Variational and Potential Methods for a Class of Linear Hyperbolic Evolutionary Processes

Springer Monographs in Mathematics Igor Chudinovich Christian Constanda Variational and Potential Methods for a Clas...

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Springer Monographs in Mathematics

Igor Chudinovich

Christian Constanda

Variational and Potential Methods for a Class of Linear Hyperbolic Evolutionary Processes

Igor Chudinovich, MS, PhD, DSc Professor of Mathematics, Faculty of Mechanical, Electrical, and Electronic Engineering The University of Guanajuato, Salamanca, GTO, Mexico Christian Constanda, MS, PhD, DSc Charles W. Oliphant Professor of Mathematical Sciences, The University of Tulsa 600 South College Avenue, Tulsa, Oklahoma 74104, USA

British Library Cataloguing in Publication Data Chudinovich, Igor Variational and potential methods for a class of linear hyperbolic evolutionary processes. – (Springer monographs in mathematics) 1. Plates (Engineering) – Mathematical models 2. Boundary element methods 3. Differential equations, Hyperbolic 4. Differential equations, Linear I. Title II. Constanda, C. (Christian) 515.3′535 ISBN 1852338881 Library of Congress Cataloging-in-Publication Data CIP data available.

Mathematics Subject Classification (2000): 35C15; 35D05; 35E05; 35L15; 35L20; 35Q72; 45F15; 74H20; 74H25; 74K20 Apart from any fair dealing for the purposes of research or private study, or criticism or review, as permitted under the Copyright, Designs and Patents Act 1988, this publication may only be reproduced, stored or transmitted, in any form or by any means, with the prior permission in writing of the publishers, or in the case of reprographic reproduction in accordance with the terms of licences issued by the Copyright Licensing Agency. Enquiries concerning reproduction outside those terms should be sent to the publishers. Springer Monographs in Mathematics ISSN 1439-7382 ISBN 1-85233-888-1 Springer is a part of Springer Science+Business Media springeronline.com © Springer-Verlag London Limited 2005 Printed in the United States of America The use of registered names, trademarks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant laws and regulations and therefore free for general use. The publisher makes no representation, express of implied, with regard to the accuracy of the information contained in this book and cannot accept any legal responsibility or liability for any errors or omissions that may be made. Typesetting: Camera-ready by the authors 12/3830-543210 Printer on acid-free paper SPIN 10991749

For Olga and Lia and the younger generation Genia and Dan

Preface

Variational and boundary integral equation techniques are two of the most useful methods for solving time-dependent problems described by systems of equations of the form ∂2u = Au, ∂t2 where u = u(x, t) is a vector-valued function, x is a point in a domain in R2 or R3 , and A is a linear elliptic differential operator. To facilitate a better understanding of these two types of methods, below we propose to illustrate their mechanisms in action on a specific mathematical model rather than in a more impersonal abstract setting. For this purpose, we have chosen the hyperbolic system of partial differential equations governing the nonstationary bending of elastic plates with transverse shear deformation. The reason for our choice is twofold. On the one hand, in a certain sense this is a “hybrid” system, consisting of three equations for three unknown functions in only two independent variables, which makes it more unusual—and thereby more interesting to the analyst—than other systems arising in solid mechanics. On the other hand, this particular plate model has received very little attention compared to the so-called classical one, based on Kirchhoff’s simplifying hypotheses, although, as acknowledged by practitioners, it represents a substantial refinement of the latter and therefore needs a rigorous discussion of the existence, uniqueness, and continuous dependence of its solution on the data before any construction of numerical approximation algorithms can be contemplated. The first part of our analysis is conducted by means of a procedure that is close in both nature and detail to a variational method, and which, for this reason, we also call variational. Once the results have been established in the general setting of Sobolev spaces, we carry out the second part of the study by seeking useful, closed-form integral representations of the solutions in terms of dynamic (retarded) plate potentials.

VIII

Preface

The problems discussed in this book include those with Dirichlet and Neumann boundary conditions (corresponding, in particular, to the clamped-edge and free-edge plate), with elastic (Robin), mixed, and combined displacementtraction (simply supported edge) boundary data, transmission (contact) problems, problems for plates with homogeneous inclusions, plates with cracks, and plates on a generalized elastic foundation. For each of them, the variational version is formulated and its solvability is examined in spaces of distributions; subsequently, the solutions are found in the form of time-dependent single-layer and double-layer potentials with distributional densities that satisfy nonstationary integral equations. The analysis technique consists in using the Laplace transformation to reduce the original problems to boundary value problems depending on the transformation parameter, and on establishing estimates for the solutions of the latter that allow conclusions to be drawn about the existence and properties of the solutions to the given initialboundary value problems. The transformed problems are solved by means of specially constructed algebras of singular integral operators defined by the boundary values of the transformed potentials. The distributional setting has the advantage over the classical one in that it enables the method to be applied in less smooth domains—for example, in regions with corners and cuts. Furthermore, Sobolev-type norms are particularly useful in the global error analysis of numerical schemes, but such analysis falls outside the scope of this book and we do not pursue it. To the authors’ knowledge, this is the first time that so many typical initial-boundary value problems have been considered in the same book for a model in conjunction with both variational and boundary integral equation methods. The text provides full details of the proofs and is aimed at researchers interested in the use of applied analysis as a tool for investigating mathematical models in mechanics. The presentation assumes no specialized knowledge beyond a basic understanding of functional analysis and Sobolev spaces. We want to emphasize that the book does not intend to explain the mechanical background of plate theory. Details of that nature and a fuller discussion of the limitations of the model that we have chosen as our object of study can be found in the article J.R. Cho and J.T. Oden, A priori modeling error estimate of hierarchical models for elasticity problems for plate and shell-like structures, Math. Comput. Modelling 23 (1996), 117–133. Ours is a purely mathematical that aims to acquaint the interested reader with two of the most powerful and general techniques of solution for this type of linear problem. We reiterate that the theory of bending of plates with transverse shear deformation has been selected merely as an application vehicle because of its unusual features and lack of previous strict mathematical treatment. The book is a natural complement to our earlier monograph [7], where we investigated the corresponding equilibrium problems.

Preface

IX

Some of the results discussed below have already been announced in concise form in the literature (see [4]–[6] and [8]). The authors would like to acknowledge help and support received from various quarters during the preparation of this book. I.C. wishes to thank his former colleagues in the Mathematical Physics and Computational Mathematics section of the Department of Mathematics and Mechanics at Kharkov National University, and his current colleagues in the Department of Mechanical, Electrical, and Electronic Engineering of the University of Guanajuato in Salamanca—in particular, Drs. Igor Chueshov, Arturo Lara Lopez, and René Jaime Rivas, for playing an instrumental role in arranging his move to Mexico. C.C. wishes to thank Dr. Bill Coberly and his other colleagues at the University of Tulsa for a departmental atmosphere that has proved highly conducive to the writing of mathematical books. Last but by no means least, we would like to place on record the debt of gratitude that we owe our wives, sine quibus non, who have guided us wisely, patiently, and selflessly, by word and by deed, to exciting and challenging new shores. Igor Chudinovich

Christian Constanda

Professor of Mathematics University of Guanajuato Salamanca, Mexico

C.W. Oliphant Professor of Mathematical Sciences University of Tulsa, USA April 2004

Contents

Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

vii

Formulation of the Problems and Their Nonstationary Boundary Integral Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 The Initial-Boundary Value Problems . . . . . . . . . . . . . . . . . . . . . 1.2 A Matrix of Fundamental Solutions . . . . . . . . . . . . . . . . . . . . . . . 1.3 Time-dependent Plate Potentials . . . . . . . . . . . . . . . . . . . . . . . . . 1.4 Nonstationary Boundary Integral Equations . . . . . . . . . . . . . . .

1 1 8 13 16

2

Problems with Dirichlet Boundary Conditions . . . . . . . . . . . . 2.1 Function Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Solvability of the Transformed Problems . . . . . . . . . . . . . . . . . . 2.3 Solvability of the Time-dependent Problems . . . . . . . . . . . . . . .

19 19 21 28

3

Problems with Neumann Boundary Conditions . . . . . . . . . . . 3.1 The Poincaré–Steklov Operators . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Solvability of the Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

37 37 42

4

Boundary Integral Equations for Problems with Dirichlet and Neumann Boundary Conditions . . . . . . . . . . . . . . . . . . . . . . 4.1 Time-dependent Potentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Nonstationary Boundary Integral Equations . . . . . . . . . . . . . . . 4.3 The Direct Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

43 43 51 53

Transmission Problems and Multiply Connected Plates . . . 5.1 Infinite Plate with a Finite Inclusion . . . . . . . . . . . . . . . . . . . . . . 5.2 Multiply Connected Finite Plate . . . . . . . . . . . . . . . . . . . . . . . . . 5.3 Finite Plate with an Inclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . .

57 57 64 75

5

XII

Contents

6

Plate Weakened by a Crack . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1 Formulation and Solvability of the Problems . . . . . . . . . . . . . . . 6.2 The Poincaré–Steklov Operator . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3 Time-dependent Potentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4 Infinite Plate with a Crack . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.5 Finite Plate with a Crack . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

7

Initial-Boundary Value Problems with Other Types of Boundary Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 7.1 Mixed Boundary Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 7.2 Combined Boundary Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . 108 7.3 Elastic Boundary Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114

8

Boundary Integral Equations for Plates on a Generalized Elastic Foundation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1 Formulation and Solvability of the Problems . . . . . . . . . . . . . . . 8.2 A Matrix of Fundamental Solutions . . . . . . . . . . . . . . . . . . . . . . . 8.3 Properties of the Boundary Operators . . . . . . . . . . . . . . . . . . . . . 8.4 Solvability of the Boundary Equations . . . . . . . . . . . . . . . . . . . .

119 119 121 126 127

Problems with Nonhomogeneous Equations and Nonhomogeneous Initial Conditions . . . . . . . . . . . . . . . . . 9.1 The Time-dependent Area Potential . . . . . . . . . . . . . . . . . . . . . . 9.2 The Nonhomogeneous Equation of Motion . . . . . . . . . . . . . . . . . 9.3 Initial Potentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

129 129 131 133

9

A

81 81 85 87 91 93

The Fourier and Laplace Transforms of Distributions . . . . . 139

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147

1 Formulation of the Problems and Their Nonstationary Boundary Integral Equations

1.1 The Initial-Boundary Value Problems All problem statements in this chapter are formal; rigorous versions will be presented after the introduction of the necessary function spaces. Below we consider initial-boundary value problems for the time-dependent homogeneous equations of the model with homogeneous initial data. In Chapter 9, we indicate how the general case can be reduced to the homogeneous one. By an elastic plate we understand an elastic body that occupies a region S¯ × [−h0 /2, h0 /2] in R3 , where S is a domain in R2 bounded by a simple closed curve ∂S and 0 < h0 = const diam S is called the thickness. Throughout the book we use the following notation and conventions. Unless otherwise specified, Greek and Latin subscripts and superscripts in all formulas take the values 1, 2 and 1, 2, 3, respectively, and summation over repeated indices is adopted. The standard inner product in R3 is (a, b) = ai bi . A generic point in R2 referred to a Cartesian system of coordinates in the middle plane x3 = 0 of the plate is written as x = (x1 , x2 ). X = (x, t), where t is the time variable. Partial derivatives are denoted by ∂α = ∂/∂xα and ∂t = ∂/∂t . A superscript T denotes matrix transposition. A superscript ∗ denotes conjugation and transposition of a complex matrix. The columns of a matrix M are denoted by M (i) . Both matrix-valued functions and scalar functions are simply referred to as functions. If Y is a space of scalar functions and g is a matrix-valued function, then g ∈ Y means that each entry of g belongs to Y. A three-component vector q = (q1 , q2 , q3 )T may be written alternatively as q = (¯ q T , q3 )T , where q¯ = (q1 , q2 )T . S + is the finite domain enclosed by ∂S, and S − = R2 \ (S + ∪ ∂S).

2

1 Formulation of the Problems

The boundary ∂S is a C 2 -curve with a uniquely defined outward (with respect to S + ) normal n = (n1 , n2 )T . We write G = S × (0, ∞),

G± = S ± × (0, ∞), +

Γ = ∂S × (0, ∞).

−

If ϕ is a smooth function defined in S (S ), then ϕ+ (ϕ− ) denotes the limiting value (if it exists) of ϕ as its argument tends to ∂S from within S + (S − ). If ϕ is not smooth but has a trace on ∂S, then the latter is denoted by γ + ϕ (γ − ϕ). Since there is no danger of ambiguity, the notation remains the same for functions defined in G+ (G− ) and their limiting values (traces) on the boundary Γ. The operators of restriction from R2 (or S + ∪S − ) to S ± , or from R2 ×(0, ∞) (or G+ ∪ G− ) to G± , are denoted by π ± . Operators of extension from ∂S to S ± , or from Γ to G± , are denoted by ± l , respectively. ∆ is the Laplacian and δij is the Kronecker delta. L and L−1 are, respectively, the Laplace transformation with respect to t, and its inverse. The Laplace transform of a function u(x, t) is denoted by u ˆ(x, p), where p is the transformation parameter. Other notation will be introduced as the need arises. Suppose that the material is homogeneous and isotropic, of density ρ and Lamé constants λ and µ, which satisfy the inequalities [9] λ + µ > 0,

µ > 0,

ρ > 0.

(1.1)

If we denote by tij , εij , vi , and fi , respectively, the components of the stress tensor, deformation tensor, displacement vector, and body force vector, then the behavior of the plate as a three-dimensional elastic body under prescribed initial and boundary conditions is governed by three main groups of equations, namely (see [14] and [17]), the kinematic formulas εij = 12 (∂i vj + ∂j vi );

(1.2)

the stress-strain relations (generalized Hooke’s law)

the equations of motion

tij = λεkk δij + 2µεij ;

(1.3)

∂j tij + fi = ρ∂t2 vi .

(1.4)

In addition, ti = tij nj are the components of the stress vector on ∂S.

1.1 The Initial-Boundary Value Problems

3

The model of bending of plates with transverse shear deformation that we intend to study here postulates a displacement field of the form vα (x, x3 , t) = x3 uα (X),

(1.5)

v3 (x, x3 , t) = u3 (X). This assumption is valid only for plates whose ratio of thickness to diameter falls within a certain range (see the Preface). Expressions (1.5) and the geometry of the plate suggest a way of simplifying equations (1.2)–(1.4). This is done by means of a well-known procedure that involves the use of the averaging operators Iα and Jα , α = 0, 1, defined by α x3 =h0 /2 (Iα g)(X) = h−1 x3 g(x, x3 , t) x =−h /2 , 0 3

(Jα g)(X) =

h−1 0

h 0 /2

0

xα 3 g(x, x3 , t) dx3 .

−h0 /2

Specifically, setting Nαβ = J1 tαβ , N3α = J0 t3α , qα = J1 fα + I1 tα3 , q3 = J0 f3 + I0 t33 , h2 = h20 /12, system (1.4) yields the plate equations of motion ∂β Nαβ − N3α + qα = ρh2 ∂t2 uα , ∂α N3α + q3 = ρ∂t2 u3 .

(1.6)

Also, from (1.2), (1.3), and (1.5), we obtain the plate constitutive relations Nαβ = h2 λ(∂γ uγ )δαβ + µ(∂α uβ + ∂β uα ) ,

(1.7)

N3α = µ(∂α u3 + uα ). Finally, substituting (1.7) into (1.6) leads to the alternative equations of motion B(∂t2 u)(X) + (Au)(X) = q(X), X ∈ G+ or X ∈ G− , (1.8) where

B = diag{ρh2 , ρh2 , ρ},

4


A is the matrix differential operator with entries [9] Aαα = −h2 µ∆ − h2 (λ + µ)∂α2 − µ (α not summed), A33 = −µ∆, A12 = A21 = −h2 (λ + µ)∂1 ∂2 , Aα3 = −A3α = µ∂α , and

u = (¯ uT , u3 )T , q = (¯ q T , q3 )T .

It is easily verified that, under conditions (1.1), A is a strongly elliptic operator and satisfies G˚ arding’s inequality [18]. The quantities Nαβ and Nα3 are the averages across the thickness of the plate of the bending and twisting moments with respect to the middle plane x3 = 0, and of the transverse shear forces [9]; qα and q3 are combinations of the body moments and forces and of the moments and forces acting on the faces x3 = ±h0 /2. Similarly, setting Nα = J1 tα , N3 = J0 t3 , we obtain N1 = h2 (λ∂α uα + 2µ∂1 u1 )n1 + µ(∂1 u2 + ∂2 u1 )n2 , N2 = h2 µ(∂1 u2 + ∂2 u1 )n1 + (λ∂α uα + 2µ∂2 u2 )n2 , N3 = µ(∂α u3 + uα )nα , which can be written as Ni = (T u)i , where T is the matrix boundary operator with entries T11 = h2 (λ + 2µ)n1 ∂1 + µn2 ∂2 , T22 = h2 (λ + 2µ)n2 ∂2 + µn1 ∂1 , T33 = µnα ∂α , T12 = h2 (λn1 ∂2 + µn2 ∂1 ), T21 = h2 (µn1 ∂2 + λn2 ∂1 ), T3α = µnα , Tα3 = 0. From what has been said above, it is obvious that T u is the vector of the averaged moments and shear force acting on the lateral part ∂S × (−h0 /2, h0 /2)


5

of the boundary. The vector u is referred to as the displacement vector since it characterizes the latter uniquely in terms of the assumption (1.5). In Chapters 2–8, we deal almost exclusively with the homogeneous equation (1.8), that is, B(∂t2 u)(X) + (Au)(X) = 0,

X ∈ G+ or X ∈ G− .

(1.9)

To (1.9) we adjoin appropriate boundary conditions and homogeneous initial conditions. The functions occurring on the right-hand side in all the boundary conditions below are prescribed. The symbolic name of each problem that we consider starts with a “D” to indicate that it is a dynamic problem. The remaining letters are fairly obvious initials related to the problem type and/or boundary condition type. Thus, the classical interior and exterior problems (DD± ) with Dirichlet boundary conditions consist, respectively, in finding functions u ∈ C 2 (G± ) ∩ ¯ ± ) such that C 1 (G B(∂t2 u)(X) + (Au)(X) = 0,

X ∈ G± ,

u(x, 0+) = (∂t u)(x, 0+) = 0,

x ∈ S±,

u± (X) = f (X),

X ∈ Γ.

In the interior and exterior initial boundary-value problems (DN± ) with ¯ ± ) of Neumann boundary conditions, we seek solutions u ∈ C 2 (G± ) ∩ C 1 (G B(∂t2 u)(X) + (Au)(X) = 0,

X ∈ G± ,

u(x, 0+) = (∂t u)(x, 0+) = 0,

x ∈ S±,

(T u)± (X) = g(X),

X ∈ Γ.

Consider two open arcs ∂S1 and ∂S2 of ∂S such that mes(∂Sα ) > 0, ∂S 1 ∪ ∂S 2 = ∂S, ∂S1 ∩ ∂S2 = ∅. The interior and exterior initial-value problems (DM± ) with mixed boundary ¯ ± ) satisfying conditions consist in finding u ∈ C 2 (G± ) ∩ C 1 (G B(∂t2 u)(X) + (Au)(X) = 0,

X ∈ G± ,

u(x, 0+) = (∂t u)(x, 0+) = 0,

x ∈ S±,

u± (X) = f (X), (T u)± (X) = g(X),

X ∈ ∂S1 × (0, ∞), X ∈ ∂S2 × (0, ∞).

6


In the interior and exterior initial-boundary value problems (DC± 1 ) with combined boundary conditions of the first kind, we look for u ∈ C 2 (G± ) ∩ ¯ ± ) such that C 1 (G B(∂t2 u)(X) + (Au)(X) = 0,

X ∈ G± ,

u(x, 0+) = (∂t u)(x, 0+) = 0,

x ∈ S±,

u± 3 (X) = f3 (X),

X ∈ Γ,

(T u)± α (X) = gα (X),

X ∈ Γ.

If the boundary conditions are of the second kind, then the solution u ∈ ¯ ± ) satisfies C 2 (G± ) ∩ C 1 (G B(∂t2 u)(X) + (Au)(X) = 0,

X ∈ G± ,

u(x, 0+) = (∂t u)(x, 0+) = 0,

x ∈ S±,

u± α (X) = fα (X), (T u)± 3 (X)

X ∈ Γ, X ∈ Γ.

= g3 (X),

If the regions S ± × [−h0 /2, h0 /2] are occupied by two different elastic materials with Lamé constants λ± , µ± and densities ρ± , respectively, then the initial-boundary value problem (DT) with transmission (contact) boundary ¯ ± ) such conditions consists in finding a pair of functions u± ∈ C 2 (G± ) ∩ C 1 (G that B± (∂t2 u± )(X) + (A± u± )(X) = 0, X ∈ G± , u± (x, 0+) = (∂t u± )(x, 0+) = 0, − u+ + (X) − u− (X) = f (X),

x ∈ S±,

X ∈ Γ,

(T+ u+ )+ (X) − (T− u− )− (X) = g(X),

X ∈ Γ,

where A± , B± , and T± have the obvious meaning. Consider an open arc ∂S0 of ∂S that models a crack, and let Ω = R2 \ ∂S 0 , ∂S1 = ∂S \ ∂S 0 , G = Ω × (0, ∞), Γi = ∂Si × (0, ∞),

i = 0, 1.

¯ k = 0, 1, 2, . . . , if the restrictions u± of u to G± are, We write u ∈ C k (G), ¯ ± ) and the limiting values on Γ1 of u+ and all its respectively, of class C k (G derivatives up to the order k coincide with those of u− . (These values may differ on Γ0 .) In the initial-boundary value problem (DKD) with Dirichlet ¯ satisfying boundary conditions, we seek u ∈ C 2 (G) ∩ C 1 (G)


B(∂t2 u)(X) + (Au)(X) = 0,

X ∈ G,

u(x, 0+) = (∂t u)(x, 0+) = 0,

x ∈ Ω,

u+ + (X)

+

= f (X),

X ∈ Γ0 ,

− u− − (X) = f (X),

X ∈ Γ0 .

7

The problem (DKN) with Neumann boundary conditions consists in find¯ such that ing u ∈ C 2 (G) ∩ C 1 (G) B(∂t2 u)(X) + (Au)(X) = 0,

X ∈ G,

u(x, 0+) = (∂t u)(x, 0+) = 0,

x ∈ Ω,

+

+

(T u+ ) (X) = g (X),

X ∈ Γ0 ,

(T u− )− (X) = g − (X),

X ∈ Γ0 .

Let K be a (3 × 3)-matrix of the form ¯ K 0 , K= 0 k33 ¯ = h2 (kαβ ) is positive definite. In where k33 > 0 and the (2 × 2)-matrix K the interior and exterior initial-boundary value problems (DD± K ) for a plate on an elastic foundation with Dirichlet boundary conditions, we look for u ∈ ¯ ± ) such that C 2 (G± ) ∩ C 1 (G B(∂t2 u)(X) + (Au)(X) + Ku(X) = 0, u(x, 0+) = (∂t u)(x, 0+) = 0, u± (X) = f (X),

X ∈ G± ,

x ∈ S±,

X ∈ Γ.

The corresponding problems (DN± K ) with Neumann boundary conditions con¯ ± ) satisfying sist in finding functions u ∈ C 2 (G± ) ∩ C 1 (G B(∂t2 u)(X) + (Au)(X) + Ku(X) = 0, u(x, 0+) = (∂t u)(x, 0+) = 0, (T u)± (X) = g(X),

X ∈ G± ,

x ∈ S±,

X ∈ Γ.

Throughout what follows, we work frequently with the Laplace transforms of vector-valued functions u(X) = u(x, t), t ∈ R, which vanish for t < 0; that is, ∞ u ˆ(x, p) = Lu(x, t) = e−pt u(x, t) dt. 0

8


This equality is understood either in the classical or in the distributional sense (see [2], [16], and the Appendix), as the case may be. To simplify the notation, functions or distributions in spaces of transforms will not carry a superposed hat in their symbols when they occur in the general analysis of such spaces; the hat will be added only when there is explicit mention that they are the Laplace transforms of solutions or data occurring in time-dependent problems, or densities of transformed nonstationary (retarded) plate potentials. We give a brief indication of the technique used to solve the initial-boundary value problems listed above. Roughly speaking, we adopt the following procedure. (i) Applying the Laplace transformation in (1.9) and taking the homogeneous initial conditions into account, we obtain the equation Bp2 u ˆ(x, p) + (Aˆ u)(x, p) = 0,

x ∈ S + or x ∈ S − .

Doing the same to the various boundary conditions, we arrive at boundary value problems whose solutions depend on p. (ii) We prove the unique solvability of the boundary value problems constructed in (i) for any value of p in the complex half-plane Cκ = {p = σ + iτ ∈ C : σ > κ},

κ > 0,

and derive estimates that show how the solutions depend on the complex parameter p. (iii) Using Parseval’s equality, we then return to the spaces of originals and prove the existence of weak solutions to the given initial-boundary value problems. A similar scheme will also be used to study the solvability of the associated time-dependent boundary integral equations.

1.2 A Matrix of Fundamental Solutions To construct single-layer and double-layer potentials in the dynamic case, we need a matrix of fundamental solutions D(X) = D(x, t) for (1.9), which vanishes for t < 0. In fact, it turns out that we only need the Laplace ˆ transform D(x, p) of such a matrix because, as shown in [15], for this type of problem numerical methods can be designed that are based solely on estimates ˆ and do not require explicit knowledge of D itself. for D ˆ Consequently, we seek a (3 × 3)-matrix D(x, p) defined for p ∈ C0 , which has polynomial growth as p → ∞ and satisfies ˆ ˆ Bp2 D(x, p) + (AD)(x, p) = δ(x)I,

(1.10)

1.2 A Matrix of Fundamental Solutions

9

where δ is the Dirac distribution and I is the identity (3 × 3)-matrix. To ˆ find D(x, p), we apply the (generalized) Fourier transformation with respect to x ∈ R2 in (1.10) and arrive at ˜ p) + A(ξ)D ˜ (ξ, p) = I, Bp2 D(ξ, or ˜ p) = I, Θ(ξ, p)D(ξ, (1.11) ˜ p) is the Fourier transform of D(x, ˆ where D(ξ, p) and Θ(ξ, p) is the (3 × 3)matrix of elements Θαβ (ξ, p) = h2 (λ + µ)ξα ξβ + δαβ (ρh2 p2 + µ + h2 µ|ξ|2 ), Θ33 (ξ, p) = ρp2 + µ|ξ|2 ,

(1.12)

Θα3 (ξ, p) = −Θ3α (ξ, p) = iµξα . A straightforward calculation yields det Θ(ξ, p)

= h4 µ2 (λ + 2µ)|ξ|6 + h2 µ ρp2 h2 (2λ + 5µ) + µ(λ + 2µ) |ξ|4 + ρp2 h2 (λ + 4µ)(ρp2 h2 + µ)|ξ|2 + ρp2 (ρp2 h2 + µ)2 .

From this it follows that det Θ(ξ, p) is invariant with respect to rotations in R2 ; that is, det Θ(ξ, p) depends only on |ξ|. Then we take ξ1 = |ξ| and ξ2 = 0 in (1.12) and find that det Θ(ξ, p) = R(|ξ|2 , p) 2 2 0 −iµ|ξ| ρp h + µ + h2 (λ + 2µ)|ξ|2 = 0 ρp2 h2 + µ + h2 µ|ξ|2 0 iµ|ξ| 0 ρp2 + µ|ξ|2 = (h2 µ|ξ|2 + ρp2 h2 + µ) h2 µ(λ + 2µ)|ξ|4 + ρp2 h2 (λ + 3µ)|ξ|2 + ρp2 (ρp2 h2 + µ) . Let |ξ|2 = s. We denote the roots of the equation R(s, p) = 0 by si = −χ2i and choose χi so that Re χi ≥ 0. It is easy to check that −1 χ21,2 = p 2hµ(λ + 2µ) 1/2

× ρph(λ + 3µ) ± ρ2 p2 h2 (λ + µ)2 − 4ρµ2 (λ + 2µ) , χ23 = (h2 µ)−1 (ρp2 h2 + µ).

10


1.1 Lemma. (i) The equation R(s, p) = 0 does not have a triple root for any p ∈ C0 . (ii) The equation R(s, p) = 0 has a double root if and only if −1 −1 1/2 p = 2µ h(λ + µ) ρ (λ + 2µ) . In this case, χ21 = χ22 . (iii) Re χi > 0 for every p ∈ C0 . Proof. (i) Suppose that χ21 = χ22 = χ23 . Then the explicit expressions of the roots show that χ21 = χ22 implies that ρ2 p2 h2 (λ + µ)2 − 4ρµ2 (λ + 2µ) = 0; hence,

−1 −1 1/2 p = 2µ h(λ + µ) ρ (λ + 2µ) .

Now the equality χ21 = χ23 yields λ + µ = 0, which contradicts (1.1). (ii) If χ21 = χ23 or χ22 = χ23 , then, as a straightforward calculation shows, we obtain p2 = −µ(ρh2 )−1 , which is impossible, since Re p > 0. (iii) First, we note that χi = 0. If Re χi = 0 for some i, then χ2i < 0; ˜ 2 = −χ2 is a positive root of the equation R(s, p) = 0. We take hence, s˜ = |ξ| i ˜ ˜ ξ = (|ξ|, 0) and denote by g = (g1 , g2 , g3 )T a nonzero solution of the system of ˜ p)g = 0. Multiplying this equality by g ∗ , we linear algebraic equations Θ(ξ, obtain ˜ 2 )|g3 |2 (ρh2 p2 + µ)(|g1 |2 + |g2 |2 ) + (ρp2 + µ|ξ| ˜ Re(g ∗ ig1 ) = 0; ˜ 2 (λ + 2µ)|g1 |2 + µ|g2 |2 − 2µ|ξ| + h2 |ξ| 3 consequently, p2 ∈ R and p2 > 0. Since ˜ Re(g ∗ ig1 ) ≥ −µ(|g1 |2 + |ξ| ˜ 2 |g3 |2 ), 2µ|ξ| 3 it follows that ρp2 h2 (|g1 |2 + |g2 |2 ) + ρp2 |g3 |2

˜ 2 (λ + 2µ)|g1 |2 + µ|g2 |2 ≤ 0. + h2 |ξ|

This contradiction completes the proof. We write

˜ p) = det Θ(ξ, p) −1 . Ψ(ξ,


11

By (1.11) and (1.12), ˜ 11 (ξ, p) = h2 µ2 |ξ|4 + h2 µ(λ + µ)|ξ|2 ξ 2 + 2ρp2 h2 µ|ξ|2 + µ2 ξ 2 D 2 1 2 2 2 2 2 2 ˜ p), + ρp h (λ + µ)ξ2 + ρp (ρp h + µ) Ψ(ξ, 2 2 4 ˜ 22 (ξ, p) = h µ |ξ| + h2 µ(λ + µ)|ξ|2 ξ12 + 2ρp2 h2 µ|ξ|2 + µ2 ξ22 D ˜ p), + ρp2 h2 (λ + µ)ξ12 + ρp2 (ρp2 h2 + µ) Ψ(ξ, 4 ˜ 33 (ξ, p) = h µ(λ + 2µ)|ξ|4 + h2 (ρp2 h2 + µ)(λ + 3µ)|ξ|2 D ˜ p), + (ρp2 h2 + µ)2 Ψ(ξ, 2 ˜ 21 (ξ, p) = −ξ1 ξ2 h µ(λ + µ)|ξ|2 ˜ 12 (ξ, p) = D D ˜ p), + ρp2 h2 (λ + µ) − µ2 Ψ(ξ, ˜ 3α (ξ, p) = iµξα (h2 µ|ξ|2 + ρp2 h2 + µ)Ψ(ξ, ˜ p). ˜ α3 (ξ, p) = −D D ˆ Hence, the elements of the matrix of fundamental solutions D(x, p) are ˆ 11 (x, p) = h2 µ2 ∆2 + h2 µ(λ + µ)∆∂ 2 − 2ρp2 h2 µ∆ − µ2 ∂ 2 D 2 1 2 2 2 2 2 2 − ρp h (λ + µ)∂2 + ρp (ρp h + µ) Ψ(x, p), 2 2 2 ˆ 22 (x, p) = h µ ∆ + h2 µ(λ + µ)∆∂ 2 − 2ρp2 h2 µ∆ − µ2 ∂ 2 D 1 2 − ρp2 h2 (λ + µ)∂12 + ρp2 (ρp2 h2 + µ) Ψ(x, p), ˆ 33 (x, p) = h4 µ(λ + 2µ)∆2 − h2 (ρp2 h2 + µ)(λ + 3µ)∆ D + (ρp2 h2 + µ)2 Ψ(x, p), ˆ 21 (x, p) = − h2 µ(λ + µ)∆ ˆ 12 (x, p) = D D + ρp2 h2 (λ + µ) − µ2 ∂1 ∂2 Ψ(x, p),

(1.13)

ˆ α3 (x, p) = −D ˆ 3α (x, p) = µ(h2 µ∆ − ρp2 h2 − µ)∂α Ψ(x, p), D ˜ p); that is, where Ψ(x, p) is the inverse Fourier transform of Ψ(ξ, Ψ(x, p) = (4π 2 )−1

˜ p) dξ. e−i(x,ξ) Ψ(ξ,

R2

By Lemma 1.1, there are two possible cases. In the first one, the roots of the equation R(s, p) = 0 are simple. Direct calculation shows that in this case −1 Ψ(x, p) = 2πh4 µ2 (λ + 2µ) ci K0 (χi |x|),

(1.14)

12


where K0 is the modified Bessel function of order zero [1] and −1 c1 = (χ21 − χ22 )(χ21 − χ23 ) , −1 c2 = (χ22 − χ23 )(χ22 − χ21 ) , −1 c3 = (χ23 − χ21 )(χ23 − χ22 ) .

(1.15)

In the second case, the equation R(s, p) = 0 has a double root: χ21 = χ22 = χ23 . Then −1 Ψ(x, p) = 2πh4 µ2 (λ + 2µ) c˜1 K0 (χ1 |x|)

+ c˜2 (2χ1 )−1 |x|K1 (χ1 |x|) + c˜3 K0 (χ3 |x|) ,

(1.16)

where K1 is the modified Bessel function of order one and c˜1 = −˜ c3 = −(χ23 − χ21 )−2 ,

(1.17)

c˜2 = (χ23 − χ21 )−1 .

1.2 Lemma. For any p ∈ C0 , the function Ψ(x, p) can be represented in the neighborhood of x = 0 in the form −1 4 Ψ(x, p) = − 128πh4 µ2 (λ + 2µ) |x| ln |x| + O(|x|6 ln |x|) + Ψ0 (x, p),

(1.18)

where Ψ0 (x, p) is an infinitely differentiable function; in addition, Ψ(x, p) → 0 exponentially as |x| → ∞. Proof. In the case of simple roots, from (1.14) and the asymptotic behavior of the modified Bessel function K0 (z) as z → 0 [1] it follows that −1 ln |x| Ψ(x, p) = − 2πh4 µ2 (λ + 2µ) 1 2 2 + c1 + c2 + c3 + ci 4 χi |x| +

4 1 4 64 χi |x|

ln |x|

+ O(|x|6 ln |x|) + Ψ0 (x, p). Using (1.15), it is easy to verify that c1 + c2 + c3 = 0, ci χ2i = 0, ci χ4i = 1; therefore, (1.18) holds.

1.3 Time-dependent Plate Potentials

13

In the case of a double root, (1.16) implies that −1 Ψ(x, p) = − 2πh4 µ2 (λ + 2µ) ln |x| 1 4 × c˜1 1 + 14 χ21 |x|2 + 64 χ1 |x|4 − 14 c˜1 |x|2 1 + 18 χ21 |x|2 1 4 + c˜3 1 + 14 χ23 |x|2 + 64 χ3 |x|4 ln |x| + O(|x|6 ln |x|) + Ψ0 (x, p), and we immediately regain (1.18) from (1.17). The last assertion follows from the fact that Re χi > 0 and the asymptotic behavior of Kn (z), n = 0, 1, as z → ∞ in such a way that Re z ≥ κ > 0. 1.3 Corollary. For any p ∈ C0 , the elements of the matrix of fundamental ˆ solutions D(x, p) can be represented in the neighborhood of x = 0 in the form ˆ αβ (x, p) = 4πµ(λ + 2µ)h2 −1 D × (λ + µ)xα xβ |x|−2 − (λ + 3µ)δαβ ln |x| ˆ 0,αβ (x, p), + O(|x|2 ln |x|) + D ˆ 33 (x, p) = −(2πh2 µ)−1 ln |x| D

(1.19)

ˆ 0,33 (x, p), + O(|x|2 ln |x|) + D ˆ α3 (x, p) = −D ˆ 3α (x, p) = − 4πh2 (λ + 2µ) −1 xα ln |x| D ˆ 0,α3 (x, p), + O(|x|2 ln |x|) + D ˆ 0,ij (x, p) are infinitely differentiable functions. where D The proof of this assertion follows from (1.13) and (1.18). 1.4 Remark. Representation (1.19) shows that for any p ∈ C0 , the asymptotic ˆ behavior of D(x, p) in the neighborhood of x = 0 coincides with that of the matrix of fundamental solutions D(x) for the equilibrium equation [10].

1.3 Time-dependent Plate Potentials Let α, β ∈ C 2 (∂S ×R) be three-component vector-valued functions with com¯ First, we introduce single-layer and double-layer potentials pact support in Γ. in terms of Laplace transforms. Thus, if α ˆ (x, p) = Lα(x, t),

ˆ p) = Lβ(x, t), β(x,

14


then for every p ∈ C0 we define a single-layer potential Vp α ˆ of density α ˆ by ˆ − y, p)ˆ D(x α(y, p) dsy ,

(Vp α ˆ )(x, p) =

x ∈ R2 ,

∂S

and a double-layer potential Wp βˆ of density βˆ by ˆ (Wp β)(x, p) =

ˆ p) dsy Pˆ (x − y, p)β(y, ∂S

=

ˆ p), Ty D ˆ (j) (y − x, p) ej dsy , β(y,

x ∈ S+ ∪ S−,

∂S

where

ˆ − x, p) T , Pˆ (x − y, p) = Ty D(y

Ty is the moment-force boundary operator acting with respect to the point y, and ej is the jth coordinate unit vector in R3 . By Remark 1.4, for any fixed p ∈ C0 the boundary properties of both potentials coincide with those of the corresponding single-layer and doublelayer potentials in the equilibrium case [10]. Below we list the most significant of these properties. (i) The single-layer potential Vp α ˆ satisfies the equation ˆ(x, p) + (Aˆ u)(x, p) = 0, Bp2 u

x ∈ S+ ∪ S−.

(1.20)

ˆ is continuous in R2 ; in particular, the direct value (Vp α ˆ )0 on ∂S (ii) Vp α of the corresponding weakly singular integral is given by ˆ )0 (x, p) = (Vp α ˆ )+ (x, p) = (Vp α ˆ )− (x, p), (Vp α

x ∈ ∂S.

(iii) There hold the jump formulas ˆ )± (x, p) = ± 12 α ˆ (x, p) + (T Vp α ˆ )0 (x, p), (T Vp α

x ∈ ∂S,

ˆ )0 is the direct value of the corresponding singular integral on where (T Vp α ∂S. (iv) The double-layer potential Wp βˆ satisfies (1.20) and can be extended by continuity from S ± to S¯± , respectively. These extensions are of class C 1,β (S¯± ) for any β ∈ (0, 1). (v) There hold the jump formulas ˆ ± (x, p) = ∓ 1 β(x, ˆ p) + (Wp β) ˆ 0 (x, p), (Wp β) 2

x ∈ ∂S,

ˆ 0 is the direct value of the corresponding singular integral on ∂S. where (Wp β)

1.3 Time-dependent Plate Potentials

15

(vi) There holds the equality ˆ + (x, p) = (T Wp β) ˆ − (x, p), (T Wp β)

x ∈ ∂S.

This equality enables us to introduce the notation ˆ ˆ + (x, p) = (T Wp β) ˆ − (x, p), (Np β)(x, p) = (T Wp β)

x ∈ ∂S.

