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
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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
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16 23 29 34 34
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40
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43 60 60
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73
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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. Poincar´e, 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 H¨older-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− .
Chapter 1. Lp -theory of Boundary Integral Equations
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)
Chapter 1. Lp -theory of Boundary Integral Equations
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)
1.2. Continuity of boundary integral operators
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 C<∞ (1.29) |Q| Q |Q| Q for any segment Q ⊂ R. Theorem 1.2.5. (Muckenhoupt [31]) The operator M obeys the estimate (1.28) for any f ∈ Lp,w (R) with 1 < p < ∞ if and only if w ∈ (Ap ). Next we consider the Hilbert operator given by 1 f (t) dt. (Hf )(x) = lim ε→+0 π |x−t|>ε 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<|x|<2r
(1.33)
p If w ∈ (Ap ) or w− p−1 ∈ (Ap ), p = p−1 , then there exists a constant Cp independent of f and such that |Tm f |p w Cp |f |p w . (1.34) 1
R
R
8
Chapter 1. Lp -theory of Boundary Integral Equations
Below we need an inequality for norms of conjugate harmonic functions. Given f ∈ Lp (−π, π), we introduce the Poisson integral π 1 (1 − r2 )f (t) iθ U (re ) = dt, r < 1, θ ∈ [−π, π) . (1.35) 2π −π 1 + r2 − 2r cos(θ − t) stand for the harmonic conjugate of U , satisfying the condition U (0) = 0. Let U For a 2π-periodic function f we put π f (θ − t) − f (θ + t) 1 dt . (1.36) f (θ) = lim U (reiθ ) = lim r→1 π ε→0 ε 2 tan t/2 Theorem 1.2.8. (M. Riesz) Let f ∈ Lp (−π, π), 1 < p < ∞. Then f (t) exists almost everywhere, f ∈ Lp (−π, π), and there exists a constant Kp , depending only on p, such that f p Kp f p . (1.37) In what follows we often use the next assertion. Consider the integral operator T : (T f )(x) = K(x, y)f (y)dy R
with the kernel K(x, y) subject to |K(x, y)| c
1 1 , |x − y| (1 + |x + y|J )
(1.38)
where J ≥ 0 and the constant c does not depend on x, y ∈ (0, ∞). Lemma 1.2.9. Let T : Lp (R) → Lp (R), 1 < p < ∞, be bounded and let 0 < α + p−1 < J + 1. Then T acts continuously in Lp,α (R). Proof. We may assume that f ≥ 0 and supp f ⊂ (0, ∞). The characteristic function of the set E ⊂ R will be denoted by χE (x). One verifies directly that ⎧ χ(0,|x|) (y) χ(|x|,∞) (y) ⎪ ⎪ ⎪ + if x < 0 , ⎪ 2 J/2 ⎪ |x|(1 + x ) y(1 + y 2 )J/2 ⎪ ⎪ ⎪ ⎨ 2J+1 χ 1 1 x (0,x/2) (y) if 0 < y < , J 2 )J/2 1 + |x − y| |x − y| ⎪ 2 x(1 + x ⎪ ⎪ ⎪ ⎪ J+1 ⎪2 χ(2x,∞) (y) ⎪ ⎪ if y > 2x . ⎩ y(1 + y 2 )J/2 By Corollaries 1.2.3 and 1.2.4, it suffices to show the boundedness of the operator T0 defined by 2x K(x, y)f (y)dy (T0 f )(x) = x/2
1.2. Continuity of boundary integral operators
9
in Lp,α (0, ∞) with an arbitrary α ∈ R. We put L(x, y) =
1 (1 + x2 )α/2 1 1 − , x, y > 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)
Chapter 1. Lp -theory of Boundary Integral Equations
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
1.2. Continuity of boundary integral operators
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
Chapter 1. Lp -theory of Boundary Integral Equations
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.
1.2. Continuity of boundary integral operators
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
Chapter 1. Lp -theory of Boundary Integral Equations
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)
1.2. Continuity of boundary integral operators
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)
Chapter 1. Lp -theory of Boundary Integral Equations
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
1.2. Continuity of boundary integral operators
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)
Chapter 1. Lp -theory of Boundary Integral Equations
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)
1.2. Continuity of boundary integral operators
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|
Chapter 1. Lp -theory of Boundary Integral Equations
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.
1.2. Continuity of boundary integral operators
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).
Chapter 1. Lp -theory of Boundary Integral Equations
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)
1.2. Continuity of boundary integral operators
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)
∂ 1 dsq log ∂nq |z − q|
∂ 1 dsq log ∂nq |z − q|
∂ 1 dsq log ∂nq |z − 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)
Chapter 1. Lp -theory of Boundary Integral Equations
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
1.2. Continuity of boundary integral operators
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
Chapter 1. Lp -theory of Boundary Integral Equations
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
1.2. Continuity of boundary integral operators
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)
Chapter 1. Lp -theory of Boundary Integral Equations
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)
1.2. Continuity of boundary integral operators
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 ∂nq |z − q|
∂ 1 dsq = log Ik (z) . ∂nq |z − q| 6
σ(q)
Γ\(Γ+ ∪Γ− )
k=1
Chapter 1. Lp -theory of Boundary Integral Equations
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|<δ/2}) + I3 Lp,β−μ (Γ∩{|q|<δ/2}) c σ Lp,β+1 (Γ) results from Hardy’s inequality (1.22). Now we estimate I4 . We can assume that z ∈ Γ+ . In the sequel, we make use of the inequality (1.57)
∂
ρ(x) 1 2 −1/2
+ 1 + (κ log (u)) +
∂nq 2 2 |z − q| (x − u) + (ρ(x))
x2μ+1 c xμ−1 + , q = u + iv ∈ Γc− (x), (x − u)2 + x2μ+1 where ρ(x) = κ+ (x) − κ− (x). Making the change of variables τ = u−μ , ξ = x−μ , we arrive at (1+ε)x δ/2 p(β−μ) x 0
x2μ+1 σ− (u) du (x − u)2 + x2μ+2
p dx
(1−ε)x
∞ c
ξ p(1−α) R
(δ/2)−μ
μ|σ− (τ −1/μ )| dτ (ξ − τ )2 + μ2 τ
(1.97)
p dξ ,
where β + p−1 = μ(α − p−1 ). It follows from Lemma 1.2.9 that the right-hand side of (1.97) is dominated by ∞ c
τ
−pα
|σ− (τ
−1/μ
δ )| dτ c
(δ/2)−μ
p
|σ− (u)|p up(β+1) du .
0
Thus it is sufficient to estimate (1+ε)x
πσ− (x) −
ρ(x) σ− (u) du . (x − u)2 + (ρ(x))2
(1−ε)x
We make the change of variables u = h(τ ), x = h(ξ), where h is specified by (1.39).
1.2. Continuity of boundary integral operators
31
By (1.51) we have (1+ε)x
δ/2
p(β−μ)
πσ x (x) −
− 0
p
ρ(x) σ− (u)
dx du
2 2 (x − u) + (ρ(x))
(1.98)
(1−ε)x
∞ c h−1 (δ/2)
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/2)
where I(ξ) admits the estimate c |I(ξ)| ξ
ξ |σ− (h(τ ))| h−1 (δ/2)
dτ +c τ
δ |σ− (h(τ ))|
dτ +c τ2
δ
h−1 (δ/2)
ξ
|σ− (h(τ ))| dτ . (ξ − τ )2 + 1 τ
It follows from Lemma 1.2.9 and Hardy’s inequality (1.22) that the last integral in (1.98) does not exceed
p ∞
δ
p(1−α)
d
c dτ = c |σ− (u)|p up(β+1) du .
dτ σ− (h(τ )) τ 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(1 + ξ −2 ) up to a real factor, it follows from the boundedness of the Hilbert transform in Lp,1−α (R) (see Theorem 1.2.6) and from Lemma 1.2.9 that the first integral on the right-hand side in (1.98) does not exceed c σ p . Lp,β+1 (Γ)
We turn to the integral I5 . Obviously, ∂ 1 1 1 = −Re cos(n cos(n log q , 0x) + Im q , 0y) . ∂nq |q − z| q−z q−z
(1.99)
Here (n q , 0x) and (n q , 0y) are the angles between the normal vector nq and the coordinate axes. For q ∈ Γr+ (x) ∪ Γr− (x) and z ∈ Γ ∩ {|q| < δ/2} we represent the first term in (1.99) in the form m zk z m+1 −Re cos(n cos( n , 0x) − Re q q , 0x). q k+1 q m+1 (q − z) k=0
Chapter 1. Lp -theory of Boundary Integral Equations
32
It is clear that
Re
z m+1 m+1 q (q − z)
m+1 μ−m−2 cos(n u . q , 0x) c x
Now we estimate the second term in (1.99). To this end, we use the equality m zk 1 z m+1 + Im Re q k+1 q m+1 q−z
Im (q − z)−1 = Im
(1.100)
k=0
+ Re
1 1 1 1 + Re m+1 Im z m+1 Re . Re z m+1 Im q m+1 q−z q q−z
Since
Im z k Re q −k−1 = O xμ u−1
and
Re z k Im q −k−1 = xk Im q −k−1 + O xμ u−1 ,
we arrive at Im
m m zk = xk Im q −k−1 + O xμ u−1 . k+1 q
k=0
k=0
The second term in (1.100) does not exceed c xm uμ−m−2 . Noting that
Im (q − z)−1 c (uμ−1 + xμ u−1 ), we obtain for the third term in (1.100)
Re q −m−1 Re z m+1 Im (q − z)−1 c (xm+1 uμ−m−2 + xμ u−1 ) . The last term in (1.100) satisfies
Re q −m−1 Im z m+1 Re (q − z)−1 c xμ u−1 . Thus we have, for q ∈ Γr+ (x) ∪ Γr− (x), m ∂ 1 1 1 = log xk −Re k+1 cos(n q , 0x) + Im k+1 cos(n q , 0y) + I(q, z), ∂nq |q − z| q q k=0
where
|I(q, z)| c (xm+1 uμ−m−2 + xμ u−1 ).
Now, set m = [μ − β − 1/p]. Then Γr+ (x)∪Γr− (x)
∂ 1 dsq = log c(k) (σ)xk + (R1 σ)(z), ∂nq |z − q| m
σ(q)
k=0
1.2. Continuity of boundary integral operators
where
σ(q) −Re
c(k) (σ) =
1 q k+1
Γ+ ∪Γ−
33
cos(n q , 0x) + Im
cos(n q , 0y) dsq
1 q k+1
and (R1 σ)(z) admits the estimate R1 σ
Lp,β−μ (Γ∩{|q|<δ/2})
c σ
Lp,β+1 (Γ)
.
It follows from Hardy’s inequality (1.22) that c(k) (σ), k = 1, . . . , m, are linear continuous functionals in Lp,β−μ (Γ). It is clear that I6 is represented in the form m
c(k) xk + (R2 σ)(z),
k=0
where
m k=0
|c(k) | + R2 σL
p,β−μ (Γ∩{|q|<δ/2})
c σ L
p,β+1 (Γ)
.
