θ$ < θ0 Cut from an Infinitely Long Cylinder with Dirichlet Boundary Conditions 1. 0 ≤ r < a 2 ∞ ∞ ( 2e−z /4κt ( Js (αn r)Js (αn ρ) −κα2n t g= 2 √ sin(sθ) sin(sθ $ ) e , [Js$ (αn a)]2 a θ0 πκt m=1 n=1

where αn is the nth positive root of Js (αa) = 0, and s = mπ/θ 0 . 2. a < r < ∞

" ∞ 2 ∞ 2 e−z /4κt ( Us (αr)Us (αρ) $ g= √ sin(sθ) sin(sθ ) e−κα t 2 α dα , Js (αa) + Ys2 (αa) θ0 πκt m=1 0

where Us (αr) = Js (αr)Ys (αa) − Js (αa)Ys (αr), and s = mπ/θ0 . 3. a < r < b g=

2 ∞ π 2 e−z /4κt ( √ sin(sθ) sin(sθ$ ) 2θ0 πκt m=1

( J 2 (αn b)Us (αn r)Us (αn ρ) 2 × α2n s 2 e−καn t , Js (αn a) − Js2 (αn b) n=1

where Us (αr) = Js (αr)Ys (αa) − Js (αa)Ys (αr), αn is the nth positive root of Us (αb) = 0, and s = mπ/θ0 .

lim |v(r, θ, z, t)| < ∞,

lim v(r, θ, z, t) → 0,

r→∞

r→0

2

v(r, θ, 0, t) = − and

2

e−(R +ζ )/[4a (t−τ )] H(t − τ ), 8a3 [π(t − τ )]3/2 2

v(r, θ, L, t) = −

2

(5.3.75)

2

(5.3.76)

2

e−[R +(L−ζ) ]/[4a (t−τ )] H(t − τ ), 8a3 [π(t − τ )]3/2

(5.3.77)

The Heat Equation

339

and the initial condition that v(r, θ, z, 0) = 0,

0 < r < ∞,

0 ≤ θ ≤ 2π,

0 < z < L.

(5.3.78)

Next, we take the Laplace transform of Equation 5.3.74 through Equation 5.3.78, ∂ 2V 1 ∂V 1 ∂ 2V ∂2V + + + − q 2 V = 0, (5.3.79) ∂r2 r ∂r r2 ∂θ2 ∂z 2 where q 2 = s/a2 . The boundary conditions are lim |V (r, θ, z, s)| < ∞,

r→0

and

lim V (r, θ, z, s) → 0,

r→∞

√ 2 2 e−sτ −q R +ζ . V (r, θ, 0, s) = − 4πa2 R2 + ζ 2 " e−sτ ∞ −ζ √ξ2 +q2 J0 (ξR) . =− e ξ dξ , 4πa2 0 ξ2 + q2 √ 2 2 e−sτ −q R +(L−ζ) . V (r, θ, L, s) = − 2πa2 R2 + (L − ζ)2 " e−sτ ∞ −(L−ζ)√ξ2 +q2 J0 (ξR) . =− e ξ dξ . 2πa2 0 ξ2 + q2

(5.3.80)

(5.3.81) (5.3.82)

(5.3.83) (5.3.84)

We will shortly show why we introduced these integral representations.38 Using Hankel transforms, the solution to Equation 5.3.79 through Equation 5.3.84 can be expressed e−sτ V (r, θ, z, s) = − 4πa2

"

∞

J (ξR) .0 ξdξ ξ2 + q2 0 . ! −ζ √ξ2 +q2 e sinh[(L − z) ξ 2 + q 2 ] . × sinh(L ξ 2 + q 2 ) √2 2 . ) e−(L−ζ) ξ +q sinh[z ξ 2 + q 2 ] . + sinh(L ξ 2 + q 2 )

(5.3.85)

38 Watson, G. N., 1966: A Treatise on the Theory of Bessel Functions. Cambridge University Press, Section 13.47, Equation 2.

340

Green’s Functions with Applications

for 0 < z < L. Consequently, G(r, θ, z, s|ρ,θ $ , ζ , τ)

" e−sτ ∞ −|z−ζ|√ξ2 +q2 J0 (ξR) . = V (r, θ, z, s) + e ξ d ξ (5.3.86) 2πa2 0 ξ2 + q2 . . " e−sτ ∞ sinh[(L − ζ) ξ 2 + q 2 ] sinh(z ξ 2 + q 2 ) . . = J0 (ξR) ξ d ξ 2πa2 0 ξ 2 + q 2 sinh(L ξ 2 + q 2 ) (5.3.87) . . −sτ " ∞i 2 2 2 2 e sinh[(L − ζ) q − ξ ] sinh(z q − ξ ) . . = K0 (ξR) ξ d ξ 2 2 2π ia −∞i q 2 − ξ 2 sinh(L q 2 − ξ 2 ) (5.3.88)

for 0 < z < L. Evaluating.Equation 5.3.88 by the residue theorem where simple poles occur at ξm = q 2 + n2 π 2 /L2 , we find that F 3 G , - * ∞ e−sτ ( nπζ nπz + s n2 π 2 $ G(r, θ, z, s|ρ,θ , ζ , τ) = sin sin K0 R + . πa2 L n=1 L L a2 L2 (5.3.89) Using tables, we invert Equation 5.3.89 and obtain , - * ∞ H(t − τ ) ( nπζ nπz + g(r, θ, z, t|ρ,θ , ζ , τ) = sin sin 2πa2 L n=1 L L $

×

e−R

2

(5.3.90)

/[4a2 (t−τ )]−a2 n2 π 2 (t−τ )/L2

t−τ

.

Tables 5.3.3 through 5.3.6 list the Green’s functions39 for the heat equation involving various three-dimensional cylindrical geometries. 5.4 HEAT EQUATION WITHIN A SPHERE Let us find the Green’s function for the radially symmetric heat equation within a sphere of radius b. Mathematically, we must solve ∂g a2 ∂ 2 (rg) δ(r − ρ)δ(t − τ ) − = , ∂t r ∂r2 4πr2

0 < r, ρ < b,

0 < t, τ ,

(5.4.1)

with the boundary conditions limr→0 |g(r, t|ρ,τ )| < ∞, g(b, t|ρ,τ ) = 0, 0 < t, and the initial condition that g(r, 0|ρ,τ ) = 0, 0 < r < b. 39 Given in Carslaw, H. S., and J. C. Jaeger, 1940: The determination of Green’s function for the equation of conduction of heat in cylindrical coordinates by the Laplace transformation. J. London Math. Soc., 15, 273–281.

The Heat Equation

341

Table 5.3.7: Green’s Functions for the Three-Dimensional Heat Conduction Equation ∂g −κ ∂t

,

∂ 2 g 1 ∂g 1 ∂2g ∂2g + + 2 2+ 2 2 ∂r r ∂r r ∂θ ∂z

-

=

δ(r − ρ)δ(θ − θ $ )δ(z − ζ)δ(t) 2πr

within the Wedge 0 < θ, θ$ < θ0 , 0 < z, ζ < L with Dirichlet Boundary Conditions 1. 0 ≤ r < ∞

, - , ∞ 4 ( −κk2 π2 t/L2 kπz kπζ g= e sin sin Lθ0 L L k=1 < ∞ > " ∞ ( $ −κα2 t × sin(sθ) sin(sθ ) αJs (αρ)Js (αr)e dα , m=1

0

where s = mπ/θ 0 . 2. 0 ≤ r < a g=

, - , ∞ 8 ( −κk2 π2 t/L2 kπz kπζ e sin sin a2 Lθ0 L L k=1 < ∞ > ∞ ( ( 2 J (α r)J (α ρ) s n s n × sin(sθ) sin(sθ$ ) e−καn t , $ (α a)]2 [J n s m=1 n=1

where αn is the nth positive root of Js (αa) = 0, and s = mπ/θ 0 . 3. a < r < ∞ , - , ∞ 4 ( −κk2 π2 t/L2 kπz kπζ g= e sin sin Lθ0 L L k=1 < ∞ > " ∞ ( Us (αr)Us (αρ) −κα2 t $ × sin(sθ) sin(sθ ) α 2 e dα , Js (αa) + Ys2 (αa) 0 m=1 where Us (αr) = Js (αr)Ys (αa) − Js (αa)Ys (αr), and s = mπ/θ 0 . 4. a < r < b , - , ∞ 2π 2 ( −κk2 π2 t/L2 kπz kπζ g= e sin sin Lθ0 L L k=1 < ∞ > ∞ 2 ( ( 2 J (α b)U (α r)U (α ρ) n s n s n × sin(sθ) sin(sθ$ ) α2n s 2 e−καn t , Js (αn a) − Js2 (αn b) m=1 n=1 where Us (αr) = Js (αr)Ys (αa) − Js (αa)Ys (αr), αn is the nth positive root of Us (αb) = 0, and s = mπ/θ0 .

342

Green’s Functions with Applications

We begin by introducing the dependent variable u(r, t|ρ,τ ) = rg(r, t |ρ,τ ) so that Equation 5.4.1 becomes ∂u ∂2u δ(r − ρ)δ(t − τ ) − a2 2 = , ∂t ∂r 4πr

0 < r, ρ < b,

0 < t, τ ,

(5.4.2)

with the boundary conditions u(0, t|ρ,τ ) = u(b, t|ρ,τ ) = 0, and the initial condition that u(r, 0|ρ,τ ) = 0. Taking the Laplace transform of Equation 5.4.2, we obtain d2 U s e−sτ − 2U = − δ(r − ρ). (5.4.3) 2 dr a 4πa2 r Let us expand δ(r − ρ)/r in the Fourier sine series ∞ * nπr + δ(r − ρ) ( = Bn sin , r b n=1

(5.4.4)

where Bn =

2 b

"

b

0

* nπr + * nπρ + δ(r − ρ) 2 sin dr = sin . r b bρ b

(5.4.5)

Therefore, we rewrite Equation 5.4.3 as ∞ d2 U s e−sτ ( * nπρ + * nπr + − U = − sin sin , dr2 a2 2πa2 bρ n=1 b b

(5.4.6)

and U (r, s|ρ,τ ) =

∞ * nπρ + * nπr + e−sτ ( 1 sin sin . 2πbρ n=1 s + a2 n2 π 2 /b2 b b

(5.4.7)

Because this particular solution also satisfies the boundary conditions, we do not require any homogeneous solutions so that the sum of the particular and homogeneous solutions satisfies the boundary conditions. Indeed, this is why we choose to expand the delta function in terms of the eigenfunction sin(nπr/b). Taking the inverse of Equation 5.4.7, using the second shifting theorem and substituting for u(r, t|ρ,τ ), we finally obtain g(r, t|ρ,τ ) =

∞ H(t − τ ) ( * nπρ + * nπr + −a2 n2 π 2 (t−τ )/b2 sin sin e . 2πbrρ n=1 b b

(5.4.8)

For the case of a hollow sphere α < r < β, the Green’s function can be found by introducing the new independent variable x = r − α into the governing partial differential equation ∂g a2 ∂ 2 (rg) δ(r − ρ)δ(t − τ ) − = , 2 ∂t r ∂r 4πr2

α < r, ρ < β

0 ,< t, τ ,

(5.4.9)

The Heat Equation

343

with the boundary conditions g(α, t|ρ,τ ) = g(β, t|ρ,τ ) = 0, 0 < t, and the initial condition that g(r, 0|ρ,τ ) = 0, α < r < β, Therefore, the Green’s function in this particular case is % & % & ∞ H(t − τ ) ( nπ(ρ − α) nπ(r − α) g(r, t|ρ,τ ) = sin sin 2π(β − α)rρ n=1 β−α β−α % 2 2 2 & a n π (t − τ ) × exp − . (5.4.10) (β − α)2 Finally we find the Green’s function for a sphere that varies in the two spatial dimensions r and µ = cos(θ). The governing equation is , 2 % & ∂g ∂ g 2 ∂g a2 ∂ δ(r − ρ)δ(µ − µ$ )δ(t − τ ) 2 ∂g − a2 + − (1 − µ ) = , ∂t ∂r2 r ∂r r2 ∂µ ∂µ 2πr2 (1 − µ2 )1/2 (5.4.11) where 0 < r, ρ < b, µ$ = cos(θ $ ), and 0 < t, τ, with the boundary conditions limr→0 |g(r, µ, t|ρ, µ$ , τ )| < ∞, g(b, µ, t|ρ, µ$ , τ ) = 0, and the initial condition that g(r, µ, 0|ρ, µ$ , τ ) = 0. Our analysis begins by taking the Laplace transform of Equation 5.4.11. This yields % & ∂ 2 G 2 ∂G 1 ∂ δ(r − ρ)δ(µ − µ$ ) −sτ 2 ∂G $ + + (1 − µ ) − s G = − e , ∂r2 r ∂r r2 ∂µ ∂µ 2πr2 (1 − µ2 )1/2 (5.4.12) √ where s$ = s/a2 . Let U (r, µ, s|ρ, µ$ , τ ) = r G(r, µ, s|ρ, µ$ , τ ). Then % & ∂ 2 U 1 ∂U U 1 ∂ δ(r − ρ)δ(µ − µ$ ) −sτ 2 ∂U $ + − + (1 − µ ) − s U = − e . ∂r2 r ∂r 4r2 r2 ∂µ ∂µ 2πr3/2 (1 − µ$2 )1/2 (5.4.13) Since ∞ ( # $ δ(µ − µ$ ) . = n + 12 Pn (µ)Pn (µ$ ), (5.4.14) $2 1−µ n=0 where Pn (·) is the Legendre polynomial of order n, we have U (r, µ, s|ρ, µ$ , τ ) = e−sτ

∞ ( #

n+

n=0

1 2

$

Un (r|ρ)Pn (µ)Pn (µ$ ).

(5.4.15)

Substituting Equation 5.4.14 and Equation 5.4.15 into Equation 5.4.13, # $2 n + 12 d2 U n 1 dUn δ(r − ρ) + − U n − s$ U n = − (5.4.16) 2 2 dr r dr r 2πρ3/2 with Un (b|ρ) = 0. Solving Equation 5.4.16, we find that Un (r|ρ) =

∞ ( Jn+ 12 (λmn r)Jn+ 12 (λmn ρ) 1 , 4 52 2 1/2 πb ρ $ + λ2 ] m=1 J $ 1 (λmn b) [s mn n+ 2

(5.4.17)

344

Green’s Functions with Applications

where λmn is the mth root of Jn+ 12 (λb) = 0. Substituting Equation 5.4.17 into Equation 5.4.15 and inverting the Laplace transform, we obtain g(r, θ, t|ρ,θ $ , τ ) =

∞ $ a2 H(t − τ ) ( # n + 12 Pn (µ)Pn (µ$ ) √ πb2 rρ n=0

    ∞ ( 1 1 Jn+ 2 (λmn r)Jn+ 2 (λmn ρ) −a2 λ2 (t−τ )  mn × e , 4 52   $ m=1  Jn+ 1 (λmn b)

(5.4.18)

2

where µ = cos(θ), and µ$ = cos(θ$ ). 5.5 PRODUCT SOLUTION

So far, we used integral transforms, eigenfunction expansions and the method of images to find Green’s functions for the heat equation. We conclude this chapter by showing how products of Green’s functions from onedimensional problems can be multiplied together to give Green’s functions for two and three dimensions in rectangular and cylindrical coordinate systems. The method fails in spherical coordinates. Consider the Green’s function in a two-dimensional rectangular coordinate system ∂g ∂ 2g ∂2g − − 2 = δ(x − ξ)δ(y − η)δ(t − τ ) 2 ∂t ∂x ∂y

(5.5.1)

over some domain with linear boundary conditions and g(x, y, 0|ξ, η,τ ) = 0. We showed in the introductory remarks that we can find the Green’s function by solving the initial-value problem ∂u ∂ 2 u ∂ 2 u − 2 − 2 =0 ∂t ∂x ∂y

(5.5.2)

over the same domain with u(x, y,τ ) = δ(x−ξ)δ(y−η) and the same boundary conditions. We solve Equation 5.5.2 by assuming that u(x, y, t) = u1 (x, t)u2 (y, t). By direct substitution, u2

,

∂u1 ∂ 2 u1 − ∂t ∂x2

-

+ u1

,

∂u2 ∂ 2 u2 − ∂t ∂y 2

-

= 0.

(5.5.3)

Equation 5.5.3 is satisfied if ∂u1 ∂ 2 u1 − = 0, ∂t ∂x2

(5.5.4)

The Heat Equation

345

and

∂u2 ∂ 2 u2 − = 0. ∂t ∂y 2

(5.5.5)

Next, consider the initial conditions. Direct substitution yields u(x, y,τ ) = u1 (x,τ )u2 (y,τ ) = δ(x − ξ)δ(y − η).

(5.5.6)

Equation 5.5.6 is satisfied if we choose u1 (x,τ ) = δ(x − ξ), and u2 (y,τ ) = δ(y − η). Finally, we must examine the boundary conditions. For our problem, there are four boundaries: x = xi , and y = yi , where i = 1, 2. For the case when x is held constant, the most general boundary condition is ' ' ' ∂u ' ki '' + hi u'' = 0, (5.5.7) ∂x x=xi x=xi or

u2 or

F

G ' ' ' ∂u1 '' ki + hi u1 '' = 0, ∂x 'x=xi x=xi

ki

' ' ' ∂u1 '' ' + h u = 0, i ' ' ∂x x=xi x=xi

(5.5.8)

(5.5.9)

where we assumed that hi is constant. Similarly, at the other two boundaries, ' ' ' ∂u2 '' ' ki + h u = 0. (5.5.10) i 2' ' ∂y y=yi y=yi

Consequently, we may find u(x, y, t) by solving Equation 5.5.4 and Equation 5.5.9 with u1 (x,τ ) = δ(x − ξ), as well as Equation 5.5.5 and Equation 5.5.10 with u2 (y,τ ) = δ(y − η). This corresponds to finding the Green’s functions for the problems ∂g1 ∂ 2 g1 − = δ(x − ξ)δ(t − τ ), (5.5.11) ∂t ∂x2 with the boundary conditions ' ' ' ∂g1 '' ' ki + h g = 0, (5.5.12) i 1' ' ∂x x=xi x=xi and

∂g2 ∂ 2 g2 − = δ(y − η)δ(t − τ ), ∂t ∂y 2

(5.5.13)

with the boundary conditions ' ' ' ∂g2 '' ki + hi g2 '' = 0. ' ∂y y=yi y=yi

(5.5.14)

346

Green’s Functions with Applications

Therefore, g(x, y, t|ξ, η,τ ) = u(x, y, t)H(t − τ ) = g1 (x, t|ξ,τ )g2 (y, t|η,τ ). In general, one-dimensional Green’s functions multiply in rectangular coordinates to give multidimensional Green’s functions regardless of whether the domain is unlimited or we have Dirichlet, Neumann or Robin boundary conditions. Does this hold for cylindrical coordinates? It does, but in a more limited sense. Let us denote the one-dimensional Green’s functions in each direction by gr , gϕ and gz . Then the multi-dimensional Green’s function grz , gϕz and grϕz are given by grz = gr gz , gϕz = gϕ gz , and grϕz = grϕ gz , if we are dealing with free-space Green’s functions or have Dirichlet or Neumann boundary conditions. Note that grϕ cannot be found by product solution. We can also include Robin boundary conditions if we require that h does not vary along any of the boundaries. • Example 5.5.1 Let us find the Green’s function for , 2 ∂g ∂ g ∂2g − a2 + = δ(x − ξ)δ(y − η)δ(t − τ ), ∂t ∂x2 ∂y 2

(5.5.15)

where −∞ < x, ξ < ∞, 0 < y, η < ∞, and 0 < t, τ, subject to the boundary conditions lim g(x, y, t|ξ, η,τ ) → 0,

|x|→∞

lim g(x, y, t|ξ, η,τ ) → 0,

y→∞

(5.5.16)

and gy (x, 0, t|ξ, η,τ ) − h g(x, 0, t|ξ, η,τ ) = 0,

(5.5.17)

with the initial condition that g(x, y, 0|ξ, η,τ ) = 0. Using the method of product solutions, we express the Green’s function as a product of two Green’s functions g(x, y, t|ξ, η,τ ) = g1 (x, t|ξ,τ )g2 (y, t|η,τ ),

(5.5.18)

where g1 (x, t|ξ,τ ) is given by the one-dimensional problem ∂g1 ∂ 2 g1 − a2 2 = δ(x − ξ)δ(t − τ ), ∂t ∂x

−∞ < x, ξ < ∞,

0 < t, τ , (5.5.19)

with the boundary conditions lim|x|→∞ g1 (x, t|ξ,τ ) → 0, and the initial condition that g1 (x, 0|ξ,τ ) = 0, while g2 (y, t|η,τ ) is given by the one-dimensional problem ∂g2 ∂ 2 g2 − a2 2 = δ(y − η)δ(t − τ ), ∂t ∂y

0 < y, η < ∞,

0 < t, τ ,

(5.5.20)

The Heat Equation

347

with the boundary conditions lim g2 (y, t|η,τ ) → 0,

y→∞

g2 y (0, t|η,τ ) − h g2 (0, t|η,τ ) = 0,

(5.5.21)

and the initial condition that g2 (x, 0|η,τ ) = 0. The solution to Equation 5.5.19 through Equation 5.5.21 is the free-space Green’s function given by Equation 5.1.7. We found the Green’s function g2 (y, t|η,τ ) in Example 5.1.2. Therefore, ! % & H(t − τ ) (x − ξ)2 + (y + η)2 g(x, y, t|ξ, η,τ ) = exp − (5.5.22) 4πa2 (t − τ ) 4a2 (t − τ ) % & (x − ξ)2 + (y − η)2 + exp − 4a2 (t − τ ) % & % & ) " ∞ (x − ξ)2 (y + η + χ)2 − 2h exp − 2 exp −hχ − dχ . 4a (t − τ ) 0 4a2 (t − τ ) A trivial extension of our two-dimensional analysis to three dimensions is of considerable importance in the design of semi-conductors.40 The problem is now , 2 ∂g ∂ g ∂2g ∂2g 2 −a + 2 + 2 = δ(x − ξ)δ(y − η)δ(z − ζ)δ(t − τ ), (5.5.23) ∂t ∂x2 ∂y ∂z where −∞ < x, ξ < ∞, −∞ < y, η < ∞, 0 < z, ζ < ∞, and 0 < t, τ, subject to the boundary conditions lim g(x, y, z, t|ξ, η, ζ,τ ) → 0,

|x|→∞

lim g(x, y, z, t|ξ, η, ζ,τ ) → 0,

|y|→∞

lim g(x, y, z, t|ξ, η, ζ,τ ) → 0,

z→∞

(5.5.24) (5.5.25)

and gz (x, y, 0, t|ξ, η, ζ,τ ) − h g(x, y, 0, t|ξ, η, ζ,τ ) = 0,

(5.5.26)

with the initial condition that g(x, y, z, 0|ξ, η, ζ,τ ) = 0. If we repeat our analysis, the Green’s function becomes ! % & H(t − τ ) (x − ξ)2 + (y − η)2 + (z + ζ)2 g(x, y, z, t|ξ, η, ζ,τ ) = exp − 3/2 4a2 (t − τ ) [4πa2 (t − τ )] % & (x − ξ)2 + (y − η)2 + (z − ζ)2 + exp − 4a2 (t − τ ) % & 2 2 (x − ξ) + (y − η) − 2h exp − (5.5.27) 4a2 (t − τ ) % & ) " ∞ (z + ζ + χ)2 × exp −hχ − dχ . 4a2 (t − τ ) 0 40 See Van Roosbroeck, W., 1955: Injected current carrier transport in a semi-infinite semiconductor and the determination of lifetimes and surface recombination velocities. J. Appl. Phys., 26, 380–391.

348

Green’s Functions with Applications # "

• Example 5.5.2 A similar problem to the previous one occurs in the study of semiconductors.41 The difference here is that the domain is finite in the z-direction. Therefore, the governing equations are ∂g ∂2g ∂ 2g ∂ 2g − − 2 − 2 = δ(x − ξ)δ(y − η)δ(z − ζ)δ(t − τ ), 2 ∂t ∂x ∂y ∂z

(5.5.28)

where −∞ < x, ξ < ∞, −∞ < y, η < ∞, 0 < z, ζ < L, and 0 < t, τ, subject to the boundary conditions lim g(x, y, z, t|ξ, η, ζ,τ ) → 0,

|x|→∞

lim g(x, y, z, t|ξ, η, ζ,τ ) → 0,

|y|→∞

(5.5.29)

gz (x, y, 0, t|ξ, η, ζ,τ ) − h g(x, y, 0, t|ξ, η, ζ,τ ) = 0,

(5.5.30)

gz (x, y, L, t|ξ, η, ζ,τ ) + h g(x, y, L, t|ξ, η, ζ,τ ) = 0,

(5.5.31)

and with the initial condition that g(x, y, z, 0|ξ, η, ζ,τ ) = 0. Using the method of product solutions, we express the Green’s function as a product of three Green’s functions g(x, y, z, t|ξ, η, ζ,τ ) = g1 (x, t|ξ,τ )g2 (y, t|η,τ )g3 (z, t|ζ,τ ).

(5.5.32)

The product solutions g1 (x, t|ξ,τ ) and g2 (y, t|η,τ ) are given by the onedimensional problems: ∂g1 ∂ 2 g1 − = δ(x − ξ)δ(t − τ ), ∂t ∂x2

−∞ < x, ξ < ∞,

0 < t, τ , (5.5.33)

with the boundary conditions lim|x|→∞ g1 (x, t|ξ,τ ) → 0 and the initial condition that g1 (x, 0|ξ,τ ) = 0, and ∂g2 ∂ 2 g2 − = δ(y − η)δ(t − τ ), ∂t ∂y 2

−∞ < y, η < ∞,

0 < t, τ , (5.5.34)

with the boundary conditions lim|y|→∞ g2 (y, t|η,τ ) → 0, and the initial condition that g2 (y, 0|η,τ ) = 0. On the other hand, for g3 (z, t|ζ,τ ) we have the boundary-value problem ∂g3 ∂ 2 g3 − = δ(z − ζ)δ(t − τ ), ∂t ∂z 2

0 < z, ζ < L,

0 < t, τ ,

(5.5.35)

41 Visvanathan, S., and J. F. Battey, 1954: Some problems in the diffusion of minority carriers in a semiconductor. J. Appl. Phys., 25, 99–102.

The Heat Equation

349

with the boundary conditions g3 z (0, t|ζ,τ ) − h g3 (0, t|ζ,τ ) = 0,

(5.5.36)

g3 z (L, t|ζ,τ ) + h g3 (L, t|ζ,τ ) = 0,

(5.5.37)

and and the initial condition that g3 (z, 0|ζ,τ ) = 0. The solutions to Equation 5.5.33 through Equation 5.5.37 are % & H(t − τ ) (x − ξ)2 + (y − η)2 g1 (x, t|ξ,τ )g2 (y, t|η,τ ) = exp − , (5.5.38) 4π(t − τ ) 4(t − τ ) and g3 (z, t|ζ,τ ) = 2H(t − τ )

∞ ( [kn cos(kn z) + h sin(kn z)][kn cos(kn ζ) + h sin(kn ζ)]

L(kn2 + h2 ) + 2h

n=1

2

× e−kn (t−τ ) ,

(5.5.39)

where kn is the nth solution of tan(kL) = 2kh/(k 2 − h2 ). The solution for g3 (z, t|ζ,τ ) was found using Laplace transforms. Therefore, the Green’s function is % & H(t − τ ) (x − ξ)2 + (y − η)2 g(x, y, z, t|ξ, η, ζ,τ ) = exp − 2π(t − τ ) 4(t − τ ) ∞ ( [kn cos(kn z) + h sin(kn z)][kn cos(kn ζ) + h sin(kn ζ)] × L(kn2 + h2 ) + 2h n=1 2

× e−kn (t−τ ) .

(5.5.40) # "

• Example 5.5.3 Using the product solution method, we find in this example the Green’s function42 for % , & ∂g 1 ∂ ∂g ∂ 2g δ(r − ρ)δ(z − ζ)δ(t − τ ) − a2 r + 2 = , (5.5.41) ∂t r ∂r ∂r ∂z 2πr where 0 < r, ρ < ∞, 0 < z, ζ < ∞, and 0 < t, τ. The boundary conditions are lim |g(r, z, t|ρ, ζ,τ )| < ∞,

r→0

lim g(r, z, t|ρ, ζ,τ )| → 0,

r→∞

(5.5.42)

42 For an example of its use, see Nelson, D. J., and B. Vick, 1995: Thermal limitations in optical recording. IEEE Trans. Comp., Packag., and Manufact. Technol.—Part A, 18, 521–526.

350

Green’s Functions with Applications gz (r, 0, t|ρ, ζ,τ ) = 0,

and

lim g(r, z, t|ρ, ζ,τ ) → 0.

z→∞

(5.5.43)

We begin by finding a product solution of the form

where

g(r, z, t|ρ, ζ,τ ) = g1 (r, t|ρ,τ )g2 (z, t|ζ,τ ),

(5.5.44)

, ∂g1 a2 ∂ ∂g1 δ(r − ρ)δ(t − τ ) − r = , ∂t r ∂r ∂r 2πr

(5.5.45)

with lim |g1 (r, t|ρ,τ )| < ∞,

and

r→0

and

lim g1 (r, t|ρ,τ ) → 0,

r→∞

∂g2 ∂ 2 g2 − a2 2 = δ(z − ζ)δ(t − τ ), ∂t ∂z

with

∂g2 (0, t|ζ,τ ) = 0, ∂z

and

lim g2 (z, t|ζ,τ ) → 0.

z→∞

(5.5.46)

(5.5.47)

(5.5.48)

As you will show in Problem 9, g1 (r, t|ρ,τ ) =

% & % & H(t − τ ) r2 + ρ2 rρ exp − I . 0 4πa2 (t − τ ) 4a2 (t − τ ) 2a2 (t − τ )

(5.5.49)

Using the method of images, we have that g2 (z, t|ζ,τ ) = .

! % & % &) (z − ζ)2 (z + ζ)2 exp − 2 + exp − 2 . 4a (t − τ ) 4a (t − τ ) 4πa2 (t − τ ) (5.5.50) H(t − τ )

Therefore,

% & % & H(t − τ ) r 2 + ρ2 rρ exp − I 0 4a2 (t − τ ) 2a2 (t − τ ) [4πa2 (t − τ )]3/2 ! % & % &) (z − ζ)2 (z + ζ)2 × exp − 2 + exp − 2 . (5.5.51) 4a (t − τ ) 4a (t − τ )

g(r, z, t|ρ, ζ,τ ) =

5.6 ABSOLUTE AND CONVECTIVE INSTABILITY One of the striking features of dynamical flows is the propensity of some of them to be unstable to small perturbations. This is especially true of fluids where the study of hydrodynamic instabilities dates back to the nineteenth century and involves such luminaries as Helmholtz, Kelvin, Rayleigh and G. I. Taylor.

The Heat Equation

351

Figure 5.6.1: Plot of u(x, t) given by Equation 5.6.5 as a function of distance x and time t when a = 12 and b = 1. This is an example of an absolutely unstable solution.

The analysis begins by deriving a set of linearized equations that describe the response of a simple basic state that satisfies the governing equations to small perturbations. The partial differential equation ∂u ∂ 2u ∂u = +a + u, 2 ∂t ∂x ∂x

−∞ < x < ∞,

0 < t,

(5.6.1)

where a is real, is prototypical43 of the perturbation equations that exhibit instability. Because of its homogeneity in space and time, solutions to Equation 5.6.1 can be expressed as a superposition of normal modes, namely u(x, t) = Aei(kx−λt) .

(5.6.2)

Upon substituting Equation 5.6.2 into Equation 5.6.1, we find that the relationship between the frequency λ and wavenumber k is governed by λ = −ak + i(1 − k 2 ).

(5.6.3)

In classical stability calculations, instability occurs if for some real wavenumber k a complex λ with positive 5(λ) is obtained from the dispersion equation and we have temporal growth of spatially periodic disturbances of infinite extent. For our particular example, instability occurs when −1 < k < 1, and 5(λ) = 1 − k 2 > 0. 43 This is a simplified version of the linearized Ginzburg-Landau equation. In the case of fluids, the Ginzburg-Landau equation governs the amplitude of perturbations in a Stewartson-Stuart [Stewartson, K., and J. T. Stuart, 1971: A non-linear instability theory for a wave system in plane Poiseuille flow. J. Fluid Mech., 48, 529–545. See their equation (4.9).] stability calculation of a fluid flow.

352

Green’s Functions with Applications

Figure 5.6.2: Same as Figure 5.6.1 except that a = 4. This is an example of a convectively unstable solution.

During the 1950s, plasma physicists further refined this concept of stability. This was necessary because most initial perturbations are not monochromatic and their presence, even if the normal modes were unstable, would not ensure instability, because they might simply propagate away, leaving nothing. In this sense, the system would be stable in the sense that the amplitude in a fixed, local region eventually decays. To illustrate this, consider an initial pulse that consists of the Gaussian distribution u(x, 0) = e−bx

2

/4

,

b > 0.

The exact solution of Equation 5.6.1 in this case is % & et b(x + at)2 u(x, t) = √ exp − . 4(1 + bt) 1 + bt

What happens as t → ∞? At x = 0, % & et ba2 t2 u(0, t) = √ exp − 4(1 + bt) 1 + bt ! ) # $2 1 t 1 = √ exp 1 + bt 1 − 14 a2 . 1 + bt 1 + bt

(5.6.4)

(5.6.5)

(5.6.6) (5.6.7)

In the limit t → ∞, u(0, t) will grow in time without limit if |a| < 2. This is also true of any other fixed point in the absolute frame of reference. Such an instability is called an absolute or globalized instability. See Figure 5.6.1. This type of instability is generally catastrophic, growing exponentially in time at all points in space leading to immediate and dramatic changes in the flow. On the other hand, if |a| > 2, u(0, t) decays exponentially to zero and we have absolute stability.

The Heat Equation

353

If we now introduce a moving reference frame with velocity −a, then Equation 5.6.5 grows exponentially in time for any real a. Consequently, if the convection speed |a| is large enough, the instability in the absolute reference frame is localized in space around the moving point x = −at. In this case, the group velocity of the disturbance is greater than the speed at which it spreads. This type of instability is called a convective or localized instability.44 See Figure 5.6.2. Because a convective or localized instability can convect away from a region of interest, it may be possible to control. The fact that a flow may not be absolutely unstable but convectively unstable has had a profound effect on our concept of what stability and instability are. The application of Green’s function to this problem yields a clean and concise understanding of this phenomenon because of the simple nature of the forcing. Consider the Green’s function associated with the partial differential equation, Equation 5.6.1, ∂g ∂ 2 g ∂g − −a −g = δ(x− ξ)δ(t − τ ), ∂t ∂x2 ∂x

−∞ < x, ξ < ∞,

0 < t, τ . (5.6.8)

Our study of instability will focus on this two-dimensional partial differential equation, which is invariant with respect to the time displacement and uniform in the spatial dimension.45 Upon applying Laplace transforms in time and Fourier transforms in space, the joint transform G(k, s|ξ,τ ) is (s + k 2 − iak − 1)G(k, s|ξ,τ ) = e−ikξ−sτ ,

(5.6.9)

and 1 g(x, t|ξ,τ ) = 4π 2 i

"

c+∞i " ∞

c−∞i

−∞

eik(x−ξ)+s(t−τ ) dk ds. s + k 2 − iak − 1

(5.6.10)

To compute g(x, t|ξ,τ ), we first invert the Laplace transform in Equation 5.6.10, which yields " H(t − τ ) ∞ ik(x−ξ)−(k2 −iak−1)(t−τ ) g(x, t|ξ,τ ) = e dk. (5.6.11) 2π −∞ Performing the k-integration, we obtain g(x, t|ξ,τ ) =

44

% & 2 e(t−τ )−a (t−τ )/4−a(x−ξ)/2 (x − ξ)2 . exp − H(t − τ ). (5.6.12) 4(t − τ ) 2 π(t − τ )

Some authors prefer spatial instability because it frees the phenomenon from its association with just fluids.

45 Equations that are nonuniform in space have been explored by Brevdo, L., and T. J. Bridges, 1996: Absolute and convective instabilities of spatially periodic flows. Phil. Trans. R. Soc. Lond., A354, 1027–1064; Huerre, P., and P. A. Monkewitz, 1990: Local and global instabilities in spatial developing flows. Ann. Rev. Fluid Mech., 22, 473–537.

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Green’s Functions with Applications

Once again, we see that if |a| > 2, the solution will decay to zero for any fixed value of x as t → ∞. Conversely, if |a| < 2, there is exponential growth for any fixed value of x and we have absolute instability. So far, we have shown that the Green’s function formulation provides a straightforward method for determining the stability of Equation 5.6.1. At this point, we would like to extend this method so that it applies to the general problem when we cannot write g(x, t|ξ,τ ) in an analytic form, as we just did. The remaining portions of this section is devoted to developing analytic tools for the treatment of this general case. We begin by noting that any linear, constant coefficient partial differential equation that varies in x and t has the Green’s function46 "

1 g(x, t|ξ,τ ) = 4π 2 i or g(x, t|ξ,τ ) =

1 4π 2

"

c+∞i

−∞+iσ

∞

eik(x−ξ)+s(t−τ ) dk ds, D(k, s)

(5.6.13)

eik(x−ξ)−iω(t−τ ) dk dω, D(k,ω )

(5.6.14)

−∞

c−∞i

∞+iσ

"

"

∞

−∞

where we have rewritten the Laplace inversion in Equation 5.6.14 as an inversion of a Fourier transform by letting s = −ωi. The contour in the complex frequency plane, commonly called the Bromwich contour, is a straight horizontal line located above all of the singularities of the integrand as dictated by causality, which requires that g(x, t|ξ,τ ) = 0 for all x when t < τ . Hence, the σ in Equation 5.6.14 is greater than the imaginary part of any ω given by D(k,ω ) = 0. The path in the complex wavenumber plane is initially taken along the real axis. For simplicity, let us assume that Equation 5.6.14 has a single discrete temporal mode ω(k). Then the Green’s function g(x, t|ξ,τ ) can be formally obtained from a residue calculation in the ω-plane at ω = ω(k). Carrying out the calculation, we have that i g(x, t|ξ,τ ) = − H(t − τ ) 2π

"

∞ −∞

ei[k(x−ξ)−ω(k)(t−τ )] dk. Dω [k,ω (k)]

(5.6.15)

We can quickly check Equation 5.6.15 by using our earlier example. In this case, Dω [k,ω (k)] = −i, and ω(k) = −ak + i(1 − k 2 ). Substituting into Equation 5.6.15, we recover Equation 5.6.11. Ideally, we would like to evaluate Equation 5.6.15 exactly as we did Equation 5.6.11. Unfortunately, most of the time we cannot and must pursue another avenue of attack. Presently, we choose to compute the Fourier integral 46 See Barlow, N. S., B. T. Helenbrook, S. P. Lin, and S. J. Weinstein, 2010: An interpretation of absolutely and convectively unstable waves using series solutions. Wave Motions, 47, 564–582.

The Heat Equation

355

by the method of steepest descent47 for large t and fixed (x − ξ)/(t − τ ). Although the exact steepest descent path depends upon the particular form of ω(k), we presently assume that there is only a single stationary point k∗ for the phase of the integrand where ∂ω(k∗ )/∂k = (x − ξ)/(t − τ ). Under this condition, the original contour of integration along the real k-axis can be deformed into a steepest descent path passing through the saddle point k∗ . Performing the calculation, eπi/4 ei[k∗ (x−ξ)−ω(k∗ )(t−τ )] . g(x, t|ξ,τ ) ∼ − √ . 2π Dω [k∗ , ω(k∗ )] ω $$ (k∗ )(t − τ )

(5.6.16)

For instance, in our present example, ω $$ (k∗ ) = −2i, ai i k∗ = + 2 2 and ω(k∗ ) = i −

,

x−ξ t−τ

a2 i i + 4 4

,

-

,

x−ξ t−τ

(5.6.17) -2

.

(5.6.18)

Upon substituting these results into Equation 5.6.16, the asymptotic evaluation of Equation 5.6.15 equals the exact solution, Equation 5.6.12; this is generally not true. Although Equation 5.6.16 is only the asymptotic evaluation of Equation 5.6.15, that is sufficient for our stability analysis. Examining Equation 5.6.16 more closely, we see that the Green’s function behaves like a wave packet in the (x, t)-plane. Along each ray (x− ξ)/(t− τ ) within the packet, the response is dominated by a specific complex wavenumber k∗ such that its real group velocity satisfies ωk (k∗ ) = (x−ξ)/(t−τ ). Along each ray, the temporal growth rate is 5 [ω(k∗ ) − (x − ξ)k∗ /(t − τ )]. Let us return to Equation 5.6.15. As k varies from −∞ to ∞, 5[ω(k)] also varies. Let us denote by kmax the wavenumber where 5[ω(k)] has its maximum value, ωi,max = 5[ω(kmax )]. Clearly, for the system to be unstable, 5[ω(k)] > 0 at k = kmax and we arrive at the following criteria for linear instability: • If ωi,max > 0, then the system is linearly unstable; • If ωi,max < 0, then the system is linearly stable. 47

Lingwood [Lingwood, R. J., 1997: On the application of the Briggs’ and steepestdescent methods to a boundary-layer flow. Stud. Appl. Math., 98, 213–254.] provided examples of saddle points through which the steepest descent path cannot be made to pass. For this reason, care must be taken to establish the topography of the phase function. Otherwise, calculations of the wave packet evolution can be seriously flawed and incorrect conclusions can be drawn about the convective or absolute instability of the flow.

356

Green’s Functions with Applications

Although we now have criteria for instability, how do we discern an absolute instability from a convective instability? Recall that the difference between them involves the behavior of a fixed point x − ξ as t − τ → ∞: absolute instability occurs when the solution grows with time, while convective instability occurs when the solution damps. Consequently, if we examine Equation 5.6.16 along the ray (x−ξ)/(t−τ ) = 0 as t−τ → ∞, we can develop a criterion. Let us denote the k∗ for the (x − ξ)/(t − τ ) = 0 ray by k0 and the corresponding ω(k∗ ) by ω0 = ω(k0 ). Then the absolute growth rate is ω0,i = 5[ω(k0 )]. From Equation 5.6.16 it follows that • If ω0,i > 0, the system is absolutely unstable; • if ω0,i < 0, the system is convectively unstable. We may illustrate these concepts with our earlier example involving Equation 5.6.8. Because ω(k) = −ak + i(1 − k 2 ), kmax = 0 and ωi,max = 1. The system is unstable for any value of a. Turning to the criteria for convective/absolute instability, a quick calculation shows that k0 = ai/2 from Equation 5.6.17 since (x − ξ)/(t − τ ) = 0; ω0 = ω(k0 ) = i(1 − a2 /4) and ω0,i = 1 − a2 /4 from Equation 5.6.18. Therefore, if |a| > 2, we have convective instability; if |a| < 2, we have absolute instability. This method of determining absolute/convective instability from ω0,i may not be precise enough when ω(k) is particularly complicated. For example, ω(k) may be multivalued so that we must consider branch cuts and Riemann surfaces. To formulate alternative criteria, consider the case where the temporal mode ω(k) exhibits a single second-order algebraic branch point ω0 with only two spatial branches k+ (ω) and k− (ω). The example that we have been using is just such a case because ω − ω0 = −i(k − k0 )2 and ω0 , the absolute frequency, is an algebraic branch point of the function k(ω) on the ω-plane; k0 is a saddle point of ω(k). Let us now follow the loci of k+ (ω) and k− (ω) as ω travels along the Bromwich contour and we lower the contour to the real axis of the ω-plane. We do this so that we can discover the asymptotic response of the inverse Laplace transform; the response in the limit t → ∞ is governed by the highest singularity in D(k,ω ) in the ω-plane. When the Bromwich contour is high enough, the spatial branches k+ and k− are located on opposite sides of the real k-axis. See the solid lines in Figure 5.6.3. As the Bromwich integral is displaced downward, k+ and k− move toward each other and the original path along the real k-axis in the Fourier inversion must be deformed to remain between them. This simultaneous deformation of Bromwich and Fourier contours cannot proceed further if the Bromwich contour touches ω0 and Fourier contour becomes pinched between the branches k+ (ω) and k− (ω). See the dotted lines in Figure 5.6.3. Pinching occurs precisely at the point k0 where the group velocity equals zero. If the corresponding ω0 is located in the upper

The Heat Equation

357

ki

i

k+ kr r k_

i

ki

k+ kr

r k_

Figure 5.6.3: The loci of k+ and k− for our sample problem as ((ω) varies for various fixed )(ω). As )(ω) decreases (first the solid line, then the dashed line, and finally the dotted line), k+ and k− move toward each other, finally pinching at a single point. In the top frame, the case for absolute instability, a = −1. In the bottom frame, the case for convective instability, a = −4.

half of the ω-plane, as is the case for the top frame shown in Figure 5.6.3, the instability is absolute; otherwise, it is convective. This graphical method of determining absolute or convective instability is commonly called the Briggs’ criterion.48 • Example 5.6.1: Absolute instability Consider the one-dimensional Klein-Gordon equation ∂ 2g 2 ∂ 2g −c −βg = δ(x−ξ)δ(t−τ ), ∂t2 ∂x2

−∞ < x, ξ < ∞,

0 < t, τ , (5.6.19)

which we studied in Example 4.1.1. Applying the asymptotic approximation for the Bessel function to Equation 4.1.31, we find that √ g(x, t|ξ,τ ) ∼ e β t , β 0,> (5.6.20) and

48

2.