ˆ Since D(x, p) has polynomial growth with respect to p ∈ Cκ , κ > 0, we can define the single-layer and double-layer time-dependent (retarded) potentials ˆ respectively; V α and W β as the inverse Laplace transforms of Vp α ˆ and Wp β, more precisely, given that the inverse transform of a product of two transforms is the convolution of the corresponding two originals, we have (V α)(X) = (L−1 Vp α ˆ )(X) ∞ = D(x − y, t − τ )α(y, τ ) dsy dτ, 0 ∂S

ˆ (W β)(X) = (L−1 Wp β)(X) ∞ = P (x − y, t − τ )β(y, τ ) dsy dτ, 0 ∂S

where P (x − y, t) is the inverse Laplace transform of Pˆ (x − y, p); that is, P (x − y, t − τ )β(y, τ ) = β(y, τ ), Ty D(j) (y − x, t − τ ) ej . ˆ and Wp βˆ it follows that, at least for smooth From the properties of Vp α densities α and β vanishing for t < 0 and with compact support with respect to the time variable, the retarded potentials have the following properties. (i) V α and W β satisfy the equation B(∂t2 u)(X) + (Au)(X) = 0,

X ∈ G+ ∪ G− .

(ii) V α and W β satisfy the initial conditions u(x, 0+) = (∂t u)(x, 0+) = 0,

x ∈ S+ ∪ S−.

(iii) For X ∈ Γ, there hold the jump formulas (V α)+ (X) = (V α)− (X), (T V α)+ (X) − (T V α)− (X) = α(X), (W β)+ (X) − (W β)− (X) = −β(X), (T W β)+ (X) = (T W β)− (X).

16


After we introduce appropriate Sobolev-type spaces and study the properties of the boundary operators generated by the potentials in these spaces, we show that the above jump formulas hold, in fact, for much wider classes of densities α and β if the limiting values of the potentials are understood as traces on Γ.

1.4 Nonstationary Boundary Integral Equations We intend to seek integral representations for the solutions of all the initialboundary value problems listed in §1.1. As a typical example, here we discuss such representations for (DD± ) and (DN± ). Suppose that we want to represent the solutions of (DD± ) in the form u(X) = (V α)(X),

X ∈ G± ,

(1.21)

where α is an unknown density. Since, as indicated in §1.3, V α satisfies both the equation of motion and the initial conditions, u needs to satisfy only the Dirichlet boundary conditions, which means that α must be a solution of the nonstationary boundary integral equation (V α)(X) = f (X),

X ∈ Γ.

(1.22)

The kernel D(X − Y ) of this equation has a retarded time argument. For every fixed value of this argument, the kernel is weakly singular with respect to the space variables. If we now seek the solutions of (DD± ) in the form u(X) = (W β)(X),

X ∈ G± ,

(1.23)

then, according to the properties of W β, the unknown density β must satisfy the corresponding boundary integral equation (W β)± (X) = f (X),

X ∈ Γ,

(1.24)

whose kernel again has a retarded time argument but is singular with respect to the space variables. In the initial-boundary value problems (DN± ), the representations (1.21) and (1.23) yield, respectively, the nonstationary boundary integral equations (T V α)± (X) = g(X), (N β)(X) = g(X),

X ∈ Γ, X ∈ Γ,

(1.25) (1.26)

ˆ The kernel of (1.25) has the where N β is the inverse Laplace transform of Np β. same structure as that of (1.24), while the kernel of (1.26) is hypersingular with respect to the space variables. Both kernels have a retarded time argument.

1.4 Nonstationary Boundary Integral Equations

17

Comparing the representations (1.21) and (1.23) with the corresponding ones in the equilibrium problems [10], we see that here we have no additive rigid displacements. This remark is valid for the integral representations of the solutions to all the initial-boundary value problems that we are considering. In what follows, we apply the Laplace transformation with respect to time in the nonstationary boundary integral equations and reduce them to new boundary integral equations that depend on the transformation parameter p. The main difficulty in the study of the latter resides in establishing estimates for their solutions that illustrate the nature of the dependence of these solutions on p. Once such estimates are obtained, we can prove the unique solvability of the time-dependent boundary equations.

2 Problems with Dirichlet Boundary Conditions

2.1 Function Spaces In what follows, we use the same notation for norms and inner products on spaces of scalar and vector-valued functions, in accordance with the convention adopted in §1.1. We also point out that all these function spaces are complex. We start by introducing spaces of functions that depend on a complex parameter p; their properties, studied in [2], are listed below without proof. Let m ∈ R, p ∈ C, k ∈ R, and S ⊂ R2 . Hm (R2 ), Hm (S), and Hm (∂S) are the standard Sobolev spaces whose elements are defined on R2 , S, and the boundary ∂S of S, respectively. Hm,p (R2 ) is the space that coincides as a set with Hm (R2 ) but is equipped with the norm 1/2 (1 + |p|2 + |ξ|2 )m |˜ u(ξ)|2 dξ , um,p = R2

where u ˜ is the (distributional) Fourier transform of the three-component distribution u ∈ S (R2 ) (see the Appendix). Clearly, for any fixed p ∈ C, the norms on Hm,p (R2 ) and Hm (R2 ) are equivalent. ¯ ˚m,p (S) is the subspace of all u ∈ Hm,p (R2 ) such that supp u ⊂ S. H Hm,p (S) is the space of the restrictions to S of all v ∈ Hm,p (R2 ). The norm of u ∈ Hm,p (S) is defined by um,p;S =

inf

v∈Hm,p (R2 ): v|S =u

vm,p .

The inner products in L2 (R2 ), L2 (S), and L2 (∂S) are denoted by (· , ·)0 , (· , ·)0;S , and (· , ·)0;∂S , respectively. H−m,p (R2 ) is the dual of Hm,p (R2 ) with respect to the duality generated by (· , ·)0 . ˚m,p (S). H−m,p (S) is the dual of H

20


H1/2,p (∂S) is the space of the traces on ∂S of all the elements of H1,p (S). It coincides as a set with H1/2 (∂S) but is equipped with the norm f 1/2,p;∂S =

inf

u∈H1,p (S):γu=f

u1,p;S .

Here γ is the trace operator, which maps H1,p (S) continuously to H1/2,p (∂S). We mention that the continuity of γ is uniform with respect to p ∈ C; that is, γu1/2,p;∂S ≤ cu1,p;S , where c does not depend on p ∈ C [2]. We recall that the trace operators corresponding to the interior and exterior domains S ± are denoted by γ ± . H−1/2,p (∂S) is the dual of H1/2,p (∂S) with respect to the duality generated by (· , ·)0;∂S . Next, l+ and l− are extension operators that, in the context of our function spaces, map H1/2,p (∂S) to H1,p (S + ) and H1,p (S − ) continuously and uniformly with respect to p ∈ C. ˆ(x, p), x ∈ S, p ∈ Cκ , such that the mapping Hm,k,κ (S) is the space of all u U (p) = u ˆ(· , p) is holomorphic from Cκ to Hm (S) and ˆ u2m,k,κ;S

∞ = sup

σ>κ −∞

(1 + |p|2 )k U (p)2m,p;S dτ < ∞,

p = σ + iτ.

(2.1)

Formula (2.1) defines the norm on Hm,k,κ (S). It is readily seen from its definition that U (p) ∈ Hm,p (S) for any p ∈ Cκ . In what follows, we write u ˆ(x, p) if we want to emphasize that this is an element of Hm,p (S), and U (p) when we want to regard it as a mapping from Cκ to Hm (S). H±1/2,k,κ (∂S) are introduced similarly; that is, these spaces consist of all fˆ(x, p), x ∈ ∂S, p ∈ Cκ , such that the corresponding mapping F (p) = fˆ(· , p) is holomorphic from Cκ to H±1/2 (∂S) and fˆ2±1/2,k,κ;∂S = sup

∞

σ>κ −∞

(1 + |p|2 )k F (p)2±1/2,p;∂S dτ < ∞.

Once again, the above equality defines the norms on these spaces and, as above, we interpret fˆ(x, p) as an element of H±1/2,p (∂S) and F (p) as a mapping from Cκ to H±1/2 (∂S). Hm,k,κ (G) and H±1/2,k,κ (Γ) consist, respectively, of the inverse Laplace transforms u and f of the elements u ˆ and fˆ of Hm,k,κ (S) and H±1/2,k,κ (∂S); these spaces are equipped with the norms um,k,κ;G = ˆ um,k,κ;S , f ±1/2,k,κ;Γ = fˆ±1/2,k,κ;∂S .

(2.2)

2.2 Solvability of the Transformed Problems

21

By the Paley–Wiener theorem and Parseval’s equality [12], for a nonnegative integer k the spaces H1,k,κ (G) consist of all three-component distributions u defined on S × R that vanish for t < 0 and are such that

e−2κt |(∂xα ∂tαt +k u)(x, t)|2 dx dt < ∞, (2.3) G

|α|+αt ≤1

where α is a two-component multi-index, αt is a nonnegative integer, and ∂xα is the partial differentiation operator acting with respect to the space variables. The norm on H1,k,κ (G) defined by (2.3) is equivalent to (2.2). A similar remark is also valid for H1/2,k,κ (Γ). In what follows, we relax the terminology and refer to the elements of all of the above spaces as “functions” instead of “distributions” or “generalized functions”, since the former term, although technically incorrect, is more familiar to the nonspecialist reader. ¯ ± ) are the spaces of infinitely differentiable functions with Finally, C0∞ (G ¯ ± , respectively. compact support in G

2.2 Solvability of the Transformed Problems In this section, we discuss the problems (D± p ) obtained after applying the Laplace transformation with respect to the time variable in the original problems (DD± ). Our aim is to establish the unique solvability of (D± p ) for every p ∈ C0 and derive certain estimates for their solutions. We use two different approaches to solve these problems. The first one is based on the Fredholm Alternative and works well in the case of interior domains. Unfortunately, the same cannot be done in exterior problems because here the operators occurring in the corresponding functional equations lose their compactness in the natural spaces where such problems are set. This makes it necessary for us to modify the method when we deal with exterior domains. The transformed problems that we consider here are more general in that they include the contribution of the body forces and moments q occurring on the right-hand side in the equation of motion (1.8). Thus, the classical version of these more general problems (D± ˆ ∈ C 2 (S ± ) ∩ C(S¯± ) p ) consists in finding u such that Bp2 u ˆ(x, p) + (Aˆ u)(x, p) = qˆ(x, p), x ∈ S ± , (2.4) u ˆ± (x, p) = fˆ(x, p), x ∈ ∂S, where u ˆ(x, p) = Lu(x, t),

qˆ(x, p) = Lq(x, t),

fˆ(x, p) = Lf (x, t).

In order to simplify the notation, and since there is no danger of ambiguity, throughout this section we omit the hat from the symbols of functions that

22


depend on x and p, it being understood that, unless otherwise stipulated, we are carrying out our arguments in spaces of Laplace transforms. As usual, to derive the variational version of (D± p ), we multiply (2.4) termwise by v ∗ ∈ C0∞ (S ± ) and integrate the result over S ± ; thus, we obtain p2 (Bu, v)0;S ± + a± (u, v) = (q, v)0;S ± , (2.5) where (Bu, v)0;S ± = (B 1/2 u, B 1/2 v)0;S ± , a± (u, v) = 2 E(u, v ∗T ) dx, S±

and E(u, v) is the sesquilinear form defined by the internal energy density; that is [9], 2E(u, v) = h2 E0 (u, v) + h2 µ(∂2 u1 + ∂1 u2 )(∂2 v¯1 + ∂1 v¯2 ) + µ[(u1 + ∂1 u3 )(¯ v1 + ∂1 v¯3 ) + (u2 + ∂2 u3 )(¯ v2 + ∂2 v¯3 )], E0 (u, v) = (λ + 2µ) (∂1 u1 )(∂1 v¯1 ) + (∂2 u2 )(∂2 v¯2 ) + λ (∂1 u1 )(∂2 v¯2 ) + (∂2 u2 )(∂1 v¯1 ) . It is obvious that a± (u, v) = a± (v, u),

(2.6)

where the superposed bar denotes complex conjugation. Equation (2.5) indicates that the variational version of problems (2.4) should consist in finding u ∈ H1,p (S ± ) such that p2 (B 1/2 u, B 1/2 v)0;S ± + a± (u, v) = (q, v)0;S ±

˚1,p (S ± ), ∀v ∈ H

γ ± u = f.

(2.7)

Throughout what follows, we denote by the same symbol c all positive constants occurring in estimates, which are independent of the functions in these estimates and of p ∈ Cκ , but may depend on κ. ¯ κ , κ > 0, 2.1 Theorem. For any f ∈ H1/2,p (∂S) and q ∈ H−1,p (S + ), p ∈ C + + problem (Dp ) has a unique weak solution u ∈ H1,p (S ) and u1,p;S + ≤ c|p|(q−1,p;S + + f 1/2,p;∂S ).

(2.8)

Proof. First we consider (D+ p ) with homogeneous boundary conditions, which ˚1,p (S + ) such that consists in finding u0 ∈ H p2 (B 1/2 u0 , B 1/2 v)0;S + + a+ (u0 , v) = (q, v)0;S +

˚1,p (S + ). ∀v ∈ H

(2.9)


23

Repeating the proof of Lemma 2.3 in [7], we can show that a+ (u, v) is coercive ˚1,p (S + )]2 . Since the form is also continuous on this space, we conclude on [H that for any q ∈ H−1,p (S + ), the variational equation a+ (u0 , v) = (q, v)0;S +

˚1,p (S + ) ∀v ∈ H

(2.10)

˚1,p (S + ), which satisfies the estimate has a unique solution u0 ∈ H u0 1 ≤ cq−1;S + . On the other hand, since a+ (u0 , v) defines a bounded antilinear (conjugate ˚1,p (S + ) for any u0 ∈ H ˚1,p (S + ), it can be written in linear) functional on H + the form (2.10) with some q ∈ H−1,p (S ). Let A be the operator defined by ˚1,p (S + ) with q ∈ H−1,p (S + ) as described a+ (u0 , v), which associates u0 ∈ H above; thus, a+ (u0 , v) = (Au0 , v)0;S +

˚1 (S + ). ∀u0 , v ∈ H

˚1 (S + ) to H−1 (S + ). Additionally, from (2.6) it A is a homeomorphism from H follows that A is self-adjoint in the sense that (Au0 , v)0;S + = (u0 , Av)0;S +

˚1,p (S + ). ∀u0 , v ∈ H

˚1,p (S + ), This is easily verified, since for any u0 , v ∈ H (Au0 , v)0;S + = a+ (u0 , v) = a+ (v, u0 ) = (Av, u0 )0;S + = (u0 , Av)0;S + . Equation (2.9) can now be written in the form p2 Bu0 + Au0 = q.

(2.11)

Applying A−1 on both sides in (2.11), we arrive at the equivalent equation p2 A−1 Bu0 + u0 = A−1 q

(2.12)

˚1,p (S + ). We denote by B0 the restriction of A−1 B in the Banach space H ˚1 (S + ) and claim that B0 is compact on H ˚1 (S + ). Let from H−1 (S + ) to H ∞ + + ˚ {un }n=1 be a weakly convergent sequence in H1 (S ). Since S is a bounded domain, Rellich’s lemma implies that the set {un }∞ n=1 is strongly compact in L2 (S + ). Therefore, there is a subsequence {unj }∞ j=1 that converges strongly in L2 (S + ), consequently, also in H−1 (S + ). Because A−1 B is continuous from ˚1 (S + ), the sequence {B0 un }∞ is strongly convergent in the H−1 (S + ) to H j j=1 ˚1 (S + ), which proves that B0 is a compact operator. space H

24


While studying the solvability of (2.12), we are under the conditions of the Fredholm Alternative, so (2.12)—hence, also (2.11)—has a unique solution ˚1 (S + ) if and only if the homogeneous equation u0 ∈ H p2 A−1 Bu0 + u0 = 0

(2.13)

has only the trivial solution. In turn, (2.13) can be rewritten in the equivalent form p2 Bu0 + Au0 = 0, or ˚1 (S + ). p2 (B 1/2 u0 , B 1/2 v)0;S + + a+ (u0 , v) = 0 ∀v ∈ H

(2.14)

Taking v = u0 in (2.14), writing p = σ + iτ , and separating the real and imaginary parts, we arrive at (σ 2 − τ 2 )B 1/2 u0 20;S + + a+ (u0 , u0 ) = 0,

(2.15)

2στ B 1/2 u0 20;S + = 0.

(2.16)

If τ = 0, then from (2.16) it follows that B 1/2 u0 = 0; hence, u0 = 0. If τ = 0, then the equality u0 = 0 follows from (2.15). So (2.11) is uniquely solvable in ˚1 (S + ) for any q ∈ H−1 (S + ). H We now establish estimate (2.8). Taking v = u0 in (2.9) and separating the real and imaginary parts, we obtain (σ 2 − τ 2 )B 1/2 u0 20;S + + a+ (u0 , u0 ) = Re(q, u0 )0;S + ,

(2.17)

2στ B 1/2 u0 20;S + = Im(q, u0 )0;S + .

(2.18)

Multiplying (2.18) by σ −1 τ and adding the new equality to (2.17), we find that |p|2 B 1/2 u0 20;S + + a+ (u0 , u0 ) = Re(q, u0 )0;S + + σ −1 τ Im(q, u0 )0;S + = σ −1 Re{¯ p(q, u0 )0;S + }. ¯ κ , it follows that σ ≥ κ. Taking into account the inequality Because p ∈ C a+ (u0 , u0 ) ≥ cu0 21 , we obtain u0 21,p;S + ≤ c|p||(q, u0 )0;S + |, from which u0 1,p;S + ≤ c|p|q−1,p;S + .


25

We now return to the full problem (2.7). Let w = l+ f ∈ H1 (S + ). We recall that since the extension operator l+ is continuous (uniformly with respect to p ∈ C), we have w1,p;S + ≤ cf 1/2,p;∂S . (2.19) ˚1,p (S + ) satisfies Representing u in the form u = w + u0 , we see that u0 ∈ H the equation p2 (B 1/2 u0 , B 1/2 v) + a+ (u0 , v) = (q, v)0;S + − p2 (B 1/2 w, B 1/2 v)0;S + − a+ (w, v) ˚1,p (S + ). ∀v ∈ H

(2.20)

We claim that the form p2 (B 1/2 w, B 1/2 v)0;S + + a+ (w, v) defines a bounded ˚1,p (S + ); true, antilinear (conjugate linear) functional on H |p2 (B 1/2 w, B 1/2 v)0;S + + a+ (w, v)| ≤ c(|p|2 w0;S + v0;S + + w1;S + v1 ) ≤ cw1,p;S + v1,p

˚1,p (S + ). ∀v ∈ H

Consequently, there is q˜ ∈ H−1,p (S + ) such that p2 (B 1/2 w, B 1/2 v)0;S + + a+ (w, v) = (˜ q , v)0;S +

˚1,p (S + ) ∀v ∈ H

and ˜ q −1,p;S + ≤ cw1,p;S + ≤ cf 1/2,p;∂S . Equation (2.20) takes the form p2 (B 1/2 u0 , B 1/2 v)0;S + + a+ (u0 , v) = (q − q˜, v)0;S +

˚1,p (S + ). ∀v ∈ H

As we already know, the latter problem is uniquely solvable and its solution satisfies the estimate u0 1,p ≤ c|p|(q−1,p;S + + ˜ q −1,p;S + ) ≤ c|p|(q−1,p;S + + f 1/2,p;∂S ).

(2.21)

Combining (2.21) and (2.19), we arrive at (2.8). ˚1,p (S + ) is a If u1 and u2 are two solutions of (2.7), then u0 = u1 − u2 ∈ H solution of (2.14); hence, u0 = 0, and the theorem is proved. As was mentioned above, we cannot repeat this proof in the case of S − because Rellich’s lemma is not valid for an unbounded domain. Consequently, we need to modify our approach.

26


¯ κ , κ > 0, 2.2 Theorem. For any f ∈ H1/2,p (∂S) and q ∈ H−1,p (S − ), p ∈ C − problem (D− ) has a unique solution u ∈ H (S ) and 1,p p u1,p;S − ≤ c|p|(q−1,p;S − + f 1/2,p;∂S ). Proof. Again, first we assume that f = 0. In this case we seek u0 ∈ H1,p (S − ) such that p2 (B 1/2 u0 , B 1/2 v)0;S − + a− (u0 , v) = (q, v)0;S −

˚1,p (S − ). ∀v ∈ H

(2.22)

To prove the unique solvability of (2.22), we consider an auxiliary variational ˚1 (S − ) such that problem that consists in finding u0 ∈ H 1 2

κ2 (B 1/2 u0 , B 1/2 v)0;S − + a− (u0 , v) = (q, v)0;S −

˚1 (S − ), ∀v ∈ H

(2.23)

where q ∈ H−1 (S − ) is prescribed. Repeating the proof of Lemma 2.2 in [7], we find that there is a constant c > 0 such that a− (u, u) + u20;S − ≥ cu21;S −

∀u ∈ H1 (S − ).

(2.24)

From (2.24) it follows that the form a−,κ (u, v) =

1 2

κ2 (B 1/2 u, B 1/2 v)0;S − + a− (u, v),

˚1 (S − ) 2 , is coercive on this space. The Lax–Milgram which is continuous on H ˚1 (S − ) for any lemma then implies that (2.23) has a unique solution u0 ∈ H − − ˚1 (S ), the form a−,κ (u0 , v) q ∈ H−1 (S ). On the other hand, for any u0 ∈ H ˚1 (S − ); theregenerates a bounded antilinear (conjugate linear) functional on H fore, it can be written in the form (2.23). This enables us to define an operator Aκ through the equality ˚1 (S − ), (Aκ u0 , v)0;S − = a−,κ (u0 , v) ∀v ∈ H ˚1 (S − ) to H−1 (S − ). Equation (2.23) can which is a homeomorphism from H be rewritten as Aκ u0 = q. In turn, (2.22) can be written in the form A κ u 0 + p2 −

1 2

κ2 Bu0 = q.

(2.25)

Applying A−1 κ on both sides in (2.25), we arrive at the equivalent equation u 0 + p2 −

1 2

−1 κ2 A−1 κ Bu0 = Aκ q.

(2.26)

Setting B 1/2 u0 = ub , we again rewrite (2.26) in the equivalent form u b + p2 −

1 2

1/2 κ2 B 1/2 A−1 ub = B 1/2 A−1 κ B κ q.

(2.27)


27

˚1 (S − ). If ub ∈ H ˚1 (S − ) is its solution, then Equation (2.27) is solvable in H ub is, at the same time, the solution of (2.27) in L2 (S − ). Conversely, let ub ∈ L2 (S − ) be a solution of (2.27). Since 1/2 ˚1 (S − ), ub ∈ H B 1/2 A−1 κ B

˚ − B 1/2 A−1 κ q ∈ H1 (S ),

˚1 (S − ). This means that problem (2.27) in H ˚1 (S − ) is it follows that ub ∈ H 2 − equivalent to itself in L (S ). We now study the properties of the restriction 1/2 from H1 (S − ) to L2 (S − ). Bκ of B 1/2 A−1 κ B 2 − Let q, ψ ∈ L (S ) be arbitrary, and let 1/2 q, u0 = A−1 κ B

1/2 v = A−1 ψ. κ B

From the definition of Aκ it follows that a−,κ (u0 , v) = (B 1/2 q, v)0;S − , a−,κ (v, u0 ) = (B 1/2 ψ, u0 )0;S − ; hence, (B 1/2 q, v)0;S − = (B 1/2 ψ, u0 )0;S − = (u0 , B 1/2 ψ)0;S − .

(2.28)

In turn, (2.28) can be rewritten as 1/2 1/2 q, B 1/2 ψ)0;S− = (B 1/2 q, A−1 ψ)0;S − , (A−1 κ B κ B

or (Bκ q, ψ)0;S − = (q, Bκ ψ)0;S −

∀q, ψ ∈ L2 (S − ).

(2.29)

By (2.29), Bκ is a symmetric operator on L2 (S − ); therefore, it is self-adjoint (as a symmetric operator defined on the whole of a Hilbert space [3]). 1/2 Once again, let q ∈ L2 (S − ) and u0 = A−1 q; then κ B 1/2 q, B 1/2 q)0;S − (Bκ q, q)0;S − = (A−1 κ B

= (u0 , Aκ u0 )0;S − = a−,κ (u0 , u0 ) ≥ 0, which means that Bκ is nonnegative. Since the spectrum of a self-adjoint nonnegative operator lies on the half-line [0, ∞) in the complex plane, every point of C \ [0, ∞) is a regular point for Bκ [3]. All that remains to do now is to remark that for any p ∈ Cκ , 1 2

κ2 − p2

−1

∈ / [0, ∞).

This implies that (2.27) is uniquely solvable for any q ∈ H−1 (S − ). Con˚1,p (S − ) for any sequently, equation (2.22) has a unique solution u0 ∈ H q ∈ H−1,p (S − ). To complete the proof, we repeat the last part of the proof of Theorem 2.1, replacing the extension operator l+ by l− .

28


2.3 Solvability of the Time-dependent Problems We start with the variational version of problems (DD± ) for the nonhomogeneous equation of motion. The classical formulation of (DD+ ) asks for a ¯ + ) such that function u ∈ C 2 (G+ ) ∩ C 1 (G B(∂t2 u)(X) + (Au)(X) = q(X), u(x, 0+) = (∂t u)(x, 0+) = 0, u+ (X) = f (X),

X ∈ G+ , x ∈ S+,

(2.30)

X ∈ Γ.

¯ + ) is such that Multiplying the first equation (2.30) by v ∗ , where v ∈ C0∞ (G + + v = 0, integrating the new equality over S with respect to x and over [0, ∞) with respect to t, and taking into account the initial data for u and the boundary value of v, we arrive at ∞

a+ (u, v) − (B 1/2 ∂t u, B 1/2 ∂t v)0;S + dt =

0

∞ (q, v)0;S + dt.

(2.31)

0

¯ + ) satisfies (2.31) for any v ∈ C ∞ (G ¯ + ) such Conversely, if u ∈ C 2 (G+ ) ∩ C 1 (G 0 + + + that v = 0, u(x, 0+) = 0, x ∈ S , and u = f , then, integrating by parts in (2.31), we find that u is the solution of (2.30). Equation (2.31) suggests that the variational problem (DD+ ) should consist in finding u ∈ H1,0,κ (G+ ) that satisfies ∞ 0

a+ (u, v) − (B 1/2 ∂t u, B 1/2 ∂t v)0;S + dt ∞ = (q, v)0;S + dt

¯ + ), v + = 0, ∀v ∈ C0∞ (G

(2.32)

0

γ + u = f, where q and f are prescribed. Similarly, the variational problem (DD− ) consists in finding u ∈ H1,0,κ (G− ) that satisfies ∞ 0

a− (u, v) − (B 1/2 ∂t u, B 1/2 ∂t v)0;S − dt ∞ = (q, v)0;S − dt,

¯ − ), v − = 0, ∀v ∈ C0∞ (G

0

γ − u = f.

(2.33)

2.3 Solvability of the Time-dependent Problems

29

2.3 Theorem. For any q ∈ H−1,1,κ (G+ ) and f ∈ H1/2,1,κ (Γ), κ > 0, problems (2.32) and (2.33) have unique solutions u ∈ H1,0,κ (G± ). If q ∈ H−1,k,κ (G± ) and f ∈ H1/2,k,κ (Γ), k ∈ R, then u ∈ H1,k−1,κ (G± ) and u1,k−1,κ;G± ≤ c(q−1,k,κ;G± + f 1/2,k,κ;Γ ).

(2.34)

Proof. We prove the assertion for (DD+ ); the case of (DD− ) is treated similarly. Let u ˆ ∈ H1,p (S + ) be the (weak) solution of the problem p2 B u ˆ(x, p) + (Aˆ u)(x, p) = qˆ(x, p), x ∈ S + , γ+u ˆ(x, p) = fˆ(x, p), x ∈ ∂S,

(2.35)

obtained by applying the Laplace transformation in (DD+ ) with a nonhomogeneous equation of motion. The existence of u ˆ was proved in the previous section. For simplicity, in what follows we write u ˆ(· , p) = U (p),

qˆ(· , p) = Q(p),

fˆ(· , p) = F (p),

and regard U , Q, and F as functions from Cκ to H1 (S + ), H−1 (S + ), and H1/2 (∂S), respectively. We claim that the inverse Laplace transform u of u ˆ belongs to the space H1,k−1,κ (G+ ) if q ∈ H−1,k,κ (G+ ) and f ∈ H1/2,k,κ (Γ). To show this, first we verify that U is holomorphic from Cκ to H1 (S + ). Let p0 ∈ Cκ , and let KR (p0 ) be a circle with center at p0 and radius R (to be specified later), and ¯ R (p0 ) ⊂ Cκ . We recall that the solution U (p0 ) of the problem such that K p20 BU (p0 ) + (AU )(p0 ) = Q(p0 ), γ + U (p0 ) = F (p0 ) satisfies the estimates U (p0 )1;S + ≤ c|p0 |(Q(p0 )−1,p0 ;S + + F (p0 )1/2,p0 ;∂S ) ≤ c(Q(p0 )−1;S + + F (p0 )1/2;∂S ). Rewriting (2.35) in the form p20 BU (p) + (AU )(p) = Q(p) − (p2 − p20 )BU (p), +

γ U (p) = F (p), we see that U (p)1;S + ≤ c(Q(p)−1;S + + F (p)1/2;∂S + |p2 − p20 |U (p)−1;S + ).

(2.36)

30


Since U (p)−1;S + ≤ U (p)1;S + , it follows that for p satisfying c|p2 − p20 | ≤

1 2

(2.37)

we have U (p)1;S + ≤ c(Q(p)−1;S + + F (p)1/2;∂S ).

(2.38)

¯ R (p0 ) ⊂ Cκ and (2.37) holds for p ∈ KR (p0 ). Since We choose R > 0 so that K Q and F are holomorphic from Cκ to H−1 (S + ) and H1/2 (∂S), respectively, ¯ R (p0 ). Estimate (2.38) shows that they are bounded in these spaces for p ∈ K + ¯ U is also bounded in H1 (S ) for p ∈ KR (p0 ). By (2.36), p20 B U (p) − U (p0 ) + A U (p) − U (p0 ) = Q(p) − Q(p0 ) − (p2 − p20 )BU (p), γ + U (p) − γ + U (p0 ) = F (p) − F (p0 ); hence, for p ∈ KR (p0 ), U (p) − U (p0 )1;S + ≤ c Q(p) − Q(p0 )−1;S + + F (p) − F (p0 )1/2;∂S + |p2 − p20 |BU (p)−1;S + ). As U (p) is bounded in H1 (S + )—hence, also in H−1 (S + )—for p ∈ KR (p0 ), it follows that lim U (p) − U (p0 )1;S + = 0, p→p0

which means that U is continuous from Cκ to H1 (S + ) at p0 . Finally, let V ∈ H1 (S + ) be the solution of the problem p20 BV − AV = Q (p0 ) − 2p0 U (p0 ), γ + V = F (p0 ); then the function W (p) = (p − p0 )−1 U (p) − U (p0 ) − V ∈ H1 (S + ) satisfies p20 BW (p) + (AW )(p)

= (p − p0 )−1 Q(p) − Q(p0 ) − Q (p0 ) − B (p + p0 )U (p) − 2p0 U (p0 ) , γ + W = (p − p0 )−1 F (p) − F (p0 ) − F (p0 ).


Next,

31

Q(p) − Q(p0 ) W (p)1;S + ≤ c − Q (p ) 0 p − p0

−1;S +

F (p) − F (p0 ) + − F (p ) 0 p − p0 1/2;∂S + (p + p0 )U (p) − 2p0 U (p0 )−1;S + . Since U is continuous at p0 , lim W (p)1;S + = 0,

p→p0

which means that U (p0 ) exists and U (p0 ) = V . The arbitrariness of p0 in Cκ implies that the mapping U is holomorphic from Cκ to H1 (S + ). Given that U (p)1,p;S + ≤ c|p|(Q(p)−1,p;S + + F (p)1/2,p;∂S ), we have u21,k−1,κ;G+

∞ = sup

σ>κ −∞

(1 + |p|2 )k−1 U (p)21,p;S + dτ

∞ ≤ c sup

σ>κ −∞

(1 + |p|2 )k (Q(p)2−1,p;S + + F (p)21/2,p;∂S ) dτ

≤ c(q2−1,k,κ;G+ + f 21/2,k,κ;Γ ),

(2.39)

where p = σ + iτ . This confirms (2.34). To complete the proof of the theorem, we need to check that u is the only solution of (2.32). We recall [11] that any two functions f1 (t) and f2 (t) such that ∞

e−2κν t |fν (t)|2 dt < ∞,

ν = 1, 2,

0

satisfy Parseval’s equality ∞

−(κ1 +κ2 )t

e 0

1 f1 (t)f2 (t) dt = 2π

∞ F1 (κ1 + iτ )F2 (κ2 + iτ ) dτ, −∞

(2.40)

32


¯ + ) be such where Fν (p) are the Laplace transforms of fν (t). Let v ∈ C0∞ (G that γ + v = 0. We make the notation ˚1 (S + ), v(· , p) = V (p), v(x, 0) = v0 (x), v0 ∈ H p = σ + iτ, p∗ = −σ + iτ,

v0 (·) = V0 ,

choose any σ > κ and fix it, then take κ1 = σ and κ2 = −σ in (2.40) and find that the Laplace transform of ∂t v at the point p∗ is p∗ V (p∗ ) − V0 , and that ∞

a+ (u, v) − (B 1/2 ∂t u, B 1/2 ∂t v)0;S + − (q, v)0;S + dt

0 −1

∞

= (2π)

a+ U (p), V (p∗ ) − B 1/2 pU (p), B 1/2 (p∗ V (p∗ ) − V0 ) 0;S +

−∞

− Q(p), V (p∗ ) 0;S + dτ.

(2.41)

Since U (p) is a weak solution of (2.4), it follows that a+ U (p), W + p2 B 1/2 U (p), B 1/2 W 0;S + = Q(p), W 0;S +

˚1 (S + ). ∀W ∈ H

(2.42)

Taking W = V (p∗ ) − (p∗ )−1 V0 in (2.42), we obtain a+ U (p), V (p∗ ) − B 1/2 pU (p), B 1/2 (p∗ V (p∗ ) − V0 ) 0;S +

− Q(p), V (p∗ ) 0;S +

= a+ U (p), (p∗ )−1 V0 − Q(p), (p∗ )−1 V0 0;S + = p−1 Q(p), V0 0;S + − p−1 a+ U (p), V0 . Therefore, (2.41) takes the form ∞

a+ (u, v) − (B 1/2 ∂t u, B 1/2 ∂t v)0;S + − (q, v)0;S + dt

0 −1

∞

= (2π)

p−1 Q(p), V0 0;S + − a+ U (p), V0 dτ.

(2.43)

−∞

We claim that the right-hand side in (2.43) vanishes. First, we remark that ˆ q 2−1;S + ≤ (1 + |p|2 )ˆ q 2−1,p;S + ∀ˆ q ∈ H−1 (S + ). (2.44)


33

For if qˆ ∈ H−1 (R2 ) and q˜ is its Fourier transform, then ˆ q 2−1

=

(1 + |ξ|2 )−1 |˜ q (ξ)|2 dξ

R2

2

≤ (1 + |p| )

(1 + |p|2 )−1 (1 + |ξ|2 )−1 |˜ q (ξ)|2 dξ

R2 2

≤ (1 + |p| )

(1 + |p|2 + |ξ|2 )−1 |˜ q (ξ)|2 dξ

R2

= (1 + |p|2 )ˆ q 2−1,p .

(2.45)

Inequality (2.44) follows from (2.45) and the definition of the norms on the spaces H−1 (S + ) and H−1,p (S + ). By (2.44), ∞

2

|(Q(p), V0 )0;S + | dτ ≤

V0 21

−∞

∞

Q(p)2−1;S + dτ

−∞

≤ V0 21

∞

(1 + |p|2 )Q(p)2−1,p;S + dτ

−∞

≤ V0 21 q2−1,1,κ;G+ < ∞. Consequently, the function ϕ(t) = (q, v0 )0;S + satisfies ∞

e−2σt |ϕ(t)|2 dt < ∞,

0

so ϕ ∈ L1loc (0, ∞), which implies that t ϕ(λ) dλ

ψ(t) = 0

is continuous for t ∈ [0, ∞); in particular, ψ(0) = 0. We have 0 = ψ(0) = (2π)

−1

∞

−∞

p−1 Q(p), V0 0;S + dτ.

34


Next, ∞

2

|a+ (U (p), V0 )| dτ ≤ −∞

cV0 21

∞

U (p)21,p;S + dτ

−∞

≤ cV0 21 u21,0,κ;G+ , and the above arguments yield ∞

p−1 a+ (U (p), V0 ) dτ = 0.

−∞

Equality (2.43) now leads to ∞

a+ (u, v) − (B 1/2 ∂t u, B 1/2 ∂t v)0;S + dt =

0

∞ (q, v)0;S + dt, 0

which means that u is a weak solution of (2.32). To prove that this solution is unique, suppose that u ∈ H1,0,κ (G+ ), γ + u = 0, satisfies ∞

a+ (u, v) − (B 1/2 ∂t u, B 1/2 ∂t v)0;S + dt = 0

0

¯ + ), γ + v = 0. ∀v ∈ C0∞ (G

(2.46)

We fix an arbitrary T > 0. It is obvious that ˚1 (S + )); u ∈ H1 ((0, T ); L2 (S + )) ∩ L2 ((0, T ); H that is, U (t) = u(· , t), regarded as a vector-valued function from (0, T ) to L2 (S + ), belongs to H1 (0, T ) and T

U (t)20;S + + U (t)20;S + dt < ∞.

(2.47)

0

˚1 (S + ), belongs to The same function, regarded as a mapping from (0, T ) to H 2 L (0, T ) and T 0

U (t)21 dt < ∞.

(2.48)


35

˚1 (S + )) From (2.47) and (2.48) we see that H1 ((0, T ); L2 (S + )) ∩ L2 ((0, T ); H can be equipped with the norm u21;G+ T

T

=

|∂ α+αt u(x, t)|2 dx dt,

0 S + |α|+αt ≤1

+ where G+ T = S × (0, T ). We now construct the function Z(t) = z(· , t), where

⎧ ⎨ z(x, t) =

⎩

T − u(x, τ ) dτ,

t ≤ T,

0,

t > T.

t

(2.49)

Clearly, the restriction of Z to (0, T ) belongs to H1 ((0, T ); L2 (S + ))∩L2 ((0, T ); ˚1 (S + )) and that z can be approximated with any accuracy in the norm H ¯ + ) such that γ + v = 0. Hence, we · 1;G+ by means of elements v ∈ C0∞ (G T may set v = z in (2.46) and obtain T

a+ (u, z) − (B 1/2 ∂t u, B 1/2 ∂t z)0;S + dt = 0.

0

We remark that

Z (t) = (∂t z)(· , t) =

u(· , t), t < T , 0, t > T,

(2.50)

and rewrite (2.46) in the form T

a+ (∂t z, z) − (B 1/2 ∂t u, B 1/2 u)0;S + dt = 0,

0

or

T 0

d a+ (z, z) − B 1/2 u20;S + dt = 0. dt

(2.51)

Since U , regarded as a mapping from (0, T ) to L2 (S + ), belongs to H1 (0, T ), it is absolutely continuous on [0, T ]; hence, T 0

d B 1/2 u20;S + dt = B 1/2 U (T )20;S + − B 1/2 U (0)20;S + dt = B 1/2 U (T )20;S + .

(2.52)

36


˚1 (S + ), From (2.49) and (2.50), it follows that Z, as a mapping from (0, T ) to H belongs to H1 (0, T ); hence, Z is absolutely continuous on [0, T ] and T 0

d a+ (z, z) dt = a+ (Z(T ), Z(T )) − a+ (Z(0), Z(0)) dt = −a+ (Z(0), Z(0)).

(2.53)

Formulas (2.51)–(2.53) now imply that a+ (Z(0), Z(0)) + B 1/2 U (T )20;S + = 0; therefore, u(T ) = u(· , T ) = 0 for any T > 0, which completes the proof.

3 Problems with Neumann Boundary Conditions

3.1 The Poincar´ e–Steklov Operators In this section we study the properties of the dynamic analogs of the Poincaré– Steklov operators introduced in [7]. We begin by considering these operators in Sobolev spaces with a parameter. Let f ∈ H1/2,p (∂S), p ∈ C0 , and let u ∈ H1,p (S ± ) be the (unique) solutions of the problems ˚1,p (S ± ), p2 (B 1/2 u, B 1/2 v)0;S ± + a± (u, v) = 0 ∀v ∈ H γ ± u = f.

