Stationary Oscillations of Elastic Plates: A Boundary Integral Equation Analysis

Gavin R. Thomson • Christian Constanda

Stationary Oscillations of Elastic Plates A Boundary Integral Equation Analysis

Gavin R. Thomson A.C.C.A. 89 Hydepark Street Glasgow, G3 8BW UK [email protected]

Christian Constanda The Charles W. Oliphant Professor of Mathematical Sciences Department of Mathematical and Computer Sciences The University of Tulsa 800 S. Tucker Drive Tulsa, OK 74104-9700 USA [email protected]

ISBN 978-0-8176-8240-8 e-ISBN 978-0-8176-8241-5 DOI 10.1007/978-0-8176-8241-5 Springer New York Dordrecht Heidelberg London Library of Congress Control Number: 2011931291 Mathematics Subject Classification (2010): 31A10, 74K20, 74H45 © Springer Science+Business Media, LLC 2011 All rights reserved. This work may not be translated or copied in whole or in part without the written permission of the publisher (Springer Science+Business Media, LLC, 233 Spring Street, New York, NY 10013, USA), except for brief excerpts in connection with reviews or scholarly analysis. Use in connection with any form of information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed is forbidden. The use in this publication of trade names, trademarks, service marks, and similar terms, even if they are not identified as such, is not to be taken as an expression of opinion as to whether or not they are subject to proprietary rights. Printed on acid-free paper Springer is part of Springer Science+Business Media (www.birkhauser-science.com)

For my parents (GT) For Lia (CC)

Preface

Many important problems in mathematical physics can be modeled by means of elliptic partial differential equations or systems. Such equations arise in the study of, for example, steady-state heat conduction (the Laplace equation), acoustics (the Helmholtz equation), elasticity (the Lam´e system), and electromagnetism (the Maxwell system). An important tool for investigating boundary value problems associated with equations of this type is the boundary integral equation technique, which relies on the derivation of Fredholm or quasi-Fredholm integral equations over the boundary of the region of interest and leads to a very convenient representation of the solution. The kernels of the ensuing integral equations are expressed in terms of a two-point (scalar or matrix) function that is, in fact, a fundamental solution of the governing linear differential operator. Boundary integral equation methods are extremely useful for a variety of reasons. First, they reduce the problem from one involving an unbounded partial differential operator to one with an integral operator, making it much more appealing from an analytic perspective; second, the methods are very general in that they can be applied to any linear second-order elliptic boundary value problem with constant coefficients; and third, the methods are attractive from a numerical point of view because they yield closed-form solutions and, therefore, lend themselves readily to boundary element treatment. Boundary integral equation methods come in many versions. Thus, the classical indirect approach seeks the solution in an appropriate form that is chosen a priori. This method is ‘indirect’ in the sense that the unknown function in the corresponding integral equation has no physical significance, being merely a convenient mathematical abstraction. By contrast, in the direct methods the unknown function in the integral formulation is an actual physical quantity. For example, in elasticity the solution of the integral equation may represent the displacement or the moment/stress on the boundary of the elastic body. Another main class of boundary integral equation methods makes use of modified fundamental solutions. This approach was developed to address problems of existence of nonunique solutions to the integral equations derived by the classical vii

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Preface

techniques. In certain instances, integral equations formulated for boundary value problems known to have at most one solution may themselves admit multiple solutions. Intuitively, this should not be the case. For this reason, we consider ways of modifying the standard fundamental solution so that it leads to uniquely solvable integral equations. This book investigates an elliptic system of equations arising in the theory of elasticity which characterizes the stationary oscillations of thin elastic plates. The system is obtained by assuming a Mindlin-type form (also known as Kirchhoff’s kinematic hypothesis) for the displacements. Approximate theories describing the bending of plates are important because they reduce the equations of classical three-dimensional elasticity to a system involving only two independent space variables, while highlighting the important bending characteristics of the elastic structure. Such theories have been used successfully in many practical engineering applications. The Mindlin-type model differs from the classical Kirchhoff theory in that it accounts for transverse shear deformation as well, thereby offering additional useful information to practitioners. Boundary integral equation methods have been widely used in the study of various elliptic systems arising in the theory of elasticity and beyond, such as equilibrium and dynamic problems in the process of deformation of two- and threedimensional elastic bodies, and the equilibrium and time-dependent bending of elastic and thermoelastic plates with transverse shear deformation. Although the equations governing the stationary oscillations of Mindlin-type plates are related in a certain way to the equilibrium equations, the two systems display very different characteristics. The main difference is the presence in the oscillatory case of so-called eigenfrequencies. These are values of the oscillation frequency for which the main homogeneous boundary value problems in a bounded domain have nonzero solutions. The book will show how such difficulties can be resolved and how the problems in question can be reduced to uniquely solvable integral equations. Here is a brief description of the contents. Chapter 1 presents a derivation of the system of equations modeling the stationary oscillations of elastic plates with transverse shear deformation. A fundamental integral formula, analogous to Green’s second identity from potential theory, is also deduced. The aim in Chapter 2 is to define the generalized single-layer and double-layer plate potentials and to describe their essential properties. All subsequent discussion of boundary value problems relies on the boundary properties of these integral functions. In order to construct the potentials, a suitable matrix of fundamental solutions is made available and its behavior, together with the behavior of a so-called matrix of singular solutions, near the boundary of the plate is investigated. Chapter 3 deals with the setting up and smoothness properties of a particular solution to the inhomogeneous system obtained in Chapter 1. In Chapter 4 we introduce the Dirichlet and Neumann problems associated with the governing system of equations. It is natural to discuss these two problems together because they are intrinsically linked in the analysis of their solvability. Radi-

Preface

ix

ation conditions that ensure the uniqueness of the solutions of the exterior problems are given, which are then shown to be satisfied by the potential functions defined in Chapter 2. The bulk of the chapter is concerned with establishing integral representations for the regular solutions of the system, to be used later as a starting point in the direct boundary integral equation method. The presence of eigenfrequencies in the interior problems, which makes the system of stationary oscillations so different from the corresponding equilibrium system, is investigated in Chapter 5. The proof of the existence of eigenfrequencies relies, however, on the relationship between the two systems. Chapter 6 is concerned with the solvability of the boundary value problems mentioned above. This issue is approached through a classical indirect formulation that results in quasi-Fredholm integral equations of the second kind. Unfortunately, owing to the existence of eigenfrequencies in the interior problems, the solvability of the latter is not always guaranteed. Furthermore, the connection between the solvability of the Dirichlet and Neumann problems leads to difficulties regarding the unique solvability of the integral equations for the exterior problems as well, an effect that, given the available uniqueness results, is not expected. The application of the direct boundary integral equation method is the subject of Chapter 7, where a coupled pair of equations for each problem—one a quasiFredholm second-kind equation and one an equation of the first kind—is obtained. Their analysis is simplified through the use of composition formulas relating various boundary integral operators of interest. It is shown that, as physically expected, each pair of equations for the exterior problems admits exactly one solution. A composite equation consisting of a linear combination of the first-kind and second-kind equations is also studied. In Chapter 8 a theory of modified integral equations is developed. This is motivated by the need for uniquely solvable equations from which the solutions of the exterior boundary value problems can then be constructed. An indirect method is employed, where the solutions are postulated in the form of modified potentials that lead to quasi-Fredholm second-kind equations. Two different types of modification are considered, with existence and uniqueness results proved for each. The chapter concludes with a look at how uniquely solvable first-kind equations can be derived (again, by an indirect method). The Robin boundary value problems are introduced in Chapter 9. After the question of uniqueness of solution has been investigated in three separate cases, integral equation methods analogous to those used in Chapters 6–8 are also constructed for these problems. Chapter 10 considers a fourth type of fundamental boundary value problem associated with the stationary oscillations of thin elastic plates, namely, the transmission problem. The existence of the solution is proved by means of an indirect method after some regularization of the operators involved. A more refined method of solution is then described, based on a direct method in conjunction with a modified fundamental solution. Finally, in Chapter 11 the null field method is examined. Though, strictly speaking, this is not an integral equation method, it is closely connected to much of the

x

Preface

work from the preceding chapters. Facts concerning the unique solvability of the null field equations and the completeness of certain sets of functions are presented. Brief announcements of the some of the results discussed in the book can be found in [58]–[69]. The methodology and results presented in this monograph should prove useful to applied mathematicians, scientists, and engineers engaged in the research of oscillatory phenomena and other similar models, as well as to graduate students in those disciplines. We would like to express our sincerest thanks to Tom Grasso and Ben Cronin at Birkh¨auser Boston for their highly professional and efficient handling of this project. Gavin R. Thomson A.C.C.A. Glasgow, UK

Christian Constanda The Charles W. Oliphant Professor of Mathematical Sciences The University of Tulsa Tulsa, Oklahoma, USA May 2011

Contents

1

The Mathematical Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1

2

Layer Potentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 2.1 Fundamental and Singular Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 2.2 Order of Singularity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 2.3 Properties of the Potentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

3

The Nonhomogeneous System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 The Newtonian Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Smoothness Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.1 Existence of the First-Order Derivatives . . . . . . . . . . . . . . . . . 3.2.2 H¨older Continuity of the First-Order Derivatives . . . . . . . . . . 3.2.3 Existence of the Second-Order Derivatives . . . . . . . . . . . . . . . 3.2.4 H¨older Continuity of the Second-Order Derivatives . . . . . . . . 3.3 A Particular Solution of the System . . . . . . . . . . . . . . . . . . . . . . . . . . .

23 23 24 24 27 30 34 43

4

The Question of Uniqueness for the Exterior Problems . . . . . . . . . . . . . 4.1 Radiation Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 The Boundary Value Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Representation Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

47 47 52 53

5

The Eigenfrequency Spectra of the Interior Problems . . . . . . . . . . . . . . 61 5.1 The Green’s Tensor for the Interior Dirichlet Problem . . . . . . . . . . . . 62 5.2 The Green’s Tensor for the Interior Neumann Problem . . . . . . . . . . . 67

6

The Question of Solvability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1 Boundary Integral Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 Boundary Integral Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3 The Interior Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4 The Exterior Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

75 76 76 81 83

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Contents

7

Direct Boundary Equation Formulation . . . . . . . . . . . . . . . . . . . . . . . . . . 87 7.1 Composition of the Boundary Operators . . . . . . . . . . . . . . . . . . . . . . . 88 7.2 Boundary Integral Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 7.3 The Interior Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90 7.4 The Eigensolutions of the Interior Problems . . . . . . . . . . . . . . . . . . . . 94 7.5 The Exterior Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96 7.6 Composite Integral Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

8

Modified Fundamental Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 8.1 The Matrix of Fundamental Solutions in Terms of Wavefunctions . . 104 8.2 The Ursell Modification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108 8.2.1 Construction of M ω (x, y) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113 8.2.2 The Invertibility of Pm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118 8.2.3 Convergence of the Infinite Series . . . . . . . . . . . . . . . . . . . . . . 127 8.3 The Jones Modification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133 8.3.1 Orthogonality Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135 8.3.2 The Exterior Neumann Problem . . . . . . . . . . . . . . . . . . . . . . . . 138 8.3.3 The Exterior Dirichlet Problem . . . . . . . . . . . . . . . . . . . . . . . . . 142 8.3.4 Modification with a Finite Series . . . . . . . . . . . . . . . . . . . . . . . 143 8.4 Equations of the First Kind . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147 8.4.1 Composition of the Modified Boundary Operators . . . . . . . . . 147 8.4.2 The Exterior Dirichlet Problem . . . . . . . . . . . . . . . . . . . . . . . . . 149 8.4.3 The Exterior Neumann Problem . . . . . . . . . . . . . . . . . . . . . . . . 151

9

Problems with Robin Boundary Conditions . . . . . . . . . . . . . . . . . . . . . . . 153 9.1 Uniqueness Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154 9.2 Indirect Boundary Equation Formulation . . . . . . . . . . . . . . . . . . . . . . . 159 9.2.1 The Interior Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159 9.2.2 The Exterior Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162 9.3 Direct Boundary Equation Formulation . . . . . . . . . . . . . . . . . . . . . . . . 167 9.3.1 The Interior Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167 9.3.2 The Exterior Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168 9.4 Composite Integral Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172 9.5 Modified Fundamental Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173 9.5.1 The Ursell Modification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173 9.5.2 The Jones Modification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174

10 The Transmission Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177 10.1 The Indirect Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178 10.2 The Direct Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187 11 The Null Field Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191 11.1 Derivation of the Null Field Equations . . . . . . . . . . . . . . . . . . . . . . . . . 192 11.2 The Question of Solvability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194 11.2.1 The Exterior Dirichlet Problem . . . . . . . . . . . . . . . . . . . . . . . . . 195 11.2.2 The Exterior Neumann Problem . . . . . . . . . . . . . . . . . . . . . . . . 196

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11.2.3 The Exterior Robin Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . 196 11.2.4 The Transmission Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197 11.3 Complete, Linearly Independent Sets of Wavefunctions . . . . . . . . . . . 201 A

Proof of Lemma 8.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205

B

The Matrix Um

(σ )

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229

Chapter 1

The Mathematical Model

In this chapter we derive the system characterizing the stationary oscillations of thin elastic plates with transverse shear deformation proposed in [56]. We assume that the body forces have a time-harmonic form, which we substitute into the full classical three-dimensional elasticity model to obtain a time-independent system. The so-called kinematic hypothesis is then applied and, by averaging over the thickness of the plate, we arrive at the desired system of equations. The boundary moment– stress operator is also defined. The chapter concludes with the reciprocity relation, which connects the solutions of the system in a bounded domain with the values of the displacements and of the moments and stresses on the boundary. Unless stated otherwise, throughout what follows Latin and Greek indices take the values 1, 2, 3 and 1, 2, respectively, and the convention of summation over repeated indices is understood. Also, a superscript T denotes matrix transposition. Consider a three-dimensional homogeneous and isotropic elastic body, and let ti j = t ji be the internal stresses, V j the displacements, F j the body forces, λ and μ the Lam´e constants of the material, and ρ the (constant) mass density. If x = (x 1 , x2 , x3 )T is a generic point in R 3 , then the equations of motion are [56] ti j,i (x,t) + Fj (x,t) = ρ

∂ 2V j (x,t), ∂t2

(1.1)

with the constitutive relations written as ti j (x,t) = λ δi jVk,k (x,t) + μ (Vi, j + V j,i )(x,t),

(1.2)

where (. . .),i = ∂ (. . .)/∂ xi and δi j is the Kronecker delta. The components of the resultant stress vector t in a direction n = (n 1 , n2 , n3 )T are ti = ti j n j . If F is of the form

  F(x,t) = Re f (x)e−iω t ,

G.R. Thomson and C. Constanda, Stationary Oscillations of Elastic Plates, A Boundary Integral Equation Analysis, DOI 10.1007/978-0-8176-8241-5_1, © Springer Science+Business Media, LLC 2011

(1.3) 1

2

1 The Mathematical Model

where f is a complex-valued vector function and ω ∈ R, and if the boundary conditions are separable in the same way with respect to the space and time variables, then the body performs stationary oscillations of frequency ω , and its expected displacements are of the form   V (x,t) = Re v(x)e−iω t . (1.4) Substituting (1.3) and (1.4) in (1.1) and (1.2) yields ti j,i (x) + f j (x) + ρω 2v j (x) = 0,

(1.5)

ti j (x) = λ δi j vk,k (x) + μ (vi, j + v j,i )(x).

(1.6)

These are the equations of stationary oscillations in classical elasticity [40]. Suppose that the body is an elastic plate occupying a region S¯ × [−h0 /2, h0 /2] in R3 , where S is a domain in R 2 bounded by a simple, closed C 2 -curve ∂ S and h 0 , 0 < h0  diam S, is the constant thickness of the plate (see Figure 1). The bounded domain enclosed by ∂ S is denoted by S + , and we write S − = R2 \(S+ ∪ ∂ S). x3

x3 = h0 /2 S−

∂S

S+

x2 x3 = −h0 /2

x1 Fig. 1. Geometric configuration of a plate. We assume a Mindlin-type displacement field; that is, vα (x1 , x2 , x3 ) = x3 uα (x1 , x2 ), v3 (x1 , x2 , x3 ) = u3 (x1 , x2 ),

(1.7)

which ensures that this model takes into account the effects of transverse shear forces [14]. Consider the averaging operators I α −1 and Jα −1 defined by

1 The Mathematical Model

3

(Iα −1 g)(x1 , x2 ) =

x =h0 /2 1  α −1 x g(x1 , x2 , x3 ) x3 =−h , 3 0 /2 h0 3

1 (Jα −1 g)(x1 , x2 ) = h0

h0 /2

xα3 −1 g(x1 , x2 , x3 ) dx3 .

−h0 /2

Substituting the kinematic assumption (1.7) in (1.5) and writing Nαβ = J1 tαβ , N3α = J0 t3α , Hα = −(J1 fα + I1t3α ), H3 = −(J0 f3 + I0 t33 ), we arrive at the system of equations Nαβ ,β − N3α + ρω 2h2 uα = Hα , N3β ,β + ρω 2u3 = H3 ,

(1.8)

where h2 = h20 /12. It can be shown that N3α , Nαα (α not summed) and N12 = N21 are, respectively, the averaged transverse shear forces and averaged bending and twisting moments with respect to the middle plane, acting on the face of a vertical cross-sectional element of the plate perpendicular to the x α -axis [14]. Similarly, Hα and H3 are related to the averaged body forces and moments and resultant averaged forces and moments acting on the faces x 3 = ±h0 /2. Using the same averaging procedure and taking (1.7) into account, we bring the constitutive relations (1.6) to the form   Nαβ = h2 λ uγ ,γ δαβ + μ (uα ,β + uβ ,α ) , (1.9) N3α = μ (uα + u3,α ). In view of (1.9), system (1.8) can now be rewritten as Aω (∂x )u(x) = A(∂ /∂ x1 , ∂ /∂ x2 )u(x) ⎛ 2 2 ⎞ ρω h 0 0 + ⎝ 0 ρω 2 h2 0 ⎠ u(x) = H(x), 0 0 ρω 2 where A(ξ1 , ξ2 ) is the matrix ⎛ 2 ⎞ h μΔ + h2 (λ + μ )ξ12 − μ h2 (λ + μ )ξ1 ξ2 −μξ1 ⎝ h2 (λ + μ )ξ1 ξ2 h2 μΔ + h2(λ + μ )ξ22 − μ −μξ2 ⎠ , μξ1 μξ2 μΔ

(1.10)

(1.11)

4

1 The Mathematical Model

u = (u1 , u2 , u3 )T , H = (H1 , H2 , H3 )T , and Δ = ξ12 + ξ22 . System (1.10) was derived in [56]. The matrix operator A(∂ x ) defined by (1.11) arises in the equilibrium bending of plates (see [14]). The components of the moments and stress on the boundary of the plate are Nα = Nαβ νβ , N3 = N3β νβ ,

(1.12)

where ν = (ν1 , ν2 )T is the unit outward normal to ∂ S in the middle plane of the plate. By (1.9), equations (1.12) may be written in terms of u as N = T (∂x )u(x). Here, N = (N1 , N2 , N3 )T and T (∂x ) = T (∂ /∂ x1 , ∂ /∂ x2 ), where T (ξ1 , ξ2 ) is the matrix ⎛ ⎞ h2 [(λ + 2μ )ν1ξ1 + μν2 ξ2 ] h2 (μν2 ξ1 + λ ν1 ξ2 ) 0 ⎜ ⎟ h2 [μν1 ξ1 + (λ + 2μ )ν2 ξ2 ] 0 ⎠ . (1.13) ⎝ h2 (λ ν2 ξ1 + μν1 ξ2 ) μν1 μν2 μνα ξα A solution u of (1.10) is called regular in S ± if u ∈ C 2 (S± ) ∩C1 (S¯± ). In what follows we assume that

λ + μ > 0, μ > 0, 1 μ ω2 > 2 . h ρ

(1.14) (1.15)

Clearly, (1.15) indicates that we are concentrating on high-frequency oscillations. Theorem 1.1. The system of equations (1.10) is elliptic. Proof. Consider the matrix ⎛ ⎞ h2 μΔ + h2 (λ + μ )ξ12 h2 (λ + μ )ξ1 ξ2 0 ⎜ ⎟ Aω h2 (λ + μ )ξ1ξ2 h2 μΔ + h2(λ + μ )ξ22 0 ⎠ , 0 (ξ ) = ⎝ 0 0 μΔ which corresponds to the second-order derivatives in (1.10). Then    4 2 2 2 2 2 det Aω 0 (ξ ) = h μΔ μΔ + (λ + μ )ξ1 μΔ + (λ + μ )ξ2 − (λ + μ ) ξ1 ξ2   = h4 μΔ μ 2 Δ 2 + μ (λ + μ )Δ 2 = h4 μ 2 (λ + 2 μ )Δ 3 = h4 μ 2 (λ + 2 μ )(ξ12 + ξ22 )3 . ω By (1.14), det A ω 0 (ξ ) > 0 for ξ = 0, so A 0 (ξ ) is invertible; therefore, (1.10) is an elliptic system [47].  

1 The Mathematical Model

5

The next assertion is known as the reciprocity relation. Theorem 1.2. If u, v ∈ C 2 (S+ ) ∩C1 (S¯+ ), then     vT Aω u − uTAω v da = vT Tu − uTT v ds. S+

∂S

Proof. The corresponding reciprocity relation for the equilibrium bending of plates [14] states that if u, v ∈ C 2 (S+ ) ∩C1 (S¯+ ), then  S+



  T vT Au − uTAv da = v Tu − uTT v ds. ∂S

Consequently,   vT Au + ρω 2(h2 u1 v1 + h2 u2 v2 + u3v3 ) S+

 − uT Av − ρω 2(h2 u1 v1 + h2 u2 v2 + u3 v3 ) da =



(vT Tu − uTT v) ds,

∂S

and the required formula follows from (1.10).   Remark 1.1. (i) Theorem 1.1 enables us to replace the discussion of the solvability of system (1.10) in the (two-dimensional) domain S by the analysis of the solvability of some related integral equations on the (one-dimensional) boundary curve ∂ S. This is a distinct advantage since integral operators have ‘better’ properties than their differential counterparts. (ii) The reciprocity relation (Theorem 1.2) is used in the construction of representation formulas (Chapter 4) and in the proof of existence of solutions by means of direct methods (Chapters 7 and 8).

Chapter 2

Layer Potentials

We solve boundary value problems associated with system (1.10) (or, rather, its homogeneous version) by means of potential-type functions with a suitably chosen kernel. In this chapter we construct a matrix of fundamental solutions for the operator Aω (∂x ), which we can then use to define generalized single-layer and doublelayer plate potentials. The method used is analogous to the one employed in [14] to construct the corresponding matrix for the operator A(∂ x ) defined by (1.11), which occurs in the study of the equilibrium bending of plates. Later, we write the matrix of fundamental solutions in a form that allows us to decompose it into an infinite series of so-called wavefunctions. This form, constructed in Theorem 2.1, is similar to the corresponding matrix in the theory of plane elastodynamics [48] with the added complication of certain computational constants. In Section 2.2 we investigate the singularities of the matrix of fundamental solutions and of its associated matrix of singular solutions. Thus, in Theorems 2.3 and 2.4 we find that these singularities coincide with those of the corresponding matrices from equilibrium plate theory. Therefore, the single-layer and double-layer potentials introduced in Section 2.3 behave in the same way as the potentials considered in [14]. The important properties of these functions, used extensively in the subsequent analysis, are contained in Theorems 2.5–2.7.

2.1 Fundamental and Singular Solutions We construct a matrix of fundamental solutions for the operator A ω (∂x ) using the method described in [13]. If A ω ∗ (ξ ) is the adjoint of the matrix A ω (ξ ), then

where B satisfies

u(x) = Aω ∗ (∂x )B(x),

(2.1)

(det Aω ) (∂x )B(x) = H(x).

(2.2)

From (1.10) and (1.11) it follows that G.R. Thomson and C. Constanda, Stationary Oscillations of Elastic Plates, A Boundary Integral Equation Analysis, DOI 10.1007/978-0-8176-8241-5_2, © Springer Science+Business Media, LLC 2011

7

8

2 Layer Potentials

det Aω (ξ )     = μξ1 μ 2 ξ1 (h2 Δ − 1) + ρω 2h2 μξ1 + μξ2 μ 2 ξ2 (h2 Δ − 1) + ρω 2h2 μξ2   + (μΔ + ρω 2 ) h4 μ (λ + 2μ )Δ 2 + h2 (λ + 3μ )(ρω 2 h2 − μ )Δ + (ρω 2h2 − μ )2 = μ 3 (ξ12 + ξ22 )(h2 Δ − 1) + ρω 2h2 μ 2 (ξ12 + ξ22)   + h4 μ 2 (λ + 2μ )Δ 3 + h2 μ (λ + 3 μ )(ρω 2 h2 − μ ) + ρω 2h4 μ (λ + 2μ ) Δ 2   + μ (ρω 2 h2 − μ )2 + ρω 2h2 (λ + 3μ )(ρω 2 h2 − μ ) Δ + ρω 2 (ρω 2 h2 − μ )2   = h4 μ 2 (λ + 2μ )Δ 3 + ρω 2 h4 μ (2λ + 5 μ ) − h2 μ 2 (λ + 2 μ ) Δ 2 + ρω 2h2 (ρω 2 h2 − μ )(λ + 4μ )Δ + ρω 2 (ρω 2 h2 − μ )2 . Factoring this expression, we obtain

 λ + 3μ 2 μ det Aω (ξ ) = h4 μ 2 (λ + 2μ ) Δ 2 + k Δ+ k2 k32 (Δ + k32 ), λ + 2μ λ + 2μ where k2 =

ρω 2 μ

and k32 = k2 − hence,

1 ; h2

(2.3)

(2.4)

det Aω (ξ ) = h4 μ 2 (λ + 2μ )(Δ + k12 )(Δ + k22 )(Δ + k32 ),

(2.5)

λ + 3μ 2 k , λ + 2μ μ k12 k22 = k2 k32 . λ + 2μ

(2.6)

where k12 + k22 =

Without loss of generality, we assume that k 12 ≥ k22 . Using (2.4) and (2.6), we find that k 12 , k22 , and k32 are connected by the equality   h2 (k12 − k32 )(k22 − k32 ) = h2 k12 k22 − k32 (k12 + k22 ) + k34

 μ λ + 3μ 2 2 2 2 2 4 =h k k3 − k k3 + k3 λ + 2μ λ + 2μ = h2 (−k2 k32 + k34 )

 1 = h2 k32 − k2 + k2 − 2 , h from which h2 (k12 − k32 )(k22 − k32 ) = −k32 .

(2.7)

2.1 Fundamental and Singular Solutions

9

We claim that, under assumptions (1.14) and (1.15), k 12 , k22 , and k32 are real, strictly positive, and distinct. First, k 12 and k22 are the two roots of the equation x2 +

λ + 3μ 2 μ k x+ k2 k32 = 0. λ + 2μ λ + 2μ

The discriminant of this quadratic is (λ + 3μ )2 4 4μ k − k2 k32 (λ + 2μ )2 λ + 2μ 

 k2 1 2 2 2 2 = k ( λ + 6 λ μ + 9 μ ) − 4 μ ( λ + 2 μ ) k − (λ + 2μ )2 h2   k2 4μ (λ + 2μ ) = k2 (λ 2 + 2λ μ + μ 2 ) + 2 (λ + 2μ ) h2   k2 4 μ (λ + 2μ ) = k2 (λ + μ )2 + > 0. (λ + 2μ )2 h2 Consequently, k 12 and k22 are real and distinct. Also, by (2.4), assumption (1.15) implies that k32 > 0. By (2.6) and (1.14), this means that k12 + k22 > 0,

k12 k22 > 0.

Hence, k12 and k22 are strictly positive. Finally, from (2.7) and the fact that k 32 = 0 it follows that k12 = k32 and k22 = k32 . Replacing, in turn, each component of H by −δ (|x − y|), where δ is the Dirac delta distribution, and setting the other two equal to zero, from (2.1) and (2.2) we obtain the matrix of fundamental solutions   D ω (x, y) = Aω ∗ (∂x ) t(x, y)E3 , (2.8) where, by (2.2) and (2.5), t(x, y) is a solution of h4 μ 2 (λ + 2μ )(Δ + k12 )(Δ + k22 )(Δ + k32 )t(x, y) = −δ (|x − y|).

(2.9)

We seek t(x, y) in the form t(x, y) =

3

(1)

∑ b j H0

(k j |x − y|),

(2.10)

j=1 (1)

where H0 is the Hankel function of the first kind of order zero and the b j are constants to be determined from (2.9). This Hankel function is a fundamental solution of the Helmholtz operator and satisfies [75] (1)

(Δ + k2j )H0 (k j |x − y|) = 4iδ (|x − y|).

(2.11)

10

2 Layer Potentials

From (2.10) and (2.11) we find that   (1) (Δ + k32 )t(x, y) = b1 4iδ (|x − y|) + (k32 − k12 )H0 (k1 |x − y|)   (1) + b2 4iδ (|x − y|) + (k32 − k22 )H0 (k2 |x − y|) + 4ib3 δ (|x − y|). To eliminate the Dirac distribution in this equation, we require that b1 + b2 + b3 = 0.

(2.12)

Now, if (2.12) is satisfied, we see that (Δ + k22 )(Δ + k32 )t(x, y)   (1) = b1 (k32 − k12 ) 4iδ (|x − y|) + (k22 − k12 )H0 (k1 |x − y|) + 4ib2(k32 − k22 )δ (|x − y|), so we must have (k32 − k12 )b1 + (k32 − k22 )b2 = 0.

(2.13)

Since t(x, y) satisfies (2.9), we obtain −

1 h4 μ 2 ( λ

+ 2μ )

δ (|x − y|) = (Δ + k12 )(Δ + k22 )(Δ + k32 )t(x, y) = 4ib1 (k12 − k22 )(k12 − k32 )δ (|x − y|),

from which we deduce that b1 =

4h4 μ 2 (λ

i . + 2 μ )(k12 − k22 )(k12 − k32 )

(2.14)

Substituting this into (2.12) and (2.13) yields b2 = b3 =

4h4 μ 2 (λ

i , + 2 μ )(k22 − k12 )(k22 − k32 )

i . 4h4 μ 2 (λ + 2 μ )(k32 − k12 )(k32 − k22 )

(2.15)

The formula for b 3 can be simplified by means of (2.7): b3 = −

i 4h2 μ 2 (λ

+ 2μ )k32

.

(2.16)

These constants are well defined since the k 2j are distinct. To calculate the matrix of fundamental solutions D ω (x, y) using (2.8), we first need to compute A ω ∗ (∂x ). From (1.10) and (1.11) it follows that

2.1 Fundamental and Singular Solutions

 2  h μΔ + h2(λ + μ )ξ 2 − μ + ρω 2h2 − μξ2  ω∗ 2   A11 (ξ ) =  μξ2 μΔ + ρω 2  = h2 μ 2 Δ Δ + h2 μ (λ + μ )Δ ξ22 + (ρω 2 h2 − μ )μΔ + ρω 2h2 μΔ + ρω 2h2 (λ + μ )ξ22 + ρω 2(ρω 2 h2 − μ ) + μ 2ξ22 = h2 μ 2 Δ Δ + h2 μ (λ + μ )Δ Δ − h2 μ (λ + μ )Δ ξ12 − μ 2 ξ12 + 2ρω 2h2 μΔ + ρω 2h2 (λ + μ )Δ − ρω 2h2 (λ + μ )ξ12 + ρω 2(ρω 2 h2 − μ ) = h2 μ (λ + 2 μ )Δ Δ − h2 μ (λ + μ )Δ ξ12 + ρω 2 h2 (λ + 3μ )Δ  − μ 2 + ρω 2 h2 (λ + μ ) ξ12 + ρω 2(ρω 2 h2 − μ ). In the same way, ∗ 2 2 2 2 2 Aω 22 (ξ ) = h μ (λ + 2μ )Δ Δ − h μ (λ + μ )Δ ξ2 + ρω h (λ + 3μ )Δ  2 − μ + ρω 2h2 (λ + μ ) ξ22 + ρω 2 (ρω 2 h2 − μ ).

Next,  2  h (λ + μ )ξ1 ξ2 −μξ1  ∗  Aω ( ξ ) = − 12  μξ2 μΔ + ρω 2   2  h (λ + μ )ξ1 ξ2 −μξ2  ∗ = −  = Aω 21 (ξ ) μξ1 μΔ + ρω 2   = −h2 μ (λ + μ )Δ ξ1 ξ2 − μ 2 + ρω 2 h2 (λ + μ ) ξ1 ξ2 ,    h2 (λ + μ )ξ1 ξ2 − μξ1  ω∗  A13 (ξ ) =  2 h μΔ + h2 (λ + μ )ξ22 − μ + ρω 2 h2 − μξ2   2  h (λ + μ )ξ1 ξ2 h2 μΔ + h2 (λ + μ )ξ 2 − μ + ρω 2 h2  2   = −Aω ∗ (ξ ) = − 31  μξ1 μξ2 = −h2 μ (λ + μ )ξ1 ξ22   + h2 μ 2 Δ ξ1 + h2 μ (λ + μ )ξ1 ξ22 + (ρω 2 h2 − μ )μξ1 = h2 μ 2 Δ ξ1 + μ (ρω 2 h2 − μ )ξ1 ,  2  h μΔ + h2 (λ + μ )ξ 2 − μ + ρω 2 h2 −μξ1  ω∗ 1   = −Aω ∗ (ξ ) A23 (ξ ) = −  32 h2 (λ + μ )ξ1 ξ2 −μξ2  = h2 μ 2 Δ ξ2 + μ (ρω 2 h2 − μ )ξ2 . Finally, ∗ Aω 33 (ξ )  2  h μΔ + h2 (λ + μ )ξ 2 − μ + ρω 2 h2  h2 (λ + μ )ξ1 ξ2 1  =  2 2 2 2 2 2 h (λ + μ )ξ1 ξ2 h μΔ + h (λ + μ )ξ2 − μ + ρω h 

11

12

2 Layer Potentials

= h4 μ 2 Δ Δ + h4 μ (λ + μ )Δ ξ22 + h2 μ (ρω 2 h2 − μ )Δ + h4 μ (λ + μ )Δ ξ12 + h4 (λ + μ )2 ξ12 ξ22 + h2 (λ + μ )(ρω 2 h2 − μ )ξ12 + h2 μ (ρω 2 h2 − μ )Δ + h2 (λ + μ )(ρω 2 h2 − μ )ξ22 + (ρω 2 h2 − μ )2 − h4 (λ + μ )2 ξ12 ξ22 = h4 μ (λ + 2μ )Δ Δ + h2 (λ + 3μ )(ρω 2h2 − μ )Δ + (ρω 2 h2 − μ )2 . Therefore, the elements of the adjoint of the matrix A ω (ξ ) are ∗ 2 2 2 2 Aω αβ (ξ ) = h μ (λ + 2 μ )δαβ Δ Δ − h μ (λ + μ )Δ ξα ξβ + ρω h (λ + 3μ )δαβ Δ  − ρω 2 h2 (λ + μ ) + μ 2 ξα ξβ + ρω 2(ρω 2 h2 − μ )δαβ , (2.17) ∗ ω∗ 2 2 2 2 Aω α 3 (ξ ) = −A3α (ξ ) = h μ Δ ξα + μ (ρω h − μ )ξα ,

(2.18)

∗ 4 2 2 2 2 2 2 Aω 33 (ξ ) = h μ (λ + 2μ )Δ Δ + h (λ + 3μ )(ρω h − μ )Δ + (ρω h − μ ) . (2.19)

Theorem 2.1. The elements of the matrix of fundamental solutions D ω (x, y) are ω Dαβ (x, y)

 i ∂2 ∂2 (1) (1) = 2 2 − α12 H0 (k1 |x − y|) − α22 H0 (k2 |x − y|) ∂ x ∂ x ∂ x ∂ x 4h μ k3 α α β β  ∂2 (1) (1) + H0 (k3 |x − y|) + δαβ k32 H0 (k3 |x − y|) , (2.20) ∂ xα ∂ xβ

Dαω3 (x, y) = −D3ωα (x, y)   i ∂ ∂ (1) (1) = 2 2 −γ H0 (k1 |x − y|) + γ H0 (k2 |x − y|) , ∂ xα ∂ xα 4h μ k3 ω D33 (x, y) =

i 4h2 μ k32

 2 (1)  (1) β1 H0 (k1 |x − y|) + β22H0 (k2 |x − y|) ,

(2.21)

(2.22)

where

α12 =

k22 − μ  k32 , k22 − k12

γ= β12 = with

k12 − μ  k32 , k12 − k22

μ  k32 , 2 k1 − k22

h2 k32 (k12 − μ k32 ) , k12 − k22

μ =

α22 =

β22 = μ . λ + 2μ

(2.23) (2.24)

h2 k32 (k22 − μ  k32 ) , k22 − k12

(2.25)

(2.26)

2.1 Fundamental and Singular Solutions

13

Proof. First, we rewrite the elements of the adjoint matrix in a more convenient form. Using (2.3) and (2.4), from (2.17) we obtain  2  ∗ 2 2 2 2 2 Aω αβ (ξ ) = δαβ h μ (λ + 2μ )Δ Δ + ρω h (λ + 3μ )Δ + ρω (ρω h − μ )   − h2 μ (λ + μ )Δ + ρω 2 h2 (λ + μ ) + μ 2 ξα ξβ   λ + 3μ 2 μ = δαβ h2 μ (λ + 2μ ) Δ Δ + k Δ+ k2 k32 λ + 2μ λ + 2μ − h2 μ (λ + μ )(Δ + k2 )ξα ξβ − μ 2 ξα ξβ , so, using (2.4) and (2.6), we arrive at ∗ 2 2 2 Aω αβ (ξ ) = δαβ h μ (λ + 2μ )(Δ + k1 )(Δ + k2 )

− h2 μ (λ + μ )(Δ + k32 )ξα ξβ − μ (λ + 2μ )ξα ξβ . Similarly, from (2.18) and (2.19), ∗ ω∗ Aω α 3 (ξ ) = −A3α (ξ )

= h2 μ 2 Δ ξα + μ (ρω 2h2 − μ )ξα

 ρω 2 h2 − μ 2 2 =h μ Δ+ ξα = h2 μ 2 (Δ + k32 )ξα , h2 μ ∗ 4 2 2 2 2 2 2 Aω 33 (ξ ) = h μ (λ + 2 μ )Δ Δ + h (λ + 3μ )(ρω h − μ )Δ + (ρω h − μ )

= h4 μ (λ + 2 μ )(Δ + μ  k32 )(Δ + k32 ). By the definition of D ω (x, y), ω ∗ Dαβ (x, y) = Aω αβ (∂x )t(x, y) (1)

= δαβ h2 μ (λ + 2 μ )(k12 − k32 )(k22 − k32 )b3 H0 (k3 |x − y|)  ∂2 (1) − h2 μ (λ + μ ) (k32 − k12 )b1 H (k1 |x − y|) ∂ xα ∂ xβ 0 + (k32 − k22 )b2  − μ (λ + 2μ ) b1

∂2 (1) H (k2 |x − y|) ∂ xα ∂ xβ 0



∂2 ∂2 (1) (1) H0 (k1 |x − y|) + b2 H (k2 |x − y|) ∂ xα ∂ xβ ∂ xα ∂ xβ 0  ∂2 (1) + b3 H (k3 |x − y|) . ∂ xα ∂ xβ 0

Taking (2.14)–(2.16) into account, we find that

14

2 Layer Potentials

i i ∂2 (1) (1) H0 (k3 |x − y|) + 2 2 H (k3 |x − y|) 2 4h μ 4h μ k3 ∂ xα ∂ xβ 0   i h2 μ (λ + μ )(k12 − k32 ) − μ (λ + 2 μ ) ∂2 (1) + H (k1 |x − y|) 2 2 2 2 4 2 4h μ (λ + 2 μ )(k1 − k2 )(k1 − k3 ) ∂ xα ∂ xβ 0   i h2 μ (λ + μ )(k22 − k32 ) − μ (λ + 2 μ ) ∂2 (1) + H (k2 |x − y|) 2 2 2 2 4 2 4h μ (λ + 2 μ )(k2 − k1 )(k2 − k3 ) ∂ xα ∂ xβ 0  i ∂2 ∂2 (1) (1) = 2 2 − α12 H0 (k1 |x − y|) − α22 H (k2 |x − y|) ∂ xα ∂ xβ ∂ xα ∂ xβ 0 4h μ k3  ∂2 (1) 2 (1) + H (k3 |x − y|) + δαβ k3 H0 (k3 |x − y|) , ∂ xα ∂ xβ 0

ω Dαβ (x, y) = δαβ

where

  k32 λ +μ 1 = 2 − , k2 − k12 λ + 2μ h2 (k12 − k32 )   k2 λ +μ 1 α22 = 2 3 2 − 2 2 . k1 − k2 λ + 2μ h (k2 − k32 )

α12

Now, using (2.7) and (2.26), we deduce that   k32 k22 − k32 2  α1 = 2 1−μ + k2 − k12 k32 =

k22 − μ  k32 1 2  2 2 2 (k − μ k + k − k ) = , 3 2 3 k22 − k12 3 k22 − k12

as required. Similarly, it can be shown that

α22 =

k12 − μ  k32 . k12 − k22

Next, ∗ Dαω3 (x, y) = −D3ωα (x, y) = Aω α 3 (∂x )t(x, y)

=

ih2 μ 2 (k32 − k12 ) ∂ (1) H0 (k1 |x − y|) 2 2 2 2 4 2 ∂ x 4h μ (λ + 2μ )(k1 − k2 )(k1 − k3 ) α

ih2 μ 2 (k32 − k22 ) ∂ (1) H (k2 |x − y|) 2 2 2 2 4 2 4h μ (λ + 2μ )(k2 − k1 )(k2 − k3 ) ∂ xα 0   i ∂ ∂ (1) (1) = 2 2 −γ H (k1 |x − y|) + γ H (k2 |x − y|) , ∂ xα 0 ∂ xα 0 4h μ k3 +

where γ is defined by (2.24).

2.1 Fundamental and Singular Solutions

15

Finally, ω ∗ D33 (x, y) = Aω 33 (∂x )t(x, y)

=

ih4 μ (λ + 2μ )(μ  k32 − k12 )(k32 − k12 ) (1) H (k1 |x − y|) 4h4 μ 2 (λ + 2 μ )(k12 − k22 )(k12 − k32 ) 0

ih4 μ (λ + 2μ )(μ k32 − k22 )(k32 − k22 ) (1) H (k2 |x − y|) 4h4 μ 2 (λ + 2μ )(k22 − k12 )(k22 − k32 ) 0   i (1) (1) = 2 2 β12 H0 (k1 |x − y|) + β22H0 (k2 |x − y|) , 4h μ k3 +

where β12 and β22 are defined by (2.25). This completes the proof.   The representation of D ω (x, y) given in Theorem 2.1 is sufficient for our purposes. There is no need to calculate explicitly the first-order and second-order derivatives of the Hankel functions in (2.20)–(2.22). To simplify the notation, we write r = |x − y|. Since

and

∂ ∂ f (r) = − f (r) ∂ xα ∂ yα ∂2 ∂2 f (r) = f (r), ∂ xα ∂ xβ ∂ yα ∂ yβ

from (2.20)–(2.22) it is easily seen that  T D ω (x, y) = D ω (y, x) .

(2.27)

We introduce the matrix of singular solutions  T P ω (x, y) = T (∂y )D ω (y, x) .

(2.28)

Theorem 2.2. Each column of D ω (x, y) and P ω (x, y) satisfies the homogeneous system Aω (∂x )u(x) = 0 at all points x ∈ R 2 , x = y. Proof. From (2.8), (2.5), and (2.9) we see that for x = y,   Aω (∂x )D ω (x, y) = Aω (∂x )Aω ∗ (∂x ) t(x, y)E3   = (det Aω )(∂x )t(x, y) E3 = 0. Using (2.28) and (2.27), we find that for x = y,   ω ω ω Aω ik (∂x )Pk j (x, y) = Aik (∂x ) T (∂y )D (y, x) jk ω = Aω ik (∂x )T jm (∂y )Dmk (y, x) ω = T jm (∂y )Aω ik (∂x )Dkm (x, y) = 0,

which proves the assertion.  

16

2 Layer Potentials

2.2 Order of Singularity The investigation of the singularities of D ω and P ω plays an important role in the study of the behavior of the single-layer and double-layer plate potentials. It is known [1] that, as ξ → 0,

 2i 1 2 1 4 (1) H0 (ξ ) = 1 − ξ + ξ − · · · ln ξ ; π 4 64 so, as r → 0, from (2.10) we see that t(x, y) = t(r) =

3

(1)

∑ b j H0

(k j r)

j=1

 1 2 2 1 4 4 b 1 − k r + k r − · · · ln(k j r) j ∑ 4 j 64 j j=1

 2i 3 1 1 = ∑ b j 1 − k2j r2 + k4j r4 ln r + d + O(r6 ln r), π j=1 4 64 =

3

2i π

where d is a constant. From (2.14) and (2.15) it follows that 3

3

3

j=1

j=1

j=1

i

∑ b j = ∑ b j k2j = 0, ∑ b j k4j = 4h4 μ 2(λ + 2μ ) ,

which leads to t(r) = c1 r4 ln r + d + O(r 6 ln r), where c1 = −

1 . 128π h 4 μ 2 (λ + 2μ )

(2.29) (2.30)

Theorem 2.3. As r → 0,

 (xα − yα )(xβ − yβ ) 1 D ω (x, y) = ln r d1 Eγγ − E33 + d2 Eαβ 2π μ r2 + C + D˜ ω (x, y),

(2.31)

where C is a constant (3 × 3)-matrix, d1 = −

λ + 3μ , + 2μ )

4π h2 μ (λ

λ +μ d2 = , 4π h 2 μ ( λ + 2 μ ) and D˜ ω (x, y) is a (3 × 3)-matrix whose elements are O(r ln r) as r → 0.

(2.32)

2.2 Order of Singularity

17

Proof. By (2.8), (2.17), (2.29), and (2.30), we find that ω ∗ Dαβ (x, y) = Aω αβ (∂x )t(x, y)

∂2 Δ (c1 r4 ln r) ∂ xα ∂ xβ   ∂2 + ρω 2 h2 (λ + 3μ )δαβ Δ (c1 r4 ln r) − ρω 2 h2 (λ + μ ) + μ 2 (c1 r4 lnr) ∂ xα ∂ xβ

= h2 μ (λ + 2μ )δαβ Δ Δ (c1 r4 ln r) − h2 μ (λ + μ )

 + ρω 2 (ρω 2 h2 − μ )δαβ c1 r4 ln r +Cαβ + O(r2 ln r)

= 64c1h2 μ (λ + 2 μ )δαβ ln r − c1 h2 μ (λ + μ )

∂2 (16r 2 lnr + 8r 2 ) ∂ xα ∂ xβ

 + Cαβ + O(r2 ln r);

hence, ω Dαβ (x, y) = 64c1 h2 μ (λ + 2μ )δαβ ln r   (xα − yα )(xβ − yβ ) − c1 h2 μ (λ + μ ) 32 + 32 δ ln r αβ r2

+ Cαβ + O(r2 ln r) =−

(xα − yα )(xβ − yβ ) λ + 3μ λ +μ δ lnr + 4π h2 μ (λ + 2 μ ) αβ 4π h2 μ (λ + 2μ ) r2 + Cαβ + O(r2 ln r).

Also, using (2.18) and (2.19), we see that ∗ Dαω3 (x, y) = −D3ωα (x, y) = Aω α 3 (∂x )t(x, y)

∂ ∂ Δ (c1 r4 ln r) + μ (ρω 2h2 − μ ) (c1 r4 ln r) ∂ xα ∂ xα +Cα 3 + O(r3 ln r) = Cα 3 + O(r ln r) = h2 μ 2

and ω ∗ D33 (x, y) = Aω 33 (∂x )t(x, y)

= h4 μ (λ + 2μ )Δ Δ (c1 r4 ln r) + h2 (λ + 3μ )(ρω 2 h2 − μ )Δ (c1 r4 ln r)  + (ρω 2 h2 − μ )2 c1 r4 ln r + C33 + O(r2 lnr)

= 64c1 h4 μ (λ + 2 μ ) lnr + C33 + O(r2 lnr) 1 =− ln r +C33 + O(r2 ln r). 2π μ The assertion is proved.  

18

2 Layer Potentials

We introduce the alternating symbol ε αβ = β − α for α , β ∈ {1, 2}. ∂ ∂ In what follows, and denote the derivatives in the normal and tan∂ ν (y) ∂ s(y) gential directions, respectively. Theorem 2.4. As r → 0,      1 ∂ ∂ ω  P (x, y) = − μ εαβ lnr Eαβ + ln r E3 2π ∂ s(y) ∂ ν (y)   ∂ (xα − yα )(xβ − yβ )   − (λ + μ )εαγ Eγβ ∂ s(y) r2    1 ∂  − (xα − yα ) ln r E3α 2 ∂ ν (y)     1 ∂  1  + εαβ (xα − yα ) ln r λ E3β + 2 Eβ 3 2 ∂ s(y) h   1 1 + να (y) 2 Eα 3 + 2(λ  + μ  )E3α 2 h + D + P˜ ω (x, y), (2.33) where D is a constant (3 × 3)-matrix,

λ =

λ , (λ + 2μ )

μ  is defined by (2.26), and P˜ ω (x, y) is a (3 × 3)-matrix whose elements are O(r ln r) as r → 0. Proof. From (2.28)–(2.30) we obtain ω ω ω ω P11 (x, y) = T11 (∂y )D11 (y, x) + T12 (∂y )D21 (y, x) + T13 (∂y )D31 (y, x)     2 ω∗ 2 ω∗ = h (λ + 2μ )ν1 A11 (∂y )t(y, x) ,1 +h μν2 A11 (∂y )t(y, x) ,2  ∗   ω∗  2 + h2 μν2 Aω 21 (∂y )t(y, x) ,1 +h λ ν1 A21 (∂y )t(y, x) ,2   = h2 (λ + 2μ )ν1 h2 μ (λ + 2μ )Δ Δ t,1 −h2 μ (λ + μ )Δ t,111   + h2 μν2 h2 μ (λ + 2μ )Δ Δ t,2 −h2 μ (λ + μ )Δ t,112   + h2 μν2 − h2 μ (λ + μ )Δ t,112   + h2 λ ν1 − h2 μ (λ + μ )Δ t,122 + D11 + O(r ln r)

= h4 μ (λ + μ )(λ + 2μ )ν1 Δ Δ t,1 +h4 μ 2 (λ + 2μ )ν1Δ Δ t,1 + h4 μ 2 (λ + 2μ )ν2 Δ Δ t,2 −2h4 μ 2 (λ + μ )ν1 Δ t,111 − 2h4 μ 2 (λ + μ )ν2 Δ t,112 −h4 λ μ (λ + μ )ν1 Δ t,111 − h4 λ μ (λ + μ )ν1 Δ t,122 +D11 + O(r ln r); so,

2.2 Order of Singularity

19

ω P11 (x, y)

= h4 μ 2 (λ + 2μ )

∂ Δ Δ t + 2h4 μ 2 (λ + μ )ν1 Δ t,111 ∂ ν (y)

− 2h4 μ 2 (λ + μ )ν1 Δ t,111 +2h4 μ 2 (λ + μ )ν1 Δ t,122 −2h4 μ 2 (λ + μ )ν2 Δ t,112 + D11 + O(r ln r)

∂ ∂ Δ Δ t + 2h4 μ 2 (λ + μ ) Δ t,12 +D11 + O(r ln r) ∂ ν (y) ∂ s(y) ∂ ∂ (x1 − y1)(x2 − y2 ) = 64h4 μ 2 (λ + 2μ )c1 ln r + 64h4 μ 2 (λ + μ )c1 ∂ ν (y) ∂ s(y) r2 + D11 + O(r ln r), = h4 μ 2 (λ + 2μ )

which means that ω P11 (x, y) = −

1 ∂ 1 ∂ (x1 − y1 )(x2 − y2 ) ln r − (λ  + μ  ) 2π ∂ ν (y) 2π ∂ s(y) r2 + D11 + O(r ln r).

Analogous manipulation yields ω P12 (x, y) = −h4 λ μ 2

∂ ∂ Δ t,11 +h4 μ 2 (λ + 2μ ) Δ t,22 +D12 + O(r lnr); ∂ s(y) ∂ s(y)

hence,   λ ∂ (x2 − y2 )2 − + lnr 4π (λ + 2μ ) ∂ s(y) r2   1 ∂ (x2 − y2 )2 − + lnr + D12 + O(r ln r) 4π ∂ s(y) r2 1 ∂ 1 ∂ (x2 − y2 )2 = − μ ln r − (λ  + μ  ) 2π ∂ s(y) 2π ∂ s(y) r2 + D12 + O(r ln r).

ω P12 (x, y) =

Similarly, we find that

∂ Δ t,2 +D13 + O(r2 ln r) ∂ s(y)  1 ∂  1 =− (x2 − y2 ) ln r − ν1 (y) + D13 + O(r2 ln r), 2 4π h ∂ s(y) 4π h 2 ∂ ∂ ω P21 (x, y) = −h4 μ 2 (λ + 2μ ) Δ t,11 +h4 λ μ 2 Δ t,22 +D21 + O(r lnr) ∂ s(y) ∂ s(y) 1  ∂ 1 ∂ (x1 − y1 )2 = μ ln r + (λ  + μ  ) + D21 + O(r ln r), 2π ∂ s(y) 2π ∂ s(y) r2 ω P13 (x, y) = h2 μ 2 (λ + 2μ )

20

2 Layer Potentials

∂ ∂ Δ Δ t − 2h4 μ 2 (λ + μ ) Δ t,12 ∂ ν (y) ∂ s(y) + D22 + O(r lnr) 1 ∂ 1 ∂ (x1 − y1 )(x2 − y2 ) =− lnr + (λ  + μ  ) 2π ∂ ν (y) 2π ∂ s(y) r2 + D22 + O(r lnr),

ω P22 (x, y) = h4 μ 2 (λ + 2μ )

∂ Δ t,1 +D23 + O(r2 lnr) ∂ s(y)  1 ∂  1 = (x1 − y1 ) ln r − ν2 (y) + D23 + O(r2 ln r), 2 4π h ∂ s(y) 4π h 2

ω P23 (x, y) = −h2 μ 2 (λ + 2μ )

∂ ∂ Δ t,1 +h4 λ μ 2 Δ t,2 +D31 + O(r2 ln r) ∂ ν (y) ∂ s(y)   1 ∂  1  ∂  =− (x1 − y1 ) ln r − λ (x2 − y2 ) ln r 4π ∂ ν (y) 4π ∂ s(y) 1 − (λ  + μ  )ν1 (y) + D31 + O(r2 ln r), 2π ∂ ∂ ω P32 (x, y) = −h4 λ μ 2 Δ t,1 +h4 μ 2 (λ + 2μ ) Δ t,2 +D32 + O(r2 ln r) ∂ s(y) ∂ ν (y)  1  1  ∂  ∂  = λ (x1 − y1 ) ln r − (x2 − y2 ) ln r 4π ∂ s(y) 4π ∂ ν (y) 1 − (λ  + μ  )ν2 (y) + D32 + O(r2 ln r), 2π ∂ ω P33 (x, y) = h4 μ 2 (λ + 2μ ) Δ Δ t + D33 + O(r lnr) ∂ ν (y) 1 ∂ =− lnr + D33 + O(r ln r), 2π ∂ ν (y) ω P31 (x, y) = h4 μ 2 (λ + 2μ )

as required.   Remark 2.1. Expansions (2.31) and (2.33) of D ω (x, y) and P ω (x, y), respectively, for y close to x coincide with those of the corresponding matrices arising in the equilibrium bending of plates, which can be found in [14].

2.3 Properties of the Potentials We define the single-layer potential (V ω ϕ )(x) =

 ∂S

D ω (x, y)ϕ (y) ds(y)

(2.34)

2.3 Properties of the Potentials

and the double-layer potential (W ω ϕ )(x) =

21



P ω (x, y)ϕ (y) ds(y),

(2.35)

∂S

where ϕ is a (3 × 1)-vector function known as the density. The properties of these potentials play an important role in formulating suitable integral equations for various boundary value problems. We are especially concerned with the behavior of (V ω ϕ )(x) and (W ω ϕ )(x) as x approaches ∂ S. This requires a very detailed analysis based on the expansions (2.31) and (2.33). As was noted in Remark 2.1, the singularities of D ω and P ω are exactly the same as those occurring in the corresponding matrices in the equilibrium bending of plates. The behavior of the corresponding potentials is thoroughly investigated in [14], and many relevant properties are simply quoted from there. We denote by C 0,α (X) the vector space of H¨older continuous functions (with index α ) on X, and by C 1,α (X) the subspace of C 1 (X) of functions whose firstorder derivatives belong to C 0,α (X ). In the sequel, when we mention that a function ϕ ∈ C 0,α (X ), α ∈ (0, 1), satisfies a certain equation or condition, we mean that ϕ ∈ C 0,α (X) for all α in the interval (0, 1). The next three assertions follow from Theorem 2.2 and the results in [14]. Theorem 2.5. (i) If ϕ ∈ C(∂ S), then V ω ϕ and W ω ϕ are analytic and satisfy Aω (∂x )u(x) = 0 in S + ∪ S− . (ii) If ϕ ∈ C 0,α (∂ S), α ∈ (0, 1), then the direct values V0ω ϕ and W0ω ϕ of V ω ϕ and W ω ϕ on ∂ S exist (the latter in the sense of principal value). Theorem 2.6. If ϕ ∈ C 0,α (∂ S), α ∈ (0, 1), then the functions   V ω + (ϕ ) = (V ω ϕ )S¯+ , V ω − (ϕ ) = (V ω ϕ )S¯−

(2.36)

are of class C ∞ (S+ ) ∩C 1,α (S¯+ ) and C ∞ (S− ) ∩C 1,α (S¯− ), respectively, and  T V ω + (ϕ ) = W0ω ∗ + 12 I ϕ on ∂ S, (2.37)  ω∗ 1 ω− T V (ϕ ) = W0 − 2 I ϕ on ∂ S, (2.38) where W0ω ∗ is the adjoint of W0ω and I is the identity operator. Theorem 2.7. If ϕ ∈ C 1,α (∂ S), α ∈ (0, 1), then the functions   (W ω ϕ )S+ in S+ , ω+ W (ϕ ) =  ω 1 W0 − 2 I ϕ on ∂ S,   (W ω ϕ )S− in S− , ω− W (ϕ ) =  ω 1 W0 + 2 I ϕ on ∂ S,

(2.39) (2.40)

are of class C ∞ (S+ ) ∩C 1,α (S¯+ ) and C ∞ (S− ) ∩C 1,α (S¯− ), respectively, and T W ω + (ϕ ) = T W ω − (ϕ )

on ∂ S.

(2.41)

22

2 Layer Potentials

Remark 2.2. Theorems 2.5–2.7 are used in Chapters 6–11 to rigorously justify the construction of regular solutions of the fundamental boundary value problems in S + and S − in the form of layer potentials.

Chapter 3

The Nonhomogeneous System

System (1.10) is not straightforward to analyze because of the nonhomogeneous term on the right-hand side. In this chapter we construct a particular solution of the system in terms of a domain potential whose kernel is the matrix of fundamental solutions for the operator A ω (∂x ) introduced in Chapter 2. It turns out that the results are also applicable to the system A(∂ x )u(x) = 0 that characterizes the equilibrium bending of thin elastic plates with transverse shear deformation. This is due to the fact that the analytic argument is based on the singularities of the matrix of fundamental solutions as r → 0, which, as was noted in Remark 2.1, coincide with those of the corresponding matrix for the operator A(∂ x ). The results pertaining to the equilibrium system are required later when we prove the existence of nonzero solutions of homogeneous boundary value problems associated with stationary plate oscillations. We start by considering the continuity of the first-order derivatives of the domain potential (see Theorems 3.1 and 3.2). In Theorems 3.3–3.5 we investigate the behavior of the second-order derivatives. Finally, in Theorem 3.6 these results are combined to show that the potential is indeed a sufficiently regular solution of the nonhomogeneous system.

3.1 The Newtonian Potential We intend to construct a particular solution of the system B(∂x )u(x) + f (x) = 0 in S + or S− . Here, the matrix operator B(∂ x ) represents either A ω (∂x ), defined by (1.10), or A(∂x ) = A(∂ /∂ x1 , ∂ /∂ x2 ), given by (1.11). Consider the integral K(x) =



k(x, y) f (y) da(y),

(3.1)

S

G.R. Thomson and C. Constanda, Stationary Oscillations of Elastic Plates, A Boundary Integral Equation Analysis, DOI 10.1007/978-0-8176-8241-5_3, © Springer Science+Business Media, LLC 2011

23

24

3 The Nonhomogeneous System

where S denotes S + or S − and k(x, y) is either D ω (x, y) or D(x, y), the matrix of fundamental solutions for the operator A(∂ x ) introduced in [14]. Such functions are known as Newtonian potentials (see, for example, [27]). For simplicity, we consider only the case of the interior domain S + . The case of the exterior domain is treated in exactly the same way, with additional restrictions on f (x) as |x| → ∞ to ensure that K(x) exists as an improper integral. From Theorem 2.3 and Remark 2.1 it follows that, as r → 0,

 (xα − yα )(xβ − yβ ) 1 k(x, y) = ln r d1 Eγγ − E33 + d2 Eαβ 2π μ r2 + C + O(r lnr),

(3.2)

where C is a constant (3 × 3)-matrix and d 1 and d2 are defined by (2.32). Throughout this chapter we differentiate the series expansion (3.2) to obtain the expansions of the derivatives of the matrices of fundamental solutions. This operation is justified by the properties of the series expansions for the Hankel functions. In what follows, σ (a, r) is the disk with the center at a, radius r, and circular boundary ∂ σ (a, r). We introduce the integrals Iαβ =



(xα − yα )(xβ − yβ ) ds(y) = πδαβ ,

(3.3)

∂ σ (x,1)

Iαβ ρη =



(xα − yα )(xβ − yβ )(xρ − yρ )(xη − yη ) ds(y)

∂ σ (x,1)

=

π (δ δρη + δαρ δβ η + δαη δβ ρ ). 4 αβ

(3.4)

3.2 Smoothness Properties The function f needs to satisfy certain restrictions to ensure that the domain integral K(x) has the necessary smoothness.