Finally, m (πσ − T σ)(z) = ±π σ(z+ ) − σ(z− ) + c(k) xk + (Rσ)(z), z ∈ Γ+ , k=0
where
m k=0
|c(k) | + RσL
p,β−μ (Γ∩{|q|<δ/2})
with (Rσ)(z) =
4
c σ L
p,β+1 (Γ)
,
Ik (z) + (R1 σ)(z) + (R2 σ)(z) .
k=2
The continuity of the operator (∂/∂s)(πI − T ) : L1p,β+1 (Γ) → Lp,β+1 (Γ) is proved in Theorem 1.2.14. Thus we obtain the boundedness of (πI − T ) : L1p,β+1 (Γ) −→ M1,+ p,β (Γ).
Changing the direction of the normal n, we obtain the following assertion from Theorem 1.2.18. Corollary 1.2.19. Let Ω+ have an outward peak, and let 0 < β + p−1 < min{μ, 1}. Then the operator
is continuous.
L1p,β+1 (Γ) σ −→ πI + T σ ∈ M1,+ p,β (Γ)
Chapter 1. Lp -theory of Boundary Integral Equations
34
1.3 Dirichlet and Neumann problems for a domain with peak In this section we obtain estimates of solutions to the auxiliary Dirichlet problem (1.7) and the Neumann problem (1.8) in domains with inward and outward peak. They relate normal derivatives of solutions and their conjugate functions under various assumptions on boundary data.
1.3.1 Dirichlet problem for a domain with outward peak Let G ⊂ R2 be a domain with C 2 -boundary such that the set {(τ, ν) ∈ G : τ 0} has compact closure and {(τ, ν) ∈ G : τ > 0} = {(τ, ν) : τ > 0, |ν| < 1}. As usual, by C0∞ (G) we mean the space of infinitely differentiable functions with compact supports in G. We denote by Wpk (G), k = 0, 1, 2, p ∈ (1, ∞), the Sobolev ˚ pk (G) the space of functions in Lp (G) with derivatives up to order k in Lp (G). By W (∂G) be the space completion of C0∞ (G) in Wpk (G) will be denoted. Let Wp of traces on ∂G of functions in Wpk (G). We introduce also the space Wp−1 (G) of distributions on G with the finite norm k−1/p
ϕWp−1 (G) = inf
3
ϕk Lp (G) ,
k=1
where the infimum is taken over all representations ϕ = ϕ0 + (∂/∂τ )ϕ1 + (∂/∂ν)ϕ2 with ϕj ∈ Lp (G), j = 0, 1, 2. k (G) if Let α ∈ R. We say that ϕ ∈ Wp,α
(1 + τ 2 )α/2 ϕ ∈ Wpk (G),
k = −1, 0, 1, . . . ,
and define the norm 2 α/2 ϕWpk (G) . ϕWp,α k (G) = (1 + τ ) k−1/p
k The spaces Wp,α (∂G) and Wp,α (∂G) are introduced in the same way. By Lp,α (∂G) we denote the space of functions with the finite norm
ϕLp,α (∂G) = (1 + τ 2 )α/2 ϕLp (∂G) . We shall make use of the same definitions for the strip Π = {(τ, ν) : τ ∈ R, |ν| < 1}.
1.3. Dirichlet and Neumann problems for a domain with peak
35
The following lemma is a particular case of more general results of the article [11]. Lemma 1.3.1. The operator k k−2 k−1/p Wp,α (G) u → {Δu, u|∂G } ∈ Wp,α (G) × Wp,α (∂G)
performs an isomorphism for every real α and k = 1, 2. The same is true for k = 1, 2, . . . if G is replaced by the strip Π. We need Kellogg’s theorem (see [10]) on boundary properties of a conformal mapping of the unit disk onto a bounded domain with smooth boundary. Theorem 1.3.2 (Kellogg). Let S be a closed Jordan curve in the complex z-plane. Suppose that S has a tangent at any point of S. Further, let the slope of the tangent τ (s), considered as a function of the arc s, is H¨ older continuous of order α, 0 < α < 1. Let w = ω(z) be a conformal mapping of the disk |z| < 1 onto the interior of S. Then ω is defined at every point z0 of the unit circumference by the relation lim ω (z) = ω (z0 ), z→z0
where z → z0 at any Stolz angle with vertex at z0 . Moreover, ω is continuous at the closed disk |z| 1 and for |z| = 1,
iθ
ω (e ) − ω (eiθ ) c|θ − θ |α and
log ω (eiθ ) − log ω (eiθ ) c|θ − θ |α .
Here we recall a well-known theorem (see [5], Ch. IX, Sect. 5) to be used in Chapter 2. Theorem 1.3.3 (Privalov). Let f (ζ) = u(r, θ) + iv(r, ζ), ζ = reiθ , be holomorphic in the disk |ζ| < 1 and let u(r, θ) be continuous in |ζ| 1. If u(1, θ) satisfies the H¨ older condition |u(1, θ) − u(1, θ )| c|θ − θ |α with 0 < α < 1, then f satisfies the H¨ older condition |f (ζ) − f (ζ )| c|ζ − ζ |α in the disk |ζ| 1.
Chapter 1. Lp -theory of Boundary Integral Equations
36
In the first proposition of the present section we show that the normal derivative of a solution of the problem (1.7) with boundary data in N1,− p,β (Γ) for a domain Ω+ with outward peak belongs to the space Np,β (Γ). The functions in the space N1,− p,β (Γ) have a simple characterization under a conformal mapping of a domain with outward peak onto a strip. Therefore the principal part of the proof concerns estimates in a strip. −1 Proposition 1.3.4. Let ϕ belong to space N1,− < min{μ, 1}. p,β (Γ), where 0 < β + p + Then there exists a harmonic extension u of ϕ onto Ω with normal derivative ∂u/∂n in Np,β (Γ) satisfying
∂u/∂n Np,β (Γ) c ϕ N1,− (Γ) .
(1.101)
∂u/∂n Lp,β+1 (Γ) c ϕ N1,− (Γ) .
(1.102)
p,β
In particular, p,β
Proof. We start with the case when ϕ ∈ N1,− p,β (Γ) vanishes on Γ ∩ {|q| < δ/2}. Let θ be a conformal mapping of the unit disk D onto Ω+ . We introduce the harmonic extension F of the continuous function ϕ◦ θ onto D. The normal derivative ∂F/∂n on ∂D has the form π ∂F s−t d 1 (ζ) = ϕ(θ(eit )) cot dt, ζ = eis . ∂n 2π −π dt 2 Hence u = F ◦ θ−1 satisfies ∂u/∂n Lp,γ+1 (Γ) c1 ∂F/∂n Lp (∂D) c2 (ϕ ◦ θ) Lp (∂D) c3 ϕ N1,− p,γ (Γ) for every real γ. Since
(∂u/∂n)ds = 0, Γ
∂u/∂n is represented in the form ∂v/∂s, where the function v satisfies v Lp,β−μ (Γ) + v Lp,β+1 (Γ) c ϕ N1,− (Γ) . p,β
It remains to prove (1.101) for ϕ ∈ N1,− p,β (Γ) vanishing outside Γ+ ∪ Γ− . Let z = ω(τ + iν) denote a conformal mapping of G onto Ω+ such that ω(∞) = 0 and ω(∂G ∩ {(x, y) : x > 1}) ⊃ Γ+ ∪ Γ− . By Theorem 1.3.2, a conformal mapping of the unit disk onto a domain with C 2 -boundary has a continuous nonvanishing derivative up to the boundary. The function z → exp(Az −μ + B)
1.3. Dirichlet and Neumann problems for a domain with peak
37
with correctly chosen real A and B maps Ω+ onto a bounded domain with C 2 boundary. Hence, ω and ω −1 obey the relations ω(τ + iν) = c (τ + iν)−1/μ (1 + o(1)), |ω (τ + iν)| c |τ + iν|−1−1/μ
(1.103)
as τ → ∞, τ + iν ∈ G, and
ω −1 (z) = c z −μ (1 + o(1)), | ω −1 (z)| c |z|−1−μ
(1.104)
as x → 0, z = x + iy ∈ Ω+ . We introduce the functions Φ± on R by Φ± (τ ) = ϕ(ω(τ ± i)) for τ ≥ 0 and Φ± (τ ) = 0 for τ < 0. The estimates hold: c1 ϕN1,− (Γ) Φ+ − Φ− Lp,1−α (R) + p,β
±
Φ± Lp,−α (R) + Φ± Lp,1−α (R)
c2 ϕN1,− (Γ)
(1.105)
p,β
with α defined by β + p−1 = μ(α − p−1 ). Let r be a measurable function on (0, ∞) subject to |r(τ )| 1. We have ∞
p
|Φ± (τ ) − Φ± (τ + r(τ ))| τ (1−α)p dτ
(1.106)
0
p τ +1 ∞ ∞ c1 |Φ± (ν)|dν τ (1−α)p dτ c2 |Φ± (τ )|p τ (1−α)p dτ. τ −1
0
0
For z, z1 ∈ Ω+ with |z1 − z| xμ+1 the distance between ω −1 (z1 ) and ω −1 (z) is bounded by (1.104). This fact and (1.106) imply the left-hand inequality in (1.105). Let h be a measurable function on [0, δ] such that |h(x)| xμ+1 . By Theorem 1.2.5 on the boundedness of the Hardy–Littlewood maximal operator in weighted Lp -spaces we have δ |ϕ(x) − ϕ(x + h(x))|p x(β−μ)p dx 0
δ
c
1
μ+1 x+x
|t ϕ (t)|dt
xμ+1 0
x−xμ+1
p
δ xβp dx c
(1.107) |ϕ (t)|p t(β+1)p dt.
0
Using (1.103), we obtain that for ζ, ζ1 ∈ G with |ζ − ζ1 | < 1 the distance between ω(ζ) and ω(ζ1 ) does not exceed c xμ+1 . This property and (1.107) show that the right-hand inequality in (1.105) holds.