(3.1)

Also, let ϕ be an arbitrary element of H1/2,p (∂S), and let w ∈ H1,p (S ± ) be such that γ ± w = ϕ. For every p ∈ C0 , we define a pair of Poincaré–Steklov operators Tp± on H1/2,p (∂S) by means of the equality (Tp± f, ϕ)0;∂S = ± p2 (B 1/2 u, B 1/2 w)0;S ± + a± (u, w) .

(3.2)

It is obvious that (3.2) defines Tp± correctly. For if w1 , w2 ∈ H1,p (S ± ) are such that γ ± w1 = γ ± w2 = ϕ, ˚1,p (S ± ) and, by (3.1), then v = w1 − w2 ∈ H p2 (B 1/2 u, B 1/2 w1 )0;S ± + a± (u, w1 ) = p2 B 1/2 u, B 1/2 (v + w2 ) 0;S ± + a± (u, v + w2 ) = p2 (B 1/2 u, B 1/2 w2 )0;S ± + a± (u, w2 ).

38


3.1 Lemma. For any p ∈ C0 , the operators Tp± are homeomorphisms from ¯ κ , κ > 0, H1/2,p (∂S) to H−1/2,p (∂S), and for any f ∈ H1/2,p (∂S), p ∈ C Tp± f −1/2,p;∂S ≤ c|p|f 1/2,p;∂S ,

(3.3)

f 1/2,p;∂S ≤ c|p|Tp± f −1/2,p;∂S .

(3.4)

Proof. Let f, ϕ ∈ H1/2,p (∂S), let w = l± ϕ, and let u ∈ H1,p (S ± ) be the solutions of (3.1) in S ± . By (3.2), |(Tp± f, ϕ)0;∂S |2 ≤ c u21,p;S ± v21,p;S ± + a± (u, u)a± (w, w) ≤ cu21,p;S ± ϕ21/2,p;∂S ; therefore, Tp± f ∈ H−1/2,p (∂S) and Tp± f −1/2,p;∂S ≤ cu1,p;S ± .

(3.5)

Taking ϕ = f and w = u in (3.1), we obtain p2 B 1/2 u20;S ± + a± (u, u) = ±(Tp± f, f )0;∂S , where p = σ + iτ . Next, we separate the real and imaginary parts above to find that (σ 2 − τ 2 )B 1/2 u20;S ± + a± (u, u) = ± Re(Tp± f, f )0;∂S ,

(3.6)

2στ B 1/2 u20;S ± = ± Im(Tp± f, f )0;∂S .

(3.7)

Multiplying (3.7) by σ −1 τ and adding the result to (3.6), we arrive at the equalities

|p|2 B 1/2 u20;S ± + a± (u, u) = ±σ −1 Re p¯(Tp± f, f )0;∂S .

(3.8)

Since, as shown in [7], a± (u, u) + u20;S ± ≥ cu21;S ± , we easily deduce from (3.8) that

u21,p;S ± ≤ ±cσ −3 Re p¯(Tp± f, f )0;∂S ;

(3.9)

u21,p;S ± ≤ c|p|Tp± f −1/2,p;∂S f 1/2,p;∂S ,

(3.10)

¯ κ, hence, for p ∈ C

3.1 The Poincaré–Steklov Operators

39

and the trace theorem implies that u1,p;S ± ≤ c|p|Tp± f −1/2,p;∂S .

(3.11)

From (3.5) and (3.10) it follows that Tp± f 2−1/2,p;∂S ≤ c|p|Tp± f −1/2,p;∂S f 1/2,p;∂S ; therefore, Tp± f −1/2,p;∂S ≤ c|p|f 1/2,p;∂S . This proves the continuity of Tp± from H1/2,p (∂S) to H−1/2,p (∂S), and (3.3). We now show that the inverse operators (Tp± )−1 exist. By the trace theorem and (3.11), f 1/2,p;∂S ≤ cu1,p;S ± ≤ c|p|Tp± f −1/2,p;∂S , so (Tp± )−1 are continuous from H−1/2,p;∂S to H1/2,p (∂S) and (3.4) holds. Suppose that the ranges of Tp± are not dense in H−1/2,p (∂S). Then there is a nonzero ϕ ∈ H1/2,p (∂S) such that (Tp± f, ϕ)0;∂S = 0 for all f ∈ H1/2,p (∂S). Taking f = ϕ, we see that (Tp± ϕ, ϕ)0;∂S = 0. If z is the solution of either of the problems (3.1) with boundary value ϕ, then (3.8) implies that z = 0; consequently, ϕ = 0. This contradiction completes the proof. At this stage, we can define operators Tˆ ± and (Tˆ ± )−1 on the elements F (p) = fˆ(· , p) and G(p) = gˆ(· , p) of H1/2,k,κ (∂S) and H−1/2,k,κ (∂S), k ∈ R, respectively, by setting (Tˆ ± F )(p) = (Tp± fˆ)(· , p), ± −1 (Tˆ ) G (p) = (Tp± )−1 gˆ (· , p). Finally, we return to the spaces of originals and define the Poincaré– Steklov operators T ± and (T ± )−1 on the elements f ∈ H1/2,k,κ (Γ) and g ∈ H−1/2,k,κ (Γ), k ∈ R, by means of the equalities T ± f = L−1 Tˆ ± Lf, (T ± )−1 g = L−1 (Tˆ ± )−1 Lg. 3.2 Theorem. For any κ > 0 and k ∈ R, the operators T ± are continuous and injective from H1/2,k,κ (Γ) to H−1/2,k−1,κ (Γ), and their ranges are dense in H−1/2,k−1,κ (Γ). Their inverses (T ± )−1 , extended by continuity from the ranges of T ± to H−1/2,k,κ (Γ), are continuous and injective from H−1/2,k,κ (Γ) to H1/2,k−1,κ (Γ) for any k ∈ R, and their ranges are dense in the corresponding spaces.

40


Proof. Let f ∈ H1/2,k,κ (Γ), g ∈ H−1/2,k,κ (Γ), Lf (x, t) = fˆ(x, p), Lg(x, t) = gˆ(x, p). First, we show that if F (p) = fˆ(· , p) is holomorphic from Cκ to H1/2 (∂S), then Φ(p) = (Tp± fˆ)(· , p) is holomorphic from Cκ to the space H−1/2 (∂S). In the proof of Theorem 2.3 we established that if F (p) is holomorphic from Cκ to H1/2 (∂S), then U (p) = u ˆ(· , p), where u ˆ(x, p) is the solution of either of the problems (3.1) with boundary data fˆ(x, p), is holomorphic from Cκ to H1 (S ± ). We now take any ϕ ∈ H1/2 (∂S) and construct w = l± ϕ. Let p0 ∈ Cκ . By (3.2),

Φ(p) − Φ(p0 ), ϕ 0;∂S = ± p2 B 1/2 U (p) − p20 B 1/2 U (p0 ), B 1/2 w 0;S ± + a± U (p) − U (p0 ), w = ± (p2 − p20 ) B 1/2 U (p), B 1/2 w 0;S ± + p20 B 1/2 U (p) − U (p0 ) , B 1/2 w 0;S ± + a± U (p) − U (p0 ), w ;

therefore,

Φ(p) − Φ(p0 ) ,ϕ p − p0 0;∂S 1/2 = ± (p + p0 ) B U (p), B 1/2 w 0;S ± U (p) − U (p0 ) 2 1/2 U (p) − U (p0 ) 1/2 + p0 B ,B w + a± ,w . p − p0 p − p0 0;S ±

¯ R (p0 ) ⊂ Let KR (p0 ) be a circle with the center at p0 and radius R such that K Cκ . Since ¯ R (p0 ), ˆ u1,p;S ± ≤ cˆ u1;S ± , p ∈ K where c = const > 0 is independent of p, it follows that lim

p→p0

Φ(p) − Φ(p0 ) ,ϕ p − p0 0;∂S 1/2 = ± 2p0 B U (p0 ), B 1/2 w 0;S ± + p2 B 1/2 U (p0 ), B 1/2 w 0;S ± + a± U (p0 ), w .

3.1 The Poincaré–Steklov Operators

41

This shows that Φ(p) is weakly holomorphic from Cκ to H−1/2 (∂S). According to Dunford’s theorem [18], any weakly holomorphic mapping is holomorphic in the strong sense, so our assertion is proved. Next, by (3.3) with p = σ + iτ and the definition of T ± , T ± f 2−1/2,k−1,κ;Γ ∞ = sup

σ>κ −∞

(1 + |p|2 )k−1 (Tp± F )(p)2−1/2,p;∂S dτ

∞ ≤ c sup

σ>κ −∞

(1 + |p|2 )k F (p)21/2,p;∂S dτ

= cf 21/2,k,κ;Γ . We now go over to the case of (T ± )−1 . For any p ∈ C0 , the Neumann boundary value problems Bp2 u ˆ(x, p) + (Aˆ u)(x, p) = 0, (T u ˆ)± (x, p) = gˆ(x, p),

x ∈ S±,

x ∈ ∂S,

admit a variational version that consists in finding u ˆ ∈ H1,p (S ± ) such that p2 (B 1/2 u ˆ, B 1/2 v)0;S ± + a± (ˆ u, v) = ±(ˆ g , γ ± v)0;∂S

∀v ∈ H1,p (S ± ),

(3.12)

where gˆ ∈ H−1/2,p (∂S) is prescribed. Let fˆ = (Tp± )−1 gˆ ∈ H1/2,p (∂S), and let u ˆ ∈ H1,p (S ± ) be the solutions of problems (3.1). Since gˆ = Tp± fˆ, from (3.2) it follows that u ˆ are the (unique) solutions of problems (3.12). In addition, (3.11) implies that ˆ u1,p;S ± ≤ c|p|ˆ g −1/2,p;∂S . Taking this inequality into account and repeating the proof of Theorem 2.3, we conclude that if G(p) = gˆ(· , p) is holomorphic from Cκ to H−1/2 (∂S), then the solutions U (p) = u ˆ(· , p) of (3.12) are holomorphic from Cκ to H1 (S ± ), ± respectively. Since the trace operators from H1 (S ± ) to ± γ−1 are continuous ± H1/2 (∂S), we deduce that F (p) = (Tp ) G (p) = γ U (p) are holomorphic from Cκ to H1/2 (∂S). Also, from (3.4) it follows that (T ± )−1 are continuous from H−1/2,k,κ (Γ) to H1/2,k−1,κ (Γ) for any k ∈ R.

42


To complete the proof, we only need to show that the ranges of T ± and (T ± )−1 are dense in the corresponding spaces. Let Hk± be the ranges of the operators T ± : H1/2,k,κ (Γ) → H−1/2,k−1,κ (Γ). We take any g ∈ H−1/2,k+1,κ (Γ) and set f = (T ± )−1 g ∈ H1/2,k,κ (Γ). Since g = T ± (T ± )−1 g, we have H−1/2,k+1,κ (Γ) ⊂ Hk± . It is obvious that for any k ∈ R, the space H−1/2,k+1,κ (Γ) is dense in H−1/2,k−1,κ (Γ); hence, Hk± are dense in H−1/2,k−1,κ (Γ). The case of (T ± )−1 is treated similarly.

3.2 Solvability of the Problems Consider problems (DN± ), whose classical formulation was given in §1.1. Their variational versions consist in finding u ∈ H1,0,κ (G± ) such that, for g prescribed on Γ, ∞ a± (u, v) − (B 1/2 ∂t u, B 1/2 ∂t v)0;S ± dt ∞ = ± (g, v ± )0;∂S dt

0

¯ ± ). ∀v ∈ C0∞ (G

(3.13)

0

3.3 Theorem. For any g ∈ H−1/2,1,κ (Γ), κ > 0, problems (3.13) have unique solutions u ∈ H1,0,κ (G± ). If g ∈ H−1/2,k,κ (Γ), k ∈ R, then these solutions belong to H1,k−1,κ (G± ), respectively, and u1,k−1,κ;G± ≤ cg−1/2,k,κ;Γ .

(3.14)

¯ κ , be the solutions of problems (3.12). It has already Proof. Let u ˆ(x, p), p ∈ C been shown that U (p) = u ˆ(· , p) are holomorphic from Cκ to H1 (S ± ). This and (3.11) imply that u = L−1 U ∈ H1,k−1,κ (G± ), and that (3.14) holds. The proof that u satisfies (3.13) is similar to that of the analogous assertion for (DD± ). Consequently, we only need to check the uniqueness of the solution to (3.13). Let u1 and u2 be two solutions of (3.13). Then u = u1 − u2 ∈ H1,0,κ (G± ) satisfies ∞ ¯ ± ). a± (u, v) − (B 1/2 ∂t u, B 1/2 ∂t v)0;S ± dt = 0 ∀v ∈ C0∞ (G 0

˚1 (S ± ), we Repeating the proof of Theorem 2.3 with H1 (S ± ) instead of H conclude that u = 0, as required. The more general problems (DN± ) for the nonhomogeneous equation of motion can be reduced to the above by means of an appropriate substitution (see Chapter 9).

4 Boundary Integral Equations for Problems with Dirichlet and Neumann Boundary Conditions

4.1 Time-dependent Potentials We recall that the single-layer (retarded) potential V α of density α defined on ∂S × R was introduced in §1.3 as ∞ D(x − y, t − τ )α(y, τ ) dsy dτ.

(V α)(X) = −∞ ∂S

If α(X) = 0 for t < 0, then the Laplace transform of this potential is ˆ − y, p)ˆ D(x α(y, p) dsy .

(Vp α ˆ )(x, p) = ∂S

ˆ ˆ T (−x, p), the above equality can also be We remark that, since D(x, p) = D written in the form ˆ (j) (y − x, p) ej dsy , (Vp α ˆ )(x, p) = α ˆ (y, p), D ∂S

where ej are the coordinate unit vectors. The main properties of Vp α ˆ for α ˆ ∈ C 2 (∂S) are listed in §1.3. On the basis of these properties, we can define a boundary operator Vp,0 by setting Vp,0 α ˆ = (Vp α ˆ )0 = γ ± π ± Vp α ˆ,

p ∈ C0 .

4.1 Lemma. For any p ∈ C0 , the operator Vp,0 , extended by continuity from C 2 (∂S) to H−1/2,p (∂S), is a homeomorphism from H−1/2,p (∂S) to ¯ κ, H1/2,p (∂S), and for any α ˆ ∈ H−1/2,p (∂S), p ∈ C

44

4 Boundary Integral Equations

Vp,0 α ˆ 1/2,p;∂S ≤ c|p|ˆ α−1/2,p;∂S ,

(4.1)

ˆ α−1/2,p;∂S ≤ c|p|Vp,0 α ˆ 1/2,p;∂S ,

(4.2)

ˆ 1,p;S + + π − Vp α ˆ 1,p;S − ≤ c|p|ˆ α−1/2,p;∂S . π + Vp α

(4.3)

Proof. Let α ˆ ∈ C 2 (∂S). From the properties of Vp α ˆ it follows that Vp,0 α ˆ∈ H1/2,p (∂S) and π ± Vp α ˆ ∈ H1,p (S ± ); consequently, ˆ 21/2,p;∂S ≤ cπ + Vp α ˆ 21,p;S + Vp,0 α ≤ c(π + Vp α ˆ 21,p;S + + π − Vp α ˆ 21,p;S − ).

(4.4)

ˆ are the solutions of (3.1) with f = Vp,0 α ˆ in S ± , respectively, Since π ± Vp α ¯ κ, (3.9) implies that for p ∈ C π + Vp α ˆ 21,p;S + + π − Vp α ˆ 21,p;S −

≤ c Re p¯ (Tp+ − Tp− )Vp,0 α ˆ , Vp,0 α ˆ 0;∂S . ˆ takes We remark that the jump formula for the single-layer potential Vp α the form (Tp+ − Tp− )Vp,0 α ˆ=α ˆ; (4.5) hence,

π + Vp α ˆ 21,p;S + + π − Vp α ˆ 21,p;S − ≤ c|p||(ˆ α, Vp,0 α ˆ )0;∂S |.

(4.6)

By (4.4) and (4.6), ˆ 21/2,p;∂S ≤ c|p|Vp,0 α ˆ 1/2,p;∂S ˆ α−1/2,p;∂S , Vp,0 α so

Vp,0 α ˆ 1/2,p;∂S ≤ c|p|ˆ α−1/2,p;∂S

∀ˆ α ∈ C 2 (∂S).

(4.7)

Furthermore, from (4.7) it follows that ˆ 1,p;S + + π − Vp α ˆ 1,p;S − ≤ c|p|ˆ α−1/2,p;∂S . π + Vp α Inequality (4.7) shows that Vp,0 can be extended by continuity to H−1/2,p (∂S), that the extended operator is continuous from the space H−1/2,p (∂S) to H1/2,p (∂S), that (4.1) holds because the operators Tp± are continuous from H1/2,p (∂S) to H−1/2,p (∂S), and that (4.5) remains valid for all densities α ˆ ∈ H−1/2,p (∂S). From the above estimates it follows that (4.3) is also valid for any α ˆ ∈ H−1/2,p (∂S). The equality ˆ = γ − π − Vp α ˆ = Vp,0 α ˆ, γ + π + Vp α

4.1 Time-dependent Potentials

45

which holds for smooth densities, also holds for any α ˆ ∈ H−1/2,p (∂S). By (4.5), Vp,0 is injective from H−1/2,p (∂S) to H1/2,p (∂S) and −1 = Tp+ − Tp− . Vp,0 −1 is continuous from H1/2,p (∂S) to This formula and (3.3) imply that Vp,0 H−1/2,p (∂S), and that (4.2) holds. To complete the proof, we need to check that the range of Vp,0 is dense in H1/2,p (∂S). If we assume the opposite, then there is a nonzero ϕ ∈ H−1/2,p (∂S) such that

(Vp,0 α ˆ , ϕ)0;∂S = 0 ∀ˆ α ∈ H−1/2,p (∂S). Taking α ˆ = ϕ, from (4.6) we see that π ± Vp ϕ = 0; therefore, Vp,0 ϕ = γ + π + Vp ϕ = γ − π − Vp ϕ = 0, so ϕ = 0. This contradiction proves the assertion. Before returning to the spaces of originals, we need an auxiliary result. We consider the problem of finding u ∈ H1,p (R2 ) such that p2 (B 1/2 u, B 1/2 v)0 + a(u, v) = (q, v)0 + (g, γv)0;∂S where

∀v ∈ H1,p (R2 ),

(4.8)

a(u, v) = 2

E(u, v) dx,

R2

γv is the trace of v on ∂S, and q ∈ H−1,p (R2 ) and g ∈ H−1/2,p (∂S) are prescribed. 4.2 Lemma. For any given q ∈ H−1,p (R2 ) and g ∈ H−1/2,p (∂S), p ∈ C0 , ¯ κ , κ > 0, problem (4.8) has a unique solution u ∈ H1,p (R2 ), and for any p ∈ C u1,p ≤ c|p|(q−1,p + g−1/2,p;∂S ).

(4.9)

Proof. First, we remark that (g, γv)0;∂S defines a bounded antilinear (conjugate linear) functional on H1,p (R2 ) since, by the trace theorem, |(g, γv)0;∂S | ≤ g−1/2,p;∂S γv1/2,p;∂S ≤ cg−1/2,p;∂S v1,p , q −1,p ≤ cg−1/2,p;∂S and so there is q˜ ∈ H−1,p (R2 ) such that ˜ q , v)0 (g, γv)0;∂S = (˜

∀v ∈ H1,p (R2 ).

46


Writing Q = q + q˜ ∈ H−1,p (R2 ), we bring (4.8) to the form p2 (B 1/2 u, B 1/2 v)0 + a(u, v) = (Q, v)0

∀v ∈ H1,p (R2 ).

(4.10)

We now fix κ > 0 and repeat the corresponding part of the proof of Theorem 2.2. Thus, we introduce the form aκ (u, v) = 12 κ2 (B 1/2 u, B 1/2 v)0 + a(u, v), which, clearly, is continuous, symmetric, and coercive on [H1,p (R2 )]2 . This form defines a self-adjoint nonnegative operator Aκ by (Aκ u, v)0 = aκ (u, v) ∀u, v ∈ H1,p (R2 ). Then we rewrite (4.10) in the form Aκ u + (p2 − 12 κ2 )Bu = Q and prove the unique solvability of this equation in H1,p (R2 ), just as we did in Theorem 2.2. Taking v = u in (4.10) and separating the real and imaginary parts, we obtain

|p|2 B 1/2 u20 + a(u, u) = σ −1 Re p¯(Q, u)0 ; therefore, u1,p ≤ c|p|Q−1,p ≤ c|p|(q−1,p + g−1/2,p;∂S ), which proves (4.9). We now define a pair of operators Vˆ0 and Vˆ0−1 on the elements α ˜ (p) = ˆ α ˆ (· , p) and F (p) = f (· , p), p ∈ Cκ , of H−1/2,k,κ (∂S) and H1/2,k,κ (∂S), respectively, by ˜ )(p) = (Vp,0 α ˆ )(· , p), (Vˆ0 α −1 ˆ (Vˆ0−1 F )(p) = (Vp,0 f )(· , p).

We also define the single-layer potential Vˆ α ˜ by ˆ )(x, p), (Vˆ α ˜ )(x, p) = (Vp α

x ∈ R2 , p ∈ Cκ .

Returning to the spaces of originals, we define operators V0 and V0−1 and the single-layer potential V α by setting V0 α = L−1 Vˆ0 Lα, V0−1 f = L−1 Vˆ0−1 Lf, (V α)(X) = (L−1 Vˆ α ˜ )(X),

X ∈ R2 × R.


47

4.3 Theorem. For any κ > 0 and k ∈ R, the operator V0 is continuous and injective from H−1/2,k,κ (Γ) to H1/2,k−1,κ (Γ), and its range is dense in H1/2,k−1,κ (Γ). The inverse V0−1 , extended by continuity from the range of V0 , is continuous and injective from H1/2,k,κ (Γ) to H−1/2,k−1,κ (Γ) for any k ∈ R, and its range is dense in H−1/2,k−1,κ (Γ). In addition, for any α ∈ H−1/2,k,κ (Γ), π + V α1,k−1,κ;G+ + π − V α1,k−1,κ;G− ≤ cα−1/2,k,κ;Γ .

(4.11)

Proof. Let α ˜ (p) ∈ H−1/2,k,κ (∂S), p ∈ Cκ . Since Vp α ˆ is the solution of (4.8) with q = 0,

g = (Tp+ − Tp− )Vp,0 α ˆ=α ˆ,

from (4.9) it follows that π + Vp α ˆ 1,p;S + + π − Vp α ˆ 1,p;S − ≤ c|p|ˆ α−1/2,p;∂S . We claim that π ± Vp α ˆ (· , p) are holomorphic mappings from Cκ to H1 (S ± ), respectively. To show this, we choose any p0 ∈ Cκ , consider a circle KR (p0 ) ¯ R (p0 ) ⊂ Cκ , and rewrite (4.8) with the center at p0 and radius R such that K as p20 (B 1/2 Vp α ˆ , B 1/2 v)0 + a(Vp α ˆ , v) = −(p2 − p20 )(B 1/2 Vp α ˆ , B 1/2 v)0 + (ˆ α, γv)0;∂S

∀v ∈ H1,p (R2 ).

Taking (4.9) into account and repeating the proof of Theorem 3.2, we conclude, successively, that Vp α ˆ (· , p) is a bounded, continuous, and holomorphic mapping from Cκ to H1 (R2 ), and that, consequently, π ± Vp α ˆ (· , p) are holomorphic mappings from Cκ to H1 (S ± ); therefore, Vp,0 α ˆ = γ + π + Vp α ˆ = γ − π − Vp α ˆ is holomorphic from Cκ to H1/2 (∂S). Since V0 α21/2,k−1,κ;Γ

∞ = sup

σ>κ −∞

(1 + |p|2 )k−1 Vp,0 α ˆ 21/2,p;∂S dτ

∞ ≤ c sup

σ>κ −∞

(1 + |p|2 )k ˆ α2−1/2,p;∂S dτ

= α2−1/2,k,κ;Γ ,

p = σ + iτ,

48


the mapping V0 : H−1/2,k,κ (Γ) → H1/2,k−1,κ (Γ) is continuous for any k ∈ R. If ˆ = 0, then (Vˆ0 α ˜ )(p) = 0, α(p) ˜ =α ˆ (·, p), which means that (Vp,0 α ˆ )(· , p) = 0 V0 α for every p ∈ Cκ ; hence, α ˆ = 0. The continuity of V0−1 follows from the equality V0−1 = T + − T − and Theorem 3.2. Finally, the statement that the ranges of V0 and V0−1 are dense in the corresponding spaces follows from the fact that H±1/2,k+1,κ (Γ) are dense in H±1/2,k−1,κ (Γ), respectively. Inequality (4.11) is obtained from (4.3), and the assertion is proved. We now go over to the properties of the (retarded) double-layer potential W β with density β defined on Γ. This potential was introduced in §1.3 for smooth densities as ∞ P (x − y, t − τ )β(y, τ ) dsy dτ

(W β)(X) = 0 ∂S ∞

=

β(y, τ ), (Ty D(j) )(y − x, t − τ ) ej dsy dτ,

0 ∂S

and its Laplace transform is ˆ ˆ p) dsy (Wp β)(x, p) = Pˆ (x − y, p)β(y, ∂S

=

ˆ p), (Ty D ˆ (j) )(y − x, p) ej dsy . β(y,

∂S

4.4 Lemma. For any p ∈ C0 and βˆ ∈ C 2 (∂S), the double-layer potential Wp βˆ can be written in the form ⎧ ˆ ⎨ (Vp Tp− β)(x, p), x ∈ S + , ˆ p) = (Wp β)(x, (4.12) ⎩ (V T + β)(x, ˆ p), x ∈ S − . p p ˆ (j) (y − x, p) as a function of Proof. Let x ∈ S + be fixed, and consider D − ¯ y ∈ S . Also, let f, ϕ ∈ H1/2 (∂S), and let u and v be the solutions of the exterior problem (3.1) with boundary data f and ϕ, respectively. By the definition of T − , ¯ 0;∂S = p2 (B 1/2 u, B 1/2 v¯)0;S − + a− (u, v¯), −(Tp− f, ϕ) −(T − ϕ, f¯)0;∂S = p2 (B 1/2 v, B 1/2 , u ¯)0;S − + a− (v, u ¯). p


49

Since the right-hand sides above are equal, we deduce that ¯ 0;∂S = (Tp− ϕ, f¯)0;∂S , (Tp− f, ϕ) or

(Tp− f )(y), ϕ(y) dsy =

∂S

f (y), (Tp− ϕ)(y) dsy .

∂S

ˆ p) and ϕ(y) = D ˆ (j) (y − x, p) in this equality, we find that Taking f (y) = β(y, − ˆ ˆ (j) (y − x, p) dsy (Tp β)(y, p), D ∂S ˆ p), (T − D ˆ (j) )(y − x, p) dsy . β(y, = p ∂S

Multiplying both sides above by ej and summing over j from 1 to 3, we obtain the first equality (4.12). The second one is derived analogously. From the properties of the double-layer potential with a smooth density it follows that π ± Wp βˆ ∈ H1,p (S ± ). Consequently, we can define the operators Wp± of the limiting values on ∂S of Wp βˆ with βˆ ∈ C 2 (∂S) by ˆ Wp± βˆ = γ ± π ± Wp β. 4.5 Lemma. For any p ∈ C0 , the double-layer potential Wp βˆ and the operators Wp± of its limiting values can be extended by continuity from C 2 (∂S) to H1/2,p (∂S). The extended operators Wp± are homeomorphisms from the space H1/2,p (∂S) to H1/2,p (∂S), and the extended functions π ± Wp βˆ are continuous from H1/2,p (∂S) to H1,p (S ± ), respectively. In addition, for any βˆ ∈ ¯ κ , κ > 0, H1/2,p (∂S), p ∈ C ˆ 1/2,p;∂S , ˆ 1,p;S + + π − Wp β ˆ 1,p;S − ≤ c|p|2 β π + Wp β ˆ 1/2,p;∂S Wp± β

2

(4.13)

ˆ 1/2,p;∂S , ≤ c|p| β

(4.14)

ˆ 1/2,p;∂S . ˆ 1/2,p;∂S ≤ c|p|2 W ± β β p

(4.15)

This assertion and inequalities (4.13)–(4.15) follow from Lemmas 3.1 and 4.1. ˜ ˆ ± )−1 on the elements β(p) ˆ ± and (W = We now define pairs of operators W ˆ , p) and F (p) = fˆ(· , p), p ∈ Cκ , of H1/2,k,κ (∂S) by β(·

˜ ˆ , p), ˆ ± β)(p) (W = (Wp± β)(· ˆ ± )−1 F (p) = (W ± )−1 fˆ (· , p), (W p

50


ˆ β˜ by and the double-layer potential W ˜ ˆ ˆ β)(x, (W p) = (Wp β)(x, p),

x ∈ S ± , p ∈ Cκ .

Returning to the spaces of originals, we define operators W ± and (W ± )−1 and the double-layer potential W β by setting ˆ ± Lβ, W ± β = L−1 W ˆ ± )−1 Lf, (W ± )−1 f = L−1 (W ˜ ˆ β)(X), (W β)(X) = (L−1 W

X ∈ G+ ∪ G− .

4.6 Theorem. For any κ > 0 and k ∈ R, the operators W ± are continuous and injective from H1/2,k,κ (Γ) to H1/2,k−2,κ (Γ). Their inverses (W ± )−1 , extended by continuity from the ranges of W ± , respectively, to H1/2,k,κ (Γ), are continuous and injective operators from H1/2,k,κ (Γ) to H1/2,k−2,κ (Γ) for any k ∈ R, and their ranges are dense in H1/2,k−2,κ (Γ). In addition, for any β ∈ H1/2,k,κ (Γ), π + W β1,k−2,κ;G+ + π − W β1,k−2,κ;G− ≤ cβ1/2,k,κ;Γ . Proof. The assertion follows from the equalities (W β)(X) =

(V T − β)(X),

X ∈ G+ ,

(V T + β)(X),

X ∈ G− ,

W ± β = V0 T ∓ β, and Theorems 3.2 and 4.3. 4.7 Lemma. For any β ∈ H1/2,k,κ (Γ), κ > 0, k ∈ R, W + β − W − β = −β,

(4.16)

T + W + β = T − W − β.

(4.17)

Proof. Let p ∈ Cκ . By (4.5), Wp+ βˆ − Wp− βˆ = Vp,0 (Tp− − Tp+ )βˆ = −βˆ

(4.18)

and Tp+ Wp+ βˆ = Tp+ Vp,0 Tp− βˆ = Tp− βˆ + Tp− Vp,0 Tp− βˆ ˆ = Tp− Vp,0 Tp+ βˆ = Tp− Wp− β.

(4.19)

4.2 Nonstationary Boundary Integral Equations

51

It is now easy to see that (4.16) and (4.17) follow from (4.18) and (4.19), respectively. We write Np = Tp+ Wp+ = Tp− Wp− and use the above procedure to define ˆ, N ˆ −1 and N, N −1 . the operators N 4.8 Theorem. For any κ > 0 and k ∈ R, the operator N is continuous and injective from H1/2,k,κ (Γ) to H−1/2,k−3,κ (Γ), and its range is dense in H−1/2,k−3,κ (Γ). Its inverse N −1 , extended by continuity from the range of N to H−1/2,k,κ (Γ), is continuous and injective from the space H−1/2,k,κ (Γ) to H1/2,k−1,κ (Γ) for any k ∈ R, and its range is dense in H1/2,k−1,κ (Γ). Proof. The assertion concerning N follows from Theorems 3.2 and 4.6. Alternatively, it can be established from the estimate ˆ 1/2,p;∂S , ˆ −1/2,p;∂S ≤ c|p|3 β Np β which is obtained from (3.3) and (4.14). ¯ κ . By (4.18), (4.19), and the trace theorem, Let βˆ ∈ H1/2,p (∂S) and p ∈ C −ˆ +ˆ 2 ˆ 2 β 1/2,p;∂S = Wp β − Wp β1/2,p;∂S −ˆ 2 ˆ 2 ≤ c(Wp+ β 1/2,p;∂S + Wp β1/2.p;∂S )

ˆ 2 + + π − Wp β ˆ 2 −) ≤ c(π + Wp β 1,p;S 1,p;S ˆ Np β) ˆ 0;∂S | ≤ c|p||(Wp+ βˆ − Wp− β, ˆ 1/2,p;∂S ; ˆ −1/2,p;∂S β ≤ c|p|Np β therefore, ˆ 1/2,p;∂S ≤ c|p|Np β ˆ −1/2,p;∂S . β This inequality and Theorems 3.2 and 4.6 complete the proof.

4.2 Nonstationary Boundary Integral Equations Problems (DD± ) consist in solving B(∂t2 u)(X) + (Au)(X) = 0,

X ∈ G± ,

u(x, 0+) = (∂t u)(x, 0+) = 0,

x ∈ S±,

u± (X) = f (X),

X ∈ Γ,

where f is prescribed on Γ. We seek their solutions in the form u(X) = (V α)(X),

X ∈ G+ or X ∈ G− ,

(4.20)

52


where α is an unknown density defined on ∂S × R, which is zero for t < 0. Alternatively, we can write u = π ± V α. This representation yields the system of nonstationary boundary integral equations V0 α = f, (4.21) where V0 is the boundary operator defined by the single-layer potential (see §4.1). On the other hand, the representation u(X) = (W β)(X), or, which is the same,

X ∈ G+ or x ∈ G− ,

(4.22)

u = π ± W β,

of the solutions of these problems yields the boundary systems W ± β = f,

(4.23)

where W ± are the operators of the limiting values on Γ of the double-layer potential and β is an unknown density defined on ∂S × R and vanishing for t < 0. 4.9 Theorem. For any f ∈ H1/2,k,κ (Γ), κ > 0, k ∈ R, systems (4.21) and (4.23) have unique solutions α ∈ H−1/2,k−1,κ (Γ) and β ∈ H1/2,k−2,κ (Γ), in which case the functions u defined by (4.20) or (4.22) belong to H1,k−1,κ (G± ). If k ≥ 1, then these functions are the weak solutions of problems (DD± ), respectively. Proof. The assertion concerning the solvability of (4.21) and (4.23) follows from Theorems 4.3 and 4.6. Now let α be the solution of (4.21) with f ∈ H1/2,k,κ (Γ). Then π ± Vp α ˆ∈ H1,p (S ± ) for any p ∈ Cκ and are infinitely differentiable with respect to x ∈ S ± . A straightforward calculation shows that these functions satisfy p2 B 1/2 (π ± Vp α ˆ ), B 1/2 v + a± (u, v) = 0 ∀v ∈ C0∞ (S ± ), γ ± π ± Vp α ˆ = Vp,0 α ˆ = fˆ.

(4.24)

˚1,p (S ± ). Furthermore, by (4.6), It is clear that (4.24) also hold for all v ∈ H −1 ˆ π + Vp α ˆ 21,p;S + + π − Vp α ˆ 21,p;S − ≤ c|p||(fˆ, Vp,0 f )0;∂S |,

so (4.2) implies that ˆ 1,p;S + + π − Vp α ˆ 1,p;S − ≤ c|p|fˆ1/2,p;∂S . π + Vp α

4.3 The Direct Method

53

Taking this inequality into account and repeating the proof of Theorem 2.3, we arrive at the required assertion. The case of (4.23) is treated similarly. We now represent the solutions of problems (DN± ) in the form (4.20) or (4.22). These representations yield, respectively, the systems of nonstationary boundary integral equations T ± V0 α = g,

(4.25)

N β = g.

(4.26)

4.10 Theorem. For any g ∈ H−1/2,k,κ (Γ), κ > 0, k ∈ R, systems (4.25) and (4.26) have unique solutions α ∈ H−1/2,k−2,κ (Γ) and β ∈ H1/2,k−1,κ (Γ), in which case the functions u defined by (4.20) or (4.22) belong to the space H1,k−1,κ (G± ). If k ≥ 1, then these functions are the weak solutions of problems (DN± ), respectively. The proof of this assertion is a repeat of that of Theorem 4.9, use being made of Theorems 3.2, 4.3, and 4.8.

4.3 The Direct Method We begin by establishing the dynamic analog of the third Green’s formula (or Somigliana representation formula, as is known in elasticity theory). Let u ∈ H1,k,κ (G+ ), k ∈ R, κ > 0, be the weak solution of problem (DD+ ) B(∂ 2 u)(X) + (Au)(X) = 0,

X ∈ G+ ,

u(x, 0+) = (∂t u)(x, 0+) = 0,

x ∈ S+,

+

(γ u)(X) = f (X),

(4.27)

X ∈ Γ,

where f ∈ H1/2,k,κ (Γ). Then for any p ∈ Cκ , u ˆ(x, p) = Lu(x, t) is the weak solution of the problem Bp2 u ˆ(x, p) + (Aˆ u)(x, p) = 0, x ∈ S + , (γ + u ˆ)(x, p) = fˆ(x, p), x ∈ ∂S, where fˆ ∈ H1/2,p (∂S). We now consider the function w = π + Vp Tp+ fˆ − π + Wp fˆ, which belongs to H1,p (S + ). Given that γ + w = Vp,0 Tp+ fˆ − Wp+ fˆ = Vp,0 (Tp+ − Tp− )fˆ = fˆ,

(4.28)

54


w is also a weak solution of (4.28). Since (4.28) has a unique solution in H1,p (S + ), we conclude that u ˆ(x, p) = (π + Vp Tp+ fˆ)(x, p) − (π + Wp fˆ)(x, p),

x ∈ S+.

Returning to the spaces of originals, we obtain u(X) = (π + V T + f )(X) − (π + W f )(X),

X ∈ G+ .

(4.29)

From this it follows that f = V0 T + f − W + f . Setting α = T + f and taking into account that f + W + f = W − f , we arrive at V0 α = W − f.

(4.30)

This nonstationary boundary integral equation for the density α has a direct physical meaning, since α is the boundary moment-force field. Solving (4.30), we obtain the solution of (4.27) as u(X) = (π + V α)(X) − (π + W f )(X),

X ∈ G+ .

In the case of the exterior problem (DD− ), it is not difficult to show that representation (4.29) takes the form u(X) = −(π − V T − f )(X) + (π − W f )(X),

X ∈ G− ,

(4.31)

which leads to the system of boundary integral equations V0 α = W + f

(4.32)

for the density α = T − f . 4.11 Theorem. For any f ∈ H1/2,k,κ (Γ), κ > 0, k ∈ R, systems (4.30) and (4.32) have unique solutions α ∈ H−1/2,k−1,κ (Γ), in which case u = ±(π ± V α − π − W f ) ∈ H1,k−1,κ (G± ). If k ≥ 1, then these functions u are the weak solutions of problems (DD± ), respectively. The assertion is proved by using the representations α = T ± f and the properties of the boundary operators defined in §4.1. Going over to (DN± ), we remark that, since here the boundary momentforce vector T ± γ ± u = g is known, we write the representations (4.29) and (4.31) as (4.33) u(X) = ± (π ± V g)(X) − (π ± W β)(X) , X ∈ G± ,

4.3 The Direct Method

55

where β = γ ± u is the unknown displacement field on the boundary. Since γ ± u = ±(V0 g−W ± β), we arrive at the systems of boundary integral equations N β = T ∓ V0 g.

(4.34)

4.12 Theorem. For any g ∈ H−1/2,k,κ (Γ), κ > 0, k ∈ R, systems (4.34) have unique solutions β ∈ H1/2,k−1,κ (Γ), in which case the functions u defined by (4.33) belong to H1,k−1,κ (G± ). If k ≥ 1, then these functions are the weak solutions of problems (DN± ), respectively. The assertion follows from the representations β = (T ± )−1 g and the properties of the boundary operators discussed above.

5 Transmission Problems and Multiply Connected Plates

5.1 Infinite Plate with a Finite Inclusion Suppose that S + and S − are the middle plane domains of plates with different Lamé constants λ+ , µ+ and λ− , µ− , different densities ρ+ , ρ− , and different thickness parameters h+ , h− . Let B+ and B− be the corresponding diagonal matrices B. Also, let A± , a± , and T± be, respectively, the matrix differential operators, internal energy bilinear forms, and boundary moment-force operators associated with S ± . If u ∈ H1,k,κ (R2 × (0, ∞)), then we make the notation u± = π ± u ∈ H1,k,κ (G± ),

u = {u+ , u− }.