3.2.1 Existence of the First-Order Derivatives The first assertion addresses the differentiability of the Newtonian potential K. Theorem 3.1. If f ∈ L∞ (S+ ), then and

∂ K(x) = ∂ xα

∂ K(x) exists at each point x ∈ S + (x ∈ ∂ S) ∂ xα



S+

∂ k(x, y) f (y) da(y). ∂ xα

3.2 Smoothness Properties

25

Proof. Consider the function 

Kδ (x) =

kδ (x, y) f (y) da(y),

S+

where

⎧ ⎪ ⎪ ⎨0,     kδ (x, y) = 12 sin π δ −1 r − 32 + 1 k(x, y), ⎪ ⎪ ⎩k(x, y),

0 ≤ r ≤ δ,

δ < r < 2δ , 2δ ≤ r < ∞.

It is clear that kδ (x, y) has first-order derivatives with respect to x ∈ S + (x ∈ ∂ S). ∂ Hence, K (x) obviously exists at each point x ∈ S + (x ∈ ∂ S) and ∂ xα δ

∂ K (x) = ∂ xα δ

 S+

∂ k (x, y) f (y) da(y), ∂ xα δ

where the kernel of the integral satisfies

∂ k (x, y) = 0, 0 ≤ r < δ , ∂ xα δ      ∂ π xα − yα kδ (x, y) = cos π δ −1 r − 32 k(x, y) ∂ xα 2δ r      ∂ + 12 sin π δ −1 r − 32 + 1 k(x, y), ∂ xα ∂ ∂ k (x, y) = k(x, y), 2δ < r < ∞. ∂ xα δ ∂ xα

δ < r < 2δ ,

Next,   Kδ (x) − K(x)       =  kδ (x, y) − k(x, y) f (y) da(y) S+

  =  −



k(x, y) f (y) da(y)

S+ ∩σ (x,δ )

1 + 2 ≤  f ∞



S+ ∩σ (x,2δ )\S+ ∩σ (x,δ )



σ (x,δ )



    −1   3 sin π δ r − 2 − 1 k(x, y) f (y) da(y)

|k(x, y)| da(y) +  f ∞



σ (x,2δ )\σ (x,δ )

|k(x, y)| da(y) → 0,

26

3 The Nonhomogeneous System

uniformly as δ → 0, since k(x, y) has only a logarithmic singularity; that is, its behavior is similar to δ

r ln r dr =

1 2

r2 lnr − 14 r2

δ 0

→ 0 as δ → 0.

0

Also,      ∂ ∂   k (x, y) f (y) da(y) − k(x, y) f (y) da(y)  ∂ xα δ  ∂ xα + + S S    ∂ =  − k(x, y) f (y) da(y) ∂ xα S+ ∩σ (x,δ )     π xα − yα + cos π (δ −1 r − 32 ) k(x, y) 2δ r S+ ∩σ (x,2δ )\S+ ∩σ (x,δ )       1 ∂ −1 3 + sin π (δ r − 2 ) − 1 k(x, y) f (y) da(y), 2 ∂ xα which means that      ∂ ∂    ∂ xα kδ (x, y) f (y) da(y) − ∂ xα k(x, y) f (y) da(y) S+ S+      ∂ k(x, y) da(y) ≤  f ∞  ∂ xα  σ (x,δ )     xα − yα  π   da(y) +  f ∞ k(x, y)   2δ r σ (x,2δ )\σ (x,δ )     ∂    da(y) +  f ∞ k(x, y)  ∂ xα  σ (x,2δ )\σ (x,δ )      xα − yα  1 π   da(y) ≤ c f ∞ da(y) +  f ∞ k(x, y)   r 2δ r σ (x,δ )

+ c f ∞



σ (x,2δ )\σ (x,δ )

σ (x,2δ )\σ (x,δ )

1 da(y) → 0, r

uniformly as δ → 0. The second integral tends to zero because, as δ → 0, it is of the same order as 2δ 2δ 1 1 1 2 r ln r dr = r ln r − 14 r2 δ → 0. 2 δ δ δ

The assertion now follows from a well-known theorem of real analysis.  

3.2 Smoothness Properties

27

3.2.2 H¨older Continuity of the First-Order Derivatives Before discussing the H¨older continuity of the derivatives of K, we require a preliminary result. Lemma 3.1. If β ∈ (0, 1], then there is a constant c such that |x − x |β ln

1 ≤ c|x − x |α , |x − x |

0 < α < β ≤ 1,

x , x ∈ R2 .

Proof. Consider the function h defined by 1 γ t + t lnt, 1−γ

h(t) =

(3.5)

where γ ∈ (0, 1) and t ∈ [0, ∞). We can write h in the form h(t) = t γ g(t), with g(t) =

(3.6)

1 + t 1−γ lnt. 1−γ

Obviously, h(0) = 0. We now show that g(t) > 0 on the interval 0 ≤ t < ∞. The function g has a turning point if and only if 0 = g (t) = (1 − γ )t −γ lnt + t −γ , which occurs when

t = e1/(γ −1) .

Differentiating g a second time yields g (t) = −γ (1 − γ )t −γ −1 lnt + (1 − γ )t −γ −1 − γ t −γ −1 ; so, since γ < 1,  1 g e1/(γ −1) = −γ (1 − γ )e(γ +1)/(1−γ ) γ −1 + (1 − γ )e(γ +1)/(1−γ ) − γ e(γ +1)/(1−γ ) = (1 − γ )e(γ +1)/(1−γ ) > 0. Also,

 g e1/(γ −1) =

1 1 1 + e−1 = 1−γ γ −1 1−γ

1−

1 e

 > 0.

Hence, g has a minimum turning point at t = e 1/(γ −1) , and g(e1/(γ −1) ) > 0. Since this is the only turning point of the function, g(t) > 0 on the interval 0 ≤ t < ∞. Therefore, h(t) given by (3.6) has exactly one root, which occurs at t = 0. Now,

28

3 The Nonhomogeneous System

since

 h e1/(γ −1) =

 1 γ /(γ −1) 1 e 1− > 0, 1−γ e

we find that h(t) ≥ 0 on the interval 0 ≤ t < ∞. Replacing t by |x  − x | in (3.5) yields 1 |x − x |γ + |x − x | ln |x − x | ≥ 0 1−γ for any γ ∈ (0, 1). Next, we set c =

1 and find that 1−γ |x − x | ln

1 ≤ c|x − x |γ , |x − x |

which implies that |x − x |β ln

1 ≤ c|x − x |α , |x − x |

0 < α < β ≤ 1.

This proves the assertion.   Theorem 3.2. If f ∈ L∞ (S+ ), then K ∈ C 1,α (∂ S), α ∈ (0, 1). Proof. By Theorem 3.1, L(x) =

∂ K(x) = ∂ xη



l(x, y) f (y) da(y)

on ∂ S,

S+

 ∂ k(x, y). By (3.2), l(x, y) = O r−1 as r → 0. ∂ xη Let x , x ∈ ∂ S and ξ = |x − x |. Then

where l(x, y) =

L(x ) − L(x ) = J1 (x , x ) + J2 (x , x ) + J3(x , x ), where 

J1 (x , x ) =

l(x , y) f (y) da(y),

S+ ∩σ (x ,2ξ )



J2 (x , x ) = −

l(x , y) f (y) da(y),

S+ ∩σ (x ,2ξ )

J3 (x , x ) =





 l(x , y) − l(x , y) f (y) da(y).

S+ \σ (x ,2ξ )

From the singularity of l(x, y) it follows that

(3.7)

3.2 Smoothness Properties

29



|J1 | ≤

|l(x , y)|| f (y)| da(y)

S+ ∩σ (x ,2ξ )

≤ c1  f ∞



1

σ (x ,2ξ )

which means that

|x − y|

2ξ

da(y) = 2π c1  f ∞ [ρ ]0 ,

|J1 | ≤ c2 |x − x |.

(3.8)

Similarly, 

|J2 | ≤

|l(x , y)|| f (y)| da(y)

S+ ∩σ (x ,2ξ )

≤ c1  f ∞



σ (x ,3ξ )

so

1 3ξ da(y) = 2π c1  f ∞ [ρ ]0 , |x − y|

|J2 | ≤ c3 |x − x |.

(3.9)

The estimate of J3 is not so straightforward. We start with the inequality |J3 | ≤



|l(x , y) − l(x , y)| | f (y)| da(y).

S+ \σ (x ,2ξ )

By the Mean Value Theorem,    ∂  |l(x , y) − l(x , y)| ≤ |xβ − xβ |  l(x , y) ∂ xβ ≤ c4 |x − x | |x − y|−2 , where x lies between x and x . Also, for y ∈ S + \σ (x , 2ξ ), |x − y| ≥ |x − y| − |x − x | > |x − y| − 12 |x − y| = 12 |x − y|. Therefore, when y ∈ S + \σ (x , 2ξ ), we have |l(x , y) − l(x , y)| ≤ c5 |x − x | |x − y|−2 , which implies that |J3 | ≤ c5  f ∞ |x − x |



S+ \σ (x ,2ξ )

1 da(y); |x − y|2

so, by extending the domain of integration, we see that

30

3 The Nonhomogeneous System



|J3 | ≤ c5  f ∞ |x − x |

σ (x ,M)\σ (x ,2ξ )

1 da(y), |x − y|2

where M is the largest distance between any two points in S¯+ . Consequently,

 M     |J3 | ≤ 2π c5  f ∞ |x − x | ln − ln|x − x | 2 1     = c6 |x − x | + c7 |x − x | ln  , |x − x | and, using Lemma 3.1, we now conclude that |J3 | ≤ c6 |x − x | + c8 |x − x |α , hence,

α ∈ (0, 1);

|J3 | ≤ c9 |x − x |α .

(3.10)

Combining (3.7)–(3.10) yields the required result; that is, |L(x ) − L(x )| ≤ c|x − x |α ,

α ∈ (0, 1),

where c is a constant independent of x  and x .  

3.2.3 Existence of the Second-Order Derivatives The investigation of the second-order derivatives of K is based on another auxiliary assertion. Theorem 3.3. If f ∈ C 0,α (S+ ), α ∈ (0, 1], then the integral M(x) =



m(x, y) f (y) da(y),

S+

∂2 k(x, y), exists in the sense of principal value at every point ∂ xα ∂ xβ x ∈ S+ and is given by the formula where m(x, y) =

M(x) =

 S+

m(x, y)[ f (y) − f (x)] da(y)   + m(x, y) da(y) S+ \σ (x,δ )

+ lim



ε →0 σ (x,δ )\σ (x,ε )

where δ > ε and σ (x, δ ) ⊆ S + .

 m(x, y) da(y) f (x),

(3.11)

3.2 Smoothness Properties

31

Proof. The principal value of the integral is computed as 

lim

ε →0 S+ \σ (x,ε )

m(x, y) f (y) da(y)





= lim

ε →0

m(x, y)[ f (y) − f (x)] da(y)

S+ \σ (x,ε )



+



m(x, y) da(y) +

S+ \σ (x,δ )



 m(x, y) da(y) f (x) ,



σ (x,δ )\σ (x,ε )

where δ is chosen so that δ > ε and σ (x, δ ) ⊆ S + . The first integral on the right-hand side of this expression converges since the integrand is O(r α −2 ), α ∈ (0, 1]; hence, 

lim

ε →0 S+ \σ (x,ε )

=



S+

+

m(x, y) f (y) da(y)

m(x, y)[ f (y) − f (x)] da(y) 



m(x, y) da(y) + lim

S+ \σ (x,δ )



ε →0 σ (x,δ )\σ (x,ε )

 m(x, y) da(y) f (x).

(3.12)

We consider the third integral and show that its limit as ε → 0 exists. By (3.2),

∂ kρη (x, y) = d1 δρη r−2 (xβ − yβ ) ∂ xβ    + d2 r−2 δρβ (xη − yη ) + δηβ (xρ − yρ )

 − 2r −4 (xρ − yρ )(xη − yη )(xβ − yβ ) + O(lnr)

and

∂2 kρη (x, y) ∂ xα ∂ xβ   = d1 δρη δαβ r−2 − 2r −4 (xα − yα )(xβ − yβ )   + d2 r−2 (δρβ δαη + δηβ δαρ ) − 2r−4 δρβ (xα − yα )(xη − yη ) + δηβ (xα − yα )(xρ − yρ )

  − 2d2 r−4 δαρ (xη − yη )(xβ − yβ ) + δαη (xρ − yρ )(xβ − yβ )



 + δαβ (xρ − yρ )(xη − yη )  − 4r −6 (xα − yα )(xβ − yβ )(xρ − yρ )(xη − yη ) + O(r−1 ).

32

3 The Nonhomogeneous System

From (3.2) we also deduce that

∂2 kρ 3 (x, y) = O(r−1 ) ∂ xα ∂ xβ and

∂ 1 −2 k33 (x, y) = − r (xβ − yβ ) + O(r lnr), ∂ xβ 2π μ

which leads to

∂2 k33 (x, y) ∂ xα ∂ xβ  1  =− δαβ r−2 − 2r−4 (xα − yα )(xβ − yβ ) + O(lnr); 2π μ therefore, 

mρη (x, y) da(y)

σ (x,δ )\σ (x,ε )

= ln

δ d1 (2πδρη δαβ − 2δρη Iαβ ) + 2π d2(δρβ δαη + δηβ δαρ ) ε  − 2d2(δρβ Iαη + δηβ Iαρ + δαρ Iηβ + δαη Iρβ + δαβ Iρη ) + 8d2 Iαβ ρη 

+

O(r−1 ) da(y).

σ (x,δ )\σ (x,ε )

Taking (3.3) and (3.4) into account, we find that 

mρη (x, y) da(y)

σ (x,δ )\σ (x,ε )

= ln

δ 2π d1 (δρη δαβ − δρη δαβ ) + 2π d2(δρβ δαη + δηβ δαρ ) ε − 2π d2(2δρβ δαη + 2δηβ δαρ + δαβ δρη )  + 2π d2(δαβ δρη + δαρ δβ η + δαη δβ ρ ) +



O(r−1 ) da(y);

σ (x,δ )\σ (x,ε )

hence,

 σ (x,δ )\σ (x,ε )

from which we see that

mρη (x, y) da(y) =

 σ (x,δ )\σ (x,ε )

O(r−1 ) da(y),

3.2 Smoothness Properties

33



lim

ε →0 σ (x,δ )\σ (x,ε )

mρη (x, y) da(y) exists.

Similarly, 



mρ 3 (x, y) da(y) =

σ (x,δ )\σ (x,ε )

σ (x,δ )\σ (x,ε )



=

∂2 kρ 3 (x, y) da(y) ∂ xα ∂ xβ O(r−1 ) da(y),

σ (x,δ )\σ (x,ε )



so lim

ε →0 σ (x,δ )\σ (x,ε )

mρ 3 (x, y) da(y) exists.

Finally, 

m33 (x, y) da(y)

σ (x,δ )\σ (x,ε )



=

σ (x,δ )\σ (x,ε )

δ = ln ε By (3.3),



∂2 k33 (x, y) da(y) ∂ xα ∂ xβ

1 1 − δαβ + I μ π μ αβ

m33 (x, y) da(y) =

σ (x,δ )\σ (x,ε )

 +



O(ln r) da(y).

σ (x,δ )\σ (x,ε )



O(ln r) da(y), σ (x,δ )\σ (x,ε )

which implies that



lim

ε →0 σ (x,δ )\σ (x,ε )

m33 (x, y) da(y) exists.

Therefore, we have shown that 

lim

ε →0 σ (x,δ )\σ (x,ε )

m(x, y) da(y) exists;

hence, from (3.12) we conclude that 

lim

ε →0 S+ \σ (x,ε )

and the assertion is proved.  

m(x, y) da(y) exists,

34

3 The Nonhomogeneous System

Remark 3.1. Since both 

k(x, y) f (y) da(y) and

S+



∂ k(x, y) f (y) da(y) ∂ xα

S+

exist as improper integrals (provided that f is bounded), their principal values obviously exist and coincide with the improper integrals. Thus, we may write 

k(x, y) f (y) da(y)

S+



=

  k(x, y) f (y) − f (x) da(y)

S+





+





k(x, y) da(y) + lim

ε →0 σ (x,δ )\σ (x,ε )

S+ \σ (x,δ )

k(x, y) da(y) f (x)

and  S+

∂ k(x, y) f (y) da(y) ∂ xα =



  ∂ k(x, y) f (y) − f (x) da(y) ∂ xα S+   ∂ + k(x, y) da(y) + lim ε →0 ∂ xα S+ \σ (x,δ )



σ (x,δ )\σ (x,ε )

 ∂ k(x, y) da(y) f (x). ∂ xα

3.2.4 H¨older Continuity of the Second-Order Derivatives In the next assertion, S 0 denotes a bounded domain in R 2 such that S0 ⊂ S+ . Theorem 3.4. If f ∈ C 0,β (S+ ), β ∈ (0, 1), then M ∈ C 0,α (S0 ), where 0 < α < β < 1 and M(x) defined by (3.11) is understood in the sense of principal value. Proof. Let ρ > 0 be the minimum distance between ∂ S and the boundary of S 0 . By Theorem 3.3, M(x) exists for x ∈ S 0 in the sense that M(x) =

 S+









m(x, y) f (y) − f (x) da(y) +

S+ \σ (x,ρ )



+ lim

 m(x, y) da(y) f (x) 

ε →0 σ (x,ρ )\σ (x,ε )

 m(x, y) da(y) f (x).

Since x ∈ S0 , the disk σ (x, ρ ) is contained entirely within S + .

3.2 Smoothness Properties

35

We have already shown in the proof of Theorem 3.3 that the limit of the third integral on the right-hand side exists. Since m is a function of x − y and we are integrating over an annular region with the center at x, this integral is a constant matrix depending on ρ ; that is, it is independent of x. For simplicity, from now on we consider m and f to be scalar functions instead of a matrix-valued function and a vector-valued function, respectively. The function M can be written in the form M(x) = M1 (x) + M2 (x) + M3 (x),

(3.13)

where M1 (x) =



  m(x, y) f (y) − f (x) da(y),

S+



M2 (x) = f (x)

m(x, y) da(y),

S+ \σ (x,ρ )







M3 (x) = lim

ε →0 σ (x,ρ )\σ (x,ε )

m(x, y) da(y) f (x) = cρ f (x).

Let x , x ∈ S0 be such that 1 ξ = |x − x | < ρ . 2 First, we have M1 (x ) − M1 (x ) = J1 (x , x ) + J2(x , x ) + J3 (x , x ) + J4(x , x ), where 

J1 (x , x ) =



  m(x , y) − m(x , y) f (y) − f (x ) da(y)

S+ \σ (x ,ρ )



J2 (x , x ) =

m(x , y) da(y),

  m(x , y) f (y) − f (x ) da(y) 

σ (x ,2ξ )

− 

J3 (x , x ) = f (x ) − f (x ) J4 (x , x ) =



S+ \σ (x ,ρ )





  + f (x ) − f (x )

 σ (x ,ρ )\σ (x ,2ξ )





  m(x , y) f (y) − f (x ) da(y),

σ (x ,2ξ )



m(x , y) da(y),

σ (x ,ρ )\σ (x ,2ξ )



  m(x , y) − m(x , y) f (y) − f (x ) da(y).

(3.14)

36

3 The Nonhomogeneous System

We now estimate these integrals. Thus, 

|J1 (x , x )| ≤

|m(x , y) − m(x , y)| | f (y) − f (x )| da(y)

S+ \σ (x ,ρ )

+ | f (x ) − f (x )|



|m(x , y)| da(y).

S+ \σ (x ,ρ )

By the Mean Value Theorem, 



|m(x , y) − m(x , y)| ≤

|xα

   ∂   m(x , y) ∂ xα

− xα | 

≤ c1 |x − x | |x − y|−3 , where x lies between x and x . Also, for y ∈ S + \σ (x , ρ ), |x − y| ≥ |x − y| − |x − x | > |x − y| − 12 |x − y| = 12 |x − y|, since |x − x | < |x − x | < 12 |x − y|. Therefore, when y ∈ S + \σ (x , ρ ), we have |m(x , y) − m(x , y)| ≤ c2 |x − x | |x − y|−3 ,

(3.15)

from which |J1 (x , x )| ≤ c3 |x − x |



|x − y|−3 |x − y|β da(y)

S+ \σ (x ,ρ )

+ c4 |x − x |β



|x − y|−2 da(y).

S+ \σ (x ,ρ )

These integrals do not pose a problem since x  lies outside the domain of integration and |x − y| is bounded in S + \σ (x , ρ ); hence, |J1 (x , x )| ≤ c5 |x − x |β ,

(3.16)

where c5 is a constant depending on ρ . Estimating J2 , we arrive at 

|J2 (x , x )| ≤

|m(x , y)| | f (y) − f (x )| da(y)

σ (x ,2ξ )

+ ≤ c6



σ (x ,2ξ )



|m(x , y)| | f (y) − f (x )| da(y)

σ (x ,2ξ )

|x − y|β −2 da(y) + c6



σ (x ,2ξ )

|x − y|β −2 da(y).

3.2 Smoothness Properties

37

2ξ

ξ x x

Fig. 2. The disks σ (x  , 2ξ ) and σ (x , 3ξ ). From Figure 2 we see that σ (x  , 2ξ ) ⊂ σ (x , 3ξ ); so, 

|J2 (x , x )| ≤ c6

|x − y|β −2 da(y) + c6

σ (x ,2ξ )

= 2π c 6

 2ξ

|x − y|β −2 da(y)

σ (x ,3ξ )

ϖ

β −1

dϖ +

0

=



3ξ

ϖ

β −1

 dϖ

0

 2π  β  c6 2 |x − x |β + 3β |x − x |β , β

which means that

|J2 (x , x )| ≤ c7 |x − x |β .

Now, by Lemma 3.1,



|J3 (x , x )| ≤ | f (x ) − f (x )|

(3.17)

|m(x , y)| da(y)

σ (x ,ρ )\σ (x ,2ξ )



≤ c8 |x − x |β

|x − y|−2 da(y)

σ (x ,ρ )\σ (x ,2ξ )  β



= 2π c8 |x − x |

ρ 2ξ

 1 ρ 1   β d ϖ = 2π c8 |x − x | ln + ln  ϖ 2 |x − x |

 β

≤ c9 |x − x | + c10 |x − x |α ≤ c11 |x − x |α , 

where 0 < α < β < 1 and c 11 depends on ρ . We go over to J4 . From the assumption on f it follows that 

|J4 (x , x )| ≤

|m(x , y) − m(x , y)| | f (y) − f (x )| da(y)

σ (x ,ρ )\σ (x ,2ξ )

≤ c12



σ (x ,ρ )\σ (x ,2ξ )

|m(x , y) − m(x , y)| |x − y|β da(y).

(3.18)

38

3 The Nonhomogeneous System

It can be shown that (3.15) holds for y ∈ σ (x  , ρ )\σ (x , 2ξ ), so 

|J4 (x , x )| ≤ c13 |x − x |

|x − y|−3 |x − y|β da(y).

σ (x ,ρ )\σ (x ,2ξ )

For y ∈ σ (x , ρ )\σ (x , 2ξ ), |x − y| ≤ |x − x | + |x − y| ≤ 12 |x − y| + |x − y| = 32 |x − y|; therefore, since β < 1, 

|J4 (x , x )| ≤ c14 |x − x |

|x − y|β −3 da(y)

σ (x ,ρ )\σ (x ,2ξ ) 



= 2π c14 |x − x |

ρ

ϖ β −2 d ϖ = 2π c14 |x − x |

2ξ

= which yields

1  β −1 2ξ ϖ ρ 1−β

  2π c14 |x − x | 2β −1 |x − x |β −1 − ρ β −1 , 1−β |J4 (x , x )| ≤ c15 |x − x |β ,

(3.19)

where c15 depends on ρ . Estimates (3.16)–(3.19) together with (3.14) now lead to the conclusion that for |x − x | < 12 ρ , |M1 (x ) − M1(x )| ≤ c16 |x − x |α , (3.20) where 0 < α < β < 1 and c 16 depends on ρ . Next, M2 (x ) − M2 (x ) = J5 (x , x ) + J6(x , x ) + J7 (x , x ) + J8(x , x ), where



J5 (x , x ) = [ f (x ) − f (x )]

S+ \σ (x ,ρ )



J6 (x , x ) = f (x )

m(x , y) da(y),



 m(x , y) − m(x , y) da(y),

S+ \σ (x ,ρ )



J7 (x , x ) = f (x )

m(x , y) da(y),

σ (x ,ρ )\σ (x ,ρ )

J8 (x , x ) = − f (x )



σ (x ,ρ )\σ (x ,ρ )

m(x , y) da(y).

(3.21)

3.2 Smoothness Properties

39

Estimating J5 , we find that 

|J5 (x , x )| ≤ | f (x ) − f (x )|

|m(x , y)| da(y)

S+ \σ (x ,ρ )



≤ c17 |x − x |β

|x − y|−2 da(y).

S+ \σ (x ,ρ )

Since x is outside the domain of integration, this yields |J5 (x , x )| ≤ c18 |x − x |β ,

(3.22)

where c18 depends on ρ . With (3.15) in mind, we see that 

|J6 (x , x )| ≤ | f (x )|

|m(x , y) − m(x , y)| da(y)

S+ \σ (x ,ρ )



 ≤ c2 sup | f (x)| |x − x | x∈S0

hence,



|x − y|−3 da(y);

S+ \σ (x ,ρ )

|J6 (x , x )| ≤ c19 |x − x |,

(3.23)

where c19 depends on ρ . For J7 we have 

|J7 (x , x )| ≤ | f (x )|

|m(x , y)| da(y)

σ (x ,ρ )\σ (x ,ρ )

≤ c20



|x − y|−2 da(y).

σ (x ,ρ )\σ (x ,ρ )

Extending the domain of integration leads to 

|J7 (x , x )| ≤ c20

|x − y|−2 da(y)

σ (x ,ρ +|x −x |)\σ (x ,ρ −|x −x |)

= 2π c20

  ρ +|x  −x |

ρ −|x −x |

1 dϖ ϖ

   ≤ 2π c20 ρ + |x − x | − ρ − |x − x | sup = 4π c20

|x − x | . ρ − |x − x |

1 ϖ



40

3 The Nonhomogeneous System

Since

1 1 ρ − |x − x | > ρ − ρ = ρ , 2 2

we arrive at |J7 (x , x )| ≤ that is,

8π c20 |x − x |; ρ

|J7 (x , x )| ≤ c21 |x − x |,

(3.24)

where c21 depends on ρ . Similarly, noting that |x  − y| ≥ |x − y| for y ∈ σ (x , ρ )\σ (x , ρ ), we find that 

|J8 (x , x )| ≤ | f (x )|

|m(x , y)| da(y)

σ (x ,ρ )\σ (x ,ρ )



≤ c22

|x − y|−2 da(y)

σ (x ,ρ )\σ (x ,ρ )



≤ c22

|x − y|−2 da(y)

σ (x ,ρ )\σ (x ,ρ )



≤ c22

|x − y|−2 da(y)

σ (x ,ρ +|x −x |)\σ (x ,ρ −|x −x |)

= 2π c22

  ρ +|x  −x |

ρ −|x −x |

1 dϖ . ϖ

Just as in the case of J7 , the above inequality yields |J8 (x , x )| ≤ c23 |x − x |,

(3.25)

where c23 depends on ρ . Combining (3.22)–(3.25) and also taking (3.21) into account, we see that for |x − x | < 12 ρ , |M2 (x ) − M2 (x )| ≤ c24 |x − x |β , (3.26) where c24 depends on ρ . Finally, it is immediately seen that |M3 (x ) − M3 (x )| ≤ |cρ ||x − x |β .

(3.27)

Thus, from (3.13), (3.20), (3.26), and (3.27) we conclude that for |x  − x | < 12 ρ , |M(x ) − M(x )| ≤ c25 |x − x |α , where 0 < α < β < 1 and c 25 depends on ρ .

3.2 Smoothness Properties

41

We can easily show that M is bounded: |M(x)| ≤



|m(x, y)| | f (y) − f (x)| da(y)

S+

+ | f (x)|



|m(x, y)| da(y) + |cρ || f (x)|

S+ \σ (x,ρ )





≤ c26 |x − y|β −2 da(y) + c27 S+

|x − y|−2 da(y) + |cρ | | f (x)| ≤ N.

S+ \σ (x,ρ )

Hence, for |x  − x | ≥ 12 ρ ,

α |x − x |α 2 ≤ 2N |x − x |α , |x − x |α ρ

|M(x ) − M(x )| ≤ 2N = 2N and we conclude that for all x  , x ∈ S0 ,

|M(x ) − M(x )| ≤ c|x − x |α , where 0 < α < β < 1 and c depends on ρ .   Theorem 3.5. If f ∈ C 0,β (S+ ), β ∈ (0, 1), then K ∈ C 2,α (Ω ), where Ω is an arbitrary domain in R2 , the closure of which lies in S + , and 0 < α < β < 1. Additionally,

∂2 K(x) = γ (α , β ) f (x) + M(x), ∂ xα ∂ xβ

(3.28)

where M is defined by (3.11) and γ (α , β ) is a symmetric (3 × 3)-matrix with elements

γρη (α , β ) = π d1 δαβ δρη + 12 π d2 (δαρ δβ η + δαη δβ ρ − δαβ δρη ), γρ 3 (α , β ) = γ3ρ (α , β ) = 0, 1 γ33 (α , β ) = − δ . 2μ αβ Proof. By Theorem 3.1 with k(x, y) = k(x − y),

∂2 ∂2 K(x) = ∂ xα ∂ xβ ∂ xα ∂ xβ =

∂ ∂ xα

 S+

According to Theorem 2.8 in [40], p. 184,



k(x − y) f (y) da(y)

S+

∂ k(x − y) f (y) da(y). ∂ xβ

(3.29) (3.30) (3.31)

42

3 The Nonhomogeneous System

∂ ∂ xα

 S+ \σ (x,δ )

∂ k(x − y) f (y) da(y) ∂ xβ 

=

S+ \σ (x,δ )

−

∂2 k(x − y) f (y) da(y) ∂ xα ∂ xβ



∂ σ (x,δ )

yα − xα ∂ k(x − y) f (y) ds(y). r ∂ xβ

(3.32)

We transform the second integral on the right-hand side of (3.32) as follows: 

−

∂ σ (x,δ )

yα − xα ∂ k(x − y) f (y) ds(y) r ∂ xβ 

=δ

(xα − yα )

∂ σ (x,1)

∂ k(δ (x − y)) f (x + δ (y − x)) ds(y). ∂ xβ

The first two components of the integral around ∂ σ (x, 1) are

δ



(xα − yα )

∂ σ (x,1)



+δ =

∂ kρη (δ (x − y)) fη (x + δ (y − x))ds(y) ∂ xβ (xα − yα )

∂ σ (x,1)



∂ kρ 3 (δ (x − y)) f3 (x + δ (y − x))ds(y) ∂ xβ

  (xα − yα ) d1 δρη (xβ − yβ ) + d2 δρβ (xη − yη ) + δηβ (xρ − yρ )   ∂ σ (x,1) − 2(xρ − yρ )(xη − yη )(xβ − yβ ) + O(δ ln δ )

+



× fη (x + δ (y − x)) ds(y) (xα − yα )(O(δ ln δ )) f 3 (x + δ (y − x)) ds(y).

∂ σ (x,1)

As δ → 0 in the above expression, from (3.3) and (3.4) it follows that the first two components are   fη (x) d1 δρη Iαβ + d2 (δβ η Iαρ + δβ ρ Iαη − 2Iαβ ρη )  = fη (x) π d1 δρη δαβ   + π d2 δβ η δαρ + δβ ρ δαη − 12 (δαβ δρη + δαρ δβ η + δαη δβ ρ ) = γρη (α , β ) fη (x) + γρ 3 (α , β ) f3 (x). The third component of the integral around ∂ σ (x, 1) is found analogously. First, we see that

3.3 A Particular Solution of the System



δ

(xα − yα )

∂ σ (x,1)



+δ =

∂ k3μ (δ (x − y)) f μ (x + δ (y − x))ds(y) ∂ xβ

(xα − yα )

∂ σ (x,1)



43

∂ k33 (δ (x − y)) f3 (x + δ (y − x)) ds(y) ∂ xβ

(xα − yα )(O(δ ln δ )) f μ (x + δ (y − x)) ds(y)

∂ σ (x,1)

+



∂ σ (x,1)



 1 2 (xα − yα ) − (x − yβ ) + O(δ ln δ ) f3 (x + δ (y − x)) ds(y). 2π μ β

As δ → 0, from (3.3) we deduce that the third component is −

1 1 I f3 (x) = − δ f3 (x) = γ3μ (α , β ) f μ (x) + γ33 (α , β ) f3 (x). 2π μ αβ 2μ αβ

Equality (3.28) is now obtained by letting δ → 0 in (3.32). By Theorem 3.3, M(x) exists in the sense of principal value and, by Theorem 3.4, M ∈ C 0,α (Ω ), which completes the proof.  

3.3 A Particular Solution of the System The next assertion is the main result of this chapter. Theorem 3.6. If f ∈ C 0,β (S+ ), β ∈ (0, 1), then K(x) defined by (3.1) is a regular solution in S + of the system B(∂x )u(x) + f (x) = 0, where B(∂x ) = A(∂x ) and k(x, y) = D(x, y) or B(∂ x ) = Aω (∂x ) and k(x, y) = D ω (x, y). Proof. The regularity of K has been shown in the proofs of Theorems 3.2 and 3.5. Let B(∂x ) = Aω (∂x ) and k(x, y) = D ω (x, y). The case where B(∂ x ) = A(∂x ) and k(x, y) = D(x, y) follows from this one with ω = 0. By (1.10) and (1.11), 

 Aω (∂x )K(x) α = h2 μ

2 ∂2 ∂2 2 K (x) + h ( λ + μ ) Kρ (x) α ∑ 2 ρ =1 ∂ xα ∂ xρ β =1 ∂ xβ 2

∑

+ (ρω 2 h2 − μ )Kα (x) − μ By Theorem 3.5,

∂ K3 (x). ∂ xα

44

3 The Nonhomogeneous System

∂2 Kα (x) = ∂ x2β

  S+

  ∂2 ω D (x, y) f (y) − f (x) da(y) ∂ x2β α





+

S+ \σ (x,δ )





+ lim

ε →0

∂2 Kρ (x) ∂ xα ∂ xρ   = S+

 ∂2 ω D (x, y) da(y) fk (x) ∂ x2β α k

σ (x,δ )\σ (x,ε )

   ∂2 ω D (x, y) da(y) fk (x) + γ (β , β ) f (x) α , αk 2 ∂ xβ

 ∂2 D ω (x, y) f (y) − f (x) ∂ xα ∂ xρ





+

S+ \σ (x,δ )

 ε →0

da(y) ρ

 ∂2 ω D (x, y) da(y) fk (x) ∂ xα ∂ xρ ρ k 

+ lim



σ (x,δ )\σ (x,ε )

   ∂2 ω D (x, y) da(y) fk (x) + γ (α , ρ ) f (x) ρ . ∂ xα ∂ xρ ρ k

Also, Remark 3.1 and Theorem 3.1 imply that Kα (x) =





  D ω (x, y) f (y) − f (x) α da(y)

S+

 +



S+ \σ (x,δ )



+ lim

ε →0

∂ K3 (x) = ∂ xα

  S+

 Dαωk (x, y) da(y) fk (x) 

 Dαωk (x, y) da(y) fk (x),

σ (x,δ )\σ (x,ε )

  ∂ D ω (x, y) f (y) − f (x) da(y) ∂ xα 3    ∂ ω + D (x, y) da(y) fk (x) ∂ xα 3k S+ \σ (x,δ )

 + lim

ε →0

consequently,



σ (x,δ )\σ (x,ε )

 ∂ ω D3k (x, y) da(y) fk (x); ∂ xα

3.3 A Particular Solution of the System



 Aω (∂x )K(x) α = h2 μ

2

∑



β =1

45

2    γ (β , β ) f (x) α + h2 (λ + μ ) ∑ γ (α , ρ ) f (x) ρ

ρ =1

    + Aω (∂x )D ω (x, y) f (y) − f (x) α da(y) S+



+



S+ \σ (x,δ )



+ lim

ε →0



Aω (∂x )D ω (x, y)





 da(y) fk (x) αk

  ω  A (∂x )D ω (x, y) α k da(y) fk (x).

σ (x,δ )\σ (x,ε )

Next, 



  Aω (∂x )D ω (x, y) f (y) − f (x) α da(y)

S+

=

  S+

=−

  − δ (|x − y|)E3 f (y) − f (x) α da(y)



δ (|x − y|) fα (y) da(y) +

S+



 δ (|x − y|) da(y) fα (x)

S+

= − fα (x) + fα (x) = 0. By Theorem 2.2, the integrals over S + \σ (x, δ ) and σ (x, δ )\σ (x, ε ) are also zero. Therefore, using (3.29) and (3.30), we obtain  ω  A (∂x )K(x) α =

=

2

∑

β =1 2

∑

β =1

 2 h μγαη (β , β ) fη (x) + h2 μγα 3 (β , β ) f3 (x) + h2(λ + μ )γβ μ (α , β ) f μ (x) + h2 (λ + μ )γβ 3 (α , β ) f3 (x)  2   h μ π d1 δαη + 12 π d2 (2δαβ δηβ − δαη ) fη (x)    + h2(λ + μ ) π d1 δαβ δβ μ + 12 π d2 δα μ f μ (x) ;

so, by (2.32),  ω   A (∂x )K(x) α = h2 μ (2π d1 − π d2) + h2(λ + μ )π d2

 + h2 μπ d2 + h2 (λ + μ )π d1 fα (x)   = h2 π (λ + 3μ )d1 + (λ + μ )d2 fα (x)   1 = (λ + μ )2 − (λ + 3μ )2 fα (x), 4 μ (λ + 2 μ )

which means that

 ω  A (∂x )K(x) α + fα (x) = 0.



46

3 The Nonhomogeneous System

Similarly,  ω  A (∂x )K(x) 3 = μ which leads to  ω  A (∂x )K(x) 3 =μ

2

∑

β =1



2

∑

β =1

 ∂ ∂2 K (x) + 2 K3 (x) + ρω 2K3 (x), ∂ xβ β ∂ xβ

    ω   γ (β , β ) f (x) 3 + A (∂x )D ω (x, y) f (y) − f (x) 3 da(y)



S+



+



S+ \σ (x,δ )



+ lim

ε →0

ω

ω

A (∂x )D (x, y)







 3k

da(y) fk (x)

  A (∂x )D (x, y) 3k da(y) fk (x). ω

ω

σ (x,δ )\σ (x,ε )

As before, the integrals vanish, and (3.30) and (3.31) yield 

 Aω (∂x )K(x) 3 = μ =μ

2

∑

β =1 2

∑

β =1

hence, as required.  



γ3α (β , β ) fα (x) + γ33 (β , β ) f3 (x)

−

1 2μ

 f3 (x);

 ω  A (∂x )K(x) 3 + f3 (x) = 0,



Chapter 4

The Question of Uniqueness for the Exterior Problems

This chapter is concerned with two questions regarding the solutions of the homogeneous system (4.1). First, we state various results established in [24] on the radiation conditions imposed on certain solutions of (4.1) to guarantee the unique solvability of a number of boundary value problems associated with the system. In Theorem 4.4 we show that the single-layer and double-layer potentials introduced in Chapter 2 satisfy the radiation conditions, which means that seeking solutions in terms of these functions does not lead to problems of nonuniqueness. Second, we derive representation formulas for the solutions of (4.1) in both the interior and exterior domains. The interior representation formula (Theorem 4.5) is obtained straightforwardly. The exterior representation formula (Theorem 4.7) requires substantially more work; it is based on far-field estimates for the matrix of fundamental solutions that are deduced in Theorem 4.6 from the fact that each column of the matrix satisfies the radiation conditions. The representation formulas are essential in the design of integral equations based on the so-called direct boundary integral equation method, which is discussed in a later chapter.

4.1 Radiation Conditions Using the result of Theorem 3.6, we can reduce system (1.10) to the corresponding homogeneous one: Aω (∂x )u(x) = 0. (4.1) Suppose that we have found a solution u of (4.1). Then the function U(x) = u(x) −



D ω (x, y)H(y) da(y)

S

satisfies (1.10), so it suffices for us to concentrate on solving system (4.1). Obviously, the given boundary condition would have to be modified in the same way. G.R. Thomson and C. Constanda, Stationary Oscillations of Elastic Plates, A Boundary Integral Equation Analysis, DOI 10.1007/978-0-8176-8241-5_4, © Springer Science+Business Media, LLC 2011

47

48

4 The Question of Uniqueness for the Exterior Problems

In this section we state the radiation conditions derived in [24], which guarantee that each of the exterior Dirichlet and Neumann problems for the system of equations (4.1) admits at most one solution. In the Dirichlet problem, the vector characterizing the displacement is prescribed on ∂ S, whereas in the Neumann problem, the moment–stress vector is given on the boundary. The question of uniqueness of solution for exterior boundary value problems in classical three-dimensional elasticity is discussed in [40], where the unknown displacement vector u is shown to have the form u = u(p) + u(s) , with the two terms satisfying, respectively, the equations (Δ + k2p)u(p) = 0,

curl u(p) = 0,

(Δ + ks2 )u(s) = 0,

divu(s) = 0,

where k2p =

ρω 2 , λ + 2μ

ks2 =

ρω 2 . μ

The two vectors u (p) and u(s) are known as the potential and solenoidal components of u. Owing to the complicated nature of system (4.1), splitting up the function u into terms that satisfy Helmholtz equations is not so simple. The following assertion was proved in [24]. ¯ of (4.1) in S can be written in the Theorem 4.1. Any solution u ∈ C 2 (S) ∩ C 1 (S) form u = U (1) +U (2) +U (3) + u(1) + u(2) , (4.2) where

(Δ + k2j )U ( j) = 0, (Δ + kβ2 )u(β ) = 0, ( j)

U3 = 0, u

(β )

=

(β ) (0, 0, u3 )T ,

(4.3) (4.4)

with the k2j defined by (2.4), (2.6), and (2.3). If w = (u 1 , u2 , 0)T , then the above functions are computed by the formulas U (1) = U (2) =

1 h2 μ k32 (k22 − k12 ) 1 h2 μ k32 (k12 − k22 )

  (Δ + k22 ) − h2 (λ + 2 μ ) graddivw + μ gradu 3 ,   (Δ + k12 ) − h2 (λ + 2 μ ) graddivw + μ gradu 3 , U (3) =

1 curlcurl w, k32

4.1 Radiation Conditions

49

1 (Δ + k22 )u3 , k22 − k12 1 (2) u3 = 2 (Δ + k12 )u3 . k1 − k22 (1)

u3 =

Since each of the functions in the representation (4.2) of a regular solution u of (4.1) satisfies a Helmholtz equation, it seems reasonable to impose Sommerfeldtype radiation conditions (see [57]) on them (cf. [40]). Specifically, we require that, as R → ∞, U ( j) = O(R−1/2 ),

(4.5)

( j)

∂U − ik jU ( j) = O(R−3/2 ), ∂R (β ) u3 = O(R−1/2 ), (β ) ∂ u3

∂R

(β )

− ikβ u3 = O(R−3/2 ).

(4.6) (4.7) (4.8)

Using conditions (4.5)–(4.8), we derive some useful far-field estimates for the (β ) and u3 . We denote by ∂ KR the circle with the center at the origin and radius R and write y = (y1 , y2 , 0)T .

U ( j)

Theorem 4.2. If y ∈ ∂ KR and u ∈ C 2 (S− ) ∩ C1 (S¯− ) satisfies conditions (4.5)–(4.8) and is a solution of (4.1) in S − , then, as R = |y| → ∞,

T y U (3) (y) = O(R−3/2 ), R

 k2 − kβ2 (β ) y U (β ) (y) = i u3 (y) + O(R−3/2 ), kβ R   T (∂y )U (β ) (y) − ih2 (λ + 2 μ )kβ U (β )(y) α = O(R−3/2 ),   T (∂y )U (3) (y) − ih2 μ k3U (3) (y) α = O(R−3/2 ),   T (∂y )u(β ) (y) − iμ kβ u(β ) (y) 3 = O(R−3/2 ),   T (∂y )U (3) (y) 3 = O(R−3/2 ).

(4.9) (4.10) (4.11) (4.12) (4.13) (4.14)

Proof. Estimates (4.9)–(4.12) are established in [24]. Using (1.13), (4.4), and (4.8), we find that  (β )  (β ) (β ) Tu − i μ kβ u(β ) 3 = T33 u3 − iμ kβ u3

(β ) (β )  ∂ u3 ∂ u3 (β ) = μ ν1 + ν2 − i μ k β u3 ∂ y1 ∂ y2

(β )  ∂ u3 (β ) =μ − ikβ u3 = O(R−3/2 ). ∂R

50

4 The Question of Uniqueness for the Exterior Problems

Finally, from (1.13) and (4.9), 

TU (3)

  (3) (3) (3) (3) = T U + T U = μ ν U + ν U 31 32 1 2 1 2 1 2 3

T y =μ U (3) = O(R−3/2 ).   R



Some of these estimates, together with Theorem 1.2, are used to prove the main result of this section [24]. Theorem 4.3. If conditions (4.5)–(4.8) hold, then each of the exterior Dirichlet and Neumann problems has at most one regular solution in S − . Definition 4.1. B ω denotes the class of functions satisfying (4.1) in S − and the radiation conditions (4.5)–(4.8) as R → ∞. Theorem 4.4. If ϕ ∈ C(∂ S), then V ω ϕ and W ω ϕ defined by (2.34) and (2.35), respectively, belong to B ω . Proof. From (2.34), (V ω ϕ )(x) =

 ∂S

=

 ∂S

D ω (x, y)ϕ (y) ds(y) ⎛ ω ⎞ ⎛ ω ⎞ D11 (x, y) D12 (x, y)  ⎜ ω ⎟ ⎜ ω ⎟ (x, y)⎠ ϕ2 (y) ds(y) ⎝D21 (x, y)⎠ ϕ1 (y) ds(y) + ⎝D22 ω (x, y) ∂S D31 D ω (x, y) ⎛ ω32 ⎞ D13 (x, y)  ⎜ ω ⎟ + ⎝D23 (x, y)⎠ ϕ3 (y) ds(y). ω (x, y) ∂S D33

By Theorem 2.2, the columns of D ω (x, y) satisfy (4.1). It remains to show that they satisfy the radiation conditions as R → ∞. Consider the first column of D ω (x, y). This can be written in the form (4.2); that is, ⎛ ω ⎞ D11 (x, y) ⎜ ω ⎟ ⎝D21 (x, y)⎠ = U (1) +U (2) +U (3) + u(1) + u(2) , ω (x, y) D31

where, by (2.20) and (2.21),

⎛

⎞ ∂ 2 (1) H (k1 |x − y|) ⎟ ⎜ ∂ x21 0 ⎜ ⎟ i ⎜ ⎟ U (1) = 2 2 ⎜ 2 ⎟, (1) ⎟ 4h μ k3 ⎜−α 2 ∂ ⎝ 1 ∂ x ∂ x H0 (k1 |x − y|)⎠ 1 2 0 − α12

4.1 Radiation Conditions

51

⎛

⎞ 2 2 ∂ H (1) (k |x − y|) − α 2 2 ⎜ ⎟ ∂ x21 0 ⎜ ⎟ i (2) ⎜ 2 ⎟, U = 2 2⎜ 2 ∂ (1) 4h μ k3 ⎝−α2 H0 (k2 |x − y|)⎟ ⎠ ∂ x1 ∂ x2 0 ⎛ 2 ⎞ ∂ (1) 2H (1) (k |x − y|) H (k |x − y|) + k 3 3 3 0 ⎜ ∂ x2 0 ⎟ ⎜ 1 ⎟ i (3) ⎜ 2 ⎟, U = 2 2⎜ ∂ (1) ⎟ 4h μ k3 ⎝ H0 (k3 |x − y|) ⎠ ∂ x1 ∂ x2 0 ⎛ ⎞ 0 i ⎜ ⎟ 0 u(1) = 2 2 ⎝ ⎠, ∂ (1) 4h μ k3 γ H0 (k1 |x − y|) ∂ x1 ⎛ ⎞ 0 i ⎜ ⎟ 0 u(2) = 2 2 ⎝ ⎠. ∂ (1) 4h μ k3 −γ H0 (k2 |x − y|) ∂ x1 Explicitly, as R → ∞, (xα − yα )(xβ − yβ ) 2 (1) ∂2 (1) H0 (k j |x − y|) = − k j H0 (k j |x − y|) ∂ xα ∂ xβ r2 (xα − yα )(xβ − yβ ) (1) +2 k j H1 (k j r) r3 1 (1) − δαβ k j H1 (k j r). r Since [1]

(1 − i) 1 ik j r (1) H0 (k j |x − y|) =  e + O(R−3/2 ), π k j R1/2 (1) H1 (k j r)

and

= O(R

−1/2

(4.15)

),

rη = Rη + O(Rη −1 ),

(4.16)

we see that (xα − yα )(xβ − yβ ) 2 (1) ∂2 (1) H (k j |x − y|) = − k j H0 (k j |x − y|) + O(R−3/2). ∂ xα ∂ xβ 0 r2 Then, using (4.15) and (4.16), we arrive at

∂2 (1 − i) 1 ik j r (1) H (k j |x − y|) = − fαβ (θ )k2j  e + O(R−3/2 ), ∂ xα ∂ xβ 0 π k j R1/2 where

52

4 The Question of Uniqueness for the Exterior Problems

⎧ 2 ⎪ ⎨cos θ fαβ (θ ) = cos θ sin θ ⎪ ⎩ 2 sin θ

if α = β = 1, if α = β , if α = β = 2;

consequently,

 ∂ ∂2 (1) − ik j H (k j |x − y|), ∂R ∂ xα ∂ xβ 0   (1 − i) 1 1 = − fαβ (θ )k2j  ik j 1/2 − ik j 1/2 + O(R−3/2 ) eik j r + O(R−3/2 ) R R πkj = O(R−3/2 ). In the same way, it can be shown that

 ∂ (1) − ik j H0 (k j |x − y|) = O(R−3/2 ) ∂R ∂ (1) H (k j |x − y|) = O(R−1/2 ), ∂ x1 0

 ∂ ∂ (1) − ik j H (k j |x − y|) = O(R−3/2 ), ∂R ∂ x1 0 which means that the first column of D ω (x, y) satisfies the radiation conditions (4.5)–(4.8). The same is verified analogously for the other two columns. Hence, V ω ϕ ∈ Bω . The double-layer potential W ω ϕ is dealt with in a similar manner.   Remark 4.1. Later on, we seek solutions to boundary value problems associated with (4.1) in the form of single-layer and double-layer potentials. From Theorems 4.3 and 4.4 we see that in the case of the exterior domain there will be at most one solution.

4.2 The Boundary Value Problems Let P, Q, R, and S be (3 × 1)-vector functions defined on ∂ S. We consider four fundamental boundary value problems for the high-frequency stationary oscillations of thin elastic plates. Interior Dirichlet problem: (Dω + ) Find u ∈ C 2 (S+ ) ∩C1 (S¯+ ) that satisfies (4.1) in S + and u(x) = P(x),

x ∈ ∂ S.

Interior Neumann problem: (Nω + ) Find u ∈ C 2 (S+ ) ∩C1 (S¯+ ) that satisfies (4.1) in S + and T (∂x )u(x) = Q(x),

x ∈ ∂ S.

4.3 Representation Formulas

53

Exterior Dirichlet problem: (Dω − ) Find u ∈ C 2 (S− ) ∩C1 (S¯− ) ∩ B ω that satisfies (4.1) in S − and u(x) = R(x),

x ∈ ∂ S.

Exterior Neumann problem: (Nω − ) Find u ∈ C 2 (S− ) ∩C1 (S¯− ) ∩ B ω that satisfies (4.1) in S − and T (∂x )u(x) = S (x),

x ∈ ∂ S.

We denote by (D 0ω ± ) or (N0ω ± ) the corresponding problems with homogeneous boundary conditions. Remark 4.2. In the Dirichlet problems, we need u to be C 1 (∂ S) since the boundary integral equations are formulated by means of representation formulas that require the computation of the moments and stress on the boundary.

4.3 Representation Formulas We may derive representation formulas for regular solutions of (4.1) in S + and S − in terms of the single-layer and double-layer potentials. We start with the interior case and construct the so-called Somigliana formula. Theorem 4.5. If u ∈ C 2 (S+ ) ∩C1 (S¯+ ) is a solution of (4.1) in S + , then ⎧ ⎪ x ∈ S+ , ⎨u(x), V ω (Tu|∂ S ) (x) − W ω (u|∂ S ) (x) = 12 u(x), x ∈ ∂ S, ⎪ ⎩ 0, x ∈ S− . Proof. Let x ∈ S + , and choose δ > 0 so that σ (x, δ ) lies entirely within S + (see Figure 3).

∂ σ (x, δ ) x S+

∂S

Fig. 3. The disk σ (x, δ ) and its boundary ∂ σ (x, δ ).

54

4 The Question of Uniqueness for the Exterior Problems

By Theorems 2.2 and 1.2 in S + \σ (x, δ ), with v replaced in turn by each column of D ω (x, y),   ω  D (x, y)T (∂y )u(y) − P ω (x, y)u(y) ds(y) ∂S



=



 D ω (x, y)T (∂y )u(y) − P ω (x, y)u(y) ds(y). (4.17)

∂ σ (x,δ )

Since u ∈ C 1 (∂ σ (x, δ )), from (2.31) it follows that 

D ω (x, y)T (∂y )u(y) ds(y) = O(δ ln δ ).

(4.18)

∂ σ (x,δ )

Using (2.33), we see that for y ∈ ∂ σ (x, δ ), P ω (x, y) = O(δ −1 ),

(4.19)

and, by the Mean Value Theorem, u(x) − u(y) = O(δ ).

(4.20)

Also, since  ∂ σ (x,δ )

∂ ln r ds(y) = ∂ s(y) 

 ∂ σ (x,δ )

∂ σ (x,δ )



we find that

∂ (xα − yα )(xβ − yβ ) ds(y) = 0, ∂ s(y) r2

∂ lnr ds(y) = 2π , ∂ ν (y)

P ω (x, y) ds(y) = −E3 + O(δ ln δ ).

(4.21)

∂ σ (x,δ )

Consequently, from (4.19)–(4.21), 

P ω (x, y)u(y) ds(y)

∂ σ (x,δ )

=



ω

P (x, y) [u(y) − u(x)] ds(y) +

∂ σ (x,δ )

 

ω



P (x, y) ds(y) u(x)

∂ σ (x,δ )

= −u(x) + O(δ ). Letting δ → 0 in (4.17) and taking (4.18) into account, we conclude that for x ∈ S + , V ω (Tu|∂ S ) (x) − W ω (u|∂ S ) (x) = u(x).

4.3 Representation Formulas

55

Suppose now that x ∈ ∂ S. We apply Theorem 1.2 to the columns of D ω (x, y) in and denote by ∂ φ (x, δ ) the part of ∂ σ (x, δ ) lying inside S + . It can be shown that the length of ∂ φ (x, δ ) is πδ + O(δ 2 ); so, S+ \ (σ (x, δ ) ∩ S+ ),  ∂ φ (x,δ )

∂ ln r ds(y) = ∂ ν (y)

which means that



π +O(  δ)

(cos2 θ + sin2 θ ) d θ = π + O(δ ),

0

P ω (x, y) ds(y) = − 12 E3 + O(δ ln δ ).

∂ φ (x,δ )

Following the argument for the case x ∈ S + , it is easy to see that V ω (Tu|∂ S ) (x) − W ω (u|∂ S ) (x) = 12 u(x). Finally, when x is in the exterior region S − , a direct application of Theorem 1.2 yields V ω (Tu|∂ S ) (x) − W ω (u|∂ S ) (x) = 0, and the theorem is proved.   Before deriving an exterior representation formula, we need to establish certain estimates for the matrix D ω (y, x) as |y| → ∞. Since, by Theorem 2.2, for x = y each of the columns of the matrix D ω (y, x) is a solution of (4.1) with respect to the variable y, we may use the result of Theorem 4.1 to write D ω (y, x) in the form D ω (y, x) = Γ (1) (y, x) + Γ (2) (y, x) + Γ (3) (y, x) + γ (1) (y, x) + γ (2) (y, x), where

( j)

Γ3k = 0,

(β )

γα k = 0.

x = y,

(4.22) (4.23)

 (β ) (β ) (β ) T (β ) Let γ3 = γ31 , γ32 , γ33 . In the proof of Theorem 4.4 we showed that, as R = |y| → ∞, Γ ( j) = O(R−1/2 ), (4.24)

∂Γ ( j) − ik j Γ ( j) = O(R−3/2 ), ∂R (β ) γ3 = O(R−1/2 ), (β ) ∂ γ3

∂R

(β )

− ikβ γ3

= O(R−3/2 ).

(4.25) (4.26) (4.27)

As before, KR is the disk of radius R with the center at the origin, ∂ K R is its boundary, and y = (y 1 , y2 , 0)T .

56

4 The Question of Uniqueness for the Exterior Problems

Theorem 4.6. If y ∈ ∂ KR and u ∈ C 2 (S− ) ∩ C1 (S¯− ) ∩ B ω is a solution of (4.1) in S− , then, as R = |y| → ∞,

 (k2 − kβ2 ) y  (β ) T (β ) Γ (y, x) = i γ3 (y, x) + O(R−3/2 ), (4.28) kβ R   (4.29) T (∂y )Γ (β ) (y, x) − ih2 (λ + 2μ )kβ Γ (β ) (y, x) α k = O(R−3/2 ),   (3) 2 (3) −3/2 T (∂y )Γ (y, x) − ih μ k3 Γ (y, x) α k = O(R ), (4.30)   (4.31) T (∂y )γ (β ) (y, x) − iμ kβ γ (β ) (y, x) 3k = O(R−3/2 ),   (3) −3/2 ), (4.32) T (∂y )Γ (y, x) 3k = O(R  (β ) T (3) Γ (y, x) U (y) = O(R−2 ), (4.33) T (β )  (3) −2 Γ (y, x) U (y) = O(R ), (4.34) where U (β ) and U (3) are the first three terms in the decomposition (4.2). Estimates (4.28)–(4.32) are established in the same way as (4.10)–(4.14) if use is made of (4.24)–(4.27). The proofs of (4.33) and (4.34) are similar to that of the T  estimate U (β ) U (3) = O(R−2 ) derived in [24]. The results of Theorem 4.6 are used in the proof of the next assertion, which constructs the exterior Somigliana representation formula. Theorem 4.7. If u ∈ C 2 (S− ) ∩C1 (S¯− ) ∩ B ω is a solution of (4.1) in S − , then ⎧ ⎪ x ∈ S+ , ⎨0, − V ω (Tu|∂ S )(x) + W ω (u|∂ S )(x) = 12 u(x), x ∈ ∂ S, ⎪ ⎩ u(x), x ∈ S− . Proof. Let R be sufficiently large so that ∂ S lies entirely within the disk K R centered at the origin and of radius R (see Figure 4). S− ∩ KR

∂ KR S+

∂S

Fig. 4. The disk KR and its boundary ∂ KR .

4.3 Representation Formulas

57

Applying Theorem 4.5 in the region S − ∩ KR , we find that −V ω (Tu|∂ S )(x) +W ω (u|∂ S )(x)    T  + D ω (x, y)T (∂y )u(y) − T (∂y )D ω (y, x) u(y) ds(y) ∂ KR

⎧ ⎪ x ∈ R2 \(S¯− ∩ K¯ R ), ⎨0, 1 = 2 u(x), x ∈ ∂ S ∪ ∂ KR , ⎪ ⎩ u(x), x ∈ S − ∩ KR .