Chapter 1. Lp -theory of Boundary Integral Equations
38
We introduce the bounded harmonic function Φ(+) on the strip Π taking the same value (Φ+ (τ ) + Φ− (τ ))/2 at the points (τ, 1) and (τ, −1) on ∂Π. The Fourier transform of Φ(+) (τ, ν) with respect to τ is given by + (ξ) + Φ − (ξ) cosh (νξ) (cosh ξ)−1 , c Φ ± denotes the Fourier transform of Φ± . Therefore, the Fourier transform where Φ of (∂/∂n)Φ(+) (τ, ±1) is equal to (ξ) tanh ξ . (ξ) + Φ ci Φ + − Hence ∂Φ(+) (τ, ±1) = c ∂n
R
−1 π d (Φ+ (t) + Φ− (t)) sinh (τ − t) dt . dt 2
By Lemma 1.2.9, ∂Φ(+) /∂n Lp,1−α (∂Π) c (Φ+ + Φ− ) Lp,1−α (∂Π) . We write the Fourier transform of (∂/∂n)Φ(+) (τ, ν) with respect to τ in the form (ξ) + Φ (ξ) sinh ξ . c iξ Φ + − ξ cosh ξ Clearly, the function (∂/∂n)Φ(+) (τ, ±1) can be represented as (∂/∂s)Y (−) (τ, ±1), where Y (−) (τ, +1) = −Y (−) (τ, −1). By Theorem 1.2.7 on the Fourier multipliers in weighted Lp -spaces, it follows that Y (−) Lp,1−α (∂Π) + (∂/∂τ )Y (−) Lp,1−α (∂Π) c (∂/∂τ )(Φ+ + Φ− ) Lp,1−α (∂Π) . Let Φ(−) be the bounded harmonic function on Π with the values (Φ+ (τ ) − Φ− (τ ))/2
and (Φ− (τ ) − Φ+ (τ ))/2
at the points (τ, 1) and (τ, −1) on ∂Π. The Fourier transform of Φ(−) (τ, ν) with respect to τ has the form − (ξ) sinh(νξ) (sinh ξ)−1 . + (ξ) − Φ c Φ
1.3. Dirichlet and Neumann problems for a domain with peak
39
Therefore, the Fourier transform of (∂/∂n)Φ(−) (τ, ±1) equals − (ξ) cosh ξ (sinh ξ)−1 + (ξ) − Φ ± cξ Φ ξ d d +i Φ+ (ξ) − Φ− (ξ) tanh ξ = ±c Φ+ (ξ) − Φ− (ξ) sinh ξ dt dt and hence
−2 π (Φ+ (t) − Φ− (t)) cosh (τ − t) dt 2 R dΦ− dΦ+ −1 (t) − (t) (sinh π(τ − t)) dt . ± c2 dt dt R
∂Φ(−) (τ, ±1) = ± c1 ∂n
By Lemma 1.2.9, 1 ∂Φ(−) /∂n Lp,1−α (∂Π) c Φ+ − Φ− Wp,1−α (∂Π) .
It is clear that (∂/∂n)Φ(−) can be written as (∂/∂s)Y (+) (τ, ±1), where Y (+) (τ, +1) = Y (+) (τ, −1). According to Theorem 1.2.7, the estimate 1 Y (+) Lp,−α (∂Π) + (∂/∂τ )Y (+) Lp,1−α (∂Π) c (Φ+ − Φ− ) Wp,1−α (∂Π)
holds. Let χ ∈ C ∞ (R) be equal to 1 for t > 1 and vanish for t < 0, and let Ψ = Δ(χ(Φ(−) + Φ(+) )) . Using Lemma 1.3.1, we have Ψ Lp (Π) c Φ(−) W 1−1/p (∂Π) + Φ(+) W 1−1/p (∂Π) . p,−α
p,−α
1−1/p
(1.108)
By the inclusion Wp1 (∂Π) ⊂ Wp (∂Π), the right-hand side of (1.108) has the majorant (+) (+) 1 c Φ(−) Wp,1−α + Φ + dΦ /dt Lp,−α (∂Π) Lp,1−α (∂Π) . (∂Π) Applying Lemma 1.3.1 with k = 2, we obtain that the Dirichlet problem ΔZ = −Ψ in G, Z = 0 on ∂G has a solution satisfying 2 (G) c2 Ψ L (G) ∂Z/∂n Lp,γ (∂G) c1 Z Wp,γ p
40
Chapter 1. Lp -theory of Boundary Integral Equations
for every real γ. Since
(∂Z/∂n)ds = 0 , ∂G
the function ∂Z/∂n can be represented in the form ∂X/∂s, where X satisfies X Lp,1−α (Γ) + X Lp,1−α (Γ) Φ± Lp,−α (R) + Φ± Lp,1−α (R) . c Φ+ − Φ− Lp,1−α (R) + ±
We set F = Z + χ(Φ(−) + Φ(+) ) . By (1.105), the function u = F ◦ ω −1 is the required harmonic extension of ϕ onto Ω+ . Since 1 ∈ N1,− p,β (Γ), the next proposition follows from the Cauchy-Riemann conditions. Proposition 1.3.5. Let ψ belong to Np,β (Γ), where 0 < β + p−1 < 1 . Then the Neumann problem (1.8) in Ω+ with boundary data ψ has a solution v in N1,− p,β (Γ) satisfying v N1,− (Γ) c ψ Np,β (Γ) . p,β
1.3.2 Boundary value problems for a domain with inward peak In this section we obtain estimates for solutions of the Dirichlet problem (1.7) and the Neumann problem (1.8) in domains with inward peak. In the first proposition we prove that the normal derivative of the harmonic extension of a boundary function from L1p,β+1 (Γ) belongs to the space Lp,β+1 (Γ). Proposition 1.3.6. Let ϕ belong to L1p,β+1 (Γ) with 0 < β + p−1 < 1
and
β + p−1 = 1/2 .
Then there exists a harmonic extension g of ϕ on Ω+ with the normal derivative in Lp,β+1 (Γ) satisfying ∂g/∂nLp,β+1(Γ) cϕL1p,β+1 (Γ) .
(1.109)
Proof. We represent ϕ ∈ L1p,β+1 (Γ) as the sum ϕ = ϕ1 +ϕ2 , where supp ϕ1 ⊂ Γ+ ∪ Γ− and ϕ2 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} 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
Chapter 1. Lp -theory of Boundary Integral Equations
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
1.3. Dirichlet and Neumann problems for a domain with peak
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)
Chapter 1. Lp -theory of Boundary Integral Equations
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 H¨older 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 H¨older 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.
1.3. Dirichlet and Neumann problems for a domain with peak
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
Chapter 1. Lp -theory of Boundary Integral Equations
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− ,
1.3. Dirichlet and Neumann problems for a domain with peak
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.
Chapter 1. Lp -theory of Boundary Integral Equations
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
<β+
1 p
R
τ 2 dτ d Φ(τ ) , dτ ξ−τ
(1.129)
< 1, satisfies (∂/∂ξ)HL
p,2β+1+1/p (R)
c (d/dξ)ΦL
p,2β+1+1/p (R)
.
(1.130)
Let n0 be the integer subject to the inequalities n0 − 1 2(μ − β − p−1 ) < n0 and let m be the largest integer satisfying 2m n0 .
(1.131)
1.3. Dirichlet and Neumann problems for a domain with peak
The function H(−) , defined by H(−) (ξ) =
1 H(ξ) − H(−ξ) = 2 π
R
49
ξ − τ
d (+)
Φ (τ ) log
dτ , dτ ξ
can be written in the form n 0 −1 ξ n0 Φ(+) (τ ) ξk Φ(+) (τ ) dτ . dτ − n π R τ 0 (ξ − τ ) π R τ k+1 k=k0
Here Φ(+) (ξ) = (Φ(ξ) + Φ(−ξ))/2, and k0 = 0 if 0 < β + 1/p < 1/2 and k0 = −1 if 1/2 < β+1/p < 1. It follows from (1.131) and the boundedness of the Hilbert transform in weighted Lp -spaces (see Theorem 1.2.6) that the Lp,2β−2μ+1/p -norm of ξ n0 Φ(+) (τ ) dτ n π R τ 0 (ξ − τ ) does not exceed Φ(+) Lp,2β−2μ+1/p (R) . Therefore, by (1.130) and (1.123) we obtain that the function h(z) = H ◦ θ0−1 (z) can be represented in the form m
ar (ϕ)Re z r−1/2 + h# (z)
(1.132)
r=r0
for z ∈ Ω+ situated in a small neighborhood of the peak, where r0 = 1, if 0 < β + 1/p < 1/2, and r0 = 0, if 1/2 < β + 1/p < 1. Here ar (ϕ) = Φ(+) (τ )τ −2r dτ , r0 r m , R
# are linear continuous functionals in N1,+ belongs to N1,− p,β (Γ), and h p,β (Γ) and satisfies h# N1,− (Γ+ ∪Γ− ) c ϕN1,+ (Γ) . p,β
∞
p,β
Now let κ ∈ C (R ) be equal to 1 for |z| < δ/2 and vanish for |z| > 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)| .
Chapter 1. Lp -theory of Boundary Integral Equations
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)
1.3. Dirichlet and Neumann problems for a domain with peak
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).
Chapter 1. Lp -theory of Boundary Integral Equations
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)
1.3. Dirichlet and Neumann problems for a domain with peak
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)
Chapter 1. Lp -theory of Boundary Integral Equations
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
1.3. Dirichlet and Neumann problems for a domain with peak
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)
Chapter 1. Lp -theory of Boundary Integral Equations
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)
1.3. Dirichlet and Neumann problems for a domain with peak
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
Chapter 1. Lp -theory of Boundary Integral Equations
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
1.3. Dirichlet and Neumann problems for a domain with peak
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
Chapter 1. Lp -theory of Boundary Integral Equations
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)
Chapter 1. Lp -theory of Boundary Integral Equations
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<|q|<2|z|}
2|z| c
∗ |σ± (v)| log
|z|/2
2 c|z|
1
+ 1 1 − v/|z|
∗ |σ± (|z|u)| log
1
+ 1 du |1 − u|
(1.162)
1/2
2 1/p ∗ c|z| |σ± (|z|u)|p u(β+1)p du = c|z|−β−1/p . 1/2
By (1.159), (1.161) and (1.162) it follows that there is a constant c such that
−1 z
dsq
c |z|−β−p . σ(q) log
z−q Γ+ ∪Γ−
Lemma 1.4.2. The double layer potential ∂ 1 dsq log W σ(z) = σ(q) ∂nq |z − q|
(1.163)
Γ
with density σ ∈
L1p,β+1 (Γ)
admits the estimate −1
|W σ(z)| |z|−β−p
.
(1.164)
1.4. Integral equations of the Dirichlet and the Neumann problems
63
Proof. Given z ∈ Ω− , we write W σ as ∂ z W σ(z) = − σ(q) arg dsq . ∂sq z−q Γ
The branch of the function arg is chosen by the condition arg 1 = 0. Using the estimate z arg = O(|q|) z−q for a fixed z and q → 0 and the inclusion σ ∈ L1p,β+1 (Γ), we obtain lim σ(q+ ) arg
q+ →0
z z = lim σ(q− ) arg = 0. q= →0 z − q+ z − q−
Integrating by parts, we find
W σ(z) = − Γ
∂ 1 dsq . σ(q) arg ∂sq z−q
Hence (1.164) follows from Lemma 1.4.1. Next, suppose z ∈ Ω+ . Given a fixed z0 ∈ Ω+ , we make the change of variables z z0 q z0 w = ω(z) = and p = ω(p) = z0 − z z0 − q in (1.163). Since q−z =
z02 (p − w) , (z0 + p)(z0 + w)
we have ∂ ∂ ∂ 1 |w| dsq = + log log log |p + z0 | dsp . ∂nq |z − q| ∂np |w − p| ∂np Let γ denote the image of Γ. Similarly to Γ, the curve γ has a peak at the origin. The image of Ω+ under the mapping ω is the exterior of γ. Clearly, σ(ω −1 (p)) ∈ L1p,β+1 (γ) and the function ∂ log |p + z0 | ∂np is bounded on γ. Hence it suffices to consider the potential ∂ |w| dsp Wγ σ(w) = σ(ω(p)) log ∂np |w − p| γ
in the exterior of γ. As shown above, −1
|Wγ σ(w)| c|w|−β−p
.