Similarly, for the Laplace transform u ˆ ∈ H1,p (R2 ) of u we write u ˆ = {ˆ u+ , u ˆ− },

u ˆ± = π ± u ˆ ∈ H1,p (S ± ).

In §1.1 we stated that the classical transmission (contact) problem (DT) consists in finding ¯ + ), u+ ∈ C 2 (G+ ) ∩ C 1 (G

¯−) u− ∈ C 2 (G− ) ∩ C 1 (G

satisfying B+ (∂t2 u+ )(X) + (A+ u+ )(X) = 0,

X ∈ G+ ,

B− (∂t2 u− )(X) + (A− u− )(X) = 0,

X ∈ G− ,

u+ (x, 0+) = (∂t u+ )(x, 0+) = 0,

x ∈ S+,

u− (x, 0+) = (∂t u− )(x, 0+) = 0,

x ∈ S−,

− u+ + (X) − u− (X) = f (X),

X ∈ Γ,

(T+ u+ )+ (X) − (T− u− )− (X) = g(X),

X ∈ Γ.

(5.1)

58

5 Transmission Problems

Let v ∈ C0∞ (R2 × [0, ∞)), and let v± = π ± v. Multiplying the equations ∗ ∗ of motion for the two plates by v+ and v− , integrating over G+ and G− , respectively, and adding the resulting equalities, we arrive at the variational equation ∞ 0

a+ (u+ , v+ ) + a− (u− , v− ) 1/2 1/2 1/2 1/2 − (B+ ∂t u+ , B+ ∂t v+ )0;S + − (B− ∂t u− , B− ∂t v− )0;S − dt ∞ = (g, v0 )0;∂S dt ∀v ∈ C0∞ (R2 × [0, ∞)),

(5.2)

0 + − where v0 = v+ = v− . Equality (5.2) suggests that the variational version of problem (DT) in the more general case of the nonhomogeneous equation of motion should consist in finding u = {u+ , u− }, u± ∈ H1,0,κ (G± ), such that

∞ 0

a+ (u+ , v+ ) + a− (u− , v− ) 1/2 1/2 1/2 1/2 − (B+ ∂t u+ , B+ ∂t v+ )0;S + − (B− ∂t u− , B− ∂t v− )0;S − dt ∞ = (q, v)0 + (g, v0 )0;∂S dt ∀v ∈ C0∞ (R2 × [0, ∞)), 0

(5.3)

γ + u+ − γ − u− = f,

where q ∈ H−1,k,κ (R2 × (0, ∞)), f ∈ H1/2,k,κ (Γ), and g ∈ H−1/2,k,κ (Γ) are prescribed. 5.1 Theorem. For any q ∈ H−1,1,κ (R2 × (0, ∞)), f ∈ H1/2,1,κ (Γ), and g ∈ H−1/2,1,κ (Γ), κ > 0, problem (5.3) has a unique solution u = {u+ , u− }, where u± ∈ H1,0,κ (G± ). If q ∈ H−1,k,κ (R2 × (0, ∞)), f ∈ H1/2,k,κ (Γ), and g ∈ H−1/2,k,κ (Γ), k ∈ R, then u± ∈ H1,k−1,κ (G± ) and u+ 1,k−1,κ;G+ + u− 1,k−1,κ;G− ≤ c(q−1,k,κ;R2 ×(0,∞) + f 1/2,k,κ;Γ + g−1/2,k,κ;Γ ).

(5.4)

Proof. We begin by rewriting (5.3) in terms of Laplace transforms. Then, with the usual notation u ˆ = Lu,

qˆ = Lq,

fˆ = Lf,

gˆ = Lg,

(DT) turns into the problem of finding u ˆ ∈ H1,p (R2 ), where u ˆ(x, p) = {ˆ u+ (x, p), u ˆ− (x, p)},

u ˆ± ∈ H1,p (S ± ),

p ∈ Cκ ,

5.1 Infinite Plate with a Finite Inclusion

59

such that a+ (ˆ u+ , v+ ) + a− (ˆ u− , v− ) 1/2

1/2

1/2

1/2

+ p2 (B+ u ˆ+ , B+ v+ )0;S + + p2 (B− u ˆ− , B− v− )0;S − = (ˆ q , v)0 + (ˆ g , γv)0;∂S

∀v ∈ H1,p (R2 ),

(5.5)

γ+u ˆ+ − γ − u ˆ− = fˆ. Here, v± = π ± v and γv is the trace of v on ∂S. First, we consider the case fˆ = 0. Then (5.5) reduces to the problem a+ (ˆ u+ , v+ ) + a− (ˆ u− , v− ) 1/2

1/2

1/2

1/2

+ p2 (B+ u ˆ+ , B+ v+ )0;S + + p2 (B− u ˆ− , B− v− )0;S − = (ˆ q , v)0 + (ˆ g , γv)0;∂S

∀v ∈ H1,p (R2 ). (5.6)

Since gˆ ∈ H−1/2,p (∂S), the form (ˆ g , γv)0;∂S defines a bounded antilinear (conjugate linear) functional on H1,p (R2 ); hence, it can be expressed as (ˆ g , γv)0;∂S = (˜ q , v)0

∀v ∈ H1,p (R2 ),

where q˜ ∈ H−1,p (R2 ) and ˜ q −1,p ≤ cˆ g −1/2,p;∂S . Writing Q = qˆ + q˜, we arrive at the problem a+ (ˆ u+ , v+ ) + a− (ˆ u− , v− ) 1/2

1/2

1/2

1/2

+ p2 (B+ u ˆ+ , B+ v+ )0;S + + p2 (B− u ˆ− , B− v− )0;S − = (Q, v)0

∀v ∈ H1,p (R2 ). (5.7)

Next, we introduce a bilinear form aκ (u, v) on [H1,p (R2 )]2 by aκ (u, v) =

1 2

1/2 1/2 1/2 1/2 κ2 (B+ u+ , B+ v+ )0;S + + (B− u− , B− v− )0;S − + a+ (u+ , v+ ) + a− (u− , v− ).

It is obvious that aκ (u, v) is continuous and coercive on [H1,p (R2 )]2 ; therefore, it generates a self-adjoint nonnegative operator Aκ through the equality (Aκ u, v)0 = aκ (u, v) ∀u, v ∈ H1,p (R2 ). Aκ is a homeomorphism from H1,p (R2 ) to H−1,p (R2 ). We can now write (5.7) in the form ˜u ˆ = Q, (5.8) Aκ u ˆ + p2 − 12 κ2 B

60


˜u where B ˆ = (B+ u ˆ+ , B− u ˆ− ) ∈ H−1,p (R2 ). In turn, (5.8) can be written as

or

u ˆ + p2 −

1 2

˜ ˆ = A−1 κ2 A−1 κ Bu κ Q,

u b + p2 −

1 2

˜ 1/2 A−1 Q, κ2 Bκ ub = B κ

(5.9)

˜ 1/2 u ˜ 1/2 A−1 B ˜ 1/2 . Following the scheme used in the where ub = B ˆ and Bκ = B κ proof of Theorem 2.2, we show that (5.9), regarded as an equation in H1,p (R2 ), is equivalent to itself in L2 (R2 ), and that Bκ is self-adjoint and nonnegative on L2 (R2 ). Consequently, (5.9) is uniquely solvable, so (5.7)—hence, also (5.6)—have a unique solution u ˆ ∈ H1,p (R2 ) for any Q ∈ H−1,p (R2 ). Taking v = u ˆ in (5.7) and separating the real and imaginary parts, we obtain ˆ u21,p ≤ c|p||(Q, u ˆ)0 |; therefore, ˆ u1,p ≤ c|p|Q−1,p ≤ c|p|(ˆ q −1,p + ˆ g −1/2,p;∂S ).

(5.10)

In the general case we construct w+ = l+ fˆ ∈ H1,p (S + ),

w = {w+ , 0},

then seek a solution of (5.5) of the form u ˆ = w+u ˜. Clearly, u ˜ ∈ H1,p (R2 ) satisfies a+ (˜ u+ , v+ ) + a− (˜ u− , v− ) 1/2

1/2

1/2

1/2

+ p2 (B+ u ˜+ , B+ v+ )0;S + + p2 (B− u ˜− , B− v− )0;S − = (ˆ q , v)0 + (ˆ g , γv)0;∂S − a+ (w+ , v+ ) 1/2

1/2

− p2 (B+ w+ , B+ v+ )0;S +

∀v ∈ H1,p (R2 ). (5.11)

Since the extension operator l+ is continuous (uniformly with respect to the parameter p ∈ C) from H1/2,p (∂S) to H1,p (S + ), we have l+ fˆ1,p;S + ≤ cfˆ1/2,p;∂S , which leads to 1/2

1/2

|a+ (w+ , v+ ) + p2 (B+ w+ , B+ v+ )0;S + | ≤ cw+ 1,p;S + v+ 1,p;S + ≤ cfˆ1/2,p;∂S v1,p .


61

Therefore, (5.11) has a solution u ˜ ∈ H1,p (R2 ) and, by (5.10), ˜ u1,p ≤ c|p|(ˆ q −1,p + fˆ1/2,p;∂S + ˆ g −1/2,p;∂S ). This implies that (5.5) also has a solution u ˆ, which satisfies u− 1,p;S − ˆ u+ 1,p;S + + ˆ ≤ c|p|(ˆ q −1,p + fˆ1/2,p;∂S + ˆ g −1/2,p;∂S ).

(5.12)

ˆ = {U ˆ+ , U ˆ− }, where U ˆ± (p) = We now use (5.12) to show successively that U u ˆ± (· , p), is bounded, continuous, and holomorphic from Cκ to H1 (S + ) × H1 (S − ). This and (5.12) prove that problem (DT) has a solution, which satisfies estimate (5.4). To show that this solution is unique, suppose that u ∈ H1,0,κ (R2 × (0, ∞)) satisfies ∞ 0

1/2 1/2 a+ (u+ , v+ ) + a− (u− , v− ) − (B+ ∂t u+ , B+ ∂t v+ )0;S + 1/2 1/2 − (B− ∂t u− , B− ∂t v− )0;S − dt = 0 ∀v ∈ C0∞ (R2 × [0, ∞)).

˚1 (S + ), Repeating the relevant part of the proof of Theorem 2.3 with L2 (S + ), H 2 2 2 and a+ (u, v) replaced by L (R ), H1 (R ), and a+ (u+ , v+ ) + a− (u− , v− ), we arrive at the desired conclusion. Having established the existence of a unique weak solution to our problem, we turn our attention to the representation of this solution in terms of various combinations of time-dependent (retarded) single-layer and double-layer potentials and prove the unique solvability of the corresponding systems of nonstationary boundary integral equations. Let V± α± and W± β± be the retarded single-layer and double-layer potentials constructed for the plates corresponding to S ± , respectively. The boundary operators generated by these potentials are denoted by V+,0 , V−,0 , W++ , W−− , and N+ , N− . The Poincaré–Steklov operators for S ± are denoted by T ± . Similar notation is used for their Laplace transform counterparts, with the only difference that the latter carry a subscript p. First we represent the (weak) solution u = {u+ , u− } of (5.1) in the form u+ (X) = (V+ α+ )(X),

X ∈ G+ ,

u− (X) = (V− α− )(X),

X ∈ G− ,

(5.13)

where α± are unknown densities defined on Γ. This yields the system of nonstationary boundary equations V+,0 α+ − V−,0 α− = f, T++ V+,0 α+

− T−− V−,0 α− = g.

(5.14)

62


The second representation is u+ (X) = (W+ β+ )(X),

X ∈ G+ ,

u− (X) = (W− β− )(X),

X ∈ G− ,

(5.15)

where β± are unknown densities defined on Γ. Representation (5.15) yields the system of boundary integral equations W++ β+ − W−− β− = f, N+ β+ − N− β− = g.

(5.16)

In the third case we seek the solution as u+ (X) = (V+ α+ )(X),

X ∈ G+ ,

u− (X) = (W− β− )(X),

X ∈ G− ,

(5.17)

which leads to the system V+,0 α+ − W−− β− = f, T++ V+,0 α+ − N− β = g.

(5.18)

Finally, seeking the solution in the form u+ (X) = (W+ β+ )(X),

X ∈ G+ ,

u− (X) = (V− α− )(X),

X ∈ G− ,

(5.19)

we arrive at the system W++ β+ − V−,0 α− = f, N+ β+ − T−− V−,0 α− = g.

(5.20)

5.2 Theorem. For any f ∈ H1/2,k,κ (Γ) and g ∈ H−1/2,k,κ (Γ), κ > 0, systems (5.14), (5.16), (5.18), and (5.20) have unique solutions α± ∈ H−1/2,k−2,κ (Γ),

β± ∈ H1/2,k−2,κ (Γ).

In each case, the corresponding function u defined by (5.13), (5.15), (5.17), or (5.19) belongs to H1,k−1,κ (G+ ) × H1,k−1,κ (G− ). If k ≥ 1, then u is the (weak ) solution of problem (DT). Proof. We rewrite all the systems of boundary equations in terms of their Laplace transforms. For p ∈ Cκ , systems (5.14), (5.16), (5.18), and (5.20)


63

take, respectively, the form Vp,+,0 α ˆ + − Vp,−,0 α ˆ − = fˆ, + − Tp,+ Vp,+,0 α ˆ + − Tp,− Vp,−,0 α ˆ − = gˆ, + ˆ − ˆ Wp,+ β+ − Wp,− β− = fˆ,

Np,+ βˆ+ − Np,− βˆ− = gˆ, − ˆ Vp,+,0 α ˆ + − Wp,− β− = fˆ, + Tp,+ Vp,+,0 α ˆ + − Np,− βˆ− = gˆ, + ˆ ˆ − = fˆ, β+ − Vp,−,0 α Wp,+ − Np,+ βˆ+ − Tp,− Vp,−,0 α ˆ − = gˆ.

(5.21)

(5.22)

(5.23)

(5.24)

Next, we consider the solution u ˆ(x, p) = {ˆ u+ (x, p), u ˆ− (x, p)} of problem (5.5) for any p ∈ Cκ . In §5.1 it was shown that u− 1,p;S − ≤ c|p| fˆ1/2,p;∂S + ˆ g −1/2,p;∂S . ˆ u+ 1,p;S + + ˆ

(5.25)

We write fˆ± = γ ± u ˆ± . By the trace theorem and (5.25), fˆ+ 1/2,p;∂S + fˆ− 1/2,p;∂S ≤ c|p| fˆ1/2,p;∂S + ˆ g −1/2,p;∂S .

(5.26)

+ ˆ − ˆ We also write gˆ+ = Tp,+ f+ and gˆ− = Tp,− f− . By (5.25) and (3.5),

ˆ g+ −1/2,p;∂S + ˆ g− −1/2,p;∂S ≤ c ˆ u+ 1,p;S + + ˆ u− 1,p;S − ≤ c|p| fˆ1/2,p;∂S + ˆ g −1/2,p;∂S .

(5.27)

−1 ˆ We now consider system (5.21). Let α ˆ ± = Vp,± f± . Then (5.26) shows that

ˆ α+ −1/2,p;∂S + ˆ α− −1/2,p;∂S ≤ c|p| fˆ+ 1/2,p;∂S + fˆ− 1/2,p;∂S ≤ c|p|2 fˆ1/2,p;∂S + ˆ g −1/2,p;∂S .

(5.28)

Clearly, u ˆ+ = Vp,+ α ˆ + and u ˆ− = Vp,− α ˆ − (hence, also α ˆ + and α ˆ − ) satisfy (5.21). The assertion concerning (5.14) follows from (5.28) and Theorem 5.1.

64


−1 −1 In the case of (5.22), we take βˆ+ = Np,+ gˆ+ and βˆ− = Np,− gˆ− . From the −1 properties of the operators (Np,± ) and (5.27) it follows that

βˆ+ 1/2,p;∂S + βˆ− 1/2,p;∂S ≤ c|p| ˆ g+ −1/2,p;∂S + ˆ g− −1/2,p;∂S ≤ c|p|2 fˆ1/2,p;∂S + ˆ g −1/2,p;∂S .

(5.29)

The desired statement now follows from (5.29) and Theorem 5.1. For systems (5.23) and (5.24) we set −1 ˆ f+ , α ˆ + = Vp,+,0 −1 βˆ+ = Np,+ gˆ+ ,

−1 βˆ− = Np,− gˆ− , −1 ˆ α ˆ − = Vp,−,0 f− ,

respectively. The proof is completed by repeating the above arguments, with the obvious changes.

5.2 Multiply Connected Finite Plate Consider a plate with Lamé constants λ+ , µ+ and density ρ+ , and characterized by the thickness parameter h+ and a middle plane domain S + whose boundary ∂S consists of two simple closed C 2 -curves ∂S1 and ∂S2 such that ∂S1 lies strictly inside the finite domain enclosed by ∂S2 . The unit normals on ∂S1 and ∂S2 are directed outwards with respect to S + . Let Si− be the domain interior to ∂S1 and Se− the domain exterior to ∂S2 . For k ∈ R, κ > 0, and p ∈ C0 , we define the spaces H±1/2,p (∂S) = H±1/2,p (∂S1 ) × H±1/2,p (∂S2 ), H±1/2,k,κ (∂S) = H±1/2,k,κ (∂S1 ) × H±1/2,k,κ (∂S2 ), H±1/2,k,κ (Γ) = H±1/2,k,κ (Γ1 ) × H±1/2,k,κ (Γ2 ), where Γν = ∂Sν × (0, ∞), ν = 1, 2,

Γ = Γ1 ∪ Γ2 .

We also write L2 (∂S) = L2 (∂S1 ) × L2 (∂S2 ). 2 If f = {f1 , f2 } and ϕ = {ϕ1 , ϕ2 } are elements of L2 (∂S) , then their L2 (∂S)inner product is defined as (f, ϕ)0;∂S = (f1 , ϕ1 )0;∂S1 + (f2 , ϕ2 )0;∂S2 .

5.2 Multiply Connected Finite Plate

65

We denote by γ + = {γ1+ , γ2+ } the trace operator that maps H1,p (S + ) continuously onto H1/2,p (∂S). Thus, for u ∈ H1,p (S + ), we write γ + u = {γ1+ u, γ2+ u}, where γν+ u are the traces of u on ∂Sν , ν = 1, 2. According to the notational convention adopted in §1.1, if u is a function in a space of originals, then its trace on Γ is also denoted by γ + u = (γ1+ u, γ2+ u). For any k ∈ R, the trace operator γ + : H1,k,κ (G+ ) → H1/2,k,κ (Γ) is continuous. Finally, we introduce extension operators lν+ , ν = 1, 2, such that for any fν ∈ H1/2,p (∂Sν ), lν+ fν ∈ H1,p (S + ),

+ γ3−ν lν+ fν = 0,

ν = 1, 2 (not summed).

If f = {f1 , f2 } ∈ H1/2,p (∂S), then we set l+ f = l1+ f1 + l2+ f2 . It is obvious that γ + l+ f = f and that l+ is continuous (uniformly with respect to p) from H1/2,p (∂S) to H1,p (S + ). We now formulate three initial-boundary value problems for the plate with middle section S + . The classical problem (DMCD) consists in finding u ∈ ¯ + ) such that C 2 (G+ ) ∩ C 1 (G B+ (∂t2 u)(X) + (A+ u)(X) = 0, u(x, 0+) = (∂t u)(x, 0+) = 0,

X ∈ G+ , x ∈ S+,

u+ (X) = f1 (X),

X ∈ Γ1 ,

u+ (X) = f2 (X),

X ∈ Γ2 ,

where f1 and f2 are prescribed. Consequently, in the variational version of (DMCD) we seek u ∈ H1,0,κ (G+ ) satisfying ∞

1/2 1/2 a+ (u, v) − (B+ ∂t u, B+ ∂t v)0;S + dt = 0

0 +

¯ + ), v + = 0, ∀v ∈ C0∞ (G

γ u = f, where a+ and B+ have the obvious meaning. ¯+) The classical problem (DMCN) consists in finding u ∈ C 2 (G+ ) ∩ C 1 (G such that B+ (∂t2 u)(X) + (A+ u)(X) = 0, u(x, 0+) = (∂t u)(x, 0+) = 0,

X ∈ G+ , x ∈ S+,

(T+ u)+ (X) = g1 (X),

X ∈ Γ1 ,

(T+ u)+ (X) = g2 (X),

X ∈ Γ2 ,

66


where g1 and g2 are prescribed. In its corresponding variational version, we seek u ∈ H1,0,κ (G+ ) satisfying ∞ 1/2 1/2 a+ (u, v) − (B+ ∂t u, B+ ∂t v)0;S + dt 0

∞ = (g, v + )0;∂S dt

¯ + ), ∀v ∈ C0∞ (G

0

where v + = {v1+ , v2+ } and g = {g1 , g2 }. ¯+) The classical problem (DMCM) consists in finding u ∈ C 2 (G+ ) ∩ C 1 (G such that B+ (∂t2 u)(X) + (A+ u)(X) = 0, X ∈ G+ , u(x, 0+) = (∂t u)(x, 0+) = 0, (T+ u)+ (X) = g1 (X), +

u (X) = f2 (X),

x ∈ S+,

X ∈ Γ1 , X ∈ Γ2 ,

where g1 and f2 are prescribed. In the variational version of this problem, we seek u ∈ H1,0,κ (G+ ) satisfying ∞ 1/2 1/2 a+ (u, v) − (B+ ∂t u, B+ ∂t v1+ )0;S + dt 0

∞ = (g1 , v1+ )0;∂S1 dt

¯ + ), v + = 0, ∀v ∈ C0∞ (G 2

0

γ2+ u = f2 . 5.3 Theorem. For any fν ∈ H1/2,1,κ (Γν ) and gν ∈ H−1/2,1,κ (Γν ), ν = 1, 2, κ > 0, problems (DMCD), (DMCN), and (DMCM) have unique solutions u ∈ H1,0,κ (G+ ). If fν ∈ H1/2,k,κ (Γν ) and gν ∈ H−1/2,k,κ (Γν ), k ∈ R, then each of these solutions u satisfies the corresponding estimate u1,k−1,κ;G+ ≤ c f1 1/2,k,κ;Γ1 + f2 1/2,k,κ;Γ2 , u1,k−1,κ;G+ ≤ c g1 −1/2,k,κ;Γ1 + g2 −1/2,k,κ;Γ2 , u1,k−1,κ;G+ ≤ c g1 −1/2,k,κ;Γ1 + f2 1/2,k,κ;Γ2 . Proof. Going over to Laplace transforms in problems (DMCD), (DMCN), and (DMCM), we arrive at new problems for every p ∈ Cκ . Thus, in (MCDp ), we seek u ˆ ∈ H1,p (S + ) such that 1/2 1/2 ˚1,p (S + ), p2 (B+ u ˆ, B+ v)0;S + + a+ (ˆ u, v) = 0 ∀v ∈ H γ+u ˆ = fˆ.

(5.30)


67

In (MCNp ), we seek u ˆ ∈ H1,p (S + ) such that 1/2

1/2

p2 (B+ u ˆ, B+ v)0;S + + a+ (ˆ u, v)

= (ˆ g , γ + v)0;∂S

∀v ∈ H1,p (S + ),

(5.31)

where γ + v = {γ1+ v, γ2+ v} is the trace of v on ∂S. ˆ ∈ H1,p (S + ) such that Finally, in (MCMp ), we seek u 1/2

1/2

p2 (B+ u ˆ, B+ v)0;S + + a+ (ˆ u, v) ˚1,p (S + , ∂S1 ), = (ˆ g1 , γ1+ v)0;∂S1 ∀v ∈ H γ+u ˆ = fˆ2 ,

(5.32)

2

˚1,p (S + , ∂S1 ) is the subspace of all u in H1,p (S + ) such that γ + u = 0. where H 2 We represent the solution of (MCDp ) in the form u ˆ = w + l+ fˆ. It is ˚1,p (S + ) satisfies obvious that w ∈ H 1/2

1/2

p2 (B+ w, B+ v)0;S + + a+ (w, v) = (Q1 , v)0;S +

˚1,p (S + ), ∀v ∈ H

(5.33)

˚1,p (S + ) by where Q1 ∈ H−1,p (S + ) is defined for v ∈ H 1/2 1/2 (Q1 , v)0;S + = − p2 (B+ l+ fˆ, B+ v)0;S + + a+ (l+ fˆ, v) . We see that |(Q1 , v)0;S + | ≤ cl+ fˆ1,p;S + v1,p ≤ cfˆ1/2,p;∂S v1,p ; hence, Q1 −1,p;S + ≤ cfˆ1/2,p;∂S . The unique solvability of (5.33) and the estimates w1,p ≤ c|p|Q1 −1,p;S + ≤ c|p|fˆ1/2,p;∂S are established by the method used in the proof of Theorem 2.1. Therefore, (5.30) is uniquely solvable in H1,p (S − ) and ˆ u1,p;S + ≤ c|p|fˆ1/2,p;∂S .

(5.34)

In the case of equation (5.31) we remark that for gˆ ∈ H−1/2,p (∂S), the form (ˆ g , γ + v)0;∂S defines a bounded antilinear (conjugate linear) functional on the space H1,p (S + ), so it can be written as (ˆ g , γ + v)0;∂S = (Q2 , v)0;S +

∀v ∈ H1,p (S + ).

68


Since

|(Q2 , v)0;S + | ≤ cˆ g −1/2,p;∂S γ + v1/2,p;∂S ≤ cˆ g −1/2,p;∂S v1,p;S + ,

˚−1,p (S + ) satisfies it follows that Q2 ∈ H Q2 −1,p ≤ cˆ g −1/2,p;∂S . We then write (5.31) in the form a+,κ (ˆ u, v) + (p2 −

1 2

1/2

1/2

κ2 )(B+ u ˆ, B+ v)0;S + = (Q2 , v)0;S +

∀v ∈ H1,p (S + ),

(5.35)

2 where the bilinear form a+,κ is defined on H1,p (S + ) by a+,κ (u, v) =

1 2

1/2

1/2

κ2 (B+ u, B+ v)0;S + + a+ (u, v).

2 It is obvious that a+,κ is continuous and coercive on H1,p (S + ) ; hence, it generates an operator Aκ that is a homeomorphism from H1,p (S + ) to ˚−1,p (S + ), by means of the equality H (Aκ u, v)0;S + = a+,κ (u, v) ∀u, v ∈ H1,p (S + ). We now rewrite (5.35) in the form Aκ u ˆ + p2 −

1 2

κ2 B+ u ˆ = Q2 .

(5.36)

Applying the Fredholm Alternative, we prove the unique solvability of (5.36)— hence, of (5.31)—in H1,p (S + ) for any gˆ ∈ H−1/2,p (∂S), p ∈ Cκ . Separating the real and imaginary parts in (5.35), we obtain ˆ u1,p;S + ≤ c|p|Q2 −1,p ≤ c|p|ˆ g −1/2,p;∂S .

(5.37)

The solution of (5.32) is sought in the form u ˆ = w + l2+ fˆ2 , where w ∈ ˚−1,p (S + , ∂S2 ) be the dual of H ˚1,p (S + , ∂S1 ) with respect ˚1,p (S + , ∂S1 ). Let H H to the duality generated by the inner product in L2 (S + ). It is obvious that ˚−1,p (S + , ∂S2 ) is the subspace of H−1,p (S + ∪ S¯− ) that consists of all q with H i ˚−1,p (S + , ∂S2 ) by q supp q ⊂ S¯+ . We denote the norm of q ∈ H ¯− . −1,p;S + ∪S i

We see that (ˆ g1 , γ1+ v)0;∂S1 defines a bounded antilinear (conjugate linear) ˚1,p (S + , ∂S1 ); hence, it can be expressed as functional on H ˜ 3 , v)0;S + (ˆ g1 , γ1+ v)0;∂S1 = (Q

˚1,p (S + , ∂S1 ), ∀v ∈ H


69

˜3 ∈ H ˚−1,p (S + , ∂S2 ) and where Q ˜3 Q g1 −1/2,p;∂S1 . ¯− ≤ cˆ −1,p;S + ∪S i

We write (5.32) as 1/2

1/2

p2 (B+ w, B+ v)0;S + + a+ (w, v) ˜ 3 , v)0;S + − p2 (B 1/2 l+ fˆ2 , B 1/2 v)0;S + − a+ (l+ fˆ2 , v) = (Q + 2 + 2 ˚1,p (S + , ∂S1 ). ∀v ∈ H

(5.38)

Since 1/2

1/2

p2 (B+ l2+ fˆ2 , B+ v)0;S + + a+ (l2+ fˆ2 , v) = (q, v)0;S + , where q ∈ H−1,p (S + , ∂S2 ) and q−1,p;S + ∪S¯− ≤ cfˆ2 1/2,p;∂S2 , i

˜ 3 − q = Q3 ∈ H ˚−1,p (S + , ∂S2 ) and see that we make the notation Q g1 −1/2,p;∂S1 . Q3 −1,p;S + ∪S¯− ≤ c fˆ2 1/2,p;∂S2 + ˆ i

Then (5.38) becomes 1/2

1/2

p2 (B+ w, B+ v) + a+ (w, v) = (Q3 , v)0;S +

˚1,p (S + , ∂S1 ). ∀v ∈ H

(5.39)

Taking into account the continuity and coerciveness of the form a+ (u, v) on ˚1,p (S + , ∂S1 )]2 , we adapt the proof of Theorem 2.1 to our situation and [H ˚1,p (S + , ∂S1 ) and that conclude that (5.39) has a unique solution w ∈ H w1,p;S + ∪S¯− ≤ c|p| fˆ2 1/2,p;∂S2 + ˆ g1 −1/2,p;∂S1 ). i

Consequently, (5.32) has a unique solution u ˆ ∈ H1,p (S + ) and ˆ u1,p;S + ≤ c|p| fˆ2 1/2,p;∂S2 + ˆ g1 −1/2,p;∂S1 .

(5.40)

To arrive at the desired assertion, we use (5.34), (5.37), and (5.40), and repeat the arguments in the proof of Theorem 2.3 with the obvious changes. We now construct an algebra of boundary operators in the case of multiply connected plates. We perform this construction in Sobolev spaces with a

70


parameter; the corresponding assertions in spaces of originals can then be easily formulated. In what follows, we use the notation a−,i (u, v) = E(u, v) dx, a−,e (u, v) = E(u, v) dx, Si−

Se−

where the bilinear form E is for the plate with the Lamé coefficients λ+ and µ+ . Let f = {f1 , f2 } and ϕ = {ϕ1 , ϕ2 } be any elements of H1/2,p (∂S), let u+ ∈ H1,p (S + ) be the solution of (5.30) with boundary data f , and suppose (i) (e) that u− ∈ H1,p (Si− ) and u− ∈ H1,p (Se− ) satisfy 1/2 (i) 1/2 (i) (i) (i) (i) ˚1,p (S − ), p2 (B+ u− , B+ v− )0;S − + a−,i (u− , v− ) = 0 ∀v− ∈ H i i

(i)

γ1− u− = f1 and 1/2 (e) 1/2 (e) (e) (e) (e) ˚1,p (S − ), p2 (B+ u− , B+ v− ) + a−,e (u− , v− ) = 0 ∀v− ∈ H e (e)

γ2− u− = f2 , where γ1− and γ2− are the trace operators that map H1,p (Si− ) and H1,p (Se− ) (i) continuously onto H1/2,p (∂S1 ) and H1/2,p (∂S2 ), respectively. Also, let v− (e)

(i)

and v− be elements of H1,p (Si− ) and H1,p (Se− ) such that γ1− v− = ϕ1 and (e) γ2− v− = ϕ2 . We define the Poincaré–Steklov operators Tp± by + + f, ϕ1 )0;∂S1 + (Tp,2 f, ϕ2 )0;∂S2 (Tp+ f, ϕ)0;∂S = (Tp,1 1/2

1/2

= p2 (B+ u+ , B+ v+ )0;S + + a+ (u+ , v+ ),

(5.41)

where v+ ∈ H1,p (S + ) and γ + v+ = ϕ, and − − (Tp− f, ϕ)0;∂S = (Tp,1 f1 , ϕ1 )0;∂S1 + (Tp,2 f2 , ϕ2 )0;∂S2

1/2 (i) 1/2 (i) (i) (i) = − p2 (B+ u− , B+ v− )0;S − + a−,i (u− , v− ) i

+p

2

1/2 (e) 1/2 (e) (B+ u− , B+ v− )0;Se−

(e) (e) + a−,e (u− , v− ) .

5.4 Lemma. For any p ∈ C0 , the operators Tp± are homeomorphisms from ¯ κ , k > 0, H1/2,p (∂S) to H−1/2,p (∂S), and for any f ∈ H1/2,p (∂S), p ∈ C Tp± f −1/2,p;∂S ≤ c|p|f 1/2,p;∂S ,

(5.42)

c|p|Tp± f −1/2,p;∂S .

(5.43)

f 1/2,p;∂S ≤


71

Proof. The assertion concerning (5.42) and (5.43) for T − has already been proved in Lemma 3.1. Taking ϕ = f and v+ = u+ in (5.41), we obtain 1/2

p2 B+ u+ 20;S + + a+ (u+ , u+ ) = (Tp+ f, f )0;∂S .

(5.44)

Separating the real and imaginary parts in (5.44), we arrive at 1/2

|p|2 B+ u+ 20;S + + a+ (u+ , u+ ) = σ −1 Re {¯ p(Tp+ f, f )0;∂S },

p = σ + iτ,

and then, finally, at u+ 21,p;S + ≤ c|p||(Tp+ f, f )0;∂S |

¯ κ. ∀p ∈ C

The rest of the argument is a verbatim repeat of the proof of Lemma 3.1. We define the single-layer potential Vp α ˆ of density α ˆ ∈ H−1/2,p (∂S), ˆ 2 }, by setting α ˆ = {ˆ α1 , α ˆ )(x, p) = (Vp,1 α ˆ 1 )(x, p) + (Vp,2 α ˆ 2 )(x, p), (Vp α where

x ∈ R2 ,

ˆ − y, p)ˆ D(x αν (y, p) dsy ,

(Vp,ν α ˆ ν )(x, p) =

ν = 1, 2 (not summed),

∂Sν

and define a boundary operator Vp,0 by ˆ = {(Vp,0 α ˆ )1 , (Vp,0 α ˆ )2 }, Vp,0 α (Vp,0 α ˆ )ν (x, p) = (Vp α ˆ )(x, p),

x ∈ ∂Sν ,

ν = 1, 2.

The second equality above can also be written as (Vp,0 α ˆ )ν = γν+ π + Vp α ˆ. 5.5 Lemma. For any p ∈ C0 , the operator Vp,0 is a homeomorphism from ¯ κ , κ > 0, H−1/2,p (∂S) to H1/2,p (∂S), and for any α ˆ ∈ H−1/2,p (∂S), p ∈ C Vp,0 α ˆ 1/2,p;∂S ≤ c|p|ˆ α−1/2,p;∂S , ˆ α−1/2,p;∂S ≤ c|p|Vp,0 α ˆ 1/2,p;∂S .

(5.45)

Proof. We remark that, as follows from the properties of single-layer potentials, for any α ˆ ∈ H−1/2,p (∂S) we have the jump formula (Tp+ − Tp− )Vp,0 α ˆ=α ˆ. The argument now continues as in the proof of Lemma 4.1.

72


Let πi− and πe− be the operators of restriction from R2 (or R2 \ ∂S) to Si− and Se− , respectively. We define the double-layer potential Wp βˆ of density βˆ ∈ H1/2,p (∂S), βˆ = {βˆ1 , βˆ2 }, by ˆ (Wp β)(x, p) = (Wp,1 βˆ1 )(x, p) + (Wp,2 βˆ2 )(x, p), where

x ∈ R2 \ ∂S,

(Wp,ν βˆν )(x, p) =

Pˆ (x − y, p)βˆν (y, p) dsy ,

ν = 1, 2 (not summed),

∂Sν

and the boundary operators Wp± of its limiting values on ∂S by ˆ Wp± βˆ = γ ± (Wp β), where, for a function u defined on R2 (or on R2 \ ∂S), γ + u = {γ1+ π + u, γ2+ π + u},

γ − u = {γ1− πi− u, γ2− πe− u}.

5.6 Lemma. For any p ∈ C0 , the operators Wp± are homeomorphisms from ¯ κ , κ > 0, H1/2,p (∂S) to H1/2,p (∂S), and for any βˆ ∈ H1/2,p (∂S), p ∈ C ˆ 1/2,p;∂S , ˆ 1/2,p;∂S ≤ c|p|2 β Wp± β ˆ 1/2,p;∂S . ˆ 1/2,p;∂S ≤ c|p|2 W ± β β p Proof. The assertion follows from the equalities ⎧ ˆ ⎨ (Vp Tp− β)(x, p), x ∈ S + , ˆ (Wp β)(x, p) = ⎩ (V T + β)(x, ˆ p), x ∈ Si− ∪ Se− , p p which are proved as in Lemmas 4.4, 5.4, and 5.5. We now define an operator Np by setting Np = Tp+ Wp+ = Tp− Wp− . 5.7 Lemma. For every p ∈ C0 , the operator Np is a homeomorphism from ¯ κ , κ > 0, H1/2,p (∂S) to H−1/2,p (∂S), and for any βˆ ∈ H1/2,p (∂S), p ∈ C ˆ −1/2,p;∂S ≤ c|p|3 β ˆ 1/2,p;∂S , Np β ˆ −1/2,p;∂S . ˆ 1/2,p;∂S ≤ c|p|Np β β This assertion is proved in exactly the same way as Theorem 4.8.

(5.46)


73

Let f = {f1 , f2 } ∈ H1/2,p (∂S),

g = {g1 , g2 } = Tp+ f ∈ H−1/2,p (∂S).

We define a pair of boundary operators χ1 and χ2 by χ1 f = {g1 , f2 } ∈ H−1/2,p (∂S1 ) × H1/2,p (∂S2 ), χ2 g = {g1 , f2 } ∈ H−1/2,p (∂S1 ) × H1/2,p (∂S2 ). 5.8 Lemma. For any p ∈ C0 , the operator χ1 is a homeomorphism from H1/2,p (∂S) to H−1/2,p (∂S1 )×H1/2,p (∂S2 ), while χ2 is a homeomorphism from H−1/2,p (∂S) to H−1/2,p (∂S1 ) × H1/2,p (∂S2 ), and for any p ∈ Cκ , κ > 0, χ1 f H−1,2,p (∂S1 )×H1/2,p (∂S2 ) ≤ c|p|f 1/2,p;∂S ,

(5.47)

f 1/2,p;∂S ≤ c|p|χ1 f H−1/2,p (∂S1 )×H1/2,p (∂S2 ) ,

(5.48)

χ2 gH−1/2,p (∂S1 )×H1/2,p (∂S2 ) ≤ c|p|g−1/2,p;∂S ,

(5.49)

g−1/2,p;∂S ≤ c|p|χ2 gH−1/2,p (∂S1 )×H1/2,p (∂S2 ) .

(5.50)

Proof. The continuity of χν , ν = 1, 2, follows from Lemma 5.4. Estimates (5.47) and (5.49) follow from (5.42) and (5.43), respectively. Let {g1 , f2 } ∈ H−1/2,p (∂S1 ) × H1/2,p (∂S2 ), and let u ∈ H1,p (S + ) be the (unique) solution of (5.32) with boundary data {g1 , f2 }. It is obvious that f = {γ1+ u, f2 } satisfies χ1 f = {g1 , f2 }. Since (5.48) follows from (5.40), the statement for χ1 is proved. The function g = Tp+ f satisfies χ2 g = {g1 , f2 }, so χ2 is a homeomorphism from H−1/2,p (∂S) to H−1/2,p (∂S1 ) × H1/2,p (∂S2 ). To complete the proof, we need to show that (5.50) holds. Again, let u be the solution of (5.32) with boundary data {g1 , f2 }, f = {γ1+ u, f2 }, and g = Tp+ f . By the definition of the operators Tp+ , |(Tp+ f, ϕ)0;∂S | ≤ cu1,p;S + ϕ1/2,p;∂S ; hence, g−1/2,p;∂S ≤ cu1,p;S + , and (5.50) follows from (5.40). We introduce the time-dependent (or retarded) single-layer and doublelayer potentials V α and W β and their corresponding boundary operators V0 , W ± , T ± , and N in the spaces of originals in the usual way, and represent the solution u of problem (DMCD) in the form u(X) = (V α)(X),

X ∈ G+ ,

(5.51)

u(X) = (W β)(X),

X ∈ G+ .