*. + g(x, t|ξ,τ ) ∼ t−1/2 cos −β t ,

β

0,
0. If c2 k 2 > β, then the system in stable. On the other hand, if c2√ k 2 < β, then the system is unstable; the maximum growth rate is ωi,max = β when kmax = 0. This agrees with Equation 5.6.20. To determine whether this instability is absolute √ or convective we note that k0 = 0 since ∂ω/∂k = 0 there. Hence, ω0 = i β. Because ω0,i > 0, the system is absolutely unstable. # " • Example 5.6.2: The Ginzburg-Landau equation The time-dependent, generalized Ginzburg-Landau equation is a simplified version of the dynamical equations for fluid mechanical systems such as Rayleigh-B´enard convection, Taylor-Couette flow, plane Poiseuille flow and wind-generated water waves. If a, b, and c denote complex constants with 3(b) ≥ 0, the Ginzburg-Landau equation is ∂u ∂u ∂2u + vg − au − b 2 + c|u|2 u = 0, ∂t ∂x ∂x

(5.6.22)

where vg denotes the group velocity. Consider the free-space Green’s function for the linearized form of Equation 5.6.22: ∂g ∂g ∂ 2g + vg − ag − b 2 = δ(x − ξ)δ(t − τ ). ∂t ∂x ∂x

(5.6.23)

The last term in Equation 5.6.22 disappears because we linearized about u = 0. The Green’s function is ! ) ea(t−τ ) H(t − τ ) [x − ξ − vg (t − τ )]2 . g(x, t|ξ,τ ) = exp − . (5.6.24) 4b(t − τ ) 2 πb(t − τ )

In the development of our criteria for absolute and convective instability, we focused on the dispersion relationship. We will do so momentarily. Before that, we will first examine an alternative definition of these instabilities given by Bers.49 They are based more on physical intuition. 49 Bers, A., 1983: Space-time evolution of plasma instabilities—absolute and convective. Basic Plasma Physics, Vol. I , A. A. Galeev and R. N. Sudan, Eds., North-Holland, 451– 517.

The Heat Equation

359

Bers states that in plasma problems there are three different types of stability possible. The first type is defined by lim |g(x, t|ξ,τ )| → ∞

(5.6.25)

t→∞

for an arbitrary, fixed value of x. This condition corresponds to absolute instability. Checking Equation 5.6.24, Equation 5.6.25 will be satisfied if F

vg2 3 a− 4b

G

> 0,

or

3(a) −

vg2 3(b) > 0. 4|b|2

(5.6.26)

Because 3(b) > 0, absolute instability occurs here if 3(a) is sufficiently large. The second type of stability is defined by lim g(Ξ+ vt, t|ξ,τ ) → 0

(5.6.27)

t→∞

for some v and for arbitrary fixed value of Ξ. This condition corresponds to absolute stability and disturbances will die out in any frame of reference. From Equation 5.6.24, this will occur when %

& (v − vg )2 3 a− < 0. 4b

(5.6.28)

Therefore, regardless of the values of v and vg , the flow will be absolutely stable if 3(a) < 0. Finally, the third type of stability is dictated by lim g(x, t|ξ,τ ) → 0,

t→∞

and

lim |g(Ξ+ vt, t|ξ,τ )| → ∞

t→∞

(5.6.29)

for some v and for arbitrary fixed values of x and Ξ, respectively. For the present problem, this occurs for 0 < 3(a) < vg2 3(b)/(4|b|2 ). This condition corresponds to convective or spatial instability. In this case, the disturbance damps as t → ∞ at any given stationary point but a moving frame of reference may be found in which the disturbance grows. As the disturbance convects away, the system returns to its original, undisturbed state. Let us now determine the stability criterion from our pulse analysis. The temporal growth rate along any ray (x − ξ)/(t − τ ) is 5 [ω(k∗ ) − (x − ξ)k∗ /(t − τ )] ,

(5.6.30)

where k∗ satisfies ωk (k∗ ) = (x − ξ)/(t − τ ). In our current problem, k∗ = −

vg i i + 2b 2b

,

x−ξ t−τ

-

,

(5.6.31)

360

Green’s Functions with Applications

and : , & , -2 ; vg2 x−ξ vg x − ξ 1 x−ξ 3(b) 5 ω(k∗ ) − k∗ = 3(a) + − − . t−τ 2 t−τ 4 4 t−τ |b|2 (5.6.32) Maximizing Equation 5.6.32 with respect to (x − ξ)/(t − τ ) gives (x − ξ)/(t − τ) = vg . Along this ray [(x−ξ)/(t−τ )]max there is maximum temporal growth that equals 3(a). Thus, the system becomes convectively unstable whenever 3(a) > 0, the same result that we obtained earlier. In this case, there is a reference frame moving at the group velocity in which a localized disturbance will appear to grow although in the stationary frame (x − ξ)/(t − τ ) = 0, it decays to zero. For 3(a) > 0, there exists a pair of values, [(x − ξ)/(t − τ )]slow and [(x − ξ)/(t − τ )]f ast , on either side of [(x − ξ)/(t − τ )]max for which Equation 5.6.32 returns to zero. These values of (x − ξ)/(t − τ ) represent fronts of the disturbance along which the system is marginally stable. Eventually, for some value of a, [(x − ξ)/(t − τ )]slow = 0 and the system becomes absolutely unstable. From Equation 5.6.32 this occurs when %

3(a) −

vg2 3(b) = 0. 4|b|2

(5.6.33)

When Equation 5.6.33 is satisfied, the growth rate is simply 5[ω(k∗ )] for any fixed x − ξ as t → ∞. Any localized disturbance will grow exponentially with time at its point of origin even though the disturbance itself may propagate away as a wave packet. On the other hand, for 0 < 3(a) < vg2 3(b)/(4|b|2 ), the disturbance propagates away quickly enough relative to its growth and spreading so that the point of origin settles back to its undisturbed state. In summary, we see that as the value of 3(a) changes, so does the behavior of the solution. For 3(a) < 0, the flow is absolutely stable. As soon as 3(a) > 0, the flow becomes convectively or spatially unstable. Eventually, for large enough and positive 3(a), the flow is absolutely unstable. Problems

1. Following the derivation in the introduction of this chapter, show that the solution to the one-dimensional heat equation ∂u ∂2u − a2 2 = q(x, t), ∂t ∂x

a < x < b,

0 < t,

The Heat Equation

361

is given by "

u(x, t) =

0

+a

2

"

+

"

t+

"

b

q(ξ,τ )g(x, t|ξ,τ ) dξdτ

a

0

b

t+ %

∂u(ξ,τ ) ∂g(x, t|ξ,τ ) g(x, t|ξ,τ ) − u(ξ,τ ) ∂ξ ∂ξ

&ξ=b

dτ

ξ=a

u(ξ, 0)g(x, t|ξ, 0) dξ, a

where g(x, t|ξ,τ ) is the one-dimensional Green’s function governed by ∂g ∂2g − a2 2 = δ(x − ξ)δ(t − τ ), ∂t ∂x

a < x, ξ < b,

0 < t, τ .

2. Following the derivation in the introduction of this chapter, show that the solution to the radial heat equation , ∂u a2 ∂ ∂u − r = q(r, t), a < r < b, 0 < t, ∂t r ∂r ∂r is given by u(r, t) =

"

+a +

t+ 0

2

"

"

0

b

"

b

q(ρ,τ )g(r, t|ρ,τ ) ρ d ρd τ

a

t+ %

∂u(ρ,τ ) ∂g(r, t|ρ,τ ) ρ g(r, t|ρ,τ ) − ρ u(ρ,τ ) ∂ρ ∂ρ

&ρ=b

dτ

ρ=a

u(ρ, 0)g(r, t|ρ, 0) ρ dρ ,

a

where g(r, t|ρ,τ ) is the one-dimensional Green’s function governed by , ∂g a2 ∂ ∂g δ(r − ρ) − r = δ(t − τ ), a < r, ρ < b, 0 < t, τ . ∂t r ∂r ∂r r Unlimited Domains 3. Find the free-space Green’s function for the linearized Ginzburg-Landau equation50 ∂g ∂g ∂ 2g +v − ag − b 2 = δ(x − ξ)δ(t − τ ), ∂t ∂x ∂x

−∞ < x, ξ < ∞, 0 < t, τ ,

50 See Deissler, R. J., 1985: Noise-sustained structure, intermittency, and the GinzburgLandau equation. J. Stat. Phys., 40, 371–395.

362

Green’s Functions with Applications

with b > 0. Step 1 : Taking the Laplace transform of the partial differential equation, show that it reduces to the ordinary differential equation b

d2 G dG −v + aG − sG = −δ(x − ξ)e−sτ . 2 dx dx

Step 2 : Using Fourier transforms, show that G(x, s|ξ,τ ) =

e−sτ 2π

"

∞ −∞

eik(x−ξ) dk, s + ikv + bk 2 − a

or g(x, t|ξ,τ ) =

ea(t−τ ) H(t − τ ) π

"

∞ 0

2

e−b(t−τ )k cos{[x − ξ − v(t − τ )]k} dk.

Step 3 : Evaluate the second integral and show that ! ) ea(t−τ ) H(t − τ ) [x − ξ − v(t − τ )]2 . g(x, t|ξ,τ ) = exp − . 4b(t − τ ) 2 πb(t − τ )

4. Show that the free-space Green’s function51 governed by the equation ∂g ∂ 2g ∂(xg) − − = δ(x − ξ)δ(t − τ ) ∂t ∂x2 ∂x is given by g(x, t|ξ,τ ) = .

, H(t − τ ) {x − ξ exp[−(t − τ )]}2 exp − . 2{1 − exp[−2(t − τ )]} 2π{1 − exp[−2(t − τ )]}

The figure captioned Problem 4 illustrates this Green’s function as functions of x and t − τ with ξ = 2. 5. Find the free-space Green’s function52 governed by ∂g ∂ 2g ∂g − a2 2 + b sin(ωt + ϕ) = δ(x − ξ)δ(t − τ ), −∞ < x, ξ < ∞, 0 < t, τ , ∂t ∂x ∂x 51 See Mahajan, S. M., P. M. Valanju, and W. L. Rowan, 1992: Closed, form invariant solutions of convective-diffusive systems with applications to impurity transport. Phys. Fluids B, 4, 2495–2498. 52 See Zon, B. A., 1974: Forced diffusion in an alternating external field. Sov. Phys. J., 17, 1594–1595.

The Heat Equation

363

Problem 4

subject to the boundary conditions lim|x|→∞ g(x, t|ξ,τ ) → 0, 0 < t, and the initial condition that g(x, 0|ξ,τ ) = 0, −∞ < x < ∞. Step 1 : Taking the Fourier transform of the partial differential equation, show that the Fourier transform of g(x, t|ξ,τ ) is governed by the ordinary differential equation dG + a2 k 2 G + ikb sin(ωt + ϕ)G = e−ikξ δ(t − τ ). dt Step 2 : Show that ! G(k, t|ξ,τ ) = H(t − τ ) exp − ikξ − a2 k 2 (t − τ )

) ikb + [cos(ωt + ϕ) − cos(ωτ + ϕ)] . ω

Step 3 : Show that g(x, t|ξ,τ ) equals ! " H(t − τ ) ∞ g(x, t|ξ,τ ) = exp ik(x − ξ) − a2 k 2 (t − τ ) 2π −∞ ) ikb + [cos(ωt + ϕ) − cos(ωτ + ϕ)] dk ω H(t − τ ) =. 4πa2 (t − τ ) % & {x − ξ + b[cos(ωt + ϕ) − cos(ωτ + ϕ)]/ω}2 × exp − . 4a2 (t − τ )

364

Green’s Functions with Applications

Problem 6

6. Show that the free-space Green’s function51 governed by the equation , ∂g ∂ ∂g ∂(xg) − x − = δ(x − ξ)δ(t − τ ), 0 < x, ξ, t, τ , ∂t ∂x ∂x ∂x is given by g(x, t|ξ,τ ) = e−x H(t − τ )

∞ (

e−n(t−τ )Ln (ξ)Ln (x)

n=0

! ) H(t − τ ) x + ξ exp[−(t − τ )] = exp − 1 − exp[−(t − τ )] 1 − exp[−(t − τ )] < . > xξ exp[−(t − τ )] × I0 2 , 1 − exp[−(t − τ )]

where Ln (·) denotes a Laguerre polynomial of order n. The figure captioned Problem 6 illustrates this Green’s function as functions of x and t − τ with ξ = 2. 7. Find the free-space Green’s function53 governed by ∂ 2g ∂2g ∂g + U (t) 2 + a + bg = δ(x − ξ)δ(t − τ ), ∂x∂t ∂x ∂x

−∞ < x, ξ < ∞, 0 < t, τ ,

with b > 0, subject to the boundary conditions lim|x|→∞ g(x, t|ξ,τ ) → 0, 0 < t, and the initial condition that g(x, 0|ξ,τ ) = 0, −∞ < x < ∞. 53 See Kozlov, V. F., 1980: Formation of a Rossby wave under the action of disturbances in a nonstationary barotropic oceanic flow. Izv. Atmos. Oceanic Phys., 16, 275–279.

The Heat Equation

365

Step 1 : Taking the Fourier transform of the governing equation, show that it becomes dG ik − k 2 U (t)G + iakG + bG = e−ikξ δ(t − τ ), dt with G(k, 0|ξ,τ ) = 0. Step 2 : Defining s(t) = G(k, t|ξ,τ ) =

/t 0

U (t$ ) dt$ , show that

e−ikξ H(t − τ ) exp{−ik[s(t) − s(τ )] − a(t − τ ) − b(t − τ )/(ik)}. ik

Step 3 : Taking the inverse Fourier transform, show that e−a(t−τ ) H(t − τ ) π

"

∞

dk sin[ηk + b(t − τ )/k] k 0 . −a(t−τ ) =e H(t − τ )H(η)J0 [2 ηb(t − τ ) ],

g(x, t|ξ,τ ) =

where η = x − ξ − s(t) + s(τ ).

8. Find the Green’s function54 governed by ∂ 2g ∂2g ∂g ∂g +a −b −c = −δ(x−ξ)δ(t−τ ), ∂x2 ∂x∂t ∂t ∂x

−∞ < x, ξ < ∞,

0 < t, τ ,

with the boundary conditions lim|x|→∞ g(x, t|ξ,τ ) → 0, 0 < t, and the initial condition that g(x, 0|ξ,τ ) = 0, −∞ < x < ∞, where a > 0. Step 1 : Taking the Laplace transform of the partial differential equation and boundary conditions, show that d2 G dG + (as − c) − bsG = −δ(x − ξ)e−sτ 2 dx dx with lim|x|→∞ G(x, s|ξ,τ ) → 0. Step 2 : Solve for G(x, s|ξ,τ ) and show that G(x, s|ξ,τ ) =

4 a 5 . e−sτ −(as−c)(x−ξ)/2 . exp − |x − ξ| (s + α)2 − σ 2 , 2 a (s + α)2 − σ 2

where α = (2b − ac)/a2 and σ 2 = α2 − c2 /a2 .

54 See Liu, Y., and D. Gidaspow, 1981: Solids mixing in fluidized beds–A hydrodynamic approach. Chem. Engng. Sci., 36, 539–547.

366

Green’s Functions with Applications

Step 3 : Invert G(x, s|ξ,τ ) and show that g(x, t|ξ,τ ) =

R 4 5S 1 c(x−ξ)/2 a e exp −α t − τ − (x − ξ) a 2 4 . 5 2 × I0 σ (t − τ ) − a(t − τ )(x − ξ) H[t − τ − a(x − ξ)/2].

9. Find the free-space Green’s function55 governed by ∂g − a2 ∂t

,

∂ 2g 1 ∂g + 2 ∂r r ∂r

-

=

δ(r − ρ)δ(t − τ ) , r

0 < r, ρ < ∞,

0 < t, τ ,

subject to the boundary conditions lim |g(r, t|ρ,τ )| < ∞,

lim g(r, t|ρ,τ ) → 0,

r→∞

r→0

0 < t,

and the initial condition that g(r, 0|ρ,τ ) = 0, 0 < r < ∞. Step 1 : Taking the Hankel transform of the governing equation, show that it becomes dG + a2 k 2 G = J0 (kρ)δ(t − τ ), dt with G(k, 0|ρ,τ ) = 0, where G(k, t|ρ,τ ) =

"

∞

g(r, t|ρ,τ )J0 (kr) r dr.

0

Step 2 : Use Laplace transforms and show that 2 2

G(k, t|ρ,τ ) = J0 (kρ)e−a

k (t−τ )

H(t − τ ).

Step 3 : Take the inverse of the Hankel transform and show that "

∞

2

J0 (kρ)J0 (kr)e−k (t−τ ) k dk 0 % & % & H(t − τ ) r2 + ρ2 rρ = 2 exp − 2 I0 . 2a (t − τ ) 4a (t − τ ) 2a2 (t − τ )

g(r, t|ρ,τ ) = H(t − τ )

The figure captioned Problem 9 illustrates this Green’s function as functions of r and a2 (t − τ ) with ρ = 0.3. 55 See Golitsyn, G. S., and N. N. Romanova, 1970: Heat conduction in a rarefied and inhomogeneous atmosphere. Geomag. Aeronomy, 10, 80–85.

The Heat Equation

367

Problem 9

10. Find the free-space Green’s function56 governed by % +& ∂g 1 ∂ 1/2 ∂ * n+1/2 = r 3κr g , 0 < r < ∞, ∂t r ∂r ∂r

0 < t,

subject to the boundary conditions lim g(r, t|ρ, 0) → 0,

r→0

lim g(r, t|ρ, 0) → 0,

r→∞

0 < t,

and the initial condition that g(r, 0|ρ, 0) = δ(r − ρ), 0 < r < ∞. Step 1 : Show that the product solution g = rp R(r)e−λt are solutions of the heat equation provided % , -& # $ $ λ 2−n 2 $$ 3 1 r R + 2p + 2n + 2 rR + (p + n) r +p+n+ 2 R = 0, 3κ where p = n − 14 .

Step 2 : Setting λ = 3κk 2 , 0 = (4 − 2n)−1 , and y = r(1−n/2) /(1 − n/2), show that R(r) = r−2n [A(k)J4 (ky) + B(k)Y4 (ky)] . 56 See Tanaka, T., 2011: Exact time-dependent solutions for the thin accretion disc equation: Boundary conditions at finite radius. Mon. Not. R. Astron. Soc., 410, 1007– 1017.

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Green’s Functions with Applications

[If 0 is not an integer, then we replace Y4 (ky) with J−4 (ky).] Therefore, " ∞ 2 g(r, t|ρ, 0) = r−n−1/4 [A(k)J4 (ky) + B(k)Y4 (ky)] e−3κk t dk. 0

Step 3 : Use the boundary condition at r = 0 to show that B(k) = 0. Step 4 : Finally show that " ∞ * 2 n + ρ5/4 g(r, t|ρ, 0) = 1 − J4 (ky $ )J4 (ky)e−3νk t k dk 2 rn+1/4 0 % & % & r−9/4 ρ5/4 2(ρ/r)1−n/2 1 + (ρ/r)2−n = (2 − n) I4 exp − , τ (r) τ (r) τ (r) where τ (r) = 12(1 − n/2)2 rn−2 κt and y $ = ρ(1−n/2) /(1 − n/2). 11. Using the free-space Green’s function problem: ∂g ∂2g 1 ∂g δ(r − ρ)δ(t − τ ) − 2 − + λg = , ∂t ∂r r ∂r r

0 ≤ r, ρ< ∞,

0 < t, τ ,

subject to the boundary conditions lim |g(r, t|ρ,τ )| < ∞,

r→0

lim g(r, t|ρ,τ ) → 0,

0 < t,

r→∞

and the initial condition that g(r, 0|ρ,τ ) = 0, 0 ≤ r < ∞, show that the solution57 to ∂u ∂ 2 u 1 ∂u − 2 − + λu = S(r), ∂t ∂r r ∂r

0 ≤ r < ∞,

0 < t,

subject to the boundary conditions limr→0 |u(r, t)| < ∞, limr→∞ u(r, t) → 0, 0 < t, and the initial condition u(r, 0) is " ∞ * rρ + 2 2 1 u(r, t) = e−r /(4t)−λt u(ρ, 0)e−ρ /(4t) I0 ρdρ 2t 2t 0 " ∞ " ∞ 5 k J0 (kρ)J0 (kr) 4 −(k2 +λ)t + ρS(ρ) 1 − e dk dρ. k2 + λ 0 0 Step 1 : Taking the Laplace transform of the Green’s function problem, show that d2 G 1 dG δ(r − ρ) −sτ + − (s + λ)G = − e , dr2 r dr r

0 ≤ r, ρ< ∞,

57 For an application, see Hassan, M. H. A., 1988: Ion distribution functions during ion cyclotron resonance heating at the fundamental frequency. Phys. Fluids, 31, 596–601.

The Heat Equation

369

with limr→0 |G(r, s|ρ,τ )| < ∞ and limr→∞ G(r, s|ρ,τ ) → 0. Step 2 : Using Hankel transforms, show that " ∞ J0 (kρ)J0 (kr) −sτ G(r, s|ρ,τ ) = e k dk. s + λ + k2 0 Step 3 : Taking the inverse Laplace transform, show that the Green’s function is " ∞ 1 2 g(r, t|ρ,τ ) = H(t − τ ) J0 (kρ)J0 (kr) exp −(k 2 + λ)(t − τ ) k dk. 0

Step 4 : Using Problem 2, show that u(r, t) =

"

0

Note:

t+

"

∞

0

S(ρ)g(r, t|ρ,τ )ρ d ρd τ +

"

∞

u(ρ, 0)g(r, t|ρ, 0)ρ dρ .

0

∂g ∂2g 1 ∂g δ(r − ρ)δ(t − τ ) + 2+ − λg = − , ∂τ ∂ρ ρ ∂ρ ρ

and −

∂u ∂ 2 u 1 ∂u + 2 + − λu = −S(ρ). ∂τ ∂ρ ρ ∂ρ

Step 5 : Substituting the Green’s function found in Step 3 into the expression for u(r, t) that you found in Step 4, find the final answer given above by evaluating the integrals as far as possible. 12. Find the free-space Green’s function58 governed by , 1 ∂g 1 ∂ ∂g = 2 r2+β + δ(r − 1)δ(t), 0 < r < ∞, r ∂t r ∂r ∂r

0 < t,

where the solution remains finite over the entire interval and is initially given by g(r, 0|1, 0) = 0. Step 1 : Take the Laplace transform of the partial differential equation and show that it equals , 1 d s 2+β dG r − G = −δ(r − 1), 0 < r < ∞, r2 dr dr r with limr→0 G(r, s|1, 0) → 0, and limr→∞ G(r, s|1, 0) → 0. 58 See Toptygin, I. N., 1973: Direct and inverse problem on cosmic-ray propagation in interplanetary space. Geomagnet. Aeronomy, 13, 181–186.

370

Green’s Functions with Applications

Step 2 : Show that Step 1 may be rewritten , 1 d s 2+β dGi r − Gi = 0, i = 1, 2, 2 r dr dr r

0 < r < ∞,

with limr→0 |G1 (r, s|1, 0)| < ∞, and limr→∞ G2 (r, s|1, 0) → 0, where G1 (r, s| 1, 0) is the Laplace transform of the Green’s function in the domain 0 < r < 1 while G2 (r, s|1, 0) is the Laplace transform of the Green’s function in the domain 1 < r < ∞. Step 3 : Show that the solution in each domain is % & 2 √ (1−β)/2 −(1+β)r/2 G1 (r, s|1, 0) = Ae Iν sr , 1−β

0 < r < 1,

and G2 (r, s|1, 0) = Be−(1+β)r/2 Kν

%

& 2 √ (1−β)/2 sr , 1−β

if β &= 1; otherwise, G1 (r, s|1, 0) = C and

r

√

s+1

r √

G2 (r, s|1, 0) = D

r−

1 < r < ∞,

,

0 < r < 1,

,

1 < r < ∞,

s+1

r

where ν = (1 + β)/(1 − β). Step 4 : Show that G1 (1, s|1, 0) = G2 (1, s|1, 0). Step 5 : By integrating the Laplace transformed version of the governing equation from 1− to 1+ , where r = 1− and r = 1+ are points just below and above r = 1, respectively, obtain the condition that G$2 (1+ , s|1, 0) − G$1 (1− , s|1, 0) = −1. Step 6 : Show that the solutions that satisfy the transformed partial differential equation and the boundary conditions are , - % & √ 2 2 √ 2 G1 (r, s|1, 0) = r−(1+β)/2 Kν s Iν r(1−β)/2 s , 1−β 1−β 1−β for 0 < r < 1, and G2 (r, s|1, 0) =

, % & √ 2 2 √ 2 r−(1+β)/2 Iν s Kν r(1−β)/2 s , 1−β 1−β 1−β

The Heat Equation

371

for 1 < r < ∞, if β &= 1;

√

1 r s+1 G1 (r, s|1, 0) = √ , 2 s+1 r and

0 < r < 1,

√

1 r− s+1 G2 (r, s|1, 0) = √ , r 2 s+1

1 < r < ∞,

if β = 1. Step 7 : Using a very good table12 of Laplace transforms, show that the Green’s function is % (1−β)/2 & % & t−1 −(1+β)/2 2r 1 + r1−β g(r, t|1, 0) = r Iν exp − , 1−β (1 − β)2 t (1 − β)2 t if β &= 1; otherwise,

% & ln2 (r) g(r, t|1, 0) = exp −t − , 4t 2r πt 1 √

if β = 1. 13. Find the free-space Green’s function55 governed by ∂g ∂ 2g − xm 2 = δ(x − ξ)δ(t − τ ), ∂t ∂x

0 < x, ξ < ∞,

0 < t, τ ,

subject to the boundary conditions lim |g(x, t|ξ,τ )| < ∞,

x→0

lim g(x, t|ξ,τ ) → 0,

x→∞

0 < t,

and the initial condition that g(x, 0|ξ,τ ) = 0, 0 < x < ∞, with m > 0. Step 1 : For the moment, let us assume that m &= 2; the case of m = 2 is treated later. Show that the original differential equation can be rewritten as , ∂R 1 ∂ 2 R 1 ∂R R δ(r − ρ)δ(t − τ ) − + − = , ∂t 4 ∂r2 r ∂r a2 r 2 (aρ)1/a

√ where 0 < r, ρ < ∞, 0 < t, τ, R(x, t|ξ,τ ) = g(x, t|ξ,τ )/ x, r = xa/2 /a, ρ = ξ a/2 /a, and a = |2 − m|. Step 2 : Taking the Hankel transform of the equation in Step 1, show that it becomes ρJ1/a (kρ)δ(t − τ ) dR k 2 + R= , 0 < t, dt 4 (aρ)1/a

372

Green’s Functions with Applications

Problem 13

with R(k, 0|ρ,τ ) = 0, where R(k, t|ρ,τ ) =

"

∞

R(r, t|ρ,τ ) J1/a (kr) r dr.

0

Step 3 : Use Laplace transforms and show that R(k, t|ρ,τ ) =

ρ J1/a (kρ) −k2 (t−τ )/4 e H(t − τ ). (aρ)1/a

Step 4 : Take the inverse of the Hankel transform and show that , a/2 √ " ∞ , a/2 2 ξ a/2 x kξ kx √ H(t − τ ) J1/a g(x, t|ξ,τ ) = J1/a e−k (t−τ )/4 k dk a a a ξ 0 % & % & √ 4ξ (a−1)/2 x ξ a + xa 2(aξ)a/2 = exp − 2 I1/a 2 H(t − τ ). a(t − τ ) a (t − τ ) a (t − τ ) Step 5 : Consider now the case m = 0. Show that the original differential equation can be rewritten as ∂R ∂ 2 R R − + = e−ρ/2 δ(r − ρ)δ(t − τ ), ∂t ∂r2 4 √ where −∞ < r, ρ < ∞, 0 < t, τ, R(x, t|ξ,τ ) = g(x, t|ξ,τ )/ x, r = ln(x), and ρ = ln(ξ).

The Heat Equation

373

Step 6 : Taking the Fourier transform of the equation in Step 5, show that it becomes dR # 2 1 $ + k + 4 R = e−ρ/2 e−ikρ δ(t − τ ), 0 < t, dt with R(k, 0|ρ,τ ) = 0, where " ∞ 1 R(r, t|ρ,τ ) = R(k, t|ρ,τ ) eikr dk. 2π −∞ Step 7 : Use Laplace transforms and show that R(k, t|ρ,τ ) = e−ρ/2 e−ikρ e−(k

2

+ 14 )(t−τ )

H(t − τ ).

Step 8 : Take the inverse of the Fourier transform and show that 3 " 2 1 H(t − τ ) x ∞ g(x, t|ξ,τ ) = cos {k [ln(x) − ln(ξ)]} e−(k + 4 )(t−τ ) dk π ξ 0 % & H(t − τ ) ln2 (x/ξ) ln(ξ/x) t − τ = . exp − − − 4(t − τ ) 2 4 2 π(t − τ ) < 1 (t−τ ) 2 > 2 ln ξe /x H(t − τ ) = . exp − , 4(t − τ ) 2 π(t − τ )

in agreement with Zhukova and Saichev.59 The figure captioned Problem 12 illustrates this Green’s function as x and t − τ vary when ξ = 0.5. 14. With the advent of financial instruments known as derivatives, partial differential equations entered into finance in a big way. During a study involving options on actively managed funds, Hyer et al.60 found the free-space Green’s function governed by ∂g ∂2g − − δ(x)g = δ(x − ξ), ∂t ∂x2

−∞ < x, ξ < ∞,

0 < t,

subject to the boundary condition lim|x|→∞ g(x, t|ξ) → 0, 0 < t, and the initial condition that g(x, 0|ξ) = 0, −∞ < x < ∞. Let us retrace their analysis. Step 1 : The interesting aspect of this problem is the presence of the delta function as a coefficient in the partial differential equation. Noting that the partial differential equation can be written ∂g ∂ 2g − = f (ξ, t)δ(x) + δ(x − ξ) = g(0, t|ξ)δ(x) + δ(x − ξ), ∂t ∂x2 59 Zhukova, I. S., and A. I. Saichev, 1997: Two-point statistical properties of a passive tracer in a turbulent compressible medium. Radiophys. Quantum Electron., 40, 682–693. 60

Hyer, T., A. Lipton-Lifschifz, and D. Pugachevsky, 1997: Passport to success. Risk , 10(9), 127–131.

374

Green’s Functions with Applications

show that 2

g(x, t|ξ) =

e−(x−ξ) /(4t) √ + 2 πt

"

t

0

2

e−x /[4(t−τ )] . f (ξ,τ ) dτ, 2 π(t − τ )

using joint Fourier and Laplace transforms.

Step 2 : Setting x = 0, show that the solution in Step 1 reduces to the integral equation " t 2 e−ξ /(4t) f (ξ,τ ) √ . f (ξ, t) = + dτ. 2 πt 0 2 π(t − τ ) Step 3 : Use Laplace transforms and show that

√ , exp[−ξ 2 /(4t)] 1 t/4−|ξ|/2 |ξ| t √ f (ξ, t) = + 4e erfc √ − , 2 2 πt 2 t and √ , 2 e−(x−ξ) /(4t) |x| + |ξ| t 1 t/4−(|x|+|ξ|)/2 √ √ g(x, t|ξ) = + 4e erfc − , 2 2 πt 2 t where erfc(·) is the complementary error function. Hint: Don’t evaluate the convolution integral directly but use the convolution theorem and tables. 15. Find the free-space Green’s function for ∂g − a2 ∂t

,

∂ 2g ∂2g ∂g + −b ∂x2 ∂y 2 ∂x

-

= δ(x − ξ)δ(y − η)δ(t − τ ),

subject to the boundary conditions lim|x|,|y|→∞ g(x, y, t|ξ, η,τ ) → 0, 0 < t, and the initial condition that g(x, y, 0|ξ, η,τ ) = 0, −∞ < x, y < ∞. Step 1 : Apply Laplace transforms to the partial differential equation and show that it becomes ∂ 2G ∂ 2G ∂G s δ(x − ξ)δ(y − η) −sτ + −b − 2G = − e , 2 2 ∂x ∂y ∂x a a2 with lim|x|,|y|→∞ G(x, y, s|ξ, η,τ ) → 0. Step 2 : Taking the Fourier transform in both the x and y directions, show that the partial differential equation in Step 1 reduces to G(k, 0, s|ξ, η,τ ) =

e−ikξ−i4η−sτ , + 02 + ibk) + s

a2 (k 2

The Heat Equation

375

where G(k, 0, s|ξ, η,τ ) denotes the double Fourier transform of the Laplace transform G(x, y, s |ξ, η,τ ). Step 3 : Following the example of Problem 8 in Chapter 6, show that : ; 3 2 . eb(x−ξ)/2−sτ b s G(x, y, s|ξ, η,τ ) = K0 (x − ξ)2 + (y − η)2 + 2 . 2πa2 4 a Step 4 : Using tables, show that g(x, y, t|ξ, η,τ ) =

! ) H(t − τ ) [(x − ξ) − a2 b(t − τ )]2 + (y − η)2 exp − . 4πa2 (t − τ ) 4a2 (t − τ )

16. Find the free-space Green’s function61 governed by , ∂ ∂ 2g ∂ 2g ∂g + +β = δ(x − ξ)δ(y − η)δ(t − τ ), ∂t ∂x2 ∂y 2 ∂x with −∞ < x, y, ξ, η < ∞, and 0 < t, τ, where the boundary conditions are lim g(x, y, t|ξ, η,τ ) → 0

|x|→∞

lim g(x, y, t|ξ, η,τ ) → 0,

|y|→∞

0 < t,

and the initial condition is g(x, y, 0|ξ, η,τ ) = 0, −∞ < x, y < ∞. Step 1 : Take the Laplace transform of the partial differential equation and show that , 2 ∂ G ∂ 2G ∂G s + +β = δ(x − ξ)δ(y − η)e−sτ . ∂x2 ∂y 2 ∂x Step 2 : Setting G(x, y, s|ξ, η,τ ) = e−βx/(2s) U (x, y, s|ξ, η,τ ), show that the partial differential equation in Step 1 reduces to ∂ 2U ∂2U β2 + − U = eβξ/(2s)−sτ δ(x − ξ)δ(y − η). ∂x2 ∂y 2 4s2 Step 3 : Show that , e−(x−ξ)β/(2s)−sτ βr G(x, y, s|ξ, η,τ ) = − K0 , 2s 2s

where r =

. (x − ξ)2 + (y − η)2 . Hint: See Problem 8 in the next chapter.

61 See Veronis, G., 1958: On the transient response of a β-plane ocean. J. Oceanogr. Soc. Japan, 14, 1–5.

376

Green’s Functions with Applications

Problem 16

Step 4 : Use the convolution theorem and a good set of tables to show that . . " t K0 [2 βr(t − ζ)] cos[ 2(x + r − ξ)β(ζ − τ )] . g(x, y, t|ξ, η,τ ) = −H(t−τ ) dζ π (t − ζ)(ζ − τ ) τ .

Step 5 : Let u − τ = (t − τ ) sin2 (ϕ), α2 = βr(t − τ ), γ = cos(θ/2), and x − ξ = r cos(θ). Using these variables and the integrals "

π/2

cos[a sin(x)] cos[b cos(x)] dx =

0

and K0 (α) =

"

∞

0

+ π *. 2 J0 a + b2 , 2

cos(αz) √ dz, z2 + 1

show that g(x, y, t|ξ, η,τ ) can be rewritten as

g(x, y, t|ξ, η,τ ) = −H(t − τ )

"

∞ 0

* . + J0 2α z 2 + γ 2 √ dz. z2 + 1

With β(t − τ ) = 10, the figure captioned Problem 16 illustrates the integral portion of g(x, y, t|ξ, η,τ ).

The Heat Equation

377

17. Find the free-space Green’s function62 governed by ∂g − a2 ∂t

,

∂ 2g 1 ∂g ∂ 2g + + 2 2 ∂r r ∂r ∂z

-

+ bg = δ(r − ρ)δ(z − ζ)δ(t − τ ),

where a and b are positive constants. Step 1 : Using Hankel transforms, show that the governing partial differential equation becomes ∂G − a2 ∂t

,

∂ 2G 2 − k G + bG = ρJ0 (kρ) δ(z − ζ) δ(t − τ ), ∂z 2

where G(k, z, t|ρ, ζ,τ ) =

"

∞

g(r, z, t|ρ, ζ,τ )J0 (kr)r dr.

0

Step 2 : Using Fourier transforms, show that the partial differential equation in Step 1 can be transformed into dG − a2 (k 2 + m2 )G + bG = ρJ0 (kρ)e−imζ δ(t − τ ), dt where G(k, m, t|ρ, ζ,τ ) =

"

∞

G(k, z, t|ρ, ζ,τ )e−imz dz.

−∞

Step 3 : Using Laplace transforms, show that the solution to the ordinary differential equations in Step 2 is G(k, m, t|ρ, ζ,τ ) = ρH(t − τ )J0 (kρ)e−imζ e−[b+a

2

(k2 +m2 )](t−τ )

H(t − τ ).

Step 4 : Taking the inverse Fourier transforms, show that % & ρ H(t − τ )J0 (kρ) (z − ζ)2 2 2 . G(k, z, t|ρ, ζ,τ ) = exp − 2 − (b + a k )(t − τ ) . 4a (t − τ ) 2 πa2 (t − τ )

Step 5 : Taking the inverse Hankel transform, show that

% & ρe−b(t−τ ) H(t − τ ) (z − ζ)2 + r2 + ρ2 g(r, z, t|ρ, ζ,τ ) = √ exp − 4a2 (t − τ ) 2π[2a2 (t − τ )]3/2 % & ρr × I0 . 2a2 (t − τ ) 62 Vyas, R., and M. L. Rustgi, 1992: Green’s function solution to the tissue bioheat equation. Med. Phys., 19, 1319–1324.

378

Green’s Functions with Applications

18. Find the free-space Green’s function for the three-dimensional heat equation63 , 2 ∂g ∂ g ∂2g ∂2g 2 −a + 2 + 2 = a2 δ(x − ξ)δ(y − η)δ(z − ζ)δ(t − τ ). ∂t ∂x2 ∂y ∂z Step 1 : Apply Laplace transforms to the partial differential equation and show that it becomes ∂ 2G ∂2G ∂ 2G s + + − 2 G = −δ(x − ξ)δ(y − η)δ(z − ζ)e−sτ . ∂x2 ∂y 2 ∂z 2 a Step 2 : Taking the Fourier transform in both the x and y directions, show that the partial differential equation in Step 1 reduces to the ordinary differential equation d2 G − (κ2 + s/a2 )G = −δ(z − ζ)e−ikξ−i4η−sτ , dz 2 where k and 0 are the Fourier transform parameters in the x and y directions, respectively, κ2 = k 2 + 02 , and G(k, 0, z, s|ξ, η, ζ,τ ) denotes the double Fourier transform of G(x, y, z, s|ξ, η, ζ,τ ). Step 3 : Show that the solution to Step 2 is G(k, 0, z, s|ξ, η, ζ,τ ) = where γ =

.

e−γ|z−ζ|−ikξ−i4η−sτ , 2γ

κ2 + s/a2 .

Step 4 : Show that G(x, y, z, s|ξ, η, ζ,τ ) = = =

e−sτ 8π 2 −sτ

e 4π

"

∞

−∞

"

0

∞

" e

∞ −∞

eik(x−ξ)+i4(y−η)−|z−ζ|γ dk d0 γ √ 2 2

−|z−ζ|

.

κ +s/a

κ2 + s/a2

J0 (κr) κ d κ

√ exp(−R s/a − sτ ) , 4πR

where k = κ cos(θ), 0 = κ sin(θ), x − ξ = r cos(β), y − η = r sin(β), and R2 = (x − ξ)2 + (y − η)2 + (z − ζ)2 . Hint: Use integral tables. Step 5 : Complete the problem by using transform tables and show that % & H(t − τ ) (x − ξ)2 + (y − η)2 + (z − ζ)2 g(x, y, z, t|ξ, η, ζ,τ ) = exp − . 3/2 4a2 (t − τ ) [4πa2 (t − τ )] 63 See Bowler, J. R., 1999: Time domain half-space dyadic Green’s functions for eddycurrent calculations. J. Appl. Phys., 86, 6494–6500.

The Heat Equation

379 Semi-Infinite Domains

19. Use Green’s functions to show that the solution64 to ∂u ∂ 2u = a2 2 , ∂t ∂x

0 < x, t,

subject to the boundary conditions u(0, t) = 0, limx→∞ u(x, t) → 0, 0 < t, and the initial condition that u(x, 0) = f (x), 0 < x < ∞, is 2

2

e−x /(4a √ u(x, t) = a πt

t)

"

∞

f (τ ) sinh

0

* xτ + 2 2 e−τ /(4a t) dτ. 2a2 t

20. Find the Green’s function65 governed by ∂g ∂ 2g − a2 2 + βg = δ(x − ξ)δ(t − τ ), ∂t ∂x

0 < x, t, ξ, τ ,

subject to the boundary conditions g(0, t|ξ,τ ) = 0,

lim g(x, t|ξ,τ ) → 0,

x→∞

0 < t,

and the initial condition that g(x, 0|ξ,τ ) = 0, 0 < x < ∞. Step 1 : Taking the Laplace transform of the governing equation and boundary conditions, show that it becomes the ordinary differential equation d2 G s + β δ(x − ξ) −sτ − G=− e , 2 2 dx a a2

0 < x, ξ < ∞,

with G(0, s|ξ,τ ) = 0, and limx→∞ G(x, s|ξ,τ ) → 0. Step 2 : Consider the similar problem of d2 G s + β δ(x − ξ) −sτ − G=− e , dx2 a2 a2

−∞ < x, ξ < ∞,

with lim|x|→∞ G(x, s|ξ,τ ) → 0. Show that the solution to this problem is G(x, s|ξ,τ ) =

√ 1 $ √ e−sτ −|x−ξ| s /a , $ 2a s

64

See Gilev, S. D., and T. Yu. Mikha˘ılova, 1996: Current wave in shock compression of matter in a magnetic field. Tech. Phys., 41, 407–411. 65 See Khantadze, A. G., and B. Ya. Chekhoskvili, 1975: Effect of plasma flow on the nighttime F region. Geomagnet. Aeron., 15, 440–441; Darling, R. B., 1995: A full dynamic model for pn-junction diode switching transients. IEEE Trans. Electron. Devices, 42, 969–976.