Chapter 1. Lp -theory of Boundary Integral Equations
64
Therefore, in a neighborhood of the peak, the estimate −1
|W σ(z)| c|z|−β−p
holds for z in Ω+ . Outside a neighborhood of the origin, W σ is a bounded function. Thus (1.163) holds in Ω+ . Theorem 1.4.3. Let Ω+ have either outward or inward peak and let 0 < β + p−1 < min{μ, 1}, β + p−1 = 1/2. Then the operator V
Lp,β+1 (Γ) × R {σ, C} −→ V σ + C ∈ N1,− p,β (Γ)
(1.165)
is surjective. Proof. For the single layer potential it does not matter whether the peak is inward or outward. Therefore, it suffices to get the result for the case of an outward peak. Let ϕ ∈ N1,− p,β (Γ) and let ϕ = 0 in a neighborhood of the peak. We introduce the + harmonic extensions H i and H e of ϕ onto Ω− constructed 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 H e = O |z|−1/2 , z ∈ Ω− , it follows that |z| ∂H i ∂H e 1 − log dsq + H e (∞). ϕ(z) = 2π ∂n ∂n |z − q| Γ
Hence the pair {σ, C} with σ = (2π)−1 (∂H i /∂n − ∂H e /∂n) ∈ Lp,β+1 (Γ) and C = H e (∞) is a solution of (1.11). −1 , there exists Now let ϕ be an arbitrary function in N1,− p,β (Γ). Since β > −p a sequence {ϕn }n≥1 of smooth functions on Γ, vanishing near the peak, and converging to ϕ in N1,− p,β (Γ). By σn we denote the solution of (1.11) just constructed with the right-hand side ϕn . By Propositions 1.3.4 and 1.3.6, {σn } converges in Lp,β+1 (Γ) to a limit σ. The harmonic extension Hne of ϕn onto Ω− satisfies |z| ∂ 1 1 ∂Hne e dsq = Hne (∞) (q) log log − Hn (q) 2π ∂nq |z − q| ∂nq |z − q| Γ
for z ∈ Ω (see (1.75)). Therefore, Proposition 1.3.6 guarantees the convergence of Hne (∞) to a certain real number C. Since the operator V : Lp,β+1 (Γ) → N1,− p,β (Γ) is continuous, it follows by passage to the limit that +
ϕ=Vσ+C.
1.4. Integral equations of the Dirichlet and the Neumann problems
65
Consequently, the equation (1.11) is solvable in Lp,β+1 (Γ) × R for every ϕ ∈ N1,− p,β (Γ) and ˙ N1,− p,β (Γ) ⊂ V (Lp,β+1 (Γ))+R. The converse inclusion was proved in Theorem 1.2.13.
The next assertion concerns the dimension of the kernel of the operator V. Theorem 1.4.4. Let Ω+ have either outward or inward peak and let the operator V be defined by (1.165). Then a) ker V = {0} for 0 < β + p−1 < 1/2; b) dim ker V = 1 for 1/2 < β + p−1 < 1. Proof. Let Ω+ have an outward peak and let {σ, C} belong to ker V. The potential V σ equals −C on Γ\{0} and by Lemma 1.4.1 admits the estimate |V σ(z)| −1 c |z|−β−p , z = 0. Therefore, V σ = −C on 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−q
The branch of arg is chosen so that arg 1 = 0. Let ζ = γ(z) be a conformal mapping of Ω− onto R2+ with γ(0) = 0. The function W (ζ) = (V ◦γ −1 )(1/ζ) is holomorphic on the lower half-plane, continuous up to the boundary, and Im W = 0 on the real axis. The holomorphic extension Wext of W onto the upper half-plane is an entire function with imaginary part satisfying −1 |Im Wext (ζ)| c |ζ|2(β+p ) . By the Schwarz integral formula, which restores an analytic function f (z) in the disk |z| < r by the boundary values of its real part: f (z) =
1 π
0
2π
Re f (reiθ )
reiθ + z dθ + ic, reiθ − z
(1.166)
we conclude that Wext has the same order of the growth at infinity. Since β +p−1 < 1, it follows that Wext is a linear function of ζ. Therefore, V σ = c0 + c1 Im (1/γ) . From the jump formula for (∂/∂n)V σ (see (1.3)) we have σ(z) = −
c1 ∂ 1 Im ∼ c x−3/2 , as z → 0, z ∈ Γ\{O}. 2π ∂n γ(z)
Chapter 1. Lp -theory of Boundary Integral Equations
66
By the integral representation (1.75) for the harmonic function Im (1/γ) on Ω− and the limit relation (1.5) for the double layer potential, we obtain ∂ |z| 1 1 1 log dsq − Im Im π ∂nq γ(q) |z − q| γ(∞) Γ (1.167) 1 1 ∂ 1 1 = Im log dsq − Im π γ(q) ∂nq |z − q| γ(z) Γ
for z ∈ Γ\{O}. Since Im (1/γ(z)) = 0 on Γ\{O}, it follows that the zeros of V have the form 1 ∂ 1 1 c Im , c Im , c ∈ R. π ∂n γ γ(∞) We note that Im (1/γ) is defined modulo a constant positive factor so that dim ker V is at most one. Since (∂/∂n)Im (1/γ) ∈ Lp,β+1 (Γ) only for β+p−1 > 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+ .
1.4. Integral equations of the Dirichlet and the Neumann problems
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.
Chapter 1. Lp -theory of Boundary Integral Equations
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
1.4. Integral equations of the Dirichlet and the Neumann problems
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
Chapter 1. Lp -theory of Boundary Integral Equations
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+ ,
1.4. Integral equations of the Dirichlet and the Neumann problems
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
Chapter 1. Lp -theory of Boundary Integral Equations
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.
1.4. Integral equations of the Dirichlet and the Neumann problems
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
Chapter 1. Lp -theory of Boundary Integral Equations
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
∂ |z| dsq , log ∂nz |z − q|
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.
1.4. Integral equations of the Dirichlet and the Neumann problems
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.
Chapter 1. Lp -theory of Boundary Integral Equations
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
1.4. Integral equations of the Dirichlet and the Neumann problems
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).
Chapter 1. Lp -theory of Boundary Integral Equations
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)
1.4. Integral equations of the Dirichlet and the Neumann problems
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.
Chapter 1. Lp -theory of Boundary Integral Equations
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),
1.4. Integral equations of the Dirichlet and the Neumann problems
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
Chapter 1. Lp -theory of Boundary Integral Equations
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
1.4. Integral equations of the Dirichlet and the Neumann problems
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 .
Chapter 1. Lp -theory of Boundary Integral Equations
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
1.4. Integral equations of the Dirichlet and the Neumann problems
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
Chapter 1. Lp -theory of Boundary Integral Equations
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
1.4. Integral equations of the Dirichlet and the Neumann problems
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
Chapter 1. Lp -theory of Boundary Integral Equations
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|
1.4. Integral equations of the Dirichlet and the Neumann problems
89
It follows from Theorem 1.2.18 that ψ ∈ M1,+ p,β (Γ). Since d ψ(z) = πσ(z) + ds
σ(q) Γ
∂ |z| dsq , log ∂nz |z − q|
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 .
Chapter 1. Lp -theory of Boundary Integral Equations
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
1.4. Integral equations of the Dirichlet and the Neumann problems
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 ◦
Chapter 1. Lp -theory of Boundary Integral Equations
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.
1.4. Integral equations of the Dirichlet and the Neumann problems
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 .
Chapter 1. Lp -theory of Boundary Integral Equations
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).
Chapter 1. Lp -theory of Boundary Integral Equations
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.
1.5. Direct method of integral equations
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
Chapter 1. Lp -theory of Boundary Integral Equations
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
1.5. Direct method of integral equations
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
Chapter 1. Lp -theory of Boundary Integral Equations
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 H¨older 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)
2.1. Direct method of integral equations
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
Chapter 2. Boundary Integral Equations in H¨ older Spaces
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.
2.1. Direct method of integral equations
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
Chapter 2. Boundary Integral Equations in H¨ older Spaces
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)
2.2. Direct method of integral equations
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
Chapter 2. Boundary Integral Equations in H¨ older Spaces
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| <
1 min(|x − t|, |y − t|), 2
(2.28)
|K(x, t) − K(y, t)|dt c,
R
d
K(x, t)dt
c (1 + |x|)−1 .
dx R
Then the integral operator T with the kernel K(x, t) continuously maps Eβα (R) with 0 < α < 1 and −1 < β < J + 1, J 0 into itself. The proof of this theorem follows the same pattern as that of Theorem 2.2.1. The next two assertions concern the continuity of two particular integral operators. Lemma 2.2.4. The integral operator T defined by ∞ 1 T g(τ ) = ν −1/2 1/2 g(ν) dν ν + τ 1/2 0 acts continuously in the space Λα β (1, ∞) if 0 < β < 1/2 and 0 < α < 1. Proof. Suppose g Λβα (1,∞) = 1. Letting 0 < x < y and 2|x − y| < |x|, we have √ √
1
| y − x| 1 1 1
√ −√ √ = √ √ √ √ √
√ √ ν x+ ν y+ ν ν ( x + ν)( y + ν) √ √ | y − x| 1 √ √ √ , c ν ( x + ν)2 because x and y are equivalent and x/2 < y < 2x. Any function g from ∧α β (1, ∞) −α obeys the estimate |g(x)| cx . Hence, for 0 < β < 1/2, ∞ √ 1 1 √ √ √ √ t−β dt |T g(x) − T g(y)| | y − x| t ( x + t)2 0 √ ∞ √ | y − x| −β 1 1 (2.29) √ √ √ t−β dt x =c x t (1 + t)2 0 |x − y| −β x c x−α−β |x − y|α c x
2.2. Direct method of integral equations
109
and
∞
1 1 √ √ √ t−β dt t ( x + t) 0 ∞ 1 1 −β √ √ t−β dt = c x−β . =x t (1 + t) 0
|T g(x)|
(2.30)
The result follows by (2.29)) and (2.30). Lemma 2.2.5. The integral operator T , T g(τ ) = τ −1/2
∞
0
τ 1/2
1 g(ν) dν, + ν 1/2
α acts continuously from Λα β (1, ∞) into Λβ (1, ∞) if 1/2 < β < 1 and 0 < α < 1.