(5.52)

or

74


Representations (5.51) and (5.52) yield, respectively, the systems of nonstationary boundary integral equations V0 α = f

(5.53)

W + β = f.

(5.54)

and

5.9 Theorem. For any f ∈ H1/2,k,κ (Γ), κ > 0, k ∈ R, systems (5.53) and (5.54) have unique solutions α ∈ H−1/2,k−1,κ (Γ) and β ∈ H1/2,k−2,κ (Γ), respectively, in which case u defined by (5.51) or (5.52) belongs to H1,k−1,κ (G+ ). If k ≥ 1, then u is the weak solution of problem (DMCD). This assertion is proved by repeating the arguments in Theorem 4.9 and using Lemmas 5.5 and 5.6. If we seek the solution u of (DMCN) in the form (5.51) or (5.52), then we arrive at the systems of boundary equations T + V0 α = g

(5.55)

N β = g.

(5.56)

and

5.10 Theorem. For any g ∈ H−1/2,k,κ (Γ), κ > 0, k ∈ R, systems (5.55) and (5.56) have unique solutions α ∈ H−1/2,k−2,κ (Γ) and β ∈ H1/2,k−1κ (Γ), in which case u defined by (5.51) or (5.52) belongs to H1,k−1,κ (G+ ). If k ≥ 1, then u is the weak solution of problem (DMCN). This assertion is proved just like Theorem 4.10, use being made of Lemmas 5.4, 5.5, and 5.7. Finally, we seek the solution u of (DMCM) in the form (5.51) or (5.52) and arrive at the systems of boundary equations (T + V0 α)1 = g1 ,

(V0 α)2 = f2

(5.57)

and (N β)1 = g1 ,

(W + β)2 = f2 ,

(5.58)

respectively. 5.11 Theorem. For any {g1 , f2 } ∈ H−1/2,k,κ (Γ) × H1/2,k,κ (Γ), κ > 0, k ∈ R, systems (5.57) and (5.58) have unique solutions α ∈ H−1/2,k−2,κ (Γ) and β ∈ H1/2,k−2,κ (Γ), in which case u defined by (5.51) or (5.52) belongs to H1,k−1,κ (G+ ). If k ≥ 1, then u is the weak solution of problem (DMCM).

5.3 Finite Plate with an Inclusion

75

Proof. In terms of Laplace transforms, (5.57) and (5.58) become (Tp+ Vp,0 α ˆ )1 = gˆ1 ,

(Vp,0 α ˆ )2 = fˆ2

and ˆ 1 = gˆ1 , (Np β)

ˆ 2 = fˆ2 , (Wp+ β)

or, equivalently, ˆ = {ˆ g1 , fˆ2 } χ1 Vp,0 α

(5.59)

χ2 Np βˆ = {ˆ g1 , fˆ2 }.

(5.60)

and

By Lemmas 5.5, 5.7, and 5.8, systems (5.59) and (5.60) have unique solutions −1 −1 (5.61) α ˆ = Vp,0 χ1 {ˆ g1 , fˆ2 } ∈ H−1/2,p (∂S) and βˆ = Np−1 χ−1 g1 , fˆ2 } ∈ H1/2,p (∂S). 2 {ˆ

(5.62)

In turn, by (5.61), (5.62), (5.45), (5.46), (5.48), and (5.50), ˆ α−1/2,p;∂S ≤ c|p|2 {ˆ g1 −1/2,p;∂S + fˆ2 1/2,p;∂S },

(5.63)

ˆ 1/2,p;∂S ≤ c|p|2 {ˆ β g1 −1/2,p;∂S + fˆ2 1/2,p;∂S }.

(5.64)

The rest of the proof now follows from (5.63), (5.64), and Theorem 5.3.

5.3 Finite Plate with an Inclusion Keeping the notation used in §5.2, below we consider an initial-boundary value problem for a piecewise homogeneous finite plate occupying the domain S = S¯i− ∪ S + . To simplify the symbols, we write S − instead of Si− . We assume that the plates with middle sections S + and S − have Lamé constants λ+ , µ+ and λ− , µ− , densities ρ+ , ρ− , and thickness parameters h+ , h− . If u is a function defined on S, then we write u = {u+ , u− }, where u± are the restrictions of u from S to S ± . The same notation is used if u is defined in the domain G. Let (DTD) be the initial-boundary value problem with transmission conditions across ∂S1 and Dirichlet condition on ∂S2 which consists in finding ¯ + ), u+ ∈ C 2 (G+ ) ∩ C 1 (G such that

¯−) u− ∈ C 2 (G− ) ∩ C 1 (G

76


B+ (∂t2 u+ )(X) + (A+ u+ )(X) = 0,

X ∈ G+ ,

B− (∂t2 u− )(X) + (A− u− )(X) = 0,

X ∈ G− ,

u+ (x, 0+) = (∂t u+ )(x, 0+) = 0,

x ∈ S+,

u− (x, 0+) = (∂t u− )(x, 0+) = 0,

x ∈ S−,

− u+ + (X) − u− (X) = f1 (X), +

X ∈ Γ1 ,

−

(T+ u+ ) (X) − (T− u− ) (X) = g1 (X), u+ + (X) = f2 (X),

X ∈ Γ1 ,

X ∈ Γ2 ,

where f1 , f2 , and g1 are prescribed. Proceeding as in §5.1, we readily see that in the variational version of (DTD) we seek u = {u+ , u− }, u± ∈ H1,0,κ (G± ), satisfying ∞ a+ (u+ , v+ ) + a− (u− , v− ) 0

1/2 1/2 1/2 1/2 − (B+ ∂t u+ , B+ ∂t v+ )0;S + − (B− ∂t u− , B− ∂t v− )0;S − dt ∞ = (g1 , v0 )0;∂S1 dt

∀v ∈ C0∞ (S × [0, ∞)),

(5.65)

0

γ1+ u+

− γ1− u− = f1 ,

γ2+ u+ = f2 ,

+ )1 = where the trace operators γ1± and γ2+ have the obvious meaning, v0 = (v+ − v− , and f = {f1 , f2 } ∈ H1/2,k,κ (Γ) and g1 ∈ H−1/2,k,κ (Γ1 ) are prescribed.

5.12 Theorem. For any f ∈ H1/2,1,κ (Γ) and g1 ∈ H−1/2,1,κ (Γ1 ), κ > 0, problem (5.65) has a unique solution u = {u+ , u− }, where u± ∈ H1,0,κ (G± ). If f ∈ H1/2,k,κ (Γ) and g1 ∈ H−1/2,k,κ (Γ1 ), k ∈ R, then u± ∈ H1,k−1,κ (G± ) and u+ 1,k−1,κ;G+ + u− 1,k−1,κ;G− ≤ c(f 1/2,k,κ;Γ + g1 −1/2,k,κ;Γ1 ). The assertion is proved exactly as Theorem 5.1, with H1,p (R2 ) replaced ˚1,p (S) and (5.12) written for this case, where it takes the form by H ˆ u+ 1,p;S + + ˆ u− 1,p;S − ≤ c|p|{fˆ1/2,p;∂S + ˆ g1 −1/2,p;∂S1 }.

(5.66)

We now represent the solution of (5.65) in terms of single-layer potentials. Let V− α− be the time-dependent (retarded) single-layer potential defined by ∞ D− (x − y, t − τ )α− (y, τ ) dsy dτ,

(V− α− )(X) = ∂S1 0

X ∈ G− ,


77

where α− is an unknown density defined on Γ1 and D− is the matrix of fundamental solutions of the equation of motion for the plate corresponding to S − . We also consider the single-layer potential V+ α+ defined by ∞ D+ (x − y, t − τ )α+ (y, τ ) dsy dτ,

(V+ α+ )(X) =

X ∈ G+ ,

∂S 0

where α+ = {α+,1 , α+,2 }, α+,1 and α+,2 are unknown densities defined on Γ1 and Γ2 , respectively, and D+ is the matrix of fundamental solutions for the plate with middle plane section S + . The properties of V+ α+ were studied in §5.2. The boundary operators generated by these potentials are denoted by V+,0 and V−,0 , where V+,0 α+ = {(V+,0 α+ )1 , (V+,0 α+ )2 }. Analogous notation W++ and W−− is used for the double-layer potentials W+ β+ and W− β− , where W++ β+ = {(W++ β+ )1 , (W++ β+ )2 } are the boundary operators generated by the limiting values of W+ β+ and W− β− on Γ. Finally, we use the symbols T++ and T−− for the corresponding Poincaré–Steklov operators, and N+ and N− for the operators of the momentforce boundary values generated by the double-layer potentials. In the spaces of Laplace transforms, all the above symbols acquire an extra subscript p. We seek the solution u = {u+ , u− } of (5.65) in the form u+ (X) = (V+ α+ )(X),

X ∈ G+ ,

u− (X) = (V− α− )(X),

X ∈ G− .

(5.67)

Representation (5.67) yields the system of boundary integral equations (V+,0 α+ )1 − V−,0 α− = f1 , (T++ V+,0 α+ )1 − T−− V−,0 α− = g1 ,

(5.68)

(V+,0 α+ )2 = f2 . 5.13 Theorem. For any f ∈ H1/2,k,κ (Γ) and g1 ∈ H−1/2,k,κ (Γ1 ), where κ > 0 and k ∈ R, system (5.68) has a unique solution α+ ∈ H−1/2,k−2,κ (Γ), α− ∈ H−1/2,k−2,κ (Γ1 ), in which case u defined by (5.67) belongs to the space H1,k−1,κ (G+ )×H1,k−1,κ (G− ). If k ≥ 1, then u is the weak solution of problem (DTD).

78


Proof. In terms of Laplace transforms, (5.68) takes the form ˆ + )1 − Vp,−,0 α ˆ − = fˆ1 , (Vp,+,0 α + − (Tp,+ Vp,+,0 α ˆ + )1 − Tp,− Vp,−,0 α ˆ − = gˆ1 ,

(5.69)

(Vp,+,0 α ˆ + )2 = fˆ2 . Next, (5.65) turns into the problem of finding uˆ = {ˆ u+ , u ˆ− }, where u ˆ± ∈ H1,p (S ± ), such that 1/2

1/2

1/2

1/2

ˆ+ , B+ v+ )0;S + + p2 (B− u ˆ− , B− v− )0;S − p2 (B+ u + a+ (ˆ u+ , v+ ) + a− (ˆ u− , v− ) = (ˆ g1 , v0 )0;∂S1 γ1+ u ˆ+ − γ1− u ˆ− = fˆ1 ,

˚1,p (S), ∀v ∈ H

(5.70)

γ2+ u ˆ+ = fˆ2 .

We recall that the solution u ˆ of (5.70) satisfies (5.66). We write γ1+ u ˆ+ = fˆ+ , − ˆ γ1 u ˆ− = f− and solve the systems of equations Vp,+,0 α ˆ + = {fˆ+ , fˆ2 }, Vp,−,0 α ˆ − = fˆ− .

(5.71)

From the properties of Vp,+,0 and Vp,−,0 it follows that (5.71) has a unique solution {ˆ α+ , α ˆ − } for any p ∈ Cκ , and that ˆ α+ −1/2,p;∂S ≤ c|p|{fˆ+ , fˆ2 }1/2,p;∂S , ˆ α− −1/2,p;∂S1 ≤ c|p|fˆ− 1/2,p;∂S1 . The trace theorem and (5.66) imply that ˆ α+ −1/2,p;∂S ≤ c|p|ˆ u+ 1,p;S + ≤ c|p|2 (fˆ1/2,p;∂S + ˆ g1 −1/2,p;∂S1 ),

(5.72)

ˆ α− −1/2,p;∂S1 ≤ c|p|ˆ u− 1,p;S − ≤ c|p|2 (fˆ1/2,p;∂S + ˆ g1 −1/2,p;∂S1 ).

(5.73)

Remarking that {ˆ α+ , α ˆ − } satisfies (5.69) and taking (5.72) and (5.73) into account, we complete the proof by means of the standard scheme used in the case of the analogous assertions in the preceding theorems. If we now seek the solution of (5.65) in the form u+ (X) = (W+ β+ )(X), u− (X) = (W− β− )(X),

X ∈ G+ , −

X∈G ,

(5.74)


79

then we arrive at the system of boundary equations (W++ β+ )1 − W−− β− = f1 , (N+ β+ )1 − N− β− = g1 , (W++ β+ )2

(5.75)

= f2 .

5.14 Theorem. For any f ∈ H1/2,k,κ (Γ) and g1 ∈ H−1/2,k,κ (Γ1 ), where κ > 0 and k ∈ R, system (5.75) has a unique solution β+ ∈ H1/2,k−2,κ (Γ), β− ∈ H1/2,k−2,κ (Γ1 ), in which case u defined by (5.74) belongs to the space H1,k−1,κ (G+ )×H1,k−1,κ (G− ). If k ≥ 1, then u is the weak solution of problem (DTD). Proof. In terms of Laplace transforms, (5.75) becomes + ˆ − ˆ (Wp,+ β+ )1 − Wp,− β− = fˆ1 ,

(Np,+ βˆ+ )1 − Np,− βˆ− = gˆ1 ,

(5.76)

+ ˆ (Wp,+ β+ )2 = fˆ2 .

Once more, let u ˆ be the solution of (5.70), and let + gˆ+ = Tp,+ {fˆ+ , fˆ2 },

− ˆ gˆ− = Tp,− f− ,

where fˆ+ = γ1+ u ˆ+ and fˆ− = γ1− u ˆ− . It is obvious that {βˆ+ , βˆ− }, where −1 βˆ± = Np,± gˆ± ,

(5.77)

−1 is a solution of (5.76). The properties of Np,± and (5.77) imply that for any p ∈ Cκ , g+ −1/2,p;∂S , βˆ+ 1/2,p;∂S ≤ c|p|ˆ

g− −1/2,p;∂S1 . βˆ− 1/2,p;∂S1 ≤ c|p|ˆ Since u+ 1,p;S + , ˆ g+ −1/2,p;∂S ≤ cˆ ˆ g− −1/2,p;∂S1 ≤ cˆ u− 1,p;S − , from (5.66) it follows that g1 −1/2,p;∂S1 ). βˆ+ 1/2,p;∂S + βˆ− 1/2,p;∂S1 ≤ c|p|2 (fˆ1/2,p;∂S + ˆ The rest of the proof is now completed in the usual way.

6 Plate Weakened by a Crack

6.1 Formulation and Solvability of the Problems Let ∂S0 be an open arc of a simple closed C 2 -curve ∂S in R2 , and let ∂S1 = ∂S \ ∂S 0 ,

S = R2 \ ∂S 0 .

We denote by πi , i = 0, 1, the operators of restriction from ∂S to ∂Si . As usual, S ± are the interior and exterior domains bounded by ∂S. We make the notation G = S × (0, ∞), Γ = ∂S × (0, ∞),

G± = S ± × (0, ∞),

Γi = ∂Si × (0, ∞), i = 0, 1.

The operators of restriction from S (or R2 ) to S ± , or from G (or R2 × (0, ∞)) to G± , are denoted by π ± . If u is a function defined on S, then we write u± = π ± u,

u = (u+ , u− ).

Let γi± = πi γ ± , i = 0, 1, be the operators of trace on ∂Si (or Γi ) for functions defined on S ± (or G± ). H1,p (S), p ∈ C, is the space of all u(x, p) = (u+ (x, p), u− (x, p)) such that u± ∈ H1,p (S ± ) and γ1+ u+ = γ1− u− . This space is equipped with the norm u1,p;S = u+ 1,p;S + + u− 1,p;S − . ˚−1,p (S) is the dual of H1,p (S) with respect to the duality generated by H the inner product (· , ·)0;S in L2 (S). ˚1,p (S) is the subspace of H1,p (R2 ) that consists of all u such that γ + u+ = H 0 − γ0 u− = 0.

82


˚1,p (S) with respect to the duality generated by H−1,p (S) is the dual of H (· , ·)0;S . ˚1/2,p (∂S0 ) is the subspace of all functions f ∈ H1/2,p (∂S) such that H supp f ∈ ∂S 0 . ˚1/2,p (∂S0 ) with respect to the duality generH−1/2,p (∂S0 ) is the dual of H ated by the inner product (· , ·)0;∂S0 in L2 (∂S0 ). H1/2,p (∂S0 ) is the space of the restrictions ϕ from ∂S to ∂S0 of all f ∈ H1/2,p (∂S), equipped with the norm ϕ1/2,p;∂S0 =

inf

f ∈H1/2,p (∂S): π0 f =ϕ

f 1/2,p;∂S .

˚−1/2,p (∂S0 ) is the dual of H1/2,p (∂S0 ) with respect to the duality generH ated by (· , ·)0;∂S0 . ˚±1/2,k,κ (Γ0 ), k ∈ R, κ > 0, and the H±1,k,κ (G), H±1/2,k,κ (Γ0 ), and H norms on these spaces, are now introduced in the usual way. We remark that the trace operators γ0± map H1,k,κ (G± ) continuously onto H1/2,k,κ (Γ0 ). We ˚1/2,k,κ (Γ0 ), also remark that if u = (u+ , u− ) ∈ H1,k,κ (G), then γ0+ u − γ0− u ∈ H where γ0± u = γ0± u± . In what follows, we denote by aS (u, v) the energy bilinear form constructed for the plate with middle plane domain S. In the variational version of the initial-boundary value problem (DKD), we seek u ∈ H1,0,κ (G) satisfying ∞

aS (u, v) − (B 1/2 ∂t u, B 1/2 ∂t v)0;S dt = 0

0

∀v ∈ C0∞ (S × [0, ∞)), γ0+ u

= f+ ,

γ0− u

(6.1)

= f− ,

where f+ and f− are prescribed on Γ0 . 6.1 Theorem. For any given f+ , f− ∈ H1/2,1,κ (Γ0 ), κ > 0, such that δf = ˚1/2,1,κ (Γ0 ), problem (6.1) has a unique solution u ∈ H1,0,κ (G). f+ − f− ∈ H ˚1/2,k,κ (Γ0 ), k ∈ R, then u ∈ H1,k−1,κ (G) If f+ , f− ∈ H1/2,k,κ (Γ0 ) and δf ∈ H and u1,k−1,κ;G ≤ c f+ 1/2,k,κ;Γ0 + δf 1/2,k,κ;Γ . Proof. Going over to Laplace transforms in (6.1), we arrive at a problem that consists in finding u ∈ H1,p (S) such that ˚1,p (S), p2 (B 1/2 u, B 1/2 v)0;S + aS (u, v) = 0 ∀v ∈ H γ0+ u = f+ ,

γ0− u = f− .

(6.2)

6.1 Formulation and Solvability of the Problems

83

˚1,p (S) First, we study an auxiliary problem, in which we want to find w ∈ H such that p2 (B 1/2 w, B 1/2 v)0;S + aS (w, v) = (q, v)0;S

˚1,p (S), ∀v ∈ H

(6.3)

where q ∈ H−1,p (S). The unique solvability of (6.3) is proved by means of standard arguments. In Theorem 2.1 it was shown that the form aS (w, v) ˚1,p (S)]2 , that, consequently, it is symmetric, continuous, and coercive on [H ˚1,p (S) through the equality defines a self-adjoint operator A on H ˚1,p (S), (Au, v)0;S = aS (u, v) ∀u, v ∈ H ˚1,p (S) to H−1,p (S). We now rewrite and that A is a homeomorphism from H (6.3) as Aw + p2 Bw = q, or ws + p2 Bws = B 1/2 A−1 q,

(6.4)

where ws = B 1/2 w and B = B 1/2 A−1 B 1/2 . Next, we verify that equation ˚1,p (S) is equivalent to itself in L2 (S), and that B is a self-adjoint (6.4) in H nonnegative operator on L2 (S). This proves that (6.4)—hence, also (6.3)— ˚1,p (S) for any q ∈ H−1,p (S). Setting v = w in have a unique solution w ∈ H (6.3) and separating the real and imaginary parts, we obtain the estimate w1,p ≤ c|p|q−1,p;S .

(6.5)

Let l0 be an extension operator from ∂S0 to ∂S, which maps H1/2,p (∂S0 ) continuously to H1/2,p (∂S); that is, l0 f+ 1/2,p;∂S ≤ cf+ 1/2,p;∂S0

∀f+ ∈ H1/2,p (∂S0 ).

Let F+ = l0 f+ , and let F− be an extension of f− from ∂S0 to ∂S such that π1 F+ = π1 F− . We take u0 = (l+ F+ , l− F− ) ∈ H1,p (S) and seek a solution u ˚1,p (S) satisfies of (6.2) in the form u = u0 + w. Clearly, w ∈ H p2 (B 1/2 w, B 1/2 v)0;S + aS (w, v) ˚1,p (S). (6.6) = −p2 (B 1/2 u0 , B 1/2 v)0;S − aS (u0 , v) ∀v ∈ H ˚1,p (S), Since for any v ∈ H |p2 (B 1/2 u0 , B 1/2 v)0;S + aS (u0 , v)| ≤ cu0 1,p;S v1,p ≤ c F+ 1/2,p;∂S + F− 1/2,p;∂S v1,p ≤ c f+ 1/2,p;∂S0 + δf 1/2,p;∂S v1,p ,

84


we can write the right-hand side in (6.6) as (q, v)0;S , where q ∈ H−1,p (S) and q−1,p;S ≤ c f+ 1/2,p;∂S0 + δf 1/2,p;∂S . Therefore, (6.2) has a unique solution u ∈ H1,p (S) and, as follows from (6.5), u1,p;S ≤ c|p| f+ 1/2,p;∂S0 + δf 1/2,p;∂S . (6.7) Using (6.7), we complete the proof in the standard way. ¯ be the space of all functions with compact support in G ¯ which Let C0∞ (G) ¯ + ) and C ∞ (G ¯ − ) and are such that their limiting values, belong to both C0∞ (G 0 and the limiting values of all their derivatives, on Γ1 from inside G+ and G− coincide. In the variational version of problem (DKN), we seek u ∈ H1,0,κ (G) that satisfies ∞

aS (u, v) − (B 1/2 ∂t u, B 1/2 ∂t v)0;S dt

0

∞ =

(g+ , γ0+ v)0;∂S0 − (g− , γ0− v)0;∂S0 dt

0

¯ ∀v ∈ C0∞ (G),

(6.8)

where g± are prescribed on Γ0 . 6.2 Theorem. For any given g+ , g− ∈ H−1/2,1,κ (Γ0 ), κ > 0, such that δg = ˚−1/2,1,κ (Γ0 ), problem (6.8) has a unique solution u ∈ H1,0,κ (G). If g+ −g− ∈ H ˚−1/2,k,κ (Γ0 ), k ∈ R, then u ∈ H1,k−1,κ (G) g+ , g− ∈ H−1/2,k,κ (Γ0 ) and δg ∈ H and u1,k−1,κ;G ≤ c δg−1/2,k,κ;Γ + g− −1/2,k,κ;Γ0 . Proof. In terms of Laplace transforms, (6.8) becomes p2 (B 1/2 u, B 1/2 v)0;S + aS (u, v) = (g+ , γ0+ v)0;∂S0 − (g− , γ0− v)0;∂S0

∀v ∈ H1,p (S),

(6.9)

or, which is the same, p2 (B 1/2 u, B 1/2 v)0;S + aS (u, v) = (δg, γ0+ v)0;∂S0 + (g− , γ0+ v − γ0− v)0;∂S0 Since for any v ∈ H1,p (S), ˚1/2,p (∂S0 ), γ0+ v − γ0− v ∈ H

∀v ∈ H1,p (S).

6.2 The Poincaré–Steklov Operator

85

it follows that |(δg, γ0+ v)0;∂S0 + (g− , γ0+ v − γ0− v)0;∂S0 | ≤ c δg−1/2,p;∂S γ0+ v1/2,p;∂S0 + g− −1/2,p;∂S0 γ0+ v − γ0− v1/2,p;∂S ≤ c δg−1/2,p;∂S + g− −1/2,p;∂S0 v1,p;S ,

so we can write (6.9) as p2 (B 1/2 u, B 1/2 v)0;S + aS (u, v) = (Q, v)0;S

∀v ∈ H1,p (S),

(6.10)

˚−1,p (S) and where Q ∈ H Q−1,p ≤ c δg−1/2,p;∂S + g− −1/2,p;∂S0 . To prove that (6.10) is uniquely solvable, we consider the bilinear form aκ,S (u, v) =

1 2

κ2 (B 1/2 u, B 1/2 v)0;S + aS (u, v).

This form is symmetric, continuous, and coercive on [H1,p (S)]2 ; consequently, it defines a self-adjoint operator Aκ , which is a homeomorphism from H1,p (S) ˚−1,p (S). Repeating standard arguments, we show that (6.9) is uniquely to H solvable in H1,p (S) and that u1,p;S ≤ c|p| δg−1/2,p;∂S + g− −1/2,p;∂S0 .

(6.11)

Taking (6.11) into account, we complete the proof by following the usual procedure.

6.2 The Poincar´ e–Steklov Operator In this section we introduce the Poincaré–Steklov operator T and study its properties. We start by defining this operator in Sobolev spaces with a parameter. Let ˚1/2,p (∂S0 ), f˜ = {f+ , δf } ∈ H1/2,p (∂S0 ) × H ˚−1/2,p (∂S0 ) × H−1/2,p (∂S0 ). g˜ = {δg, g− } ∈ H We write

f˜, g˜0;∂S0 = (f+ , δg)0;∂S0 + (δf, g− )0;∂S0 .

86


˚1/2,p (∂S0 ) and H ˚−1/2,p (∂S0 ) × The norms of f˜ and g˜ in H1/2,p (∂S0 ) × H H−1/2,p (∂S0 ), respectively, are defined by f˜1/2,p;∂S0 ;1/2,p;∂S = f+ 1/2,p;∂S0 + δf 1/2,p;∂S , ˜ g −1/2,p;∂S;−1/2,p;∂S0 = δg−1/2,p;∂S + g− −1/2,p;∂S0 . ˚1/2,p (∂S0 ), ϕ˜ = {ϕ+ , δϕ}, let u ∈ H1,p (S) be Let f˜, ϕ˜ ∈ H1/2,p (∂S0 ) × H the solution of (6.2) with f− = f+ − δf , and let v ∈ H1,p (S) be such that γ0+ v = ϕ+ and γ0+ v − γ0− v = δϕ. For every p ∈ C0 we define the Poincaré– Steklov operator Tp by means of the equality Tp f˜, g˜0;∂S0 = p2 (B 1/2 u, B 1/2 v)0;S + aS (u, v).

(6.12)

It is easy to convince ourselves that (6.12) defines Tp correctly; that is, its definition does not depend on the choice of v. 6.3 Lemma. For any p ∈ C0 , the operator Tp is a homeomorphism from the ˚1/2,p (∂S0 ) to its dual H ˚−1/2,p (∂S0 ) × H−1/2,p (∂S0 ), and space H1/2,p (∂S0 ) × H ¯ κ , κ > 0, ˚1/2,p (∂S0 ), p ∈ C for any f˜ ∈ H1/2,p (∂S0 ) × H Tp f˜−1/2,p;∂S;−1/2,p;∂S0 ≤ c|p|f˜1/2,p;∂S0 ;1/2,p;∂S ,

(6.13)

f˜1/2,p;∂S0 ;1/2,p;∂S ≤ c|p|Tp f˜−1/2,p;∂S;−1/2,p;∂S0 .

(6.14)

Proof. Let ϕ˜+ = l0 ϕ+ , and let ϕ˜− ∈ H1/2,p (∂S) be an extension of ϕ− from ˚1/2,p (∂S0 ). We take v± = l± ϕ˜± . ∂S0 to ∂S such that ϕ˜+ − ϕ˜− = δ ϕ˜ ∈ H Clearly, v = (v+ , v− ) ∈ H1,p (S) and v1,p;S ≤ cϕ ˜ 1/2,p;∂S0 ;1/2,p;∂S . By (6.12) and (6.15), ˜ 0;∂S0 | ≤ cu1,p;S v1,p;S |Tp f˜, ϕ ≤ cu1,p;S ϕ ˜ 1/2,p;∂S0 ;1/2,p;∂S ; ˚−1/2,p (∂S0 ) × H−1/2,p (∂S0 ) and therefore, Tp f˜ ∈ H Tp f˜−1/2,p;∂S;−1/2,p;∂S0 ≤ cu1,p;S . Formula (6.7) now implies that Tp f˜−1/2,p;∂S;−1/2,p;∂S0 ≤ c|p|f˜1/2,p;∂S0 ;1/2,p;∂S .

(6.15)


87

Taking in ϕ˜ = f˜ and v = u in (6.12) and separating the real and imaginary ¯ κ, parts, we find that for any p ∈ C u21,p;S ≤ c|p||Tp f˜, f˜0;∂S0 |;

(6.16)

u1,p;S ≤ c|p|Tp f˜−1/2,p;∂S;−1/2,p;∂S0 .

(6.17)

hence,

By the trace theorem and (6.17), f˜1/2,p;∂S0 ;1/2,p;∂S ≤ cu1,p;S ≤ c|p|Tp f˜−1/2,p;∂S;−1/2,p;∂S0 , which proves (6.13) and (6.14). ˚−1/2,p (∂S0 ) × H−1/2,p (∂S0 ), then there If the range of Tp is not dense in H ˚1/2,p (∂S0 ) such that is a nonzero ϕ˜ ∈ H1/2,p (∂S0 ) × H ˚1/2,p (∂S0 ). ˜ 0;∂S0 = 0 ∀f˜ ∈ H1/2,p (∂S0 ) × H Tp f˜, ϕ

(6.18)

Taking f˜ = ϕ˜ in (6.18), we use (6.16) and the trace theorem to conclude that ϕ˜ = 0. This contradiction completes the proof. We now introduce the operators Tˆ , Tˆ −1 and T , T −1 in the usual way (see §3.1). 6.4 Theorem. For any κ > 0 and k ∈ R, the operator T is continuous ˚1/2,k,κ (Γ0 ) to H ˚−1/2,k−1,κ (Γ0 ) × and injective from the space H1/2,k,κ (Γ0 ) × H ˚ H−1/2,k−1,κ (Γ0 ), and its range is dense in H−1/2,k−1,κ (Γ0 )×H−1/2,k−1,κ (Γ0 ). The inverse operator T −1 , extended by continuity from the range of T to ˚−1/2,k,κ (Γ0 ) × H−1/2,k,κ (Γ0 ), is continuous and injective from the space H ˚1/2,k−1,κ (Γ0 ) for any ˚−1/2,k,κ (Γ0 ) × H−1/2,k,κ (Γ0 ) to H1/2,k−1,κ (Γ0 ) × H H ˚ k ∈ R, and its range is dense in H1/2,k−1,κ (Γ0 ) × H1/2,k−1,κ (Γ0 ). This assertion is proved in the same way as those of a similar nature in the preceding chapters.

6.3 Time-dependent Potentials ˚−1/2,p (∂S0 ), p ∈ C0 , and let Vp α Let α ˆ ∈ H ˆ be the single-layer potential of ˆ ∈ H1,p (S) and that the boundary operator density α. ˆ It is obvious that Vp α Vp,0 generated by Vp α ˆ on ∂S0 through the equality Vp,0 α ˆ = γ0+ Vp α ˆ = γ0− Vp α ˆ

88


˚−1/2,p (∂S0 ) to H1/2,p (∂S0 ). Let is continuous from H ˆ. f± = γ0± Vp α From the properties of single-layer potentials it follows that f+ = Vp,0 α ˆ,

δf = f+ − f− = 0.

For this reason, in what follows we do not distinguish between the space ˚1/2,p (∂S0 ) consisting of all H1/2,p (∂S0 ) and the subspace of H1/2,p (∂S0 ) × H the elements of the form {f+ , 0}, f+ ∈ H1/2,p (∂S0 ). 6.5 Lemma. For any p ∈ C0 , the operator Vp,0 is a homeomorphism from ¯ κ, ˚−1/2,p (∂S0 ) to H1/2,p (∂S0 ), and for any function α ˚−1/2,p (∂S0 ), p ∈ C H ˆ∈H κ > 0, Vp,0 α ˆ 1/2,p;∂S0 ≤ c|p|ˆ α−1/2,p;∂S ,

(6.19)

ˆ α−1/2,p;∂S ≤ c|p|Vp,0 α ˆ 1/2,p;∂S0 .

(6.20)

Proof. Estimate (6.19) follows from Lemma 4.1. By the jump formula, ˆ = {ˆ α, π0 Tp− γ − π − Vp α ˆ} Tp Vp,0 α

˚−1/2,p (∂S0 ), ∀ˆ α∈H

(6.21)

where, as before, Tp± are the Poincaré–Steklov operators defined in relation to the domains S ± ; hence, ˆ α−1/2,p;∂S ≤ Tp Vp,0 α ˆ −1/2,p;∂S;−1/2,p;∂S0 ≤ c|p|Vp,0 α ˆ 1/2,p;∂S0 . If the range of Vp,0 is not dense in H1/2,p (∂S0 ), then we can find a nonzero ˚−1/2,p (∂S0 ) such that ψ∈H ˚−1/2.p (∂S0 ). ˆ , ψ)0;∂S0 = 0 ∀ˆ α∈H (Vp,0 α Taking α ˆ = ψ, from (6.16) with u = Vp ψ, (6.21), and the equality Vp,0 α ˆ = {Vp,0 α ˆ , 0} we see that Vp ψ21,p;S ≤ c|p||Tp Vp,0 ψ, Vp,0 ψ0;∂S0 | = 0; therefore, Vp ψ = 0, so Vp,0 ψ = 0, which implies that ψ = 0. This contradiction completes the proof. We now define the time-dependent (retarded) single-layer potential V α and its corresponding boundary operator V0 in the usual way.


89

6.6 Theorem. For any κ > 0 and k ∈ R, the operator V0 is continuous ˚−1/2,k,κ (Γ0 ) to H1/2,k−1,κ (Γ0 ), and its range is dense and injective from H in H1/2,k−1,κ (Γ0 ). The inverse V0−1 , extended by continuity from the range ˚−1/2,k−1,κ (Γ0 ) for of V0 , is continuous and injective from H1/2,k,κ (Γ0 ) to H ˚ any k ∈ R, and its range is dense in H−1/2,k−1,κ (Γ0 ). In addition, for any ˚−1/2,k,κ (Γ0 ), α∈H V α1,k−1,κ;G ≤ cα−1/2,k,κ;Γ .

(6.22)

Proof. The assertions concerning V0 and V0−1 follow from Lemma 6.5, in particular, from (6.19) and (6.20). ˚−1/2,p (∂S0 ). By (6.16) with u = Vp α Let α ˆ∈H ˆ and f˜ = {Vp,0 α ˆ , 0}, ˆ 21,p;S ≤ c|p||(ˆ α, Vp α ˆ )0;∂S0 | ≤ c|p|2 ˆ α2−1/2,p;∂S ; Vp α hence, Vp α ˆ 1,p;S ≤ c|p|ˆ α−1/2,p;∂S ,

(6.23)

and (6.22) follows from (6.23). ˚1/2,p (∂S0 ), Next, we introduce the double-layer potential Wp βˆ for βˆ ∈ H ± define its corresponding boundary operators Wp,0 on ∂S0 by ± = π0 Wp± , Wp,0

and introduce a boundary operator Wp,0 by setting + ˆ + ˆ − ˆ + ˆ ˆ Wp,0 βˆ = {Wp,0 β, Wp,0 β − Wp,0 β} = {Wp,0 β, −β},

ˆ By Lemma 4.5, for any where we have used the jump property of Wp β. p ∈ Cκ , κ > 0, ˆ 1/2,p;∂S ˆ 1/2,p;∂S ;1/2,p;∂S = W + β ˆ 1/2,p;∂S + β Wp,0 β p,0 0 0 ˆ 1/2,p;∂S . ≤ c|p|2 β We now define an operator Np,0 by writing Np,0 = Tp Wp,0 . ˆ for any βˆ ∈ H ˚1/2,p (∂S0 ), where It is obvious that Np,0 βˆ = {0, π0 Np β} Np = Tp+ Wp+ = Tp− Wp− .

(6.24)

90


In what follows we do not distinguish between the space H−1/2,p (∂S0 ) and ˚−1/2,p (∂S0 ) × H−1/2,p (∂S0 ) consisting of all the elements of the subspace of H the form {0, g− }, where g− ∈ H−1/2,p (∂S0 ). 6.7 Lemma. For any p ∈ C0 , the operator Wp,0 is continuous from the ˚1/2,p (∂S0 ) to H1/2,p (∂S0 ) × H ˚1/2,p (∂S0 ), while the operator Np,0 is space H ˚ a homeomorphism from H1/2,p (∂S0 ) to H−1/2,p (∂S0 ). In addition, for any ¯ κ , κ > 0, ˚1/2,p (∂S0 ), p ∈ C βˆ ∈ H ˆ 1/2,p;∂S , ˆ 1/2,p;∂S ;1/2,p;∂S ≤ c|p|2 β Wp,0 β 0

(6.25)

ˆ 1/2,p;∂S , ˆ −1/2,p;∂S ≤ c|p|3 β Np,0 β 0

(6.26)

ˆ 1/2,p;∂S ≤ c|p|Np,0 β ˆ −1/2,p;∂S . β 0

(6.27)

Proof. Estimate (6.25) has already been proved. Inequality (6.26) follows from (6.24), (6.25), and (6.13). To prove (6.27), we see that for any density ˚1/2,p (∂S0 ), βˆ ∈ H − + ˆ 2 ˆ ˆ 2 ˆ 2 β 1/2,p;∂S = γ0 Wp β − γ0 Wp β1/2,p;∂S ≤ cWp β1,p;S .

(6.28)

ˆ By (6.28) and (6.16) with u = Wp β, ˆ 2 ˆ ˆ ˆ ˆ β 1/2,p;∂S ≤ c|p||(Np,0 β, β)0;∂S0 | ≤ c|p|Np,0 β−1/2,p;∂S0 β1/2,p;∂S , which proves (6.27). If the range of Np,0 is not dense in H−1/2,p (∂S0 ), then there is a nonzero ˆ ψ)0;∂S = 0 for any βˆ ∈ H ˚1/2,p (∂S0 ). ˚1/2,p (∂S0 ) such that (Np,0 β, ψ ∈ H 0 Taking βˆ = ψ, we use (6.16) with u = Wp ψ and find that Wp ψ21,p;S ≤ c|p||(Np,0 ψ, ψ)0;∂S0 | = 0; therefore, Wp ψ = 0, so ψ = γ0− Wp ψ − γ0+ Wp ψ = 0. This contradiction proves the lemma. We now introduce the time-dependent (retarded) double-layer potential W β and its corresponding boundary operators W0 , N0 , and N0−1 in the usual way.

6.4 Infinite Plate with a Crack

91

6.8 Theorem. For any κ > 0 and k ∈ R, the operator W0 is continuous from ˚1/2,k,κ (Γ0 ) to H1/2,k−2,κ (Γ0 ) × H ˚1/2,k,κ (Γ0 ), while the operator N0 is conH ˚ tinuous and injective from H1/2,k,κ (Γ0 ) to H−1/2,k−3,κ (Γ0 ), and its range is dense in H−1/2,k−3,κ (Γ0 ). The inverse N0−1 , extended by continuity from the ˚1/2,k−1,κ (Γ0 ) range of N0 , is continuous and injective from H−1/2,k,κ (Γ0 ) to H ˚1/2,k−1,κ (Γ0 ). In addition, for any for any k ∈ R, and its range is dense in H ˚ β ∈ H1/2,k,κ (Γ0 ), W β1,k−2,κ;G ≤ cβ1/2,k,κ;Γ . Proof. The assertion follows from Lemma 6.7 and the Laplace transform estimate 4 ˆ 2 ˆ 2 ˆ ˆ Wp β 1,p;S ≤ c|p||(Np,0 β, β)0;∂S0 | ≤ c|p| β1/2,p;∂S .

6.4 Infinite Plate with a Crack We seek the solution of problem (DKD) in the form u(X) = (V α)(X) + (W β)(X),

X ∈ G,

(6.29)

where α and β are unknown densities defined on ∂S0 . Representation (6.29) yields the system of boundary equations V0 α + W0 β = f˜ = {f+ , δf }.