380

Green’s Functions with Applications

where s$ = s + β. Step 3 : Using the method of images and the results from Step 2, show that the solution to Step 1 is G(x, s|ξ,τ ) =

5 √ e−sτ 4 −|x−ξ|√s$ /a $ √ e − e−(x+ξ) s /a . 2a s$

Step 4 : Using the shifting theorems and tables, show that g(x, t|ξ,τ ) =

S 2 2 e−β(t−τ ) H(t − τ ) R −(x−ξ)2 /[4a2 (t−τ )] . e − e−(x+ξ) /[4a (t−τ )] . 2a π(t − τ )

21. Find the Green’s function66 for

∂g ∂2g − 2 = δ(x − ξ)δ(t − τ ), ∂t ∂x

0 < x, t, ξ, τ ,

subject to the boundary conditions g(0, t|ξ,τ ) = κ

"

t 0

gx (0, t$ |ξ,τ ) dt$ ,

and

lim g(x, t|ξ,τ ) → 0,

x→∞

and the initial condition that g(x, 0|ξ,τ ) = 0, 0 < x < ∞. The interesting aspect of this problem is the presence of the integral in the boundary condition. Step 1 : Taking the Laplace transform of the governing equation and boundary conditions, show that it becomes the ordinary differential equation d2 G − sG = −δ(x − ξ)e−sτ , dx2

0 < x, ξ < ∞,

with G(0, s|ξ,τ ) = κG$ (0, s|ξ,τ )/s, and limx→∞ G(x, s|ξ,τ ) → 0. Step 2 : Show that the solution to the ordinary differential equation in Step 1 is √ √ √ e−|x−ξ| s−sτ e−(x+ξ) s−sτ κ e−(x+ξ) s−sτ √ √ √ G(x, s|ξ,τ ) = − + √ . 2 s 2 s s(κ + s ) Step 3 : Taking the inverse Laplace transform, show that ! % & % &) H(t − τ ) (x − ξ)2 (x + ξ)2 g(x, t|ξ,τ ) = . exp − − exp − 4(t − τ ) 4(t − τ ) 4π(t − τ ) % & x + ξ + 2κ(t − τ ) √ + κH(t − τ )eκ[x+ξ+κ(t−τ )] erfc , 2 t−τ 66

See Agmon, N., 1984: Diffusion with back reaction. J. Chem. Phys., 81, 2811–2817.

The Heat Equation

381

Problem 21

where erfc(·) is the complementary error function. The figure captioned Problem 21 illustrates this Green’s function when κ = −0.1, and ξ = 2. 22. Find the Green’s function for the linearized Ginzburg-Landau equation ∂g ∂g ∂ 2g +v − ag − b 2 = δ(x − ξ)δ(t − τ ), ∂t ∂x ∂x

0 < x, ξ < ∞,

0 < t, τ ,

with b > 0, subject to the boundary conditions g(0, t|ξ,τ ) = 0,

lim g(x, t|ξ,τ ) → 0,

x→∞

0 < t,

and the initial condition that g(x, 0|ξ,τ ) = 0, 0 < x < ∞. Step 1 : Taking the Laplace transform of the partial differential equation, show that it reduces to b

d2 G dG −v + aG − sG = −δ(x − ξ)e−sτ , 2 dx dx

with G(0, s|ξ,τ ) = 0, and limx→∞ G(x, s|ξ,τ ) → 0. Step 2 : Setting G(x, s|ξ,τ ) = evx/(2b) U (x, s|ξ,τ ), show that the ordinary differential equation in Step 1 becomes b

d2 U v2 − U + aU − sU = −δ(x − ξ)e−vξ/(2b)−sτ , dx2 4b

382

Green’s Functions with Applications

with U (0, s|ξ,τ ) = 0, and limx→∞ U (x, s|ξ,τ ) → 0. Step 3 : Using Fourier sine transforms, show that % &" ∞ 2 v(x − ξ) sin(kξ) sin(kx) G(x, s|ξ,τ ) = exp − sτ dk. π 2b s + bk 2 + v 2 /(4b) − a 0 Step 4 : Invert the Laplace transform and show that % & 2 2 v(x − ξ) g(x, t|ξ,τ ) = exp H(t − τ )ea(t−τ )−v (t−τ )/(4b) π 2b " ∞ 2 × e−b(t−τ )k sin(kξ) sin(kx) dk 0

% & 2 ea(t−τ )−v (t−τ )/(4b) v(x − ξ) . = exp H(t − τ ) 2b 2 πb(t − τ ) ! % & % &) (x − ξ)2 (x + ξ)2 × exp − − exp − . 4b(t − τ ) 4b(t − τ )

23. Find the Green’s function67 governed by ∂g ∂ 2g ∂g − − = δ(x − ξ)δ(t − τ ), 2 ∂t ∂x ∂x

0 < x, t, ξ, τ ,

subject to the boundary conditions gx (0, t|ξ,τ ) + (1 − h)g(0, t|ξ,τ ) = 0,

lim g(x, t|ξ,τ ) → 0,

x→∞

0 < t,

and the initial condition that g(x, 0|ξ,τ ) = 0, 0 < x < ∞. Step 1 : Taking the Laplace transform of the governing equation and boundary conditions, show that it becomes the ordinary differential equation d2 G dG + − sG = −δ(x − ξ)e−sτ , dx2 dx

0 < x, ξ < ∞,

with G$ (0, s|ξ,τ ) + (1 − h)G(0, s|ξ,τ ) = 0, and limx→∞ G(x, s|ξ,τ ) → 0. Step 2 : Show that the solution to the ordinary differential equation in Step 1 is √ 1 √ 1 e−(x−ξ)/2−|x−ξ| s+ 4 −sτ e−(x−ξ)/2−(x+ξ) s+ 4 −sτ = = G(x, s|ξ,τ ) = − 2 s + 14 2 s + 14 √ 1 e−(x−ξ)/2−(x+ξ) s+ 4 −sτ = + . h − 12 + s + 14 67 See Kim, S.-H., S. A. Korpela, and A. Chait, 1998: Axial segregation in unsteady diffusion-dominated solidification of a binary alloy. J. Cryst. Growth, 183, 490–496.

The Heat Equation

383

Step 3 : Taking the inverse Laplace transform, show that # $ g(x, t|ξ,τ ) = H(t − τ ) 12 − h e−(1−h)[x+h(t−τ )]+hξ % & # $√ x+ξ × erfc √ − 12 − h t − τ 2 t−τ H(t − τ ) −(x−ξ)/2−(t−τ )/4 + . e 2 π(t − τ ) R S 2 2 × e−(x+ξ) /[4(t−τ )] + e−(x−ξ) /[4(t−τ )] , where erfc(·) is the complementary error function.

24. Find the Green’s function68 for , ∂g ∂ ∂g xα − x1−α = xα δ(x − ξ)δ(t − τ ), ∂t ∂x ∂x

0 < x, t, ξ, τ ,

where 0 ≤ α ≤ 1, subject to the boundary conditions gx (0, t|ξ,τ ) = 0, limx→∞ g(x, t|ξ,τ ) → 0, 0 < t, and the initial condition that g(x, 0|ξ,τ ) = 0, 0 < x < ∞. Step 1 : Taking the Laplace transform of the governing equation and boundary conditions, show that it becomes the ordinary differential equation d2 G 1 − α dG + − sx2α−1 G = −x2α−1 δ(x − ξ)e−sτ , dx2 x dx

0 < x, ξ < ∞,

with G$ (0, s|ξ,τ ) = 0, and limx→∞ G(x, s|ξ,τ ) → 0. Step 2 : Show that the solution to the ordinary differential equation in Step 1 is % & 2 2 √ (1+2α)/2 α/2 −sτ G(x, s|ξ,τ ) = (xξ) e I−α/(1+2α) s x< 1 + 2α 1 + 2α % & 2 √ (1+2α)/2 × Kα/(1+2α) s x> . 1 + 2α Step 3 : Find the inverse√Laplace transform by applying Bromwich’s integral with the branch cut for s taken along the negative axis of the s-plane. Show that 1 + 2α g(x, t|ξ,τ ) = (xξ)α/2 H(t − τ ) 2 " ∞ 4 5 4 5 × J−α/(1+2α) ηξ (1+2α)/2 J−α/(1+2α) ηx(1+2α)/2 0 1 2 × exp −(1 + 2α)2 (t − τ )η 2 /4 η dη . 68 Similar to a problem solved by Smith, F. B., 1957: The diffusion of smoke from a continuous elevated point-source into a turbulent atmosphere. J. Fluid Mech., 2, 49–76.

384

Green’s Functions with Applications

Step 4 : Using integral tables, evaluate the integral and show that % & (xξ)α/2 H(t − τ ) x1+2α + ξ 1+2α g(x, t|ξ,τ ) = exp − (1 + 2α)(t − τ ) (1 + 2α)2 (t − τ ) % & 2(xξ)(1+2α)/2 × I−α/(1+2α) . (1 + 2α)2 (t − τ ) 25. Find the Green’s function governed by , ∂g 1 ∂ ∂g ∂2g ∂g δ(r − ρ)δ(z − ζ)δ(t − τ ) − r − 2 +2 = , ∂t r ∂r ∂r ∂z ∂z 2πr where 0 < r, ρ < 1, |z − ζ| < ∞, and 0 < t, τ, with the boundary conditions gr (0, z, t|ρ, ζ,τ ) = gr (1, z, t|ρ, ζ,τ ) = 0, and lim|z|→∞ g(r, z, t|ρ, ζ,τ ) → 0, and the initial condition that g(r, z, 0|ρ, ζ,τ ) = 0. Step 1 : Taking the Laplace transform of the partial differential equation, show that it becomes , 1 ∂ ∂G ∂2G ∂G δ(r − ρ)δ(z − ζ) −sτ sG − r − +2 = e , 2 r ∂r ∂r ∂z ∂z 2πr with the boundary conditions Gr (0, z, s|ρ, ζ,τ ) = Gr (1, z, s|ρ, ζ,τ ) = 0, and lim|z|→∞ G(r, z, s|ρ, ζ,τ ) → 0. Step 2 : Assuming that G(r, z, s|ρ, ζ,τ ) can be expressed in terms of the expansion ∞ ( G(r, z, s|ρ, ζ,τ ) = Zn (z)J0 (kn r), n=1

where kn is the nth zero of J1 (k) = 0, show that Zn (z) is governed by $ d2 Zn dZn # J0 (kn ρ) −2 − s + kn2 Zn = − δ(z − ζ)e−sτ . 2 dz dz π J02 (kn )

Why did we choose this particular expansion? Step 3 : Show that Zn (z) =

√ 2 ez−ζ−sτ J0 (kn ρ) −|z−ζ| s+kn +1 . e . 2π J02 (kn ) s + kn2 + 1

Step 4 : Finish up the problem by inverting the Laplace transform and show that 2 ∞ ( ez−ζ−(z−ζ) /[4(t−τ )] J0 (kn ρ)J0 (kn r) −(1+kn2 )(t−τ ) . g(r, z, t|ρ, ζ,τ ) = H(t−τ ) e . J02 (kn ) 2π π(t − τ ) n=1

The Heat Equation

385

Problem 25

The figure captioned Problem 25 illustrates this Green’s function (multiplied by 100) as functions of r and z − ζ when t − τ = 0.2, and ρ = 0.3. 26. Use the Green’s function given by Equation 5.1.10 to solve the initialboundary-value problem69 ∂u i ∂2u = , ∂t 4a ∂x2

0 < x < ∞,

0 < a, t

subject to the boundary conditions u(0, t) = 0, limx→∞ u(x, t) → 0, 0 < t, and the initial condition that u(x, 0) = f (x), 0 < x < ∞. Finite Domains 27. Use Equation 5.2.13 to construct the Green’s function for the one-dimensional heat equation gt − gxx = δ(x − ξ)δ(t − τ ) for 0 < x, ξ < L, 0 < t, τ, with the initial condition that g(x, 0|ξ,τ ) = 0, 0 < x < L, and the boundary conditions that g(0, t|ξ,τ ) = gx (L, t|ξ,τ ) = 0 for 0 < t. 28. Use Equation 5.2.13 to construct the Green’s function70 for the one-dimensional heat equation gt − gxx = δ(x − ξ)δ(t − τ ) for 0 < x, ξ < L, 0 < t, τ, 69 Puga, A., and B. N. Miller, 2013: Solution of the quantum initial value problem with transparent boundary conditions. Am. J. Phys., 81, 50–56. 70

For an application, see J. Liu and Y. X. Zhou, op. cit.

386

Green’s Functions with Applications

with the initial condition that g(x, 0|ξ,τ ) = 0, 0 < x < L, and the boundary conditions that gx (0, t|ξ,τ ) = g(L, t|ξ,τ ) = 0 for 0 < t. 29. Use Equation 5.2.13 to construct the Green’s function71 for the one-dimensional heat equation gt − gxx = δ(x − ξ)δ(t − τ ) for 0 < x, ξ < L, 0 < t, τ, with the initial condition that g(x, 0|ξ,τ ) = 0, 0 < x < L, and the boundary conditions that gx (0, t|ξ,τ ) = gx (L, t|ξ,τ ) = 0 for 0 < t. 30. Use Equation 5.2.13 to construct the Green’s function72 for the one-dimensional heat equation gt − gxx = δ(x − ξ)δ(t − τ ) for 0 < x, ξ < L, 0 < t, τ, with the initial condition that g(x, 0|ξ,τ ) = 0, 0 < x < L, and the boundary conditions that gx (0, t|ξ,τ ) − h g(0, t|ξ,τ ) = g(L, t|ξ,τ ) = 0 for 0 < t. 31. Use Problem 1 and the Green’s function given by Equation 5.2.21 to find the solution to the heat equation ut = uxx for 0 < x < L, 0 < t, with the initial data u(x, 0) = 1, 0 < x < L, and the boundary conditions u(0, t) = e−t and u(L, t) = 0 when 0 < t. 32. Use Problem 1 and the Green’s function that you found in Problem 27 to find the solution to the heat equation ut = uxx for 0 < x < L, 0 < t, with the initial data u(x, 0) = 1, 0 < x < L, and the boundary conditions u(0, t) = sin(t) and ux (L, t) = 0 when 0 < t. 33. Use Problem 1 and the Green’s function that you found in Problem 29 to find the solution to the heat equation ut = uxx for 0 < x < L, 0 < t, with the initial data u(x, 0) = 1, 0 < x < L, and the boundary conditions ux (0, t) = 1 and ux (L, t) = 0 when 0 < t. 34. Consider the Green’s function problem73 ∂g ∂ 2g = , ∂t ∂x2

0 < x < 1,

0 < t,

subject to the boundary conditions gx (0, t|0, 0+) = 0, gx (1, t|0, 0+ ) = δ(t−0+ ), 0 < t, and the initial condition that g(x, 0|0, 0+) = 0, 0 < x < L. Step 1 : Show that

√ cosh(x s ) √ . G(x, s|0, 0 ) = √ s sinh( s ) +

71

For an application, see Deng, Z.-S., and J. Liu, 2002: Analytical study on bioheat transfer problems with spatial or transient heating on skin surface or inside biological bodies. J. Biomech. Eng., 124, 638–649. 72

Ibid.

73 See Bazant, M. Z., K. Thornton, and A. Ajdari, 2004: Diffuse-charge dynamics in electrochemical systems. Phys. Review, Ser. E , 70, Art. No. 021506.

The Heat Equation

387

Step 2 : Invert G(x, s|0, 0+ ) and show that % & ∞ ( 1 (x − 2m + 1)2 √ g(x, t|0, 0 ) = exp − . 4t πt m=−∞ +

Step 3 : Consider the Green’s function problem ∂g ∂2g = , ∂t ∂x2

0 < x < 1,

0 < t,

subject to the boundary conditions gx (0, t|1− , 0) = gx (1, t|1− , 0) = 0, 0 < t, and the initial condition that g(x, 0|1− , 0) = δ(x − 1− ), 0 < x < L. Show that this Green’s function problem is equivalent to the original problem. Hint: Find G(x, s|1− , 0) and show that it is the same as that given in Step 1. Step 4 : Use the Green’s function problem stated in the previous step and the method of images to obtain the solution stated in Step 2. 35. Find the Green’s function for ∂g ∂2g − a2 2 + a2 k 2 g = δ(x − ξ)δ(t − τ ), ∂t ∂x

0 < x, ξ < L,

0 < t, τ ,

subject to the boundary conditions g(0, t|ξ,τ ) = gx (L, t|ξ,τ ) = 0, 0 < t, and the initial condition that g(x, 0|ξ,τ ) = 0, 0 < x < L, where a and k are real constants. 36. Find the Green’s function74 for ∂g ∂2g − + Aσe σt g = δ(x − ξ)δ(t − τ ), ∂t ∂x2

0 < x, ξ < 1,

0 < t, τ ,

subject to the boundary conditions gx (0, t|ξ,τ ) = gx (1, t|ξ,τ ) + h g(1, t|ξ,τ ) = 0,

0 < t,

and the initial condition that g(x, 0|ξ,τ ) = 0, 0 < x < 1. Step 1 : Show that the delta function and Green’s function can be expressed by the eigenfunction expansions: δ(x − ξ) = 2

∞ (

kn2 + h2 cos(kn ξ) cos(kn x), k 2 + h2 + h n=1 n

74 See Srikumar, A., T. G. Stanford, and J. W. Weidner, 1998: Linear sweep voltammetry in flooded porous electrodes at low sweep rates. J. Electroanal. Chem., 458, 161–173.

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Green’s Functions with Applications

and g(x, t|ξ,τ ) =

∞ (

gn (t|τ ) cos(kn ξ) cos(kn x),

n=1

where kn is the nth root of k tan(k) = h and dgn k 2 + h2 + kn2 gn + Aσe σt gn = 2 2 n 2 δ(t − τ ). dt kn + h + h Hint: Use the dispersion relationship kn tan(kn ) = h to show that (kn2 + h2 ) sin2 (kn ) = h2 . Step 2 : Show that the equation for gn (t|τ ) can be rewritten # $ 2 # $ d1 k 2 + h2 exp kn2 t + Aeσt gn = 2 2 n 2 exp kn2 t + Aeσt δ(t − τ ). dt kn + h + h

Step 3 : Integrating the equation in Step 2 from 0 to t, show that gn (t|τ ) = 2

1 # $2 kn2 + h2 exp −kn2 (t − τ ) − A eσt − eστ H(t − τ ). 2 2 kn + h + h

37. Find the Green’s function75 for

∂g ∂2g − a2 2 = δ(x − ξ)δ(t − τ ), ∂t ∂x

0 < x, ξ < L,

0 < t, τ ,

subject to the boundary conditions a2 gx (0, t|ξ,τ ) − h g(0, t|ξ,τ ) = −αϕ(t), and a2 gx (L, t|ξ,τ ) + h g(L, t|ξ,τ ) = αψ(t), where and

ϕ$ (t) + α ϕ(t) = h g(0, t|ξ,τ ),

ϕ(0) = 0,

ψ $ (t) + α ψ(t) = h g(L, t|ξ,τ ),

ψ(0) = 0.

The initial condition is g(x, 0|ξ,τ ) = 0 for 0 < x < L. Step 1 : Take the Laplace transform of the partial differential equation and boundary conditions and show that we obtain the transformed set sG − a2

d2 G = δ(x − ξ)e−sτ , dx2

75 See Kim, H., and K. J. Shin, 2000: On the diffusion-influenced reversible trapping problem in one dimension. J. Chem. Phys., 112, 8312–8317.

The Heat Equation with

389

a2 G$ (0, s|ξ,τ ) = a2 qX(q)G(0, s|ξ,τ ),

and

a2 G$ (L, s|ξ,τ ) = −a2 qX(q)G(L, s|ξ,τ ),

where q 2 = s/a2 , and X(q) = qh/(a2 q 2 + α). Step 2 : Show that

e−sτ [cosh(qx< ) + X(q) sinh(qx< )] (q, L) × {cosh[a(L − x> )] + X(q) sinh[q(L − x> )]} ,

G(x, s|ξ,τ ) =

a2 qY

where Y (q, L) = [1 + X 2 (q)] sinh(qL) + 2X(q) cosh(qL). Step 3 : Show that G(x, s|ξ,τ ) has singularities at q = 0 and qn , where qn is the nth root of Y (q, L) = 0 and n = 1, 2, 3, . . . Step 4 : Conclude the problem by showing that g(x, t|ξ,τ ) =

% 2 2 & ∞ ( H(t − τ ) W (qn , x)W (qn , ξ) a βn (t − τ ) +2H(t−τ ) exp − , L + 2h/α Z(qn , L) L2 n=1

where W (q, x) = cosh(qx) + X(q) sinh(qx), Z(q, L) = L[1 − X 2 (q)] + 2X $ (q), and qn = βn i/L. 38. Find the Green’s function76 governed by ∂g ∂ 2g b ∂g − − = δ(x − ξ)δ(t − τ ), 2 ∂t ∂x x ∂x

b < 1,

0 < x, ξ < 1,

0 < t, τ ,

subject to the boundary conditions g(0, t|ξ,τ ) = g(1, t|ξ,τ ) = 0, 0 < t, and the initial condition that g(x, 0|ξ,τ ) = 0, 0 < x < 1. Step 1 : Consider the Sturm-Liouville problem , d b dϕ x + λxb ϕ = 0, ϕ(0) = ϕ(1) = 0. dx dx Show that the orthonormal eigenfunctions to this Sturm-Liouville problem are . . √ ϕn (x) = 2 xν Jν (x λn )/|Jν+1 ( λn )|,

76 See Chan, C. Y., and B. M. Wong, 1995: Existence of classical solutions for singular parabolic problems. Q. Appl. Math., 53, 201–213.

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Green’s Functions with Applications

√ where ν = (1 − b)/2, and λn is the nth root of Jν ( λ ) = 0. Step 2 : Assuming that g(x, t|ξ,τ ) can be written as g(x, t|ξ,τ ) =

∞ (

an (t)ϕn (x),

n=1

show that an (t) is governed by a$n (t) + λn an (t) = ξ b ϕn (ξ)δ(t − τ ). Step 3 : Show that an (t) = ξ b ϕn (ξ)e−λn (t−τ ) H(t − τ ). Step 4 : Show that g(x, t|ξ,τ ) = H(t − τ )

∞ (

ξ b ϕn (ξ)ϕn (x)e−λn (t−τ ) .

n=1

39. Consider the boundary-value problem77 , ∂u a2 ∂ ∂u − r = f (r, t), ∂t r ∂r ∂r

0 ≤ r < 1,

0 < t,

with the boundary conditions limr→0 |u(r, t)| < ∞, ur (1, t) = 0, 0 < t, and the initial condition that u(r, 0) = 0, 0 ≤ r < 1. Let us solve this problem using Green’s functions. Step 1 : Show that the corresponding Green’s function problem is , ∂g a2 ∂ ∂g δ(r − ρ)δ(t − τ ) − r = , 0 ≤ r < 1, 0 < t, ∂t r ∂r ∂r 2πr with the boundary conditions limr→0 |g(r, t|ρ,τ )| < ∞, gr (1, t|ρ,τ ) = 0, 0 < t, and the initial condition that g(r, 0|ρ,τ ) = 0, 0 ≤ r < 1. Step 2 : Because ∞ δ(r − ρ) 1 1 ( J0 (kn ρ)J0 (kn r) = + , 2πr π π n=1 J02 (kn )

0 ≤ r < 1,

where kn is the nth root of J1 (k) = 0, show that our Green’s function problem reduces to dG0 = δ(t − τ ), G0 (0|τ ) = 0, dt 77 See Ahmad, B., and R. Pratap, 2011: Analytical evaluation of squeeze film forces in a CMUT with sealed air-filled cavity. IEEE Sens. J., 11, 2426–2431. See also Section 4.3.2 in Ma, K. Q., J. Liu, S. H. Xiang, K. W. Xie, and Y. X. Zhou, 2009: Study of thawing behavior of liquid metal used as computer chip coolant. Int. J. Therm. Sci., 48, 964–974.

The Heat Equation and

391

dGn − a2 kn2 Gn = δ(t − τ ), dt

Gn (0|τ ) = 0,

where n = 1, 2, 3, . . ., and ∞ G0 (t|τ ) 1( J0 (kn ρ)J0 (kn r) g(r, t|ρ,τ ) = + Gn (t|τ ) . π π n=1 J02 (kn )

Step 3 : Show that 2 2 kn (t−τ )

Gn (t|τ ) = e−a

G0 (t|τ ) = H(t − τ ),

H(t − τ ),

and g(r, t|ρ,τ ) =

∞ H(t − τ ) H(t − τ ) ( −a2 kn2 (t−τ ) J0 (kn ρ)J0 (kn r) + e . π π J02 (kn ) n=1

Step 4 : Show that the solution to our boundary-value problem is u(r, t) = 2π

" t" 0

1

f (ρ,τ )g(r, t|ρ,τ ) ρ d ρdτ . 0

40. Use Green’s function to show that the solution to the heat conduction problem78 , ∂u a2 ∂ ∂u = r , 0 ≤ r < b, 0 < t, ∂t r ∂r ∂r subject to the boundary conditions limr→0 |u(r, t)| < ∞, ur (b, t)+h u(b, t) = 0, 0 < t, and the initial condition that u(r, 0) =

!

0, T0 ,

0 ≤ r < a, a < r < b,

where h denotes the relative heat transfer coefficient, is u(r, t) =

∞ 2T0 ( −a2 ηn2 t J0 (rηn )[bJ1 (bηn ) − aJ1 (aηn )] e , b2 n=1 ηn [J02 (bηn ) + J12 (bηn )]

where ηn is the nth root of ηJ1 (bη) − h J0 (bη) = 0. Hint: Use Equation 5.3.9 and the results from Problem 2. Watch out for a factor of 2π. 78 For an application, see Mack, W., and M. Pl¨ ochl, 2000: Transient heating of a rotating elastic-plastic shrink fit. Int. J. Engng. Sci., 38, 921–938.

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Green’s Functions with Applications

41. Find the Green’s function79 governed by ∂g − a2 ∂t

,

∂ 2g ∂2g + 2 2 ∂x ∂y

-

= δ(x − ξ)δ(y − η)δ(t − τ ),

where 0 < x, ξ < L, and 0 < y, η < h, with the boundary conditions g(0, y, t|ξ, η,τ ) = g(L, y, t|ξ, η,τ ) = 0, gy (x, 0, t|ξ, η,τ ) − α0 g(x, 0, t|ξ, η,τ ) = 0, and gy (x, h, t|ξ, η,τ ) + α1 g(x, h, t|ξ, η,τ ) = 0, and the initial condition that g(x, y, 0|ξ, η,τ ) = 0. Step 1 : Taking the Laplace transform of the partial differential equation, show that it becomes , 2 ∂ G ∂ 2G sG − a2 + = δ(x − ξ)δ(y − η)e−sτ . ∂x2 ∂y 2 Step 2 : Assuming that G(x, y, s|ξ, η,τ ) can be expressed in terms of the expansion , - * ∞ 2 ( mπξ mπx + G(x, y, s|ξ, η,τ ) = Gm (y|η) sin sin , L m=1 L L show that Gm (y|η) is governed by d2 Gm − dy 2

,

s m2 π 2 + 2 a L2

-

Gm = −

δ(y − η) −sτ e , a2

with G$m (0|η) − α0 Gm (0|η) = 0, and G$m (h|η) + α1 Gm (h|η) = 0, Why have we chosen this particular eigenfunction expansion? Step 3 : Consider the eigenfunctions ϕn (y) = βn cos(βn y) + α0 sin(βn y),

n = 1, 2, 3, . . .

Show that they satisfy the differential equation ϕ$$n + βn2 ϕn = 0, with the boundary conditions ϕ$ (0) − α0 ϕn (0) = 0 and ϕ$ (h) + α1 ϕn (h) = 0, if βn is chosen so that (α0 α1 − βn2 ) + (α0 + α1 )βn cot(βn h) = 0. 79 See Kidawa-Kukla, J., 1997: Vibration of a beam induced by harmonic motion of a heat source. J. Sound Vibr., 205, 213–222.

The Heat Equation

393

Step 4 : Show that the solution to the ordinary differential equation in Step 2 is ∞ ( 2 ϕn (η)ϕn (y) Gm (y|η) = e−sτ , 2 h q (s + a2 βn2 + a2 m2 π 2 /L2 ) n=1 n where

qn2 = α20 + βn2 +

(α0 + α1 )(α0 α1 + βn2 ) . h(α21 + βn2 )

Hint: You will need to show that [(α0 α1 − βn2 )2 + (α0 + α1 )2 βn2 ] sin2 (βn h) = (α0 + α1 )2 βn2 . Step 5 : Taking the inverse Laplace transform, , ∞ ( ∞ ( 4 ϕn (η)ϕn (y) mπξ g(x, y, t|ξ, η,τ ) = H(t − τ ) sin hL qn2 L m=1 n=1 % , & * mπx + m2 π 2 2 × sin exp −a2 + β (t − τ ) . n L L2 42. Find the Green’s function governed by , 2 ∂g ∂ g 1 ∂g ∂2g ∂g δ(r − ρ)δ(z − ζ)δ(t − τ ) − a2 + − b + = , 2 2 ∂t ∂r r ∂r ∂z ∂z 2πr where 0 ≤ r, ρ< 1, 0 < z, ζ < 1, and 0 < t, τ, with the boundary conditions lim |g(r, z, t|ρ, ζ,τ )| < ∞,

r→0

gr (1, z, t|ρ, ζ,τ ) = 0,

0 < z < 1,

gz (r, 0, t|ρ, ζ,τ )−bg(r, 0, t|ρ, ζ,τ ) = 0, gz (r, 1, t|ρ, ζ,τ ) = 0,

0 < t,

0 ≤ r < 1, 0 < t,

and the initial condition that g(r, z, 0|ρ, ζ,τ ) = 0, 0 ≤ r < 1, 0 < z < 1. Step 1 : Introducing the new dependent variable g(r, z, t|ρ, ζ,τ ) = exp[(z − ζ)/(2b)]v(r, z, t|ρ, ζ,τ ), show that the problem becomes , 2 ∂v ∂ v 1 ∂v ∂ 2v 1 δ(r − ρ)δ(z − ζ)δ(t − τ ) − a2 + −b 2 + v = , 2 ∂t ∂r r ∂r ∂z 4b 2πr with the boundary conditions lim |v(r, z, t|ρ, ζ,τ )| < ∞,

r→0

vr (1, z, t|ρ, ζ,τ ) = 0,

0 < z < 1,

0 < t,

2bvz (r, 0, t|ρ, ζ,τ ) − v(r, 0, t|ρ, ζ,τ ) = 0,

0 ≤ r < 1,

0 < t,

2bvz (r, 1, t|ρ, ζ,τ ) + v(r, 1, t|ρ, ζ,τ ) = 0,

0 ≤ r < 1,

0 < t,

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Green’s Functions with Applications

and the initial condition that v(r, z, 0|ρ, ζ,τ ) = 0, 0 ≤ r < 1, 0 < z < 1. Step 2 : Show that the corresponding eigenfunction expansions for the delta functions are δ(z − ζ) =

∞ (

m=1

Am sin[αm (z − 12 )],

Am =

2(1 + 4b2 α2m ) sin[αm (ζ − 12 )] , 1 + 4b + 4b2 α2m

∞ δ(r − ρ) 1 1( = + Bn J0 (kn r), 2πr π π n=1

Bn =

J0 (βn ρ) , J02 (βn )

so that the eigenfunction expansions for the Green’s function is v(r, z, t|ρ, ζ,τ ) =

∞ 1 ( Am v0m (t|τ ) sin[αm (z − 12 )] π m=1

∞ ∞ 1(( + vnm (t|τ )Am Bn J0 (βn r) sin[αm (z − 12 )] π n=1 m=1

with

dg0m 1 + bα2m g0m + g0m = δ(t − τ ), dt 4b

g0m (0|τ ) = 0,

and dgnm 1 + bα2m gnm + a2 βn2 gnm + gnm = δ(t − τ ), dt 4b

gnm (0|τ ) = 0,

where αm is the mth root of 4b (αm /2) = − tan(αm /2) and βn is the nth root of J1 (β). Hint: For the eigenfunction expansion in z, consider the SturmLiouville problem: $$ Zm + α2m Zm = 0,

$ 2bZm − Zm (0) = 0,

$ 2bZm (1) + Zm (1) = 0.

Step 3 : Show that % , & 1 2 g0m (t|τ ) = exp − bαm + (t − τ ) H(t − τ ) 4b and

% , & 1 gnm (t|τ ) = exp − bα2m + a2 βn2 + (t − τ ) H(t − τ ) 4b

43. Find the Green’s function80 governed by , 2 ∂g ∂ g ∂2g ∂2g − a2 + + = δ(x − ξ)δ(y − η)δ(z − ζ)δ(t − τ ), ∂t ∂x2 ∂y 2 ∂z 2 80

Deng and Liu, op. cit.

The Heat Equation

395

where 0 < x, ξ < L x , 0 < y, η < L y , 0 < z, ζ < L z , and 0 < t, τ, with the boundary conditions gx (0, y, z, t|ξ, η, ζ,τ ) = g(Lx , y, z, t|ξ, η, ζ,τ ) = 0, gy (x, 0, z, t|ξ, η, ζ,τ ) = gy (x, Ly , z, t|ξ, η, ζ,τ ) = 0, gz (x, y, 0, t|ξ, η, ζ,τ ) = gz (x, y, Lz , t|ξ, η, ζ,τ ) = 0, and the initial condition that g(x, y, z, 0|ξ, η, ζ,τ ) = 0. 44. Redo81 Problem 43 except that the boundary conditions along the x boundaries are gx (0, y, z, t|ξ, η, ζ,τ ) = h g(0, y, z, t|ξ, η, ζ,τ ),

g(Lx , y, z, t|ξ, η, ζ,τ ) = 0.

45. Redo82 Problem 43 except that the boundary conditions are now gx (0, y, z, t|ξ, η, ζ,τ ) = 0,

gx (Lx , y, z, t|ξ, η, ζ,τ ) = −h g(Lx , y, z, t|ξ, η, ζ,τ ),

gy (x, 0, z, t|ξ, η, ζ,τ ) = 0,

gy (x, Ly , z, t|ξ, η, ζ,τ ) = −h g(x, Ly , z, t|ξ, η, ζ,τ ),

and gz (x, y, 0, t|ξ, η, ζ,τ ) = 0,

gz (x, y, Lz , t|ξ, η, ζ,τ ) = −h g(x, y, Lz , t|ξ, η, ζ,τ ),

and the initial condition that g(x, y, z, 0|ξ, η, ζ,τ ) = 0.

81

Ibid.

82 Li, F.-F., J. Liu, and K. Yue, 2009: Exact analytical solution to three-dimensional phase change heat transfer problems in biological tissues subject to freezing. Appl. Math. Mech.-Engl. Ed., 30, 63–72.

Chapter 6 Green’s Functions for the Helmholtz Equation In the previous chapters, we sought solutions to the heat and wave equations via Green’s functions. In this chapter, we turn to the reduced wave equation ∇2 u + λu = −f (r). (6.0.1) Equation 6.0.1, generally known as Helmholtz’s equation, includes the special case of Poisson’s equation when λ = 0. Poisson’s equation has a special place in the theory of Green’s functions because George Green invented his technique for its solution. The reduced wave equation arises during the solution of the harmonically forced wave equation1 by separation of variables. In one spatial dimension, the problem is ∂ 2u 1 ∂2u − 2 2 = −f (x)e−iωt . (6.0.2) 2 ∂x c ∂t Equation 6.0.2 occurs, for example, in the mathematical analysis of a stretched string over some interval subject to an external, harmonic forcing. Assuming 1 See, for example, Graff, K. F., 1991: Wave Motion in Elastic Solids. Dover Publications, Inc., Section 1.4.

397

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Green’s Functions with Applications

that u(x, t) is bounded everywhere, we seek solutions of the form u(x, t) = y(x)e−iωt . Upon substituting this solution into Equation 6.0.2, we obtain the ordinary differential equation y $$ + k02 y = −f (x),

(6.0.3)

where k02 = ω 2 /c2 . This is an example of the one-dimensional Helmholtz equation. Similar considerations hold as we include more spatial dimensions. For example, in three spatial dimensions we have ∂2g ∂ 2g ∂ 2g + 2 + 2 + λg = −δ(x − ξ)δ(y − η)δ(z − ζ), 2 ∂x ∂y ∂z

(6.0.4)

or in vectorial form ∇2 g(r|r0 ) + λg(r|r0 ) = −δ(r − r0 ).

(6.0.5)

In this notation r0 , the source variable, denotes the position of the point source (ξ, η,ζ ) and appears in the partial differential equation as a parameter. On the other hand, r, the field variable, is the variable with respect to which differentiation is carried out. Consider now the equations for the Green’s function for two different source points r1 and r2 : ∇2 g(r|r1 ) + λg(r|r1 ) = −δ(r − r1 ),

(6.0.6)

∇2 g(r|r2 ) + λg(r|r2 ) = −δ(r − r2 ).

(6.0.7)

and Multiplying Equation 6.0.6 by g(r|r2 ) and Equation 6.0.7 by g(r|r1 ), subtracting and employing Green’s second formula, we find "" ⊂⊃ [g(r|r2 )∇g(r|r1 ) − g(r|r1 )∇g(r|r2 )] · n dS = g(r2 |r1 ) − g(r1 |r2 ), (6.0.8) S

where the closed surface S is a piece-wise smooth surface that includes the points r1 and r2 . To proceed further, we need to discuss the boundary conditions. The most common ones are • the Dirichlet boundary condition, where g vanishes on the boundary, • the Neumann boundary condition, where the normal gradient of g vanishes on the boundary ∇g · n = 0, and • the Robin boundary condition, which is the linear combination of the Dirichlet and Neumann conditions.

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399

When any of these conditions hold, the surface integral vanishes and we obtain g(r2 |r1 ) = g(r1 |r2 ). Because r1 and r2 are two arbitrary points inside the region, we obtain our first important result that g(r|r0 ) = g(r0 |r),

(6.0.9)

which establishes the symmetry or reciprocity of the Green’s function under the interchange of field and source variables. Let us now use Green’s functions to solve the Helmholtz equation ∇2 u(r) + λu(r) = −f (r),

(6.0.10)

where f (r) is a specified scalar function of the field variables. If we multiply Equation 6.0.10 by g(r|r0 ) and Equation 6.0.5 by u(r), subtract and integrate, we find that u(r0 ) =

"""

V

1

2 g(r|r0 )∇2 u(r) − u(r)∇2 g(r|r0 ) dV +

"""

f (r)g(r|r0 ) dV. V

(6.0.11) Applying Green’s second formula to the first volume integral, Equation 6.0.11 becomes u(r0 ) =

"""

V

& "" % f (r)g(r|r0 ) dV − ⊂⊃ u(r)∇g(r|r0 ) − g(r|r0 )∇u(r) · n dS. S

(6.0.12) Since r0 is an arbitrary point inside of volume V , we denote it in general by r. Furthermore, the variable r is now merely a dummy integration variable that we now denote by r0 . Finally, to emphasize the fact that the gradient operation is occurring with the variable r0 , we write the gradient operator as ∇0 . Upon making these substitutions and using the symmetry condition, we have that u(r) =

"""

V

"" f (r0 )g(r|r0 ) dV0 − ⊂⊃ u(r0 ) [∇0 g(r0 |r) · n] dS0 S

0 "" 0 + ⊂⊃ g(r|r0 ) [∇0 u(r0 ) · n] dS0 .

(6.0.13)

S0

Equation 6.0.13 shows that the solution of Helmholtz’s equation depends upon the sources inside the volume V and values of u(r) and ∇u(r) on the enclosing surface. On the other hand, we must still find the particular Green’s function for a given problem; this Green’s function depends directly upon the boundary conditions. At this point, we work out several special cases.

400

Green’s Functions with Applications

1. Nonhomogeneous Helmholtz equation and homogeneous Dirichlet boundary conditions In this case, let us assume that we can find a Green’s function that also satisfies the same Dirichlet boundary conditions as u(r). Once the Green’s function is found, then Equation 6.0.13 reduces to """ u(r) = f (r0 )g(r|r0 ) dV0 . (6.0.14) V0

A possible source of difficulty would be the nonexistence of the Green’s function. From our experience in Chapter 3, we know that this will occur if λ equals one of the eigenvalues of the corresponding homogeneous problem. An example of this occurs in acoustics when the Green’s function for the Helmholtz equation does not exist at resonance. 2. Homogeneous Helmholtz equation and nonhomogeneous Dirichlet boundary conditions In this particular case f (r) = 0. For convenience, let us use the Green’s function from the previous example so that g(r|r0 ) equals zero along the boundary. Under these conditions, Equation 6.0.13 becomes "" ∂g(r0 |r) u(r) = ⊂⊃ u(r0 ) dS0 , (6.0.15) ∂n0 S0 where n0 is the unit, inwardly pointing normal to the boundary. Consequently, the solution is determined once we compute the normal gradient of the Green’s function along the boundary. 3. Nonhomogeneous Helmholtz equation and homogeneous Neumann boundary conditions If we require that u(r) satisfies the nonhomogeneous Helmholtz equation with homogeneous Neumann boundary conditions, then the governing equations are ∇2 u(r) + λu(r) = −f (r),

and

∂u = 0, ∂n

on S.

(6.0.16)

Integrating Equation 6.0.16 and using Gauss’s divergence theorem, we obtain "" """ """ ⊂⊃ ∇u · n dS + λ u(r) dV = − f (r) dV. (6.0.17) S

V

V

Because the first integral in Equation 6.0.17 must vanish in the case of homogeneous Neumann boundary conditions, this equation cannot be satisfied if λ = 0 unless """ f (r) dV = 0. (6.0.18) V

The Helmholtz Equation

401

A physical interpretation of Equation 6.0.18 is as follows: Consider the physical process of steady-state heat conduction within a finite region V . The temperature u is given by Poisson’s equation ∇2 u = −f , where f is proportional to the density of the heat sources and sinks. The boundary condition ∂u/∂n = 0 implies that there is no heat exchange across the boundary. Consequently, no steady-state temperature distribution can exist unless the heat sources are balanced by heat sinks. This balance of heat sources and sinks is given by Equation 6.0.18. This same problem can be solved using Green’s functions. Equation 6.0.13 suggests that it would be convenient to choose a Green’s function that also satisfies the homogeneous boundary condition ∂g/∂n = 0 on S. However, the differential equation governing the Green’s function /// is ∇2 g + λg = −δ(r − r0 ). Integrating this equation over the volume leads to λ V g(r|r0 ) dV = −1. This cannot be true if λ = 0. On the other hand, we just showed that solutions to/// Poisson’s equation with homogeneous Neumann boundary conditions exist if V f (r) dV = 0. This contradiction forces us to abandon the homogeneous Neumann boundary condition for another. Let us try ∂g/∂n =//C, a constant. To find C, we integrate ∇2 g = −δ(r − r0 ) and find that C ⊂⊃S dS = −1 or C = −1/As , where As is the area of the surface S. A disadvantage of using this nonhomogeneous Neumann boundary condition is the loss of the symmetry condition g(r|r0 ) = g(r0 |r). However, retracing our steps in the // symmetry proof, we see that if we impose the additional constraint that ⊂⊃S g(r|r0 ) dS = 0, then symmetry still holds. Equation 6.0.13 still applies and yields """ "" 1 u(r) = f (r0 )g(r|r0 ) dV0 + ⊂⊃ u(r0 ) dS0 . (6.0.19) A s S0 V0 While the last term cannot be evaluated unless u(r) is known, it is a constant. Therefore, Equation 6.0.19 yields u(r) up to an arbitrary constant. For most physical problems, this is adequate. We omit the proof that a Green’s function g(r|r0 ) that satisfies all of our conditions can be found. 6.1 FREE-SPACE GREEN’S FUNCTIONS FOR HELMHOLTZ’S AND POISSON’S EQUATIONS Before we plunge into the solution of the various shades of the Helmholtz equation, let us focus on the simple problem of finding the Green’s function for the Helmholtz equation ∇2 g(r|r0 ) + k02 g(r|r0 ) = −δ(r − r0 )

(6.1.1)

when there are no boundaries present. The primary use of this free-space Green’s function 2 is as a particular solution to the Green’s function problem. 2 In electromagnetic theory, a free-space Green’s function is the particular solution of the differential equation valid over a domain of infinite extent, where the Green’s function remains bounded as we approach infinity, or satisfies a radiation condition there.