Proof. Let g Λαβ (1,∞) = 1. Similarly to the previous lemma, we assume that 0 < x < y and 2|x − y| < |x|. We have √ √ | y − x| y−x √ √ c 3/2 ; x y x √ √ | y − x| 1 √ √ √ √ √ x ( x + t)( y + t) √ √ 1 | y − x| √ . c√ √ x ( x + t)2
1 1
√ − √ = x y
1
1 1
√ −√ √ =
√ √ x x+ t y+ t
For 1/2 < β < 1 these inequalities imply
1 1 1
∞
√ |g(t)| dt |T g(x) − T g(y)| √ − √
√ x y 0 x+ t ∞
1 1 1
√ − √ √ |g(t)| dt +√
√ y 0 ( x + t) ( y + t) √ √ | y − x| ∞ 1 √ t−β dt √ √ (2.31) √ x y x+ t 0 ∞ √ 1 1 √ √ √ √ t−β dt + √ | y − x| √ y ( x + t)( y + t) 0 ∞ ∞ 1 1 |y − x| |y − x| √ t−β dt + √ t−β dt c 3/2 √ √ x x x+ t ( x + t)2 0 0 |y − x| −β x c x−α−β |x − y|α c x
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Chapter 2. Boundary Integral Equations in H¨ older Spaces
and
∞
1 √ t−β dt √ ( x + t) 0 ∞ 1 √ t−β dt = c x−β . = x−β (1 + t) 0
1 |T g(x)| √ x
(2.32)
The inclusion of T g into Λα β (1, ∞) follows by (2.31) and (2.32).
2.2.2 Continuity of the operator πI − T The main results of this section are theorems on the continuity of the mapping α ˙ ˙ (Lodd )1,α 2−β (Γ)+R σ −→ πσ − T σ ∈ Λ−β (Γ)+R
(2.33)
on a contour Γ with an outward peak and of the mapping α ˙ (Lodd )1,α 2−β (Γ) σ −→ πσ − T σ ∈ Λ−β (Γ)+R
(2.34)
on a contour Γ with an inward peak. We assume that ϕ ∈ Λα β (Γ± ∩ {x < δ}), if the restrictions ϕ+ and ϕ− of ϕ to Γ+ and Γ− , considered as functions of x, belong to Λα β (0, δ). We fix a positive ε, ε < 1/2, such that |κ± (x) − κ∓ (u)| ≥ c u2 for all u with |x − u| < 2εx. By Γ ± (x), Γc± (x) and Γr± (x) we mean the parts of the arcs Γ± with abscissas [0, (1 + ε)−1 x], [(1 + ε)−1 x, (1 − ε)−1 x] and [(1 − ε)−1 x, δ]. In the next assertion we collect some estimates to be used in the proof of continuity of the operator (2.33). Lemma 2.2.6. Suppose that z, z1 , z2 ∈ Γ+ . Then a) b)
∂ |q|
log
c < ∞ on Γ ± (x) ∪ Γc+ (x),
∂nq |q − z|
∂ x |q|
on Γr± (x), u = Re q; log
c ∂nq |q − z| u
x = Re z;
and, for |x1 − x2 | (ε/2)|x1 |, c) d) e)
∂ |z1 − z2 | |z1 − q|
log
c
∂nq |z2 − q| |z1 |
∂ |z |z − q|
1 1 − z2 | log
c
∂nq |z2 − q| |z1 − q|
∂
− q| |z |z
1 1 − z2 | log
c ∂nq |z2 − q| |q|
on
Γ ± (x1 ),
on
Γc+ (x1 );
on
Γr± (x1 ) .
x1 = Re z1 ;
2.2. Direct method of integral equations
111
Proof. We have
∂ 1/2 | − κ± (u) − κ± (u)(x − u) + κ+ (x) | 1
log (u)2 =
1 + κ±
∂nq |q − z| |q − z|2 on Γ± . The right-hand side does not exceed c
(x − u)2 c<∞ |q − z|2
on Γ + (x) ∪ Γc+ (x) and is dominated by c
|κ+ (x) − κ− (x)| x2 (x − u)2 c<∞ + = c 1 + |q − z|2 |q − z|2 |q − z|2
on Γ − (x). Therefore, a) follows. To obtain the assertion b), we use the equality
∂ 1/2 1 ∂ 1 |q|
∂
log + log log (u)2 = Re − κ± (u)
1 + κ±
∂nq |q − z| ∂u 1 − z/q ∂v 1 − z/q 1 z 1 z − κ± (u)Re − Im . q−z q q−z q The absolute value of the first term on the right-hand side does not exceed x c u
z
c , |q − z| q u because |q − z| |q|(1 − |z/q|) c u, while the absolute value of the second term is majorized by
1 z
1
1
1
Re + Re
Im
Im z Re + Re z Im
q−z q q−z q q u2 2 1 x x x c + +x c . (u − x)2 u u − x u u These estimates imply b). Next we show that c), d), and e) hold for q ∈ Γ+ . If the angle (z1 − q, z2 − q) at a point q does not exceed π/2, we use the inequality
∂ − q, nq ) cos (z2 − q, nq )
|z2 − q|
cos (z1
− log
=
∂nq |z1 − q| |q − z1 | |q − z2 |
1
1
=
− cos (z2 − q, nq )
|q − z1 | |q − z2 |
1
cos (z1 − q, nq ) − cos (z2 − q, nq )
+ |q − z1 | − q, nq )| c 1 |z2 − z1 | | cos (z2 + | sin (z1 − q, z2 − q)| . |q − z1 | |q − z2 | |q − z1 | 2
112
Chapter 2. Boundary Integral Equations in H¨ older Spaces
Otherwise, we replace z2 by the point z˜2 , which is centrally symmetric with respect to q. Then we have
∂ |z2 − q|
∂ |˜ z2 − q|
log log
=
∂nq |z1 − q| ∂nq |z1 − q|
− q, nq )| c 1 |z2 − z1 | | cos (z2 + | sin (z1 − q, z˜2 − q)| . |q − z1 | |q − z2 | |q − z1 | 2
In both cases, the sine of the half-angle is dominated by
sin (z1 − q, z2 − q) = sin (z1 − q, z˜2 − q)
(x1 − u)(κ+ (x2 ) − κ+ (u)) − (x2 − u)(κ+ (x1 ) − κ+ (u))
. = |q − z1 ||q − z2 |
(2.35)
Let us write the numerator in (2.35) as (x1 − u)(κ+ (x2 ) − κ+ (u)) − (x2 − u)(κ+ (x1 ) − κ+ (u)) κ (u) − κ (x ) κ (x ) − κ (x ) + + 2 + 1 + 2 = (x2 − x1 )(u − x2 ) . − u − x2 x1 − x2 We take the Taylor expansion of
κ+ (u)−κ+ (x2 ) u−x2
(2.36)
in a neighborhood of u = x1
(c1 ) κ+ (x1 ) − κ+ (x2 ) κ+ κ+ (c1 ) − κ+ (x2 ) κ+ (u) − κ+ (x2 ) (u − x1 ) , = + − u − x2 x1 − x2 c1 − x2 (c1 − x2 )2
where c1 is taken between u and x1 . The expression in square brackets on the right-hand side of (2.36) does not exceed |κ+ (c1 )(c1 − x2 ) − κ+ (c1 ) − κ(x2 )| 1 = |κ (c2 )| c < ∞ , 2 (c1 − x2 ) 2
where c2 is taken between c1 and x2 and the constant c is independent of u, x1 and x2 in (0, δ). This proves that | sin (z1 − q, z2 − q)| c |z1 − z2 | . This inequality and a) imply
∂ |z1 − z2 | |z2 − q|
log
c
∂nq |z1 − q| |q − z1 |
(2.37)
for q ∈ Γ+ . On Γr− (x1 ) ∪ Γ − (x1 ) we have | sin (z1 − q, z2 − q)| =
|(x1 − u)(κ+ (x2 ) − κ− (u)) − (x2 − u)(κ+ (x1 ) − κ− (u)) . |q − z1 | |q − z2 |
2.2. Direct method of integral equations
113
The numerator does not exceed
(x1 −u)(κ+ (x2 )−κ+ (u))+(x2 −u)(κ+ (x1 )−κ+ (u)) + (κ− (u)−κ+ (u))(x2 −x1 ) . As shown above, the first term is dominated by c|(x2 − x1 )(x1 − u)(x2 − u)|. The second term obeys the estimate |(κ− (u) − κ+ (u))(x2 − x1 )| c u2 |x2 − x1 | c|(x2 − x1 )(x1 − u)(x2 − u)| , because u ε−1 |x1 − u| and u ε−1 |x2 − u| on Γr− (x1 ) ∪ Γ − (x1 ). Hence | sin (z1 − q, z2 − q)| c |z1 − z2 | and thus
∂ |z1 − z2 | |z2 − q|
log
c
∂nq |z1 − q| |q − z1 |
Γr− (x1 ) ∪ Γ − (x1 ) .
on
(2.38)
Now, c), d), and e) follow from (2.37) and (2.38).
Theorem 2.2.7. Let Ω+ have an outward peak and let 0 < α, β < 1. Then the operator α ˙ ˙ (Lodd )1,α (2.39) 2−β +R σ −→ πσ − T σ ∈ Λ−β (Γ)+R is continuous. Proof. Let κ ∈ C0∞ ({|z| < δ}) and κ(0) = 0. It suffices to show that κ πσ − T σ + T σ(0) Λα−β (Γ) c σ(Lodd )1,α (Γ) . 2−β
A function σ from (Lodd )1,α 2−β (Γ) can be represented as the sum σ = σ1 + σ2 , where σ1 belongs to Λ1,α and satisfies σ1 (z+ ) + σ2 (z− ) = 0, and σ2 is an Λα −β (Γ). 2−β (i) Let σ ∈ Λ1,α 2−β (Γ) and let σ(z+ ) + σ(z− ) = 0. Let us limit consideration to z ∈ Γ+ . We express (πI − T )σ(z) + T σ(0), z = x + iκ+ (x) ∈ Γ+ in the form ∂ |q| ∂ |q| dsq − dsq − σ(q) log σ(q) log ∂nq |z − q| ∂nq |z − q| Γ\(Γ+ ∪Γ− )
− Γc+ (x)
σ(q)
∂ |q| dsq − log ∂nq |z − q|
− Γr+ (x)∪Γr− (x)
Γ+ (x)∪Γ− (x)
σ(q) Γc− (x)
∂ |q| dsq = log Ik (z) . ∂nq |z − q| 6
σ(q)
∂ |q| dsq + πσ(z− ) log ∂nq |z − q|
k=1
114
Chapter 2. Boundary Integral Equations in H¨ older Spaces
Clearly, I1 belongs to Λα −β (Γ+ ∩ {x < δ/2}) and satisfies I1 Λα−β (Γ+ ∩{x<δ/2}) c σ Λα1−β (Γ) c σ Λ1,α
2−β (Γ)
c σ(Lodd )1,α
2−β (Γ)
.