(6.30)

˚1/2,k,κ (Γ0 ), κ > 0, k ∈ R, 6.9 Theorem. For any f˜ ∈ H1/2,k,κ (Γ0 ) × H ˚1/2,k,κ (Γ0 ), ˚ system (6.30) has a unique solution {α, β} ∈ H−1/2,k−1,κ (Γ0 ) × H in which case u defined by (6.29) belongs to H1,k−1,κ (G). If k ≥ 1, then u is the weak solution of problem (DKD). Proof. In terms of Laplace transforms, (6.30) takes the form Vp,0 α ˆ + Wp,0 βˆ = fˆ = {fˆ+ , δ fˆ},

(6.31)

where fˆ(x, p) = Lf˜(x, t). Since the second “component” of the left-hand side ˆ we have βˆ = −δ fˆ. By Lemma 6.3, (6.31) is equivalent to in (6.31) is −β, Tp Vp,0 α ˆ + Np,0 βˆ = Tp fˆ = {δˆ g , gˆ− }.

(6.32)

92


The first “component” of the left-hand side in (6.32) is α; ˆ hence, α ˆ = δˆ g. ˆ ˆ Consequently, (6.31) has a unique solution {ˆ α, β} = {δˆ g , −δ f } and, by (6.13), ˆ α−1/2,p;∂S ≤ c|p|fˆ1/2,p;∂S0 ;1/2,p;∂S ,

(6.33)

ˆ 1/2,p;∂S ≤ cfˆ1/2,p;∂S ;1/2,p;∂S . β (6.34) 0 ˆ ∈H ˚1/2,p (∂S0 ) is the solution of (6.31), then, by ˚−1/2,p (∂S0 ) × H If {ˆ α, β} (6.16) with u = Vp α ˆ + Wp βˆ and (6.13), ˆ 2 ˆ ˆ Vp α ˆ + Wp β 1,p;S ≤ c|p||Tp f , f 0;∂S0 | ≤ c|p|2 fˆ21/2,p;∂S0 ;1/2,p;∂S .

(6.35)

Using (6.33)–(6.35), we complete the proof in the standard way. We now represent the solution of problem (DKN) in the form (6.29) and arrive at the system of boundary equations T V0 α + N0 β = g˜ = {δg, g− }.

(6.36)

˚−1/2,k,κ (Γ0 ) × H−1/2,k,κ (Γ0 ), κ > 0, k ∈ R, 6.10 Theorem. For any g˜ ∈ H ˚1/2,k−1,κ (Γ0 ), ˚−1/2,k,κ (Γ0 ) × H system (6.36) has a unique solution {α, β} ∈ H in which case u defined by (6.29) belongs to H1,k−1,κ (G). If k ≥ 1, then u is the weak solution of problem (DKN). Proof. Going over to Laplace transforms, we reduce (6.36) to the system Tp Vp,0 α ˆ + Np,0 βˆ = gˆ = {δˆ g , gˆ− },

(6.37)

where gˆ(x, p) = L˜ g (x, t). Comparing the first “components” on both sides of (6.37), we see that α ˆ = δˆ g. System (6.37) is equivalent to Vp,0 α ˆ + Wp,0 βˆ = Tp−1 gˆ = {fˆ+ , δ fˆ}.

(6.38)

From (6.38) it follows that βˆ = −δ fˆ. Using (6.14), we obtain g −1/2,p;∂S;−1/2,p;∂S0 , ˆ α−1/2,p;∂S ≤ ˆ

(6.39)

ˆ 1/2,p;∂S ≤ c|p|ˆ g −1/2,p;∂S;−1/2,p;∂S0 . β

(6.40)

6.5 Finite Plate with a Crack

93

ˆ ∈H ˚−1/2,p (∂S0 ) × H ˚1/2,p (∂S0 ) is the solution of (6.37), then, by If {ˆ α, β} ˆ ˆ + Wp β and (6.14), (6.16) with u = Vp α ˆ 2 Vp α ˆ + Wp β g , Tp−1 gˆ0;∂S0 | 1,p;S ≤ c|p||ˆ ≤ c|p|2 ˆ g 2−1/2,p;∂S;−1/2,p;∂S0 .

(6.41)

Estimates (6.39)–(6.41) enable us to complete the proof by following the usual procedure.

6.5 Finite Plate with a Crack Let S˜ be a finite domain whose boundary is a simple closed C 2 -curve ∂S2 , and ˜ The arc ∂S0 , which models a crack, is assumed to let ∂S0 be an open arc in S. ˜ We denote be part of a simple closed C 2 -curve ∂S that lies strictly inside S. by Si+ the domain interior to ∂S and by S − the (infinite) domain exterior to ∂S2 . We also write S = S˜ \ ∂S 0 , ∂S1 = ∂S \ ∂S 0 , ˜ = S˜ × (0, ∞), G G+ i

=

Si+

× (0, ∞),

G+ e

=

Se+

Se+ = S˜ \ S¯i+ , ∂ S˘ = ∂S0 ∪ ∂S2 , G = S × (0, ∞), × (0, ∞),

G− = S − × (0, ∞),

˘ = ∂ S˘ × (0, ∞). Γ ˜ S, The operators of restriction from R2 (or Si+ ∪ Se+ ∪ S − ) to Si+ , Se+ , S, ˜ , π, and π − , respectively. and S − are denoted by πi+ , πe+ , π + + We use the symbols γν,i , γν,e , ν = 0, 1, for the operators of trace on ∂Sν + + + − + (Γν ) from inside Si (Γi ) and Se (Γ+ e ), and γ2,e and γ2 for the operators of + + − − trace on ∂S2 (Γ2 ) from inside Se (Γe ) and S (Γ ). H1,p (S), p ∈ C, is the space of all functions u(x, p) = {u+,i , u+,e } such that + + u+,i ∈ H1,p (Si+ ), u+,e ∈ H1,p (Se+ ), γ1,i u+,i = γ1,e u+,e , equipped with the norm u1,p;S = u+,i 1,p;S + + u+,e 1,p;Se+ . i

˜ and u+,i To simplify the notation, if u is a function defined on S (or S) + + and u+,e are its restrictions to Si and Se , then we write + + γν,i u+,i = γν,i u,

+ + γν,e u+,e = γν,e u,

ν = 0, 1,

+ + γ2,e u+,e = γ2,e u.

94


˜ consisting of all u such that ˚1,p (S) is the subspace of H1,p (S) H + + + γ0,i u = γ0,e u = γ2,e u = 0.

˘ = H1/2,p (∂S0 ) × H ˚1/2,p (∂S0 ) × H1/2,p (∂S2 ) is the space of H1/2,p (∂ S) functions of the form f˘ = {f˜, f2 },

f˜ = {f+,i , δf },

δf = f+,i − f+,e .

˘ =H ˚−1/2,p (∂S0 )×H−1/2,p (∂S0 )×H−1/2,p (∂S2 ), with elements H−1/2,p (∂ S) g˘ = {˜ g , g2 },

g˜ = {δg, g+,e },

δg = g+,i − g+,e

˘ with respect to the duality generated by the L2 (∂ S)˘ is the dual of H1/2,p (∂ S) inner product f˘, g˘ ˘ = f˜, g˜0;∂S + (f2 , g2 )0;∂S . 0;∂ S

0

2

˚±1/2,k,κ (Γ0 ), equipped with the norms The spaces H±1/2,k,κ (Γ0 ) and H ˘ and f+ ±1/2,k,κ;Γ0 and δf ±1/2,k,κ;Γ , as well as H1,k,κ (G) and H±1/2,k,κ (Γ) ˘ their norms u1,k,κ;G and f ±1/2,k,κ;Γ˘ , are now defined in the usual way, with reference to the above spaces and with the same convention regarding the notation for the various traces of functions defined on G. In this section we consider only one initial-boundary value problem and only one integral representation for its solution. The other problems and solution representations are treated analogously. Thus, the variational version of problem (DKD) consists in finding a function u ∈ H1,0,κ (G) such that ∞

aS (u, v) − (B 1/2 ∂t u, B 1/2 ∂t v)0;S dt = 0

0

∀v ∈ C0∞ (S × [0, ∞)), + γ0,i u = f+,i ,

− γ0,e u = f+,e ,

+ γ2,e u = f2 ,

where f+,i and f+,e are prescribed on Γ0 and f2 is prescribed on Γ2 .

(6.42)


95

˘ κ > 0, where 6.11 Theorem. For any given f˘ = {f˜, f2 } ∈ H1/2,1,κ (Γ), ˜ f = {f+,i , δf } and δf = f+,i − f+,e , problem (6.42) has a unique solution ˘ k ∈ R, then u ∈ H1,k−1,κ (G) and u ∈ H1,0,κ (G). If f˘ ∈ H1/2,k,κ (Γ), u1,k−1,κ;G ≤ c f+,i 1/2,k,κ;Γ0 + δf 1/2,k,κ;Γ + f2 1/2,k,κ;Γ2 . The proof is a repeat of that of Theorem 6.1, with the obvious changes. ˘ p ∈ C0 , and let Let f˘ = {f+,i , δf, f2 }, ϕ˘ = {ϕ+,i , δϕ, ϕ2 } ∈ H1/2,p (∂ S), u ∈ H1,p (S) be the solution of the variational problem ˚1,p (S), aS (u, v) + p2 (B 1/2 u, B 1/2 v)0;S = 0 ∀v ∈ H + γ0,i u = f+,i ,

+ γ0,e u = f+,e ,

+ γ2,e u = f2 ,

where v is any element of H1,p (S) such that + γ0,i v = ϕ+,i ,

+ γ0,e v = ϕ+,e ,

+ γ2,e v = ϕ2 .

For any p ∈ C0 , we define the Poincaré–Steklov operator Tp+ by means of the equality Tp+ f˘, ϕ ˘ 0;∂ S˘ = p2 (B 1/2 u, B 1/2 v)0;S + aS (u, v). 6.12 Lemma. For any p ∈ C0 , the operator Tp+ is a homeomorphism from ¯ κ , κ > 0, ˘ to H−1/2,p (∂ S), ˘ and for any f˘ ∈ H1/2,p (∂ S), ˘ p∈C H1/2,p (∂ S) Tp+ f˘−1/2,p;∂ S˘ ≤ c|p|f˘1/2,p;∂ S˘ , f˘1/2,p;∂ S˘ ≤ c|p|Tp+ f˘−1/2,p;∂ S˘ , ˘ where · 1/2,p;∂ S˘ and · −1/2,p;∂ S˘ are, respectively, the norms on H1/2,p (∂ S) ˘ and H−1/2,p (∂ S). This assertion is proved just like Lemma 6.3. In what follows, we use the symbol Tp− for the Poincaré–Steklov operator constructed in relation to the domain S − . ˚−1/2,p (∂S0 ) and α ˆ 2 ∈ H−1/2,p (∂S2 ). For any p ∈ C0 , we Let α ˆ0 ∈ H ˚−1/2,p (∂S0 )× ˆ of density α ˆ = {ˆ α0 , α ˆ2} ∈ H define the single-layer potential Vp α H−1/2,p (∂S2 ) by ˆ )(x, p) = (Vp(0) α ˆ 0 )(x, p) + (Vp(2) α ˆ 2 )(x, p), (Vp α (j)

x ∈ R2 ,

where Vp α ˆ j , j = 0, 2, are the single-layer potentials of densities α ˆ j constructed in terms of the parts ∂Sj of ∂S, respectively.

96


+ + Since γ0,i Vp α ˆ = γ0,e Vp α ˆ , the corresponding boundary operator Vp,0 is defined by

Vp,0 α ˆ = {(Vp α ˆ )0 , 0, (Vp α ˆ )2 }, where + + (Vp α ˆ )0 = γ0,i Vp α ˆ = γ0,e Vp α ˆ,

+ (Vp α ˆ )2 = γ2,e Vp α ˆ.

˚−1/2,p (∂S0 ) × H−1/2,p (∂S2 ) Vp,0 is continuous as a mapping from the space H to H1/2,p (∂S0 ) × H1/2,p (∂S2 ). Writing

Tp+ Vp,0 α ˆ = δ(Tp+ Vp,0 α ˆ ), (Tp+ Vp,0 α ˆ )+,e , (Tp+ Vp,0 α ˆ )2 , we immediately see that the jump formula leads to α ˆ 0 = δ(Tp+ Vp,0 α ˆ ),

α ˆ 2 = (Tp+ Vp,0 α ˆ )2 − Tp− (Vp,0 α ˆ )2 ,

(6.43)

where, according to the above notation, + (Vp,0 α ˆ )2 = (Vp α ˆ )2 = γ2,e Vp α ˆ.

In what follows, we identify H1/2,p (∂S0 ) × H1/2,p (∂S2 ) with the subspace ˘ of all functions of the form f˘ = {f+,i , 0, f2 } with f+,i ∈ of H1/2,p (∂ S) ˚−1/2,p (∂S0 )× ˆ∈H H1/2,p (∂S0 ) and f2 ∈ H1/2,p (∂S2 ). We define the norm of α H−1/2,p (∂S2 ) by ˆ α−1/2,p;∂S;−1/2,p;∂S2 = ˆ α0 −1/2,p;∂S + ˆ α2 −1/2,p;∂S2 . 6.13 Lemma. For any p ∈ C0 , the operator Vp,0 is a homeomorphism from ˚−1/2,p (∂S0 ) × H−1/2,p (∂S2 ) to H1/2,p (∂S0 ) × H1/2,p (∂S2 ), and for the space H ¯ κ , κ > 0, ˚−1/2,p (∂S0 ) × H−1/2,p (∂S2 ), p ∈ C any α ˆ∈H Vp,0 α ˆ 1/2,p;∂ S˘ ≤ c|p|ˆ α−1/2,p;∂S;−1/2,p;∂S2 ,

(6.44)

ˆ α−1/2,p;∂S;−1/2,p;∂S2 ≤ c|p|Vp,0 α ˆ 1/2,p;∂ S˘ .

(6.45)

˚−1/2,p (∂S0 ) × H−1/2,p (∂S2 ). By the trace theorem, Proof. Let α ˆ∈H ˆ 21/2,p;∂ S˘ ≤ c πVp α ˆ 21,p;S + π − Vp α ˆ 21,p;S − . Vp,0 α Since in our case

πVp α ˆ 21,p;S ≤ c Re p¯Tp+ Vp,0 α ˆ , Vp,0 α ˆ 0;∂ S˘ ,

π − Vp α ˆ 21,p;S − ≤ c Re p¯Tp− (Vp,0 α ˆ )2 , (Vp,0 α ˆ )2 0;∂S2 ,

(6.46)


97

from (6.46) and (6.43) it follows that

Vp,0 α ˆ 21/2,p;∂ S˘ ≤ c|p| |(ˆ α0 , (Vp α ˆ )0 )0;∂S0 | + |(ˆ α2 , (Vp α ˆ )2 )0;∂S2 | ˆ 1/2,p;∂ S˘ , ≤ c|p|ˆ α−1/2,p;∂S;−1/2,p;∂S2 Vp,0 α

(6.47)

which implies that (6.44) holds. Estimate (6.45) follows from equalities (6.43) and the properties of the Poincaré–Steklov operators. If the range of Vp,0 is not dense in H1/2,p (∂S0 ) × H1/2,p (∂S2 ), then there ˚−1/2,p (∂S0 ) × H−1/2,p (∂S2 ) such that is a nonzero ψ = {ψ0 , ψ2 } ∈ H

ˆ )0 , ψ0 (Vp α

0;∂S0

+ (Vp α ˆ )2 , ψ2 0;∂S2 = 0 ˚−1/2,p (∂S0 ) × H−1/2,p (∂S2 ). ∀ˆ α∈H

(6.48)

Taking α ˆ = ψ in (6.48), we see that, by (6.47) and (6.48), Vp ψ = 0. This implies that Vp,0 ψ = 0; hence, ψ = 0, and the assertion is proved. In the usual way, we now construct the time-dependent single-layer po˚−1/2,k,κ (Γ0 ) × H−1/2,k,κ (Γ2 ) and its correspondtential V α of density α ∈ H ing boundary operator V0 , and the double-layer potential W0 β of density ˚1/2,k,κ (Γ0 ). β∈H ˆ Let (Wp,0 β)(x, p) be the Laplace transform of (W0 β)(X). We define the ˘ p,0 acting on densities βˆ ∈ H ˚1/2,p (∂S0 ) by setting boundary operator W ˆ −β, ˆ γ + Wp,0 β}. ˆ ˘ p,0 βˆ = {γ + Wp,0 β, W 2,e 0,i ˚1/2,p (∂S0 ) to H1/2,p (∂ S) ˘ ˘ p,0 is continuous from H By (6.25) and Theorem 6.8, W ˆ ¯ ˚ and for any β ∈ H1/2,p (∂S0 ), p ∈ Cκ , κ > 0, 2 ˆ ˆ ˘ p,0 β W ˘ ≤ c|p| β−1/2,p;∂S . 1/2,p;∂ S

The corresponding boundary operator in the spaces of originals is denoted by ˘ 0. the symbol W We seek the solution of (6.42) in the form u(X) = (V α)(X) + (W0 β)(X),

X ∈ G,

(6.49)

where α and β are unknown densities. Representation (6.49) leads to the system of boundary equations ˘ 0 β = f˘, V0 α + W where

(6.50)

98


f˘ = {f˜, f2 },

f˜ = {f+,i , δf },

δf = f+,i − f+,e .

˘ k ∈ R, κ > 0, system 6.14 Theorem. For any given f˘ ∈ H1/2,k,κ (Γ), ˚−1/2,k−1,κ (Γ0 ) × H−1/2,k−1,κ (Γ2 ) × (6.50) has a unique solution {α, β} ∈ H ˚ H1/2,k,κ (Γ0 ), in which case u defined by (6.49) belongs to H1,k−1,κ (G). If k ≥ 1, then u is the weak solution of problem (DKD). Proof. In terms of Laplace transforms, (6.50) takes the form Vp,0 α ˆ + Wp,0 βˆ = fˆ,

(6.51)

where fˆ = {fˆ+,i , δ fˆ, fˆ2 } is the transform of f . Comparing the second “components” on both the sides in (6.51), we see that βˆ = −δ fˆ.

(6.52)

By Lemma 6.12, (6.51) is equivalent to the equation ˆ + Tp+ Wp,0 βˆ = gˆ, Tp+ Vp,0 α where

(6.53)

gˆ = {δˆ g , gˆ+,e , gˆ2 } = Tp+ fˆ.

Since the first “component” on the left-hand side in (6.53) is equal to α ˆ 0 , it follows that α ˆ 0 = δˆ g. (6.54) Remarking that in our case the second equation (6.43) is written as ˆ 2 − T − (Vp,0 α ˆ 2, α ˆ 2 = (Tp+ Vp,0 α ˆ + Tp+ Wp,0 β) ˆ + Wp,0 β) p we find that

α ˆ 2 = (Tp+ fˆ)2 − Tp− fˆ2 .

(6.55)

By (6.52), (6.54), (6.55), and Lemma 6.12, ˆ 1/2,p;∂S = δ fˆ1/2,p;∂S , β ˆ α0 −1/2,p;∂S ≤ c|p|fˆ1/2,p;∂ S˘ ,

(6.56)

ˆ α2 −1/2,p;∂S2 ≤ c|p|fˆ1/2,p;∂ S˘ . At the same time, we have +ˆ ˆ 2 ˆ 2 ˆ 2 Vp,0 α ˆ + Wp,0 β ˘ | ≤ c|p| f 1/2,p;∂ S ˘. 1,p;S ≤ c|p||Tp f , f 0;∂ S

(6.57)

Taking (6.56) and (6.57) into account, we now complete the proof by following the standard procedure.

7 Initial-Boundary Value Problems with Other Types of Boundary Conditions

7.1 Mixed Boundary Conditions As mentioned in §1.1, here we assume that the closed boundary curve ∂S consists of two parts ∂S1 and ∂S2 such that mes ∂Sν > 0,

ν = 1, 2,

∂S1 ∩ ∂S2 = ∅,

∂S = ∂S 1 ∪ ∂S 2 ,

and write Γν = ∂Sν × (0, ∞). We denote by πν , ν = 1, 2, the operators of restriction from ∂S to ∂Sν (or from Γ to Γν ). The operators of trace on ∂Sν (Γν ) for functions defined on S ± (G± ) are denoted by γν± . We remark, in particular, that γν± map H1,k,κ (G± ) continuously onto H1/2,k,κ (Γν ) for any k ∈ R. The notation regarding the single-layer and double-layer potentials and the boundary operators generated by them is the same as in Chapter 4. In this section we consider the initial-boundary value problems (DM± ) with mixed boundary data; in other words, we assume that the displacement field is prescribed on Γ1 and that the moments and shear force are prescribed on Γ2 . ¯±) The classical problems (DM± ) consist in finding u ∈ C 2 (G± ) ∩ C 1 (G that satisfy, respectively, B(∂t2 u)(X) + (Au)(X) = 0,

X ∈ G± ,

u(x, 0+) = (∂t u)(x, 0+) = 0,

x ∈ S±,

u± (X) = f1 (X), (T u)± (X) = g2 (X),

X ∈ Γ1 , X ∈ Γ2 .

100

7 Other Types of Boundary Conditions

We call u ∈ H1,0,κ (G± ) weak solutions of (DM± ) if they satisfy ∞ 0

a± (u, v) − (B 1/2 ∂t u, B 1/2 ∂t v)0;S ± dt ∞ = ± (g2 , γ2± v ± )0;∂S2 dt 0

¯ ± ), γ ± v ± = 0, ∀v ∈ C0∞ (G 1

(7.1)

γ1± u = f1 .

7.1 Theorem. For any f1 ∈ H1/2,1,κ (Γ1 ) and g2 ∈ H−1/2,1,κ (Γ2 ), κ > 0, problems (7.1) have unique solutions u ∈ H1,0,κ (G± ). If f1 ∈ H1/2,k,κ (Γ1 ) and g2 ∈ H−1/2,k,κ (Γ2 ), k ∈ R, then u ∈ H1,k−1,κ (G± ) and u1,k−1,κ;G± ≤ c f1 1/2,k,κ;Γ1 + g2 −1/2,k,κ;Γ2 . Proof. We prove the statement for problem (DM+ ); the case of (DM− ) is treated similarly. We begin by rewriting (7.1) in terms of Laplace transforms. ˚1,p (S + , ∂Sν ) be the subspace in H1,p (S + ) of all u such that Let H π3−ν γ + u = 0,

ν = 1, 2;

˚1/2,p (∂Sν ). In the transform domain, (7.1) turns into the that is, γ + u ∈ H ) of seeking u ˆ ∈ H1,p (S + ) such that for any p ∈ Cκ , problem (M+ p ˆ, B 1/2 v)0;S + + a+ (ˆ u, v) p2 (B 1/2 u = (ˆ g2 , γ2+ v)0;∂S2

˚1,p (S + , ∂S2 ), ∀v ∈ H

(7.2)

γ1+ u ˆ = fˆ1 , where gˆ2 and fˆ1 are the Laplace transforms of g2 and f1 , respectively. Since ˚1,p (S + , ∂S2 ), for any v ∈ H g2 −1/2,p;∂S2 γ + v1/2,p;∂S |(ˆ g2 , γ2+ v)0;∂S2 | ≤ ˆ ≤ cˆ g2 −1/2,p;∂S2 v1,p;S + , the form (ˆ g2 , γ2+ v)0;∂S2 defines a bounded antilinear (conjugate linear) func˚1,p (S + , ∂S2 ); hence, it can be written as tional on H (ˆ g2 , γ2+ v)0;∂S2 = (q2 , v)0;S +

˚1,p (S + , ∂S2 ), ∀v ∈ H

(7.3)

˚1,p (S + , ∂S2 ) of H ˚1,p (S + , ∂S2 ) and where q2 belongs to the dual H q2 [H g2 −1/2,p;∂S2 . ˚1,p (S + ,∂S2 )] ≤ cˆ

(7.4)

7.1 Mixed Boundary Conditions

101

˚1,p (S + , ∂S2 ) be the solution of our problem with fˆ1 = 0. First, let u ˆ0 ∈ H Taking (7.3) into account, in this case we rewrite (7.2) in the form p2 (B 1/2 u ˆ0 , B 1/2 v)0;S + + a+ (ˆ u0 , v) ˚1,p (S + , ∂S2 ). ∀v ∈ H

= (q2 , v)0;S +

(7.5)

The unique solvability of (7.5) and the estimate ˆ u0 1,p;S + ≤ c|p|q2 [H ˚1,p (S + ,∂S2 )]

(7.6)

are established by standard arguments. Let lν , ν = 1, 2, be extension operators that map H1/2,p (∂Sν ) continuously to H1/2,p (∂S). In the general problem, we consider a function w = l+ l1 f1 ∈ H1,p (S + ), remark that w1,p;S + ≤ cfˆ1 1/2,p;∂S1 ,

(7.7)

and seek a solution u ˆ of (7.2) of the form u ˆ = u ˆ0 + w. ˆ ˚1,p (S + , ∂S2 ) satisfies H

Clearly, u ˆ0 ∈

p2 (B 1/2 u ˆ0 , B 1/2 v)0;S + + a+ (ˆ u0 , v) = (q2 , v)0;S + − p2 (B 1/2 w, B 1/2 v)0;S + − a+ (w, v) ˚1,p (S + , ∂S2 ). ∀v ∈ H

(7.8)

By (7.7), |p2 (B 1/2 w, B 1/2 v)0;S + + a+ (w, v)| ≤ cw1,p;S + v1,p;S + ≤ cf1 1/2,p;∂S1 v1,p;S + ; hence, p2 (B 1/2 w, B 1/2 v)0;S + + a+ (w, v) = (q1 , v)0;S +

˚1,p (S + , ∂S2 ), ∀v ∈ H

˚1,p (S + , ∂S2 ) and where q1 ∈ H ˆ q1 [H ˚1,p (S + ,∂S2 )] ≤ cf1 1/2,p;∂S1 .

(7.9)

Equation (7.8) now takes the form p2 (B 1/2 u ˆ0 , B 1/2 v)0;S + + a+ (ˆ u0 , v) = (q2 − q1 , v)0;S +

˚1,p (S + , ∂S2 ). ∀v ∈ H

(7.10)

102


˚1,p (S + , ∂S2 ). We have already shown that (7.10) has a unique solution u ˆ0 ∈ H By (7.6), (7.4), and (7.9), ˆ u0 1,p;S + ≤ c|p|q1 − q2 [H ˚1,p (S + ,∂S2 )] ≤ c|p| fˆ1 1/2,p;∂S1 + ˆ g2 −1/2,p;∂S2 ; therefore, u ˆ=u ˆ0 + w satisfies

ˆ u1,p;S + ≤ c|p| fˆ1 1/2,p;∂S1 + ˆ g2 −1/2,p;∂S2 .

(7.11)

The proof is now completed by means of standard arguments. Next, we consider suitable integral representations for the solutions of (DM± ). With this aim, for p ∈ C0 , we introduce three pairs of boundary integral operators and study their properties. ± Let f ∈ H1/2,p (∂S), and let ν, τ = 1, 2, ν = τ . We define operators πp,ντ by ± πp,ντ f = {πν f, πτ Tp± f }, where Tp± are the Poincaré–Steklov operators constructed in terms of the domains S ± . Let g ∈ H−1/2,p (∂S). For the same values of ν and τ as above, the ± operators θp,ντ are defined by ± ± θp,ντ g = πp,ντ (Tp± )−1 g.

Finally, for {fν , gτ } ∈ H1/2,p (∂Sν ) × H−1/2,p (∂Sτ ), we define operators ρ± p,τ ν by ± ± −1 ρ± {fν , gτ }. p,τ ν {fν , gτ } = πp,τ ν (πp,ντ ) The norm of {fν , gτ } in H1/2,p (∂Sν ) × H−1/2,p (∂Sτ ) is {fν , gτ }1/2,p;∂Sν ;−1/2,p;∂Sτ = fν 1/2,p;∂Sν + gτ −1/2,p;∂Sτ . ± 7.2 Lemma. (i) The operators πp,ντ are homeomorphisms from H1/2,p (∂S) ¯ κ , κ > 0, to H1/2,p (∂Sν ) × H−1/2,p (∂Sτ ), and for any f ∈ H1/2,p (∂S), p ∈ C ± πp,ντ f 1/2,p;∂Sν ;−1/2,p;∂Sτ ≤ c|p|f 1/2,p;∂S ,

(7.12)

± f 1/2,p;∂Sν ;−1/2,p;∂Sτ . f 1/2,p;∂S ≤ c|p|πp,ντ

(7.13)

± (ii) The operators θp,ντ are homeomorphisms from H−1/2,p (∂S) to the ¯ κ, space H1/2,p (∂Sν ) × H−1/2,p (∂Sτ ), and for any g ∈ H−1/2,p (∂S), p ∈ C κ > 0, ± θp,ντ g1/2,p;∂Sν ;−1/2,p;∂Sτ ≤ c|p|g−1/2,p;∂S ,

(7.14)

± g1/2,p;∂Sν ;−1/2,p;∂Sτ . g−1/2,p;∂S ≤ c|p|θp,ντ

(7.15)


103

(iii) The operators ρ± p,ντ are homeomorphisms from the space H1/2,p (∂Sν )× H−1/2,p (∂Sτ ) to H1/2,p (∂Sτ ) × H−1/2,p (∂Sν ), and for any element {fν , gτ } ∈ ¯ κ , κ > 0, H1/2,p (∂Sν ) × H−1/2,p (∂S), p ∈ C ρ± p,ντ {fν , gτ }1/2,p;∂Sτ ;−1/2,p;∂Sν ≤ c|p|{fν , gτ }1/2,p;∂Sν ;−1/2,p;∂Sτ ,

(7.16)

{fν , gτ }1/2,p;∂Sν ;−1/2,p;∂Sτ ≤ c|p|ρ± p,ντ {fν , gτ }1/2,p;∂Sτ ;−1/2,p;∂Sν .

(7.17)

± and (7.12) follow from Lemma 3.1. Let Proof. (i) The continuity of πp,ντ ± u ∈ H1,p (S ) be the unique solutions of the problems

p2 (B 1/2 u, B 1/2 v)0;S ± + a± (u, v) = ±(gτ , γτ± v)0;∂Sτ

˚1,p (S ± , ∂Sτ ), ∀v ∈ H

(7.18)

γν± u = fν , where {fν , gτ } ∈ H1/2,p (∂Sν ) × H−1/2,p (∂Sτ ). By (7.11), u1,p;S ± ≤ c|p|{fν , gτ }1/2,p;∂Sν ;−1/2,p;∂Sτ .

(7.19)

± ± If f = γ ± u ∈ H1/2,p (∂S), then πp,ντ f = {fν , gτ }; hence, πp,ντ are surjective. The trace theorem and (7.19) imply that (7.13) holds. ± (ii) By Lemma 3.1 and (i) above, the operators θp,ντ are homeomorphisms from H−1/2,p (∂S) to H1/2,p (∂Sν ) × H−1/2,p (∂Sτ ). Let u ∈ H1,p (S ± ) be the unique solutions of the problems

p2 (B 1/2 u, B 1/2 v)0;S ± + a± (u, v) = ±(g, γ ± v)0;∂S

∀v ∈ H1,p (S ± ),

where g ∈ H−1/2,p (∂S). By (3.12), u1,p;S ± ≤ c|p|g−1/2,p;∂S . If f = (Tp± )−1 g, then f 1/2,p;∂S ≤ c|p|g−1/2,p;∂S . We have πν f 1/2,p;∂Sν + πτ g−1/2,p;∂Sτ ≤ f 1/2,p;∂S + g−1/2,p;∂S ≤ c|p|g−1/2,p;∂S , which proves (7.14). Now let u be the solution of (7.2) with boundary data f1 and g2 , and let γ ± u = f and Tp± f = g. From (3.5) and (7.11) it follows that (7.15) holds. (iii) The definition of the ρ± p,ντ and statement (i) above imply that these operators are homeomorphisms from the space H1/2,p (∂Sν ) × H−1/2,p (∂Sτ )

104


to H1/2,p (∂Sτ ) × H−1/2,p (∂Sν ). As above, let u be the solution of (7.2) with boundary data f1 and g2 , and let f = γ ± u and g = Tp± f . Then πν g−1/2,p;∂Sν + πτ f 1/2,p;∂Sτ ≤ g−1/2,p;∂S + f 1/2,p;∂S ≤ cu1,p;S ± ≤ c|p|{fν , gτ }1/2,p;∂Sν ;−1/2,p;∂Sτ , which proves (7.16). Estimate (7.17) is established similarly. We now consider four representations for the solutions of problems (DM± ) in terms of time-dependent (retarded) plate potentials and demonstrate the unique solvability of the corresponding systems of boundary equations. First, we seek these solutions in the form u(X) = (V α)(X),

X ∈ G± ,

(7.20)

or u = π ± V α, where V α is the single-layer potential of unknown density α. This leads to the system of boundary equations (V α)(X) = f1 (X), (T ± V α)(X) = g2 (X), or π1 V0 α = f1 ,

X ∈ Γ1 , X ∈ Γ2 ,

(7.21)

π2 T ± V0 α = g2 ,

where V0 is the boundary operator generated by V α. 7.3 Theorem. For any given f1 ∈ H1/2,k,κ (Γ1 ) and g2 ∈ H−1/2,k,κ (Γ2 ), κ > 0, k ∈ R, systems (7.21) have unique solutions α ∈ H−1/2,k−2,κ (Γ), in which case the functions u defined by (7.20) belong, respectively, to the spaces H1,k−1,κ (G± ). If k ≥ 1, then these functions are the weak solutions of problems (DM± ). Proof. For simplicity, we prove this only for (DM+ ); the case of (DM− ) is treated similarly. In terms of Laplace transforms, (7.21) becomes π1 Vp,0 α ˆ = fˆ1 , or

π2 Tp+ Vp,0 α ˆ = gˆ2 ,

+ πp,12 Vp,0 α ˆ = {fˆ1 , gˆ2 }.

(7.22)

By Lemmas 7.2 and 4.1, system (7.22) has a unique solution α ˆ ∈ H−1/2,p (∂S), which, for any p ∈ Cκ , satisfies ˆ α−1/2,p;∂S ≤ c|p|2 {fˆ1 , gˆ2 }1/2,p;∂S1 ;−1/2,p;∂S2 .

(7.23)

Taking (7.23) and Theorem 7.1 into account, we now complete the proof by following the standard procedure.


105

The second solution representation is u(X) = (W β)(X),

X ∈ G± ,

(7.24)

or u = π ± W β, where W β is the double-layer potential of unknown density β. This leads to the system of boundary integral equations (W ± β)(X) = f1 (X), (N β)(X) = g2 (X),

X ∈ Γ1 , X ∈ Γ2 ,

(7.25)

or π1 W ± β = f1 ,

π2 N β = g2 ,

where W ± are the boundary operators generated by W β and N = T +W + = T −W −. 7.4 Theorem. For any given f1 ∈ H1/2,k,κ (Γ1 ) and g2 ∈ H−1/2,k,κ (Γ2 ), κ > 0, k ∈ R, systems (7.25) have unique solutions β ∈ H1/2,k−2,κ (Γ), in which case the functions u defined by (7.24) belong, respectively, to the spaces H1,k−1,κ (G± ). If k ≥ 1, then these functions are the weak solutions of problems (DM± ). Proof. Once again, for simplicity we discuss only the interior problem. In terms of Laplace transforms, system (7.25) for (DM+ ) takes the form π1 Wp+ βˆ = fˆ1 ,

π2 Np βˆ = gˆ2 ,

or + πp,12 Wp+ βˆ = {fˆ1 , gˆ2 }.

(7.26)

ˆ Then W + βˆ = (T + )−1 gˆ, and (7.26) can be written as Let gˆ = Np β. p p + θp,12 gˆ = {fˆ1 , gˆ2 }.

(7.27)

By Lemma 7.2 and Theorem 4.8, equation (7.27)—hence, also (7.26)—has a unique solution βˆ ∈ H1/2,p (∂S), which satisfies ˆ 1/2,p;∂S ≤ c|p|2 {fˆ1 , gˆ2 }1/2,p;∂S ;−1/2,p;∂S . β 1 2

(7.28)

We now complete the proof in the usual way, making use of Theorem 7.1 and (7.28). The third integral representation is u(X) = (V α1 )(X) + (W β2 )(X), or u = π ± (V α1 + W β2 ),

X ∈ G± ,

(7.29)

106


˚−1/2,k,κ (Γ1 ) and β2 ∈ H ˚1/2,k,κ (Γ2 ) are unknown densities. This where α1 ∈ H yields the systems of boundary equations (V0 α1 )(X) + (W ± β2 )(X) = f1 (X),

X ∈ Γ1 ,

±

(T V0 α1 )(X) + (N β2 )(X) = g2 (X),

(7.30)

X ∈ Γ2 ,

or π1 (V0 α1 + W ± β2 ) = f1 ,

π2 (T ± V0 α1 + N β2 ) = g2 .

7.5 Theorem. For any f1 ∈ H1/2,k,κ (Γ1 ) and g2 ∈ H−1/2,k,κ (Γ2 ), κ > 0, ˚−1/2,k−1,κ (Γ1 ) × k ∈ R, systems (7.30) have unique solutions {α1 , β2 } ∈ H ˚1/2,k−1,κ (Γ2 ), in which case the functions u defined by (7.29) belong, respecH tively, to H1,k−1,κ (G± ). If k ≥ 1, then these functions are the weak solutions of problems (DM± ). Proof. We discuss only the interior problem; the exterior one is dealt with similarly. In terms of Laplace transforms, (7.30) for (DM+ ) becomes + (Vp,0 α ˆ 1 + Wp+ βˆ2 ) = {fˆ1 , gˆ2 }. πp,12

We claim that

(7.31)

+ −1 − α ˆ 1 = (θp,12 ) − (θp,12 )−1 {fˆ1 , gˆ2 }, − −1 + βˆ2 = (πp,12 ) − (πp,12 )−1 {fˆ1 , gˆ2 }

± is the solution of (7.31). True, from the definition of πp,12 it follows that + −1 − −1 ˆ ˚ π1 (πp,12 ) − (πp,12 ) = 0, so β2 ∈ H1/2,p (∂S2 ). Similarly, it is shown that ˚ α ˆ 1 ∈ H−1/2,p (∂S1 ). Consequently, since ± ± (θp,12 )−1 = Tp± (πp,12 )−1 ,

Vp,0 (Tp+ − Tp− ) = I,

where I is the identity operator, we have Vp,0 α ˆ 1 + Wp+ βˆ2 + −1 − −1

− + = Vp,0 (θp,12 ) − (θp,12 )−1 + Tp− (πp,12 ) − (πp,12 )−1 {fˆ1 , gˆ2 } + −1 + + = Vp,0 (θp,12 ) − Tp− (πp,12 )−1 {fˆ1 , gˆ2 } = (πp,12 )−1 {fˆ1 , gˆ2 }, which proves the assertion. If (7.31) has more than one solution, then the difference {˜ α1 , β˜2 } of any ˚1/2,p (∂S2 ) and satisfies ˚−1/2,p (∂S1 ) × H two solutions belongs to ∈ H + πp,12 (Vp,0 α ˜ 1 + Wp+ β˜2 ) = {0, 0},

so Vp,0 α ˜ 1 + Wp+ β˜2 = 0.


107

ˆ It is clear that π ± (Vp α ˜ 1 +Wp β˜2 ) are solutions of (M± ˆ2 = 0; p ) with f1 = 0 and g consequently, π ± (Vp α ˜ 1 + Wp β˜2 ) = 0. This implies that β˜2 = γ − (Vp α ˜ 1 + Wp β˜2 ) − γ + (Vp α ˜ 1 + Wp β˜2 ) = 0, ˜ 1 + Wp β˜2 ) − Tp− γ − (Vp α ˜ 1 + Wp β˜2 ) = 0, α ˜ 1 = Tp+ γ + (Vp α which proves that the solution of (7.31) is unique. The estimates ˆ α1 −1/2,p;∂S ≤ c|p|{fˆ1 , gˆ2 }1/2,p;∂S1 ;−1/2,p;∂S2 , βˆ2 1/2,p;∂S ≤ c|p|{fˆ1 , gˆ2 }1/2,p;∂S1 ;−1/2,p;∂S2 follow from (7.13) and (7.15). The proof now continues according to the standard procedure. The fourth representation is u(X) = (W β1 )(X) + (V α2 )(X),

X ∈ G± ,

(7.32)

or u = π ± (W β1 + V α2 ), ˚1/2,k,κ (Γ1 ) and α2 ∈ H ˚−1/2,k,κ (Γ2 ). This representation leads where β1 ∈ H to the boundary systems (W ± β1 )(X) + (V0 α2 )(X) = f1 (X), ±

(N β1 )(X) + (T V0 α2 )(X) = g2 (X),

X ∈ Γ1 , X ∈ Γ2 ,

(7.33)

or π1 (W ± β1 + V0 α2 ) = f1 ,

π2 (N β1 + T ± V0 α2 ) = g2 .

