Boundary Integral Equations on Contours with Peaks (Operator Theory: Advances and Applications)

Operator Theory: Advances and Applications Vol. 196 Editor: I. Gohberg Editorial Office: School of Mathematical Science...

Author: Vladimir Maz'ya | Alexander Soloviev

24 downloads 503 Views 3MB Size Report

This content was uploaded by our users and we assume good faith they have the permission to share this book. If you own the copyright to this book and it is wrongfully on our website, we offer a simple DMCA procedure to remove your content from our site. Start by pressing the button below!
Report copyright / DMCA form

DOWNLOAD PDF

Operator Theory: Advances and Applications Vol. 196

Editor: I. Gohberg Editorial Office: School of Mathematical Sciences Tel Aviv University Ramat Aviv Israel

Editorial Board: D. Alpay (Beer Sheva, Israel) J. Arazy (Haifa, Israel) A. Atzmon (Tel Aviv, Israel) J.A. Ball (Blacksburg, VA, USA) H. Bart (Rotterdam, The Netherlands) A. Ben-Artzi (Tel Aviv, Israel) H. Bercovici (Bloomington, IN, USA) A. Böttcher (Chemnitz, Germany) K. Clancey (Athens, GA, USA) R. Curto (Iowa, IA, USA) K. R. Davidson (Waterloo, ON, Canada) M. Demuth (Clausthal-Zellerfeld, Germany) A. Dijksma (Groningen, The Netherlands) R. G. Douglas (College Station, TX, USA) R. Duduchava (Tbilisi, Georgia) A. Ferreira dos Santos (Lisboa, Portugal) A.E. Frazho (West Lafayette, IN, USA) P.A. Fuhrmann (Beer Sheva, Israel) B. Gramsch (Mainz, Germany) H.G. Kaper (Argonne, IL, USA) S.T. Kuroda (Tokyo, Japan) L.E. Lerer (Haifa, Israel) B. Mityagin (Columbus, OH, USA)

V. Olshevski (Storrs, CT, USA) M. Putinar (Santa Barbara, CA, USA) A.C.M. Ran (Amsterdam, The Netherlands) L. Rodman (Williamsburg, VA, USA) J. Rovnyak (Charlottesville, VA, USA) B.-W. Schulze (Potsdam, Germany) F. Speck (Lisboa, Portugal) I.M. Spitkovsky (Williamsburg, VA, USA) S. Treil (Providence, RI, USA) C. Tretter (Bern, Switzerland) H. Upmeier (Marburg, Germany) N. Vasilevski (Mexico, D.F., Mexico) S. Verduyn Lunel (Leiden, The Netherlands) D. Voiculescu (Berkeley, CA, USA) D. Xia (Nashville, TN, USA) D. Yafaev (Rennes, France)

Honorary and Advisory Editorial Board: L.A. Coburn (Buffalo, NY, USA) H. Dym (Rehovot, Israel) C. Foias (College Station, TX, USA) J.W. Helton (San Diego, CA, USA) T. Kailath (Stanford, CA, USA) M.A. Kaashoek (Amsterdam, The Netherlands) P. Lancaster (Calgary, AB, Canada) H. Langer (Vienna, Austria) P.D. Lax (New York, NY, USA) D. Sarason (Berkeley, CA, USA) B. Silbermann (Chemnitz, Germany) H. Widom (Santa Cruz, CA, USA)

Boundary Integral Equations on Contours with Peaks Translated into English and edited by Tatyana Shaposhnikova

Vladimir G. Maz’ya Alexander A. Soloviev

Birkhäuser Basel · Boston · Berlin

Authors: Vladmir G. Maz’ya Department of Mathematical Sciences University of Liverpool Liverpool L69 7ZL UK e-mail: [email protected]

Alexander A. Soloviev Department of Mathematics Chelyabinsk State University 454021 Chelyabinsk Russia e-mail: [email protected]

and Department of Mathematics Linköping University 581 83 Linköping Sweden e-mail: [email protected]

2000 Mathematical Subject Classification: Primary: 45P05, 47B38, 47G10

Library of Congress Control Number: 2009932071

Bibliographic information published by Die Deutsche Bibliothek. Die Deutsche Bibliothek lists this publication in the Deutsche Nationalbibliografie; detailed bibliographic data is available in the Internet at http://dnb.ddb.de

ISBN 978-3-0346-0170-2 Birkhäuser Verlag AG, Basel – Boston – Berlin This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, re-use of illustrations, recitation, broadcasting, reproduction on microfilms or in other ways, and storage in data banks. For any kind of use permission of the copyright owner must be obtained.

© 2010 Birkhäuser Verlag AG Basel · Boston · Berlin P.O. Box 133, CH-4010 Basel, Switzerland Part of Springer Science+Business Media Printed on acid-free paper produced from chlorine-free pulp. TCF∞ Printed in Germany

ISBN 978-3-0346-0170-2

e-ISBN 978-3-0346-0171-9

987654321

www.birkhauser.ch

Contents

Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 Lp -theory of Boundary Integral Equations on a Contour with Peak 1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Continuity of boundary integral operators . . . . . . . . . . 1.2.1 Auxiliary assertions . . . . . . . . . . . . . . . . . . 1.2.2 Estimates for kernels of integral operators . . . . . . 1.2.3 Single and double layer potentials on contours with peak . . . . . . . . . . . . . . . . . . . . . . . . 1.2.4 Operator πI − T on a contour with outward peak . . 1.2.5 Operator πI − T on a contour with inward peak . . 1.3 Dirichlet and Neumann problems for a domain with peak . 1.3.1 Dirichlet problem for a domain with outward peak . 1.3.2 Boundary value problems for a domain with inward peak . . . . . . . . . . . . . . . . . . . . . . . 1.3.3 Auxiliary boundary value problems for a domain with peak . . . . . . . . . . . . . . . . . . . . . . . . 1.4 Integral equations of the Dirichlet and Neumann problems . 1.4.1 Integral equations of the first kind . . . . . . . . . . 1.4.2 Integral equation of the interior Dirichlet problem in a domain with outward peak . . . . . . . . . . . . 1.4.3 Integral equation of the interior Neumann problem in a domain with outward peak . . . . . . . . . . . . 1.4.4 Integral equations of the exterior Dirichlet and Neumann problems in a domain with inward peak . 1.4.5 Boundary integral equation of the Dirichlet problem in a domain with inward peak . . . . . . . . . . . . . 1.4.6 Boundary integral equation of the Neumann problem in a domain with inward peak . . . . . . . . . . . . . 1.4.7 Integral equations of the exterior Dirichlet and Neumann problems for domain with outward peak .

ix

. . . .

. . . .

. . . .

1 5 5 9

. . . . .

. . . . .

. . . . .

16 23 29 34 34

. . .

40

. . . . . . . . .

43 60 60

. . .

68

. . .

73

. . .

78

. . .

80

. . .

85

. . .

92

vi

Contents

1.5

Direct method of integral equations of the Neumann and Dirichlet problems . . . . . . . . . . . . . .

2 Boundary Integral Equations in H¨ older Spaces on a Contour with Peak 2.1 Weighted H¨ older spaces . . . . . . . . . . . . . . . . . . . . . . . 2.2 Boundedness of integral operators . . . . . . . . . . . . . . . . . . 2.2.1 Integral operators in weighted H¨ older spaces . . . . . . . . 2.2.2 Continuity of the operator πI − T . . . . . . . . . . . . . 2.2.3 Continuity of the operator πI + S . . . . . . . . . . . . . 2.2.4 Continuity of the operator V . . . . . . . . . . . . . . . . 2.3 Dirichlet and Neumann problems in a strip . . . . . . . . . . . . 2.4 Boundary integral equations of the Dirichlet and Neumann problems in domains with outward peak . . . . . . . . 2.4.1 Auxiliary assertions . . . . . . . . . . . . . . . . . . . . . 2.4.2 Integral equation of the Dirichlet problem on a contour with outward peak . . . . . . . . . . . . . . 2.4.3 Integral equation of the Neumann problem on a contour with outward peak . . . . . . . . . . . . . . 2.5 Boundary integral equations of the Dirichlet and Neumann problems in domains with inward peak . . . . . . . . . . . . . . . 2.5.1 Integral equation of the Dirichlet problem on a contour with inward peak . . . . . . . . . . . . . . . 2.5.2 Integral equation of the Neumann problem on contour with inward peak . . . . . . . . . . . . . . . . 2.6 Integral equation of the first kind on a contour with peak . . . . 2.7 Appendices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Appendix A: Proof of Theorem 2.2.1 . . . . . . . . . . . . . . . . Appendix B: Proof of Corollary 2.2.2 . . . . . . . . . . . . . . . . Appendix C: To proof of Theorem 2.2.7 . . . . . . . . . . . . . . Appendix D: To proof of Theorem 2.2.9 . . . . . . . . . . . . . . Appendix E: To proof of Theorem 2.2.12 . . . . . . . . . . . . . . 3 Asymptotic Formulae for Solutions of Boundary Integral Equations Near Peaks 3.1 Preliminary facts . . . . . . . . . . . . . . . . . . . . . . 3.1.1 Asymptotics of a conformal mapping of a domain with outward peak onto a strip . . . . . . . . . . 3.1.2 Asymptotics of a conformal mapping of a domain with inward peak onto the upper half-plane . . . 3.2 The Dirichlet and Neumann problems in domains with peaks . . . . . . . . . . . . . . . . . . . 3.3 Integral equations of the Dirichlet problem . . . . . . . .

95 103 106 106 110 120 127 134 151 151 157 166 175 175 179 187 193 193 198 200 205 207

. . . . . 219 . . . . . 219 . . . . . 223 . . . . . 226 . . . . . 235

Contents

3.4

3.5

vii

3.3.1 Homogeneous integral equation of the problem D(i) . . . 3.3.2 Solvability of the integral equation of the problem D(i) . 3.3.3 Integral equation of the problem D(e) . . . . . . . . . . Integral equations of the Neumann problem . . . . . . . . . . . 3.4.1 Homogeneous integral equation of the problem N (e) . . 3.4.2 Solvability of the integral equation of the problem N (e) 3.4.3 Integral equation of the problem N (i) . . . . . . . . . . Appendices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Appendix A: Counterexample . . . . . . . . . . . . . . . . . . . Appendix B: Proof of Lemma 3.2.1 . . . . . . . . . . . . . . . . Appendix C: Proof of Lemma 3.2.2 . . . . . . . . . . . . . . . . Appendix D: Proof of Lemma 3.2.3 . . . . . . . . . . . . . . . . Appendix E: Proof of Lemma 3.2.4 . . . . . . . . . . . . . . . .

4 Integral Equations of Plane Elasticity in Domains with Peak 4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Boundary value problems of elasticity . . . . . . . . . . . . . 4.2.1 Asymptotic behavior of solutions to the problem D(i) . 4.2.2 Asymptotic behavior of solutions to the problem N (e) 4.2.3 Properties of solutions to the problem D(e) . . . . . . 4.2.4 Uniqueness theorems . . . . . . . . . . . . . . . . . . . 4.3 Integral equations on a contour with inward peak . . . . . . . 4.3.1 Integral equations of the problem D(i) . . . . . . . . . 4.3.2 Integral equation of the problem N (e) . . . . . . . . . 4.4 Integral equations on a contour with outward peak . . . . . . 4.4.1 Integral equation of the problem D(i) . . . . . . . . . . 4.4.2 Integral equation of the problem N (e) . . . . . . . . .

. . . . . . . . . . . .

. . . . . . . . . . . . .

235 244 247 248 248 250 253 254 254 256 259 265 270

. . . . . . . . . . . .

273 281 281 293 310 312 313 313 323 326 326 331

Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 335 List of Symbols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 339 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 341

Preface An equation of the form

K(x, y)ϕ(y)dν(y) = f (x), x ∈ X,

λϕ(x) −

(1)

X

is called a linear integral equation. Here (X, ν) is a space with σ-finite measure ν and λ is a complex parameter, K and f are given complex-valued functions. The function K is called the kernel and f is the right-hand side. The equation is of the first kind if λ = 0 and of the second kind if λ = 0. Integral equations have attracted a lot of attention since 1877 when C. Neumann reduced the Dirichlet problem for the Laplace equation to an integral equation and solved the latter using the method of successive approximations. Pioneering results in application of integral equations in the theory of harmonic functions were obtained by H. Poincaré, G. Robin, O. H¨ older, A.M. Lyapunov, V.A. Steklov, and I. Fredholm. Further development of the method of boundary integral equations is due to T. Carleman, G. Radon, G. Giraud, N.I. Muskhelishvili, S.G. Mikhlin, A.P. Calderon, A. Zygmund and others. A classical application of integral equations for solving the Dirichlet and Neumann boundary value problems for the Laplace equation is as follows. Solutions of boundary value problems are sought in the form of the double layer potential W σ and of the single layer potential V σ. In the case of the internal Dirichlet problem and the external Neumann problem, the densities of corresponding potentials obey the integral equation −πσ + W σ = g (2) and

∂ Vτ =h (3) ∂n respectively, where ∂/∂n is the derivative with respect to the outward normal to the contour. Henceforth, integral equations of the theory of logarithmic potentials will be called boundary integral equations and abbreviated as BIE. The equations (2) and (3) for domains with nonzero angles at the boundary, that is without cusps, were studied in various function spaces by means of the theory of Fredholm operators. (For a historic survey and bibliography see [16].) −πσ +

x

Preface

However, as Radon showed in [34], zero angles at the contour prevent the use of Fredholm’s theory for equations (2) and (3), when they are considered in the space of continuous functions and its dual. Radon defined the Fredholm radius of an operator T in a Banach space B as the supremum of R such that the Fredholm theorem holds for each of the equations σ − λT = ϕ and τ − λT ∗ τ = ψ (ϕ ∈ B, ψ ∈ B ∗ ) for all |λ| < R. Here T ∗ is the operator adjoint of T and it acts in the dual space B ∗ . In the case of a boundary curve Γ with bounded rotation without zero angles, Radon proved that the Fredholm radius of the logarithmic double layer potential in the space C(Γ) is greater than 1. In the presence of a cusp, the Fredholm radius is equal to 1. This was, probably, the reason why equations on contours with cusps have not been studied since then. For the curves Γ from the above class, a necessary and sufficient condition of Fredholm properties of the operator in (2) acting in Lp (Γ) was found by Shelepov (see [35]), who developed an earlier work by Lopatinskii (see [15]). In this book we construct a theory of boundary integral equations for plane domains with a finite number of cusps at the boundary. We use another approach introduced in [16] and based on representations of solutions to integral equations of potential theory in terms of solutions to auxiliary exterior and interior boundary value problems. This method allows us to obtain information on inverse operators of boundary integral equations from theorems on inverse operators of boundary value problems. In this way, the theory of elliptic boundary value problems is used to get solvability results for boundary integral equations. The theory developed in the present book is based on our joint work which started in 1991 and goes back to the papers [19]–[29]. Recently, the papers [3], [4] dealing with BIEs on curves with peaks appeared, where conditions of the solvability in a weighted Lp -space were found. The method used in [3], [4] reduces the boundary value problem for the Laplace operator to the Riemann–Hilbert problem for analytic functions on a unit disk. The book consists of four chapters. The first three chapters are devoted to the study of the solvability of boundary integral equations for planar domains with peaks at the boundary in certain classes of functions with weak singularity. We also describe kernels of the operator in these equations. In the first chapter we construct a theory of BIEs of the first and second kind in weighted Lp -spaces. Pairs of function spaces are found, which serve as domains and targets for the operators just mentioned. Furthermore, we describe the kernels of the operators of integral equations and conditions of their triviality. The content of this chapter follows the articles [21]–[25]. The second chapter concerns BIEs in a weighted Hölder-type space for a contour with peak. We prove solvability theorems, describe kernels of operators

Preface

xi

and find conditions for the triviality of kernels. The content of this chapter is based on [26]. In the third chapter we find asymptotic formulas for solutions of BIEs. Such formulas were found in [19], [20] and [27]. We limit consideration to the case of contours with peaks, having the first-order tangency, in order to simplify the presentation. Similarly to harmonic potentials, the theory of elastic potentials in domains with smooth boundary is well developed (see [12], [33]). In the case of piecewise smooth boundaries without zero angles, theorems on the unique solvability of boundary integral equations of elasticity theory were obtained in [16], where the method outlined above, avoiding Fredholm theorems and the theory of singular operators, was used. We apply the same approach to integral equations of plane elasticity on contours with peaks. In our proofs we use the complex form of representation of solutions to equations of elasticity theory due to G.V. Kolosov. His method was developed by Muskhelishvili (see [32]). Since even for smooth functions on the right-hand side these integral equations generally do not have solutions in classes of integrable functions, we study certain modified integral equations. Chapter four is devoted to solvability and uniqueness theorems in classes of functions with weak singularity as well as to the asymptotic behavior of solutions near peaks. The material follows the authors’ paper [28] (see also [29]). A more detailed description of the content can be found in the introduction to each chapter. Now we comment on the structure of the book. Each chapter contains several sections and most of the sections have subsections. The presentation within each chapter is independent of the content of other chapters. We use double numeration of subsections. For example, 2.3 means Subsection 3 of Section 2. Within each section the numeration of theorems, propositions, lemmas, and corollaries is independent. For instance, Proposition 4.3 means Proposition 3 in Section 4.

Chapter 1

Lp-theory of Boundary Integral Equations on a Contour with Peak In this chapter we are concerned with two topics: solvability of boundary integral equations and Fredholm property of operators generated by these equations in Lp spaces. We find under what conditions and in which spaces we have solvability and Fredholm theorems and we describe the cases when the Fredholm property fails. We arrange the material in the following way. In the Introduction, we collect information on the single and double layer potentials and describe the domains and function spaces. Section 2 contains auxiliary facts from harmonic analysis. We also study the continuity of single and double potentials, as well as boundary integral operators. The auxiliary boundary value problems are treated in Section 3, whereas the integral equations of the Dirichlet and Neumann problems for the Laplace operators are studied in Sections 4 and 5.

1.1 Introduction Single and double layer potentials. By Ω+ we denote a bounded simply connected domain with a boundary Γ of the class C 2 . We put Ω− = R2 \Ω+ . A function u± in the class C 2 (Ω± ) is said to be harmonic in Ω± , if it satisfies the Laplace equation Δu± = 0 in Ω± and u− obeys the condition u− (x) = O(1) as |x| → ∞ . Given any z ∈ R2 \Γ, the integrals τ (q) log (V τ )(z) = Γ

1 dsq |z − q|

(1.1)

Chapter 1. Lp -theory of Boundary Integral Equations

2

and (W σ)(z) =

σ(q) Γ

∂ 1 dsq log ∂nq |z − q|

(1.2)

are called single and double layer potentials with densities τ and σ respectively. The potentials (1.2) and (1.1) are usually called logarithmic. Let n denote the unit vector of the normal to the boundary curve, directed inward Ω− . If Γ is smooth at z0 , the normal derivatives of the single layer potential, as z → z0 ∈ Γ for z in Ω− τ) ∂(V τ ) or in Ω+ , will be denoted by ∂(V ∂n− (z0 ) and ∂n+ (z0 ) respectively, while (Sτ )(z0 ) stands for the direct value of the normal derivative: ∂ 1 dsq . Sτ (z0 ) = τ (q) log ∂nz |z − q| Γ We recall that the single layer potential is continuous when one crosses Γ, unlike its normal derivative which has the jump ∂(V τ ) ∂(V τ ) (z0 ) − (z0 ) = 2πτ (z0 ) . ∂n+ ∂n−

(1.3)

The functions ∂(V τ )/∂n± and Sτ are related through the equality ∂(V τ ) (z0 ) = ±πτ (z0 ) + (Sτ )(z0 ) . ∂n±

(1.4)

The potential V τ is a harmonic function in Ω+ and in general nonharmonic in Ω− because of the logarithmic growth at infinity. The boundedness of V τ in Ω− holds if τ ds = 0. Γ

The direct value of the double layer potential at z0 ∈ Γ will be denoted by (T σ)(z0 ), and its limit values from Ω+ and Ω− will be W+ (z0 ) and W− (z0 ). They are related by (W± σ)(z0 ) = ∓πσ(z0 ) + (T σ)(z0 ) . (1.5) The jump formula for the double layer potential is (W− σ)(z0 ) − (W+ σ)(z0 ) = 2πσ(z0 ) .

(1.6)

The potential W σ is a harmonic function in Ω± . Dirichlet and Neumann boundary value problems and corresponding boundary integral equations. By classical solutions of the Dirichlet problems in Ω± one means harmonic functions u± ∈ C(Ω± ) such that u± = ϕ± on Γ , where ϕ± are given functions.

(1.7)

1.1. Introduction

3

By classical solutions of the Neumann problems in Ω± , one means harmonic functions u± ∈ C(Ω± ) satisfying the conditions ∂u± /∂n = ψ ± on Γ .

(1.8)

The necessary conditions for solvability of the Neumann problems in Ω± are ψ ± (q)dsq = 0 . (1.9) Γ

Relations (1.4) and (1.5) reduce the solution of the Dirichlet and Neumann problems for the Laplace equation to boundary integral equations. We are looking for a solution of the Dirichlet problem in the form of the potential (1.1) with density σ ± . By (1.5), the density σ ± is a solution of the equation πσ ± ∓ T σ ± = ∓ϕ± . (1.10) There is another approach to solving the Dirichlet problem when its solution is represented in the form u(z) = V σ ± (z) + C ± , where C ± is constant and V σ ± is defined by (1.1). Both σ ± and C ± are to be found from the equation V σ ± + C ± = ϕ± on Γ .

(1.11)

To reduce the Neumann problems in Ω± to integral equations, let us look for their solutions in the form of the potential (1.2) with unknown densities τ ± . The potential V τ provides a solution of the boundary value problem if τ ± satisfy the equations πτ ± ± Sτ ± = ±ψ ± . (1.12) Sets of points and function spaces. In this chapter we are interested in a class of domains with cusps. Note that the arcs of the boundary may have different orders of tangency near a cusp. Let Γ\{O} belong to the class C 2 . We say that O is an outward (inward) peak if Ω+ (the complementary domain Ω− ) is given near O by the inequalities κ− (x) < y < κ+ (x), 0 < x < δ, where x−μ−1 κ± (x) ∈ C 2 [0, δ],

lim x−μ−1 κ± (x) = α±

x→+0

(1.13)

with μ > 0 and α+ > α− . By Γ± we denote the arcs {(x, κ± (x)) : x ∈ [0, δ]}. Points on Γ+ and Γ− with equal abscissas will be denoted by q+ and q− .


4

We introduce the space N1,− p,β (Γ) of absolutely continuous functions ϕ on Γ\{O} with the finite norm ϕN1,− (Γ) = |ϕ(q+ ) − ϕ(q− )|p |q|p(β−μ) dsq p,β

Γ+ ∪Γ−

|(∂/∂s)ϕ(q)| |q| p

+

p(β+1)

dsq +

Γ

1/p |ϕ(q)| |q| dsq . p

(1.14)

pβ

Γ

By Np,β (Γ) we denote the space of functions on Γ\{O} represented in the form ϕ = (d/ds)ψ, where ψ ∈ N1,− p,β (Γ) and ψ(z0 ) = 0 for a fixed point z0 ∈ Γ\{O}. A norm on Np,β (Γ) is defined by ϕN

p,β (Γ)

= ψN1,− (Γ) .

(1.15)

p,β

If |q|β ϕ ∈ Lp (Γ), we say that ϕ belongs to Lp,β (Γ). We define the norm in this space by ϕLp,β (Γ) = |q|β ϕLp (Γ) . (1.16) Let L1p,β (Γ) be the space of absolutely continuous functions on Γ \ {O} with the finite norm ϕ L1p,β (Γ) = (∂/∂s)ϕ Lp,β (Γ) + ϕLp,β−1 (Γ) . (1.17) It is an easy exercise to check the density, in the space L1p,β (Γ), of the set of smooth functions on Γ vanishing near O. We introduce the space N1,+ p,β (Γ) of absolutely continuous functions ϕ on Γ \ {O} with the finite norms 1/p ϕN1,+ (Γ) = |ϕ(q+ ) + ϕ(q− )|p |q|p(β−μ) dsq + ϕ L1p,β+1 (Γ) . (1.18) p,β

Γ+ ∪Γ−

Let P(Γ) denote the space of restrictions to Γ \ {O} of functions of the form P (z) =

m

t(k) Re z k

k=0

with real numbers t(k) , where m = [μ − β − p−1 ].

(1.19)

A norm of P is defined by P P(Γ) =

m k=0

| t(k) | .

(1.20)

1.2. Continuity of boundary integral operators

5

1,+ By M1,+ p,β (Γ) we denote the direct sum of Np,β (Γ) and P(Γ). Let Mp,β (Γ) denote the space of functions on Γ \ {O} represented in the form ϕ = (d/ds)ψ, where ψ ∈ M1,+ p,β (Γ) and ψ(z0 ) = 0 for a fixed point z0 ∈ Γ \ {O}. A norm on Mp,β (Γ) is defined by ϕM (Γ) = ψM1,+ (Γ) . (1.21) p,β

p,β

1.2 Continuity of boundary integral operators 1.2.1 Auxiliary assertions In this subsection we collect some facts of the measure and integral theory used in the sequel. All of them but for the last lemma are well known. Lemma 1.2.1. (Hardy’s inequality [7]). If f (x) ≥ 0, then ∞

−r

x 0

p p F (x) dx |r − 1|

p

∞

x−r (xf (x))p dx

(1.22)

0

where p > 1, r = 1 and x F (x) =

∞ f (t)dt for r > 1, and F (x) =

f (t)dt for r < 1 . x

0

We introduce the space Lp,α (R) of measurable functions on R, with the finite norm ϕ L (R) = (1 + x2 )α/2 ϕ L (R) . p,α

p

Theorem 1.2.2. (Muckenhoupt [31]). Let 1 p ∞. In order that there exists a constant C independent of f and such that ∞

x |w(x)

0

1/p ∞ 1/p f (t)dt| dx C |v(x)f (x)|p dx , p

0

(1.23)

0

it is necessary and sufficient that ∞ 1/p r 1/p p B = sup |w(x)| dx |v(x)|−p dx < ∞, r>0

where p = p/(p − 1).

r

0

(1.24)


6

In particular, we have Corollary 1.2.3. The integral operator 1 1 f (x) → (1 + x2 )J/2 x

x f (y)dy, x ∈ (0, ∞), 0

acts continuously in Lp,α (0, ∞) if α + p−1 < J + 1. Proof. One can check directly that the functions w(x) = (1 + x2 )(α−J)/2 x−1 and v(x) = (1 + x2 )α/2 satisfy (1.24) if α + p−1 < J + 1.

Corollary 1.2.4. The integral operator ∞ f (x) →

1 f (y) dy, x ∈ (0, ∞), (1 + y 2 )J/2 y

(1.25)

x

acts continuously in Lp,α (0, ∞) if α + p−1 > −J. Proof. By Theorem 1.2.2,

∞∞

0

1 f (y)

dy g(x)dx

(1 + y 2 )J/2 y

x

∞

= f (y)

1 1 (1 + y 2 )J/2 y

0

y

g(x)dxdy

0

y p ∞ p−1 p−1 α−J p 2 − 2 1 (1 + y ) g(x)dx dy f Lp,α y 0

f Lp,α gL

0 p ,α p−1

provided α + p−1 > −J. The result follows.

Let f be locally integrable on R. The Hardy maximal function is defined by 1 (M f )(x) = sup |f (y)|dy, x ∈ R, (1.26) Qx |Q| Q where the supremum is taken over all segments Q in R, containing x. The operator M : f → M f is called the Hardy–Littlewood maximal operator. According to the classical theorem (cf. [38]), the operator M is bounded in Lp (R), i.e., M f p Cp f p , 1 < p ∞ .

(1.27)


7

Let Lp,w (R) be the weighted space with the norm 1/p f p,w = |f (x)|p w(x)dx , R

where w is a nonnegative function. Let us describe the class of functions w for which |(M f )(x)|p w(x)dx C |f |p w(x)dx, f ∈ Lp,w (R) . (1.28) R

R

We introduce the so-called (Ap )-condition. A nonnegative measurable function w given on R satisfies the (Ap )-condition if there exists a constant C such that p 1 1 1 w w− p−1 Cε x − t

(1.30)

We shall use weighted estimates for H obtained in [8]. Theorem 1.2.6. Let 1 < p < ∞. The inequality p |Hf | w Cp |f |p w R

(1.31)

R

holds for any f ∈ Lp,w (R) with a constant Cp depending only on p, if and only if w ∈ (Ap ). In the sequel, we also use the following theorem on Fourier multipliers in weighted Lp -spaces (cf. [38]). Let m be a bounded measurable function given on R. Further, let Tm be the linear transform defined on Lp ∩ L2 (R), (Tm f )(x) = m(x)f(x) ,

(1.32)

where f is the Fourier transform on R. Theorem 1.2.7. Let m be a bounded function given on R and let

∂ 1

sup

m(x) dx < ∞ . r ∂x r>0 r 0 . 1 + |x − y|J |x − y| (1 + y 2 )α/2

Obviously,

α 1

(1 + y 2 )α/2 − (1 + x2 )α/2

|α|(1 + ξ 2 ) 2 −1 |ξ| |L(x, y)| = ,

|x − y|

(1 + y 2 )α/2 (1 + y 2 )α/2 where the real ξ is placed between x and y. Since 1/2 < ξ/y < 2, one has α

(1 + ξ 2 ) 2 −1 = O(1) . α (1 + y 2 ) 2 −1 Therefore, L(x, y) ∼ x/(1 + x2 ) for x/2 < y < 2x. By Minkowski’s inequality, the operator L defined by 2x Lg(x) = L(x, y)g(y)dy x/2

acts continuously in Lp (R+ ). Now let f ∈ Lp,α (R). We put g(y) = (1 + y 2 )α/2 f (y). The continuity of T in Lp,α (R+ ) results from the estimate p p T f |(Lg)(x)| dx + |(T0 g)(x)|p dx Lp,α (R)

R

R

Cp,α g

p Lp (R)

= Cp,α f p

Lp,α (R)

where the constant Cp,α does not depend on f .

,

1.2.2 Estimates for kernels of integral operators Here we obtain estimates for the kernels of the single and double layer potentials on contours with peaks. Let (u) = κ+ (u) − κ− (u) with u ∈ [0, δ]. We define the function h by the equation δ dν = τ, τ > 0 . (1.39) (ν) h(τ )

The function q given by

δ q(u) = u

dν (ν)


10

is strictly decreasing and continuously differentiable three times. The function 1/ is not integrable on (0, δ). Hence the inverse function u = h(τ ) = q −1 (τ ) is defined on (0, ∞), strictly decreasing on (0, δ) and three times continuously differentiable. Furthermore, 1 1 = + O(ν −μ ) as ν → +0 . (1.40) (ν) αν μ+1 Integrating (1.40) over (u, δ), we obtain τ=

1 −μ u + O(1) as u → +0 . αμ

(1.41)

It follows from (1.41) that τ ∼ (αμ)−1 u−μ as u → 0. Hence, u ∼ (αμτ )−1/μ as τ → ∞. We move the term O(1) in (1.41) to the left-hand side and raise both sides to the power −1/μ. Then h(τ ) =(αμτ )−1/μ (1 + O(1))−1/μ =(αμτ )−1/μ + O(τ −1−1/μ ) .

(1.42)

In order to obtain a representation of h (τ ) as τ → ∞, we differentiate (1.39), h (τ ) = −(h(τ )),

(1.43)

and plug (1.42) into (1.43) instead of h(τ ). Then h (τ ) = − α((αμτ )−1/μ + O(τ −1−1/μ ))μ+1 + O(τ −1−2/μ ) = − α−1/μ (μτ )−1−1/μ + O(τ −1−1/μ−1/κ ) ,

(1.44)

where κ = max{μ, 1}. A representation of h can be derived by differentiating (1.43), and plugging in (1.42) and (1.44) instead of h(τ ) and h (τ ). This results in the asymptotic formula h (τ ) = (αμ)−1/μ

1 1 (1 + ) τ −2−1/μ + O(τ −2−1/μ−1κ ) . μ μ

(1.45)

The next step is the equality h (τ ) = −(αμ)−1/μ

1 1 1 (1 + ) (2 + ) τ −3−1/μ + O(τ −3−1/μ−1κ ) . μ μ μ

(1.46)

Therefore, the asymptotic representation (1.42) of h(τ ) is differentiable three times. Lemma 1.2.10. Let (u) = κ+ (u) − κ− (u) and let the function h be defined by (1.39). If |1 − ξ/τ | < ε, where ε is a sufficiently small number; then

(h(τ )) (h(ξ) − h(τ ))

τ −ξ

c. (1.47) −

(h(ξ) − h(τ ))2 + ((h(τ )))2 2 (τ − ξ) + 1 τ


11

Proof. Decompose the function α(ξ, η) =

h(ξ) − h(τ ) ξ−τ

in powers of ξ − τ : α(ξ, τ ) = h (τ ) + h (τ )(ξ − τ ) + O(τ −3−1/μ )(ξ − τ )2 . Dividing by h (ξ), and using (1.44), we obtain (ξ − τ )2 h (τ ) α(ξ, τ ) = 1 + (ξ − τ ) + O . h (τ ) h (τ ) τ2 The ratio h (τ )/h (τ ) is of the same order as τ −1 as τ → ∞. Hence,

By (1.43) and (1.48),

ξ − η α(ξ, η) =1+O . h (τ ) τ

(1.48)

ξ − η (h(ξ)) = −1 + O . α(ξ, η) τ

(1.49)

Represent the difference (h(τ )) (h(ξ) − h(τ )) τ −ξ − (h(ξ) − h(τ ))2 + ((h(τ )))2 (τ − ξ)2 + 1 in the form (h(τ ))

ξ−τ α(ξ, τ ) (ξ − τ )2 + 1 (ξ − τ )(1 − ((h(τ ))/α(ξ, τ ))2 ) (h(τ )) . + α(ξ, τ ) (ξ − τ )2 + 1 (ξ − τ )2 + ((h(τ ))/α(ξ, τ ))2 −1

Now, using (1.49), one can see that every term in (1.50) is O(τ −1 ).

(1.50)

Lemma 1.2.11. Let (u) = κ+ (u) − κ− (u) and let the function h be specified by (1.39). If |1 − ξ/τ | < ε, where ε is a sufficiently small number, then

(h(ξ))(h(τ )) 1

(h(ξ) − h(τ ))2 + ((h(ξ)))2 − (ξ − τ )2 + 1

(1.51) c 1 1 . + τ (ξ − τ )2 + 1 ξ Proof. We use Taylor’s formula and decompose α(ξ, η) =

h(ξ) − h(η) ξ−η


12

in powers of ξ − η: α(ξ, η) = h (τ ) + h (τ )(ξ − τ ) + O(τ −3−1/μ )(ξ − τ )2 and in powers of η − ξ: α(ξ, η) = h (ξ) + h (ξ)(τ − ξ) + O(ξ −3−1/μ )(τ − ξ)2 . Multiplying these formulae and dividing by h (τ )h (ξ), we arrive at (ξ − τ )2 h (ξ)h (τ ) − h (τ )h (ξ) α(ξ, τ )2 = 1 + (ξ − τ ) + O . h (ξ)h (τ ) h (ξ)h (τ ) ξ2 By the Lagrange theorem, h (ξ)h (τ ) − h (τ )h (ξ) h (τ ) h (ξ) h (z)h (z) − h (z)2 = − = (ξ − τ ) , h (ξ)h (τ ) h (τ ) h (ξ) h (z)2 where z is situated between ξ and τ . Since the factor in front of ξ − τ is of the same order as ξ −2 (see (1.46), (1.45) and (1.44)), the equality (ξ − τ )2 α(ξ, τ )2 = 1 + O h (ξ)h (τ ) ξ2 holds. Using (1.43), we find (ξ − τ )2 (h(ξ))(h(τ )) . = 1 + O α(ξ, τ )2 ξ2

(1.52)

The difference

(h(ξ))(h(τ )) 1 − (h(ξ) − h(τ ))2 + (h(ξ))2 (ξ − τ )2 + 1 admits the representation 2 1 − (h(ξ))/α(ξ, τ ) (h(ξ))(h(τ )) 2 (α(ξ.τ ))2 (ξ − τ )2 + 1 (ξ − τ )2 + (h(ξ))/α(ξ, τ ) 1 − ((h(ξ))(h(τ )))/(α(ξ, τ ))2 . − (ξ − τ )2 + 1 By (1.52) and (1.49), we deduce that (1.53) does not exceed c 1 1 . + 2 τ (ξ − τ ) + 1 ξ

(1.53)

We choose ε > 0 so small that, for a certain constant c > 0 and for all u ∈ [0, δ] such that |x − u| < εx, the estimate |κ± (x) − κ∓ (u)| ≥ c uμ+1

(1.54)

holds. In the sequel, the arcs of Γ± , which can be projected on the segments [0, (1 − ε)x], [(1 − ε)x, (1 + ε)x] and [(1 + ε)x, δ], will be denoted by Γ ± (x), Γc± (x), and Γr± (x), respectively.


13

Lemma 1.2.12. Let z = x + κ+ (x) ∈ Γ+ and ρ(u) = κ+ (u) − κ− (u). (a) There is a constant c independent of z, such that, for all q = u + iκ+ (u) ∈ Γc+ (x), the inequality

∂

u |z| 2 −1/2

c x2μ−1 (1.55)

∂sz log |z − q| − (1 + (κ+ (x)) ) x (u − x)

holds. (b) There is a constant c independent of z, such that

∂

u−x |z| 2 −1/2 u

c

∂sz log |z − q| − (1 + (κ+ (x)) ) 2 2 x (u − x) + ((u))

x for all q = u + iκ− (u) ∈ Γc− (x). (c) There is a constant c independent of z, such that

∂ (x) 1 2 −1/2

− (1 + (κ log (u)) ) +

∂nq 2 2 |z − q| (x − u) + (x)

x2μ+1 c xμ−1 + (x − u)2 + x2μ+2 for all q = u + iκ− (u) ∈ Γc− (x). (d) There is a constant c independent of z, such that

∂ (u) |z| 2 −1/2 u

c

− (1 + (κ log (u)) ) +

∂nz 2 2 |z − q| x (x − u) + (u) x

(1.56)

(1.57)

(1.58)

for all q = u + iκ− (u) ∈ Γc− (x). Proof. (a) Our starting point is the equality ∂ |z| q = (1 + (κ+ . log (x))2 )−1/2 Re (1 + iκ+ (x)) ∂sz |z − q| z(q − z) The left-hand side of (1.55) does not exceed

u q

Re (1 + iκ (x))

. − +

z(q − z) x(u − x)

Using

(1.59)

(1.60)

1 1 κ (x) 1 + Re (1 + iκ+ − = Im (1 + iκ+ (x)) (x)) z x x z

and 1 (u) − κ+ (x) 1 κ+ 1 − = Im (1 + iκ+ , (x)) (x)) Re (1 + iκ+ q−z u−x u−x q−z


14

we find

∂

u |z| 2 −1/2

− (1 + (κ log (x)) ) +

∂sz |z − q| x (u − x)

|κ+ (x) − x κ+ (x)| |κ+ (x)| 3 x |κ+ (u) − κ+ (x) − κ+ (x)(u − x)| + |κ+ (x) − κ+ (u)| . 3 |x − u|

(1.61)

By the Taylor formula, the right-hand side of (1.61) is dominated by c x2μ−1 . Now, we prove (b). Set z ∗ = x + iκ+ (u). One can check that 1 1 u u−x − − (x)) Re (1 + iκ+ z z−q x (u − x)2 + ((u))2 q 1 u 1 − Re = Re (1 + iκ+ (x)) zq−z x q − z∗ q u 1 − (x)) = Re (1 + iκ+ z x q−z 1 u 1 − + Re (1 + iκ+ (x)) x q−z q − z∗ 1 u . (x)) i Im + Re (1 + iκ+ x q − z∗

(1.62)

Let the three terms on the right-hand side of (1.62) be denoted by I1 , I2 , and I3 . We have

q 1 1 u

1 c

|I1 | −

c uμ c uμ μ+1 , (1.63) z x |z − q| |z − q| u x because |z − q| ≥ |Im (z − q)| ≥ c uμ+1 ,

1 (u − x) uμ+1

Im

x (q − z)(q − z ∗ ) (u − x)2 u2μ+2 c c , x ((u − x)2 + ((x))2 )2 x

|I2 |

and

1

u2μ+2 c u c

|I3 | = κ+ (x) Im .

x q − z∗ x (u − x)2 + ((x))2 x

Now, (1.56) follows by (1.62), (1.63), (1.64) and (1.65). (c) To prove (1.57), we note that |κ− (x) − κ− (u) − κ− (u)(x − u)| c xμ−1 (x − u)2 .

(1.64)

(1.65)


15

Hence

1+

(u − x)κ− (u) − κ− (u) − κ+ (x) ∂ 1 = log 2 ∂nq |z − q| (x − u)2 + κ− (u) − κ+ (x) μ−1 κ+ (x) − κ− (x) . = 2 + O x (x − u)2 + κ− (u) − κ+ (x)

1/2 (κ+ (u)2

(1.66)

Taking into account that |κ− (x) − κ− (u)| c xμ |x − u|, we find κ+ (x) − κ− (x) 2 (x − u)2 + κ− (u) − κ+ (x) ρ(x) x2μ+1 = + O 2 (x − u)2 + x2μ+2 (x − u)2 + ρ(x)

(1.67)

which together with (1.66) implies (1.57). (d) In view of the relations ∂ |z| q (1 + (κ+ log (x))2 )−1/2 = Im (1 + i κ+ (x)) ∂nz |z − q| z (z − q) and Im

1 (u) = q − z∗ (x − u)2 + (u)2

with z ∗ = x + iκ+ (u), the left-hand side of (1.58) can be written as

u q

+ Im (x))

Im (1 + iκ+ ∗ z(z − q) x(q − z )

q u q

= Im + Im + κ (x)Re

+ z(z − q) x(q − z ∗ ) z(z − q)

q 1

u 1

q

− Im + |κ (x)|

= K1 + K2 .

Re + z q−z x q − z∗ z(z − q)

(1.68)

Since |κ+ (x) − κ− (u)| > c xμ+1 , the term K2 obeys the estimate

(x)|

|κ+

q c xμ+1 c

.

2 2 z(z − q) x (x − u) + (κ+ (x) − κ− (u)) x

(1.69)

Next, we majorize the term K1 in (1.68):

q 1 u 1

−

Im z q − z x q − z∗

q u 1

u 1 1

− − Im

+ Im

= J1 + J2 . z x q−z x q − z q − z∗

(1.70)


16

We have

q 1 u

1 xμ c

c c xμ μ+1 = |J1 | −

z x |q − z| |q − z| x x

for |q − z| ≥ |Im (q − z)| ≥ c xμ+1 . The second term J2 in (1.70) satisfies

u

z − z∗ u

1

|κ (u) − κ (x)| J2 = Im

Re + + x (q − z)(q − z ∗ ) x (q − z)(q − z ∗ ) 1 |u − x|xμ+1 1 . 2 2μ+2 x (x − u) + x x

(1.71)

(1.72)

Then (1.58) follows from (1.68)–(1.72).