(4.35)

By (2.27), (4.2), and (4.22), the integrand of the integral around ∂ K R can be written as  ω T  T D (y, x) T (∂y )u(y) − T (∂y )D ω (y, x) u(y)  T  = Γ (1) + Γ (2) + Γ (3) TU (1) − ih2 (λ + 2μ )k1U (1)  + TU (2) − ih2 (λ + 2μ )k2U (2) + TU (3) − ih2 μ k3U (3) + Tu(1) + Tu(2)  T  + γ (1) + γ (2) TU (1) + TU (2) + TU (3)  + Tu(1) − iμ k1 u(1) + Tu(2) − iμ k2 u(2)  − T Γ (1) − ih2 (λ + 2 μ )k1Γ (1) + T Γ (2) − ih2(λ + 2μ )k2Γ (2) T + T Γ (3) − ih2 μ k3 Γ (3) + T γ (1) + T γ (2)   × U (1) +U (2) + U (3)  T − T Γ (1) + T Γ (2) + T Γ (3) + T γ (1) − iμ k1 γ (1) + T γ (2) − i μ k2γ (2)   × u(1) + u(2)  T  T  + ih2(λ + 2μ )(k2 − k1 ) Γ (1) U (2) − Γ (2) U (1)  T  T  + iμ (k2 − k1 ) γ (1) u(2) − γ (2) u(1)  T  T  + ih2(λ + 2μ )k1 Γ (3) U (1) − Γ (1) U (3)  T  T  + ih2(λ + 2μ )k2 Γ (3) U (2) − Γ (2) U (3)  T  T   + ih2 μ k3 Γ (1) + Γ (2) U (3) − Γ (3) U (1) + U (2) . Taking formulas (1.13), (4.3)–(4.5), (4.7), (4.11)–(4.14), (4.23), (4.24), (4.26), and (4.29)–(4.34) into account, we deduce that  ω T  T D (y, x) T (∂y )u(y) − T (∂y )D ω (y, x) u(y)  T    T   = γ (1) + γ (2) TU (1) + TU (2) − T Γ (1) + T Γ (2) u(1) + u(2)  T  T  + ih2 (λ + 2 μ )(k2 − k1 ) Γ (1) U (2) − Γ (2) U (1)  T  T  + iμ (k2 − k1 ) γ (1) u(2) − γ (2) u(1) + O(R−2 ),

58

4 The Question of Uniqueness for the Exterior Problems

so (4.10) and (4.28) imply that  ω T  T D (y, x) T (∂y )u(y) − T (∂y )D ω (y, x) u(y)  (1) (1) (2) (2)  = μ (y/R)T U (1) γ3 + U (2) γ3 +U (1)γ3 + U (2) γ3  (1)  T T T T  (2)  (1)  (2)  − μ u3 Γ (1) + u3 Γ (1) + u3 Γ (2) + u3 Γ (2) (y/R) + ih2 (λ + 2μ )(k2 − k1 )k1−1 k2−1 (k2 − k12 )(k2 − k22 )  (1) (2) (2) (1)  × u3 γ3 − u3 γ3 (y/R)T (y/R)  (2) (1) (1) (2)  + iμ (k2 − k1) u3 γ3 − u3 γ3 + O(R−2). Since (y/R)T (y/R) = 1, we find that [D ω (y, x)] T (∂y )u(y) − [T (∂y )D ω (y, x)] u(y)  (1) (1) (2) (1) = iμ k1−1 (k2 − k12 )u3 γ3 + k2−1 (k2 − k22 )u3 γ3 T

T

(2) (2) 

(1) (2)

+ k1−1 (k2 − k12 )u3 γ3 + k2−1 (k2 − k22 )u3 γ3

 (1) (1) (2) (1) − iμ k1−1 (k2 − k12 )u3 γ3 + k1−1 (k2 − k12 )u3 γ3

(1) (2) (2) (2)  + k2−1 (k2 − k22 )u3 γ3 + k2−1 (k2 − k22 )u3 γ3  (1) (2) (2) (1)  + ih2 (λ + 2μ )(k2 − k1 )k1−1 k2−1 (k2 − k12 )(k2 − k22 ) u3 γ3 − u3 γ3  (2) (1) (1) (2)  + iμ (k2 − k1 ) u3 γ3 − u3 γ3 + O(R−2)  = ik1−1 k2−1 h2 (λ + 2μ )(k2 − k1 )(k2 − k12 )(k2 − k22 )   + μ k2 (k2 − k12 ) − k1 (k2 − k22 ) − k1 k2 (k2 − k1 )  (1) (2) (2) (1)  × u3 γ3 − u3 γ3 + O(R−2 ),

from which we obtain 

T  T D ω (y, x) T (∂y )u(y) − T (∂y )D ω (y, x) u(y)   = i(k2 − k1 )k1−1 k2−1 h2 (λ + 2μ )(k2 − k12 )(k2 − k22 ) + μ k2 (1) (2)

(2) (1)

× (u3 γ3 − u3 γ3 ) + O(R−2 ). Using (2.6), we now see that h2 (λ + 2 μ )(k2 − k12 )(k2 − k22 ) + μ k2   = h2 (λ + 2μ ) k4 − (k12 + k22 )k2 + k12 k22 + μ k2 = h2 (λ + 2μ )k4 − h2(λ + 3μ )k4 + h2 μ k4 − μ k2 + μ k2 = 0. Consequently, the integrand is O(R −2 ), and the desired result is obtained by letting R → ∞ in (4.35).  

4.3 Representation Formulas

59

The functions V ω ± and W ω ± defined by (2.36), (2.39), and (2.40) may be used to write the results of Theorems 4.5 and 4.7 in a simpler form. Thus, the interior representation formula becomes  u in S¯+ , ω+ ω+ V (Tu|∂ S ) − W (u|∂ S ) = (4.36) 0 in S− . Similarly, the exterior representation formula can be expressed as  0 in S + , − V ω − (Tu|∂ S ) + W ω − (u|∂ S ) = u in S¯− .

(4.37)

Chapter 5

The Eigenfrequency Spectra of the Interior Problems

As we saw in the previous chapter, each of the exterior boundary value problems has at most one regular solution. Unfortunately, this is not true for (D ω + ) and (Nω + ). + ω+ We prove that (Dω 0 ) and (N0 ) have nonzero solutions by establishing their equivalence to certain integral equations that are known to have such solutions. To do so, we construct the kernels of these integral equations—the so-called Green’s tensors. Throughout the chapter we exploit the close connection between system (4.1) and the corresponding homogeneous system governing the equilibrium bending of plates. In Section 5.1 we prove the existence of the Green’s tensor for the interior Dirichlet problem. This is straightforward because the corresponding interior Dirichlet problem for the equilibrium system is uniquely solvable (Theorem 5.2). We then show that the Green’s tensor is symmetric and that it can be used as the basis for an interior representation formula (Theorem 5.3). In Theorem 5.4 we use the representation formula and results on the Newtonian potential from Chapter 3 to show the equivalence of the homogeneous boundary value problem to its associated integral equation. Theorem 5.5 contains the main result of the section: the existence of eigenfrequencies. The Green’s tensor for the interior Neumann problem (N ω + ) is considered in Section 5.2. Its existence is not so easily proved because the corresponding boundary value problem for the equilibrium system is not always solvable. For this reason, it is necessary to assume that the Green’s tensor takes a rather complicated form, chosen so that the tensor satisfies solvability conditions while remaining symmetric. Once the question of solvability has been settled, the discussion follows the pattern adopted in the preceding section, encompassing the representation formula (Theorem 5.8), the equivalence of the homogeneous boundary value problem to its associated integral equation (Theorem 5.9), and the existence of eigenfrequencies (Theorem 5.10). The proof of the existence of eigenfrequency spectra for the homogeneous prob+ ω+ lems (Dω 0 ) and (N0 ) requires several results from the study of the equilibrium bending of plates. These results, established in [14], are stated wherever they need to be used. G.R. Thomson and C. Constanda, Stationary Oscillations of Elastic Plates, A Boundary Integral Equation Analysis, DOI 10.1007/978-0-8176-8241-5_5, © Springer Science+Business Media, LLC 2011

61

62

5 The Eigenfrequency Spectra of the Interior Problems

5.1 The Green’s Tensor for the Interior Dirichlet Problem Consider the system of equations A(∂x )u(x) = 0,

(5.1)

where the matrix operator A(∂ x ) is defined by (1.11). This system characterizes the displacements in the equilibrium bending of plates discussed in [14], where the next two assertions were also proved. Theorem 5.1. If u, v ∈ C 2 (S+ ) ∩C1 (S¯+ ), then 

(vT Au − uTAv) da =

S+



(vT Tu − uTT v) ds.

∂S

Theorem 5.2. The interior Dirichlet problem for system (5.1) has a unique regular solution for boundary data of class C 1,α (∂ S), α ∈ (0, 1). The Green’s tensor for the interior Dirichlet problem is a (3 × 3)-matrix G 1 (x, y) such that A(∂x )G1 (x, y) = 0, G1 (x, y) = 0,

x ∈ S+ , y ∈ S+ , x = y, +

x ∈ ∂ S, y ∈ S ,

(5.2) (5.3)

and which in the domain S + is represented by G1 (x, y) = D(x, y) − v1(x, y),

(5.4)

where D(x, y) is the matrix of fundamental solutions for the operator A(∂ x ) (see [14]) and the columns of v 1 (x, y) are regular solutions of (5.1) in S + . By (5.2)–(5.4), the construction of G 1 (x, y) is therefore reduced to solving the equilibrium interior Dirichlet problems (i)

A(∂x )v1 (x, y) = 0, (i) v1 (x, y)

x ∈ S + , y ∈ S+ ,

(i)

= D (x, y),

(5.5)

+

x ∈ ∂ S, y ∈ S ,

where i = 1, 2, 3. Unfortunately, the columns of D(x, y) are not in C 1,α (∂ S). However, since D(x, y) is a function of x − y (see [14]), we could consider the boundary value problems (i)

A(∂z )v1 (z) = 0, (i)

v1 (z) = D(i) (z),

z ∈ S+ , z ∈ ∂ S,

i = 1, 2, 3, where D (i) (z) = D(i) (x, 0). The columns of D(z) are in C ∞ (∂ S), so these boundary value problems are solvable. Replacing z by x − y in the solutions yields the required columns of v 1 (x, y).

5.1 The Green’s Tensor for the Interior Dirichlet Problem

63

We now verify the symmetry property of the Green’s tensor. From Theorem 5.1 (i) ( j) with u = G1 (x, x1 ) and v = G1 (x, x2 ) and (5.3) we see that   (i)

T  ( j) T  ( j) (i) G1 (x, x1 ) A(∂x )G1 (x, x2 ) − G1 (x, x2 ) A(∂x )G1 (x, x1 ) da(x) = 0.

S+

Using (5.4), (5.5), and the properties of the matrix of fundamental solutions, we arrive at       G1 (x, x1 ) ji δ (|x − x2 |) da(x) = G1 (x, x2 ) i j δ (|x − x1 |) da(x); S+

S+

    G1 (x2 , x1 ) ji = G1 (x1 , x2 ) i j ,

hence,

 T G1 (x, y) = G1 (y, x) .

which means that

(5.6)

Theorem 5.3. If u ∈ C 2 (S+ ) ∩C1 (S¯+ ), then it can be represented in the form u(x) = −

 

 T T (∂y )G1 (y, x) u(y) ds(y) − G1 (x, y)A(∂y )u(y) da(y),

x ∈ S+.

S+

∂S

(5.7)

Proof. We proceed as in the case of Theorem 4.5. Let x ∈ S + , and let δ > 0 be sufficiently small so that σ (x, δ ) ⊂ S + . By Theorem 5.1 in S + \σ (x, δ ) with v replaced in turn by each column of G 1 (y, x),     T T G1 (y, x) A(∂y )u(y) da(y) − G1 (y, x) A(∂y )u(y) da(y) S+

σ (x,δ )

=

 



−

uT (y)A(∂y )G1 (y, x) da(y)

S+ \σ (x,δ )

T

 G1 (y, x) T (∂y )u(y) − u (y)T (∂y )G1 (y, x) ds(y)

∂S

−



T

 T  G1 (y, x) T (∂y )u(y) − uT(y)T (∂y )G1 (y, x) ds(y).

∂ σ (x,δ )

From (5.2), (5.3), and (5.6) it follows that 

G1 (x, y)A(∂y )u(y) da(y) −

S+

=−

 



G1 (x, y)A(∂y )u(y) da(y)

σ (x,δ )

T

T (∂y )G1 (y, x) u(y) ds(y) −

∂S



G1 (x, y)T (∂y )u(y) ds(y)

∂ σ (x,δ )

+



∂ σ (x,δ )



T T (∂y )G1 (y, x) u(y) ds(y).

(5.8)

64

5 The Eigenfrequency Spectra of the Interior Problems

Since u is regular in S + , we have Au ∈ C(S + ); so, by (5.4) and (2.31), 

G1 (x, y)A(∂y )u(y) da(y) =

σ (x,δ )



  D(x, y) − v1 (x, y) A(∂y )u(y) da(y)

σ (x,δ )

=O

δ

 r ln r dr ,

0

as δ → 0. Also, following the argument in the proof of Theorem 4.5, we see that, as δ → 0,  

G1 (x, y)T (∂y )u(y) ds(y) = O(δ ln δ ),

∂ σ (x,δ )

 T T (∂y )G1 (y, x) u(y) ds(y) = −u(x) + O(δ ).

∂ σ (x,δ )

Letting δ → 0 in (5.8), we arrive at (5.7).   We now turn our attention to the interior Dirichlet problem for stationary oscillations. In view of (1.10), the system of equations (4.1) may be written in the form A(∂x )u(x) + ω 2H u(x) = 0,

(5.9)

⎛ 2 ⎞ ρh 0 0 H = ⎝ 0 ρ h2 0 ⎠ . 0 0 ρ

(5.10)

where

+ Theorem 5.4. The homogeneous problem (D ω 0 ) is equivalent to the integral equation  u(x) − ω 2 H G1 (x, y)u(y) da(y) = 0, x ∈ S + , (5.11) S+

where H is defined by (5.10). + Proof. Let u be a solution of (D ω 0 ). Since u| ∂ S = 0, from (5.7) we obtain

u(x) = −



G1 (x, y)A(∂y )u(y) da(y),

S+

which, by (5.9), becomes u(x) − ω 2 H

 S+

as required.

G1 (x, y)u(y) da(y) = 0,

5.1 The Green’s Tensor for the Interior Dirichlet Problem

65

Suppose now that u satisfies (5.11). Then, by (5.3), u|∂ S = 0, and, using (5.4), Theorem 3.6, and the regularity of v 1 , we find that A(∂x )u(x) = ω 2 H A(∂x )



G1 (x, y)u(y) da(y)

S+



= ω 2 H A(∂x )

D(x, y)u(y) da(y)

S+



− ω 2 H A(∂x ) = −ω 2 H u(x) − ω 2H



v1 (x, y)u(y) da(y)

S+

A(∂x )v1 (x, y)u(y) da(y).

S+

By (5.5) and (5.9),

Aω (∂x )u(x) = 0,

+ which means that u is a solution of (D ω  0 ). 

Definition 5.1. A value of ω for which the homogeneous Dirichlet (Neumann) + problem has a nonzero regular solution u is called an eigenfrequency of (D ω 0 ) ω+ ((N0 )), and the corresponding nonzero solution u is called an eigensolution. + Theorem 5.5. (Dω 0 ) has infinitely many real eigenfrequencies, which are the values of ω for which the integral equation (5.11) has nonzero solutions. + Proof. By Theorem 5.4, (D ω 0 ) is equivalent to (5.11). In [26] it is shown that there is an infinite set of values of ω 2 , all real, for which (5.11) has nonzero solutions. This is based on the fact that G 1 (x, y) is a symmetric, nondegenerate kernel. To verify that the eigenfrequencies are real, we must show that ω 2 ≥ 0 when (5.11) has nonzero solutions. According to the so-called Betti formula for the equilibrium bending of plates (see [14]), if u, v ∈ C 2 (S+ ) ∩C1 (S¯+ ), then

    vT Au + 2E(u, v) da = vT Tu ds, S+

∂S

where E(u, v) =

1 2

 2   h λ uα ,α vβ ,β + μ uα ,β (vα ,β + vβ ,α ) + μ (uα + u3,α )(vα + v3,α ) .

+ Suppose that u is a solution of (D ω 0 ) or, equivalently, of (5.11). From the Betti formula with v = u¯ and (5.9) it follows that

66

5 The Eigenfrequency Spectra of the Interior Problems





 − ρω 2 (h2 |u1 |2 + h2 |u2 |2 + |u3 |2 ) + 2E(u, u) ¯ da = 0.

(5.12)

S+

Since  2   h λ uα ,α u¯β ,β + μ uα ,β (u¯α ,β + u¯β ,α ) + μ (uα + u3,α )(u¯α + u¯3,α )    = 12 h2 (λ + μ )| divu|2 + μ |u1,2 + u2,1 |2 + μ |u1,1 − u2,2 |2  + μ |u1 + u3,1 |2 + μ |u2 + u3,2 |2 ≥ 0,

E(u, u) ¯ =

1 2

+ from (5.12) we deduce that ω 2 ≥ 0; therefore, the eigenfrequencies of (D ω 0 ) are real.   + Representation (5.11) of the regular solutions of (D ω 0 ) leads to the next assertion. + 0,α (∂ S), α ∈ (0, 1). Theorem 5.6. If u is a regular solution of (D ω 0 ), then Tu ∈ C

∂u ∈ C 0,α (∂ S). Since u is a regular solution of ∂ xη + (Dω 0 ), it is also a solution of (5.11); that is, Proof. It suffices to show that

u(x) = ω 2 H



G1 (x, y)u(y) da(y),

x ∈ S+;

S+

hence,

∂ ∂ u(x) = ω 2 H ∂ xη ∂ xη = ω 2H

∂ ∂ xη



G1 (x, y)u(y) da(y)

S+

 

 D(x, y) − v1 (x, y) u(y) da(y),

x ∈ S+ .

S+

In view of Theorem 3.1 and the regularity of v 1 , we can differentiate under the integral sign to obtain

∂ u(x) = ω 2 H ∂ xη

 S+

 ∂  D(x, y) − v1 (x, y) u(y) da(y), ∂ xη

Since u is regular in S + , it follows that

∂ u(x) = ω 2 H ∂ xη

 S+

∂u ∈ C(S¯+ ); so, ∂ xη

 ∂  D(x, y) − v1 (x, y) u(y) da(y), ∂ xη

From (3.2) we now see that, as r → 0,

x ∈ S+.

x ∈ ∂ S.

5.2 The Green’s Tensor for the Interior Neumann Problem

67

 ∂  D(x, y) − v1 (x, y) = O(r−1 ); ∂ xη consequently, arguing as in the proof of Theorem 3.2, we arrive at

∂u ∈ C 0,α (∂ S), ∂ xη

α ∈ (0, 1),

which concludes the proof.  

5.2 The Green’s Tensor for the Interior Neumann Problem Owing to the solvability conditions required for the interior Neumann problem associated with system (5.1), the construction of the Green’s tensor for (N ω + ) is not so straightforward. Consider the matrix ⎛ ⎞ 1 0 0 F˜ = ⎝ 0 1 0⎠ . −x1 −x2 1 The columns F˜ (i) of F˜ form a basis for the space of rigid displacements and are also linearly independent eigensolutions of the interior Neumann problem associated with (5.1) [14]. Let F be the (3 × 3)-matrix whose columns are the vectors obtained by orthonor 3 malizing the set F˜ (i) i=1 in S+ ; that is,   T F (i) (x) F ( j) (x) da(x) = δi j ,

i, j = 1, 2, 3.

(5.13)

S+

The Green’s tensor for (N ω + ) is a (3 × 3)-matrix G 2 (x, y) satisfying A(∂x )G2 (x, y) =



3

∑ F (k) (x) F (k) (y)

T

,

x ∈ S+ , y ∈ S+ , x = y,

(5.14)

k=1

T (∂x )G2 (x, y) = 0,

x ∈ ∂ S, y ∈ S + .

(5.15)

In S+ , G2 (x, y) is represented as G2 (x, y) = Q(x, y) − v2(x, y), where Q(x, y) is the (3 × 3)-matrix Q(x, y) 3



k=1

S+

= D(x, y) − ∑ F (k) (x)



T D(y, ξ )F (k) (ξ ) da(ξ )

(5.16)

68

5 The Eigenfrequency Spectra of the Interior Problems



3

−∑

k=1

+

S+



3

∑

D(x, ξ )F (k) (ξ ) da(ξ )

F

k,r=1

(k)

 

F (k) (y)

T

    T  T (k) (r) (x) F (ξ ) D(ξ , η )F (η ) da(η ) da(ξ ) F (r) (y) S+ S+

(5.17) and each column of v 2 (x, y) is a regular solution of (5.1) in S + . Next, since [14] A(∂x )D(x, y) = 0,

x ∈ S + , x = y,

A(∂x )F (k) (x) = 0,

x ∈ S+ ,

from (5.17) it follows that     T (k) A(∂x )Q(x, y) = − ∑ A(∂x ) D(x, ξ )F (ξ ) da(ξ ) F (k) (y) , 3

k=1

S+

where x, y ∈ S + , x = y. By Theorem 3.6, A(∂x )Q(x, y) =

3



∑ F (k) (x) F (k) (y)

T

x, y ∈ S+ , x = y.

,

(5.18)

k=1

From (5.16), (5.18), and the fact that A(∂ x )v2 (x, y) = 0, we deduce that A(∂x )G2 (x, y) =

3



T

∑ F (k) (x) F (k) (y)

x, y ∈ S+ , x = y,

,

k=1

as required. To find the matrix v 2 (x, y), we must therefore solve the boundary value problems (i)

A(∂x )v2 (x, y) = 0,

x ∈ S + , y ∈ S+ ,

(i)

T (∂x )v2 (x, y) = T (∂x )Q(i) (x, y),

x ∈ ∂ S, y ∈ S + ,

(5.19)

where i = 1, 2, 3. The next assertion was established in [14]. Theorem 5.7. Boundary value problems (5.19) are solvable if and only if   T F (k) (x) T (∂x )Q(i) (x, y) ds(x) = 0,

i, k = 1, 2, 3.

(5.20)

∂S

Their solutions are unique up to a linear combination of the columns of F. We now show that conditions (5.20) are satisfied by our choice of matrix Q(x, y). Since [14] T (∂x )F (k) (x) = 0, k = 1, 2, 3,

5.2 The Green’s Tensor for the Interior Neumann Problem

69

from (5.17) we obtain T (∂x )Q(i) (x, y) = T (∂x )D(i) (x, y) 3



k=1

S+

− ∑ T (∂x )

(k)

D(x, ξ )F (k) (ξ )Fi (y) da(ξ ),

i = 1, 2, 3.

Conditions (5.20) may now be written as   T F (r) (x) T (∂x )D(i) (x, y) ds(x) ∂S

3

−∑

 



T (k) F (r) (x) T (∂x )D(x, ξ )F (k) (ξ )Fi (y) ds(x) da(ξ ) = 0, (5.21)

k=1 + S ∂S

where r, i = 1, 2, 3. Here we have used Theorem 3.1 to take the T operator inside the integral in the second term. Equations (5.21) can be expressed in the form  

T T (∂x )D(i) (x, y) F (r) (x) ds(x)

∂S 3

−∑

    T   T (r) (k) (k) F (ξ )Fi (y) T (∂x )D(x, ξ ) F (x) ds(x) da(ξ ) = 0,

k=1 + S

(5.22)

∂S

where r, i = 1, 2, 3. Since each column of F satisfies (5.1) and T F = 0, we use the interior representation formula for the equilibrium bending of plates (see [14]) to find that, for ξ ∈ S + ,   T F (r) (ξ ) = − T (∂x )D(x, ξ ) F (r) (x) ds(x), r = 1, 2, 3. (5.23) ∂S

In view of (5.23), we can bring (5.22) to the form 3

(r)

−Fi (y) + ∑



 (k) T (k) F (ξ )Fi (y) F (r) (ξ ) da(ξ ) = 0,

r, i = 1, 2, 3;

  T F (k) (ξ ) F (r) (ξ ) da(ξ ) = 0,

r, i = 1, 2, 3;

k=1 + S

that is, 3

(r)

(k)

−Fi (y) + ∑ Fi (y) k=1

S+

so, taking the orthonormalization condition (5.13) into account, we see that the solvability conditions are automatically satisfied. By Theorem 5.7, (5.19) are solvable for each i and their solutions are unique to within three additive vectors of rigid displacement. Because of this arbitrariness, we can impose the normalizing conditions   T (i) F (k) (x) v2 (x, y) da(x) = 0, S+

i, k = 1, 2, 3.

(5.24)

70

5 The Eigenfrequency Spectra of the Interior Problems

The symmetry of the Green’s tensor is proved by means of an auxiliary formula. The ith column of the matrix Q(x, y) is Q(i) (x, y) 3

= D(i) (x, y) − ∑ F (k) (x) 

3

−∑

k=1

S+

 (k) D(x, ξ )F (k) (ξ ) da(ξ ) Fi (y)

S+

     T (r) F (k) (x) F (k) (ξ ) D(ξ , η )F (r) (η ) da(η ) da(ξ ) Fi (y);

3

∑

+

k=1

  T F (k) (ξ ) D(i) (ξ , y) da(ξ )

k,r=1

S + S+

hence, for i, m = 1, 2, 3,   T F (m) (x) Q(i) (x, y) da(x) S+

=





T F (m) (x) D(i) (x, y) da(x)

S+ 3

−∑

    T T F (m) (x) F (k) (x) F (k) (ξ ) D(i) (ξ , y) da(ξ ) da(x)

k=1 + S 3

−∑

S+

    T  (k) (m) (k) F (x) D(x, ξ )F (ξ ) da(ξ ) da(x)Fi (y)

k=1 + S

+

3

∑

S+



k,r=1



S+

×

T F (m) (x) F (k) (x)

    T (r) F (k) (ξ ) D(ξ , η )F (r) (η ) da(η ) da(ξ ) da(x) Fi (y).

S + S+

Using (5.13), we arrive at   T F (m) (x) Q(i) (x, y) da(x) S+

=

    T T F (m) (x) D(i) (x, y) da(x) − F (m) (ξ ) D(i) (ξ , y) da(ξ ) S+

S+

  3   T  (k) − F (m) (x) D(x, ξ )F (k) (ξ ) da(ξ ) da(x)Fi (y)

∑

k=1 + S 3

+∑

S+

   T (r) F (m) (ξ ) D(ξ , η )F (r) (η ) da(η ) da(ξ )Fi (y);

r=1 + + S S

5.2 The Green’s Tensor for the Interior Neumann Problem

that is,

71

  T F (m) (x) Q(i) (x, y) da(x) = 0,

i, m = 1, 2, 3.

(5.25)

S+

Consequently, from (5.16), (5.24), and (5.25) we find that   T (i) F (k) (x) G2 (x, y) da(x) = 0,

i, k = 1, 2, 3.

(5.26)

S+

We can now prove that the Green’s tensor for (N ω + ) is symmetric. Using (5.15) (i) ( j) and Theorem 5.1 with u = G 2 (x, x1 ) and v = G2 (x, x2 ), we see that   (i)

T  ( j) T  ( j) (i) G2 (x, x1 ) A(∂x )G2 (x, x2 ) − G2 (x, x2 ) A(∂x )G2 (x, x1 ) da(x) = 0.

S+

By (5.14), (5.16), and (5.17),      (i) T − G2 (x, x1 ) ji δ (|x − x2 |) + G2 (x, x1 ) S+

=

  S+

 (k) (k) F (x)F (x ) da(x) 2 ∑ j 3

k=1

    ( j) T 3 (k) (k) − G2 (x, x2 ) i j δ (|x − x1 |) + G2 (x, x2 ) ∑ F (x)Fi (x1 ) da(x); k=1

that is,  T 3    (k) T (i) (k) − G2 (x2 , x1 ) ji + ∑ Fj (x2 ) F (x) G2 (x, x1 ) da(x) k=1

S+

 T    (k) T ( j) (k) = − G2 (x1 , x2 ) i j + ∑ Fi (x1 ) F (x) G2 (x, x2 ) da(x) . 3

k=1

By (5.26),

S+

    G2 (x2 , x1 ) ji = G2 (x1 , x2 ) i j ,

so

 T G2 (x, y) = G2 (y, x) .

(5.27)

We now derive a representation formula analogous to Theorem 5.3, which makes use of G2 (x, y). Theorem 5.8. If u ∈ C 2 (S+ ) ∩C1 (S¯+ ), then it can be represented in the form u(x) =



G2 (x, y)T (∂y )u(y) ds(y) −

∂S

+



S+



G2 (x, y)A(∂y )u(y) da(y)

S+

3  T uT (y) ∑ F (k) (y) F (k) (x) da(y), k=1

x ∈ S+.

(5.28)

72

5 The Eigenfrequency Spectra of the Interior Problems

Proof. As in the case of Theorem 5.3, we apply Theorem 5.1 in S + \σ (x, δ ), with v replaced in turn by each column of G 2 (y, x). This leads to  

 T G2 (y, x) A(∂y )u(y) da(y) −

S+

−  



T G2 (y, x) A(∂y )u(y) da(y)

σ (x,δ )



uT (y)A(∂y )G2 (y, x) da(y)

S+ \σ (x,δ )

T  G2 (y, x) T (∂y )u(y) − uT(y)T (∂y )G2 (y, x) ds(y)

=

∂S

−



 T  G2 (y, x) T (∂y )u(y) − uT(y)T (∂y )G2 (y, x) ds(y).

∂ σ (x,δ )

Taking (5.14), (5.15), and (5.27) into account, we arrive at 

G2 (x, y)A(∂y )u(y) da(y) −

S+

−





S+

3  T uT (y) ∑ F (k) (y) F (k) (x) da(y) k=1

G2 (x, y)A(∂y )u(y) da(y) +

σ (x,δ )

=





G2 (x, y)T (∂y )u(y) ds(y) −

∂S



3  T uT (y) ∑ F (k) (y) F (k) (x) da(y)

σ (x,δ )

k=1

G2 (x, y)T (∂y )u(y) ds(y)

∂ σ (x,δ )

+





T T (∂y )G2 (y, x) u(y) ds(y).

∂ σ (x,δ )

Arguing as before, we find that 

lim

δ →0 σ (x,δ )



lim

δ →0 σ (x,δ )

G2 (x, y)A(∂y )u(y) da(y) = 0, 3  T uT (y) ∑ F (k) (y) F (k) (x) da(y) = 0,



lim

δ →0 ∂ σ (x,δ )



lim

δ →0 ∂ σ (x,δ )

k=1

G2 (x, y)T (∂y )u(y) ds(y) = 0, 

T T (∂y )G2 (y, x) u(y) ds(y) = −u(x),

and (5.28) is now obtained by letting δ → 0 in (5.29).  

(5.29)

5.2 The Green’s Tensor for the Interior Neumann Problem

73

+ Theorem 5.9. The homogeneous problem (N ω 0 ) is equivalent to the integral equation 

u(x) − ω 2 H

G2 (x, y)u(y) da(y) = 0,

x ∈ S +,

(5.30)

S+

where H is defined by (5.10). + (k) Proof. First, suppose that u is a solution of (N ω 0 ). Since the vectors F , k = 1, 2, 3, are eigensolutions of the interior Neumann problem for (5.1), applying Theorem 5.1 with v = F (k) yields

  T F (k) (y) A(∂y )u(y) da(y) = 0,

k = 1, 2, 3;

S+

so, by (5.9),  

T u(y) F (k) (y) da(y) = 0,

− ω 2H

k = 1, 2, 3.

S+ + It is already known that ω = 0 is an eigenfrequency of (N ω 0 ) [14]. If we assume that ω = 0, then, since H is nonsingular, we have

 

T u(y) F (k) (y) da(y) = 0,

k = 1, 2, 3.

(5.31)

S+

Next, by (5.31) and the boundary condition Tu| ∂ S = 0, equality (5.28) implies that u(x) = −



G2 (x, y)A(∂y )u(y) da(y),

S+

which leads to u(x) − ω 2 H



G2 (x, y)u(y) da(y) = 0,

x ∈ S +,

S+

as required. Suppose now that u is a solution of (5.30); then A(∂x )u(x) = ω 2 H A(∂x )



G2 (x, y)u(y) da(y).

S+

Using (5.16), (5.17), the smoothness of v 2 (x, y) and F (k) (x), k = 1, 2, 3, and the fact that A(∂x )v2 (x, y) = A(∂x )F (k) (x) = 0 in S + , for each k we arrive at

74

5 The Eigenfrequency Spectra of the Interior Problems

A(∂x )u(x) = ω 2 H A(∂x )



D(x, y)u(y) da(y)

S+

− ω 2H

3

∑

 

k=1 + S

  T D(x, ξ )F (k) (ξ ) da(ξ ) F (k) (y) u(y) da(y).



A(∂x )

S+

By Theorem 3.6, A(∂x )u(x) = −ω 2 H u(x) + ω 2 H

  3 S+

that is,

  (k) T (k) F (x) F (y) u(y) da(y); ∑

k=1

3

Aω (∂x )u(x) = ω 2 H

∑ F (k) (x)

k=1

  T F (k) (y) u(y) da(y).

(5.32)

S+

Since u is a solution of (5.30), it follows that      T T F (k) (y) u(y) da(y) = ω 2 H F (k) (y) G2 (y, η )u(η ) da(η ) da(y); S+

S+ S +

hence, (5.26) yields

  T F (k) (y) u(y) da(y) = 0, S+

so (5.32) becomes

Aω (∂x )u(x) = 0.

Also, by Theorem 3.1 and the boundary property (5.15), from (5.28) we see that T (∂x )u(x)|∂ S = T (∂x ) =

 S+



S+

3  T uT (y) ∑ F (k) (y) F (k) (x) da(y) k=1



3

∑ T (∂x )F (k) (x) F (k) (y)

T

u(y) da(y) = 0;

k=1

+ therefore, u is a solution of (N ω 0 ).

 

In view of Theorem 5.9, we are able to formulate an assertion concerning the + eigenfrequencies of (N ω 0 ), which is proved just like Theorem 5.5. + Theorem 5.10. (Nω 0 ) has infinitely many real eigenfrequencies, which coincide with the values of ω for which the integral equation (5.30) has nonzero solutions. + ω+ Remark 5.1. The sets of infinitely many real eigenfrequencies of (D ω 0 ) and (N 0 ) identified above play an important role in the subsequent discussion of the existence and uniqueness of solution for both the interior and exterior problems. As will be seen in Chapter 6, the presence of these sets makes the oscillation problems fundamentally different from those in the equilibrium case and is responsible for their attending mathematical difficulties.

Chapter 6

The Question of Solvability

In this chapter we prove the existence of regular solutions to the interior and exterior Dirichlet and Neumann boundary value problems introduced in Section 4.2. This is done by means of an indirect boundary integral equation method in which the solution of each problem is sought in the form of a particular potential. Such a method has been used successfully to prove the solvability of a variety of important boundary value problems, for example, in connection with the Helmholtz equation (see [11]) and stationary oscillations in classical three-dimensional elasticity (see [40]). The above models have many characteristics in common with our system of equations, an important one being the presence of eigenfrequencies. The integral equations characterizing each boundary value problem are derived in Section 6.1 by substituting a suitable form of solution in the relevant boundary condition. The aim is to obtain quasi-Fredholm integral equations of the second kind, for which an existence theory can be constructed. Specifically, given the boundary properties of the potentials mentioned in Theorems 2.6 and 2.7, we seek the solutions of the Dirichlet problems as double-layer potentials and those of the Neumann problems as single-layer potentials. Theorems 6.4 and 6.5 indicate the link between the eigenfrequencies of the homogeneous interior problems and the occurrence of eigenfunctions (see Definition 6.1) of the boundary integral operators arising from the integral equation formulation. These assertions are of great importance because they signal the existence of multiple solutions of the integral equations for certain values of the oscillation frequency. The solvability of the interior problems is discussed in Theorems 6.6–6.9. As expected, existence cannot be established when ω is an eigenfrequency of the corresponding homogeneous problem. In the case of the exterior problems, the existence of solutions for all values of the oscillation frequency is proved in Theorems 6.10 and 6.11. As the analysis shows, although the indirect boundary integral equation technique is successful in proving solvability, it is inadequate when it comes to the construction of the solutions because the integral equations involved may have multiple solutions. G.R. Thomson and C. Constanda, Stationary Oscillations of Elastic Plates, A Boundary Integral Equation Analysis, DOI 10.1007/978-0-8176-8241-5_6, © Springer Science+Business Media, LLC 2011

75

76

6 The Question of Solvability

6.1 Boundary Integral Equations We seek the solutions of (D ω + ), (Nω + ), (Dω − ), and (Nω − ) in the form W ω + (ϕ ), V ω + (ϕ ), W ω − (ϕ ), and V ω − (ϕ ), respectively. Taking Theorem 2.5(i), (2.37)– (2.40), and the relevant boundary condition into account, we arrive at (W0ω − 12 I )ϕ = P,

(6.1)

(W0ω ∗ + 12 I )ϕ = Q,

(6.2)

(W0ω + 12 I )ϕ = R,

(6.3)

(W0ω ∗ − 12 I )ϕ = S .

(6.4)

Though (6.1)–(6.4) are systems of singular integral equations, it can be shown [49] that Fredholm’s theorems are applicable to them. The next assertions are proved exactly as Theorems 2.39 and 2.38 in [14]. Theorem 6.1. The Fredholm alternative holds for the pairs of equations (W0ω + 12 I )ϕ = f ,

(W0ω ∗ + 12 I )ψ = g

and

(W0ω − 12 I )ϕ = p, (W0ω ∗ − 12 I )ψ = q  in the dual system C 0,α (∂ S),C 0,α (∂ S) , α ∈ (0, 1), with the bilinear form (ϕ , ψ ) =



ϕ T (y)ψ (y) ds(y).

(6.5)

∂S

Theorem 6.2. If h ∈ C 1,α (∂ S), α ∈ (0, 1), then any solution ϕ ∈ C 0,α (∂ S) of  ω 1 W0 ± 2 I ϕ = h, is of class C 1,α (∂ S).

6.2 Boundary Integral Operators In this section we show the connection between the eigenfrequencies of the interior problems and the presence of eigenfunctions of the boundary integral operators of interest. Definition 6.1. Let Lω be an operator depending on a parameter ω . If there is a value of ω such that the equation L ω ϕ = 0 has a nonzero solution ϕ , then ϕ is called an eigenfunction of L ω .

6.2 Boundary Integral Operators

77

A typical example is the Sturm–Liouville operator, whose eigenfunctions are widely used in the theory of partial differential equations (see [25]). Theorem 6.3. (i) V

ω + (ϕ )

= 0 in S¯+ if and only if ϕ = 0 on ∂ S.

(ii) W ω + (ϕ ) = 0 in S¯+ if and only if ϕ = 0 on ∂ S. Proof. (i) If ϕ = 0 on ∂ S, then it is obvious that V ω + (ϕ ) = 0 in S¯+ . Suppose that V ω + (ϕ ) = 0 in S¯+ . By the continuity of the single-layer potential, − ω V − (ϕ ) = 0 on ∂ S, and we deduce that V ω − (ϕ ) is a solution of (Dω 0 ). Since ω − ω ω − − ¯ V (ϕ ) ∈ B , by Theorem 4.3, V (ϕ ) = 0 in S . Consequently, from (2.37) and (2.38), ϕ = T V ω + (ϕ ) − T V ω − (ϕ ) = 0 on ∂ S. (ii) As before, if ϕ = 0 on ∂ S, we immediately conclude that W ω + (ϕ ) = 0 in S¯+ follows immediately. If W ω + (ϕ ) = 0 in S¯+ , then, by (2.41), T W ω + (ϕ ) = T W ω − (ϕ ) = 0 on ∂ S. − ω − (ϕ ) = 0 in S¯− Hence, W ω − (ϕ ) is a solution of (N ω 0 ). So, by Theorem 4.3, W and from (2.39) and (2.40) we find that

ϕ = W ω − (ϕ ) − W ω + (ϕ ) = 0 on ∂ S, which completes the proof.   + Theorem 6.4. The parameter ω is an eigenfrequency of (D ω 0 ) of multiplicity n if and only if W0ω − 12 I has n linearly independent C 1,α -eigenfunctions, α ∈ (0, 1), and W0ω ∗ − 12 I has n linearly independent C 0,α -eigenfunctions. The linearly independent C 0,α -eigenfunctions of W0ω ∗ − 12 I are the boundary values of the images + of the linearly independent eigensolutions of (D ω 0 ) under the boundary moment– stress operator. + Proof. Let ω be an eigenfrequency of (D ω 0 ) of multiplicity n with corresponding (k) linearly independent eigensolutions, say, u , k = 1, 2, . . . , n. Since u(k) |∂ S = 0, k = 1, 2, . . . , n, applying T to (4.36) and taking (2.37) into account, we see that   Tu(k) |∂ S = W0ω ∗ + 12 I Tu(k) |∂ S , k = 1, 2, . . . , n,

 ω ∗ 1  (k) W0 − 2 I Tu |∂ S = 0, k = 1, 2, . . . , n.  (k) n The set Tu |∂ S k=1 is linearly independent. To see this, suppose that there are constants ck , k = 1, 2, . . . , n, not all zero, such that so

n

∑ ck Tu(k) |∂ S = 0

k=1

and consider the function u(x) =

n

∑ ck u(k) (x),

k=1

x ∈ S+ .

(6.6)

78

6 The Question of Solvability

+ Since each u (k) is an eigensolution of (D ω 0 ), we have

u|∂ S = 0, and, by (6.6), Tu|∂ S = 0. Applying the interior representation yields u = 0 in S + , which, by the  (k) formula n (4.36) + linear independence of the set u (x) k=1 in S , means that ck = 0, k = 1, 2, . . . , n, contradicting our assumption.  n Consequently, Tu(k) |∂ S k=1 is a set of n linearly independent eigenfunctions for W0ω ∗ − 12 I , which, by Theorem 5.6, are of class C 0,α (∂ S), α ∈ (0, 1). We now show that W0ω ∗ − 12 I has exactly n linearly independent C 0,α -eigenfunctions. Let ϕ ∈ C 0,α (∂ S) be an arbitrary eigenfunction of W 0ω ∗ − 12 I and consider the single-layer potential V ω − (ϕ ). From (2.38) we see that V ω − (ϕ ) is a solution − of (Nω 0 ), so, by Theorem 4.3, V ω − (ϕ ) = 0 in S − .

(6.7)

By the continuity of the single-layer potential, 0 = V ω − (ϕ )|∂ S = V0ω ϕ = V ω + (ϕ )|∂ S , which means that V

ω + (ϕ )

(6.8)

+ is a solution of (Dω 0 ). Therefore, we can write

V ω + (ϕ ) =

n

∑ ak u(k)

in S+ ,

(6.9)

k=1

where the ak , k = 1, 2, . . . , n, are constants. From (6.7)–(6.9) we obtain T V ω − (ϕ ) = 0 on ∂ S and T V ω + (ϕ ) =

n

∑ ak Tu(k)

on ∂ S,

k=1

so, using (2.37) and (2.38), we find that

ϕ = T V ω + (ϕ ) − T V ω − (ϕ ) =

n

∑ ak Tu(k)|∂ S .

k=1

 n Hence, Tu(k) |∂ S k=1 is a complete set of linearly independent C 0,α -eigenfunctions of W0ω ∗ − 12 I . According to the Fredholm alternative, W 0ω − 12 I also has exactly n linearly independent C 0,α -eigenfunctions. That these eigenfunctions are of class C 1,α (∂ S) follows from Theorem 6.2.

6.2 Boundary Integral Operators

79

Conversely, let ω be such that W0ω − 12 I has n linearly independent C 1,α -eigen + functions ϕ (k) . Then, by (2.39), W ω + ϕ (k) is a solution of (Dω 0 ) for each k. We now show that the W ω + (ϕ (k) ), k = 1, 2, . . . , n, are nonzero and linearly independent. The former property follows immediately from Theorem 6.3(ii). Suppose that the W ω + (ϕ (k) ) are linearly dependent. Then there are constants b k , k = 1, 2, . . . , n, not all zero, such that n  ∑ bk W ω + ϕ (k) = 0 in S+. k=1

From the linearity of the double-layer potential we see that

n  W ω + ∑ bk ϕ (k) = 0 in S + , k=1

so, by Theorem 6.3(ii), n

∑ bk ϕ (k) = 0

on ∂ S.

k=1

Thus, the set  ω +bk =(k)0, kn = 1, 2, . . . , n, which contradicts our assumption. Therefore, ω+ W ϕ is linearly independent, which means that (D ) has at least n 0 k=1 linearly independent eigensolutions. Suppose that there are m linearly independent eigensolutions, with m > n. Then, by the first part of the proof, the operator W 0ω − 12 I has exactly m linearly independent C 1,α -eigenfunctions. Since we have assumed that W 0ω − 12 I has n linearly independent C 1,α -eigenfunctions, it follows that m = n. + Consequently, ω is an eigenfrequency of (D ω  0 ) of multiplicity n.  + Theorem 6.5. The parameter ω is an eigenfrequency of (N ω 0 ) of multiplicity n if 1 ω 1, α and only if W0 + 2 I has n linearly independent C -eigenfunctions, α ∈ (0, 1), and W0ω ∗ + 12 I has n linearly independent C 0,α -eigenfunctions. The linearly independent C 1,α -eigenfunctions of W0ω + 12 I are the boundary values of the linearly + independent eigensolutions of (N ω 0 ).  (k) n + Proof. Suppose that ω is an eigenfrequency of (N ω be a lin0 ), and let u k=1 early independent set of the corresponding eigensolutions. Following the argument used in the proof of Theorem 6.4, it is easily seen that the boundary values of these eigensolutions are also linearly independent. Since Tu(k) |∂ S = 0, k = 1, 2, . . . , n, from (4.36) and (2.39) we deduce that

 u(k) |∂ S = −(W0ω − 12 I ) u(k) |∂ S , that is,

 (W0ω + 12 I ) u(k) |∂ S = 0,

k = 1, 2, . . . , n;

k = 1, 2, . . . , n,

so W0ω + 12 I has at least n linearly independent eigenfunctions given by the bound+ ary values of the linearly independent eigensolutions of (N ω 0 ). Each of these eigen-

80

6 The Question of Solvability

functions belongs to C 0,α (∂ S) since u(k) ∈ C 1 (S¯+ ) for each k. The assertion that the eigenfunctions of W0ω + 12 I are in C 1,α (∂ S) follows from Theorem 6.2. We now show that W0ω + 12 I has exactly n linearly independent C 0,α -eigenfunctions. Suppose that there are m such eigenfunctions, where m > n. Then, by the Fredholm alternative, W0ω ∗ + 12 I also has m linearly independent C 0,α eigenfunctions ϕ (k) . Consider the single-layer potentials V ω + (ϕ (k) ),

k = 1, 2, . . . , m.

By (2.37),  T V ω + ϕ (k) = (W0ω ∗ + 12 I )ϕ (k) = 0,

k = 1, 2, . . . , m,

+ so each V ω + (ϕ (k) ) is a solution of (Nω 0 ). Suppose that the single-layer poω + (k) tentials V (ϕ ) are linearly dependent in S + ; that is, there are constants a k , k = 1, 2, . . . , m, not all zero, such that m

∑ ak V ω +



ϕ (k) = 0 in S + .

k=1

By the linearity of the single-layer potentials, we obtain

m  V ω + ∑ ak ϕ (k) = 0 in S+ , k=1

so, by Theorem 6.3(i), m

∑ ak ϕ (k) = 0

k=1

on ∂ S.

 m Since the set ϕ (k) k=1 is linearly independent, it follows that a k = 0, k = 1, 2, . . . , m,   m which contradicts our assumption. Therefore, the set V ω + ϕ (k) k=1 is linearly independent. Also, since the ϕ (k) , k = 1,  2, . . . , m, are nonzero, from Theorem 6.3 it follows  that so are the functions V ω + ϕ (k) , k = 1, 2, . . . , m. Consequently, the V ω + ϕ (k) , + k = 1, 2, . . . , m, are linearly independent eigensolutions of (N ω 0 ). But we have asω+ sumed that (N0 ) has only n linearly independent eigensolutions, so m = n. Then, by the Fredholm alternative, W0ω ∗ + 12 I also has exactly n linearly independent C 0,α eigenfunctions. Conversely, suppose that the operator W 0ω ∗ + 12 I has n linearly independent C 0,α eigenfunctions ϕ (k) , k = 1, 2, . . . , n, for a particular value of the parameter ω . As above, the single-layer potentials V ω + ϕ (k) , k = 1, 2, . . . , n, are linearly indepen+ ω+ dent eigensolutions of (N ω 0 ), so ω is an eigenfrequency of (N 0 ) of multiplicω+ ity not less than n. Suppose that (N 0 ) has m linearly independent eigensolutions, where m > n. Then, from what was said above, W 0ω ∗ + 12 I has exactly m linearly independent C 0,α -eigenfunctions, which contradicts our assumption. Hence, ω is + an eigenfrequency of (N ω  0 ) of multiplicity n. 

6.3 The Interior Problems

81

6.3 The Interior Problems The assertions below discuss the solvability of (D ω + ) and (Nω + ). + ω + ) has a unique Theorem 6.6. If ω is not an eigenfrequency of (D ω 0 ), then (D 1, α regular solution for any P ∈ C (∂ S), α ∈ (0, 1), which can be represented in the form u = W ω + (ϕ ), with ϕ ∈ C 1,α (∂ S). + Proof. Suppose that ω is not an eigenfrequency of (D ω 0 ). Then, by Theorem 6.4, the equation (W0ω − 12 I )ϕ = 0

has only the zero solution; so, according to the Fredholm alternative, (6.1) has a unique solution ϕ ∈ C 0,α (∂ S). Since P ∈ C 1,α (∂ S), from Theorem 6.2 we find that ϕ ∈ C 1,α (∂ S); hence, using (2.39), we see that u = W ω + (ϕ ) is a regular solution of + (Dω + ). Because ω is not an eigenfrequency of (D ω  0 ), this solution is unique.  + ω + ) has regular solutions Theorem 6.7. If ω is an eigenfrequency of (D ω 0 ), then (D if and only if



P T (y)T (∂y )u(k) (y) ds(y) = 0,

k = 1, 2, . . . , n,

∂S

 n where P ∈ C 1,α (∂ S), α ∈ (0, 1), and u(k) k=1 is a complete set of linearly inde+ pendent eigensolutions of (D ω 0 ). Each of these solutions can be represented in the ω + form u = W (ϕ ), with ϕ ∈ C 1,α (∂ S). + Proof. Let ω be an eigenfrequency of (D ω 0 ) of multiplicity n. By Theorem 6.4, (k) 1, α there are nonzero functions ϕ ∈ C (∂ S), k = 1, 2, . . . , n, and ψ (k) ∈ C 0,α (∂ S), k = 1, 2, . . . , n, such that

and

(W0ω − 12 I )ϕ (k) = 0,

k = 1, 2, . . . , n,

(W0ω ∗ − 12 I )ψ (k) = 0,

k = 1, 2, . . . , n.

By the Fredholm alternative, (6.1) is solvable if and only if  P, ψ (k) = 0, k = 1, 2, . . . , n; that is, by (6.5), 

P T (y)ψ (k) (y) ds(y) = 0,

k = 1, 2, . . . , n.

∂S

Also, by Theorem 6.4,

ψ (k) = Tu(k) |∂ S ,

k = 1, 2, . . . , n,

82

6 The Question of Solvability

 n + where u(k) k=1 is a complete set of linearly independent eigensolutions of (D ω 0 ). Therefore, we may rewrite the solvability conditions as 

P T (y)T (∂y )u(k) (y) ds(y) = 0,

k = 1, 2, . . . , n.

∂S

If these conditions aresatisfied, then, in view of (2.39), the regular solutions of (Dω + ) are given by W ω + η (k) , where the η (k) , k = 1, 2, . . . , n, are linearly indepen+ dent C 1,α -solutions of (6.1). Since ω is an eigenfrequency of (D ω 0 ), these solutions are not unique.   + ω + ) has a unique Theorem 6.8. If ω is not an eigenfrequency of (N ω 0 ), then (N 0, α regular solution for any Q ∈ C (∂ S), α ∈ (0, 1), which can be represented in the form u = V ω + (ϕ ), with ϕ ∈ C 0,α (∂ S). + Proof. By Theorem 6.5, since ω is not an eigenfrequency of (N ω 0 ), the equation

(W0ω ∗ + 12 I )ϕ = 0 has only the zero solution. By the Fredholm alternative, since Q ∈ C 0,α (∂ S), it follows that (6.2) has a unique solution ϕ ∈ C 0,α (∂ S), and from (2.37) we see that u = V ω + (ϕ ) is a regular solution of (N ω + ), which is unique because of the assumption on ω .   + ω + ) has regular solutions Theorem 6.9. If ω is an eigenfrequency of (N ω 0 ), then (N if and only if 

Q T (y)u(k) (y) ds(y) = 0,

k = 1, 2, . . . , n,

∂S

 n where Q α ∈ (0, 1), and u(k) k=1 is a complete set of linearly inde+ pendent eigensolutions of (N ω 0 ). Each of these solutions can be represented in the ω + form u = V (ϕ ), with ϕ ∈ C 0,α (∂ S). ∈ C 0,α (∂ S),

+ Proof. If ω is an eigenfrequency of (N ω 0 ) of multiplicity n, then, according to Theorem 6.5, there are linearly independent sets of functions ϕ (k) ∈ C 0,α (∂ S), k = 1, 2, . . . , n, and ψ (k) ∈ C 1,α (∂ S), k = 1, 2, . . . , n, respectively, such that

and

(W0ω ∗ + 12 I )ϕ (k) = 0,

k = 1, 2, . . . , n,

(W0ω + 12 I )ψ (k) = 0,

k = 1, 2, . . . , n.

The Fredholm alternative guarantees that (6.2) is solvable if and only if 

 Q, ψ (k) = Q T (y)ψ (k) (y) ds(y) = 0, ∂S

By Theorem 6.5,

k = 1, 2, . . . , n.

6.4 The Exterior Problems

83

ψ (k) = u(k) |∂ S , k = 1, 2, . . . , n,  (k) n + where u is a complete set of linearly independent eigensolutions of (N ω 0 ); k=1 therefore, the solvability conditions become 

Q T (y)u(k) (y) ds(y) = 0,

k = 1, 2, . . . , n.

∂S ω+ If these conditions  (k) hold, then,(k)by (2.37), the regular solutions of (N ) 0,are ω + given by V η , where the η , k = 1, 2, . . . , n, are linearly independent C α solutions of (6.2).  

6.4 The Exterior Problems Similar solvability results can also be established for (D ω − ) and (Nω − ). The only difference here is that, by Theorem 4.3, each of these problems has a unique solution for every value of the oscillation frequency ω . Theorem 6.10. For any R ∈ C 1,α (∂ S), α ∈ (0, 1), and any ω , (D ω − ) has a unique + regular solution. If ω is not an eigenfrequency of (N ω 0 ), then the solution can be ω − 1, α represented in the form u = W (ϕ ), with ϕ ∈ C (∂ S). If ω is an eigenfrequency + of (Nω 0 ), then the solution can be represented in the form n  u = W ω − (ϕ ) + ∑ ck V ω − ψ˜ (k) , k=1

where the ck , k = 1, 2, . . . , n, are constants, ϕ ∈ C 1,α (∂ S), and ψ˜ (k) ∈ C 0,α (∂ S), k = 1, 2, . . . , n, are linearly independent eigenfunctions of W 0ω ∗ + 12 I . + Proof. If ω is not an eigenfrequency of (N ω 0 ), then, by Theorem 6.5, the equation  ω 1 W0 + 2 I ϕ = 0

has only the zero solution. Since R ∈ C 1,α (∂ S), from Theorem 6.2 and the Fredholm alternative we see that (6.3) has a unique solution ϕ ∈ C 1,α (∂ S). Then, by (2.40), u = W ω − (ϕ ) is a regular solution of (D ω − ) which, according to Theorem 4.3, is unique. + Suppose now that ω is an eigenfrequency of (N ω 0 ) of multiplicity n. By Theorem 1 ω 1, α 6.5, W0 + 2 I has n linearly independent C -eigenfunctions and W0ω ∗ + 12 I has n linearly independent C 0,α -eigenfunctions. We consider such eigenfunctions ϕ (k) and ψ (k) , k = 1, 2, . . . , n, respectively. By the Fredholm alternative, (6.3) is solvable if and only if   R, ψ (k) = R T (y)ψ (k) (y) ds(y) = 0, k = 1, 2, . . . , n. ∂S

84

6 The Question of Solvability

 n  n Without loss of generality, we assume that the sets ϕ (k) k=1 and ψ (k) k=1 have been biorthonormalized by means of the procedure described in [40]; that is, 

  (m) T (n) ϕ (m) , ψ (n) = ϕ (y) ψ (y) ds(y) = δmn .

(6.10)

∂S

Consider the function n

˜ R(y) = R(y) − ∑



l=1

(l)



R (y)ψ (y) ds(y) ϕ (l) (y). T

(6.11)

∂S

By our assumption, R ∈ C 1,α (∂ S) and ϕ (l) ∈ C 1,α (∂ S) for all l = 1, 2, . . . , n, so R˜ ∈ C 1,α (∂ S). From (6.10) and (6.11) it follows that 

R˜ T (y)ψ (k) (y) ds(y) = 0,

k = 1, 2, . . . , n,

∂S

which ensures the solvability of the equation  ω 1 ˜ W0 + 2 I ϕ = R.

(6.12)

 n Since ψ (k) k=1 is a linearly independent set of C 0,α -eigenfunctions of the op erator W0ω ∗ + 12 I , the single-layer potentials V ω + ψ (k) , k = 1, 2, . . . , n, are linearly + independent eigensolutions of (N ω shown in the proof of Theorem 6.5. By 0 ),as is  n (k) Theorem 6.5, we can construct a set ψ˜ of linear combinations of the ψ (k) k=1 satisfying V0ω ψ˜ (k) = ϕ (k) , k = 1, 2, . . . , n. Let ϕ ∈ C 1,α (∂ S) be any solution of (6.12). Then the function  n   u = W ω − (ϕ ) + ∑ R T (y)ψ (k) (y) ds(y) V ω − ψ˜ (k) k=1

∂S

is a solution of (Dω − ), whose uniqueness is guaranteed by Theorem 4.3.   Theorem 6.11. For any S ∈ C 0,α (∂ S), α ∈ (0, 1), and any ω , (N ω − ) has a unique + regular solution. If ω is not an eigenfrequency of (D ω 0 ), then the solution can be ω − 0, α represented in the form u = V (ϕ ), with ϕ ∈ C (∂ S). If ω is an eigenfrequency + of (Dω ), then the solution can be represented in the form 0 n  u = V ω − (ϕ ) + ∑ ck W ω − ψ˜ (k) , k=1

where the ck , k = 1, 2, . . . , n, are constants, ϕ ∈ C 0,α (∂ S), and ψ˜ (k) ∈ C 1,α (∂ S), k = 1, 2, . . . , n, are linearly independent eigenfunctions of W 0ω − 12 I .

6.4 The Exterior Problems

85

+ Proof. First, suppose that ω is not an eigenfrequency of (D ω 0 ). Then, by Theorem 6.4, the equation  ω∗ 1 W0 − 2 I ϕ = 0

has only the zero solution, so, according to the Fredholm alternative, (6.4) has a unique solution ϕ ∈ C 0,α (∂ S). Consequently, by (2.38), u = V ω − (ϕ ) is a regular solution of (Nω − ) which, by Theorem 4.3, is unique. + If ω is an eigenfrequency of (D ω ) of multiplicity n, then, by Theorem 6.4, there  (k) n  (k) n 0 are sets ϕ and ψ of linearly independent C 0,α -eigenfunctions of k=1 k=1 1 ω ∗ W0 − 2 I and linearly independent C 1,α -eigenfunctions of W0ω − 12 I , respectively. By the Fredholm alternative, (6.4) is solvable if and only if 

 S , ψ (k) = S T (y)ψ (k) (y) ds(y) = 0,

k = 1, 2, . . . , n.

∂S

As in the proof of Theorem 6.10, we may assume that the two sets of eigenfunctions have been biorthonormalized (see [40]), which means that 

  (m) T (n) ϕ (m) , ψ (n) = ϕ (y) ψ (y) ds(y) = δmn .

(6.13)

∂S

We introduce the function n

S˜(y) = S (y) − ∑



l=1

S (y)ψ T

(l)



ds(y) ϕ (l) (y),

(6.14)

∂S

which is of class C 0,α (∂ S) since S ∈ C 0,α (∂ S) and ϕ (l) ∈ C 0,α (∂ S), l = 1, 2, . . . , n. From (6.13) and (6.14) it is easily seen that 

S˜T (y)ψ (k) (y) ds(y) = 0,

k = 1, 2, . . . , n,

∂S

which means that the equation  ω∗ 1 W0 − 2 I ϕ = S˜

(6.15)

has C 0,α -solutions. As in the proof of Theorem 6.4, we can show that the double-layer potentials + W ω + (ψ (k) ), k = 1, 2, . . . , n, are linearly independent eigensolutions of (N ω 0 ). Using  (k) n the result of Theorem 6.4, we can construct a set ψ˜ of linear combinations k=1 (k) of the ψ satisfying  T W ω + (ψ˜ (k) )∂ S = ϕ (k) , k = 1, 2, . . . , n. Taking (2.41) into account, we obtain  T W ω − (ψ˜ (k) ) = ϕ (k) , ∂S

k = 1, 2, . . . , n.