(2.40)
The required estimates for I2 and I3 , I2 Λα−β (Γ+ ∩{x<δ/2}) +I3 Λα−β (Γ+ ∩{x<δ/2}) c σ Λα1−β (Γ) c σ(Lodd )1,α
2−β (Γ)
,
(2.41)
result from the inequalities in Lemma 2.2.6. Let us show this for I3 , which is a more difficult case. By the estimate a) in Lemma 2.2.6, |I3 (z)| c
(1−ε)−1 x
(1+ε)−1 x
σ+ (u)du cσΛα1−β (Γ)
cσ(Lodd )1,α
2−β (Γ)
(1−ε)−1 x
uβ−1 du
(1+ε)−1 x
(2.42)
β
x .
Suppose that |x1 − x2 | 2ε x1 and let x1 < x2 . We represent the difference I3 (z1 ) − I3 (z2 ) in the form ∂ |q| dsq σ(q) log c c ∂n |z q 1 − q| Γ+ (x1 )\Γ+ (x2 ) ∂ |q| dsq + σ(q) log ∂nq |z2 − q| Γc+ (x2 )\Γc+ (x1 ) ∂ |z2 − q| dsq . + σ(q) log & c c ∂n |z q 1 − q| Γ+ (x2 ) Γ+ (x1 ) The first and the second integrals are estimated in the same way. By the inequality b) from Lemma 2.2.6 we have for the first integral
σ(q) Γc+ (x1 )\Γc+ (x2 )
∂ |q|
dsq c σΛα1−β (Γ) log ∂nq |z1 − q|
c σΛα1−β (Γ) |x1 −
x2 |xβ−1 1
c σ(Lodd )1,α
2−β
(Γ) |x1
−
(1+ε)−1 x2
(1+ε)−1 x
x2 |α xβ−α 2
uβ−1 du
1
.
Next, we estimate the remaining third integral:
∂ |z2 − q|
ds σ(q) log
q
& ∂nq |z1 − q| Γc+ (x1 ) Γc+ (x2 ) ' (1−ε)−1 x1 (x1 +x2 )/2 du du c |x1 − x2 | + |σ+ (u)| |σ+ (u)| x2 − u u − x1 (x1 +x2 )/2 (1+ε)−1 x2 = {I31 + I32 } .
2.2. Direct method of integral equations
115
For I31 we obtain I31 c σΛα1−β (Γ) |x1 − x2 |
(x1 +x2 )/2
uβ−1
(1+ε)−1 x2
du x2 − u
(x1 +x2 )/2
du x 2−u 2 x − x
x2
2 1 1−α 1 + log c σ(Lodd )1,α (Γ) |x1 − x2 |α x2 β−α 2−β x2 x2 − x1 α β−α c σ(Lodd )1,α (Γ) |x1 − x2 | x2 .
c σΛα1−β (Γ) |x1 − x2 |x2
β−1
(1+ε)−1 x
2−β
The term I32 is estimated in a similar way: I32 c σΛα1−β (Γ) |x1 − x2 |
(1−ε)−1 x1
(x1 +x2 )/2
uβ−1
du x2 − u
(1−ε)−1 x1
du (x1 +x2 )/2 u − x1 x − x
x1
2 1 1−α 1 + log c σ(Lodd )1,α (Γ) |x1 − x2 |α x1 β−α 2−β x1 x2 − x1 c σ(Lodd )1,α (Γ) |x1 − x2 |α x1 β−α . c σΛα1−β (Γ) |x1 − x2 |x1 β−1
2−β
Let S± stand for the image of Γ± under the mapping w = 1/z, z ∈ Γ± . Since 2 1 + κ±x(x) 1 1 w κ± (x) = , = =1−i 1 κ (x) ± u x + iκ± (x) Re x+iκ± (x) x 1+i x
it follows that Im w = −
κ± (x) u. x
(2.43)
Using the relations u x = Re w x = x Re
2 1 1 = κ± (x) 2 = 1 + O x , x + iκ± (x) 1+ x
which can be differentiated twice, we find u=
1 1 1 1 1 2 + O(x), ux = − 2 + O(1), ux = 2 + O , and x = + O 3 . x x x x u u
Hence we get the equality Im w = − which can be differentiated twice.
κ± (x) + O x2 , x2
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Chapter 2. Boundary Integral Equations in H¨ older Spaces
Thus, the curves S± are graphs of the functions v = ϕ± (u), u + iv = w, admitting the representations 1 1 , ϕ± (u) = O 2 . ϕ± (u) = −α± + o(1), ϕ± (u) = O u u Using the change of variables p = 1/q, w = 1/z in I4 , we can write the integral over Γc− (x) as ∂ 1 dsp . σ(1/p) log ∂np |p − w| S c (w− )
Here w stands for the image of z, w− denotes the image of z− and S c (w− ) is the image of Γc− (x). The increment ΔS c (w) of the function arg(p − w) along S c (w− ) admits the representation ∂ ∂ 1 1 dsp = π+O Re . ΔS c (w) = arg(p−w)dsp = − log ∂sp ∂np |p − w| w S c (w− )
S c (w− )
Considered as a function of Re w− , the product σ(1/w− ) [π−ΔS c (w)] belongs to Λα β (2/δ, ∞) and obeys the required estimate. Therefore, it suffices to consider ∂ 1 − dsp . σ(1/p) − σ(1/w− ) log (2.44) ∂np |p − w| S c (w− )
By w and wr we denote the end points of the arc S c (w− ), which are images of the end points (1 + ε)−1 x + iκ− ((1 + ε)−1 x) and (1 − ε)−1 x + iκ− ((1 − ε)−1 x) of the arc Γc− (x). Further, let S (w− ) and S r (w− ) be the arcs on S c (w− ), connecting w , w− and w− , wr . We write the integral in (2.44) as 1 ∂ 1 p−w −σ dsp σ arg p w− ∂sp w − w S (w− ) 1 ∂ 1 p−w −σ dsp . σ arg + p w− ∂sp wr − w S r (w− ) π We choose the branch of arg by fixing its range as (− 3π 2 , 2 ]. Integrating by parts, we find (∂/∂sp )σ(1/p) arg(p − w)dsp S c (w− )
− arg(wr − w)
(∂/∂sp )σ(1/p)dsp
S r (w
−)
− arg(w − w) S (w− )
(∂/∂sp )σ(1/p)dsp .
(2.45)
2.2. Direct method of integral equations
117
The function arg(p − w), p ∈ S c (w− ), obeys (2.26) (see Appendix C(i)). By Theorem 2.2.1, the norm of the first integral in (2.45), considered as a function of Re w, in the space Λα β (2/δ, ∞) does not exceed c σ (Lodd )1,α (Γ) . 2−β
The two remaining integrals in (2.45) are estimated in the same way. Consider, for example, the last one. Since Re (wr − w) = εu + O the expression
1 u
,
1 1 1 = +O 3 Re (wr − w) εu u
is comparable with 1/u. In other words, there are constants c1 and c2 such that c1
1 1 1 c2 , u Re (wr − w) u
while Im (wr − w) is comparable with 1. Hence
Im (w − w)
c
r
. Re (wr − w) u Using this estimate, we obtain
arg(wr − w)
S r (w− )
(∂/∂sp )σ(1/p)dsp
Im (w − w) γ(w− ) r (∂/∂t)σ− (1/t)dt Re (wr − w) γ(wr ) c γ(w− ) 2−β −2 t t dt = cσ(Lodd )1,α u−β , σ Λα2−β 2−β u γ(wr ) c arctan
where γ(wr ) and γ(w− ) are comparable with (1 − ε)u and u, respectively. In order to estimate the increment of the last term in (2.45) at the points w1 and w2 , with |w1 − w2 | (ε/2) min{|w1 |, |w2 |} we represent it as [arg(w1r − w1 ) − arg(w2r − w2 )] (∂/∂sp )σ(1/p)dsp S r (w1− ) (∂/∂sp )σ(1/p)dsp = I41 + I42 . + arg(w2r − w2 ) − S r (w1− )
S r (w2− )
To be specific, let u1 = Re w(1) < u2 = Re w(2) . In view of ∂ Re wr − w = O(1) ∂u
and
1 ∂ Im wr − w = O , ∂u u
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Chapter 2. Boundary Integral Equations in H¨ older Spaces
by the mean value theorem we have, on Γc (w− ), |Re w1r − w1 − Re w2r − w2 | c|u1 − u2 | and
|u1 − u2 | . |Im w1r − w1 − Im w2r − w2 | c u1
Hence,
Im (w − w ) Im (w − w )
1r 1 2r 2
− | arg(w1r − w1 ) − arg(w2r − w2 )|
Re (w1r − w1 ) Re (w2r − w2 )
1 1
c
−
|Im (w1r − w1 )| Re (w1r − w1 ) Re (w2r − w2 )
|u − u | 1
1 2 (Im (w1r − w1 ) − Im (w2r − w2 ) +
. Re (w2r − w2 ) u21 As a result we obtain |I41 | c σ Λα2−β
|u1 − u2 | −β u1 c σ(Lodd )1,α |u1 − u2 |α u−β−α . 1 2−β u1
To estimate I42 we use the inequality − S r (w1− )
c σ
Λα 2−β
S r (w2− )
|(∂/∂sp )σ(1/p)|dsp
γ(w2r )
γ(w2− )
+ γ(w1r )
u−β du
γ(w1− )
c σ(Lodd )1,α |u2 − u1 |u−β , 2−β
where γ(wir ) and γ(wi− ) are comparable with (1 − ε)ui and ui , respectively, i = 1, 2. Taking into account that c | arg w2r − w2 ) | , u2 we conclude |I42 | c σ(Lodd )1,α
2−β
|u2 − u1 | −β u1 c σ(Lodd )1,α |u2 − u1 |α u−β−α . 1 2−β u1
Thus, the last two terms in (2.45), considered as functions of Re w, belong to Λα β (2/δ, ∞) and their norms do not exceed c σ(Lodd )1,α (Γ) . 2−β
Finally, consider I6 . By b) and e) from Lemma 2.2.6
∂ |z| |q|
, q ∈ Γr± (x), log
c
∂nq |z − q| |q|
x = Re z ,
(2.46)
2.2. Direct method of integral equations
and
119
∂ |z1 − z2 | |z2 − q|
, log
c ∂nq |z1 − q| |q|
(2.47)
where q ∈ Γr± (x1 ), z1 , z2 ∈ Γr+ (x1 ) and |x1 − x2 | 2ε x1 . Hence the norm of I6 in Λα −β (Γ+ ∩ {x < δ/2}) is dominated by σ (Lodd )1,α (Γ) . To be precise, 2−β
I6 Λα−β (Γ+ ∩{δ/2}) c σ Λα1−β (Γ) c σ (Lodd )1,α
2−β
(Γ)
.
(2.48)
As a result we obtain πσ − T σ + T σ(0)Λα−β (Γ) c σ Λ1,α
2−β (Γ)
c σ (Lodd )1,α
2−β (Γ)
.