1.2.3 Single and double layer potentials on contours with peak In the rest of this chapter, instead of the single layer potential (1.1), we consider the potential |z| (V τ )(z) = dsq τ (q) log (1.73) |z − q| Γ on a contour Γ with peak. We keep the notation (Sτ )(z), z ∈ Γ\{0}, for the direct value of the normal derivative of V τ . Note that the limit relations (1.4) and the jump formula (1.3) remain valid on Γ\{0}. If functions u± , harmonic in Ω± , belong to the classes C(Ω± ) and have continuous normal derivatives, it follows that u+ (z) =

1 (V u+ )(z) − (W u+ )(z), 2π

z ∈ Ω+ ,

(1.74)

and u− (z) =

1 (W u− )(z) − (V u− )(z) + u− (∞), 2π

z ∈ Ω− .

(1.75)

Using the estimates obtained in Sect. 1.2.2, we can prove the boundedness of the potential V in a pair of spaces on Γ. Theorem 1.2.13. Let Ω+ have either an inward or an outward peak and let 0 < β + p−1 < min{μ, 1}. Then the operator V : Lp,β+1 (Γ) → N1,− p,β (Γ) is continuous. Proof. Let us choose a positive δ to be so small that δ δ z : |z| < ∩ Γ = z : |z| < ∩ (Γ+ ∪ Γ− ) . 2 2


17

We introduce a function χ ∈ C0∞ (R2 ) such that supp χ ⊂ {z : |z| < δ} and χ = 1 in a small neighborhood of the origin. We represent V σ(z) in the form |z| |z| dsq + dsq χ(q)σ(q) log (1 − χ(q))σ(q) log |z − q| |z − q| Γ+ ∪Γ−

Γ\(Γ+ ∪Γ− )

=V∗ σ(z) + V0 σ(z) . It is enough to check the continuity of the operator V∗ : σ Lp,β+1 → V∗ σ(z) ∈ N1,− p,β (Γ). We start by showing that there is a constant c, independent of σ, and such that ∂ V∗ σ c σLp,β+1 (Γ) . ∂sz Lp,β+1 (Γ) Using the relation (1.59), we find ⎧ u

⎪ ⎨ c 2 on Γ+ (x) ∪ Γ− (x),

∂

|z| x

∂sz log |z − q| ⎪ 1 ⎩c on Γr+ (x) ∪ Γr− (x) . x

(1.76)

Setting σ(u + iκ± (u)) = σ± (u), we obtain by (1.59) that

I1 =

Γ+ ∪ Γ− (x)

c ∂ |z|

dsq 2 σ(q) log ∂sz |z − q| x

(1−ε)x

(|σ+ (u)| + |σ− (u)|)udu 0

and

I2 =

Γr+ ∪ Γr− (x)

c ∂ |z|

dsq σ(q) log ∂sz |z − q| x

δ (|σ+ (u)| + |σ− (u)|)du , (1+ε)x

where z = x + iy, q = u + iv. By Hardy’s inequality (1.22), I1 Lp,β+1 (Γ) + I2 Lp,β+1 (Γ) c σLp,β+1 (Γ) . In what follows, we may assume that z ∈ Γ+ . In order to estimate the integral over Γc+ (x), we use (1.55) to conclude (1+ε)x

c

∂ |z| du

c

dsq

log σ+ (u) u

+

σ(q) ∂sz |z − q| x u−x x Γc+

(1−ε)x

(1+ε)x

|σ+ (u)|du . (1−ε)x

(1.77)


18

The required estimate for the left-hand side follows by Hardy’s inequality (1.22) and the boundedness of the Hilbert transform in weighted Lp -spaces (see Theorem 1.2.6). It follows from (1.56) that

∂ |z|

dsq

σ(q) log

∂sz |z − q| Γc− (x)

(1+ε)x

c c

x−u

σ− (u)u du

+ x

(x − u)2 + (ρ(u))2 x (1−ε)x

(1+ε)x

|σ− (u)|du . (1−ε)x

The second integral on the right-hand side is estimated in Lp,β+1 (0, δ) by Hardy’s inequality (1.22). It remains to consider the first integral. We make the change of variables x = h(ξ), u = h(τ ), where h is the function defined in (1.39). By (1.47) we have δ 0

xp(β+1)

p

x−u u

σ− (u) du

dx x (x − u)2 + (ρ(u))2

(1+ε)x

(1−ε)x

∞

ξ

γr (ξ)

p

ξ−τ h(τ )

dτ σ− (h(τ ))

dξ 2

τ (ξ − τ ) + 1

p(1−α)

h−1 (δ)

γ (ξ)

∞ ξ

+c

γr (ξ)

p

h(τ )

dτ dξ , |σ− (h(τ ))|

ξ

τ

p(1−α) 1

h−1 (δ)

γ (ξ)

where μ(α − p−1 ) = β + p−1 and the limits of integration γ (ξ) and γr (ξ) are comparable with ξ. According to Theorem 1.2.8 and Hardy’s inequality (1.22), the two last integrals do not exceed

∞ |σ− (h(τ ))|

p

c

h(τ ) τ

p

δ τ

p(1−α)

h−1 (δ)

|σ− (u)|p up(β+1) du.

dτ = c 0

Therefore, the norm in Lp,β+1 (Γ) of the integral ∂ |z| dsq σ(q) log c ∂s |z − q| z Γ− (x) is dominated by c σLp,β+1 (Γ) . Thus, (∂/∂s) V∗ σ L

p,β+1 (Γ)

c σLp,β+1 (Γ) .

(1.78)


19

We turn to the estimate of V∗ σ(z+ )−V∗ σ(z− ) =

+ Γ+

Γ−

1 − q/z−

dsq = I+ + I−

σ(q)log

1 − q/z+

(1.79)

in Lp,β−μ (Γ+ ∪ Γ− ). Clearly, it is sufficient to treat I+ . We represent it as the sum

1 − q/z−

dsq . σ(q)log

+ + (1.80) 1 − q/z+

Γc+ (x)

Γ+ (x)

Γr+ (x)

−1 −1 Since |z+ − z− | c xμ−1 , the inequality

1 − q/z−

xμ u

1 − q/z+ 1 + c |x − u|

holds. Hence,

1 − q/z−

ds σ(q)log

q

1 − q/z+

+

Γ+ (x)

Γr+ (x) (1−ε)x

cx

δ

|σ+ (u)|udu + c x

μ−1

|σ+ (u)|du .

μ

0

(1+ε)x

This and Hardy’s inequality (1.22) imply the required estimate in Lp,β−μ (Γ+ ∪Γ− ) for the left-hand side. Now we need to consider the integral over Γc+ (x) in (1.80). We obtain for q ∈ Γc+ (x): u−x z+ − z− q − z− z+ − z− 1− 1− . = 1+ q − z+ u−x q − z+ (u − x) + (z+ − z− ) Since Re (z+ − z− ) = 0, we have

z+ − z−

(u − x) + (z+ − z− ) 1. Taking into account that 1−

u−x κ+ (u) − κ+ (x) u − x =i = O(xμ ) q − z+ u−x q − z+

and z+ /z− = 1 + O(xμ ), we deduce

μ+1

1 − q/z−

(1 + c xμ ) 1 + c x .

1 − q/z+

|u − x|


20

Hence,

1 − q/z−

dsq

σ(q)log

1 − q/z+

Γc+ (x)

(1+ε)x

cx

cxμ+1 du = J1 + J2 . |σ+ (u)|log 1 + |u − x|

(1+ε)x

|σ+ (u)|du + c

μ (1−ε)x

(1−ε)x

The integral J1 can be estimated in Lp,β−μ (0, δ) by Hardy’s inequality (1.22). Making the change ξ = x−μ , τ = u−μ in J2 , we see that δ

p xp(β−μ) J2 (x) dx

0

∞

∞

c

ξ p(1−α) 1

−1/μ −1−1/μ |σ+ (τ )|τ log 1 +

1

c |ξ − τ |2

p

dτ

dξ .

It follows from Lemma 1.2.9 that the right-hand side is majorized by ∞ |σ+ (τ

c

−1/μ

p −p(β+1)/μ −1−1/μ

)| τ

τ

δ |σ+ (u)|p up(β+1) du .

dτ = c

1

0

Hence we obtain the required estimate for the norm of I+ Lp,β−μ (Γ), and we arrive at the inequality

(see (1.79))

in

1/p

|V∗ σ(z+ ) − V∗ σ(z− )| |z| p

p(β−μ)

dsz

cσLp,β+1 (Γ) .

Γ+ ∩Γ−

This along with (1.78) implies the boundedness of V∗ : Lp,β+1 (Γ) → N1,− p,β (Γ).

Using the estimates obtained in Section 1.2.2, we show that the direct value of the double layer potential T is a continuous operator in L1p,β+1 (Γ). Theorem 1.2.14. Let Ω+ have either inward or outward peak and let 0 < β + p−1 < min{μ, 1}. Then the operator T : L1p,β+1 (Γ) σ −→ T σ ∈ L1p,β+1 (Γ) is continuous.


21

Proof. Since σ ∈ Lp,β (Γ), we have qσ(q) → 0 as q → 0 on Γ± . Therefore, ∂ ∂ |z| T σ(z) = − σ (q) dsq log ∂s ∂nz |z − q| Γ

and it suffices to verify the continuity of the operator T∗ : Lp,β+1 (Γ) → Lp,β+1 (Γ) given by ∂ |z| dsq . σ (q) log T∗ σ(z) = ∂nz |z − q| Γ+ ∪Γ−

We represent T∗ σ(z) in the form ∂ |z| dsq + I(z), σ (q) + log ∂nz |z − q| Γc− (x)

Γc+ (x)

where the last term admits the estimate c |I(z)| 2 x

x

(|σ+ (u)| + |σ− (u)|)udu

0

c + x

δ

(|σ+ (u)| + |σ− (u)|)du , z ∈ Γ+ ∪ Γ− .

x

It follows from Hardy’s inequality (1.22) that IL

p,β+1 (Γ+ ∪Γ− )

c σ L

p,β+1 (Γ)

.

In the sequel we assume that z ∈ Γ+ . Since

∂ |z|

1

log

= Im (1 + iκ (x))

∂nz |z − q| z−q

κ(u) − κ(x) − κ (x)(u − x) κ(x) − κ (x)(x)

+

cxμ−1 |z − q|2 |z|2

(1.81)

for q ∈ Γc+ (x), we obtain

Γc+ (x)

c ∂ |z| σ (q) log dsq

∂nz |z − q| x

(1+ε)x

|σ+ (u)|du .

(1−ε)x

Now the required estimate for the Lp,β+1 (Γ)-norm of the left-hand side is a consequence of Hardy’s inequality (1.22).


22

The following estimate is derived from (1.58),

∂

ρ(u) |z| 2 −1/2

c,

∂nz log |z − q| − 1 + (κ− (x)) 2 2 (x − u) + (ρ(u)) x where q ∈ Γc− (x). This implies

∂ |z|

dsq

σ (q) log

∂nz |z − q| Γc− (x)

(1+ε)x

c x

|σ− (u)|du

(1+ε)x

+

(1−ε)x

|σ− (u)| ρ(u) du = J1 + J2 . (x − u)2 + (ρ(u))2

(1−ε)x

The integral J1 is estimated in Lp,β+1 (0, δ) by Hardy’s inequality (1.22). Further, making the change of variables τ = u−μ , ξ = x−μ in J2 , we obtain δ xp(β+1) |J2 (x)|p dx 0

∞

∞

c

ξ δ −μ

p(1−α) δ −μ

|σ− (τ −1/μ )| −1−1/μ τ dτ (ξ − τ )2 + 1

p

1/p dξ

.

It follows from Lemma 1.2.9 that the integral on the right-hand side is majorized by ∞ δ −1/μ p −p(β+1)/μ −1−1/μ c |σ− (τ )| τ τ dτ = c |σ− (u)|p up(β+1) du . δ −μ

0

Thus T∗ σL

p,β+1 (Γ)

c σL1

p,β+1 (Γ)

.

By (1.82) we arrive at the boundedness of T : L1p,β+1 (Γ) → L1p,β+1 (Γ). In passing, we have proved the following assertion. Theorem 1.2.15. Let Ω+ have either inward or outward peak and let 0 < β + p−1 < min{μ, 1}. Then the operator S : Lp,β+1 (Γ) σ − −→ Sσ ∈ Lp,β+1 (Γ) is continuous.

(1.82)


23

1.2.4 Operator πI − T on a contour with outward peak Let T be the operator defined by (1.5). In this subsection, we prove a property of πI − T which is stronger than the continuity of T in L1p,β+1 (cf. Theorem 1.2.14). Theorem 1.2.16. Let Ω+ have an outward peak and let 0 < β + p−1 < min{μ, 1} . Then the operator (πI − T ) : L1p,β+1 (Γ) → N1,− p,β (Γ) is continuous. Proof. First, we estimate the norm of (πI − T )σ(z+ ) − (πI − T )σ(z− ) in Lp,β−μ (Γ+ ∪ Γ− ). To this end we represent (πI − T )σ(z) in the form ∂ 1 σ(q) log dsq π[σ(z+ ) + σ(z− )] − ∂nq |q| Γr+ (x)∪Γr− (x)

−

σ(q)

k=1

Γr+ (x)∪Γr− (x)

−

σ(q) Γc∓ (x)

σ(q)

Γ+ (x)∪Γ− (x)

− Γr+ (x)∪Γr− (x)

1 ∂ Re k ∂nq

∂ 1 log ∂nq |z − q|

−

N −1

k z dsq − q

σ(q)

Γ\(Γ+ ∪Γ− )

dsq + πσ(z∓ ) −

Γc± (x)

σ(q)




k N −1 8 |q| ∂ 1 z − Re log dsq = σ(q) Ik (z), ∂nq |z − q| k q k=1

k=1

where z = x + iκ± (x) ∈ Γ± . It is clear that I1 N1,− (Γ) = I1 L1p,β+1 (Γ) cσL1p,β+1 (Γ) . p,β

Since 1 1 ∂ log c , ∂n |q| |q|

(1.83)


24

we have

δ

|σ(τ )| dτ, z ∈ Γ+ ∪ Γ− . τ

|I2 (z)| c (1−ε)x

By Hardy’s inequality (1.22), we obtain I2

N1,− p,β (Γ)

= I2 L1p,β+1 (Γ) cσL1p,β+1 (Γ) .

(1.84)

Now we estimate I3 . We have Re

∂ ∂nq

k ∂ ∂ z = Re z k Im q −k − Im z k Re q −k , q ∂nq ∂nq

where q = u + iκ± (u) ∈ Γ± . Since the inequalities and |(∂/∂nq )q −k | c u−k−1

|Im z k | c xk+μ hold, it follows that

∂ xk+μ Re q −k | c k+1 . ∂nq u

|Im z k We have

Re z k = (Re z)k −

k−1

Im z k−r (Re z)r−1 Im z .

r=1

Hence,

Re z k − (Re z)k c xk+μ .

Therefore,

k

xk+μ ∂ ∂ z

− (Re z)k Im q −k c k+1 , q ∈ Γr+ (x) ∪ Γr− (x),

Re

∂nq q ∂nq u which implies the identity σ(q) Γr+ (x)∪Γr− (x)

∂ Re ∂nq

k z dsq = xk q

σ(q) Γr+ (x)∪Γr− (x)

∂ Im q −k dsq + (Rσ)(z) . ∂nq

The last term on the right-hand side admits the estimate δ |(Rσ)(z)| c x

μ (1−ε)x

|σ(τ )| dτ z ∈ Γ+ ∪ Γ− . τ


25

By Hardy’s inequality (1.22), we obtain |z|p(β−μ) |I3 (z+ ) − I3 (z− )|p dsz c σ p

Lp,β+1 (Γ)

Γ+ ∪Γ−

.

(1.85)

Using Taylor’s decomposition of 1 ∂ , q ∈ Γ\(Γ+ ∪ Γ− ), log ∂nq |z − q| for z → 0, we represent I4 in the form [μ]

c(k) xk + (Rσ)(z),

(1.86)

k=0

where RσLp,β−μ (Γ+ ∪Γ− ) cσL1p,β+1 (Γ) . It follows from (1.86) that |z|p(β−μ) |I4 (z+ ) − I4 (z− )|p dsz σL1p,β+1 (Γ) .

(1.87)

Γ+ ∪Γ−

Thus, the required estimate for I4 (z+ ) − I4 (z− ) holds. In what follows, it is sufficient to consider the case z ∈ Γ+ . In order to estimate I5 , we make use of the inequality (1.57). By the change of variables τ = u−μ , ξ = x−μ and by setting σ(u + iκ± (u)) = σ± (u), u ∈ [0, δ], we obtain (1+ε)x

δ p(β−μ)

x 0

p dx

(1−ε)x

∞ c

ξ p(1−α) R

δ −μ −1

σ− (u) x2μ+1 du (x − u)2 + x2μ+2 −1/μ

μ |σ− (τ )| dτ (ξ − τ )2 + μ2 τ

(1.88)

p dξ ,

−1

where β + p = μ(α − p ). It follows from Lemma 1.2.9 that the right-hand side in (1.88) is majorized by ∞ τ

−pα

|σ− (τ

−1/μ

δ )| dτ p

δ −μ

|σ− (u)|p up(β+1) du .

0

Thus, it suffices to estimate (1+ε)x

πσ− (x) − (1−ε)x

σ− (u)ρ(x) du . (x − u)2 + (ρ(x))2


26

Making the change of variables u = h(τ ), x = h(ξ), where h is the function in (1.39), we obtain by (1.51) that γr (ξ)

πσ− (h(ξ)) −

σ− (h(τ ))ρ(h(ξ))h (τ ) dτ (h(ξ) − h(τ ))2 + (ρ(h(ξ)))2

γ (ξ)

= πσ− (h(ξ)) − R

σ− (h(τ )) dτ + I(ξ). (ξ − τ )2 + 1

Here the limits of integration γ (ξ) = h−1 ((1 + ε)x)) and γr (ξ) = h−1 (1 − ε)x) are comparable with ξ and the last term admits the estimate ξ

c |I(ξ)| ξ

dτ |σ− (h(τ ))| +c τ

h−1 (δ)

∞

dτ |σ− (h(τ ))| 2 + c τ

∞

h−1 (δ)

ξ

|σ− (h(τ ))| dτ . (ξ − τ )2 + 1 τ

Therefore, δ

(1+ε)x

πσ (x) − −

p(β−μ)

x 0

(1−ε)x

∞ h−1 (δ)

p

σ− (u)ρ(x) du

dx 2 2 (x − u) + (ρ(x))

(1.89)

p

∞

σ− (h(τ )) p(1−α)

dτ dξ + ξ ξ p(1−α) |I(ξ)|p dξ .

πσ− (h(ξ)) − (ξ − τ )2 + 1

R

h−1 (δ)

According to Lemma 1.2.9 and Hardy’s inequality (1.22), the last integral does not exceed

p ∞

δ

d

σ− (h(τ )) τ p(1−α) dτ = c |σ− (u)|p up(β+1) du .

dτ

h−1 (δ)

0

The Fourier transform of πσ− (h(ξ)) − R

σ− (h(τ )) dτ (ξ − τ )2 + 1

equals −πi sign(ν) (σ ◦ h) (ν)

1 − exp(−|ν|) . |ν|

Since the function (1−exp(−|ν|))|ν|−1 is the Fourier transform of log(ξ −2 +1) up to a real factor, it follows from the boundedness of the Hilbert transform in weighted


27

Lp -spaces (see Theorem 1.2.6) and from Lemma 1.2.9 that the first integral on the right-hand side in (1.89) does not exceed c σ p . Thus, the estimate Lp,β+1 (Γ)

I5 Lp,β−μ (Γ+ ∪Γ− ) c σ p

(1.90)

Lp,β+1 (Γ)

holds. We pass to the estimates for I6 and I7 . For q ∈ Γc+ (x) ∪ Γ + (x) ∪ Γ − (x) the inequality

∂ 1

μ−1

∂nq log |z − q| c x

(1.91)

holds. Indeed,

∂

1 1 2 1/2

1 + (κ log (u) +

|z − q|2

∂nq |z − q|

x

κ+ (τ )(x − u)dτ c xμ−1

u

if q ∈ Γc+ (x), and

∂

1 2 1/2

∂nq log |z − q| 1 + (κ+ (u) uμ + xμ 1 |κ± (u) − κ+ (x)| |κ± (u)| + c |z − q| |q − z| x if q ∈ Γ ± (x). Hence, (1.91) follows. Therefore, (1+ε)x

|I6 (z)| + |I7 (z)| c x

(|σ+ (u)| + |σ− (u)|)du .

μ−1 0

By Hardy’s inequality (1.22), we obtain I6 L

p,β−μ (Γ+ ∪Γ− )

+ I7 L

p,β−μ (Γ+ ∪Γ− )

c σ L

p,β+1 (Γ)

.

It remains to consider I8 . We have k m 1 z ∂ 1 − Re log ∂nq 1 − z/q k q k=1

= −Re

m+1

(z/q) q−z

cos(n q , 0x) + Im

(z/q)m+1 cos(n q , 0y) , q−z

(1.92)


28

where (n q , 0x) and (n q , 0y) are the angles between the normal vector nq and the coordinate axes. The first term on the right-hand side does not exceed c xm+1 uμ−m−2 . We represent the function Im (z/q)m+1 (q − z)−1 in the form m+1 1 z Im Re q q−z m 1 zk 1 m+1 m−k . Re + Im z Re m+1 (Re z) + Im m+1 (Re z) q q q−z k=1

Since |Im (q − z)−1 | c uμ−1 on Γr+ (x) ∪ Γr− (x), we obtain

m+1

z 1

Im

c xm+1 uμ−m−2 .

Re

q q − z

We have

(1.93)

m

zk

Re m+1 (Re z)m−k c xm+μ+1 u−m−1

Im z

q

(1.94)

k=1

Im 1 c uμ−m−1 .

m+1 q

and

(1.95)

It follows from the estimates (1.93), (1.94) and (1.95) that

m+1

Im (z/q)

c uμ−m−2 xm+1 on Γr+ (x) ∪ Γr− (x).

q−z

Therefore,

δ |I8 (z)| c x

(|σ+ (u)| + |σ− (u)|) uμ−m−2 du .

m+1 x

Now set m = [μ]. By Hardy’s inequality (1.22), the integral on the right-hand side belongs to Lp,β−μ (0, δ) and the norm in Lp,β−μ (Γ+ ∪ Γ− ) of I8 is estimated by c σ L . Therefore and by (1.83), (1.84), (1.85), (1.87), (1.90) and (1.92) it (Γ) p,β+1

follows that

(πI − T )σ(z+ ) − (πI − T )σ(z− ) p |z|p(β−μ) dsz c σ p

Lp,β+1 (Γ)

Γ+ ∪Γ−

We turn to the operator (∂/∂s)(πI − T ) : L1p,β+1 (Γ) → Lp,β+1 (Γ).

.

(1.96)


Since

∂ (πI − T )σ(z) = πσ (z) + ∂s

29

σ (q)

Γ

∂ |z| dsq , log ∂nz |z − q|

the estimate

∂ c σ Lp,β+1 (Γ) (πI − T )σ ∂s Lp,β+1 (Γ) follows from Theorem 1.2.15. Corollary 1.2.17. Let Ω

+

have an inward peak and let 0 < β + p−1 < min{μ, 1}.

Then the operator (πI + T ) : L1p,β+1 (Γ) −→ N1,− p,β (Γ) is continuous.

1.2.5 Operator πI − T on a contour with inward peak On contours with inward peaks, the operator πI − T has better properties than T . We show, for instance, that the operator πI − T maps L1p,β+1 (Γ) continuously 1 into M1,+ p,β (Γ) where as T is continuous in Lp,β+1 (Γ) (Theorem 1.2.14). Theorem 1.2.18. Let Ω+ have an inward peak and let 0 < β + p−1 < min{μ, 1}. Then the operator L1p,β+1 (Γ) σ −→ (πI − T )σ ∈ M1,+ p,β (Γ) is continuous. Proof. Let us estimate the norm of (πI − T )σ(z+ ) + (πI − T )σ(z− ) in Lp,β−μ (Γ ∩ {|q| < δ/2}). We represent (πI − T )σ(z) for z = x + iκ± (x) ∈ Γ± in the form ∂ 1 dsq ± π[σ(z+ ) − σ(z− )] − σ(q) log ∂nq |z − q| Γ+ (x)∪Γ− (x)

−

σ(q) Γc± (x)

∂ 1 dsq + πσ(z∓ ) − log ∂nq |z − q|

−

σ(q) Γr+ (x)∪Γr− (x)

∂ 1 dsq − log ∂nq |z − q|

σ(q)

Γc∓ (x)


∂ 1 dsq = log Ik (z) . ∂nq |z − q| 6

σ(q)

Γ\(Γ+ ∪Γ− )

k=1


30

For any q ∈ Γc+ (x) ∪ Γ + (x) ∪Γ − (x), the inequality (1.91) holds. Therefore, (1+ε)x

|I2 (z)| + |I3 (z)| c x

(|σ+ (u)| + |σ− (u)|)du .

μ−1 0

Hence the estimate I2 Lp,β−μ (Γ∩{|q| 0} γ(0) = 0. By Kellogg’s conformal mapping theorem (see Theorem 1.3.2) we have γ −1 (ζ) = c ζ 2 (1 + o(1)), | γ −1 (ζ)| c |ζ|

1.3. Dirichlet and Neumann problems for a domain with peak

41

as ζ → 0, ζ ∈ R2+ , and γ(z) = c z 1/2 (1 + o(1)), |γ (z)| c |z|−1/2 as z → 0, z ∈ Ω+ . We introduce the function Φ1 = ϕ1 ◦ γ −1 . Since γ −1 (ζ) ∼ c ζ 2 as ζ → 0, it follows that Φ1 ∈ Lp,1+2β+1/p (R). A harmonic extension of Φ1 on R2+ has the form ∞ 1 G1 (ξ, η) = − Φ1 (τ ) Im (log(1 − τ /ζ))dτ π −∞

for 0 < β + p

−1

< 1/2 and

1 G1 (ξ, η) = − π

∞

Φ1 (τ ) Im (log(1 − τ /ζ) + τ /ζ)dτ

−∞

for 1/2 < β + p−1 < 1. The branch of log is chosen to ensure log(1 + i0) = 0. In the case 0 < β + p−1 < 1/2 we obtain ∂G1 1 (ξ, 0) = − ∂η πξ

∞ −∞

Φ1 (τ )τ

dτ . ξ−τ

If 1/2 < β + p−1 < 1, then ∂G1 1 (ξ, 0) = − 2 ∂η πξ

∞ −∞

Φ1 (τ )τ 2

dτ . ξ−τ

By Theorem 1.2.6 on the boundedness of the Hilbert transform in weighted Lp spaces we have ∂G1 ( · , 0) ∈ Lp,1+2β+1/p (R). ∂η Since, in addition, (∂G1 /∂τ )(ξ, 0) = O(ξ −2 ) as |ξ| → ∞ , the normal derivative (∂g1 /∂n) of g1 = G1 ◦ γ belongs to Lp,β+1 (Γ) and satisfies ∂g1 /∂nLp,β+1 (Γ) cϕ1 L1p,β+1 (Γ) . Let λ be a conformal mapping of the unit disk D onto Ω+ and let G2 be the harmonic extension of ϕ2 ◦ λ on D. From the representation 1 ∂G2 (ζ) = ∂n 2π

π −π

s−t d ϕ2 (λ(eit )) cot dt, ζ = eit , dt 2


42

it follows that g2 = G2 ◦ λ−1 satisfies ∂g2 /∂nLp,β+1 (Γ) c1 ∂G2 /∂nLp (∂D) c2 d(ϕ2 ◦ λ)/dtLp (∂D) c3 ϕ2 L1p,β+1 (Γ) . Thus (1.109) holds for the harmonic extension g = g1 + g2 of ϕ.

In the next proposition we show that solutions of the Neumann boundary value problem belong to L1p,β+1 (Γ) if boundary data belong to Lp,β+1 (Γ). Proposition 1.3.7. Let ψ belong to Lp,β+1 (Γ) with 0 < β + p−1 < 1,

β + p−1 = 1/2 .

Then there exists a solution g of the Neumann problem (1.8) such that g ∈ L1p,β+1 (Γ) and gL1 c ψL . (1.110) (Γ) (Γ) p,β+1

p,β+1

Proof. We represent ψ ∈ Lp,β+1 (Γ) as the sum ψ = ψ1 + ψ2 , where ψ1 and ψ2 belong to Lp,β+1 (Γ) and supp ψ1 ⊂ Γ+ ∪ Γ− , ψ2 is equal to zero in a neighborhood of the peak. By ζ = γ(z) we denote the conformal mapping of Ω+ onto the upper half-plane R2+ = {ζ = ξ + iη, η > 0} with γ(0) = 0. It follows from Kellogg’s conformal mapping Theorem 1.3.2 that γ −1 (ζ) = c ζ 2 (1 + o(1)), | γ −1 (ζ)| c |ζ| as ζ → 0, ζ ∈ R2+ , (1.111) and (1.112) γ(z) = c z 1/2 (1 + o(1)), |γ (z)| c |z|−1/2 as z → 0, z ∈ Ω+ . Let Ψ1 be equal to (ψ1 ◦ γ −1 )| γ −1 |. We notice from estimates (1.111) for γ −1 and γ −1 that Ψ1 belongs to Lp,1+2β+1/p (R). We represent the solution of the Neumann problem in R2+ with boundary data Ψ1 in the form 1 G1 (ξ, η) = − π

∞ Ψ1 (τ ) Re (log(1 − τ /ζ))dτ −∞

for 0 < β + p−1 < 1/2 and 1 G1 (ξ, η) = − π

∞ Ψ1 (τ ) Re (log(1 − τ /ζ) + τ /ζ)dτ −∞

−1

for 1/2 < β+p < 1. The branch of log is chosen here to ensure log(1 + i0) = 0. In the case 0 < β + p−1 < 1/2 we obtain 1 ∂G1 (ξ, 0) = − ∂ξ πξ

∞ Ψ1 (τ )τ −∞

dτ . ξ−τ


43

If 1/2 < β + p−1 < 1, then 1 ∂G1 (ξ, 0) = − 2 ∂ξ πξ

∞ Ψ1 (τ )τ 2 −∞

dτ . ξ−τ

From Theorem 1.2.6 on the boundedness of the Hilbert transform in weighted Lp -spaces we have ∂G1 ( · , 0) ∈ Lp,1+2β+1/p (R). ∂ξ Since, in addition, (∂G1 /∂ξ)(ξ, 0) = O(ξ −2 ) as |ξ| → ∞, the derivative (∂g1 /∂s) of g1 = G1 ◦ γ belongs to Lp,β+1 (Γ) and satisfies ∂g1 /∂sLp,β+1 (Γ) cψ1 Lp,β+1 (Γ) . Let λ be a conformal mapping of the unit disc D onto Ω+ and let G2 the solution of the Neumann problem (1.8) in D with the boundary data (ψ2 ◦ λ)|λ |. From the representation ∂G2 1 (ζ) = ∂s 2π

π

ψ2 (λ(eit ))|λ (eit )| cot

−π

s−t dt, ζ = eis , 2

it follows that g2 = G2 ◦ λ−1 satisfies ∂g2 /∂sLp,β+1 (Γ) c1 ∂G2 /∂sLp (∂D) c2 (ψ2 ◦ λ)|λ |Lp (∂D) c3 ψ2 Lp,β+1 (Γ) . Thus (1.110) holds for the solution g = g1 + g2 of the Neumann problem (1.8) in Ω+ .

1.3.3 Auxiliary boundary value problems for a domain with peak The concluding principal proposition in this section concerns a representation of the harmonic conjugate for a harmonic extension of a function in N1,+ p,β (Γ). Before passing to this result, we state an auxiliary assertion which is proved at the end of the section. Lemma 1.3.8. Let Φ be integrable on R and let Φ(ξ) = |ξ|2μ (β± + Λ(ξ)) as ξ → ±0,

(1.113)

where Λ(ξ) satisfies the H¨ older condition |Λ(ξ1 ) − Λ(ξ2 )| c |ξ1 − ξ2 |γ , 0 < γ < min{2μ, 1},

(1.114)


44

near the peak and Λ(0) = 0. Then the Hilbert transform HΦ(ξ) of Φ(ξ) is represented in the form [2μ]

for [2μ] = 2μ c(k) ξ k + c(±) |ξ|2μ + c(m+1) ξ m+1 + O ξ 2μ+γ

k=0

and

(1.115)

[2μ]−1

c(k) ξ k + c(m) ξ m log |ξ| + c(±) |ξ|2μ + O ξ 2μ+γ

for [2μ] = 2μ

k=0

with γ < γ. we denote the image of Ω+ under the mapping u + iv = (x + iy)1/2 . By D near the origin is the graph of the function κ(u) such that The boundary ∂ D 1 κ(u) = ± α± |u|2μ+1 1 + O u2 min{2μ,1} , u → ±0. 2

(1.116)

This asymptotic decomposition can be differentiated twice. normalized by θ(0) = 0. Let θ denote a conformal mapping of R2+ onto D, According to Kellogg’s conformal mapping Theorem 1.3.2, we have = ξ + ψ(ξ), ξ ∈ R, Re θ(ξ)

(1.117)

where ψ(0) = 0 and dψ/dξ satisfies the Hölder condition |(dψ/dξ)(ξ1 ) − (dψ/dξ)(ξ2 )| c |ξ1 − ξ2 |γ , 0 < γ < min{2μ, 1}. Substituting (1.117) into (1.116), we obtain d = |ξ|2μ β± + Λ(ξ) , ξ → ±0, Im θ(ξ) dξ where Λ is subject to the Hölder condition (1.114) with a small exponent γ and Λ(0) = 0. belongs to the Hardy space H 1 in the upper half-plane The derivative of θ(ξ) R2+ . Therefore, d = H d Im θ (ξ) + c, Re θ(ξ) (1.118) dξ dξ where H denotes the Hilbert transform.


45

By Lemma 1.3.8, ⎧ [2μ]−1 ⎪ ⎪ ⎪ ⎪ a(k) ξ k + a([2μ]) ξ [2μ] log |ξ| + a(±) |ξ|2μ ⎪ ⎪ ⎪ ⎪ k=0 ⎪ ⎪ ⎪ ⎨ + O ξ 2μ+γ if 2μ ∈ N, d Re θ(ξ) = [2μ] ⎪ dξ ⎪ ⎪ ⎪ ⎪ a(k) ξ k + a(±) |ξ|2μ + a([2μ]+1) ξ [2μ]+1 ⎪ ⎪ ⎪ ⎪ k=0 ⎪ ⎪ ⎩ + O ξ 2μ+γ if 2μ ∈ N, as ξ → ±0. Hence

= θ(ξ)

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨

[2μ] k=1

b(k) ξ k + b([2μ]+1) ξ [2μ]+1 log |ξ| + b(±) |ξ|2μ+1 + O ξ 2μ+1+γ if 2μ ∈ N,

[2μ]+1 ⎪ ⎪ ⎪ ⎪ ⎪ b(k) ξ k + b(±) |ξ|2μ+1 + b([2μ]+2) ξ [2μ]+2 ⎪ ⎪ ⎪ ⎪ k=1 ⎪ ⎪ ⎩ + O ξ 2μ+1+γ if 2μ ∈ N.

The inverse mapping θ−1 (z) restricted to Γ± has the form ξ=

[2μ]

k=1 (±) μ+1/2

+β

1 x

(1.119)

(±1)k β (k) xk/2 + β (±) xμ+1/2 + o (xμ+1/2 ),

(1.120)

(±1)k β (k) xk/2 + (±1)[2μ]+1 β ([2μ]+1) x([2μ]+1)/2 log x

+ o (xμ+1/2 ),

if 2μ ∈ N, and

[2μ]+1

ξ=

k=1

if 2μ ∈ N. Here β (k) , k = 1, . . . , [2μ] + 1, are real coefficients. We notice that there exists a function of the form ⎧ [2μ] ⎪ ⎪ ⎪ ζ + ⎪ a(k) ζ k if 2μ ∈ N, ⎪ ⎨ k=2 d0 (ζ) = [2μ]+1 ⎪ ⎪ ⎪ ⎪ ⎪ a(k) ζ k if 2μ ∈ N, ζ + ⎩ k=2


46

defined on R2+ and satisfying (θ ◦ d0 )(ζ) = ζ + γ(ζ) . Here γ(ζ) is a holomorphic function on a neighborhood of ζ = 0 in R2+ taking real values near ξ = 0 in R and having the representation b[2μ]+1 ξ [2μ]+1 log |ξ| + O |ξ|2μ+1 if 2μ ∈ N, γ(ξ) = O |ξ|2μ+1 if 2μ ∈ / N, where ξ = Re ζ. Then the function d0 (ζ) + γ(ζ) if 2μ ∈ N , d(ζ) = if 2μ ∈ / N, d0 (ζ) satisfies

θ ◦ d (ξ) = ξ + O |ξ|2μ+1 .

It is clear that

2 θ0 = θ ◦ d

(1.121) R2+

is a conformal mapping of a neighborhood of ζ = 0 in onto a neighborhood of the peak in Ω+ which admits the representation x := Re θ0 (ξ) = ξ 2 + O |ξ|2μ+2 as ξ → ±0 . (1.122) The inverse mapping θ0−1 satisfies

ξ := Re θ0−1 (z) = ±x1/2 + O xμ+1/2 on Γ± .

By diminishing δ in the definition of Γ± , we can assume that θ0 is defined on Γ+ ∪ Γ − . Let n0 be an integer subject to the inequalities n0 − 1 2(μ − β − p−1 ) < n0 and let m be the largest integer satisfying 2m n0 . One can check directly that m = [μ − β − 1/p]. Proposition 1.3.9. Let Ω+ have an inward peak and let ϕ ∈ N1,+ p,β (Γ), where 0 < β + p−1 < min{μ, 1}, β + p−1 = 2−1 and 2(μ − β − p−1 ) ∈ / N. Then there exists a harmonic extension of ϕ onto Ω− with normal derivative in the space Lp,β+1 (Γ) such that the conjugate function g with g(z0 ) = 0 at a fixed point z0 ∈ Γ \ {O} has the representation m r=r0

cr (ϕ)Re z r−1/2 + g # (z), z ∈ Ω− ,


47

where ck (ϕ) are linear continuous functionals in N1,+ p,β (Γ), r0 =

0 1

if 1/2 < β + 1/p < 1 , if 0 < β + 1/p < 1/2 ,

and g # satisfies g # N1,− (Γ) c ϕN1,+ (Γ) p,β

p,β

with c independent of ϕ. Proof. (i) Let ϕ ∈ N1,+ p,β (Γ) vanish outside Γ+ ∪Γ− . We extend the function Φ(τ ) = (ϕ ◦ θ0 )(τ ) by zero outside a small neighborhood of O. We first prove the estimates c1 ϕN1,+ (Γ) Φ(+) L p,β

p,2β−2μ+1/p (R)

+ (dΦ/dτ )L

p,2β+1+1/p (R)

+ ΦL

c2 ϕN1,+ (Γ) ,

p,2β+1/p (R)

(1.123)

p,β

where Φ(+) (ξ) = (Φ(ξ) + Φ(−ξ))/2. Let r(ξ) be a measurable function on (0, ∞) subject to |r(ξ)| |ξ|2μ+1 . We choose ∈ [0, 1] such that /2 < β + p−1 < ( + 1)/2 . Then, from the boundedness of the Hardy–Littlewood maximal operator in a weighted Lp -space (see Theorem 1.2.5), we obtain R

|Φ(ξ) − Φ(ξ + r(ξ))|p |ξ|2βp−2μp+1 dξ

c

R

1

|ξ|2μ+1

R

1+

τ (dΦ/dτ )(τ ) dτ

p |ξ|2βp− p+1 dξ

(1.124)

ξ−c|ξ|2μ+1

c

2μ+1 ξ+c|ξ|

|(dΦ/dτ )(ξ)|p |ξ|2βp+p+1 dξ .

For z ∈ Γ+ we have |θ0−1 (z) + θ0−1 (z− )| c |ξ|2μ+1 . This inequality and (1.124) show that the left-hand inequality in (1.123) follows.


48

Let h be a measurable function on [0, δ] such that |h(x)| xμ+1 . Similarly to (1.124), we have δ

δ |ϕ(x) − ϕ(x + h(x))|p x(β−μ)p dx c 0

1

p

x+cx μ+1

t |(dϕ/dt)(t)|dt

xμ+1

xβ p dx

x−cxμ+1

0

δ c

|(dϕ/dt)(t)|p x(β+1)p dx .

(1.125)

0

Using (1.122), we obtain that for ξ in a small neighborhood of the origin the distance between θ0 (ξ) and θ0 (−ξ) does not exceed c xμ+1 , which together with (1.125) implies the right-hand inequality in (1.123). We introduce a function H by dΦ 1 ζ −τ H(ζ) = (τ ) Re log dτ , ζ = ξ + iη ∈ R2+ , (1.126) π R dτ ζ if 0 < β +

1 p

< 12 , and H(ζ) =

1 π

R

dΦ ζ −τ τ (τ ) Re log + dτ, dτ ζ ζ

(1.127)

< β + p1 < 1. From the inequality for the norm of the Hilbert transform (see Theorem 1.2.6) in the space Lp,2β+1+p−1 (R), it follows that the function given by 1 d τ dτ (∂/∂ξ)H(ξ) = Φ(τ ) , (1.128) πξ R dτ ξ−τ if

1 2

if 0 < β +

1 p

< 12 , and by 1 (∂/∂ξ)H(ξ) = 2 πξ

if

1 2

δ. We extend κh by zero outside a small neighborhood of O and set 2

ψ1 (z) = −Δ(κh)(z), z ∈ Ω+ , ϕ1 (z) = (∂/∂s)ϕ(z) − (∂/∂n)(κh)(z), z ∈ Γ. Since (∂/∂η)H = (∂/∂ξ)Φ ∈ Lp,2β+1+1/p (R), it follows that ∂h/∂n belongs to Lp,β+1 (Γ+ ∪ Γ− ). We consider the boundary value problem ΔF (ζ) = Q(ζ), ζ ∈ R2+ , (∂/∂n)F (ξ + i0) = T (ξ), ξ ∈ R,

(1.133)

where Q(ζ) = (ψ1 ◦ θ)(ζ) |θ (ζ)|2

and T (ξ) = (ϕ1 ◦ θ)(ξ + i0) |θ (ξ + i0)| .


50

For 0 < β + 1/p < 1/2 by (1.126) we find c

τ (dΦ/dτ )(τ )

dτ , |grad H(ζ)|

|ζ| R ζ −τ

Φ(τ )

|H(ζ)|

dτ , R ζ −τ

(1.134)

Since the functions Q and T vanish outside {ζ ∈ R2+ : r−1 < |ζ| < r} for some positive r, by (1.134) and Theorems 1.2.5 and 1.2.6 on the boundedness of the Hardy–Littlewood maximal operator and the Hilbert transform in weighted Lp spaces we obtain T Lp (R) + QLp (R2+ ) c dΦ/dτ Lp,2β+1+1/p(R) + ΦLp,2β+1/p(R) .