402

Green’s Functions with Applications

• One-dimensional Helmholtz equation Clearly, the simplest problem is the one-dimensional case. Let us find the Green’s function for g $$ + k02 g = −δ(x − ξ),

−∞ < x, ξ < ∞.

(6.1.2)

If we solve Equation 6.1.2 by piecing together homogeneous solutions, then

for x < ξ while

g(x|ξ) = Ae−ik0 (x−ξ) + Beik0 (x−ξ) ,

(6.1.3)

g(x|ξ) = Ce−ik0 (x−ξ) + Deik0 (x−ξ) ,

(6.1.4)

for ξ < x. Let us examine the solution, Equation 6.1.3, more closely. The solution represents two propagating waves. Because x < ξ, the first term is a wave propagating out to infinity, while the second term gives a wave propagating in from infinity. This is seen most clearly by including the e−iωt term into Equation 6.1.3 or g(x|ξ)e−iωt = Ae−ik0 (x−ξ)−iωt + Beik0 (x−ξ)−iωt .

(6.1.5)

Because we have a source only at x = ξ, solutions that represent waves originating at infinity are nonphysical and we must discard them. This requirement that there are only outwardly propagating wave solutions is commonly called Sommerfeld’s radiation condition.3 Similar considerations hold for Equation 6.1.4 and we must take C = 0. To evaluate A and D, we use the continuity conditions on the Green’s function: g(ξ + |ξ) = g(ξ − |ξ),

and g $ (ξ + |ξ) − g $ (ξ − |ξ) = −1,

(6.1.6)

or A = D,

and

ik0 D + ik0 A = −1.

(6.1.7)

Therefore,

g(x|ξ) =

i ik0 |x−ξ| e . 2k0

(6.1.8)

3 Sommerfeld, A., 1912: Die Greensche Funktion der Schwingungsgleichung. Jahresber. Deutsch. Math.-Verein., 21, 309–353.

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403

iy (b) Contour for x


Figure 6.1.1: Contour used to evaluate Equation 6.1.10.

We can also solve Equation 6.1.2 by Fourier transforms. Assuming that the Fourier transform of g(x|ξ) exists and denoting it by G(k|ξ), we find that G(k|ξ) = and

1 g(x|ξ) = 2π

"

e−ikξ , k 2 − k02 ∞

−∞

eik(x−ξ) dk. k 2 − k02

(6.1.9)

(6.1.10)

Immediately we see that there is a problem with the singularities lying on the path of integration at k = ±k0 . How do we avoid them? There are four possible ways that we might circumvent the singularities. One of them is shown in Figure 6.1.1. Applying Jordan’s lemma to close the line integral along the real axis (as shown in Figure 6.1.1),

1 g(x|ξ) = 2π

0

C

eiz(x−ξ) dz. z 2 − k02

(6.1.11)

For x < ξ, %

& eiz(x−ξ) i −ik0 (x−ξ) g(x|ξ) = −i Res 2 ; −k0 = e , z − k02 2k0 while g(x|ξ) = i Res

%

& eiz(x−ξ) i ik0 (x−ξ) ; k e , 0 = z 2 − k02 2k0

(6.1.12)

(6.1.13)

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Green’s Functions with Applications

for x > ξ. A quick check shows that these solutions agree with Equation 6.1.8. If we try the three other possible paths around the singularities, we obtain incorrect solutions. • Example 6.1.1 During their development of a uniform integral representation for geometric optics solutions near caustics, Hazak et al.4 found the Green’s function for the one-dimensional Helmholtz equation: d2 g + (B − Ax)g = δ(x − ξ), dx2

−∞ < x, ξ < ∞.

(6.1.14)

A quick check shows that the Green’s function has the solution g(x|ξ) = CAi[−z(x> )] {Bi[−z(x< )] + iAi[−z(x< )]} ,

(6.1.15)

where z(x) = (B − Ax)/A2/3 and Ai(·) and Bi(·) are Airy functions. This solution assumes that B/A >ξ . The first term on the right side of Equation 6.1.15 represents a damped wave as x → ∞ while the bracketed term is a leftward propagating wave as x → −∞ if we assume that the time dependence is eiωt . Because 'ξ + ' $ g (x|ξ)'' = 1, (6.1.16) − ξ

we find that C = −π/A1/3 since the Wronskian W [Ai(z), Bi(z)] = 1/π. Therefore, the final answer is g(x|ξ) = −πAi[−z(x> )] {Bi[−z(x< )] + iAi[−z(x< )]} /A1/3 ,

(6.1.17) # "

• One-dimensional Poisson equation There is no solution for the domain −∞ < x < ∞. • Two-dimensional Helmholtz equation At this point, we have found two forms of the free-space Green’s function for the one-dimensional Helmholtz equation. The first form is the analytic 4 Hazak, G., L. Friedland, and I. B. Bernstein, 1984: A uniform integral representation for geometric optics solutions near caustics. Phys. Fluids, 27, 129–132.

The Helmholtz Equation

405

Table 6.1.1: Free-Space Green’s Functions for the Poisson and Helmholtz Equations Dimension

Poisson Eq.

One

no solution

Two

g(x, y|ξ,η ) = −

Three

g(x, y, z|ξ, η,ζ ) =

r=

.

(x − ξ)2 + (y − η)2 ,

Helmholtz Eq. g(x|ξ) =

ln(r) 2π 1 4πR

R=

.

i ik0 |x−ξ| e 2k0

g(x, y|ξ,η ) =

i (1) H (k0 r) 4 0

g(x, y, z|ξ, η,ζ ) =

eik0 R 4πR

(x − ξ)2 + (y − η)2 + (z − ζ)2

Note: For the Helmholtz equation, we have taken the temporal forcing to be e−iωt and k0 = ω/c.

solution, Equation 6.1.8, while the second is the integral representation, Equation 6.1.10, where the line integration along the real axis is shown in Figure 6.1.1. In the case of two dimensions, the Green’s function5 for the Helmholtz equation symmetric about the point (ξ,η ) is the solution of the equation d2 g 1 dg δ(r) + + k02 g = − , 2 dr r dr 2πr

(6.1.18)

. where r = (x − ξ)2 + (y − η)2 . The homogeneous form of Equation 6.1.18 is Bessel’s differential equation of order zero. Consequently, the general solution in terms of Hankel functions is (1)

(2)

g(r|r0 ) = A H0 (k0 r) + B H0 (k0 r).

(6.1.19)

Why have we chosen to use Hankel functions rather than J0 (·) and Y0 (·)? As we argued earlier, solutions to the Helmholtz equation must represent outwardly propagating waves (the Sommerfeld radiation condition). If we again assume that the temporal behavior is e−iωt and use the asymptotic expressions for Hankel functions, Equation 2.3.24 and Equation 2.3.25, we see (1) that H0 (k0 r) represents outwardly propagating waves and B = 0. 5 For an alternative derivation, see Graff, K. F., 1991: Wave Motion in Elastic Solids. Dover Publications, Inc., pp. 284–285.

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Green’s Functions with Applications

What is the value of A? Integrating Equation 6.1.18 over a small circle around the point r = 0 and taking the limit as the radius of the circle vanishes, A = i/4 and

g(r|r0 ) =

i (1) H (k0 r). 4 0

(6.1.20)

If a real function is needed, then the free-space Green’s function equals the Neumann function Y0 (k0 r) divided by −4. As we did in the one-dimensional case, let us solve the two-dimensional Helmholtz equation by Fourier transforms.6 The governing equation is ∂2g ∂ 2g + 2 + k02 g = −δ(x − ξ)δ(y − η), 2 ∂x ∂y

−∞ < x, y, ξ, η < ∞.

(6.1.21)

Taking the Fourier transform of Equation 6.1.21, we find that d2 G + 02 G = −δ(y − η)e−ikξ , dy 2

(6.1.22)

where 02 = k02 − k 2 , and k is the transform variable in the x direction. Using the results from the one-dimensional Helmholtz equation, G(k, y|ξ,η ) =

i i4|y−η|−ikξ e , 20

(6.1.23)

with 3(0) and 5(0) ≥ 0. These choices ensure outwardly radiating waves if the temporal behavior is e−iωt . Taking the inverse of Equation 6.1.23, we obtain " ∞ i 1 i4|y−η|+ik(x−ξ) g(x, y|ξ,η ) = e dk (6.1.24) 4π −∞ 0 " ∞ i cos[k(x − ξ)]ei4|y−η| = dk. (6.1.25) 2π 0 0 As an interesting aside, we can use Equation 6.1.20 and Equation 6.1.25 to obtain an integral property involving Hankel functions. Because there is only one unique free-space Green’s function for the two-dimensional Helmholtz 6 Gunda, R., S. M. Vijayakar, R. Singh, and J. E. Farstad, 1998: Harmonic Green’s functions of a semi-infinite plate with clamped or free edges. J. Acoust. Soc. Am., 103, 888–899. See Section I.D.

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407

equation, these equations must be equivalent. Setting these two representations equal to each other, we obtain the Cartesian decomposition of the zero order Hankel function of the first kind √ 2 2 " 1 ∞ eik(x−ξ)−i k0 −k |y−η| (1) . H0 (k0 r) = dk, (6.1.26) π −∞ k02 − k 2

where the imaginary part of the radical must be zero or negative and 5(k0 ) ≥ 0. In Section 4.5 we showed how the free-space Green’s function for the three-dimensional Helmholtz equation can be written as a superposition of cylindrical waves, Equation 4.5.29. Here we found that the free-space Green’s function for the two-dimensional Helmholtz equation can be written as a superposition of plane waves,7 Equation 6.1.25. Why are such decompositions useful? Imagine a situation where we wish to use the free-space Green’s function but we need to evaluate it along some straight edge. In that case, this equation would be crucial to our analysis. This ability to re-express the solutions of Helmholtz’s equation in one coordinate system in terms of another coordinate system is further examined in Appendix A. We can also solve the planar Helmholtz equation in terms of polar coordinates , 1 ∂ ∂g 1 ∂ 2g δ(r − ρ)δ(θ − θ$ ) r + 2 2 + k02 g = − , (6.1.27) r ∂r ∂r r ∂θ r where 0 < r, ρ < ∞, and 0 ≤ θ,θ $ ≤ 2π. Because δ(θ − θ $ ) =

∞ ∞ 1 1( 1 ( + cos[n(θ − θ$ )] = cos[n(θ − θ$ )], 2π π n=1 2π n=−∞

(6.1.28)

we assume a solution of the form g(r,θ |ρ,θ $ ) =

∞ (

n=−∞

gn (r|ρ) cos[n(θ − θ $ )].

(6.1.29)

Substituting Equation 6.1.28 and Equation 6.1.29 into Equation 6.1.27 and simplifying, we find that , 1 d dgn n2 δ(r − ρ) r − 2 gn + k02 gn = − . r dr dr r 2πr

(6.1.30)

7 This result has been generalized by Cincotti, G., F. Gori, M. Santarsiero, F. Frezza, F. Furn` o, and G. Schettini, 1993: Plane wave expansion of cylindrical functions. Optics Comm., 95, 192–198.

408

Green’s Functions with Applications

The homogeneous solution of Equation 6.1.30 is < AJ|n| (k0 r), 0 ≤ r ≤ ρ, gn (r|ρ) = (1) BH|n| (k0 r), ρ ≤ r < ∞.

(6.1.31)

This solution possesses the properties of remaining finite as r → 0 and corresponding to a radiating wave solution in the limit of r → ∞. If we now use the requirements that the solution is continuous at r = ρ, and that 'r=ρ+ dgn (r|ρ) '' 1 ρ =− , (6.1.32) ' dr 2π r=ρ− where ρ+ and ρ− denote points slightly greater and less than ρ, then gn (r|ρ) =

i (1) H (k0 r> )J|n| (k0 r< ), 4 |n|

(6.1.33)

so that g(r,θ |ρ,θ $ ) =

∞ i ( (1) cos[n(θ − θ$ )]H|n| (k0 r> )J|n| (k0 r< ). 4 n=−∞

(6.1.34)

Watanabe and Payton8 developed an alternative representation of the free-space Green’s function. We begin by rotating the coordinate system so that θ$ = 0 and find that g(r,θ |ρ, 0) =

∞ i ( (1) H (k0 r> )J|n| (k0 r< )einθ . 4 n=−∞ |n|

(6.1.35)

Next, we introduce the integral formula9 " 4. 5 1 π In (a)Kn (b) = K0 a2 + b2 − 2ab cos(ϕ) cos(nϕ) dϕ (6.1.36) π 0 " 4. 5 sin(nπ) ∞ − K0 a2 + b2 + 2ab cosh(ξ) e−nξ dξ. π 0 Because10

* + * + πi In ae−πi/2 Kn be−πi/2 = Jn (a)Hn(1) (b), 2

(6.1.37)

8 Watanabe, K., and R. G. Payton, 2001: Source of a time-harmonic SH wave in a cylindrically orthotropic elastic solid. Geophys. J. Int., 145, 709–713. Because they assume a temporal dependence eiωt rather than e−iωt , their radial dependence involved (2) (1) Hν (·) rather than Hν (·). 9

Oberhettinger, F., 1958: On the diffraction and reflection of waves and pulses by wedges and corners. J. Res. NBS, Ser. D , 61, 343–365. 10 Lebedev, N. N., 1972: Special Functions and Their Applications. Dover, Equation 5.7.4 and Equation 5.7.5.

The Helmholtz Equation

409

we can rewrite Equation 6.1.36 as " 5 1 π (1) 4. 2 Jn (a)Hn(1) (b) = H0 a + b2 − 2ab cos(ϕ) cos(nϕ) dϕ (6.1.38) π 0 " 5 sin(nπ) ∞ (1) 4. 2 − H0 a + b2 + 2ab cosh(ξ) e−nξ dξ. π 0 Let

z1 =

. ρ2 + r2 − 2ρr cos(ϕ)

and z2 =

Then, Equation 6.1.35 becomes g(r,θ |ρ, 0) =

i 4π

!"

and

(1)

H0 (k0 z1 )

0

− Since

π

"

0

∞

.

∞ (

ρ2 + r2 + 2ρr cosh(ξ). (6.1.39)

cos(|n|ϕ)einθ dϕ

n=−∞ ∞ (

(1) H0 (k0 z2 )

sin(|n|π)e

inθ −|n|ξ

n=−∞

∞ ∞ ( 1 ( + cos(nx) = π δ(x − 2mπ), 2 n=1 m=−∞

2

∞ (

sin(nx)e−ny =

n=1

sin(x) , cosh(y) − cos(x)

e

(6.1.40) ) dξ .

(6.1.41)

(6.1.42)

we can rewrite Equation 6.1.40 as ∞ ( 4π g(r,θ |ρ, 0) = H(π − |θ + 2mπ|) i m=−∞ 4 . 5 (1) × H0 k0 ρ2 + r2 − 2ρr cos(θ + 2mπ) " ∞ 4 . 5 1 (1) − H0 k0 ρ2 + r2 + 2ρr cosh(ξ) (6.1.43) 2π 0 % & sin(π − θ) sin(π + θ) × + dξ. cosh(ξ) − cos(π − θ) cosh(ξ) − cos(π + θ)

The summation term in Equation 6.1.43 arises from a series of fictitious source points located at r = a and θ = −2mπ. The integral term are internally circular diffracted waves originating from the origin. A version of this problem that arose in diffraction theory11 involves solving Equation 6.1.27 over the wedge 0 ≤ r, ρ< ∞ and 0 < θ, θ$ < β. Here 11 Davies, H., 1906: On the solution of problems in diffraction by the aid of contour integration. Philos. Mag., Ser. 6 , 12, 63–67.

410

Green’s Functions with Applications

g(r, 0|ρ,θ $ ) = g(r,β |ρ,θ $ ) = 0. Because of these boundary conditions, we choose to write the delta and Green’s function as , - , ∞ 2( nπθ$ nπθ sin sin β n=1 β β

(6.1.44)

, - , ∞ 2( nπθ $ nπθ g(r,θ |ρ,θ ) = gn (r|ρ) sin sin β n=1 β β

(6.1.45)

, 1 d dgn n2 π 2 δ(r − ρ) r − 2 2 gn + k02 gn = − . r dr dr r β r

(6.1.46)

δ(θ − θ$ ) = and

$

with

The homogeneous solution of Equation 6.1.46 is gn (r|ρ) =

!

AJnπ/β (k0 r), BKnπ/β (k0 r),

0 ≤ r ≤ ρ, ρ ≤ r < ∞.

(6.1.47)

This solution possesses the properties of remaining finite as r → 0 and corresponding to a radiating wave solution in the limit of r → ∞. If we now use the requirements that the solution is continuous at r = ρ, and that 'r=ρ+ dgn (r|ρ) '' ρ = −1, dr 'r=ρ−

(6.1.48)

where ρ+ and ρ− denote points slightly greater and less than ρ, then gn (r|ρ) = Jnπ/β (k0 r< )Knπ/β (k0 r> ),

(6.1.49)

so that , - , ∞ 2( nπθ$ nπθ g(r,θ |ρ,θ ) = Jnπ/β (k0 r< )Knπ/β (k0 r> ) sin sin , β n=1 β β (6.1.50) since Kν$ (z)Jν (z) − Jν$ Kν (z) = −1/z. $

• Example 6.1.2 Let us find the free-space Green’s function for , 1 ∂ ∂g ∂2g δ(r − ρ)δ(z − ζ) r + 2 + k02 g = − , r ∂r ∂r ∂z r where 0 < r, ρ < ∞, and −∞ < z, ζ < ∞.

(6.1.51)

The Helmholtz Equation

411

Taking the Fourier transform of Equation 6.1.51 with respect find that , 1 d dG δ(r − ρ) −ikζ r + κ2 G = − e , r dr dr r . where κ = k02 − k 2 with the 5(κ) ≥ 0. The homogeneous solution tion 6.1.52 is ! AJ0 (κr), 0 ≤ r ≤ρ, G(r, k|ρ,ζ ) = (1) BH0 (κr), ρ ≤ r < ∞.

to z, we (6.1.52) of Equa-

(6.1.53)

This solution possesses the properties of remaining finite as r → 0 and corresponding to a radiating wave solution in the limit of r → ∞. If we now use the requirements that the solution is continuous at r = ρ so that G(ρ− , k|ρ,ζ ) = G(ρ+ , k|ρ,ζ ) and that 'r=ρ+ dG(r, k|ρ,ζ ) '' 1 ' − = −ρ, dr r=ρ

(6.1.54)

where ρ+ and ρ− denote points slightly greater and less than ρ, the coefficients A and B are determined by , -, - , (1) J0 (κρ) −H0 (κρ) A 0 = . (6.1.55) (1) B 1/ρ κJ0$ (κρ) −κH $ 0 (κρ) Because the Wronskian determinant is 4 5 2i (1) W J0 (κρ), H0 (κρ) = , πκρ A=

πi (1) H (κρ), 2 0

Therefore, G(r, k|ρ,ζ ) = and g(r, z|ρ,ζ ) =

i 4

"

∞ −∞

and B =

πi J0 (κρ). 2

πi (1) J0 (κr< )H0 (κr> ), 2 (1)

J0 (κr< )H0 (κr> )eik(z−ζ) dk.

(6.1.56) (6.1.57)

(6.1.58)

(6.1.59) # "

• Two-dimensional Laplace equation In this subsection, we find the free-space Green’s function for Poisson’s equation in two dimensions. This Green’s function is governed by , 1 ∂ ∂g 1 ∂ 2g δ(r − ρ)δ(θ − θ $ ) r + 2 2 =− . (6.1.60) r ∂r ∂r r ∂θ r

412

Green’s Functions with Applications

If we now choose our . coordinate system so that the origin is located at the point source, r = (x − ξ)2 + (y − η)2 and ρ = 0. Multiplying both sides of this simplified Equation 6.1.60 by r dr d θ and integrating over a circle of radius 2, we obtain −1 on the right side from the surface integration over the delta functions. On the left side, ' " 2π ∂g '' r ' dθ = −1. (6.1.61) ∂r r=' 0

The Green’s function g(r,θ |0, θ$ ) = − ln(r)/(2π) satisfies Equation 6.1.61. To find an alternative form of the free-space Green’s function when the point of excitation and the origin of the coordinate system do not coincide, we first note that ∞ 1 ( in(θ−θ $ ) δ(θ − θ$ ) = e . (6.1.62) 2π n=−∞ This suggests that the Green’s function should be of the form g(r,θ |ρ,θ $ ) =

∞ (

$

gn (r|ρ)ein(θ−θ ) .

(6.1.63)

n=−∞

Substituting Equation 6.1.62 and Equation 6.1.63 into Equation 6.1.60, we obtain the ordinary differential equation , 1 d dgn n2 δ(r − ρ) r − 2 gn = − . (6.1.64) r dr dr r 2πr The homogeneous solution to Equation 6.1.64 is ! a, 0 ≤ r ≤ ρ, g0 (r|ρ) = b ln(r), ρ ≤ r < ∞, and gn (r|ρ) =

!

c (r/ρ)n , d (ρ/r)n ,

0 ≤ r ≤ ρ, ρ ≤ r < ∞,

(6.1.65)

(6.1.66)

if n &= 0. At r = ρ, the gn ’s must be continuous, in which case, a = b ln(ρ), On the other hand,

or a=−

ln(ρ) , 2π

and

c = d.

'r=ρ+ dgn '' 1 ρ =− , dr 'r=ρ− 2π

b=−

1 , 2π

and c = d =

(6.1.67)

(6.1.68) 1 . 4πn

(6.1.69)

The Helmholtz Equation

413

Therefore, g(r,θ |ρ,θ $ ) = −

, -n ∞ ln(r> ) 1 ( 1 r< + cos[n(θ − θ $ )]. 2π 2π n=1 n r>

(6.1.70)

We can simplify Equation 6.1.70 by noting that ∞ ( 1 2 ρn cos[n(θ − θ$ )] ln 1 + ρ2 − 2ρ cos(θ − θ$ ) = −2 , n n=1

(6.1.71)

if |ρ| < 1. Applying this relationship to Equation 6.1.70, we find that g(r,θ |ρ,θ $ ) = −

1 2 1 ln r2 + ρ2 − 2rρ cos(θ − θ$ ) . 4π

Note that when ρ = 0 we recover g(r,θ |0, θ$ ) = − ln(r)/(2π). A similar free-space problem involves solving , 1 ∂ ∂g ∂2g δ(r − ρ)δ(z − ζ) r + 2 =− , r ∂r ∂r ∂z 2πr

(6.1.72)

(6.1.73)

over the domain 0 ≤ r, ρ< ∞ and −∞ < z, ζ < ∞, with the boundary conditions lim |g(r, z|ρ,ζ )| < ∞,

r→0

lim g(r, z|ρ,ζ ) → 0,

r→∞

−∞ < z < ∞,

(6.1.74)

and lim g(r, z|ρ,ζ ) → 0,

|z|→∞

0 ≤ r < ∞.

Introducing the Hankel transform " ∞ G(k, z) = g(r, z)J0 (kr) r dr,

(6.1.75)

(6.1.76)

0

Equation 6.1.73 becomes d2 G J0 (kρ) − k2 G = − δ(z − ζ). dz 2 2π

(6.1.77)

subject to the boundary condition lim|z|→∞ G(k, z|ρ,ζ ) → 0. The solution to Equation 6.1.77 is e−k|z−ζ| G(k, z|ρ,ζ ) = . (6.1.78) 4πk Consequently, the free-space Green’s function is " ∞ 1 g(r, z|ρ,ζ ) = e−k|z−ζ| J0 (kρ)J0 (kr) dk. (6.1.79) 4π 0

414

Green’s Functions with Applications • Three-dimensional Helmholtz equation

Finally, we treat the three-dimensional Helmholtz equation. In particular, we now prove that

g(r|r0 ) =

eik0 R , 4πR

(6.1.80)

where R = |r − r0 |. This particular solution is the three-dimensional freespace Green’s function. We begin by noting that Equation 6.1.80 satisfies ∇2 g(r|r0 )+k02 g(r|r0 ) = 0 everywhere except at the point r0 . Next, we integrate both sides of this equation over a sphere of very small radius 2, centered at r0 . On the right side, we obtain −1 from the volume integration over the delta functions. The integration over the left side consists of two integrals. The integral involving k02 g(r|r0 ) vanishes because it tends to zero as 23 as 2 → 0. The volume integral of ∇2 g(r|r0 ) can be replaced by a closed surface integral of ∇g(r|r0 ) · n by Gauss’s divergence theorem. This radial gradient, centered at r0 , equals ∂g(r|r0 )/∂r = [−1/(4πR2 ) + ik0 /(4πR)]eik0 R , which is constant over the spherical surface R = 2. Thus, the closed surface integral equals ∂g(r|r0 )/∂r times the surface area 4π22 of the sphere. As 2 → 0, this integral approaches the value −1, which also equals the right side, verifying our solution for g(r|r0 ). We chose eik0 R /(4πR) rather than e−ik0 R /(4πR) for the Green’s function because the former represents an outwardly radiating solution. • Three-dimensional Poisson equation Because Helmholtz’s equation becomes Poisson’s equation if k0 = 0, 1/(4πR) is the free-space Green’s function for the three-dimensional Poisson equation. • Example 6.1.3 Let us show that the three-dimensional, free-space Green’s function12 for Poisson’s equation in cylindrical coordinates is given by " ∞ 1 $ g(r, θ, z|ρ,θ , ζ) = e−k|z−ζ| J0 (kR) dk (6.1.81) 4π 0 12 Dougall, J., 1900: The determination of Green’s functions by means of cylindrical and spherical harmonics. Proc. Edinburgh Math. Soc., 18, 33–83.

The Helmholtz Equation =

1 4π

415 "

∞

e−k|z−ζ|

0

∞ (

Jn (kρ)Jn (kr) cos[n(θ − θ$ )] dk,

n=−∞

(6.1.82)

. where R = r2 + ρ2 − 2rρ cos(θ − θ$ ). To prove these relationships, we begin with , 1 ∂ ∂g ∂2g 1 ∂ 2g δ(r − ρ)δ(z − ζ)δ(θ − θ $ ) r + 2 + 2 2 =− . r ∂r ∂r ∂z r ∂θ r

(6.1.83)

Because δ(θ − θ$ ) can be re-expressed as Equation 6.1.28, we write the Green’s function as g(r, θ, z|ρ,θ $ , ζ) =

∞ (

n=−∞

gn (r, z|ρ,ζ ) cos[n(θ − θ $ )].

(6.1.84)

Substituting Equation 6.1.28 and Equation 6.1.84 into Equation 6.1.83 and simplifying, we find that , 1 ∂ ∂gn ∂ 2 gn n2 δ(r − ρ)δ(z − ζ) r + − 2 gn = − . (6.1.85) 2 r ∂r ∂r ∂z r 2πr Taking the Hankel transform of Equation 6.1.85, we obtain d2 Gn Jn (kρ)δ(z − ζ) − k 2 Gn = − . dz 2 2π

(6.1.86)

The solution to Equation 6.1.86 is Gn (k, z|ρ,ζ ) =

Jn (kρ) exp(−k|z − ζ|) . 4πk

(6.1.87)

Taking the inverse of Equation 6.1.87 and substituting the inverse into Equation 6.1.84, we obtain Equation 6.1.82. To obtain Equation 6.1.81, we recall the addition theorem for Bessel functions: J0

4.

ζ2

+

z2

5

− 2ζz cos(ϕ) =

∞ (

Jn (ζ)Jn (z) cos(nϕ).

(6.1.88)

n=−∞

Setting ζ = kρ, z = kr, and ϕ = θ − θ$ in Equation 6.1.88 and then substituting this modified version of Equation 6.1.88 into Equation 6.1.82, we obtain Equation 6.1.81. An alternative to Equation 6.1.82 can be derived starting with Equation 6.1.85. However, at this point, we take the Fourier cosine transform with respect to z: " 1 ∞ f (z) = F (k) cos[k(z − ζ)] dk, (6.1.89) π 0

416

Green’s Functions with Applications

so that the transformed Equation 6.1.85 is , - , 1 d dGn n2 δ(r − ρ) r − k 2 + 2 Gn = − . r dr dr r 2πr

(6.1.90)

Applying the techniques of Section 3.3, we solve Equation 6.1.90 and find that Gn (r|ρ) =

In (kr< )Kn (kr> ) . 4π

(6.1.91)

Therefore, g(r, θ, z|ρ,θ $ , ζ) =

=

∞ " ∞ $ 1 ( ein(θ−θ ) cos[k(z − ζ)] 2 2π n=−∞ 0 ∞ (

1 4π 2n=−∞

"

∞

(6.1.92)

× In (kr< )Kn (kr> ) dk $

ein(θ−θ )+ik(z−ζ) In (|k|r< )Kn (|k|r> ) dk.

−∞

(6.1.93)

Finally, Murashima13 derived the integral representation 1 g(r, θ, z|ρ,θ , ζ) = 3 π $

"

∞ 0

provided that 0 < |θ − θ$ | < 2π.

"

0

∞

cos[k(z − ζ)] cosh[η(π − |θ − θ$ |)] × Kiη (kr)Kiη (kρ) dη dk,

(6.1.94)

# "

• Example 6.1.4 As an example of how we can re-express a Green’s function in one coordinate system (spherical coordinates) in terms of another one (cylindrical coordinates), let us show that %" k0 ∞ ( eik0 R iξJ n (rξ)Jn (ρξ) i√k2 −ξ2 |z−ζ| 0 . = cos[n(θ − θ$ )] e dξ R k02 − ξ 2 0 n=−∞ & " ∞ ξJn (rξ)Jn (ρξ) −√ξ2 −k2 |z−ζ| 0 . + e dξ . ξ 2 − k02 k0 (6.1.95) 13

Murashima, S., 1973: Neumann functions for Laplace’s equation for a circular cylinder of finite length. Jap. J. Appl. Phys., 12, 1232–1243. See his Equation (21).

The Helmholtz Equation

417

Table 6.1.2: Free-Space Green’s Function for the Axisymmetric Modified Helmholtz Equation , 1 ∂ ∂g ∂ 2g n2 δ(r − ρ)δ(z − ζ) r + 2 − k02 g − 2 g = − r ∂r ∂r ∂z r 2πr k0 0

k0

si

Free-Space Green’s Function 1 4π

"

1 g= 4π

"

g=

∞

Jn (rξ)Jn (ρξ)e−ξ|z−ζ| dξ

0

0

∞

.

ξ ξ 2 + k02

√2 2 Jn (rξ)Jn (ρξ)e− ξ +k0 |z−ζ| dξ

" s 1 iξJ n (rξ)Jn (ρξ) i√s2 −ξ2 |z−ζ| . g= e dξ 4π 0 s2 − ξ 2 " ∞ 1 ξJn (rξ)Jn (ρξ) −√ξ2 −s2 |z−ζ| . + e dξ 4π s ξ 2 − s2

Consider the modified Helmholtz equation , 1 ∂ ∂g ∂2g 1 ∂ 2g δ(r − ρ)δ(z − ζ)δ(θ − θ$ ) r + 2 + 2 2 − k02 g = − . (6.1.96) r ∂r ∂r ∂z r ∂θ r Because δ(θ − θ $ ) is given by Equation 6.1.28, we assume a solution of the form ∞ ( g(r, θ, z|ρ,θ $ , ζ) = gn (r, z|ρ,ζ ) cos[n(θ − θ $ )]. (6.1.97) n=−∞

Substituting Equation 6.1.28 and Equation 6.1.97 into Equation 6.1.96 and simplifying, we find that , 1 ∂ ∂gn ∂ 2 gn n2 δ(r − ρ)δ(z − ζ) r + − gn − k02 gn = − . (6.1.98) 2 2 r ∂r ∂r ∂z r 2πr Next, we take the Hankel transform " ∞ Gn (k, z|ρ,ζ ) = gn (r, z|ρ,ζ )Jn (kr) r dr

(6.1.99)

0

of Equation 6.1.98 and obtain $ d2 Gn # 2 Jn (kρ)δ(z − ζ) − k0 + k 2 Gn = − . 2 dz 2π

(6.1.100)

418

Green’s Functions with Applications The solution to Equation 6.1.100 is . Jn (kρ) exp(−|z − ζ| k02 + k 2 ) . Gn (k, z|ρ,ζ ) = . 4π k02 + k 2

Taking the inverse of Equation 6.1.101, we obtain " ∞ √ 1 ξ −|z−ζ| k02 +ξ2 . gn (r, z|ρ,ζ ) = J dξ. n (rξ)Jn (ρξ)e 4π 0 k02 + ξ 2

(6.1.101)

(6.1.102)

Equation 6.1.102 serves as the free-space Green’s function for the axisymmetric modified Helmholtz equation, Equation 6.1.98. Additional variations are listed in Table 6.1.2. Lu14 developed equivalent expressions when n = 0 and 1 in terms of (1) a power series of spherical Hankel functions and (2) a sum of a singular term involving a Legendre function of degree 12 of the second kind and a regular term. The final result follows from the substitution of Equation 6.1.102 into Equation 6.1.97 or g(r,θ, z|ρ,θ $ , ζ) (6.1.103) :" ; ∞ ∞ √ ( 1 ξJ (rξ)J (ρξ) 2 2 n n . = cos[n(θ − θ $ )] e−|z−ζ| k0 +ξ dξ . 4π n=−∞ k02 + ξ 2 0

Because e−k0 R /(4πR), where R2 = r2 + ρ2 − 2rρ cos(θ − θ$ ) + (z − ζ)2 , is also a free-space Green’s function, we find that :" ; ∞ ∞ √ 2 2 ( e−k0 R ξJ (rξ)J (ρξ) n n . = cos[n(θ − θ$ )] e−|z−ζ| k0 +ξ dξ . R k02 + ξ 2 0 n=−∞ (6.1.104) Equation 6.1.95 follows by replacing k0 with −ik0 . In the case when there is no θ dependence and ρ = 0, a similar derivation leads to the famous Lamb or Sommerfeld integral: eik0 R = R +

"

"

k0

0 ∞

k0

iξJ n (rξ)Jn (ρξ) i√k2 −ξ2 |z−ζ| 0 . e dξ k02 − ξ 2 ξJn (rξ)Jn (ρξ) −√ξ2 −k2 |z−ζ| 0 . e dξ, ξ 2 − k02

(6.1.105)

where R2 now equals r2 + (z − ζ)2 .

14 Lu, Y., 1998: The elastodynamic Green’s function for a torsional ring source. J. Appl. Mech., 65, 566–568 and Lu, Y., 1999: The full-space elastodynamic Green’s function for time-harmonic radial and axial ring sources in a homogeneous isotropic medium. J. Appl. Mech., 66, 639–645.

The Helmholtz Equation

419

For the special case k0 = 0, " ∞ ∞ ( 1 = cos[n(θ − θ$ )] Jn (rξ)Jn (ρξ)e−|z−ζ|ξ dξ. R n=−∞ 0

Since the integral15 " ∞ e−at Jm (bt)Jm (ct) dt = 0

1 √ Qm− 1 2 π bc

,

a2 + b2 + c2 2bc

-

(6.1.106)

,

(6.1.107)

where Qm− 1 (·) is the half-integer degree Legendre function of the second kind, 2 Cohl and Tohline16 found that % 2 & ∞ ( 1 1 r + ρ2 + (z − ζ)2 $ = √ cos[n(θ − θ )]Qn− 1 (6.1.108) 2 R π ρr n=−∞ 2rρ % 2 & ∞ 1 ( r + ρ2 + (z − ζ)2 = √ 2n cos[n(θ − θ$ )]Qn− 12 , (6.1.109) π ρr n=0 2rρ

since Qn− 1 (χ) = Q−n− 1 (χ), where 20 = 1 and 2n = 2 if n > 0. Equations 2 2 6.1.108 and Equation 6.1.109 are significantly more compact and easier to numerically evaluate than Equation 6.1.106. Conway and Cohl17 found the Fourier expansion for a free-space Green’s function for Helmholtz’s equation in cylindrical coordinates. Their motivation sprang from the fact that the source often appears as a circular ring. These Fourier coefficients consist of a series that involves associate Legendre functions, Bessel and Hankel functions and a hypergeometric function.

• Example 6.1.5

# "

We can use the free-space Green’s function e−k0 R /(4πR), where R2 = r + ρ2 − 2rρ cos(θ − θ $ ) + (z − ζ)2 , for Equation 6.1.96 to generate additional axisymmetric Green’s functions by simply eliminating the θ dependence by integration. We listed several solutions in Table 6.1.3. Tezer-Sezgin and Dost18 give additional Green’s functions for Helmholtz-type equations. 2

# " 15 Gradshteyn, I. S., and I. M. Ryzhik, 1965: Tables of Integrals, Series, and Products. Academic Press, Formula 6.612, Number 3. 16 Cohl, H. S., and J. E. Tohline, 1999: A compact cylindrical Green’s function expansion for the solution of potential problems. Astrophys. J., 527, 86–101. 17

Conway, J. T., and H. S. Cohl, 2010: Exact Fourier expansion in cylindrical coordinates for the three-dimensional Helmholtz Green function. Zeit. Angew. Math. Phys., 61, 425–443. 18 Tezer-Sezgin, M., and S. Dost, 1993: On the fundamental solutions of the axisymmetric Helmholtz-type equations. Appl. Math. Modelling, 17, 47–51.

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Green’s Functions with Applications

Table 6.1.3: Free-Space Green’s Function for Some Axisymmetric Helmholtz-like Equations: LHS = −δ(r − ρ)δ(z − ζ) LHS

Free-Space Green’s Function

, 1 ∂ ∂g ∂ 2g r + 2 r ∂r ∂r ∂z , 1 ∂ ∂g ∂ 2g r + 2 − k02 g r ∂r ∂r ∂z , 1 ∂ ∂g ∂ 2g n2 g r + 2− 2 r ∂r ∂r ∂z r , 2 1 ∂ ∂g ∂ g g r + 2 − 2 + k02 g r ∂r ∂r ∂z r , 1∂ ∂g ∂ 2g g r + 2 − 2 − k02 g r ∂r ∂r ∂z r

ρ g= 4π ρ g= 4π g= g= g=

ρ 4π ρ 4π ρ 4π

"

2π

0

"

2π −k0 R

e

R

0

"

cos(nθ) dθ R

2π

cos(θ) −ik0 R e dθ R

2π

cos(θ) −k0 R e dθ R

0

"

dθ

2π

0

"

1 dθ R

0

R2 = r2 + ρ2 − 2rρ cos(θ) + (z − ζ)2 See Tsuchimoto, M., K. Miya, T. Honma, and H. Igarashi, 1990: Fundamental solutions of the axisymmetric Helmholtz-type equations. Appl. Math. Modelling, 14, 605–611.

• Example 6.1.6 Let us solve19 ∂ 2g ∂2g ∂2g + + + 2Aδ(z) ∂x2 ∂y 2 ∂z 2

,

-' ∂ 2g ∂ 2 g '' + 2 ' = −δ(x)δ(y)δ(z), (6.1.110) ∂x2 ∂y z=0

for the domain −∞ < x, y, z < ∞. We begin by introducing the Fourier transform in three dimensions: " ∞" ∞" ∞ 1 Z 0, m|0, 0, 0)ei(kx+4y+mz) dk d0dm. g(x, y, z|0, 0, 0) = G(k, 8π 3 −∞ −∞ −∞ (6.1.111) Substituting Equation 6.1.111 into Equation 6.1.110, we obtain " A 2 ∞ Z 2 2 Z (κ + m )G(k, 0, m|0, 0, 0) + κ G(k, 0,η |0, 0, 0) dη = 1, (6.1.112) π −∞ 19 Dick, R., 2003: Hamiltonians and Green’s functions which interpolate between two and three dimensions. Int. J. Theoret. Phys., 42, 569–581.

The Helmholtz Equation

421

where κ2 = k 2 + 02 . Because the integral is independent of m, we have Z 0, m|0, 0, 0) = G(k,

Because

1 π

"

∞

−∞

F (κ) . + m2

κ2

dη 1 = , κ2 + η 2 κ

(6.1.113)

(6.1.114)

we find that 1 . (1 + Aκ)(κ2 + m2 )

Z 0, m|0, 0, 0) = G(k,

(6.1.115)

Taking the inverse of Equation 6.1.115, G(k, 0, z|0, 0, 0) =

exp(−κ|z|) . 2κ(1 + Aκ)

(6.1.116)

Therefore, " ∞ " 2π κ[ir cos(ϕ)−|z|] 1 e dϕdκ 8π 2 0 1 + Aκ 0 " ∞ 1 exp(−κ|z|) = J0 (κr) dκ. 4π 0 1 + Aκ

g(x, y, z|0, 0, 0) =

(6.1.117) (6.1.118) # "

• Example 6.1.7: Free-space Green’s function for the biharmonic equation An equation similar to Laplace’s equation is the biharmonic equation ∇4 g =

∂ 4g ∂ 4g ∂4g + 2 2 2 + 4 = δ(x − ξ)δ(y − η). 4 ∂x ∂x ∂y ∂y

(6.1.119)

Let us find its free-space Green’s function. We start by taking the Fourier transform in both directions. If we denote the transform variables in the x and y directions by k and 0, respectively, we find that 1 g(x, y|ξ,η ) = 4π 2

"

∞ −∞

"

∞

−∞

eik(x−ξ)+i4(y−η) dk d0. (k 2 + 02 )2

(6.1.120)

If we try to evaluate Equation 6.1.120 directly, we fail because the integral diverges. To avoid this difficulty, we first compute gxy (x, y|ξ,η ) = −

1 4π 2

"

∞ −∞

"

∞

−∞

k 0 eik(x−ξ)+i4(y−η) dk d0. (k 2 + 02 )2

(6.1.121)

422

Green’s Functions with Applications

Then 1 gxy (x, y|ξ,η ) = 2π 2 i

"

0

∞

!"

∞

−∞

) k eik(x−ξ) dk 0 sin[0(y − η)] d0 (k 2 + 02 )2 (6.1.122)

" (x − ξ) ∞ sin[0(y − η)] e−|x−ξ|4 d0 4π 0 1 (x − ξ)(y − η) = . 4π (x − ξ)2 + (y − η)2 =

(6.1.123) (6.1.124)

Integrating twice yields 5 S 2 R 4. 1 1 (x − ξ)2 + (y − η)2 ln (x − ξ)2 + (y − η)2 − 1 8π + f (x − ξ) + g(y − η), (6.1.125)

g(x, y|ξ,η ) =

where f (·) and g(·) are arbitrary, differentiable functions. Actually, only that portion of Equation 6.1.125 that contains the logarithm is the free-space Green’s function because it contains the singular nature of the function. Hayes et al.20 found the free-space Green’s function for the slightly different equation: , 4 ∂g ∂ g ∂ 4g ∂4g +τ + 2 2 2 + 4 = δ(x − ξ)δ(y − η). (6.1.126) ∂x ∂x4 ∂x ∂y ∂y 6.2 METHOD OF IMAGES One of the earliest methods for computing Green’s functions was the method of images. Drawing from the strong analog between electrostatics and Green’s functions, Lord Kelvin introduced this method of adding additional free-space Green’s functions to the original one whose singularity occurs at (ξ, η,ζ ). The singularities associated with these additional free-space Green’s functions must lie outside of the domain of interest because the Green’s function must satisfy Laplace’s equation everywhere within the domain except at (ξ, η,ζ ). From practical considerations the method of images works best for very symmetrical boundaries, such as half-planes, cylinders, spheres and intersecting planes with an angle π/m, where m is an integer. A simple example is the Green’s function for the half-space z > 0 with the boundary condition g(x, y, 0|ξ, η,ζ ) = 0. With the source lying at (ξ, η,ζ ), let us introduce an image source at (ξ, η, −ζ). Note how this image is located outside of the half-space. The Green’s function for the half-space is g(x, y, z|ξ, η,ζ ) =

1 1 − , r r$

(6.2.1)

20 Hayes, M., S. B. G. O’Brien, and J. H. Lammers, 2000: Green’s function for steady flow over a small two-dimensional topography. Phys. Fluids, 12, 2845–2858.