7.6 Theorem. For any f1 ∈ H1/2,k,κ (Γ1 ) and g2 ∈ H−1/2,k,κ (Γ2 ), κ > 0, ˚1/2,k−2,κ (Γ1 ) × k ∈ R, systems (7.33) have unique solutions {β1 , α2 } ∈ H ˚−1/2,k−2,κ (Γ2 ), in which case the functions u defined by (7.32) belong, reH spectively, to H1,k−1,κ (G± ). If k ≥ 1, then these functions are the weak solutions of problems (DM± ). Proof. As for the preceding assertions, we consider only the interior problem. In the Laplace transform domain, (7.33) for (DM+ ) becomes + (Wp+ βˆ1 + Vp,0 α ˆ 2 ) = {fˆ1 , gˆ2 }. πp,12

(7.34)

108


We claim that − −1 + + ) ρp,21 − (πp,12 )−1 {fˆ1 , gˆ2 }, βˆ1 = (πp,12 + −1 − ˆ ˆ2 } α ˆ 2 = (θp,12 ) − (θp,12 )−1 ρ+ p,21 {f1 , g is the solution of (7.34). True, we have ˆ2 Wp+ βˆ1 + Vp,0 α − −1 + + = Vp,0 Tp− (πp,12 ) ρp,21 − (πp,12 )−1

+ − ˆ ˆ2 } + (θp,12 )−1 − (θp,12 )−1 ρ+ p,21 {f1 , g

+ + + = Vp,0 Tp+ (πp,12 )−1 − Tp− (πp,12 )−1 {fˆ1 , gˆ2 } = (πp,12 )−1 {fˆ1 , gˆ2 ,

so (7.34) holds. Since

− −1 + + π2 (πp,12 ) ρp,21 − (πp,12 )−1 ]{fˆ1 , gˆ2 } = 0,

˚p,1/2 (∂S1 ). Similarly, we verify that α ˚−1/2,p (∂S). ˆ∈H it follows that βˆ1 ∈ H To establish the unique solvability of (7.34), we repeat the arguments in the proof of Theorem 7.5, with the obvious changes. By (7.13), (7.15), and (7.16), for any p ∈ Cκ , βˆ1 1/2,p;∂S ≤ c|p|2 {fˆ1 , gˆ2 }1/2,p;∂S1 ;−1/2,p;∂S2 , ˆ α2 −1/2,p;∂S ≤ c|p|2 {fˆ1 , gˆ2 }1/2,p;∂S1 ;−1/2,p;∂S2 . The rest of the proof is now completed according to the standard scheme.

7.2 Combined Boundary Conditions In the classical dynamic problems (DC± 1 ) with combined boundary conditions, ¯ ± ) that satisfy we seek functions u = (¯ u, u3 ) ∈ C 2 (G± ) ∩ C 1 (G B(∂t2 u)(X) + (Au)(X) = 0,

X ∈ G± ,

u(x, 0+) = (∂t u)(x, 0+) = 0,

x ∈ S±,

(T u)± (X) = g¯(X),

u± 3 (X) = f3 (X),

X ∈ Γ,

where v¯ = (v1 , v2 ) and the subscript 3 indicates the third component of the corresponding vector-valued function. We denote by γ¯ ± and γ3± the trace operators acting on functions u ∈ H1,p (S ± ), p ∈ C0 , according to the formulas γ¯ ± u = γ ± u,

γ3± u = (γ ± u)3 .

7.2 Combined Boundary Conditions

109

¯ 1,p (S ± ) and H (3) (S ± ) are the subspaces of H1,p (S ± ) that consist of all H 1,p u such that γ3± u = 0 and γ¯ ± u = 0, respectively. ¯ ±1/2,p (∂S) and H (3) (∂S) are defined just as H±1/2,p (∂S) but consist H ±1/2,p of two-component vector-valued functions and scalar functions, respectively. For simplicity, the norms on all these spaces are denoted by · ±1/2,p;∂S . ¯ ±1/2,k,κ (Γ) and H (3) H (Γ), κ > 0, k ∈ R, and their norms ·±1/2,k,κ;Γ ±1/2,k,κ

are introduced by analogy with the above. The trace operators in the spaces of originals which correspond to γ¯ ± and ± γ3 are denoted by the same symbols. ± The variational versions of (DC± 1 ) consist in finding u ∈ H1,0,κ (G ) that satisfy ∞

a± (u, v) − (B 1/2 ∂t u, B 1/2 ∂t v)0;S ± dt

0

∞ = ± (¯ g , v¯± )0;∂S dt 0

7.7 Theorem.

¯ ± ), γ ± v ± = 0, ∀v ∈ C0∞ (G 3

(7.35)

γ3± u = f3 .

(3) ¯ −1/2,1,κ (Γ), κ > 0, For any f3 ∈ H1/2,1,κ (Γ) and g¯ ∈ H (3)

problems (7.35) have unique solutions u ∈ H1,0,κ (G± ). If f3 ∈ H1/2,k,κ (Γ) ¯ −1/2,k,κ (Γ), k ∈ R, then u ∈ H1,k−1,κ (G± ) and and g¯ ∈ H g −1/2,k,κ;Γ . u1,k−1,κ;G± ≤ c f3 1/2,k,κ;Γ + ¯

(7.36)

The proof of this assertion is based on the application of the usual procedure, so we omit it. Let p ∈ C0 , and let u ∈ H1,p (S ± ) be the solutions of the problems p2 (B 1/2 u, B 1/2 v)0;S ± + a± (u, v) = 0,

˚1,p (S ± ), ∀v ∈ H

γ ± u = f,

(7.37)

± where f ∈ H1/2,p (∂S). We define operators πp,3 on H1/2,p (∂S) by ± f = {Tp± f , f3 }, πp,3

where Tp± are the Poincaré–Steklov operators constructed with respect to the ± domains S ± . The operators θp,3 are defined on the space H−1/2,p (∂S) by ± ± = πp,3 (Tp± )−1 . θp,3

110


± ± 7.8 Lemma. For any number p ∈ C0 , the operators πp,3 and θp,3 are homeomorphisms from the spaces H1/2,p (∂S) and H−1/2,p (∂S), respectively, to ¯ −1/2,p (∂S) × H (3) (∂S), and for any f ∈ H1/2,p (∂S) and g ∈ H−1/2,p (∂S), H 1/2,p ¯ κ , κ > 0, p∈C ± πp,3 f H¯

(3) −1/2,p (∂S)×H1/2,p (∂S)

≤ c|p|f 1/2,p;∂S ,

(7.38)

,

(7.39)

± f H¯ f 1/2,p;∂S ≤ c|p|πp,3

(3) −1/2,p (∂S)×H1/2,p (∂S)

± gH¯ θp,3

(3) −1/2,p (∂S)×H1/2,p (∂S)

≤ c|p|g−1/2,p;∂S ,

(7.40)

.

(7.41)

± gH¯ g−1/2,p;∂S ≤ c|p|θp,3

(3) −1/2,p (∂S)×H1/2,p (∂S)

± Proof. The continuity of πp,3 and (7.38) follow from Lemma 3.1. Let u ∈ ± H1,p (S ) be the (unique) solutions of the problems

p2 (B 1/2 u, B 1/2 v)0;S ± + a± (u, v) = ±(¯ g , γ ± v¯)0;∂S γ3± u By (7.36),

¯ 1,p (S ± ), ∀v ∈ H

(7.42)

= f3 .

u1,p;S ± ≤ c|p| f3 1/2,p;∂S + ¯ g −1/2,p;∂S .

(7.43)

The trace theorem and (9.3) imply that (7.39) holds, which proves the state± . ment concerning πp,3 ± ± , the operators θp,3 are homeoBy Lemma 3.1 and the properties of πp,3 (3) ¯ −1/2,p (∂S) × H morphisms from H−1/2,p (∂S) to H (∂S). Estimate (7.40) 1/2,p

follows from Lemma 3.1. If u is the solution of (7.42), then formulas (7.36), written in terms of Laplace transforms (but with the hat omitted from the transform symbol), and (3.5) imply that (7.41) holds. We represent the solutions of problems (DC± 1 ) in the form u(X) = (V α)(X),

X ∈ G± ,

(7.44)

or u = π ± V α. This leads to the systems of boundary equations T ± V0 α = g¯,

(V0 α)3 = f3 . (3)

(7.45)

¯ −1/2,k,κ (Γ), κ > 0, 7.9 Theorem. For any given f3 ∈ H1/2,k,κ (Γ) and g¯ ∈ H k ∈ R, systems (7.45) have unique solutions α ∈ H−1/2,k−2,κ (Γ), in which case the functions u defined by (7.44) belong, respectively, to H1,k−1,κ (G± ). If k ≥ 1, then these functions are the weak solutions of problems (DC± 1 ).


111

Proof. In terms of Laplace transforms, (7.45) becomes Tp± Vp,0 α ˆ = g¯ˆ,

(Vp,0 α ˆ )3 = fˆ3 ,

or ± πp,3 Vp,0 α ˆ = {g¯ˆ, gˆ3 }.

(7.46)

By Lemmas 7.8 and 4.1, equation (7.46) is uniquely solvable, and its solution satisfies ˆ α−1/2,p;∂S ≤ c|p|2 {g¯ˆ, fˆ3 }H¯ . (3) (7.47) (∂S) −1/2,p (∂S)×H 1/2,p

Taking (7.47) and (7.36) into account, we complete the proof in the usual way. We now represent the solutions of our problems in the form X ∈ G± ,

u(X) = (W β)(X),

(7.48)

or u = π ± W β, and arrive at the systems of boundary equations N β = g¯,

(W ± β)3 = f3 .

(7.49)

(3) ¯ −1/2,k,κ (Γ), κ > 0, 7.10 Theorem. For any f3 ∈ H1/2,k,κ (Γ) and g¯ ∈ H k ∈ R, systems (7.49) have unique solutions β ∈ H1/2,k−2,κ (Γ), in which case the functions u defined by (7.48) belong, respectively, to H1,k−1,κ (G± ). If k ≥ 1, then these functions are the weak solutions of problems (DC± 1 ).

Proof. Going over to Laplace transforms, we bring (7.49) to the form Np βˆ = g¯ˆ,

ˆ 3 = fˆ3 , (Wp± β)

or ± θp,3 Np βˆ = {g¯ˆ, fˆ3 }.

(7.50)

By Lemma 7.8 and Theorem 4.8, (7.50) has a unique solution βˆ ∈ H1/2,p (∂S), which satisfies ˆ 1/2,p;∂S ≤ c|p|2 {g¯ˆ, fˆ3 } ¯ β H

(3) −1/2,p (∂S)×H1/2,p (∂S)

.

(7.51)

Using (7.51) and (7.36), we now complete the proof according to the standard scheme.

112


The classical interior and exterior initial-boundary value problems (DC± 2) ¯ ± ) such that consist in finding u ∈ C 2 (G± ) ∩ C 1 (G B(∂t2 u)(X) + (Au)(X) = 0,

X ∈ G± ,

u(x, 0+) = (∂t u)(x, 0+) = 0,

x ∈ S±,

u ¯± (X) = f¯(X),

(T u)± 3 (X) = g3 (X),

X ∈ Γ,

where f¯ and g3 are prescribed on Γ. We call u ∈ H1,0,κ (G± ) weak solutions of (DC± 2 ) if ∞

a± (u, v) − (B 1/2 ∂t u, B 1/2 ∂t v)0;S ± dt

0

=±

∞ (g3 , v3± )0;∂S dt 0

¯ ± ), γ¯ ± v ± = 0, ∀v ∈ C0∞ (G

(7.52)

γ¯ ± u = f¯.

¯ 1/2,1,κ (Γ) and g3 ∈ H (3) 7.11 Theorem. For any f¯ ∈ H −1/2,1,κ (Γ), κ > 0, ¯ 1/2,k,κ (Γ) and problems (7.52) have unique solutions u ∈ H1,0,κ (G± ). If f¯ ∈ H (3)

g3 ∈ H−1/2,k,κ (Γ), k ∈ R, then u ∈ H1,k−1,κ (G± ) and u1,k−1,κ;G± ≤ c f¯1/2,k,κ;Γ + g3 −1/2,k,κ;Γ .

(7.53)

Once again, since the proof of this assertion is carried out according to the standard procedure, we omit it. Let u ∈ H1,p (S ± ) be the solutions of (7.37). We define operators π ¯p± on ± ¯ H1/2,p (∂S) and θp on H−1/2,p (∂S) by π ¯p± f = {f¯, (Tp± f )3 }, θ¯p± = π ¯p± (Tp± )−1 . ¯p± and θ¯p± are home7.12 Lemma. For any number p ∈ C0 , the operators π ¯ 1/2,p (∂S) × omorphisms from H1/2,p (∂S) and H−1/2,p (∂S), respectively, to H (3) ¯ κ, H−1/2,p (∂S), and for any f ∈ H1/2,p (∂S) and g ∈ H−1/2,p (∂S), p ∈ C κ > 0, ¯ πp± f H¯

(3) 1/2,p (∂S)×H−1/2,p (∂S)

πp± f H¯ f 1/2,p;∂S ≤ c|p|¯

≤ c|p|f 1/2,p;∂S ,

(3) 1/2,p (∂S)×H−1/2,p (∂S)

θ¯p± gH¯

(3) 1/2,p (∂S)×H−1/2,p (∂S)

g−1/2,p;∂S ≤ c|p|θ¯p± gH¯

,

≤ c|p|g−1/2,p;∂S ,

(3) 1/2,p (∂S)×H−1/2,p (∂S)

.


113

The proof of this assertion is a repeat of that of Lemma 7.3, with the obvious changes. We now seek the solutions of (DC± 2 ) in the form (7.44). This yields the systems of boundary equations V0 α = f¯,

(T ± V0 α)3 = g3 .

(7.54)

¯ 1/2,k,κ (Γ) and g3 ∈ H (3) 7.13 Theorem. For any given f¯ ∈ H −1/2,k,κ (Γ), κ > 0, k ∈ R, systems (7.54) have unique solutions α ∈ H−1/2,k−2,κ (Γ), in which case the functions u defined by (7.44) belong, respectively, to the spaces H1,k−1,κ (G± ). If k ≥ 1, then these functions are the weak solutions of problems (DC± 2 ). Proof. In terms of Laplace transforms, (7.54) is written as ¯ Vp,0 α ˆ = fˆ,

(Tp± Vp,0 α ˆ )3 = gˆ3 ,

or ¯ π ¯p± Vp,0 α ˆ = {fˆ, gˆ3 }.

(7.55)

By Lemmas 7.12 and 4.1, (7.55) have unique solutions α ˆ ∈ H−1/2,p (∂S), which satisfy ¯ ˆ α−1/2,p;∂S ≤ c|p|2 {fˆ, gˆ3 }H¯ . (3) (7.56) 1/2,p (∂S)×H−1/2,p (∂S) Estimates (7.56) and (7.53) now enable us to complete the proof by following the usual procedure. Representation (7.48) for the solutions (DC± 2 ) leads to the systems of boundary equations W ± β = f¯,

(N β)3 = g3 .

(7.57)

¯ 1/2,k,κ (Γ) and g3 ∈ H (3) 7.14 Theorem. For any f¯ ∈ H −1/2,k,κ (Γ), κ > 0, k ∈ R, systems (7.57) have unique solutions β ∈ H1/2,k−2,κ (Γ), in which case the functions u defined by (7.48) belong, respectively, to H1,k−1,κ (G± ). If k ≥ 1, then these functions are the weak solutions of problems (DC± 2 ). Proof. In the Laplace transform domain, (7.57) become ¯ Wp± βˆ = fˆ,

ˆ 3 = gˆ3 , (Np β)

or ¯ θ¯p± Np βˆ = {fˆ, gˆ3 }.

(7.58)

114


By Lemma 7.12 and Theorem 4.8, systems (7.58) have unique solutions β ∈ H1/2,p (∂S), which satisfy ˆ, gˆ3 } ¯ ˆ 1/2,p;∂S ≤ c|p|2 {f¯ β H

(3) 1/2,p (∂S)×H−1/2,p (∂S)

.

(7.59)

Inequalities (7.59) and (7.53) are now used to complete the proof be means of the standard scheme.

7.3 Elastic Boundary Conditions Let χ be a self-adjoint (3 × 3)-matrix-valued function defined on ∂S and such that (i) the operator of multiplication by χ is continuous from H1/2 (∂S) to H−1/2 (∂S); (ii) there is a χ0 = const > 0 such that (χf, f )0;∂S ≥ χ0 f 21/2

∀f ∈ H1/2 (∂S).

It is easy to see that any constant, positive definite (3 × 3)-matrix satisfies these conditions. In the classical dynamic problems (DR± ) with elastic (Robin-type) bound¯ ± ) such that ary conditions, we seek a function u ∈ C 2 (G± ) ∩ C 1 (G B(∂t2 u)(X) + (Au)(X) = 0,

X ∈ G± ,

u(x, 0+) = (∂t u)(x, 0+) = 0,

x ∈ S±,

(T u)± (X) ± χ(x)u± (X) = g(X),

X ∈ Γ, x ∈ ∂S,

where g is prescribed on Γ. The variational versions of (DR± ) consist in finding u ∈ H1,0,κ (G± ) such that ∞ 0

a± (u, v) − (B 1/2 ∂t u, B 1/2 ∂t v)0;S ± + (χu, v ± )0;∂S dt ∞ = ± (g, v ± )0;∂S dt

¯ ± ). ∀v ∈ C0∞ (G

(7.60)

0

7.15 Theorem. For any g ∈ H−1/2,1,κ (Γ), κ > 0, problems (7.60) have unique solutions u ∈ H1,0,κ (G± ). If g ∈ H−1/2,k,κ (Γ), k ∈ R, then u ∈ H1,k−1,κ (G± ) and u1,k−1,κ;G± ≤ cg−1/2,k,κ;Γ . This assertion is proved just like Theorem 3.3, with the obvious changes.

7.3 Elastic Boundary Conditions

115

Let f ∈ H1/2,p (∂S), and let u ∈ H1,p (S ± ) be the solutions of the problems ˚1,p (S ± ), p2 (B 1/2 u, B 1/2 v)0;S ± + a± (u, v) = 0 ∀v ∈ H γ ± u = f.

(7.61)

± We define operators Tp,χ by means of the equality ± (Tp,χ f, ϕ)0;∂S

= ± a± (u, v) + p2 (B 1/2 u, B 1/2 , v)0;S ± + (χf, ϕ)0;∂S ,

(7.62)

where ϕ ∈ H1/2,p (∂S) is arbitrary and v ∈ H1,p (S ± ) is such that γ ± v = ϕ. ± are homeomorphisms from 7.16 Lemma. For any p ∈ C0 , the operators Tp,χ ¯ κ , κ > 0, H1/2,p (∂S) to H−1/2,p (∂S), and for any f ∈ H1/2,p (∂S), p ∈ C ± Tp,χ f −1/2,p;∂S ≤ c|p|f 1/2,p;∂S ,

(7.63)

± f −1/2,p;∂S . f 1/2,p;∂S ≤ c|p|Tp,χ

(7.64)

Proof. By (7.62), ± f = Tp± f ± χf. Tp,χ

Lemma 3.1 and the inequality χf −1/2,p;∂S ≤ cf 1/2,p;∂S ± imply that Tp,χ are continuous from H1/2,p (∂S) to H−1/2,p (∂S) and that (7.63) holds. From (7.62) it also follows that

|p|2 B 1/2 u20;S ± + a± (u, u) + (χf, f )0;∂S ± = ±σ −1 Re{¯ p(Tp,χ f, f )0;∂S },

σ = Re p;

hence, ± p(Tp,χ f, f )0;∂S }. u21,p;S ± ≤ ±cσ −1 Re{¯

By the trace theorem and (7.65), ± f, f )0;∂S | f 21/2,p;∂S ≤ c|p|(Tp,χ ± ≤ c|p|Tp,χ f −1/2,p;∂S f 1/2,p;∂S ,

which proves (7.64).

(7.65)

116


± If the ranges of Tp,χ are not dense in H−1/2,p (∂S), then there is a nonzero ψ ∈ H1/2,p (∂S) such that ± (Tp,χ f, ψ)0;∂S = 0 ∀f ∈ H1/2,p (∂S).

(7.66)

We take f = ψ in (7.66) and construct the corresponding solution u ∈ H1,p (S ± ) of (7.61). By (7.65) and (7.66), u = 0; therefore, ψ = γ ± u = 0. This contradiction proves the lemma. We now represent the solutions of problems (DR± ) in the form u(X) = (V α)(X),

X ∈ G± ,

(7.67)

or u = π ± V α. This yields the systems of boundary equations Tχ± V0 α = g,

(7.68)

where Tχ± are the boundary operators in the spaces of originals constructed ± from Tp,χ in the usual way. 7.17 Theorem. For any g ∈ H−1/2,k,κ (Γ), κ > 0, k ∈ R, systems (7.68) have unique solutions α ∈ H−1/2,k−2,κ (Γ), in which case the functions u defined by (7.67) belong, respectively, to H1,k−1,κ (G± ). If k > 1, then these functions are the weak solutions of problems (DR± ). Proof. Going over to Laplace transforms, for any p ∈ Cκ we obtain the system of boundary equations ± Vp,0 α ˆ = gˆ. Tp,χ (7.69) By Lemmas 4.1 and 7.16, system (7.69) is uniquely solvable for any gˆ ∈ H−1/2,p (∂S), and its solution α ˆ ∈ H−1/2,p (∂S) satisfies ˆ α−1/2,p;∂S ≤ c|p|2 ˆ g −1/2,p;∂S .

(7.70)

Estimate (7.70) and Theorem 7.15 now enable us to complete the proof in the usual way. We introduce the operators ± ± Np,χ = Tp,χ Wp± . ± 7.18 Lemma. (i) For any p ∈ C0 , the operators Np,χ are homeomorphisms from H1/2,p (∂S) to H−1/2,p (∂S), and for any function βˆ ∈ H1/2,p (∂S), with ¯ κ , κ > 0, p∈C ± ˆ ˆ 1/2,p;∂S , β−1/2,p;∂S ≤ c|p|3 β Np,χ

ˆ −1/2,p;∂S . ˆ 1/2,p;∂S ≤ c|p|3 N ± β β p,χ

7.3 Elastic Boundary Conditions

117

¯ κ , κ0 > 0, (ii) If χ is such that for p ∈ C 0 χw−1/2,p;∂S ≤ cw−1/2,p;∂S

∀w ∈ H−1/2,p;∂S ,

(7.71)

¯ κ∗ , then there is κ∗ > 0 such that for any p ∈ C ˆ 1/2,p;∂S ≤ c|p|N ± β ˆ −1/2,p;∂S . β p,χ

(7.72)

Proof. The first statement follows from Lemmas 4.5 and 7.16. Suppose now that (7.71) holds, and let βˆ ∈ H1/2,p (∂S). By the trace theorem, −ˆ 2 ˆ 2 Wp+ β 1/2,p;∂S + Wp β1/2,p;∂S ˆ 2 ≤ c π + Wp β

1,p;S +

ˆ 2 − + π − Wp β 1,p;S

ˆ β) ˆ 0;∂S }. p(Np β, ≤ cσ −1 Re{¯ Clearly,

± ˆ ˆ β = Np βˆ ± χWp± β, Np,χ

so, −ˆ 2 ˆ 2 Wp+ β 1/2,p;∂S + Wp β1/2,p;∂S

ˆ β) ˆ 0;∂S | + |p||(χW ± β, ˆ β) ˆ 0;∂S | . ≤ σ −1 |p||(N ± β, p,χ

p

¯ κ . Then, by (7.71), Let p ∈ C 0 ˆ −1/2,p;∂S ≤ cβ ˆ 1/2,p;∂S ˆ −1/2,p;∂S ≤ c|p|β |p|χβ and −ˆ 2 ˆ 2 Wp+ β 1/2,p;∂S + Wp β1/2,p;∂S ± ˆ ˆ ˆ 1/2,p,∂S ; ˆ 1/2,p;∂S β ≤ cσ −1 |p||(Npχ β, β)0;∂S | + cσ −1 Wp± β

therefore, −ˆ 2 ˆ 2 Wp+ β 1/2,p;∂S + Wp β1/2,p;∂S ± ˆ ˆ ˆ 2 ≤ cσ −1 |(Np,χ β, β)0;∂S | + cσ −2 β 1/2,p;∂S .

The jump formula βˆ = Wp− βˆ − Wp+ βˆ implies that +ˆ 2 −ˆ 2 ˆ 2 β 1/2,p;∂S ≤ c(Wp β1/2,p;∂S + Wp β1/2,p;∂S ) ± ˆ ˆ 1/2,p;∂S + cσ −2 β ˆ 2 ≤ cσ −1 Np,χ β−1/2,p;∂S β 1/2,p;∂S . (7.73)

118


By (7.73), there is κ∗ > 0 such that for σ ≥ κ∗ , −1 ± ˆ ˆ 2 ˆ 1/2,p;∂S , β Np,χ β−1/2,p;∂S β 1/2,p;∂S ≤ cσ

which proves (7.72). We now seek the solutions of problems (DR± ) in the form u(X) = (W β)(X),

X ∈ G± ,

(7.74)

or u = π ± W β. This leads to the systems of boundary equations Nχ± β = g,

(7.75)

where Nχ± are the boundary operators in the spaces of originals which cor± respond to Np,χ . 7.19 Theorem. For any g ∈ H−1/2,k,κ (Γ), κ > 0, k ∈ R, systems (7.75) have unique solutions β ∈ H1/2,k−3,κ (Γ). If χ satisfies (7.71), then there is κ∗ > 0 such that β ∈ H1/2,k−1,κ (Γ) for κ ≥ κ∗ , in which case the functions u defined by (7.74) belong, respectively, to H1,k−1,κ (Γ). If k ≥ 1, then these functions are the weak solutions of problems (DR± ). Proof. In the Laplace transform domain, (7.75) becomes ± ˆ Np,χ β = gˆ.

The assertion now follows from Lemma 7.18 and Theorem 7.15.

8 Boundary Integral Equations for Plates on a Generalized Elastic Foundation

8.1 Formulation and Solvability of the Problems In the case of bending of a plate that lies on a generalized elastic foundation, the equations of motion (1.4) are replaced by ∂j tij − (Kv)i + fi = ρ∂t2 vi ,

i = 1, 2, 3,

where K = (kij ) is a constant, symmetric, positive definite (3 × 3)-matrix. The averaging procedure described in §1.1 leads to B(∂t2 u)(X) + (Au)(X) + (Ku)(X) = q(X), X ∈ G = S × (0, ∞); here

K=

¯ K 0

0 k33

¯ = h2 K

,

k11 k21

k12 k22

(8.1)

.

¯ and K are positive definite matrices. Clearly, K For simplicity, we restrict our attention to only two types of boundary conditions; the corresponding theory can easily be developed for all the initialboundary value problems investigated earlier. The variational problems (DD± K ) with Dirichlet boundary conditions and zero initial conditions for the homogeneous equation of motion consist in finding u ∈ H1,0,κ (G± ) such that ∞

a± (u, v) + (Ku, v)0;S ± − (B 1/2 ∂t u, B 1/2 ∂t v)0;S ± dt = 0

0 ±

γ u = f, where f is prescribed on Γ.

¯ ± ), v ± = 0, ∀v ∈ C0∞ (G

(8.2)

120

8 Plates on a Generalized Elastic Foundation

Obviously, in terms of Laplace transforms, (8.2) turns into the problem of seeking the (weak) solutions u ˆ ∈ H1,p (S ± ) of problems (D± K,p ) p2 B u ˆ(x, p) + (Aˆ u)(x, p) + Kˆ u(x, p) = 0, γ±u ˆ(x, p) = fˆ(x, p),

x ∈ S±,

x ∈ ∂S.

¯ κ , κ > 0, problems (D± ) have 8.1 Lemma. For any fˆ ∈ H1/2,p (∂S), p ∈ C K,p unique solutions u ˆ ∈ H1,p (S ± ) and ˆ u1,p;S ± ≤ c|p|fˆ1/2,p;∂S . This assertion is proved just like Theorems 2.1 and 2.2. 8.2 Theorem. For any f ∈ H1/2,1,κ (Γ), κ > 0, problems (8.2) have unique solutions u ∈ H1,0,κ (G± ). If f ∈ H1/2,k,κ (Γ), k ∈ R, then u ∈ H1,k−1,κ (Γ± ) and u1,k−1,κ;G± ≤ cf 1/2,k,κ;Γ . The proof of this statement is similar to that of Theorem 2.3. In problems (DN± K ) for the homogeneous equation of motion with Neumann boundary conditions and zero initial conditions, we want to find u ∈ H1,0,κ (G± ) such that ∞ 0

a± (u, v) + (Ku, v)0;S ± − (B 1/2 ∂t u, B 1/2 ∂t v)0;S ± dt ∞ = ± (g, v ± )0;∂S dt

¯ ± ), ∀v ∈ C0∞ (G

(8.3)

0

where g is prescribed on Γ. When the Laplace transformation is applied to (8.3), we arrive at problems (N± ˆ ∈ H1,p (S ± ) such that K,p ), which consist in finding (weak) solutions u p2 B u ˆ(x, p) + (Aˆ u)(x, p) + Kˆ u(x, p) = 0, (T u ˆ)± (x, p) = gˆ(x, p),

x ∈ S±,

x ∈ ∂S.

The next two assertions are stated without proof because they are fully analogous to their corresponding counterparts in the case K = 0. ¯ κ , κ > 0, problems (N± ) 8.3 Lemma. For any gˆ ∈ H−1/2,p (∂S), p ∈ C K,p ± have unique solutions u ˆ ∈ H1,p (S ) and ˆ u1,p;S ± ≤ c|p|ˆ g −1/2,p;∂S .


121

8.4 Theorem. For any g ∈ H−1/2,1,κ (Γ), κ > 0, problems (8.3) have unique solutions u ∈ H1,0,κ (G± ). If g ∈ H−1/2,k,κ (Γ), k ∈ R, then u ∈ H1,k−1,κ (G± ) and u1,k−1,κ;G± ≤ cg−1/2,k,κ;Γ . For simplicity, in what follows we consider the case when K = diag{h2 k0 , h2 k0 , k3 },

k0 , k3 = const > 0;

however, all the results obtained below remain valid for a general case constant, symmetric, and positive definite matrix K.

8.2 A Matrix of Fundamental Solutions The construction of single-layer and double-layer plate potentials requires the knowledge of a matrix of fundamental solutions DK (x, t) for equation (8.1), ˆ K (x, p). The latter is a (3 × 3)-matrix such that or of its Laplace transform D ˆ K (x, p) + (AD ˆ K )(x, p) + KD ˆ K (x, p) = δ(x)I, x ∈ R2 , p2 B D

(8.4)

where δ is the Dirac delta distribution and I is the identity (3 × 3)-matrix. Going over to Fourier transforms in (8.4), we arrive at ˜ K (ξ, p) = I, ˜ K (ξ, p) + A(ξ)D ˜ K (ξ, p) + KD p2 B D or ˜ K (ξ, p) = I, ΨK (ξ, p)D

(8.5)

where the entries of the (3 × 3)-matrix ΨK (ξ, p) are ΨK,αβ (ξ, p) = h2 (λ + µ)ξα ξβ + δαβ h2 (k0 + ρp2 ) + µ + h2 µ|ξ|2 , ΨK,33 (ξ, p) = k3 + ρp2 + µ|ξ|2 ,

(8.6)

ΨK,α3 (ξ, p) = −ΨK,3α (ξ, p) = −iµξα . To obtain the inverse matrix Ψ−1 K (ξ, p), we first need to compute det ΨK (ξ, p). Direct calculation shows that det ΨK (ξ, p) is a function of |ξ|2 alone; in other words, it is rotation-invariant in R2 . Therefore, we may take ξ1 = |ξ|, to find that

ξ2 = 0

122


det ΨK (ξ, p) = RK (|ξ|2 , p) = h2 µ|ξ|2 + h2 (k0 + ρp2 ) + µ × h2 µ(λ + µ)|ξ|4 + h2 (k3 + ρp2 )(λ + 2µ) + (k0 + ρp2 )µ |ξ|2

+ (k3 + ρp2 ) h2 (k0 + ρp2 ) + µ . The roots tj = −χ2j , j = 1, 2, 3, of the equation RK (t, p) = 0 are −1 χ21,2 = 2hµ(λ + 2µ) × h (k3 + ρp2 )(λ + 2µ) + (k0 + ρp2 )µ 2 ± h2 (k3 + ρp2 )(λ + 2µ) − (k0 + ρp2 )µ

1/2

, − 4(k3 + ρp2 )µ2 (λ + 2µ)

χ23 = (h2 µ)−1 h2 (k0 + ρp2 ) + µ .

We choose χj so that Re χj ≥ 0, j = 1, 2, 3. 8.5 Lemma. (i) The equation RK (t, p) = 0 does not have a triple root for any p ∈ C0 . (ii) The equation RK (t, p) = 0 has a double root only for finitely many values of p ∈ C0 . (iii) Re χj > 0, j = 1, 2, 3. Proof. (i) It is not difficult to see that χ21 = χ22 = χ23 only if k0 + ρp2 = −

2λµ(λ + 2µ) 2h2 (λ + µ)2

or k0 + ρp2 = −

µ , h2

and that either of these equalities leads immediately to a contradiction. (ii) The validity of the second assertion is obvious. (iii) First, we remark that χi = 0. True, if χi = 0 for some i ∈ {1, 2, 3}, then µ ρp2 = − 2 + k0 , h which contradicts the fact that p ∈ C0 . ˜ 2 = −χ2 is a posiIf Re χi = 0 for some i, then χ2i < 0; hence, t˜ = |ξ| i ˜ 0) and consider a tive root of the equation RK (t, p) = 0. We take ξ˜ = (|ξ|,


123

nonzero solution g = (g1 , g2 , g3 )T of the system of linear algebraic equations ˜ p)g = 0. Multiplying this equality by g ∗ , we obtain ΨK (ξ, 2 2 h (ρp + k0 ) + µ (|g1 |2 + |g2 |2 ) ˜ 2 )|g3 |2 + h2 |ξ| ˜ 2 (λ + 2µ)|g1 |2 + µ|g2 |2 + (ρp2 + k3 + µ|ξ| ˜ Re(g ∗ ig1 ) = 0; − 2µ|ξ| 3 therefore, p2 ∈ R and p2 > 0. Since ˜ Re(g ∗ ig1 ) ≥ −µ(|g1 |2 + |ξ| ˜ 2 |g3 |2 ), 2µ|ξ| 3 we obtain (ρp2 + k0 )h2 (|g1 |2 + |g2 |2 ) + (ρp2 + k3 )|g3 |2 ˜ 2 (λ + 2µ)|g1 |2 + µ|g2 |2 ≤ 0. + h2 |ξ| This contradiction completes the proof. If we write

˜ K (ξ, p) = det ΨK (ξ, p) −1 , Ψ

then, by (8.5) and (8.6), ˜ K,11 (ξ, p) = h2 µ2 |ξ|4 + h2 µ(λ + µ)|ξ|2 ξ 2 D 2 + µh2 (k0 + k3 + 2ρp2 )|ξ|2 + µ2 ξ12 + h2 (λ + µ)(k3 + ρp2 )ξ22 ˜ K (ξ, p), + h2 (k0 + p2 ) + µ (k3 + ρp2 ) Ψ

˜ K,22 (ξ, p) = h2 µ2 |ξ|4 + h2 µ(λ + µ)|ξ|2 ξ12 D

+ µh2 (k0 + k3 + 2ρp2 )|ξ|2 + µ2 ξ22 + h2 (λ + µ)(k3 + ρp2 )ξ12 ˜ K (ξ, p), + h2 (k0 + ρp2 ) + µ (k3 + ρp2 ) Ψ 4 ˜ K,33 (ξ, p) = h µ(λ + µ)|ξ|4 + h2 (λ + 3µ) h2 (k0 + ρp2 ) + µ |ξ|2 D ˜ K (ξ, p), + h2 (k0 + ρp2 ) + µ)2 Ψ ˜ K,21 (ξ, p) ˜ K,12 (ξ, p) = D D = −ξ1 ξ2 h2 µ(λ + µ)|ξ|2

˜ K (ξ, p), + h2 (λ + µ)(k3 + ρp2 ) − µ2 Ψ

˜ 3α,K (ξ, p) ˜ K,α3 (ξ, p) = −D D ˜ K (ξ, p). = iµξα h2 µ|ξ|2 + h2 (k0 + ρp2 ) + µ Ψ

124


Consequently, applying the inverse Laplace transformation, we find that the ˆ K (x, p) are entries of D ˆ K,11 (x, p) = h2 µ2 ∆2 + h2 µ(λ + µ)∆∂22 D − µh2 (k0 + k3 + 2ρp2 )∆ − µ2 ∂12 − h2 (λ + µ)(k3 + ρp2 )∂22 + h2 (k0 + ρp2 ) + µ k3 ΨK (x, p),

ˆ K,22 (x, p) = h2 µ2 ∆2 + h2 µ(λ + µ)∆∂12 D − h2 µ(k0 + k3 + 2ρp2 )∆

− µ2 ∂22 − h2 (λ + µ)(k3 + ρp2 )∂12 + h2 (k0 + ρp2 ) + µ (k3 + ρp2 ) ΨK (x, p), (8.7) 4 ˆ K,33 (x, p) = h µ(λ + 2µ)∆2 − h2 (λ + 3µ) h2 (k0 + ρp2 ) + µ ∆ D 2 + h2 (k0 + ρp2 ) + µ ΨK (x, p), ˆ K,12 (x, p) = D ˆ K,21 (x, p) D = − h2 µ(λ + µ)∆

+ h2 (λ + µ)(k3 + ρp2 ) − µ2 ∂1 ∂2 ΨK (x, p),

ˆ K,α3 (x, p) = −D ˆ K,3α (x, p) D 2 = µ h µ∆ − h2 k0 + ρp2 ) − µ ∂α ΨK (x, p), where ΨK (x, p) = (4π 2 )−1

˜ K (ξ, p)dξ e−i(x,ξ) Ψ

R2

˜ K (ξ, p). is the inverse Fourier transform of Ψ There are two possible cases. If the equation RK (t, p) = 0 has distinct roots, then we easily convince ourselves that −1 ΨK (x, p) = 2πh4 µ2 (λ + 2µ) ci K0 (χi |x|), where

c1 = [(χ21 − χ23 )(χ21 − χ22 )]−1 , c2 = [(χ21 − χ22 )(χ22 − χ23 )]−1 , c3 = [(χ21 − χ23 )(χ22 − χ23 )]−1 .

If, on the other hand, RK (t, p) = 0 has a double root, say, χ21 = χ22 = χ23 ,


125

then −1 ΨK (x, p) = 2πh4 µ2 (λ + 2µ) × c˜1 K0 (χ1 |x|) + c˜3 K0 (χ3 |x|)

+ c˜2 (2χ1 )−1 |x|K1 (χ1 |x|) ,

where c˜1 = −˜ c3 = −(χ23 − χ21 ),

c˜2 = (χ23 − χ21 )−1 .

8.6 Lemma. For any p ∈ C0 , the function ΨK (x, p) can be represented near x = 0 in the form −1 4 ΨK (x, p) = − 128πh4 µ2 (λ + 2µ) |x| ln |x| + O(|x|6 ln |x|) + ΨK,0 (x, p),

(8.8)

where ΨK,0 (x, p) is infinitely differentiable. In addition, ΨK (x, p) → 0 exponentially as |x| → ∞. This assertion is proved just like Lemma 1.2. 8.7 Corollary. For any p ∈ C0 , the elements of the matrix of fundamental ˆ K (x, p) can be represented in the neighborhood of x = 0 in the form solutions D ˆ K,αβ (x, p) = 4πµ(λ + 2µ)h2 −1 D × (λ + µ)xα xβ |x|−2 − (λ + 3µ)δαβ ln |x| ˆ K,0,αβ (x, p), + O(|x|2 ln |x|) + D ˆ K,33 (x, p) = −(2πh2 µ)−1 ln |x| D

(8.9)

ˆ K,0,33 (x, p), + O(|x|2 ln |x|) + D ˆ K,α3 (x, p) = −D ˆ 3αK (x, p) = − 4πh2 (λ + 2µ) −1 xα ln |x| D ˆ K,0,α3 (x, p), + O(|x|2 ln |x|) + D ˆ K,0,ij (x, p) are infinitely differentiable functions. where D The proof of this assertion follows from (8.7) and (8.8). 8.8 Remark. Equalities (8.9) show that for any p ∈ C0 , the asymptotic behavˆ K (x, p) near x = 0 coincides with that of the matrix of fundamental ior of D solutions D(x) in the equilibrium case [10].