(1.135)

In order to prove (1.135) for 1/2 < β + 1/p < 1, we use the estimates c

τ 2 (dΦ(+) /dτ )(τ )

(−) dτ , |grad H (ζ)| 2

|ζ| R ζ −τ 1

τ Φ(+) (τ )

dτ , |H(−) (ζ)|

|ζ| R ζ − τ which hold in view of (1.127). A solution of (1.133) is given by F (ζ) = T (u)G(u, ζ)du − Q(w)G(w, ζ)dudv, w = u + iv, R

R2+

with Green’s function

w w

1

log 1 − 1− . G(w, ζ) = 2π ζ ζ

We write F on R in the form F (ξ) = t−1 (ϕ) log |ξ| + t0 (ϕ)

1 ξ

ξ

1

T (u) log 1 − du − Q(w) log 1 − dudv , + π u π w R

R2+

where t−1 (ϕ) = − and

1 t0 (ϕ) = π

R

1 π

T (u)du + R

1 T (u) log |u|du − π

1 π

Q(w)dudv R2+

Q(w) log |w|dudv . R2+

(1.136)


Hence t−1 (ϕ) 1 ∂F (ξ) − = ∂ξ ξ π

R

T (u) 1 du − ξ−u π

R2+

51

(ξ − u)Q(w) dudv . |ξ − w|2

By the boundedness of the Hilbert transform in Lp (R) and the Minkowski inequality, we find that (∂F /∂ξ)(ξ) − t−1 (ϕ) ξ −1 ∈ Lp (R) (1.137) and that the Lp -norm of this function does not exceed T Lp (R) + QLp(R2+ ) . It is clear that in a neighborhood of infinity (∂F /∂ξ)(ξ) = R∞ (ξ) ξ −2 , where

(1.138)

|R∞ (ξ)| c T Lp (R) + QLp(R2+ ) for large |ξ|.

We set f = F ◦ θ. From (1.137) and (1.138) it follows that ∂f /∂s belongs to Lp,β+1 (Γ) and satisfies ∂f /∂sLp,β+1(Γ) c ϕL1p,β+1 (Γ) .

(1.139)

By Taylor’s decomposition of the integral terms in (1.136), we obtain F (ξ) = t−1 (ϕ) log |ξ| + t0 (ϕ) +

n 0 −1

tk (ϕ)ξ k + |ξ|n0 Rn0 (ξ) ,

(1.140)

k=1

where and

|tk (ϕ)| c T Lp (R) + QLp(R2+ ) , k = −1, . . . , n0 − 1, |Rn0 (ξ)| c T Lp (R) + QLp(R2+ )

for small |ξ|. Taking into account the asymptotic representations (1.119) and (1.120) of θ−1 and the inequality 2(μ − β − p−1 ) < n0 , we find from (1.139) and (1.140) that f can be written in the form f (z) =

m

bk (ϕ)Re z k−1/2 + f # (z), z ∈ Ω+ ,

(1.141)

k=1

where f # ∈ N1,− p,β (Γ), and bk (ϕ), k = 1, . . . , m, are linear combinations of the coefficients t (ϕ) in (1.140).


52

According to (1.132) and (1.141), the function g = κh + f is harmonic in Ω+ and can be represented as m

g(z) =

ck (ϕ)Re z k−1/2 + g # (z), z ∈ Ω− ,

k=1

with ck (ϕ) = ak (ϕ) + bk (ϕ). Moreover, m

|c(k) | + g # N1,− (Γ) c ϕN1,+ (Γ) p,β

k=1

p,β

and (g ◦ θ)(∞) = 0 by definition of g. Owing to (∂/∂s)g ∈ Lp,β+1 (Γ), one of the functions conjugate to −g is a harmonic extension of ϕ onto Ω+ with normal derivative in Lp,β+1 (Γ). (ii) Now let ϕ belong to N1,+ p,β (Γ) and let ϕ vanish on Γ ∩ {|q| < δ/2}. We introduce the function Φ(ξ) = (ϕ ◦ θ)(1/ξ), ξ ∈ R, which vanishes outside a certain interval. Set 1 G(ζ) = P.V. π

∞

ζ dτ, ζ ∈ R2+ . τ (ζ − τ )

Φ(τ )Re

−∞

(1.142)

It is clear that one of the conjugate functions G is a harmonic extension of −Φ onto R2+ . It follows from the boundedness of the Hilbert transform in Lp -spaces that g = G 1/θ−1 belongs to L1p,β+1 (Γ) and satisfies g L1p,β+1 (Γ) c ϕ N1,+ (Γ) . p,β

(1.143)

Further, we represent G on R in the form 1 G(ξ) = P.V. π +

Φ(τ )τ

−1

dτ +

R

1 πξ n0 −1

where

R

k=1

1 πξ k

n0 −1

Φ(τ ) τ ξ−τ

1 tk (ϕ) = π

and G # (ξ) =

n 0 −1

1 πξ n0 −1

dτ =

Φ(τ )τ k−1 dτ R

n 0 −1

tk (ϕ)ξ k + G # (ξ),

k=0

R

Φ(τ )τ k−1 dτ

(1.144)

R

Φ(τ ) τ n0 −1 dτ, ξ ∈ R . ξ−τ

(1.145)


53

Since −1/p < 2μ − 2β − 3/p − n0 + 1 < 1 − 1/p, we have n 0 −1 k=0

|tk (ϕ)| + G # L

p,2μ−2β−3/p (R)

c ΦL

p,2μ−2β−3/p (R)

.

Hence, it follows from (1.143) that g is represented as g(z) =

m

ck (ϕ)Re z k−1/2 + g # (z), z ∈ Ω+ ,

k=1

where g # ∈ N1,− p,β (Γ), and ck (ϕ), k = 1, . . . , m, are linear combinations of the coefficients t (ϕ), = 1, . . . , n0 − 1 in (1.144). The coefficients ck (ϕ) and the function g # satisfy m

|ck (ϕ)| + g # N1,− (Γ) c ϕN1,+ (Γ) p,β

k=1

p,β

−1 and the conjugate function g = G(1/θ ) is a harmonic extension of ϕ onto Ω+ with the normal derivative in Lp,β+1 (Γ).

Proof of Lemma 1.3.8. We represent HΦ(ξ) for a positive ξ in the form 1 HΦ(ξ) = π

∞ −∞

1 dt = Φ(t) ξ−t π

2ξ

1 + π =

dt Φ(ξ − t) − Φ(ξ + t) t

0

2 dt + Φ(−t) ξ+t π

0 4

ξ

∞ Φ

(odd)

2ξ

tdt 2 (t) 2 + ξ ξ − t2 π

∞ Φ(odd) (t) 2ξ

dt ξ 2 − t2

Ik (ξ),

k=1

where Φ(odd) (ξ) =

1 1 Φ(ξ) − Φ(−ξ) , Φ(odd) (ξ) = Φ(ξ) + Φ(−ξ) . 2 2

Let us write I1 (ξ) as 1 I1 (ξ) = π

ξ [(ξ − t)

2μ

0

+

1 π

1 dt − (ξ + t) ] + t π

ξ [(ξ − t)2μ − (ξ + t)2μ ]Λ(ξ − t)

2μ

dt t

0

ξ (ξ + t)2μ [Λ(ξ − t) − Λ(ξ + t)] 0

dt t

(1.146)


54

and consider each of the three terms. We have 1 π

ξ [(ξ − t)2μ − (ξ + t)2μ ]

1 dt = ξ 2μ t π

0

1 [(1 − t)2μ − (1 + t)2μ ]

dt = c ξ 2μ , t

0

ξ 1

dt

[(ξ − t)2μ − (ξ + t)2μ ]Λ(ξ − t)

π t 0

ξ

2μ+γ

1 π

1 [(1 − t)2μ − (1 + t)2μ ](1 − t)γ

dt , t

0

because |Λ(ξ(1 − t))| c |ξ|γ |1 − t|γ , and 1

π

ξ (ξ + t)2μ [Λ(ξ − t) − Λ(ξ − t)]

dt

t

0

ξ c

dt = c ξ 2μ+γ (ξ + t) t t

0

Therefore,

1

2μ γ

(1 + t)2μ tγ

dt . t

0

I1 (ξ) = c ξ 2μ + O ξ 2μ+γ .

(1.147)

Since Φ(ξ − t) − Φ(ξ + t) = β+ ξ 2μ−1 t + O(ξ 2μ tγ + ξ 2μ−1+γ t) for 0 < τ < ξ, it is clear that I2 (ξ) = c ξ 2μ + O ξ 2μ+γ .

(1.148)

Let us find an asymptotic representation of I3 (ξ). We use the identity ξ2

t 1 ξ2 ξ 2m ξ 2m+2 . = − − 3 − · · · − 2m+1 − 2m+1 2 2 −t t t t t (ξ − t2 )

It suffices to consider the integral 1 2ξ

m

1

k=0

2ξ

tdt Φ(−t) 2 =− ξ 2k 2 ξ −t

Φ(−t) dt + ξ 2(m+1) t2k+1

1 2ξ

Φ(−t) dt . t2 − ξ 2 t2m+1


55

Suppose that μ is noninteger and m < μ < m + 1. For all k subject to k < μ, the function t−2k−1 Φ(−t) is integrable in a neighborhood of the origin. Therefore, 1 ξ

1

Φ(−t) dt = ξ 2k t2k+1

2k

Φ(−t) dt − ξ 2k t2k+1

0

2ξ

= ck ξ

1 2

−ξ

2k

2k

2ξ

0 2μ

t

2μ−2k−1

2μ−2k−1+γ

+O t

dt

(1.149)

+ O ξ 2μ+γ .

we have

1 ξ

Φ(−t) dt t2k+1

0

= ck ξ 2k + c ξ For m < μ m +

2ξ

Φ(−t) dt − ξ2 )

2m+2

t2m+1 (t2 2ξ

1 =ξ

t2μ−2m−1 dt + ξ 2m+2 (t2 − ξ 2 )

2m+2 2ξ

=ξ

2μ

∞

−

2

= cξ

2μ

∞

+O ξ

2ξ

O t2μ−2m−1+γ dt t2 − ξ 2

t2μ−2m−1 dt + ξ 2μ+γ t2 − 1

1/ξ

1

2μ+γ

(1.150)

1/ξ 2μ−2m−1+γ O t dt t2 − 1 2

,

because 2m + 2 ≥ 2μ + 1 > 2μ + γ . Hence, for m < μ m +

(1.151)

1 2,

I3 (ξ) =

m

c( ) ξ 2 + c ξ 2μ + O ξ 2μ+γ .

=0

Now, suppose that μ is noninteger and m + 1 ξ

2m+2 2ξ

t2μ−2m−1 dt = ξ 2μ t2 − ξ 2

∞ 2

t2μ−2m−3 dt + ξ 2μ+2

= cξ

+ cm+1 ξ

2m+2

+O ξ

2m+4

∞

t2μ−2m−1 dt t2 − 1

1/μ

∞

1/ξ 2μ

< μ < m + 1. Then

t2μ−2m−1 dt − ξ 2μ t2 − 1

∞ = c ξ 2μ − ξ 2μ

1 2

1/ξ

O t2μ−2m−5 dt

= c ξ 2μ + cm+1 ξ 2m+2 + O ξ 2μ+γ ,

(1.152)


56

owing to (1.148). Furthermore, since 1 1 1 ξ2 = + , 2 2 2 4 t −ξ t t 1 − ξ 2 /t2 we have 1 ξ

t2μ Λ(−t) dt t2m+1 (t2 − ξ 2 )

2m+2 2ξ

1 = ξ 2m+2

2μ

t Λ(−t) dt + ξ 2m+4 t2m+3

2ξ

1 2ξ

(1.153) 2μ

t Λ(−t) 1 dt . t2m+5 1 − ξ 2 /t2

If 2μ + γ 2m + 2, we see that, for γ < γ, 1 ξ

2m+2

t2μ Λ(−t) dt = ξ 2μ+γ 2m+3 t

1/ξ O t2μ−2m+γ −3 dt = O ξ 2μ+γ . 2

2ξ

In the case 2μ + γ > 2m + 2, 1 ξ

2m+2

t2μ Λ(−t) dt = ξ 2m+2 t2m+3

1

t2μ Λ(−t) dt + ξ 2m+2 t2m+3

0

2ξ

= cξ

2m+2

+O ξ

2μ+γ

2ξ 0

.

The second term on the right-hand side of (1.153) is O ξ 2μ + γ. Hence, for m + 12 < μ < m + 1, I3 (ξ) =

m

O t2μ+γ−2m−3 dt (1.154)

2μ+γ

, since 2m + 4 >

c( ) ξ 2 + O ξ 2μ+γ ,

=0

where γ < γ. Suppose that μ is integer and μ = m. Then 1 ξ

2m

Φ(−t) dt = ξ 2m t2m+1

2ξ

1

1 + O tγ−1 dt = ξ 2m log ξ + c ξ 2μ + O ξ 2μ+γ t

2ξ

and 1 ξ

2m+2 2ξ

Φ(−t) dt = ξ 2m+2 t2 − ξ 2 t2m+1

1 2ξ

+ ξ 2μ+γ

1 dt + ξ 2μ t2 − ξ 2 t

1/ξ 1

dt t(t2 − 1)

1/ξ γ O t dt = cξ 2μ + O ξ 2μ+γ . 2 t −1 t 1

(1.155)


57

Thus I3 (ξ) =

m−1

c( ) ξ 2 + c ξ 2m log ξ + c(m) ξ 2m + O ξ 2μ+γ .

=0

Finally, ⎧ m−1 ⎪ ⎪ ⎪ c( ) ξ 2 + c ξ 2μ log(ξ −1 ) + c(m) ξ 2m + O ξ 2μ+γ ⎪ ⎪ ⎪ ⎪ ⎪

=0 ⎪ ⎪ ⎪ ⎪ if μ = m, ⎪ ⎪ ⎪ ⎪ m ⎪ ⎪ ⎪ ⎪ c( ) ξ 2 + c ξ 2μ + O ξ 2μ+γ ⎪ ⎨ I3 (ξ) =

=0

⎪ 1 ⎪ ⎪ if m < μ m + , ⎪ ⎪ 2 ⎪ ⎪ ⎪ m ⎪ ⎪ ⎪ ⎪ c( ) ξ 2 + c ξ 2μ + c(m) ξ 2(m+1) + O ξ 2μ+γ ⎪ ⎪ ⎪ ⎪

=0 ⎪ ⎪ ⎪ ⎪ 1 ⎪ ⎩ if m + < μ < m + 1 . 2

(1.156)

Let us show that I4 (ξ) admits the asymptotic representation ⎧ m−1 2μ+γ ⎪ ( ) 2 +1 2μ (m+1) 2m+1 ⎪ ⎪ c ξ + c ξ + c ξ + O ξ ⎪ ⎪ ⎪ ⎪

=0 ⎪ ⎪ ⎪ ⎪ 1 ⎪ ⎪ if m < μ < m + , ⎪ ⎪ 2 ⎪ ⎪ ⎪ m−1 ⎪ ⎪ ⎪ ⎪ ⎨ c( ) ξ 2 +1 + c ξ 2μ log(ξ −1 ) + c(m) ξ 2m+1 + O ξ 2μ+γ I4 (ξ) =

(1.157)

=0

⎪ ⎪ 1 ⎪ ⎪ ⎪ if μ = m + , ⎪ ⎪ 2 ⎪ ⎪ ⎪ m ⎪ ⎪ ⎪ ⎪ c( ) ξ 2 +1 + c ξ 2μ + O ξ 2μ+γ ⎪ ⎪ ⎪ ⎪

=0 ⎪ ⎪ ⎪ ⎪ 1 ⎩ if m + < μ m + 1 . 2

As in the case of I3 (ξ), it is enough to consider the integral 1 ξ 2ξ

m−1 Φ(−t) Φ(−t) dt dt 2k+1 2m+1 . Φ(−t) 2 = − ξ dt + ξ ξ − t2 t2k+2 t2 − ξ 2 t2 m 1

k=0

2ξ

1

2ξ


58

For all k subject to k < μ − 12 , the function t2k+2 Φ(−t) is integrable in a neighborhood of the origin and 1 ξ

2k+1

1

Φ(−t) dt = ξ 2k+1 t2k+2

2ξ

Φ(−t) dt − ξ 2k+1 t2k+2

0

2ξ

Φ(−t) dt t2k+2

0

2ξ

t2μ−2k−2 + O t2μ−2k−2+γ dt

= ck ξ 2k+1 − ξ 2k+1 0

= ck ξ 2k+1 + c ξ 2μ + O ξ 2μ+1 . For m < μ < m + 1 ξ

2m+1 2ξ

1 2

we have

Φ(−t) dt = ξ 2m+1 t2m (t2 − ξ 2 )

1 2ξ

t2μ−2m dt + ξ 2m+1 t2 − ξ 2

1 2ξ

O t2μ−2m+γ dt . t2 − ξ 2 (1.158)

The first term on the right-hand side of (1.158) can be represented in the form ∞ ξ

2μ 2

t2μ−2m dt − ξ 2μ t2 − 1

∞ t

1/ξ

= cξ

2μ

− cm ξ

2m+1

∞ 2μ−2m−2

+O ξ

2m+3

dt + ξ

2μ+2

O t2μ−2m−4 dt

1/ξ

= cξ

2μ

− cm ξ 2m+1 + O ξ 2μ+γ .

The last term in (1.158) is estimated under the condition 2μ + γ 2m + 1 as follows: 1 ξ

2m+1 2ξ

1/ξ 2μ−2m+γ O t2μ−2m+γ O t 2μ+γ dt dt = ξ 2 2 t −ξ t2 − 1 1

=ξ

1/ξ O t2μ−2m+γ −2 dt = O ξ 2μ+γ ,

2μ+γ

1

where γ < γ. In the case 2μ + γ > 2m + 1 we have 1 ξ

2m+1 2ξ

O t2μ−2m+γ dt t2 − ξ 2 1

=ξ

2m+1

O t 2ξ

2μ−2m+γ−2

1

dt + ξ

2m+3 2ξ

O t2μ−2m+γ−4 dt


2ξ = cm ξ

2m+1

+ξ

2m+1

59

O t2μ−2m+γ−2 dt

0

+ ξ 2μ+γ

1/ξ O t2μ−2m+γ−4 dt = cm ξ 2m+1 + O ξ 2μ+γ . 2

Thus, under the condition m < μ < m + 12 , I4 (ξ) =

m−1

c( ) ξ 2 +1 + c ξ 2μ + c(m) ξ 2m+1 + O ξ 2μ+γ , γ < γ .

=0

Let μ satisfy m + 12 < μ m + 1. We estimate the first term on the right-hand side of (1.158). We have 1 ξ

2m+1 2ξ

t2μ−2m dt t2 − ξ 2 1

=ξ

2m+1

1 t

2μ−2m−2

dt + ξ

2m+3

2ξ

2ξ

1 =ξ

2m+1

= cm ξ

2ξ t

0 2m+1

t2μ−2m−4 O 1 dt

2μ−2m−2

+ cξ

2μ

dt − ξ

+O ξ

2m+1

2m+3

t

2μ−2m−2

dt + ξ

2μ

0

1/ξ t2μ−2m−4 O 1 dt 2

= cm ξ

2m+1

+ cξ

2μ

+ O ξ 2μ+γ ,

since 2m + 3 > 2μ + γ. For the second term on the right-hand side of (1.158), the relation 1 ξ

2m+1 2ξ

1 O t2μ−2m+γ 1 2m+1 =ξ O t2μ−2m+γ−2 dt t2 − ξ 2 1 − ξ 2 /t2 2ξ

1 = ξ 2m+1

O t2μ−2m+γ−2 = cm ξ 2m+1 + O ξ 2μ+γ

2ξ

holds. Hence I4 (ξ) =

m

=0

for m + 1/2 < μ < m + 1.

c( ) ξ 2 +1 + c ξ 2μ + O ξ 2μ+γ


60

Finally, consider the case μ = m + 12 . By (1.158), 1 ξ

2m+1 2ξ

Φ(−t) dt t2m (t2 − ξ 2 ) 1

=ξ

2m+1 2ξ

t dt + ξ 2m+1 t2 − ξ 2

1

= ξ 2m+1 log |t2 − ξ 2 |

O t1+γ dt t2 − ξ 2

1 2ξ

2ξ

1 + ξ 2m+1

O tγ−1 dt = c ξ 2m+1 log ξ + cm ξ 2m+1 + O ξ 2μ+γ .

2ξ

Thus I4 (ξ) =

m−1

c( ) ξ 2 +1 + c ξ 2μ log

=0

Since 1 π

∞ −∞

1 + c(m) ξ 2m+1 + O ξ 2μ+γ . ξ

dt 1 Φ(t) =− ξ−t π

∞ Φ(−t) −∞

dt |ξ| − t

for ξ < 0, we arrive at (1.115) by (1.147), (1.148), (1.156) and (1.157).

1.4 Integral equations of the Dirichlet and Neumann problems In this section we study the boundary integral equations related to the Dirichlet problem (1.7) and the Neumann problem (1.8) on contours with peak. The proofs of all theorems here are similar and differ only in details. This is hardly surprising, since we use the same idea discussed in Section 1.1.

1.4.1 Integral equations of the first kind We start with the proof of solvability and a description of the kernel of the integral equations of the first kind (1.11) on a contour Γ. We first prove two auxiliary assertions concerning estimates of the single and double layer potentials. Lemma 1.4.1. The complex potential (VC σ)(z) =

σ(q) log Γ

z dsq z−q

1.4. Integral equations of the Dirichlet and the Neumann problems

61

with density σ ∈ Lp,β+1 (Γ) admits the estimate −1

|(VC σ)(z)| c |z|−β−p ,

z = 0 .

Proof. It suffices to estimate σ(q) log Γ+ ∪Γ−

z dsq z−q

in a small neighborhood of z = 0. Suppose that |z| > 2|q|. Then

log 1 − q c |q| . z |z| Hence

|I(z)| =

σ(q) log

Γ∩{|q|< |z| 2 }

c z

dsq z−q |z|

|σ(q)||q|dsq .

(1.159)

Γ∩{|q|< |z| 2 }

Applying H¨ older’s inequality to the functions |σ(q)||q|β+1 and |q|−β in (1.159), we find −1 |I(z)| c |z|−β−p . (1.160) For |z| < |q|/2 one has

log

z

z |q| u z

− 1 log + c log + c,

log + log z−q q q |z| |z|

where q = u + iκ± (u), u ∈ [0, δ]. Setting σ ± (u) = σ(u + iκ± (u)), 0 u δ, and σ ± (u) = 0, u > δ, we obtain

σ(q) log Γ± ∩{|q|>2|z|}

∞ = c|z|

z dsq z−q

∞

u

+ 1 du c | σ± (u)| log

|z| |z|

| σ± (u)| log u + 1 du

1

1/p ∞ −1 | σ± (|z|u)|p u(β+1)p du c|z|−β−p . c|z| 1

Now let |z| < |q| < 2|z|. Then either |1 − q/z| < 1 and log

|z| |z|

, log

|z − q| |z| − |q|

(1.161)


62

or |1 − q/z| > 1 and log Therefore,

log

|z| c. |z − q|

z

|z|

+ c.

log

z−q |z| − |q|

We introduce the change of variable

v = v(u) = |q| = u 1 +

κ± (u) 2 . u

One can take δ > 0 so small that κ± (u)

d κ± (u) ! > 0. du u

∗ In this case the function v(u) increases and 0 < c−1 8 < v (u) < c8 . Setting σ± (v) = σ(u + iκ± (u)), we obtain

q

dsq

σ(q) log

z−q Γ± ∩{|z|/2 1/2, we obtain that ker V is one-dimensional for 1/2 < β + p−1 < 1 and zero for 0 < β + p−1 < 1/2. The result for the case of an inward peak can be obtained similarly by using the identity ∂ 1 |z| 1 log dsq Im π ∂nq γ(q) |z − q| Γ 1 1 ∂ 1 1 dsq + Im , z ∈ Γ\{O}, = Im log π γ(q) ∂nq |z − q| γ(z) Γ

instead of (1.167) (see (1.74) and (1.5)).

In passing, we have proved the following characterization of the kernel of V in the case 1/2 < β + p−1 < 1. Proposition 1.4.5. a) Let Ω+ have an outward peak and 1/2 < β + p−1 < 1. Then 1 t ∂ 1 Im (out) , t Im (out) , ker V = π ∂n γ γ (∞) where t ∈ R and γ (out) is a conformal mapping of Ω− onto R2+ normalized by the conditions γ (out) (0) = 0 and γ (out) (∞) = i. b) Let Ω+ have an inward peak and 1/2 < β + p−1 < 1. Then 1 t ∂ Im (in) , 0 , ker V = π ∂n γ where t ∈ R and γ (in) is a conformal mapping Ω+ onto R2+ normalized by the conditions γ (in) (0) = 0 and γ (in) (z0 ) = i with a fixed point z0 ∈ Ω+ .


67

In the exceptional case β + p−1 = 1/2, the operator V is not Fredholm. Proposition 1.4.6. Let Ω+ have either outward or inward peak and let β + p−1 = 1/2. Then the operator V : Lp,β+1 (Γ) × R → N1,− p,β (Γ) is not Fredholm. Proof. It suffices to consider the case of an inward peak. We introduce the function H1i (z) = Im z −1/2 (log z − πi)d , z ∈ Ω+ , where −p−1 < d < 0 and the branch of log is chosen so that log(1 + i0) = 0. The trace ϕ1 of H1i to Γ\{O} has the representation 1 ϕ1 (z) = − x−1/2 Im (log x + πi)d ∓ α± xμ−1/2 Re (log x + πi)d 2 μ−1/2 d−1 , +O x (log x) which implies

∂ ϕ1 (z) = O x−3/2 (log x)d−1 ∈ Lp,β+1 (Γ). ∂s

Let χ ∈ C ∞ (R) be equal to 1 for t > 1 and vanish for t < 1/2, and let 1 ϕ2 (z) = − α± xμ−1/2 Re (log x ∓ πi)d χ(|z|/δ), z ∈ Γ\{O}. 2 Set ϕ = ϕ1 − ϕ2 on Γ\{O}. Since ϕ(z) = −x−1/2 Im (log x + πi)d + O xμ−1/2 (log x)d−1 , it follows that ϕ belongs to N1,− p,β (Γ). It is clear that we can find β < β such that ϕ2 ∈ L1p,β +1 (Γ). By Proposition 1.3.6 there exists a harmonic extension H2i of ϕ2 on Ω+ satisfying ∂H2i /∂n ∈ Lp,β +1 (Γ). Hence ∂H2i /∂n ∈ Lp,β+1 (Γ). By

∂ i ∂ H1 (z) = Re z −1/2 (log z − πi)d = O x−3/2 (log x)d ∂n ∂s and d > −1/p, we have (∂/∂n)H1i ∈ Lp,β+1 (Γ). Therefore, the function H i = i i H1 + H2 is a harmonic extension of ϕ to Ω+ and (∂/∂n)H i ∈ Lp,β+1 (Γ). Let H e be the harmonic of ϕ to Ω− constructed in Proposition extension e N 1.3.4. We have grad H = O |z| for an arbitrary integer N . Since (∂/∂n)H1i ∈ Lp,β +1 (Γ), β > β, the pair # $ (2π)−1 (∂H i /∂n − ∂H e /∂n), H e (∞) ∈ Lp,β +1 (Γ) × R is a solution of (1.11) with the right-hand side ϕ.


68

Assume that the equation (1.11) has a solution {σ, c} in Lp,β+1 (Γ) × R. Then σ ∈ Lp,β +1 (Γ) for β > β and by Proposition 1.4.5 we have ∂ 1 ∂H i ∂H e −1 − +c Im , (1.168) σ = (2π) ∂n ∂n ∂n γ where γ is a conformal mapping of Ω+ onto R2+ , γ(0) = 0. Since ∂ 1 ∂H1i Im ∼ c1 x−3/2 and ∼ c2 x−3/2 (log x)d (d < 0), ∂n γ ∂n the equality (1.168) is impossible. Hence the equation (1.11) with the right-hand side ϕ is not solvable in the space Lp,β+1 (Γ) × R. Since the set of smooth functions vanishing near the cusp is dense in N1,− p,β (Γ), β ≤ β, and since the equation (1.11) with the right-hand 1,− side in Np,β (Γ) has a solution in the space Lp,β +1 (Γ), which is embedded into Lp,β+1 (Γ), it follows that the image of Lp,β+1 (Γ) × R under the mapping V is 1,− dense in N1,− p,β (Γ). Therefore the range of V is not closed in Np,β (Γ).

1.4.2 Integral equation of the interior Dirichlet problem in a domain with outward peak Here we obtain the solvability and describe zeros of the integral equation (1.10) on the contour Γ with outward peak. Theorem 1.4.7. Let Ω+ have an outward peak and let 0 < β + p−1 < min{μ, 1}, β + p−1 = 1/2. Then the operator L1p,β+1 (Γ) σ −→ (πI − T )σ ∈ N1,− p,β (Γ)

(1.169)

is surjective. Proof. Let ϕ ∈ N1,− p,β (Γ) and let ϕ = 0 in a neighborhood of the peak. We introduce + the harmonic extensions hi and he of ϕ onto Ω− mentioned in Propositions Ω and i N + 1.3.4 and 1.3.6. Since grad h = O |z| , z ∈ Ω , for an arbitrary integer N and grad he = O |z|−1/2 , z ∈ Ω− , it follows that i ∂he |z| ∂h 1 − log dsq + he (∞), z ∈ Γ\{O}. (1.170) ϕ(z) = 2π ∂n ∂n |z − q| Γ

Let g e ∈ L1p,β+1 (Γ) be the solution of the Neumann problem in Ω− with boundary data ∂hi /∂n on Γ\{O}, g e (∞) = 0, constructed in Proposition 1.3.7. Taking


69

into account that grad g e = O |z|−1/2 , z ∈ Ω− , we obtain that the function w = g e − he + he (∞) admits the representation ∂w ∂ 1 |z| 1 − dsq , z ∈ Ω− , w(q) log (q) log w(z) = 2π ∂nq |z − q| ∂nq |z − q| Γ

(see (1.75)). This and (1.170) imply that, for all z ∈ Γ \ {O}, 1 ∂ 1 w(z) − dsq = 2(ϕ(z) − he (∞)) . w(q) log π ∂nq |z − q| Γ

Since T he (∞) = −πhe (∞), we obtain that the function σ = (2π)−1 (g e − ϕ) ∈ L1p,β+1 (Γ)

(1.171)

is a solution of (1.10). −1 Now let ϕ be an arbitrary function in N1,− , there p,β (Γ). Because of β > −p exists a sequence {ϕn }n≥1 of smooth functions on Γ\{O}, vanishing near the peak and approaching ϕ in N1,− p,β (Γ). By σn we denote the solution of (1.10) with the right-hand side ϕn given by (1.171). In view of Propositions 1.3.4 and 1.3.7, {σn } converges in L1p,β+1 (Γ) to a limit σ. Since the operator (1.169) is continuous, it follows by passage to the limit that σ is a solution of (1.10). Consequently, equation (1.10) is solvable in L1p,β+1 (Γ) for every ϕ ∈ N1,− p,β (Γ) and 1 N1,− p,β (Γ) ⊂ (πI − T )(Lp,β+1 (Γ)).

The converse inclusion was proved in Theorem 1.2.16.

The next assertion contains a description of the kernel of the operator πI −T . Theorem 1.4.8. Let Ω+ have an outward peak and let the operator πI −T be defined by (1.169). Then a) ker (πI − T ) = {0} for 0 < β + p−1 < 1/2; b) dim ker (πI − T ) = 1 for 1/2 < β + p−1 < 1 and 1 , ker (πI − T ) = c Re γ0 where c ∈ R and γ0 is the conformal mapping of Ω− onto R2+ normalized by the conditions γ0 (0) = 0, γ0 (∞) = i. Proof. Let σ belong to ker (πI − T ). Since |W σ(z)| = O |z|−N for a positive integer N and for all z ∈ Ω+ , and since the limit values of W σ(z) are equal to zero in Γ\{O}, it follows that W σ = 0 in Ω+ . Therefore, the function 1 ∂ % dsq , z ∈ R2 \ {O} , W σ(z) = σ(q) log ∂s |z − q| Γ


70

%σ(z) = C, z ∈ Ω+ . We introduce the holomorphic function is constant in Ω+ . Set W %σ(z) − C), z ∈ Ω− . W (z) = (W σ)(z) + i(W Let ζ = γ(z) be a conformal mapping of Ω− onto R2+ , γ(0) = 0. The function F (ζ) = W ◦ γ −1 (1/ζ) is holomorphic in the lower half-plane R2− = {ζ = ξ + iη, η < 0}, continuous up to its boundary, and Im F = 0 on the real axis. Since W σ in Ω− does not grow faster than a power function as z → 0, it follows that the real part of the holomorphic extension Fext of F onto C has the same property. From the Schwarz integral formula (1.166) it follows that Fext has the power growth at infinity. Therefore, there exists a polynomial P with real coefficients such that W σ(z) = Re P (1/γ(z)), z ∈ Ω− . From the jump formula (1.6) for W σ we deduce σ(z) =

1 Re P (1/γ(z)) , z ∈ Γ\{0}. 2π

Since σ ∈ L1p,β+1 (Γ) and since 1 ∈ ker (πI − T ), it follows that σ is proportional to Re (1/γ). By the integral representation (1.75) for the harmonic function Re (1/γ) on Ω− and the limit relation (1.5) for the double layer potential, we obtain 1 ∂ |z| 1 1 log dsq − Re Re π ∂nq γ(q) |z − q| γ(∞) Γ

=

1 π

Re Γ

1 ∂ 1 1 dsq − Re log γ(q) ∂nq |z − q| γ(z)

for z ∈ Γ\{O}. Since (∂/∂n)Re (1/γ) = 0 on Γ\{O}, it follows that the zeros of πI − T have the form c Re (1/γ), where c ∈ R, and γ is an arbitrary conformal mapping of Ω− onto R2+ , normalized by the conditions γ(0) = 0 and Re γ(∞) = 0. We note that the functions of the form Re (1/γ) differ in pairs by a constant positive factor so that dim ker (πI − T ) is at most one. Since Re (1/γ) ∈ L1p,p+1 (Γ) only for β + p−1 > 1/2, we obtain that ker (πI − T ) is one-dimensional for 1/2 < β + p−1 < 1 and trivial for 0 < β + p−1 < 1/2. In passing we have described the kernel of πI − T in the case 1/2 < β + p−1 < 1. The next assertion shows that the case β + p−1 = 1/2 is exceptional. Proposition 1.4.9. Let Ω+ have an outward peak and let β + p−1 = 1/2. Then the operator πI − T defined by (1.169) is not Fredholm. Proof. We introduce the function −r hi (z) = Im z −1/2 log z −1 , z = x + iy ∈ Ω+ ,


71

where r > 1/p and the branch of log is chosen to ensure log(1 + i0) = 0. The restriction ϕ of hi on Γ\{0} has the representation −r −r−1 1 + r α± xμ−1/2 log x−1 ϕ(z) = − α± xμ−1/2 log x−1 2 −r + O x2μ−1/2 log x−1 and belongs to N1,− p,β (Γ). Since −r −r = x−1/2 (log x−1 )−r + O x2μ−1/2 log x−1 , Re z −1/2 log z −1 we have −r ∂hi ∂ = − Re z −1/2 log z −1 ∂n ∂s −r −r d −1/2 x =± + O x2μ−3/2 log x−1 . log x−1 dx Let ζ = γ(z) be a conformal mapping of Ω− onto R2+ with γ(0) = 0. Using Kellogg’s Theorem 1.3.2, one can easily check that γ −1 (ζ) = aγ ξ 2 + O |ξ|2+ε , where ε is a small positive number and aγ ∈ R. This equality can be differentiated at least once. We normalize γ so that aγ = 1. Making the change of variable x = Re γ −1 (ξ + i0), we obtain d −r −r d −1/2 x |ξ|−1 log |ξ|−1 =c log x−1 ± dx dξ

−1 −r

d

+ O |ξ|ε−2 log |ξ|−1

Re γ −1 (ξ + i0) . dξ Let χ be an even function from C ∞ (R), equal to 1 for |ξ| < 1/2 and vanishing for |ξ| > 1, and let −r d −1 Ψ(ξ) = |ξ| χ(ξ). log |ξ|−1 dξ The function ∞ 1 G(ξ, η) = Ψ(τ )Re log(1 − τ /ζ) + τ /ζ dτ π −∞

is a solution of the Neumann problem in R2+ with the boundary data Ψ on R \ {O} (see Proposition 1.3.7). Moreover, for ξ = 0, 1 ∂G (ξ, 0) = − 2 ∂ξ πξ

∞ −∞

Ψ(τ )τ 2 d. ξ−τ


72

In the case ξ > 0 we represent the integral on the right-hand side as ∞ −∞

Ψ(τ )τ 2 dτ = ξ−τ

ξ

Ψ(ξ − τ )(ξ − τ )2 − Ψ(ξ + τ )(ξ + τ )2

0

2ξ +

Ψ(−τ )τ 2 dτ + ξ+τ

0

+ 2ξ

I1 (ξ) = O

∞ 2ξ

∞

We have

dτ τ

(Ψ(τ ) − Ψ(−τ )) 3 τ dτ ξ2 − τ 2

(Ψ(τ ) + Ψ(−τ )) 2 τ dτ = Ik (ξ). 2 2 ξ −τ 4

k=0

−r −r , I2 (ξ) = O log ξ −1 and I4 = 0 . log ξ −1

The integral I3 takes the form ∞ I3 (ξ) = −2

∞ Ψ(τ )τ dτ − 2ξ

2

2ξ

2ξ

Ψ(τ ) dτ . 1 − (ξ/τ )2 τ

1−r The first term on the right-hand side is asymptotically equivalent to c log ξ −1 as ξ → 0+ with r < 1 and the second term admits the estimate O ξ ε , ε > 0. Thus, ∞ 1−r Ψ(τ )τ 2 dτ ∼ c log ξ −1 as ξ → 0 + . ξ−τ −∞

Since

∞ −∞

Ψ(τ )τ 2 dτ = − ξ−τ

∞

−∞

Ψ(−τ )τ 2 dτ |ξ| − τ

for ξ < 0, we obtain 1−r ∂ G(ξ, 0) ∼ c ξ −2 log |ξ|−1 as ξ → 0. ∂ξ It is clear that (∂G/∂ξ) ∈ Lp,1+2β+1/p (R) for r < 1. The function g1 = G ◦ γ is a solution of the Neumann problem in Ω− with the boundary data ψ1 (z) = ±c1

−r −r d −1/2 x + O x−3/2+ε log x−1 , z ∈ Γ\{O}, log x−1 dx

where c1 is a real number. Hence g1 ∈ L1p,β+1 (Γ) and g1 ∈ L1p,β +1 (Γ) for β > β.


73

−r −3/2+ε log x−1 , we can Set ψ2 = (∂/∂n)hi − c−1 1 ψ1 . Since ψ2 (z) = O x choose β < β so that ψ2 ∈ Lp,β +1 (Γ). By Proposition 1.3.7, the Neumann problem in Ω− with the boundary data ψ2 has a solution g2 in L1p,β +1 (Γ). Therefore, g2 ∈ L1p,β+1 (Γ). Thus, the function σ = (2π)−1 (g1 + g2 − ϕ) ∈ L1p,β +1 (Γ) is a solution of (1.10) for β > β (see Theorem 1.4.7). Now we assume that (1.10) has a solution σ0 in L1p,β+1 (Γ). Then by Theorem 1.4.8, σ0 = σ + Re (1/γ0 ), (1.172) where γ0 is the conformal mapping of Ω− onto R2+ normalized by γ0 (0) = 0 and γ0 (∞) = i. Taking into account that 1−r , p−1 < r < 1 , Re (1/γ0 (z)) ∼ ±c1 x−1/2 and σ(z) ∼ ±c2 x−1/2 log x−1 we conclude that (1.172) is not possible. Therefore, the equation (1.10) is not solvable in L1p,β+1 (Γ). Since the set of smooth functions vanishing in a neighborhood of the peak is dense in the space 1 N1,− p,β (Γ), it follows that the image of Lp,β+1 (Γ) under the mapping πI − T is dense 1,− in N1,− p,β (Γ). Therefore, the range of πI − T is not closed in Np,β (Γ).

1.4.3 Integral equation of the interior Neumann problem in a domain with outward peak In this section we prove the solvability and describe zeros of the boundary integral equation (1.12) on a contour with outward peak. Theorem 1.4.10. Let Ω+ have an outward peak and let 0 < β + p−1 < min{μ, 1}, β + p−1 = 1/2. Then the operator (Lp,β+1 (Γ)) σ −→ (πI + S)σ ∈ N1,− p,β (Γ)

(1.173)

is surjective. Proof. Let ψ belong to N1,− p,β (Γ) and let ψ = 0 in a neighborhood of the peak. We introduce the harmonic extension hi of ψ onto Ω+ mentioned in Proposition 1.3.4. By g i we denote a conjugate harmonic function which is a solution of the Neumann problem in Ω+ with the boundary data −(d/ds)ψ on Γ\{O}. Since (∂/∂s)g i = (∂/∂n)hi ∈ Lp,β+1 (Γ), it follows that g i belongs to L1p,β+1 (Γ). e − i We also introduce the harmonic extension N h onto+Ω of g given on Γ\{O} i (see Proposition 1.3.6). Since grad h = O |z| , z ∈ Ω , for an arbitrary integer N , we have grad he = O |z|−1/2 , z ∈ Ω− . Therefore, |z| −1 dψ ∂he i + log dsq + he (∞), z ∈ Γ\{O} . g (z) = 2π ds ∂n |z − q| Γ


74

We choose the conjugate harmonic function g i so that he (∞) = 0. Set σ(z) = (2π)−1 (d/ds)ψ(z) + (∂/∂n)he (z) , z ∈ Γ\{O}.

(1.174)

−1 Since V σ(z) = O |z|−β−p , z ∈ Ω+ , (see Lemma 1.4.1), we obtain V σ(z) = −g i (z) on Ω+ . Hence σ ∈ Lp,β+1 (Γ) is a solution of the equation πσ + Sσ = dψ/ds .

(1.175)

Now, let ψ be an arbitrary function in N1,− p,β (Γ). Then there exists a sequence {ψn }n≥1 of smooth functions on Γ\{O} vanishing near the peak, which tends to ψ in N1,− p,β (Γ). By σn we denote the solution of (1.175) with the right-hand side (d/ds)ψn given by (1.174). By Propositions 1.3.4 and 1.3.7, {σn } converges to a limit σ in Lp,β+1 (Γ). Since the operator S : Lp,β+1 (Γ) → Lp,β+1 (Γ) is continuous (cf. Theorem 1.2.15), we obtain by passage to the limit that σ is a solution of (1.175). Hence N1,− p,β (Γ) ⊂ πI + S Lp,β+1 (Γ) . It remains to prove the converse inclusion. Let σ belong to Lp,β+1 (Γ) and let S ∈ L1p,β+1 (Γ) be such that (d/ds)S = σ on Γ\{O}. By ψ we denote the function S(q)

ψ(z) = πS(z) − Γ

∂ 1 dsq . log ∂nq |z − q|

According to Theorem 1.2.16, ψ ∈ N1,− p,β (Γ). Since d ψ(z) = πσ(z) + ds

σ(q) Γ

we arrive at


πI + S Lp,β+1 (Γ) ⊂ N1,− p,β (Γ) .