86

6 The Question of Solvability

We choose any solution ϕ ∈ C 0,α (∂ S) of (6.15) and consider the function  n  ω− T (k) u = V (ϕ ) + ∑ S (y)ψ (y) ds(y) W ω − (ψ˜ (k) ). k=1

∂S

This function is a solution of (N ω − ) which, by Theorem 4.3, is unique.   Even though the exterior problems are known to have unique regular solutions, we are still, for certain values of ω , faced with integral equations that may have more than one solution. This problem of nonuniqueness is built into the chosen ‘indirect’ boundary integral equation formulation and will be addressed in the next chapter.

Chapter 7

Direct Boundary Equation Formulation

To overcome the difficulties that arose in the indirect method—that is, the ill-posed nature of certain integral equations even though the boundary value problems they represent have at most one solution—in this chapter we do not assume a priori that the solutions of the boundary value problems take the form of specific layer potentials. Instead, using the Somigliana representation formulas (4.36) and (4.37), we apply the so-called direct method to obtain an integral equation of the second kind for an unknown density. These equations alone do not, however, prevent the nonuniqueness of solution since they involve the same boundary integral operators encountered in the indirect method. An equation of the first kind is also derived for each problem, and we show that, in the case of (D ω − ) and (Nω − ), the pair of second-kind and first-kind integral equations constructed in this way has a unique solution irrespective of the oscillation frequency ω . In Section 7.1 we show that the boundary operators occurring in the direct formulation are connected by four composition relations, which turn out to be extremely useful in manipulating the solutions of the corresponding integral equations. The pairs of equations characterizing the solutions of our four boundary value problems are derived in Section 7.2, and their solvability is discussed in Sections 7.3 and 7.5. In doing so, we uncover many other results connecting the eigenfrequencies of the homogeneous interior problems and the eigenfunctions of the boundary operators. In Section 7.4 we indicate how the eigensolutions of the homogeneous interior problems can be represented as both single-layer and double-layer potentials. This is essential in establishing the solvability of the pairs of equations representing the exterior problems when ω is an eigenfrequency of the ‘adjoint’ homogeneous interior problem (see Theorems 7.9 and 7.11). Finally, in Section 7.6 we show that the solutions of each of (D ω − ) and (Nω − ) can be constructed from single, uniquely solvable integral equations. These equations are obtained by taking a certain linear combination of the first-kind and second-kind equations derived in Section 7.2. A similar approach was used in [32] (see also the references cited there) to discuss the Dirichlet and Neumann problems for the three-dimensional Helmholtz equation. In what follows we simplify many of the ideas developed in [32] by making use of the composition relations for the relevant boundary operators. G.R. Thomson and C. Constanda, Stationary Oscillations of Elastic Plates, A Boundary Integral Equation Analysis, DOI 10.1007/978-0-8176-8241-5_7, © Springer Science+Business Media, LLC 2011

87

88

7 Direct Boundary Equation Formulation

7.1 Composition of the Boundary Operators We introduce an operator N 0ω : C 1,α (∂ S) → C 0,α (∂ S) by writing N0ω ϕ = T W ω + (ϕ ) = T W ω − (ϕ ).

(7.1)

The various boundary operators satisfy some useful composition formulas. Analogous formulas for the three-dimensional Helmholtz equation can be found in [51]. The procedure used below, however, is that described in [23]. Let ϕ ∈ C 1,α (∂ S) and ψ ∈ C 0,α (∂ S) be arbitrary, and consider the function u = V ω + (ψ ) − W ω + (ϕ ) in S¯+ . Then, by (2.37), (2.39), and (7.1),  u|∂ S = V0ω ψ − W0ω − 12 I ϕ ,  Tu|∂ S = W0ω ∗ + 12 I ψ − N0ω ϕ .

(7.2)

Since u is a regular solution of (4.1) in S¯+ , from (4.36) we deduce that  u|∂ S = V0ω (Tu|∂ S ) − W0ω − 12 I (u|∂ S ),  Tu|∂ S = W0ω ∗ + 12 I (Tu|∂ S ) − N0ω (u|∂ S ). If we set u|∂ S

U=

Q=

!

Tu|∂ S

,

! ϕ Φ= , ψ

 − W0ω − 12 I −N0ω

(7.3)

V0ω W0ω ∗ + 12 I

! ,

then (7.2) and (7.3) take the form U = QΦ

(7.4)

U = QU,

(7.5)

and respectively. Substituting (7.4) in (7.5) leads to (Q 2 − Q)Φ = 0; so, given the arbitrariness of Φ in C 1,α (∂ S) × C 0,α (∂ S), we have Q 2 = Q, which implies that

7.2 Boundary Integral Equations

89

W0ω − W0ω + 14 I −V0ω N0ω = −W0ω + 12 I 2

on C 1,α (∂ S),

− W0ω V0ω + 12 V0ω +V0ω W0ω ∗ + 12 V0ω = V0ω

on C 0,α (∂ S),

N0ω W0ω − 12 N0ω − W0ω ∗ N0ω − 12 N0ω = −N0ω

on C 1,α (∂ S),

− N0ω V0ω + W0ω ∗ +W0ω ∗ + 14 I = W0ω ∗ + 12 I

on C 0,α (∂ S),

2

and we arrive at the composition formulas W0ω V0ω = V0ω W0ω ∗ ω ∗2

N0ω V0ω = W0

on C 0,α (∂ S),

− 14 I

N0ω W0ω = W0ω ∗ N0ω ω2

V0ω N0ω = W0 − 14 I

on C 0,α (∂ S), on C 1,α (∂ S), on C 1,α (∂ S).

(7.6) (7.7) (7.8) (7.9)

7.2 Boundary Integral Equations We use the representation formulas (4.36) and (4.37) to reformulate (D ω + ), (Nω + ), (Dω − ), and (Nω − ) as integral equations of the second kind. Consider the interior Neumann problem. Applying the given boundary condition Tu|∂ S = Q to (4.36) yields  V0ω Q − W0ω − 12 I u|∂ S = u|∂ S . Writing ϕ = u|∂ S , we obtain the integral equation  ω 1 W0 + 2 I ϕ = V0ω Q.

(7.10)

By (4.36), a solution u of (4.1) in S + is given by u = V ω + (Q) − W ω + (ϕ ),

(7.11)

where ϕ is a solution of (7.10). Since u must also satisfy the boundary condition, we require that Tu|∂ S = T V ω + (Q) − T W ω + (ϕ ) = Q, which leads to

 N0ω ϕ = W0ω ∗ − 12 I Q.

(7.12)

Consequently, u given by (7.11) is a solution of (N ω + ) in S + if ϕ satisfies both (7.10) and (7.12). Equations corresponding to the other three boundary value problems can be derived by the same procedure. Thus, a solution of (D ω + ) in S + is given by u = V ω + (ϕ ) − W ω + (P),

(7.13)

90

7 Direct Boundary Equation Formulation

where ϕ satisfies and

 ω∗ 1 W0 − 2 I ϕ = N0ω P

(7.14)

 V0ω ϕ = W0ω + 12 I P.

(7.15)

A solution of (Dω − ) in S − is given by

where ϕ satisfies and

u = W ω − (R) − V ω − (ϕ ),

(7.16)

 ω∗ 1 W0 + 2 I ϕ = N0ω R

(7.17)

 V0ω ϕ = W0ω − 12 I R.

(7.18)

A solution of (Nω − ) in S − is given by

where ϕ satisfies and

u = W ω − (ϕ ) − V ω − (S ),

(7.19)

 ω 1 W0 − 2 I ϕ = V0ω S

(7.20)

 N0ω ϕ = W0ω ∗ + 12 I S .

(7.21)

The solvability of these pairs of equations is discussed in the following sections. For the integral equations corresponding to the interior problems we can prove existence and uniqueness results in all cases except when ω is an eigenfrequency of the problem. For (D ω − ) and (Nω − ), existence and uniqueness hold for all values of the parameter ω .

7.3 The Interior Problems First, we establish some auxiliary results concerning the eigenfunctions of the boundary integral operators. Theorem 7.1. If ϕ ∈ C 0,α (∂ S), α ∈ (0, 1), is nonzero, then V0ω ϕ = 0 if and only if ϕ is an eigenfunction of W0ω ∗ − 12 I . Proof. Suppose that V0ω ϕ = 0 and consider u = V ω − (ϕ ). Then u is a solution of − ¯− (Dω 0 ) and, by Theorem 4.3, u = 0 in S ; therefore,  0 = T V ω − (ϕ ) = W0ω ∗ − 12 I ϕ on ∂ S, as required. Suppose now that ϕ is an eigenfunction of W 0ω ∗ − 12 I and let u = V ω − (ϕ ). Since,  − by assumption, Tu| ∂ S = W0ω ∗ − 12 I ϕ = 0, we deduce that u is a solution of (N ω 0 );

7.3 The Interior Problems

so, by Theorem 4.3,

91

u = V ω − (ϕ ) = 0 in S − .

From the continuity of the single-layer potential we see that V0ω ϕ = 0, which completes the proof.   Corollary 7.1. (i) There is a nonzero function ϕ ∈ C 0,α (∂ S), α ∈ (0, 1), such that + V0ω ϕ = 0 if and only if ω is an eigenfrequency of (D ω 0 ). (ii) If ϕ is a C 0,α -eigenfunction of W0ω ∗ + 12 I , then V0ω ϕ = 0. Proof. (i) This is a direct consequence of Theorems 7.1 and 6.4. (ii) If ϕ ∈ C 0,α (∂ S) is an eigenfunction of W0ω ∗ + 12 I , then  ω∗ 1 W0 − 2 I ϕ = −ϕ = 0, so, by Theorem 7.1, V0ω ϕ = 0.   Theorem 7.2. If ϕ ∈ C 1,α (∂ S), α ∈ (0, 1), is nonzero, then N0ω ϕ = 0 if and only if ϕ is an eigenfunction of W0ω + 12 I . − Proof. Suppose that N0ω ϕ = 0. Then u = W ω − (ϕ ) is a solution of (N ω 0 ), so u = 0 − 1,α ω − − ¯ in S . Since ϕ ∈ C (∂ S), W (ϕ ) is continuous in S and we deduce that  0 = W ω − (ϕ )|∂ S = W0ω + 12 I ϕ .

If, on the other hand, ϕ is a C 1,α -eigenfunction of W0ω + 12 I , then u = W ω − (ϕ ) − ¯− is a solution of (Dω 0 ), so u = 0 in S ; hence, as required.  

Tu|∂ S = N0ω ϕ = 0,

Corollary 7.2. (i) There is a nonzero function ϕ ∈ C 1,α (∂ S), α ∈ (0, 1), such that + N0ω ϕ = 0 if and only if ω is an eigenfrequency of (N ω 0 ). (ii) If ϕ is a C 1,α -eigenfunction of W0ω − 12 I , then N0ω ϕ = 0. Proof. (i) This follows from Theorems 7.2 and 6.5. (ii) Let ϕ be a C 1,α -eigenfunction of W0ω − 12 I . Then  ω 1 W0 + 2 I ϕ = ϕ = 0, so, by Theorem 7.2, N 0ω ϕ = 0.   Using the composition relations (7.6)–(7.9) together with Corollaries 7.1(ii) and 7.2(ii), we can establish connections between the eigenfunctions of W 0ω ± 12 I and those of W0ω ∗ ± 12 I.

92

7 Direct Boundary Equation Formulation

Theorem 7.3. (i) If ϕ ∈ C 0,α (∂ S), α ∈ (0, 1), is an eigenfunction of W0ω ∗ + 12 I , then V0ω ϕ is an eigenfunction of W0ω + 12 I . (ii) If ϕ ∈ C 1,α (∂ S), α ∈ (0, 1), is an eigenfunction of W0ω − 12 I , then N0ω ϕ is an eigenfunction of W0ω ∗ − 12 I . Proof. (i) Suppose that ϕ is a C 0,α -eigenfunction of W0ω ∗ + 12 I ; that is,  ω∗ 1 W0 + 2 I ϕ = 0. By (7.6),

  0 = V0ω W0ω ∗ + 12 I ϕ = W0ω + 12 I (V0ω ϕ ).

By Corollary 7.1(ii), V0ω ϕ is nonzero and is, therefore, an eigenfunction of W 0ω + 12 I . (ii) Let ϕ be a C 1,α -eigenfunction of W0ω − 12 I . Then, by (7.8),   0 = N0ω W0ω − 12 I ϕ = W0ω ∗ − 12 I (N0ω ϕ ). Corollary 7.2(ii) guarantees that N 0ω ϕ is nonzero and is, therefore, an eigenfunction of W0ω ∗ − 12 I .   The above theorem offers an alternative derivation of the statements of Theorems 6.4 and 6.5 concerning the relationships between the eigenfunctions of W 0ω ∗ − 12 I and W0ω + 12 I and the eigensolutions of (D ω + ) and (Nω + ), respectively. Consider the integral equations (7.14) and (7.15) representing (D ω + ), in the case + when ω is not an eigenfrequency of (D ω 0 ). According to the Fredholm alternative and Theorem 6.4, (7.14) has a unique solution (assuming that N 0ω P is sufficiently smooth), so, by Corollary 7.1(i), (7.15) has at most one solution. We now show that the unique solution of (7.14) also satisfies (7.15), which means that we require only one of these integral equations to construct the solution (7.13) of (D ω + ). + 1,α (∂ S) with Theorem 7.4. If ω is not an eigenfrequency of (D ω 0 ) and P ∈ C α ∈ (0, 1), then the equations  ω∗ 1 W0 − 2 I ϕ = N0ω P

and

 V0ω ϕ = W0ω + 12 I P

have a unique solution ϕ ∈ C 0,α (∂ S) and  ω∗ 1 W0 − 2 I ϕ = N0ω P if and only if

 V0ω ϕ = W0ω + 12 I P.

Proof. If P ∈ C 1,α (∂ S), then N0ω P ∈ C 0,α (∂ S), so, by Theorem 6.4 and the Fredholm alternative, (7.14) has a unique solution ϕ ∈ C 0,α (∂ S). Using (7.6) and (7.9), we find that

7.3 The Interior Problems

93

 0 = W0ω ∗ − 12 I ϕ − N0ω P   = V0ω W0ω ∗ − 12 I ϕ − N0ω P     = W0ω − 12 I V0ω ϕ − W0ω + 12 I P . By Theorem 6.4,

 V0ω ϕ = W0ω + 12 I P;

that is, ϕ is also a solution of (7.15). According to Corollary 7.1(i), this is the only solution of (7.15). Again by (7.6) and (7.9), if ϕ is a solution of (7.15), then  0 = V0ω ϕ − W0ω + 12 I P     = W0ω − 12 I V0ω ϕ − W0ω + 12 I P   = V0ω W0ω ∗ − 12 I ϕ − N0ω P . Consequently, by Corollary 7.1(i), ϕ is a solution of (7.14), and the theorem is proved.   We can establish similar results for (Nω + ). The next assertion shows that when + ω is not an eigenfrequency of (N ω 0 ), we need only one of the equations (7.10) and (7.12) to construct the solution (7.11). + 0,α (∂ S), α ∈ (0, 1), Theorem 7.5. If ω is not an eigenfrequency of (N ω 0 ) and Q ∈ C then the equations  ω 1 W0 + 2 I ϕ = V0ω Q

and

 N0ω ϕ = W0ω ∗ − 12 I Q

have a unique solution ϕ ∈ C 1,α (∂ S) and  ω 1  W0 + 2 I ϕ = V0ω Q if and only if N0ω ϕ = W0ω ∗ − 12 I Q. Proof. If Q ∈ C 0,α (∂ S), then, by Theorem 2.6, V 0ω Q ∈ C 1,α (∂ S). Theorem 6.5 and the Fredholm alternative guarantee the existence of a solution ϕ ∈ C 0,α (∂ S) of (7.10). In fact, since the right-hand side of (7.10) is of class C 1,α (∂ S), Theorem 6.2 states that ϕ ∈ C 1,α (∂ S). From the composition relations (7.7) and (7.8) we see that  0 = W0ω + 12 I ϕ −V0ω Q   = N0ω W0ω + 12 I ϕ − V0ω Q     = W0ω ∗ + 12 I N0ω ϕ − W0ω ∗ − 12 I Q . + Since ω is not an eigenfrequency of (N ω 0 ), Theorem 6.5 yields

 N0ω ϕ = W0ω ∗ − 12 I Q.

94

7 Direct Boundary Equation Formulation

Consequently, ϕ is a solution of (7.12) which, by Corollary 7.2(i), is unique. If ϕ is a solution of (7.12), from (7.8) and (7.9) we find that  0 = N0ω ϕ − W0ω ∗ − 12 I Q     = W0ω ∗ + 12 I N0ω ϕ − W0ω ∗ − 12 I Q   = N0ω W0ω + 12 I ϕ −V0ω Q , so, by Corollary 7.2(i),

 ω 1 W0 + 2 I ϕ = V0ω Q,

which completes the proof.  

7.4 The Eigensolutions of the Interior Problems To prove the solvability of the integral equations associated with the exterior prob+ ω+ lems, we must first consider the eigensolutions of (D ω 0 ) and (N 0 ). These eigensolutions may be represented as either single-layer or double-layer potentials. + Theorem 7.6. (i) A function u is an eigensolution of problem (D ω 0 ) if and only if 1 ω + + 0, α ω ∗ u = V (ϕ ) in S¯ , where ϕ is a C -eigenfunction of W0 − 2 I . + ω + (ϕ ) in (ii) The function u is an eigensolution of (D ω 0 ) if and only if u = W 1 + 1, α ω S¯ , where ϕ is a C -eigenfunction of W0 − 2 I . + Proof. (i) Suppose that u is an eigensolution of (D ω 0 ). Representation formula (4.36) yields u = V ω + (Tu|∂ S ) in S¯+ .

By (2.37), that is,

 Tu|∂ S = T V ω + (Tu|∂ S ) = W0ω ∗ + 12 I (Tu|∂ S );  ω∗ 1 W0 − 2 I (Tu|∂ S ) = 0,

and writing ϕ = Tu| ∂ S leads to the required result. Suppose now that ϕ is a C 0,α -eigenfunction of W0ω ∗ − 12 I . Then from Theorem + 7.1 we see that V0ω ϕ = 0, so u = V ω + (ϕ ) is a solution of (Dω 0 ). Since ϕ is nonzero, by Theorem 6.3(i) so is u; therefore, u is an eigensolution of the problem. (ii) Suppose that ϕ is a C 1,α -eigenfunction of W0ω − 12 I and let u = W ω + (ϕ ). + From (2.39) we see that u| ∂ S = 0, so u is a solution of (D ω 0 ). By Theorem 6.3(ii), u ω+ is nonzero and is, therefore, an eigensolution of (D 0 ). + Let u be an eigensolution of (D ω 0 ) for an eigenfrequency of multiplicity n. Then, by Theorem 6.4, the operator W 0ω − 12 I has n linearly independent C 1,α  n eigenfunctions ϕ (k) . In the proof of Theorem 6.4 it was shown that the set u(k) k=1 , where  u(k) = W ω + ϕ (k) , k = 1, 2, . . . , n,

7.4 The Eigensolutions of the Interior Problems

95

+ is a basis for the eigenspace of (D ω 0 ). Hence, there are constants b k , k = 1, 2, . . . , n, such that n n  u = ∑ bk u(k) = ∑ bk W ω + ϕ (k) = W ω + (ϕ ), k=1

where ϕ =

n

∑ bk ϕ (k) .

k=1

 

k=1

+ Theorem 7.7. (i) The function v is an eigensolution of problem (N ω 0 ) if and only if 1 ω + + 1, α ω v = −W (ϕ ) in S , where ϕ is a C -eigenfunction of W0 + 2 I . + ω + (ϕ ) in S+ , (ii) The function v is an eigensolution of (N ω 0 ) if and only if v = V 1 0,α ω ∗ where ϕ is a C -eigenfunction of W0 + 2 I . + Proof. (i) Let v be an eigensolution of (N ω 0 ). Then, by (4.36),

v = −W ω + (v|∂ S ) in S¯+ , so, using (2.39), we see that  v|∂ S = − W0ω − 12 I (v|∂ S ), which leads to

 ω 1 W0 + 2 I ϕ = 0,

where ϕ = v|∂ S . Suppose that ϕ is a C 1,α -eigenfunction of W0ω + 12 I . Then, by Theorem 7.2, N0ω ϕ = T W ω + (ϕ ) = 0; + consequently, v = −W ω + (ϕ ) is a solution of (Nω 0 ). Since ϕ is nonzero, by Theo+ rem 6.3(ii) so is v, which implies that v is an eigensolution of (N ω 0 ). (ii) Consider the function v = V ω + (ϕ ), where ϕ is a C 0,α -eigenfunction of + ω W0 ∗ + 12 I . From (2.37) we see that v is a solution of (N ω 0 ), which, by Theorem + 6.3(i), is nonzero. Consequently, v is an eigensolution of (N ω 0 ). ω+ Suppose now that v is an eigensolution of problem (N 0 ) for an eigenfrequency of multiplicity n. In the proof of Theorem 6.5 it was shown that the functions  v(k) = V ω + ϕ (k) , k = 1, 2, . . . , n,

where the ϕ (k) are linearly independent eigenfunctions of W 0ω ∗ + 12 I , form a basis + for the eigenspace of (N ω 0 ). Hence, there are constants a k , k = 1, 2, . . . , n, such that v= where ϕ =

n

n

n

k=1

k=1

∑ ak v(k) = ∑ ak V ω +

∑ ak ϕ (k) , as required.

k=1

 

 (k) ϕ = V ω + (ϕ ),

96

7 Direct Boundary Equation Formulation

7.5 The Exterior Problems We can now show that even in the case of interior eigenfrequencies the integral equation formulations of the exterior problems have unique solutions. We start with (Dω − ) and the corresponding integral equations (7.17) and (7.18). + Theorem 7.8. If ω is not an eigenfrequency of (N ω 0 ), then the equation

 ω∗ 1 W0 + 2 I ϕ = N0ω R has a unique solution ϕ ∈ C 0,α (∂ S) for every R ∈ C 1,α (∂ S), α ∈ (0, 1). This solution also satisfies  V0ω ϕ = W0ω − 12 I R. Proof. By the Fredholm alternative and Theorem 6.5, since N 0ω R ∈ C 0,α (∂ S), equation (7.17) has a unique solution ϕ ∈ C 0,α (∂ S). Using (7.8) and (7.9), we find that  0 = W0ω ∗ + 12 I ϕ − N0ω R   ω ∗ 1  = W0ω ∗ − 12 I W0 + 2 I ϕ − N0ω R    = N0ω V0ω ϕ − W0ω − 12 I R and that, by Corollary 7.2(i),  V0ω ϕ = W0ω − 12 I R, which proves the theorem.   + Theorem 7.9. If ω is an eigenfrequency of (N ω 0 ), then the equation

 V0ω ϕ = W0ω − 12 I R has a unique solution ϕ ∈ C 0,α (∂ S) for every R ∈ C 1,α (∂ S), α ∈ (0, 1). This solution also satisfies  ω∗ 1 W0 + 2 I ϕ = N0ω R. Proof. First, we show that (7.17) is solvable; that is, that the eigenfunctions of W0ω + 12 I are orthogonal to N0ω R with respect to the bilinear form (6.5). Let u = W ω + (R) and

v = −W ω + (η ),

where η is an eigenfunction of W0ω + 12 I . By Theorem 7.7(i), v is an eigensolution + of (Nω 0 ), so T v|∂ S = 0. Applying Theorem 1.2 to u and v yields

7.5 The Exterior Problems

97



(vT Tu − uTT v) ds = 0;

∂S

that is,



vT (y) (N0ω R) (y) ds(y) = 0.

(7.22)

∂S

 Since W0ω + 12 I η = 0, from (2.39) it follows that  v|∂ S = − W0ω − 12 I η = η , so (7.22) leads to



η T (y)(N0ω R)(y) ds(y) = 0,

∂S

as required. Hence, by the Fredholm alternative, there are C 0,α -solutions, not necessarily unique, of (7.17). Let ϕ1 be any such solution. Then  ω ∗ 1  ω ∗ 1  W0 − 2 I W0 + 2 I ϕ1 − N0ω R = 0; so, using (7.7) and (7.8), we find that    N0ω V0ω ϕ1 − W0ω − 12 I R = 0. Therefore, either ϕ 1 is a solution of (7.18) or  V0ω ϕ1 = W0ω − 12 I R + η ,

(7.23)

where, according to Theorem 7.2, η is an eigenfunction of W 0ω + 12 I . By Theorem 6.5, η = v|∂ S , + where v is an eigensolution of (N ω 0 ). Consequently, Theorem 7.7(ii) implies that 0,α there is an eigenfunction ψ ∈ C (∂ S) of W0ω ∗ + 12 I such that

η = V0ω ψ . Substituting this in (7.23) yields  V0ω (ϕ1 − ψ ) = W0ω − 12 I R, so (7.18) has at least one C 0,α -solution. Since ϕ1 is a solution of (7.17) and ψ , being an eigenfunction of W 0ω ∗ + 12 I, satisfies the homogeneous version of (7.17), it follows that ϕ 1 − ψ is a solution of (7.17). As we have seen, ϕ 1 − ψ is also a solution of (7.18).

98

7 Direct Boundary Equation Formulation (0)

To show that (7.18) is uniquely solvable in C 0,α (∂ S), consider a pair ϕ 1 , ψ (0) constructed as above with R = 0; in other words, such that  ω ∗ 1 (0) W0 + 2 I (ϕ1 − ψ (0) ) = 0. (0)

Given that ϕ1 − ψ (0) is also a solution of (7.18) with R = 0, the function v=V satisfies

ω+

(0)

(ϕ1 − ψ (0))

Aω v = 0 in S + , (0)

v|∂ S = V0ω (ϕ1 − ψ (0) ) = 0,  (0) T v|∂ S = W0ω ∗ + 12 I (ϕ1 − ψ (0) ) = 0, so, by the interior representation formula (4.36), V

ω+

(0)

(ϕ1 − ψ (0)) = v = V

ω+

(T v|∂ S ) − W

ω+

(v|∂ S ) = 0.

(0)

Theorem 6.3(i) now implies that ϕ 1 − ψ (0) = 0, which means that (7.18) has a unique solution.   We remark that the above result is stronger than the corresponding result derived in [32] for the three-dimensional Helmholtz equation. There, it was shown only that the pair of equations has a unique solution, whereas Theorem 7.9 proclaims that the equation of the first kind has a unique solution which also satisfies the second-kind equation. Similar results can be proved for the integral equation formulation of (N ω − ). + Theorem 7.10. If ω is not an eigenfrequency of (D ω 0 ), then the equation  ω 1 W0 − 2 I ϕ = V0ω S

has a unique solution ϕ ∈ C 1,α (∂ S) for every S ∈ C 0,α (∂ S), α ∈ (0, 1). This solution also satisfies  N0ω ϕ = W0ω ∗ + 12 I S . Proof. By Theorem 6.4 and the Fredholm alternative, (7.20) has a unique solution ϕ ∈ C 0,α (∂ S). Since, by Theorem 2.6, V 0ω S ∈ C 1,α (∂ S), using Theorem 6.2 we conclude that ϕ ∈ C 1,α (∂ S). The composition relations (7.6) and (7.9) lead to  0 = W0ω − 12 I ϕ − V0ω S    = W0ω + 12 I W0ω − 12 I ϕ − V0ω S    = V0ω N0ω ϕ − W0ω ∗ + 12 I S ,

7.5 The Exterior Problems

99

which, in view of Corollary 7.2(i), yields  N0ω ϕ = W0ω ∗ + 12 I S , as required.   + Theorem 7.11. If ω is an eigenfrequency of (D ω 0 ), then the equation

 N0ω ϕ = W0ω ∗ + 12 I S has a unique solution ϕ ∈ C 1,α (∂ S) for every S ∈ C 0,α (∂ S), α ∈ (0, 1). This solution also satisfies  ω 1 W0 − 2 I ϕ = V0ω S . Proof. Let η be an eigenfunction of W 0ω ∗ − 12 I . since, by Theorem 7.1, V0ω η = 0, we have 

η T (y) (V0ω S ) (y) ds(y) =

∂S



η T (y)

∂S

=

 

∂S

 η (y)D ω (y, x) ds(y) S (x) ds(x)

∂S

 

∂S

=

D ω (y, x)S (x) ds(x) ds(y)

T

∂S

=



 

ω

D (x, y)η (y) ds(y)

T S (x) ds(x)

∂S

T (V0ω η )(x) S (x) ds(x) = 0.

∂S

Consequently, there are solutions, not necessarily unique, of (7.20). By Theorem 6.2, since V0ω S ∈ C 1,α (∂ S), these solutions are of class C 1,α (∂ S). Suppose that ϕ1 is any such solution. Then  ω 1  ω 1  W0 + 2 I W0 − 2 I ϕ1 − V0ω S = 0, so, using (7.6) and (7.9), we deduce that    V0ω N0ω ϕ1 − W0ω ∗ + 12 I S = 0. Hence, either ϕ1 is a solution of (7.21) or  N0ω ϕ1 = W0ω ∗ + 12 I S + η ,

(7.24)

where, by Theorem 7.1, η is an eigenfunction of W 0ω ∗ − 12 I . According to Theorem 6.4, η = Tu|∂ S , + 1,α (∂ S) where u is an eigensolution of (D ω 0 ). By Theorem 7.6(ii), there is ψ ∈ C such that

100

7 Direct Boundary Equation Formulation

η = T W ω + (ψ )|∂ S = N0ω ψ .

(7.25)

Substituting (7.25) in (7.24) leads to  N0ω (ϕ1 − ψ ) = W0ω ∗ + 12 I S , so the existence of at least one solution of (7.21) is guaranteed. An argument similar to that used in the proof of Theorem 7.9 can now be applied, according to which ϕ 1 − ψ is a solution of both (7.20) and (7.21). (0) Let ϕ1 , ψ (0) be a pair constructed as above with S = 0. Then the function w=W

ω+

(0)

(ϕ1 − ψ (0) )

satisfies the homogeneous versions of (7.20) and (7.21), so Aω w = 0 in S+ ,  (0) w|∂ S = W0ω − 12 I (ϕ1 − ψ (0) ) = 0, (0)

Tw|∂ S = N0ω (ϕ1 − ψ (0) ) = 0. By the interior representation formula (4.36), W

ω+

(0)

(ϕ1 − ψ (0) ) = w = V

ω+

(Tw|∂ S ) − W

ω+

(w|∂ S ) = 0,

(0)

and from Theorem 6.3(ii) we conclude that ϕ 1 − ψ (0) = 0; therefore, (7.21) has a unique solution.   Again, we remark that through the application of the composition relations we have proved a stronger result than in the case of the three-dimensional Helmholtz equation [32].

7.6 Composite Integral Equations In the preceding section we showed that there is a unique function satisfying a coupled pair of integral equations representing each of the exterior problems, irrespective of the oscillation frequency. We now consider forming single composite integral equations by taking a linear combination of the equations in each of the coupled pairs. This approach was used in [8] to study the exterior Neumann problem for the three-dimensional Helmholtz equation. The integral equations arising in problems of time-harmonic waves in three-dimensional elastodynamics have been investigated by the same method in [30]. Let Λ = (Λi j ) be a constant (3 × 3)-matrix whose entries are such that

Λi j = Λ¯ ji

for i = j

(7.26)

7.6 Composite Integral Equations

101

and either Im(Λ11 ), Im(Λ22 ), Im(Λ33 ) > 0

(7.27)

Im(Λ11 ), Im(Λ22 ), Im(Λ33 ) < 0.

(7.28)

or We consider a linear combination of (7.17) and (7.18). Theorem 7.12. If (7.26) and either (7.27) or (7.28) hold, then the equation  ω∗ 1  W0 + 2 I ϕ + Λ V0ω ϕ = N0ω R + Λ W0ω − 12 I R

(7.29)

has a unique solution ϕ ∈ C 0,α (∂ S) for every R ∈ C 1,α (∂ S), α ∈ (0, 1). Proof. We already know from Theorems 7.8 and 7.9 that (7.29) has a C 0,α -solution. It remains to show that this solution is the only one. Consider the corresponding homogeneous equation  ω∗ 1 W0 + 2 I ϕ + Λ V0ω ϕ = 0. (7.30) Let ϕ be any solution of (7.30) and define u = V ω + (ϕ ). Then, by (7.30), Tu|∂ S + Λ u|∂ S = 0.

(7.31)

Using Theorem 1.2 with v = u, ¯ we find that 

(u¯ T Tu − uTT u) ¯ ds = 0,

∂S

which, in view of (7.31) and (7.26), leads to 0=



(u Λ¯ u¯ − u¯ TΛ u) ds =



(uiΛ¯ i j u¯ j − u¯ j Λ ji ui ) ds

T

∂S

= −2i

 

∂S

 Im(Λ11 )|u1 |2 + Im(Λ22 )|u2 |2 + Im(Λ33 )|u3 |2 ds.

∂S

From either (7.27) or (7.28) we deduce that

so (7.31) implies that ω − (ϕ )

u|∂ S = V0ω ϕ = 0,

(7.32)

 Tu|∂ S = W0ω ∗ + 12 I ϕ = 0.

(7.33)

By (7.32), U = V is a solution of that U = 0 in S¯− . Consequently,

− (D ω 0 );

hence, using Theorem 4.3, we see

102

7 Direct Boundary Equation Formulation

 0 = TU|∂ S = W0ω ∗ − 12 I ϕ , which, in view of (7.33), means that ϕ = 0. Thus, the homogeneous equation (7.30) has only the zero solution, so the solution of (7.29) is unique.   If the solution ϕ of (7.29) is found, then u given by (7.16) is the unique regular solution of (Dω − ). A similar result can be established for (N ω − ). Theorem 7.13. If (7.26) and either (7.27) or (7.28) hold, then the equation   Λ W0ω − 12 I ϕ + N0ω ϕ = Λ V0ω S + W0ω ∗ + 12 I S

(7.34)

has a unique solution ϕ ∈ C 1,α (∂ S) for every S ∈ C 0,α (∂ S), α ∈ (0, 1). Proof. The existence of at least one C 1,α -solution of (7.34) is guaranteed by Theorems 7.10 and 7.11. Consider a solution ϕ of the corresponding homogeneous equation  Λ W0ω − 12 I ϕ + N0ω ϕ = 0 (7.35) and let

u = W ω + (ϕ ).

Then (7.35) yields

Λ u|∂ S + Tu|∂ S = 0. Arguing exactly as in the proof of Theorem 7.12, we arrive at  u|∂ S = W0ω − 12 I ϕ = 0 and

Tu|∂ S = N0ω ϕ = 0.

(7.36) (7.37)

− (N ω 0 ),

From (7.37) we see that U = W ω − (ϕ ) is a solution of Theorem 4.3, U = 0 in S¯− . In particular,  0 = U|∂ S = W0ω + 12 I ϕ ,

which means that, by

so, taking (7.36) into account, we conclude that ϕ = 0, which proves that the solution of (7.34) is unique.   The unique regular solution of (N ω − ) in S− is given by (7.19), where ϕ is the unique solution of (7.34).

Chapter 8

Modified Fundamental Solutions

As we have seen in Chapters 6 and 7, the integral equations of the second kind that arise in the classical direct and indirect boundary integral formulations may have multiple solutions for certain values of the oscillation frequency ω . In Chapter 7 it was shown that, for the exterior problems, a unique solution does exist if we operate with a special pair of integral equations. The question is, can the matrix of fundamental solutions D ω (x, y) be modified so that we may obtain a single secondkind equation for each of the exterior problems, which has a unique solution for all values of ω ? Modified fundamental solutions in connection with exterior boundary value problems for the Helmholtz equation have been the subject of much investigation. In [71] it is suggested that a fundamental solution be used which satisfies a Robin-type condition on a circle inside the interior region. The analysis there is restricted to the two-dimensional case, although it is remarked that the ideas involved can be generalized to three dimensions. The two-dimensional Neumann problem is solved in [28] through the addition of a finite series of radiating circular wavefunctions, which eliminates nonuniqueness for a range of wavenumbers that depends on the number of terms in the series. Indication is also given on how the three-dimensional case can be handled with similar treatment. A key proof in the theory developed in [28] was later simplified in [72], but only in the context of the two-dimensional exterior Neumann problem. A systematic account of the Jones modification [28], incorporating the Ursell simplification [72], applied to the exterior Dirichlet and Neumann problems in three dimensions is given in [33]. A general survey of modified fundamental solutions in the theory of the Helmholtz equation can be found in [53] (or [52]). The ideas expounded in [28] and [72] have also been considered for problems in elastodynamics. Thus, the three-dimensional case is investigated in [29] by means of an infinite series of spherical wavefunctions. In two dimensions, a Jones-type modification with a finite series is operated in [37]. A much more general finite series is used for the two-dimensional exterior Neumann problem in [7], where a matrix of fundamental solutions is constructed based on the Ursell modification. G.R. Thomson and C. Constanda, Stationary Oscillations of Elastic Plates, A Boundary Integral Equation Analysis, DOI 10.1007/978-0-8176-8241-5_8, © Springer Science+Business Media, LLC 2011

103

104

8 Modified Fundamental Solutions

In this chapter we attempt to apply the preceding ideas to the exterior boundary value problems of stationary oscillations of thin elastic plates. The need for modified integral equations stems from the existence of nonzero solutions to interior problems with homogeneous boundary data. Our aim is to construct functions that, when they satisfy (4.1) in some bounded region contained in S + and homogeneous conditions on ∂ S, are forced to vanish in the entire bounded region. Both modifications described here require the matrix of fundamental solutions to be written as an infinite series of certain wavefunctions. This is done for D ω (x, y) in Section 8.1. We investigate the Ursell-type modification in Section 8.2. The theory is based on functions that satisfy a dissipative condition on a circle interior to the boundary ∂ S (see Theorem 8.1). A crucial role is played by Lemma 8.1, which is overlooked in other work using this type of modification (cf. [7]). We then define the modified matrix and show that it is symmetric (see Theorem 8.2). This symmetry is necessary since the solution of (D ω − ) is sought as a modified double-layer potential (cf. Theorem 2.2). In Theorems 8.3 and 8.4 we prove the unique solvability of the modified integral equations. The remaining work is concerned with the actual construction of the modified matrix. The main result here is Theorem 8.6, which guarantees the computability of the modification. Finally, in Subsection 8.2.3 we discuss the convergence of the infinite series representing the modified matrix. An alternative type of modification is considered in Section 8.3. This takes the form of a certain infinite series with a number of arbitrary constants that can be chosen in an appropriate way. The unique solvability of the modified integral equations is proved in Theorems 8.10 and 8.11. The corresponding modification with a finite series is investigated in Subsection 8.3.4. In the final section we obtain well-posed first-kind equations for the solution of the exterior boundary value problems by means of modified potentials. Modified composition relations are deduced from a modified exterior representation formula derived in Theorem 8.13. These relations are then used to establish the equivalence of our first-kind equations to certain uniquely solvable equations of the second kind (see Theorems 8.15 and 8.17).

8.1 The Matrix of Fundamental Solutions in Terms of Wavefunctions First, we need to write the original matrix of fundamental solutions in terms of certain wavefunctions, in a form similar to that given in [74] for the two-dimensional elastodynamic case. This is not straightforward because of the complicated nature of the matrix D ω (x, y) constructed in Theorem 2.1. We notice some important connections between the constants α 12 , α22 , γ , β12 , and 2 β2 arising in the definition of D ω (x, y). Thus, by (2.23) and (2.25),

β12 = h2 k32 α22 ,

β22 = h2 k32 α12 .

(8.1)

8.1 The Matrix of Fundamental Solutions in Terms of Wavefunctions

105

Also, (2.23), (2.26), (2.6), and (2.4) imply that h2 k32 α12 α22 =

h2 k32 (μ  k32 − k22 )(k12 − μ  k32 ) (k12 − k22 )2

 h2 k32  2 − μ  k34 − k12 k22 + μ  k32 (k12 + k22 ) 2 2 2 (k1 − k2 )   h2 k 2 μ2 μ λ + 3μ 2 2 4 2 2 = 2 32 2 − k − k k + μ k k 3 3 (λ + 2 μ )2 3 λ + 2μ (λ + 2 μ )2 (k1 − k2 ) =

  h2 k34 2 2 2 2 − μ k − μ ( λ + 2 μ )k + μ ( λ + 3 μ )k 3 2 2 (λ + 2μ )2(k1 − k2 )2 

  h2 k34 1 2 2 2 2 = −μ k − 2 +μ k h (λ + 2μ )2(k12 − k22 )2 =

=

μ 2 k34 ; (λ + 2μ )2(k12 − k22 )2

so, by (2.24), h2 k32 α12 α22 = γ 2 .

(8.2)

α 12

This equality indicates that and α22 are both either strictly positive or strictly negative. From (2.7) we deduce that k 12 > k32 , since it was assumed that k12 > k22 . Also, taking (1.14) into account, we see that k 32 > μ  k32 . Consequently, k12 − μ  k32 > 0,

k12 − k22 > 0,

so, by (2.23), we have α 22 > 0, which in turn implies that α 12 > 0. We introduce the Neumann factor ε m defined by  1, m = 0, εm = 2, m ≥ 1, (σ )

where m is a nonnegative integer. We also consider E m (θ ), where  cos mθ , σ = 1, (σ ) Em ( θ ) = sin mθ , σ = 2. Let (Rx , θx ) and (Ry , θy ) be the polar coordinates of x and y, respectively. Then, as shown in [48], p. 827, the Hankel function of the first kind and order zero can be written for R x < Ry in the form (1)

H0 (k j r) =

∞

∑ εm

m=0



(1)

Jm (k j Rx )(cos mθx )Hm (k j Ry )(cos mθy )

 (1) + Jm (k j Rx )(sin mθx )Hm (k j Ry )(sin mθy ) ,

(8.3)

106

8 Modified Fundamental Solutions

where Jm is the usual Bessel function. Let √ (1) (σ ) (σ ) (σ ) εm Hm (k1 Rx )Em (θx ), φˆm (x) = εm Jm (k1 Rx )Em (θx ), √ √ (σ ) (1) (σ ) (σ ) (σ ) υm (x) = εm Hm (k2 Rx )Em (θx ), υˆm (x) = εm Jm (k2 Rx )Em (θx ), √ √ (σ ) (1) (σ ) (σ ) (σ ) ψm (x) = εm Hm (k3 Rx )Em (θx ), ψˆ m (x) = εm Jm (k3 Rx )Em (θx ); (σ )

φm (x) =

√

then, using (8.3), we find that for R x < Ry , (1)

H0 (k1 r) = (1)

H0 (k2 r) = (1)

H0 (k3 r) =

∞

2

(σ )

(x)φm (y),

(σ )

(x)υm (y),

∑ ∑ φˆm

m=0 σ =1 ∞

2

∑ ∑ υˆm

m=0 σ =1 ∞

2

(σ )

∑ ∑ ψˆ m

m=0 σ =1

(σ )

(σ )

(8.4)

(σ )

(x)ψm (y).

We now use series (8.4) and the fact that

∂ ∂ (1) (1) H0 (k j r) = − H (k j r) ∂ xα ∂ yα 0 to write D ω (x, y) in terms of these new functions. By (2.20),  i ∂ 2 (1) ∂ 2 (1) ω D11 (x, y) = 2 2 − α12 2 H0 (k1 |x − y|) − α22 2 H0 (k2 |x − y|) 4h μ k3 ∂ x1 ∂ x1 − =

∂ 2 (1) H (k3 |x − y|) ∂ x22 0



 i ∂2 ∂2 (1) (1) α12 H0 (k1 |x − y|) + α22 H (k2 |x − y|) 2 2 ∂ x1 ∂ y1 ∂ x1 ∂ y1 0 4h μ k3 +

 ∂2 (1) H0 (k3 |x − y|) ; ∂ x2 ∂ y2

hence, ω D11 (x, y) =

  ∞ 2  i ∂ ˆ (σ ) ∂ (σ ) 2 α φ (x) φ (y) m m ∑∑ ∂ y1 4h2 μ k32 m=0 σ =1 1 ∂ x1

  ∂ (σ ) ∂ (σ ) 2 + α2 υˆ m (x) υm (y) ∂ x1 ∂ y1

  ∂ ∂ (σ ) (σ ) + ψˆ m (x) ψm (y) . ∂ x2 ∂ y2

(8.5)

8.1 The Matrix of Fundamental Solutions in Terms of Wavefunctions

107

In the same way, ω D12 (x, y) =

ω D21 (x, y) =

ω D22 (x, y) =

ω D13 (x, y) =

ω D23 (x, y) =

ω D31 (x, y) =

ω D32 (x, y) =

  ∞ 2  i ∂ ˆ (σ ) ∂ (σ ) 2 α φ (x) φ (y) m m ∑∑ ∂ y2 4h2 μ k32 m=0 σ =1 1 ∂ x1

  ∂ (σ ) ∂ (σ ) 2 ˆ + α2 υm (x) υm (y) ∂ x1 ∂ y2

  ∂ ∂ (σ ) (σ ) ˆ + ψm (x) − ψm (y) , ∂ x2 ∂ y1

  ∞ 2  i ∂ ˆ (σ ) ∂ (σ ) 2 φm (x) φm (y) ∑∑ α ∂ y1 4h2 μ k32 m=0 σ =1 1 ∂ x2

  ∂ (σ ) ∂ (σ ) + α22 υˆm (x) υm (y) ∂ x2 ∂ y1

  ∂ ∂ (σ ) (σ ) + − ψˆ m (x) ψm (y) , ∂ x1 ∂ y2 

  ∞ 2 i ∂ ˆ (σ ) ∂ (σ ) 2 φm (x) φm (y) ∑∑ α ∂ y2 4h2 μ k32 m=0 σ =1 1 ∂ x2

  ∂ (σ ) ∂ (σ ) 2 + α2 υˆm (x) υm (y) ∂ x2 ∂ y2

  ∂ ∂ (σ ) (σ ) + − ψˆ m (x) − ψm (y) , ∂ x1 ∂ y1

 ∞ 2  i ∂ ˆ (σ ) (σ ) − γ φ (x) φm (y) m ∑∑ ∂ x1 4h2 μ k32 m=0 σ =1

  ∂ (σ ) (σ ) +γ υˆm (x) υm (y) , ∂ x1

 ∞ 2  i ∂ ˆ (σ ) (σ ) ∑ ∑ − γ ∂ x2 φm (x) φm (y) 4h2 μ k32 m=0 σ =1

  ∂ (σ ) (σ ) +γ υˆm (x) υm (y) , ∂ x2 

 ∞ 2 i ∂ (σ ) (σ ) ˆ ∑ ∑ − γ φm (x) ∂ y1 φm (y) 4h2 μ k32 m=0 σ =1

 ∂ (σ ) (σ ) + γ υˆ m (x) υm (y) , ∂ y1 

 ∞ 2 i ∂ (σ ) (σ ) ˆ − γ φ (x) φ (y) m m ∑∑ ∂ y2 4h2 μ k32 m=0 σ =1

 ∂ (σ ) (σ ) + γ υˆ m (x) υm (y) . ∂ y2

(8.6)

(8.7)

(8.8)

(8.9)

(8.10)

(8.11)

(8.12)

108

8 Modified Fundamental Solutions

Finally, by (2.22) and (8.1), ω D33 (x, y) =

∞

i

2

∑∑

4h2 μ k32 m=0 σ =1

 2 2 2 (σ ) (σ ) h k3 α2 φˆm (x)φm (y)

 (σ ) (σ ) + h2 k32 α12 υˆm (x)υm (y) .

We introduce the wavefunctions (cf. [7] and [74]) ⎛ ⎞ ∂ (σ ) α φ (x) ⎜ 1 ∂ x1 m ⎟ ⎜ ⎟ ⎜ ⎟ (σ ) Φm (x) = ⎜ α ∂ φ (σ ) (x) ⎟ , ⎜ 1 ∂x m ⎟ ⎝ 2 ⎠ (σ ) −hk3 α2 φm (x) ⎛ ⎞ ∂ (σ ) ⎜α2 ∂ x1 υm (x)⎟ ⎜ ⎟ ⎜ ⎟ (σ ) ϒm (x) = ⎜α ∂ υ (σ ) (x)⎟ , ⎜ 2 ∂x m ⎟ ⎝ 2 ⎠ (σ ) hk3 α1 υm (x) ⎛ ⎞ ∂ (σ ) ψm (x) ⎜ ∂ x2 ⎟ ⎜ ⎟ (σ ) ⎜ ⎟. Ψm (x) = ⎜ ∂ (σ ) ⎟ − ψ (x) ⎝ ∂ x1 m ⎠ 0

(8.13)

(8.14)

(8.15)

(8.16)

(σ ) (σ ) (σ ) (σ ) The vectors Φˆ m , ϒˆm , and Ψˆm are defined in the same way as above, with φ m , (σ ) (σ ) (σ ) (σ ) (σ ) υm , and ψm replaced by φˆm , υˆm , and ψˆ m , respectively. Next, using (8.2) and (8.14)–(8.16), we can write (8.5)–(8.13) for R x < Ry in the form

D ω (x, y) =

i

∞

2

∑∑

4h2 μ k32 m=0 σ =1

 (σ )  (σ )  T  (σ ) T (σ ) Φˆ m (x) Φm (y) + ϒˆm (x) ϒm (y)  (σ )  T  (σ ) + Ψˆm (x) Ψm (y) . (8.17)

Although this is similar in form to the expansion in the plane elastodynamic case [7], here the wavefunctions have a more complicated structure.

8.2 The Ursell Modification We construct a modified matrix of fundamental solutions that satisfies a Robin-type (dissipative) condition on a circle inside ∂ S. Modifications of this kind were first considered in [71] in connection with the Helmholtz equation; the ideas involved

8.2 The Ursell Modification

109

were later developed in [7] to deal with an exterior boundary value problem in plane elastodynamics. Let D be the annular region bounded externally and internally by simple, smooth, closed curves ∂ S 1 and ∂ S2 , respectively. Lemma 8.1. If u is an analytic solution of (4.1) in D ∪ ∂ S 2 and u = Tu = 0 on ∂ S 2 , then u = 0 in D. The proof of this assertion involves lengthy calculations and is deferred to Appendix A. Theorem 8.1. Let u be an analytic solution of (4.1) in D ∪ ∂ S 2 satisfying u = 0 or Tu = 0 on ∂ S 1 , Tu + Ku = 0 on ∂ S2 ,

(8.18) (8.19)

where K is a constant (3 × 3)-matrix whose elements are such that Ki j = K¯ ji

for i = j

(8.20)

and either Im(K11 ), Im(K22 ), Im(K33 ) > 0

(8.21)

Im(K11 ), Im(K22 ), Im(K33 ) < 0.

(8.22)

or Then u = 0 in D for all ω . Proof. Applying the reciprocity relation to u and u¯ in D and taking (8.18)–(8.20) into account, we arrive at 0=



(uT T u¯ − u¯ TTu) ds −

∂ S1

=





(uT T u¯ − u¯ TTu) ds

∂ S2

(uT K¯ u¯ − u¯ TKu) ds =

∂ S2

= −2i

 



(ui K¯i j u¯ j − u¯ j K ji ui ) ds

∂ S2

Im(K11 )|u1 |2 + Im(K22 )|u2 |2 + Im(K33 )|u3 |2 ] ds.

∂ S2

Either (8.21) or (8.22) ensures that u = 0 on ∂ S 2 , and from (8.19) we have Tu = 0 on ∂ S2 . Applying Lemma 8.1, we now deduce that u = 0 in D.   Without loss of generality, we assume that the origin lies in S + . Let ∂ σ (a) be the circle centered at the origin and of radius a, which is chosen so that ∂ σ (a) lies

110

8 Modified Fundamental Solutions

entirely within ∂ S, and let S a− be the region R > a. We consider a modified matrix of fundamental solutions of the form ω DM (x, y) = D ω (x, y) + M ω (x, y),

(8.23)

where the columns of M ω (x, y) are regular solutions of A ω u = 0 in Sa− ∪ ∂ σ (a) with respect to x and satisfy the radiation conditions as R = |x| → ∞. We also require that ω ω T (∂x )DM (x, y) + KDM (x, y) = 0,

x ∈ ∂ σ (a),

(8.24)

⎛ ⎞ h2 μκ 0 0 ⎜ ⎟ K = ⎝ 0 h2 μκ 0 ⎠ ; 0 0 μκ

where

here

κ = |κ |eiδ ,

(8.25)

0 < δ < π.

(8.26)

We remark that K satisfies the conditions of Theorem 8.1. The condition 0 < δ < π in (8.26) is needed at a later stage. We can show that this particular modified matrix of fundamental solutions is symmetric. Theorem 8.2. If (8.24) is satisfied with K of the form (8.25), then  ω T ω DM (x, y) = DM (y, x) . ω (i)

Proof. We apply the reciprocity relation in the region S a− ∩ KR with u = DM (x, x1 ) ω ( j) and v = DM (x, x2 ); then 



T ω (i) ω ( j) DM (x, x1 ) Aω (∂x )DM (x, x2 )

S− a ∩KR

 ω ( j) T  ω (i) − DM (x, x2 ) Aω (∂x )DM (x, x1 ) da(x) =

  ω (i)

T ω ( j) DM (x, x1 ) T (∂x )DM (x, x2 )

∂ KR

−

 ω ( j) T  ω (i) − DM (x, x2 ) T (∂x )DM (x, x1 ) ds(x) 

∂ σ (a)

 ω (i) T ω ( j) DM (x, x1 ) T (∂x )DM (x, x2 )  ω ( j) T  ω (i) − DM (x, x2 ) T (∂x )DM (x, x1 ) ds(x).

(8.27)

ω (x, y) satisfies the radiation The integral around ∂ K R tends to zero as R → ∞ since D M conditions (see the proof of Theorem 4.7). Also, since K is a diagonal matrix, the boundary condition (8.24) implies that

8.2 The Ursell Modification





T ω (i) ω ( j) DM (x, x1 ) T (∂x )DM (x, x2 )  ω ( j) T  ω (i) − DM (x, x2 ) T (∂x )DM (x, x1 ) ds(x)  ω (i) T   ω ( j) DM (x, x1 ) − KDM (x, x2 )

∂ σ (a)



=

111

 ω ( j) T   ω (i) − DM (x, x2 ) − KDM (x, x1 ) ds(x)

∂ σ (a)

=K



 ω (i) T ω ( j) DM (x, x1 ) DM (x, x2 )  ω ( j) T ω (i)  − DM (x, x2 ) DM (x, x1 ) ds(x) = 0.

∂ σ (a)

Therefore, letting R → ∞ in (8.27), we obtain   ω (i)

T ω ( j) DM (x, x1 ) Aω (∂x )DM (x, x2 )  ω ( j) T  ω (i) − DM (x, x2 ) Aω (∂x )DM (x, x1 ) da(x) = 0.

S− a

From the definition of D ω (x, y) and the fact that A ω (∂x )M ω (x, y) = 0 in Sa− , it follows that     ω   ω DM (x, x1 ) ji δ (|x − x2 |) − DM (x, x2 ) i j δ (|x − x1 |) da(x) = 0; S− a

hence,



ω DM (x2 , x1 )

 ji

 ω  = DM (x1 , x2 ) i j ,

as required.   The symmetry of the modified matrix is not discussed in [7], since for the exterior Neumann problem only the single-layer potential is required. Consider the modified single-layer and double-layer potentials (VMω ϕ )(x) (WMω ϕ )(x) =

=

 



ω DM (x, y)ϕ (y) ds(y),

∂S

T ω T (∂y )DM (y, x) ϕ (y) ds(y),

∂S ω (x, y) is given by (8.23) and satisfies (8.24). Since M ω (x, y) is regular where DM − in Sa , these potentials behave in the same way as the corresponding unmodified potentials when x ∈ ∂ S. Consequently, we may define the functions

VMω + (ϕ ) = VMω (ϕ )|S¯+ , which are such that

VMω − (ϕ ) = VMω (ϕ )|S¯− ,

112

8 Modified Fundamental Solutions

 ω∗ 1 T VMω + (ϕ ) = WM0 + I ϕ  ω ∗ 21 ω− T VM (ϕ ) = WM0 − 2 I ϕ and the functions

 W ω (ϕ )| + =  Mω S1 WM0 − 2 I ϕ  W ω (ϕ )| − WMω − (ϕ ) =  Mω S1 WM0 + 2 I ϕ WMω + (ϕ )

with

T WMω + (ϕ ) = T WMω − (ϕ )

on ∂ S, on ∂ S,

(8.28)

in S+ , on ∂ S, in S− , on ∂ S, on ∂ S.

(8.29)

The modified potentials are also of class B ω since the columns of M ω (x, y) are required to satisfy the radiation conditions as R x → ∞. Owing to the fact that M ω (x, y) may have singularities in the region interior to the circle ∂ σ (a), we cannot say that V Mω + (ϕ ) and WMω + (ϕ ) satisfy Aω u = 0 in all of S+ . They do, however, satisfy the system in S a− ∩ S+ . Using Theorem 8.1 and the modified potentials, we can formulate boundary integral equations of the second kind representing (D ω − ) and (Nω − ), each of which is uniquely solvable for every value of the oscillation frequency ω . Theorem 8.3. If R ∈ C 1,α (∂ S), α ∈ (0, 1), the unique solution of (D ω − ) is given by u = WMω − (ϕ ), (8.30) where ϕ ∈ C 1,α (∂ S) is the unique solution of  ω WM0 + 12 I ϕ = R.

(8.31)

Proof. Seeking the solution of (D ω − ) in the form (8.30) leads to the integral equation (8.31) for the unknown density ϕ . Consider the homogeneous adjoint equation to (8.31), namely,  ω∗ 1 WM0 + 2 I ψ = 0. (8.32) Let v1 = VMω + (ψ ), where ψ satisfies (8.32). Then T v 1 |∂ S = 0, so, by Theorem 8.1, v1 = 0 in Sa− ∩ S+ . By continuity, ω v1 |∂ S = VM0 ψ = 0, − which implies that v2 = VMω − (ψ ) is a solution of (Dω 0 ). By the uniqueness property, − ¯ v2 = 0 in S , so  ω∗ 1 T v2 |∂ S = T VMω − (ψ )|∂ S = WM0 − 2 I ψ = 0. (8.33)

Combining (8.32) and (8.33) yields ψ = 0. Therefore, according to the Fredholm alternative, (8.31) has a unique solution ϕ ∈ C 1,α (∂ S).  

8.2 The Ursell Modification

113

Theorem 8.4. If S ∈ C 0,α (∂ S), α ∈ (0, 1), then the unique solution of (N ω − ) is given by u = VMω − (ϕ ), (8.34) where ϕ ∈ C 0,α (∂ S) is the unique solution of  ω∗ 1 WM0 − 2 I ϕ = S .

(8.35)

Proof. If the solution of (N ω − ) is sought in the form (8.34), then the integral equation (8.35) follows from (8.28). As in the previous theorem, we consider the homogeneous adjoint equation to (8.35); that is,  ω WM0 − 12 I ψ = 0. (8.36) We define

v1 = WMω + (ψ ),

where ψ is a solution of (8.36). Then v 1 |∂ S = 0, so Theorem 8.1 implies that v 1 = 0 in Sa− ∩ S+ . Consequently, T v1 |∂ S = T WMω + (ψ )|∂ S = 0, and, by (8.29),

T WMω − (ψ )|∂ S = 0.

− Hence, v2 = WMω − (ψ ) is a solution of (N ω 0 ), and the uniqueness property leads to − ¯ the conclusion that v 2 = 0 in S . In particular,  ω v2 |∂ S = WM0 + 12 I ψ = 0, (8.37)

so from (8.36) and (8.37) we deduce that ψ = 0. As in the proof of Theorem 8.3, the assertion now follows from the Fredholm alternative.  

8.2.1 Construction of M ω (x, y) We assume M ω (x, y) to be of the form M ω (x, y) =

(σ )

∞ 2   (σ )  T  (σ ) T i (σ ) (σ ) Φm (x) Am (y) + ϒm (x) Bm (y) ∑ ∑ 4h2 μ k32 m=0 σ =1  (σ )  T  (σ ) + Ψm (x) Cm (y) , (8.38)

(σ )

(σ )

where Am , Bm , and Cm are (3 × 1)-vectors to be determined via (8.24). It is easily seen that the columns of M ω (x, y) satisfy the system of equations ω A u = 0 in R2 \{0}. The fact that the columns of the matrix satisfy the radiation

114

8 Modified Fundamental Solutions

conditions as R x → ∞ is shown in the same way as in the proof of Theorem 4.4 for D ω (x, y). ω (x, y) explicitly for R < R and using (8.23), (8.17), and (8.38), we Writing DM x y have ω DM (x, y)

=

i

∞

2

∑∑



4h2 μ k32 m=0 σ =1

 (σ )  T  (σ ) T (σ ) (σ ) Φˆ m (x) Φm (y) + ϒˆm (x) ϒm (y)  (σ )  T  (σ ) T (σ ) (σ ) + Ψˆm (x) Ψm (y) + Φm (x) Am (y)  (σ ) T  (σ )  T  (σ ) (σ ) + ϒm (x) Bm (y) + Ψm (x) Cm (y) .