(2.49)
(ii) Now, suppose that σ ∈ Λα −β (Γ). It was shown in part (i) that the integral ∂ |q| dsq σ(q) log ∂nq |z − q| Γ+
belongs to Λα −β (Γ+ ∩ {x < δ/2}) and its norm does not exceed c σΛα1−β (Γ) cσΛα−β (Γ) . Let us represent
σ(q) Γ−
∂ |q| dsq , z ∈ Γ+ , log ∂nq |z − q|
as the sum of integrals over the arcs Γ − (x), Γc− (x) and Γr− (x), denoted by J , J c and J r , respectively. Using (2.41) and (2.48), we show that J and J r belong to Λα −β (Γ+ ∩ {x < δ/2}) and satisfy J Λα−β (Γ+ ∩{x<δ/2}) + J r Λα−β (Γ+ ∩{x<δ/2}) c σΛα−β (Γ) . After the change of variables q = 1/p, z = 1/w the integral J c takes the form (1+ε) ∂ 1 ∂ dsp = − arg(p − w)du , (2.50) σ(1/p) log σ∼ (u) ∂np |w − p| ∂u (1−ε)x S c (w− )
where σ∼ (u) = σ(1/p), p ∈ S c (w− ). Since (∂/∂u) log |p−w| satisfies the conditions (2.28) (see Appendix C(ii)), it follows by Theorem 2.2.3 that the integral (2.50), considered as a function of Re p, belongs to Λα β (2/δ, ∞) and its norm in this space does not exceed c σΛα−β (Γ) . Thus, πσ − T σ + T σ(0)Λα−β (Γ) c σΛα−β (Γ) . The continuity of the operator (2.49) results by this inequality and (2.39).
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Chapter 2. Boundary Integral Equations in H¨ older Spaces
Theorem 2.2.8. Let Ω+ have an inward peak and let 0 < α, β < 1. Then the operator α ˙ (Lodd )1,α (2.51) 2−β (Γ) σ −→ πσ − T σ ∈ Λ−β (Γ)+R is continuous. Proof. It suffices to show that κ πσ − T σ + T σ(0) Λα−β (Γ) c σ(Lodd )1,α
2−β (Γ)
,
where κ is the same as in Theorem 2.2.7. With this aim in view, we represent (πI − W )σ + W σ(0) on Γ+ in the form ∂ |q| dsq σ(q) log π[σ(z+ ) − σ(z− )] − ∂nq |z − q| Γ\(Γ+ ∪Γ− )
−
σ(q) Γ+ (x)∪Γ− (x)
+ πσ(z− ) − − Γr+ (x)∪Γr− (x)
∂ |q| dsq − log ∂nq |z − q| σ(q)
Γc− (x)
σ(q) Γc+ (x)
∂ |q| dsq log ∂nq |z − q|
∂ |q| dsq log ∂nq |z − q|
∂ |q| dsq = σ(q) log Ik (z) . ∂nq |z − q| 6
k=1
Clearly, I1 belongs to Λα −β (Γ+ ∩ {x < δ/2}) and obeys the required estimate. It was proved in Theorem 2.2.7 that I2 − I6 belong to Λα −β (Γ+ ∩ {x < δ/2}) and their norms in this space are dominated by c σ (Lodd )1,α (Γ) . 2−β
2.2.3 Continuity of the operator πI + S We introduce the linear hull D1 of (∂/∂s)Re z, z ∈ Γ \ {O}. Theorem 2.2.9. Let Ω+ have an outward peak and let 0 < α, β < 1. Then the operator α ˙ (Lodd )1,α (2.52) 2−β (Γ) σ −→ πσ + Sσ ∈ Λ−β (Γ)+D1 is continuous. We collect auxiliary estimates, to be used in the proof of this theorem, in the following assertion similar to Lemma 2.2.6. Lemma 2.2.10. For any z, z1 , z2 ∈ Γ+ the inequalities hold:
∂ |q|
log a)
c < ∞ on Γ ± (x) ∪ Γc+ (x), x = Re z;
∂nz |q − z|
∂ 1
x 1
+ Im c on Γr± (x), u = Re q; b) log
∂nz |q − z| q u
2.2. Direct method of integral equations
and for |x1 − x2 | c) d) e)
ε 2
121
min{|x1 |, |x2 |},
∂ |z1 − z2 | |z1 − q|
log
c
∂nz |z2 − q| |z1 |
∂ |z1 − z2 | |z1 − q|
log
c
∂nz |z2 − q| |z1 − q|
∂ |z1 − z2 | |z1 − q|
log
c
∂nz |z2 − q| |q|
on
Γ ± (x1 ),
on
Γc+ (x1 );
on
Γr± (x1 ) .
x1 = Re z1 ;
Proof. We have
∂ 1/2 | − κ (u) + κ (x)(u − x) + κ (x) | 1
± +
± log (x)2 = on Γ± .
1 + κ±
∂nz |q − z| |q − z|2 The right-hand side does not exceed c
(x − u)2 c < ∞ on Γ + (x) ∪ Γc+ (x) . |q − z|2
Hence c
(x − u)2 |κ+ (u) − κ− (u)| x2 c < ∞ on Γ − (x) , + = c 1 + |q − z|2 |q − z|2 |q − z|2
because |q − z| (1 + ε)−1 x. Thus, a) follows. To obtain b), which is valid on Γr± (x), we start with the relation 1/2 1 1 ∂ 1 + κ± log (u)2 + Im ∂nz |q − z| q 1 ∂ 1 1 ∂ log + log + Im (x) = Re − κ± ∂x z − q ∂y z−q q z 1 − Im . = −κ± (x)Re z−q q(z − q) The absolute value of the first term on the right-hand side does not exceed x cx c , |q − z| u because |q − z| |q|(1 − | zq |) c u. The absolute value of the second term is dominated by
1 z
1
1
1
Re + Re
Im
Im z Re + Re z Im
q−z q q−z q q u2 x 1 x c (x + u) c . + (u − x)2 u−x u u These estimates imply b).
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Chapter 2. Boundary Integral Equations in H¨ older Spaces
Let us show c), d) and e) for q ∈ Γ+ . Assuming |(z1 − q, z2 − q)| π/2, we have
∂ − q, nz1 ) cos (z2 − q, nz2 )
|z2 − q|
cos (z1
− log
=
∂nz |z1 − q| |q − z1 | |q − z2 |
1
1
− cos (z2 =
− q, nz2 )
|q − z1 | |q − z2 |
1
cos (z1 − q, nz1 ) − cos (z2 − q, nz2 )
(2.53) + |q − z1 | − q, nz2 )| |z2 − z1 | | cos (z2 |q − z1 | |q − z2 | 1 c | sin (z1 − + q, z2 − q)| + | sin (n z1 , nz2 )| , |q − z1 | 2 because
cos (z1 − q, nz1 ) − cos (z2 − q, nz2 )
cos (z1 − q, nz1 ) − cos (z2 − q, nz1 ) + cos (z2 − q, nz1 ) − cos (z2 − q, nz2 )
1 1 q, z2 − q)| + c | sin (n c | sin (z1 − z , nz2 )| , 2 2 1 and
1 q, z2 − q)| | sin (z1 − q, z2 − q)| . | sin (z1 − 2 q, z2 − q)| > π/2. By q˜ we denote a point which is centrally Suppose that |(z1 − symmetric to q with respect to z2 . Since |z2 − q| = |z2 − q˜|, we arrive at the inequality
∂ − q, nz2 )| |z2 − q˜|
|z2 − z1 | | cos (z2
log
∂nz |z1 − q| |q − z1 | |˜ q − z2 | c 1 | sin (z1 − + q, z2 − q˜)| + | sin (n , n )| z z2 |q − z1 | 2 1 in the same way as we did in (2.53). However, now we have |(z1 − q, z2 − q˜)| π/2, hence 1 q, z2 − q˜)| | sin (z1 − q, z2 − q˜)| = | sin (z1 − q, z2 − q)|. | sin (z1 − 2 Next, we use the equality
sin (z1 − q, z2 − q)
(x1 − u)(κ+ (x2 ) − κ+ (u)) − (x2 − u)(κ+ (x1 ) − κ+ (u))
. = |q − z1 ||q − z2 |
(2.54)
2.2. Direct method of integral equations
123
Let us write the numerator in (2.54) as (x1 − u)(κ+ (x2 ) − κ+ (u)) − (x2 − u)(κ+ (x1 ) − κ+ (u)) κ (u) − κ (x ) κ (x ) − κ (x ) + + 2 + 1 + 2 = (x2 − x1 )(u − x2 ) . − u − x2 x1 − x2 We develop
(2.55)
κ+ (u) − κ+ (x2 ) u − x2
by the Taylor formula in a neighborhood of u = x1 , (c1 ) κ+ (u) − κ+ (x2 ) κ+ (x1 ) − κ+ (x2 ) κ+ κ+ (c1 ) − κ+ (x2 ) (u − x1 ), = + + u − x2 x1 − x2 c1 − x2 (c1 − x2 )2
with c1 between u and x1 . The expression in square brackets in (2.55) does not exceed |κ+ (c1 )(c1 − x2 ) + κ+ (c1 ) − κ(x2 )| 1 = |κ (c2 )| c < ∞ 2 (c1 − x2 ) 2 with c2 between c1 and x2 and with c independent of u, x1 and x2 in (0, δ). This proves the inequality | sin (z1 − q, z2 − q)| c |z1 − z2 | . By our choice of δ > 0 we have 1 | sin (n z , nz2 )| | sin (nz1 , nz2 )| c |κ+ (x1 ) − κ+ (x2 | c |z1 − z2 | . 2 1 Unifying the above estimates with a), we conclude
∂ |z1 − z2 | |z2 − q|
log
c ∂nz |z1 − q| |q − z1 |
for q ∈ Γ+ .
(2.56)
On Γr− (x1 ) ∪ Γ − (x1 ) we have |(x1 − u)(κ+ (x2 ) − κ− (u)) − (x2 − u)(κ+ (x1 ) − κ− (u)) . q, z2 − q)| = | sin (z1 − |q − z1 | |q − z2 | The numerator in the last ratio is dominated by |(x1 − u)(κ+ (x2 ) − κ+ (u)) + (x2 − u)(κ+ (x1 ) − κ+ (u))|q − z1 | + |(κ− (u) − κ+ (u))(x2 − x1 )| . As shown above, the first term does not exceed c |(x2 − x1 )(x1 − u)(x2 − u)|.