Next we show that the kernel of the operator πI + S is nontrivial for 1/2 < β + p−1 < 1. Theorem 1.4.11. Let Ω+ have an outward peak and let the operator πI + S be defined by (1.173). Then a) ker (πI + S) = {0} for 0 < β + p−1 < 1/2; b) dim ker (πI + S) = 1 for 1/2 < β + p−1 < 1 and ∂ 1 Im , ker (πI + S) = c ∂n γ0 where c ∈ R and γ0 is the conformal mapping of Ω− onto R2+ normalized by the conditions γ0 (0) = 0 and γ0 (∞) = i.


75

Proof. Let σ belong to ker (πI + S). Since −1 |(V σ)(z)| = O |z|−β−p , z ∈ Ω+ , (see Lemma 1.4.1) and V σ is a solution of the Neumann problem with zero boundary data, it follows that V σ = C in Ω+ . We introduce the holomorphic function V (z) = V"σ(z) − i (V σ)(z) − C , z ∈ Ω− , where V"σ(z) =

σ(q) arg Γ

z dsq , z ∈ Ω− . z−q

By Lemma 1.4.1 we have |V"σ(z)| c |z|−(β+p

−1

)

.

Let ζ = γ(z) be a conformal mapping of Ω− on R2+ , γ(0) = 0. The function F (ζ) = V ◦ γ −1 (1/ζ) is holomorphic in the lower half-plane, continuous up to the boundary, and Im F = 0 on ∂R2− . The holomorphic extension Fext of F onto the upper half-plane is an entire function with the real part satisfying −1

|Re Fext (ζ)| c |ζ|2(β+p

)

.

By the Schwarz integral formula (1.166) we see that Fext has the same order of growth at infinity. Since β + p−1 < 1, it follows that Fext is a linear function. Therefore, (V σ)(z) = c0 + c1 Im (1/γ(z)), z ∈ Ω− . The jump formula (1.3) for the single layer potential implies σ(z) =

∂ 1 1 c1 Im . 2π ∂n γ(z)

By the integral representation (1.75) of the harmonic function Im (1/γ(z)) on Ω− we have |z| 1 1 1 ∂ log Im dsq − Im 2π ∂nq γ(q) |z − q| γ(∞) Γ

=

1 2π

Im Γ

1 ∂ 1 1 dsq − Im . log γ(q) ∂nq |z − q| γ(z)

Since Im (1/γ) = 0 on Γ\{O}, we obtain from the limit relation for the single layer potential that zeros of πI + S are the functions c (∂/∂n)Im (1/γ), where c ∈ R, and γ is an arbitrary conformal mapping of Ω+ onto R2+ , normalized by γ(0) = 0.


76

We note that functions of the form (∂/∂n)Im (1/γ) differ only by a positive factor. Hence dim ker (πI + S) is at most 1. Since (∂/∂n)Im (1/γ(z)) ∈ Lp,β+1 (Γ) only for β + p−1 > 1/2, we conclude that ker (πI + S) is one-dimensional for 1/2 < β + p−1 < 1 and trivial for 0 < β + p−1 < 1/2. In passing we have described ker (πI + S) in the case 1/2 < β + p−1 < 1. Now we show that the operator πI + S is not Fredholm for β + p−1 = 1/2. Proposition 1.4.12. Let Ω+ have an outward peak and let β + p−1 = 1/2. Then the operator πI + S defined by (1.173) is not Fredholm. Proof. We introduce the function −r , z ∈ Ω+ , hi (z) = Im z −1/2 log z −1 where r > 1/p. The restriction ψ of hi on Γ\{O} belongs to N1,− p,β (Γ) and has the asymptotic representation −r −r−1 1 ψ(z) = − α± xμ−1/2 log x−1 + rα± xμ−1/2 log x−1 2 −r . + O x2μ−1/2 log x−1 −r The conjugate function h i (z) = Re z −1/2 log z −1 is a solution of the Neumann problem in Ω+ with the boundary data (d/ds)ψ and has the form −r −r h i (z) = x−1/2 log x−1 , z ∈ Γ\{O} . + O x2μ−1/2 log x−1 Let ζ = γ(z) be a conformal mapping of Ω− onto R2+ , γ(0) = 0. We have γ −1 (ζ) = aγ ξ 2 + O |ξ|2+ε , where aγ ∈ R and ε is a small positive number. This equality can be differentiated at least once. We normalize γ so that aγ = 1. Making the change of variable x = Re γ −1 (ξ + i0), we obtain −r −r −r . x−1/2 log x−1 = c |ξ|−1 log |ξ|−1 + O |ξ|ε−1 log |ξ|−1 Let χ ∈ C ∞ (R) be equal to 1 for |ξ| < 1/2 and vanish for |ξ| > 1, and let −r Φ(ξ) = |ξ|−1 log |ξ|−1 χ(ξ). The function of the form 1 G1 (ξ, η) = − π

∞ −∞

Φ (τ )Im log(1 − τ /ξ) + τ /ξ dτ


77

is a harmonic extension of Φ on R2+ . Hence 1 ∂G1 (ξ, 0) = − 2 ∂η πξ

∞

Φ (τ )τ 2

−∞

dτ , ξ = 0 . ξ−τ

Similarly to Proposition 1.4.9, one proves that 1−r ∂G1 (ξ, 0) ∼ c ξ −2 log |ξ|−1 as ξ → 0, ∂η if 1/p < r < 1. Therefore, ∂G1 /∂η does not belong to Lp,1+2β+1/p (R). The trace φ1 of the harmonic function g1 = G1 ◦ γ on Γ\{O} has the asymptotic representation −r −r φ1 (z) = c1 x−1/2 log x−1 , + O x−1/2+ε log x−1 where c1 is a real number. It is clear that ∂g1 /∂n ∈ Lp,β+1 (Γ) and ∂g1 /∂n ∈ Lp,β +1 (Γ) for β > β. Setting φ2 = hi − c−1 1 φ1 on Γ\{O}, we have −r . φ2 (z) = O x−1/2+ε log x−1 According to Proposition 1.3.6, there exists a harmonic extension g2 on Ω− of φ2 satisfying ∂g2 /∂n ∈ Lp,β +1 (Γ) for some β < β. It follows that ∂g2 /∂n ∈ Lp,β+1 (Γ). The function σ = (2π)−1 (d/ds)ψ − (∂/∂n)(g1 + g2 ) ∈ Lp,β +1 (Γ) is a solution of (1.175) for β > β (see Theorem 1.4.10). Now, we assume that the equation (1.175) has a solution σ0 ∈ Lp,β+1 (Γ). By Theorem 1.4.11, 1 ∂ σ0 = σ + Im , (1.176) ∂n γ0 where γ0 is the conformal mapping Ω− onto R2+ normalized by γ0 (0) = 0 and γ0 (∞) = i. Taking into account that 1−r ∂ 1 Im ∼ c x−3/2 and σ(z) ∼ c x−3/2 log x−1 , ∂n γ we obtain that (1.176) is not possible. Thus, (1.175) is not solvable in Lp,β+1 (Γ). Since the set of smooth functions vanishing in a neighborhood of the peak is dense in N1,− p,β (Γ), it follows that the image of Lp,β+1 (Γ) under the mapping πI + S is dense in N1,− p,β (Γ). Therefore, the 1,− range of πI + S is not closed in Np,β (Γ).


78

1.4.4 Integral equations of the exterior Dirichlet and Neumann problems in a domain with inward peak Analogous results hold for boundary integral equations of the exterior Dirichlet and Neumann problems for a domain Ω+ with inward peak. The solution of the exterior Dirichlet problem with boundary data ϕ ∈ N1,− p,β (Γ) is sought in the form ∂ 1 + 1 dsq , z ∈ Ω− . log (1.177) W ext σ(z) = σ(q) ∂nq |z − q| Γ

Then the density σ satisfies the equation (πI + T ext )σ = ϕ on Γ\{O},

(1.178)

where T extσ is the value of the potential W ext σ on Γ\{O}. We represent a solution of the Neumann problem in Ω− with the boundary data ϕ ∈ Np,β (Γ) in the form of the single layer potential V σ. This leads to the equation (πI − S)σ = −ϕ on Γ\{O} (1.179) for the density σ. The proofs of the next two assertions on invertibility of the operators πI + T ext and πI − S are very much similar to those of Theorems 1.4.7 and 1.4.11. We only indicate the differences. Theorem 1.4.13. Let Ω+ have an inward peak and let 0 < β+p−1 < min{μ, 1}, β+ p−1 = 1/2. Then the operator L1p,β+1 (Γ) σ −→ (πI + T ext )σ ∈ N1,− p,β (Γ)

(1.180)

is surjective. Proof. The only change to be made in the proof of Theorem 1.4.7 is that the solution g i ∈ L1p,β+1 (Γ) of the Neumann problem in Ω+ with the boundary data (∂/∂n)he (see Proposition 1.3.7) should be chosen to satisfy g i ds = ϕds − 2πhe (∞), Γ

Γ

− where h is the harmonic extension of ϕ ∈ N1,− p,β (Γ) on Ω described in Proposition 1.3.4. The solution of (1.178) is given by e

σ = (2π)−1 (ϕ − g i ) ∈ L1p,β+1 (Γ) . +

Theorem 1.4.14. Let Ω have an inward peak and let 0 < β + p β + p−1 = 1/2. Then the operator (Lp,β+1 (Γ)) σ −→ πI − S σ ∈ Np,β (Γ) is surjective.

−1

< min{μ, 1}, (1.181)


79

Proof. The only difference with the proof of Theorem 1.4.13 consists in the follow− ing. We introduce the harmonic extension he of ψ ∈ N1,− bounded p,β (Γ) onto Ω e at infinity, which is described in Proposition 1.3.4. By g we denote the conjugate function vanishing at infinity. Let hi be the harmonic extension of g e on Ω+ mentioned in Proposition 1.3.6. Then 1 ∂hi dψ − σ= 2π ∂n ds is a solution of the equation πσ − Sσ = dψ/ds .

The next two assertions concern the kernels of the operators πI + T ext and πI − S. Their proofs duplicate those of Theorem 1.4.8 and 1.4.11 and therefore we restrict ourselves to the statements. Theorem 1.4.15. Let Ω+ have an inward peak and let the operator πI + T ext be defined by (1.180). Then a) ker (πI + T ext ) = {0} for 0 < β + p−1 < 1/2; b) dim ker (πI + T ext) = 1 for 1/2 < β + p−1 < 1 and 1 , ker (πI + T ext ) = c Re γ0 where c ∈ R and γ0 is the conformal mapping of Ω+ onto R2+ normalized by the conditions 1 γ0 (0) = 0, Re ds = 0 , Im γ0 (z0 ) = 1 , γ 0 Γ where z0 is a fixed point in Ω+ . Theorem 1.4.16. Let Ω+ have an inward peak and let the operator πI −S be defined by (1.181). Then a) ker (πI − S) = {0} for 0 < β + p−1 < 1/2; b) dim ker (πI − S) = 1 for 1/2 < β + p−1 < 1 and ∂ 1 Im ker (πI − S) = c , ∂n γ0 where γ0 is the conformal mapping of Ω+ onto R2+ normalized by the conditions γ(0) = 0 and γ(z0 ) = i with a fixed point z0 in Ω+ . The next proposition can be proved in the same way as Propositions 1.4.9 and 1.4.12. Proposition 1.4.17. Let Ω+ have an inward peak and let β + p−1 = 1/2. Then the operators πI + T ext and πI − S defined by (1.178) and (1.179) are not Fredholm.


80

1.4.5 Boundary integral equation of the Dirichlet problem in a domain with inward peak In this and further subsections of Section 1.4 we study boundary integral equations for the Dirichlet and Neumann problems in domains with inward peaks. We consider modified boundary integral equations, since the equations (1.10) and (1.12) do not have solutions in spaces of integrable functions. We look for a solution of the Dirichlet problem in the form m u(z) = W σ (z) − t(k) Ik (z),

z = x + iy ∈ Ω+ ,

k=1

where W σ is the double layer potential ∂ 1 dsq , W σ (z) = σ(q) log ∂nq |z − q| Γ

t(1) , . . . , t(m) are real numbers and Ik (z) = Im z k−1/2 . Similarly to Proposition 1.3.9, we assume that m is equal to [μ − β − 1/p]. We are looking for a function σ and a vector t = t(1) , . . . , t(m) satisfying the equation πσ − T σ +

m

t(k) Ik = −ϕ on Γ \ {O} ,

(1.182)

k=1

where (T σ)(z), as before, is the value of the potential W σ at the point z ∈ Γ\{O}. Theorem 1.4.18. Let Ω+ have an inward peak and let / N. 0 < β + p−1 < min{μ, 1}, β + p−1 = 2−1 and 2(μ − β − p−1 ) ∈ Then the operator L1p,β+1 (Γ) × Rm (σ, t) −→ πσ − T σ +

m

t(k) Ik ∈ M1,+ p,β (Γ) ,

(1.183)

k=1

where Ik (z) = Im z k−1/2 , is surjective. Proof. (i) Let ϕ ∈ N1,+ p,β (Γ) and let ϕ = 0 in a neighborhood of the peak. We consider the harmonic extension hi of ϕ onto Ω+ and its conjugate function g i , normalized by the condition g(z0 ) = 0, z0 ∈ Γ \ {O}, which were introduced in Proposition 1.3.9. By Proposition 1.3.9, (∂/∂s)g i ∈ Lp,β+1 (Γ) and there exist real numbers c(k) (ϕ), k = 1, . . . , m, such that the function g0i (z) = g i (z) −

m k=1

c(k) (ϕ)Rk (z),


81

(k) where Rk (z) = Re z k−1/2 , belongs to N1,− (ϕ), k = p,β (Γ). The coefficients c i 1, . . . , m, and the function g0 (z) satisfy m

|c(k) (ϕ)|+ g0i N1,− (Γ) c ϕ N1,+ (Γ) . p,β

k=1

The function hi0 (z)

i

= h (z) +

m

p,β

(1.184)

c(k) (ϕ)Ik (z), z ∈ Ω+ ,

k=1

is a harmonic extension of ϕ+

m

c(k) (ϕ)Ik ∈ N1,+ p,β (Γ) .

k=1

Let he0 be a harmonic extension of hi0 onto Ω− subject to grad he0 (z) = O |z|−1/2 . This relation and the estimate hi0 (z) = O |z|−1/2 , z ∈ Ω+ , show that hi0 (z)

1 = 2π

i ∂he0 |z| ∂h0 − log dsq + he0 (∞), z ∈ Γ \ {O} . ∂n ∂n |z − q|

(1.185)

Γ

According to Proposition 1.3.4, the Dirichlet problem in Ω− with the boundary data g0i has a solution f e such that (∂/∂n)f e ∈ Lp,β+1 (Γ) and satisfies (∂/∂n)f e Lp,β+1 (Γ) c g0i N1,− (Γ) p,β

(1.186)

Let g e denote the harmonic function conjugate of f e and vanishing at infinity. We have (∂/∂s)g e = (∂/∂n)f e ∈ Lp,β+1 (Γ) . Since (∂/∂n)g e = −(∂/∂s)f e = −(∂/∂s)g0i = −(∂/∂n)hi0 , on Γ \ {O}, it follows that g e belongs to L1p,β+1 (Γ) and satisfies the Neumann problem in Ω− with the boundary data −(∂/∂n)hi0 . By the integral representation (1.75) of a harmonic function in Ω− and by (1.186) we arrive at the estimate g e L1p,β+1 (Γ) c g0i N1,− (Γ) . p,β

(1.187)

Since grad g e = O |z|−μ−1/2 , z ∈ Ω− , it follows that the function w = −g e − he0 + he0 (∞) admits the representation ∂w ∂ 1 1 |z| − dsq , z ∈ Ω− . w(q) w(z) = log (q) log 2π ∂nq |z − q| ∂nq |z − q| Γ


82

By the limit relation for the double layer potential and by (1.185) we find w − π −1 T w = −2 hi0 − he0 (∞) , in Γ \ {O}. Since T 1 = −π, we obtain that the function m c(k) (ϕ)Ik ∈ L1p,β+1 (Γ) σ = −(2π)−1 g e + ϕ + k=1

satisfies πσ(z) − T σ(z) = −ϕ(z) −

m

c(k) (ϕ)Ik (z), z ∈ Γ \ {O} .

k=1

We set t(k) = c(k) (ϕ), k = 1, . . . , m. From (1.184) and (1.187) it follows that m

|t(k) |+ σ L1p,β+1 (Γ) c ϕ N1,+ (Γ) . p,β

k=1

(1.188)

(ii) Let ϕ be an arbitrary function in N1,+ p,β (Γ). There exists a sequence {ϕr }r≥1 of smooth functions on Γ \ {O} vanishing near the peak, which tends m 1 be the solution of (1.178) with the to ϕ ∈ N1,+ p,β (Γ). Let (σr , tr ) ∈ Lp,β+1 (Γ) × R right-hand side −ϕr which is constructed as in (i). According to (1.188), the sequence {(σr , tr )}r≥1 converges to a limit (σ, t) in L1p,β+1 (Γ) × Rm . Since the operator T : L1p,β+1 (Γ) → L1p,β+1 (Γ) is continuous (see Theorem 1.2.14), it follows by passage to the limit that πσ − T σ +

m

t(k) Ik = −ϕ .

(1.189)

k=1

Consequently, the equation (1.182) is solvable in L1p,β+1 (Γ) × Rm for every ϕ ∈ N1,+ p,β (Γ). (iii) Let us turn to the case ϕ(z) = Re z k , z ∈ Γ \ {O}. We take the function hi (z) = Re z k as the harmonic extension of ϕ onto Ω+ . It is clear that the conjugate harmonic function g i (z) = Im z k belongs to N1,− p,β (Γ). According to Proposition 1.3.4, g i admits a harmonic extension f e onto Ω− such that (∂/∂n)f e ∈ Lp,β+1 (Γ). Let g e be the harmonic function conjugate to f e and vanishing at infinity. We have ∂g e ∂hi =− on Γ \ {O} and g e ∈ L1p,β+1 (Γ) . ∂n ∂n


83

Using the same argument as in (i), we prove that the pair (σ, 0), where σ(z) = −(2π)−1 g e (z) + Re z k , belongs to L1p,β+1 (Γ) × Rm and satisfies (1.182). This fact and (1.189) imply m (k) L1p,β+1 (Γ) × Rm . (Γ) ⊂ πI − T + t I M1,+ k p,β k=1

The converse inclusion was proved in Theorem 1.2.18.

Next we show that the kernel of the operator (1.183) has zero dimension. Theorem 1.4.19. Let Ω+ have an inward peak. Then the operator (1.183) is injective for 0 < β + p−1 < min{μ, 1}. Proof. Let (σ, t) ∈ L1p,β+1 (Γ) × Rm be an element of m t(k) Ik . ker πI − T + k=0

Then the harmonic function (W σ)(z) +

m

t(k) Ik (z) , z ∈ Ω+ ,

k=1

vanishes on the contour Γ \ {O}. By (W σ) we denote an arbitrary function con+ jugate of W σ in Ω . We introduce the holomorphic function W (z) = −(W σ)(z) + i(W σ)(z) +

m

t(k) z k−1/2 , z ∈ Ω+ .

k=1

Let ζ = γ(z) be a conformal mapping of Ω+ onto R2+ with γ(0) = 0. The function F (z) = W ◦ γ −1 (1/ζ) is holomorphic in the lower half-plane R2− = {ζ = ξ + iη, η < 0}, continuous up to the boundary, and Im F = 0 on the real axis. We notice that the function W σ admits the estimate |(W σ)(z)| c |z|−N for an integer N . Therefore, the holomorphic extension Fext of F onto C is an entire function with the real part satisfying |Re Fext (ζ)| c |ζ|2N .


84

It follows from the Schwarz integral formula (1.166) that Fext has the same order of growth at infinity as Re Fext . In particular, there exists a polynomial P with real coefficients such that %σ)(z) + −(W

m

t(k) Rk (z) = Re P (1/γ(z)), z ∈ Ω+ .

k=1

Since %σ)(z) = (W

Γ

dσ dσ |z| (q) log dsq = V (z), z ∈ Ω+ , ds |z − q| ds

the equality m dσ (z) = Re P (1/γ(z)) − t(k) Rk (z), z ∈ Γ \ {O} , V ds

(1.190)

k=1

holds. Taking into account that V (dσ/ds) ∈ N1,− p,β (Γ) (see Theorem 1.2.13), we obtain that the right-hand side in (1.190) belongs to N1,− p,β (Γ). However, since m − 1/2 + β − μ < −1/p, the functions Rk , k = 1, . . . , m, and positive integer (k) powers of 1/γ do not belong to N1,− , p,β (Γ). Hence, it follows that the coefficients t k = 1, . . . , m, are equal to zero and the polynomial P is a constant. Therefore, %σ and W σ are constants in Ω+ . W − By W − σ we denote the harmonic function conjugate of (W σ)(z), z ∈ Ω , %σ on Γ \ {O}. Since W which equals W − σ admits the estimate −N as z → 0 |(W − σ)(z)| c |z| − for an integer N and is constant on Γ \ {O}, it follows that W − σ = const in Ω . − Therefore, W σ is constant in Ω . From the jump formula (1.6) for W σ we obtain that σ = const on Γ \ {O}. Since a nonzero constant does not satisfy the homogeneous equation (1.182), we conclude that σ equals zero.

We show in the next assertion that the range of the operator (1.183) is not −1 closed in M1,+ ) ∈ N. p,β (Γ) if 2(μ − β − p Proposition 1.4.20. Let Ω+ have an inward peak, and let 0 < β + p−1 < min{μ, 1}, β + p−1 = 2−1 , 2(μ − β − p−1 ) ∈ N. Then the operator (1.183) is not Fredholm. Proof. Let

−γ Φ(ξ) = |ξ|n0 −1 − log |ξ|

in a small neighborhood of the origin and let supp Φ be in the domain of the mapping θ0 introduced in (1.121). Further, let γ be such that 1/p < γ < 1. By ϕ we denote the function Φ ◦ θ0−1 ∈ N1,+ p,β (Γ) and introduce the harmonic extension


85

hi of ϕ onto Ω+ constructed in Proposition 1.3.9. Let g i be the harmonic function conjugate of hi from Proposition 1.3.9. We have g i (z) =

m

c(k) Rk (z) + g # (z).

k=1

Here

g # (z) = c Re z n0 /2−1/2 (log z)−γ+1 + g0# (z),

e # − where g0# ∈ N1,− p,β (Γ). The harmonic extension f of g to Ω described in Proposition 1.3.4 has the form zz −μ+n0 /2−1/2 −γ+1 0 f e (z) = c1 Im log z z0 − z zz n0 /2−1/2 −γ+1 0 + f0e (z), + c2 Re log z z0 − z

where c1 , c2 ∈ R, z0 is a fixed point of Ω+ and ∂f0e /∂n ∈ Lp,β+1 (Γ). The function g e , g e (∞) = 0, conjugate of f e , has the representation −1

g e (z) ∼ c x−β−p

−γ+1 log x .

It is clear that g e ∈ / L1p,β+1 (Γ) and g e ∈ L1p,β +1 (Γ) for β > β. By Theorem 1.4.18, the pair (σ, t), where t = (c(1) , . . . , c(m) ) and m c(k) Ik on Γ \ {O}, σ = −(2π)−1 g e + ϕ + k=1

is the solution of (1.182) in L1p,β +1 (Γ) × Rm for β > β. From Theorem 1.4.19 it follows that (1.182) is not solvable in L1p,β+1 (Γ) × Rm . According to Theorem 1.4.18, the equation (1.182) with the right-hand side in m 1 N1,+ p,β (Γ), β < β, is solvable in Lp,β +1 (Γ) × R . Since the set of smooth functions m 1 vanishing near the peak is dense in N1,+ is embedded p,β (Γ) and Lp,β +1 (Γ) × R m to L1p,β+1 (Γ) × R , it follows that the range of operator (1.183) is not closed in M1,+ p,β (Γ).

1.4.6 Boundary integral equation of the Neumann problem in a domain with inward peak Let us look for a solution of the Neumann problem (1.8) in the form m u(z) = V σ (z) + t(k) Rk (z), z ∈ Ω+ , k=1


86

where V σ is defined by (1.73) and Rk (z) = Re z k−1/2 . As in Proposition 1.3.9, we put m = [μ − β − 1/p]. Then the function σ and the vector t = t(1) , . . . , t(m) satisfy the equation πσ + Sσ +

m

t(k)

k=1

where

σ(q)

(Sσ)(z) = Γ

∂ Rk = ϕ, on Γ \ {O} , ∂n

(1.191)

∂ |z| dsq , z ∈ Γ \ {O}. log ∂nz |z − q|

In this section we prove the unique solvability of (1.191) on a contour Γ with inward peak. Theorem 1.4.21. Let Ω+ have an inward peak and let / N. 0 < β + p−1 < min{μ, 1}, β + p−1 = 2−1 and 2(μ − β − p−1 ) ∈ Then the operator

Lp,β+1 (Γ) × Rm (σ, t) −→ πσ + Sσ +

m k=1

t(k)

∂ Rk ∈ Mp,β (Γ) ∂n

(1.192)

with Rk (z) = Re z k−1/2 is surjective. i Proof. (i) Let ψ ∈ N1,+ p,β (Γ) and let ψ = 0 in a neighborhood of the peak. By h we + denote the harmonic extension of ψ onto Ω which is introduced in Proposition 1.3.9. Let g i be the function conjugate of hi and normalized by the condition g i (z0 ) = 0 with z0 ∈ Γ \ {O}. By Proposition 1.3.9, g i belongs to L1p,β+1 (Γ) and there exist real numbers c(k) (ψ), k = 1, . . . , m, such that

g i (z) = −

m

c(k) (ψ)Rk (z) + g0i (z), z ∈ Ω+ .

k=1

These coefficients c(k) (ψ) and the function g0i satisfy m

|c(k) (ψ)|+ g0i N1,− (Γ) c ψ N1,+ (Γ) . p,β

k=1

p,β

(1.193)

Since ∂g i /∂n = −∂hi /∂s = −dψ/ds , the function −g0i satisfies the Neumann problem in Ω+ with the boundary data dψ (k) ∂ − Rk ∈ Mp,β (Γ) . c ds ∂n m

k=1


87

By Proposition 1.3.4, the Dirichlet problem in Ω− with the boundary data −g0i has a solution he such that (∂/∂n)he ∈ Lp,β+1 (Γ) and satisfies (∂/∂n)he Lp,β+1 (Γ) c g0i N1,− (Γ) . p,β

(1.194)

From the equality ∂he 1 |z0 | 1 ∂ 1 e dsq − dsq , (q) log he (q) log h (∞) = 2π ∂nq |z0 − q| 2π ∂nq |z0 − q| Γ

Γ

+

where z0 is a fixed point in Ω , and from (1.194) we obtain that the linear funci tional g i → he (∞) is continuous in N1,− p,β (Γ). Therefore, we can choose g so that e i h (∞) = 0 and the inequality (1.194) remains valid. Since grad g0 = O |z|−1/2 and grad he = O |z|−μ−1/2 , it follows that i ∂he |z| ∂g0 1 (q) + (q) log dsq . g0i (z) = 2π ∂n ∂n |z − q| Γ

Set m 1 d ∂ ∂ e ψ(z) − h (z) , z ∈ Γ \ {0} . (1.195) c(k) (ψ) Rk (z) − 2π ds ∂n ∂n k=1 Taking into account that V σ(z) = O |z|−(β+1/p) , z = 0, as well as that the functions g0i (z), z ∈ Ω+ , and he (z), z ∈ Ω− ,

σ(z) =

are bounded, we have g0i (z) + V σ(z) = c Im (1/γ(z)), z ∈ Ω+ , and

he (z) + V σ(z) = 0, z ∈ Ω− ,

where γ(z) is a conformal mapping of Ω+ onto R2+ with γ(0) = 0. From the jump formula (1.3) for (∂/∂n)V σ we obtain c

∂ 1 Im = 0, z ∈ Γ \ {0}. ∂n γ(z)

Since (∂/∂n)Im (1/γ(z)) ∼ (1/2)x−3/2 as x → 0, we have c = 0. Thus, V σ = −g0i on Ω+ . The limit relation for the normal derivative of the single layer potential implies d ∂ ψ(z) − c(k) (ψ) Rk (z), z ∈ Γ \ {0}. ds ∂n m

πσ(z) + (Sσ)(z) =

k=1

We set t(k) = c(k) (ψ), k = 1, . . . , m. Hence it follows that the pair (σ, t), where t = (t(1) , . . . , t(m) ) and σ is defined by (1.195), belongs to Lp,β+1 (Γ) × Rm and


88

satisfies (1.191). From (1.193) and (1.194) we conclude that σ Lp,β+1 (Γ) +

m

|c(k) (ψ)| ψ N1,+ (Γ) . p,β

k=1

(1.196)

Now let ψ be an arbitrary function in N1,+ p,β (Γ). There exits a sequence {ψr }r≥1 of smooth functions on Γ \ {0}, which vanishes in a neighborhood of the peak and m tends to ψ ∈ N1,+ be the solution of (1.191) p,β (Γ). Let (σr , tr ) ∈ Lp,β+1 (Γ) × R with the right-hand side (d/ds)ψr . According to (1.196), the sequence {(σr , tr )}r≥1 converges in Lp,β+1 (Γ)× Rm to a limit (σ, t). Since the operator S : Lp,β+1 (Γ) −→ Lp,β+1 (Γ) is continuous (see Theorem 1.2.15), it follows by passage to the limit that m

πσ + Sσ +

k=1

t(k)

∂ d Rk = ψ. ∂n ds

(1.197)

Consequently, the equation (1.191) is solvable in Lp,β+1 (Γ) × Rm for every ψ ∈ N1,+ p,β (Γ). (ii) Let us turn to the case ψ(z) = Re z k , z ∈ Γ \ {0}. We take the function h (z) = Re z k as a harmonic extension of ψ onto Ω+ . The conjugate function i g i (z) = −Im z k belongs to N1,− p,β (Γ). By Proposition 1.3.4, the function −g has the harmonic extension he on Ω− such that (∂/∂n)he ∈ Lp,β+1 (Γ). Set i

σ(z) =

∂ e 1 d ψ(z) − h (z) , z ∈ Γ \ {O}. 2π ds ∂n

Then the pair (σ, 0) ∈ Lp,β+1 (Γ) × Rm is a solution of (1.191) with the right-hand side (d/ds)ψ (see (i)). This and (1.197) imply m ∂ Mp,β (Γ) ⊂ πI + S + t(k) Rk Lp,β+1 (Γ) × Rm . ∂n k=1

(iii) It remains to prove the converse embedding. Clearly, m

t(k)

k=1

∂ Rk ⊂ Mp,β (Γ) ∂n

for any t ∈ Rm . Now, let σ belong to Lp,β+1 (Γ) and let a function S ∈ L1p,β+1 (Γ) be defined by (d/ds)S = σ on Γ \ {O}. By ψ we denote the function ψ(z) = πS(z) −

S(q) Γ

∂ |z| dsq . log ∂nq |z − q|


89

It follows from Theorem 1.2.18 that ψ ∈ M1,+ p,β (Γ). Since d ψ(z) = πσ(z) + ds

σ(q) Γ


we obtain that the image of Lp,β+1 (Γ) under the mapping (1.192) is the space Mp,β (Γ). Next we show that the homogeneous equation (1.191) has only the trivial solution in the space Lp,β+1 (Γ) × Rm . Theorem 1.4.22. Let Ω+ have an inward peak. Then the operator (1.192) is injective provided 0 < β + p−1 < min{μ, 1}. Proof. Let (σ, t) be an element of Lp,β+1 (Γ) × Rm , where t = (t(1) , . . . , t(m) ), and let (σ, t) belong to m ∂ ker πI + S + (1.198) t(k) Rk . ∂n k=1

Then the harmonic function v(z) = V σ(z) +

m

t(k) Rk (z), z ∈ Ω− ,

k=1

has zero Neumann boundary data on Γ \ {O}. Since |v(z)| |z|−β−1/p

(1.199)

(see Lemma 1.4.1), we obtain by the integral representation (1.75) for the harmonic function v(z) and the limit relation (1.5) for the double layer potential that ∂ 1 πv(z) + v(q) dsq = 0, z ∈ Γ \ {O}. log ∂nq |z − q| Γ

Thus v is a solution of the homogeneous integral equation of the Dirichlet problem in Ω− . The double layer potential (W v)(z), z ∈ Ω− , grows not faster than a power function as z → 0. Since the limit values of W v vanish on Γ \ {O}, it follows that %v is constant (W v)(z) = 0, z ∈ Ω− . Therefore, an arbitrary conjugate function W − % in Ω . We set W v = C. Let W+ v be defined by W+ v = W v in Ω+ and let W + v be a conjugate function such that W v = C on Γ \ {O}. We introduce the holomorphic function + + W (z) = (W+ v)(z) + i(W + v − C), z ∈ Ω .


90

Let ζ = γ(z) be a conformal mapping of Ω+ onto R2+ , γ(0) = 0. The function F (ζ) = (W ◦ γ −1 )(1/ζ) is holomorphic in the lower half-plane R2− , continuous up to the boundary, and Im F = 0 on ∂R2− . The holomorphic extension Fext of F to C is the entire function, which does not grow faster than a power function as ζ → ∞. Hence it follows that W (z) = P (1/γ(z)), where P is a polynomial with real coefficients. Therefore, (W+ v)(z) =

c(k) Re (1/γ(z))−k , z ∈ Ω+ .

k=0

From the jump formula (1.6) for W v we obtain v(z) = −(2π)−1

c(k) Re (1/γ(z))−k , z ∈ Γ \ {O}.

k=0

By (1.199) we have v(z) = −(2π)−1 c(0) + c(1) Re (1/γ(z)) on Γ \ {O}. Therefore, (V σ)(z)+

m

t(k) Rk (z)

k=1

= − (2π)−1 c(0) − (2π)−1 c(1) Re (1/γ(z)) + c Im (1/γ(z)), z ∈ Ω+ . By hek , k = 1, . . . , m and he0 we denote harmonic extensions of Rk and Re (1/γ) onto Ω− which do not grow faster than a power function as z → 0. Since Vσ+

m

t(k) hek + (2π)−1 c(1) he0 + (2π)−1 c(0)

k=1

vanishes on Γ \ {O}, we have (V σ)(z) = −

m

t(k) hek (z) − (2π)−1 c(1) he0 (z) − (2π)−1 c(0) , z ∈ Ω− .

k=1

From the jump formula (1.3) for the normal derivative of V σ it follows that 2πσ(z) = − +

m k=1 m k=1

t(k)

∂ 1 ∂ Rk (z) + c Im ∂n ∂n γ(z) (1.200)

t

(k)

∂ e ∂ h (z) + (2π)−1 c(1) he0 (z), z ∈ Γ \ {O}, ∂n k ∂n


91

where ∂ Rk (z) ∼ ak α± |z|k+μ−3/2 , ∂n 1 ∂ Im ∼ a0 |z|−3/2 , ∂n γ(z) ∂ e h (z) ∼ ±bk |z|k−μ−3/2 , ∂n k

k = 1, . . . , m,

k = 0, 1, . . . , m .

Here ak and bk , k = 0, 1, . . . , m, are real coefficients. Since m β − p−1 , we have k − μ − 1/2 + β < p−1 for k = 0, 1, . . . , m . This means that the function (∂/∂n)hek does not belong to Lp,β+1 (Γ) for k = 0, 1, . . . , m. Therefore, the coefficients t(1) , . . . , t(m) and c(1) are equal to zero. It follows from (1.198) and (1.200) that 1 ∂ σ ∈ ker πI + S and σ = (2π)−1 c Im , on Γ \ {O}. ∂n γ

(1.201)

By the integral representation (1.74) for the harmonic function Im (1/γ) on Ω+ we have ∂ |z| 1 1 log dsq Im 2π ∂nq γ(q) |z − q| Γ 1 1 ∂ |z| 1 dsq + Im = Im log 2π γ(q) ∂nq |z − q| γ(z) Γ

for z ∈ Ω+ . Since Im (1/γ(z)) = 0 on Γ \ {O}, we obtain from the limit relation (1.4) for the single layer potential that πσ −Sσ = 0 on Γ\{O}. However, according to (1.201), we have πσ + Sσ = 0 on Γ \ {O}. Hence we conclude that σ = 0. Now we show that the operator (1.192) fails to be Fredholm if one of the conditions in Theorem 1.4.21 is violated, namely if (μ − β − p−1 ) ∈ / N. Proposition 1.4.23. Let Ω+ have an inward peak, and let 0 < β + p−1 < min{μ, 1}, β + p−1 = 2−1 and 2(μ − β − p−1 ) ∈ N. Then the operator (1.192) is not Fredholm. Proof. Let

−γ Ψ(ξ) = |ξ|n0 −1 − log |ξ|

in a small neighborhood of the origin and let suppΨ be in the domain of the mapping θ0 introduced in (1.121). We assume that 1/p < γ < 1 and set ψ = Ψ ◦


92

i + θ0−1 ∈ N1,+ p,β (Γ). By h we denote the harmonic extension of ψ onto Ω constructed i in Proposition 1.3.9. Let g be the conjugate function from Proposition 1.3.9. We have m g i (z) = c(k) Rk (z) + g # (z). k=1

Here

g # (z) = c Re z n0 /2−1/2 (log z)−γ+1 + g0# (z),

where g0# ∈ N1,− p,β (Γ).

By he we denote the harmonic extension of −g # onto Ω− from Proposition 1.3.4. We have zz −μ+n0 /2−1/2 −γ+1 0 log z z0 − z zz n0 /2−1/2 −γ+1 0 + he0 (z), + c2 Re log z z0 − z

he (z) = c1 Im

where z0 is a fixed point of Ω+ and ∂he0 /∂n ∈ Lp,β+1 (Γ). Since −γ−1 −1 ∂ e h (z) ∼ c x−β−p −1 log x , ∂n / Lp,β+1 (Γ) and ∂he /∂n ∈ Lp,β +1 (Γ) for β > β. it follows that ∂he /∂n ∈ By Theorem 1.4.21, the pair (σ, t), where t = (c(1) , . . . , c(m ) and m c(k) ∂Rk /∂n − ∂he ∂n on Γ \ {O}, σ = (2π)−1 dψ/ds − k=1

belongs to Lp,β +1 (Γ)×Rm for β > β and satisfies (1.197). It results from Theorem 1.4.22 that the same equation is not solvable in Lp,β+1 (Γ) × Rm . The equation (1.197) is solvable in Lp,β +1 (Γ) × Rm with β < β. Since the set of smooth functions vanishing near the peak is dense in N1,+ p,β (Γ) and since Lp,β +1 (Γ) × Rm is embedded to Lp,β+1 (Γ) × Rm , we obtain that the range of the operator (1.192) is not closed in Mp,β (Γ).

1.4.7 Integral equations of the exterior Dirichlet and Neumann problems for domain with outward peak Now we briefly discuss the integral equations mentioned in the title of this subsection. The proofs are similar to those of the corresponding results relating the interior problems for a domain with inward peak, which were proved in Section 1.4.4 and 1.4.5.


Ω

−

93

Let Ω+ have an outward peak. The solution of the Dirichlet problem (1.7) in is sought in the form m u(z) = W ext σ (z) + t(k) Ikext (z), z ∈ Ω− . k=1

Here

ext W σ (z) =

Γ

and Ikext (z) = Re

∂ 1 + 1 dsq σ(q) log ∂nq |z − q| zz k−1/2 0 , z ∈ Ω− , z0 − z

where z0 is a fixed point in Ω+ . The density σ and the vector t(1) , . . . , t(m) satisfy the equation m ext πσ + T σ + t(k) Ikext = ϕ on z ∈ Γ \ {O}, (1.202) k=1

where (T

ext

σ)(z) is the value of the potential W ext σ at a point z ∈ Γ \ {O}.

Theorem 1.4.24. Let Ω+ have an outward peak and let / N. 0 < β + p−1 < min{μ, 1}, β + p−1 = 2−1 and 2(μ − β − p−1 ) ∈ Then the operator L1p,β+1 (Γ) × Rm (σ, t) −→ πσ + T extσ +

m k=1

t(k) Ikext ∈ M1,+ p,β (Γ)

(1.203)

is surjective. − Proof. Let he be the harmonic extension of ϕ ∈ N1,+ constructed in p,β (Γ) on Ω Proposition 1.3.9, and let g e be the conjugate function vanishing at a fixed point on Γ \ {O}. By Proposition 1.3.9, there exist real numbers c(k) , k = 1, . . . , m, such that m e ge = c(k) Rext k + g0 , k=1

where g0e ∈

N1,− p,β (Γ)

and Rext k (z) = Re

We set he0 = he +

zz k−1/2 0 . z0 − z

m k=1

t(k) Ikext .


94

The only change to be made in the proof of Theorem 1.4.18 is that the solution g i of the Neumann problem on Ω+ with the boundary data ∂he0 /∂n should be chosen to satisfy he0 ds − 2he0 (∞) .

g i ds = Γ

Γ (1)

Then the pair (σ, t), where t = (c , . . . , c(m) ) and m σ = (2π)−1 ϕ + c(k) Ikext − g i , k=1

m

× R of the equation (1.202). The case ϕ ∈ P(Γ) is is a solution in considered as in Theorem 1.4.18. L1p,β+1 (Γ)

We represent the solution of the Neumann problem (1.8) in the form u(z) = (V σ)(z) −

m

− t(k) Rext k (z), z ∈ Ω ,

k=1

where V σ is the single layer potential. The density σ and the vector t = t(1) , . . . , t(m) satisfy πσ − Sσ +

m k=1

t(k)

∂ ext R = −ϕ on Γ \ {O} . ∂n k

(1.204)

Theorem 1.4.25. Let Ω+ have an outward peak and let 0 < β + p−1 < min{μ, 1}, β + p−1 = 2−1 and 2(μ − β − p−1 ) ∈ / N. Then the operator Lp,β+1 (Γ) × Rm (σ, t) −→ πσ − Sσ +

m k=1

t(k)

∂ ext R ∈ Mp,β (Γ) (1.205) ∂n k

is surjective. − Proof. Let he be a harmonic extension of ψ ∈ N1,+ and let g e be a p,β (Γ) onto Ω conjugate function constructed in Proposition 1.3.9. Then there exist real numbers c(k) such that m e ge = c(k) Rext k + g0 , k=1

N1,− p,β (Γ).

where ∈ We choose g0e to satisfy g0e (∞) = 0. Here the function g e plays the same role as g i in the proof of Theorem 1.4.21. Now we use the same argument as in Theorem 1.4.21. By hi we denote a harmonic extension of g0e onto Ω+ such that (∂/∂n)hi ∈ Lp,β+1 (Γ) (see Proposition 1.3.4). Then the pair (σ, t), where t = (c(1) , . . . , c(m) ) and m σ = (2π)−1 (∂/∂n)hi − (d/ds)ψ + ∈ Lp,β+1 (Γ) , c(k) (∂/∂n)Rext k g0e

k=1

solves the equation (1.204).

1.5. Direct method of integral equations

95

The case ϕ ∈ P(Γ) is considered in the same way as in Theorem 1.4.21; one should only replace g i , hi , and he by g e , he , and hi respectively. The following two theorems can be proved in the same way as Theorems 1.4.19 and 1.4.22. Theorem 1.4.26. Let Ω+ have an outward peak. Then the operator (1.203) is injective for 0 < β + p−1 < min{μ, 1}. Theorem 1.4.27. Let Ω+ have an outward peak. Then the operator (1.205) is injective for 0 < β + p−1 < min{μ, 1}. The proof of the following Proposition is essentially the same as those of Propositions 1.4.20 (the case of operator (1.203)) and 1.4.23 (the case of operator (1.205)). Proposition 1.4.28. Let Ω+ have an outward peak, and let 0 < β + p−1 < min{μ, 1}, β + p−1 = 2−1 and 2(μ − β − p−1 ) ∈ N. Then the operators (1.203) and (1.205) are not Fredholm.