The Helmholtz Equation

423

where r2 = (x − ξ)2 + (y − η)2 + (z − ζ)2 ,

(6.2.2)

(r$ )2 = (x − ξ)2 + (y − η)2 + (z + ζ)2 .

(6.2.3)

and A quick check shows that g(x, y, 0|ξ, η,ζ ) = 0. • Example 6.2.1 Let us find the Green’s function for the domain between the planes located at x = ±a/2 where g(±a/2, y, z|ξ, η,ζ ) = 0. Consider the Green’s function g(x, y, z|ξ, η,ζ ) =

∞ (

(−1)n

=

. 2 {x − [na + (−1)n ξ]} + (y − η)2 + (z − ζ)2 (6.2.4) This Green’s functions consists of an infinite of positive images located at x = na + ξ and an infinite number of negative images located at x = na − ξ. It can be shown that it is the correct one for our problem because (1) it satisfies Laplace equation everywhere for −a/2 < x < a/2 except at the singularity point (ξ, η,ζ ) and the boundary conditions. To prove the last point, the summation in Equation 6.2.4 when x = ±a/2 is broken up into 4 new summations: (1) n = 2m with m = 0, 1, 2, . . ., (2) n = 2m − 1 with m = 1, 2, 3, . . ., (3) n = −1 − 2m with m = 0, 1, 2, . . . and (4) n = −2m with m = 1, 2, 3, . . .. The sum from (1) cancels the sum from (3) while the sum from (2) cancels the sum from (4). n=−∞

# " • Example 6.2.2 Let us use the method of images to solve ∂2g ∂ 2g + = δ(x − ξ)δ(y − η), ∂x2 ∂y 2

−∞ < x, ξ < ∞,

0 < y, η < 1, (6.2.5)

subject to the boundary conditions g(x, 0|ξ,η ) = g(x, 1|ξ,η ) = 0.

(6.2.6)

Consider the summation g(x, y|ξ,η ) = −

∞ 1 ( ln |r − r0 |, 2π n=−∞

(6.2.7)

424

Green’s Functions with Applications

where |r − r0 | = (x − ξ)2 + (y − ηn )2 , and ηn =

R n + η, n + 1 + η,

n even, n odd.

(6.2.8) (6.2.9)

The construction is very similar to the previous example. The free-space Green’s function for Laplace’s equation is − ln(r)/(2π). We then create a series of sources and sinks at (ξ,η n ) so that the Green’s function equals zero along y = 0 and 1. Hakim and Keyser21 illustrate the use of this Green’s function in atmospheric dynamics. # " • Example 6.2.3 In Section 6.7 we will show that the Green’s function for Laplace’s equation inside an infinitely long tube of radius a is g(r, θ, z|ρ,θ $ , ζ) =

∞ ∞ ( ( 1 Jn (knm ρ)Jn (knm r) cos[n(θ − θ$ )] 2πa2 n=−∞ m=1 knm J $ 2n (knm a)

× e−knm |z−ζ| ,

(6.2.10)

where knm is the m root of Jn (k) = 0. This Green’s function satisfies the boundary condition g(a, θ, z|ρ,θ $ , ζ) = 0. Let us use the method of images and Equation 6.2.10 to give the Green’s function for Laplace’s equation within a cylindrical pill box. Thus, this Green’s function must satisfy the additional boundary conditions g(r, θ, 0|ρ,θ $ , ζ) = g(r, θ, L|ρ,θ $ , ζ) = 0.

(6.2.11)

Let us focus on the z dependence for a moment. Drawing upon the previous example, the Green’s function will vanish at z = 0 and z = L if we introduce positive images at z = 2nL + ζ and negative images at z = 2nL − ζ, where n has all integral values from −∞ to ∞. If z < ζ, the z-dependence must be ∞ (

n=0

e−knm (2nL+ζ−z) +

∞ (

n=1

e−knm (2nL−ζ+z) − −

∞ (

∞ (

e−knm (2nL−ζ−z)

n=1

e−knm (2nL+ζ+z)

n=0

21 Hakim, G. J., and D. Keyser, 2001: Canonical frontal circulation patterns in terms of Green’s functions for the Sawyer-Eliassen equation. Quart. J. R. Meteor. Soc., 127, 1795–1814.

The Helmholtz Equation

425 F

= 2 e−knm ζ

=2

: #

∞ (

n=0

e−2nknm L − eknm ζ

× sinh (knm z) e−knm ζ − eknm ζ

∞ $(

∞ (

e−2nknm L

n=1

e−2nknm L + eknm ζ

n=0

× sinh (knm z) sinh[knm (L − ζ)] sinh(knm z) =2 . sinh(knm L)

G

(6.2.12) ; (6.2.13) (6.2.14)

For z > ζ, we substitute L − z for z and L − ζ for ζ in Equation 6.2.14. Therefore, the Green’s function for Laplace’s equation for a cylindrical pill box is g(r, θ, z|ρ,θ $ , ζ) =

∞ ∞ 1 ( ( Jn (knm ρ)Jn (knm r) cos[n(θ − θ$ )] πa2 n=−∞ m=1 J $ 2n (knm a)

×

sinh[knm (L − z< )] sinh(knm z> ) . knm sinh(knm L)

(6.2.15)

This is in agreement with Equation 6.7.14 which is obtained in a different manner. # " • Example 6.2.4: Green’s function for a quarter plane Let us find the Green’s function for the planar Poisson equation ∂ 2g ∂2g + = −δ(x − ξ)δ(y − η), ∂x2 ∂y 2

(6.2.16)

for the quarter space 0 < x, y, with the boundary conditions g(0, y|ξ,η ) = gy (x, 0|ξ,η ) = 0,

(6.2.17)

and lim g(x, y|ξ,η ) → 0.

x,y→∞

(6.2.18)

One method for solving Equation 6.2.16 through Equation 6.2.18 is to use Fourier sine transforms in the x direction. Taking the transform " ∞ G(k, y|ξ,η ) = g(x, y|ξ,η ) sin(kx) dx (6.2.19) 0

of these equations, they become the ordinary differential equation d2 G − k 2 G = − sin(kξ)δ(y − η), dy 2

(6.2.20)

426

Green’s Functions with Applications

Figure 6.2.1: Configuration of images to find the Green’s function for the planar Poisson equation with the boundary conditions g(0, y|ξ,η ) = gy (x, 0|ξ,η ) = 0. The positive images are filled circles while the negative images are open circles.

with G$ (k, 0|ξ,η ) = 0, and limy→∞ G(k, y|ξ,η ) → 0. The solution to Equation 6.2.20 is 5 sin(kξ) 4 −k|y−η| G(k, y|ξ,η ) = e + e−k(y+η) . (6.2.21) 2k Therefore,

g(x, y|ξ,η ) =

1 π

"

∞ 0

4 5 dk e−k|y−η| + e−k(y+η) sin(kξ) sin(kx) . k

(6.2.22)

Performing the integration and simplifying, ! ) 1 [(x + ξ)2 + (y − η)2 ][(x + ξ)2 + (y + η)2 ] g(x, y|ξ,η ) = ln . (6.2.23) 4π [(x − ξ)2 + (y − η)2 ][(x − ξ)2 + (y + η)2 ] Now let us do this problem by the method of images. To satisfy the boundary conditions, we must introduce three images, which are shown in Figure 6.2.1. The filled circles denote positive images while the open circles are negative images. Using the free-space Green’s function for each image and simplifying, we obtain Equation 6.2.23. This solution is shown in Figure 6.2.2. Note that the point x = y = 1 has not been plotted. # "

The Helmholtz Equation

427

Figure 6.2.2: The Green’s function for the planar Poisson equation over the quarter plane 0 < x, y, with the boundary conditions g(0, y|ξ,η ) = gy (x, 0|ξ,η ) = 0 when ξ = η = 1.

• Example 6.2.5: Sommerfeld’s method In the late 1890s A. Sommerfeld22 developed a technique using integration on the complex plane to extend the method of images to several other useful geometries in three dimensions. In this example we will illustrate his technique by constructing the Green’s function for a conducting half-space.23 Davis and Reitz24 applied this technique to compute the Green’s function for a space defined by two intersecting half-planes where the angle between the planes lies between 0 and 2π. Sommerfeld’s generalization of the method of images involves mapping the original problem into one on an unbounded and unbranched space. Here Davis and Reitz applied Sommerfeld’s technique to a semi-infinite conducting sheet and mapped it into a Riemann double space. The method of images is then applied on this new domain. The details are as follows: Consider a conducting half-plane that coincides with the xz-plane where x ≥ 0 and |z| < ∞. If the xy-plane is the horizontal plane, then the conducting half-space is a semi-infinite ribbon (x > 0) of infinite length. Using cylindrical coordinates, x = r cos(ϕ) and y = r sin(ϕ). An important aspect of Sommerfeld’s method is his generalization of the concepts of the Riemann surface, branch cut and branch point to three 22 Sommerfeld, A., 1897: Uber ¨ verzweigte Potentiale im Raum. Proc. London Math. Soc., Ser. 2, 28, 395–429. 23 Davis, L. C., and J. R. Reitz, 1971: Solution to potential problems near a conducting semi-infinite sheet or conducting disk. Am. J. Phys., 39, 1255–1265. 24 Davis, L. C., and J. R. Reitz, 1975: Solution of potential problems near the corner of a conductor. J. Math. Phys., 16, 1219–1226.

428

Green’s Functions with Applications

dimensions. If this were a two-dimensional problem, the line x > 0, y = 0 would be the branch cut and x = y = 0 would be the branch point. One Riemann surface (the physical one) is defined by 0 < ϕ < 2π while a second Riemann surface occurs for 2π < ϕ < 4π. In the three dimension case the branch cut becomes a branch membrane (the ribbon) and the branch point becomes a branch line (the z-axis). If the source point is located at (ρ,θ $ , ζ), the free-space Green’s function is 1/(4πR), where R2 = r2 + ρ2 − 2ρr cos(θ − θ$ ) + (z − ζ)2 .

(6.2.24)

To obtain the Green’s function g for the Riemann space, Sommerfeld generalized R to be a function of a complex parameter α and wrote " 1 f (α) g= dα, (6.2.25) 4π C R where the integral is taken over a suitable path in the α-plane. Through a suitable choice for f (α) we will transform the single-valued Green’s function 1/(4πR) into a double-valued one as follows: We begin by replacing θ $ by the complex variable α and denote the modified R by Rα , or 2 Rα = r2 + ρ2 − 2ρr cos(θ − α) + (z − ζ)2 .

(6.2.26)

Next, we multiply 1/(4πRα ) by a function f1 (α) which has a simple pole at α = θ $ and integrate around some contour C which encloses this pole. The function f1 (α) must be periodic (with period 2π) in both α and θ $ . Let us try f1 (α) =

i exp(iα) . [exp(iα) − exp(iθ$ )]

From Cauchy’s integral formula, we have 0 1 1 f1 (α) = dα. R 2πi C Rα

(6.2.27)

(6.2.28)

Let us now deform the path of integration in such a manner that we do not pass over any singularity of the integrand. The branch points are located 2 at Rα = 0 or cos(θ − α) = cos(iα1 ), where cos(iα1 ) = [r2 + ρ2 + (z − ζ)2 ]/(2ρr).

(6.2.29)

Thus, the branch points are given by α = θ + 2kπ ± iα1 , where k is an integer. The poles of the integrand are located at α = θ + 2kπ. We also introduce branch cuts in the α-plane along lines parallel to the imaginary axis from the branch points to α = θ + 2kπ ± ∞i. See Figure 6.2.3. Consequently

The Helmholtz Equation

429

A C2

θ = ια 1

C1 0

θ

θ

2π θ = −ια 1

Figure 6.2.3: Integration contours used in an illustration of Sommerfeld’s extension of the method of images.

we can deform the initial contour C into a new one Γ which consists of two loops enclosing the branch cuts and a rectangle of width 2π and arbitrary height. The contribution to the integral from one vertical side of the rectangle cancels the other vertical side because of the periodicity of the integrand. Furthermore, if the height of the rectangle is made very large, the contribution from the top and bottom of the rectangle vanishes because 1/Rα → 0 as 5(α) → ∞. We are left with two contours around the branch cuts, taken in the appropriate sense and extending to 5(α) = ±∞. Let us call this contour A. We now proceed to the solution for the Riemann double space. We will first obtain a solution for a Riemann space of n windings and then set n = 2. Consider the following fn (α) that is periodic in α and θ$ with period 2nπ and has a simple pole at α = θ$ : fn (α) = The function

i exp(iα/n) . n exp(iα/n) − exp(iθ $ /n)

1 g= 8π 2 i

"

A

fn (α) dα Rα

(6.2.30)

(6.2.31)

430

Green’s Functions with Applications

is the Green’s function of an n-fold Riemann space. The branch points of the integrand are the same as when n = 1, but the poles are now located at α = θ$ + 2knπ. Sommerfeld proved that g has the following properties: (1) In the entire Riemann space it is uniquely defined, finite, and continuous, except in the vicinity of the source point where it becomes infinite as 1/R; (2) it satisfies Laplace’s equation everywhere in Riemann space except at the source point and along the branch cuts; (3) it vanishes at infinity; and (4) in ordinary space it is multivalued with a separate branch for each winding of Riemann space. Let us evaluate Equation 6.2.31 when n = 2. Setting α = ϕ ± iβ, depending upon which of the two contours we are located, β runs from ∞ to α1 and back again in both cases. Furthermore, the two halves of the contour along the branch cut give the same value. Therefore, "

$

g(r, θ, z|ρ,θ , ζ) =

8π

1 √ 2

∞

sinh(β/2) . dβ, 8π 2ρr α1 [cosh(β/2) − τ ] cosh(β) − cosh(α1 ) (6.2.32) where τ = cos[(θ − θ $ )/2]. Introducing ξ = cosh(β/2) and σ = cosh(α1 /2), Equation 6.2.32 becomes g(r, θ, z|ρ,θ $ , ζ) =

1 2√

ρr

"

∞

σ

%3 & dξ 1 σ+τ . = 2 arctan . 2π R σ−τ (ξ − τ ) ξ 2 − σ 2 (6.2.33)

6.3 TWO-DIMENSIONAL POISSON’S EQUATION OVER RECTANGULAR AND CIRCULAR DOMAINS Consider the two-dimensional Poisson equation ∂2u ∂2u + 2 = −f (x, y). ∂x2 ∂y

(6.3.1)

This equation arises in equilibrium problems, such as the static deflection of a rectangular membrane. In that case, f (x, y) represents the external load per unit area, divided by the tension in the membrane. The solution u(x, y) must satisfy certain boundary conditions. For the present, let us choose u(0, y) = u(a, y) = 0, and u(x, 0) = u(x, b) = 0. • Rectangular Area To find the Green’s function for Equation 6.3.1 we must solve the partial differential equation ∂ 2g ∂ 2g + = −δ(x − ξ)δ(y − η), ∂x2 ∂y 2

0 < x, ξ < a,

0 < y, η < b,

(6.3.2)

The Helmholtz Equation

431

subject to the boundary conditions g(0, y|ξ,η ) = g(a, y|ξ,η ) = g(x, 0|ξ,η ) = g(x, b|ξ,η ) = 0.

(6.3.3)

From Equation 6.0.14, "

u(x, y) =

a 0

"

b

g(x, y|ξ,η )f (ξ,η ) dηd ξ.

(6.3.4)

0

One approach to finding the Green’s function is to expand it in terms of the eigenfunctions ϕ(x, y) of the differential equation ∂ 2ϕ ∂2ϕ + = −λϕ, ∂x2 ∂y 2

(6.3.5)

and the boundary conditions, Equation 6.3.3. The eigenvalues are n2 π 2 m2 π 2 + , (6.3.6) 2 a b2 where n = 1, 2, 3, . . ., m = 1, 2, 3, . . ., and the corresponding eigenfunctions are * nπx + * mπy + ϕnm (x, y) = sin sin . (6.3.7) a b Therefore, we seek g(x, y|ξ,η ) in the form ∞ ( ∞ * nπx + * mπy + ( g(x, y|ξ,η ) = Anm sin sin . (6.3.8) a b n=1 m=1 λnm =

Because the delta functions can be written , - * ∞ ∞ 4 (( nπξ mπη + * nπx + * mπy + δ(x − ξ)δ(y − η) = sin sin sin sin , ab n=1 m=1 a b a b (6.3.9) we find that , 2 2 , - * n π m2 π 2 4 nπξ mπη + + A = sin sin , (6.3.10) nm a2 b2 ab a b after substituting Equation 6.3.8 and Equation 6.3.9 into the partial differential equation, Equation 6.3.2, and setting the corresponding harmonics equal to each other. Therefore, the biharmonic bilinear formula for the Green’s function of Poisson’s equation is

g(x, y|ξ,η ) =

∞ ( ∞ (

4 ab n=1 m=1

sin

* nπx + a

, - * nπξ mπy + * mπη + sin sin sin a b b . n2 π 2 /a2 + m2 π 2 /b2

(6.3.11)

432

Green’s Functions with Applications

Thus, solutions to Poisson’s equation can now be written as u(x, y) =

* nπx + * mπy + anm sin sin , n2 π 2 /a2 + m2 π 2 /b2 a b n=1 m=1 ∞ ( ∞ (

(6.3.12)

where anm are the Fourier coefficients for the function f (x, y): anm =

4 ab

"

0

a" b

f (x, y) sin

0

* nπx + a

sin

* mπy + b

dy dx.

(6.3.13)

We save as Problem 36 the derivation of the Green’s function with Neumann boundary conditions. Another form of the Green’s function can be obtained by considering each direction separately. To satisfy the boundary conditions along the edges y = 0 and y = b, we write the Green’s function as the Fourier series g(x, y|ξ,η ) =

∞ (

Gm (x|ξ) sin

m=1

* mπy + b

,

(6.3.14)

where the coefficients Gm (x|ξ) are left as undetermined functions of x, ξ, and η. Substituting this series into the partial differential equation for g, multiplying by 2 sin(nπy/b)/b, and integrating over y, we find that * nπη + d2 Gn n2 π 2 2 − G = − sin δ(x − ξ). n dx2 b2 b b

(6.3.15)

This differential equation shows that the expansion coefficients Gn (x|ξ) are one-dimensional Green’s functions; we can find them, as we did in Chapter 3, by piecing together homogeneous solutions to Equation 6.3.15 that are valid over various intervals. For the region 0 ≤ x ≤ ξ, the solution to Equation 6.3.15 that vanishes at x = 0 is * nπx + Gn (x|ξ) = An sinh , (6.3.16) b where An is presently arbitrary. The corresponding solution for ξ ≤ x ≤ a is Gn (x|ξ) = Bn sinh

%

& nπ(a − x) . b

(6.3.17)

Note that this solution vanishes at x = a. Because the Green’s function must be continuous at x = ξ, An sinh

,

nπξ b

-

% & nπ(a − ξ) = Bn sinh . b

(6.3.18)

The Helmholtz Equation

433

Figure 6.3.1: Equation 6.3.22 or Equation 6.3.23 for the planar Poisson equation over a rectangular area with Dirichlet boundary conditions on all sides when a = b and ξ/b = η/b = 0.3.

On the other hand, the appropriate jump discontinuity of G$n (x|ξ) yields −

% & , * nπη + nπ nπ(a − ξ) nπ nπξ 2 Bn cosh − An cosh = − sin . (6.3.19) b b b b b b

Solving for An and Bn , An =

* nπη + sinh[nπ(a − ξ)/b] 2 sin , nπ b sinh(nπa/b)

(6.3.20)

* nπη + sinh(nπξ/b) 2 sin . nπ b sinh(nπa/b)

(6.3.21)

and Bn =

This yields the Green’s function25

∞ * nπη + * nπy + 2 ( sinh[nπ(a − x> )/b] sinh(nπx< /b) sin sin . π n=1 n sinh(nπa/b) b b (6.3.22) Figure 6.3.1 illustrates Equation 6.3.22 in the case of a square domain with ξ/b = η/b = 0.3.

g(x, y|ξ,η ) =

25 For an application of this Green’s function in studying a foil-less diode, see Chen, J., and R. V. Lovelace, 1978: Beam generation in foil-less diodes. Phys. Fluids, 21, 1623–1633.

434

Green’s Functions with Applications

If we began with a Fourier expansion in the y direction, we would have obtained , - * ∞ 2 ( sinh[mπ(b − y> )/a] sinh(mπy < /a) mπξ mπx + g(x, y|ξ,η ) = sin sin . π m=1 m sinh(mπb/a) a a (6.3.23) • Circular Disk Let us now find the Green’s function for Poisson’s equation for the circular domain r ≤ 1 with Dirichlet boundary conditions. We begin by writing the Green’s function as a sum of the free-space Green’s function gp plus a homogeneous solution gH such that ∇2 gH = 0. From separation of variables it is easily shown that gH =

∞

a0 ( n + r [an cos(nθ) + bn sin(nθ)] , 2 n=1

(6.3.24)

while the free-space Green’s function, Equation 6.1.72, is gp = −

1 2 1 ln r2 + ρ2 − 2rρ cos(θ − θ$ ) . 4π

(6.3.25)

Along r = 1, it follows that gH = −gp = Next, we note that

1 2 1 ln 1 + ρ2 − 2ρ cos(θ − θ $ ) . 4π

R 4 5S 1 2 $ $ ln 1 + ρ2 − 2ρ cos(θ − θ$ ) = ln 1 + ρ2 − ρ ei(θ−θ ) + e−i(θ−θ ) R4 54 5S $ $ = ln 1 − ρei(θ−θ ) 1 − ρe−i(θ−θ ) 4 5 4 5 $ $ = ln 1 − ρei(θ−θ ) + ln 1 − ρe−i(θ−θ ) 4 5 $ $ = − ρei(θ−θ ) + 12 ρ2 e2i(θ−θ ) + · · · 4 5 $ $ − ρe−i(θ−θ ) + 12 ρ2 e−2i(θ−θ ) + · · · =−2

∞ ( ρn cos[n(θ − θ $ )] , n n=1

(6.3.26)

(6.3.27) (6.3.28) (6.3.29)

(6.3.30) (6.3.31)

provided |ρ| < 1. Therefore, when r = 1, we have −2

∞ ∞ ( ρn cos[n(θ − θ$ )] a0 ( = + an cos(nθ) + bn sin(nθ) n 2 n=1 n=1

(6.3.32)

The Helmholtz Equation

435

Figure 6.3.2: Equation 6.3.36 for the planar √ Poisson equation over a unit disk with a Dirichlet boundary condition when ρ = 0.55 2, and θ ! = 5π/4.

so that a0 = 0,

an =

ρn cos(nθ$ ), 2πn

bn =

ρn sin(nθ $ ). 2πn

(6.3.33)

Consequently, ∞ 1 ( (rρ)n cos[n(θ − θ$ )] 2π n=1 n 1 2 1 =− ln 1 + (rρ)2 − 2rρ cos(θ − θ$ ) , 4π

gH =

and

(6.3.34) (6.3.35)

1 2 1 ln r2 + ρ2 − 2rρ cos(θ − θ$ ) 4π 1 2 1 + ln 1 + (rρ)2 − 2rρ cos(θ − θ$ ) (6.3.36) 4π 1 2 1 =− ln r2 + ρ2 − 2rρ cos(θ − θ$ ) 4π 1 2 1 1 + ln r2 + (1/ρ)2 − (2r/ρ) cos(θ − θ$ ) + ln(ρ2 ). (6.3.37) 4π 4π

g(r,θ |ρ,θ $ ) = −

Equation 6.3.37 shows that the homogeneous portion of the Green’s function behaves (within an additive constant) in the same manner as the free-space Green’s function except that its source is located at (1/ρ,θ $ ). Because 1/ρ is the inverse of ρ, the relationship between (ρ,θ $ ) and (1/ρ,θ $ ) is called inversion

436

Green’s Functions with Applications

with respect to the circle.26 We will see a similar inversion pattern when we find the Green’s function for Poisson’s equation with a Dirichlet boundary condition on a sphere of radius a. We can apply Equation 6.3.36 to find the solution to Laplace’s equation with the Dirichlet boundary condition u(1, θ). Using Equation 6.0.13 with f (r0 ) = 0 and ∇0 g(r0 |r) =

' ∂g '' 1 1 − r2 ˆ r=− ˆ r, ' 2 ∂ρ ρ=1 2π 1 + r − 2r cos(θ − θ$ )

(6.3.38)

we obtain Poisson’s integral formula 1 u(r,θ ) = 2π

"

2π

0

1 − r2 u(1, θ$ ) dθ $ . 1 + r2 − 2r cos(θ − θ$ )

(6.3.39)

An alternative to Equation 6.3.36 can be found by solving , 1 ∂ ∂g 1 ∂ 2g δ(r − ρ)δ(θ − θ $ ) r + 2 2 =− , r ∂r ∂r r ∂θ r

(6.3.40)

where 0 < r, ρ < a, and 0 ≤ θ,θ $ ≤ 2π, with the boundary conditions lim |g(r,θ |ρ,θ $ )| < ∞,

r→0

g(a,θ |ρ,θ $ ) = 0,

0 ≤ θ,θ $ ≤ 2π.

(6.3.41)

The Green’s function must be periodic in θ. We begin by expressing δ(θ − θ$ ) by the Fourier cosine series δ(θ − θ $ ) =

∞ ∞ 1 1( 1 ( + cos[n(θ − θ$ )] = cos[n(θ − θ$ )]. 2π π n=1 2π n=−∞

(6.3.42)

Thus, the form of the Green’s function is g(r,θ |ρ,θ $ ) =

∞ (

n=−∞

gn (r|ρ) cos[n(θ − θ $ )].

(6.3.43)

Substituting Equation 6.3.42 and Equation 6.3.43 into Equation 6.3.40 and simplifying, we find that , 1 d dgn n2 δ(r − ρ) r − 2 gn = − . r dr dr r 2πr

(6.3.44)

26 Thomson, W., 1845: Extrait d’une lettre de M. William Thomson ` a M. Liouville. J. Math. Pures Appl. 10, 364–367; Thomson, W., 1847: Extraits de deux lettres address´ees a M. Liouville. J. Math. Pures Appl., 12, 256–264. `

The Helmholtz Equation Because

437

∞ δ(r − ρ) 1 ( Jn (knm ρ)Jn (knm r) = , 2πr πa2 m=1 J $ 2n (knm a)

(6.3.45)

where knm is the mth root of Jn (knm a) = 0, we take gn (r|ρ) =

∞ (

Anm Jn (knm r) .

(6.3.46)

m=1

Substituting Equation 6.3.45 and Equation 6.3.46 into Equation 6.3.44, we obtain 1 Jn (knm ρ) 2 knm Anm = , (6.3.47) πa2 J $ 2n (knm a) after using the differential equation that governs Bessel functions of order n and the first kind. The Green’s function follows by substituting Equation 6.3.46 and Equation 6.3.47 into Equation 6.3.43 or

g(r,θ |ρ,θ $ ) =

∞ ∞ 1 ( ( Jn (knm ρ)Jn (knm r) cos[n(θ − θ$ )]. 2 J $ 2 (k πa2 n=−∞ m=1 knm n nm a)

(6.3.48) • Infinitely long, axisymmetric cylinder Let us find the Green’s function for the exterior of an infinitely long, axisymmetric cylinder,27 where the Green’s function equals zero along the boundary r = a. This problem is governed by the partial differential equation ∂ 2g 1 ∂g ∂ 2g δ(r − ρ)δ(z − ζ) + + 2 =− , 2 ∂r r ∂r ∂z 2πr

(6.3.49)

where a < r, ρ < ∞, and −∞ < z, ζ < ∞, with the boundary conditions lim g(r, z|ρ,ζ ) → 0,

|z|→∞

a < r < ∞,

(6.3.50)

and g(a, L|ρ,ζ ) = 0,

lim g(r, z|ρ,ζ ) → 0,

r→∞

|z| < ∞.

(6.3.51)

27 See Marr-Lyon, M. J., D. B. Thiessen, F. J. Blonigen, and P. L. Marston, 2000: Stabilization of electrically conducting capillary bridges using feedback control of radial electrostatic stresses and the shapes of extended bridges. Phys. Fluids, 12, 986–995.

438

Green’s Functions with Applications We begin by taking the Fourier transform of Equation 6.3.49, which yields d2 G 1 dG δ(r − ρ) −ikζ + − k2 G = − e , 2 dr r dr 2πr

a < r < ∞,

(6.3.52)

with G(a|ρ) = 0, and limr→∞ G(r|ρ) → 0. Because the homogeneous solutions of Equation 6.3.52 are I0 (kr) and K0 (kr), a Green’s function that satisfies this equation and the boundary conditions is G(r|ρ) = A[I0 (ka)K0 (|k|r< ) − K0 (|k|a)I0 (kr< )]K0 (|k|r> ).

(6.3.53)

Equation 6.3.53 enjoys the additional property of being continuous at the point r = ρ. To compute A, we use the condition that the Green’s function must satisfy the jump condition 'r=ρ+ dG '' e−ikζ = − . dr 'r=ρ− 2πρ

(6.3.54)

Upon substituting Equation 6.3.53 into Equation 6.3.54 and using the Wronskian relationship, we find that 2πAK0 (|k|a) = −e−ikζ . Therefore, G(r, k|ρ,ζ ) =

[K0 (|k|a)I0 (kr< ) − I0 (ka)K0 (|k|r< )]K0 (|k|r> ) , 2πK0 (|k|a)

(6.3.55)

and "

∞

[K0 (|k|a)I0 (kr< ) − I0 (ka)K0 (|k|r< )]K0 (|k|r> ) ik(z−ζ) e . K0 (|k|a) −∞ (6.3.56) A problem that is very similar to Equation 6.3.49 through Equation 6.3.51 involves solving Equation 6.3.49 in the interior of a cylinder of radius a with the boundary conditions 1 g(r, z|ρ,ζ ) = 4π 2

dk

lim g(r, z|ρ,ζ ) → 0,

|z|→∞

0 < r < a,

(6.3.57)

and lim |g(r, z|ρ,ζ )| < ∞,

r→0

g(a, z|ρ,ζ ) = 0,

|z| < ∞.

(6.3.58)

We begin as before and take the Fourier transform of Equation 6.3.49 with respect to z. This leads to Equation 6.3.52. The solution to Equation 6.3.52 in the present case is G(r|ρ) = A[I0 (ka)K0 (|k|r> ) − K0 (|k|a)I0 (kr> )]I0 (kr< )

(6.3.59)

The Helmholtz Equation

439

with 2πAI0 (ka) = e−ikζ . Therefore, " ∞ 1 [I0 (ka)K0 (|k|r> ) − K0 (|k|a)I0 (kr> )]I0 (kr< ) ik(z−ζ) g(r, z|ρ,ζ ) = dk e . 2 4π −∞ I0 (ka) (6.3.60) We can evaluate Equation 6.3.60 by closing28 the line integration in this integral with a semicircle of infinite radius as dictated by Jordan’s lemma and applying the residue theorem. This yields g(r, z|ρ,ζ ) =

∞ 1 ( J0 (kn ρ/a)J0 (kn r/a) −kn |z−ζ|/a e , 2πa n=1 kn J12 (kn )

(6.3.61)

where kn is the nth root of J0 (k) = 0. • Infinitely and semi-infinitely long cylinder, axisymmetric cylinder A problem that is similar to the previous one consists of solving the partial differential equation29 ∂ 2g 1 ∂g ∂ 2g δ(r − ρ)δ(z − ζ) + + =− , ∂r2 r ∂r ∂z 2 2πr

(6.3.62)

where 0 ≤ r, ρ < a, and −∞ < z, ζ < ∞, subject to the boundary conditions lim |g(r, z|ρ,ζ )| < ∞,

r→0

g(a, z|ρ,ζ ) = 0,

−∞ < z < ∞,

(6.3.63)

and lim g(r, z|ρ,ζ ) → 0,

|z|→∞

0 ≤ r < a.

(6.3.64)

We begin by noting that since ∞ δ(r − ρ) 2 ( J0 (kn ρ)J0 (kn r) = 2 , r a n=1 J12 (kn a)

(6.3.65)

where kn is nth zero of J0 (ka) = 0, we can express the Green’s function via the expansion g(r, z|ρ,ζ ) =

∞ (

n=1

Gn (z|ζ)

J0 (kn ρ)J0 (kn r) . J12 (kn a)

(6.3.66)

28 See Allen, C. K., N. Brown, and M. Reiser, 1994: Image effects for bunched beams in axisymmetric systems. Particle Accel., 45, 149–165. 29 See Bouwkamp, C. J., and N. G. de Bruijn, 1947: The electrostatic field of a point charge inside a cylinder, in connection with wave guide theory. J. Appl. Phys., 18, 562–577.

440

Green’s Functions with Applications

Substituting Equation 6.3.65 and Equation 6.3.66 into Equation 6.3.62, we obtain the ordinary differential equation d2 Gn δ(z − ζ) − kn2 Gn = − , 2 dz πa2

−∞ < z < ∞,

(6.3.67)

for Gn (z|ζ). The boundary conditions on Gn (z|ζ) are lim|z|→∞ Gn (z|ζ) → 0. The solution to Equation 6.3.67 (except for the point z = ζ) that satisfies the boundary conditions is Gn (z|ζ) = Ae−kn |z−ζ| .

(6.3.68)

To compute A, we integrate Equation 6.3.67 and find that 'z=ζ + dGn '' 1 = − 2. dz 'z=ζ − πa

(6.3.69)

Upon substituting Equation 6.3.68 into Equation 6.3.69 and simplifying, we find that 2πa2 kn A = 1. Therefore, g(r, z|ρ,ζ ) =

∞ 1 ( −kn |z−ζ| J0 (kn ρ)J0 (kn r) e . 2πa2 n=1 kn J12 (kn a)

(6.3.70)

Let us now introduce a plane30 at z = 0 so that z, ζ> 0. If we require that the Green’s function vanishes along z = 0, Equation 6.3.64 now reads g(r, 0|ρ,ζ ) = 0,

lim g(r, z|ρ,ζ ) → 0,

z→∞

0 ≤ r < a.

(6.3.71)

The z-dependence of Green’s function becomes Gn (z|ζ) = A sinh(kn z< )e−kn z> ,

(6.3.72)

and πa2 kn A = 1. Consequently, the Green’s function within a semi-infinite cylinder of radius a where the Green’s function equals zero along the boundary is ∞ 1 ( J0 (kn ρ)J0 (kn r) g(r, z|ρ,ζ ) = sinh(kn z< )e−kn z> . (6.3.73) πa2 n=1 kn J12 (kn a)

30

See Chen and Lovelace, op. cit.

The Helmholtz Equation

441

• Solid cylinder with Dirichlet conditions In the same vein let us find the Green’s function for an annular region r1 < r < r2 of finite depth L where the Green’s function equals zero along the boundary. Mathematically,31 this is equivalent to solving ∂ 2g 1 ∂g ∂ 2g δ(r − ρ)δ(z − ζ) + + 2 =− , 2 ∂r r ∂r ∂z 2πr

(6.3.74)

where r1 < r, ρ < r2 , and 0 < z, ζ < L, with the boundary conditions g(r1 , z|ρ,ζ ) = g(r2 , z|ρ,ζ ) = 0,

0 < z < L,

(6.3.75)

r1 < r < r2 .

(6.3.76)

and g(r, 0|ρ,ζ ) = g(r, L|ρ,ζ ) = 0, We begin by noting that since δ(z − ζ) =

, - * ∞ 2 ( nπζ nπz + sin sin , L n=1 L L

we can express the Green’s function via the expansion , - * ∞ ( nπζ nπz + g(r, z|ρ,ζ ) = Gn (r|ρ) sin sin . L L n=1

(6.3.77)

(6.3.78)

Substituting Equation 6.3.77 and Equation 6.3.78 into Equation 6.3.74, we obtain the ordinary differential equation d2 Gn 1 dGn n2 π 2 δ(r − ρ) + − Gn = − , dr2 r dr L2 Lπr

r1 < r < r2 ,

(6.3.79)

for Gn (r|ρ). Because the homogeneous solutions of Equation 6.3.79 are I0 (nπr /L) and K0 (nπr/L), solutions to this equation that satisfy the boundary conditions are  A[K0 (nπr/L)I0 (nπr1 /L) − K0 (nπr 1 /L)I0 (nπr/L)],   r1 ≤ r ≤ ρ, Gn (r|ρ) = B[K (nπr/L)I (nπr /L) − K (nπr  0 0 2 0 2 /L)I0 (nπr/L)], (6.3.80)  ρ ≤ r ≤ r2 . To compute A and B, we first use the condition that the Green’s function must be continuous from r = ρ− to r = ρ+ , or

A[K0 (nπρ/L)I0 (nπr1 /L) − K0 (nπr 1 /L)I0 (nπρ/L)] (6.3.81) = B[K0 (nπρ/L)I0 (nπr2 /L) − K0 (nπr2 /L)I0 (nπρ/L)]. 31 A generalization of a problem from Oliveira, J. C., and C. H. Amon, 1998: Pressurebased semi-analytical solution of radial Stokes flows. J. Math. Anal. Appl., 217, 95–114.

442

Green’s Functions with Applications

Figure 6.3.3: Equation 6.3.84 [times ρπ2 ] for the planar Poisson equation within an annulus of finite height with Dirichlet boundary conditions on all sides when r1 /L = 1, r2 /L = 2, ζ/L = 1/π, and ρ/L = 1 + 1/π.

On the other hand, integrating Equation 6.3.79 from r = ρ− to r = ρ+ yields the jump condition 'r=ρ+ dGn '' 1 =− , (6.3.82) dr 'r=ρ− πρL or A[K1 (nπρ/L)I0 (nπr1 /L) + K0 (nπr1 /L)I1 (nπρ/L)]

(6.3.83) 1 − B[K1 (nπρ/L)I0 (nπr2 /L) + K0 (nπr 2 /L)I1 (nπρ/L)] = − 2 . nπ ρ

Solving for A and B and substituting into Equation 6.3.80, and then Equation 6.3.78, we obtain , - * ∞ ( nπζ nπz + R1 (r< )R2 (r> ) g(r, z|ρ,ζ ) = − sin sin , (6.3.84) L L nπ 2 ρCn n=1

where

Rm (r) = K0 (nπr/L)I0 (nπrm /L) − K0 (nπr m /L)I0 (nπr/L),

(6.3.85)

Cn = [K0 (nπρ/L)I1 (nπρ/L) + I0 (nπρ/L)K1 (nπρ/L)]

(6.3.86)

and × [K0 (nπr1 /L)I0 (nπr2 /L) − I0 (nπr1 /L)K0 (nπr2 /L)]. Figure 6.3.3 illustrates Equation 6.3.84.

The Helmholtz Equation

443

The special case32 of r1 = 0 follows if we replace Equation 6.3.75 with lim |g(r, z|ρ,ζ )| < ∞,

r→0

g(r2 , z|ρ,ζ ) = 0,

0 < z < L.

(6.3.87)

Equation 6.3.76 through Equation 6.3.79 still hold with r1 = 0. A solution that satisfies the boundary conditions is now  AI0 (nπr/L), 0 ≤ r ≤ ρ,  Gn (r|ρ) = B[K0 (nπr/L)I0 (nπr2 /L) (6.3.88)  −K0 (nπr2 /L)I0 (nπr/L)], ρ ≤ r ≤ r2 .

To compute A and B, we use the fact that Gn (ρ+ , z|ρ,ζ ) = Gn (ρ− , z|ρ,ζ ) and Equation 6.3.82. Solving for A and B and substituting them into Equation 6.3.78 and Equation 6.3.88, we find that g(r, z|ρ,ζ ) = −

, - * ∞ 1 ( nπζ nπz + sin sin πL n=1 L L % I0 (nπr/L)I0 (nπρ/L)K0 (nπr2 /L) × I0 (nπr2 /L)

(6.3.89)

& − I0 (nπr< /L)K0 (nπr> /L) .

• Solid cylinder with Neumann conditions An interesting Green’s function problem involving Laplace’s equation arises when the boundary conditions are Neumann along all of the boundaries. The governing equations are ∂ 2 g 1 ∂g ∂ 2 g δ(r − ρ)δ(z − ζ) 1 + + =− + 2, ∂r2 r ∂r ∂z 2 r πa

(6.3.90)

where 0 ≤ r, ρ < a, and 0 < z, ζ < L, with the boundary conditions lim |g(r, z|ρ,ζ )| < ∞,

r→0

gr (a, z|ρ,ζ ) = 0,

0 < z < L,

(6.3.91)

and gz (r, 0|ρ,ζ ) = gz (r, L|ρ,ζ ) = 0,

0 ≤ r < a.

(6.3.92)

Note the last term on the right side of Equation 6.3.90. It occurs because the elliptic operator on the left side has a null eigenvalue in its spectrum. From the general theory of linear inhomogeneous equations, a solution to this 32 See Mukkavilli, S., L. L. Tavlarides, and C. V. Wittmann, 1987: Integral method of analysis for chemical reaction in a nonisothermal finite cylindrical catalyst pellet–I. Dirichlet problem. Chem. Engng. Sci., 42, 27–33.

444

Green’s Functions with Applications

equation will exist only if the source term (the right side) is orthogonal to the null space of the operator. This constant term is necessary so that this condition is satisfied. Let us introduce the normalized eigenfunctions ψn (r) = An J0 (kn r),

n = 1, 2, . . . ,

(6.3.93)

where J0$ (kn a) = −J1 (kn a) = 0 and A2n = 2/[a2 J02 (kn a)]. Then, we can write the Green’s functions as g(r, z|ρ,ζ ) = G0 (z|ζ) +

∞ (

Gn (z|ζ)ψn (ρ)ψn (r).

(6.3.94)

n=1

Substituting Equation 6.3.94 into Equation 6.3.90, we find that G$$0 +

∞ ( #

n=1

$ δ(r − ρ)δ(z − ζ) 1 G$$n − kn2 Gn ψn (ρ)ψn (r) = − + 2 , (6.3.95) 2πr πa L

or π 2 aG$$0 = −δ(z − ζ) + "

1 , L

L

G0 (z|ζ) dz = 0, 0

and G$$n − kn2 Gn = −δ(z − ζ),

G$0 (0|ζ) = G$0 (L|ζ) = 0, 'ζ + '

G$0 (z|ζ)''

ζ−

=−

1 , πa2

G$n (0|ζ) = G$n (L|ζ) = 0.