(8.39)

For simplicity, from now on we write (R, θ ) instead of (R x , θx ) and omit the (1) superscript (1) in the symbol of the Hankel function H m (k j R). We want to satisfy condition (8.24) on the circle R = a. This happens if   (σ )  T  (σ )  (σ ) T (σ ) (σ )  (σ )  T Φm + K Φm R=a Am (y) + T ϒm + Kϒm R=a Bm (y) σ =1   (σ )  T  (σ ) (σ )  + T Ψm + KΨm R=a Cm (y) 2

∑

=−

  (σ ) T (σ ) (σ )  T Φˆ m + K Φˆ m R=a Φm (y) σ =1  (σ )  (σ )  T (σ )  + T ϒˆm + Kϒˆm R=a ϒm (y)   (σ )  T  (σ ) (σ )  + T Ψˆm + KΨˆm  Ψm (y) 2

∑

R=a

for m = 0, 1, 2, . . .. This can be written as the (3 × 3)-systems of algebraic equations 2

(σ )

∑ Um

σ =1

(σ )

Xm = −

(σ ) (σ ) Ym ,

σ =1

m = 0, 1, 2, . . . ,

(8.40)

⎛ (σ ) ⎞ Φm (y) ⎜ (σ ) ⎟ ⎜ (σ ) ⎟ (σ ) (σ ) ⎜ ⎟ ⎜ Xm = ⎝Bm (y)⎠ , Ym = ⎝ϒm (y) ⎟ ⎠, (σ ) (σ ) Cm (y) Ψm (y) ⎛ ⎞ (σ ) (σ )   (σ ) (σ )   (σ ) (σ )  T Φm + K Φm 1 T ϒm + Kϒm 1 T Ψm + KΨm 1  ⎜ ⎟ (σ ) (σ ) (σ )   (σ ) (σ )   (σ ) (σ )  ⎟  Um = ⎜ , ⎝ T Φm + K Φm 2 T ϒm + Kϒm 2 T Ψm + KΨm 2 ⎠   (σ ) (σ )   (σ ) (σ )   (σ ) (σ )  T Φm + K Φm 3 T ϒm + Kϒm 3 T Ψm + KΨm 3 R=a ⎛ ⎞ (σ ) (σ )   (σ ) (σ )   (σ ) (σ )  T Φˆ m + K Φˆ m 1 T ϒˆm + Kϒˆm 1 T Ψˆm + KΨˆm 1  ⎜   (σ )    ⎟ (σ ) ˆ ˆ (σ ) ˆ (σ ) ˆ (σ ) ⎟  ˆ (σ ) ˆ (σ ) Uˆ m = ⎜ , ⎝ T Φm + K Φm 2 T ϒm + Kϒm 2 T Ψm + KΨm 2 ⎠    (σ )     ( σ ) ( σ ) ( σ ) ( σ ) ( σ )  T Φˆ m + K Φˆ m T ϒˆm + Kϒˆm T Ψˆm + KΨˆm

where

⎛

2

∑ Uˆ m

(σ )

Am (y)

3

⎞

3

3

R=a

8.2 The Ursell Modification

115

with σ = 1, 2 and m = 0, 1, 2, . . . . (σ ) The elements of the matrix U m are computed explicitly in Appendix B. The ( σ ) elements of Uˆ m are obtained by replacing the Hankel functions H m by the Bessel functions Jm in the expressions given below. We have  (σ ) (σ )   T Φm + K Φm 1 R=a  √ (σ ) (σ ) 2 = εm h α1 − λ k12 Hm (k1 a)(cos θ )Em (θ ) + 2μ k12 Hm (k1 a)(cos θ )Em (θ ) 2μ mk1 (−1)σ Hm (k1 a)(sin θ )Em3−σ (θ ) a 2μ m + 2 (−1)σ Hm (k1 a)(sin θ )Em3−σ (θ ) a −

(σ )

+ μκ k1Hm (k1 a)(cos θ )Em (θ ) −

 μκ m (−1)σ Hm (k1 a)(sin θ )Em3−σ (θ ) , a

where the ‘prime’ symbol denotes differentiation with respect to the argument. Similarly,  (σ ) (σ )   T Φm + K Φm 2 R=a  √ (σ ) (σ ) = εm h2 α1 − λ k12 Hm (k1 a)(sin θ )Em (θ ) + 2μ k12 Hm (k1 a)(sin θ )Em (θ ) 2μ mk1 (−1)σ Hm (k1 a)(cos θ )Em3−σ (θ ) a 2μ m − 2 (−1)σ Hm (k1 a)(cos θ )Em3−σ (θ ) a (σ ) + μκ k1 Hm (k1 a)(sin θ )Em (θ ) +

 (σ ) (σ )   T Φm + K Φm 3 R=a =

√

 μκ m σ 3−σ + (−1) Hm (k1 a)(cos θ )Em (θ ) , a

  (σ ) (σ ) εm μ (α1 − hk3α2 )k1 Hm (k1 a)Em (θ ) − μκ hk3α2 Hm (k1 a)Em (θ ) .

Also,  (σ ) (σ )   T ϒm + Kϒm 1 R=a  √ (σ ) (σ ) 2 = εm h α2 − λ k22 Hm (k2 a)(cos θ )Em (θ ) + 2μ k22 Hm (k2 a)(cos θ )Em (θ ) −

2μ mk2 (−1)σ Hm (k2 a)(sin θ )Em3−σ (θ ) a

116

8 Modified Fundamental Solutions

2μ m (−1)σ Hm (k2 a)(sin θ )Em3−σ (θ ) a2 (σ ) + μκ k2Hm (k2 a)(cos θ )Em (θ ) +

 μκ m σ 3−σ − (−1) Hm (k2 a)(sin θ )Em (θ ) , a

 (σ ) (σ )   T ϒm + Kϒm 2 R=a  √ (σ ) (σ ) 2 = εm h α2 − λ k22 Hm (k2 a)(sin θ )Em (θ ) + 2μ k22 Hm (k2 a)(sin θ )Em (θ ) 2μ mk2 (−1)σ Hm (k2 a)(cos θ )Em3−σ (θ ) a 2μ m − 2 (−1)σ Hm (k2 a)(cos θ )Em3−σ (θ ) a +

(σ )

+ μκ k2 Hm (k2 a)(sin θ )Em (θ ) +

 μκ m (−1)σ Hm (k2 a)(cos θ )Em3−σ (θ ) , a

 (σ ) (σ )   T ϒm + Kϒm 3 R=a  √  (σ ) (σ ) = εm μ (α2 + hk3α1 )k2 Hm (k2 a)Em (θ ) + μκ hk3α1 Hm (k2 a)Em (θ ) . Finally, 



(σ ) (σ )   T Ψm + KΨm 1 R=a  √ k3 (σ ) (σ ) 2 = εm h μ k32 Hm (k3 a)(sin θ )Em (θ ) − Hm (k3 a)(sin θ )Em (θ ) a 2mk3 + (−1)σ Hm (k3 a)(cos θ )Em3−σ (θ ) a 2m − 2 (−1)σ Hm (k3 a)(cos θ )Em3−σ (θ ) a m2 (σ ) (σ ) + 2 Hm (k3 a)(sin θ )Em (θ ) + κ k3Hm (k3 a)(sin θ )Em (θ ) a  κm + (−1)σ Hm (k3 a)(cos θ )Em3−σ (θ ) , a

 (σ ) (σ )   T Ψm + KΨm 2  R=a  √ k3 (σ ) (σ ) = εm h2 μ − k32 Hm (k3 a)(cos θ )Em (θ ) + Hm (k3 a)(cos θ )Em (θ ) a 2mk3 + (−1)σ Hm (k3 a)(sin θ )Em3−σ (θ ) a

8.2 The Ursell Modification

117

2m (−1)σ Hm (k3 a)(sin θ )Em3−σ (θ ) a2 m2 (σ ) (σ ) − 2 Hm (k3 a)(cos θ )Em (θ ) − κ k3Hm (k3 a)(cos θ )Em (θ ) a  κm σ 3−σ + (−1) Hm (k3 a)(sin θ )Em (θ ) , a  √ μm (σ ) (σ )   T Ψm + KΨm 3 R=a = εm (−1)σ Hm (k3 a)Em3−σ (θ ). a −

Let



 λ 2 2 2 2  2  am = α1 − k1 a Hm (k1 a) + 2k1 a Hm (k1 a) + κ k1a Hm (k1 a) , μ   λ 2 2 2 2  2  bm = α2 − k2 a Hm (k2 a) + 2k2 a Hm (k2 a) + κ k2a Hm (k2 a) , μ cm = −2mk3 aHm (k3 a) − m(κ a − 2)Hm(k3 a),   dm = α1 2mk1 aHm (k1 a) + m(κ a − 2)Hm(k1 a) ,   em = α2 2mk2 aHm (k2 a) + m(κ a − 2)Hm(k2 a) , fm = −k32 a2 Hm (k3 a) − k3 a(κ − a)Hm (k3 a) − m2 Hm (k3 a), gm = (α1 − hk3α2 )k1 aHm (k1 a) − κ hk3α2 aHm (k1 a), hm = (α2 + hk3α1 )k2 aHm (k2 a) + κ hk3α1 aHm (k2 a), im = −mHm (k3 a).

We define aˆm , bˆ m , . . . , ıˆm similarly, with Hm replaced by Jm . Using the above, we can show that system (8.40) is equivalent to ⎛

(1)

Am

⎞

⎛

(2)

Am

⎞

⎜ (1) ⎟ ⎜ (2) ⎟ ⎟ ⎜ ⎟ Rm (θ )Pm ⎜ ⎝ Bm ⎠ + Sm (θ )Pm ⎝Bm ⎠ (2) (1) −Cm Cm

⎛ (2) ⎞ Φm ⎜ (1) ⎟ ⎜ (2) ⎟ ⎟ ⎟ ˆ ⎜ = −Rm (θ )Pˆm ⎜ ⎝ ϒm ⎠ − Sm (θ )Pm ⎝ϒm ⎠ , (2) (1) −Ψm Ψm

where

and

⎛

⎞ am bm cm ⎜ ⎟ Pm = ⎝dm em fm ⎠ , gm hm i m

⎛

(1)

Φm

⎞

⎛ ⎞ aˆm bˆ m cˆm ⎜ ⎟ Pˆm = ⎝dˆm eˆm fˆm ⎠ , gˆm hˆ m ıˆm

118

8 Modified Fundamental Solutions

√

⎛

cos θ cosmθ sin θ sin mθ

0

⎞

εm h2 μ ⎜ ⎟ 0 ⎝ sin θ cos mθ − cos θ sin mθ ⎠, a2 2 0 0 (a/h ) cos mθ ⎛ ⎞ cos θ sin m θ − sin θ cos m θ 0 √ εm h2 μ ⎜ ⎟ 0 S m (θ ) = ⎝ sin θ sin mθ cos θ cos mθ ⎠. a2 2 0 0 (a/h ) sin mθ R m (θ ) =

From the orthogonality of the trigonometric functions we see that it suffices to (σ ) (σ ) (σ ) choose Am , Bm , and Cm so that ⎛ ⎞ ⎛ ⎞ (σ ) (σ ) Am (y) Φm (y) ⎜ ⎟ ⎜ ⎟ (σ ) ⎟ = −Pˆm ⎜ ϒ (σ ) (y) ⎟ Pm ⎜ Bm (y) m ⎝ ⎠ ⎝ ⎠ σ 3− σ σ 3− σ (−1) Cm (y) (−1) Ψm (y) ω (x, y) defined by (8.39) satisfies for σ = 1, 2 and m = 0, 1, 2, . . . . Consequently, D M (σ ) (σ ) (σ ) condition (8.24) on the circle R = a if the vectors A m , Bm , and Cm are chosen so that ⎛ ⎞ ⎛ ⎞ (σ ) (σ ) Am (y) Φm (y) ⎜ ⎟ ⎜ ⎟ (σ ) (σ ) ⎜ ⎟ = −Pm−1 Pˆm ⎜ ⎟. B (y) ϒ (y) m m ⎝ ⎠ ⎝ ⎠ (−1)σ Cm3−σ (y) (−1)σ Ψm3−σ (y)

The assertion that Pm−1 exists if (8.26) holds is proved below.

8.2.2 The Invertibility of Pm Before proving the existence of Pm−1 for m ≥ 0, we establish certain orthogonality (σ ) (σ ) (σ ) properties of the wavefunctions Φ m , ϒm , and Ψm . First, we must consider the asymptotic behavior of these functions as R → ∞. In [1], p. 364, it is shown that, as R → ∞, " 2 i(k j R−πm ) Hm (k j R) = e + O(R−3/2 ), (8.41) πk jR where

πm =

π (2m + 1). 4

For the moment we consider m ≥ 1, which means that ε m = 2. Let

8.2 The Ursell Modification

119

⎛ ⎞ cos θ ⎜ ⎟ eR = ⎝ sin θ ⎠ , 0

⎛ ⎞ − sin θ ⎜ ⎟ eθ = ⎝ cos θ ⎠ , 0

⎛ ⎞ 0 e3 = ⎝0⎠ . 1

We remark that eTR eθ = eTR e3 = eTθ e3 = 0

(8.42)

eTR eR = eTθ eθ = eT3 e3 = 1.

(8.43)

and

(σ )

From the definition of Φ m and (8.41) we find that (σ )

Φm (x)

⎛  (σ ) ∂  1 ∂ ⎜α1 (cos θ ) ∂ R Hm (k1 R) Em (θ ) − (sin θ ) R Hm (k1 R) ∂ θ ⎜ √ ⎜ ⎜  = 2 ⎜α (sin θ ) ∂ H (k R) E (σ ) (θ ) + (cos θ ) 1 H (k R) ∂ m m 1 m 1 ⎜ 1 ∂R R ∂θ ⎜ ⎝ (σ )



⎞

(σ ) Em ( θ )

⎟ ⎟ ⎟  (σ ) ⎟ Em ( θ ) ⎟ ⎟ ⎟ ⎠

−hk3 α2 Hm (k1 R)Em (θ )

#  ⎛  ⎞ ∂ 2 i(k1 R−πm ) (σ ) −3/2 e Em (θ ) + O(R )⎟ ⎜α1 (cos θ ) ∂ R π k1 R ⎜ ⎟ ⎜  ⎟ #

 ⎜ ⎟ √ ⎜ ∂ 2 i(k1 R−πm ) ⎟ (σ ) −3/2 = 2 ⎜ α1 (sin θ ) e Em (θ ) + O(R )⎟, ⎜ ⎟ ∂ R π k R 1 ⎜ ⎟ ⎜ ⎟ # ⎝ ⎠ 2 i(k1 R−πm ) (σ ) −3/2 −hk3 α2 e Em (θ ) + O(R ) π k1 R so (σ )

1 (σ ) ei(k1 R−πm ) Em (θ )eR π k1 R 1 (σ ) − 2hk3 α2 √ ei(k1 R−πm ) Em (θ )e3 + O(R−3/2 ). π k1 R

Φm (x) = 2α1 ik1 √

(8.44)

Similarly, 1 (σ ) (σ ) ϒm (x) = 2α2 ik2 √ ei(k2 R−πm ) Em (θ )eR π k2 R 1 (σ ) + 2hk3 α1 √ ei(k2 R−πm ) Em (θ )e3 + O(R−3/2 ). π k2 R Also,

(8.45)

120

8 Modified Fundamental Solutions (σ )

Ψm (x)

⎛

⎞ (σ ) ∂  1 ∂  (σ ) (sin θ ) H (k R) E ( θ ) + (cos θ ) H (k R) E ( θ ) m m m 3 m 3 ⎜ ⎟ ∂R R ∂θ √ ⎜ ⎟ ⎜ ⎟   ∂ 1 ∂ = 2⎜ (σ ) (σ ) ⎟ − (cos θ ) H (k R) E ( θ ) + (sin θ ) H (k R) E ( θ ) m m m 3 m 3 ⎝ ⎠ ∂R R ∂θ 0

#  ⎛ ⎞ ∂ 2 i(k3 R−πm ) (σ ) −3/2 ) (sin θ ) e E ( θ ) + O(R m ⎜ ⎟ ∂R π k3 R ⎟ √ ⎜ #  ⎜ ⎟ = 2⎜ ∂ 2 i(k3 R−πm ) (σ ) ⎟; ⎜ − (cos θ ) e Em (θ ) + O(R−3/2 )⎟ ⎝ ⎠ ∂ R π k3 R 0

that is, (σ )

Ψm (x) = −2ik3 √

1 (σ ) ei(k3 R−πm ) Em (θ )eθ + O(R−3/2 ). π k3 R

(8.46)

We also need to know the behavior of the images under the operator T of the wavefunctions on a circle centered at the origin and of radius R, as R → ∞. We start by deriving two useful formulas. By (2.23) and (2.26),  2  h (λ + 2μ )k12 − h2 μ k32 α12 = =

h2 k22 − k12

  (λ + 2 μ )k12 − μ k32 (k22 − μ  k32 )

   h2 (λ + 2 μ )k12 − μ k32 (λ + 2μ )k22 − μ k32 . (λ + 2μ )(k22 − k12 )

Therefore, using (2.6) and (2.4), we obtain  2  h (λ + 2μ )k12 − h2 μ k32 α12   h2 (λ + 2μ )2 k12 k22 − μ (λ + 2μ )k32 (k12 + k22 ) + μ 2k34 (λ + 2μ )(k22 − k12 )   h2 k32 μ μ  k32 μ2 2 2 2 2 = μ ( λ + 2 μ )k − μ ( λ + 3 μ )k + μ k − = , h2 (λ + 2μ )(k22 − k12 ) k12 − k22 =

so, by (2.24), h2 (λ + 2μ )k12 α12 = h2 μ k32 α12 + μγ , which, in view of (8.2), implies that h2 (λ + 2μ )k12 α12 = h2 μ k32 α12 + h μ k3α1 α2 ; that is, h2 (λ + 2μ )k12 α1 = h2 μ k32 α1 + hμ k3 α2 .

(8.47)

8.2 The Ursell Modification

121

In just the same way, we can show that h2 (λ + 2μ )k22 α2 = h2 μ k32 α2 − hμ k3 α1 .

(8.48)

Next, on the circle of radius R we find that, as R → ∞,   (σ ) T (∂x )Φm (x) 1  √   (σ ) = 2 h2 α1 λ (cos θ )Δ Hm (k1 R)Em (θ )

    ∂2 ∂2 (σ ) + 2μ (cos θ ) 2 + (sin θ ) Hm (k1 R)Em (θ ) ∂ x1 ∂ x2 ∂x #1  √ 2 i(k1 R−πm ) (σ ) = 2 h2 α1 − λ k12 (cos θ ) e Em (θ ) π k1 R #    ∂2 2 i(k1 R−πm) (σ ) + 2μ (cos θ ) 2 e Em (θ ) + O(R−3/2 ) ∂R π k1 R 1 (σ ) = −2h2 (λ + 2μ )k12 α1 √ ei(k1 R−πm ) (cos θ )Em (θ ) + O(R−3/2 ). π k1 R Similarly, 

 1 (σ ) (σ ) T (∂x )Φm (x) 2 = −2h2 (λ + 2 μ )k12 α1 √ ei(k1 R−πm ) (sin θ )Em (θ ) π k1 R + O(R−3/2 ).

Also,   (σ ) T (∂x )Φm (x) 3

 √  ∂ ∂  (σ ) = 2 (α1 − hk3 α2 )μ (cos θ ) + (sin θ ) Hm (k1 R)Em (θ ) ∂ x1 ∂ x2

#  √ ∂ 2 i(k1 R−πm ) (σ ) = 2 (α1 − hk3 α2 )μ e Em (θ ) + O(R−3/2) ∂R π k1 R 1 (σ ) = 2(α1 − hk3 α2 )ik1 μ √ ei(k1 R−πm ) Em (θ ) + O(R−3/2 ). π k1 R Consequently, taking (8.47) into account, we see that  1 (σ ) (σ ) T (∂x )Φm (x) = −2 h2 μ k32 α1 + hμ k3 α2 √ ei(k1 R−πm ) Em (θ )eR π k1 R 1 (σ ) + 2(α1 − hk3 α2 )ik1 μ √ ei(k1 R−πm ) Em (θ )e3 π k1 R + O(R−3/2 ). (8.49) Analogously, this time using (8.48), we obtain

122

8 Modified Fundamental Solutions

1 (σ ) ei(k2 R−πm ) Em (θ )eR π k2 R 1 (σ ) + 2(α2 + hk3 α1 )ik2 μ √ ei(k2 R−πm) Em (θ )e3 π k2 R + O(R−3/2 ). (8.50)

(σ )

T (∂x )ϒm (x) = −2(h2 μ k32 α2 − hμ k3 α1 ) √

Finally,   (σ ) T (∂x )Ψm (x) 1

√ = 2 h2 μ 2(cos θ )

  ∂2 ∂2 ∂2  (σ ) + (sin θ ) 2 − (sin θ ) 2 Hm (k3 R)Em (θ ) ∂ x1 ∂ x2 ∂ x2 ∂ x1

#  2 √ ∂ 2 i(k3 R−πm ) (σ ) = 2 h2 μ (sin θ ) 2 e Em (θ ) + O(R−3/2 ) ∂R π k3 R = −2h2 μ k32 √

1 (σ ) ei(k3 R−πm ) (sin θ )Em (θ ) + O(R−3/2 ), π k3 R

  1 (σ ) (σ ) T (∂x )Ψm (x) 2 = 2h2 μ k32 √ ei(k3 R−πm ) (cos θ )Em (θ ) + O(R−3/2 ) π k3 R and 

 (σ ) T (∂x )Ψm (x) 3

  ∂ ∂  (σ ) = 2 μ (cos θ ) − (sin θ ) Hm (k3 R)Em (θ ) ∂ x2 ∂ x1

#  √ 1 ∂ 2 i(k3 R−πm ) (σ ) = 2μ e Em (θ ) + O(R−3/2 ) R ∂θ π k3 R √

= O(R−3/2 ); hence, (σ )

T (∂x )Ψm (x) = 2h2 μ k32 √

1 (σ ) ei(k3 R−πm ) Em (θ )eθ + O(R−3/2 ). π k3 R

(8.51)

We recall the well-known orthogonality relations 2π

(σ )

(σ )

(1)

(1)

Em (θ )En (θ ) d θ = πδmn,

m = 0,

(8.52)

0

2π

E0 (θ )E0 (θ ) d θ = 2π ,

(8.53)

0

2π

(σ )

Em (θ )En3−σ (θ ) d θ = 0

0

(8.54)

8.2 The Ursell Modification

123

and define (σ 1)

χm

(σ )

(x) = Φm (x),

(σ 2)

χm

(σ )

(x) = ϒm (x),

(σ 3)

χm

(σ )

(x) = Ψm (x).

(8.55)

Theorem 8.5. If ∂ C is any closed curve that contains the origin in its interior, then, for m ≥ 1,   (σ j) T

χ¯ m

(ν k)

T χn

 (ν k) T (σ j)  − χn T χ¯ m ds = 8ih2 μ k32 δmn δσ ν δ jk .

(8.56)

∂C

This equality also holds for m = 0, σ = 1. Proof. Let ∂ KR be the circle with the center at the origin and radius R, with R large enough so that ∂ C is contained entirely within ∂ K R . Applying Theorem 1.2 in the (σ j) (ν k) region between ∂ KR and ∂ C, with u = χ¯ m and v = χn , we find that 



 (ν k) T (σ j)  (σ j) T (ν k) χ¯ m T χn − χn T χ¯ m ds

∂C

=

  (σ j) T

χ¯ m

(ν k)

T χn

 (ν k) T (σ j) − χn T χ¯ m ds. (8.57)

∂ KR

We consider the integral around ∂ K R . It is clear that this integral vanishes as R → ∞ when m = n or σ = ν , because of (8.52) and (8.54), respectively. From (8.44)– (8.46), (8.49)–(8.51), and (8.42) it is also easy to see that the integral vanishes as R → ∞ when j = 1, k = 3 and j = 2, k = 3. By (8.44), (8.45), (8.49), (8.50), (8.42), (8.43), and (8.52), for j = 1, k = 2 we have    ¯ (σ ) T

Φm

∂ KR

=√

(σ )  (σ ) T ¯ (σ )  T ϒm − ϒm T Φm ds

 4i ei(k2 −k1 )R (h2 μ k32 α2 − hμ k3 α1 )α1 k1 − hk3 α2 (α2 + hk3 α1 )k2 μ k1 k2  + (h2 μ k32 α1 + hμ k3 α2 )α2 k2 + hk3 α1 (α1 − hk3α2 )k1 μ + O(R−1 )

= O(R−1 ). Obviously, the complex conjugate of the integral is also O(R −1 ). This takes care of the case j = 2, k = 1. We now investigate what happens when m = n, σ = ν , and k = j. First,    ¯ (σ ) T

Φm

∂ KR

(σ )  (σ ) T ¯ (σ )  T Φm − Φm T Φm ds   = 8i α1 (h2 μ k32 α1 + h μ k3α2 ) − hμ k3 α2 (α1 − hk3 α2 ) + O(R−1)

= 8ih2 μ k32 (α12 + α22 ) + O(R−1 ); so, since α12 + α22 = 1, we deduce that

124

8 Modified Fundamental Solutions





(σ ) T (σ )  (σ )  T ¯ (σ )  Φ¯ m T Φm − Φm T Φm ds = 8ih2 μ k32 + O(R−1).

∂ KR

Similarly,    ¯ (σ ) T

ϒm

∂ KR

(σ )

T ϒm

 (σ ) T (σ )  − ϒm T ϒ¯m ds

  = 8i α2 (h2 μ k32 α2 − h μ k3α1 ) + hμ k3 α1 (α2 + hk3 α1 ) + O(R−1) = 8ih2 μ k32 + O(R−1 )

and

   ¯ (σ ) T

Ψm

(σ )  (σ ) T ¯ (σ )  T Ψm − Ψm T Ψm ds = 8ih2 μ k32 + O(R−1 ).

∂ KR

Consequently, letting R → ∞ in (8.57), we obtain (8.56). We remark that we get the same result when m = 0 and σ = 1 since in this case εm = 1 and, by (8.53), integrating the angular parts of the wavefunctions yields 2π , as opposed to the value π when ε m = 2.   Orthogonality relations for the corresponding wavefunctions arising in the theory of the three-dimensional Helmholtz equation and of two-dimensional elastodynamics are given in [53]. However, in [50] it is not mentioned that the results may not necessarily hold when m = n = 0. Theorem 8.6. Pm−1 exists for m ≥ 0. Proof. First, we deal with the case m ≥ 1. Consider the system of equations 

 (1) (1) (1)  (1)  T Φm + K Φm R=a z1 + T ϒm + Kϒm R=a z2  (2) (2)  + T Ψm + KΨm R=a z3 = 0. (8.58)

This can be written as TU + KU = 0 on R = a, where

(1)

(1)

(2)

U = Φm z1 + ϒm z2 + Ψm z3 . By (8.59) and (8.26), on R = a we have U¯ T TU − U TT U¯ = −h2 μκ |U1 |2 − h2 μκ |U2 |2 − μκ |U3 |2 + h2 μ κ¯ |U1 |2 + h2 μ κ¯ |U2 |2 + μ κ¯ |U3 |2  = −2iμ |κ | sin δ h2 |U1 |2 + h2|U2 |2 + |U3 |2 , so, taking (8.60) into account, we find that

(8.59) (8.60)

8.2 The Ursell Modification



125

(1) (1) (2) T  (1) (1) (2)  Φ¯ m z¯1 + ϒ¯m z¯2 + Ψ¯m z¯3 T Φm z1 + T ϒm z2 + T Ψm z3  (1) (1) (2) T  (1) (1) (2)  − Φm z1 + ϒm z2 + Ψm z3 T Φ¯ m z¯1 + T ϒ¯m z¯2 + T Ψ¯m z¯3  = −2iμ |κ | sin δ h2 |U1 |2 + h2 |U2 |2 + |U3 |2 . (8.61)

Integrating (8.61) around ∂ σ (a) and using (8.56) yields  8ih2 μ k32 |z1 |2 + |z2 |2 + |z3 |2

   = −2i μ |κ | sin δ h2 |U1 |2 ds + h2 |U2 |2 ds + ∂ σ (a)

∂ σ (a)

 |U3 | ds , 2

∂ σ (a)

which, since 0 < δ < π , implies that z1 = z2 = z3 = 0. Therefore, we have shown that the system (8.58) or, equivalently,

Θm Z = 0, where Z = (z1 , z2 , z3 )T and ⎛ (1) (1)  T Φm + K Φm 1 ⎜ (1) (1)  Θm = ⎜ ⎝ T Φm + K Φm 2  (1) (1)  T Φm + K Φm 3



(1)  1  (1) (1)  T ϒm + Kϒm 2  (1) (1)  T ϒm + Kϒm 3 (1)

T ϒm + Kϒm

⎞  1  ⎟    (2) (2)  T Ψm + KΨm 2 ⎟ ⎠    (2) (2)  T Ψm + KΨm 

(2)

(2) 

T Ψm + KΨm

3

,

R=a

has only the zero solution; hence, det Θ m = 0. Explicitly, we have 0 = det Θm   

√ 2 3 cos θ cos(mθ ) sin θ sin(mθ ) 0  am bm −cm   a 2h μ  0  dm em − fm  = 2  sin θ cos(mθ ) − cos θ sin(mθ ) h a2  0 0 cos(mθ ) gm hm −im  √ h4 μ 3 = 2 2 5 cos2 (mθ ) sin(mθ )(det Pm ). a Since m = 0, we conclude that det Pm = 0, so Pm−1 exists for m ≥ 1. When m = 0, the matrix reduces to ⎛ ⎞ a0 b 0 0 ⎜ ⎟ P0 = ⎝ 0 0 f 0 ⎠ , g0 h 0 0 from which we see that detP0 = − f0 (a0 h0 − b0 g0 ).

126

8 Modified Fundamental Solutions

Consider the system   (1) (1) (1)  (1)  T Φ0 + K Φ0 R=a z1 + T ϒ0 + Kϒ0 R=a z2 = 0.

(8.62)

This is not an over-determined system since the second row is equal to the first row multiplied by tan θ . Consequently, (8.62) is equivalent to !    (1) (1) (1)   (1)   T Φ0 + K Φ0 1 R=a T ϒ0 + Kϒ0 1 R=a z1 = 0. (8.63)   (1) (1) (1)   (1)   z2 T Φ + KΦ T ϒ + Kϒ 0

0

0

3 R=a

0

3 R=a

Equations (8.62) can be written as TU + KU = 0 on R = a, where

(1)

(1)

U = Φ0 z1 + ϒ0 z2 . Following the argument for the case m ≥ 1, we arrive at  8ih2 μ k32 |z1 |2 + |z2 |2

 = −2i μ |κ |(sin δ ) h2

|U1 | ds + h 2

∂ σ (a)

 2

∂ σ (a)

|U2 | ds +



2

 |U3 | ds , 2

∂ σ (a)

from which we conclude that z1 = z2 = 0. Hence, (8.63) has only the zero solution, so   T Φ (1) + K Φ (1)    0 1 R=a 0 =  0(1)    T Φ + K Φ (1)  0

0

3 R=a

   h2 μ      a2 cos θ 0  a0 b0  = μ  g0 h0   0   a =

  (1) (1)   T ϒ0 + Kϒ0 1 R=a    (1) (1)   T ϒ0 + Kϒ0 3 R=a 

h2 μ 2 (cos θ )(a0 h0 − b0g0 ), a3

which implies that a0 h0 − b0g0 = 0. It remains to show that f 0 = 0. First, we remark that H0 (k3 a) = −H1 (k3 a), 1 1  H0 (k3 a) = −H1 (k3 a) = −H0 (k3 a) + H1 (k3 a) = −H0 (k3 a) − H (k3 a), k3 a k3 a 0

8.2 The Ursell Modification

127

which means that

  1  f0 = −k32 a2 − H0(k3 a) − H0 (k3 a) − k3 a(κ − a)H0 (k3 a) k3 a = k32 a2 H0 (k3 a) − k3 a(κ − a − 1)H0 (k3 a).

Suppose that f 0 = 0; that is, k3 aH0 (k3 a) = (κ − a − 1)H0 (k3 a).

(8.64)

Then we also have f¯0 = 0, which leads to k3 aH¯0 (k3 a) = (κ¯ − a − 1)H¯0 (k3 a).

(8.65)

By (8.64), (8.65), and (8.26),   k3 a H0 (k3 a)H¯ 0 (k3 a) − H¯ 0 (k3 a)H0 (k3 a) = (κ¯ − a − 1)|H0 (k3 a)|2 − (κ − a − 1)|H0 (k3 a)|2 = (κ¯ − κ )|H0 (k3 a)|2 = −2i|κ | sin δ |H0 (k3 a)|2 . Also, as is shown in [71], 4i H0 (k3 a)H¯ 0 (k3 a) − H¯ 0 (k3 a)H0 (k3 a) = , π k3 a so

 2 4i = −2i|κ | sin δ H0 (k3 a) ; π

that is,

 2 2 + π |κ | sin δ H0 (k3 a) = 0.

But this is impossible since, owing to the fact that 0 < δ < π , the left-hand side is strictly positive. Hence, f 0 = 0, which implies that det P0 = 0, and we conclude that P0−1 exists.  

8.2.3 Convergence of the Infinite Series We now establish conditions under which the infinite series (8.38) converges. Writing M ω (x, y) explicitly, we have M ω (x, y) =

∞ 2   (σ )  T  (σ )  T i (1) (σ ) (2) (σ ) βm Φm (x) Φm (y) + βm Φm (x) ϒm (y) ∑ ∑ 2 2 4h μ k3 m=0 σ =1

128

8 Modified Fundamental Solutions

 T (3) (σ ) + (−1)σ βm Φm (x) Ψm3−σ (y)  (σ ) T  (σ ) T (4) (σ ) (5) (σ ) + βm ϒm (x) Φm (y) + βm ϒm (x) ϒm (y)  T (6) (σ ) + (−1)σ βm ϒm (x) Ψm3−σ (y)  T (7) (σ ) + (−1)σ βm Ψm (x) Φm3−σ (y)  T (8) (σ ) + (−1)σ βm Ψm (x) ϒm3−σ (y)  (σ )  T  (9) (σ ) + βm Ψm (x) Ψm (y) , (r)

where βm , r = 1, 2, . . . , 9, are constants. We restrict our attention to one part of this series, namely,     ∞ 2 iα 2 ∂ (1) (σ ) MRωx,1Ry (x, y) = 2 1 2 ∑ ∑ βm (cos θx ) Hm (k1 Rx ) Em (θx ) ∂ Rx 2h μ k3 m=0 σ =1     ∂ (σ ) × (cos θy ) Hm (k1 Ry ) Em (θy ) . ∂ Ry This can be written as MRωx,1Ry (x, y) = where

∞

∑ um ,

m=0

 um = vm cos θx cos θy cos m(θx − θy ) ,

with vm =

   iα12 ∂ ∂ (1) β H (k R ) H (k R ) . m m x m y 1 1 ∂ Rx ∂ Ry 2h2 μ k32

From (8.66) we see that |u m | ≤ |vm |. If we can show that ∞

by the comparison test, the series

(8.66)

(8.67)

∞

∑ |vm | is convergent, then,

m=0

∑ um is absolutely convergent.

m=0

We have (1)

βm = −[Pm−1Pˆm ]11 = −[Pm−1 ]11 [Pˆm ]11 − [Pm−1 ]12 [Pˆm ]21 − [Pm−1]13 [Pˆm]31 . The inverse of the matrix Pm is ⎛ ⎞ em im − fm hm cm hm − bmim bm fm − cmem 1 ⎜ ⎟ Pm−1 = ⎝ fm gm − dmim am im − cmgm cm dm − am fm ⎠ , det Pm dm hm − emgm bm gm − am hm am em − bmdm so 1  (1) βm = − (em im − fm hm )aˆm + (cm hm − bm im )dˆm detPm  + (bm fm − cmem )gˆm .

(8.68)

8.2 The Ursell Modification

129

As m → ∞, we have [1]

m 1 ez √ Jm (z) ∼ , 2m 2π m



 1 e ez m−1 1 m ez m Jm (z) ∼ √ =√ , 2π m 2 2m 2π m z 2m

 m−2 1 e2 1 ez  √ Jm (z) ∼ 1− 4 m 2m 2π m

 m

 1 m2 1 ez 1 m2 ez m =√ 1− ∼√ . m 2m 2π m z2 2π m z2 2m

(8.69)

Also, 1 Hm (z) = Jm (z) + iYm (z) ∼ √ 2π m

ez 2m

#

m −i

2 πm

ez 2m

−m

,

which means that #

−m 2 ez Hm (z) ∼ −i , π m 2m # #



 2 e ez −m−1 2 m ez −m  Hm (z) ∼ i =i , π m 2 2m π m z 2m # #

 −m−2

 2 e2 1 ez 2 m2 ez −m Hm (z) ∼ −i 1+ ∼ −i . πm 4 m 2m π m z2 2m

Using (8.70), we find that # am ∼ 2α1 k12 a2 Hm (k1 a) ∼ # = −2iα1 m

2

2 πm

−2iα1 k12 a2

ek1 a 2m

−m

2 m2 π m k12 a2

ek1 a 2m

−m

, #

 2 ek2 a −m −2iα2 m , π m 2m #

 2 1 ek3 a −m cm ∼ −2mk3 aHm (k3 a) ∼ −2im2 k3 a π m k3 a 2m #

−m 2 ek3 a = −2im2 , π m 2m #

 2 1 ek1 a −m  2 dm ∼ 2α1 mk1 aHm (k1 a) ∼ 2iα1 m k1 a π m k1 a 2m #

−m 2 ek1 a = 2iα1 m2 , π m 2m bm ∼ 2α2 k22 a2 Hm (k2 a) ∼

2

(8.70)

130

8 Modified Fundamental Solutions

#

 2 ek2 a −m em ∼ 2α2 mk2 aHm (k2 a) ∼ 2iα2 m2 , π m 2m # #

−m

−m 2 m2 ek3 a 2 ek3 a 2 2  2 2 2 fm ∼ −k3 a Hm (k3 a) ∼ ik3 a = im , π m k32 a2 2m π m 2m #

 2 m ek1 a −m  gm ∼ (α1 − hk3 α2 )k1 aHm (k1 a) ∼ i(α1 − hk3 α2 )k1 a π m k1 a 2m #

−m 2 ek1 a = im(α1 − hk3 α2 ) , π m 2m #

 2 ek2 a −m  hm ∼ (α2 + hk3 α1 )k2 aHm (k2 a) ∼ im(α2 + hk3 α1 ) , π m 2m #

−m 2 ek3 a im = −mHm (k3 a) ∼ im . π m 2m From the above equalities we deduce the asymptotic behavior of certain terms of interest. Thus, as m → ∞, em im − fm hm # 

−m  #

−m  2 ek2 a 2 ek3 a 2 ∼ 2iα2 m im π m 2m π m 2m #  #

−m 

  2 ek3 a 2 ek2 a −m 2 − im im(α2 + hk3α1 ) , π m 2m π m 2m

 2 3 e2 k2 k3 a2 −m ∼ m (−2α2 + α2 + hk3 α1 ) πm 4m2

2  2m2 e k2 k3 a2 −m =− (α2 − hk3α1 ) , π 4m2 d m im − f m g m # 

  #

  2 ek1 a −m 2 ek3 a −m 2 ∼ 2iα1 m im π m 2m π m 2m # # 

−m 

−m  2 ek3 a 2 ek1 a 2 − im im(α1 − hk3α2 ) π m 2m π m 2m

2 −m 2 2 3 e k1 k3 a = m (−2α1 + α1 − hk3 α2 ) πm 4m2

2  2m2 e k1 k3 a2 −m =− (α1 + hk3α2 ) , π 4m2 d m h m − e m gm # # 

 

  2 ek1 a −m 2 ek2 a −m 2 ∼ 2iα1 m im(α2 + hk3 α1 ) π m 2m π m 2m

8.2 The Ursell Modification

 − 2iα2 m2

#

2 πm

131

ek2 a 2m

−m  im(α1 − hk3 α2 )

#

2 πm

ek1 a 2m

−m 

  4 3 e2 k1 k2 a2 −m  m − α1 (α2 + hk3α1 ) + α2 (α1 − hk3α2 ) 2 πm 4m

2  2 4hk3m e k1 k2 a2 −m =− , π 4m2 =

c m hm − b m i m # # 

−m 

−m  2 ek3 a 2 ek2 a 2 ∼ − 2im im(α2 + hk3α1 ) π m 2m π m 2m # 

−m  #

  2 ek2 a 2 ek3 a −m 2 − − 2iα2 m im π m 2m π m 2m

2 −m

 2 4 3 e k2 k3 a 4hk3 α1 m2 e2 k2 k3 a2 −m = m (α2 + hk3 α1 − α2 ) = , πm 4m2 π 4m2 b m f m − cm em # 

  #

  2 ek2 a −m 2 ek3 a −m 2 2 ∼ − 2iα2 m im π m 2m π m 2m # # 

−m 

−m  2 ek3 a 2 ek2 a 2 2 − − 2im 2iα2 m π m 2m π m 2m

2 

 4 4 e k2 k3 a2 −m 4α2 m3 e2 k2 k3 a2 −m = m (α2 − 2α2 ) = − . πm 4m2 π 4m2 These results imply that det Pm = am (em im − fm hm ) − bm(dm im − fm gm ) + cm(dm hm − emgm ) # 

 

2   2 ek1 a −m 2m2 e k2 k3 a2 −m 2 ∼ − 2iα1 m − (α2 − hk3 α1 ) π m 2m π 4m2 # 

−m 

2   2 2 ek2 a 2m e k1 k3 a2 −m − − 2iα2 m2 − (α1 + hk3 α2 ) π m 2m π 4m2 # 

−m 

  2 ek3 a 4hk3 m2 e2 k1 k2 a2 −m + − 2im2 − π m 2m π 4m2 #

3 −m 4 3   4im 2 e k1 k2 k3 a = α1 (α2 − hk3 α1 ) − α2 (α1 + hk3 α2 ) + 2hk3 , 3 π πm 8m so

4ihk3 m4 det Pm ∼ π

#

2 πm

e3 k1 k2 k3 a3 8m3

−m

.

132

8 Modified Fundamental Solutions (1)

Before investigating the asymptotic behavior of β m , we need to know what happens to aˆ m , dˆm , and gˆm as m → ∞. By (8.69),

m 1 m2 ek1 a aˆm ∼ 2α1 k12 a2 Jm (k1 a) ∼ 2α1 k12 a2 √ 2π m k12 a2 2m

m 1 ek1 a = 2 α 1 m2 √ , 2π m 2m

 1 m ek1 a m  ˆ dm ∼ 2α1 mk1 aJm (k1 a) ∼ 2α1 mk1 a √ 2π m k1 a 2m

m 1 ek1 a = 2 α 1 m2 √ , 2π m 2m

 1 m ek1 a m  gˆm ∼ (α1 − hk3 α2 )k1 aJm (k1 a) ∼ (α1 − hk3 α2 )k1 a √ 2π m k1 a 2m

m m ek1 a = (α1 − hk3 α2 ) √ ; 2π m 2m hence, by (8.68), (1) βm

#

 π π m e3 k1 k2 k3 a3 m ∼− 4ihk3 m4 2 8m3 

2  

  2m2 e k2 k3 a2 −m 1 ek1 a m 2 × − (α2 − hk3 α1 ) 2 α1 m √ π 4m2 2π m 2m 

2 −m 

m  2 2 4hk3 α1 m e k2 k3 a 1 ek1 a + 2 α 1 m2 √ 2 π 4m 2π m 2m 

2 −m 

  3 2 4 α2 m e k2 k3 a m ek1 a m √ + − ( α − hk α ) 1 3 2 π 4m2 2π m 2m #

3 

 m π π m e k1 k2 k3 a3 4m4 2mk1 m 1 √ =− 4ihk3 m4 2 8m3 π ek2 k3 a 2π m   2 × − α1 (α2 − hk3 α1 ) + 2hk3α1 − α2 (α1 − hk3α2 ) ,

which means that (1) βm

Also,

2α1 α2 − 2hk3α12 − hk3 ∼ 2ihk3

e2 k12 a2 4m2

m .

#

m ∂ im 2 2m  Hm (k1 Rx ) = k1 Hm (k1 Rx ) ∼ , ∂ Rx Rx π m ek1 Rx #

 ∂ im 2 2m m Hm (k1 Ry ) ∼ . ∂ Ry Ry π m ek1 Ry

8.3 The Jones Modification

133

Putting all these results together and using (8.67), we obtain iα 2 2α1 α2 − 2hk3α12 − hk3 vm ∼ − 2 1 2 2ihk3 2h μ k3

 m2 Rx Ry

 m 2 4m2 × ; πm e2 k12 Rx Ry

e2 k12 a2 4m2

that is, (2hk3 α12 + hk3 − 2α1 α2 )α12 vm ∼ 2π h3 μ k33

m

m Rx Ry



a2 Rx Ry

m .

(8.71)

∞

Theorem 8.7. The series

∑ |vm | converges in the region R x Ry > a2.

m=0

Proof. By (8.71), vm+1 ∼ vm Therefore,

m+1 m



a2 Rx R y



1 = 1+ m



  2  vm+1  = a , lim   m→∞ vm Rx Ry ∞

so, by the ratio test,

 a2 . Rx Ry

∑ |vm | converges provided that R x Ry > a2 .

m=0

As was remarked earlier, this result implies that

 

∞

∑ um is absolutely convergent

m=0

in the region R x Ry > a2 . The rest of the series (8.38) is handled in the same manner. We can now formulate the following assertion. Theorem 8.8. The infinite series (8.38) for M ω (x, y) is absolutely convergent in the region Rx Ry > a2 . Since Ry > a, we remark that R x Ry > a2 implies that Rx ≥ a, so the series is absolutely convergent in our region of interest.

8.3 The Jones Modification We now describe a modification analogous to the one introduced in [28] for the Helmholtz equation. However, instead of adding a finite series to the matrix of fundamental solutions, as was done in [28], here we consider an infinite series. This has the advantage that it leads to uniquely solvable integral equations for all values of ω instead of only for a range of values. We give conditions that ensure the convergence

134

8 Modified Fundamental Solutions

of this infinite series and indicate how the results are adapted to the case of a finite series. The modified matrix of fundamental solutions D Lω (x, y) is defined by DLω (x, y) = D ω (x, y) + Lω (x, y),

(8.72)

where Lω (x, y) =

∞ 2   (σ )  T  (σ ) T i (σ ) (σ ) (σ ) (σ ) am Φm (x) Φm (y) + bm ϒm (x) ϒm (y) ∑ ∑ 2 2 4h μ k3 m=0 σ =1  (σ ) T (σ ) (σ ) + cm Ψm (x) Ψm (y)  (σ )  T  (σ )  T (σ )  (σ ) (σ ) + dm Φm (x) ϒm (y) + ϒm (x) Φm (y)  (σ )  T + em (−1)σ Φm (x) Ψm3−σ (y)  T (σ ) − Ψm (x) Φm3−σ (y)  (σ )  T + fm (−1)σ ϒm (x) Ψm3−σ (y)  T  (σ ) − Ψm (x) ϒm3−σ (y) ; (8.73)

(σ )

(σ )

(σ )

(σ )

here, am , bm , cm , dm , em , and f m are arbitrary constants. The matrix L ω (x, y) is chosen in this way so that it takes the same form as M ω (x, y) constructed in Section 8.2. The reason for this will be mentioned later. (σ ) It is not necessary to include the terms with coefficients d m , em , and f m in series (σ ) (8.73). They are present there only for the sake of generality. If d m = em = fm = 0, σ = 1, 2, m = 0, 1, 2, . . . , then our modification is analogous to that considered in [29] for three-dimensional elastodynamics. Let σ (b) be the disk with the center at the origin and radius b, which is sufficiently small so that σ (b) ⊂ S + . We assume that, as m → ∞,

2 2 2 m

2 2 2 m e k1 b e k2 b (σ ) (σ ) am ∼ , b ∼ , m 4m2 4m2

2 2 2 m

2  e k3 b e k1 k2 b2 m (σ ) (σ ) (8.74) cm ∼ , d ∼ , m 4m2 4m2

2 

2  e k1 k3 b2 m e k2 k3 b2 m em ∼ , f ∼ . m 4m2 4m2 Following the argument developed in Section 8.2.3, we can show that, if the constants satisfy (8.74) as m → ∞, then the infinite series for L ω (x, y) defined by (8.73) is absolutely convergent in the region R x Ry > b2 . Therefore, the series is also absolutely convergent in the region R x ≥ b (since this is a subset of R x Ry > b2 ).

8.3 The Jones Modification

135

From (8.73) it is easily seen that  T Lω (x, y) = Lω (y, x) ; hence,

 T DLω (x, y) = DLω (y, x) .

The columns of the modification L ω (x, y) are regular solutions of A ω u = 0 with respect to x in R2 \{0} and satisfy the radiation conditions as R = |x| → ∞. We introduce the modified single-layer and double-layer potentials VLω (ϕ ) = WLω (ϕ ) =

 



DLω (x, y)ϕ (y) ds(y),

∂S

T T (∂y )DLω (y, x) ϕ (y) ds(y),

∂S

where DLω (x, y) is defined by (8.72). As in the Ursell modification, we define VLω + (ϕ ) = VLω (ϕ )|S¯+ , VLω − (ϕ ) = VLω (ϕ )|S¯− , which satisfy  ω∗ 1 T VLω + (ϕ ) = WL0 + 2I ϕ  ω∗ T VLω − (ϕ ) = WL0 − 12 I ϕ

on ∂ S, on ∂ S.

We also define WLω + (ϕ )

 WLω (ϕ )|S+ =  ω − 1I ϕ WL0 2

WLω − (ϕ )

 WLω (ϕ )|S− =  ω + 1I ϕ WL0 2

in S+ , on ∂ S, in S− , on ∂ S,

which satisfy T WLω + (ϕ ) = T WLω − (ϕ )

on ∂ S.

8.3.1 Orthogonality Relations Some further orthogonality relations, in addition to those mentioned in Theorem 8.5, are needed.

136

8 Modified Fundamental Solutions

  Since Jm (z) = Re Hm (z) when the argument of the Bessel function is real, from (8.44)–(8.46) and (8.49)–(8.51) we deduce that, for m ≥ 1, as R → ∞, 1 (σ ) (σ ) Φˆ m (x) = −2α1 k1 √ sin(k1 R − πm )Em (θ )eR π k1 R 1 (σ ) − 2hk3α2 √ cos(k1 R − πm )Em (θ )e3 + O(R−3/2 ), π k1 R 1 (σ ) (σ ) ϒˆm (x) = −2α2 k2 √ sin(k2 R − πm )Em (θ )eR π k2 R 1 (σ ) + 2hk3α1 √ cos(k2 R − πm )Em (θ )e3 + O(R−3/2 ), π k2 R 1 (σ ) (σ ) Ψˆm (x) = 2k3 √ sin(k3 R − πm)Em (θ )eθ + O(R−3/2 ), π k3 R

(8.75)

and that, on the circle with center at the origin and radius R, as R → ∞, (σ ) T (∂x )Φˆ m (x)

= −2(h2 μ k32 α1 + h μ k3α2 ) √ − 2(α1 − hk3 α2 )k1 μ √

1 (σ ) cos(k1 R − πm)Em (θ )eR π k1 R

1 (σ ) sin(k1 R − πm)Em (θ )e3 + O(R−3/2 ), π k1 R

(σ ) T (∂x )ϒˆm (x)

(8.76)

1 (σ ) = −2(h2 μ k32 α2 − h μ k3α1 ) √ cos(k2 R − πm)Em (θ )eR π k2 R 1 (σ ) − 2(α2 + hk3 α1 )k2 μ √ sin(k2 R − πm)Em (θ )e3 + O(R−3/2 ), π k2 R 1 (σ ) (σ ) T (∂x )Ψˆm (x) = 2h2 μ k32 √ cos(k3 R − πm)Em (θ )eθ + O(R−3/2 ). π k3 R By analogy with (8.55), we define (σ 1) (σ ) χˆ m (x) = Φˆ m (x),

(σ 2) (σ ) χˆ m (x) = ϒˆm (x),

(σ 3) (σ ) χˆ m (x) = Ψˆm (x).

(8.77)

Theorem 8.9. If ∂ C is any closed curve that contains the origin in its interior, then, for m ≥ 1, 



(σ j) T (ν k)  (ν k) T (σ j)  χˆ m T χˆ n − χˆ n T χˆ m ds = 0,

∂C

  (σ j) T

χˆ m

(ν k)

T χn

 (ν k) T (σ j)  − χn T χˆ m ds = 4ih2 μ k32 δmn δσ ν δ jk .

(8.78) (8.79)

∂C

Equality (8.78) actually holds for all m, while (8.79) is also true for m = 0, σ = 1.

8.3 The Jones Modification

137

Proof. Equation (8.78) is the result of a straightforward application of the reci(σ j) (ν k) procity relation in the region interior to ∂ C, with u = χˆ m and v = χˆ n , since both these functions are well behaved everywhere. To derive (8.79), we follow the procedure used in the proof of Theorem 8.5. Basically, we have to evaluate   (σ )  T (σ )  (σ ) T ˆ (σ )  lim Φˆ m T ϒm − ϒm T Φm ds, R→∞ ∂ KR

lim

   ˆ (σ ) T

lim

   ˆ (σ ) T

lim

   ˆ (σ ) T

lim

   ˆ (σ ) T

R→∞ ∂ KR

R→∞ ∂ KR

R→∞ ∂ KR

R→∞ ∂ KR

ϒm

 (σ ) T (σ )  (σ ) T Φm − Φm T ϒˆm ds,

Φm

ϒm

Ψm

 (σ )  T (σ ) (σ )  T Φm − Φm T Φˆ m ds,

 (σ )  T (σ )  (σ ) T ϒm − ϒm T ϒˆm ds,  (σ )  T (σ ) (σ )  T Ψm − Ψm T Ψˆm ds,

making use of (8.44)–(8.46), (8.49)–(8.51), (8.75), and (8.76). Thus, we find that    ˆ (σ ) T

Φm

 (σ ) T (σ ) (σ )  T ϒm − ϒm T Φˆ m ds

∂ KR

4 ei(k2 R−πm) k1 k2   × α1 k1 (h2 μ k32 α2 − hμ k3 α1 ) + hk3α1 (α1 − hk3 α2 )k1 μ sin(k1 R − πm)   + − hk3 α2 (α2 + hk3α1 )k2 μ + α2 k2 (h2 μ k32 α1 + hμ k3 α2 )  × i cos(k1 R − πm ) + O(R−1)

=√

= O(R−1 ),    ˆ (σ ) T

ϒm

 (σ )  T (σ )  (σ ) T Φm − Φm T ϒˆm ds

∂ KR

4 ei(k1 R−πm) k1 k2   × α2 k2 (h2 μ k32 α1 + hμ k3 α2 ) − hk3α2 (α2 + hk3 α1 )k2 μ sin(k2 R − πm)   + hk3 α1 (α1 − hk3 α2 )k1 μ + α1 k1 (h2 μ k32 α2 − hμ k3 α1 )  × i cos(k2 R − πm ) + O(R−1)

=√

= O(R−1 ),

138

8 Modified Fundamental Solutions





(σ ) T (σ )  (σ )  T ˆ (σ )  Φˆ m T Φm − Φm T Φm ds

∂ KR

  = 4ei(k1 R−πm ) α1 (h2 μ k32 α1 + hμ k3 α2 ) − hk3 α2 μ (α1 − hk3 α2 )   × i cos(k1 R − πm) + sin(k1 R − πm ) + O(R−1) = 4ih2 μ k32 (α12 + α22 ) + O(R−1) = 4ih2 μ k32 + O(R−1 ).

(σ ) (σ ) The behavior of the integral involving ϒˆm and ϒm is similar to the one above. Finally,



 (σ ) T (σ )  (σ ) T ˆ (σ )  Ψˆm T Ψm − Ψm T Ψm ds

∂ KR

  = 4h2 μ k32 i cos(k3 R − πm ) + sin(k3 R − πm) ei(k3 R−πm) + O(R−1 ) = 4ih2 μ k32 + O(R−1).

The assertion is now established by reasoning as in the proof of Theorem 8.5.  

8.3.2 The Exterior Neumann Problem We seek the solution of (N ω − ) in the form v = V Lω − (ϕ ), which leads to the integral equation  ω∗ 1 WL0 − 2 I ϕ = S . (8.80) To prove that (8.80) has a unique solution, we must show that  ω∗ 1 WL0 − 2 I ϕ = 0

(8.81)

has only the zero solution. Let u = VLω + (ϕ ) and v = VLω − (ϕ ), where ϕ satisfies (8.81). Then T v| ∂ S = 0, so, by the uniqueness property, v = 0 in S¯− ; in particular,

ω v|∂ S = VL0 ϕ = u|∂ S = 0.

Consider the constants (σ )

Am = (σ )

Bm =

i 2 4h μ k32 i 2 4h μ k32

  (σ )

T Φm (y) ϕ (y) ds(y),

∂S

  (σ )

T ϒm (y) ϕ (y) ds(y),

∂S

8.3 The Jones Modification (σ )

139

Cm =

i 4h2 μ k32

  (σ )

T Ψm (y) ϕ (y) ds(y)

∂S

and let Rmin be the minimum distance from the origin to ∂ S. Then, using (8.72), (8.17), and (8.73), we find that for b ≤ R x < Rmin , u(x) =

∞

 (σ ) (σ ) (σ ) (σ ) (σ ) (σ ) Am Φˆ m (x) + Bm ϒˆm (x) + Cm Ψˆm (x) m=0 σ =1   (σ ) (σ ) (σ ) (σ ) (σ ) + am Am + dm Bm + (−1)σ emCm3−σ Φm (x)  (σ ) (σ )  (σ ) (σ ) (σ ) + bm Bm + dm Am + (−1)σ fmCm3−σ ϒm (x)  (σ ) (σ )  (σ )  σ σ + cm Cm − (−1)σ em A3− − (−1)σ fm B3− Ψm (x) . m m 2

∑∑

(8.82)

We apply the reciprocity relation to u and u¯ in S + \σ (b). Since u|∂ S = u| ¯ ∂ S = 0, we arrive at  (uT T u¯ − u¯T Tu) ds = 0.

(8.83)

∂ σ (b) (σ j) Substituting (8.82) in (8.83), taking into account the fact that χˆ m is real when its argument is real, and using (8.56), (8.78), and (8.79), we find that ∞

1

2

∑∑

m=0 σ =1

2

 (σ )  (σ ) ¯ (σ ) (σ ) (σ ) a¯m Am + d¯m B¯ m + (−1)σ e¯mC¯m3−σ

Am

 (σ )  ¯ (σ ) ¯ (σ ) (σ ) (σ ) bm Bm + d¯m A¯ m + (−1)σ f¯mC¯m3−σ

+ 12 Bm

(σ )  (σ ) ¯ (σ ) σ c¯m Cm − (−1)σ e¯m A¯ 3− m

 σ − (−1)σ f¯m B¯ 3− m  (σ )  (σ ) (σ ) (σ ) (σ ) + 12 A¯ m am Am + dm Bm + (−1)σ emCm3−σ  (σ ) (σ )  (σ ) (σ ) + am Am + dm Bm + (−1)σ emCm3−σ  (σ ) (σ )  (σ ) (σ ) × a¯m A¯ m + d¯m B¯ m + (−1)σ e¯mC¯m3−σ  (σ )  (σ ) (σ ) (σ ) (σ ) + 12 B¯ m bm Bm + dm Am + (−1)σ fmCm3−σ  (σ ) (σ )  (σ ) (σ ) + bm Bm + dm Am + (−1)σ fmCm3−σ  (σ ) (σ )  (σ ) (σ ) × b¯ m B¯ m + d¯m A¯ m + (−1)σ f¯mC¯m3−σ  (σ )  (σ ) (σ ) σ σ + 12 C¯m cm Cm − (−1)σ em A3− − (−1)σ fm B3− m m  (σ ) (σ )  σ σ + cm Cm − (−1)σ em A3− − (−1)σ fm B3− m m  (σ ) (σ )  σ σ × c¯m C¯m − (−1)σ e¯m A¯ 3− − (−1)σ f¯m B¯ 3− = 0. m m + 12 Cm

We consider two different ways of rearranging (8.84). First, we may write ∞

2

∑∑

m=0 σ =1



(σ )

|Am |2

1

(σ ) 1 2 a¯m + 2

(σ )

(σ )

(σ )

am + |am |2 + |dm |2 + |em|2



(8.84)

140

8 Modified Fundamental Solutions (σ )

1

¯ (σ ) 2 bm + (σ )  (σ ) + |Cm |2 12 c¯m + + |Bm |2

(σ ) ¯(σ ) 1 (σ ) ¯(σ ) ¯ (σ ) (σ ) ¯ 2 dm + am dm + 2 dm + bm dm + em f m (σ ) (σ )  (σ ) (σ ) (σ ) (σ ) (σ ) (σ ) + A¯ m Bm 12 d¯m + 12 dm + a¯m dm + bm d¯m + e¯m fm  (σ ) (σ ) (σ ) (σ ) + (−1)σ Am C¯m3−σ 12 e¯m + am e¯m + dm f¯m + 12 em + c¯m em (σ ) (σ ) + Am B¯ m

1

(σ ) 2 (σ ) 2 2 1 (σ ) 2 bm + |bm | + |dm | + | f m | (σ ) 2 2 2 1 (σ ) 2 cm + |cm | + |em | + | f m |

(σ ) (σ ) 1 ¯(σ ) 2 e¯m + 2 em + a¯ m em + dm f m + cm e¯m  (σ ) (σ ) (σ ) (σ ) + (−1)σ Bm C¯m3−σ 12 f¯m + dm e¯m + bm f¯m + 12 fm + c¯m fm (σ ) + (−1)σ A¯ m Cm3−σ

(σ )

+ (−1)σ B¯ m Cm3−σ

1 1

(σ ) ¯  1 ¯(σ ) ¯ (σ ) ¯ 2 f m + dm em + 2 f m + bm f m + cm f m

= 0;

therefore,  (σ ) 2  (σ ) 1 2 1 (σ ) |Am | am + 2  − 4 + |dm |2 + |em |2 m=0 σ =1 2 (σ )  (σ ) (σ ) + |Bm |2 bm + 12  − 14 + |dm |2 + | fm |2 2 (σ )  (σ ) + |Cm |2 cm + 12  − 14 + |em|2 + | fm |2  (σ ) (σ )  (σ ) 1 (σ )  (σ ) 1 (σ )  + 2 Re Am B¯ m am + 2 d¯m + b¯ m + 2 dm + em f¯m  (σ )  (σ ) + 2(−1)σ Re Am C¯m3−σ am + 12 e¯m  (σ ) (σ )  + c¯m + 12 em + dm f¯m  (σ )  (σ ) + 2(−1)σ Re Bm C¯m3−σ bm + 12 f¯m  (σ ) (σ )  + c¯m + 12 fm + dm e¯m = 0. ∞

2

∑∑

If we choose the constants so that, for σ = 1, 2 and m = 0, 1, 2, . . . , 

and either

or

(σ )  (σ ) (σ ) (σ ) am + 12 d¯m + b¯ m + 12 dm + em f¯m = 0,  (σ ) 1  (σ ) (σ ) am + 2 e¯m + c¯m + 12 em + dm f¯m = 0,  (σ ) 1  (σ ) (σ ) bm + 2 f¯m + c¯m + 12 fm + dm e¯m = 0

(8.85)

 (σ ) 1 2 am +  + |dm(σ )|2 + |em |2 > 1 , 2 4  (σ ) 1 2 bm +  + |dm(σ )|2 + | fm |2 > 1 , 2 4  (σ ) 1 2 cm +  + |em |2 + | fm |2 > 1 2 4

(8.86)

 (σ ) 1 2 am +  + |dm(σ )|2 + |em |2 < 1 , 2 4  (σ ) 1 2 bm +  + |dm(σ )|2 + | fm |2 < 1 , 2 4  (σ ) 1 2 2 2 1 cm +  + |em | + | fm | < , 2 4

(8.87)

8.3 The Jones Modification

then

(σ )

141 (σ )

(σ )

σ = 1, 2, m = 0, 1, 2, . . . .