1.5 Direct method of integral equations of the Neumann and Dirichlet problems Here we consider the so-called direct method when solutions of integral equations are represented explicitly by the solutions of the Dirichlet and Neumann boundary value problems (see details in [16]). In this case integral equations can be obtained from the integral representation for a harmonic function 1 ∂u 1 u(z) = V (z) − W u (z), z ∈ Ω+ , 2π ∂n 2π where V is the single layer potential (1.73) and W is the double layer potential (1.2). Using the continuity of the single layer potential and the limit relation (1.5) for the double layer potential, we obtain ∂u πu(z) = V (z) − T u (z), z ∈ Γ \ {O} , (1.206) ∂n where T u (z) is the direct value of W u at the point z on Γ \ {O}. Clearly, (1.206) can be interpreted as either an integral equation of the first kind for the normal derivative with prescribed values of u on Γ \ {O} or an integral equation of the second kind for u|Γ\{O} with given ∂u/∂n. We study both equations. Let us introduce the double layer potential in Ω− as in (1.177): ∂ ext 1 W σ (z) = σ(q) log + 1 dsq , z ∈ Ω− . ∂nq |z − q| Γ

The value of this potential at the point z ∈ Γ \ {O} will be denoted by T extσ (z).


96

We start by proving the existence and uniqueness of a solution to the equation (1.207) at the boundary of a domain with outward peak. Theorem 1.5.1. Let Ω+ have an outward peak, and let 0 < β + p−1 < min{μ, 1}. Then, for any ψ ∈ Np,β (Γ), the integral equation πσ + T ext σ = V ψ

(1.207)

has a unique solution σ in N1,− p,β (Γ), satisfying σ ds = 0. Γ

Proof. Let ψ belong to C0∞ (Γ \ {O}). By h we denote a solution of the Neumann problem in Ω+ with the boundary data ψ as in Proposition 1.3.5. From the integral representation of the harmonic function h in Ω+ (1.74) and the limit relation (1.5) for the double layer potential we obtain 1 ∂ 1 1 |z| dsq = ψ(q)dsq . h(z) + h(q) log log (1.208) π ∂nq |z − q| π |z − q| Γ

Γ

We choose h so that

h(q)dsq = 0 . Γ

According to Proposition 1.3.5, h belongs to N1,− p,β (Γ). From (1.208) it follows that σ = h ∈ N1,− p,β (Γ) is a solution of (1.207). Now, let ψ be an arbitrary function in Np,β (Γ). For β > −p−1 there exists a sequence {ψn }n≥1 of smooth functions on Γ \ {O} vanishing near the peak and approximating ψ in Np,β (Γ). By σn we denote the constructed solution of (1.207) with ψn in the right-hand side which is unique by Theorem 1.4.26. Since the operator Lp,β+1 (Γ) ψ − −→ V ψ ∈ N1,− p,β (Γ) is continuous (see Theorem 1.2.13), it follows that {V ψn } converges to a limit V ψ in L1p,β+1 (Γ). According to Proposition 1.3.5, the sequence {σn } converges to a limit σ in L1p,β+1 (Γ). Since the operator W

1 L1p,β+1 (Γ) σ − −→ (πI + T ext)σ ∈ M1,+ p,β (Γ) ⊂ Lp,β+1 (Γ)

is continuous (see Corollary 1.2.19), we obtain that σ is a solution of (1.207) by passing to the limit. From Theorem 1.4.26 it follows that the kernel of W in L1p,β+1 (Γ) is trivial. Therefore, the equation (1.207) is uniquely solvable in N1,− p,β (Γ). It results from Proposition 1.5.2 that Theorem 1.5.1 is not a corollary of theorems in the preceding subsections.


97

Proposition 1.5.2. Under the assumptions of Theorem 1.5.1 with / N, β + p−1 = 2−1 , and ψ ∈ Np,β (Γ) 2(μ − β − p−1 ) ∈ 1,+ the function V ψ belongs to N1,− p,β (Γ) ∩ Mp,β (Γ) and satisfies the orthogonality conditions ∂ 1 (V ψ)Re 2k−1 ds = 0, k = 1, . . . , m , ∂s ζ Γ

where ζ is the conformal mapping of Ω− onto R2+ subject to ζ(0) = 0, Re ζ(∞) = 0 and Re (1/ζ(z)) = ±x−1/2 + O(1). Proof. By Theorem 1.2.13 we have V ψ ∈ N1,− p,β (Γ), because ψ ∈ Lp,β+1 (Γ), Therefore, Vψ =ϕ+

and V ψ ∈ M1,+ p,β (Γ). m

d(k) Re z k ,

k=0

N1,+ p,β (Γ). According −

where ϕ ∈ tension of ϕ onto Ω the representation

to Proposition 1.3.9, there exists a harmonic exsuch that the conjugate function g satisfying g(∞) = 0 has m

c(k) (ϕ)Re z k−1/2 + g # (z),

k=1 1,− # where c (ϕ) are linear continuous functionals in N1,+ p,β (Γ) and g ∈ Np,β (Γ). We 1−2k apply the Green formula to the harmonic functions g and Re ζ in Ω− ∩ {|z| < ε}. Passing to the limit as ε → 0, we obtain ∂g 1 1 Re 2k−1 ds = ck ϕs Re 2k−1 ds. c(k) (ϕ) = ck ∂n ζ ζ (k)

Γ

Let (σ, t) ∈

L1p,β+1 (Γ)

×R

m

Γ

be a solution of the equation

(πI + T ext )σ +

m

t(k) Ik = V ψ

k=1

with the right-hand side from M1,+ p,β (Γ). As it is shown in Theorem 1.4.24, the components t(k) are equal to c(k) (ϕ), k = 1, . . . , m. Since the operator W is surjective (see Theorem 1.4.24) and since 1 ∂ Im z k Re 2k−1 ds = 0 , ∂n ζ Γ

it follows that V ψ ∈

N1,− p,β (Γ)

(k) ∩ M1,+ (V ψ) = 0, k = 1, . . . , m. p,β (Γ) and c


98

The next assertion concerns the equation (1.209) on a contour with inward peak. Theorem 1.5.3. Let Ω+ have an inward peak, and let 0 < β + p−1 < min{μ, 1} and β + p−1 = 1/2. Then, for any ψ ∈ L1p,β+1 (Γ), the boundary integral equation πσ + T extσ = V ψ

(1.209)

has a solution σ ∈ L1p,β+1 (Γ) satisfying σ ds = 0.

(1.210)

Γ

The homogeneous equation (1.209) has only the trivial solution for 0 < β + p−1 < 1/2 and a one-dimensional space of solutions for 1/2 < β + p−1 < 1 given by t Re

1 , γ0

where t ∈ R and γ0 is the conformal mapping of Ω+ onto R2+ subject to γ0 (0) = 0,

Re Γ

1 ds = 0 and Im γ0 (z0 ) = 1 γ0

with a fixed point z0 ∈ Ω+ . Proof. Let ψ ∈ L1p,β+1 (Γ). Then V ψ ∈ N1,− p,β (Γ) (see Theorem 1.2.13). According to Theorem 1.4.13, the equation (1.209) is solvable in L1p,β+1 (Γ). Since the harmonic extension of V ψ vanishes at infinity, we have (1.210). By Theorem 1.4.15, the set of solutions to the homogeneous equation (1.209) in L1p,β+1 (Γ) is onedimensional for 1/2 < β + p−1 < 1 and trivial for 0 < β + p−1 < 1/2. The set of solutions to the homogeneous equation (1.209) is described in Theorem 1.4.15. We conclude this subsection with a study of equation (1.211) on contours with inward and outward peak. Theorem 1.5.4. Let Ω+ have an outward peak and let 0 < β + p−1 < min{μ, 1}. Then the boundary integral equation V γ = πϕ + T ϕ

(1.211)

has a solution γ ∈ Np,β (Γ) for every ϕ ∈ N1,− p,β (Γ). This solution is unique for β + p−1 = 2−1 . Proof. Let ϕ ∈ C0∞ (Γ \ {O}). By u we denote the bounded harmonic extension of ϕ onto Ω+ constructed in Proposition 1.3.4. Owing to the integral representation


99

(1.74) of the harmonic function u on Ω+ and the limit relation (1.4) for the single layer potential, we obtain |z| ∂u ∂ 1 (q)dsq = πϕ(z) + ϕ(q) dsq log log (1.212) |z − q| ∂n ∂nq |z − q| Γ

Γ

for z ∈ Γ \ {O} Since ∂u/∂n belongs to Np,β (Γ), it follows from (1.212) that γ = ∂u/∂n satisfies (1.211). Now, let ϕ be an arbitrary function in N1,− p,β (Γ). There exists a sequence {ϕn }n≥1 of smooth functions on Γ \ {O} vanishing near the peak and converging to ϕ in N1,− p,β (Γ). By γn we denote the constructed solution of (1.211) with ϕn instead of ϕ in the right-hand side. Since the operator (πI + T ) : L1p,β+1 (Γ) → M1,+ p,β (Γ) is continuous (see Theorem 1.4.24), we obtain by passing to the limit that {πϕn + T ϕn }n≥1 converges in L1p,β+1 (Γ). In view of Proposition 1.3.4, the sequence {∂un/∂n}, where un is the bounded extension of ϕn onto Ω+ , converges in Lp,β+1 (Γ). According to Theorem 1.2.13, the operator V

−→ V γ ∈ N1,− Lp,β+1 (Γ) γ − p,β (Γ) is continuous. Then, passing to the limit in the equation V γn = πϕn + T ϕn , we obtain that γ is a solution of (1.211). For Ω+ with an outward peak Theorem 1.5.1 implies that ker V is trivial provided β + p−1 = 2−1 . Therefore, the solution of (1.211) just obtained is unique. Theorem 1.5.5. Let Ω+ have an inward peak, and let 0 < β + p−1 < min{μ, 1}, β + p−1 = 1/2. Then the integral equation V γ = πϕ + T ϕ

(1.213)

has a solution γ ∈ Lp,β+1 (Γ) for every ϕ ∈ L1p,β+1 (Γ). This solution is unique for 0 < β + p−1 < 1/2 and the homogeneous equation (1.213) has a one-dimensional space of solutions for 1/2 < β + p−1 < 1 given by ∂ 1 Im (in) , t ∂n γ where t ∈ R and γ (in) is the conformal mapping of Ω+ onto R2+ subject to the conditions γ (in) (0) = 0 and γ (in) (z0 ) = i with a fixed point z0 ∈ Ω+ . Proof. Let ϕ ∈ L1p,β+1 (Γ). According to Corollary 1.2.17, the function (πI + T )ϕ − belongs to N1,− p,β (Γ) and its harmonic extension onto Ω , ∂ 1 dsq , z ∈ Ω− , W σ (z) = σ(q) log ∂nq |z − q| Γ

100


vanishes at infinity. The range of the operator Lp,β+1 (Γ) γ − −→ V γ ∈ N1,− p,β (Γ) − consists of the elements of N1,− p,β (Γ) with harmonic extensions to Ω vanishing at infinity (see Theorem 1.4.3). Therefore, (1.213) has a solution in Lp,β+1 (Γ). The homogeneous equation (1.213) has only the zero solution in Lp,β+1 (Γ) for 0 < β + p−1 < 1/2 and the one-dimensional space of solutions for 1/2 < β + p−1 < 1. The set of solutions to the homogeneous equation (1.213) is described in Proposition 1.4.5.

Chapter 2

Boundary Integral Equations in Hölder Spaces on a Contour with Peak In this chapter we are concerned with the solvability in H¨ older spaces and description of kernels of boundary integral equations of the Dirichlet problem Δu = 0 in Ω+ , u|Γ = ϕ,

(2.1)

Δu = 0 in Ω+ , (∂u/∂n)|Γ = ψ,

(2.2)

and the Neumann problem

in a plane bounded simply connected domain Ω+ with a peak at the boundary Γ. Here and elsewhere we assume the normal n to be outward. Another assumption is that the vertex of the peak is placed at the origin. A possible approach to solution of the Dirichlet problem (2.1) is to represent u in the form u(z) = V σ(z) + C, z ∈ Ω+ , where C ∈ R and V is the single layer potential |z| dsq . V σ(z) = σ(q) log |z − q| Γ The density σ and the constant C are found by the equation V σ + C = ϕ on Γ .

(2.3)

Another approach to solution of the same Dirichlet problem is to look for u as the double layer potential u(z) = W σ (z) , z = x + iy ∈ Ω+ ,

102

Chapter 2. Boundary Integral Equations in H¨ older Spaces

if Ω+ has an outward peak, and in the form u(z) = W σ (z) + t I(z) , if Ω+ has an inward peak, where I(z) = Im z 1/2 and t is a real number. Here, as in Chapter 1, ∂ 1 dsq , z ∈ W σ (z) = σ(q) log / Γ. ∂nq |z − q| Γ

In the case of an outward peak, σ is found from the equation πσ − T σ = −ϕ on Γ \ {O},

(2.4)

where (T σ)(z) is the value of (W σ)(z) at the boundary point z. In the case of an inward peak the pair (σ, t) satisfies the equation πσ − T σ + tI = −ϕ

(2.5)

on Γ\ {O}. A solution of the Neumann problem (2.2) in a domain with an outward peak is required in the form u(z) = V σ (z), z ∈ Ω+ , where V σ is the single layer potential V σ (z) = σ(q) log Γ

1 / Γ. dsq , z ∈ |z − q|

In the case of an inward peak we set u(z) = V σ (z) + t R(z), z ∈ Ω+ , where

α+ − α− 3/2 α+ + α− 3/2 1/2 z (log z − πi) + i z , z ∈ Ω+ . (2.6) + R(z) = Re z 4π 4

Here the branches of the analytic functions z 1/2 , z 3/2 and log z are chosen in such a way that Im z 1/2 > 0, Im z 3/2 < 0 and Im log z = π at the points z = x+i0 ∈ Ω+ , x < 0. Then σ satisfies πσ + Sσ = ψ (2.7) on Γ \ {O} in the case of an outward peak and the pair (σ, t) is a solution of πσ + Sσ + t

∂ R = ψ on Γ \ {O} ∂n

(2.8)


103

in the case of an inward peak. Here and elsewhere ∂ 1 dsq , z ∈ Γ \ {O}. (Sσ)(z) = σ(q) log ∂nz |z − q| Γ

The necessity of introducing additional terms in the potential representations of solutions to the boundary value problems in domains with inward peak will be justified later in the proof of solvability of integral equations.

2.1 Weighted H¨ older spaces Let Ω+ be a plane simply connected domain with compact closure, bounded by a contour Γ with a peak z = 0. We assume that the curve Γ\{O} belongs to the class C 2 . We say that the peak is outward (inward) if Ω+ (the complementary domain Ω− ) is given near the origin O by the inequalities κ− (x) < y < κ+ (x), 0 < x < δ, where x−2 κ± (x) ∈ C 2 (0, δ),

lim x−2 κ± (x) = α±

x→+0

with α+ > α− . By Γ± we denote the arcs {(x, κ± (x)) : x ∈ [0, δ]}. The points on Γ+ and Γ− which have equal abscissas will be denoted by q+ and q− . Further, we assume δ to be so small that for any z1 and z2 which are both on Γ+ or on Γ− , |(n z1 , nz2 )| < π/2 . older space of functions on (0, δ) with By Λ1,α γ (0, δ) we denote the weighted H¨ the norm |uγ+α φ (u) − v γ+α φ (v)| φΛ1,α = sup γ (0,δ) |u − v|α u,v∈(0,δ) (2.9) + sup uγ |φ (u)| + sup uγ−1 |φ(u)|. u∈(0,δ)

u∈(0,δ)

Another weighted H¨ older space Λα γ (0, δ) is endowed with the norm φΛαγ (0,δ) =

|uγ+α φ(u) − v γ+α φ(v)| + sup uγ |φ(u)| . |u − v|α u,v∈(0,δ) u∈(0,δ) sup

(2.10)

We introduce two more norms on Λα γ (0, δ): |ϕ(x) − ϕ(y)| + sup |x|γ |ϕ(x)| |x − y|α x∈(0,δ)

(2.11)

|xγ+α ϕ(x) − y γ+α ϕ(y)| + sup |x|γ |ϕ(x)| .g. |x − y|α 2|x−y|min{|x|,|y|} x∈(0,δ)

(2.12)

sup 2|x−y|min{|x|,|y|}

|x|γ+α

and sup

In the next lemma and elsewhere, by c we denote different positive constants.

104


Lemma 2.1.1. Suppose 0 < α < 1 and β ∈ R. The norms (2.10), (2.11) and (2.12) in Λα −β (0, δ) are equivalent. Proof. We start with the equivalence of (2.11) and (2.12). Suppose that 2|x − y| < min{|x|, |y|}, with 0 < x < y δ. We observe that |x|α−β

|xα−β ϕ(x) − y α−β ϕ(y)| |ϕ(x) − ϕ(y)| |x − y|α |x − y|α |xα−β − y α−β | β + y −β |ϕ(y)| y . |x − y|α

We estimate the second term on the right-hand side of (2.13):

1

|xα−β − y α−β | β

β

α−β−1 y = c y [y + τ (x − y)] dτ

|x − y|1−α |x − y|α 0

1 x − y !α−β−1

x − y 1−α

1+τ = c

dτ

. y y 0 Under the above assumptions we have x − y

3 1

1+τ 2 y 2

(2.13)

(2.14)

x − y 1

. and

y 2

Hence (2.14) implies |xα−β − y α−β | β y c, |x − y|α which together with (2.13) gives |x|α−β

|ϕ(x) − ϕ(y)| |xα−β ϕ(x) − y α−β ϕ(y)| + c y −β |ϕ(y)| . |x − y|α |x − y|α

Therefore, the norm (2.11) is estimated by the norm (2.12). On the other hand, |xα−β ϕ(x) − y α−β ϕ(y)| |xα−β − y α−β | α−β |ϕ(x) − ϕ(y)| |x| + |ϕ(y)| |x − y|α |x − y|α |x − y|α |ϕ(x) − ϕ(y)| |x|α−β + c y −β |ϕ(y)| . |x − y|α Thus, the norm (2.12) is dominated by (2.11) up to a constant factor. To complete the proof, we estimate the norm (2.10) by the norm (2.12). To this end we assume that x < y and |x − y| > x/2. Then |x − y| > y/3 and |xα−β ϕ(x) − y α−β ϕ(y)| xα−β |ϕ(x)| y α−β |ϕ(y)| + α |x − y| |x − y|α |x − y|α −β c x |ϕ(x)| + y −β |ϕ(y)| . Hence the equivalence of the norms (2.10) and (2.12) follows. The lemma is proved.


105

α We say that φ ∈ Λ1,α Λγ (Γ± ) , if φ± = φ |Γ± , as a function of the γ (Γ± ) Λα local variable x, belongs to Λ1,α γ (0, δ) γ (0, δ) . We introduce the space Nβ1,α (Γ) of functions which are continuous on Γ\{O} and have the finite norm ϕ N 1,α (Γ) = ϕ+ − ϕ− Λαβ−1 (0,d) + ϕ Λ1,α

β+1 (Γ)

β

.

1,α We define Λodd 2−β (Γ) as the space of functions with the norm φ(Λodd )1,α

2−β (Γ)

= φ+ − φ− Λ1,α

2−β (Γ+ ∪Γ− )

+ φ+ + φ− Λα−β (Γ+ ∪Γ− ) + φΛα (Γ\(Γ+ ∪Γ− )) . 1,β The space Λeven 2−α (Γ) is endowed with the norm φ(Λeven )1,α

2−β (Γ)

= φ+ + φ− Λ1,α

2−β (Γ+ ∪Γ− )

+ φ+ − φ− Λα−β (Γ+ ∪Γ− ) + φΛα (Γ\(Γ+ ∪Γ− )) . 1,α Further, we introduce the space Lodd 2−β (Γ) of functions φ on Γ such that φ = 1,α φ1 + φ2 with φ1 ∈ Λodd 2−β (Γ) and φ2 ∈ Λα −β (Γ). We put 1,α (2.15) = inf φ1 (Λodd )1,α + φ2 Λα−β . φ Lodd

2−β

(Γ)

2−β

1,α In a similar way, one can define the space Leven 2−β (Γ). Finally, by Λα γ (1, ∞) we denote the space of continuous functions on [1, ∞) satisfying the condition φΛαγ (1,∞) =

|uγ+α φ(u) − v γ+α φ(v)| |u − v|α u,v∈(1,∞) sup

+

sup |uγ φ(u)| < ∞.

(2.16)

u∈(1,∞)

Here without loss of generality we may assume that δ = 1. We present two simple properties of the above spaces. Lemma 2.1.2. The norm (2.16) in Λα γ (1, ∞) is equivalent to φΛαγ (1,∞) =

sup u,v∈[1,∞) 2|u−v|min(u,v)

|u|γ+α

|φ(u) − φ(v)| + sup |uγ φ(u)|. |u − v|α u∈[1,∞)

(2.17)

Proof. Note that the inequality 2|x − y| < min |x|,|y| with x = 1/u and y = 1/v, u, v ∈ (1, ∞) holds if and only if 2|u − v| < min |u|, |v| . The rest of the proof duplicates the proof of Lemma 2.1.1 .

106


Lemma 2.1.3. The operator S : Sφ = φ ◦ s is the bijection from Λα β (1, ∞) onto Λα (0, 1), where by s we denote the mapping s(u) = 1/u of [1, ∞) onto (0, 1] . −β Proof. Let 2|x − y| < min(|x|, |y|), and let x = 1/u and y = 1/v. Then |x|α−β

|ϕ(x) − ϕ(y)| |ϕ(s(u)) − ϕ(s(v))| = |u|β |v|α α |x − y| |u − v|α |ϕ(s(u)) − ϕ(s(v))| c|u|α+β , |u − v|α

(2.18)

because 23 v < u < 32 v. The inequality converse to (2.18) is proved in a similar way. α Thus S performs a bijection of Λα β (1, ∞) onto a Λ−β (0, 1). Now, we give simple assertions to be used in the sequel, omitting their proofs. Lemma 2.1.4. If σ ∈ Λ1,α 2−β (Γ), then

σ± (u) u

∈ Λα 2−β (0, 1) and

σ (u) ± c σ Λ1,α (Γ) . α 2−β u Λ2−β (0,1)

(2.19)

γ α Lemma 2.1.5. If σ ∈ Λα β (Γ) and γ ∈ R, then u σ± (u) ∈ Λβ−γ (0, 1) and

uγ σ± (u) Λαβ−γ (0,1) c σ Λαβ (Γ) .

(2.20)

α Lemma 2.1.6. If σ ∈ Λ1,α 2−β (Γ), then σ ∈ Λ1−β (Γ) and

σ Λα1−β (Γ) c σ Λ1,α

2−β (Γ)

.

(2.21)

α Lemma 2.1.7. If σ ∈ Λ1,α 2−β (Γ), then uσ± (u) ∈ Λ−β (0, 1) and

uσ± (u) Λα−β (0,1) c σ Λ1,α

2−β (Γ)

.

(2.22)

1 Lemma 2.1.8. If σ ∈ Λα β (Γ), μ ∈ C (0, 1) and if the derivative μ is bounded, then α μσ± ∈ Λβ (0, 1). Moreover,

μσ± Λαβ (0,1) c σ Λαβ (Γ) .

(2.23)

2.2 Boundedness of integral operators 2.2.1 Integral operators in weighted H¨ older spaces Let Eβ1,α (R) denote the space of functions on R, continuously differentiable and subject to: |(1 + u2 )(β+α)/2 ϕ (u) − (1 + v 2 )(β+α)/2 ϕ (v)| |u − v|α u,v∈R

ϕE 1,α (R) = sup β

+ sup |(1 + u2 )β/2 ϕ (u)| + sup |(1 + u2 )(β−1)/2 ϕ(u)| < ∞. u∈R

u∈R

(2.24)


107

The space Eβα (R) is defined as the set of functions on R, which are continuous and have the finite norm: |(1 + u2 )(β+α)/2 ϕ(u) − (1 + v 2 )(β+α)/2 ϕ(v)| |u − v|α u,v∈R

ϕEβα (R) = sup

(2.25)

+ sup |(1 + u2 )β/2 ϕ(u)| < ∞. u∈R

The spaces Eβ1,α (0, ∞) and Eβα (0, ∞) are introduced in a similar way. The next assertion contains sufficient conditions for the boundedness of singular integral operators in weighted H¨ older spaces. Theorem 2.2.1. Let the function K(x, t) with x, t ∈ R obey the conditions: 1 < ∞; |x − t|(1 + |x − t|J ) |x − y| (b) |K(x, t) − K(y, t)| c |x − t|2 (1 + |x − t|J ) 1 for |x − y| < min(|x − t|, |y − t|) ; 2 (a) |K(x, t)| c

(2.26)

|K(x, x + t) + K(x, x − t)|dt c ;

R

d

(d)

K(x, t)dt

c |x|−1 . dx R (c)

Suppose that 0 < α < 1 and 0 β < J + 1, J 0. Then the integral operator T with the kernel K(x, t) is a continuous mapping of Eβα (R) onto itself. The property (c) implies existence of the integral K(x, t)dt R

in the sense of its principal value. We assume that this integral, considered as a function of x, is differentiable and that its derivative satisfies the condition (d). While proving the boundedness of I −T and I +S, we need a stronger variant of Theorem 2.2.1 for H¨ older spaces with the negative exponent β in the weight, which holds under an additional condition on the kernel K(x, t). Corollary 2.2.2. Let the function K(x, t) vanish outside the set {(x, t) : |x − t| < x/2} and let the conditions (a), (c), (d) in Theorem 2.2.1 hold together with (b ) |K(x, t) − K(y, t)| c

|x − y| 1 for |x − y| < min(|x − t|, |y − t|) (2.27) |x − t|2 2

on the intersection of supports of K(x, t) and K(y, t). Then the integral operator T with the kernel K(x, t) is continuous in the space Eβα (R) with 0 < α < 1 and −1 < β < 1.

108


For the proof of Theorem 2.2.1 and Corollary 2.2.2 below see Appendix A and Appendix B. Theorem 2.2.3. Suppose that the function K(x, t) satisfies the conditions: (a) |K(x, t)|(1 + |x − t|J+1 ) c < ∞, (b) |K(x, t) − K(y, t)| c (c) (d)

|x − y| (1 + |x − t|J+2 ) for |x − y|

(k) are linearly independent functions with the Here p = 2(n + [ν] + 1), and U properties: (k) = 0 on ∂Λ ∩ {t > log(1/ε}, (k) = 0 in Λε = Λ ∩ {t > log(1/ε}, U ΔU

3.5. Appendices

267

" (1) ∈ W 3 (Λε ), where β ∈ (n + [ν] + 1, n + [ν] + 3/2). Making (k) ∈ W 3 (Λε ), W U 2,β 2,β the reverse change of variables, we obtain (2)

v˜

(r, θ) =

p

u ˜(k) (r, θ) + w ˜ (1) (r, θ) ,

k=1

where (k) (log(1/r), θ) and ∇ w u ˜(k) (r, θ) = U ˜(1) (r, θ) = O rn+[ν]+1− for = 0, 1 . The same argument as in the proof of Lemma 3.2.2 implies the equality p

p

u ˜(k) (r, θ) =

αm v˜m (r, θ),

αm ∈ R,

m=1

k=1

where

m+[ν]

v˜m (r, θ) = Re

Pm/2,k (log z)z m/2+k + O |z|n+[ν] ,

m = 1, . . . , p .

k=1

The functions v˜m are harmonic in Ω+ and vanish on Γ± . Thus v˜ = vñ + v˜(1) + v˜(2) is harmonic in Ω+ and v˜ = H on Γ. Let v be the conjugate function, taken with the minus sign. Then v satisfies ∂v ∂˜ v ∂H = = =h ∂n ∂s ∂s

on Γ

and has the required representation with ϕ(ext) (z) = iϕ n+1 (z) n

and

ψn(ext) (z) = i

p

αm (˜ vm + ivm )(z).

m=1

Let ζ c stand for the conformal mapping of Ω− onto the upper half-plane satisfying ζ(0) = 0. Applying the same argument as in the proof of Lemma 3.2.2, we find −1+ε 1 aθ = √ + O |z| 2 . c ζ z In particular, Re

−1+ε 1 = aθ r−1/2 cos(ϕ/2) + O r 2 , ζ c (z)

0 < ε < 1.

To find the coefficient r0 in the expansion of the complex potential ψ, we apply the Green formula to the solution v (e) of the problem N (e) and to the function

268

Chapter 3. Asymptotic Formulae for Solutions of Integral Equations

Re (1/ζ c (z) in Ω− \{|z| < ε}: ∂ 1 1 ∂ (e) v (e) (z) Re c (3.79) − Re c v (z) ds ∂n ζ (z) ζ (z) ∂n Γε ∂ 1 ∂ (e) 1 =− v (e) (z) Re c ds + Re c v (z)ds , ∂n ζ (z) ζ (z) ∂n {r=ε}∩Ω− {r=ε}∩Ω− where Γε = ∂Ω− \{|z| < ε}. In the case ν = /2, ∈ Z, we write, using polar coordinates (r, ϕ): v (e) (z) = A11 rν+1 cos(ν + 1)ϕ + A12 rν+1 sin(ν + 1)ϕ + A21 rν+2 cos(ν + 2)ϕ + A22 rν+2 sin(ν + 2)ϕ + Re r0 r1/2 cos ϕ/2 − Im r0 r1/2 sin ϕ/2 + o(r) . Putting this expansion of v (e) into the right-hand side of (3.79), we find 1 ∂ v (e) (z) Re c ds ∂n ζ (z) − {r=ε}∩Ω 2π 2 1 ν+k−1/2 =− Ak1 ε cos((ν + k)ϕ) cos(ϕ/2)dϕ 2 0 k=1 2π + Ak2 εν+k−1/2 sin((ν + k)ϕ) cos(ϕ/2)dϕ aθ − Re r0 2

0 2π

cos2 (ϕ/2)dϕ +

0

(1)

(1)

= A1 εν+1/2 + A2 εν+3/2 − and

aθ Im r0 2

2π

sin(ϕ/2) cos(ϕ/2)dϕ + o(1) 0

π aθ Re r0 + o(1) 2

∂ (e) v (z)ds ∂n {r=ε}∩Ω− 2π 2 = (ν + k) Ak1 εν+k−1/2 cos((ν + k)ϕ) cos(ϕ/2)dϕ Re

1

ζ c (z)

k=1

+ Ak2 εν+k−1/2

sin((ν + k)ϕ) cos(ϕ/2)dϕ

0

2π aθ cos (ϕ/2)dϕ − Im r0 sin(ϕ/2) cos(ϕ/2)dϕ + o(1) 2 0 0 π (2) (2) = A1 εν+1/2 + A2 εν+3/2 + aθ Re r0 + o(1) , 2 aθ − Re r0 2

0 2π

2π

2

(i)

where Aki and Ak can be simply expressed by β1 , β2 . Noting that (∂/∂n)Re 1/ζ c (z) = 0 on Γ\{z = 0} ,

3.5. Appendices

269

we obtain 1 ∂ (e) v (z)ds Re c ζ (z) ∂n Γε 1 ∂ 1 ∂ (e) v (z)ds = v (e) (z) Re c ds − Re c ∂n ζ (z) ζ (z) ∂n − − {r=ε}∩Ω {r=ε}∩Ω (1) (1) (2) (2) = εν+1/2 A1 − A1 + εν+3/2 A2 − A2 − πaθ Re r0 + o(1) . Passing to the limit as ε → 0, we find

1

1 ∂ (e) v (z)ds . Re c Re r0 = − πaθ Γε ζ (z) ∂n

(3.80)

We use the same argument for ν ∈ Z, ν = −1/2, −3/2. Now let us consider the case ν = −1/2. Since Im β1,1 = 0 and Re β1,0 = 0, we have v (e) (z) = − Im β1,0 r1/2 sin(ϕ/2) + Re β1,1 r1/2 log r cos(ϕ/2) − Re β1,1 r1/2 ϕ sin(ϕ/2) + Re 0 r1/2 cos(ϕ/2) − Im r0 r1/2 sin(ϕ/2) + o(r1/2 ) . Then v (e) (z) {r=ε}∩Ω−

and

1 ∂ 1 1 1 ds = − πRe β1,1 log ε − πRe β1,1 − πRe r0 + o(1) ∂n ζ c (z) 2 2 2

{r=ε}∩Ω−

∂ (e) 1 1 v (z) c ds = πRe β1,1 log ε + πRe β1,1 ∂n ζ (z) 2 1 1 − πRe β1,1 + πRe r0 + o(1) . 2 2

By (3.79) we have

1 ∂ (e) v (z)ds = πaθ Re β11 + πaθ Re r0 . −

c (z) ∂n ζ Γ Thus

1

1 ∂ (e) v (z)ds − Re β11 Re r0 = − πaθ Γ ζ c (z) ∂n

(+) (−) 1

h 0 − h0 1 ∂ (e) v . =− (z)ds − πaθ Γ ζ c (z) ∂n π

In the case ν = −3/2 we have v (e) (z) = Re β1,0 z −1/2 + β1,1 z −1/2 log z + β2,0 z 1/2 + β2,1 z 1/2 log z

! + β2,2 z 1/2 log2 z + β2,3 z 1/2 log3 z + r0 z 1/2 + o(|z|1/2 ) .

270


We may take Re β2,0 = 0. Applying the same argument as for ν = −1/2, we conclude that

1

1 ∂ (e) v (z)ds − Re β2,1 . Re r0 = − Re c πaθ Γ ζ (z) ∂n It remains to find β2,1 .

Appendix E: Proof of Lemma 3.2.4 , ( ) un (z) = Re χ(z)ϕn+3 (z) ,

Let

z ∈ Ω− ,

, ( )

where ϕn+3 are given in (3.43) for ν = l/2 and in (3.44) for ν = l/2, and χ is a smooth cut-off function supported near the peak. We choose the coefficients of , ( ) ϕn+3 in such a way that hn (z) = (h − ∂un /∂n)(z) = O(|z|n+ν+2 ). Suppose that the peak is at the point z = 0. We introduce the function H(z) = ± h(s)ds, z ∈ Γ± (0) , 0z

where the integral is taken along the shortest arc of the curve Γ connecting 0 and z. We have H(z) = ±

n k=1

0 h(±) (±) xk+ν+1 + T k (log x)xk/2 + O(xγ+1 ) k+ν +1

n

k=1

for ν = /2 and (±) (±) h(±) h0,1 ν+1 h0,1 ν+1 0,0 − x x ± log x ν + 1 (ν + 1)2 ν +1 n0 n (±) (±) + P k+1 (log x)xk+ν+1 + T k (log x)xk/2 + O(xγ+1 )

H(z) = ±

k=1

k=1 ,

, ( )

( )

, ( )

n+3 (z), where ϕ n+3 obeys the for ν = /2 ∈ Z. We look for ϕn+3 in the form iϕ relation , ( ) H(z) − Re ϕ n+3 (z) = O |z|n+ν+3 near 0. We construct it in the same way as in proof of Lemma 3.2.1 to obtain , ( )

ϕ n+3 (z) =

n+3 k=0

α k z k+ν +

n 0 +3 k=1

k (log z)z −1+k/2 U

(3.81)

3.5. Appendices

271

for ν = /2 and , ( )

ϕ n+3 (z) = z ν

2

n+3 k+1 (log z)z k+ν α 0,r (log z)r + Q

r=0

+

k=1 n 0 +3

(3.82) k+1 (log z)z U

k+ν

k=1

for ν = /2. In a small neighborhood of the origin, we have , , ( ) ( ) n+3 (z) = (∂/∂s) H(z) − Re ϕ n+3 (z) . hn (z) = h(z) − (∂/∂n)Re iϕ

The coefficient α 0 in (3.81) is α 0 = −i

(+)

(−)

h 0 + h0 2 (0) − κ (0)) . ν(ν + 1) (κ+ −

The coefficients α 0,r in (3.82) are found separately for ν = 0 and ν = 0. We have (+)

α 0,1 = − i α 0,0 = − i

(−)

h0,1 + h0,1 2 0,2 = 0, (0)) , α ν(ν + 1) (κ+ (0) − κ− (+) (−) h +h 2 0,0

0,0

(0) − κ (0)) ν (ν + 1)(κ+ −

(+) (−) (+) (−) h0,1 + h0,1 h0,1 + h0,1 1 − 2 (ν + 1) (κ+ (0) − κ− (0)) ν(ν + 1) (κ+ (0) − κ− (0)) (+) (−) (+) (−) h0,0 + h0,0 h0,1 + h0,1 2 2ν + 1 = −i − ν (ν + 1)(κ+ (0) − κ− (0)) ν(ν + 1) (κ+ (0) − κ− (0))

−

for ν = 0 and (+)

α 0,0 = 0, α 0,1 = −2i

(+)

α 0,2

(−)

(+)

(−)

h0,1 + h0,1 h0,0 + h0,0 + 2i (0) − κ (0)) , (ν + 1)(κ+ (0) − κ− (0)) (ν + 1)2 (κ+ − (−)

h0,1 + h0,1 = −i (0) − κ (0)) (ν + 1)(κ+ −

for ν = 0. We express the harmonic extension of , ! ( ) u n (z) = H(z) − Re χ(z)ϕ n+3 (z)

into Ω− in the form u (1) + u (2) , where u (1) = u n on Γ\{O}, ∇k v˜(2) (z) = O(|z|n+ν+3−k ), k = 0, 1, 2, 3,

272


in a neighborhood of the origin and u (2) is a solution of the boundary value problem Δ u(2) = −Δ u(1) ,

˚ 1 (Ω− ), v˜(2) ∈ W 2

u (2) = O(1)

as z → ∞ .

(3.83)

Making the change of variable ζ =1−

z0 , z

ζ = ξ + iη,

where z0 is a fixed point of Ω+ , we rewrite the equation in (3.83) in the form (2) = F (1) ΔU

in Λ,

where the curvilinear strip Λ is the image of Ω+ and F (1) obeys the relations ∇k F (1) (z) = O |ζ|−n−ν−k−3 , k = 0, 1. (2) Applying the same arguments as in the proof of Lemma 3.2.1, we show that U 1 ˚ belongs to W2 (Λ) and (2) (ξ, η) = O |ξ|−n−[ν] k = 0, 1. ∇k U Making the inverse change of variable, we find the solution u (2) of the problem 3 + (3.83) in W2 (Ω ). The Sobolev embedding theorem (see [17], Theorem 1.4.5(e)) implies that u (2) is differentiable at least once and ∇ u(2) (z) = O |z|n+[ν]−1 , z → 0 . (1) + u (2) is harmonic in Ω+ , u = H on Γ and Hence the function u =u n + u , ( ) u = Re ϕ n+3 (z) + O |z|n+[ν] ,

z ∈ Ω− ,

in a neighborhood of z = 0. Then the function u, equal to the conjugate function of u taken with the minus sign, satisfies ∂ u ∂H ∂u = = =h ∂n ∂s ∂s , ( )

on Γ , ( )

and has the required representation with ϕn (z) = iϕ n+1 (z).

Chapter 4

Integral Equations of Plane Elasticity in Domains with Peak 4.1 Introduction Boundary integral equations provide an effective tool in the study and solution of boundary value problems in elasticity theory. Boundary value problems for the Lamé system can be reduced to a system of integral equations for which one gets results similar to those given in the previous chapters. In order to describe the stress and strain state of a body in plane elasticity, one uses the displacement vector u(x, y) = (u1 (x, y), u2 (x, y)) and the stress tensor with components σxx , σyy and τxy , which are considered as functions of the complex variables z = x + iy and z = x − iy. Here x and y are Cartesian coordinates of the initial position of points of an elastic body, whose displacement is the vector u(z, z). In the absence of volume forces, the components of the stress tensor satisfy the equilibrium equations ∂σxx ∂τxy + = 0, ∂x ∂y (4.1) ∂σyy ∂τxy + = 0. ∂y ∂x The stress tensor and the displacement vector are related by the linear relations (Hook’s law): ∂u1 , ∂x ∂u2 σyy = λ div u + 2μ , ∂y ∂u ∂u2 1 + . τxy = μ ∂y ∂x

σxx = λ div u + 2μ

(4.2)

274

Chapter 4. Integral Equations of Plane Elasticity in Domains with Peak

Inserting these expressions into (4.1), one arrives at the homogeneous Lamé system μΔu + (λ + μ)∇div u = 0 ,

(4.3)

where λ and μ are physical constants related to the media. Called the Lamé constants, λ and μ are subject to the conditions: μ > 0, λ > −2μ/3. In what follows, by Ω+ we mean a planar simply connected domain bounded by a closed piecewise smooth curve with a peak at the origin O. Suppose that either Ω+ or its complement Ω− is described in Cartesian coordinates x, y near O by the inequalities κ− (x) < y < κ+ (x), 0 < x < δ, where κ± are C ∞ -functions on [0, δ] satisfying

κ± (0) = κ± (0) = 0 and κ+ (0) > κ− (0). In the first case, we say that O is an outward peak and in the second one, O is an inward peak. The matrix with elements 1 λ + 3μ 1 0 log Γ(z, ζ) = 4πμ(λ + 2μ) |z − ζ| 0 1 1 λ+μ (x − ξ)(y − η) (x − ξ)2 + (y − η)2 λ + 3μ |z − ζ|2 (x − ξ)(y − η) is the fundamental solution of the Lamé system. Applying the operator on the left-hand side of (4.3) to the kth column of this matrix, we obtain δ(z − ζ)e(k) . Here e(k) is the unit vector in the direction of the kth axis of the Cartesian system and δ(z − ζ) is the Dirac measure concentrated at ζ. The matrix Γ is called the Kelvin-Somigliana tensor. Boundary conditions h = (h1 , h2 ) for stresses at the boundary Γ = ∂Ω+ have the form h1 = σxx cos(n, 0x) + τxy cos(n , 0y), 0x) + σyy cos(n , 0y) , h2 = τxy cos(n, where n is an outward normal to Γ. Putting (4.2) into the boundary conditions we arrive at the matrix differential operator T (∂ζ , nζ )u : = 2μ∂u/∂n + λn div u + μn × rot u , called the stress operator; T (∂ζ , nζ ) u is the stress vector in the direction nζ . We consider the interior and exterior first boundary value problems Δ∗ u : = μΔu + (λ + μ)∇div u = 0 in Ω+ , Δ∗ u = 0 in Ω− ,

u=g

on Γ ,

u=g

u(z) = O(1)

on Γ ;

as |z| → ∞ ;

(D(i) ) (D(e) )

4.1. Introduction

275

and the interior and exterior second boundary value problems Δ∗ u : = 0 Δ∗ u = 0

in Ω− ,

in Ω+ , Tu = h

Tu = h on Γ ,

(N (i) )

on Γ ; u(z) = o(1) as

|z| → ∞ (N (e) )

for the displacement u = (u1 , u2 ). Henceforth we shall not distinguish between a displacement u = (u1 , u2 ) and a complex displacement u = u1 + iu2 . The equilibrium equations (4.1) are satisfied by introducing the Airy function F (x, y) by the formulae σxx =

∂2F , ∂y 2

σyy =

∂2F , ∂x2

τxy = −

∂2F . ∂x∂y

(4.4)

Thus solving the plane elasticity problems can be reduced to the search for the Airy function which satisfies the biharmonic equation ΔΔF = 0. The function F can be expressed via two analytic functions ϕ and χ in the following way (see [32]) # $ F (x, y) = Re zϕ(z) + χ(z) . (4.5) This formula was obtained by Goursat and named after him. The stress components σxx , σyy , τxy and the displacement components u1 , u2 are related to the stress functions ϕ and ψ = ∂χ/∂z by Kolosov’s formulae (see [32]) σxx + σyy = 4Re {ϕ (z)}, σxx − σyy + 2iτxy = −2[zϕ (z) + ψ(z)],

(4.6)

2μ(u1 + iu2 ) = κϕ(z) − zϕ (z) − ψ(z) . Here κ = (λ + 3μ)(λ + μ)−1 . The formula (4.5) enables one to reduce the basic boundary value problems of the plane elasticity theory to the problem of finding two analytic functions ϕ and ψ. The first boundary value problem, when the displacements u = u1 + iu2 are given on the boundary contour, reduces to looking for functions ϕ and ψ, satisfying the boundary condition κϕ(z) − zϕ (z) − ψ(z) = 2μu(z),

z ∈ Γ.