(6.3.96) (6.3.97)

(6.3.98)

Solving Equation 6.3.96 and Equation 6.3.97, G0 (z|ζ) =

L z2 + ζ 2 z> + − 2. 3πa2 2πa2 L πa

(6.3.99)

On the other hand, the equation governing Gn (z|ζ) is G$$n

−

kn2 Gn

, ∞ * mπz + 1 ( mπζ =− 2m cos cos , L m=0 L L

(6.3.100)

where 20 = 1 and 2m = 2 for m > 0. The right side of Equation 6.3.100 is the representation of the delta function as a Fourier cosine series. Solving Equation 6.3.100, we have that Gn (z|ζ) =

∞ 1 ( cos(mπζ) cos(mπz) 2m . L m=0 kn2 + m2 π 2 /L2

(6.3.101)

The Helmholtz Equation

445

Therefore, the generalized Green’s function is ∞ ∞ L z2 + ζ2 z> 1 (( 2m ψn (ρ)ψn (r) + − + . 3πa2 2πa2 L πa2 πa2 L n=1 m=0 (kn2 + m2 π 2 /L2 )J02 (kn a) (6.3.102) • Solid cylinder with Robin’s conditions

g(r, z|ρ,ζ ) =

A problem33 that is similar to the previous one is ∂ 2g 1 ∂g ∂ 2g δ(r − ρ)δ(z − ζ) + + =− , ∂r2 r ∂r ∂z 2 r

(6.3.103)

where 0 ≤ r, ρ < r2 , and 0 < z, ζ < L, with the boundary conditions lim |g(r, z|ρ,ζ )| < ∞,

r→0

gr (r2 , z|ρ,ζ ) + a g(r2 , z|ρ,ζ ) = 0,

gz (r, 0|ρ,ζ ) − a g(r, 0|ρ,ζ ) = 0,

0 ≤ r < r2 ,

gz (r, L|ρ,ζ ) + a g(r, L|ρ,ζ ) = 0,

0 ≤ r < r2 ,

0 < z < L, (6.3.104) (6.3.105)

and (6.3.106)

where a > 0. We begin by introducing the eigenfunctions Zn (z) = cos[βn (2z − L)], where βn is the nth solution of β tan(βL) = a/2 and n = 1, 2, 3, . . .. This choice is useful because Zn$ (0) − a Zn (0) = 0 and Zn$ (L) + a Zn (L) = 0. The eigenfunction representation for the (Dirac) delta function is δ(z − ζ) =

∞ ( 1 cos[βn (2ζ − L)] cos[βn (2z − L)], δ n=1 n

where δn =

% & L 2 1+ sin2 (βn L) . 2 aL

(6.3.107)

(6.3.108)

Consequently, the Green’s function is given by ∞ ( 1 g(r, z|ρ,ζ ) = Gn (r|ρ) cos[βn (2ζ − L)] cos[βn (2z − L)]. δ n=1 n

(6.3.109)

This Green’s function satisfies the boundary conditions, Equation 6.3.105 and Equation 6.3.106. 33 See Mukkavilli, S., L. L. Tavlarides, and Ch. V. Wittmann, 1987: Integral method of analysis for chemical reaction in a nonisothermal finite cylindrical catalyst pellet–II. Robin problem. Chem. Engng. Sci., 42, 35-40.

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Green’s Functions with Applications

Substituting Equation 6.3.107 and Equation 6.3.109 into Equation 6.3. 103, we obtain the ordinary differential equation d2 Gn 1 dGn δ(r − ρ) + − 4βn2 Gn = − , 2 dr r dr r

0 ≤ r < r2 ,

(6.3.110)

for Gn (r|ρ). A solution to Equation 6.3.110 which is continuous at r = ρ is Gn (r|ρ) = AI0 (2βn r< ) [I0 (2βn r> ) − αK0 (2βn r> )] .

(6.3.111)

To satisfy the boundary condition G$n (r2 |ρ) + a Gn (r2 |ρ) = 0, α=

aI0 (2βn r2 ) + 2βn I1 (2βn r2 ) . aK0 (2βn r2 ) − 2βn K1 (2βn r2 )

(6.3.112)

Finally, to find A we use the condition that 'r=ρ+ dGn '' 1 =− . ' dr r=ρ− ρ

(6.3.113)

This yields αA = −1. Consequently, the Green’s function for this problem is g(r, z|ρ,ζ ) = −

∞ 1( 1 cos[βn (2ζ − L)] cos[βn (2z − L)] α n=1 δn

× I0 (2βn r< ) [I0 (2βn r> ) − αK0 (2βn r> )] .

(6.3.114)

• Elliptic coordinates During the computation of the potential around a dielectric cylinder, Martin et al.34 found the Green’s function in elliptic coordinates. The transformation from ξ,η to x, y is . . x = aξη, y = a ξ2 − 1 1 − η2 , (6.3.115)

where 1 ≤ ξ < ∞ and −1 ≤ η < 1. The Green’s function in the present case is given by % & 2 1 ∂ 2g ∂g ∂g 2 2 ∂ g (ξ − 1) 2 + ξ + (1 − η ) 2 − η a2 (ξ 2 − η 2 ∂ξ ∂ξ ∂η ∂η 1 =− δ(ξ − ξ $ )δ(η − η $ ), (6.3.116) hξ hη 34 Martin, J. L., I. Marin-Enriquez, R. Riera, and R. P´ erez-Enriquez, 2007: On the potential of an infinite dielectric cylinder and a line of charge: Green’s function in an elliptic coordinate approach. Rev. Mex. Fis., 53, 41–47.

The Helmholtz Equation with hξ = a

T

447

ξ2 − η2 , ξ2 − 1

hη = a

T

ξ2 − η2 . 1 − η2

The quantity 2a gives the interfocal distance. Because ∞ ( Tm (η)Tm (η $ ) δ(η − η $ ) = bm . 1 − η2 m=0

with

"

1

−1

and bm

Tm (η)Tm$ (η) . dη = am δm,m$ , 1 − η2

1 = = am

!

1/π, 2/π,

m = 0, m &= 0,

(6.3.117)

(6.3.118)

(6.3.119)

(6.3.120)

we assume that the Green’s function is given by the orthogonal expansion: g(ξ,η |ξ $ , η$ ) =

∞ (

gm (ξ|ξ $ , η $ )Tm (η),

(6.3.121)

m=0

where Tm (·) is the Chebyshev polynomial of the first kind of order m. Substituting Equation 6.3.121 into Equation 6.3.116, we obtain the ordinary differential equation %

(ξ 2 − 1)

& . d2 d 2 + ξ − m gm = − ξ 2 − 1bm Tm (η $ )δ(ξ − ξ $ ) 2 dξ dξ

(6.3.122)

for gm . The homogeneous solution to Equation 6.3.122 is gm (ξ|ξ $ , η$ ) = ATm (η) + BSm (η),

(6.3.123)

where Sm (·) is the Chebyshev functions of the second kind. To evaluate A and B, we have the conditions that g(ξ + |ξ $ , η $ ) = g(ξ − |ξ $ , η $ ) and

bm Tm (η$ ) g $ (ξ + |ξ $ , η $ ) − g $ (ξ − |ξ $ , η $ ) = − . . ξ $2 − 1

(6.3.124)

(6.3.125)

Applying these conditions and substituting the results into Equation 6.3.125, we obtain the final solution g(ξ,η |ξ $ , η $ ) =

∞ (

m=0

Sm (η> )Tm (η< )

Tm (η)Tm (η $ ) . am

(6.3.126)

448

Green’s Functions with Applications

• Green’s functions via complex variables It is well known that the real and imaginary parts of any analytic function satisfy not only the Cauchy-Riemann but also Laplace’s equation. Can we use this fact to find the Green’s function for Laplace equation? For the biharmonic equation, Dean35 applied complex variables to find the Green’s functions when there is an elliptic hole in the plane. Consider the complex function % & % & 1 1 log = − ln |f (z; z0)| + i arg (6.3.127) f (z; z0 ) f (z; z0 ) = g(x, y|ξ,η ) − iθ(x, y|ξ,η ), (6.3.128) where f is an analytic function within the domain D surrounded by the boundary C with f (z0 ; z0 ) = 0, z = x + iy, and z0 = ξ + iη. Of course, the point z0 lies within D. If we define g(ξ,η |ξ,η ) as the limit of g(x, y|ξ,η ) as z → z0 , we have that g(ξ,η |ξ,η ) = ∞. Next, we note that any point in D other than ξ,η , ∇2 g = 0 because g is the real part of a harmonic function. Furthermore, g(x, y|ξ,η ) = g(ξ,η |x, y). Finally, if g(x, y|ξ,η ) satisfies certain homogeneous boundary conditions, then g satisfies all of the conditions required of a twodimensional Green’s function for Poisson’s equation in the closed domain D. Because g(x, y|ξ,η ) is analytic at each point on the boundary C, we have from the Cauchy-Riemann relationships that ∂g ∂θ =− , ∂x ∂y

and

∂g ∂θ = . ∂y ∂x

(6.3.129)

We next observe that ∂g ∂g ∂g ∂g = cos(n, y) + cos(n, x) = , ∂s ∂y ∂x ∂n

(6.3.130)

where s and n denote any pair of mutually perpendicular directions. For example, s = −y and n = x, or s = x and n = y. The cosine terms are the directional cosines. From Green’s identity, we have that 0 1 ∂g(ζ|z0 ) u(z0 ) = u(ζ) ds, (6.3.131) 2π C ∂n or

1 u(z) = 2π

0

C

u ˜(ζ)

∂g(ζ|z0 ) ds, ∂n

(6.3.132)

where u ˜(ζ) denotes any real function defined and continuous on the boundary C of D and we have written g(x, y|ξ,η ) using complex variables as g(z|z0 ). 35 Dean, W. R., 1954: Note on the Green’s function of an elastic plate. Proc. Cambridge Philos. Soc., 50, 623–627.

The Helmholtz Equation

449

Equations 6.3.131 and 6.3.132 are the complex variable form of Green’s formula. It expresses the value of a harmonic function u at any point z0 in the domain D in terms of the values taken by this function on the boundary of D. • Example 6.3.1: Poisson’s integral formula Consider the function f (z; z0 ) = Then,

R(z − z0 ) , R2 − zz0∗

|z| < R.

' ' ' R(z − z0 ) ' ' '. g(z|z0 ) = − ln ' 2 R − zz0∗ '

(6.3.133)

(6.3.134)

Let z0 = reiϕ and ζ = Reiψ , a point on the boundary |z| = R. Then ds = R d ψ and ∂g ∂ arg[f (ζ; z0 )] 1 ∂ arg[f (ζ; z0 )] = = . (6.3.135) ∂n ∂n R ∂ψ Here arg denotes a branch of the argument that is continuous in a neighborhood surrounding the boundary point ζ. If the Green’s function satisfies a Dirichlet boundary condition, g(ζ|z0 ) = 0 on C and we have that ∂ arg[f (ζ; z0 )] ∂ log[f (ζ; z0 )] = −i , ∂ψ ∂ψ

(6.3.136)

and ! % &) ∂g ∂ log[f (ζ; z0 )] i ∂ Reiψ − reiϕ = −i =− log ∂n ∂ψ R ∂ψ R − rei(ψ−ϕ) 2 2 1 R −r = . R R2 − 2Rr cos(ψ − ϕ) + r2

(6.3.137) (6.3.138)

Substituting this result in Equation 6.3.131 and writing u(Reiψ ) for u(ζ), we obtain " 2π 1 R2 − r2 u(reiϕ ) = u(Reiψ ) 2 dψ. (6.3.139) 2π 0 R − 2Rr cos(ψ − ϕ) + r2 This result is known as Poisson’s integral formula; it gives the solution to Dirichlet’s problem for a disc. A similar formula can be obtained for the upper half-plane y > 0. Using f (z; z0 ) =

z − z0 , z − z0∗

5(z) > 0,

(6.3.140)

450

Green’s Functions with Applications

in place of Equation 6.3.133 and proceeding as before, we obtain Poisson’s integral formula for the upper half-plane, " 1 ∞ y u(x, y) = u(ξ) dξ. (6.3.141) π −∞ (x − ξ)2 + y 2 Here u(ξ) denotes the prescribed function giving the boundary values of the harmonic function on the real axis; it is assumed that u(ξ) is such that it has only a finite number of finite discontinuities along the x axis and the integral in Equation 6.3.141 converges. 6.4 TWO-DIMENSIONAL HELMHOLTZ EQUATION OVER RECTANGULAR AND CIRCULAR DOMAINS One of the classic problems of mathematical physics is the forced vibrations of a membrane over a rectangular region. Because the edges are usually clamped, the solution equals zero there. After the time dependence is removed, we are left with the two-dimensional Helmholtz equation ∂2u ∂2u + 2 + k02 u = −f (x, y). ∂x2 ∂y

(6.4.1)

In this section, we obtain the Green’s function for Equation 6.4.1. • Rectangular Area The problem to be solved is ∂2g ∂ 2g + 2 + k02 g = −δ(x − ξ)δ(y − η), 2 ∂x ∂y

(6.4.2)

where 0 < x, ξ < a, and 0 < y, η < b, subject to the boundary conditions that g(0, y|ξ,η ) = g(a, y|ξ,η ) = g(x, 0|ξ,η ) = g(x, b|ξ,η ) = 0.

(6.4.3)

We use the same technique to solve Equation 6.4.2 as we did in Section 6.3 by assuming that the Green’s function has the form g(x, y|ξ,η ) =

∞ (

m=1

Gm (x|ξ) sin

* mπy + b

,

(6.4.4)

where the coefficients Gm (x|ξ) are undetermined functions of x, ξ, and η. Substituting this series into Equation 6.4.2, multiplying by 2 sin(nπy/b) /b, and integrating over y, we find that , 2 2 * nπη + d2 Gn n π 2 2 − − k δ(x − ξ). (6.4.5) 0 Gn = − sin 2 2 dx b b b

The Helmholtz Equation

451

The first method for solving Equation 6.4.5 involves writing δ(x − ξ) =

, - * ∞ 2 ( mπξ mπx + sin sin , a m=1 a a

(6.4.6)

∞ * mπx + 2 ( anm sin . a m=1 a

(6.4.7)

and Gn (x|ξ) =

Upon substituting Equation 6.4.6 and Equation 6.4.7 into Equation 6.4.5, we obtain ∞ , * mπx + ( m2 π 2 n2 π 2 k02 − 2 − 2 anm sin a b a m=1 , - * ∞ * nπη + 4 ( mπξ mπx + =− sin sin sin . (6.4.8) ab m=1 b a a Matching the harmonics, anm =

4 sin(mπξ/a) sin(nπη/b) , ab(m2 π 2 /a2 + n2 π 2 /b2 − k02 )

(6.4.9)

and the bilinear form of the Green’s function 36 is

g(x, y|ξ,η ) =

∞

∞

4 ( ( sin(mπξ/a) sin(nπη/b) sin(mπx/a) sin(nπy/b) . ab n=1m=1 m2 π 2 /a2 + n2 π 2 /b2 − k02

(6.4.10) The bilinear form of the Green’s function for the two-dimensional Helmholtz equation with Neumann boundary conditions and the three-dimensional Helmholtz equation with Dirichlet and Neumann boundary conditions are left as Problems 42, 45, and 47, respectively. As in the previous section, we can construct an alternative to the bilinear form of the Green’s function, Equation 6.4.10, by writing Equation 6.4.5 as * nπη + d2 Gn 2 2 − k G = − sin δ(x − ξ), n n dx2 b b

(6.4.11)

36 For an application, see Section 4 of Rasband, S. N., and L. Turner, 1981: Solutions of the Helmholtz equation with boundary conditions for force-free magnetic fields. Phys. Fluids, 24, 931–935.

452

Green’s Functions with Applications

Figure 6.4.1: Equation 6.4.10 for Helmholtz’s equation over a rectangular region with a Dirichlet boundary condition on the sides when a = b, k0 a = 10, and ξ/a = η/a = 0.35.

where kn2 = n2 π 2 /b2 − k02 . The homogeneous solution to Equation 6.4.11 is now ! An sinh(kn x), 0 ≤ x ≤ ξ, Gn (x|ξ) = (6.4.12) Bn sinh[kn (a − x)], ξ ≤ x ≤ a.

This solution satisfies the boundary conditions at both end points. Because Gn (x|ξ) must be continuous at x = ξ, An sinh(kn ξ) = Bn sinh[kn (a − ξ)]. On the other hand, the jump discontinuity involving

(6.4.13)

G$n (x|ξ)

yields * + 2 nπη −kn Bn cosh[kn (a − ξ)] − kn An cosh(kn ξ) = − sin . (6.4.14) b b Solving for An and Bn , * nπη + sinh[k (a − ξ)] 2 n An = sin , (6.4.15) bkn b sinh(kn a) and

* nπη + sinh(k ξ) 2 n sin . bkn b sinh(kn a) This yields the Green’s function Bn =

g(x, y|ξ,η ) =

(6.4.16)

N * nπη + * nπy + 2 ( sin[κn (a − x> )] sin(κn x< ) sin sin b n=1 κn sin(κn a) b b

∞ * nπη + * nπy + 2 ( sinh[kn (a − x> )] sinh(kn x< ) + sin sin . b kn sinh(kn a) b b n=N +1

(6.4.17)

The Helmholtz Equation

453

Here N denotes the largest value of n such that kn2 < 0, and κ2n = k02 − n2 π 2 /b2 . If we began with a Fourier expansion in the y direction, we would have obtained , - * M 2 ( sin[κm (b − y> )] sin(κm y< ) mπξ mπx + g(x, y|ξ,η ) = sin sin a m=1 κm sin(κm b) a a , ∞ * mπx + 2 ( sinh[km (b − y> )] sinh(km y< ) mπξ + sin sin , a km sinh(km b) a a m=M +1

(6.4.18)

2 2 where M denotes the largest value of m such that km < 0, km = m2 π 2 /a2 −k02 , 2 2 2 2 2 and κm = k0 − m π /a .

• Circular Disk In this subsection, we find the Green’s function for the Helmholtz equation when the domain consists of the circular region 0 < r < a. The Green’s function is governed by the partial differential equation , 1 ∂ ∂g 1 ∂ 2g δ(r − ρ)δ(θ − θ$ ) r + 2 2 + k02 g = − , r ∂r ∂r r ∂θ r

(6.4.19)

where 0 < r, ρ < a, and 0 ≤ θ,θ $ ≤ 2π, with the boundary conditions lim |g(r,θ |ρ,θ $ )| < ∞,

r→0

g(a,θ |ρ,θ $ ) = 0,

0 ≤ θ,θ $ ≤ 2π.

(6.4.20)

The Green’s function is periodic in θ. We begin by noting that δ(θ − θ $ ) =

∞ ∞ 1 1( 1 ( + cos[n(θ − θ$ )] = cos[n(θ − θ$ )]. 2π π n=1 2π n=−∞

(6.4.21)

Therefore, the solution has the form g(r,θ |ρ,θ $ ) =

∞ (

n=−∞

gn (r|ρ) cos[n(θ − θ $ )].

(6.4.22)

Substituting Equation 6.4.21 and Equation 6.4.22 into Equation 6.4.19 and simplifying, we find that , 1 d dgn n2 δ(r − ρ) r − 2 gn + k02 gn = − . r dr dr r 2πr

(6.4.23)

454

Green’s Functions with Applications

Figure 6.4.2: Equation 6.4.27 for Helmholtz’s equation within a√ circular disk with a Dirichlet boundary condition on the rim when k0 a = 10, ρ/a = 0.35 2, and θ ! = π/4.

The solution to Equation 6.4.23 is the Fourier-Bessel series , ∞ ( knm r gn (r|ρ) = Anm Jn , a m=1

(6.4.24)

where knm is the mth root of Jn (k) = 0. Upon substituting Equation 6.4.24 into Equation 6.4.23 and solving for Anm , we have that , " a 1 knm r 2 2 2 (k0 − knm /a )Anm = − 2 $ 2 δ(r − ρ)Jn dr, (6.4.25) a πa J n (knm ) 0 or Jn (knm ρ/a) Anm = . (6.4.26) 2 π(knm − k02 a2 )J $ 2n (knm ) Thus, the Green’s function37 is

∞ ∞ 1 ( ( Jn (knm ρ/a)Jn (knm r/a) g(r,θ |ρ,θ ) = cos[n(θ − θ$ )]. 2 − k 2 a2 )J $ 2 (k π n=−∞ m=1 (knm n nm ) 0 $

(6.4.27) 37 Sommerfeld, A., 1912: Die Greensche Funktion der Schwingungsgleichung. Jahresber. Deutschen Math.-Vereinung, 21, 309–353. For an example of its use, see Zhang, D. R., and C. F. Foo, 1999: Fields analysis in a solid magnetic toroidal core with circular cross section based on Green’s function. IEEE Trans. Magnetics, 35, 3760–3762.

The Helmholtz Equation

455

An alternative to Equation 6.4.27 can be obtained as follows: As we noted earlier, a particular solution to Equation 6.4.19 is −Y0 (k0 R)/4, where R2 = r2 + ρ2 − 2rρ cos(θ − θ $ ). Consequently, the most general solution to Equation 6.4.19 consists of this particular solution plus a homogeneous solution: g(r,θ |ρ,θ $ ) = − 14 Y0 (k0 R) +

∞ (

n=−∞

An Jn (k0 r) cos[n(θ − θ $ )].

(6.4.28)

Along r = a, g(a,θ |ρ,θ $ ) = 0, and 1 Y 4 0

∞ 4 . 5 ( k0 a2 + ρ2 − 2aρ cos(θ − θ $ ) = An Jn (k0 a) cos[n(θ − θ$ )]. n=−∞

(6.4.29)

However, from the addition theorem of Bessel functions, ∞ 4 . 5 ( Y0 k0 a2 + ρ2 − 2aρ cos(θ − θ$ ) = Jn (k0 ρ)Yn (k0 a) cos[n(θ − θ$ )]. n=−∞

(6.4.30)

Therefore, An =

Jn (k0 ρ)Yn (k0 a) 4Jn (k0 a)

(6.4.31)

so that 4 . 5 g(r,θ |ρ,θ $ ) = − 41 Y0 k0 r2 + ρ2 − 2rρ cos(θ − θ$ ) +

1 4

∞ ( Jn (k0 ρ)Yn (k0 a) Jn (k0 r) cos[n(θ − θ$ )]. J (k a) n 0 n=−∞

(6.4.32)

Again, we can use the addition theorem and find that g(r,θ |ρ,θ $ ) = − 14

∞ (

n=−∞

cos[n(θ − θ$ )]

% & Yn (k0 a) × Jn (k0 r< ) Yn (k0 r> ) − Jn (k0 r> ) . Jn (k0 a)

(6.4.33)

Let us now turn to the case of Neumann boundary conditions. In a manner38 similar to that used to derive Equation 6.4.27, we find that the 38

See Okoshi, T., T. Takeuchi, and J.-P. Hsu, 1975: Planar 3-dB hybrid circuit. Electron. Commun. Japan, 58-B(8), 80–90.

456

Green’s Functions with Applications

Green’s function for Helmholtz’s equation with Neumann boundary conditions is ∞ ∞ 1 2 ( ( Jn (knm ρ)Jn (knm r) cos[n(θ − θ $ )] + , 2 )(k 2 − k 2 )J 2 (k πa2 k02 π n=0 m=1 2n (a2 − n2 /knm nm 0 n nm a) (6.4.34) where knm in the mth root of Jn$ (knm a) = 0, 20 = 2, and 2n = 1 for n > 0. Equation 6.4.34 has two disadvantages. First, we must compute two power series as well as the Bessel functions. Second, we must find the eigenvalue knm . Alhargan and Judah39 showed how both of these difficulties can be eliminated by summing the inner power series exactly using Mittag-Leffler’s expansion theory. This yields

g(r,θ |ρ,θ $ ) = −

∞ 1 ( Jn (k0 r< )Fn (k0 r> ) cos[n(θ − θ$ )] 2 n=0 2n Jn$ (k0 a) 4 . 5 = − 41 Y0 k0 r2 + ρ2 − 2rρ cos(θ − θ$ )

g(r,θ |ρ,θ $ ) =

+ where

1 2

∞ ( Jn (k0 ρ)Yn$ (k0 a) Jn (k0 r) cos[n(θ − θ$ )], $ (k a) 2 J n 0 n n=−∞

Fn (kr) = Yn$ (ka)Jn (kr) − Jn$ (ka)Yn (kr).

(6.4.35)

(6.4.36)

(6.4.37)

McIver40 applied Equation 6.4.36 to understand how water waves propagate through an array of cylindrical structures. • Circular Sector As the previous examples showed, Green’s functions for various twodimensional domains can be expressed in terms of the orthonormal eigenfunction expansion ( un (r)un (r0 ) g(r|r0 ) = , (6.4.38) kn2 − k02 n

where the eigenfunction un satisfies the differential equation and homogeneous boundary conditions and kn is the eigenvalue, respectively, of the Helmholtz equation ∇2 un + kn2 un = 0. (6.4.39) Similar considerations were developed in Section 3.4. 39

Alhargan, F. A., and S. R. Judah, 1991: Reduced form of the Green’s function for disks and annular rings. IEEE Trans. Microwave Theory Tech., MMT-39, 601–604. 40 McIver, P., 2000: Water-wave propagation through an infinite array of cylindrical structures. J. Fluid Mech., 424, 101–125.

The Helmholtz Equation

457

In this subsection, we use Equation 6.4.38 to immediately find the Green’s function for various circular sectors.41 We begin by computing the Green’s function for a circular sector defined over the region 0 < r < a and 0 < θ < α. We will find the Green’s function with Dirichlet conditions along the boundary. Consider the product solution: umγ (r,θ ) = Jγ (kmγ r) sin(γθ),

(6.4.40)

where γ = nπ/α, and n is any positive integer. A quick check shows that Equation 6.4.40 satisfies the homogeneous Helmholtz equation and boundary conditions if Jγ (kmγ a) = 0. Here the term kmγ a is the mth zero of the Bessel function Jγ (·). Equation 6.4.40 does not form an orthogonal set unless γ is an integer or α = π/0, where 0 is a positive integer. In that case, "

a 0

2 Jγ2 (kmγ r) r dr = 12 a2 Jγ+1 (kmγ a).

(6.4.41)

Employing Equation 6.4.41 to normalize the eigenfunctions and then using Equation 6.4.38, we immediately obtain the Green’s function g(r,θ |ρ,θ $ ) =

∞ ∞ 40 ( ( Jγ (kmγ ρ)Jγ (kmγ r) sin(γθ) sin(γθ$ ) . 2 − k 2 )J 2 (k πa2 n=1 m=1 2n (kmγ 0 γ+1 mγ a)

(6.4.42)

Similar considerations hold if we wish to find the Green’s function for Helmholtz’s equation with a Neumann condition. The product solution42 in this case is umγ (r,θ ) = Jγ (kmγ r) cos(γθ), (6.4.43) where n = 0, 1, 2, . . . and kmγ a is the mth root of Jγ$ (·). In addition to these roots, which we will number m = 1, 2, . . ., we have the additional solution k00 = 0 when n = 0. The orthogonality condition here is "

a 0

Jν2 (kmν r) r dr

=

 

1 2 a , 2

11 2

2

2 a2 − ν 2 /kmν Jν2 (kmν a),

m = ν = 0, (6.4.44) otherwise,

41 A summary is given in Lozyanoy, V. I., I. V. Petrusenko, I. G. Prokhoda, and V. P. Prudkiy, 1984: Green’s function of simple regions in a polar coordinate system for solving electromagnetic problems. Radio Engng. Electron. Phys., 29(6), 8–13. 42 Several useful product solutions that satisfy Neumann boundary conditions are summarized in Lo, Y. T., D. Solomon, and W. F. Richards, 1979: Theory and experiment on microstrip antennas. IEEE Trans. Antennas Propagat., AP-27, 137–145.

458

Green’s Functions with Applications

where ν = n0. Note that ν is an nonnegative integer. Performing the same analysis43 as above, we now have that g(r,θ |ρ,θ ) = −

∞ ∞ 20 40 ( ( Jν (kmν ρ)Jν (kmν r) cos(νθ) cos(νθ$ ) + . 2 )(k 2 − k 2 )J 2 (k πa2 k02 π n=0 m=1 2ν (a2 − n2 02 /kmν mν 0 ν mν a) (6.4.45)

• Annular rings In the case of an annular ring defined by a < r, ρ < b, the mutually orthogonal eigenfunctions that satisfy Dirichlet conditions along r = a and r = b are ! cos(nθ), umn(r,θ ) = [Yn (kmn a)Jn (kmn r) − Jn (kmn a)Yn (kmn r)] sin(nθ), (6.4.46) where m = 1, 2, 3, . . . and n = 0, 1, 2, . . . The eigenvalues kmn are computed from Jn (kmn a) Jn (kmn b) = . (6.4.47) Yn (kmn a) Yn (kmn b) Because "

b

a

[Yn (kmn a)Jn (kmn r) − Jn (kmn a)Yn (kmn r)]2 r dr ! 1 = 2 b2 [Yn (kmn a)Jn$ (kmn b) − Jn (kmn a)Yn$ (kmn b)]2 − a2 [Yn (kmn a)Jn$ (kmn a) − Jn (kmn a)Yn$ (kmn b)]2

(6.4.48) )

and using the normalizing factors for cos(nθ) and sin(nθ), we obtain the Green’s function ∞ ∞ 2(( Fmn (ρ)Fmn (r) cos[n(θ − θ$ )] g(r,θ |ρ,θ ) = 2 , 2 $2 (b) − a2 F $2 (a)] (k 2 π n=0 m=1 2n [b Fmn mn nm − k0 ) $

(6.4.49)

where Fmn (r) = Yn (kmn a)Jn (kmn r) − Jn (kmn a)Yn (kmn r),

(6.4.50)

43 See Okoshi, T., and T. Miyoshi, 1972: The planar circuit—An approach to microwave integrated circuitry. IEEE Trans. Microwave Theory Tech., MTT-20, 245–252; Chadha, R., and K. C. Gupta, 1981: Green’s functions for circular sectors, annular rings, and annular sectors in planar microwave circuits. IEEE Trans. Microwave Theory Tech., MTT-29, 68–71.

The Helmholtz Equation

459

Figure 6.4.3: Equation 6.4.42 for Helmholtz’s equation within a circular sector√ with a Dirichlet boundary condition on the rim when k0 a = 10, α = π, ρ/a = 0.35 2, and θ ! = π/4.

20 = 2, and 2n = 1 for n > 0. When there are Neumann boundary conditions, the eigenfunctions are now umn(r,θ ) = [Yn$ (kmn a)Jn (kmn r) − Jn$ (kmn a)Yn (kmn r)]

!

cos(nθ), sin(nθ). (6.4.51)

The eigenvalues kmn are given by the solution of Jn$ (kmn a) Jn$ (kmn b) = , Yn$ (kmn a) Yn$ (kmn b)

(6.4.52)

and the orthogonality condition becomes "

a

b

[Yn$ (kmn a)Jn (kmn r) − Jn$ (kmn a)Yn (kmn r)]2 r dr ! # $ $ 2 1 = 2 b2 − n2 /kmn [Yn (kmn a)Jn (kmn b) − Jn$ (kmn a)Yn (kmn b)]2 (6.4.53) ) # 2 $ $ 2 2 $ 2 − a − n /kmn [Yn (kmn a)Jn (kmn a) − Jn (kmn a)Yn (kmn a)] .

We have the additional eigenvalue k00 = 0 when n = 0.

460

Green’s Functions with Applications Proceeding as before, we obtain the Green’s function44 1 − a2 )k02 ∞ ∞ 2(( 4 + π n=0 m=1 2n (b2 −

g(r,θ |ρ,θ $ ) = −

π(b2

×

Fmn (ρ)Fmn (r) n2 2 (b) )Fmn 2 knm

− (a2 −

5

n2 2 (a) )Fmn 2 knm

cos[n(θ − θ $ )] , 2 − k2 kmn 0

(6.4.54)

where Fmn (r) = Yn$ (kmn a)Jn (kmn r) − Jn$ (kmn a)Yn (kmn r).

(6.4.55)

One of the disadvantages of Equation 6.4.54 is the evaluation of a double series as well as the Bessel functions. We can reduce this double Fourier series to a single one using Mittag-Leffler’s expansion theorem.39 This yields g(r,θ |ρ,θ $ ) = where

∞ 1 ( Fn (r< , a)Fn (r> , b) cos[n(θ − θ $ )], 2 n=0 2n Fn$ (b, a)

(6.4.56)

Fn (r, a) = Yn$ (k0 a)Jn (k0 r) − Jn$ (k0 a)Yn (k0 r),

(6.4.57)

Fn$ (b, a) = Yn$ (k0 a)Jn$ (k0 b) − Jn$ (k0 a)Yn$ (k0 b).

(6.4.58)

and A problem45 that is similar to Equation 6.4.19 with Neumann boundary conditions is , 1 ∂ ∂g 1 ∂ 2g δ(r − ρ) r + 2 2 − k02 g = − δ(θ − θ$ ), (6.4.59) r ∂r ∂r r ∂θ r a < r, ρ < b and 0 ≤ θ,θ $ < 2π, with the boundary conditions that gr (a,θ |ρ,θ $ ) = gr (b,θ |ρ,θ $ ),

0 ≤ θ,θ $ < 2π.

(6.4.60)

The Green’s function is periodic in θ. We begin by expanding δ(θ − θ$ ) as a Fourier cosine series: δ(θ − θ$ ) =

∞ 1 ( cos[n(θ − θ $ )]. 2π n=−∞

(6.4.61)

44 As given in Chadha, R., and K. C. Gupta, 1981: Green’s functions for circular sectors, annular rings, and annular sectors in planar microwave circuits. IEEE Trans. Microwave Theory Tech., MMT-29, 68–71. 45 Weston, V. H., 1980: Stability effects of an external conducting open sheet on magnetically confined plasmas. Phys. Fluids, 23, 1388–1395.

The Helmholtz Equation

461

Then the Green’s function can be written g(r,θ |ρ,θ $ ) = where

∞ (

n=−∞

gn (r|ρ) cos[n(θ − θ $ )],

, 1 d dgn n2 δ(r − ρ) r − 2 gn − k02 gn = − . r dr dr r 2πr

(6.4.62)

(6.4.63)

The solution gn (r|ρ) is then gn (r|ρ) = AR< (r)R> (r), where R< (r) = Kn (k0 r)In$ (k0 a) − In (k0 r)Kn$ (k0 a),

(6.4.64)

R> (r) = Kn (k0 r)In$ (k0 b) − In (k0 r)Kn$ (k0 b).

(6.4.65)

and $ $ Note that R< (a) = R> (b) = 0. To determine A, we integrate Equation 6.4.63 − + over the interval (ρ , ρ ) and find that

'ρ+ dgn '' 1 =− . dr 'ρ− 2πρ

(6.4.66)

Substituting gn (r|ρ) into Equation 6.4.66 and simplifying, we obtain A [In$ (k0 b)Kn$ (k0 a) − In$ (k0 a)Kn$ (k0 b)] = −

1 . 2π

(6.4.67)

Upon substituting gn (r|ρ) into Equation 6.4.62, the final result is g(r,θ |ρ,θ $ ) = − =−

∞ 1 ( R< (r)R> (r) cos[n(θ − θ$ )] $ 2π n=−∞ In (k0 b)Kn$ (k0 a) − In$ (k0 a)Kn$ (k0 b)

∞ 1( R< (r)R> (r) cos[n(θ − θ$ )] . π n=0 2n [In$ (k0 b)Kn$ (k0 a) − In$ (k0 a)Kn$ (k0 b)]

(6.4.68) (6.4.69)

We also can find an approximate expression for k0 2 1 by using the asymptotic expression for the Bessel functions. The result is g(r,θ |ρ,θ $ ) ≈ const − −

[(r2 + ρ2 )/(2b2 ) + (a/b)2 ln(a/r< ) + ln(b/r> )] 2π [1 − (a/b)2 ] ∞ n ( [(a/r< ) + (r< /a)n ](b/r> )n + (r> /b)n ]

n=1

2nπ [1 − (a/b)2n ]

× (a/b)n cos[n(θ − θ$ )].

(6.4.70)

462

Green’s Functions with Applications

r=b

( , ) =

r=a

(r, )

=0 Figure 6.4.4: The geometry of the annular sector.

• Annular sectors The Green’s functions for annular sectors (see Figure 6.4.4), where a < r, ρ < b, and 0 < θ, θ$ < α, are found following the same technique as we used for annular rings. Here, the mutually orthogonal eigenfunctions that satisfy Dirichlet conditions along its edges are umν (r,θ ) = [Yν (kmν a)Jν (kmν r) − Jν (kmν a)Yν (kmν r)] sin(νθ),

(6.4.71)

where ν = nπ/α, m = 1, 2, 3, . . ., and n = 1, 2, 3, . . . The eigenvalues kmν are computed from Jν (kmν a) Jν (kmν b) = . (6.4.72) Yν (kmν a) Yν (kmν b) Proceeding as we did for the annular rings, we obtain the Green’s function g(r,θ |ρ,θ $ ) =

∞ ∞ 2 ( ( Fmν (ρ)Fmν (r) sin(νθ) sin(νθ$ ) , $2 (b) − a2 F $2 (a)] (k 2 − k 2 ) π n=1 m=1 [b2 Fmν mν mν 0

(6.4.73)

where Fmν (r) = Yν (kmν a)Jν (kmν r) − Jν (kmν a)Yν (kmν r).

(6.4.74)

In the case of Neumann boundary conditions, the eigenfunctions are umν (r,θ ) = [Yν$ (kmν a)Jν (kmν r) − Jν$ (kmν a)Yν (kmν r)] cos(νθ).

(6.4.75)

The eigenvalues kmν are found by solving Jν$ (kmν a) J $ (kmν b) = ν$ . $ Yν (kmν a) Yν (kmν b)

(6.4.76)

The Helmholtz Equation

463

We have the additional eigenvalue k00 = 0 when n = 0. Proceeding as before, we obtain the Green’s function44 20 π(b2 − a2 )k02 ∞ ∞ 40 ( ( 4 + π n=0 m=1 2 (b2 − ν

g(r,θ |ρ,θ $ ) = −

×

Fmν (ρ)Fmν (r) n2 42 2 (b) )Fmν 2 kmν

− (a2 −

5

n2 42 2 (a) )Fmν 2 kmν

cos(νθ) cos(νθ$ ) , 2 − k2 kmν 0

(6.4.77)

where Fmν (r) = Yν$ (kmν a)Jν (kmν r) − Jν$ (kmν a)Yν (kmν r).

(6.4.78)

In the derivation of Equation 6.4.73 and Equation 6.4.77, Chadha and Gupta required that π/α must be a positive integer. However, Alhargan and Judah46 showed that this condition is unnecessary. An alternative expression to Equation 6.4.73 is found by first noting that δ(θ − θ$ ) =

, - , ∞ 2( nπθ$ nπθ sin sin , α n=1 α α

(6.4.79)

where 0 < α < 2π. From the form of Equation 6.4.79 we anticipate that g(r,θ |ρ,θ $ ) =

, - , ∞ 2( nπθ $ nπθ Gn (r|ρ) sin sin , α n=1 α α

(6.4.80)

where Gn (r|ρ) satisfies the equation , 1 d dGn ν 2 Gn δ(r − ρ) r − + k02 Gn = − , r dr dr r2 r

(6.4.81)

and ν = nπ/α. Note that Equation 6.4.80 satisfies the Dirichlet boundary conditions if Gn (a|ρ) = Gn (b|ρ) = 0. Solutions to Equation 6.4.81 that satisfy these boundary conditions and the condition that Gn (ρ− |ρ) = Gn (ρ+ |ρ) are Gn (r|ρ) = An Fν (k0 a, k0 r< )Fν (k0 b, k0 r> ),

(6.4.82)

Fν (x, y) = Jν (x)Yν (y) − Yν (x)Jν (y).

(6.4.83)

where

46 Alhargan, F. A., and S. R. Judah, 1994: Circular and annular sector planar components of arbitrary angle for N -way power dividers/combiners. IEEE Trans. Microwave Theory Tech., MTT-42, 1617–1623.

464

Green’s Functions with Applications

Figure 6.4.5: Equation 6.4.86 for Helmholtz’s equation within an annular sector 0 < θ < √π with a Dirichlet boundary conditions on all of the sides when k0 a = 1, k0 b = 2, ρ = 2, and θ ! = π/4.

The coefficient An is determined by the condition that 'r=ρ+ dGn '' 1 =− . ' dr r=ρ− ρ

This leads to Gn (r|ρ) =

πFν (k0 a, k0 r< )Fν (k0 b, k0 r> ) , 2Fν (k0 b, k0 a)

(6.4.84)

(6.4.85)

and , - , ∞ π ( Fν (k0 a, k0 r< )Fν (k0 b, k0 r> ) nπθ$ nπθ g(r,θ |ρ,θ ) = sin sin . α n=1 Fν (k0 b, k0 a) α α (6.4.86) We can also find an alternative to Equation 6.4.86, namely $

g(r,θ |ρ,θ $ ) = where

, - , ∞ π ( Fν (k0 a, k0 r< )Fν (k0 b, k0 r> ) nπθ $ nπθ cos cos , α n=0 2n Fν$ (k0 b, k0 a) α α (6.4.87) Fν (x, y) = Yν$ (x)Jν (y) − Jν$ (x)Yν (y), Fν$ (x, y) = Yν$ (x)Jν$ (y) − Jν$ (x)Yν$ (y),

20 = 2, and 2n = 1 for n > 0.

(6.4.88) (6.4.89)

The Helmholtz Equation

465

a ( , )

Figure 6.4.6: The location of the various image sources used to find the Green’s function within an isosceles right-angled triangular region with Neumann boundary conditions.

• Triangular section Deriving Green’s functions for non-rectangular and non-cylindrical domains is difficult. Chadha and Gupta47 developed an ingenious technique for triangular shaped regions with Neumann boundary conditions. We illustrate their technique for an isosceles right-angled triangle. The interested reader is referred to their paper for 30◦ -60◦ right-angle and equilateral triangles. The problem hinges on devising an eigenfunction expansion to represent δ(x − ξ)δ(y − η) for the region shown in Figure 6.4.6. Along each of the three boundaries drawn as solid lines, we want the derivative of this expansion to vanish. Why? The reason lies in the commonly employed technique of using the same eigenfunctions to represent the Green’s function and the delta function. How did they find this expansion? Chadha and Gupta introduced an infinite number of virtual sources whose basic pattern is shown in Figure 6.4.6 and is repeated an infinite number of times in both the x and y directions. Then they used the method of images. Chadha and Gupta found that δ(x − ξ)δ(y − η) =

∞ ∞ 1 ( ( U (ξ,η )U (x, y), 2a2 m=−∞ n=−∞

(6.4.90)

47 Chadha, R., and K. C. Gupta, 1980: Green’s functions for triangular segments in planar microwave circuits. IEEE Trans. Microwave Theory Tech., MTT-28, 1139–1143.

466

Green’s Functions with Applications

Figure 6.4.7: Equation 6.4.94 for Helmholtz’s equation over an isosceles, right-angled triangular region with a Neumann boundary condition on each side when k0 a = 10 and x/a = y/a = 0.25.

where * mπx + * nπy + * nπx + * mπy + U (x, y) = cos cos + (−1)m+n cos cos . (6.4.91) a a a a The first term in Equation 6.4.91 represents the inner four sources, while the second term represents the outer four sources. Having found the eigenfunction expansion for the delta functions, the corresponding Green’s function is g(x, y|ξ,η ) =

∞ ∞ 1 ( ( Amn U (x, y). 2a2 m=−∞ n=−∞

(6.4.92)

Recall that these eigenfunctions were chosen to satisfy the Neumann boundary conditions along the boundaries. Therefore, we only have to find Amn so that Equation 6.4.92 satisfies the differential equation. Substituting Equation 6.4.90 and Equation 6.4.92 into Equation 6.4.2, we have that % ) ∞ ! * mπ +2 * nπ +2 & ( Amn k02 − − + U (ξ,η ) U (x, y) = 0. a a m=−∞ n=−∞ (6.4.93) Because the bracketed term must vanish for any x and y, we finally obtain ∞ (

g(x, y|ξ,η ) =

∞ ∞ 1 ( ( U (ξ,η )U (x, y) . 2 m=−∞ n=−∞ (m2 + n2 )π 2 − k02 a2

(6.4.94)

The Helmholtz Equation

467

In a similar manner, Terras and Swanson48 found all of the domains bounded by linear planes for which the method of images produces Green’s functions for Laplace’s equation with Dirichlet boundary conditions. 6.5 POISSON’S AND HELMHOLTZ’S EQUATIONS ON A RECTANGULAR STRIP In this section, we continue our study of the two-dimensional Poisson and Helmholtz equations. Our particular focus will be on domains that are of infinite extent in the y direction but are of finite length in the x direction. • Poisson’s equation We begin by finding the Green’s function for Poisson’s equation on the semi-infinite strip 0 < x, ξ < L and 0 < y, η < ∞ where we solve ∂ 2g ∂2g + = −δ(x − ξ)δ(y − η), ∂x2 ∂y 2

(6.5.1)

subject to the boundary conditions g(0, y|ξ,η ) = g(L, y|ξ,η ) = g(x, 0|ξ,η ) = 0,

(6.5.2)

lim g(x, y|ξ,η ) → 0.