Am = Bm = Cm = 0,

(8.88)

Inequalities (8.86) and (8.87) are similar to those obtained for the Helmholtz equation (see [72]) and for the two-dimensional elastodynamic system (see [7]). There is another set of restrictions on the constants which ensures that (8.88) holds. By (8.84), ∞

 (σ ) 2  1 (σ ) 1 (σ ) (σ )  (σ ) (σ ) |Am | 2 a¯m + 2 am + |Bm |2 12 b¯ m + 12 bm m=0 σ =1 (σ )  (σ ) (σ ) + |Cm |2 12 c¯m + 12 cm (σ ) (σ ) (σ ) (σ )  (σ ) (σ ) + (Am B¯ m + A¯ m Bm ) 12 d¯m + 12 dm  (σ ) (σ ) + (−1)σ (Am C¯m3−σ + A¯ m Cm3−σ ) 12 e¯m + 12 em  (σ ) (σ ) + (−1)σ (Bm C¯m3−σ + B¯ m Cm3−σ ) 12 f¯m + 12 fm 2

∑∑

(σ ) (σ )

(σ ) (σ )

(σ ) (σ )

(σ ) (σ )

+ |am Am + dm Bm + (−1)σ emCm3−σ |2 + |bm Bm + dm Am + (−1)σ fmCm3−σ |2

 (σ ) (σ ) σ σ 2 + |cm Cm − (−1)σ em A3− − (−1)σ fm B3− = 0; m m | hence, ∞

2

∑∑

m=0 σ =1



(σ )

(σ )

(σ )

(σ )

(σ )

(σ )

Re(am )|Am |2 + Re(bm )|Bm |2 + Re(cm )|Cm |2 (σ ) (σ ) (σ ) + 2 Re(dm ) Re(Am B¯ m ) (σ ) + 2(−1)σ Re(em ) Re(Am C¯m3−σ ) (σ ) + 2(−1)σ Re( fm ) Re(Bm C¯m3−σ ) (σ ) (σ )

(σ ) (σ )

(σ ) (σ )

(σ ) (σ )

+ |am Am + dm Bm + (−1)σ emCm3−σ |2 + |bm Bm + dm Am + (−1)σ fmCm3−σ |2  (σ ) (σ ) σ σ 2 + |cm Cm − (−1)σ em A3− − (−1)σ fm B3− = 0. m m | Consequently, if for σ = 1, 2 and m = 0, 1, 2, . . . we have (σ )

(σ )

(σ )

Re(am ), Re(bm ), Re(cm ) > 0, (σ )

Re(dm ) = Re(em ) = Re( fm ) = 0, then (8.88) holds. Suppose that (8.88) holds. Then, by (8.82), u(x) = 0,

b ≤ Rx < Rmin .

(8.89)

142

8 Modified Fundamental Solutions

In view of the analyticity of the function u in S + \σ (b), from (8.89) we deduce that u = 0 in S + \σ (b), so

 ω∗ 1 0 = Tu|∂ S = WL0 + 2 I ϕ,

which, by (8.81), implies that ϕ = 0. Therefore, under certain smoothness conditions on S , (8.80) has a unique solution. Theorem 8.10. If S ∈ C 0,α (∂ S), α ∈ (0, 1), and the constants in L ω (x, y) are chosen so that (8.88) holds, then the integral equation  ω∗ 1 WL0 − 2 I ϕ = S has a unique solution ϕ ∈ C 0,α (∂ S). The matrix Lω (x, y) was chosen to be of the same form as M ω (x, y) because the Ursell modification may be just a special case of the Jones modification; that is, the constants arising in M ω (x, y) may satisfy, for example, (8.85) and (8.87). In [47] it is shown that this is indeed the case for the modifications used in connection with the two-dimensional Helmholtz equation.

8.3.3 The Exterior Dirichlet Problem The treatment of (D ω − ) is similar to that of (N ω − ). The solution is sought as a modified double-layer potential v = W Lω − (ϕ ). This leads to the equation  ω 1 WL0 + 2 I ϕ = R (8.90) for the unknown density ϕ . We show that the corresponding homogeneous equation  ω 1 WL0 + 2 I ϕ = 0 (8.91) has only the zero solution. Suppose that ϕ is a solution of (8.91) and let u = W Lω + (ϕ ) and v = WLω − (ϕ ). Then v|∂ S = 0 and, by the uniqueness of the solution of (D ω − ), we see that v = 0 in S − ; therefore,

0 = T v|∂ S = T WLω − (ϕ )|∂ S = T WLω + (ϕ )|∂ S = Tu|∂ S .

Let (σ ) A˜ m =

i 4h2 μ k32

 

T (σ ) T (∂y )Φm (y) ϕ (y) ds(y),

∂S

(8.92)

8.3 The Jones Modification (σ ) B˜ m =

(σ ) C˜m =

Using the fact that

143

i 4h2 μ k32 i 4h2 μ k32

 

T (σ ) T (∂y )ϒm (y) ϕ (y) ds(y),

∂S

 

T (σ ) T (∂y )Ψm (y) ϕ (y) ds(y).

∂S

 T DLω (y, x) = DLω (x, y) , (σ )

(σ )

(σ )

we can write u(x) in the same way as (8.82) with A m , Bm , and Cm replaced by (σ ) (σ ) (σ ) A˜ m , B˜ m , and C˜m , respectively. The analysis then proceeds in exactly the same way as in the previous subsection, with the use of boundary condition (8.92) when the reciprocity relation is applied. Under the same conditions as in Subsection 8.3.2, we find that (σ ) (σ ) (σ ) A˜ m = B˜ m = C˜m = 0,

σ = 1, 2, m = 0, 1, 2, . . . .

(8.93)

Arguing as before, we see that if (8.93) holds, then u = 0 in S + \σ (b). By continuity,

 ω 1 0 = u|∂ S = WL0 − 2 I ϕ,

which, in view of (8.91), implies that ϕ = 0. Consequently, (8.90) has a unique solution. Theorem 8.11. If R ∈ C 1,α (∂ S), α ∈ (0, 1), and the constants in L ω (x, y) are chosen so that (8.93) holds, then the integral equation  ω 1 WL0 + 2 I ϕ = R has a unique solution ϕ ∈ C 1,α (∂ S). A drawback of this approach is the requirement that the arbitrary constants must satisfy (8.74) for the infinite series to converge.

8.3.4 Modification with a Finite Series The exterior Neumann problem in acoustics was studied in [28] by means of a modification based on a finite series of wavefunctions. Similarly, in [7], the corresponding matrix of fundamental solutions in two-dimensional elastodynamics was modified by adding to it a finite series of the relevant wavefunctions. We investigate an analogous approach for the exterior boundary value problems of plate oscillations. As

144

8 Modified Fundamental Solutions

in [7], this method leads to uniquely solvable integral equations for only a range of values of the oscillation frequency. Let ω DL (x, y) = D ω (x, y) + L ω (x, y), (8.94) where L ω (x, y) =

M 2   (σ )  T  (σ ) T i (σ ) (σ ) (σ ) (σ ) am Φm (x) Φm (y) + bm ϒm (x) ϒm (y) ∑ ∑ 2 2 4h μ k3 m=0 σ =1  (σ ) T (σ ) (σ ) + cm Ψm (x) Ψm (y)  (σ )  T  (σ )  T (σ )  (σ ) (σ ) + dm Φm (x) ϒm (y) + ϒm (x) Φm (y)  (σ )  T + em (−1)σ Φm (x) Ψm3−σ (y)  T (σ ) − Ψm (x) Φm3−σ (y)  (σ )  T + fm (−1)σ ϒm (x) Ψm3−σ (y)  T  (σ ) − Ψm (x) ϒm3−σ (y) . (8.95) (σ )

(σ )

The nonnegative integer M is at our disposal along with the constants a m , bm , (σ ) (σ ) cm , dm , em , and f m . Later on, we specify conditions on these constants to ensure the unique solvability of our modified integral equations. Consider the exterior Neumann problem. We adopt the analysis procedure followed in Subsection 8.3.2 and seek the solution in the form v = VLω − (ϕ ). The unknown density must then satisfy the integral equation  ω∗ 1 WL 0 − 2 I ϕ = S .

(8.96)

We show that the homogeneous equation  ω∗ 1 WL 0 − 2 I ϕ = 0

(8.97)

has only the zero solution. As in Subsection 8.3.2, we set u = V Lω + (ϕ ) and v = VLω − (ϕ ), where ϕ is a solution of (8.97), and deduce that v|∂ S = VLω 0 ϕ = u|∂ S = 0. Using (8.94), (8.17), and (8.95), for R x < Rmin we may write u(x) =

∞

2

∑ ∑

m=M+1 σ =1

 (σ ) (σ )  (σ ) (σ ) (σ ) (σ ) Am Φˆ m (x) + Bm ϒˆm (x) +Cm Ψˆm (x)

8.3 The Jones Modification

+

(σ )

M

2

∑∑

m=0 σ =1

(σ )

145

 (σ ) (σ ) (σ ) (σ ) (σ ) (σ ) Am Φˆ m (x) + Bm ϒˆm (x) + Cm Ψˆm (x)  (σ ) (σ )  (σ ) (σ ) (σ ) + am Am + dm Bm + (−1)σ emCm3−σ Φm (x)  (σ ) (σ )  (σ ) (σ ) (σ ) + bm Bm + dm Am + (−1)σ fmCm3−σ ϒm (x)  (σ ) (σ )  (σ )  σ σ + cm Cm − (−1)σ em A3− − (−1)σ fm B3− Ψm (x) , m m (8.98) (σ )

where Am , Bm , and Cm are defined as in Subsection 8.3.2. Let ε be small enough so that the disk σ (ε ) is contained entirely within the interior domain S + . Applying the reciprocity relation to u and its complex conjugate u¯ in S + \σ (ε ) and taking into account the fact that u| ∂ S = u| ¯ ∂ S = 0, we arrive at 

 T u T u¯ − u¯ TTu ds = 0.

(8.99)

∂ σ (ε )

As in Subsection 8.3.2, substituting (8.98) in (8.99) leads to  (σ ) 2  (σ ) 1 2 1 (σ ) |Am | am + 2  − 4 + |dm |2 + |em |2 m=0 σ =1 2 (σ )  (σ ) (σ ) + |Bm |2 bm + 12  − 14 + |dm |2 + | fm |2 2 (σ )  (σ ) + |Cm |2 cm + 12  − 14 + |em|2 + | fm |2  (σ ) (σ )  (σ ) 1 (σ )  (σ ) 1 (σ )  + 2 Re Am B¯ m am + 2 d¯m + b¯ m + 2 dm + em f¯m  (σ )  (σ ) + 2(−1)σ Re Am C¯m3−σ am + 12 e¯m  (σ ) (σ )  + c¯m + 12 em + dm f¯m  (σ )  (σ ) + 2(−1)σ Re Bm C¯m3−σ bm + 12 f¯m  (σ ) (σ )  + c¯m + 12 fm + dm e¯m =0 M

2

∑∑

or, equivalently, M

2

∑∑

m=0 σ =1



(σ )

(σ )

(σ )

(σ )

(σ )

(σ )

Re(am )|Am |2 + Re(bm )|Bm |2 + Re(cm )|Cm |2 (σ ) (σ ) (σ ) + 2 Re(dm ) Re(Am B¯ m ) (σ ) (σ ) + 2(−1)σ Re(em ) Re(Am C¯m3−σ ) + 2(−1)σ Re( fm ) Re(Bm C¯m3−σ ) (σ ) (σ )

(σ ) (σ )

(σ ) (σ )

(σ ) (σ )

+ |am Am + dm Bm + (−1)σ emCm3−σ |2 + |bm Bm + dm Am + (−1)σ fmCm3−σ |2

 (σ ) (σ ) σ σ 2 + |cm Cm − (−1)σ em A3− − (−1)σ fm B3− = 0. m m |

146

8 Modified Fundamental Solutions

The constants can then be chosen to ensure that (σ )

(σ )

(σ )

Am = Bm = Cm = 0,

σ = 1, 2, m = 0, 1, 2, . . . , M.

(8.100)

From (8.98) we see that, if (8.100) holds, then for R x < Rmin , u(x) =

∞

2

∑ ∑

m=M+1 σ =1

 (σ ) (σ )  (σ ) (σ ) (σ ) (σ ) Am Φˆ m (x) + Bm ϒˆm (x) +Cm Ψˆm (x) .

(8.101)

Theorem 8.12. If (8.100) holds, then the function u has a zero in R at the origin, of order at least M − 1. Proof. By (8.101), u(x) = Next,

2

∑

σ =1

 (σ ) (σ )  (σ ) (σ ) (σ ) (σ ) AM+1 Φˆ M+1 (x) + BM+1ϒˆM+1 (x) +CM+1ΨˆM+1 (x) + · · · .

⎛ ⎞ ∂  (σ ) α J (k R)E ( θ ) M+1 1 M+1 ⎜ 1 ⎟ √ ⎜ ∂ x1 ⎟ ( σ )  ⎟ . ∂ Φˆ M+1 (x) = 2 ⎜ ( σ ) ⎜α1 ⎟ ⎝ ∂ x2 JM+1 (k1 R)EM+1 (θ ) ⎠ (σ ) −hk3 α2 JM+1 (k1 R)EM+1 (θ )

It is easy to show that  ∂  (σ ) JM+1 (k1 R)EM+1 (θ ) ∂ x1   (σ ) (σ ) = 12 k1 JM (k1 R)EM (θ ) − JM+2 (k1 R)EM+2 (θ ) ,  ∂  (σ ) JM+1 (k1 R)EM+1 (θ ) ∂ x2   3−σ 3−σ = 12 (−1)σ k1 JM (k1 R)EM (θ ) + JM+2 (k1 R)EM+2 (θ ) , which means that

  ⎞ (σ ) (σ ) α1 k1 JM (k1 R)EM (θ ) − JM+2(k1 R)EM+2 (θ )  ⎟ 1 ⎜ (σ ) 3−σ 3−σ Φˆ M+1 (x) = √ ⎜ (−1)σ α1 k1 JM (k1 R)EM (θ ) + JM+2 (k1 R)EM+2 (θ ) ⎟ ⎝ ⎠. 2 (σ ) −2hk3 α2 JM+1 (k1 R)EM+1 (θ )

Also,

⎛

2l  1 l ∂l J (k R) = k J (k R) + βm JM−l+m (k1 R), M 1 1 1 M−l ∑ 2 ∂ Rl m=2

where the βm are constants. Since J0 (0) = 1

8.4 Equations of the First Kind

147

and Jn (0) = 0, we find that



n ≥ 1,

  ∂l  J (k R) = 0 if M − l ≥ 1. M 1  l ∂R R=0

(σ ) Consequently, Φˆ M+1 (x) has a zero in R at the origin, of order at least M − 1. Re(σ ) (σ ) peating this argument for ϒˆ (x) and Φˆ (x) leads to the desired result.   M+1

M+1

From Theorem 8.12 we see that, by increasing the value of M, we increase the order of the zero in R of u at the origin and, hence, strengthen the constraints on the function. As in [7], we can argue that this increases—or, rather, does not decrease— the lowest value of the oscillation frequency, ω (M), say, for which u can possibly − be an eigensolution of (D ω 0 ). Unfortunately, there is no way of knowing how many terms to take in the series (8.95) to avoid a particular eigenfrequency. All we can claim is that we have a greater chance of eliminating eigenfrequencies as we increase M. Suppose that ω < ω (M) and that (8.100) holds. Then u(x) = 0 in S + , which, as in Subsection 8.3.2, implies that (8.97) has only the zero solution. Consequently, if S ∈ C 0,α (∂ S), then (8.96) has a unique C 0,α -solution. The exterior Dirichlet problem can be treated analogously.

8.4 Equations of the First Kind Using the modified potentials discussed in the preceding subsections, we can derive Fredholm integral equations of the first kind for (D ω − ) and (N ω − ) which have unique solutions. In [15]–[20], an extensive investigation by means of modified potentials can be found of equations of the first kind arising in the Dirichlet boundary value problems in two-dimensional elastostatics and equilibrium bending of elastic plates with transverse shear deformation. We concentrate on the Ursell modification, although all of the results remain valid for the Jones modification provided that the conditions of Theorems 8.10 and 8.11 are satisfied.

8.4.1 Composition of the Modified Boundary Operators Consider the modification matrix M ω (y, x) defined by (8.38) and let K R be a disk centered at the origin, whose radius R is large enough so that S + is contained entirely

148

8 Modified Fundamental Solutions

within ∂ KR . We apply the reciprocity relation in S − ∩ KR to the columns of M ω (y, x) and a function u(x) that satisfies A ω u = 0 in S − and the radiation conditions as |x| → ∞; thus,    T  uT (y)T (∂y )M ω (y, x) − M ω (y, x) T (∂y )u(y) ds(y) ∂S

   T  uT (y)T (∂y )M ω (y, x) − M ω (y, x) T (∂y )u(y) ds(y).

=

(8.102)

∂ KR

As in the proof of Theorem 4.7, the integral around ∂ K R vanishes as R → ∞, so, taking the symmetry of M ω (y, x) into account, we find that (8.102) yields  

T  T (∂y )M ω (y, x) u(y) − M ω (x, y)T (∂y )u(y) ds(y) = 0.

(8.103)

∂S

Theorem 8.13. If u ∈ C 2 (S− ) ∩C1 (S¯− ) ∩ Bω is a solution of (4.1) in S − , then u can be represented in the form u = WMω − (u|∂ S ) − VMω − (Tu|∂ S ) in S¯− .

(8.104)

Proof. By the exterior Somigliana formula, u = W ω − (u|∂ S ) − V ω − (Tu|∂ S )   T  = T (∂y )D ω (y, x) u(y) − D ω (x, y)T (∂y )u(y) ds(y). ∂S

Adding (8.103) to the above equation gives  

 T T (∂y ) D ω (y, x) + M ω (y, x) u(y)    ∂S − D ω (x, y) + M ω (x, y) T (∂y )u(y) ds(y)   T  ω ω = T (∂y )DM (y, x) u(y) − DM (x, y)T (∂y )u(y) ds(y)

u=

∂S

= WMω − (u|∂ S ) − VMω − (Tu|∂ S ), as required.   Composition formulas similar to those derived for the unmodified boundary integral operators may now be obtained. We introduce the function u = WMω − (ϕ ) − VMω − (ψ ) in S¯− , where ϕ ∈ C 1,α (∂ S) and ψ ∈ C 0,α (∂ S) are arbitrary. Then

8.4 Equations of the First Kind

149

 ω ω u|∂ S = WM0 + 12 I ϕ − VM0 ψ  ω ω∗ Tu|∂ S = NM0 ϕ − WM0 − 12 I ψ . Also, since u satisfies the conditions of Theorem 8.13, we can write  ω ω u|∂ S = WM0 + 12 I (u|∂ S ) −VM0 (Tu|∂ S )  ω ω∗ 1 Tu|∂ S = NM0 (u|∂ S ) − WM0 − 2 I (Tu|∂ S ).

(8.105)

(8.106)

We can express (8.105) and (8.106), respectively, in the form U = Q Φ and

U = Q U,

 u|∂ S U= , Tu|∂ S

where

and 

Q =

 ω WM0 + 12 I ω NM0

 ϕ Φ= ψ ω −VM0

!

 ω∗ 1 . − WM0 − 2I

Arguing as in the unmodified case, we arrive at the equality Q  = Q , 2

from which we obtain the composition relations ω ω ω ω∗ WM0 VM0 = VM0 WM0

on C 0,α (∂ S),

ω ω ω∗ NM0 VM0 = WM0 − 14 I 2

ω ω NM0 WM0

on C 0,α (∂ S),

(8.107)

ω∗ ω = WM0 NM0

on C 1,α (∂ S),

(8.108)

ω2

on C 1,α (∂ S).

(8.109)

ω ω VM0 NM0 = WM0 − 14 I

8.4.2 The Exterior Dirichlet Problem We want to solve (Dω − ) via an integral equation of the first kind. Using the modified matrix of fundamental solutions, we are able to overcome the problem of nonuniqueness that affects the corresponding unmodified equations (see Chapter 7). Seeking the solution of (D ω − ) in the form u = V Mω − (ϕ ) leads to the integral equation ω VM0 ϕ = R. (8.110) In the proofs of Theorems 8.3 and 8.4 it was shown that each of the four equations

150

8 Modified Fundamental Solutions

 ω WM0 ± 12 I ϕ = 0,

 ω∗ 1 WM0 ± 2 I ϕ = 0,

has only the zero solution. ω ϕ = 0, then ϕ = 0. Theorem 8.14. If NM0

Proof. Consider the function v = W Mω − (ϕ ). Then ω T v|∂ S = NM0 ϕ = 0,

so, by the uniqueness of the solution of (N ω − ), v = 0 in S¯− . Consequently,  ω 0 = WMω − (ϕ )|∂ S = WM0 + 12 I ϕ , from which it follows that, since this equation has only the zero solution, ϕ = 0.   Theorem 8.15. If R ∈ C 1,α (∂ S), α ∈ (0, 1), then the equation ω VM0 ϕ =R

has a unique C 0,α -solution. Also, ϕ ∈ C 0,α (∂ S) is a solution of this equation if and only if  ω ∗2 1 ω WM0 − 4 I ϕ = NM0 R. (8.111) Proof. Suppose that ϕ ∈ C 0,α (∂ S) is a solution of (8.110). Then ω ω ω NM0 (VM0 ϕ ) = NM0 R,

so, by (8.107),

 ω ∗2 1 ω WM0 − 4 I ϕ = NM0 R.

Conversely, let ϕ be a solution of (8.111). Then, again by (8.107),  ω ∗2 1 ω ω ω NM0 R = WM0 − 4 I ϕ = NM0 (VM0 ϕ ); hence,

ω ω NM0 (VM0 − R) = 0.

From Theorem 8.14 we deduce that ω VM0 ϕ = R.

It remains to show that (8.111) has a unique C 0,α -solution. We can rewrite (8.111) in the form  ω∗ 1 ω WM0 + I ψ = NM0 R, (8.112)  ω ∗ 21 WM0 − 2 I ϕ = ψ . (8.113) By the Fredholm alternative, (8.112) has a unique solution ψ ∈ C 0,α (∂ S). Similarly, (8.113) has a unique solution ϕ ∈ C 0,α (∂ S), and the theorem is proved.  

8.4 Equations of the First Kind

151

8.4.3 The Exterior Neumann Problem If the solution of (N ω − ) is postulated in the form of a modified double-layer potential, then we need to solve the equation ω NM0 ϕ =S

(8.114)

for the unknown density ϕ . ω ϕ = 0, then ϕ = 0. Theorem 8.16. If VM0

Proof. Let

v = VMω − (ϕ ).

By the uniqueness property, since v| ∂ S = 0, we have v = 0 in S¯− . Hence,  ω∗ 1 0 = T VMω − (ϕ )|∂ S = WM0 − 2I ϕ, and we deduce that ϕ = 0.   Theorem 8.17. If S ∈ C 0,α (∂ S), α ∈ (0, 1), then the equation ω NM0 ϕ =S

has a unique C 1,α -solution. Also, ϕ ∈ C 1,α (∂ S) is a solution of this equation if and only if  ω2 1 ω WM0 − 4 I ϕ = VM0 S. (8.115) Proof. If ϕ ∈ C 1,α (∂ S) satisfies (8.114), then ω ω ω VM0 (NM0 ϕ ) = VM0 S,

so, using (8.109), we arrive at  ω2 1 ω WM0 − 4 I ϕ = VM0 S. On the other hand, if ϕ ∈ C 1,α (∂ S) is a solution of (8.115), then, taking (8.109) into account, we have  ω2 1 ω ω ω VM0 S = WM0 − 4 I ϕ = VM0 (NM0 ϕ ), which implies that

ω ω VM0 (NM0 ϕ − S ) = 0;

therefore, Theorem 8.16 yields ω NM0 ϕ =S.

The fact that (8.115) has a unique C 1,α -solution is shown in the same way as in the proof of Theorem 8.15.  

152

8 Modified Fundamental Solutions

Remark 8.1. As shown above, the question of nonuniqueness can be resolved if certain suitable modifications of the matrix of fundamental solutions are operated. The solutions supplied by the methods used in Chapters 6–11 are smooth because they are generated by smooth data functions prescribed on a smooth boundary curve. Loss of smoothness in the data and/or the boundary would make it necessary to conduct the analysis in a distributional setting that would reveal the existence of weak solutions. Such a discussion, which is beyond the scope of this book, can be found in the case of static and dynamic deformation of plates in [9] and [10], respectively.

Chapter 9

Problems with Robin Boundary Conditions

In this chapter we consider the third type of fundamental boundary value problems, namely, problems with Robin boundary conditions, where a linear combination of the stresses and displacements is prescribed on ∂ S. The exterior Robin problem for the three-dimensional Helmholtz equation has been studied in [3]–[5], [6], and [41] by means of various integral equation methods. In the theory of equilibrium bending of thin plates with transverse shear deformation, the interior and exterior Robin problems have been investigated in [55] and [21]. A discussion of the corresponding boundary value problems for plane strain can be found in [54] and [22]. The question of uniqueness of solution for both the interior and exterior Robin boundary value problems is addressed in Section 9.1, where we examine, in particular, the role of the matrix that connects the stresses and displacements in the boundary conditions. In Theorems 9.1 and 9.5 we show that, depending on the properties of this matrix, the interior problem may have either a unique solution or countably many eigensolutions. The existence of an eigenfrequency spectrum is proved by the same technique used in the case of the interior Dirichlet and Neumann problems in Chapter 5. Also, Theorem 9.2 indicates when the exterior Robin problem has at most one solution. In Section 9.2 we use an indirect boundary integral equation formulation, analogous to that in Chapter 6, to prove the existence of solutions to the boundary value problems. As in Chapter 6, this leads to difficulties generated by integral equations with nonunique solutions. The direct method used in Section 9.3 circumvents this drawback, at least in the exterior case, by providing pairs of equations (cf. Chapter 7), the solvability of which is discussed in Theorems 9.13–9.15 and 9.17. In Theorem 9.18 (Section 9.4) we show that the solution of the exterior problem can be found from a uniquely solvable equation formed by combining the two equations derived in Section 9.3. Finally, in Section 9.5 the modified theory developed in Chapter 8 to deal with (Dω − ) and (Nω − ) is extended to the exterior Robin problem. G.R. Thomson and C. Constanda, Stationary Oscillations of Elastic Plates, A Boundary Integral Equation Analysis, DOI 10.1007/978-0-8176-8241-5_9, © Springer Science+Business Media, LLC 2011

153

154

9 Problems with Robin Boundary Conditions

9.1 Uniqueness Properties Let F and G be (3 × 1)-vector functions defined on ∂ S, and let σ ∈ C 1,α (∂ S), α ∈ (0, 1), be a symmetric (3 × 3)-matrix function. The interior and exterior Robin problems are defined as follows: (Rω + ) Find u ∈ C 2 (S+ ) ∩C1 (S¯+ ) that satisfies (4.1) in S + and T (∂x )u(x) + σ (x)u(x) = F (x),

x ∈ ∂ S.

(9.1)

(Rω − ) Find u ∈ C 2 (S− ) ∩C1 (S¯− ) ∩ B ω that satisfies (4.1) in S − and T (∂x )u(x) + σ (x)u(x) = G (x),

x ∈ ∂ S.

ω− + We denote the corresponding homogeneous problems by (R ω 0 ) and (R0 ), respectively. Later on, we place further restrictions on the symmetric matrix σ when we ascertain whether or not the above boundary value problems have unique solutions.

Theorem 9.1. If Im(σ ) is either positive definite or negative definite, then (R ω + ) has at most one regular solution. + Proof. Let u be a regular solution of (R ω 0 ). Then, applying the reciprocity relation to u and u, ¯ we arrive at  (u¯ T Tu − uTT u) ¯ ds = 0. (9.2)

∂S

Let v1 = Re(u|∂ S ), v2 = Im(u|∂ S ), σ1 = Re(σ ), and σ2 = Im(σ ). Since Tu|∂ S + σ u|∂ S = 0, on ∂ S we have u¯ T Tu − uTT u¯ = −(v1 − iv2 )T (σ1 + iσ2 )(v1 + iv2 ) + (v1 + iv2 )T (σ1 − iσ2 )(v1 − iv2 ) = −2i(vT1 σ2 v1 + vT2 σ2 v2 − vT2σ1 v1 + vT1σ1 v2 ). Owing to the symmetry of σ 1 , (u¯ T Tu − uTT u)| ¯ ∂ S = −2i(vT1 σ2 v1 + vT2 σ2 v2 ), so (9.2) leads to



(9.3)

(vT1 σ2 v1 + vT2 σ2 v2 ) ds = 0.

∂S

By assumption, σ2 is either positive definite or negative definite; hence, v 1 = v2 = 0. Consequently, u|∂ S = 0, (9.4)

9.1 Uniqueness Properties

155

and from the boundary condition we deduce that Tu|∂ S = 0.

(9.5)

Substituting (9.4) and (9.5) in the interior representation formula (4.36) yields u = 0 in S¯+ , from which we conclude that (R ω + ) can have at most one solution under the given conditions on σ .   Theorem 9.2. If Im(σ ) is positive semidefinite, then (R ω − ) has at most one regular solution. Proof. Applying the reciprocity relation to u and u¯ in S − ∩ KR , where u is a regular − solution of (Rω 0 ) and KR denotes the usual disk, we see that 

(u¯ T Tu − uTT u) ¯ ds =

∂ KR



(u¯ T Tu − uTT u) ¯ ds.

(9.6)

∂S

In [24] it is shown that on ∂ KR , u¯T Tu − uTT u¯  = 2i h2 (λ + 2μ )(k1 |U (1) |2 + k2 |U (2) |2 ) + h2 μ k3 |U (3) |2  (1) (2) + μ k2 (k1−1 |u3 |2 + k2−1 |u3 |2 ) + O(R−2 ), which, in view of (9.3), means that, letting R → ∞ in (9.6), we have

   h2 (λ + 2μ ) k1 lim |U (1) |2 ds + k2 lim |U (2) |2 ds R→∞ ∂ KR

+ h2 μ k32 lim

R→∞ ∂ KR



R→∞ ∂ KR

|U (3) |2 ds

   (1) (2) + μ k2 k1−1 lim |u3 |2 ds + k2−1 lim |u3 |2 ds =−



R→∞ ∂ KR

R→∞ ∂ KR

(vT1 σ2 v1 + vT2 σ2 v2 ) ds,

(9.7)

∂S

where v1 , v2 , σ1 , and σ2 are as in the proof of Theorem 9.1. Since σ 2 is assumed to be positive semidefinite, from (9.7) we deduce that 

lim

R→∞ ∂ KR

|U ( j) |2 ds = lim



R→∞ ∂ KR

(β )

|u3 |2 ds = 0.

156

9 Problems with Robin Boundary Conditions

An application of Rellich’s lemma [57] now yields (β )

U ( j) = u3 = 0 in S − , as required.   We show that if Re(σ ) is positive definite and Im(σ ) = 0, the homogeneous interior Robin problem has eigenfrequencies. The Green’s tensor for (R ω + ) is a (3 × 3)-matrix G 3 (x, y) that satisfies A(∂x )G3 (x, y) = 0,

x, y ∈ S + , x = y,

T (∂x )G3 (x, y) + σ (x)G3 (x, y) = 0,

x ∈ ∂ S, y ∈ S + .

(9.8)

In S+ , G3 (x, y) is given by G3 (x, y) = D(x, y) − v3(x, y), where D(x, y) is a matrix of fundamental solutions for the operator A(∂ x ) (see [14]) and the columns of v 3 (x, y) are regular solutions of (5.1) in S + . To construct G 3 (x, y), we must solve the boundary value problems (i)

A(∂x )v3 (x, y) = 0, (i)

x, y ∈ S + ,

(i)

T (∂x )v3 (x, y) + σ (x)v3 (x, y) = T (∂x )D(i) (x, y) + σ (x)D(i) (x, y),

x ∈ ∂ S, y ∈ S +

for i = 1, 2, 3. The solvability of the interior Robin problem associated with (5.1) has been proved in [55] and [21] for a real positive definite matrix σ . Hence, the above arguments guarantee the existence of the Green’s tensor for (R ω + ). The Green’s tensor is symmetric. To verify this, we apply the reciprocity relation (i) ( j) to u(x) = G3 (x, x1 ) and v(x) = G3 (x, x2 ) and take the boundary condition (9.8) into account:   (i)

T  ( j) T  ( j) (i) G3 (x, x1 ) A(∂x )G3 (x, x2 ) − G3 (x, x2 ) A(∂x )G3 (x, x1 ) da(x)

S+

=

  ( j)

T  (i) T  (i) ( j) G3 (x, x2 ) σ (x)G3 (x, x1 ) − G3 (x, x1 ) σ (x)G3 (x, x2 ) ds(x).

∂S

The integral around ∂ S vanishes since  ( j) T  (i) T (i) ( j) G3 (x, x2 ) σ (x)G3 (x, x1 ) − G3 (x, x1 ) σ (x)G3 (x, x2 )  T ( j)  (i) T (i) ( j) = σ (x)G3 (x, x1 ) G3 (x, x2 ) − G3 (x, x1 ) σ (x)G3 (x, x2 )  (i) T  (i) T ( j) ( j) = G3 (x, x1 ) σ (x)G3 (x, x2 ) − G3 (x, x1 ) σ (x)G3 (x, x2 ) = 0,

9.1 Uniqueness Properties

157

where we have used the symmetry of σ (x). Arguing as in the case of the Green’s tensor for (Dω + ), we now find that  T G3 (x, y) = G3 (y, x) .

(9.9)

Theorem 9.3. If u ∈ C 2 (S+ ) ∩C1 (S¯+ ), then for x ∈ S + it can be represented in the form u(x) =



G3 (x, y)T (∂y )u(y) ds(y) +

∂S



G3 (x, y)σ (y)u(y) ds(y)

∂S

−



G3 (x, y)A(∂y )u(y) da(y). (9.10)

S+

Proof. Following the argument in the proof of Theorem 5.3 and recalling the symmetry property (9.9), we obtain 

G3 (x, y)A(∂y )u(y) da(y) −

S+

=



G3 (x, y)A(∂y )u(y) da(y)

σ (x,δ )



G3 (x, y)T (∂y )u(y) ds(y) −

∂S

−

 

T T (∂y )G3 (y, x) u(y) ds(y)

∂S



G3 (x, y)T (∂y )u(y) ds(y) +

∂ σ (x,δ )



 T T (∂y )G3 (y, x) u(y) ds(y).

∂ σ (x,δ )

Letting δ → 0 yields 

G3 (x, y)A(∂y )u(y) da(y)

S+

=



G3 (x, y)T (∂y )u(y) ds(y) −

∂S

 

T T (∂y )G3 (y, x) u(y) ds(y) − u(x),

∂S

and the assertion follows from an application of the boundary condition (9.8).   + Theorem 9.4. The homogeneous problem (R ω 0 ) is equivalent to the integral equation  u(x) − ω 2 H G3 (x, y)u(y) da(y) = 0, x ∈ S + , (9.11) S+

where H is defined by (5.10). + Proof. First, suppose that u is a solution of (R ω 0 ). Then

Tu|∂ S = −σ u|∂ S , which, when substituted in (9.10), leads to

158

9 Problems with Robin Boundary Conditions

u(x) = −



G3 (x, y)A(∂y )u(y) da(y).

S+

By (5.9), A(∂y )u(y) = −ω 2 H u(y), hence, u(x) − ω 2 H



y ∈ S+;

G3 (x, y)u(y) da(y) = 0,

S+

as required. Conversely, if u satisfies (9.11), then, by Theorem 3.1 and (9.8), on ∂ S we have T (∂x )u(x) = ω 2 H



T (∂x )G3 (x, y)u(y) da(y)

S+

= −ω 2 H σ (x)



G3 (x, y)u(y) da(y),

S+

so, since H is a diagonal matrix, T (∂x )u(x) + σ (x)ω 2 H



G3 (x, y)u(y) da(y) = 0;

S+

that is, T (∂x )u(x) + σ (x)u(x) = 0, The fact that

x ∈ ∂ S.

Aω (∂x )u(x) = 0

+ is shown as in the proof of Theorem 5.4; hence, u is a solution of (R ω 0 ).

 

We can now state a result analogous to Theorems 5.5 and 5.10, concerning the + existence of eigenfrequencies of (R ω 0 ). + Theorem 9.5. If Re(σ ) is positive definite and Im(σ ) = 0, then (R ω 0 ) has countably many real eigenfrequencies. These eigenfrequencies correspond to the values of ω for which the integral equation (9.11) has nonzero solutions.

Proof. As in the case of Theorem 5.5, it follows that there are countably many real values of ω 2 for which (9.11) has nonzero solutions. To prove that the eigenfrequen+ cies are real, we remark that if u is a solution of (R ω 0 ), then   S+

 − ρω 2 (h2 |u1 |2 + h2 |u2 |2 + |u3 |2 ) + 2E(u, u) ¯ da =−



∂S

(vT1 σ v1 + vT2σ v2 ) ds,

9.2 Indirect Boundary Equation Formulation

159

where v1 and v2 are as in the proof of Theorem 9.1. Since σ is positive definite, we deduce that ω 2 > 0, so the eigenfrequencies are real.   In view of Theorems 9.1, 9.2, and 9.5, we consider three separate cases for the matrix σ , namely, (C1) Re(σ ) is positive definite and Im(σ ) = 0. (C2) Im(σ ) is either positive definite or negative definite. (C3) Im(σ ) is positive semidefinite.

9.2 Indirect Boundary Equation Formulation We solve the Robin boundary value problems by following a classical indirect boundary integral equation approach.

9.2.1 The Interior Problem If we assume that the solution of (R ω + ) is of the form u = V we obtain the integral equation  ω∗ W0 + σ V0ω + 12 I ϕ = F .

ω+

(ϕ ), then from (9.1) (9.12)

We remark that the integral operator in (9.12) has the same singular part as the operator W0ω ∗ + 12 I studied in Chapter 6. Consequently, the index of this equation is zero, which means that the Fredholm alternative is applicable. We start with the case (C1). + Theorem 9.6. If (C1) holds, then ω is an eigenfrequency of (R ω 0 ) of multiplicity 1 n if and only if W0ω + V0ω σ + 2 I has n linearly independent C 1,α -eigenfunctions and W0ω ∗ + σ V0ω + 12 I has n linearly independent C 0,α -eigenfunctions. The linearly independent C 1,α -eigenfunctions of W0ω + V0ω σ + 12 I coincide with the boundary + values of the linearly independent eigensolutions of (R ω 0 ). + Proof. Let ω be an eigenfrequency of (R ω 0 ) of multiplicity n with corresponding linearly independent eigensolutions u (k) , k = 1, 2, . . . , n. By the interior representation formula (4.36),

u(k) |∂ S = V0ω (Tu(k) |∂ S ) − (W0ω − 12 I )(u(k) |∂ S ), so, since we have

Tu(k) |∂ S + σ u(k) |∂ S = 0,

k = 1, 2, . . . , n,

k = 1, 2, . . . , n,

160

9 Problems with Robin Boundary Conditions

u(k) |∂ S = −V0ω (σ u(k) |∂ S ) − (W0ω − 12 I )(u(k) |∂ S ),

k = 1, 2, . . . , n;

that is,

 ω W0 + V0ω σ + 12 I (u(k) |∂ S ) = 0, k = 1, 2, . . . , n.  n We now show that the set u(k) |∂ S k=1 is linearly independent. Suppose that this is not the case. Then there are constants a k , k = 1, 2, . . . , n, not all zero, such that n

∑ ak u(k) |∂ S = 0.

(9.13)

k=1

Let u(x) =

n

∑ ak u(k)(x),

x ∈ S+ .

k=1

+ Since each u (k) is an eigensolution of (R ω 0 ), it follows that

Tu|∂ S + σ u|∂ S = 0. But by (9.13), u|∂ S = 0,

(9.14)

Tu|∂ S = 0.

(9.15)

so + Substituting (9.14) and (9.15) in (4.36)  (k) n yields+ u = 0 in S , which, because of the linear independence of the set u in S , implies that ak = 0, k = 1, 2, . . . , n. k=1 This contradicts our assumption, so the boundary values of the linearly independent + eigensolutions of (R ω 0 ) are linearly independent. Consequently, W0ω + V0ω σ + 12 I has at least n linearly independent C 1,α -eigenfunctions. We now show that W0ω + V0ω σ + 12 I has exactly n linearly independent C 1,α eigenfunctions. Suppose that there are m, m > n, linearly independent C 1,α -eigenfunctions. The Fredholm alternative then the existence of a set of m linearly independent  guarantees m C 0,α -eigenfunctions, ϕ (k) k=1 , say, of the operator W0ω ∗ + σ V0ω + 12 I. We construct  m the single-layer potentials V ω + (ϕ (k) ) k=1 . Then

T V ω + (ϕ (k) ) + σ V ω + (ϕ (k) ) = 0 on ∂ S,

k = 1, 2, . . . , m,

+ which implies that V ω + (ϕ (k) ) is a solution of (Rω 0 ) for each k. By Theorem 6.3(i), + ω + (k) each V (ϕ ) is also nonzero and is therefore an eigensolution of (R ω 0 ). The  ω + (k) m + fact that V (ϕ ) k=1 is linearly independent in S is shown in exactly the  m same way as in the proof of Theorem 6.5. Hence, the set V ω + (ϕ (k) ) k=1 con+ sists of linearly independent eigensolutions of (R ω 0 ). Since we have assumed that ω+ (R0 ) has n linearly independent eigensolutions, we deduce that m = n. Conse-

9.2 Indirect Boundary Equation Formulation

161

quently, W0ω +V0ω σ + 12 I and W0ω ∗ + σ V0ω + 12 I have exactly n linearly independent eigenfunctions of class C 1,α (∂ S) and C 0,α (∂ S), respectively. On the other hand, suppose that ω is such that the operator W 0ω ∗ + σ V0ω + 12 I  n has a set of n linearly independent C 0,α -eigenfunctions ϕ (k) k=1 . The above argument shows that the single-layer potentials formed with these eigenfunctions as + densities are linearly independent eigensolutions of (R ω 0 ). Consequently, ω is an ω+ + eigenfrequency of (R 0 ) of multiplicity not less than n. But if (R ω 0 ) has m linearly independent eigensolutions, where m > n, then from the first part of the proof it follows that W0ω ∗ + σ V0ω + 12 I has exactly m linearly independent C 0,α -eigenfunctions, which contradicts our assumption. Thus, we deduce that ω is an eigenfrequency of + (Rω  0 ) of multiplicity n.  + ω+) Theorem 9.7. If (C1) holds and ω is not an eigenfrequency of (R ω 0 ), then (R 0, α has a unique regular solution for any F ∈ C (∂ S), α ∈ (0, 1), which can be represented in the form u = V ω + (ϕ ), with ϕ ∈ C 0,α (∂ S). + Proof. If ω is not an eigenfrequency of (R ω 0 ), then, by Theorem 9.6, the integral equation  ω∗ W0 + σ V0ω + 12 I ϕ = 0

has only the zero solution. Therefore, by the Fredholm alternative, (9.12) has a unique solution ϕ ∈ C 0,α (∂ S).   + ω + ) has Theorem 9.8. If (C1) holds and ω is an eigenfrequency of (R ω 0 ), then (R regular solutions if and only if



F T (y)u(k) (y) ds(y) = 0,

k = 1, 2, . . . , n,

∂S

 n where F ∈ C 0,α (∂ S), α ∈ (0, 1), and u(k) k=1 is a complete set of linearly inde+ pendent eigensolutions of (R ω 0 ). Each of these solutions can be represented in the ω + form u = V (ϕ ), with ϕ ∈ C 0,α (∂ S). + Proof. If ω is an eigenfrequency of (R ω 0 ) of multiplicity n, then, by Theorem 9.6,

 ω W0 + V0ω σ + 12 I (u(k) |∂ S ) = 0,

k = 1, 2, . . . , n,

(9.16)

+ where the u(k) are the linearly independent eigensolutions of (R ω 0 ). Since (9.16) is the homogeneous adjoint equation to (9.12), according to the Fredholm alternative, (9.12) is solvable if and only if

(F , u(k) |∂ S ) = 0, that is,



k = 1, 2, . . . , n;

F T (y)u(k) (y) ds(y) = 0,

∂S

which completes the proof.  

k = 1, 2, . . . , n,

162

9 Problems with Robin Boundary Conditions

We now turn our attention to the case (C2). This differs from (C1) in that here + (Rω 0 ) has no eigenfrequencies. Theorem 9.9. If (C2) holds, then the equations  ω∗ W0 + σ V0ω + 12 I ϕ = 0 and

(9.17)

 ω W0 +V0ω σ + 12 I ϕ = 0

(9.18)

have only the zero solution. Proof. Let ϕ be a solution of (9.17) and consider the function u = V ω + (ϕ ). Then Tu|∂ S + σ u|∂ S = 0, + ¯+ so u is a solution of (R ω 0 ). By Theorem 9.1, u = 0 in S ; in particular,

and

u|∂ S = V0ω ϕ = 0

(9.19)

 Tu|∂ S = W0ω ∗ + 12 I ϕ = 0.

(9.20)

From (9.19) we see that the function v = V ω − (ϕ ) − ¯− is a solution of (Dω 0 ). By the uniqueness property, v = 0 in S , so  T v|∂ S = W0ω ∗ − 12 I ϕ = 0.

(9.21)

Combining (9.20) and (9.21) yields ϕ = 0. Since (9.18) is the homogeneous adjoint equation to (9.17), by the Fredholm alternative, it must also have only the zero solution.   Theorem 9.10. If (C2) holds, then for any F ∈ C 0,α (∂ S), α ∈ (0, 1), and any value of ω , (Rω + ) has a unique regular solution, which can be represented in the form u = V ω + (ϕ ) with ϕ ∈ C 0,α (∂ S). The assertion follows immediately from (9.12), Theorem 9.9, and the Fredholm alternative.

9.2.2 The Exterior Problem If the solution of (R ω − ) is sought in the form u = V the equation

ω − (ϕ ),

then we need to solve

9.2 Indirect Boundary Equation Formulation

163

 ω∗ W0 + σ V0ω − 12 I ϕ = G

(9.22)

for the unknown density ϕ . The integral operator in (9.22) has the same singular part as W0ω ∗ − 12 I , so the Fredholm alternative is valid. + Theorem 9.11. If (C3) holds, then ω is an eigenfrequency of (D ω 0 ) of multiplicity n if and only if W0ω ∗ + σ V0ω − 12 I has n linearly independent C 0,α -eigenfunctions and W0ω + V0ω σ − 12 I has n linearly independent C 1,α -eigenfunctions. The linearly independent C 0,α -eigenfunctions of W0ω ∗ + σ V0ω − 12 I are the images under the the boundary moment–stress operator of the linearly independent eigensolutions + of (Dω 0 ). + Proof. If ω is an eigenfrequency of (D ω 0 ) of multiplicity n, then, by Theorem 6.4,

 ω∗ 1 W0 − 2 I (Tu(k) |∂ S ) = 0,

k = 1, 2, . . . , n,

(9.23)

+ where the u (k) are linearly independent eigensolutions of (D ω 0 ). Also, Theorem 7.1 implies that V0ω (Tu(k) |∂ S ) = 0, k = 1, 2, . . . , n, (9.24)

so, by (9.23) and (9.24),  ω∗ W0 + σ V0ω − 12 I (Tu(k) |∂ S ) = 0,

k = 1, 2, . . . , n;

that is, W0ω ∗ + σ V0ω − 12 I has at least n linearly independent C 0,α -eigenfunctions. Suppose now that this operator has m > n such eigenfunctions. Then for each k = 1, 2, . . . , m, the function V ω − (ϕ (k) ) ∈ B ω satisfies Aω V ω − (ϕ (k) ) = 0 in S− and  T V ω − (ϕ (k) )|∂ S + σ V ω − (ϕ (k) )|∂ S = W0ω ∗ + σ V0ω − 12 I ϕ (k) = 0, − so V ω − (ϕ (k) ) is a solution of (Rω 0 ); hence, by Theorem 9.2,

V ω − (ϕ (k) ) = 0, which implies that Then

V0ω (ϕ (k) ) = 0.  ω ∗ 1 (k) W0 − 2 I ϕ = 0;

that is, W0ω ∗ − 12 I has m linearly independent eigenfunctions. By Theorem 6.4, + (Dω 0 ) has m linearly independent eigensolutions, which contradicts our assumption. Consequently, the operator W 0ω ∗ + σ V0ω − 12 I has exactly n linearly independent C 0,α -eigenfunctions. By the Fredholm alternative, its adjoint W 0ω + V0ω σ − 12 I also has exactly n linearly independent C 0,α -eigenfunctions. That they are of class C 1,α (∂ S) follows from a result similar to Theorem 6.2.

164

9 Problems with Robin Boundary Conditions

Suppose now that ω is such that W0ω ∗ + σ V0ω − 12 I has exactly n linearly independent C 0,α -eigenfunctions ϕ (k) . Then, as was seen above,  ω∗ W0 + σ V0ω − 12 I ϕ (k) = 0, k = 1, 2, . . . , n, (9.25) so the single-layer potentials V the uniqueness property,

ω − (ϕ (k) ),

− k = 1, 2, . . . , n, are solutions of (R ω 0 ). By

V ω − (ϕ (k) ) = 0 in S¯− , in particular,

V0ω ϕ (k) = 0,

k = 1, 2, . . . , n;

k = 1, 2, . . . , n.

Hence, (9.25) becomes (W0ω ∗ − 12 I )ϕ (k) = 0,

k = 1, 2, . . . , n,

and the assertion follows from Theorem 6.4.   Lemma 9.1. Suppose that (C3) holds and let ϕ ∈ C 1,α (∂ S), α ∈ (0, 1). Then W ω + (ϕ ) + V ω + (σ ϕ ) = 0 in S¯+ if and only if ϕ = 0. Proof. Trivially, if ϕ = 0, then W Consider the function

ω + (ϕ ) + V ω + (σ ϕ )

= 0 in S¯+ .

u = W ω + (ϕ ) + V ω + (σ ϕ ) and suppose that u vanishes identically in S¯+ . In particular, u| ∂ S = 0, which means that  ω W0 + V0ω σ − 12 I ϕ = 0. (9.26) Also, Tu|∂ S = 0; that is,

 N0ω ϕ + W0ω ∗ + 12 I (σ ϕ ) = 0.

(9.27)

We introduce the function v = W ω − (ϕ ) + V ω − (σ ϕ ). By (9.26) and (9.27),    T v|∂ S + σ v|∂ S = N0ω ϕ + W0ω ∗ − 12 I (σ ϕ ) + σ W0ω + 12 I ϕ + V0ω (σ ϕ ) = −σ ϕ + σ ϕ = 0, − ¯− so v is a solution of (Rω 0 ). By the uniqueness property, v = 0 in S . Then

 v|∂ S = W0ω + 12 I ϕ + V0ω (σ ϕ ) = 0, which, in view of (9.26), implies that ϕ = 0.  

9.2 Indirect Boundary Equation Formulation

165

Theorem 9.12. If (C3) holds, then for any G ∈ C 0,α (∂ S), α ∈ (0, 1), and any value + of ω , (Rω − ) has a unique regular solution. If ω is not an eigenfrequency of (D ω 0 ), ω − 0,α then the solution can be represented in the form u = V (ϕ ), ϕ ∈ C (∂ S). If ω + is an eigenfrequency of (D ω 0 ), then the solution can be represented in the form n   u = V ω − (ϕ ) + ∑ ck W ω − (ψ˜ (k) ) + V ω − (σ ψ˜ (k) ) , k=1

where the ck , k = 1, 2, . . . , n, are constants, ϕ ∈ C 0,α (∂ S), and the ψ˜ (k) ∈ C 1,α (∂ S), k = 1, 2, . . . , n, are linearly independent solutions of (9.26). + Proof. If ω is not an eigenfrequency of (D ω 0 ), then, by Theorem 9.11 and the Fredholm alternative, (9.22) has a unique solution ϕ ∈ C 0,α (∂ S), so u = V ω − (ϕ ) is the unique regular solution of (R ω − ). + Now let ω be an eigenfrequency of (D ω 0 ) of multiplicity n. By Theorem 9.11, 0, α there are n linearly independent C -eigenfunctions of W0ω ∗ + σ V0ω − 12 I and n linearly independent C 1,α -eigenfunctions of W0ω + V0ω σ − 12 I. We denote them by ϕ (k) and ψ (k) , respectively. Without loss of generality, we may assume that these two sets have been biorthonormalized [43]; that is,



[ϕ (m) ]T ψ (n) ds = δmn .

(9.28)

∂S

We introduce the function n

G˜ = G − ∑

l=1



G ψ T

(l)



ds ϕ (l)

(9.29)

∂S

and the related integral equation  ω∗ W0 + σ V0ω − 12 I ϕ = G˜.

(9.30)

By the Fredholm alternative, (9.30) is solvable if and only if 

G˜T ψ (k) ds = 0,

k = 1, 2, . . . , n.

(9.31)

∂S

Using (9.29) and (9.28), we easily see that (9.31) is indeed satisfied, so (9.30) has at least one solution in C 0,α (∂ S). Consider the functions v(k) = W ω + (ψ (k) ) + V ω + (σ ψ (k) ),

k = 1, 2, . . . , n.

We show that the v(k) , k = 1, 2, . . . , n, are n linearly independent eigensolutions of + (k) are eigenfunctions of W ω + V ω σ − 1 I, we find that (Dω 0 ). Since the ψ 0 0 2  v(k) |∂ S = W0ω − 12 I ψ (k) + V0ω (σ ψ (k) ) = 0,

k = 1, 2, . . . , n.

166

9 Problems with Robin Boundary Conditions

By Lemma 9.1, each v (k) is nonzero since ψ (k) = 0, so each v(k) is an eigensolu+ (k) are linearly dependent in S + ; that is, there are tion of (Dω 0 ). Suppose that the v constants ak , k = 1, 2, . . . , n, not all zero, such that n

∑ ak



 W ω + (ψ (k) ) + V ω + (σ ψ (k) ) = 0 in S+ .

k=1

By the linearity of the potentials, W ω + (ψ ) + V ω + (σ ψ ) = 0 in S + , where

ψ=

n

∑ ak ψ (k) .

k=1

Then from Lemma 9.1 it follows that n

∑ ak ψ (k) = 0.

k=1

Since ψ (k) are linearly independent, we must have a k = 0, k = 1, 2, . . . , n. Hence,  (k) the n + v is the set of linearly independent eigensolutions of (D ω ). k=1  (k) n 0 Theorem 9.11 implies that we can construct a set ψ˜ , where each ψ˜ (k) is k=1 a linear combination of the functions ψ (k) , k = 1, 2, . . . , n, chosen to satisfy   T W ω + (ψ˜ (k) ) + V ω + (σ ψ˜ (k) ) ∂ S = ϕ (k) , k = 1, 2, . . . , n, which means that  N0ω ψ˜ (k) + W0ω ∗ + 12 I (σ ψ˜ (k) ) = ϕ (k) ,

k = 1, 2, . . . , n.

(9.32)

The unique solution of (R ω − ) is then given by  n    ω− T (k) u = V (ϕ ) + ∑ G ψ ds W ω − (ψ˜ (k) ) + V ω − (σ ψ˜ (k) ) , k=1

∂S

where ϕ ∈ C 0,α (∂ S) is any solution of (9.30). To see this, we note that, by (9.30), (9.29), (9.32), and the fact that the ψ˜ (k) are eigenfunctions of W0ω + V0ω σ − 12 I,  Tu|∂ S + σ u|∂ S = W0ω ∗ + σ V0ω − 12 I ϕ  n    T (k) +∑ G ψ ds N0ω ψ˜ (k) + W0ω ∗ − 12 I (σ ψ˜ (k) ) k=1

  + σ W0ω + 12 I ψ˜ (k) + V0ω (σ ψ˜ (k) )   n  n  T (l) (l) T (k) =G −∑ G ψ ds ϕ + ∑ G ψ ds ϕ (k) = G . l=1

The assertion is proved.  

∂S

∂S

k=1

∂S

9.3 Direct Boundary Equation Formulation

167

9.3 Direct Boundary Equation Formulation In this section we derive integral equations for (R ω + ) and (Rω − ) based on the representation formulas (4.36) and (4.37).

9.3.1 The Interior Problem Substituting the boundary condition Tu| ∂ S = F − σ u|∂ S in (4.36) leads to u = V ω + (F ) − V ω + (σ u|∂ S ) − W ω + (u|∂ S ); hence,

 u|∂ S = V0ω F − V0ω (σ u|∂ S ) − W0ω − 12 I (u|∂ S ).

Writing ϕ = u|∂ S , we arrive at the integral equation  ω W0 +V0ω σ + 12 I )ϕ = V0ω F .

(9.33)

Suppose that ϕ satisfies (9.33). Then the function u = V ω + (F ) − V ω + (σ ϕ ) − W ω + (ϕ )

(9.34)

is a solution of (Rω + ) if Tu|∂ S + σ u|∂ S = F . Consequently, we must have  ω∗ 1  W0 + 2 I F − W0ω ∗ + 12 I (σ ϕ ) − N0ω ϕ   + σ V0ω F − V0ω (σ ϕ ) − (W0ω − 12 I )ϕ = F , so, since ϕ is a solution of (9.33), we require that  ω∗ 1  W0 − 2 I (σ ϕ ) + N0ω ϕ = W0ω ∗ − 12 I F .

(9.35)

Therefore, u given by (9.34) is a regular solution of (R ω + ) provided that ϕ satisfies both (9.33) and (9.35). We consider cases (C1) and (C2). + Theorem 9.13. If assumption (C1) holds, ω is not an eigenfrequency of (R ω 0 ), and 0, α 1, F ∈ C (∂ S), α ∈ (0, 1), then equation (9.33) has a unique solution ϕ ∈ C α (∂ S) that also satisfies (9.35) and, thus, generates the unique solution of (R ω + ).

Proof. By Theorem 9.6 and the Fredholm alternative, (9.33) has a unique solution ϕ ∈ C 0,α (∂ S). Since, by Theorem 2.6, the first two terms on the left-hand side and the one on the right-hand side in (9.33) belong to C 1,α (∂ S), it follows that ϕ ∈ C 1,α (∂ S).

168

9 Problems with Robin Boundary Conditions

Consider this unique solution of (9.33). Applying N 0ω to (9.33) yields  0 = W0ω +V0ω σ + 12 I ϕ −V0ω F   = N0ω W0ω + V0ω σ + 12 I ϕ − V0ω F , so, by the composition formulas (7.7) and (7.8),  ω ∗ 1  ω ∗ 1   W0 + 2 I W0 − 2 I (σ ϕ ) + N0ω ϕ − W0ω ∗ − 12 I F = 0.