The boundary condition for the second boundary value problem is ϕ(z) + zϕ (z) + ψ(z) = g(z) , where g(z) is a line integral of stresses applied at the boundary Γ. To be more precise, g(z) = i hds + const , z ∈ Γ . (0z)

276


Here (0z) stands for the arc on Γ connecting the points 0 and z. A classical method for solving the first and the second boundary value problems of elasticity theory consists in representation of their solutions in the form of the double layer potential ∗ {T (∂ζ , nζ ) Γ (z, ζ)} σ(ζ)dsζ , (4.7) W σ(z) = Γ

where

1 0

0 1 2(λ + μ) ∂ 1 (x − ξ)(y − η) (x − ξ)2 + log 2 2 (x − ξ)(y − η) (y − η) μ|z − ζ| ∂nζ |z − ζ| ' 1 0 1 ∂ , z = (x, y), ζ = (ξ, η), + log −1 0 ∂sζ |z − ζ|

$∗ # T (∂ζ , nζ )Γ(z, ζ) =

μ 2π(λ + 2μ)

and the single layer potential V τ (z) = Γ(z, ζ)τ (ζ)dsζ , z = (x, y) ∈ Ω+ or Ω− .

(4.8)

(4.9)

Γ

Recall that the vector-valued functions u = (u1 , u2 ) and v = (v1 , v2 ) from the class C 2 (Ω+ ) obey the first Betti formula ∗ vΔ u + E(v, u) dω = vT uds , (4.10) Ω+

Γ

where E(v, u) = λ div v div u + μ

2 ∂vi ∂ui ∂uj + ∂xj ∂xj ∂xi i,j=1

and T is the stress operator. Replacing u = (u1 , u2 ) by v = (v1 , v2 ) and vice versa in (4.10) and subtracting the resulting equality from (4.10), one finds (vΔ∗ u − uΔ∗ v)dω = (vT u − uT v)ds , (4.11) Ω+

Γ

which is called the second Betti formula. Letting u = (u1 , u2 ) and v = (v1 , v2 ) be equal and satisfying the Lamé equation (4.3), we arrive at E(u, u)dω = uT u ds , (4.12) Ω+

which is called the third Betti formula.

Γ

4.1. Introduction

277

Replacing v(z) by Γ(z, ζ) in (4.11), we conclude that the solution u(z) of the Lamé system in the class C 2 (Ω+ ) has the representation u(z) = −W u(z) + V (T u)(z),

z ∈ Ω+ ,

(4.13)

while the solution u(z) in the class C 2 (Ω− ) admits the representation u(z) = W u(z) − V (T u)(z) + u(∞), Using the equalities

⎧ ⎪ ⎨−1, W 1 = −1/2, ⎪ ⎩ 0,

z ∈ Ω− .

(4.14)

z ∈ Ω+ , z ∈ Γ, z ∈ Ω− ,

one proves that the limit values of the double layer potential are given by 1 lim W σ(z) = − σ(z0 ) + W σ(z0 ), 2

z→z0 ∈Γ z∈Ω+

lim W σ(z) =

z→z0 ∈Γ z∈Ω−

1 σ(z0 ) + W σ(z0 ) . 2

(4.15)

The single layer potential has the limit values lim T V τ (z) =

z→z0 ∈Γ z∈Ω+

1 τ (z0 ) + T (V τ )(z0 ), 2

1 lim T V τ (z) = − τ (z0 ) + T (V τ )(z0 ) . z→z0 ∈Γ 2 −

(4.16)

z∈Ω

Then, for the problems D(i) and N (e) , the densities of corresponding potentials can be found from the 2 × 2 systems of boundary integral equations −2−1 σ + W σ = g

(4.17)

−2−1 τ + T V τ = h .

(4.18)

and

Under certain general conditions on g in (4.17), there exist solutions u+ and u− of the problems D(i) and D(e) in Ω+ and Ω− with the boundary data g satisfying g(z) = lim Γ(z, ζ) T (∂ζ , nζ )u+ (ζ) − T (∂ζ , nζ )u− (ζ) dsζ + u− (∞) (4.19) ε→0 {Γ:|ζ|>ε}

on Γ \ {O}. Let v − denote a solution of N (e) in Ω− , vanishing at infinity, with the boundary data T u+ on Γ \ {O}.

278


We can choose v − so that, for w = v − − u− + u− (∞)

on Γ \ {O} ,

the equality w(z) − 2 lim

ε→0 {Γ:|ζ|>ε}

{T (∂ζ , nζ ) Γ (z, ζ)}∗ w(ζ)dsζ = −2ϕ(z) + 2u− (∞)

(4.20)

holds with z ∈ Γ\{O}. Solutions of the equations (4.17) and (4.18) are constructed by means of (4.19) and (4.20). Hence the function σ = v− − g is a solution of (4.17). A solution of (4.18) can be obtained as follows. Take a solution v − of N (e) , vanishing at infinity, with the boundary data h and a solution u+ of D(i) equal to v − on Γ \ {O}. Under sufficiently general assumptions on h we can select v − and u+ in such a way that the density τ = T u+ − h satisfies (4.18). We introduce the class Nν (ν > −1) of infinitely differentiable vector-valued functions h on Γ\{O} admitting the representations: h± (x) = xν q± (x) on the arcs Γ± = {(x, κ± (x)) : x ∈ (0, δ]}, where the vector-valued functions q± belong to C ∞ [0, δ] and satisfy |q+ (0)| + |q− (0)| = 0. Let N denote the set 2 Nν N= ν>−1

and let Mβ (β > −1) be the class of differentiable vector-valued functions on Γ\{O} satisfying ∂rσ (z) = O(xβ−r ), z = x + iy = (x, y), r = 0, 1. ∂srz We introduce the class M as

2

M=

Mβ .

β>−1

For domains with outward peak we put Mext =

2 β>−1/2

Mβ .

4.1. Introduction

279

Inward peak. In general, the integral equation (4.17) has no solutions in M even if g ∈ N vanishes on Γ± . However, for a function from Nν with ν > 3, the solvability of (4.17) can be attained by changing the equation in the following way. A solution u of the problem D(i) is sought as the sum of the double layer potential with density σ and a linear combination of explicitly given functions A1 , A2 and A3 with unknown real coefficients: u(z) = W σ(z) + c1 A1 (z) + c2 A2 (z) + c3 A3 (z). The functions A1 , A2 , A3 are given by i [2κ Im z 1/2 − z −1/2 Im z] A1 (z) = 2μ (κ − 1)(α+ − α− ) [2κ Im (z 3/2 log z) − 3z 1/2 log z Im z +i 8πκμ (κ − 1)α+ 3/2 − 2z 1/2 Im z] − i z , 2μ Q [2κ Im z 1/2 + z −1/2 Im z] (4.21) 2μ (α+ − α− )(κ + 1) [2κ Im (z 3/2 log z) + 3z 1/2 log z Im z −Q 8πμκ i (κ + 1)α+ 3/2 [2κ Im z 3/2 − 3z 1/2 Im z] + Q + 2z 1/2 Im z] + z , 2μ 2μ

A2 (z) = −

(α+ − α− )(κ + 1) κ+1 Im z − [2κ Im (z 2 log z) μ 4πμκ (κ + 1)α+ 2 + 4z log z Im z + 2z Im z] + z , μ

A3 (z) = −

where κ = (λ + 3μ)/(λ + μ) and Q = [(α+ + α− ) − (α+ − α− )/2κ + 2α− ] /2 . Here and in the sequel by z ν (log z)k we mean the branch taking real values on the upper boundary of the slit along the positive part of the real axis. By the limit relation for the double layer potential we obtain −2−1 σ + W σ + c1 A1 + c2 A2 + c3 A3 = g

(4.22)

for the pair (σ, c), where c = (c1 , c2 , c3 ). We prove the uniqueness assertion for the equation (4.22) in the class of pairs (σ, c) with σ ∈ M and c ∈ R3 in Section 4.3.1. The solvability of (4.22) with the right-hand side g ∈ Nν , ν > 3, in M × R3 is proved in the same section. Moreover, in the same theorem we derive the following asymptotic formula for σ near the peak σ(z) = α(log x)2 + β log x + γ x−1/2 + O(x−ε ), z ∈ Γ, with a positive ε.

280


A solution v of the problem N (e) with the boundary data h from Nν , ν > 3, is sought in the form of the single layer potential V τ . The density τ satisfies the system of integral equations (4.18) on Γ\{O}. In Section 4.3.2 we prove that if the mean value of h on Γ vanishes, then (4.18) has a unique solution τ in the class M and this solution admits the representation τ± (z) = α± x−1/2 + O(1) on the arcs Γ± . Outward peak. We represent a solution u of the problem D(i) as the double layer potential W σ. The density σ is found from the system of integral equations (4.17). It is proved that the kernel of the integral operator in (4.17) is two-dimensional in the class M. Solutions of the homogeneous system of integral equations (4.17) are functions obtained as restrictions to Γ of solutions to the homogeneous problem N (e) . Near the peak, these displacements have the estimate O(r−1/2 ) with r standing for the distance to the peak. Thus Mext is the uniqueness class for the equation (4.17). We examine the case when (4.17) has at least two solutions in Section 4.4.1. The nonhomogeneous integral equation (4.17) is studied in the same section. We show that the solvability in M holds for all functions g from the class Nν with ν > 0. One of the solutions of (4.17) has the following representations on Γ± : σ(x) = β± xν−1 + O(1) −1/2

σ(x) = β± x

for ν = 1/2,

log x + O(1) for ν = 1/2.

The integral equation (4.18) is uniquely solvable in the class M if the righthand side h ∈ Nν with ν > 0 satisfies hds = 0, hζds = 0, Γ

Γ

where ζ is any solution of the homogeneous equation (4.17) in the class M. In order to remove the orthogonality conditions, we are looking for a solution v of the problem N (e) with the boundary data h from N as the sum of the single layer potential V τ and a linear combination of functions 1 (z), 2 (z) with unknown coefficients v(z) = V τ (z) + t1 1 (z) + t2 2 (z). The functions k (z), k = 1, 2, are defined by complex stress functions (complex potentials) ϕk (z), ψk (z): 1 κϕk (z) − zϕk (z) − ψk (z) , 2μ 1/2 1/2 zz0 zz0 3 ϕ1 (z) = , ψ1 (z) = − , z − z0 2 z − z0 k (z) =

4.2. Boundary value problems of elasticity

ϕ2 (z) = i

zz0 z − z0

1/2 ,

281

i ψ2 (z) = 2

zz0 z − z0

1/2 ,

where z0 is a fixed point in Ω+ . The boundary equation −2−1 τ + T V τ + t1 T 1 + t2 T 2 = h

(4.23)

is considered with respect to the pair (τ, t), where τ is the density of the single layer potential and t = (t1 , t2 ) is a vector in R2 . In Section 4.4.2 we prove the existence and uniqueness for solutions of (4.23). We also study the asymptotic behavior of the solutions. We prove that, for h ∈ Nν with 0 < ν < 1, the density τ has the following representations on the arcs Γ± : τ (x) = β± xν−1 + O(x−1/2 ) −1/2

τ (x) = γ± x

−1/2

τ (x) = γ± x

for 0 < ν < 1/2, −1/2

log x + β± x ν−1

+ β± x

+ O(log x)

+ O(log x)

for ν = 1/2, for 1/2 < ν < 1.

4.2 Boundary value problems of elasticity We represent densities of integral equations of elasticity theory by means of solutions of certain interior and exterior boundary value problems. The auxiliary results concerning such problems are collected in this section.

4.2.1 Asymptotic behavior of solutions to the problem D (i) We introduce some notation and cite some known facts (cf. [11], [18]) to be used in the proof of the next theorem and elsewhere. Let G be a domain in the plane R2 , = 0, 1, . . . , and let H (G) be the Sobolev space of functions given on G which is endowed with the norm

2 ∂ μ+ν 1/2

f H (G) = . (4.24)

μ ν f (x, y) dxdy G ∂x ∂y μ+ν

By H −1/2 (∂G) we denote the space of traces on ∂G of functions from H (G) with the norm g H −1/2 (∂G) = inf{ f H (G) : g = f on ∂G} . (4.25)

(G) we denote the weighted Sobolev space with the norm Let β > 0. By W2,β

⎛ ⎝

⎞1/2

∂ μ+ν

2

e2βx μ ν f (x, y) dxdy ⎠ . ∂x ∂y G

μ+ν

(4.26)

282


˚ (G) we mean the completion of C0∞ in the W (G)-norm and let W 0 = By W 2,β 2,β 2,β L2,β .

Note that the norm of f (x, y) in W2,β (G) is equivalent to the norm of

−1/2

exp(βx)f (x, y) in H (G). We supply the space W2,β (∂G) of traces on ∂G of

functions from W2,β (G) with the norm exp(βx)gH −1/2 (∂G) . Let L(∂ξ , ∂η ) stand for either a scalar differential operator of order 2m in the strip Π = {(ξ, η) : ξ ∈ R, a < η < b}, or for a matrix operator in Π whose elements are scalar differential operators Li,j (∂ξ , ∂η ) (i, j = 1, 2) of order 2m. Similarly, let B (r)(∂ξ , ∂η ) (r = 1, . . . , m) be either a scalar differential operator in Π of order μr with μr m, or a matrix differential operator in Π whose r components Bi,j (∂ξ , ∂η ) are scalar differential operators of orders μr with μr m. We assume that the boundary value problem n−m L(∂ξ , ∂η )u = f in Π, f ∈ W2,β (Π),

B (r) (∂ξ , ∂η )u = g (r) on ∂Π, g (r) ∈ W n+m−μr −1/2 (∂Π), 1 r m,

(4.27)

is elliptic. The operator U(∂ξ , ∂η ) = L(∂ξ , ∂η ), B (1) (∂ξ , ∂η ), . . . , B (m)(∂ξ , ∂η ) of the problem (4.27) continuously maps n+m (Π) W2,β

into

n−m W2,β (Π)

×

m 7

n+m−μr −1/2

W2,β

(∂Π).

r=1

We introduce the boundary value problem with a complex parameter k: L(ik, ∂η )ϕ = 0 in (a, b), B (r) (ik, ∂η )ϕ|η=a,b = 0 , 1 r m .

(4.28)

The family of problems (4.28) with complex k is related to (4.27) by the complex Fourier transform. This transform and its inverse are given by ∞ exp(−ikt)f (t)dt, k ∈ R + iβ, f(k) = (2π)−1/2 −∞ ∞ (4.29) exp(ikt)f(k)dk. f (t) = (2π)−1/2 −∞

Then the Parseval formula ∞ exp(2βt)|f (t)|2 dt = −∞

∞+iβ

−∞+iβ

|f(k)|2 dk

(4.30)


283

holds. For all k, except for certain isolated values, the problem (4.28) has only the trivial solution in W2n+m (a, b). The exceptional values of k are called eigenvalues of the operator pencil of the problem (4.28). Nonzero solution u of (4.28) is called the eigenvector, while the dimension of the space of solutions of (4.28) is referred to as the multiplicity of the eigenvalue. Elements ϕ(1) , . . . , ϕ(r−1) of W2n+m (a, b) are called generalized eigenvectors corresponding to the eigenvector ϕ(0) = ϕ, if j 1 ∂

L(ik, ∂η )ϕ(j− ) = 0 ! ∂k

on Π,

=0

j 1 (r) B (ik, ∂η )ϕ(j− ) |η=a,b = 0 , !

(4.31) r = 1, . . . , m, j = 1, . . . , r − 1 .

=0

The number r is called the length of the chain ϕ(0) , ϕ(1) , . . . , ϕ(r−1) of the eigenvector and the generalized eigenvectors. The maximal length of the chain ϕ(0) , ϕ(1) , . . . . . . , ϕ(r−1) is referred to as the multiplicity of the eigenvector ϕ(0) . The following two assertions are particular cases of more general results from [11] and [18]. Theorem 4.2.1. Suppose that there are no eigenvalues of the operator pencil of the problem (4.28) on the line Im k = β. Then the operator of the problem (4.28) performs an isomorphism n+m n−m W2,β (Π) ≈ W2,β (Π) ×

m 7

n+m−μr −1/2

W2,β

(∂Π) .

(4.32)

r=1

The solution

m ˚ 2,β ∩ W n+m (Π) u∈ W 2,β

of the problem (4.27) satisfies the estimate m B (r) u W n+m−μr −1/2 (∂Π) . u W n+m (Π) c Lu W n−m (Π) + 2,β

2,β

(4.33)

2,β

r=1

Let k0 be an eigenvalue of the operator pencil of the problem (4.28) and let ϕ(0) , ϕ(1) , . . . , ϕ(p) be the chain of the corresponding eigenvector and its generalized eigenvectors. One verifies directly that the function u = eik0 t

p (it)j j=0

j!

ϕ(p−j)

is a solution of the homogeneous problem (4.27).

284


Theorem 4.2.2. Let n−m n−m (Π) ∩ W2,β (Π), f ∈ W2,β 1 n+m−μr −1/2

g (r) ∈ W2,β

n+m−μr −1/2

(∂Π) ∩ W2,β1

(∂Π) ,

(4.34)

where β1 > β and let the lines Im k = β and Im k = β1 do not contain eigenvalues of the operator pencil in (4.28), whereas the eigenvalues k1 , . . . , kM lie in the strip n+m β < Im k < β1 . If u is a solution of the problem (4.27) from W2,β (Π), it has the form Jμ rj,μ −1 M (j) u= cj,k,μ uk,μ + u1 . μ=1 j=1 k=0

Here (j)

uk,μ = eikμ t

k (it)s

s!

s=0 −1,j)

(r

(0,j)

k = 0, . . . , rj,μ − 1, j = 1, . . . , Jμ ,

ϕ(k−s,j) , μ

and ϕμ , . . . , ϕμ j,μ is the chain of maximal length of the eigenvector and its generalized eigenvectors corresponding to the eigenvalue kμ , μ = 1, . . . , M , Jμ is the multiplicity of kμ , cj,k,μ are certain constants, and u1 is a solution of (4.27) n+m from W2,β (Π). 1 We start our study of auxiliary boundary value problems with an asymptotic representation for solutions of the problem D(i) in domains with outward peaks. Theorem 4.2.3. Let Ω+ have an outward peak. Suppose that g is an infinitely differentiable function on the curve Γ\{O} and let g have the following representations on the arcs Γ± : g± (z) =

n+1

(k+1)

Q±

(log x)xk+ν + O xn+2+ν−ε , z = x + iy, ν > −1,

k=0 (j)

where Q± are polynomials of degree j and ε is a small positive number. Suppose the above representations can be differentiated n + 2 times. Then the problem D(i) has a solution u of the form 1 κϕn (z) − zϕn (z) − ψn (z) + u0 (z), z ∈ Ω+ , u(z) = (4.35) 2μ where ∇k u0 (z) = O(|z|n−2k ) for k = 0, . . . , n and ϕn (z) = ψn (z) =

n+1 k=0 n+1

Pϕ(k+2) (log z)z ν+k−1 , (k+2)

Pψ

(log z)z ν+k−1 .

k=0

Here

(j) Pϕ

and

(j) Pψ

are polynomials of degree j.


285

Proof. (i) We are looking for a displacement vector un such that the vector-valued function gn = g − un belongs to C ∞ (Γ\{O}) and (gn )± (x) = xν (qn )± (x), where (qn )± are infinitely differentiable on [0, δ] and satisfy ∇k (qn )± (x) = O(xn+1−k−ε ) ,

k = 0, . . . , n + 2 ,

on the arcs Γ± with ε being a small positive number. To this end, we use the so-called method of complex stress functions (see [32], Ch. II). The displacement vector u is related to complex potentials ϕ and ψ as follows: 2μu(z) = κϕ(z) − zϕ (z) − ψ(z), where the functions ϕ and ψ are to be defined by the boundary data of the problem D(i) . It suffices to consider a function g(z) coinciding with A± xν (log x)m on Γ± . We shall seek ϕ and ψ in the form ϕ(z) = β0 z ν−1 (log z)m−k + ε0 z ν (log z)m , ψ(z) = γ0 z ν−1 (log z)m−k + δ0 z ν (log z)m for ν = 1. There exist β0 , γ0 , ε0 and δ0 such that κϕ(z) − zϕ (z) − ψ(z), restricted to Γ± , is equal to 2μA± xν (log x)m plus terms of the form c± xi (log x)j , admitting the estimate O(xν (log x)m−1 ). We substitute the expansions of ϕ and ψ in powers of x along Γ± into the equation 1 (κϕ(z) − zϕ (z) − ψ(z)) = g(z), z = x + iy ∈ Γ . 2μ Equalizing the coefficients for xν (log x)m and xν−1 (log x)m , we obtain the algebraic system

i(κ+ (0) − κ− (0))(ν − 1)(κβ0 + (ν − 3)β0 + γ0 ) = 4μ(A+ − A− ), κβ0 − (ν − 1)β0 − γ0 = 0

,

with respect to β0 and γ0 . Let us choose ε0 arbitrarily. Then δ0 is defined by the equation i κε0 − νε0 − δ0 = μ(A+ + A− ) − (κ+ (0) + κ− (0))(ν − 1)(κβ0 + (ν − 3)β0 + γ0 ) . 4

286


In the case ν = 1 we seek ϕ and ψ in the form ϕ(z) =

ψ(z) =

2 k=0 2

βk (log z)m+1−k + ε0 z(log z)m , γk (log z)m+1−k + δ0 z(log z)m .

k=0

The coefficients β0 and γ0 are found from the algebraic system κβ0 − γ0 = 0, i(m + 1)(κ+ (0) − κ− (0))(κβ0 − β0 + γ0 ) = 2μ(A+ − A− ) . Further, we choose ε0 arbitrarily and find δ0 from the equation i κε0 − ε0 − δ0 = (m + 1)β0 + μ(A+ + A− ) − (κ+ (0) + κ− (0))(κβ0 − β0 + γ0 ) . 4 Given β1 , we find γ1 from the equation κβ1 − γ1 = m β0 . (ii) By u(1) we denote a vector-valued function equal to gn on Γ\{O} and satisfying the estimates u(1) (z) = O(|z|n+1+ν−ε ), ∇k u(1) (z) = O(|z|n+ν−k−ε ), k = 1, . . . , n + 2 . Let the vector-valued function u(2) be a unique solution of the boundary value problem ˚21 (Ω+ ). Δ∗ u(2) = −Δ∗ u(1) in Ω+ , u(2) ∈ W (4.36) After the change of variable z = ζ −1 with ζ = ξ + iη, the equation (4.36) with respect to ξ η U (2) (ξ, η) = u(2) , − |ζ|2 |ζ|2 takes the form L(∂ξ , ∂η ) U (2) = Δ∗ U (2) + L(∂ξ , ∂η ) U (2) = F (1) in Λ, where a curvilinear semi-infinite strip Λ is the image of Ω+ , L(∂ξ , ∂η ) is the secondorder differential operator with coefficients subject to the estimate O(1/ξ) as ξ → ∞, and ∇k F (1) (ζ) = O(|ζ|−n−ν−2−k+ε ) , k = 0, . . . , n . Let ρ be a function from the class C ∞ (R) vanishing for ξ < 1 and equal to 1 for ξ > 2, and let ρr (ξ) = ρ(ξ/r). Clearly, (2) , (2) = Δ∗ U (2) + R(∂ξ , ∂η )U ξ n L(∂ξ , ∂η )ξ −n U


287

where R(∂ξ , ∂η ) is the second-order differential operator with coefficients admitting the estimate O(1/ξ) as ξ → ∞. Therefore, the boundary value problem (2) = F (2) in Λ, U (2) = 0 on ∂Λ, (2) + ρr R(∂ξ , ∂η )U Δ∗ U where F (2) (ξ, η) = ξ n F (1) (ξ, η) and ∇k F (2) (ξ, η) = O(ξ −2−ν−k+ε ), k = 0, . . . , n , ˚ 1 (Λ) for large r. It follows from the local estimate is uniquely solvable in W 2 (2) (2) n+2 (2) L (Λ) , n (Λ) + χU U (4.37) c χF W 2 W (Λ∩{ −1 2r. Using of unity and the same local estimate, we obtain that a1 partition ˚2 ∩ W n+2 (Λ2r ), where Λ2r = Λ ∩ {ξ > 2r}. U (2) − U (3) ∈ W 2 Let D(∂ξ , ∂η ) denote the differential operator Δ∗ continuously mapping 1 ˚ ∩ W n+2 (Π) into W n (Π), where W 2,β 2,β 2,β $ # Π = (ξ, η) : ξ ∈ R, −κ+ (0)/2 < η < −κ− (0)/2 . Eigenvalues of the operator pencil D(ik, ∂η ) are non-zero roots of the equation α2 k 2 = κ2 (sinh αk)2 , where α = (κ+ (0) − κ− (0))/2 and κ = (λ + 3μ)/(λ + μ). Using shift and dilation, we reduce the boundary value problem Δ∗ V = 0

in Π and V |∂Π = 0

(4.38)

to Δ∗ V = 0

in Π = {(ξ, η) : ξ ∈ R, −1 < η < 1} and V |∂Π = 0 .

(4.39)

Hence, it is enough to deal with the problem (4.39). The general solution of D(i0, ∂η )V = 0, −1 < η < 1, is a linear vector-valued function. Therefore, the only solution of the boundary value problem D(i0, ∂η )V = 0,

|η| < 1,

and V |η=±1 = 0

is trivial. Thus k = 0 is not an eigenvalue of the operator pencil D(ik, ∂η ).

(4.40)

288


Now let k = 0. One checks directly that the vector-valued functions κ (a1 + b1 )e−kη , ia1 + ib1 + ib1 η e−kη k

κ ib2 + ib2 η ekη , (a2 + b2 )ekη k are linearly independent solutions of the equation

and

(4.41)

ia2 +

(4.42)

D(ik, ∂η )V = 0 . We plug (4.41) and (4.42) into the boundary condition V |η=±1 = 0. One gets a nontrivial solution of (4.40) if the determinant Det of the system with respect to a1 , a2 , b1 , b2 vanishes. Straightforward calculations give sinh 2k 2 Det = 16 1 − κ2 . 2k Thus the real axis has no eigenvalues of D(ik, ∂η ). Since the operator D(∂ξ , ∂η ) is the “limit” operator for L(∂ξ , ∂η ), there exists β > 0 such that n+2 1 ˚2,β (Λ) ∩W U (2) − U (3) ∈ W2,β (see Theorem 7.2 [18]). Now, since U (2) = U (3) + (U (2) − U (3) ), it follows from the Sobolev embedding theorem that ∇k U (2) (ξ, η) = O(|ξ|−n ) for k = 0, . . . , n . Therefore, we find from (i) and (ii) that the function u = un + u(1) + u(2) is a solution of the problem D(i) and has the required representation (4.35) with u0 = u(1) + u(2) . Corollary 4.2.4. Let g have the following representations on the arcs Γ± : g± (x) =

n+1

(k)

q± xk+ν + O(xn+2+ν ), ν > −1,

k=0 (k)

with real coefficients q± . Then the functions ϕn and ψn in (4.35) have the form ϕn (z) = β0 z

−1

+ (β1,0 + β1,1 log z) +

n+1

βk z k−1 ,

k=2

ψn (z) = γ0 z −1 + (γ1,0 + γ1,1 log z) +

n+1 k=2

γk z k−1


289

for ν = 0, ϕn (z) = β0,0 + β0,1 log z +

n+1

βk z k , ψn (z) = γ0,0 + γ0,1 log z +

k=1

n+1

γk z k

k=1

for ν = 1, and ϕn (z) =

n+1

βk z k+ν−1 , ψn (z) =

k=0

n+1

γk z k+ν−1

k=0

otherwise. Next we find an asymptotic representation for solutions of D(i) in domains with inward peaks. Theorem 4.2.5. Let Ω+ have an inward peak. Suppose g is an infinitely differentiable function on the curve Γ\{O} and its restrictions to the arcs Γ± have the representations g± (z) =

n+1

(k+1)

Q±

(log x)xk+ν + O(xn+ν+2−ε ), ν > −1,

k=0 (j) Q±

where are polynomials of degree j and ε is a small positive number. Suppose that these representations can be differentiated n + 2 times. Then the problem D(i) has a solution of the form u(z) =

1 κ(ϕn (z) + ϕ∗ (z))−z(ϕn (z) + ϕ∗ (z)) 2μ ! −(ψn (z) + ψ∗ (z)) + u0 (z) ,

(4.43)

where ∇ u0 (z) = O(|z|n+[ν]+1− −ε ), = 1, . . . , n. The complex potentials ϕn , ψn , ϕ∗ and ψ∗ are: ϕn (z) = ϕ∗ (z) =

n

Pϕ(k+2) (log z)z k+ν , ψn (z) =

k=0 p m=1

(j) Pϕ ,

n

(k+2)

Pψ

(log z)z k+ν ,

k=0

Rϕ,m (log z)z m/2 , ψ∗ (z) =

p

Rψ,m (log z)z m/2 .

m=1

(j) Pψ

Here are polynomials of degree j, Rϕ,m , Rψ,m are polynomials of degree [(m − 1)/2], and p = 2(n + [ν] + 1). Proof. (i) We are looking for a displacement vector un such that the vector-valued function gn = g − un on Γ\{O}

290


belongs to C ∞ (Γ\{O}) and ∇k (gn )± (z) = O(xn+ν+3−k )

for k = 1, . . . , n + 2 .

We use the method of complex stress functions described in [32]. It suffices to take g(z) equal to A± xν (log x)m on Γ± . As in the proof of Theorem 4.2.3, we introduce the potentials ϕ(z) = βm z ν (log z)m and ψ(z) = γm z ν (log z)m for ν = /2, ∈ Z, such that κϕ(z) − zϕ (z) − ψ(z) on Γ± is the sum of 2μA± xν (log x)m and the terms of the form c± xi (log x)j , admitting the estimate O(xν (log x)m−1 ). We substitute the expansions of ϕ and ψ in powers of x along Γ± into the equation 1 (κϕ(z) − zϕ (z) − ψ(z)) = g(z), z = x + iy ∈ Γ . 2μ The coefficients βm and γm are found from the algebraic system κβm − νβm − γm = 2μA+ e4iπν κβm − νβm − γm = 2μe2iπν A− . If ν = /2, ∈ Z, we seek the functions ϕ and ψ in the form ϕ(z) = βm,1 (log z)m+1 + βm,0 (log z)m z ν , ψ(z) = γm,1 (log z)m+1 + γm,0 (log z)m z ν . The values βm,1 and γm,1 are found from the system ⎧ ⎪ ⎨ κβm,1 − νβm,1 − γm,1 = 0, A+ − (−1)m A− ⎪ . ⎩ κβm,1 + νβm,1 + γm,1 = iμ π(m + 1) Finally, we choose βm,0 arbitrarily. Then γm,0 is defined by the equation κβm,0 − νβm,0 − γm,0 = 2μA+ + (m + 1)βm,1 . (ii) Let u(1) be a vector-valued function equal to gn on Γ\{O} and admitting the estimates u(1) (z) = O(|z|n+ν+3 ) and ∇k u(1) (z) = O(|z|n+ν+2−k ), k = 1, . . . , n + 2, (2)

(2)

in a neighborhood of the peak. By u(2) = (u1 , u2 ) we denote a unique solution of the Dirichlet problem ˚21 (Ω+ ) . Δ∗ u(2) = −Δ∗ u(1) in Ω+ , u(2) ∈ W


291

Let Λ be the image of Ω+ under the mapping (r, θ) → (t, θ), where (r, θ) are polar coordinates of (x, y) and t = log(1/r). The vector-valued function U (2) (t, θ) with the components u1 (e−t , θ) cos θ + u2 (e−t , θ) sin θ and u2 (e−t , θ) cos θ − u1 (e−t , θ) sin θ, (2)

(2)

(2)

(2)

is a solution of the equation ˚ 21 (Λ) , Δ∗ U (2) + KU (2) = F (1) in W where F (1) (t, θ) = O(e−(n+ν+2)t ). Here K is the first-order differential operator −(λ + 2μ) −(λ + 3μ)(∂/∂θ) . K= (λ + 3μ)(∂/∂θ) −μ From the local estimate

U (2) W n+2 (Λ∩{ −1 0 , since the necessary condition for solvability of the problem (4.47) is violated. Therefore, the strip 0 < Im z < β, where β ∈ (n + [ν] + 1, n + [ν] + 3/2), contains p = 2(n + [ν] + 1) eigenvalues of D(ik, ∂θ ). n (Λ), it follows that U (2) admits the representation (see Since F (1) ∈ W2,β Theorem 4.2.2) p U (2) = ck V (k) + W (1) , k=1 (k) (k) (V1 , V2 ) (k)

where V (k) = are linearly independent vector-valued functions satisfying (Δ∗ + K)V = 0 in ΛR = Λ ∩ {t > R} and vanishing on ∂Λ ∩ {t > R}, n+2 n+2 V (k) ∈ / W2,β (ΛR ) and W (1) ∈ W2,β (ΛR ). Making the inverse change of variable t = − log r, we obtain u(2) (r, θ) =

p k=1

ck v (k) (r, θ) + w(1) (r, θ) ,


293

where v (k) (r, θ) = V (k) (log(1/r), θ) · eiθ and ∇ w(1) (r, θ) = O(rn+[ν]+1− ) for = 1, . . . , n. The functions ϕ(z) = z k/2 , ψ(z) = Az k/2 with A = (k/2 − κ), and the functions ϕ(z) = iz k/2 , ψ(z) = iAz k/2 with A = (κ + k/2) obey the equations ! 1 κϕ(z) − zϕ (z) − ψ(z) = O(xk/2+1 ) , z ∈ Γ± . μ Duplicating the argument used at the beginning of the proof of the theorem, we can construct functions ϕ∗ (z) and ψ∗ (z) such that p 1 κϕ∗ (z) − zϕ∗ (z) − ψ∗ (z) + w(2) (z) , ck v (k) (z) = 2μ k=1

where ∇ w(2) (z) = O(|z|n+[ν]+1− −ε ) for = 1, . . . , n and ϕ∗ (z) = ε1 z 1/2 + ε2 z + (ε3,0 + ε3,1 log z)z 3/2 + · · · + Rϕ, p−1 (log z)z n+[ν]+1/2 , ψ∗ (z) = δ1 z 1/2 + δ2 z + (δ3,0 + δ3,1 log z)z 3/2 + · · · + Rψ, p−1 (log z)z n+[ν]+1/2 . It follows from (i) and (ii) that the vector-valued function u = un +u(1) +u(2) is the required solution of the problem D(i) with u0 = u(1) + w(1) + w(2) . Corollary 4.2.6. Let g have the asymptotic representations g± (z) =

n+1

(k,1)

(α±

(k,0)

log x + α±

)xk+ν + O(xn+ν+2 )

k=0 (k,i)

on the arcs Γ± for ν = m/2, m ∈ Z, where α± functions ϕn and ψn in (4.43) have the form ϕn (z) = ψn (z) =

n

are real numbers. Then the

(β (k,1) log z + β (k,0) )z k+ν ,

k=0 n

(γ (k,1) log z + γ (k,0) )z k+ν

k=0

with β

(k,i)

, γ

(k,i)

∈ C.

4.2.2 Asymptotic behavior of solutions to the problem N (e) We introduce the weighted Sobolev space W k,ω (Ω− ) with the inner product ∂ μ+ν f1 ∂ μ+ν f2 (f1 , f2 )k,ω := ω −2k+2(μ+ν) μ ν μ ν dxdy, ∂x ∂y ∂x ∂y μ+νk Ω−

˚ k,ω (Ω− ) we denote the completion of C ∞ (Ω− ) where ω(z) = (1+x2 +y 2 )1/2 . By W 0 in the W k,ω (Ω− ) norm.

294


The required assertion on solvability of the problem Δ2 f = g ˚ 2,ω (Ω− ) is a consequence of the following Lax-Milgram theorem and the in W lemma below. Theorem 4.2.7. (Lax-Milgram [14]) Let V be a Hilbert space and let B(u, v) denote a bilinear form: V × V → R with the properties (i) |B(u, v)| c1 uv, u, v ∈ V, (ii) B(u, u) c2 u2, u ∈ V , where c1 , c2 are positive constants. Then, for any g in the dual space V ∗ , there exists a unique u ∈ V such that B(u, v) = g(v), v ∈ V . The mapping L : V ∗ → V defined by the equality Lg = u is a linear continuous operator with continuous inverse. Besides, L 1/c2 and L−1 c1 . In the next lemma we use the notation D1 = ∂/∂x, D2 = ∂/∂y, x1 = x, x2 = y . Lemma 4.2.8. (see [9]) The uniquality (ω

−k

Ω−

f ) dx1 dx2 ck 2

2 i=1

Ω−

(ω −k+1 Di f )2 dx1 dx2 ,

˚ k,ω (Ω− ) with a constant ck depending on k. holds for any f ∈ W Proof. We put ωr (x1 , x2 ) = (r + (x21 + x22 )1/2 ), r 1 . Integrating by parts, we obtain Ω−

ωr−2k f 2 dx1 dx2

=−

2 i=1

=k

Ω−

−

Ω−

Ω−

1 = 2 i=1 2

Ω−

ωr−2k f 2 Di xi dx1 dx2

! ωr−k f ωr−k Di f − kf ωr−k−1 Di ωr xi dx1 dx2

ωr−2k f 2

2

xi Di ωr ωr−1 dx1 dx2

i=1

ωr−k f

ωr−k

2 i=1

xi Di f dx1 dx2 ,

(4.48)


295

where f ∈ C0∞ (Ω− ). Hence, Ω−

2 (f ωr−k )2 1 − k xi Di ωr ωr−1 dx1 dx2 i=1

=−

Ω−

ωr−k f ωr−k

2

xi Di f dx1 dx2 .

i=1

We choose r to satisfy kωr−1 1/2. Since xi ωr−1 1, it follows that (f ωr−k )2 dx1 dx2 Ω−

2 4

Ω−

Ω−

(f ωr−k )2 dx1 dx2 (f ωr−k )2 dx1 dx2

1/2

2 Ω−

2 1/2 i=1

xi Di f

2

ωr−2k dx1 dx2

i=1

Ω−

(wr−k+1 Di f )2 dx1 dx2

1/2

1/2

.

Thus we arrive at the inequality Ω−

(ωr−k f )2 dx1 dx2

16

2 i=1

Ω−

(ωr−k+1 Di f )2 dx1 dx2 .

The result follows from the estimate r−1 (r + (x21 + x22 )1/2 )−1 (1 + (x21 + x22 ))1/2 1 .

˚ 2,ω (Ω− ) Theorem 4.2.9. Let ω 2 g ∈ L2 (Ω− ). Then there exists a unique u in W such that Δ2 u = g . Our goal is to find an asymptotic representation for solutions of the Neumann problem for domains with inward peaks. Theorem 4.2.10. Let Ω have an inward peak. Suppose that h is an infinitely differentiable vector-valued function on Γ\{O} with h ds = 0 Γ

and let the restrictions of h to Γ± admit the representation h± (z) =

n−1 k=0

(k+1)

H±

(log x)xk+ν + O(xn+ν−ε ), ν > −1,

296

Chapter 4. Integral Equations of Plane Elasticity in Domains with Peak (j)

where H± are polynomials of degree j and ε is a small positive number. Let this representation be differentiable n times. Then the problem N (e) with the boundary data h has a solution v bounded at infinity, satisfying the condition V.P. T v ds = lim T v ds = 0 , (4.49) Γ

ε→0 {q∈Γ, |q|≥ε}

and, up to a linear function α + icz with real c, represented in the form 1 κϕn (z) − zϕn (z) − ψn (z) + v0 (z) . v(z) = 2μ Here

∇k v0 (z) = O |z|n−2k−1 for k = 1, . . . , n − 1 , p m ν−2 zz0 zz0 ϕn (z) = i β0,m log z0 − z z0 − z m=0 k+ν−2 n+1 zz0 zz0 + Pϕ(k+2) log , z0 − z z0 − z k=1 p m ν−2 zz0 zz0 γ0,m log ψn (z) = i z0 − z z0 − z m=0 k+ν−2 n+1 (k+2) zz0 zz0 log + Pψ , z0 − z z0 − z

(4.50)

z ∈ Ω− ,

k=1

where z0 is a fixed point of Ω+ , β0,m and γ0,m are real numbers, p = 1 if ν = (j) (j) 0, 1, 2, 3, and p = 2 otherwise, Pϕ and Pψ are polynomials of degree j. Proof. (i) We are looking for a displacement vector vn such that the traction hn = h − T vn belongs to C ∞ (Γ\{0}) and admits the estimates ∇k (hn )± (z) = O(xn+ν−k−ε ), z = x + iy, on Γ± for k = 0, . . . , n. To this end, we represent the boundary condition of the problem N (e) with the boundary data h in the Muskhelishvili form (see [32], Ch. II, Sect. 30) ϕ(z) + zϕ (z) + ψ(z) = f (z), z ∈ Γ\{0} . (4.51) Here ϕ and ψ are complex stress functions and f (z) = −i h ds + c, z ∈ Γ , (0z)

where by (0z) we denote the arc of Γ connecting 0 and z. As f in (4.51), it suffices to consider the function ±ih± xν+1 (log x)m on Γ± .


297

We begin with the case ν = 0, 1, 2, 3. As in Theorem 4.2.3, we find coefficients of ϕ˜ and ψ˜ so that the restriction ˜ of ϕ(z) ˜ + z ϕ˜ (z) + ψ(z) to Γ± is the sum of ±ih± xν+1 (log x)m and the terms of i j the form c± x (log x) , admitting the estimate O(xν+1 (log x)m−1 ). In a small neighborhood of the peak, we are looking for complex potentials ϕ˜ and ψ˜ in the form ϕ(z) ˜ = (log z)m

3

˜ β r z ν+r−2 and ψ(z) = (log z)m

r=0

3

γ r z ν+r−2 .

r=0

The coefficients β 0 , γ 0 are defined by the algebraic system β 0 + (ν − 2)β 0 + γ 0 = 0, β 0 − (ν − 4)β 0 − γ 0 = 0 . We find Re β 0 = Re γ 0 = 0,

and

(3 − ν)Im β 0 = Im γ 0 .