(6.5.3)

and y→∞

Cases when −∞ < y, η < ∞ are left as Problems 14, 15, and 16. Because the Green’s function must vanish when x = 0 and x = L, we express it and δ(x − ξ) in terms of the Fourier sine expansions δ(x − ξ) = and g(x, y|ξ,η ) =

, - * ∞ 2 ( nπξ nπx + sin sin , L n=1 L L

, - * nπξ nπx + Gn (y|η) sin sin , L L n=1 ∞ (

(6.5.4)

(6.5.5)

respectively. Substituting Equation 6.5.4 and Equation 6.5.5 into Equation 6.5.1, we find that d2 Gn n2 π 2 − Gn = −δ(y − η), dy 2 L2

0 < y, η < ∞,

(6.5.6)

48 Terras, R., and R. A. Swanson, 1980: Electrostatic image problems with plane boundaries. Am. J. Phys., 48, 526–531; Terras, R., and R. Swanson, 1980: Image methods for constructing Green’s functions and eigenfunctions for domains with plane boundaries. J. Math. Phys., 21, 2140–2153.

468

Green’s Functions with Applications

with Gn (0|η) = 0, and limy→∞ Gn (y|η) → 0. The solution of this boundaryvalue problem is Gn (y|η) =

e−nπ|y−η|/L − e−nπ(y+η)/L . nπ

(6.5.7)

The first term in Equation 6.5.7 is the particular solution to Equation 6.5.6, while the second term is a homogeneous solution, required so that Gn (0|η) = 0. Therefore, the Green’s function governed by Equation 6.5.1 through Equation 6.5.3 is , - * ∞ 1 ( e−nπ|y−η|/L − e−nπ(y+η)/L nπξ nπx + g(x, y|ξ,η ) = sin sin . π n=1 n L L (6.5.8) An alternative expression49 for Equation 6.5.8 is ∞ * nπy + * nπy + , nπξ - * nπx + 2 (1 < > g(x, y|ξ,η ) = sinh exp − sin sin . π n=1 n a a L L (6.5.9) Following Melnikov,50 we can find a closed form for the Green’s function. We begin by first rewriting Equation 6.5.9 as

g(x, y|ξ,η ) =

Since

∞ 5 1 ( 1 4 −nπ|y−η|/L e − e−nπ(y+η)/L 2π n=1 n ! % & % &) nπ(x − ξ) nπ(x + ξ) × cos − cos . L L

∞ 4. 5 ( qn cos(nα) = − ln 1 − 2q cos(α) + q 2 , n n=1

(6.5.10)

(6.5.11)

provided that |q| < 1, and 0 ≤ α ≤ 2π, we have that

!T 1 1 − 2e−π(y+η)/L cos[π(x − ξ)/L] + e−2π(y+η)/L g(x, y|ξ,η ) = ln 2π 1 − 2e−π|y−η|/L cos[π(x − ξ)/L] + e−2π|y−η|/L T ) 1 − 2e−π|y−η|/L cos[π(x + ξ)/L] + e−2π|y−η|/L × . 1 − 2e−π(y+η)/L cos[π(x + ξ)/L] + e−2π(y+η)/L

(6.5.12)

49

Chen, J., and R. V. Lovelace, 1978: Beam generation in foil-less diodes. Phys. Fluids, 21, 1623–1633. 50 Melnikov, Yu. A., 1995: Green’s Functions in Applied Mechanics. WIT Press/Computational Mechanics Publications, pp. 24–25.

The Helmholtz Equation

469

Multiplying the numerator of the first radial and the denominator of the second radial by e2π(y+η)/L , we arrive at the closed form solution % & 1 E(z + ζ ∗ )E(z − ζ ∗ ) g(x, y|ξ,η ) = ln , (6.5.13) 2π E(z − ζ)E(z + ζ) ' ' where z = y + ix, ζ = η + iξ, ζ ∗ = η − iξ, and E(z) = 'eπz/L − 1'. A similar technique can be used to find the Green’s function for Equation 6.5.1 with the boundary conditions g(0, y|ξ,η ) = g(L, y|ξ,η ) = gy (x, 0|ξ,η ) − β g(x, 0|ξ,η ) = 0,

and Equation 6.5.3, where β ≥ 0. Performing a similar analysis, we % & ∞ 1 ( 1 −nπ|y−η|/L βL − nπ −nπ(y+η)/L g(x, y|ξ,η ) = e − e π n=1 n βL + nπ , - * nπξ nπx + × sin sin L L ∞ 4 5 ( 1 1 −nπ|y−η|/L = e + e−nπ(y+η)/L π n=1 n , - * nπξ nπx + × sin sin L L , - * ∞ −nπ(y+η)/L ( 2βL e nπξ nπx + − sin sin . π n=1 n(βL + nπ) L L % & 1 E(z + ζ)E(z − ζ ∗ ) = ln 2π E(z − ζ)E(z + ζ ∗ ) , - * ∞ 2βL ( e−nπ(y+η)/L nπξ nπx + − sin sin . π n=1 n(βL + nπ) L L

(6.5.14) find that

(6.5.15)

(6.5.16)

(6.5.17)

The advantage of Equation 6.5.17 over Equation 6.5.15 is its uniform convergence. We obtain the logarithmic term in the same manner as we derived it in Equation 6.5.13. Finally we solve Equation 6.5.1 subject to the boundary conditions g(0, y|ξ,η ) = gx (L, y|ξ,η ) = 0,

(6.5.18)

gy (x, 0|ξ,η ) − β g(x, 0|ξ,η ) = 0,

(6.5.19)

and 51

with Equation 6.5.3. The Green’s function is % & 1 E1 (z + ζ ∗ )E1 (z − ζ)E2 (z + ζ)E2 (z − ζ ∗ ) g(x, y|ξ,η ) = ln 2π E1 (z + ζ)E1 (z − ζ ∗ )E2 (z − ζ)E2 (z + ζ ∗ ) ∞ 2β ( e−ν(y+η) − sin(νξ) sin(νx), (6.5.20) L n=1 ν(β + ν) 51

Ibid., p. 28.

470

Green’s Functions with Applications

Figure 6.5.1: Equation 6.5.20 for the planar Poisson equation over the semi-infinite strip 0 < x < L and 0 < y subject to the boundary conditions that g(0, y|ξ,η ) = gx (0, y|ξ,η ) = 0, and gy (x, 0|ξ,η ) − βg(x, 0|ξ,η ) = 0 when ξ/L = η/L = 0.5, and βL = 1.

where z = y' + ix, ζ = η +' iξ, ζ ∗ = η − ' πz/(2L) ' iξ, ν = (2n − 1)π/(2L), E1 (z) = 'e + 1', and E2 (z) = 'eπz/(2L) − 1'. • Example 6.5.1

During the solution of a magnetostatic boundary-value problem, Scharstein and Daniel52 found the Green’s function governed by the equation ∂ 2g ∂ 2g + = −δ(x)δ(y−η), ∂x2 ∂y 2

−∞ < x < ∞,

−π/2 < y, η < π , (6.5.21)

subject to the boundary conditions lim g(x, y|0, η) = −

|x|→∞

|x| , 3π

−π/2 < y < π ,

(6.5.22)

and gy (x, −π/2|0, η) = g(x,π |0, η) = 0,

|x| < ∞.

(6.5.23)

Because the Green’s function does not vanish as |x| → ∞, Scharstein and Daniel introduced the generalized Fourier transform53 " ∞ g(x, y|0, η) = G(k, y|0, η) cos(kx) dk (6.5.24) 0

52 Scharstein, R. W., and D. S. Daniel, 2002: Magnetostatic field of a conductor of semicircular cross section in a highly permeable half-space: Exact analytic solution for infinite permeability with perturbative correction. IEEE Trans. Magn., 38, 3594–3606. 53 See pp. 11-12 in Van der Pol, B., and H. Bremmer, 1987: Operational Calculus: Based on the Two-Sided Laplace Integral. Amer. Math. Soc., 415 pp.

The Helmholtz Equation with

471

1 δ(x) = π

"

∞

cos(kx) dk.

(6.5.25)

0

Taking the generalized Fourier transform of Equation 6.5.21, we have that d2 G 1 − k 2 G = − δ(y − η). dy 2 π

(6.5.26)

The solution to Equation 6.5.26 is G(k, y|0, η) = A(k) cosh[k(y + π/2)] −

H(y − η) sinh[k(y − η)]. kπ

(6.5.27)

The first term on the right side of Equation 6.5.27 is the homogeneous solution while the second term is the particular solution due to the delta function. Upon applying the boundary conditions, the generalized Fourier transform of the Green’s function is cosh[k(π − η)] cosh[k(y + π/2)] kπ sinh(3kπ/2) H(y − η) − sinh[k(y − η)] kπ cosh[k(π − y> )] cosh[k(y< + π/2)] = . kπ sinh(3kπ/2)

G(k, y|0, η) =

(6.5.28) (6.5.29)

We must now take the inverse Fourier transform of Equation 6.5.29. Using the residue theorem, we have that " ∞ cosh(bx) cos(ax) dx = − 12 ln{2 [cosh(2a/3) + cos(2b/3)]} , (6.5.30) x sinh(3πx/2) 0 Applying Equation 6.5.30 to Equation 6.5.29, we find that , 1 g(x, y|ξ,η ) = − ln{cosh [2(x − ξ)/3] + cos [2(y + η − π/2)/3]} 4π + ln{cosh [2(x − ξ)/3] + cos [2(|y − η| − 3π/2)/3]} −

ln(2) . 2π

-

(6.5.31)

• Helmholtz’s equation Consider the Helmholtz equation ∂2g ∂ 2g + + k02 g = −δ(x − ξ)δ(y − η), ∂x2 ∂y 2

(6.5.32)

472

Green’s Functions with Applications

where 0 < x, ξ < L, and −∞ < y, η < ∞, subject to the boundary conditions that g(0, y|ξ,η ) = g(L, y|ξ,η ) = 0, (6.5.33) and lim g(x, y|ξ,η ) → 0.

|y|→∞

We begin by introducing the Fourier transform " ∞ G(x,0 |ξ,η ) = g(x, y|ξ,η )e−i4y dy,

(6.5.34)

(6.5.35)

−∞

and g(x, y|ξ,η ) =

1 2π

"

∞

−∞

G(x,0 |ξ,η )ei4y d0,

(6.5.36)

where 3(0) > 0. We chose this form of the Fourier transform so that we have outwardly propagating waves if the time dependence is e−iωt . Equation 6.5.32 then becomes the ordinary differential equation $ d2 G # 2 + k0 − 02 G = −δ(x − ξ)e−i4η , dx2

0 < x, ξ < L,

(6.5.37)

with G(0, 0|ξ,η ) = G(L,0 |ξ,η ) = 0. To solve Equation 6.5.37, we assume that G(x,0 |ξ,η ) = with δ(x − ξ) =

* nπx + Gn sin , L n=1 ∞ (

, - * ∞ 2 ( nπξ nπx + sin sin . L n=1 L L

(6.5.38)

(6.5.39)

Substituting Equation 6.5.38 and Equation 6.5.39 into Equation 6.5.37, and matching the harmonics, , , n2 π 2 2 nπξ k02 − 02 − G = − sin e−i4η , (6.5.40) n L2 L L and g(x, y|ξ,η ) =

, - * " ∞ 1 ( nπξ nπx + ∞ ei4(y−η) sin sin d0, 2 2 πL n=1 L L −∞ 0 + κ

(6.5.41)

. where κ = n2 π 2 /L2 − k02 . We now integrate Equation 6.5.41 via contour integration. By Jordan’s lemma, we must close the line integration from (−∞, 0) to (∞, 0) with a

The Helmholtz Equation

473

semicircle of infinite radius in the upper half of the 0-plane if y > η; if y < η, then the semicircle will be in the lower half-plane. Performing the integration via the residue theorem, we obtain √ 2 2 2 2 , - * ∞ 1 ( nπξ nπx + e− n π /L −k0 |y−η| . g(x, y|ξ,η ) = sin sin . L n=1 L L n2 π 2 /L2 − k02

(6.5.42)

What happens if nπ/L < k0 for some n? Then

= = n2 π 2 /L2 − k02 = i k02 − n2 π 2 /L2 . We must have this particular root so that the waves decay away from the source at y = η. In summary, the Green’s function for the two-dimensional Helmholtz equation where the domain is the infinite strip 0 < x < L consists of a superposition of two types of waves. For nπ/L < k0 , we have propagating waves radiating away from the source. For sufficiently large n, then the waves become evanescent modes that are trapped near the source. Let us retrace our steps and find another form of the Green’s function by the method of images. We begin at the point where we took the Fourier transform, Equation 6.5.37. In Section 6.1 we showed that the free-space Green’s function that satisfies this equation is G(x,0 |ξ,η ) =

i ik|x−ξ|−i4y e , 2k

(6.5.43)

where k 2 = k02 − 02 . Although Equation 6.5.43 satisfies Equation 6.5.37, it does not satisfy any of the boundary conditions. On the other hand, if we modify Equation 6.5.43 to read G(x,0 |ξ,η ) =

i ik|x−ξ|−i4y i ik|x+ξ|−i4y e − e , 2k 2k

(6.5.44)

a quick check shows that Equation 6.5.44 does satisfy the boundary condition at x = 0. Although the second term appears to be a wave that radiates from the source point (−ξ,η ), it is really the reflection of a wave emanating from (ξ,η ) off the surface at x = 0. The reason that Equation 6.5.44 is not the Green’s function that we are seeking is its failure to satisfy the boundary condition at x = L. Let us modify it to read i ik|x−ξ|−i4y i ik|x+ξ|−i4y e − e 2k 2k i ik|x−ξ−2L|−i4y i ik|x+ξ−2L|−i4y + e − e . 2k 2k

G(x,0 |ξ,η ) =

(6.5.45)

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Green’s Functions with Applications

Although Equation 6.5.45 now satisfies the boundary conditions at x = L, it also violates the one at x = 0. Consequently, we must add in more free-space Green’s functions so that we again satisfy the boundary condition at x = 0. Clearly, this repeated introduction of free-space Green’s functions to satisfy both boundary conditions leads to the series: % ∞ ( i ik|x−ξ|−i4y G(x,0 |ξ,η ) = e + eik|x−ξ−2nL|−i4y 2k n=−∞ −

n&=0 ∞ (

n=−∞

& eik|x+ξ+2nL|−i4y .

(6.5.46)

Upon using Equation 6.1.26 on each term in Equation 6.5.46, we find that g(x, y|ξ,η ) =

∞ ! 4 . 5 i ( (1) H0 k0 (x − ξ − 2nL)2 + (y − η)2 (6.5.47) 4 n=−∞ 4 . 5) (1) − H0 k0 (x + ξ + 2nL)2 + (y − η)2 .

We can convert Equation 6.5.47 into Equation 6.5.42 through the use of Poisson’s summation formula. What is the physical interpretation of Equation 6.5.47? Recall that the Helmholtz equation describes the steady-state radiation of waves from the source point (ξ,η ). Consequently, this equation appears to be describing the constructive and destructive interference of waves emanating from an infinite number of sources. Actually, these additional terms are due to multiple reflections of the original wave emanating from (ξ,η ) off the boundaries. For this reason, this equation is often referred to as a multiple-reflective wave series. Our third and final method for solving Equation 6.5.32 involves separation of variables. We begin by considering the homogeneous version of this equation, namely ∂2g ∂ 2g + + k02 g = 0. (6.5.48) ∂x2 ∂y 2 Assuming that g(x, y) = X(x)Y (y), the homogeneous Helmholtz equation becomes 1 d2 X 1 d2 Y + + k02 = 0, (6.5.49) X dx2 Y dy 2 or d2 X + λx X = 0; (6.5.50) dx2 and d2 Y + λy Y = 0, (6.5.51) dy 2

The Helmholtz Equation

475

with λx + λy = k02 . Let us associate with Equation 6.5.50 and Equation 6.5.51, a corresponding one-dimensional Green’s function problem: d2 X + λx X = −δ(x − ξ), dx2

0 < x, ξ < L,

(6.5.52)

with X(0|ξ) = X(L|ξ) = 0, and d2 Y + λy Y = −δ(y − η), dy 2

−∞ < y, η < ∞,

(6.5.53)

where Y (y|η) must satisfy the radiation condition. The solution to Equation 6.5.52 follows directly from the eigenfunction method presented in Section 3.4 and equals ∞ 2 ( sin(nπξ/L) sin(nπx/L) X(x|ξ) = − . (6.5.54) L n=1 λx − n2 π 2 /L2 Because the y-domain is infinite, we employ Fourier transforms in that direction and find that √ i Y (y|η) = . ei λy *y−η| (6.5.55) 2 λy . with 5( λy ) > 0. Consider now the following Green’s function: 0 1 g(x, y|ξ,η ) = − X(x|ξ)Y (y|η) dλx (6.5.56) 2πi Cx 0 1 =− X(x|ξ)Y (y|η) dλy , (6.5.57) 2πi Cy where Cx encloses the singularities of X(x|ξ) and excludes those of Y (y|η); Cy , on the other hand, encloses only the singularities of Y (y|η). Here, the relationship λx + λy = k02 was used to eliminate λy in Equation 6.5.56 and λx in Equation 6.5.57. Furthermore, ∂ 2g ∂2g ∂ 2g ∂2g + 2 + k02 g = + λx g + 2 + λy g (6.5.58) 2 2 ∂x ∂y ∂x ∂y 0 , 2 1 ∂ ∂2 =− + λ + + λ x y X(x|ξ)Y (y|η) dλx 2πi Cx ∂x2 ∂y 2 (6.5.59) 0 1 =− [−δ(x − ξ)Y (y|η) − X(x|ξ)δ(y − η)] dλx 2πi Cx (6.5.60) 0 δ(y − η) = X(x|ξ) dλx = −δ(x − ξ)δ(y − η) (6.5.61) 2πi Cx

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Green’s Functions with Applications

x-plane

Cx 2

k20

L2

n2 L2

2

poles

pole

y-plane

k 20

Cy

n2 2 poles L2 2

k20

2

L

pole

Figure 6.5.2: The contours Cx and Cy used in finding Equation 6.5.56 and Equation 6.5.57.

by Equation 3.4.34. The integral, Equation 6.5.61, follows from [Equation 6.5.60 because Cx excludes the singularities of Y (y|η); consequently, CxY (y|η) dλx = 0 by Cauchy’s theorem. A similar demonstration proves Equation 6.5.57. If we use Equation 6.5.56, then √ 2 0 ∞ 1 iei k0 −λx |y−η| ( sin(nπξ/L) sin(nπx/L) . g(x, y|ξ,η ) = dλx 2πi Cx L k02 − λx n=1 λx − (nπ/L)2 (6.5.62)

=

∞ (

√

1 sin(nπξ/L) sin(nπx/L) − . e L n=1 (nπ/L)2 − k02

(nπ/L)2 −k02 |y−η|

, (6.5.63)

. where 5[ (nπ/L)2 − k02 ] > 0. In Figure 6.5.2 we illustrate Cx . The . poles are located at λx = (nπ/L)2 , while the branch cut associated with k02 − λx is taken along the real axis from the branch point k02 to −∞. As the figure shows, Cx encloses the poles but not the branch point k02 . If we had used Equation 6.5.57, then the solution would be , - * " ∞ ( ∞ 1 nπξ nπx + eiω|y−η| g(x, y|ξ,η ) = sin sin dω, πL −∞ n=1 L L ω 2 + n2 π 2 /L2 − k02 (6.5.64)

The Helmholtz Equation

477

which is a branch cut integral and equals an integral over the continuous spectrum of Y (y|η) rather than a sum over the discrete spectrum of X(x|ξ). Figure 6.5.2 illustrates the contour Cy , where it was assumed that π/L < k0 < 2π/L. This contour encloses the branch cut for Y (y|η) but not the poles of X(x|ξ) which occur now when λy = k02 − (nπ/L)2 . In this section, we have shown that Equation 6.5.42, Equation 6.5.47 and Equation 6.5.64 give the Green’s function for the two-dimensional Helmholtz equation over the infinite strip 0 < x < L. Which one should we use? Because the rate of convergence for each series depends upon the particular (x, y), so does the choice. For example, near the source point, Equation 6.5.49 is best because the lead term explicitly resolves the singularity at (ξ,η ) and the remaining terms in the series tend rapidly to zero. 6.6 THREE-DIMENSIONAL PROBLEMS IN A HALF-SPACE Here we find the Green’s functions for Poisson’s and Helmholtz’s equations within the half-space z > 0. • Poisson’s equation In this case, the Green’s function g(x, y, z|ξ, η,ζ ) satisfies the partial differential equation ∂ 2g ∂ 2g ∂ 2g + 2 + 2 = −δ(x − ξ)δ(y − η)δ(z − ζ) 2 ∂x ∂y ∂z

(6.6.1)

over the half-space z > 0, with the boundary condition αg(x, y, 0|ξ, η,ζ ) − β

∂g(x, y, 0|ξ, η,ζ ) = 0, ∂z

(6.6.2)

where α and β are constants. There are three possible cases: (1) the Dirichlet problem with α = 1 and β = 0, (2) the Neumann problem with α = 0 and β = −1, and (3) the Robin problem with α = h and β = 1 with h > 0. Finally, we require that the Green’s function vanishes at infinity. We begin by noting that the most general solution consists of a sum of the free-space Green’s function g1 (x, y, z|ξ, η,ζ ) =

2−1/2 1 1 (x − ξ)2 + (y − η)2 + (z − ζ)2 4π

(6.6.3)

found in Section 6.1 and a homogeneous solution to Equation 6.6.1, g2 . Consequently, from Equation 6.6.2 the boundary condition on g2 is αg2 (x, y, 0|ξ, η,ζ ) − β

∂g2 (x, y, 0|ξ, η,ζ ) ∂z

= −αg1 (x, y, 0|ξ, η,ζ ) + β

∂g1 (x, y, 0|ξ, η,ζ ) . ∂z

(6.6.4)

478

Green’s Functions with Applications

Let us now introduce an image point at (ξ, η, −ζ), which is the reflection of (ξ, η,ζ ) in the plane z = 0. Then the free-space Green’s function corresponding to this image point is g2 (x, y, z|ξ, η,ζ ) = Z

2−1/2 1 1 (x − ξ)2 + (y − η)2 + (z + ζ)2 . 4π

(6.6.5)

Note, however, that Z g2 (x, y, z|ξ, η,ζ ) is harmonic within the domain z > 0. Can we use Equation 6.6.5 to find g2 (x, y, z|ξ, η,ζ )? Yes! Because gZ2 (x, y, 0| ξ, η,ζ ) = g1 (x, y, 0|ξ, η,ζ ), g2 (x, y, z|ξ, η,ζ ) = −Z g2 (x, y, z|ξ, η,ζ ) in the case of a Dirichlet condition. Similarly, since ' ' 2 ∂Z g2 (x, y, 0|ξ, η,ζ ) (z + ζ) 1 2 2 2 −3/2 ' =− (x − ξ) + (y − η) + (z + ζ) ' (6.6.6) ∂z 4π z=0 ∂g1 (x, y, 0|ξ, η,ζ ) =− , (6.6.7) ∂z g2 (x, y, z|ξ, η,ζ ) = gZ2 (x, y, z|ξ, η,ζ ) in the case of the Neumann problem. Thus, from the method of images, the Green’s function for Poisson’s equation within the half-space z > 0 is 2−1/2 1 1 (x − ξ)2 + (y − η)2 + (z − ζ)2 4π 2−1/2 1 1 − (x − ξ)2 + (y − η)2 + (z + ζ)2 , 4π

g(x, y, z|ξ, η,ζ ) =

(6.6.8)

when we have a Dirichlet boundary condition and

2−1/2 1 1 (x − ξ)2 + (y − η)2 + (z − ζ)2 4π 2−1/2 1 1 + (x − ξ)2 + (y − η)2 + (z + ζ)2 , 4π

g(x, y, z|ξ, η,ζ ) =

(6.6.9)

when we have a Neumann boundary condition. If we try to apply this technique for a Robin boundary condition, we fail. We cannot obtain g2 (x, y, z|ξ, η,ζ ) solely in terms of an image source point at (ξ, η, −ζ). Motivated by Example 5.1.2 (recall that Poisson’s equation is the steady-state version of the three-dimensional heat equation), we introduce an entire line of image sources on the line x = ξ and y = η with z extending from z = −ζ to z = −∞. Let us try 2−1/2 1 1 (x − ξ)2 + (y − η)2 + (z + ζ)2 4π " −ζ 1 ρ(s) . + ds, (6.6.10) 2 4π −∞ (x − ξ) + (y − η)2 + (z − s)2

g2 (x, y, z|ξ, η,ζ ) =

The Helmholtz Equation

479

where ρ(s) is a presently unknown source density. When h = 0 in the Robin boundary condition, Equation 6.6.10 must reduce to the solution for the Neumann problem. Therefore, ρ(s) = 0 when h = 0. This is why we introduced the free-space Green’s function, Equation 6.6.5, in Equation 6.6.10. If ρ(s) decays sufficiently rapidly at infinity so that the integral in Equation 6.6.10 converges and that differentiation under the integral sign is permitted, then g2 is harmonic for z > 0 because all of the singular points in this equation occur in the lower half-space z < 0. Substituting Equation 6.6.10 into Equation 6.6.4. ∂g2 (x, y, 0|ξ, η,ζ ) − h g2 (x, y, 0|ξ, η,ζ ) ∂z 2−3/2 ζ 1 =− (x − ξ)2 + (y − η)2 + ζ 2 4π " −ζ 2−1/2 1 ∂ 1 − ρ(s) (x − ξ)2 + (y − η)2 + s2 ds 4π −∞ ∂s 2−1/2 h1 − (x − ξ)2 + (y − η)2 + ζ 2 4π " −ζ h ρ(s) . − ds (6.6.11) 4π −∞ (x − ξ)2 + (y − η)2 + s2 ∂g1 (x, y, 0|ξ, η,ζ ) =− + h g1 (x, y, 0|ξ, η,ζ ) (6.6.12) ∂z 2−3/2 ζ 1 =− (x − ξ)2 + (y − η)2 + ζ 2 4π 2−1/2 h1 + (x − ξ)2 + (y − η)2 + ζ 2 . (6.6.13) 4π

Because the operator ∂/∂z has the same effect as −∂/∂s at z = 0, and the use of ∂/∂s in the integral term allows us to integrate by parts, we obtain "

−ζ

−∞

ρ(s)

2−1/2 ∂ 1 (x − ξ)2 + (y − η)2 + s2 ds ∂s 1 2−1/2 = ρ(−ζ) (x − ξ)2 + (y − η)2 + ζ 2 ds " −ζ 1 2−1/2 − ρ$ (s) (x − ξ)2 + (y − η)2 + s2 ds.

(6.6.14)

−∞

Substituting Equation 6.6.14 into Equation 6.6.11, we find that %

& 2−1/2 h ρ(−ζ) 1 − − (x − ξ)2 + (y − η)2 + ζ 2 2π 4π " −ζ 1 ρ$ (s) − h ρ(s) . + ds = 0. 4π −∞ (x − ξ)2 + (y − η)2 + s2

(6.6.15)

480

Green’s Functions with Applications

Equation 6.6.15 is satisfied if we set ρ$ (s) − h ρ(s) = 0, s < −ζ, with ρ(−ζ) = −2h. The solution of the ordinary differential equation is ρ(s) = −2h eh(s+ζ) . Note that ρ(s) vanishes for h = 0 and that it decays exponentially as s → −∞. Thus, the Green’s function for Poisson’s equation within the half-space z > 0 with Robin boundary condition is 2−1/2 1 1 (x − ξ)2 + (y − η)2 + (z − ζ)2 4π 2−1/2 1 1 + (x − ξ)2 + (y − η)2 + (z + ζ)2 (6.6.16) 4π " −ζ h eh(s+ζ) . − ds. 2π −∞ (x − ξ)2 + (y − η)2 + (z − s)2

g(x, y, z|ξ, η,ζ ) =

• Example 6.6.1

Consider steady-state heat conduction within the half-space z > 0. At the surface heat radiates according to 1 ∂u(x, y, 0) − u(x, y, 0) = −f (x, y). h ∂z

(6.6.17)

Denoting the surface area byΣ , Equation 6.0.13 simplifies to ' ' "" "" ∂g '' ∂u '' u(x, y, z) = u(ξ, η, 0) dσ − g(x, y, z|ξ, η, 0) dσ. ∂ζ 'ζ=0 ∂ζ 'ζ=0 Σ Σ (6.6.18) Equation 6.6.18 can be rewritten as % & "" 1 ∂g u(x, y, z) = h u(ξ, η, 0) −g dσ h ∂ζ Σ ζ=0 % & "" 1 ∂u −h g(x, y, z|ξ, η, 0) −u dσ. (6.6.19) h ∂ζ Σ ζ=0 The first integral vanishes. Substituting Equation 6.6.16 and Equation 6.6.17, Equation 6.6.19 becomes "" h f (ξ,η ) . u(x, y, z) = dσ (6.6.20) 2 2π (x − ξ) + (y − η)2 + z 2 Σ "" " ∞ h2 e−hs . − f (ξ,η ) ds dσ. 2π (x − ξ)2 + (y − η)2 + (z + s)2 Σ 0 Although Equation 6.6.20 is the desired result, we can integrate the second integral by parts to obtain "" " ∞ h (z + s) e−hs u(x, y, z) = f (ξ,η ) dσ ds. 2 2π [(x − ξ) + (y − η)2 + (z + s)2 ]3/2 Σ 0 (6.6.21)

The Helmholtz Equation

481

This result was first obtained by Boussinesq54 although he did not use a Green’s function to find his result. •Helmholtz’s equation Let us now find the Green’s function for the three-dimensional Helmholtz equation, Equation 6.1.1, in the space z > 0 with the boundary condition that g(x, y, 0|ξ, η,ζ ) = 0. Using the method of images, the desired Green’s function is √ √ 2 2 2 2 eik0 (z−ζ) +ρ eik0 (z+ζ) +ρ . g(x, y, z|ξ, η,ζ ) = − . , (6.6.22) 4π (z − ζ)2 + ρ2 4π (z + ζ)2 + ρ2

where ρ2 = (x − ξ)2 + (y − η)2 . The first term in Equation 6.6.22 corresponds to the free-space Green’s function resulting from the point source (ξ, η,ζ ) in the region z > 0. Consequently, it satisfies the nonhomogeneous Helmholtz equation, Equation 6.0.1. On the other hand, the second term represents the free-space Green’s function for a (negative) mirror image about z = 0. Because its source point lies at (ξ, η, −ζ), it satisfies the homogeneous Helmholtz equation. Its purpose is to ensure that Equation 6.6.22 satisfies the boundary condition at z = 0. Equation 6.6.22 arises in acoustics and gives the complicated interference pattern due to the presence of a wall. We can see this by examining this equation when the distance from the origin to the point (x, y, z) is much greater than the source depth ζ. Denoting the declination angle by θ, our far-field assumption leads to the following approximations .

and .

.

(z − ζ)2 + ρ2 ≈ R − ζ sin(θ),

(6.6.23)

(z + ζ)2 + ρ2 ≈ R + ζ sin(θ),

(6.6.24)

where R = ρ2 + z 2 . See Figure 6.6.1. Assuming further that the ranges appearing in the denominator of both terms of Equation 6.6.22 can be replaced by the slant range R because the amplitude decays slowly with range, we obtain S 1 R ik0 [R−ζ sin(θ)] e − eik0 [R+ζ sin(θ)] 4πR 5 eik0 R 4 −ik0 ζ sin(θ) = e − eik0 ζ sin(θ] . 4πR

g(x, y, z|ξ, η,ζ ) =

(6.6.25) (6.6.26)

54 Boussinesq, J., 1900: Echauffement ´ permanent mais in´egal, par rayonnement, d’un mur l’´ epaisseur ind´efinie, ramen´ e au cas d’un ´echauffement analogue par contact. C. R. Acad. Sci. Paris, 131, 9–13. See his Equation (6).

482

Green’s Functions with Applications

z P ( ,z) ( , , ) R

( , ,

)

sin( )

Figure 6.6.1: The solution of the Helmholtz equation for the half-space z > 0 using the method of images.

The two exponentials can be combined to yield sine and Equation 6.6.26 becomes i g(x, y, z|ξ, η,ζ ) = − sin [k0 ζ sin(θ)] eik0 R , (6.6.27) 2πR which means that the amplitude of the Green’s function varies as ' 1 '' |g(x, y, z|ξ, η,ζ )| = sin [k0 ζ sin(θ)]'. (6.6.28) 2πR For a point source in free space we would have a spherically expanding wave with |g(x, y, z|ξ, η,ζ )| = 1/(4πR). Therefore, the presence of the reflecting surface generates a directional pattern with maxima and minima given by 1 π |g(x, y, z|ξ, η,ζ )|max = for sin(θ) = (2m − 1) , m = 1, 2, . . . 2πR 2k0 ζ (6.6.29) and π |g(x, y, z|ξ, η,ζ )|min = 0 for sin(θ) = (m−1) , m = 1, 2, . . . (6.6.30) k0 ζ This is the classic surface-image or Lloyd-mirror interference pattern.55 Note that the maximum is twice the value that a single source would have (constructive interference), while the minimum is zero (destructive interference). 55 In 1833 Humphrey Lloyd (1800–1881) published the results of an experiment that confirmed conical refraction as predicted by William Rowan Hamilton (1805–1865) from Fresnel’s wave theory of light.

The Helmholtz Equation

483

−3

| Green’s function |

10

−4

10

−5

10

0

1000

2000

3000

range

4000

5000

Figure 6.6.2: Surface-interference (Lloyd mirror) solution for a point source in a homogeneous halfspace z > 0. The solid line gives |g(x, y, 200 m|ξ, η, 25 m)| from Equation 6.6.25 . with k0 = 2π/(10 m) as a function of range ρ in meters. The dashed line is 1/[4π ρ2 + (200 m − 25 m)2 ].

Figure 6.6.2 shows the variation of the |g(x, y, z|ξ, η,ζ )| as a function of range ρ. 6.7 THREE-DIMENSIONAL POISSON’S EQUATION IN A CYLINDRICAL DOMAIN In this section, we present Green’s functions for various limited cylindrical domains. We assume Dirichlet or Neumann boundary conditions along any surface. In all cases, the governing equation is , 1 ∂ ∂g 1 ∂2g ∂ 2g δ(r − ρ)δ(θ − θ$ )δ(z − ζ) r + 2 2 + 2 =− . r ∂r ∂r r ∂θ ∂z r

(6.7.1)

We illustrate several simple geometries; Gray and Mathews56 and Dougall57 present solutions for more complicated configurations. 56 Gray, A., and G. B. Mathews, 1966: Treatise on Bessel Functions and Their Applications to Physics. Dover Publications, Chapter IX, Section 3. 57

Dougall, op. cit.

484

Green’s Functions with Applications

• Space bounded by two parallel plates In Example 6.1.3 we found the three-dimensional free-space Green’s function in cylindrical coordinates. Although this solution satisfies the differential equation, it does not satisfy most boundary conditions. For example, in the case of Dirichlet boundary conditions along the planes z = 0 and z = c, we must find a homogeneous solution to Equation 6.7.1 such that it plus the free-space Green’s function satisfy the boundary conditions.58 A quick check shows that ) " ∞! sinh(kz) −k(c−ρ) sinh[k(c − z)] −kρ V (x, y, z) = − e + e J0 (kR) dk, sinh(kc) sinh(kc) 0 (6.7.2) . where R = r2 + ρ2 − 2rρ cos(θ − θ $ ), is this homogeneous solution, since its combination with the free-space Green’s function, Equation 6.1.81, yields the Green’s function59 " ∞ 1 sinh[k(c − z> )] sinh(kz< ) g(r, θ, z|ρ,θ $ , ζ) = J0 (kR) dk, (6.7.3) 2π 0 sinh(kc) which satisfies Dirichlet boundary conditions along z = 0 and z = c. • Infinitely long tube In the case of an infinitely long tube, we must solve Equation 6.7.1 with the boundary conditions lim |g(r, θ, z|ρ,θ $ , ζ)| < ∞,

r→0

g(a, θ, z|ρ,θ $ , ζ) = 0,

(6.7.4)

for 0 ≤ θ,θ $ ≤ 2π. The Green’s function must be periodic in θ. We begin by noting that ∞ ∞ δ(r − ρ)δ(θ − θ$ ) 1 ( ( Jn (knm ρ)Jn (knm r) = cos[n(θ − θ$ )], r πa2 n=−∞ m=1 J $ 2n (knm a) (6.7.5) where knm is the mth root of Jn (ka) = 0. The form of the Green’s function is then

g(r, θ, z|ρ,θ $ , ζ) =

∞ ∞ ( (

n=−∞ m=1

Gnm (z|ζ)Jn (knm r) cos[n(θ − θ $ )].

(6.7.6)

58 Bates [Bates, J. W., 1997: On toroidal Green’s functions. J. Math. Phys., 38, 3679–3691.] used a similar technique to find the Green’s functions for Laplace’s and the biharmonic equations with Dirichlet boundary conditions along the entire boundary. 59

Dougall, op. cit.

The Helmholtz Equation

485

Substitution of Equation 6.7.5 and Equation 6.7.6 into Equation 6.7.1 yields the differential equation d2 Gnm Jn (knm ρ) 2 − knm Gnm = − 2 $ 2 δ(z − ζ). dz 2 πa J n (knm a)

(6.7.7)

The solution to Equation 6.7.7 that remains bounded as |z| → ∞ is Gnm (z|ζ) =

Jn (knm ρ) e−knm |z−ζ| . 2πa2 knm J $ 2n (knm a)

(6.7.8)

Consequently, the Green’s function for an infinitely long tube where the solution vanishes along the cylindrical side at r = a is ∞ ∞ 1 ( ( Jn (knm ρ)Jn (knm r) cos[n(θ−θ$ )]e−knm |z−ζ| . 2πa2n=−∞ m=1 knm J $ 2n (knm a) (6.7.9) For applications of this Green’s functions to electrodynamics, see Bouwkamp and de Bruijn,60 Jackson,61 Ilyenko et al.,62 Ilyenko and Yatsenko,63 and Lopes and Motta.64 A disadvantage of using Equation 6.7.9 is its slow convergence when we are near the source. Barlow65 derived an alternative and showed that

g(r, θ, z|ρ,θ $ , ζ) =

g(r, θ, z|ρ,θ $ , ζ) =

∞ 1 2 ( − (2 − δ0m ) cos[m(θ − θ$ )] (6.7.10) R πa m=0 " ∞ Km (k) × Im (kρ/a)Im (kr/a) cos[k(z − ζ)/a] dk, Im (k) 0

60 Section 8 in Bouwkamp, C. J., and N. G. de Bruijn, 1947: The electrostatic field of a point charge inside a cylinder, in connection with wave guide theory. J. Appl. Phys., 18, 562–577. 61

Page 75 in Jackson, J. D., 1962: Classical Electrodynamics. New York: Wiley, 641

pp. 62 Ilyenko, K. V., G. M. Gorbik, and T. Yu. Yatsenko, 2005: On the calculation of the force acting upon a moving charge in a cylindrical drift tube. Telecommun. Radio Engng., 63, 871–882. 63

Ilyenko, K., and T. Yatsenko, 2011: Three-dimensional model for Green’s function charged-particle-beam simulations in cylindrical geometry. IEEE Trans. Plasma Sci., 39, 659–667. 64 Lopes, D. T., and C. C. Motta, 2012: Electrostatic force between rings and discs of charge inside a grounded metallic pipe using the Green’s function technique. J. Electrost., 70, 166–173. 65 Barlow, S. E., 2003: Alternative electrostatic Green’s function for a long tube. J. Appl. Phys., 94, 6221–6222.

486

Green’s Functions with Applications

where R2 = r2 + ρ2 − 2rρ cos(θ − θ$ ) + (z − ζ)2 . The k-integrals behave well except where both r and ρ are large; they converge there but not quickly. Finally Lopes and Motta66 derived the following representation: g(r, θ, z|ρ,θ $ , ζ) =

" ∞ 1 ( im(θ−θ $ ) ∞ e Im (kr< ) cos[k(z − ζ)] dk 2π 2 m=−∞ 0 % & Km (ka) × Km (kr> ) − Im (kr> ) , (6.7.11) Im (ka)

which they used in some electrostatic calculations. • Cylindrical pill box with Dirichlet boundary conditions When the tube is of finite length and has the additional boundary conditions that g(r, θ, 0|ρ,θ $ , ζ) = g(r, θ, L|ρ,θ $ , ζ) = 0, (6.7.12) we must solve Equation 6.7.7 subject to the boundary conditions Gnm (0| ζ) = Gnm (L|ζ) = 0. The solution is Gnm (z|ζ) =

Jn (knm ρ) sinh[knm (L − z> )] sinh(knm z< ) . πa2 knm sinh(knm L)J $ 2n (knm a)

(6.7.13)

Substituting Equation 6.7.13 into Equation 6.7.6, we obtain g(r, θ, z|ρ,θ $ , ζ) =

∞ ∞ 1 ( ( Jn (knm ρ)Jn (knm r) cos[n(θ − θ$ )] πa2 n=−∞ m=1 J $ 2n (knm a)

×

sinh[knm (L − z< )] sinh(knm z> ) . knm sinh(knm L)

(6.7.14)

In Section 3.4 we showed that the Green’s function for ordinary differential equations such as Equation 6.7.7 can be written as an eigenfunction expansion. In the present case, Gnm (z|ζ) =

∞

2L Jn (knm ρ) ( sin(0πζ/L) sin(0πz/L) . 2 L 2 + 02 π 2 πa2 J $ 2n (knm a) knm

(6.7.15)

4=1

Thus, an alternative to Equation 6.7.14 is g(r, θ, z|ρ,θ $ , ζ) =

∞ ∞ ∞ 2L ( ( ( Jn (knm ρ)Jn (knm r) cos[n(θ − θ$ )] πa2 n=−∞ m=1 J $ 2n (knm a) 4=1

× 66

sin(0πζ/L) sin(0πz/L) . 2 L 2 + 02 π 2 knm

See the first half of the appendix of Lopes and Motta, op. cit.