126


8.3 Properties of the Boundary Operators Let u ∈ H1,p (S ± ) be the (unique) solutions of problems (D± K,p ) with bound± ary data f ∈ H1/2,p (∂S). We define Poincaré–Steklov operators TK,p on H1/2,p (∂S) through the equality ± f, ϕ)0;∂S = ± p2 (B 1/2 u, B 1/2 w)0,S ± + aK,± (u, w) , (TK,p where ϕ ∈ H1/2,p (∂S), w is any element of H1,p (S ± ) such that γ ± w = ϕ, and aK,± are the bilinear forms defined by aK,± (u, w) = a± (u, w) + (Ku, w)0,S ± . ± 8.9 Lemma. For any p ∈ C0 , the operators TK,p are homeomorphisms from ¯ κ , κ > 0, H1/2,p (∂S) to H−1/2,p (∂S), and for any f ∈ H1/2,p (∂S), p ∈ C ± f −1/2,p;∂S ≤ c|p|f 1/2,p;∂S , TK,p ± f 1/2,p;∂S ≤ c|p|TK,p f −1/2,p;∂S .

The proof of this assertion is identical to that of Lemma 3.1, with the appropriate changes. At this stage, we introduce the operators TˆK± , (TˆK± )−1 and TK± , (TK± )−1 in the usual way. 8.10 Theorem. For any κ > 0 and k ∈ R, the operators TK± are continuous and injective from H1/2,k,κ (Γ) to H−1/2,k−1,κ (Γ), and their ranges are dense in H−1/2,k−1,κ (Γ). The inverses (T ± )−1 , extended by continuity from the ranges of T ± to H−1/2,k,κ (Γ), are continuous and injective from H−1/2,k,κ (Γ) to H1/2,k−1,κ (Γ) for any k ∈ R, and their ranges are dense in H1/2,k−1,κ (Γ) . The assertion follows from Lemma 8.9 and the arguments used in the proof of Theorem 3.2. The time-dependent (retarded) single-layer and double-layer potentials are now defined by ∞ (VK α)(X) = DK (x − y, t − τ )α(y, τ ) dsy dτ, −∞ ∂S

and

∞ (WK β)(X) =

(j) β(y, τ ), (Ty DK )(y − x, t − τ ) ej dsy dτ.

−∞ ∂S

In view of (8.8), the boundary properties of VK α and WK β coincide with those of the corresponding plate potentials in the case K = 0.

8.4 Solvability of the Boundary Equations

127

The boundary operators VK,0 , WK± , and NK = TK+ WK+ = TK− WK− associated with the above potentials are introduced according to the scheme used in §4.1. Their properties are stated in the next three assertions, whose proofs are essentially identical to those of Theorem 4.3, 4.6, and 4.8. 8.11 Theorem. For any κ > 0 and k ∈ R, the operator VK,0 is continuous and injective from H−1/2,k,κ (Γ) to H1/2,k−1,κ (Γ), and its range is dense in H1/2,k−1,κ (Γ). The inverse (VK,0 )−1 , extended by continuity from the range of VK,0 to H1/2,k,κ (Γ), is continuous and injective from H1/2,k,κ (Γ) to H−1/2,k−1,κ (Γ) for any k ∈ R, and its range is dense in H−1/2,k−1,κ (Γ). In addition, for any α ∈ H−1/2,k,κ (Γ), π + VK α1,k−1,κ;G+ + π − VK α1,k−1,κ;G− ≤ cα−1/2,k,κ;Γ . 8.12 Theorem. For any κ > 0 and k ∈ R, the operators WK± are continuous and injective from H1/2,k,κ (Γ) to H1/2,k−2,κ (Γ), and their ranges are dense in H1/2,k−2,κ (Γ). The inverses (WK± )−1 , extended by continuity from the ranges of WK± , respectively, to H1/2,k,κ (Γ) are continuous and injective from H1/2,k,κ (Γ) to H1/2,k−2,κ (Γ) for any k ∈ R, and their ranges are dense in H1/2,k−2,κ (Γ). In addition, for any β ∈ H1/2,k,κ (Γ), π + WK β1,k−2,κ;G+ + π − WK β1,k−2,κ;G− ≤ cβ1/2,k,κ;Γ . 8.13 Theorem. For any κ > 0 and k ∈ R, the operator NK is continuous and injective from H1/2,k,κ (Γ) to H−1/2,k−3,κ (Γ), and its range is −1 dense in H−1/2,k−3,κ (Γ). The inverse NK , extended by continuity from the range of NK to H−1/2,k,κ (Γ), is continuous and injective from H−1/2,k,κ (Γ) to H1/2,k−1,κ (Γ) for any k ∈ R, and its range is dense in H1/2,k−1,κ (Γ).

8.4 Solvability of the Boundary Equations We seek the solutions u of (DD± K ) in the form or

u = π ± VK α

(8.10)

u = π ± WK β.

(8.11)

Representations (8.10) and (8.11) yield, respectively, the systems of boundary equations (8.12) VK,0 α = f and WK± β = f. (8.13) The next assertion follows from Theorems 8.11 and 8.12.

128


8.14 Theorem. For any f ∈ H1/2,k,κ (Γ), k ∈ R, κ > 0, systems (8.12) and (8.13) have unique solutions α ∈ H−1/2,k−1,κ (Γ) and β ∈ H1/2,k−2,κ (Γ), in which case the functions u defined by (8.10) or (8.11) belong to H1,k−1,κ (G± ). If k ≥ 1, then these functions are the solutions of problems (DD± K ), respectively. We now seek the solutions of (DN± K ) in the form (8.10) and (8.11). These representations lead to the systems of boundary equations and

TK± (VK,0 α) = g

(8.14)

NK β = g.

(8.15)

8.15 Theorem. For any g ∈ H−1/2,k,κ (Γ), k ∈ R, κ > 0, systems (8.14) and (8.15) have unique solutions α ∈ H−1/2,k−2,κ (Γ) and β ∈ H1/2,k−1,κ (Γ), in which case the functions u defined by (8.10) or (8.11) belong to the space H1,k−1,κ (G± ). If k ≥ 1, then these functions are the solutions of problems (DN± K ), respectively. The assertion follows from Theorems 8.10, 8.11, and 8.13. 8.16. Remark. As shown in Chapter 9, the general case with a nonhomogeneous equation of motion and nonhomogeneous initial conditions can be reduced to its homogenous counterpart by means of suitable substitutions in terms of so-called area and initial potentials.

9 Problems with Nonhomogeneous Equations and Nonhomogeneous Initial Conditions

9.1 The Time-dependent Area Potential Let G = R2 × (0, ∞). We define the area potential U q of density q ∈ C0∞ (G) by D(x − y, t − τ )q(y, τ ) dy dτ, X ∈ G, (U q)(X) = G

where D(x, t) is the matrix of fundamental solutions computed in §1.2. It is obvious that the Laplace transform U qˆ of U q with respect to the time variable has the form ˆ − y, p)ˆ (U qˆ)(x, p) = D(x q (y, p) dy, X ∈ R2 . R2

9.1 Lemma. For any p ∈ C0 , the operator U can be extended by continuity from C0∞ (R2 ) to H−1,p (R2 ). The extended operator is continuous from H−1,p (R2 ) to H1,p (R2 ), and for any qˆ ∈ H−1,p (R2 ), p ∈ Cκ , κ > 0, q −1,p . U qˆ1,p ≤ c|p|ˆ

(9.1)

˜ p) of Proof. First, we establish an estimate for the Fourier transform D(ξ, ˆ ˜ p) is the solution D(x, p) with respect to x. For any ξ ∈ R2 and p ∈ C0 , D(ξ, of the problem ˜ p) + (A(ξ)D)(ξ, ˜ Bp2 D(ξ, p) = I, where A(ξ) is the matrix ⎞ ⎛ 2 h2 (λ + µ)ξ1 ξ2 −iµξ1 h (λ + µ)ξ12 + h2 µ|ξ|2 + µ ⎜ h2 (λ + µ)ξ1 ξ2 h2 (λ + µ)ξ22 + h2 µ|ξ|2 + µ −iµξ2 ⎟ ⎠ ⎝ iµξ1 iµξ2 µ|ξ|2 and I is the identity (3 × 3)-matrix.

(9.2)

130

9 Nonhomogeneous Problems

Let Ψ be the solution of the equation p2 BΨ + A(ξ)Ψ = F

(9.3)

for some function F . Multiplying (9.3) by Ψ∗ , we obtain p2 |B 1/2 Ψ|2 + A(ξ)Ψ, Ψ)C3 = (F, Ψ)C3 ,

(9.4)

¯ i is the standard inner product in C3 . By (9.2), where (F, Ψ)C3 = Fi Ψ (A(ξ)Ψ, Ψ)C3 = h2 (λ + µ)(ξ12 |Ψ1 |2 + ξ22 |Ψ2 |2 ) + (h2 µ|ξ|2 + µ)(|Ψ1 |2 + |Ψ2 |2 ) + µ|ξ|2 |Ψ3 |2 ¯ 2) + 2h2 (λ + µ)ξ1 ξ2 Re(Ψ1 Ψ ¯3 + 2µ Re i(ξ1 Ψ1 + ξ2 Ψ2 )Ψ ≥ µ 12 h2 |ξ|2 + 1 (|Ψ1 |2 + |Ψ2 |2 ) + (|ξ|2 − 2h−2 )|Ψ3 |2 ≥ k1 (1 + |ξ|2 ) − k2 |Ψ|2

(9.5)

for some positive constants k1 and k2 . Separating the real and imaginary parts in (9.4), we arrive at |p|2 |B 1/2 Ψ|2 + A(ξ)Ψ, Ψ C3 = σ −1 Re{¯ p(F, Ψ)C3 },

p = σ + iτ.

(9.6)

By (9.5) and (9.6), for any p ∈ Cκ , κ > 0, (1 + |p|2 + |ξ|2 )|Ψ| ≤ c|p||F |.

(9.7)

˜ (j) (ξ, p) is the jth If F = ej is the jth coordinate unit vector and Ψ = D ˜ p), then, by (9.7), column of the matrix D(ξ, ˜ (j) (ξ, p)| ≤ c|p|(1 + |p|2 + |ξ|2 )−1 , |D

j = 1, 2, 3.

Consequently, for any qˆ ∈ C0∞ (R2 ), U qˆ21,p

=

qˆ)(ξ, p)|2 dξ (1 + |ξ|2 + |p|2 )|(U

R2

≤ c|p|2

(1 + |ξ|2 + |p|2 )−1 |˜ q (ξ, p)|2 dξ = c|p|2 ˆ q 2−1,p ,

R2

where q˜ is the Fourier transform of qˆ. This proves (9.1) and the assertion.

9.2 The Nonhomogeneous Equation of Motion

131

We now return to the spaces of originals and say that u ∈ H1,0,κ (G) is a weak solution of the Cauchy problem with zero initial data and right-hand side q ∈ H−1,1,κ (G) in the equation of motion if u satisfies ∞

a(u, v) − (B 1/2 ∂t u, B 1/2 ∂t v)0 dt =

0

∞ (q, v)0 dt

¯ ∀v ∈ C0∞ (G).

(9.8)

0

9.2 Theorem. For any q ∈ H−1,1,κ (G), κ > 0, equation (9.8) has a unique solution u ∈ H1,0,κ (G). If q ∈ H−1,k,κ (G), k ∈ R, then u ∈ H1,k−1,κ (G) and u1,k−1,κ;G ≤ cq−1,k,κ;G . Proof. In terms of Laplace transforms, (9.8) assumes the form p2 (B 1/2 u ˆ, B 1/2 vˆ)0 + a(ˆ u, vˆ) = (ˆ q , vˆ)0

∀ˆ v ∈ H1,p (R2 ).

(9.9)

Clearly, p2 (BU qˆ)(x, p) + (AU qˆ)(x, p) = qˆ(x, p),

x ∈ R2 ,

for any qˆ ∈ C0∞ (R2 ); hence, for any qˆ ∈ H−1,p (R2 ), (9.9) admits the solution u ˆ = U qˆ ∈ H1,p (R2 ). Separating the real and imaginary parts in (9.9) with vˆ = u ˆ, that is, in the equality p2 B 1/2 u ˆ20 + a(ˆ u, u ˆ) = (ˆ q, u ˆ )0 , we arrive at the estimate ˆ u1,p ≤ c|p|ˆ q −1,p , which shows that (9.9) is uniquely solvable. We now complete the proof by using (9.9) to establish that the mapping (U qˆ)(·, p) is holomorphic from Cκ to H1 (R2 ) and that, therefore, U q = L−1 U qˆ exists and is the desired unique solution of (9.8).

9.2 The Nonhomogeneous Equation of Motion With the notation used in Chapter 2, let q ∈ H−1,1,κ (G± ). We recall (see §2.3) that u ∈ H1,0,κ (G± ) are called weak solutions of the initial-boundary value problems (DD± ) with a nonhomogeneous equation of motion and homogeneous initial data if they satisfy

132


∞

a± (u, v) − (B 1/2 ∂t u, B 1/2 ∂t v)0;S ± dt

0

∞ = (q, v)0;S ± dt

¯ ± ), v ± = 0, ∀v ∈ C0∞ (G

(9.10)

0

γ ± u = f, where f and q are prescribed. Theorem 2.3 states conditions for the unique solvability of these problems. In what follows, we show how (DD± ) can be reduced to their simpler versions for the corresponding homogeneous equation of motion. The Laplace transform qˆ(x, p) of q(x, t) with respect to t belongs to the space H−1,p (S ± ) for any p ∈ Cκ . Using extension operators, we construct a ˆ ∈ H−1,p (R2 ) such that π ± Q ˆ = qˆ ∈ H−1,p (S ± ) and function Q ˆ −1,p ≤ cˆ Q q −1,p;S ± . ˆ = Q ∈ H−1,1,κ (G) and π ± Q ∈ H−1,1,κ (G± ). We now Obviously, L−1 Q represent the solutions of (9.10) in the form u(X) = w(X) + (U Q)(X),

X ∈ G± ,

or u = w + π ± U Q. It is clear that the functions w ∈ H1,0,κ (G± ) satisfy, respectively, the variational problems ∞ 0

a± (w, v) − (B 1/2 ∂t w, B 1/2 ∂t v)0;S ± dt ∞ = (q, v)0;S ± dt 0

∞ − 0

a± (π ± U Q, v) − (B 1/2 ∂t π ± U Q, B 1/2 ∂t v)0;S ± dt ¯ ± ), supp v ⊂ S ± × [0, ∞), ∀v ∈ C0∞ (G γ ± w = f − γ ± π ± U Q.

By Theorem 9.2, the right-hand side in the first equality above is zero; consequently, (9.10) reduce to the corresponding problems for the homogeneous equation of motion. If q ∈ H−1,k,κ (G± ) and f ∈ H1/2,k,κ (Γ), then, by Theorems 9.2 and 2.3, w ∈ H1,k−1,κ (G± ) and w1,k−1,κ;G± ≤ c(q−1,k,κ;G± + f 1/2,k,κ;Γ ).

9.3 Initial Potentials

133

In the case of the initial-boundary value problems (DN± ) with a nonzero ˚−1,k,κ (G± ) in the equation of motion and homogeneous right-hand side q ∈ H initial conditions, we again represent the weak solutions u ∈ H1,0,κ (G± ) in the form u(X) = w(X) + (U Q)(X), X ∈ G± , (9.11) or

u = w + π ± U Q,

with Q constructed as above. Arguing in the same way, we then show that representation (9.11) reduces (DN± ) to the corresponding problems for the homogeneous equation of motion, which were discussed in Chapter 3. Similar procedures enable us to reduce all the problems considered in the preceding chapters to their counterparts for the homogeneous equation of motion.

9.3 Initial Potentials In this section, we consider the Cauchy problem in G = R2 × (0, ∞) with the homogeneous equation of motion and nonhomogeneous initial conditions, and represent its solution in the form of some special functions that we call initial single-layer and double-layer potentials. The classical Cauchy problem in this case consists in finding a function ¯ such that u ∈ C 2 (G) ∩ C 1 (G) B(∂t2 u)(X) + (Au)(X) = 0, u(x, 0+) = ϕ(x),

X ∈ G,

(∂t u)(x, 0+) = ψ(x),

x ∈ R2 .

(9.12)

Let Hm,κ (G), m = 0, 1, 2, . . . , be the space of three-component vectorvalued functions u(X) = u(x, t) defined in G, equipped with the norm u2m,κ;G

=

m ∞

l=0 0

R2

e−2κt (1 + |ξ|2 )m−l |∂tl u ˜(ξ, t)|2 dξ dt,

where u ˜(ξ, t) is the Fourier transform of u(x, t) with respect to x. We say that u ∈ H1,κ (G) is a weak solution of problem (9.12) if ∞ 0

a(u, v) − (B 1/2 ∂t u, B 1/2 ∂t v)0 dt = (Bψ, γ0 v)0

¯ ∀v ∈ C0∞ (G),

γ0 u = ϕ,

where γ0 is the time trace operator defined by (γ0 u)(x, t) = u(x, 0+),

u ∈ H1,κ (G).

(9.13)

134


9.3 Lemma. The variational problem (9.13) has at most one solution. This assertion is proved exactly like the uniqueness statement in the proof of Theorem 2.3 with S + replaced by R2 . We now introduce the initial single-layer potential Vi ψ of density ψ ∈ L2 (R2 ) and the initial double-layer potential Wi ϕ of density ϕ ∈ H1 (R2 ), respectively, by (Vi ψ)(X) = D(x − y, t)Bψ(y) dy, X ∈ G, (9.14) R2

∂t D(x − y, t)Bϕ(y) dy,

(Wi ϕ)(X) =

X ∈ G.

(9.15)

R2

Going over to Fourier transforms, we see that (9.14) and (9.15) take the form ˜ ˜ (9.16) (V i ψ)(ξ, t) = D(ξ, t)B ψ(ξ), ˜ (W ˜ i ϕ)(ξ, t) = ∂t D(ξ, t)B ϕ(ξ),

(9.17)

˜ t) is the Fourier transform of the matrix of fundamental solutions where D(ξ, D(x, t) with respect to x. In what follows we use the matrix norm M 2 =

3

|Mij |2

i,j=1

for any (3 × 3)-matrix M = (Mij ). As usual, we denote by M (j) the columns of M . ¯ and l = 0, 1, 2, . . . , 9.4 Lemma. For any (ξ, t) ∈ G ˜ t) ≤ cl (1 + t)(1 + |ξ|)l−1 . ∂tl D(ξ,

(9.18)

˜ t) = θ(t)M (ξ, t), where θ is the characteristic Proof. It is obvious that D(ξ, function of the interval (0, ∞) and the columns M (j) of M satisfy B(∂t2 M (j) )(ξ, t) + A(ξ)M (j) (ξ, t) = 0, M

(j)

(ξ, 0) = 0,

(∂t M

(j)

(ξ, t) ∈ G,

)(ξ, 0) = B −1 ej .

By (9.19),

d 1/2 |B ∂t M (j) (ξ, t)|2 + A(ξ)M (j) (ξ, t), M (j) (ξ, t) C3 = 0; dt

(9.19)


135

hence, |B 1/2 ∂t M (j) (ξ, t)|2 + A(ξ)M (j) (ξ, t), M (j) (ξ, t) C3 = |B −1/2 ej |2 and

|(∂t M (j) )(ξ, t)| < c.

(9.20)

By (9.5), (1 + |ξ|)2 |M (j) (ξ, t)|2 ≤ c A(ξ)M (j) (ξ, t), M (j) (ξ, t) C3

≤ c 1 + |M (j) (ξ, t)|2 , and, since |M (j) (ξ, t)| ≤

0

t

+ |M (j) (ξ, t)|2

|(∂τ M (j) )(ξ, τ )| dτ ≤ ct,

it follows that |M (j) (ξ, t)| ≤ c(1 + t)(1 + |ξ|)−1 .

(9.21)

Inequalities (9.20) and (9.21) prove (9.18) for l = 0, 1. Suppose that (9.18) holds for l = 0, 1, . . . , k, k ≥ 1. Differentiating the equation in (9.19) with respect to t, we find that (∂tk+1 M (j) )(ξ, t) = −B −1 A(ξ)(∂tk−1 M (j) ),

(ξ, t) ∈ G;

therefore, |(∂tk+1 M (j) )(ξ, t)| ≤ c(1 + |ξ|)2 (1 + t)(1 + |ξ|)k−2 = c(1 + t)(1 + |ξ|)k , which implies that (9.18) holds for all l = 0, 1, 2, . . . . 9.5 Theorem. For any ψ ∈ L2 (R2 ), ϕ ∈ H1 (R2 ), and κ > 0, the function u = Vi ψ + W i ϕ is the (unique) weak solution of the Cauchy problem (9.13). If ψ ∈ Hk (R2 ) and ϕ ∈ Hk+1 (R2 ), k = 0, 1, 2, . . . , then u ∈ Hk+1,κ (G) and uk+1,κ;G ≤ c(ψk + ϕk+1 ). Proof. By (9.16) and (9.17), ˜ + ∂t D(ξ, ˜ t)B ϕ(ξ); ˜ t)B ψ(ξ) ˜ u ˜(ξ, t) = D(ξ,

(9.22)

136


therefore, u2k+1,κ;G =

∞ k+1

e−2κt (1 + |ξ|2 )k+1−l

l=0 0 R2 2 ˜ + ∂ l+1 D(ξ, ˜ t)B ψ(ξ) ˜ t)B ϕ(ξ)| × |∂tl D(ξ, dξ ˜ t

≤c

∞ k+1

2 ˜ ˜ t)2 |ψ(ξ)| e−2κt (1 + |ξ|2 )k+1−l ∂tl D(ξ, dξ dt

l=0 0 R2

+c

∞ k+1

2 ˜ t)2 |ϕ(ξ)| e−2κt (1 + |ξ|2 )k+1−l ∂tl+1 D(ξ, ˜ dξ dt,

l=0 0 R2

so, by (9.18), u2k+1,κ;G ∞ ≤c

e−2κt (1 + t)2 (1 + |ξ|2 )k+1−l

0 R2

2 2 ˜ × (1 + |ξ|2 )l−1 |ψ(ξ)| + (1 + |ξ|2 )l |ϕ(ξ)| ˜ } dξ dt

≤ c(ψ2k + ϕ2k+1 ), which proves (9.22). We now show that u(x, t) → ϕ(x) in H1/2 (R2 ) as t → 0+. We have u(· , t) −

ϕ21/2

=

2 (1 + |ξ|)|˜ u(ξ, t) − ϕ(ξ)| ˜ dξ

R2

2 ˜ ˜ t)2 |ψ(ξ)| ≤c (1 + |ξ|)D(ξ, dξ R2

+

2 ˜ t)B − I2 |ϕ(ξ)| (1 + |ξ|)∂t D(ξ, ˜ dξ .

R2

˜ t) → 0 and (∂t D)(ξ, ˜ Since D(ξ, t) → B −1 , as t → 0+, for any ξ ∈ R2 , from Lebesgue’s dominated convergence theorem and (9.18) it follows that lim u(·, t) − ϕ1/2 = 0.

t→0+


137

To complete the proof, it remains to verify that u satisfies the variational equation in (9.13). Let ϕ, ψ ∈ C0∞ (R2 ). By (9.22), u is a smooth function. Going over to Fourier transforms with respect to x in (9.13), integrating by parts with respect to t and taking the initial data in (9.19) into account, we conclude that u indeed satisfies the relevant variational equation. The proof of the assertion now follows from (9.22) and the fact that C0∞ (R2 ) is dense in Hk (R2 ). 9.6 Remark. The above analysis suggests the following solution procedure for problems (DD± ) with boundary data f and nonhomogeneous initial data ϕ and ψ. If ϕ ∈ Hk+1 (S ± ) and ψ ∈ Hk (S ± ) for some k = 0, 1, 2, . . . , then we use extension operators to construct Φ ∈ Hk+1 (R2 ), π ± Φ = ϕ

and

Ψ ∈ Hk (R2 ), π ± Ψ = ψ,

and represent the solutions u in the form u(X) = u0 (X) + w(X),

X ∈ G± ,

where u0 = π ± Vi Ψ + π ± Wi Φ. Clearly, w must satisfy the same equation of motion as u but with homogeneous initial data. The boundary condition for w takes the form γ ± w = f − γ ± u0 . If f − γ ± u0 ∈ H1/2,l,κ (Γ),

(9.23)

then our problems have been reduced to those considered in §2.3. It turns out that (9.23) is the compatibility condition for the initial and boundary data. A more detailed discussion of such compatibility is outside the scope of this book. The interested reader is referred, for example, to [13]. Analogous techniques can also be set up for the other problems in the preceding chapters, to reduce their general form with nonhomogeneous initial conditions to their corresponding homogeneous counterparts.

A The Fourier and Laplace Transforms of Distributions

Below we introduce the Fourier and Laplace transformations for test functions and distributions and list (without proof) some of their most important properties. A.1 Definition. A complex-valued function ϕ defined on Rm is called a test function (of rapid descent) if ϕ ∈ C ∞ (Rm ) and for any k ∈ Z+ ,

ϕk = sup (1 + |x|)k x∈Rm

|∂ α ϕ(x)| < ∞,

|α|≤k

where |x| = (x21 + · · · + x2m )1/2 . The space of all such functions is denoted by S(Rm ). A.2 Definition. The Fourier transform ϕ˜ of a test function (of rapid descent) ϕ ∈ S(Rm ) is defined by ϕ(ξ) ˜ = (Fϕ)(ξ) =

ei(x,ξ) ϕ(x) dx.

(A.1)

Rm

A.3 Theorem. The operator F is a homeomorphism from S(R2 ) to S(R2 ), and its inverse F −1 acts according to the formula ˜ = (2π)−m e−i(x,ξ) ϕ(ξ) ˜ dξ, ϕ˜ ∈ S(Rm ). ϕ(x) = (F −1 ϕ)(x) Rm

A.4 Remark. The continuity of F from S(Rm ) to S(Rm ) means that for any k ∈ Z+ , there is l(k) such that Fϕk ≤ ck ϕl(k)

∀ϕ ∈ S(Rm ).

140


A.5 Theorem. (i) For any ϕ, ψ ∈ S(Rm ), there holds Parseval’s equality ˜ dξ = (2π)m ϕ(ξ) ˜ ψ(ξ) ϕ(x)ψ(x) dx. (A.2) Rm

Rm

˜ (ii) F(ϕ ∗ ψ)(ξ) = ϕ(ξ) ˜ ψ(ξ) for any ϕ, ψ ∈ S(Rm ). 2 (iii) For any ϕ ∈ S(R ) and any multiindex α, F(∂ α ϕ)(ξ) = (−i)|α| ξ α ϕ(ξ), ˜ F(xα ϕ)(ξ) = (−i)|α| ∂ α ϕ(ξ). ˜ 2

x2

, a > 0, then

2

x2

dx

A.6 Example. If m = 1 and ϕ(x) = e−a 2

(Fe−a

x2

eixξ−a

)(ξ) = R

=

1 −ξ2 /(4a2 ) e a

2 2 1 = e−ξ /(4a ) a

R

2

e−(σ+iξ/(2a)) dσ

2

e−ζ dζ.

Im ζ=ζ/(2a)

Using Cauchy’s theorem, we can easily check that √ 2 −ζ 2 e dζ = e−ζ dζ = π; R

Im ζ=ζ/(2a)

therefore, 2

(Fe−a

x2

√ )(ξ) =

π −ξ2 /(4a2 ) . e a

A.7 Remark. Since S(Rm ) is dense in L1 (Rm ) and L2 (Rm ), the Fourier transformation can be extended by continuity to the latter spaces. Thus, if f ∈ L1 (Rm ), then Ff is simply defined by (A.1) and is continuous and bounded in Rm . 2 m If f ∈ L2 (Rm ), then we approximate f by a sequence {f }∞ n=1 in L (R ) ∩ 1 m L (R ), for example, the sequence of truncations fn (x) =

f (x), |x| < n, 0, |x| > n.

It is not difficult to show that fn → f and fñ → f˜ ∈ L2 (Rm ). Then we again define the Fourier transform of f by (A.1), understanding that equality in the above sense.


141

A.8 Theorem. If f, g ∈ L2 (Rm ), then (i) the inverse transformation F −1 acts according to the formula −1 ˜ −m f (x) = (F f )(x) = (2π) e−i(x,ξ) f˜(ξ) dξ. Rm

(ii) Parseval’s equality (A.2) holds. A.9 Definition. The (generalized) Fourier transform f˜ = Ff of a tempered distribution f ∈ S (Rm ) (an element of the dual of S(Rm )) is defined by (f˜, ϕ) ˜ = (2π)m (f, ϕ) ∀ϕ˜ ∈ S(Rm ), where ϕ = F −1 ϕ˜ and (· , ·) is the duality generated by the inner product in L2 (Rm ). A.10 Theorem. (i) The distributional operator F is a homeomorphism from S (Rm ) to S (Rm ). (ii) For any f ∈ S (Rm ) and any ϕ ∈ S(Rm ), F(f ∗ ϕ) = ϕ˜f˜. (iii) For any f ∈ S (Rm ) and any multiindex α, F(∂ α f ) = (−i)|α| ξ α f˜, F(xα f ) = (−i)|α| ∂ α f˜. A.11 Examples. (i) If δ is the Dirac delta, then, by Theorem A.3(iii), ˜ F(∂ α δ) = (−i)|α| ξ α δ. Since for any ϕ˜ ∈ S(Rm ), ˜ ϕ) (δ, ˜ = (2π)m (δ, ϕ) = (2π)m ϕ(0) =

ϕ(ξ) ˜ dξ = (1, ϕ), ˜ Rm

it follows that

F(∂ α δ) = (−i)|α| ξ α .

142


−1 generated by the function (ii) For the one-dimensional distribution x x (in the sense of principal value) we have −1

−1 , ϕ) (F x ˜ = 2π(x−1 , ϕ) = 2π lim

ε→0 R→∞ ε a. A.13 Remark. We may write the Laplace transform G of g in the form ∞ G(p) = G(ξ + iη) = 0

e−itη e−ξt g(t) dt = F (e−ξt g) (−η).

(A.3)


143

By Parseval’s equality, ∞ 2π

e−2ξt |g(t)|2 dt =

0

|G(ξ + iη)|2 dη;

R

hence, ∞

e−2at |g(t)|2 dt = sup ξ>a

0

∞

e−2ξt |g(t)|2 dt =

0

1 sup 2π ξ>a

|G(ξ + iη)|2 dη.

R

From (A.3) it now follows that if ξ > a, then −ξt

e

1 g(t) = 2π

−iηt

e R

1 G(ξ − iη) dη = 2π

eiηt G(ξ + iη) dη,

R

so the inverse Laplace transformation is defined by the formula 1 −1 g(t) = (L G)(t) = ept G(p) dp. 2πi Re p=ξ

Clearly, the convergence of these integrals is understood in the L2 -sense. Let g and h be functions of the above class with abscissas of absolute convergence a and b, respectively. The convolution u = g ∗h of these functions is defined by t u(t) = (g ∗ h)(t) = g(t − τ )h(τ ) dτ. 0

A.14 Theorem. The Laplace transform U (p) = (Lu)(p) exists, is holomorphic for ξ > max{a, b}, and U (p) = G(p)H(p), where G and H are the Laplace transforms of f and h, respectively. Next, we define the Laplace transform of a distribution. We restrict the space of test functions (of rapid descent) for m = 1 to the subspace D(R) = C0∞ (R) of all infinitely differentiable functions with compact support in R, and denote by D+ (R) the set of distributions in D (R) (the dual of D(R)) with support in [0; ∞). Let g ∈ D+ (R), and let a ∈ R be such that e−ξt g(t) belongs to S (R) for ξ > a and e−ξt g(t) does not belong to S (R) for ξ < a. The number a in

144


this case is called the abscissa of convergence of the Laplace transform of the distribution g. It is easy to show that if ξ > a, then e−ξt g(t) belongs to S (R). To define the distributional (generalized) Laplace transformation, we first introduce the distributional equivalent of the change of variable for functions, which takes us from f (x) to f (−x). Specifically, for any f ∈ S (R), we define f−x ∈ S (R) by

f−x , ϕ(x) = f, ϕ(−x)

∀ϕ ∈ D(R).

A.15 Definition. If g ∈ D+ (R) is a distribution with abscissa of convergence a ∈ R, then its (generalized) Laplace transform G(p) = (Lg)(p) is defined for ξ = Re p > a by

G(p) = (Lg)(p) = F (e−ξt g −η (p),

p = ξ + iη.

(R) with abscissas of converA.16 Theorem. If g and h are elements of D+ gence a and b and Laplace transforms G and H, respectively, then (i) G(p), p = ξ + iη, is holomorphic for ξ > a; (ii) (L(∂tl g))(p) = pl G(p); (iii) u = g ∗ h exists and

U (p) = (Lu)(p) = G(p)H(p)

for ξ = Re p > max{a, b}.

A.17 Theorem. Let g ∈ D+ (R) be a distribution with abscissa of convergence a and Laplace transform G. (i) For any ε > 0, there are positive numbers c and l such that

|G(p)| ≤ c(1 + |p|)l

for Re p = ξ ≥ a + ε;

(A.4)

that is, G(p) grows at infinity no faster than a polynomial. (ii) If G(p) is holomorphic for ξ > a and satisfies (A.4), then there is a unique distribution g ∈ D+ (R) such that (Lg)(p) = G(p)

for ξ > a.

A.18 Example. It is easy to verify that the Laplace transform of the onedimensional Dirac delta is (Lδ)(p) = 1.

References

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18.

Abramowitz, M., Stegun, I.: Handbook of Mathematical Functions. Dover (1964) Agranovich, M.S., Vishik, M.I.: Elliptic problems with a parameter and parabolic problems of a general form. Russian Math. Surveys, 19:3, 53–161 (1964) Akhiezer, N.I., Glazman, I.M.: Theory of Linear Operators in Hilbert Space, vol. 1. F. Ungar (1961–63) Chudinovich, I., Constanda, C.: Non-stationary integral equations for elastic plates. C.R. Acad. Sci. Paris Sér. I, 329, 1115–1120 (1999) Chudinovich, I., Constanda, C.: Solvability of initial-boundary value problems in bending of plates. J. Appl. Math. Phys., 51, 449–466 (2000) Chudinovich, I., Constanda, C.: The Cauchy problem in the theory of plates with transverse shear deformation. Math. Models Methods Appl. Sci., 10, 463– 477 (2000) Chudinovich, I., Constanda, C.: Variational and Potential Methods in the Theory of Bending of Plates with Transverse Shear Deformation. Chapman & Hall/CRC (2000) Chudinovich, I., Constanda, C.: Dynamic transmission problems for plates. J. Appl. Math. Phys. (ZAMP), 53, 1060–1074 (2002) Constanda, C.: A Mathematical Analysis of Bending of Plates with Transverse Shear Deformation. Longman/Wiley (1990) Constanda, C.: Direct and Indirect Boundary Integral Equation Methods. Chapman & Hall/CRC (1999) Doetsch, G.: Anleitung zum praktischen Gebrauch der Laplace-Transformation und der Z -Transformation. R. Oldenbourg (1967) Eskin, G.I.: Boundary Value Problems for Elliptic Pseudodifferential Equations. Amer. Math. Soc. (1981) Lions, J.-L., Magenes, E.: Nonhomogeneous Boundary Value Problems and Applications, vol. 1. Springer-Verlag (1972) Love, A.E.H.: A Treatise on the Mathematical Theory of Elasticity. Cambridge University Press (1959) Lubich, Ch.: On the multistep time discretization of linear initial-boundary value problems and their boundary integral equations. Numer. Math., 67, 364– 389 (1994) Mizohata, S.: The Theory of Partial Differential Equations. Cambridge University Press (1973) Sokolnikoff, I.S.: Mathematical Theory of Elasticity. McGraw-Hill (1956) Yosida, K.: Functional Analysis. Springer-Verlag (1966)

Index

Area potential, time-dependent 129 Bessel function, modified of order one 12 of order zero 12 Boundary conditions combined 6, 108 Dirichlet 5, 19, 43, 119 elastic 114 mixed 5, 99 Neumann 5, 37, 43, 120 transmission (contact) 6 integral equations, nonstationary 1, 16, 51 operators 73, 126 algebra of 69 Cauchy problem 131, 135 theorem 140 Crack 6 Dirac distribution 8 Direct method 53 Displacement vector 5 Distribution, tempered 141 Dual 19, 141 Equations, nonhomogeneous 129 Forces, transverse shear 4

Form energy bilinear 82 sesquilinear 22 Fourier transform 9, 139 distributional 19 inverse 11 transformation 8 Fredholm Alternative 21, 24, 68 Functional, bounded antilinear (conjugate linear) 23, 68 Homeomorphism 23, 68, 95, 139 Initial conditions, nonhomogeneous 129 Internal energy density 22 Jump formulas 14, 15 Lamé constants 2 Laplace transform 2, 142 of a distribution 143 transformation 2 distributional (generalized) 144 Lemma Lax–Milgram 26 Rellich’s 23 Mapping absolutely continuous 35 holomorphic 20

148

Index

Matrix boundary operator 4 of fundamental solutions 8, 77, 121 Moments, bending and twisting 4 Operators averaging 3 boundary 73, 126 algebra of 69 extension 2, 65 Poincaré–Steklov 37, 39, 61, 70, 102 restriction 2 trace 70, 109 Parseval’s equality 8, 21, 31, 140 Plate constitutive relations 3 elastic 1 equations of motion 3 nonhomogeneous 131 finite, with a crack 93 infinite, with a finite inclusion 57 middle plane of 1 multiply connected 57 on an elastic foundation 7, 119 piecewise homogeneous finite 75

potentials, time-dependent 13, 15 thickness 1 weakened by a crack 82 with transverse shear deformation 3 Potentials 8 boundary operators generated by 16 density 14 initial 133 in terms of Laplace transforms 13 time-dependent 43, 73, 87, 126 Sobolev spaces with a parameter 37, 69, 85 Solution, weak 22, 34, 74, 97, 135 Somigliana representation formula 53 Test function (of rapid descent) 139 Theorem Dunford’s 41 Lebesgue’s dominated convergence 136 Paley–Wiener 21 trace 87, 96, 103, 110 Variational equation 23 problem 94

Variational and Potential Methods for a Class of Linear Hyperbolic Evolutionary Processes

Variational and Potential Methods for a Class of Linear Hyperbolic Evolutionary Processes

Variational Methods

Lectures on Numerical Methods for Non-Linear Variational Problems

Variational Methods for Eigenvalue Approximation

Potential Function Methods for Approximately Solving Linear Programming Problems

Lectures on Numerical Methods for Non-Linear Variational Problems

Markov Processes and Potential Theory

Variational methods for eigenvalue approximation

Fencing for Conservation: Restriction of Evolutionary Potential Or a Riposte to Threatening Processes?

Markov Processes and Potential Theory

Markov processes and potential theory

Statistical Processes of Evolutionary Theory

Finite Volume Methods for Hyperbolic Problems

Combined relaxation methods for variational inequalities

Finite Volume Methods for Hyperbolic Problems

Principles and Applications of Variational Methods

Iterative Methods for Linear and Nonlinear Equations

Optimization by variational methods

Variational methods in statistics

Iterative Methods for Linear and Nonlinear Equations

Topological methods for variational problems with symmetries

Iterative Methods For Linear And Nonlinear Equations

Iterative methods for linear and nonlinear equations

Iterative methods for linear and nonlinear equations

Linear and Quasilinear Complex Equations of Hyperbolic and Mixed Types

Linear and Quasilinear Complex Equations of Hyperbolic and Mixed Types

Linear and quasilinear complex equations of hyperbolic and mixed type

Optimization by variational methods

Numerical Methods for Nonlinear Variational Problems

Variational and Potential Methods for a Class of Linear Hyperbolic Evolutionary Processes