The coefficients β 1 and γ 1 satisfy ⎧ ⎨ β 1 + (ν − 1)β 1 + γ 1 = 0, ⎩ β − (ν − 3)β − γ = (ν − 3)(ν − 2) (κ (0) + κ (0))Im β . 1 + − 0 1 1 ν −1 Hence, it follows that (ν − 3)(ν − 2) (κ+ (0) + κ− (0))Im β 0 , 4(ν − 1) Re γ 1 = −νRe β 1 and Im γ 1 = −(ν − 2)Im β 1 .

Re β 1 =

We set

Im γ 1 = Im β 1 = 0.

The coefficients Im β 2 and Im γ 2 are evaluated from the system β 2 + νβ 2 + γ 2 = A2 , β 2 − (ν − 2)β 2 − γ 2 = B2 ,

(4.52)

where 1 A2 = − iκ+ (0)κ− (0)(ν − 3)(ν − 2)Im β 0 , 2 (ν − 2)(ν − 3) h+ + h− + (κ+ (0) + κ B2 = 2 − (0))Im β 0 ν(κ+ (0) − κ− (0)) 3ν (ν − 2)(ν − 3)(ν − 4) (κ+ (0))2 + κ+ (0)κ− (0) + (κ− (0))2 Im β 0 −i 6ν (ν − 1)(ν − 2) (κ+ (0) + κ− (0))Re β 1 . +i ν

298


The system (4.52) is solvable if (ν − 2)(ν − 3)(ν + 2) (κ+ (0) − κ− (0))3 Im β 0 = −8Im (h+ + h− ). 3

(4.53)

We express Im β 0 from (4.53). Adding together the equations of the system (4.52), we find Re β 2 . Then Re γ 2 is defined from the first equation in (4.52) Re γ 2 = −(ν + 1)Re β 2 . We choose Im β 2 arbitrarily and find Im γ 2 from the equation (ν − 1)Im β 2 + Im γ 2 =

1 κ (0)κ− (0)(ν − 3)(ν − 2)Im β 0 . 2 +

Let us choose β 3 arbitrarily. Then γ 3 is defined by the equation β 3 + (ν + 1)β 3 + γ 3 =

i i (h+ − h− ) − ν(κ+ (0) + κ− (0) β 2 − (ν − 2)β 2 − γ 2 . 2 2

˜ In the exceptional cases ν = 0, 1, 2, 3 complex potentials ϕ(z) ˜ and ψ(z) it should be looked for in the form ϕ(z) ˜ = (log z)m+1 ψ˜1 (z) = (log z)m+1

2

β r,0 z ν+r−2 + (log z)m

3

r=0

r=0

2

3

γ r,0 z ν+r−2 + (log z)m

r=0

β r,1 z ν+r−2 , γ r,1 z ν+r−2 .

r=0

We choose βr,k and γr,k in such a way that the coefficients of xν−k (log x)m , k = 0, 1, 2, m = 1, . . . , p, in the decompositions of ϕn and ψn along Γ± coincide with the corresponding coefficients in the decompositions of ϕ and ψ. Once the complex stress functions ϕn and ψn have been found, the displacement vector vn is given by vn (z) = (2μ)−1 κϕn (z) − zϕ n (z) − ψn (z) . Then ∇vn = O(|z|−2 ) as |z| → ∞ and V.P. T vn ds = 0. Γ

(ii) Let fn,1 and fn,2 denote the components of the vector-valued function fn on Γ\{O} defined by hn ds + c. fn (z) = −i (0z)


We set

i An = − 2π

299

fn,1 dx + fn,2 dy Γ

and consider the function ⎛

⎞ k n+[ν]+1 zz0 z ⎠ , z ∈ Ω− , + χn (z) = An ⎝log 1 − ck z0 z0 − z k=1

where z0 is a fixed point in Ω+ and the coefficients ck are chosen to ensure χn (z) = (1) (1) O(|z|n+[ν]+2 ) as z tends to zero. Let vn denote the displacement vector vn (z) = (1) (1) (1) (1/2μ)χ (z). Then T vn = i(∂/∂s)χn . The traction hn = hn − T vn on Γ\{O} (1) (1) with the components hn1 and hn2 has zero principal vector h(1) n (z)ds Γ

and zero principal moment ! (1) (1) (x − x0 )hn2 (x, y) − (y − y0 )hn1 (x, y) ds Γ

with respect to z0 = (x0 , y0 ) ∈ Ω+ . (2) (1) We need to construct a displacement vector vn with the given stress hn on Γ. To this end, we represent the boundary conditions of this problem via the Airy function F (z), biharmonic in Ω− (see [32], Ch. II, Sect. 30). We have F = b and ∂F/∂n = d on Γ\{O}, where the functions b and d belong to C ∞ (Γ\{O}) and admit the estimates b(p) = O(|p|n+2+ν−ε ), d(p) = O(|p|n+1+ν−ε ), p ∈ Γ\{O} , (1)

because of their relations with hn : (1) (1) [(xs − x)hn1 (xs , ys ) − (ys − y)hn2 (xs , ys )]ds, F (z) = (0z) ∂F (1) (z) = − hn2 ds + C1 , ∂x (0z) ∂F (1) (z) = hn1 ds + C2 , z = (x, y) ∈ Γ\{O}. ∂y (0z) Let F1 (z), defined on Ω− , satisfy F1 (p) = b(p), (∂F1 /∂n)(p) = d(p) on Γ\{O}, and admit the estimate |∇k F1 (z)| = O(|z|n+1+ν−k−ε ), k = 0, . . . , n + 2, as z → 0 .

(4.54)

300


We can assume also that ω 2 Δ2 F1 ∈ L2 (Ω− ), where ω(z) = (1 + |z|2 )1/2 . By Theorem 4.2.9, the boundary value problem Δ2 F2 = −Δ2 F1 in Ω− , F2 = ∂F2 /∂n = 0 on Γ, ˚ 2,ω (Ω− ). We use the Kelvin transform, setting is uniquely solvable in W U (ζ) = |ζ|2 F2 (1/ζ), ζ = ξ + iη ∈ Λ, where Λ is the image of Ω− under the inversion ζ = 1/z. Hence, Δ2 U = H in Λ, where ∇k H(ζ) = O(|ζ|−n−ν−3−k+ε ). Let ρ be a function from the class C ∞ (R) vanishing for ξ < 1 and equal to 1 for ξ > 2, and let ρr (ξ) = ρ(ξ/r). One checks directly that ξ n Δ2 ξ −n Un = Δ2 Un + R(∂ξ , ∂η )Un , where R(∂ξ , ∂η ) is a third-order differential operator with coefficients admitting the estimate O(1/ξ) as ξ → ∞. Hence, the boundary value problem Δ2 Un + ρr R(∂ξ , ∂η )Un = Hn in the strip Λ, Un = 0 on ∂Λ, with Hn (ξ, η) = ξ n H(ξ, η) and ∇k Hn (ξ, η) = O(ξ −2−ν−k+ε ), k = 0, . . . , n , ˚ 1 (Λ) for large r. The local estimate is uniquely solvable in W 2 Un W n+2 (Λ∩{ −1 2r. Using a partition of unity and the local estimate (4.55), we find that the ˚ 1 ∩ W n+2 (Λ2r ) with Λ2r = Λ ∩ {ξ > 2r} for difference U − U (1) belongs to W 2 2 large r.


301

By U(∂ξ , ∂η ) we mean the operator Δ2 continuously mapping 2 ˚ ∩ W n+2 (Π) into W n−2 (Π), W 2,β 2,β 2,β where

# $ Π = (ξ, η) : ξ ∈ R, −κ+ (0)/2 < η < −κ− (0)/2 .

Let us show that the eigenvalues of the operator pencil U(ik, ∂η ) are nontrivial roots of the equation (αk)2 − (sinh αk)2 = 0 with α = (κ+ (0) − κ− (0))/2. Making a shift and a dilation in variables ξ, η, we reduce the boundary value problem ˚ 2 (Π), (4.56) U(∂ξ , ∂η )V = 0, V ∈ W 2 to the problem U(∂ξ , ∂t )V = 0,

˚22 ({(ξ, t) : ξ ∈ R, |t| < 1}) . V ∈W

The general solution of U(ik, ∂t )V = 0, |t| < 1, has the form γ1 cosh kt + γ2 sinh kt + γ3 t cosh kt + γ4 t sinh kt . We plug it into the boundary conditions V |t=±1 = (∂V /∂t)|t=±1 = 0. As a result we arrive at an algebraic system of linear equations with a nontrivial solution provided that the determinant Det of the system vanishes. One can easily see that 2 2 Det = sinh 2k − 2k . Hence the eigenvalues of the pencil U(ik, ∂t ) are roots of the transcendental equation 2 2 sinh 2k − 2k = 0 . Consider first the case k = 0. The equation U(ik, ∂t )V = 0 has a solution V = γ0 t3 + γ1 t2 + γ2 t + γ3 . Setting this into the boundary conditions

∂V

V t=±1 = = 0,

∂t t=±1 we find that γ0 = γ1 = γ2 = γ3 = 0. Therefore, k = 0 is not an eigenvalue of the pencil U(ik, ∂t ).

302


Turning back to the original problem, we conclude that the eigenvalues of the operator pencil U(ik, ∂η ), η ∈ (−κ+ (0)/2, −κ− (0)/2), are nontrivial roots of the equation (αk)2 − (sinh αk)2 = 0 with α = (κ+ (0) − κ− (0))/2. Since the real axis contains no eigenvalues of U(ik, ∂η ), there exists a positive β such that n+2 ˚ 2 (Λ ∩ {ξ > 2R}) U − Un(1) ∈ W2,β ∩W 2,β (see Theorem 4.2.2). This inclusion and the Sobolev embedding theorem imply (1) that U − Un and its derivatives up to order n have the estimate O exp(−βξ) (1) (1) as ξ → ∞. Thus U = Un + (U − Un ) admits the estimate |∇k U (ξ, η)| = O(|ξ|−n+1 )

for k = 1, . . . , n.

Therefore, the Airy function F (z), equal to F1 (z) + F2 (z), satisfies F (z) = O(|z|n−1 ) and ∇k F (z) = O(|z|n−1−2k ), k = 1, . . . , n . (2)

The displacement vector vn corresponding to F (z) has the form (2)

vn(2) (z) = α + icz + vn0 (z), (2)

where c is a real coefficient and ∇k vn0 (z) = O(|z|n−2k−2 ). Since the gradient (2) (2) ∇vn of the displacement vector vn is square integrable in a neighborhood of infinity, we have |z|=Rn

T vn(2) ds → 0

for a certain sequence {Rn }, Rn → ∞. This and the condition h ds = 0 Γ

imply

Γ

T vn(2) ds = 0.

It follows from the first Kolosov formula (4.6) ΔF (z) = 4Re ∂ϕ(2) n /∂z (z) (2) (2) (2) that Re ∂ϕn /∂z (z), where ϕn is the complex potential corresponding to vn , is square integrable in a neighborhood of infinity. Therefore, ϕ(2) n (z) = icz +

0 k=−∞

bk z −k , c ∈ R.


303

The second Kolosov formula (4.6) ! ∂2F ∂2F ∂2F (2) (z) = 2 z(ϕ(2) (z) − (z) − 2i n ) (z) + (ψn ) (z) 2 2 ∂x ∂y ∂x∂y (2)

implies that the second derivative of the complex potential ψn (z) is also square integrable in a neighborhood of infinity. Therefore, we have ψn(2) (z) =

0

ak z −k .

k=−∞ (2)

Thus, the displacement vector vn admits the representation vn(2) (z) =

1 (κ + 1)icz + O(1). 2μ

Hence, up to the rigid displacement (2μ)−1 (κ+1)icz, the constructed displacement (2) vector vn is bounded at infinity. We remove this rigid displacement term, since it satisfies the homogeneous problem N (e) . Thus the displacement vector v = (1) (2) vn + vn + vn is a solution of the problem N (e) . It has the required representation (4.50) and satisfies the condition (4.49). (1)

Corollary 4.2.11. Let the polynomials H± satisfy (1) (1) (1) (1) Im H+ (0) + H− (0) = Im (∂H+ /∂t)(0) + (∂H− /∂t)(0) = 0 . Then the solution of the problem N (e) has a finite energy integral and can be represented in the form (4.50) with p+1 m ν−1 zz0 zz0 ϕn (z) = i β1,m log z0 − z z0 − z m=0 k+ν−2 n+1 zz0 zz0 + Pϕ(k+2) log , z0 − z z0 − z k=2

m ν−1 zz0 zz0 γ1,m log ψn (z) = i z0 − z z0 − z m=0 k+ν−2 n+1 (k+2) zz0 zz0 log + Pψ , z0 − z z0 − z p+1

k=2

where β1,m , γ1,m are real numbers, p = 1 if ν = −1, 0, 1, 2, and p = 2 otherwise, (j) (j) and Pϕ , Pψ are polynomials of degree j.

304


Corollary 4.2.12. Let h have the representation h± (z) = h± x−1/2 + h± x1/2 log x + h± x1/2 + · · · (1)

(2)

(3)

on Γ± . Then the problem N (e) with the boundary data h has a solution in the class M if and only if (1)

(1)

(2)

(2)

Im (h+ + h− ) = 0, Im (h+ + h− ) = 0, 1 (1) (1) (3) (3) Re h+ α− + Re h− α+ − Im (h+ + h− ) = 0 . 2 The next theorem concerns asymptotics of solutions to the problem N (e) in a domain with an outward peak. Theorem 4.2.13. Let Ω+ have an outward peak. Suppose that h is a C ∞ function on Γ\{O} and its restrictions to the arcs Γ± have the representations h± (z) =

n−1

(k)

H± (log x)xν+k + O(xν+n ), ν > −2.

k=0 (j)

Here H± are polynomials of degree at most j. Suppose that the above representations can be differentiated n times and V.P. h ds = 0 . Γ

Then the problem N (e) with the boundary data h on Γ has a solution v bounded at infinity, admitting the representation v(z) =

1 κ (ϕn (z) + ϕ∗ (z)) − z(ϕn (z) + ϕ∗ (z)) 2μ ! − ψn (z) + ψ∗ (z) + v0 (z)

(4.57)

up to a linear function α + icz with a real coefficient c, and satisfying the condition T v ds = 0 . Γ

Here ∇k v0 (z) = O(|z|n+ν−k ) ,

k = 0, . . . , n − 1,

the complex potentials ϕn , ψn have the form k+ν n zz0 zz0 (k+1) log ϕn (z) = Pϕ , z0 − z z0 − z k=1 k+ν n zz0 zz0 (k+1) log Pψ , ψn (z) = z0 − z z0 − z k=1

4.2. Boundary value problems of elasticity (j)

where Pϕ

(j)

and Pψ

305

are polynomials of degree j, and

ϕ∗ (z) =

m k=1 k=2

ψ∗ (z) =

m k=1 k=2

k/2 zz0 zz0 Rϕ,k log , z0 − z z0 − z k/2 zz0 zz0 Rψ,k log , z0 − z z0 − z

where Rϕ,k and Rψ,k are polynomials of degree at most [k/2], and m is the largest integer not exceeding n + ν + 1. Proof. We choose a displacement vector vn such that hn = h − T vn belongs to C ∞ (Γ\{O}) and (hn )± (z) = O(xn+ν ). To this end, as in the proof of Theorem 4.2.10, we use the method of complex stress functions. It is convenient to write the boundary conditions of the problem N (e) in the Muskhelishvili form ϕ(z) + zϕ (z) + ψ(z) = f (z), z ∈ Γ .

Here f (z) = −i

h ds + c , (0z)

where (0z) stands for the arc of Γ connecting 0 and z. It suffices to consider a function f (z) of the form ±ih± xν+1 (log x)m on the arcs Γ± . We set

ν+1 m zz0 log , z − z0 ν+1 m zz0 zz0 log ψn (z) = γm z − z0 z − z0

ϕn (z) = βm

zz0 z − z0

for ν = n/2, n ∈ Z, where z0 is a fixed point in Ω+ . Let us show that there exist βm and γm such that the function ϕn (z) + zϕn (z) + ψn (z), considered on Γ± , is equal to the sum of ±ih± xν+1 (log x)m and a linear combination of the functions c± xi (log x)j , admitting the estimate O(xν+1 (log x)m−1 ). As in the proof of Theorem 4.2.5, we decompose ϕn and ψn in powers of x along Γ± . The coefficients βm and γm of xν+1 (log x)m are solutions of the algebraic system βm + (ν + 1)βm + γm = ih+ , e4iπν βm + (ν + 1)βm + γm = −ih− e2iπν .

306


In the case ν = n/2, n ∈ Z , the functions ϕn and ψn are defined by m+1 m ν+1 zz0 zz0 zz0 + βm,0 log , ϕn (z) = βm,1 log z − z0 z − z0 z − z0 m+1 m ν+1 zz0 zz0 zz0 + γm,0 log , ψn (z) = γm,1 log z − z0 z − z0 z − z0 where z0 is a fixed point in Ω+ . The coefficients βm,1 and γm,1 are found from the system ⎧ ⎪ ⎨ βm,1 + (ν + 1)βm,1 + γm,1 = 0, h+ + (−1)m h− ⎪ . ⎩ βm,1 − (ν + 1)βm,1 − γm,1 = − 2π(m + 1) Given βm,0 , we find γm,0 from the equation βm,0 + (ν + 1)βm,0 + γm,0 = ih+ + (m + 1)βm,1 . Once the complex stress functions ϕn and ψn have been found, the displacement vector vn is defined as vn (z) = (2μ)−1 κϕn (z) − zϕ n (z) − ψn (z) .

Since

h ds = 0,

V.P. Γ (0)

(0)

we have H+ = −H− for −2 < ν −1. Therefore, T vn ds = 0 . lim ε→0 {q∈Ω− , |q|=ε}

Hence we arrive at V.P.

T vn ds = 0. Γ (1)

(ii) As in Theorem 4.2.10, we can construct the displacement vector vn such that vn(1) (z) = O(|z|n+[ν]+1 ) as z → 0 (1)

(1)

(1)

and vn vanishes at infinity. Furthermore, the stress function hn = hn − T vn on Γ has zero principal vector and zero principal moment with respect to an arbitrary point z0 ∈ Ω+ . (2) (1) We find a displacement vector vn with the stress hn on Γ. To this end, we express the boundary condition of this problem by the Airy function F in Ω− . Let (1) F = b and ∂F/∂n = d on Γ. Taking into account the relation of b and d with hn (see (4.54)), we have d(z) = O(|z|n+1+ν ), b(z) = O(|z|n+2+ν ) for z ∈ Γ\{O}.


307

Let us represent the Airy function F (z) as the sum F1 (z) + F2 (z), where F1 (z) is chosen in such a way that it vanishes at infinity, satisfies the conditions F1 (z) = b(z), ∂F1 (z)/∂n = d(z) on Γ , and admits the estimate ∇k F1 (z) = O(|z|n+2+ν−k ),

k = 0, . . . , n + 2,

in a neighborhood of the origin. As F2 (z) we choose a solution of the problem Δ2 F2 = −Δ2 F1 in Ω− , F2 = ∂F2 /∂n = 0 on Γ ˚ 2,ω (Ω− ) with ω(z) = (1 + |z|2 )1/2 (see Theorem 4.2.9). in W We make the change of variable t = log(1/r), r = |z|. Let Λ be the image of the domain Ω− under the mapping (r, θ) → (t, θ), where (r, θ) are polar coordinates of (x, y). The function U (t, θ) = F2 (e−t , θ) satisfies the equation + * 2 ∂2 ∂ ∂2 ∂2 +2 + 2 + 2 U (t, θ) L(∂t , ∂θ )U (t, θ) = ∂t ∂θ ∂t2 ∂θ (4.58) = H(t, θ) in Λ ∩ {t > R} and the boundary condition U = ∂U/∂n = 0 on ∂Λ ∩ {t > R} for large R, where ∇k H(t, θ) = O(e−(n+2+ν)t ),

k = 0, . . . , n − 2,

t → ∞.

small negative β the solution U (t, θ) of the equation (4.58) belongs to For n+2 ˚ 2 (Λ ∩ {t > R}) (see Theorem 7.2 [18]). W2 ∩W 2 By U(∂t , ∂θ ) we denote the operator of the boundary value problem U(∂t , ∂θ )U = F in Π = {(t, θ) : t ∈ R, 0 < θ < 2π}, U = (∂/∂n)U = 0 on ∂Π . Now we prove that the eigenvalues of the operator pencil ˚ 2 (0, 2π) → W n−2 (0, 2π) U(ik, ∂θ ) : W2n+2 ∩ W 2 2 with a complex parameter k are the numbers k = i/2, ∈ Z, = ±2. The general solution of the equation U(ik, ∂θ )u(θ) = (ik + 2)2 +

∂2 ! ∂2 ! 2 − k u(θ) = 0 + ∂θ2 ∂θ2

with k = 0, ±i, ±2i has the form u(θ) = c1 ekθ + c2 e−kθ + c3 e−kθ e2iθ + c4 ekθ e−2iθ .

(4.59)

308


Satisfying the boundary conditions u(0) = u(2π) = 0 ,

∂u ∂u (0) = (2π) = 0 , ∂θ ∂θ

(4.60)

we arrive at a linear algebraic system of equations in unknowns c1 , c2 , c3 , c4 . The values of k, for which the determinant Det vanishes are eigenvalues of the operator pencil U(ik, ∂θ). Straightforward calculations give Det = −16(sinh 2πk)2 ,

if k = 0, i, 2i .

The multiplicity of each eigenvalue is equal to 2. The linear independent solutions of the boundary value problem U(ik, ∂θ )u(θ) = 0, θ ∈ (0, 2π), u(0) = u(2π) = 0,

∂u ∂u (0) = (2π) = 0 (4.61) ∂θ ∂θ

are u1 (θ) = − cos θ + cos(2 − )θ , 2 2 u2 (θ) = ( − 4) sin θ + sin(2 − )θ , 2 2

= 2 .

For k = ±i, the general solution of (4.59) has the form u(θ) = c1 eiθ + c2 θeiθ + c3 e−iθ + c4 θe−iθ and the determinant of the linear system does not vanish. Hence the boundary value problem (4.61) has only the trivial solution, that is k = ±i are not eigenvalues of the operator pencil U(ik, ∂θ ). For k = 0 and k = ±2i the general solution takes the form u(θ) = c1 + c2 θ + c3 e2iθ + c4 e−2iθ . After substituting it into (4.60) we conclude that, up to a constant, the only solution of (4.61) is given by u(θ) = cos 2θ − 1 . Let us show that the eigenvectors do not generate generalized eigenvectors. The first generalized eigenvector u(1) , corresponding to the eigenvector u, is a solution of the boundary value problem ∂ U(ik, ∂θ )u(θ), θ ∈ (0, 2π) , ∂k ∂u ∂u (0) = (2π) = 0 . u(0) = u(2π) = 0 , ∂θ ∂θ

U(ik, ∂θ )u(1) (θ) = −


309

A necessary condition for solvability of this problem is ∂ U(ik, ∂θ )u, u = 0 . ∂k Substituting the expression for U(ik, ∂θ ) into the last equality, we find ∂

∂u

2 U(ik, ∂θ )u, u = (k − i) 4k(k − 2i)||u||2L2 (0,2π) +

L2 (0,2π) = 0 . ∂k ∂θ Let β satisfy n + ν + 1 < β < n + ν + 3/2 and let m be the largest noninteger not exceeding n+ν+1. The operator pencil U(ik, ∂θ ) has p = 2m eigenvalues in the strip {k : −1/2 < Im k < β}. Therefore, the solution U admits the representation U=

p

cs U s + W

s=1

(see Theorem 4.2.2). Here Us are linearly independent, each Us satisfies the equation LUs = 0 in the domain Λ ∩ {t > R}, vanishes on ∂Λ ∩ {t > R}, and n+2 n+2 Us ∈ / W2,β (Λ ∩ {t > R}). In addition, W ∈ W2,β (Λ ∩ {t > R}). Making the inverse change of variable, we obtain F (r, θ) =

p

ck Uk (− log r, θ) + O(rβ )

k=1

and this equality can be differentiated n times. (2) A displacement vector vn corresponding to F (z) has the form (2)

vn(2) = α + icz + vn0 (z) , where c is a real coefficient. The functions ϕ(z) = z k/2 , ψ(z) = −(k/2 + 1)z k/2 , and ϕ(z) = iz k/2 , ψ(z) = i(k/2 − 1)z k/2 with k = 0, . . . satisfy ϕ(z) + zϕ (z) + ψ(z) = O(xk/2+1 ) , z ∈ Γ± . Duplicating the argument used at the beginning of the present proof, we construct functions ϕ∗ (z) and ψ∗ (z) of the form ϕ∗ (z) =

m k=1 k=2

and ψ∗ (z) =

m k=1 k=2

k/2 zz0 zz0 Rϕ,k log z0 − z z0 − z k/2 zz0 zz0 Rψ,k log . z0 − z z0 − z

310

Chapter 4. Integral Equations of Plane Elasticity in Domains with Peak (2)

Then the displacement vector vn0 can be written as 1 (2) κϕ∗ (z) − zϕ∗ (z) − ψ∗ (z) + w(1) (z) , vn0 (z) = 2μ

(4.62)

where w(1) (z) = O(|z|n+ν ). The equality (4.62) can be differentiated n − 1 times. As in the proof of Theorem 4.2.10, we show that T vn(2) ds = 0. Γ

(1)

(2)

Hence the displacement vector v = vn + vn + vn is a solution of the problem N (e) with the boundary data h and admits the required representation. Corollary 4.2.14. Let h be expressed by h± (z) =

n−1

(k)

h± xk+ν + O(xn+ν ), x ∈ Γ± ,

k=0

for ν > −1, ν = /2, ∈ Z. Then the functions ϕn and ψn in (4.57) have the form k+ν k+ν n n zz0 zz0 ϕn (z) = βk , ψn (z) = γk . z − z0 z − z0 k=1

k=1

4.2.3 Properties of solutions to the problem D (e) The next two assertions are related to the behavior of solutions to the Dirichlet boundary value problem for the Lamé operator on a contour near the peak under mild assumptions on boundary data. Proposition 4.2.15. Let Ω+ have an outward peak and let a vector-valued function g satisfy the conditions ∂kg (z) = O(|z|β−k ), k = 0, 1, 2, ∂sk on Γ± . Then the problem D(e) with the boundary data g has a solution u bounded at infinity and subject to

|q| ∇u(q) dsq < ∞, u(z) = O(|z|γ ) (4.63) Γ

if −1 < γ < β 0, and |∇u(q)|dsq < ∞, u(z) = O(|z|γ ) Γ

if 0 < γ < min{β, 1/2}.

(4.64)


311

Proof. We can construct a solution u satisfying either (4.63) or (4.64) (cf. Theorem 4.2.5) and belonging to W 1,ω (|z| > R) (cf. Theorem 4.2.7 and Lemma 4.2.8) for sufficiently large R. Now we check that u is bounded at infinity. By the first Kolosov formula (4.6), Re {ϕ (z)} is a linear combination of the components of ∇u. It follows that Re ϕ (z) is square integrable in a neighborhood of infinity. So, ϕ(z) has the decomposition 0

ϕ(z) = icz +

bk z −k , c ∈ R.

k=−∞

It follows from the equality u(z) = (2μ)−1 (κϕ(z) − zϕ (z) − ψ(z)) that ψ(z) is bounded in |z| > R. Therefore, ψ(z) can be presented in the form ψ(z) =

0

dk z −k .

k=−∞

Finally, we have

u(z) = i(2μ)−1 (κ + 1)cz + O(1).

Since the function ω −1 u is square integrable in |z| > R, we conclude that c = 0. Consequently, u is bounded at infinity. Proposition 4.2.16. Let Ω+ have an inward peak and let g satisfy the conditions ∂kg (z) = O(|z|β−k ), k = 0, 1, 2 , ∂sk on Γ± . Then the problem D(e) with the boundary data g has a solution u bounded at infinity and satisfying |u(q)|dsq < ∞, |q||∇u(q)|dsq < ∞. (4.65) Γ

Γ

Proof. Let u be an extension of g onto Ω− , equal to zero outside a certain disk, and satisfying Δ∗ u(1) = O(|z|β−3 ) as z → 0. We look for a function u(2) such that (1)

Δ∗ u(2) = −Δ∗ u(1) in Ω− , u(2) = 0 on Γ\{O} .

(4.66)

After the change of variable z = 1/ζ with ζ = ξ + iη, the differential equation in (4.66) takes the form Δ∗ U + L(∂ξ , ∂η )U = F in Λ, (4.67) where U (ξ, η) = u(2) (ξ/|ζ|2 , η/|ζ|2 ), Λ is the image of Ω+ and L(∂ξ , ∂η ) is the second-order differential operator with coefficients admitting the estimate O(ξ −β−1 ). Since F ∈ L2 (Λ), the problem (4.67) has a unique solution in the space W22 (Λ). Hence u = u(1) + u(2) satisfies (4.65).

312


4.2.4 Uniqueness theorems Here we collect classical uniqueness theorems for the Dirichlet and Neumann problems for the Lamé system in domains with peak to be referred to henceforth. We start with the Neumann problem. Theorem 4.2.17. Let Ω+ have an inward or outward peak. Suppose that v = (v1 , v2 ) is a solution of the problem N (e) with zero boundary data and obeys the conditions: (i) v(z) = O(1), ∇v1 (z), ∇v2 (z) = O(|z|β ) as z → 0 in Ω− with some β > −1; (ii) v(z) = O(|z|−1 ), ∇vk (z) = O(|z|−2 ), k = 1, 2, as z → ∞. Then v(z) = 0 in Ω− . Proof. The quadratic form E(v, v) in the first Betti formula (4.10) can be written as 2 μ ∂vi ∂vj 2 E(v, v) = λ(div v)2 + + . 2 i,j=1 ∂xj ∂xi Here x1 = x and x2 = y. Since T v = 0 on Γ\{O}, the third Betti formula (4.12) implies E(v, v) = 0 in Ω− . In particular, ∂v2 ∂v1 = 0, = 0, ∂x ∂y

∂v1 ∂v2 + = 0. ∂y ∂x

(4.68)

By the Lamé equation (4.3) we have Δv1 = 0 and Δv2 = 0 in Ω− . Combining this with (4.68), we find that ∂ 2 v2 /∂x2 = 0 and ∂ 2 v1 /∂y 2 = 0. Therefore, ∂v2 /∂x and ∂v1 /∂y are constants in Ω− . This implies that v1 and v2 are linear functions. Since v is bounded at infinity, we conclude that v = 0. Using quite similar arguments, we can prove a uniqueness theorem for the Dirichlet problem for the Lamé system. Theorem 4.2.18. Let Ω+ have either an inward or outward peak. Suppose that u = (u1 , u2 ) is a solution of the problem D(e) with zero boundary data. Also let u(z) − u(∞) = O(|z|−1 ), ∇uk = O(|z|−2 ), k = 1, 2, z → ∞, and u(z) = O(1),

∇vk (z) = O(|z|β ), k = 1, 2, as z → 0 in Ω− ,

where β > −1, Then u = 0 in Ω− .

4.3. Integral equations on a contour with inward peak

313

4.3 Integral equations on a contour with inward peak 4.3.1 Integral equations of the problem D (i) Our next goal will be to show that the modified homogeneous boundary equation (4.22) has only the trivial solution.

Lemma 4.3.1. Let Ω+ have an inward peak and let α = max(|κ+ (0)|, |κ− (0)|). Then, for any σ = σ (1) , σ (2) from M, 1 (1) (1) σ+ − σ− (x) 2 δ (2) dτ μ (2) + O(1) , σ+ − σ− (τ ) ± 2π(λ + 2μ) 0 x−τ 1 (2) (2) (2) σ+ − σ− (x) (W σ) (z) = ∓ 2 δ (1) dτ μ (1) + O(1) σ+ − σ− (τ ) ∓ 2π(λ + 2μ) 0 x−τ (W σ)

(1)

(z) = ∓

with the upper sign if z ∈ {(x, y) : κ+ (x) < y < αx, x ∈ (0, δ)} and with the lower sign if $ # z ∈ (x, y) : −αx2 < y < κ− (x), x ∈ (0, δ) . Proof. We use the equality 2 1 1 1 Re (z/¯ z) Im (z/¯ z) x xy = I+ z) −Re (z/¯ z) |z|2 xy y 2 2 2 Im (z/¯ and write the kernel (4.8) of the operator W in the form ⎞ ⎛ μ λ+μ λ+μ RK3 K2 + I K3 K1 + 1 ⎜ ⎟ λ + 2μ λ + 2μ (λ + 2μ) ⎠, ⎝ λ+μ λ+μ 2π − μ K + I K3 RK3 K1 − 2 λ + 2μ 2π(λ + 2μ) (λ + 2μ) where K1 τ (z) =

∂ 1 dsq , log ∂nq |z − q| Γ ∂ 1 dsq , K2 τ (z) = τ (q) log ∂sq |z − q| Γ z−q ∂ 1 dsq RK3 τ (z) = τ (q) Re log z ¯ − q ¯ ∂n |z − q| q Γ and

I K3 τ (z) =

τ (q)

τ (q) Im Γ

z−q ∂ 1 dsq . log z¯ − q¯ ∂nq |z − q|

(4.69)

314


By Lemma 3.3.1, for τ ∈ Mβ , β > −1, we have K1 τ (z) = ±π τ+ (x) − τ− (x) + O(1) and

K2 τ (z) = ∓

δ

0

du + O(1) , z = x + iy . τ+ (u) − τ− (u) x−u

Hence it suffices to show that z−q ∂ 1 dsq = O(1). τ (q) log z ¯ − q ¯ ∂n |z − q| q Γ Using the arguments and notation from the proof of Lemma 3.3.1 we arrive at the inequality z−q ∂ 1 − dsq τ (q) log z − q ∂n |z − q| q Γ± ξ + i(η(ξ) + ν) dξ = ν τ (x) + O(1) . 2 2 |ξ|<x/2 ξ − i(η(ξ) + ν) ξ + (η(ξ) + ν) We write the integral on the right-hand side as 1 1 dξ − 2 (ξ − iν)2 |ξ|<x/2 (ξ − i(η(ξ) + ν)) dξ + O(1) = J1 (x) + J2 (x) + O(1) . − 2 |ξ|<x/2 (ξ − iν) Since

1 |η(ξ)| |ξ| + |η(ξ)||η(ξ) + ν| + η(ξ)2 1

c −

(ξ − i(η(ξ) + ν))2 (ξ − iν)2 (ξ 2 + (η(ξ) + ν)2 )(ξ 2 + ν 2 ) 3 ξ 2 (|η(ξ)| + ν) |ξ| |ξ| + c c 2 2 , 2 2 2 2 2 2 2 (ξ + (η(ξ) + ν) )(ξ + ν ) (ξ + (η(ξ) + ν) )(ξ + ν ) ξ + ν2 we find that |ν τ (x) J1 (x)| c|ν τ (x)|

|ξ|<x/2

ξ2

x2 |ξ| ν dξ c log + 1 xβ+1 = O(1) . 2 2 +ν x 4ν

This along with

dξ dξ

|ν τ (x) J2 (x)| = ν τ (x) = τ (x)

ν 2 2

(ξ − iν) (ξ − iν) |ξ|<x/2 |ξ|x/2 dξ ν c |τ (x)| c xβ+1 ν|τ (x)| 2 2 x |ξ|x/2 ξ + ν completes the proof of the lemma.


315

Theorem 4.3.2. Let Ω+ have an inward peak. The boundary integral equation 2−1 σ − W σ + c1 A1 + c2 A2 + c3 A3 = 0

(4.70)

has only the trivial solution in the Cartesian product M × R3 . Proof. Let the pair (σ, c) with σ ∈ M and c ∈ R3 be a solution of the equation (4.70). Consider the displacement vector W σ(z) + c1 A1 (z) + c2 A2 (z) + c3 A3 (z), z ∈ Ω+ ,

(4.71)

with zero boundary value on Γ\{O}. We show that this vector-valued function is equal to zero in Ω+ . Let us use the change of variable t = log 1/r and denote the components of (4.71) by W1 (z), W2 (z). Then U (t, θ), with the components W1 (e−t , θ) cos θ + W2 (e−t , θ) sin θ, −W1 (e−t , θ) sin θ + W2 (e−t , θ) cos θ , is a solution of the problem L(∂t , ∂θ )U = (Δ∗ + K)U = 0 in Λ,

U = 0 on ∂Λ,

where (r, θ) are polar coordinates of (x, y) and Λ is the image of Ω+ (see Theorem (4.2.5).) Here by K we denote the first-order differential operator with constant coefficients: −λ + 2μ, −(λ + 3μ)∂/∂θ . K= (λ + 3μ)∂/∂θ −μ Since the potential W σ(z) admits at most a power growth as z → 0, there exists β < 0 such that U ∈ L2,β (Λ). By the local estimate (4.44) we obtain that 2 ˚ 1 (Λ). The eigenvalues of the operator pencil D(ik, ∂θ ) : ∩W U belongs to W2,β 2,β 2 ˚ 1 (0, 2π) → L2 (0, 2π) are the numbers i/2, where is a nonzero integer W2 ∩ W 2 (see the proof of Theorem 4.2.5). According to Theorem 4.2.1, the operator L is invertible if β = i/2, = ±1, ±2, . . . . Since U ∈ ker L, it follows from the asymptotic representation of U (see Theorem 4.2.2), that either U (t, θ) = O(e−t/2 ) or U (t, θ) grows faster than et/(2+ε) (ε > 0) as t → +∞. In the first case the energy integral of (W1 , W2 ) is finite and therefore U is equal to zero in Ω+ . Now we show that the second case is impossible. Let χ be a C ∞ -function equal to zero outside of a small neighborhood of the origin and let χ = 1 near the origin. The components of χσ will be denoted by σ (1) , σ (2) . Since the boundary values of W σ are bounded on Γ\{O}, by Lemma 4.3.1 we have δ 1 (1) μ dτ (1) (2) (2) − σ+ − σ− (x) + = O(1), σ+ − σ− (τ ) 2 2π(λ + 2μ) 0 x−τ δ 1 (2) μ dτ (2) (1) (1) − σ+ − σ− (x) − = O(1). σ+ − σ− (τ ) 2 2π(λ + 2μ) 0 x−τ

316


(1) (1) (2) (2) These equalities imply that the functions σ+ − σ− and σ+ − σ− are integrable with any degree p > 1 in a neighborhood of the origin. Therefore, W σ(z) is also integrable with any degree p > 1 near the origin along any ray {(x, y) : x = αt, y = βt, t > 0} with α < 0. Since Aj = O(1), j = 1, 2, 3, our goal is achieved. Hence we have W σ(z) + c1 A1 (z) + c2 A2 (z) + c3 A3 (z) = 0 in Ω+ . Let u− be a solution of the problem D(e) with the boundary data σ. We notice that V T u− − W u− = u− (∞) in Ω+ . (4.72) Here by V τ (z) we denote the single layer potential with density τ and with the kernel Γ(z, q) − Γ(z, 0), q ∈ Γ\{O}. Since W u− = −

3

ck Ak ,

k=1

by (4.72) we find V T u− +

3

ck Ak = u− (∞) in Ω+ .

k=1

The limit relations for the single layer potential imply 3 1 − − = 0 on Γ\{O}. T V Tu + ck Ak + u 2 k=1

Let wk− be the solution of the problem Δ∗ wk− = 0 in Ω− ,

T wk− = T Ak on Γ\{O}

constructed in the proof of Theorem 4.2.10. Since the boundary function T Ak does not satisfy the conditions of Corollary 4.2.12, it follows that the solution wk− does not belong to M. Hence V T u− (z) = −u− (z) −

3

ck wk− (z) + u0 (z) in Ω− ,

(4.73)

k=1

where u0 is the displacement vector satisfying T u0 = 0 on Γ\{O}. We substitute (4.73) into the identity W u− − V T u− = u− − u− (∞) in Ω−


and obtain W σ(z) = u0 (z) −

3

317

ck wk− (z) − u− (∞), z ∈ Ω− .

(4.74)

k=1

From the jump formula for W σ we see that σ = −u0 +

3

ck wk− + u− (∞) −

k=1

3

ck Ak on Γ\{O}.

(4.75)

k=1

Since the density σ belongs to M, it follows from (4.74) that the function w0 defined by w0 (ζ) = u0 (ζ −1 ) and its gradient grow not faster than a power function 2 in the image Λ of Ω− . Therefore, w0 ∈ W2,β (Λ) with a small negative β. Using −2 the relations ∇u0 (z) = O(|z| ) as z → ∞ and T u0 = 0 on Γ\{O}, we find from Betti’s formula (4.11) that T u0 ds = 0 for 0 < x < δ . {κ− (x) 0. We set c1 = d11 , c2 = (Qd12 + d31 )/(Q2 + 1), c3 = d22 ,

! where κ = (λ + 3μ)/(λ + μ) and Q = (α+ + α− ) − (α+ − α− )/2κ + 2α− /2. Then the displacement vector 3 + u (z) − ck Ak (z) k=1

is the sum of a linear combination of the functions U12 (z) − Q U31 (z), U21 (z), U32 (z), U41 (z), U42 (z) and a function W1 (z) admitting the estimate O(|z|2+ε ). According to Theorem 4.2.10 and Corollaries 4.2.11, 4.2.12, the problem N (e) with the boundary data 3 T u+ − ck Ak k=1

has a solution whose trace on Γ\{O} is in the class M. Now, once the vector (c1 , c2 , c3 ) is chosen, by U + and U − we denote solutions of the problems D(i) and D(e) with the data g−

3

ck Ak

on Γ\{O}.

k=1

Then T U + admits the estimate O x−1/2 on Γ\{O}. According to Proposition 4.2.16, the displacement vector U − (z) is bounded at infinity and

T U − (p) |p|dsp < ∞ . Γ

Hence (see (4.19)) we obtain g(p) −

3 k=1

ck Ak (p) = V T U + − T U − (p) + U e (∞), p ∈ Γ\{O}.


323

Let V − be a solution of the problem N (e) : V − = 0 in Ω− ,

T V − = T U + on Γ\{O}, V − (∞) = 0 .

Applying Betti’s formula (4.11) to w = V − − U − + U − (∞) in Ω− and using the jump relation for W w, we obtain w(p) − 2W w(p) = − 2V T U + − T U − = − 2g(p) + 2

3

ck Ak (p) + 2U − (∞), p ∈ Γ\{O}.

k=1

Hence the function σ(p) = w(p) − U − (∞) = V − (p) − g(p) +

3

ck Ak (p), p ∈ Γ\{O}

k=1

with chosen c1 , c2 , c3 is a solution of the equation (4.81) in the class M. By Theorems 4.2.5 and 4.2.10, σ has the required asymptotics and by Theorem 4.3.2 the obtained solution (σ, c1 , c2 , c3 ) is unique.