(9.36)

Similarly, operating with W0ω − 12 I, we have  0 = W0ω +V0ω σ + 12 I ϕ − V0ω F    = W0ω − 12 I W0ω +V0ω σ + 12 I ϕ −V0ω F , which, in view of (7.6) and (7.9), leads to    V0ω W0ω ∗ − 12 I (σ ϕ ) + N0ω ϕ − W0ω ∗ − 12 I F = 0.

(9.37)

Equations (9.36) and (9.37) together with Corollary 7.2 imply that  ω∗ 1  W0 − 2 I (σ ϕ ) + N0ω ϕ = W0ω ∗ − 12 I F , and the theorem is proved.   In fact, equations (9.33) and (9.35) are equivalent. We have already shown that (9.35) follows from (9.33). To verify the converse, we operate with W 0ω ∗ + 12 I and then with V0ω in (9.35) and use the composition formulas to find that   N0ω W0ω + V0ω σ + 12 I ϕ − V0ω F = 0,  ω 1  ω  W0 − 2 I W0 + V0ω σ + 12 I ϕ − V0ω F = 0. From Corollary 7.2(ii) we now deduce that (9.33) holds. Theorem 9.14. If (C2) holds and F ∈ C 0,α (∂ S), α ∈ (0, 1), then equation (9.33) has a unique solution ϕ ∈ C 1,α (∂ S) for any value of ω , which also satisfies (9.35). This assertion is proved in exactly the same way as Theorem 9.13, this time with the use of Theorem 9.9.

9.3.2 The Exterior Problem Applying the boundary condition Tu| ∂ S = G − σ u|∂ S to the exterior representation formula (4.37) yields u = W ω − (u|∂ S ) − V ω − (G ) + V ω − (σ u|∂ S );

9.3 Direct Boundary Equation Formulation

hence,

169

u|∂ S = (W0ω + 12 I )(u|∂ S ) − V0ω G +V0ω (σ u|∂ S ).

Writing ϕ = u|∂ S leads to

 ω W0 + V0ω σ − 12 I ϕ = V0ω G .

(9.38)

If ϕ is a solution of (9.38), then the function u = W ω − (ϕ ) + V ω − (σ ϕ ) − V ω − (G )

(9.39)

satisfies (Rω − ) provided that Tu|∂ S + σ u|∂ S = G . Therefore, we require that   N0ω ϕ + W0ω ∗ − 12 I (σ ϕ ) − W0ω ∗ − 12 I G   + σ W0ω + 12 I ϕ + V0ω (σ ϕ ) −V0ω G = G , which, since ϕ satisfies (9.38), means that  ω∗ 1  W0 + 2 I (σ ϕ ) + N0ω ϕ = W0ω ∗ + 12 I G .

(9.40)

Thus, if (C3) holds, then u defined by (9.39) is the unique regular solution of (R ω − ) provided that ϕ satisfies both (9.38) and (9.40). We claim that this pair of equations has a unique solution for all values of the oscillation frequency. + Theorem 9.15. If (C3) holds and ω is not an eigenfrequency of (D ω 0 ), then the equation  ω W0 + V0ω σ − 12 I ϕ = V0ω G

has a unique solution ϕ ∈ C 1,α (∂ S) for every G ∈ C 0,α (∂ S), α ∈ (0, 1). This solution also satisfies  ω∗ 1  W0 + 2 I (σ ϕ ) + N0ω ϕ = W0ω ∗ + 12 I G . Proof. By the Fredholm alternative and Theorem 9.11, (9.38) has a unique solution ϕ that, since V0ω G ∈ C 1,α (∂ S), is of class C 1,α (∂ S). Using (7.7) and (7.8), we find that  0 = W0ω + V0ω σ − 12 I ϕ − V0ω G   = N0ω W0ω +V0ω σ − 12 I ϕ −V0ω G      = W0ω ∗ − 12 I N0ω ϕ + W0ω ∗ + 12 I (σ ϕ ) − W0ω ∗ + 12 I G + and, since ω is not an eigenfrequency of (D ω 0 ), from Theorem 6.4 we deduce that

170

9 Problems with Robin Boundary Conditions

 ω∗ 1  W0 + 2 I (σ ϕ ) + N0ω ϕ = W0ω ∗ + 12 I G , so ϕ is also a solution of (9.40).   + Before considering the case when ω is an eigenfrequency of (D ω 0 ), we need a ω+ result concerning the eigensolutions of (D 0 ) (cf. Theorem 7.6). + Theorem 9.16. If (C3) holds, then the function u is an eigensolution of (D ω 0 ) if and only if u = W ω + (ϕ ) + V ω + (σ ϕ ) in S¯+ ,

where ϕ is a C 1,α -eigenfunction of W0ω +V0ω σ − 12 I. + Proof. Suppose that u is an eigensolution of (D ω 0 ) for an eigenfrequency ω of multiplicity n. Then, by Theorem 9.11, the operator W 0ω + V0ω σ − 12 I has n linearly independentC 1,α-eigenfunctions, say, ψ (k) . In the proof of Theorem 9.12 it is shown n that the set v(k) k=1 , where

v(k) = W ω + (ψ (k) ) + V ω + (σ ψ (k) ),

k = 1, 2, . . . , n,

+ is a basis for the eigenspace of problem (D ω 0 ). Consequently, there are constants bk , k = 1, 2, . . . , n, such that

u=

n

n

k=1

k=1

∑ bk v(k) = ∑ bk



 W ω + (ψ (k) ) + V ω + (σ ψ (k) ) = W ω + (ϕ ) + V ω + (σ ϕ ),

where

ϕ=

n

∑ bk ψ (k) .

k=1

Conversely, if ϕ is

a C 1,α -eigenfunction of W0ω

+V0ω σ − 12 I, then

 u|∂ S = W0ω − 12 I ϕ +V0ω (σ ϕ ) = 0, + so u is a solution of (D ω 0 ). From Lemma 9.1 we deduce that u is nonzero; conse+ quently, u is an eigensolution of (D ω  0 ).  + Theorem 9.17. If (C3) holds and ω is an eigenfrequency of (D ω 0 ), then equation 1, α 0, α (9.40) has a unique solution ϕ ∈ C (∂ S) for every G ∈ C (∂ S), α ∈ (0, 1). This solution also satisfies equation (9.38).

Proof. First, we demonstrate that (9.38) is solvable. Let η be an eigenfunction of W0ω ∗ + σ V0ω − 12 I. By Theorems 9.11 and 6.4, η is also an eigenfunction of the operator W0ω ∗ − 12 I . As in the proof of Theorem 7.11, we deduce that 

η TV0ω G ds = 0,

∂S

so (9.38) has solutions of class C 1,α (∂ S).

9.3 Direct Boundary Equation Formulation

171

Let ϕ1 be one of the solutions of (9.38). Then, arguing as in the proof of Theorem 9.15, we find that  ω ∗ 1  ω    W0 − 2 I N0 ϕ1 + W0ω ∗ + 12 I (σ ϕ1 ) − W0ω ∗ + 12 I G = 0. Hence, either

 ω∗ 1  W0 + 2 I (σ ϕ1 ) + N0ω ϕ1 = W0ω ∗ + 12 I G ,

which means that ϕ1 is a solution of (9.40), or  ω∗ 1  W0 + 2 I (σ ϕ1 ) + N0ω ϕ1 = W0ω ∗ + 12 I G + η ,

(9.41)

where η is an eigenfunction of W0ω ∗ − 12 I . By Theorem 6.4,

η = T v|∂ S , + where v is an eigensolution of (D ω 0 ). According to Theorem 9.16, we can write

v = W ω + (ψ ) + V ω + (σ ψ ), where ψ is an eigenfunction of W0ω + V0ω σ − 12 I. Hence,  η = T v|∂ S = N0ω ψ + W0ω ∗ + 12 I (σ ψ ), and from (9.41) we deduce that  ω ∗ 1    W0 + 2 I σ (ϕ1 − ψ ) + N0ω σ (ϕ1 − ψ ) = W0ω ∗ + 12 I G ;

(9.42)

that is, ϕ = ϕ1 − ψ , which is a solution of (9.38), is also a solution of (9.40). This proves that (9.40) is solvable. Clearly, for each chosen ϕ 1 , the pairing eigenfunction ψ of W 0ω + V0ω σ − 12 I occurring in (9.42) is uniquely determined. (0) To establish the uniqueness of the solution of (9.40), let ϕ 1 , ψ (0) be a pair such (0)

that ϕ1 − ψ (0) satisfies the homogeneous equation (9.40); that is,

 ω ∗ 1  (0) (0) W0 + 2 I σ (ϕ1 − ψ (0)) + N0ω (ϕ1 − ψ (0)) = 0.

(9.43)

The function  (0) (0) v = W ω + (ϕ1 − ψ (0) ) + V ω + σ (ϕ1 − ψ (0) ) (0)

satisfies Aω v = 0 in S + . Since ϕ1 − ψ (0) is also a solution of the homogeneous version of (9.38), in view of (9.43) we have  (0) v|∂ S = W0ω +V0ω σ − 12 I (ϕ1 − ψ (0) ) = 0,   (0) (0) T v|∂ S = W0ω ∗ + 12 I σ (ϕ1 − ψ (0) ) + N0ω (ϕ1 − ψ (0) ) = 0,

172

9 Problems with Robin Boundary Conditions

so, by the interior representation formula (4.36),  (0) (0) v = W ω + (ϕ1 − ψ (0) ) + V ω + σ (ϕ1 − ψ (0) ) = V ω + (T v|∂ S ) − W ω + (v|∂ S ) = 0. (0)

Lemma 9.1 now implies that ϕ 1 − ψ (0) = 0, which means that (9.40) has a unique solution.   From Theorems 9.15 and 9.17 we conclude that (R ω − ) is uniquely solvable for any value of ω .

9.4 Composite Integral Equation We consider the unique solvability of a linear combination of (9.38) and (9.40) (cf. Section 7.6). Theorem 9.18. If (C3) together with (7.26) and either (7.27) or (7.28) hold, then the equation  ω∗ 1  W0 + 2 I (σ ϕ ) + N0ω ϕ + Λ W0ω +V0ω σ − 12 I ϕ  = W0ω ∗ + 12 I G + Λ V0ω G

(9.44)

has a unique solution ϕ ∈ C 1,α (∂ S) for every G ∈ C 1,α (∂ S), α ∈ (0, 1). Proof. The existence of at least one C 1,α -solution of (9.44) is guaranteed by Theorems 9.15 and 9.17. Let ϕ be a solution of the corresponding homogeneous equation  ω∗ 1  W0 + 2 I (σ ϕ ) + N0ω ϕ + Λ W0ω + V0ω σ − 12 I ϕ = 0. Then the function

u = W ω + (ϕ ) + V ω + (σ ϕ )

satisfies Tu|∂ S + Λ u|∂ S = 0. Arguing as in the proof of Theorem 7.12, we arrive at Tu|∂ S = u|∂ S = 0, which, in view of (4.36) and (9.45), means that W ω + (ϕ ) + V ω + (σ ϕ ) = 0 in S¯+ , so, by Lemma 9.1, ϕ = 0. Thus, the solution of (9.44) is unique.  

(9.45)

9.5 Modified Fundamental Solutions

173

9.5 Modified Fundamental Solutions We now investigate whether or not the theory of modified potentials developed in Chapter 8 for the exterior Dirichlet and Neumann boundary value problems may be used to construct solutions of the exterior Robin problem.

9.5.1 The Ursell Modification We start with the Ursell-type modification introduced in Section 8.2. At first glance, one may expect to have to prove an assertion analogous to Theorem 8.1 with (8.18) replaced by Tu + σ u = 0 on ∂ S 1 . This, however, is not the case. Theorem 8.1, as it stands, is sufficient for our purposes since it is designed to eliminate nonuniqueness difficulties when ω is an eigen+ ω+ frequency of either (D ω 0 ) or (N0 ). As Theorem 9.11 indicates, the problem of nonuniqueness in the context of (R ω − ) arises only when ω is an eigenfrequency of + problem (D ω 0 ). Theorem 9.19. If (C3) holds and G ∈ C 0,α (∂ S), α ∈ (0, 1), then the unique regular solution of (Rω − ) is given by u = VMω − (ϕ ), (9.46) where ϕ ∈ C 0,α (∂ S) is the unique solution of  ω∗ ω WM0 + σ VM0 − 12 I ϕ = G .

(9.47)

Proof. If the solution of (R ω − ) is sought in the form (9.46), then we arrive at the integral equation (9.47) for the unknown density. We consider the homogeneous adjoint equation to (9.47), namely,  ω ω WM0 +VM0 σ − 12 I ψ = 0, (9.48) and introduce the function v1 = WMω + (ψ ) + VMω + (σ ψ ), where ψ is a solution of (9.48). Then v1 |∂ S = 0, so, according to Theorem 8.1, v1 = 0 in Sa− ∩ S+ , where Sa− is the region R > a; therefore,

174

9 Problems with Robin Boundary Conditions

 ω∗ 1 ω T v1 |∂ S = NM0 ψ + WM0 + 2 I )(σ ψ ) = 0.

(9.49)

Now consider the function v2 = WMω − (ψ ) + VMω − (σ ψ ). Then, by (9.48) and (9.49), we find that T v2 |∂ S + σ v2 |∂ S

 ω∗ 1  ω  ω ω = NM0 ψ + WM0 − 2 I (σ ψ ) + σ WM0 + 12 I ψ + VM0 (σ ψ ) = −σ ψ + σ ψ = 0,

− so v2 is a solution of (Rω 0 ). By the uniqueness property, this solution is identically − ¯ zero in S . In particular,  ω ω v2 |∂ S = WM0 + 12 I ψ + VM0 (σ ψ ) = 0,

which, combined with (9.48), yields ψ = 0. Thus, by the Fredholm alternative, (9.47) has a unique C 0,α -solution.  

9.5.2 The Jones Modification We turn our attention to the Jones-type modification discussed in Section 8.3. If the solution of (R ω − ) is assumed to be of the form u = VLω − (ϕ ), then we need to solve the integral equation  ω∗ ω WL0 + σ VL0 − 12 I ϕ = G .

(9.50)

The corresponding homogeneous equation is  ω∗ ω WL0 + σ VL0 − 12 I ϕ = 0.

(9.51)

Let ϕ be a solution of equation (9.51) and consider the functions u = V Lω + (ϕ ) and v = VLω − (ϕ ). We see that T v|∂ S + σ v|∂ S = 0. Since (Rω − ) has at most one solution, this means that v = 0 in S¯− , so

ω v|∂ S = VL0 ϕ = u|∂ S = 0.

9.5 Modified Fundamental Solutions

175

Continuing the argument in exactly the same way as in Subsection 8.4.2 and assuming that certain conditions on the constants in the definition of L ω (x, y) are satisfied, we arrive at u = 0 in S + \σ (b). Consequently,  ω∗ 1 ω 0 = Tu|∂ S + σ u|∂ S = WL0 + 2 I ϕ + σ VL0 ϕ, which, in view of (9.51), implies that ϕ = 0. Hence, by the Fredholm alternative, (9.50) has a unique solution. This result is summed up in the next assertion. Theorem 9.20. If (C3) holds, G ∈ C 0,α (∂ S), α ∈ (0, 1), and the constants in Lω (x, y) are chosen so that (8.88) is satisfied, then equation (9.50) has a unique solution ϕ ∈ C 0,α (∂ S). ω (x, y) defined by (8.94) can Results involving the finite series modification D L also be easily obtained.

Chapter 10

The Transmission Problem

We now examine the transmission problem (also referred to as the inclusion or boundary–contact problem) in which the plate consists of two elastic materials in separate subdomains with conditions specified on the interface. The analogous problem in acoustics was investigated in [38] by means of a simple indirect formulation that leads to a coupled pair of integral equations at the interface. Although the integral equations involved do not have the disadvantage of nonunique solutions previously observed for the Dirichlet, Neumann, and Robin problems, the method used in [34] and [35] is that of modified fundamental solutions. This was done with the aim of minimizing the norm of the boundary integral operators involved—a maneuver that maintains the unique solvability and is advantageous when the equations are solved numerically. Subsequently, combined direct and indirect formulations were proposed in [31], which yielded a single integral equation in a single unknown that turned out to be sufficient for constructing the solution. The three-dimensional elastodynamic transmission problem was solved in [40] by means of an indirect method. Unfortunately, this gives rise to hypersingular operators that require regularization. A similar problem is encountered in two dimensions [45]. Two-dimensional elastodynamic transmission problems with a variety of conditions across the interface, ranging from ‘welded contact’ to ‘complete debonding’, were investigated in [46]. In each case, the analysis was performed on boundary integral equations established by means of direct formulations. Similar problems arise in the case of bounded elastic obstacles surrounded by an inviscid compressible fluid [42]. As usual, we first consider the question of uniqueness of solution for the boundary value problem. The answer, given in Theorem 10.1, is based on results obtained in [24]. We then turn our attention to proving solvability. Just as we did for the Dirichlet, Neumann, and Robin problems, here we adopt an indirect method. But this classical approach produces hypersingular operators for which the Fredholm existence theory is not valid. We circumvent this by forming a combination of hypersingular operators in such a way as to eliminate the strong singularities. Singular integral equation G.R. Thomson and C. Constanda, Stationary Oscillations of Elastic Plates, A Boundary Integral Equation Analysis, DOI 10.1007/978-0-8176-8241-5_10, © Springer Science+Business Media, LLC 2011

177

178

10 The Transmission Problem

theory is then employed to deduce the validity of the Fredholm alternative. Theorem 10.2 guarantees the existence of a unique solution of the transmission problem. In Section 10.2 we consider an alternative method for solving the boundary value problem, which avoids the use of hypersingular operators. A system of equations is derived from an interior representation formula and a modified exterior representation formula (cf. Theorem 8.13). Even though this is not a Fredholm system, the existence of at least one solution is assured by the existence of a solution of the transmission problem itself established in Theorem 10.2. The unique solvability of the system is proved in Theorem 10.3.

10.1 The Indirect Method We consider an infinite plate of thickness h 0 with a bounded inclusion occupying the region S¯+ × [−h0 /2, h0 /2]. The inclusion is homogeneous and isotropic, has Lam´e constants λi and μi , and is of density ρ i . The remainder of the plate, which lies in S¯− × [−h0 /2, h0 /2], is also assumed to be homogeneous and isotropic. Its Lam´e ω constants are λe and μe and its density is ρe . Let Aω i (∂x ) and Ae (∂x ) be the operator ω A (∂x ) with the elastic constants λ i , μi , ρi and λe , μe , ρe , respectively. The boundary moment–stress operators Ti (∂x ) and Te (∂x ) are defined analogously. We assume that both sets of elastic constants satisfy the restrictions (1.14) and (1.15). The transmission problem for the stationary oscillations of thin elastic plates is formulated as follows: Find ui ∈ C2 (S+ ) ∩C1 (S¯+ ) and ue ∈ C2 (S− ) ∩C1 (S¯− ) ∩ Bω such that + Aω i ui = 0 in S , − Aω e ue = 0 in S

and u e − ui = f Te ue − Ti ui = g

on ∂ S, on ∂ S.

(10.1) (10.2)

Equations (10.1) and (10.2) are known as transmission (or interface) conditions. Theorem 10.1. The transmission problem has at most one regular solution pair {ui , ue }. Proof. Suppose that {u i , ue } is a solution of the corresponding homogeneous problem. Then ue − ui = 0 on ∂ S, Te ue − Ti ui = 0 on ∂ S. Writing the reciprocity relation for u i and u¯ i yields

(10.3) (10.4)

10.1 The Indirect Method

179



(u¯ Ti Ti ui − uTiTi u¯i ) ds = 0.

(10.5)

∂S

Similarly, for u e and u¯e in S− ∩ KR we find that 

(u¯ Te Te ue − uTeTe u¯e ) ds =

∂S



(u¯ Te Te ue − uTeTe u¯e ) ds.

(10.6)

∂ KR

Making use of the homogeneous boundary conditions and adding (10.5) and (10.6), we arrive at  (u¯ Te Te ue − uTeTe u¯e ) ds = 0. ∂ KR

Since ue satisfies the radiation conditions, we deduce that [24] ue = 0 in S¯− ; hence, ue |∂ S = Te ue |∂ S = 0. Boundary conditions (10.3) and (10.4) then imply that ui |∂ S = Ti ui |∂ S = 0, so, by the interior representation formula, ui = 0 in S¯− . Consequently, the homogeneous problem has only the zero solution, which means that the transmission problem has at most one regular solution pair.   Let Diω (x, y) and Deω (x, y) be the matrices of fundamental solutions constructed with λi , μi , ρi and λe , μe , ρe , respectively. The corresponding pairs of single-layer and double-layer potentials are defined in the obvious way, namely, (Viω ϕ )(x) =



Diω (x, y)ϕ (y) ds(y),

(Wiω ϕ )(x) =

∂S

and (Veω ϕ )(x) =





Piω (x, y)ϕ (y) ds(y),

∂S

Deω (x, y)ϕ (y) ds(y),

(Weω ϕ )(x) =

∂S



Peω (x, y)ϕ (y) ds(y).

∂S

The most obvious choice for an indirect formulation is to assume that ui = Vi ω + (ϕ ) + Wi ω + (ψ ), ue = Veω − (ϕ ) + Weω − (ψ ).

180

10 The Transmission Problem

This is analogous to the form of solution assumed in [38] for the transmission problem in acoustics. Unfortunately, this choice leads to the occurrence of the hypersinω − N ω when the boundary condition (10.2) is applied. gular operator Ne0 i0 A similar difficulty in the corresponding boundary value problem of plane elastodynamics was overcome in [45] by the construction of a linear combination of the hypersingular operators, which resulted in a singular integral operator. We solve our problem by the same procedure. We consider N0ω and investigate the singularity of the matrix T (∂ x )P ω (x, y) as r = |x − y| → 0. From (2.33) we see that, as r → 0,    ∂ ∂ ln r Eαβ − ln r E3 ∂ s(y) ∂ ν (y)   ∂ (xα − yα )(xβ − yβ ) + (1 − μ )εαγ Eγβ + O(lnr), (10.7) ∂ s(y) r2

2π P ω (x, y) = −μ  εαβ



 T where we have used the fact that λ  + μ  = 1 − μ . If we write ν (y) = ν1y , ν2y and Yα = yα − xα , then, explicitly,

∂ 1 ln r = 2 (Y1 ν1y + Y2 ν2y ), ∂ ν (y) r y

(10.8)

y

so, since τ (y) = (−ν2 , ν1 )T , we have

∂ 1 y y lnr = 2 (Y2 ν1 − Y1 ν2 ), ∂ s(y) r ∂ (x1 − y1 )2 2 = − 4 Y1Y2 (Y1 ν1y +Y2 ν2y ), 2 ∂ s(y) r r ∂ (x1 − y1 )(x2 − y2 ) 1 = 4 (Y12 −Y22 )(Y1 ν1y + Y2 ν2y ), ∂ s(y) r2 r

(10.9)

∂ (x2 − y2 )2 2 = 4 Y1Y2 (Y1 ν1y + Y2 ν2y ). ∂ s(y) r2 r Using (10.7)–(10.9), we can easily show that, as r → 0,  1 = 2 μ  (Yα νβy −Yβ ναy ) r    1 y y   + − μ δαβ + 2(μ − 1) 2 Yα Yβ (Y1 ν1 + Y2 ν2 ) + O(lnr) r

ω 2π Pαβ (x, y)

(10.10)

and

1 (Y1 ν1y + Y2 ν2y ) + O(lnr). r2 All the other terms in the matrix are O(ln r) as r → 0. ω 2π P33 (x, y) = −

(10.11)

10.1 The Indirect Method

181

We now show that y

y

Y1 ν1 +Y2 ν2 = O(r2 ) as r → 0. Let the boundary ∂ S be defined by an equation of the form  T y = ψ (s) = ψ1 (s), ψ2 (s) , where s is the arc coordinate, and let  T x = ψ (s0 ) = ψ1 (s0 ), ψ2 (s0 ) , where s0 is fixed. Expanding the function ψ α (s) in a Taylor series about the point s = s0 yields  ψα (s) = ψα (s0 ) + (s − s0 )ψα (s0 ) + 12 (s − s0 )2 ψα (s0 ) + O (s − s0 )3 (10.12) and, similarly,

 ψα (s) = ψα (s0 ) + (s − s0)ψα (s0 ) + O (s − s0 )2 .

(10.13)

From (10.12) we see that Yα = ψα (s) − ψα (s0 )

 = (s − s0 )ψα (s0 ) + 12 (s − s0 )2 ψα (s0 ) + O (s − s0 )3 ;

(10.14)

hence, r2 = Y12 +Y22 = (s − s0 )2

  2  2   ψ1 (s0 ) + ψ2 (s0 ) + O (s − s0 )3 .

(10.15)

Since   T 1  (ν1y , ν2y )T = (τ2y , −τ1y )T =  ψ (s), − ψ (s) ,     2 1 2 2 1/2 ψ1 (s) + ψ2 (s) from (10.13) and (10.14) we obtain    1 2     y y 3 2 (s − s0 ) ψ1 (s0 )ψ2 (s0 ) − ψ1 (s0 )ψ2 (s0 ) Y1 ν1 + Y2 ν2 = + O (s − s ) , 0    1/2 [ψ1 (s)]2 + [ψ2 (s)]2 so, taking (10.15) into account, we find that Y1 ν1 +Y2ν2 ψ  (s0 )ψ2 (s0 ) − ψ1 (s0 )ψ2 (s0 ) = 1 3/2 ; 2 s→s0 r 2 [ψ1 (s0 )]2 + [ψ2 (s0 )]2 y

y

lim

thus,

Y1 ν1y +Y2 ν2y = O(r2 ) as r → 0.

(10.16)

182

10 The Transmission Problem

With (10.16) in mind, from (10.10) we can show that, as r → 0,   ∂ 1 ω  y  1 2 y 2π P (x, y) = 2 μ να + 2(1 − μ ) 2 Y1 να + O(r−1 ), ∂ xα 11 r r   ∂ ω 1 y  y  1 2π P (x, y) = 2 μ ν2 + 2(1 − μ ) 2 Y1Y2 ν1 + O(r−1 ), ∂ x1 12 r r   ∂ ω 1 1 2π P12 (x, y) = 2 − μ ν1y + 2(1 − μ ) 2 Y1Y2 ν2y + O(r−1 ), ∂ x2 r r   ∂ ω 1 1 2π P21 (x, y) = 2 − μ  ν2y + 2(1 − μ ) 2 Y1Y2 ν1y + O(r−1 ), ∂ x1 r r   ∂ ω 1 1 y y 2π P21 (x, y) = 2 μ  ν1 + 2(1 − μ ) 2 Y1Y2 ν2 + O(r−1 ), ∂ x2 r r   ∂ 1 1 ω 2π P22 (x, y) = 2 μ  ναy + 2(1 − μ ) 2 Y22 ναy + O(r−1 ), ∂ xα r r

(10.17)

(10.18) (10.19)

and then, using (10.11), we find that, as r → 0, 2π

∂ 1 y ω P33 (x, y) = 2 να + O(r−1 ). ∂ xα r

From (10.17)–(10.19) we see that, as r → 0,  2π  T (∂x )P ω (x, y) 11 2 h   1 1 = λ ν1x 2 2μ  ν1y + 2(1 − μ ) 2 Y1 (Y1 ν1y + Y2 ν2y ) r r   1 1 y y + 2μν1x 2 μ  ν1 + 2(1 − μ ) 2 Y12 ν1 r r   1 1 1 + μν2x 2 2(1 − μ ) 2 Y12 ν2y + 2(1 − μ ) 2 Y1Y2 ν1y + O(r−1 ). r r r By (10.16), as r → 0, Y1Y2 ν1y = Y1Y2 ν1y + Y22 ν2y − Y22 ν2y = Y2 (Y1 ν1y +Y2 ν2y ) − (r2 − Y12 )ν2y = Y12 ν2y − r2 ν2y + O(r3 ); therefore,  π  T (∂x )P ω (x, y) 11 2 h  1 1 y y y y = 2 λ μ  ν1x ν1 + μ μ ν1x ν1 + 2μ (1 − μ ) 2 Y12 (ν1x ν1 + ν2x ν2 ) r r − μ (1 − μ



)ν2x ν2y



+ O(r−1 ).

10.1 The Indirect Method

Since

λ μ =

183

λμ (λ + 2 μ )μ − 2μ 2 = = μ (1 − 2 μ ), λ + 2μ λ + 2μ

it follows that  π  T (∂x )P ω (x, y) 11 h2 μ



 1 2 2 x y x y x y x y = (1 − μ ) 2 ν1 ν1 − ν2 ν2 + 2 Y1 ν1 ν1 + ν2 ν2 + O(r−1 ), r r 

which can be written in the form 

2    π  Y1 ω  1 x y x y −1 T ( ∂ )P (x, y) = (1 − μ ) 2 ν ν + 2 − 1 ν ν x 1 1 γ γ + O(r ). 11 h2 μ r2 r2 Proceeding in the same way for the other elements of T (∂ x )P ω (x, y), we can show that, as r → 0, 

T (∂x )P ω (x, y)

 αβ



  Yα Yβ 1 2  1 x y x y x y = h μ (1 − μ ) 2 να νβ + νβ να + 2 2 − δαβ νγ νγ π r r + O(r−1 )

and



T (∂x )P ω (x, y)

 33

=

μ 1 + O(r−1 ). 2π r2

(10.20)

(10.21)

The remaining elements of the matrix are O(r −1 ) as r → 0. In [45] it is shown that, as r → 0,

νγx νγy = 1 + O(r) and

Yα Yβ = 2δαβ − (ναx νβy + νβx ναy ) + O(r). r2 Hence, (10.20) implies that, as r → 0, 2



T (∂x )P ω (x, y)

 αβ

=

1 2 1 h μ (1 − μ ) 2 δαβ + O(r−1 ). π r

We define the constant matrices ⎛ ⎞ 1 0 0 ⎜ μα (1 − μα ) ⎟ ⎜ ⎟ 1 ⎜ ⎟ 0 0 Cα = ⎜ ⎟, ⎜ ⎟ μα (1 − μα ) ⎝ 1 ⎠ 0 0 μα

α = e, i.

(10.22)

184

10 The Transmission Problem

ω − C N ω is O(r−1 ) as r → 0, which From (10.21) and (10.22) we see that C e Ne0 i i0 qualifies it as a singular integral operator. We now discuss the solvability of the transmission problem by using an indirect boundary integral equation formulation. Suppose that ui = Vi ω + (ϕ ) +Ci Wi ω + (ψ ) (10.23)

and

ue = Veω − (ϕ ) +Ce Weω − (ψ ).

This leads to the system of singular integral equations   ω 1   ω (Ve0 −Vi0ω )ϕ + Ce We0 + 2 I −Ci Wi0ω − 12 I ψ = f ,  ω ∗ 1  ω ∗ 1  ω We0 − 2 I − Wi0 + 2 I ϕ + (Ce Ne0 −Ci Ni0ω )ψ = g

(10.24)

(10.25) (10.26)

for the unknown density functions ϕ and ψ . We claim that system (10.25), (10.26) is a quasi-Fredholm system. With the notation used in [45], let σ be the (6× 6)-symbol matrix of this system (see [45], [78], or [49] for details). From (10.10), (10.11), and (10.16) it follows that, as r → 0, ω P11 (x, y) = O(1),

ω P22 (x, y) = O(1),

ω P33 (x, y) = O(1).

Also,

y y μ  Y1 ν2 −Y2ν1 ω + O(1) = −P21 (x, y). 2π r2 As in [45], this implies that the (3 × 3)-symbol matrix for the operators W αω0 and Wαω0∗ , α = e, i, is

ω P12 (x, y) =

σ (Wαω0 ) = σ (Wαω0∗ ) = where θ = ±1 and

μα iθ J, 2

α = e, i,

⎛

⎞ 0 10 J = ⎝−1 0 0⎠ . 0 00

ω −C N ω , We denote by M the symbol matrix of the singular integral operator C e Ne0 i i0 which is not required explicitly in what follows. The symbol matrix of (10.25), (10.26) is

 0 σ1 σ= , (10.27) σ2 M

where

σ1 = 12 Ce (μe iθ J + E3) − 12 Ci (μi iθ J − E3), σ2 =

1 2

(μe iθ J − E3 ) − 12

(μi iθ J + E3).

The matrix σ is said to be regular if det(σ ) = 0.

(10.28)

10.1 The Indirect Method

185

Lemma 10.1. The symbol matrix σ is regular. Proof. From (10.27) we see that det(σ ) = det(σ1 ) det(σ2 ).

(10.29)

Explicitly, (10.28) gives ⎛ ⎞ iθ μi 1 1 iθ μe 0 ⎟ ⎜ μe (1 − μe ) + μi (1 − μ ) μe (1 − μe ) − μi (1 − μ  ) i i ⎜ ⎟ ⎜ iθ μ  ⎟  i θ μ 1 1 ⎜ e i 2σ 1 = ⎜ − + 0 ⎟ ⎟;     ⎜ μi (1 − μi ) μe (1 − μe ) μe (1 − μe ) μi (1 − μi ) ⎟ ⎝ 1 1⎠ 0 0 + μe μi therefore,

 2 1 1 1 1 1 det(σ1 ) = + + 8 μe μi μe (1 − μe ) μi (1 − μi)  2  μi μe − − . μe (1 − μe ) μi (1 − μi ) From (2.26) and (1.14) we deduce that

μα = so we have

μα λα + μα = 1− < 1, λ α + 2 μα λα + 2 μ α

α = e, i,

    μi μi μe μe   −  μe (1 − μ  ) μi (1 − μ  )  ≤ μe (1 − μ  ) + μi (1 − μ  ) e e i i 1 1 < + ; μe (1 − μe ) μi (1 − μi )

consequently, det(σ1 ) > 0. Also,

which means that Since

⎛

⎞ −2 iθ (μe − μi ) 0 −2 0 ⎠, 2σ2 = ⎝−iθ (μe − μi ) 0 0 −2   det(σ2 ) = − 14 4 − (μe − μi )2 . |μe − μi | ≤ μe + μi < 2,

we obtain det(σ2 ) < 0. Hence, from (10.29) we deduce that det(σ ) = 0, so σ is regular.  

186

10 The Transmission Problem

From the definition of the so-called index of system (10.25), (10.26) (see [49], [78], [45], and [14]) it is easily seen that, since the symbol matrix σ does not depend on the boundary curve ∂ S, the system has index zero. This means that the Fredholm alternative is applicable to it. It can be shown [45] that if ϕ , ψ ∈ C 0,α (∂ S), α ∈ (0, 1), satisfy (10.25) with f ∈ C 1,α (∂ S), then ψ is of class C 1,α (∂ S). Theorem 10.2. For any f ∈ C 1,α (∂ S) and any g ∈ C 0,α (∂ S), α ∈ (0, 1), system (10.25), (10.26) has a unique solution pair ϕ ∈ C 0,α (∂ S), ψ ∈ C 1,α (∂ S). Proof. Suppose that ϕ and ψ satisfy the homogeneous versions of (10.25), (10.26), namely,   ω 1   ω (Ve0 −Vi0ω )ϕ + Ce We0 + 2 I −Ci Wi0ω − 12 I ψ = 0, (10.30)  ω ∗ 1  ω ∗ 1  ω We0 − 2 I − Wi0 + 2 I ϕ + (Ce Ne0 − Ci Ni0ω )ψ = 0. We introduce the functions ω− u− (ϕ ) + Ce Weω − (ψ ), e = Ve ω+ u+ (ϕ ) + Ci Wi ω + (ψ ), i = Vi ω+ u+ (ϕ ) − Ce Weω + (ψ ), e = −Ve ω− u− (ϕ ) + Ci Wi ω − (ψ ). i = Vi

From (10.30) we see that and

+ Te u− e |∂ S − Ti ui |∂ S = 0.

Hence, by Theorem 10.1, and Consequently, that is,

+ u− e |∂ S − ui |∂ S = 0

¯+ u+ i = 0 in S ¯− u− e = 0 in S .

+ − − u+ i |∂ S = Ti ui |∂ S = ue |∂ S = Te ue |∂ S = 0;

 Vi0ω ϕ + Ci Wi0ω − 12 I ψ = 0,  ω∗ 1 Wi0 + 2 I ϕ + Ci Ni0ω ψ = 0,  ω 1 ω Ve0 ϕ + Ce We0 + 2 I ψ = 0,  ω∗ 1 ω We0 − 2 I ϕ + Ce Ne0 ψ = 0.

Using (10.33) and (10.31), we now obtain

(10.31) (10.32) (10.33) (10.34)

10.2 The Direct Method

 ω 1 ω u+ e |∂ S = −Ve0 ϕ − Ce We0 − 2 I ψ = ψ ,  ω 1 ω u− i |∂ S = Vi0 ϕ + Ci Wi0 + 2 I ψ = ψ ;

similarly, (10.34) and (10.32) imply that  ω∗ 1 ω Te u+ e |∂ S = − We0 + 2 I ϕ −Ce Ne0 ψ = −ϕ ,  ω∗ ω 1 Ti u− i |∂ S = Wi0 − 2 I ϕ +Ci Ni0 ψ = −ϕ . Therefore,

187

(10.35)

(10.36)

+ u− i |∂ S − ue |∂ S = 0, + Ti u− i |∂ S − Te ue |∂ S = 0,

+ so {u− i , ue } is the solution of a homogeneous transmission problem (one in which the interior and exterior elastic materials have been interchanged). Hence, u − i = 0 in + , and from (10.35) and (10.36) we deduce that ϕ = ψ = 0. ¯ S¯− and u+ = 0 in S e The assertion now follows from the Fredholm alternative.  

The existence of a unique regular solution pair for the transmission problem is, therefore, guaranteed. This pair can be represented by (10.23) and (10.24), where {ϕ , ψ } is the unique solution pair of system (10.25), (10.26). The regularization of the hypersingular operators described in this section could also be applied to the corresponding operators in the static theory of plates since the matrix of fundamental solutions has exactly the same singularities as D ω (x, y).

10.2 The Direct Method We can avoid the troublesome hypersingular operators N αω0 , α = e, i, by deriving simple integral equations based on the representation formulas. To guarantee uniqueness of solution for every oscillation frequency, we make use of a modified matrix of fundamental solutions in the exterior region. By (4.36), ui = Vi ω + (Ti ui |∂ S ) − Wi ω + (ui |∂ S ) in S¯+ (10.37) and, by (8.104),

ω− ω− ue = WeM (ue |∂ S ) − VeM (Te ue |∂ S ) in S¯− .

(10.38)

ω− ω− Here, WeM and VeM are the potentials constructed with an Ursell-type modified matrix of fundamental solutions (see Section 8.2). From (10.37) we obtain  ui |∂ S = Vi0ω (Ti ui |∂ S ) − Wi0ω − 12 I (ui |∂ S ),

so, writing ϕ = u i |∂ S and ψ = Ti ui |∂ S , we arrive at  ω 1 Wi0 + 2 I ϕ −Vi0ω ψ = 0.

(10.39)

188

10 The Transmission Problem

Similarly, using (10.1) and (10.2) in (10.38), we find that  ω  ω ω ω WeM0 − 12 I ϕ −VeM0 ψ = − WeM0 − 12 I f +VeM0 g.

(10.40)

Since we have already proved that system (10.39), (10.40) has a solution, namely, {ϕ , ψ } = {ui |∂ S , Ti ui |∂ S }, all that remains for us to show is that this is the only solution. Theorem 10.3. For any f ∈ C 1,α (∂ S) and any g ∈ C 0,α (∂ S), α ∈ (0, 1), system (10.39), (10.40) has a unique solution pair ϕ ∈ C 1,α (∂ S), ψ ∈ C 0,α (∂ S). This solution also satisfies  Ni0ω ϕ − Wi0ω ∗ − 12 I ψ = 0, (10.41)  ω∗ 1  ω∗ 1 ω ω NeM0 ϕ − WeM0 + 2 I ψ = −NeM0 f + WeM0 + 2 I g. (10.42) Proof. Consider the corresponding pair of homogeneous equations  ω 1 Wi0 + 2 I ϕ −Vi0ω ψ = 0,  ω ω WeM0 − 12 I ϕ −VeM0 ψ = 0.

(10.43) (10.44)

Let {ϕ , ψ } be a solution pair of (10.43), (10.44) and define ω+ u+ (ψ ) − Wi ω + (ϕ ), i = Vi ω− ω− u− e = WeM (ϕ ) − VeM (ψ ), ω− u− (ϕ ) − Vi ω − (ψ ), i = Wi ω+ ω+ u+ e = WeM (ϕ ) − VeM (ψ ).

Using (10.43) and (10.44), we see that  ω 1 ω u+ i |∂ S = Vi0 ψ − Wi0 − 2 I ϕ = ϕ ,  ω ω 1 u− e |∂ S = WeM0 + 2 I ϕ − VeM0 ψ = ϕ ,  ω ω 1 u− i |∂ S = Wi0 + 2 I ϕ − Vi0 ψ = 0,  ω ω 1 u+ e |∂ S = WeM0 − 2 I ϕ − VeM0 ψ = 0.

(10.45) (10.46) (10.47) (10.48)

By (10.47), u − i is a solution of a homogeneous exterior Dirichlet problem, which + − + ¯− implies that u− i = 0 in S . Also, by (10.48) and Theorem 8.1, u e = 0 in Sa ∩ S (see Section 8.2). Consequently, + Ti u− i |∂ S = Te ue |∂ S = 0;

that is,

 Ni0ω ϕ − Wi0ω ∗ − 12 I ψ = 0,  ω∗ 1 ω NeM0 ϕ − WeM0 + 2 I ψ = 0.

(10.49) (10.50)

10.2 The Direct Method

189

From (10.49) and (10.50) we now see that  ω∗ 1 ω Ti u+ i |∂ S = Wi0 + 2 I ψ − Ni0 ϕ = ψ ,  ω∗ 1 ω Te u− e |∂ S = NeM0 ϕ − WeM0 − 2 I ψ = ψ .

(10.51) (10.52)

Therefore, (10.45), (10.46), (10.51), and (10.52) imply that + u− e |∂ S − ui |∂ S = 0, + Te u− e |∂ S − Ti ui |∂ S = 0, − so {u+ i , ue } is a solution of the homogeneous transmission problem. By Theorem 10.1, ¯+ u+ i = 0 in S ,

¯− u− e = 0 in S . Hence, from (10.45) and (10.51), or (10.46) and (10.52), we deduce that − ϕ = u+ i |∂ S = ue |∂ S = 0, − ψ = Ti u+ i |∂ S = Te ue |∂ S = 0,

so (10.43) and (10.44) admit only the zero solution. Thus, system (10.39), (10.40) has a unique solution. It remains to show that the unique solution of (10.39), (10.40) also satisfies (10.41), (10.42). Let {ϕ , ψ } be this unique solution. Then    0 = Wi0ω + 12 I ϕ − Vi0ω ψ = Ni0ω Wi0ω + 12 I ϕ − Vi0ω ψ , so, using (7.7) and (7.8), we find that  ω ∗ 1  ω   Wi0 + 2 I Ni0 ϕ − Wi0ω ∗ − 12 I ψ = 0. Similarly,

(10.53)

 0 = Wi0ω + 12 I ϕ − Vi0ω ψ    = Wi0ω − 12 I Wi0ω + 12 I ϕ − Vi0ω ψ ,

and an application of (7.6) and (7.9) yields    Vi0ω Ni0ω ϕ − Wi0ω ∗ − 12 I ψ = 0. In view of (10.53) and (10.54), Corollary 7.2(ii) implies that  Ni0ω ϕ − Wi0ω ∗ − 12 I ψ = 0; that is, {ϕ , ψ } satisfies (10.41).

(10.54)

190

10 The Transmission Problem

Also,  ω ω 0 = WeM0 − 12 I (ϕ + f ) − VeM0 (ψ + g)   ω ω ω 1 = NeM0 WeM0 − 2 I (ϕ + f ) −VeM0 (ψ + g) , which, when (8.107) and (8.108) are taken into account, leads to  ω ∗ 1  ω  ω∗ 1  WeM0 − 2 I NeM0 (ϕ + f ) − WeM0 + 2 I (ψ + g) = 0. ω ∗ − 1 I has no eigenfunctions (see Theorem 8.4), so we deduce that The operator WeM0 2

 ω∗ 1  ω∗ 1 ω ω NeM0 ϕ − WeM0 + 2 I ψ = −NeM0 f + WeM0 + 2 I g, as required.   It is easily seen that if {ϕ , ψ } satisfies (10.39), (10.40) (and, hence, (10.41), (10.42)), then {u i , ue }, where ui = Vi ω + (ψ ) − Wi ω + (ϕ ),

ω− ω− ω− ω− ue = WeM (ϕ ) − VeM (ψ ) + WeM ( f ) − VeM (g),

is the unique solution of the transmission problem.

Chapter 11

The Null Field Equations

In the preceding chapters we solved various boundary value problems associated with the high-frequency stationary oscillations of thin elastic plates, by means of quasi-Fredholm boundary integral equations. We now consider the so-called null field method (also known as the T -matrix, or extended boundary condition, method) which, although not generating integral equations in the classical sense, makes use of many of the ideas that we have introduced earlier. This method was considered for the first time in [76] in connection with electromagnetic scattering, and was extended subsequently to the acoustic case (see [77]). The question of unique solvability of the null field equations of acoustics was examined in [43] and [44]. The same topic was studied in [12] by a different method. As shown in [36], there is a link between the null field method and the modified Green’s function technique developed in [33] with the aim of minimizing the norm of the boundary integral operators in the modified theory. It was then established (see [2]) that these two methods are also connected to the so-called reproducing kernel (or Bergman–Schiffer kernel). These three procedures are collectively referred to as the scattering trinity. A list of papers on the null field method can be found in the book [73]. The null field equations for the system governing high-frequency stationary oscillations of thin elastic plates are not established straightforwardly. Their derivation is discussed in Section 11.1, where Theorem 11.1 plays an important part. In Section 11.2 each of the infinite sets of equations characterizing the solutions of the exterior Dirichlet, Neumann, and Robin problems, as well as the transmission problem, is shown to have a unique solution (see Theorems 11.2–11.4 and 11.6). Before deriving the null field equations for the transmission problem, however, we need to prove an auxiliary assertion similar to Theorem 11.1 (see Theorem 11.5). Finally, results concerning the completeness and linear independence of certain sets of wavefunctions are proved in Section 11.3, where we apply many of the theorems on uniquely solvable sets of equations discussed in the preceding section. These properties have important practical value in the numerical solution of the null field equations. G.R. Thomson and C. Constanda, Stationary Oscillations of Elastic Plates, A Boundary Integral Equation Analysis, DOI 10.1007/978-0-8176-8241-5_11, © Springer Science+Business Media, LLC 2011

191

192

11 The Null Field Equations

11.1 Derivation of the Null Field Equations In acoustics, the derivation of the null field equations relies on the orthogonality of the so-called regular circular (spherical) wavefunctions over circles (spheres) centered at the origin (see [43], [44], and [36]). Unfortunately, our corresponding reg(σ j) ular circular wavefunctions χˆ m , defined by (8.77), are not mutually orthogonal. Nevertheless, we can obtain the null field equations for the system characterizing the stationary oscillations of thin elastic plates by means of an auxiliary assertion. (2 j) Since χˆ 0 = 0, j = 1, 2, 3, in all the summations that follow, the terms for m = 0 and σ = 2 are not strictly included. Theorem 11.1. If

∞

2

3

(σ j) (σ j) χˆ m (x)

∑ ∑ ∑ cm

m=0 σ =1 j=1

=0

(11.1)

in a bounded region containing the origin, then (σ j)

cm

= 0 ∀σ , j, m.

Proof. Taking the curl of the first two components of (11.1) leads to ∞

2

(σ 3)

∑ ∑ cm

m=0 σ =1

hence, −k32

∞

2

∑∑

m=0 σ =1

√

(σ )

Δ ψˆ m (x) = 0;

(σ 3)

(σ )

εm cm Jm (k3 R)Em (θ ) = 0.

(σ )

By the orthogonality of the E m (θ ) on circles centered at the origin, we find that (σ 3)

cm

= 0 ∀σ , m.

It is now easily seen that, explicitly,  √ α1 m(−1)σ (σ ) (σ ) (3−σ ) Φˆ m (x) = εm α1 k1 Jm (k1 R)Em (θ )eR + Jm (k1 R)Em (θ )eθ R (σ )



− hk3α2 Jm (k1 R)Em (θ )e3 and  √ α2 m(−1)σ (σ ) (σ ) (3−σ ) ϒˆm (x) = εm α2 k2 Jm (k2 R)Em (θ )eR + Jm (k2 R)Em (θ )eθ R

 (σ ) + hk3 α1 Jm (k2 R)Em (θ )e3 .

11.1 Derivation of the Null Field Equations

193

As R → 0, we have [1] Jm (k j R) = and

1 km Rm + O(Rm+2 ) 2mΓ (m + 1) j m

Jm (k j R) =

kmj Rm−1 + O(Rm+1 ),

2mΓ (m + 1)

which means that, as R → 0, (σ ) Φˆ m (x) =

√   m εm (σ ) (3−σ ) Rm−1 α1 k1m+1 Em (θ )eR + α1 k1m (−1)σ Em (θ )eθ m 2 Γ (m + 1) + O(Rm )

and (σ ) Ψˆm (x) =

√   m εm (σ ) (3−σ ) Rm−1 α2 k2m+1 Em (θ )eR + α2 k2m (−1)σ Em (θ )eθ m 2 Γ (m + 1) + O(Rm ).

From (11.1) and the orthogonality of the various trigonometric functions we deduce that for σ = 1, 2 and m = 0, 1, 2, . . . , (σ 1)

α1 k1m+1 cm

(σ 2)

+ α2 k2m+1 cm

+ O(R) = 0,

(σ 1) (σ 2) α1 k1m cm + α2 k2m cm + O(R) =

0.

Letting R → 0 yields the systems (σ )

BmCm = 0, where

∀σ , m,

α1 k1m+1 α2 k2m+1 Bm = α1 k1m α2 k2m

and (σ ) Cm

(σ 1)

=

cm

(σ 2)

cm

! ∀m

! ∀σ , m;

therefore, det Bm = α1 α2 k1m+1 k2m − α1 α2 k1m k2m+1 = α1 α2 k1m k2m (k1 − k2 ). Since k1 and k2 are nonzero and distinct, we deduce that det B m = 0, m = 0, 1, 2, . . . , so (σ 1) (σ 2) cm = cm = 0 ∀σ , m, and the theorem is proved.  

194

11 The Null Field Equations

By (8.17), (8.55), and (8.77), for R x < Ry the matrix of fundamental solutions may be written in the form D ω (x, y) =

∞ 2 3  (σ j) T i (σ j) χˆ m (x) χm (y) . 2 2 4h μ k3 m=0 σ =1 j=1

∑∑∑

(11.2)

If u is a regular solution of (4.1) in S − and satisfies the radiation conditions, then, by Theorem 4.7, u(x) =

 

T  T (∂y )D ω (y, x) u(y) − D ω (x, y)T (∂y )u(y) ds(y),

x ∈ S− , (11.3)

∂S

0=

 

T  T (∂y )D ω (y, x) u(y) − D ω (x, y)T (∂y )u(y) ds(y),

x ∈ S+ . (11.4)

∂S

We denote by R min and Rmax the minimum and maximum distances from the origin to ∂ S, respectively. Let σ in and σex be the disks centered at the origin with radii R min and Rmax . When x ∈ σin , we may substitute (11.2) in (11.4) to obtain ∞

2

∑∑

3

(σ j) ∑ χˆ m (x)

m=0 σ =1 j=1

 

∂S

(σ j)

T (∂y )χm

T (y) u(y)

 (σ j) T  − χm (y) T (∂y )u(y) ds(y) = 0,

x ∈ σin ,

so, by Theorem 11.1,  

(σ j)

T (∂y )χm

T  (σ j) T  (y) u(y) − χm (y) T (∂y )u(y) ds(y) = 0

∀σ , j, m. (11.5)

∂S

Equations (11.5) are called the null field equations associated with system (4.1).

11.2 The Question of Solvability We now use the infinite set of equations (11.5) to construct solutions of the exterior Dirichlet, Neumann, and Robin boundary value problems. We show that in each case this infinite set of equations has precisely one solution. We will also address the transmission problem. The unique solvability of the null field equations for the exterior Dirichlet and Neumann problems of acoustics was proved by showing that a function is a solution of the null field equations if and only if it is also a solution of a certain uniquely solvable integral equation of the second kind that involves a Jones-type modified fundamental solution (see [43] and [44]). In [12] it was shown that the same results

11.2 The Question of Solvability

195

could be obtained in a much simpler manner. It is this latter approach that we use for our problems.

11.2.1 The Exterior Dirichlet Problem We start with (Dω − ). Applying the boundary condition u| ∂ S = R to (11.5) yields   (σ j)

χm

T (y) T (∂y )u(y) ds(y)

∂S

=

 

(σ j)

T (∂y )χm

T (y) R(y) ds(y) ∀σ , j , m,

(11.6)

∂S

which is an infinite set of equations for the unknown moment–stress vector on ∂ S. Assuming that this set of equations can be solved, we compute the function u by means of (11.3). Theorem 11.2. The infinite set of equations (11.6) has a unique solution for every R ∈ C 1,α (∂ S), α ∈ (0, 1). Proof. The solvability of (11.6) follows from the solvability of the boundary value problem itself (see Theorem 6.10). We need to show that the equation   (σ j)

χm

T (y) ϕ (y) ds(y) = 0 ∀σ , j, m

(11.7)

∂S

has only the zero solution. (σ j) Suppose that (11.7) holds. Multiplying each equation by χˆ m (x), we arrive at 

 (σ j) T (σ j) χˆ m (x) χm (y) ϕ (y) ds(y) = 0 ∀σ , j, m;

∂S

hence,

 ∂S

∞ 2 3  (σ j) T i (σ j) χˆ m (x) χm (y) ϕ (y) ds(y) 2 2 4h μ k3 m=0 σ =1 j=1

∑∑∑

= 0.

(11.8)

From (8.17) and the definition of the single-layer potential we see that (11.8) leads to V ω + (ϕ ) = 0 for Rx < Rmin . Since the single-layer potential is analytic, we deduce that V ω + (ϕ ) = 0 in S¯+ . Theorem 6.3(i) then implies that ϕ = 0, and so the solution of (11.6) is unique.  

196

11 The Null Field Equations

11.2.2 The Exterior Neumann Problem Proceeding similarly for (N ω − ), we need to solve the infinite set of equations  

(σ j)

T (∂y )χm

  T T (σ j) (y) u(y) ds(y) = χm (y) S (y) ds(y)

∂S

∀σ , j, m

(11.9)

∂S

for the unknown boundary value of u. Theorem 11.3. The infinite set of equations (11.9) has a unique solution for every S ∈ C 0,α (∂ S), α ∈ (0, 1). Proof. As in the proof of Theorem 11.2, we are concerned only with the question of uniqueness. Suppose that there is a function ϕ such that  

(σ j)

T (∂y )χm

T (y) ϕ (y) ds(y) = 0 ∀σ , j, m.

(11.10)

∂S (σ j) Multiplying (11.10) by χˆ m (x) yields



 T (σ j) (σ j) χˆ m (x) T (∂y )χm (y) ϕ (y) ds(y)

∂S

=

 

(σ j)

T (∂y )χm

 (σ j) T T (y) χˆ m (x) ϕ (y) ds(y) = 0 ∀σ , j, m,

∂S

which means that    T (∂y ) ∂S

therefore,

∞ 2 3  (σ j) T i (σ j) χm (y) χˆ m (x) ∑ ∑ ∑ 2 2 4h μ k3 m=0 σ =1 j=1

T

ϕ (y) ds(y) = 0;

W ω + (ϕ ) = 0 for Rx < Rmin .

We now use standard arguments to deduce that W ω + (ϕ ) = 0 in S¯+ ; consequently, by Theorem 6.3(ii), ϕ = 0, which implies that the solution of the infinite set of equations (11.9) is unique.  

11.2.3 The Exterior Robin Problem Substituting the boundary condition Tu| ∂ S + σ u|∂ S = G in (11.5) and using the symmetry of the matrix σ , we obtain

11.2 The Question of Solvability





(σ j)

T (∂y )χm

197 (σ j)

(y) + σ (y)χm

T (y) u(y) ds(y)

∂S

=

  (σ j)

χm

T (y) G (y) ds(y)

∀σ , j, m. (11.11)

∂S

Theorem 11.4. If (C3) holds, then the infinite set of equations (11.11) has a unique solution for every G ∈ C 0,α (∂ S), α ∈ (0, 1). Proof. Consider the equality 

 T (σ j) (σ j) T (∂y )χm (y) + σ (y)χm (y) ϕ (y) ds(y) = 0 ∀σ , j, m.

(11.12)

∂S (σ j) Multiplying each equation in (11.12) by χˆ m (x), we arrive at



 T (σ j) (σ j) χˆ m (x) T (∂y )χm (y) ϕ (y) ds(y)

∂S

+



 T (σ j) (σ j) χˆ m (x) σ (y)χm (y) ϕ (y) ds(y) = 0 ∀σ , j, m;

∂S

that is,  

(σ j)

T (∂y )χm

 (σ j) T T (y) χˆ m (x) ϕ (y) ds(y)

∂S

+



 (σ j) T (σ j) χˆ m (x) χm (y) σ (y)ϕ (y) ds(y) = 0 ∀σ , j, m.

∂S

Arguing as in the proofs of Theorems 11.2 and 11.3, we conclude that W ω + (ϕ ) + V ω + (σ ϕ ) = 0 for Rx < Rmin and, since the potentials are analytic, deduce that W ω + (ϕ ) + V ω + (σ ϕ ) = 0 in S¯+ . By Lemma 9.1, we have ϕ = 0. Consequently, the solution of (11.11) is unique.  

11.2.4 The Transmission Problem The null field equations for the transmission problem are not as easily derived as those for the exterior Dirichlet, Neumann, and Robin problems. To establish them, we need an auxiliary assertion (cf. Theorem 11.1).

198

11 The Null Field Equations

Theorem 11.5. If

∞

2

3

(σ j) (σ j) χm (x)

∑ ∑ ∑ dm

m=0 σ =1 j=1

=0

(11.13)

outside a bounded region containing the origin, then (σ j)

dm

= 0 ∀σ , j, m.

Proof. As in the proof of Theorem 11.1, by taking the curl of the first two components of (11.13), we can easily deduce that (σ 3)

dm (σ β )

We introduce the functions Fm (σ 1)

Fm

= 0 ∀σ , m.

by writing

1 (σ ) (x) = √ e−i(k1 R−πm ) (hk3 α1 eR − iα2 k2 e3 )Em (θ ), R

(σ 2) Fm (x)

1 (σ ) = √ e−i(k2 R−πm ) (hk3 α2 eR + iα1 k1 e3 )Em (θ ), R

(11.14)

where eR , eθ , and e3 are the vectors defined in Section 8.2.2. From (11.14), (8.42)– (8.45), and (8.52)–(8.54) we easily see that lim

  (σ α )

R→∞ ∂ KR

Fm

T (νβ ) (x) χn (x) ds(x) = aβ δmn δσ ν δαβ ,

(11.15)

where aβ are nonzero constants and ∂ K R is the circle of radius R centered at the origin. Using the orthogonality relations (11.15), from (11.13) we obtain (σ 1)

dm

(σ 2)

= dm

=0

∀σ , m,

which completes the proof.   We can now use Theorem 11.5 to derive the null field equations for the transmission problem. By Theorem 4.5, ui (x) =



 ω  T  Di (x, y)Ti (∂y )ui (y) − Ti (∂y )Diω (y, x) ui (y) ds(y),

∂S

x ∈ S+ (11.16)

and 0=

   T  Diω (x, y)Ti (∂y )ui (y) − Ti (∂y )Diω (y, x) ui (y) ds(y), ∂S

x ∈ S− , (11.17)

11.2 The Question of Solvability

199

while Theorem 4.7 yields ue (x) =

 

T  Te (∂y )Deω (y, x) ue (y) − Deω (x, y)Te (∂y )ue (y) ds(y),

x ∈ S−

∂S

(11.18)

and  

0=

T  Te (∂y )Deω (y, x) ue (y) − Deω (x, y)Te (∂y )ue (y) ds(y),

x ∈ S+ . (11.19)

∂S

Substituting (11.2), with the obvious notational changes, in (11.19) for x ∈ σ in leads to  ∞ 2 3  T (σ j) (σ j) ∑ ∑ ∑ χˆ em (x) Te(∂y )χem (y) ue(y) m=0 σ =1 j=1  (σ j) T  ∂S − χem (y) Te (∂y )ue (y) ds(y) = 0; hence, by Theorem 11.1,  

T  (σ j) T  (σ j) Te (∂y )χem (y) ue (y) − χem (y) Te (∂y )ue (y) ds(y) = 0 ∀σ , j, m.

∂S

(11.20) Just as we did when we derived (8.17), we can easily show that for R x > Ry , D ω (x, y) =

∞ 2 3  (σ j) T i (σ j) χm (x) χˆ m (y) . ∑ ∑ ∑ 2 2 4h μ k3 m=0 σ =1 j=1

(11.21)

For x ∈ σex we may substitute (11.21) in (11.17) to obtain ∞

2

3

(σ j)

∑ ∑ ∑ χim

m=0 σ =1 j=1

(x)

  (σ j)

∂S

T χˆ im (y) Ti (∂y )ui (y)  T  (σ j) − Ti (∂y )χˆ im (y) ui (y) ds(y) = 0.