(6.7.16)

The Helmholtz Equation

487

Finally Hernandes and Assis67 derived g(r, θ, z|ρ,θ $ , ζ) =

4 L

, ( ∞

$

eim(θ−θ )

m=−∞

!( ∞

sin(kn ζ) sin(kn z)Im (kn r< )

n=1

% &)Km (kn a) × Km (kn r> ) − Im (kn r> ) , Im (kn a)

(6.7.17)

where kn = nπ/L. • Cylindrical pill box with Neumann boundary conditions An interesting Green’s function problem68 involving Laplace’s equation arises when the boundary conditions are Neumann along all of the boundaries. The governing equation is , 1 ∂ ∂g 1 ∂2g ∂ 2g δ(r − ρ)δ(θ − θ $ )δ(z − ζ) 1 r + 2 2+ 2 =− + 2 , (6.7.18) r ∂r ∂r r ∂θ ∂z r πa L where 0 ≤ r, ρ < a, 0 ≤ θ,θ $ < 2π, and 0 < z, ζ < L, with the boundary conditions lim |g(r, θ, z|ρ,θ $ , ζ)| < ∞, gr (a, θ, z|ρ,θ $ , ζ) = 0,

r→0

g(r, θ, z|ρ,θ $ , ζ) = g(r,θ + 2π, z|ρ,θ $ , ζ),

0 ≤ θ < 2π, 0 < z < L,

(6.7.19) 0 ≤ r < a, 0 < z < L, (6.7.20)

and gz (r, θ, 0|ρ,θ $ , ζ) = gz (r, θ, L|ρ,θ $ , ζ) = 0,

0 ≤ r < a, 0 ≤ θ < 2π. (6.7.21) Note the last term on the right side of Equation 6.7.18. It occurs because the elliptic operator on the left side has a null eigenvalue in its spectrum. From the general theory of linear inhomogeneous equations, a solution to this equation will exists only if the source term (the right side) is orthogonal to the null space of the operator. This constant term is necessary so that this condition is satisfied. Consider the eigenvalue problem , 1 ∂ ∂ψnm 1 ∂ 2 ψnm 2 r + 2 + knm ψnm = 0. (6.7.22) r ∂r ∂r r ∂θ2 67 Hernandes, J. A., and A. K. T. Assis, 2005: Electric potential due to an infinite conducting cylinder with internal or external point charge. J. Electrost., 63, 1115–1131. 68 See Ludwig, R., G. Leuenberger, S. Makarov, D. Apelian, 2002: Electric voltage predictions and correlation with density measurements in green-state powder metallurgy compacts. J. Nondestr. Eval., 21, 1–9.

488

Green’s Functions with Applications

The normalized eigenfunction is ψnm (r,θ ) = Anm Jm (knm r) cos(mθ),

n, m = 0, 1, 2, . . . ,

(6.7.23)

$ where Jm (knm a) = 0 and knm = 0. To find Anm , we have that

"

0

a " 2π 0

2 A2nm Jm (knm r) cos2 (mθ) r dθ dr = 1,

or A2nm =

2m πa2

H %, & m2 2 1− 2 Jm (knm a) , knm

(6.7.24)

(6.7.25)

where 20 = 1 and 2m = 2 for m > 0. We introduced these eigenfunctions because the Green’s function can be written g(r, θ, z|ρ,θ $ , ζ) = G00 (z|ζ)+

∞ ( ∞ (

Gnm (z|ζ)ψnm (ρ,θ $ )ψnm (r,θ ). (6.7.26)

n=0 m=0

Substituting Equation 6.7.26 into Equation 6.7.18, we obtain G$$00 (z|ζ) +

∞ ( ∞ ( 1

n=0 m=0

=−

2 2 G$$nm (z|ζ) − knm Gnm (z|ζ) ψnm (ρ,θ $ )ψnm (r,θ )

δ(r − ρ)δ(θ − θ$ )δ(z − ζ) 1 + 2 . r πa L

(6.7.27)

Equation 6.7.27 yields π 2 aG$$00 = −δ(z − ζ) + "

1 , L

G$00 (0|ζ) = G$00 (L|ζ) = 0,

L

ζ+

G$00 (z|ζ)|ζ − = −

G00 (z|ζ) dz = 0,

0

and

2 G$$nm (z|ζ) − knm Gnm = −δ(z − ζ),

1 ; πa2

(6.7.28) (6.7.29)

G$nm (0|ζ) = G$nm (L|ζ) = 0. (6.7.30)

The solution to Equation 6.7.28 and Equation 6.7.29 is G00 (z|ζ) =

L z2 + ζ 2 z> + − 2. 3πa2 2πa2 L πa

(6.7.31)

Let us rewrite Equation 6.7.30 as 2 G$$nm − knm Gnm = −

, , ∞ 1( jπζ jπz 2j cos cos , L j=0 L L

(6.7.32)

The Helmholtz Equation

489

where we have rewritten the delta function as a Fourier cosine series. The solution to Equation 6.7.32 is Gnm (z|ζ) =

∞

1 ( cos(jπζ) cos(jπz) 2j 2 . L j=0 knm + j 2 π 2 /L2

(6.7.33)

Consequently, the final solution is L z2 + ζ 2 z> + − 2 (6.7.34) 3πa2 2πa2 L πa ∞ ∞ ∞ 1 ((( 2j 2m ψnm (r,θ )ψnm (ρ,θ $ ) + 2 . 2 + j 2 π 2 /L2 (1 − m2 /k 2 )J 2 (k πa L n=0 m=0 j=0 knm nm m nm a)

g(r, θ, z|ρ,θ $ , ζ) =

• Wedge with Dirichlet boundary conditions A historically important Green’s function problem69 involves the solution of Equation 6.7.1 with the boundary conditions: lim |g(r, θ, z|ρ,θ $ , ζ)| < ∞,

r→0

and

g(a, θ, z|ρ,θ $ , ζ) = 0,

(6.7.35)

g(r, 0, z|ρ,θ $ , ζ) = g(r, α, z|ρ,θ $ , ζ) = 0,

(6.7.36)

lim g(r, θ, z|ρ,θ $ , ζ) → 0,

(6.7.37)

|z|→∞

where 0 ≤ r, ρ < a, 0 < θ, θ$ < α, −∞ < z < ∞, and α > 0. Consequently we are finding the Green’s function within an infinitely deep pie-shaped volume. Our analysis begins by rewriting δ(θ − θ$ ) as δ(θ − θ$ ) =

∞ 2( sin(νθ $ ) sin(νθ), α n=1

(6.7.38)

where ν = nπ/α. Consequently we expand the Green’s function as g(r, θ, z|ρ,θ $ , ζ) =

∞ (

Gn (r, z|ρ,ζ ) sin(νθ$ ) sin(νθ),

(6.7.39)

n=1

where Gn (r, z|ρ,ζ ) is given by , 1 ∂ ∂Gn ∂ 2 Gn 2 r + = − δ(r − ρ)δ(z − ζ) r ∂r ∂r ∂z 2 αρ

(6.7.40)

69 Macdonald, H. M., 1895: The electrical distribution on a conductor bounded by two spherical surfaces cutting at any angle. Proc. London Math. Soc., Ser. 1 , 26, 156–172.

490

Green’s Functions with Applications

with the boundary conditions lim |Gn (r, z|ρ,ζ )| < ∞,

r→0

Gn (a, z|ρ,ζ ) = 0,

and

lim Gn (r, z|ρ,ζ ) → 0.

|z|→∞

(6.7.41)

Next, we expand the delta function δ(r − ρ) as ∞ δ(r − ρ) 2 ( Jν (knm ρ)Jν (knm r) = 2 , r a n=1 Jν$ 2 (knm a)

(6.7.42)

where knm is the mth root of Jν (ka) = 0. Therefore, Gn (r, z|ρ,ζ ) is given by the eigenfunction expansion Gn (r, z|ρ,ζ ) =

∞ (

Gnm (z|ζ)Jν (knm r),

(6.7.43)

m=1

where

d2 Gnm 4 2 − knm Gnm = − 2 δ(z − ζ) dz 2 αa with lim|z|→∞ Gnm (z|ζ) → 0. The solution to Equation 6.7.44 is Gnm (z|ζ) =

2 Jν (knm ρ) e−knm |z−ζ| . $2 nm Jν (knm a)

αa2 k

(6.7.44)

(6.7.45)

Consequently the Green’s function is ∞ ∞ 2 ( ( Jν (knm ρ)Jν (knm r) −knm |z−ζ| e sin(νθ $ ) sin(νθ), αa2 n=1 m=1 knm Jν$ 2 (knm a) (6.7.46) A particularly interesting case follows when a → ∞. Then the summation over m becomes an integral and the Green’s function equals

g(r, θ, z|ρ,θ $ , ζ) =

g(r, θ, z|ρ,θ $ , ζ) =

" ∞ ∞ 1( sin(νθ$ ) sin(νθ) Jν (kρ)Jν (kr)e−k|z−ζ| dk. α n=1 0 (6.7.47)

• Example 6.7.1: Magnetic field in a magnetic recording device In 1977 A. Van Herk70 found the magnetic field near the side of a recording head in a magnetic recording device. To compute the magnetic scalar potential u(r, θ, z) he used Equation 6.0.13 and Equation 6.7.47 with u(ξ,θ $ , ζ) = − 21 sgn(ζ).

(6.7.48)

70 Van Herk, A., 1977: Side fringing fields and write and read crosstalk of narrow magnetic recording heads. IEEE Trans. Magnet., MAG-13, 1021–1028.

The Helmholtz Equation

491

Because the first and third integrals in Equation 6.0.13 vanish, we only have the second integral which equals π u(r, θ, z) = − 2 α

∞ (

n sin(νθ)

"

n=1,3,5,...

"

0

and

"

∞

∞

"

0

∞

"

∞

sgn(ζ)e−k|z−ζ|

−∞

Jν (ρk) dρd ζ dk. ρ

Jν (ρk) 1 dρ = , ρ ν

sgn(ζ)e−k|z−ζ| dζ =

−∞

Jν (rk)

0

× Now,

∞

(6.7.49)

(6.7.50)

* + 2 sgn(z) 1 − e−k|z| . k

(6.7.51)

Therefore, the magnetic scalar potential for this boundary forcing is " ∞ ∞ 2sgn(z) ( 1 − e−k|z| sin(νθ) Jν (rk) dk (6.7.52) α k 0 n=1,3,5,... ! ( ∞ 2sgn(z) sin(νθ) =− (6.7.53) α n n=1,3,5,... :√ ;ν ) ∞ ( r2 + z 2 − |z| sin(νθ) − r n n=1,3,5,... < > 1 2 sinh (π/α) sinh−1 (z/r) 1 = − arctan . (6.7.54) π sin(πθ/α)

u(r, θ, z) = −

# " R. W. Scharstein71 examined a similar version of this problem. If 0 ≤ r < ∞ and |z| < ∞, the governing equations are , 1 ∂ ∂g1 1 ∂ 2 g1 ∂ 2 g1 δ(r − ρ)δ(θ − θ$ )δ(z) r + 2 + = − , r ∂r ∂r r ∂θ2 ∂z 2 r

α 1) LHS(i,j-1) = LHS(i,j-1) + alpha*cos(arg2)/(dt*pi*h); end % compute the |t − τ | contribution, use power series to compute % si and Ci minus the logarithm term arg = abs(t-tau)/h; arg2 = arg*arg; sgn = 1; power s = arg; power c = arg2; fact s = 1; fact c = 2; si = power s - pi/2; Ci = big C - power c / 4; for kk = 2:10 power s = power s * arg2; power c = power c * arg2; sgn = - sgn; fact s = (2*kk-1)*(2*kk-2)*fact s; fact c = 2*kk *(2*kk-1)*fact c; si = si + sgn*power s/((2*kk-1)*fact s); Ci = Ci - sgn*power c/(2*kk*fact c); end K = K - sin(arg) * si - cos(arg) * Ci; if (j < NT) LHS(i,j) = LHS(i,j) + K*dt/(h*pi); else LHS(i,j) = LHS(i,j) + 0.5*K*dt/(h*pi); end % logarithm contribution for t − τ via Atkinson scheme arg1 = abs(t-tau)/h; arg2 = abs(t-tau+dt)/h; kk = i - j + 1; km1 = kk-1; Psi 0 = -1; Psi 1 = 0.25*(kk*kk-km1*km1);

632 if (kk ∼= 0) Psi 0 = Psi 0 Psi 1 = Psi 1 end if (km1 ∼= 0) Psi 0 = Psi 0 Psi 1 = Psi 1 end Psi 1 = Psi 1 +

Green’s Functions with Applications

+ kk*log(abs(kk)); - 0.5*kk*kk*log(abs(kk));

- km1*log(abs(km1)); + 0.50*km1*km1*log(abs(km1)); kk*Psi 0;

W0 = Psi 0 - Psi 1; W1 = Psi 1; alpha = 0.5*dt*log(dt) + dt*W0; beta = 0.5*dt*log(dt) + dt*W1; LHS(i,j) = LHS(i,j) - beta * cos(arg1) / (pi*h); if (j > 1) LHS(i,j-1) = LHS(i,j-1) - alpha * cos(arg2) / (pi*h); end end; end % compute ϕ(t) phi = LHS\RHS’ Next, A(k) is computed from Equation 7.7.8. Using Filon’s method,27 the MATLAB code is: for index = 1:NK k = index*dk; theta = k*dt; sine = sin(theta); cosine = cos(theta); alpha = (theta*theta+theta*sine*cosine-2*sine*sine) ... / (theta*theta*theta); beta = 2*(theta*(1+cosine*cosine)-2*sine*cosine) ... / (theta*theta*theta); gamma = 4*(sine-theta*cosine) / (theta*theta*theta); tau = NT*dt; even = 0.5*phi(NT)*sin(k*tau); odd = 0; for j = 1:NT-1 tau = j*dt; if (mod(j,2) == 0) even = even + phi(j)*sin(k*tau); else odd = odd + phi(j)*sin(k*tau); end; end; tau = NT*dt; last = phi(NT)*cos(k*tau); A(index) = beta*even + gamma*odd - alpha*last; A(index) = dt*A(index); 27 See Section III in Mandel, F., and R. J. Bearman, 1971: Some remarks on the numerical evaluation of Fourier and Fourier-Bessel transforms. J. Comput. Phys., 7, 637–645.

Numerical Methods

633

0.04

0.03

A(k)

0.02

0.01

0

−0.01

−0.02 0

10

20

k

30

40

50

Figure 7.7.1: The variation of A(k) with k for the parameters a = h = 1, ζ = ∆τ = 0.05.

√ 2 and

end Figure 7.7.1 illustrates A(k) as a function of k. Using numerical integration the computed A(k) was found to satisfy Equation 7.7.6 and Equation 7.7.7. Finally, the Green’s function is given by : ; 1 1 1 . g(r, z|0.ζ) = +. 4π r2 + (z − ζ)2 r2 + (z + ζ)2 & " ∞% e−kζ e−kz + A(k) − J0 (kr) dk. (7.7.26) 2π 1 + kh 0 The first two terms in Equation 7.7.26 follow by evaluating the first two terms in Equation 7.7.5. The integral must be computed numerically using Simpson’s rule. Introducing the temporary arrays temp1, temp2, and temp3 to speed up the computations, the MATLAB code is: for index = 1:NK k = index*dk; temp1(index) = (A(index)-exp(-k*zeta)/(2*pi))/(1+k*h); end for nr = 1:51 rr = 0.1*(nr-1); for index = 1:NK k = index*dk; temp2(index,nr) = besselj(0,k*rr); end; end for nz = 1:51 zz = 0.1*(nz-1);

634

Green’s Functions with Applications

0.1

g(r,z|0, )

0.075 0.05 0.025 0 0

5 1

4 2

3 3

r

2 4

1 0 5

z

Figure 7.7.2: The Green’s function √ governed by Equation 7.7.1 through Equation 7.7.4 when a = h = 1, ∆k = 0.01, and ζ = 2.

for index = 1:NK k = index*dk; temp3(index,nz) = exp(-k*zz); end; end for nz = 1:51 z(nz) = 0.1*(nz-1); for nr = 1:51 r(nr) = 0.1*(nr-1); R(nr,nz) = r(nr); Z(nr,nz) = z(nz); denom1 = 4*pi*sqrt(r(nr)*r(nr) + (z(nz)-zeta)*(z(nz)-zeta)); denom2 = 4*pi*sqrt(r(nr)*r(nr) + (z(nz)+zeta)*(z(nz)+zeta)); gf(nr,nz) = 1/denom1 + 1/denom2; sum = - 1/(2*pi); % k = 0 contribution for index = 1:NK k = index*dk; integrand = temp1(index)*temp2(index,nr)*temp3(index,nz); if (mod(index,2) == 0) sum = sum + 2*integrand; else sum = sum + 4*integrand; end; end gf(nr,nz) = gf(nr,nz + dk*sum/3; end; end Figure 7.7.2 illustrates this Green’s function.

Appendix: Relationship between Solutions of Helmholtz’s and Laplace’s Equations in Cylindrical and Spherical Coordinates In Section 6.1 we showed how solutions to the Helmholtz or scalar wave equation in one coordinate system can be re-expressed as a superposition (integral) of solutions in another coordinate system. We extend these thoughts as they apply to product solutions involving Laplace’s and Helmholtz’s equations. Consider the three-dimensional Helmholtz equation in cylindrical and spherical coordinates ∂ 2 u 1 ∂u 1 ∂2u ∂ 2u + + 2 + 2 + k02 u = 0, 2 ∂ρ ρ ∂ρ ρ ∂ϕ2 ∂z and % & ∂ 2 u 2 ∂u 1 ∂ ∂u 1 ∂ 2u + + sin(θ) + 2 2 + k02 u = 0. 2 2 ∂r r ∂r r sin(θ) ∂θ ∂θ r sin (θ) ∂ϕ2

(A.1)

(A.2)

Separation of variables1 leads to the product solutions

Zm,θ0 (ρ,ϕ , z) = eik0 z cos(θ0 ) Jm [k0 ρ sin(θ0 )]eimϕ ,

(A.3)

1 See Stratton, J. A., 1941: Electromagnetic Theory. McGraw-Hill, p. 395, Eq. (21) with h = k0 cos(θ0 ).

635

636

Green’s Functions with Applications

and m imϕ Ψm n (r, ϕ,θ ) = jn (k0 r)Pn [cos(θ)]e

(A.4)

for Equation E.1 and Equation E.2, respectively, where jn ( ) is a spherical Bessel function of order n, and Pnm (·) is an associated Legendre polynomial of the first kind, order n and type m. The first relationship is given in Stranton’s book,2 namely that

Zm,θ0 (ρ,ϕ , z) =

∞ ( (2n + 1)in−m (n − m)! m Pn [cos(θ0 )]Ψm n (r, ϕ,θ ), (n + m)! n=m

(A.5) where n and m are nonnegative integers and 0 ≤ θ0 ≤ π. To obtainΨ m n in terms of Zm,θ0 , we multiply Equation E.5 by Pnm [cos(θ0 )] sin(θ0 ) dθ0 , integrate from 0 to π and obtain that

Ψm n (r, ϕ,θ ) =

im−n 2

"

0

π

Pnm [cos(θ0 )]Zm,θ0 (ρ,ϕ , z) sin(θ0 ) dθ0 .

(A.6) 3

Let us apply Equation E.5 to establish the relationship between solutions of Laplace’s equation in cylindrical and spherical coordinates. Setting θ0 = it and k0 = 2λe−t , then 1 # $2 −2t eiλ(1+e )z Jm ρλi 1 − e−2t eimϕ , ∞ ( (2n + 1)in−m (n − m)! m et + e−t = Pn (n + m)! 2 n=m , 2λr × jn Pnm [cos(θ)]eimϕ . (A.7) et 2

Ibid., p. 413, Eq. (82).

3 Erofeenko, V. T., 1972: Relation between the basic solutions of the Helmholtz and Laplace equations in cylindrical and spherical coordinates (in Russian). Izv. Akad. Nauk BSSR, Ser. Fiz.-Mat. Nauk, No. 4, 42–46.

Appendix

637

In the limit of t → ∞, Pnm

,

et + e−t 2

-

≈

(−1)m (2n)!e(n−m)t , 2n n!(n − m)!2n−m

(A.8)

,

-

≈

2n n!(2rλ)n , (2n + 1)!ent

(A.9)

and jn so that

2λr et

eiλz Jm (iρλ)eimϕ = (−1)m

∞ (

(iλ)n rn Pnm [cos(θ)]eimϕ , (n + m)! n=m

and

Jm,λ (ρ,ϕ , z) = (−1)m

∞ (

(iλ)n Rnm (r, ϕ,θ ), (n + m)! n=m

(A.10)

(A.11)

where Jm,λ (ρ,ϕ , z) = eiλz Jm (iλρ)eimϕ ,

(A.12)

Rnm (r, ϕ,θ ) = rn Pnm [cos(θ)]eimϕ

(A.13)

and 4

are product solutions of Laplace’s equation in cylindrical and spherical coordinates, respectively. The final relationship is derived from the expression 1 (ρ2

+

1 z 2 )m+ 2

= =

1 r2m+1

√ " ∞ π # $ eiλz |λ|m Km (|λ|ρ) dλ, π2m ρm Γ m + 12 −∞

for m ≥ 0. Therefore, 1

r

(A.14)

(2m)!(−1)m sinm (θ) rm+1 2m m! (2m)!(−1)m ρm = 2m m! r2m+1 " (−1)m ∞ iλz m = e |λ| Km (|λ|ρ) dλ. π −∞

P m [cos(θ)] = m+1 m

1

(A.15)

(A.16) (A.17) (A.18)

4 See Stinson, D. C., 1976: Intermediate Mathematics of Electromagnetics. PrenticeHall, pp. 125 and 155.

638

Green’s Functions with Applications

Because n > m ≥ 0, dn−m dz n−m

!

1

r

P m [cos(θ)] m+1 m

)

1 = (−1)n−m (n − m)! n+1 Pnm [cos(θ)] r " (−1)m n−m ∞ iλz n−m m = i e λ |λ| Km (|λ|ρ) dλ. π −∞

(A.19) (A.20)

Consequently, 1

r

P m [cos(θ)] = n+1 n

(−1)n in−m π(n − m)!

"

in−m (−1)n π(n − m)!

"

∞

−∞

eiλz λn−m |λ|m Km (|λ|ρ) dλ.

(A.21)

The final result is

Φm n (r, ϕ,θ ) =

∞

λn Km,λ (ρ,ϕ , z) dλ,

(A.22)

−∞

where Km,λ (ρ,ϕ , z) = eiλz Km (λρ)eimϕ , with the modified Bessel function of the second kind ! Km (λ), λ Km (λ) = (−1)m Km (−λ), λ and

(A.23)

0, 0,


(A.24)

(A.25)

are product solutions4 of Laplace’s equation in cylindrical and spherical coordinates, respectively. Problems 1. Use Equation E.5 to establish the identity that eikr cos(θ) =

∞ (

n=0

in (2n + 1)jn (kr)Pn [cos(θ)].

Answers to Some of the Problems Chapter 3 1.

g(t|τ ) = e−k(t−τ ) H(t − τ )

2. g(t|τ ) =

1 4

3. g(t|τ ) =

1 2

4 5 e3(t−τ ) − e−(t−τ ) H(t − τ )

4 5 e−(t−τ ) − e−3(t−τ ) H(t − τ )

4. g(t|τ ) = 5.

1 2

sin[2(t − τ )] et−τ H(t − τ )

4 5 g(t|τ ) = e2(t−τ ) − et−τ H(t − τ )

6. g(t|τ ) = (t − τ )e−2(t−τ ) H(t − τ ) 639

640

Green’s Functions with Applications

7. g(t|τ ) =

1 6

4 5 e3(t−τ ) − e−3(t−τ ) H(t − τ )

8. g(t|τ ) = sin(t − τ )H(t − τ ) 9.

1 2 g(t|τ ) = et−τ − 1 H(t − τ )

18. g(x|ξ) = and g(x|ξ) = 2

(L − x> )(α + x< ) L+α

∞ ( (1 + α2 kn2 ) sin[kn (L − ξ)] sin[kn (L − x)] , [α + L(1 + α2 kn2 )]kn2 n=1

where kn is the nth root of tan(kL) = −αk. 19. g(x|ξ) = (1 + x< )(L − 1 − x> )/L and g(x|ξ) = −

∞ 2ex+ξ 2L3 ( ϕn (ξ)ϕn (x) + , e2L − 1 π 2 n=1 n2 (n2 π 2 + 1)

where ϕn (x) = sin(nπx/L) + nπ cos(nπx/L)/L. 20. g(x|ξ) =

(1 + x< )(L + 1 − x> ) 2+L

and g(x|ξ) = 2

∞ (

ϕn (ξ)ϕn (x) , (2 + L + kn2 L)kn2 n=1

where ϕn (x) = sin(kn x) + kn cos(kn x) and kn is the nth root of tan(kL) = 2k/(k 2 − 1). 21. g(x|ξ) = 22. g(x|ξ) =

a x2 + ξ 2 − x> + 3 2a

sinh(kx< ) sinh[k(L − x> )] k sinh(kL)

Answers to Some of the Problems and g(x|ξ) = 2L

23. g(x|ξ) = and g(x|ξ) =

24. g(x|ξ) = and

∞ ( sin(nπx/L) sin(nπξ/L) k 2 L2 + n2 π 2 n=1

cosh(kx< ) cosh[k(L − x> )] k sinh(kL)

∞ ( cos(nπx/L) cos(nπξ/L) 1 + 2L 2 k L k 2 L2 + n2 π 2 n=1

sinh(kx< ){k cosh[k(x> − L)] − sinh[k(x> − L)]} k sinh(kL) + k 2 cosh(kL)

∞ ( (1 + kn2 ) sin(kn ξ) sin(kn x) g(x|ξ) = 2 , (k 2 + kn2 )[1 + L(1 + kn2 )] n=1

where kn is the nth root of tan(kL) = −k. 25. g(x|ξ) =

sinh(kx< ){sinh[k(x> − L)] + k cosh[k(x> − L)]} k 2 cosh(kL) − k sinh(kL)

and g(x|ξ) = 2

∞ ( (1 + kn2 ) sin(kn ξ) sin(kn x) , (k 2 + kn2 )[L(1 + kn2 ) − 1] n=1

where kn is the nth root of tan(kL) = k. 26. g(x|ξ) = and g(x|ξ) = 2

a sinh(kx< ) − k cosh(kx< ) cosh[k(L − x> )] k[a cosh(kL) − k sinh(kL)] ∞ ( (a2 + kn2 ) cos[kn (ξ − L)] cos[kn (x − L)] , (k 2 + kn2 )[L(a2 + kn2 ) − a] n=1

where kn is the nth root of kL tan(kL) = −aL. 27. g(x|ξ) =

{sinh[k(L − x> )] − k cosh[k(L − x> )]} k[(1 + k 2 ) sinh(kL) − 2k cosh(kL)]

× [sinh(kx< ) − k cosh(kx< )]]

641

642

Green’s Functions with Applications

and g(x|ξ) = 2

∞ (

n=1

(k 2

ϕn (ξ)ϕn (x) , + kn2 )[(1 + kn2 )L − 2]

where ϕn (x) = sin(kn x) − kn cos(kn x) and tan(kn L) = 2kn /(1 − kn2 ). 31. 4 5 * . + . ea(ξ−x)/2 sinh (π − x> ) k 2 + a2 /4 sinh x< k 2 + a2 /4 * . + g(x|ξ) = . k 2 + a2 /4 sinh π k 2 + a2 /4 32.

 4 √ 5 1+4νs  exp 1+ 4ν (x − ξ) ,  1 g(x|ξ) = √ 4 √ 5 1 + 4νs   exp 1− 1+4νs (x − ξ) , 4ν

39.

g(x|ξ) =

x≤ξ x≥ξ

$ $ $ πk 2 L2 Jm (kx< )[Jm (kx> )Ym$ (kL) − Jm (kL)Ym$ (kx> )] $ (kL) 2 Jm

and g(x|ξ) = 2 $ where Jm (knm L) = 0.

40. g(r|ρ) =

∞ (

4 $ $ knm Jm (knm ξ)Jm (knm x) , 2 2 2 (k − k )(knm − m2 )Jm nm L)

2 (knm n=1

∞ * nπρ + * nπr + a ( 1 sin sin . 3 2 2π ρr n=1 n a a

43. For (a), if n = 1, g(x|ξ) = ln(x/ξ)H(x − ξ) + ln(ξ); otherwise, g(x|ξ) = For (b), if n = 1,

$ 2 1 1# 1−n x − ξ 1−n H(x − ξ) + ξ 1−n − 1 1−n

g(x|ξ)/ξ = H(ξ − x) ln(ξ/x) + ln(x) − 1/a;

Answers to Some of the Problems

643

otherwise, g(x|ξ)/ξ n =

1# 1−n $ 2 1 ξ − x1−n H(ξ − x) + x1−n − 1 − 1/a. (1 − n)

44. g(x|ξ) =

kb cosh(kx< ) − a sinh(kx< ) −kx> e k(kb − a)

46. H 4 √ √ 5 √ −|x−ξ| s −(x+ξ) s   e − R(s)e (2 s ) ,     H   √ √ √   x s+ρ−ξ s  T (s)e (2 s ) ,  H g(x|ξ) = √ √  √  ξ s+ρ−x s  T (s)e (2 s ) ,     4 5H √  √ √    e−|x−ξ| s+ρ + R(s)e(x+ξ) s+ρ (2 s + ρ ) ,

where

√ √ s+ρ− s √ R(s) = √ s+ρ+ s

and

T (s) = √

x ≥ 0, ξ ≥ 0, x ≤ 0, ξ ≥ 0, x ≥ 0, ξ ≤ 0, x ≤ 0, ξ ≤ 0,

√ 2 s √ . s+ρ+ s

47. (a) g(x|ξ) =

1 −κ|x−ξ| κ−µ e + eκ(x+ξ) , 2κ 2κ(κ + µ) g(x|ξ) =

with µ =

1 e−µx+κξ , κ+µ

x < 0,

x > 0,

√ κ2 + C.

(b) g(x|ξ) =

1 −κ|x−ξ| C e + eκ(x+ξ) , 2κ 2κ(2κ − C) g(x|ξ) =

1 e−κ(x−ξ) . 2κ − C

x > 0,

x < 0,

644

Green’s Functions with Applications

(c) * √ + * √ + π J2κ 2i C ex/2 J−2κ 2i C eξ/2 sin(2κπ) 4 * √ + * √ + * √ + * √ +5 $ $ π J−2κ 2i C J2κ 2i C + J2κ 2i C J−2κ 2i C * √ + * √ + − $ 2 sin(2κπ)J2κ 2i C J−2κ 2i C * √ + * √ + × J2κ 2i C eξ/2 J2κ 2i C ex/2

g(x|ξ) =

for x < 0,

for x > 0. 49.

* √ + * √ + J2κ 2i C eξ/2 J2κ 2i C e−x/2 * √ + * √ + g(x|ξ) = √ $ 2i C J2κ 2i C J2κ 2i C

, , x2 x3 x2 x3 1 x> g(x|ξ) = x3< − + >2 − >3 + x2< − > + >2 6 2b 3b 2 b 2b

50. g(x|ξ) = 51.

x2< * x< + x> − 2 3

$ (x> − L)x< # 2 x − 2Lx> + ξ 2 6L ∞ ( sin(nπξ/L) sin(nπx/L) 3 = 2L n4 π 4 n=1

g(x|ξ) =

52. g(x|ξ) = − 53.

+ 1 * ik|x−ξ| −k|x−ξ| ie − e 4k 3

sin[k(L − x> )] sin(kx< ) 2k 3 sin(kL) sinh[k(L − x> )] sinh(kx< ) − 2k 3 sinh(kL)

g(x|ξ) =

and g(x|ξ) = 2L3

∞ ( sin(nπξ/L) sin(nπx/L) n4 π 4 − k 4 L4 n=1

Answers to Some of the Problems Chapter 4 21. g(x, t|ξ,τ ) =

% & % & ∞ ( 4 1 (2n − 1)πξ (2n − 1)πx H(t − τ ) sin sin π 2n − 1 2L 2L n=1 % & (2n − 1)π(t − τ ) × sin . 2L

22. g(x, t|ξ,τ ) =

23.

, ∞ * nπx + ( t−τ 2 1 nπξ H(t − τ ) + H(t − τ ) cos cos L π n L L n=1 % & nπ(t − τ ) × sin . L

, , ∞ 8 ( m 2mπx 2mπt u(x, t) = sin cos . L m=1 4m2 − 1 L L

24. * nπx + sin L n=1 ! % , -& , -) nπ nπt L nπt −t × e − cos + 2 sin L 2 + n2 π 2 L L + n2 π 2 L , * πx + πt + 2 sin cos L L % & % & ∞ 4L ( 1 (2m − 1)πx (2m − 1)πt + 2 sin sin . π m=1 (2m − 1)2 L L

u(x, t) = 2

∞ (

25. % &! % &) ∞ 8L ( (−1)n+1 (2n − 1)πx (2n − 1)πt sin 1 − cos π 2 n=1 (2n − 1)2 2L 2L % & % & ∞ 8L ( (−1)n+1 (2n − 1)πx (2n − 1)πt + 2 sin cos π n=1 (2n − 1)2 2L 2L % & % & ∞ 8L ( 1 (2n − 1)πx (2n − 1)πt − 2 sin sin . π n=1 (2n − 1)2 2L 2L

u(x, t) = −

645

646 26.

Green’s Functions with Applications , -& ∞ * nπx + % t2 2L ( 1 nπt u(x, t) = 1 − − 2 cos 1 − cos . 2L π n=1 n2 L L

Chapter 5 26.

" 5 2 4iaC ∞ 4 ia(x+ξ)2 /t u(x, t) = √ e − eia(x−ξ) /t f (ξ) dξ, t 0 . where C = 12 (1 − i) 2a/π. 27.

g(x, t|ξ,τ ) =

% & % & ∞ ( 2 (2n − 1)πξ (2n − 1)πx H(t − τ ) sin sin L 2L 2L n=1 % & (2n − 1)2 π 2 (t − τ ) × exp − 4L2

28. g(x, t|ξ,τ ) =

∞ ( 2 2 H(t − τ ) cos(kn ξ) cos(kn x)e−kn (t−τ ), L n=0

where kn = (2n + 1)π/(2L). 29. H(t − τ ) < L ∞ > , * nπx + ( nπξ −n2 π 2 (t−τ )/L2 × 1+2 cos cos e L L n=1

g(x, t|ξ,τ ) =

30. g(x, t|ξ,τ ) = 2H(t − τ )

∞ (

n=1

2 2 kn (t−τ )

e−a

kn2 + h2 + h2 ) + h

L(kn2

× sin[kn (L − ξ)] sin[kn (L − x)],

where kn is the nth root of k cot(kL) = −h.

Answers to Some of the Problems 31. , -& * nπx + % n n2 π 2 t −t sin e − exp − n2 π 2 − L2 L L2 n=1 % & % & ∞ 4 ( 1 (2m − 1)πx (2m − 1)2 π 2 t + sin exp − . π m=1 2m − 1 L L2

u(x, t) = 2π

∞ (

32. % & 2n − 1 (2n − 1)πx u(x, t) = 16πL sin (2n − 1)4 π 4 + 16L4 2L n=1 ! % &) 2 2 (2n − 1) π (2n − 1)2 π 2 t × sin(t) − cos(t) − exp − 4L2 4L2 % & % & ∞ 4( 1 (2n − 1)πx (2n − 1)2 π 2 t + sin exp − . π n=1 2n − 1 2L 4L2 2

∞ (

33. u(x, t) = 1 −

, -& ∞ * nπx + % t 2L ( 1 n2 π 2 t − 2 cos 1 − exp − . L π n=1 n2 L L2

35. % & % & ∞ ( 2 (2n − 1)πξ (2n − 1)πx g(x, t|ξ,τ ) = H(t − τ ) sin sin L 2L 2L n=1 % & a2 (2n − 1)2 π 2 (t − τ ) 2 2 × exp −a k (t − τ ) − 4L2 43. g(x, y,z, t|ξ, η, ζ,τ ) =

∞ ∞ ∞ 2 2 2 2 2 H(t − τ ) ( ( ( Ci Cj e−a (kn +4j +mi )(t−τ ) cos(kn ξ) Lx L y Lz n=1 i=0 j=0

× cos(kn x) cos(0j η) cos(0j y) cos(mi ζ) cos(mi z),

where Ci = 2 − δi0 , kn = (2n − 1)π/(2Lx ), 0j = jπ/Ly , and mi = iπ/Lz .

647

648

Green’s Functions with Applications

44. g(x,y, z, t|ξ, η, ζ,τ ) ∞ ( ∞ ( ∞ ( 2 2 2 2 = Aijn e−a (kn +4j +mi )(t−τ ) sin[kn (Lx − ξ)] i=0 j=0 n=1

× sin[kn (Lx − x)] cos(0j η) cos(0j y) cos(mi ζ) cos(mi z),

where Aijn =

2H(t − τ )(kn2 + h2 )Ci Cj , [Lx (kn2 + h2 ) + h]Ly Lz

kn is the nth positive root of k cot(kLx ) = −h, Ci = 2 − δi0 , 0j = jπ/Ly , and mi = iπ/Lz . 45. g(x, y, z, t|ξ, η, ζ,τ ) ∞ ( ∞ ( ∞ ( cos(βj x) cos(γn y) cos(µk z) cos(βj ξ) cos(γn η) cos(µk ζ) = N (βj )N (γn )N (µk ) k=1 j=1 n=1 1 2 2 2 × exp −a (βj + γn2 + µ2k )(t − τ ) H(t − τ ),

where the positive constants βj , γn , µk , N (βj ), N (γn ) and N (µk ) are given by Lx (βj2 + h2 ) + h N (βj ) = , βj tan(βj Lx ) = h, 2(βj2 + h2 ) N (γn ) =

Ly (γn2 + h2 ) + h , 2(γn2 + h2 )

γn tan(γn Ly ) = h,

N (µk ) =

Lz (µ2k + h2 ) + h , 2(µ2k + h2 )

µk tan(µk Lz ) = h.

and

Chapter 6 14.

, - * ∞ ( exp (−nπ|y − η|/a) nπξ nπx + g(x, y|ξ,η ) = sin sin nπ a a n=1

15.

1 # $ 2 ∞ ( exp − n + 12 π|y − η|/a # $ g(x, y|ξ,η ) = n + 12 π n=0 :# :# $ ; $ ; n + 12 πx n + 12 πξ × cos cos a a

Answers to Some of the Problems

649

16. g(x, y|ξ,η ) =

, ∞ * nπx + |y − η| ( exp (−nπ|y − η|/a) nπξ − cos cos 2a nπ a a n=1

18. g(r,θ |ρ,θ $ ) = 23.

, - , ∞ 1 ( 1 nπ/β −nπ/β nπθ$ nπθ r< r> sin sin . π n=1 n β β

* . + 2 2 π 2 /L2 |x − ξ| ∞ exp − k + n ( 0 1 . g(x, y|ξ,η ) = 2 n=−∞ k02 L2 + n2 π 2 * nπη + * nπy + × cos cos L L

36. g(x, y|ξ,η ) =

∞ ∞ 4 (( 2nm ab n=0 m=0 n2 π 2 /a2 + m2 π 2 /b2 , * nπx + * mπy + * mπη + nπξ × cos cos cos cos , a a b b

where 200 = 0, 2n0 = 20m = 12 , and 2nm = 1 if n > 0 and m > 0. 38. g(r,θ |ρ,θ $ ) = 2 +2

ln(r< /a) ln(b/r> ) 2π ln(b/a) ∞ ( cos[n(θ − θ$ )]

2πn [1 − (a/b)2n ] n=1

39. ∞ 1 ( 1 nπ/β g(r,θ |ρ,θ $ ) = r π n=1 n


#

n n r< − a2n /r
− 2nπ/β a

G

$# n $ n 1/r> − r> /b2n .

,

nπθ$ sin β

-

, nπθ sin . β

∞ ∞ 2 (( J0 (βm r/a)J0 (βm ρ/a) 4 5 g(r, z|ρ,ζ ) = πa2 L m=1 n=1 (βm /a)2 + n2 π 2 /L2 J 2 (βm ) 1 , - * + nπζ nπz × sin sin , L L

650

Green’s Functions with Applications

where βm is the mth root of J0 (β) = 0. 43.

∞ 2 ( cos(km x< ) cos[km (x> − a)] αm b m=0 km sin(km a) * mπη + * mπy + × cos cos b b ∞ ∞ 4 (( cos(nπξ/a) cos(nπx/a) = αn αm 2 ab n=0 m=0 n2 π 2 /a2 − km * mπη + * mπy + × cos cos , b b

g(x, y|ξ,η ) =

2 where km = k02 − m2 π 2 /b2 , α0 = 12 , and αm = 1 for m > 0.

44. g(x, y, z|ξ, η,ζ ) =

∞ ∞ ∞ 8 ( ( ( sin(pπζ/c) sin(pπz/c) abcπ 2 m=1 n=1 p=1 n2 /a2 + m2 /b2 + p2 /c2 * mπη + * mπy + , nπξ - * nπx + × sin sin sin sin b b a a

46. g(x, y, z|ξ, η,ζ ) =

∞ ∞ 4 ( ( sin(kmn x< ) sin[kmn (a − x> )] bc m=0 n=0 kmn sin(kmn a) * mπη + * mπy + , nπζ - * nπz + × sin sin sin sin b b c c ∞ ∞ ∞ 8 ( ( ( sin(kπξ) sin(kπx) = 2 abc n=0 m=0 k 2 π 2 /a2 − km k=0 * mπη + * mπy + , nπζ - * nπz + × sin sin sin sin , b b c c

2 where km = k02 − m2 π 2 /b2 − n2 π 2 /c2 .

47. g(x, y, z|ξ, η,ζ ) ∞ ∞ ∞ 8 ((( cos(nπξ/a) cos(nπx/a) = αn 2 2 2 abc n=0 m=0 n π /a + m2 π 2 /b2 + k 2 π 2 /c2 − k02 k=0 , * mπη + * mπy + , kπζ kπz × sin sin sin cos , b b c c

Answers to Some of the Problems where α0 =

1 2

and αn = 1 for n > 0.

48. g(x, y, z|ξ, η,ζ ) =

∞ ∞ 4 (( cos(kmn x< ) cos[kmn (x> − a)] αn αm bc m=0 n=0 kmn sin(kmn a) , * mπη + * mπy + * nπz + nπζ × cos cos cos cos b b c c ∞ ( ∞ ( ∞ ( 8 cos(kπξ/a) cos(kπx/a) = αn αm αk 2 abc n=0 m=0 k 2 π 2 /a2 − km k=0 , * mπη + * mπy + * nπz + nπζ × cos cos cos cos . b b c c

2 Here km = k02 − m2 π 2 /b2 − n2 π 2 /c2 , α0 = 12 , and αm = 1 for m > 0.

651

Mathematics

The text then presents Green’s functions for each class of differential equation (ordinary differential, wave, heat, and Helmholtz equations) according to the number of spatial dimensions and the geometry of the domain. Detailing step-by-step methods for finding and computing Green’s functions, each chapter contains a special section devoted to topics where Green’s functions particularly are useful. For example, in the case of the wave equation, Green’s functions are beneficial in describing diffraction and waves. To aid readers in developing practical skills for finding Green’s functions, worked examples, problem sets, and illustrations from acoustics, applied mechanics, antennas, and the stability of fluids and plasmas are featured throughout the text. A new chapter on numerical methods closes the book. Included solutions and hundreds of references to the literature on the construction and use of Green’s functions make Green’s Functions with Applications, Second Edition a valuable sourcebook for practitioners as well as graduate students in the sciences and engineering.

WITH APPLICATIONS

The book opens with necessary background information: a new chapter on the historical development of the Green’s function, coverage of the Fourier and Laplace transforms, a discussion of the classical special functions of Bessel functions and Legendre polynomials, and a review of the Dirac delta function.

GREEN’S FUNCTIONS

Since publication of the first edition over a decade ago, Green’s Functions with Applications has provided applied scientists and engineers with a systematic approach to the various methods available for deriving a Green’s function. This fully revised Second Edition retains the same purpose, but has been meticulously updated to reflect the current state of the art.

Second Edition

Advances in Applied Mathematics

GREEN’S FUNCTIONS WITH APPLICATIONS Second Edition

Second Edition

About the Author

DUFFY

Dean G. Duffy received his bachelor of science in geophysics from Case Institute of Technology, Cleveland, Ohio, USA, and his doctorate of science in meteorology from the Massachusetts Institute of Technology, Cambridge, USA. He served in the US Air Force for four years as a numerical weather prediction officer. After his military service, he began a twenty-five year association with the National Aeronautics and Space Administration’s Goddard Space Flight Center, Greenbelt, Maryland, USA. Widely published, Dr. Duffy has taught courses at the US Naval Academy, Annapolis, Maryland, and the US Military Academy, West Point, New York.

DEAN G. DUFFY

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