4.3.2 Integral equation of the problem N (e) ˚21 outside the peak if We say that a function v(z), z ∈ Ω+ , belongs to the space W ∞ ˚ 1 (Ω+ ). for any C0 -function κ such that O ∈ / supp κ one has κv ∈ W 2 Lemma 4.3.4. Let Ω+ have an outward peak. If a vector-valued function v(z), ˚ 1 outside the peak, satisfies Δ∗ v = 0 in a vicinity of the z ∈ Ω+ , belongs to W 2 peak, and v(z) and ∇v(z) admit at most a power growth, then there exists β < 0 such that v(z) and ∇v(z) satisfy the estimate O exp(βRe z1 ) as z → 0. Proof. The function w defined by w(ζ) = (χv)(ζ −1 ) satisfies (Δ∗ + L)w = 0

in Λ = {ζ : ζ −1 ∈ Ω+ }

and w = 0 on ∂Λ. Here χ belongs to C0∞ (R2 ) and is equal to 1 in a neighborhood of the peak and L is a second-order differential operator with coefficients vanishing ˚ 1 (Λ). Hence, by the local estimate as ζ → ∞. For small negative β we have w ∈ W 2,β 3 1 (4.37), it follows that w ∈ W2,β ∩ W2,β (Λ). The operator D(∂ξ , ∂η ) of the boundary value problem # $ Δ∗ u = f in Π = (ξ, η) : ξ ∈ R, −κ+ (0)/2 < η < −κ− (0)/2 , u = 0 on ∂Π 2 ˚ 1 (Π) into L2,β (Π). The eigenvalues of the operator continuously maps W2,β ∩W 2,β pencil ˚21 (−κ+ (0)/2, −κ− (0)/2) → L2 (−κ+ (0)/2, −κ−(0)/2) D(ik, ∂η ) : W22 ∩ W

324


are nonzero solutions of the equation α2 k 2 = κ2 (sinh k α)2 , where α = (κ+ (0) − κ− (0))/2 and κ = (λ + 3μ)/(λ + μ). We can assume that there in the strip {k : β < Im k < −β}. Then w belongs to 3 are no1 eigenvalues ˚ W2,−β ∩ W (Λ) (see Theorem 4.2.2). The Sobolev embedding theorem implies 2,−β that w and ∇w admit the estimate O exp(βξ) as ξ → ∞. The following two theorems concern the existence of the sole trivial solution of the homogeneous equation (4.18) and the solvability of the nonhomogeneous equation. Theorem 4.3.5. Let Ω+ have an inward peak. Then the homogeneous integral equation −2−1 τ + T V τ = 0 (4.82) has only the trivial solution in the class M. Proof. Let τ ∈ M satisfy (4.82). Integrating this equation over Γ, we obtain τ ds = 0. Γ

Hence the potential v(z) = V τ (z) is a solution of the problem N (i) with the boundary data τ , that is T v(p) = τ (p), p ∈ Γ\{O}. By the Betti integral representation for v we obtain W v = 0 in Ω+ . Therefore, the density v of this double layer potential is a solution of the homogeneous integral equation of the problem D(i) , (1 − 2W ) v(p) = 0, p ∈ Γ\{O}. The restriction of v to Γ\{O} belongs to M, hence, by Theorem 4.3.2, v(p) = 0 for p ∈ Γ\{O}. Since v(z) and ∇v(z), z ∈ Ω− , vanish at infinity and admit at most a power growth as z → 0, Lemma 4.3.4 implies that v(z) and ∇v(z) satisfy the estimate O exp(βRe 1/z) with a negative β. According to the classical uniqueness theorem (see Theorem 4.2.18), we have v = 0 in Ω− . We make the change of variable t = log 1/r and by v (1) (z), v (2) (z) denote the components of the displacement vector v(z), z ∈ Ω+ . Then the vector-valued function U (t, θ) = v(e−t , θ)e−it , where (r, θ) are polar coordinates of (x, y), is a solution of the equation (Δ∗ + K)U = 0 in the image Λ of Ω+


325

satisfying U = 0 on ∂Λ. Here K is the first-order differential operator −λ + 2μ, −(λ + 3μ)∂/∂θ . K= (λ + 3μ)∂/∂θ −μ ˚ 1 (Λ) with a negative β. From the local The vector-valued function v belongs to W 2,β 2 (Λ). Consider the operator estimate (4.44) it follows that v is an element of W2,β 2 ˚ 1 (Π) → L2,β (Π) ∩W D(∂t , ∂θ ) : W2,β 2,β of the boundary value problem (Δ∗ + K)U = F in Π = {(t, θ) : t ∈ R, 0 < θ < 2π}, U = 0 on ∂Π 2 as well as the operator pencil D(ik, ∂θ ) : k ∈ C, continuously mapping W2 ∩ ˚ 21 (0, 2π) into L2 (0, 2π) and examined in Theorem 4.2.5. Then the displaceW ment vector v(z) is represented as a linear combination of linearly independent nonzero solutions of the homogeneous problem D(i) (see Theorem 4.2.2). Since the displacement vector v(z), z ∈ Ω+ , admits the estimate O(|z|β ), β > −1, one has v(z) = d1 ζ1 (z) + d2 ζ2 (z), where ζ1 and ζ2 are solutions of the homogeneous problem D(i) subject to the estimate O(|z|−1/2 ) and d1 , d2 ∈ R. From the limit relations for V τ it follows that τ (p) = d1 T ζ1 (p) + d2 T ζ2 (p), p ∈ Γ\{O} , where T ζ1 (p) = x−3/2 + O x−1/2 log x

and T ζ2 (p) = i x−3/2 + O x−1/2 log x .

Noting that the stresses T ζk , k = 1, 2, do not belong to M, the components of stresses have different orders, and τ ∈ M, we conclude that the coefficients d1 and d2 vanish. Theorem 4.3.6. Let Ω+ have an inward peak and let h belong to Nν , ν > 3, and satisfy hds = 0. Γ

Then the integral equation

−2−1 τ + T V τ = h

has a unique solution in the class M. The solution τ can be represented as τ (z) = α± x−1/2 + O(1).

326


Proof. Let v − denote a solution of the problem N (e) with the boundary data h on Γ, vanishing at infinity. By Theorem 4.2.10, v − has the representation v − (z) = (α(0) + α(1) log x)xν−2 + O(xν−1−ε ) on Γ. By Theorem 4.2.5, the displacement vector u+ , equal to v − on Γ\{O}, is given by u+ (z) =

1 κϕ(z) − zϕ (z) − ψ(z) , 2μ

where ϕ(z) and ψ(z) admit the representations ϕ(z) = βz 1/2 + O(z),

ψ(z) = γz 1/2 + O(z).

Therefore, we have T u+(z) = α± x−1/2 + O(1) on Γ\{O}. Hence

v − = V (T u+ − h) on Γ\{O}.

Let us show that the same equality holds in Ω− . We put v(z) = v − (z) − V (T u+ − h)(z), z ∈ Ω− . As in Theorem 4.2.10, we can prove that ∇v(z) admits the estimate O(|z|−2 ) ˚ 1,ω (Ω− ) with at infinity. Hence, it results by Lemma 4.3.4 that v belongs to W 2 1/2 (e) ˚ 1,ω (Ω− ) ω(z) = (1 + |z| ) . Since the problem D is uniquely solvable in W − (cf. [9]), it follows that v vanishes in Ω . Thus, the density τ = T u+ − h satisfies the integral equation of the problem N (e) . We see from Theorem 4.3.5, that every solution in the class M is unique.

4.4 Integral equations on a contour with outward peak 4.4.1 Integral equation of the problem D (i) Now we start our study of boundary integral equations on contours with outward peak. It is shown in the next assertion that the homogeneous integral equation corresponding to the exterior Neumann problem has a two-dimensional space of solutions. Theorem 4.4.1. Let Ω+ have an outward peak. The homogeneous integral equation (4.17) has a two-dimensional space of solutions in M.

4.4. Integral equations on a contour with outward peak

327

(e) Proof. By Theorem 4.2.13, the homogeneous β problem N , considered in the class of functions admitting the estimate O |z| , β > −1, has a nonzero solution vanishing at infinity and satisfying

ζ(z) = cz −1/2 + O(1). Furthermore, ζ spans a two-dimensional real linear space. From the integral representation ζ(z) = (W ζ)(z) − (V T ζ)(z), z ∈ Ω− , by the jump formula for W ζ we get ζ(p) − 2(W ζ)(p) = −2(V T ζ)(p) = 0, p ∈ Γ\{O}. Thus ζ(p) is a solution of the homogeneous integral equation (1 − 2W ) σ = 0

(4.83)

for the problem D(i) . Now let σ ∈ M be a nontrivial solution of the equation (4.83). The potential W σ and its gradient in Ω+ have a power growth as z → 0. By Lemma 4.3.4, ˚ 1 (Ω+ ). Since W σ(z) = O exp(βRe 1z ) with a negative β and therefore, W σ ∈ W 2 ˚ 1 (Ω+ ), we obtain the homogeneous problem D(i) has only the trivial solution in W 2 that W σ vanishes in Ω+ . Let u− satisfy the problem D(e) with the boundary function σ. We have V T u− − W u− = u− (∞) in Ω+ . Here, by V τ (z) we denote a single layer potential with the kernel Γ(z, q) − Γ(z, 0), q ∈ Γ\{O}. Since W u− vanishes in Ω+ , we have V T u− (z) = u− (∞), z ∈ Ω+ . The limit relations for the single layer potential imply 1 T V T u− + u− (p) = 0, p ∈ Γ\{O}. 2 Thus we have

V T u−(z) = −u− (z) + u0 (z) in Ω− ,

where u0 is a displacement vector satisfying the boundary condition T u0 (p) = 0 on Γ\{O}. From the integral representation (4.14) of u− and (4.84) we obtain W u− (z) = u0 (z) − u− (∞), z ∈ Ω− . Since the potential W u− vanishes at infinity, we have u0 (∞) = u− (∞).

(4.84)

328


The jump formula for W u− implies σ(p) = u− (p) = u0 (p) − u0 (∞),

p ∈ Γ\{O}.

Noting that σ belongs to the class M, we see by (4.84) that u0 and its gradient have at most power growth as z → 0. For any function χ ∈ C0∞ (R2 ), satisfying O∈ / supp χ, we see that χu0 belongs to W21 (Ω− ). From the estimate ∇u0 (z) = O(|z|−2 ) as z → ∞ and Betti’s formula (4.11) we obtain T u0 (z)ds = 0 , 0 < r < δ . {z∈Ω− , |z|=r}

We assume that the Airy function U , generated by the displacement vector u0 in − the domain Ω+ δ = Ω ∩ {|z| < δ}, satisfies ∂U ∂U +i = 0 on Γ± ∂x ∂y (cf. (4.76)). In particular, U is constant on Γ+ and Γ− . We make the change of variable t = log(1/r), r = |z|. Let Λ be the image of Ω+ under the mapping (r, θ) → (t, θ), where (r, θ) are polar coordinates of (x, y). δ By S+ , S− we denote the images of Γ+ and Γ− . Then W (t, θ), equal to U (e−t , θ), is the solution of the equation * + 2 ∂ ∂2 ∂2 ∂2 +2 + 2 + 2 W (t, θ) = 0 in Λ ∂t ∂θ ∂t2 ∂θ and satisfies W = c± , ∂W/∂n = 0 on S± . Moreover, the function W belongs 2 to W2,γ (Λ) with a negative γ. It follows from the local estimate (4.77) that W 3 belongs to W2,γ (Λ). Using the relations (4.2), (4.4) between u0 and U in Ω+ δ , we 2 (Λ) with γ < 0. conclude that W0 (t, θ) = (ρu0 )(e−t , θ) · e−iθ belongs W2,γ Let D(∂t , ∂θ ) denote the operator of the boundary value problem L(∂t , ∂θ )U = (Δ∗ + K)U = F in Π = {(t, θ) : t ∈ R, 0 < θ < 2π}, B(∂t , ∂θ ) U = f on ∂Π , 1/2

2 continuously mapping W2,β (Π) into L2,β (Π) × W2,β (∂Π). Here K has the form

−λ + 2μ −(λ + 3μ)(∂/∂θ) (λ + 3μ)(∂/∂θ) −μ

K=

μ(∂/∂θ) −μ (∂/∂θ) + 1 . −λ(∂/∂θ) + (λ + 2μ) (λ + 2μ)(∂/∂θ)

and B=


329

Consider the auxiliary problem with a complex parameter k L(ik, ∂θ )V (θ) = 0,

θ ∈ (0, 2π) ,

B(ik, ∂θ )V |θ=0,2π = 0 .

(4.85)

The eigenvalues of (4.85) are k = i/2, ∈ Z. In order to show that nonzero eigenvalues have multiplicity 2, we plug a linear combination of functions (4.45) into the boundary conditions B(ik, ∂θ )V = 0, θ = 0 and θ = 2π. Thus we arrive at a linear algebraic system of equations whose determinant, for k = 0, equals sinh 2πk times a constant. Therefore, W0 admits the representation W0 =

p

ck Vk + V0

k=1

(see Theorem 4.2.2), where p is the largest integer not exceeding −2γ, Vk are linear independent vector-valued functions satisfying the equation L(∂t , ∂θ )U = 0 in ΛR = Λ ∩ {t > R} and the boundary condition B(∂t , ∂θ ) U = 0 on ∂Λ ∩ {t > R}. 2 2 / W2,β (ΛR ) and V0 ∈ W2,β (ΛR ) for β ∈ (−1/2, 0). Making Moreover, we have Vk ∈ −t the inverse change of variable r = e we obtain

u0 (z) = (2μ)−1 (κϕ∗ (z) − zϕ ∗ (z) − ψ∗ (z)) + O(|z|−ε ), where ε > −1/2, and ϕ∗ (z) = ψ∗ (z) =

p k=1 p k=1

(k)

zz0 zz0 (k−p−1)/2 (k) log Rϕ , z0 − z z0 − z (k)

Rψ

log

zz0 zz0 (k−p−1)/2 . z0 − z z0 − z

(k)

Here Rϕ and Rψ are polynomials of degree [(k − 1)/2], and z0 is a fixed point in Ω+ . Since σ ∈ M, we have u0 (z) = O |z|−1/2 . Thus σ is the restriction to Γ\{O} of a solution of the homogeneous problem N (e) . The next assertion concerns the solvability of the nonhomogeneous equation (4.17) Theorem 4.4.2. Let Ω+ have an outward peak and let g belong to Nν , ν > 0. Then the equation (4.17) is solvable in the class M and the homogeneous equation has a two-dimensional space of solutions. Among solutions of (4.17) there is only one

330


with the representation σ(x) = β± xν−1 + O(1), if ν = 1/2 and

σ(x) = β± x−1/2 log x + O(log x),

if ν = 1/2. Proof. By V σ we denote the single layer potential with the kernel Γ(z, q). Let u+ , u− be solutions of the problems D(i) and D(e) . Theorem 4.2.3 implies T u+(z) = ±β0 xν−2 + O(xν−1 ), z = x + iy ∈ Ω+ . By Proposition 4.2.15, T u− is integrable on Γ and the displacement vector u− (z) is bounded at infinity. Hence g(p) = V T u+ − T u− (p) + u− (∞), p ∈ Γ\{O}. Here the single layer potential (4.9) is understood in the sense of the principal value. Let v − satisfy the problem N (e) with the boundary data T u+ . By Theorem 4.2.13 we have ν−1 v − (z) = β± x + O(1) for ν = 1/2, v − (z) = (β± log x + γ± )x−1/2 + O(log x) for ν = 1/2, z ∈ Γ\{O}.

Using the integral representation for w = v − − u− + u− (∞) in Ω− and the jump formula (4.15) for the double layer potential, we obtain w − 2W w = −2V T u+ − T u− = −2g + 2u− (∞) on Γ\{O}. Thus the displacement vector σ = w−u− (∞) = v − −g is a solution of the equation (4.17). By Theorem 4.4.1, the restrictions of solutions of the homogeneous problem N (e) to Γ\{O} (and only such functions) satisfy the homogeneous integral equation of the problem D(i) . The space of these solutions in M is two-dimensional. We choose two linear independent solutions ζR and ζI such that κ + 1 −1/2 x + O(1), μ κ + 1 −1/2 + O(1), z = x + iy ∈ Γ\{O}. ζI (z) = ∓ ix μ

ζR (z) = ∓

(4.86)

Therefore, solutions of the equation (4.17) in M have the form σ = v − − g − ( Re c)ζR + ( Im c)ζI . The complex constant c can be chosen in such a way that σ has the required asymptotics.


331

4.4.2 Integral equation of the problem N (e) As in the previous section, we start with the trivial solution of the homogeneous equation. The functions 1 and 2 used below are defined in the introduction to this chapter. We have already noted there that, in order to eliminate the conditions of solvability of the equation (4.18), i.e., the conditions hζR ds = 0, hζI ds = 0, Γ

Γ

(see (4.86)) we consider the modified boundary equation (4.23). Theorem 4.4.3. Let Ω+ have an outward peak. Then the homogeneous equation −2−1 τ + T V τ + t1 T 1 + t2 T 2 = 0

(4.87)

has only the trivial solution in the Cartesian product M × R2 . Proof. Let τ be a solution of (4.87). We define the displacement vector v(z) by v(z) = V τ (z) + t1 1 (z) + t2 2 (z), z ∈ Ω− . Integrating (4.87) over the boundary Γ, we find τ (p)dsp = 0. Γ

Therefore, the potential V τ (z) takes the form V τ (z) = {Γ(z, q) − Γ(z, 0)} τ (q)dsq . Hence we have v(z) = O(|z|−1 ), ∇v(z) = O(|z|−2 ),

as |z| → ∞.

This shows that v(z) is a solution of the problem N (e) with zero boundary data and v(z) = O(1), ∇v(z) = O(|z|β ), β > −1, as z → 0. The classical uniqueness theorem (see Theorem 4.2.17) implies v(z) = c, z ∈ Ω− . Since v(z) vanishes at infinity, it follows that V τ (z) = −t1 1 (z) − t2 2 (z), z ∈ Ω− .

332


We extend the functions k (z), z ∈ Γ, k = 1, 2, onto Ω+ as solutions of the + problem D(i) for the Lamé system. Let + 1 (z) and 2 (z) denote these extensions. The vector-valued function + 0 (z) = V τ (z) + t1 + 1 (z) + t2 2 (z), z ∈ Γ\{O} ,

is bounded in Ω+ , vanishes on Γ\{O} and its gradient ∇0 admits at most power growth. According to Lemma 4.3.4, 0 (z) satisfies 1 as z → 0 0 (z) = O exp β Re z with a negative β. We have V τ (z) = −

2

tk + k (z) + 0 (z).

1

The jump formula (4.16) for the single layer potential entails τ (z) = −

2

tk T + k (z) +

1

2

tk T k (z) + T 0 (z), z ∈ Γ.

1

+ We have T k (z) = O(|z|−1/2 ), k = 0, 1, and T + 1 (z), T 2 (z) have the order −3/2. Since the components of these functions have different singularities, it follows that t1 and t2 vanish. Thus, V τ (z) = 0 in Ω− . By Lemma 4.3.4 applied to V τ in Ω+ and by the classical uniqueness theorem, we conclude that V τ (z) = 0, z ∈ Ω+ . Thus the density τ given by

τ (z) = T (V τ )− (z) − T (V τ )+ (z), z ∈ Γ\{O},

vanishes on Γ.

Let ζR and ζI be solutions of the homogeneous problem N (e) introduced in Theorem 4.4.2. Theorem 4.4.4. Suppose that Ω+ has an outward peak. Let h belong to Nν , ν > 0, and satisfy hds = 0. Γ

Then the integral equation (4.23) has a unique solution in M×R2 and this solution can be represented as follows: τ (z) = β± xν−1 + O(x−1/2 ) for 0 < ν < 1/2, τ (z) = γ± log x + β± x−1/2 + O(log x) for ν = 1/2, τ (z) = γ± x−1/2 + β± xν−1 + O(log x) where z = x + iy ∈ Γ\{O}.

for 1/2 < ν < 1,


333

Proof. Let v − denote the solution of the problem N (e) with the boundary data h such that v − (∞) = 0. It follows from Theorem 4.2.13 that v − has the form 1 κϕ(z) − zϕ (z) − ψ(z) + O(z 2 log z), v − (z) = 2μ where ϕ(z) = δ1,0

zz0 z − z0

1/2 + δ2,0

zz0 z − z0 zz

1+ν 3/2 zz0 zz0 0 + δ3,0 log + δ3,1 + β0 , z − z0 z − z0 z − z0 1/2 zz0 zz0 + ε2,0 ψ(z) = ε1,0 z − z0 z − z0 1+ν zz 3/2 zz0 zz0 0 + ε3,0 log + ε3,1 + γ0 , z 0 ∈ Ω+ . z − z0 z − z0 z − z0 Thus v − admits the representation v − (z) = c0 ± c1 x1/2 + c2 x + d± x3/2 log x + f± x3/2 + g± x1+ν + O(x3/2 log x) on Γ, where c1 = (1 + κ)δ1,0 /2μ, c2 = (1 + κ)δ2,0 /2μ. We apply Betti’s formula (4.11) to v − and first to ζR and then to ζI in Ω+ ε = Ω− \{|z| < ε}. Taking the limit as ε → 0, we obtain 1 1 hζR ds, Im c1 = hζI ds. Re c1 = 4π Γ 4π Γ Setting t1 =

1 μ 1 + κ 2π

h ζR ds, Γ

t2 =

1 μ 1 + κ 2π

h ζI ds , Γ

we find that the problem N (e) with the boundary data h − t1 T 1 − t2 T 2 has a solution on Γ± of the form v(z) = c0 + c2 x + d± x3/2 log x + f± x3/2 + g± x1+ν + O(x3/2 log x). The solution of the problem D(i) with the given displacement c2 z on Γ admits the estimate O(z) and this estimate can be differentiated at least once. Hence, by Theorem 4.2.3, it follows that the solution u+ of the problem Δu+ = 0 in Ω+ ,

u+ = v − on Γ,

334


admits the representation u+ (z) =

1 κϕ(z) − zϕ (z) − ψ(z) + O(z log z), 2μ

which is differentiable at least once. Here ϕ(z) = β0 z ν + β1 z 1/2 , ψ(z) = γ0 z ν + γ1 z 1/2 for 0 < ν < 1/2, ϕ(z) = β0 + β1 log z z 1/2 , ψ(z) = γ0 + γ1 log z z 1/2 for ν = 1/2, ϕ(z) = β0 z 1/2 + β1 z ν , ψ(z) = γ0 z 1/2 + γ1 z ν

for

(4.88)

1/2 < ν < 1.

Therefore, the stress T u+ has the following representation on Γ± : T u+ (z) = β± xν−1 + O(x−1/2 ) for −1/2

+

T u (z) = (γ± log x + β± )x −1/2

+

T u (z) = γ± x

ν−1

+ β± x

0 < ν < 1/2,

+ O(log x) for

+ O(log x),

for

ν = 1/2,

(4.89)

1/2 < ν < 1 ,

where z = x + iy. On Γ we have v = V T u+ − h + t1 T 1 + t2 T 2 .

(4.90)

The single layer potential V T u+ is bounded in a neighborhood of the origin and its gradient admits the estimate O(|z|ν−1 ). Hence, by a classical uniqueness theorem (see Theorem 4.2.18), we obtain that (4.90) remains valid in Ω− . Consequently, the density τ = T u+ − h + t1 T 1 + t2 T 2 satisfies the integral equation of the problem N (e) and has the required asymptotics. Theorem 4.4.3 implies that the constructed solution of the equation (4.23) is unique. Unifying Theorems 4.4.3 and 4.4.4, we obtain the following result. Theorem 4.4.5. Let Ω+ have an outward peak. Furthermore, let h belong to Nν , ν > 0, and satisfy hds = 0, hζR ds = 0 and hζI ds = 0. Γ

Γ

Then the equation (4.23) is uniquely solvable in M.

Γ

Bibliography

[1] J. Hadamard, Lectures on Cauchy’s Problem in Linear Partial Differential Equations. Yale University Press, New Haven, 1922. [2] K.F. Andersen, Weighted norm inequalities for Hilbert transforms and conjugate functions of even and odd functions. Proc. Amer. Math. Soc. (1976), 56, 99–107. [3] R. Duduchava, T. Latsabidze, A. Saginashvili, Singular integral operators with the complex conjugation on curves with peaks. Integral Equation and Operator Theory (1995), 22, 1–36. [4] R. Duduchava, B. Silbermann, Boundary value problems in domain with peaks. Memoirs on Differential Equations and Mathematical Physics (2000), 21, 1–122. [5] G.M. Goluzin, Geometric Theory of Functions of a Complex Variable, “Nauka”, Moskow, 1966; English transl.: Transl. Math. Monogr., vol. 26, Amer. Math. Soc., Providence, RI, 1969. [6] H.G. Hahn, Elastizit¨ atstheorie. Grundlagen der linearen Theorie und Anwendungen auf eindimensionale, ebene und r¨ aumliche Probleme. Leitf¨ aden der Angewandten Mathematik und Mechanik, Band 62, B.G. Teubner, Stuttgart, 1985. [7] G.H. Hardy, J.E. Littlewood, G. Polya, Inequalities, Cambridge Univ. Press, 1934. [8] R.A. Hunt, B. Muckenhoupt, R.L. Wheeden, Weighted norm inequalities for the conjugate function and Hilbert transform. Trans. Amer. Math. Soc. (1973), 176, 227–251. [9] R. Janssen. Elliptic problems on unbounded domains. SIAM J.Math. Anal. (1986), 17, no. 6, 1370–1389. [10] O.D. Kellogg, On the derivatives of harmonic functions on the boundary, Trans. Amer. Math. Soc. (1931), 33, no. 2, 486–510. [11] V.A. Kondratiev, Boundary value problems for elliptic equations in domains with conical or angular points. Trudy Moskov. Mat. Ob. (1967), 16, 209–292. English transl.: Trans. Moscow Math. Soc. (1967), 16, 227–313.

336

Bibliography

[12] V.D. Kupradze, T.G. Gegelia, M.D. Basheleishvilli, T.V. Burchuladze, Three-dimensional Problems of the Mathematical Theory of Elasticity and Thermoelasticity. Nauka, Moscow, 1976. [13] D.S. Kurtz, R.L. Wheeden, Results on weighted norm inequalities for multipliers. Trans. Amer. Math. Soc. (1979), 255, 343–362. [14] P.D. Lax, A.N. Milgram, Parabolic equations. Contributions to the theory of partial differential equations. Annals of Math. Studies (1954), no. 33, 167– 190. [15] Ya.B. Lopatinskii, On certain types of singular integral equations. Teor. i prikl. matem., L’vovskii universitet (1963), II, 53–57 (Ukrainian). [16] V.G. Maz’ya, Boundary integral equations. Analysis, IV, Itogi Nauki i Tekhniki, Sovrem. Probl. Mat. Fund. Naprav., vol. 27, VINITI, Moscow, 1988, 131–228; English transl.: Encyclopaedia Math. Sci., vol. 27, SpringerVerlag, 1991, 127–222. [17] V. Maz’ya, Sobolev Spaces. Springer, 1985. [18] V.G. Maz’ya, B.A. Plamenevskii, Estimates in Lp and in H¨ older classes, and the Miranda-Agmon maximum principle for solutions of elliptic boundary value problems in domains with singular points on the boundary. Math. Nachr. (1978), 81, 25–82. English transl.: Amer. Math. Soc. Transl. (1984), 123, 1–56. [19] V.G. Maz’ya, A.A. Soloviev, On the integral equation of the Dirichlet problem in a plane domain with peaks at the boundary, Matem. Sbornik (1989), 180, no. 9, 1211–1233; English transl.: Math. USSR-Sb. Vol. (1991), 68, no. 1, 61–83. [20] V.G. Maz’ya, A.A. Soloviev, On the boundary integral equation of the Neumann problem in a domain with a peak, Trudy Leningrad. Mat. Ob. (1990), 1, 109–134; English transl.: Amer. Math. Soc. Transl. (1993), 155, no. 2, 101–127. [21] V. Maz’ya, A. Soloviev, Lp -theory of a boundary integral equation on a cuspidal contour. Applicable Analysis (1997), 65, 289–305. [22] V. Maz’ya, A. Soloviev, Lp -theory of boundary integral equation on a contour with outward peak. Integral Equations and Operator Theory (1998), 32, pp. 75–100. [23] V. Maz’ya, A. Soloviev, Lp -theory of boundary integral equation on a contour with inward peak. Zeitschrift f¨ ur Analysis und ihre Anwendungen (1998), 17, no. 3, 641–673. [24] V. Maz’ya, A. Soloviev, Lp -theory of direct boundary integral equations on a contour with peak. In: Mathematical Aspects of Boundary Element Methods (M. Bonnet, A.-M. S¨ andig and W.L. Wendland eds.), Chapman & Hall/CRC Research Notes in Mathematics 414, London, 2000, 203–214.

Bibliography

337

[25] V. Maz’ya, A. Soloviev, The direct method for boundary integral equations on a contour with peak. Georgian Mathematical J. (2003), 10, no. 3, 573–593. [26] V.G. Maz’ya, A.A. Soloviev, Integral equations of logarithmic potential theory in H¨ older spaces on a contour with peaks. Algebra i Analis (1998), 10, 5, 85–142. English transl.: St. Petersburg Math. J. (1999), 10, no. 5, 791–832. [27] V. Maz’ya, A. Soloviev, Boundary integral equations of the logarithmic potential theory for domains with peaks. Atti Accad. Naz. Lincei Cl. Sci. Fis. Mat. Natur. Rend. Lincei (1995), 9, Mat. Appl. 6, no. 4, 211–236. [28] V. Maz’ya, A. Soloviev, Boundary integral equation of plane elasticity in domains with peak. Georgian Mathematical J. (2001), 8, no. 3, 571–614. [29] V. Maz’ya, A. Soloviev, Boundary integral equation of plane elasticity in domains with peak. Addendum, Georgian Mathematical J. (2002), 9, no. 3, 403–404. [30] Y. Meyer, Continuité sur les espaces de H¨ older et de Sobolev des opérateurs définis par les intégrales singulières. In : Recent Progress in Fourier Analysis. (I. Peral J.-L. Rubio de Francia (eds.).) – Amsterdam: North-Holland, 1985. 45–172. [31] B. Muckenhoupt, Weighted norm inequalities for the Hardy maximal function. Trans. Amer. Math. Soc. (1972), 165, 207–226. [32] N.I. Muskhelishvili, Some Basic Problems of the Mathematical Theory of Elasticity. Academia Nauk USSR, Moscow, 1954. English transl.: Noordhoff, Groningen, 1963. [33] V.Z. Parton, P.I. Perlin, Integral Equations in Elasticity Theory. Nauka, Moscow, 1977. ¨ [34] J. Radon, Uber die Randwertaufgaben beim logarithmischen Potential, S.-B. Akad. Wiss. Wien Math.-Nat. Cl. Abt. 2a (1919), 128, 1123–1167. [35] V.Yu. Shelepov, On the index of an integral operator of potential type in the space Lp . Dokl. Akad. Nauk SSSR (1969), 186, no. 6, 1266–1270 (Russian). [36] V.I. Smirnov, N.A. Lebedev, Constructive Theory of Functions of a Complex Variable, M.I.T. Press, Cambridge, Massachusetts, 1968. [37] E.M. Stein, Note on singular integrals. Proc. Amer. Math. Soc. (1957), 8, 250–254. [38] E.M. Stein, Singular Integrals and Differentiability Properties of Functions, Princeton University Press, Princeton, New Jersey, 1970. [39] S. Stoilov, Theory of Functions of a Complex Variable, Edit. Acad. Rep. Pop. Romaine, Bucharest, 1954. [40] S.E. Warschawski, On conformal mapping of infinite strips. Trans. Amer. Math. Soc. (1942), 51, 280–335.

List of Symbols Symbols listed in order of appearance

Chapter 1 Ω± . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . V , S . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1, W, T ............................... κ± . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . C2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . N1,− p,β (Γ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . · N1,− (Γ) . . . . . . . . . . . . . . . . . . . . . . . . . . p,β

Chapter 2 1 2 2 3 3 4 4

R(z) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102 Λ1,α γ (0, δ) . . . . . . . . . . . . . . . . . . . . . . . . . 103 · Λ1,α . . . . . . . . . . . . . . . . . . . . . . . 103 γ (0,δ) Λα γ (0, δ) . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 · Λαγ (0,δ) . . . . . . . . . . . . . . . . . . . . . . . . 103 Λ1,α γ (Γ± ) . . . . . . . . . . . . . . . . . . . . . . . . . 105 Λα γ (Γ± ) . . . . . . . . . . . . . . . . . . . . . . . . . . . 105

Np,β (Γ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 N 1,α (Γ) . . . . . . . . . . . . . . . . . . . . . . . . . . β · N (Γ) . . . . . . . . . . . . . . . . . . . . . . . . . . 4 1,α p,β Λodd 2−β (Γ) . . . . . . . . . . . . . . . . . . . . . L1p,β (Γ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 1,β · L1p,β (Γ) . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 Λeven 2−α (Γ) . . . . . . . . . . . . . . . . . . . . · (Λodd )1,α (Γ) . . . . . . . . . . . . . . . . . . . . Lp,β (Γ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 2−β ................... · Lp,β (Γ) . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 · (Λeven )1,α 2−β (Γ) 1,α 1,+ Np,β (Γ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 Lodd (Γ) . . . . . . . . . . . . . . . . . . . . . 2−β · N1,+ (Γ) . . . . . . . . . . . . . . . . . . . . . . . . . . 4 · 1,α ................... p,β Lodd (Γ) 2−β P(Γ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 α · P(Γ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 Λγ (1, ∞) . . . . . . . . . . . . . . . . . . . . . . . . . · Λαγ (1,∞) . . . . . . . . . . . . . . . . . . . . . . . m = [μ − β − p−1 ] . . . . . . . . . . . . . . . . . . . 4 · α Λγ (1,+∞) . . . . . . . . . . . . . . . . . . . . . . . M1,+ p,β (Γ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 1,α Eβ (R) . . . . . . . . . . . . . . . . . . . . . . . . . . . Mp,β (Γ). . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 · E 1,α (R) . . . . . . . . . . . . . . . . . . . . . . . . · M (Γ) . . . . . . . . . . . . . . . . . . . . . . . . . . 5 β p,β Eβα (R) . . . . . . . . . . . . . . . . . . . . . . . . . . . . H................................... 7 · Eβα (R) . . . . . . . . . . . . . . . . . . . . . . . . . . πI − T . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

,α V . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64 C (G) (0 < α < 1) . . . . . . . . . . . . . . πI + S . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73 · C ,α (G) . . . . . . . . . . . . . . . . . . . . . . . . −1,α (G) . . . . . . . . . . . . . . . . . . . . . . . . . W ext , T ext . . . . . . . . . . . . . . . . . . . . . . . . . 78 C ext Ik . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93 · C −1,α (G) . . . . . . . . . . . . . . . . . . . . . . .

105 105 105 105 105 105 105 105 105 105 106 106 107 107 134 134 134 134

1,α Rext k . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93 Cβ (G) . . . . . . . . . . . . . . . . . . . . . . . . . . 134

340

List of Symbols

· C 1,α (G) . . . . . . . . . . . . . . . . . . . . . . . . 134 h = (h1 , h2 ) . . . . . . . . . . . . . . . . . . . . . . β Eβα (∂G) . . . . . . . . . . . . . . . . . . . . . . . . . . . 134 T (∂ζ , nζ ) . . . . . . . . . . . . . . . . . . . . . . . . . ∗ · Eβα (∂G) . . . . . . . . . . . . . . . . . . . . . . . . 134 Δ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (i) D , D(e) . . . . . . . . . . . . . . . . . . . . . . . . . Eβ1,α (∂G). . . . . . . . . . . . . . . . . . . . . . . . . . 135 N (i) N (e) . . . . . . . . . . . . . . . . . . . . . . . . . ϕE 1,α (∂G) . . . . . . . . . . . . . . . . . . . . . . . 135 β E(· , ·) . . . . . . . . . . . . . . . . . . . . . . . . . . . . Gω,θ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162 N (ν > −1) . . . . . . . . . . . . . . . . . . . . . . ν (Λ0 )α −β (Γ) . . . . . . . . . . . . . . . . . . . . . . . . 168 N . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . D1 , D2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168 M (β > −1) . . . . . . . . . . . . . . . . . . . . . β Wβ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175 M, M . . . . . . . . . . . . . . . . . . . . . . . . . . ext B0 , B1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181 A , A , A . . . . . . . . . . . . . . . . . . . . . . 1 2 3 κ ................................ Chapter 3 k (z), k = 1, 2 . . . . . . . . . . . . . . . . . . . . M, Mext . . . . . . . . . . . . . . . . . . . . . . . . . 215 H (G) . . . . . . . . . . . . . . . . . . . . . . . . . . . . N, Next . . . . . . . . . . . . . . . . . . . . . . . . . . . 215 · H (G) . . . . . . . . . . . . . . . . . . . . . . . . . D(i) , N (e) . . . . . . . . . . . . . . . . . . . . . . . . . 217 H −1/2 (∂G). . . . . . . . . . . . . . . . . . . . . . . D(e) , N (i) . . . . . . . . . . . . . . . . . . . . . . . . . 218 · H −1/2 (∂G) . . . . . . . . . . . . . . . . . . . . . n (z) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 218 W (G) . . . . . . . . . . . . . . . . . . . . . . . . . . 2,β

, ∞

˚ (G) . . . . . . . . . . . . . . . . . . . . . . . . . . −∞ · dt . . . . . . . . . . . . . . . . . . . . . . . . . . 218 W 2,β 0 W2,β

= L2,β . . . . . . . . . . . . . . . . . . . . . . L(∂ξ , ∂η ) . . . . . . . . . . . . . . . . . . . . . . . . . . σxx , σyy , τxy . . . . . . . . . . . . . . . . . . . . . . 273 B (r)(∂ξ , ∂η ) . . . . . . . . . . . . . . . . . . . . . . . λ, μ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274 U(∂ξ , ∂η ) . . . . . . . . . . . . . . . . . . . . . . . . . . Γ(z, ζ) . . . . . . . . . . . . . . . . . . . . . . . . . . . 274 ζR , ζI . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter 4

274 274 274 274 275 276 278 278 278 278 279 279 280 281 281 281 281 281 282 282 282 282 282 330

Index (Ap )-condition, 7 Betti formula first, 276, 312 second, 276, 317, 323, 328, 333 third, 276, 312 Betti integral representation, 324 Biharmonic equation, 275 Biharmonic function, 299 Boundary integral equation, x, 1–3, 34, 60, 73, 78, 80, 85, 98, 101, 151, 175, 273, 277, 315, 326 Calderon A.P., ix Carleman T., ix Classical uniqueness theorem, 312, 320, 324, 331, 334 Complex potential, 60, 239, 251, 267, 280, 285, 289, 297, 298, 302–304 Complex stress functions, 280, 285, 290, 296, 298, 305, 306 Conjugate function, 8, 34, 43, 46, 52, 53, 73, 76, 79–86, 88, 89, 92–94, 97, 142–147, 161, 170, 172, 173, 175, 176, 178, 181, 189, 242, 267, 272 Contour with peak, 1, 9, 16, 23, 29, 60, 68, 73, 86, 98, 101, 103, 110, 157, 166, 175, 179, 187, 215, 216, 254, 326 Direct method of integral equations, 95 Dirichlet P., ix Dirichlet boundary value problem, 2 Dirichlet problem, ix Domain with inward peak, 3, 40, 78, 80, 85, 92, 103, 223, 289, 295 with outward peak, 3, 34, 36, 68, 73, 92, 151, 278, 284 Double layer potential, x, 2, 254

Elastic potential, xi Elasticity problems, 275, 281 Elasticity theory, 275, 276, 281 First Betti formula, 276, 312 Fourier multiplier, 7 Fourier transform, 7 Fredholm I., ix Giraud G., ix Green formula, 97, 252, 263, 267 Hardy maximal function, 6 Hardy space, 44, 224 Hardy’s inequality, 5, 17–22, 24–28, 30, 31, 33 Hardy–Littlewood maximal operator, 6, 37, 47 Harmonic extension, 36, 40, 41, 43, 46, 52, 53, 64, 67, 68, 73, 77–79, 81, 82, 85, 86, 88, 92–94, 97–99, 135, 138, 139, 142, 144, 145, 147, 148, 150, 162, 173, 174, 176, 182, 184, 187, 189, 190, 216, 217, 244, 247, 249, 253, 258, 271 function, 1, 2, 8, 38, 66, 73, 75, 77, 81–85, 89, 91, 95–97, 99, 136, 140–143, 145, 146, 148–151, 154, 155, 161, 169, 174, 176, 181, 186, 189, 191, 219, 221, 241, 245, 246, 252, 253, 261, 264 Hilbert operator, 7 space, 294 transform, 18, 26, 31, 41, 43, 44, 48, 224 H¨ older O., ix

342 H¨ older condition, 35 H¨ older space, 101 weighted, 103 Homogeneous integral equation, 89 Integral equations of the Dirichlet problems, 68, 77, 79, 92 of the Neumann problems, 73, 77, 79, 85, 92 of plane elasticity, 273 of the Dirichlet problems, 1, 60 of the first kind, ix, 60, 187 of the Neumann problems, 1, 3, 60 of the second kind, ix, 95 Jump formula elastic double layer potential, 317, 323, 327, 328, 330 single layer potential, 332 logarithmic double layer potential, 2, 70, 84, 90, 161, 178, 182 single layer potential, 16, 65, 75, 87, 90, 169, 182, 191, 249 Kellogg O.D., 35 Kellogg theorem, 40, 42, 44, 71, 224, 263 Kelvin transform, 300, 317 Kelvin-Somigliana tensor, 274 Kolosov G.V., xi Kolosov formula, 275, 302, 311 Lamé constants, 274 equation, 312 operator, 310 system, 273, 274, 277, 312, 332 Laplace equation, 3 Laplace P.S., ix Lax-Milgram theorem, 294 Linear integral equation, ix Lopatinskii Ja. B., x Lyapunov A.M., ix

Index Mikhlin S.G., ix Minkowski inequality, 9, 51 Muckenhoupt B., 5, 7 Muskhelishvili N.I., ix Muskhelishvili form, 296, 305 Neumann C., ix Neumann boundary value problem, 2 Neumann problem, ix Parseval formula, 282 Poincaré H., ix Poisson integral, 8 Privalov I.I., 35 Radon I., ix Riemann–Hilbert problem, x Riesz M., 8 Robin G., ix Schwartz class, 135 Schwarz integral formula, 65, 70, 75, 84, 161, 169, 178, 191 Second Betti formula, 276, 317, 323, 328, 333 Shelepov V.Yu., x Single layer potential, x, 2 Sobolev space, 34 Steklov V.A., ix Third Betti formula, 276, 312 Warschawski S.E., 151 Warschawski theorem, 221 Weighted Lp -space, 7 Weighted H¨ older space, 103 Zygmund A., ix

Boundary Integral Equations on Contours with Peaks (Operator Theory: Advances and Applications)

Boundary integral equations on contours with peaks

Boundary Integral Equations on Contours with Peaks (Operator Theory: Advances and Applications)

Integral Equations and Operator Theory - Volume 62

Integral Equations and Operator Theory - Volume 70

Integral Equations and Operator Theory - Volume 68

Integral Equations and Operator Theory - Volume 65

Integral Equations and Operator Theory - Volume 67

Integral Equations and Operator Theory - Volume 69

Integral Equations and Operator Theory - Volume 66

Integral Equations and Operator Theory - Volume 58

Integral Equations and Operator Theory - Volume 61

Integral Equations and Operator Theory - Volume 63

Integral Equations and Operator Theory - Volume 57

Integral Equations and Operator Theory - Volume 59

Integral Equations and Operator Theory - Volume 64