Theorem 11.5 then implies that   (σ j)

T  T  (σ j) χˆ im (y) Ti (∂y )ui (y) − Ti (∂y )χˆ im (y) ui (y) ds(y) = 0 ∀σ , j, m.

∂S

(11.22) Applying the boundary conditions u i |∂ S = ue |∂ S − f and Ti ui |∂ S = Te ue |∂ S − g to (11.22) leads to  

T  (σ j) T  (σ j) Ti (∂y )χˆ im (y) ue (y) − χˆ im (y) Te (∂y )ue (y) ds(y)

∂S

=

 

(σ j)

Ti (∂y )χˆ im (y)

∂S

T

 (σ j) T  f (y) − χˆ im (y) g(y) ds(y) ∀σ , j, m.

(11.23)

200

11 The Null Field Equations

Equations (11.20) and (11.23) are the null field equations for the transmission problem. If these equations are solved for the unknown boundary values of u e and Te ue , then the solution pair of the problem itself is given by (11.16) and (11.18) once the boundary values of u i and Ti ui have been found from (10.1) and (10.2). Theorem 11.6. The coupled pair of infinite sets of equations (11.20) and (11.23) has a unique solution pair for every f ∈ C 1,α (∂ S) and g ∈ C 0,α (∂ S), α ∈ (0, 1). Proof. We consider the homogeneous equations  

T  (σ j) T  (σ j) Te (∂y )χem (y) ϕ (y) − χem (y) ψ (y) ds(y) = 0 ∀σ , j, m,

(11.24)

T  (σ j) T  (σ j) Ti (∂y )χˆ im (y) ϕ (y) − χˆ im (y) ψ (y) ds(y) = 0 ∀σ , j, m.

(11.25)

∂S

 

∂S (σ j) (σ j) Multiplying (11.24) by χˆ em (x) and (11.25) by χ im (x) leads to

 

 (σ j) T T (σ j) Te (∂y )χem (y) χˆ em (x) ϕ (y) ds(y)

∂S

−



 (σ j) T (σ j) χˆ em (x) χem (y) ψ (y) ds(y) = 0 ∀σ , j, m

∂S

and  

 (σ j) T T (σ j) Ti (∂y )χˆ im (y) χim (x) ϕ (y) ds(y)

∂S

−



 (σ j) T (σ j) χim (x) χˆ im (y) ψ (y) ds(y) = 0 ∀σ , j, m.

∂S

Arguing in the usual way and taking (11.2) and (11.21) into account, we deduce that Weω + (ϕ ) − Veω + (ψ ) = 0 in S¯+ , Wi Let

ω−

(ϕ ) − Vi

ω−

(ψ ) = 0 in

S¯− .

ω+ u+ (ϕ ) − Veω + (ψ ), e = We ω− u− (ϕ ) − Vi ω − (ψ ), i = Wi ω− u− (ϕ ) − Veω − (ψ ), e = We ω+ u+ (ψ ) − Wi ω + (ϕ ). i = Vi

From (11.26) and (11.27) we see that + − − u+ e |∂ S = Te ue |∂ S = ui |∂ S = Ti ui |∂ S = 0;

(11.26) (11.27)

11.3 Complete, Linearly Independent Sets of Wavefunctions

201

that is, ω ω (We0 − 12 I)ϕ − Ve0 ψ = 0, ω ω∗ Ne0 ϕ − (We0 + 12 I)ψ = 0,

(Wi0ω + 12 I)ϕ − Vi0ω ψ = 0,

(11.28)

Ni0ω ϕ − (Wi0ω ∗ − 12 I)ψ = 0. Using (11.28), we obtain  ω 1 ω u− e |∂ S = We0 + 2 I ϕ −Ve0 ψ = ϕ ,  ω ω∗ 1 Te u− e |∂ S = Ne0 ϕ − We0 − 2 I ψ = ψ ,  ω 1 ω u+ i |∂ S = Vi0 ψ − Wi0 − 2 I ϕ = ϕ ,  ω∗ 1 ω Ti u+ i |∂ S = Wi0 + 2 I ψ − Ni0 ϕ = ψ ,

(11.29)

which means that + u− e |∂ S − ui |∂ S = 0, + Te u− e |∂ S − Ti ui |∂ S = 0. + ¯− ¯+ By Theorem 10.1, we have u − e = 0 in S and ui = 0 in S , so equations (11.29) imply that ϕ = ψ = 0,

as required.  

11.3 Complete, Linearly Independent Sets of Wavefunctions One possible way of solving the infinite set of null field equations for each problem is to expand the unknown function in a series of elements of a complete set. This would generate an infinite set of algebraic equations for the unknown coefficients in the series, which, for computational purposes, would obviously need to be truncated, thereby raising the question of whether the sequence of approximate solutions obtained in this way converges to the exact solution. Below, we answer this question by considering the completeness of six particular sets that involve the (σ j) (σ j) wavefunctions χ m and χˆ m . We define the sets  (σ j)  X1 = χm : σ = 1, 2, j = 1, 2, 3, m = 0, 1, 2, . . . , (11.30)  (σ j)  X2 = T χm : σ = 1, 2, j = 1, 2, 3, m = 0, 1, 2, . . . , (11.31)  (σ j)  (σ j) X3 = T χm + σ χm : σ = 1, 2, j = 1, 2, 3, m = 0, 1, 2, . . . . (11.32)

202

11 The Null Field Equations (2 j)

As was mentioned earlier, strictly speaking, since χ 0 = 0, j = 1, 2, 3, these sets do not include the terms for m = 0 and σ = 2. Using the results of the preceding section, we show that each of X 1 , X2 , and X3 is complete in L2 (∂ S), where L2 (∂ S) is the set of complex-valued, square integrable functions on ∂ S. A set is said to be closed in L2 (∂ S) if, for an arbitrary function ϕ ∈ L 2 (∂ S), the orthogonality of ϕ to every element in the set implies that ϕ = 0 almost everywhere on ∂ S (see [70], p. 90). Sometimes such a set is also known as total or fundamental (see [39]). In [70], p. 91, it is shown that if a set is closed in L 2 (∂ S), then it is also complete in L2 (∂ S) in the mean square sense. Therefore, if f ∈ L 2 (∂ S), there are constants (σ j) am , depending on n, such that, as n → ∞, 

| f (x) − fn (x)|2 ds(x) → 0,

∂S

where fn (x) =

n

2

3

(σ j) (σ j) χm (x).

∑ ∑ ∑ am

m=0 σ =1 j=1

Theorem 11.7. X1 is complete in L2 (∂ S). Proof. Suppose that ϕ ∈ L 2 (∂ S) is orthogonal to every element in X 1 ; that is, by (11.30),   T (σ j) χm (y) ϕ¯ (y) ds(y) = 0 ∀σ , j, m. (11.33) ∂S

In [14], Theorems 2.33–2.36, it is shown that for L 2 -densities, the properties of the potentials listed in Theorems 2.5–2.7 hold almost everywhere on ∂ S. With this in mind, following the argument in the proof of Theorem 11.2 with (11.7) replaced by (11.33) leads to the conclusion that ϕ¯ = 0 almost everywhere on ∂ S. Consequently, X1 is closed—hence, complete—in L 2 (∂ S).   Similar reasoning, this time involving Theorems 11.3 and 11.4, yields the corresponding results for X2 and X3 defined by (11.31) and (11.32), respectively. Theorem 11.8. X2 is complete in L2 (∂ S). Theorem 11.9. If (C3) holds, then X 3 is complete in L2 (∂ S). The linear independence of each of these sets on ∂ S can also be established. Theorem 11.10. X1 is linearly independent on ∂ S. Proof. Suppose that the elements of X 1 are linearly dependent on ∂ S; that is, there (σ j) are constants dm , not all zero, such that

11.3 Complete, Linearly Independent Sets of Wavefunctions ∞

2

3

(σ j) (σ j) χm (x)

∑ ∑ ∑ dm

m=0 σ =1 j=1

203

x ∈ ∂ S.

= 0,

Then U defined by U(x) =

∞

2

3

(σ j) (σ j) χm (x),

∑ ∑ ∑ dm

m=0 σ =1 j=1

x ∈ S¯− ,

(11.34)

− is a solution of (Dω 0 ); hence, by the uniqueness property, ∞

2

3

(σ j) (σ j) χm (x)

∑ ∑ ∑ dm

m=0 σ =1 j=1

= 0,

x ∈ S¯− .

From Theorem 11.5 we now deduce that (σ j)

dm

= 0 ∀σ , j, m,

which contradicts our assumption.   Theorem 11.11. X2 is linearly independent on ∂ S. The proof of this assertion is similar to that of the previous one, the only change − being that U defined by (11.34) is now a solution of (N ω 0 ). Theorem 11.12. If (C3) holds, then X 3 is linearly independent on ∂ S. − Here, the proof is based on the fact that (11.34) is a solution of (R ω 0 ). For certain values of the oscillation frequency we can also prove completeness of the sets of regular wavefunctions defined by

 (σ j)  Xˆ1 = χˆ m : σ = 1, 2, j = 1, 2, 3, m = 0, 1, 2, . . . ,  (σ j)  Xˆ2 = T χˆ m : σ = 1, 2, j = 1, 2, 3, m = 0, 1, 2, . . . ,  (σ j)  (σ j) Xˆ3 = T χˆ m + σ χˆ m : σ = 1, 2, j = 1, 2, 3, m = 0, 1, 2, . . . . Again, these sets do not include the terms for m = 0 and σ = 2 since they are identically zero. + ˆ Theorem 11.13. If ω is not an eigenfrequency of (D ω 0 ), then X1 is complete in 2 L (∂ S) and linearly independent on ∂ S.

Proof. Consider ϕ ∈ L2 (∂ S) such that   (σ j)

χˆ m

T (y) ϕ¯ (y) ds(y) = 0

∀σ , j, m.

∂S (σ j)

Multiplying (11.35) by χ m

(x) and summing over all σ , j, and m leads to

(11.35)

204

11 The Null Field Equations

V ω − (ϕ¯ ) = 0 for Rx > Rmax , which, by analyticity, means that V ω − (ϕ¯ ) = 0 in S¯− .

(11.36)

+ Thus, V0ω ϕ¯ = 0, so u = V ω + (ϕ¯ ) is a solution of (Dω 0 ). Since we assumed that ω ω+ is not an eigenfrequency of (D 0 ), we deduce that

V ω + (ϕ¯ ) = 0 in S¯+ .

(11.37)

By the usual arguments, from (11.36) and (11.37) we conclude that ϕ¯ = 0 almost everywhere on ∂ S, and the completeness of Xˆ1 in L2 (∂ S) follows immediately. Suppose now that the elements in Xˆ1 are linearly dependent on ∂ S. Then there (σ j) are constants cm , not all zero, such that ∞

2

3

(σ j) (σ j) χˆ m (x)

∑ ∑ ∑ cm

m=0 σ =1 j=1

= 0,

x ∈ ∂ S.

If this is the case, the function U defined by U(x) =

∞

2

3

(σ j) (σ j) χˆ m (x),

∑ ∑ ∑ cm

m=0 σ =1 j=1

x ∈ S¯+ ,

+ ω+ is a solution of (Dω 0 ). The fact that ω is not an eigenfrequency of (D 0 ) implies that ∞

2

3

(σ j) (σ j) χˆ m (x)

∑ ∑ ∑ cm

m=0 σ =1 j=1

= 0,

x ∈ S¯+ .

An application of Theorem 11.1 now yields (σ j)

cm

= 0 ∀σ , j, m.

Consequently, Xˆ1 is linearly independent on ∂ S.

 

The corresponding completeness results for Xˆ2 and Xˆ3 are obtained after simple modifications of the proof of Theorem 11.13. + ˆ Theorem 11.14. If ω is not an eigenfrequency of (N ω 0 ), then X2 is complete in 2 L (∂ S) and linearly independent on ∂ S. + ˆ Theorem 11.15. If (C1) holds and ω is not an eigenfrequency of (R ω 0 ), then X3 is 2 complete in L (∂ S) and linearly independent on ∂ S.

Theorem 11.16. If (C2) holds, then Xˆ3 is complete in L2 (∂ S) and linearly independent on ∂ S. As already mentioned, Theorems 11.14–11.16 facilitate the numerical approximation of the solutions of the null field equations (11.5).

Appendix A

Proof of Lemma 8.1

We prove the following assertion, listed as Lemma 8.1 in Chapter 8: If u is an analytic solution of (4.1) in D ∪ ∂ S 2 and u = Tu = 0 on ∂ S 2 , then u = 0 in D. Since u is analytic in D, it can be expressed as a power series with a nonzero radius of convergence, whose coefficients are the derivatives of the components of u at the point of expansion. We claim that all the derivatives of the components of u of every order are zero on ∂ S 2 . First, we consider the derivatives of order 1. Since Tu = 0 on ∂ S 2 , we have

∂ u1 ∂ u1 ∂ u2 ∂ u2 + μν2 + μν2 + λ ν1 = 0, ∂ x1 ∂ x2 ∂ x1 ∂ x2 ∂ u1 ∂ u1 ∂ u2 ∂ u2 λ ν2 + μν1 + μν1 + (λ + 2μ ) = 0, ∂ x1 ∂ x2 ∂ x1 ∂ x2 ∂ u3 ∂ u3 ν1 u 1 + ν2 u 2 + ν1 + ν2 =0 ∂ x1 ∂ x2

(λ + 2μ )ν1

(A.1) (A.2) (A.3)

on ∂ S2 , where ν = (ν1 , ν2 )T is the unit normal to ∂ S 2 pointing into D. The fact that u = 0 on ∂ S 2 implies that

∂ u1 ∂ u1 ∂ u1 = − ν2 + ν1 = 0 on ∂ S 2 , ∂s ∂ x1 ∂ x2 ∂ u2 ∂ u2 ∂ u2 = − ν2 + ν1 = 0 on ∂ S 2 , ∂s ∂ x1 ∂ x2 ∂ u3 ∂ u3 ∂ u3 = − ν2 + ν1 = 0 on ∂ S 2 , ∂s ∂ x1 ∂ x2

(A.4) (A.5) (A.6)

so, from (A.1), (A.2), (A.4), and (A.5) it follows that L(1) d (1) = 0 on ∂ S2 , G.R. Thomson and C. Constanda, Stationary Oscillations of Elastic Plates, A Boundary Integral Equation Analysis, DOI 10.1007/978-0-8176-8241-5, © Springer Science+Business Media, LLC 2011

(A.7) 205

206

A Proof of Lemma 8.1

where

⎛

(λ + 2μ )ν1 ⎜ λ ν2 L(1) = ⎜ ⎝ −ν 2 0

and d (1) =

μν2 μν1 ν1 0

⎞ μν2 λ ν1 μν1 (λ + 2μ )ν2 ⎟ ⎟ ⎠ 0 0 − ν2 ν1

∂ u 1 ∂ u1 ∂ u2 ∂ u2 , , , ∂ x1 ∂ x2 ∂ x1 ∂ x2

T .

Next,   det L(1) = (λ + 2μ )ν1 − μν1 ν12 + (λ + 2μ )ν2 ν1 (−ν2 )   − μν2 − μν1 (−ν2 )ν1 + (λ + 2μ )ν2(−ν2 )2     + μν2 λ ν2 ν12 − μν1 (−ν2 )ν1 − λ ν1 λ ν2 ν1 (−ν2 ) − μν1 (−ν2 )2   = −μ (λ + 2 μ )(ν14 + ν24 ) + − (λ + 2μ )2 − μ 2 + (λ + μ )2 ν12 ν22 = −μ (λ + 2 μ )(ν14 + ν24 ) + (−2λ μ − 4 μ 2 )ν12 ν22 = −μ (λ + 2 μ )(ν14 + 2ν12 ν22 + ν24 ) = −μ (λ + 2 μ )(ν12 + ν22 )2 = −μ (λ + 2μ ), which, because of the restrictions (1.14) on the elastic constants, is nonzero. Hence, by (A.7), d (1) = 0 on ∂ S2 ; that is,

∂ u1 ∂ u 1 ∂ u 2 ∂ u2 = = = = 0 on ∂ S 2 . ∂ x1 ∂ x2 ∂ x1 ∂ x2 Similarly, by (A.3), the fact that u 1 = u2 = 0 on ∂ S 2 , and (A.6), for u 3 we obtain (1) (1)

L3 d3 = 0 on ∂ S2 , where (1) L3

 ν1 ν2 = , −ν2 ν1

(1) d3

=

∂ u3 ∂ u3 , ∂ x1 ∂ x2

(A.8) T .

Since (1)

detL3 = ν12 + ν22 = 1, (1)

from (A.8) we deduce that d 3 = 0 on ∂ S2 ; hence,

∂ u3 ∂ u3 = = 0 on ∂ S 2 . ∂ x1 ∂ x2 Suppose now that all the derivatives of the components of u of all orders up to and including n = 2l − 1 are zero on ∂ S 2 . We claim that this implies that the derivatives of orders n = 2l and n = 2l + 1 are also zero on ∂ S 2 . We start with the case n = 2l. Since u satisfies (4.1) in D ∪ ∂ S 2 , we have

A Proof of Lemma 8.1

h2 μΔ u1 + h2 (λ + μ ) h2 ( λ + μ )

207

∂ 2 u1 ∂ 2 u2 ∂ u3 2 + h ( λ + μ ) + h2 μ k32 u1 − μ = 0, 2 ∂ x1 ∂ x2 ∂ x1 ∂ x1

(A.9)

∂ 2 u1 ∂ 2 u2 ∂ u3 + h2 μΔ u2 + h2 (λ + μ ) 2 + h2 μ k32 u2 − μ = 0 (A.10) ∂ x1 ∂ x2 ∂ x2 ∂ x2

in D ∪ ∂ S2 . Applying Δ l−1 to (A.9) and (A.10) and recalling that the derivatives of orders n = 2l − 2 and n = 2l − 1 are assumed to be zero on ∂ S 2 , we obtain   2l





 2l  l ∂ u1 l ∂ 2l u1 l ∂ 2l u1 l ∂ u1 μ + + ···+ + 2l 2l−2 2l−2 2 2 0 ∂ x1 1 ∂ x1 ∂ x2 l − 1 ∂ x1 ∂ x2 l ∂ x2l 2   2l

 l − 1 ∂ u1 l−1 ∂ 2l u1 + (λ + μ ) + + ··· 0 1 ∂ x2l ∂ x2l−2 ∂ x22 1 1



  l−1 ∂ 2l u1 l −1 ∂ 2l u1 + + l − 2 ∂ x41 ∂ x2l−4 l − 1 ∂ x21 ∂ x2l−2 2 2  

 l −1 ∂ 2l u2 l −1 ∂ 2l u2 + (λ + μ ) + + ··· 0 1 ∂ x2l−1 ∂ x2 ∂ x2l−3 ∂ x32 1 1



  l−1 ∂ 2l u2 l −1 ∂ 2l u2 + + =0 l − 2 ∂ x31 ∂ x2l−3 l − 1 ∂ x1 ∂ x2l−1 2 2 and

 

 l−1 ∂ 2l u1 l−1 ∂ 2l u1 (λ + μ ) + + ··· 0 1 ∂ x2l−1 ∂ x2 ∂ x2l−3 ∂ x32 1 1



  l−1 ∂ 2l u1 l −1 ∂ 2l u1 + + l − 2 ∂ x31 ∂ x2l−3 l − 1 ∂ x1 ∂ x2l−1 2 2   2l





 2l  l ∂ u2 l ∂ 2l u2 l ∂ 2l u2 l ∂ u2 +μ + + ···+ + 2l 2l−2 2l−2 2 2 0 ∂ x1 1 ∂ x1 ∂ x2 l − 1 ∂ x1 ∂ x2 l ∂ x2l 2  

 l −1 ∂ 2l u2 l −1 ∂ 2l u2 + (λ + μ ) + + ··· 0 1 ∂ x2l−2 ∂ x22 ∂ x2l−4 ∂ x42 1 1



  l−1 ∂ 2l u2 l − 1 ∂ 2l u2 + + =0 l − 2 ∂ x21 ∂ x2l−2 l − 1 ∂ x2l 2 2

on ∂ S2 . Also, since the derivatives of order n = 2l − 1 are assumed to be zero on ∂ S2 , it follows that

 ∂ ∂ 2l−1 u1 ∂ 2l u1 ∂ 2l u1 = −ν2 2l + ν1 2l−1 = 0 on ∂ S2 , 2l−1 ∂ s ∂ x1 ∂ x1 ∂ x1 ∂ x2

2l−1  ∂ ∂ u1 ∂ 2l u1 ∂ 2l u1 = −ν2 2l−1 + ν1 2l−2 2 = 0 on ∂ S2 , 2l−2 ∂ s ∂ x1 ∂ x2 ∂ x1 ∂ x2 ∂ x1 ∂ x2 .. .

208

A Proof of Lemma 8.1

∂ ∂s ∂ ∂s



∂ 2l−1 u1 ∂ x2l−1 2 ∂ 2l−1 u2 ∂ x2l−1 1

 = − ν2

∂ 2l u1 ∂ 2l u1 + ν1 2l = 0 on ∂ S2 , 2l−1 ∂ x2 ∂ x1 ∂ x2

= − ν2

∂ 2l u2 ∂ 2l u2 + ν = 0 on ∂ S2 , 1 ∂ x2l ∂ x2l−1 ∂ x2 1 1



2l−1  ∂ ∂ u2 ∂ 2l u2 ∂ 2l u2 = −ν2 2l−1 + ν1 2l−2 2 = 0 on ∂ S2 , 2l−2 ∂ s ∂ x1 ∂ x2 ∂ x1 ∂ x2 ∂ x1 ∂ x2 .. .

 ∂ ∂ 2l−1 u2 ∂ 2l u2 ∂ 2l u2 = − ν2 + ν1 2l = 0 on ∂ S2 . 2l−1 2l−1 ∂ s ∂ x2 ∂ x2 ∂ x1 ∂ x2 We define

 ⎧  l l−1 ⎪ ⎪ + (λ + μ ) , 0 ≤ n ≤ l − 1, ⎨μ n n (l)

 bn = ⎪ l ⎪ ⎩μ , n = l, l

 l−1 (l) cn = (λ + μ ) , 0 ≤ n ≤ l − 1, n ⎧  l ⎪ ⎪ n = 0, ⎨μ 0 , (l)



 dn = ⎪ l l−1 ⎪ ⎩μ + (λ + μ ) , 1 ≤ n ≤ l. n n−1

(A.11)

Then we arrive at a system of 4l + 2 equations in 4l + 2 unknowns, namely, L(2l) d (2l) = 0 on ∂ S2 , where L

(2l)

(A.12)

is the matrix ⎛

(l)

b0

⎜ ⎜ 0 ⎜ ⎜ ⎜−ν2 ⎜ ⎜ 0 ⎜ ⎜ .. ⎜ . ⎜ ⎜ ⎜ 0 ⎜ ⎜ 0 ⎜ ⎜ 0 ⎜ ⎜ ⎜ 0 ⎜ ⎜ . ⎜ .. ⎜ ⎝ 0 0

(l)

(l)

0 b1 · · ·

0 bl

0

(l)

(l)

(l)

c0 ν1 − ν2 .. . 0 0 0 0 .. . 0 0

· · · cl−1 ··· 0 ··· 0 . . .. . . .. . ν1 0 0 · · · −ν 2 0 ··· 0

0 0 ν1 .. .

0 ··· .. .. . . 0 ··· 0 ···

0 d0 0 0 0 0 .. .. . .

0 0 ν1 0 0 −ν 2

0 .. .

0 .. .

0 .. .

0 0

0 0

0 0

(l)

c0

···

(l)

0 cl−1 0

⎞

⎟ (l) (l) 0 · · · dl−1 0 dl ⎟ ⎟ ⎟ 0 ··· 0 0 0 ⎟ ⎟ 0 ··· 0 0 0 ⎟ ⎟ .. .. .. .. .. ⎟ . . . . . ⎟ ⎟ ⎟ 0 ··· 0 0 0 ⎟ ⎟ 0 ··· 0 0 0 ⎟ ⎟ ν1 · · · 0 0 0 ⎟ ⎟ ⎟ .. −ν 2 . 0 0 0 ⎟ ⎟ .. . . . . .. .. ⎟ . . . . . ⎟ ⎟ 0 · · · −ν 2 ν1 0 ⎠ 0 · · · 0 −ν 2 ν1

A Proof of Lemma 8.1

209

and d

(2l)

=

∂ 2l u1 ∂ 2l u1 ∂ 2l u1 ∂ 2l u2 ∂ 2l u2 ∂ 2l u2 , , . . . , , , , . . . , ∂ x2l ∂ x2l ∂ x2l ∂ x2l ∂ x2l−1 ∂ x2 ∂ x2l−1 ∂ x2 1 2 1 2 1 1

T .

Obviously, if L(2l) is invertible, then d (2l) = 0 on ∂ S 2 . By direct computation, det L(2l)  (l)  (l) (l) (l) (l) = b0 d0 ν14l + d1 ν14l−2 (−ν2 )2 + · · · + dl−1ν12l+2 (−ν2 )2l−2 + dl ν12l (−ν2 )2l  (l)  (l) (l) (l) + b1 d0 ν14l−2 (−ν2 )2 + d1 ν14l−4 (−ν2 )4 + · · · + dl ν12l−2 (−ν2 )2l+2  (l)  (l) (l) (l) + · · · + bl−1 d0 ν12l+2 (−ν2 )2l−2 + d1 ν12l (−ν2 )2l + · · · + dl ν12 (−ν2 )4l−2  (l)  (l) (l) (l) + bl d0 ν12l (−ν2 )2l + d1 ν12l−2 (−ν2 )2l+2 + · · · + dl (−ν2 )4l  (l)  (l) (l) (l) + c0 − c0 ν14l−2 (−ν2 )2 − c1 ν14l−4 (−ν2 )4 − · · · − cl−1ν12l (−ν2 )2l  (l)  (l) (l) (l) + c1 − c0 ν14l−4 (−ν2 )4 − c1 ν14l−6 (−ν2 )6 − · · · − cl−1ν12l−2 (−ν2 )2l+2  (l)  (l) (l) (l) + · · · + cl−2 − c0 ν12l+2 (−ν2 )2l−2 − c1 ν12l (−ν2 )2l − · · · − cl−1 ν14 (−ν2 )4l−4  (l)  (l) (l) (l) + cl−1 − c0 ν12l (−ν2 )2l − c1 ν12l−2 (−ν2 )2l+2 − · · · − cl−1 ν12 (−ν2 )4l−2 , which yields (2l)

det L

(l) (l) = b0 d0 ν14l

+

l−1 r

+∑

r=1

l

∑

l−1 r

r=1

s=0

(l) (l) bs dl−s −

s=0

+∑

∑

∑

s=0

(l) (l) bs dr−s −

r−1

∑

(l) (l) cs cr−1−s



ν14l−2r ν22r

s=0 l−1

∑

(l) (l) cs cl−1−s

 ν12l ν22l

s=0 (l) (l) bl−r+s dl−s −

r−1

∑

(l) (l) cl−r+s cl−1−s

 (l) (l) ν12r ν24l−2r + bl dl ν24l .

s=0

Using (A.11), we find that detL(2l)  

 

 

 l l −1 l 4l l l l−1 = μ + (λ + μ ) μ ν1 + μ μ + (λ + μ ) ν24l 0 0 0 l l l−1

 

 l−1  r−1   l l−1 l l−1 +∑ ∑ μ + (λ + μ ) μ + (λ + μ ) s s r−s r−1−s r=1 s=0

  l −1 l−1 − (λ + μ )2 s r−1−s  

  l l−1 l + μ + (λ + μ ) μ ν14l−2r ν22r r r 0

210

A Proof of Lemma 8.1

 l−1  

 

 l l−1 l l−1 μ + ( λ + μ ) μ + ( λ + μ ) ∑ s s l−s l −1−s s=0

 

  l −1 l−1 l l − (λ + μ )2 +μ μ ν12l ν22l s l−1−s l 0 

 l−1  r−1  l l −1 +∑ ∑ μ + (λ + μ ) l −r+s l−r+s r=1 s=0  

 l l−1 × μ + (λ + μ ) l −s l−1−s

  l −1 l −1 − (λ + μ )2 l −r+s l −1−s

 

 l l l −1 +μ μ + (λ + μ ) ν12r ν24l−2r ; l l−r l −1−r +

that is, det L(2l) = μ (λ + 2μ )ν14l  r   r−1   l−1  l −1 l l l−1 + ∑ μ (λ + μ ) ∑ +∑ s r−s r−1−s r=1 s=0 s=0 s  r  l l + μ2 ∑ ν14l−2r ν22r s r − s s=0      l−1  l −1 l l l −1 + μ (λ + μ ) ∑ + s l−s s l −1−s s=0  l  l l + μ2 ∑ ν12l ν22l s l − s s=0  r   l−1  l−1 l + ∑ μ (λ + μ ) ∑ l − (r − s) r=1 s=0 l − 1 − s

    r−1 r l l−1 l l +∑ + μ2 ∑ ν12r ν24l−2r l − s l − (r − s) l − s l − (r − s) s=0 s=0 + μ (λ + 2 μ )ν24l . Next, r

∑

s=0

  r−1   l−1 l l l−1 +∑ s r−s r−1−s s=0 s

 

  r−1   r l −1 l l −1 l l l−1 = +∑ +∑ 0 r s r−s r−1−s s=1 s=0 s

A Proof of Lemma 8.1

211

    r−1   r   l l l l −1 l l l−1 +∑ − +∑ 0 r s s−1 r−s r−1−s s=1 s=0 s

 

 

  r r r−1 l l l −1 l l l −1 =∑ −∑ +∑ ; r−s r−s r−1−s s=0 s s=1 s − 1 s=0 s =

hence,

  r−1    r  l−1 l l l −1 l l +∑ =∑ , ∑ s r−s r−1−s r−s s=0 s=0 s s=0 s r

which also implies that r

∑

s=0

  r−1   l−1 l l l −1 +∑ l − 1 − s l − (r − s) l − (r − s) s=0 l − s   r l l =∑ . l − (r − s) s=0 l − s

Similarly, l−1 

∑

s=0

    l −1 l l l −1 + s l −s s l−1−s

    l−1   l −1 l l l −1 l−1 l = + +∑ 0 l 0 l −1 s l −s s=1

  l−1 l l−1 +∑ l −1−s s=1 s

    l−1     l l l l l l −1 l = + +∑ − 0 l l 0 s s−1 l −s s=1

  l−1 l l−1 +∑ l −1−s s=1 s

 

 

  l l−1 l−1 l l l −1 l l l−1 =∑ −∑ +∑ ; l −s l−s l−1−s s=0 s s=1 s − 1 s=1 s

that is, l−1 

∑

s=0

Therefore,

     l  l−1 l l l−1 l l + =∑ . s l −s s l−1−s l−s s=0 s l−1  r

detL(2l) = μ (λ + 2μ ) ∑

r=0

  l l ∑ s r − s ν14l−2r ν22r s=0

212

A Proof of Lemma 8.1

  l l ν12l ν22l s l − s s=0   l−1  r l l + μ (λ + 2 μ ) ∑ ∑ ν12r ν24l−2r , l − s l − (r − s) r=0 s=0 l

+ μ (λ + 2 μ ) ∑

which means that det L(2l)

 





  l 2l l 2l−2 2 l l 2l 2 2l−2 = μ (λ + 2 μ ) ν + ν ν2 + · · · + ν ν + ν 0 1 1 1 l−1 1 2 l 2  





  l 2l l 2l−2 2 l l 2l × ν + ν ν2 + · · · + ν 2 ν 2l−2 + ν . 0 1 1 1 l −1 1 2 l 2

From this we deduce that  2 detL(2l) = μ (λ + 2 μ ) (ν12 + ν22 )l = μ (λ + 2 μ ) = 0. Hence, by (A.12),

∂ 2l u1 ∂ 2l u1 ∂ 2l u2 ∂ 2l u2 = = · · · = = = 0 on ∂ S 2 . ∂ x2l ∂ x2l ∂ x2l−1 ∂ x2 ∂ x1 ∂ x2l−1 1 2 1 2 We now consider the derivatives of order n = 2l of u 3 . From the system of equations we find that ∂ u 1 ∂ u2 + + Δ u3 + k 2 u3 = 0 (A.13) ∂ x1 ∂ x2 in D ∪ ∂ S2 . Applying Δ l−1 to (A.13) and bearing in mind that the derivatives of the components of u of orders n = 2l − 2 and n = 2l − 1 are assumed to be zero on ∂ S 2 , we arrive at Δ l u3 = 0 on ∂ S 2 , which means that

 2l

 l ∂ u3 l ∂ 2l u3 + 2l 0 ∂ x1 1 ∂ x2l−2 ∂ x22 1



 2l l ∂ 2l u3 l ∂ u3 + ···+ + = 0 on ∂ S 2 . 2l−2 2 l − 1 ∂ x1 ∂ x2 l ∂ x2l 2 Also, on ∂ S 2 we have

 ∂ ∂ 2l−1 u3 ∂ 2l u3 ∂ 2l u3 = − ν + ν = 0, 2 1 ∂ s ∂ x2l−1 ∂ x2l ∂ x2l−1 ∂ x2 1 1 1

2l−1  ∂ ∂ u3 ∂ 2l u3 ∂ 2l u3 = −ν2 2l−1 + ν1 2l−2 2 = 0, 2l−2 ∂ s ∂ x1 ∂ x2 ∂ x1 ∂ x2 ∂ x1 ∂ x2

A Proof of Lemma 8.1

.. .

∂ ∂s

213

∂ 2l−1 u3 ∂ x2l−1 2

 = − ν2

∂ 2l u3 ∂ 2l u3 + ν = 0. 1 ∂ x2l ∂ x1 ∂ x2l−1 2 2

Hence, we end up with a system of 2l + 1 equations in 2l + 1 unknowns written in the form (2l) (2l) L3 d3 = 0 on ∂ S2 , (A.14)

 ⎛  l l 0 ⎜ 0 1 ⎜ ⎜ − ν2 ν1 0 ⎜ ⎜ 0 − ν2 ν1 ⎜ ⎜ 0 0 −ν 2 ⎜ (2l) .. .. .. L3 = ⎜ ⎜ . . . ⎜ ⎜ ⎜ 0 0 0 ⎜ ⎜ 0 0 0 ⎜ ⎝ 0 0 0 0 0 0

where

and (2l)

d3

=

0 ···

0

··· 0 ··· 0 ··· 0 . . .. . . .. 0 . ν1 0 · · · −ν 2 0 ··· 0 0 ··· 0

0 0 ν1 .. .

 l l −1 0 0 0 .. . 0 ν1 −ν 2 0

0 0 0 0 .. . 0 0 ν1 −ν 2

⎞ l l ⎟ ⎟ 0 ⎟ ⎟ 0 ⎟ ⎟ 0 ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ 0 ⎟ ⎟ 0 ⎟ ⎟ 0 ⎠ ν1

∂ 2l u3 ∂ 2l u3 ∂ 2l u3 ∂ 2l u3 , 2l−1 , ..., , 2l 2l−1 ∂ x1 ∂ x1 ∂ x2 ∂ x2l ∂ x1 ∂ x2 2

T .

It is easily seen that (2l)

det L3





 l l l 2l−1 2l−2 = ν1 ν1 + (−ν2 )(−ν2 )ν1 + (−ν2 )(−ν2 )3 ν12l−4 0 1 2



 l l 2l−3 2 + ···+ (−ν2 )(−ν2 ) ν1 + (−ν2 )(−ν2 )2l−1 ; l−1 l

that is,

(2l)

det L3

= (ν12 + ν22 )l = 1.

Therefore, (A.14) implies that

∂ 2l u3 ∂ 2l u3 ∂ 2l u3 ∂ 2l u3 = 2l−1 = ··· = = = 0 on ∂ S 2 . 2l 2l−1 ∂ x1 ∂ x2l ∂ x1 ∂ x2 ∂ x1 ∂ x2 2 We now go over to the derivatives of order n = 2l + 1. We apply Δ l−1 to (A.9) and (A.10) and take the normal derivative of each equation on ∂ S 2 , approaching the limit from D. Since the derivatives of order n = 2l have been shown to be zero on ∂ S2 , this yields

214

A Proof of Lemma 8.1 (l)

b 0 ν1

∂ 2l+1 u1 ∂ 2l+1 u1 ∂ 2l+1 u1 (l) (l) + b0 ν2 2l + b1 ν1 2l−1 2 2l+1 ∂ x1 ∂ x2 ∂ x1 ∂ x1 ∂ x2 (l)

+ · · · + b l ν1 (l)

+ c 0 ν1

2l+1 u ∂ 2l+1 u1 (l) ∂ 1 + b ν 2 l 2l 2l+1 ∂ x1 ∂ x2 ∂ x2

∂ 2l+1 u2 ∂ 2l+1 u2 ∂ 2l+1 u2 (l) (l) + c0 ν2 2l−1 2 + c1 ν1 2l−2 3 2l ∂ x1 ∂ x2 ∂ x1 ∂ x2 ∂ x1 ∂ x2 (l)

+ · · · + cl−1 ν1

∂ 2l+1 u2 ∂ 2l+1 u2 (l) + c ν = 0 on ∂ S2 2 l−1 ∂ x1 ∂ x2l ∂ x21 ∂ x2l−1 2 2

and (l)

c0 ν1

∂ 2l+1 u1 ∂ 2l+1 u1 ∂ 2l+1 u1 (l) (l) + c ν + c ν 2 1 0 1 ∂ x2l ∂ x2l−1 ∂ x22 ∂ x2l−2 ∂ x32 1 ∂ x2 1 1 (l)

+ · · · + cl−1 ν1 (l)

+ d 0 ν1

∂ 2l+1 u1 ∂ 2l+1 u1 (l) + c ν 2 l−1 2l−1 ∂ x1 ∂ x2l ∂ x21 ∂ x2 2

∂ 2l+1 u2 ∂ 2l+1 u2 ∂ 2l+1 u2 (l) (l) + d0 ν2 2l + d1 ν1 2l−1 2 2l+1 ∂ x1 ∂ x2 ∂ x1 ∂ x1 ∂ x2 (l)

+ · · · + d l ν1

2l+1 u ∂ 2l+1 u2 (l) ∂ 2 + d ν = 0 on ∂ S 2 . 2 l 2l 2l+1 ∂ x1 ∂ x2 ∂ x2

By taking the tangential derivative on ∂ S 2 of the derivatives of order n = 2l, we obtain the usual 4l +2 equations involving the derivatives of order n = 2l +1. Hence, we arrive at a system of 4l + 4 equations in 4l + 4 unknowns written in the form L(2l+1) d (2l+1) = 0 on ∂ S 2 , where L(2l+1) is the matrix ⎛ (l) ⎞ (l) (l) (l) (l) (l) b0 ν1 b0 ν2 · · · bl ν1 bl ν2 0 c0 ν1 · · · cl−1 ν2 0 ⎜ ⎟ ⎜ 0 c(l) ν1 · · · c(l) ν2 0 d (l) ν1 d (l) ν2 · · · d (l) ν1 d (l) ν2 ⎟ 0 0 0 l−1 l l ⎜ ⎟ ⎜ − ν2 ν1 · · · 0 0 0 0 ··· 0 0 ⎟ ⎜ ⎟ ⎜ .. .. .. .. .. .. .. .. ⎟ .. .. ⎜ . . . . . . . . . . ⎟ ⎜ ⎟ ⎜ ⎟ . . ⎜ 0 . ν1 0 0 0 0 ··· 0 0 ⎟ ⎜ ⎟ ⎜ 0 0 · · · − ν2 ν1 0 0 ··· 0 0 ⎟ ⎜ ⎟ ⎜ 0 0 ··· 0 0 − ν 2 ν1 · · · 0 0 ⎟ ⎜ ⎟ ⎜ ⎟ .. ⎜ 0 0 ··· 0 0 0 −ν 2 . 0 0 ⎟ ⎜ ⎟ ⎜ . .. .. .. .. .. .. . . .. ⎟ .. ⎝ .. . . . . . . . . . ⎠ 0 0 ··· 0 0 0 0 · · · −ν 2 ν1 and

(A.15)

A Proof of Lemma 8.1

d (2l+1) =

215

∂ 2l+1 u1 ∂ 2l+1 u1 ∂ 2l+1 u2 ∂ 2l+1 u2 , ..., , , ..., 2l+1 2l+1 2l+1 ∂ x1 ∂ x2 ∂ x1 ∂ x2l+1 2

T .

As before, to show that d (2l+1) = 0 on ∂ S 2 we need to prove that L (2l+1) is invertible. By direct calculation, we find that detL(2l+1) (l)  (l) (l) (l) = b0 ν1 − d0 ν1 ν14l+2 + d0 ν2 ν14l+1 (−ν2 ) − d1 ν1 ν14l (−ν2 )2 (l)

(l)

+ · · · − dl ν1 ν12l+2 (−ν2 )2l + dl ν2 ν12l+1 (−ν2 )2l+1

(l)  (l) (l) − b0 ν2 − d0 ν1 ν14l+1 (−ν2 ) + d0 ν2 ν14l (−ν2 )2 (l)

(l)

− · · · − dl ν1 ν12l+1 (−ν2 )2l+1 + dl ν2 ν12l (−ν2 )2l+2

(l)  (l) (l) + · · · + bl ν1 − d0 ν1 ν12l+2 (−ν2 )2l + d0 ν2 ν12l+1 (−ν2 )2l+1 (l)

(l)

− · · · − dl ν1 ν12 (−ν2 )4l + dl ν2 ν1 (−ν2 )4l+1

(l)  (l) (l) − bl ν2 − d0 ν1 ν12l+1 (−ν2 )2l+1 + d0 ν2 ν12l (−ν2 )2l+2 (l)

(l)

− · · · − dl ν1 ν1 (−ν2 )4l+1 + dl ν2 (−ν2 )4l+2

 





(l)  (l) (l) − c0 ν1 − c0 ν1 ν14l (−ν2 )2 + c0 ν2 ν14l−1 (−ν2 )3 (l)

(l)

− · · · − cl−1ν1 ν12l+2 (−ν2 )2l + cl−1 ν2 ν12l+1 (−ν2 )2l+1

(l)  (l) (l) + c0 ν2 − c0 ν1 ν14l−1 (−ν2 )3 + c0 ν2 ν14l−2 (−ν2 )4 (l)

(l)

− · · · − cl−1ν1 ν12l+1 (−ν2 )2l+1 + cl−1 ν2 ν12l (−ν2 )2l+2

 (l) (l) (l) − · · · − cl−1 ν1 − c0 ν1 ν12l+2 (−ν2 )2l + c0 ν2 ν12l+1 (−ν2 )2l+1 (l)

(l)

− · · · − cl−1ν1 ν14 (−ν2 )4l−2 + cl−1ν2 ν13 (−ν2 )4l−1

 



 (l) (l) (l) + cl−1 ν2 − c0 ν1 ν12l+1 (−ν2 )2l+1 + c0 ν2 ν12l (−ν2 )2l+2

 (l) (l) − · · · − cl−1ν1 ν13 (−ν2 )4l−1 + cl−1ν2 ν12 (−ν2 )4l ,

which means that detL(2l+1)  (l)  (l) (l) = − b0 d0 ν14l (ν14 + 2ν12 ν22 + ν24 ) + · · · + dl ν12l ν22l (ν14 + 2ν12 ν22 + ν24 ) (l)  (l) − b1 d0 ν14l−2 ν22 (ν14 + 2ν12 ν22 + ν24 )  (l) + · · · + dl ν12l−2 ν22l+2 (ν14 + 2ν12 ν22 + ν24 ) (l)  (l) − · · · − bl−1 d0 ν12l+2 ν22l−2 (ν14 + 2ν12 ν22 + ν24 ) (l)

+ · · · + dl ν12 ν24l−2 (ν14 + 2ν12 ν22 + ν24 )]

 (l)  (l) 2l 2l 4 (l) d0 ν1 ν2 (ν1 + 2ν12 ν22 + ν24 ) + · · · + dl ν24l (ν14 + 2ν12 ν22 + ν24 )

− bl

216

A Proof of Lemma 8.1 (l)  (l) + c0 c0 ν14l−2 ν22 (ν14 + 2ν12ν22 + ν24 )

 (l) + · · · + cl−1 ν12l ν22l (ν14 + 2ν12 ν22 + ν24 )

(l)  (l) + · · · + cl−1 c0 ν12l ν22l (ν14 + 2ν12 ν22 + ν24 )

 (l) + · · · + cl−1 ν12 ν24l−2 (ν14 + 2ν12 ν22 + ν24 ) ;

hence,

det L(2l+1) = −(ν12 + ν22 )2 detL(2l) ,

which implies that

det L(2l+1) = −μ (λ + 2μ ) = 0.

Consequently, from (A.15) we deduce that

∂ 2l+1 u1 ∂ 2l+1 u1 ∂ 2l+1 u2 ∂ 2l+1 u2 = 2l = ··· = = = 0 on ∂ S 2 . 2l+1 2l ∂ x1 ∂ x2 ∂ x1 ∂ x2 ∂ x1 ∂ x2l+1 2 It remains to show that the derivatives of order n = 2l + 1 of u 3 are zero on ∂ S 2 . Using the same procedure as for the first two components, from (A.13) we obtain the system of equations (2l+1) (2l+1) d3

L3 where

= 0 on ∂ S 2 ,

(A.16)

   ⎞ ⎛    l l l l l l ν1 ν2 ν1 · · · ν2 ν1 ν2 ⎜ 0 ⎟ 0 1 l − 1 l l ⎜ ⎟ ⎜ − ν2 ν1 0 ··· 0 0 0 ⎟ ⎜ ⎟ ⎜ 0 − ν2 ν1 · · · 0 0 0 ⎟ ⎜ ⎟ (2l+1) .. .. .. .. ⎟ L3 =⎜ .. .. ⎜ ... . . . . . . ⎟ ⎜ ⎟ ⎜ ⎟ .. ⎜ 0 . 0 0 ν1 0 0 ⎟ ⎜ ⎟ ⎝ 0 0 0 ··· −ν2 ν1 0 ⎠ 0 0 0 ··· 0 −ν 2 ν1

and (2l+1)

d3

=

∂ 2l+1 u3 ∂ 2l+1 u3 ∂ 2l+1 u3 ∂ 2l+1 u3 , , ..., , 2l+1 ∂ x2l ∂ x ∂ x1 ∂ x2l ∂ x1 ∂ x2l+1 1 2 2 2

T .

Direct computation now yields (2l+1)

det L3





 l l l = ν1 ν12l+1 − ν2 (−ν2 )ν12l + ν1 (−ν2 )2 ν12l−1 0 0 1



 l l 2l − ···+ ν1 (−ν2 ) ν1 − ν2 (−ν2 )2l+1 l l

A Proof of Lemma 8.1

217

 





  l 2l l 2l−2 2 l l 2l 2 2 2 2l−2 = (ν1 + ν2 ) ν1 + ν1 ν2 + · · · + ν1 ν2 + ν 0 1 l −1 l 2  l+1 = ν12 + ν22 = 1; therefore, (A.16) implies that

∂ 2l+1 u3 ∂ 2l+1 u3 ∂ 2l+1 u3 ∂ 2l+1 u3 = = · · · = = = 0 on ∂ S 2 . ∂ x2l ∂ x1 ∂ x2l ∂ x2l+1 ∂ x2l+1 1 ∂ x2 2 1 2 By induction, we deduce that the derivatives of the components of u of any order are zero on ∂ S 2 and, using the power series expansion of u, we conclude that u is zero in a nonempty neighborhood of any point on ∂ S 2 . Since u is an analytic solution of (4.1) throughout D, we can continue the above process to prove that u = 0 in D.

Appendix B (σ )

The Matrix Um

(σ )

We compute the elements of the matrix U m arising in Subsection 8.2.1. The ele(σ ) ments of Uˆ m are computed in the same way, with H m replaced by Jm . From (1.13), (8.14), and (8.25) we find that 

(σ ) (σ )   T Φm + K Φm 1 R=a  ∂ 2 (σ ) ∂2 (σ ) 2 = h (λ + 2μ )α1 (cos θ ) 2 φm + 2 μα1(sin θ ) φm ∂ x1 ∂ x2 ∂ x1  ∂ 2 (σ ) ∂ (σ )  + λ α1 (cos θ ) 2 φm + μκα1 φm  ∂ x1 ∂ x2 R=a    √ (σ ) = εm h2 α1 λ (cos θ )Δ Hm (k1 R)Em (θ )

   ∂2 ∂2 (σ ) + 2μ (cos θ ) 2 + (sin θ ) Hm (k1 R)Em (θ ) ∂ x1 ∂ x2 ∂ x1    ∂  (σ ) + μκ Hm (k1 R)Em (θ )  , ∂ x1 R=a

so, since (cos θ )

∂2 ∂2 ∂2 1 ∂2 1 ∂ + (sin θ ) = (cos θ ) 2 − (sin θ ) + (sin θ ) , 2 ∂ x1 ∂ x2 ∂R R ∂ R∂ θ R2 ∂θ ∂ x1

we see that  (σ ) (σ )   T Φm + K Φm 1 R=a   (σ ) √ ∂2  (σ ) 2 = εm h α1 − λ k12 Hm (k1 R)(cos θ )Em (θ ) + 2 μ (cos θ ) 2 Hm (k1 R) Em (θ ) ∂R  ∂  (σ )  2μ ∂  − (sin θ ) Hm (k1 R) Em ( θ ) R ∂R ∂θ 219

(σ )

220

B The Matrix Um

2μ ∂  (σ )  (sin θ )Hm (k1 R) Em ( θ ) R2 ∂θ  (σ ) ∂  + μκ (cos θ ) Hm (k1 R) Em (θ ) ∂R  μκ ∂  (σ )   − (sin θ )Hm (k1 R) Em (θ )  , R ∂θ R=a +

which means that  (σ ) (σ )   T Φm + K Φm 1 R=a  √ (σ ) (σ ) = εm h2 α1 − λ k12 Hm (k1 a)(cos θ )Em (θ ) + 2μ k12 Hm (k1 a)(cos θ )Em (θ ) 2μ mk1 (3−σ ) (−1)σ Hm (k1 a)(sin θ )Em (θ ) a 2μ m (3−σ ) + 2 (−1)σ Hm (k1 a)(sin θ )Em (θ ) a (σ ) + μκ k1Hm (k1 a)(cos θ )Em (θ )  μκ m (3−σ ) σ − (−1) Hm (k1 a)(sin θ )Em (θ ) . a −

Similarly,  (σ ) (σ )   T Φm + K Φm 2 R=a    √ (σ ) = εm h2 α1 λ (sin θ )Δ Hm (k1 R)Em (θ )

  ∂2 ∂2  (σ ) + 2μ cos θ + (sin θ ) 2 Hm (k1 R)Em (θ ) ∂ x1 ∂ x2 ∂x  2    ∂ (σ ) + μκ Hm (k1 R)Em (θ )  . ∂ x2 R=a Taking into account the fact that (cos θ )

∂2 ∂2 ∂2 1 ∂2 1 ∂ + (sin θ ) 2 = (sin θ ) 2 + (cos θ ) − (cos θ ) , ∂ x1 ∂ x2 ∂R R ∂ R∂ θ R2 ∂θ ∂ x2

we have  (σ ) (σ )   T Φm + K Φm 2 R=a   (σ ) √ ∂2  (σ ) 2 = εm h α1 − λ k12 Hm (k1 R)(sin θ )Em (θ ) + 2 μ (sin θ ) 2 Hm (k1 R) Em (θ ) ∂R  ∂  (σ )  2μ ∂  + (cos θ ) Hm (k1 R) Em (θ ) R ∂R ∂θ

(σ )

B The Matrix Um

221

2μ ∂  (σ )  (cos θ )Hm (k1 R) Em (θ ) R2 ∂θ  (σ ) ∂  + μκ (sin θ ) Hm (k1 R) Em (θ ) ∂R  μκ ∂  (σ )   + (cos θ )Hm (k1 R) Em ( θ )  , R ∂θ R=a −

which leads to  (σ ) (σ )   T Φm + K Φm 2 R=a  √ (σ ) (σ ) = εm h2 α1 − λ k12 Hm (k1 a)(sin θ )Em (θ ) + 2μ k12 Hm (k1 a)(sin θ )Em (θ ) 2μ mk1 (3−σ ) (−1)σ Hm (k1 a)(cos θ )Em (θ ) a 2μ m (3−σ ) − 2 (−1)σ Hm (k1 a)(cos θ )Em (θ ) a +

(σ )

+ μκ k1Hm (k1 a)(sin θ )Em (θ )

 μκ m (3−σ ) σ + (−1) Hm (k1 a)(cos θ )Em (θ ) . a Also, 

(σ ) (σ )   T Φm + K Φm 3 R=a 

  √ ∂ ∂  (σ ) = εm μ (α1 − hk3 α2 ) (cos θ ) + (sin θ ) Hm (k1 R)Em (θ ) ∂ x1 ∂ x2   (σ ) − μκ hk3α2 Hm (k1 R)Em (θ )  ; R=a

therefore, since (cos θ )

∂ ∂ ∂ + (sin θ ) = , ∂ x1 ∂ x2 ∂ R

we arrive at 

(σ ) (σ )   T Φm + K Φm 3 R=a  √  (σ ) (σ ) = εm μ (α1 − hk3 α2 )k1 Hm (k1 a)Em (θ ) − μκ hk3α2 Hm (k1 a)Em (θ ) .

Using (8.15) instead of (8.14), we can easily compute the corresponding terms (σ ) involving ϒm . From (8.16) we have

(σ )

222

B The Matrix Um

(σ ) (σ )   T Ψm + KΨm 1 R=a    √ ∂2 ∂2 ∂2  (σ ) 2 = εm h μ 2(cos θ ) + (sin θ ) 2 − (sin θ ) 2 Hm (k3 R)Em (θ ) ∂ x1 ∂ x2 ∂ x2 ∂ x1    ∂  (σ ) +κ Hm (k3 R)Em (θ )  , ∂ x2



R=a

which, given that 2(cos θ )

∂2 ∂2 ∂2 + (sin θ ) 2 − (sin θ ) 2 ∂ x1 ∂ x2 ∂ x2 ∂ x1

= (sin θ )

∂2 1 ∂ 2 ∂2 2 ∂ 1 ∂2 − (sin θ ) + (cos θ ) − (cos θ ) − (sin θ ) , ∂ R2 R ∂R R ∂ R∂ θ R2 ∂ θ R2 ∂θ2

leads to  (σ ) (σ )   T Ψm + KΨm 1 R=a   (σ )  (σ ) √ ∂2  1 ∂  = εm h2 μ (sin θ ) 2 Hm (k3 R) Em (θ ) − (sin θ ) Hm (k3 R) Em (θ ) ∂R R ∂R  ∂  (σ )  2 ∂  + (cos θ ) Hm (k3 R) Em (θ ) R ∂R ∂θ 2 ∂  (σ )  − 2 (cos θ )Hm (k3 R) Em (θ ) R ∂θ 1 ∂ 2  (σ )  − 2 (sin θ )Hm (k3 R) 2 Em (θ ) R ∂θ  (σ ) ∂  + κ (sin θ ) Hm (k3 R) Em (θ ) ∂R  κ ∂  (σ )   + (cos θ )Hm (k3 R) Em ( θ )  , R ∂θ R=a and we arrive at  (σ ) (σ )   T Ψm + KΨm 1 R=a  √ k3 (σ ) (σ ) 2 = εm h μ k32 Hm (k3 a)(sin θ )Em (θ ) − Hm (k3 a)(sin θ )Em (θ ) a 2mk3 (3−σ ) + (−1)σ Hm (k3 a)(cos θ )Em (θ ) a 2m (3−σ ) − 2 (−1)σ Hm (k3 a)(cos θ )Em (θ ) a m2 (σ ) (σ ) + 2 Hm (k3 a)(sin θ )Em (θ ) + κ k3Hm (k3 a)(sin θ )Em (θ ) a  κm (3−σ ) + (−1)σ Hm (k3 a)(cos θ )Em (θ ) . a

(σ )

B The Matrix Um

223

In the same way, (σ ) (σ )   T Ψm + KΨm 2 R=a     √ ∂2 ∂2 ∂2 (σ ) = εm h2 μ (cos θ ) 2 − (cos θ ) 2 − 2(sin θ ) Hm (k3 R)Em (θ ) ∂ x1 ∂ x2 ∂ x2 ∂ x1    ∂  (σ ) −κ Hm (k3 R)Em (θ )  , ∂ x1 R=a



where, in polar coordinates, (cos θ )

∂2 ∂2 ∂2 − (cos θ ) 2 − 2(sin θ ) 2 ∂ x1 ∂ x2 ∂ x2 ∂ x1 = −(cos θ )

∂2 1 ∂ 2 ∂2 + (cos θ ) + (sin θ ) ∂ R2 R ∂R R ∂ R∂ θ 2 ∂ 1 ∂2 − 2 (sin θ ) + 2 (cos θ ) 2 . R ∂θ R ∂θ

Therefore, 

(σ ) (σ )   T Ψm + KΨm 2 R=a   (σ )  (σ ) √ ∂2  1 ∂  2 = εm h μ − (cos θ ) 2 Hm (k3 R) Em (θ ) + (cos θ ) Hm (k3 R) Em (θ ) ∂R R ∂R  ∂  (σ )  2 ∂  + (sin θ ) Hm (k3 R) Em ( θ ) R ∂R ∂θ 2 ∂  (σ )  − 2 (sin θ )Hm (k3 R) Em ( θ ) R ∂θ 1 ∂ 2  (σ )  + 2 (cos θ )Hm (k3 R) 2 Em (θ ) R ∂θ  (σ ) ∂  − κ (cos θ ) Hm (k3 R) Em (θ ) ∂R  κ ∂  (σ )   + (sin θ )Hm (k3 R) Em (θ )  , R ∂θ R=a

from which we see that  (σ ) (σ )   T Ψm + KΨm 2 R=a  √ k3 (σ ) (σ ) = εm h2 μ − k32 Hm (k3 a)(cos θ )Em (θ ) + Hm (k3 a)(cos θ )Em (θ ) a 2mk3 (3−σ ) + (−1)σ Hm (k3 a)(sin θ )Em (θ ) a 2m (3−σ ) − 2 (−1)σ Hm (k3 a)(sin θ )Em (θ ) a

(σ )

224

B The Matrix Um

m2 (σ ) (σ ) Hm (k3 a)(cos θ )Em (θ ) − κ k3Hm (k3 a)(cos θ )Em (θ ) a2  κm (3−σ ) + (−1)σ Hm (k3 a)(sin θ )Em (θ ) . a −

Finally, since (cos θ )

∂ ∂ 1 ∂ − (sin θ ) = , ∂ x2 ∂ x1 R ∂ θ

we find that  √ μm (σ ) (σ )   (3−σ ) T Ψm + KΨm 3 R=a = εm (−1)σ Hm (k3 a)Em (θ ). a

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Index

A

F

analytic solution 109 asymptotic behavior 118, 132 averaging operators 2

far-field estimates 49 Fredholm alternative 76, 97, 150, 159, 175 frequency 2

B Bessel function 115, 136 Betti formula 65 boundary integral equations 76, 89 boundary integral operators 76 boundary value problems 52, 68 C complete set of wavefunctions 201 composition formulas 88, 149, 168 constitutive relations 1 D direct method 87, 153, 167 Dirichlet problem 52, 62, 142, 149, 195 displacements 2 double-layer potential 21, 179 modified 104, 111, 135, 151

G Green’s tensor

62, 67, 156

H Hankel function 9, 105, 114 Helmholtz equation 48, 88, 124, 141 Helmholtz operator 9 hypersingular operator 180 I index of a system 186 indirect method 75, 153, 159, 178 integral equation 73, 157, 167 composite 100, 172 of the first kind 87, 147, 149 of the second kind 75, 87, 103, 112, 194 quasi-Fredholm 75, 191 J Jones modification

E eigenfrequency 65, 77, 79, 95, 156, 163 spectrum 61, 153 eigenfunction 77 eigensolution 65, 77, 94, 170 elastodynamics, two-dimensional 109, 124, 141, 180 equations of motion 1

133, 174

K kinematic assumption

3

L Lam´e constants 1, 178 layer potentials 7, 20 logarithmic singularity 26 229

230

Index

M matrix of fundamental solutions 7, 12, 156, 194 Mindlin-type displacements 2 modified fundamental solution 103, 149, 173, 194 modified integral equations 144 modified potential 111, 135, 142 moment–stress operator 77, 163 N Neumann problem 52, 67, 138, 151, 196 Newtonian potential 23 null field equations 191

regular solution 4, 66, 81, 154, 167, 178 Rellich’s lemma 156 representation formula 53, 155, 167 Robin problem 153, 196 S single-layer potential 20, 179 modified 111, 135 singular integral equations 76 singular integral operator 184 solenoidal component 48 solvability 75 conditions 67, 69, 82 Somigliana representation formula stationary oscillations 2 symbol matrix 184

53, 56

O T orthogonality relations

135 transmission problem

177, 197

P U potential component principal value 30

48 uniqueness of solution 47, 153, 154 Ursell modification 108, 173

R W radiation conditions 47, 110, 179 reciprocity relation 5, 109, 139, 154

wavefunctions

7, 104, 120, 143, 201