124
Chapter 2. Boundary Integral Equations in H¨ older Spaces
The second term obeys the estimate |(κ− (u) − κ+ (u))(x2 − x1 )| c u2 |x2 − x1 | c |(x2 − x1 )(x1 − u)(x2 − u)| , because u ε−1 |x1 − u| and u ε−1 |x2 − u| on Γr− (x1 ) ∪ Γ − (x1 ). Thus, on Γr− (x1 ) ∪ Γ − (x1 ) we have q, z2 − q)| c |z1 − z2 | | sin (z1 − and
∂ |z1 − z2 | |z2 − q|
. log
c ∂nz |z1 − q| |q − z1 |
(2.57)
Inequalities (2.56), (2.57) imply c), d) and e). Proof of Theorem 2.2.9. We set A=
lim
Γ+ z→0
Sσ(z) .
Let us write πσ + Sσ in the form ∂ ∂ πσ + Sσ + A Re z − A Re z . ∂s ∂s It suffices to show that κ πσ+ Sσ − A satisfies κ πσ + Sσ − A
Λα −β (Γ+ )
c σ(Lodd )1,α
2−β (Γ)
on Γ+ , where κ is the same as in Theorem 2.2.7. (i) Given σ ∈ (Lodd )1,α 2−β (Γ), we represent πσ(z) + (Sσ)(z) − A with z = x + iκ+ (x) ∈ Γ+ in the form ∂ 1 1 + Im dsq π[σ(z+ ) − σ(z− )] + σ(q) log ∂nz |z − q| q
Γ\(Γ+ ∪Γ− )
+
+ Γ+ \Γr+ (x)
+
σ(q) Γc+ (x)
+ Γc− (x)
Γ− \Γr− (x)
1 σ(q) Im dsq + q
∂ 1 dsq + log ∂nz |z − q|
σ(q) Γ+ (x)∪Γ− (x)
Γr+ (x)∪Γr− (x)
∂ 1 dsq log ∂nz |z − q|
∂ 1 1 + Im dsq σ(q) log ∂nz |z − q| q
7 ∂ 1 dsq + πσ(z− ) = σ(q) log Ik (z) . ∂nz |z − q| k=1
2.2. Direct method of integral equations
125
Using estimates in Lemma 2.2.10, we check that I1 (z), I2 (z) and I3 (z) belong to Λα −β (Γ+ ∩ {x < δ/2}) and satisfy 3
Ik Λα−β (Γ+ ∩{x<δ/2}) c σΛα1−β (Γ) c σ(Lodd )1,α
2−β (Γ)
k=1
.
It is also shown in Lemma 2.2.10 that on Γ ± (x) ∪ Γc+ (x) one has
∂ ∂ |z1 − z2 | 1 |z2 − q|
c and log
∂nz log |z1 − q| c |z1 | ∂nz |z − q| for z1 , z2 ∈ Γ+ with |z1 − z2 |
ε 2
min{|z1 |, |z2 |}. These estimates imply
I4 Λα−β (Γ+ ∩{x<δ/2}) + I5 Λα−β (Γ+ ∩{x<δ/2}) c σ Λα1−β (Γ) c σ (Lodd )1,α
2−β (Γ)
.
We omit the proof of this inequality, since it does not differ from the proof of (2.41) in Theorem 2.2.7. Using inequalities b) and e) in Lemma 2.2.10, we show that I6 Λα−β (Γ+ ∩{x<δ/2}) c σ Λα1−β (Γ) c σ(Lodd )1,α (Γ) . 2−β
Let us write I7 (z) as ∂ ∂ 1 1 − dsq σ(q) log log ∂nz |z − q| ∂nq |z − q| Γc− (x)
+ Γc− (x)
∂ 1 dsq + πσ(z− ) = J1 σ (z) + J2 σ (z) . σ(q) log ∂nq |z − q|
The kernel of J1 (z) has the form ∂ 1 1 1 ∂ − = κ+ , q ∈ Γc− (x) . log log (x) − κ− (u) Re ∂nz |z − q| ∂nq |z − q| z−q By Appendix D and Theorem 2.2.1, the integral operator 1 dsq σ(q) κ− (u)Re J11 σ (z) = c z − q Γ− (x) is subject to J11 σ Λα−β (Γ+ ∩{x<δ/2}) c σ− Λα1−β (0,δ) c σ Λ1,α
2−β (Γ− )
According to Corollary 2.2.2, the operator J12 σ (z) = κ+ (x) σ(q)Re Γc− (x)
.
c 1 dsq = κ+ (x) J12 σ (z) z−q
126
Chapter 2. Boundary Integral Equations in H¨ older Spaces
satisfies the estimate J12 σ Λα−β (Γ+ ∩{x<δ/2}) c J12 σ Λα1−β (Γ+ ∩{x<δ/2}) c σ− Λα1−β (0,δ) c σ Λ1,α
2−β (Γ)
.
Thus, for J1 = J11 + J12 we have J1 Λα−β (Γ+ ∩{x<δ/2}) c σ Λα1−β (Γ) c σ(Lodd )1,α
2−β (Γ)
.
The required inequality for J2 is proved in Theorem 2.2.7. Unifying the estimates obtained, we find c σ(Lodd )1,α (Γ) . πσ + Sσ − A α (2.58) 2−β Λ−β (Γ+ ∩{x<δ/2})
(ii) Now, suppose that σ ∈ Λα −β (Γ). It has been shown in (i) that the integral 1 |q| ∂ + Im dsq , z ∈ Γ+ , σ(q) log ∂nz |z − q| q Γ−
as a function of z, belongs to Λα −β (Γ+ ∩ {x < δ/2}) and its norm is dominated by σΛα1−β (Γ) σΛα−β (Γ) . We represent the integral 1 ∂ |q| + Im dsq , σ(q) log ∂nz |z − q| q
z ∈ Γ+ ,
Γ+
as the sum of three integrals over the arcs Γ + (x), Γc+ (x) and Γr+ (x). Let them be denoted by I , I c and I r . Following the same pattern as in (i), we can show that I Λα−β (Γ+ ∩{x<δ/2}) + I r Λα−β (Γ+ ∩{x<δ/2}) cσΛα−β (Γ) . Similarly to (2.58), let us represent I c (z) as J1 (z)+ J2 (z). In (i) above we obtained the required estimate for J1 (z): J1 Λα−β (Γ+ ∩{x<δ/2}) c σΛα1−β (Γ) c σΛα−β (Γ) . It was shown in Theorem 2.2.7 that J2 belongs to Λα −β (Γ+ ∩ {x < δ/2}) and J2 Λα−β (Γ+ ∩{x<δ/2}) c σΛα−β (Γ) . Thus we arrive at πσ + Sσ − A
Λα −β (Γ+ ∩{x<δ/2})
c σΛα−β (Γ) .
This inequality together with (2.58) implies the boundedness of the operator (2.52).
2.2. Direct method of integral equations
127
Theorem 2.2.11. Let Ω+ have an inward peak and let 0 < α, β < 1. Then the operator α ˙ (Lodd )1,α (2.59) 2−β (Γ) σ −→ πσ + Sσ ∈ Λ−β (Γ)+D1 is continuous. Proof. Similarly to Theorem 2.2.9 we introduce the number A=
lim
Γ+ z→0
Sσ(z) .
It suffices to show that κ πσ + Sσ − A
Λα −β (Γ+ )
c σ(Lodd )1,α
2−β (Γ)
.
To this end we represent πσ(z) + Sσ(z) − A with z = x + iκ+ (x) ∈ Γ+ in the form ∂ 1 1 + Im dsq σ(q) log π[σ(z+ ) + σ(z− )] + ∂nz |z − q| q
Γ\(Γ+ ∪Γ− )
+
+ Γ+ \Γr+ (x)
+
σ(q) Γc+ (x)
Γ− \Γr− (x)
∂ 1 dsq + log ∂nz |z − q|
+
σ(q) Γc− (x)
1 σ(q) Im dsq + q
σ(q) Γ+ (x)∪Γ− (x)
Γr+ (x)∪Γr− (x)
∂ 1 dsq log ∂nz |z − q|
∂ 1 1 + Im dsq σ(q) log ∂nz |z − q| q
7 ∂ 1 dsq − πσ(z− ) = log Ik (z) . ∂nz |z − q| k=1
Clearly, I1 belongs to Λα −β (Γ+ ∩ {x < δ/2}) and I1 Λα−β (Γ+ ) c σ (Lodd )1,α
2−β (Γ)
.
It was proved in Theorem 2.2.9 that the terms I2 , . . . , I7 belong to Λα −β (Γ+ ∩ {x < δ/2}) and the norms in this space do not exceed c σ (Lodd )1,α
2−β (Γ)
2.2.4 Continuity of the operator V Our aim is to prove the following assertion. Theorem 2.2.12. The single layer potential Vσ(z) = σ(q) log Γ
continuously maps Λα 1+β (Γ) into
Nβ1,α (Γ).
|z| dsq |z − q|
.
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Chapter 2. Boundary Integral Equations in H¨ older Spaces
Proof. We fix a small positive ε such that, |κ± (x) − κ∓ (u)| ≥ c u2 for all u, with |x − u| < 2ε x. By Γ ± (x), Γc± (x) and Γr± (x) we denote the subarcs of Γ± with abscissas in [0, (1 + ε)−1 x], [(1 + ε)−1 x, (1 − ε)−1 x] and [(1 − ε)−1 x, δ]. ( Consider two cases: 1◦ supp σ ⊂ Γ+ Γ− and 2◦ σ = 0 in a small fixed neighborhood of O. 1◦ We have ∂ ∂ |z| dsz Vσ(z) = σ(q) log ∂sz ∂sz Γ+ ( Γ− |z − q| (2.60) ∂ |z| dsq + Iσ (z) . σ(q) = + log ∂sz |z − q| Γc+ (x) Γc− (x) We claim that the last term on the right-hand side of (2.60) obeys the inequality Iσ Λα1+β (Γ) c σ Λα1+β (Γ) .
(2.61)
Taking into account that 1 |z| 1 ∂ = ∓Re 1 − iκ± on Γ± , log (x) − ∂sz |z − q| z z−q one can easily check that the last term on the right-hand side of (2.60) admits the estimate (1−ε/2)x c δ c |σ± (u)|udu + |σ± (u)|du, (2.62) |I(z)| 2 x 0 x (1+ε/2)x where z = x + iy ∈ Γ± . Hence, # $ max |I(z)| x1+β c σ± Λα1+β (0,δ) . x∈(0,δ)
(2.63)
Let z1 , z2 belong to Γ+ or Γ− . Suppose that x1 = Rez1 < Rez2 = x2 . In the case |x1 − x2 | < 12 min{|x1 |, |x2 |} the mean value theorem implies
∂ u |z1 | |z2 |
∂
log log −
c 3 |x1 − x2 | on Γ ± (x1 ) ∪ Γ ± (x2 )
∂sz |z1 − q| ∂sz |z2 − q| x1
and
∂ ∂ |x1 − x2 | |z1 | |z2 |
− log log on Γr± (x1 ) ∪ Γr± (x2 ) .
c
∂sz |z1 − q| ∂sz |z2 − q| x21 From these inequalities and (2.62) we obtain x1+α+β | Iσ (z1 ) − Iσ (z2 )| 1 c σ± Λα1+β (0,δ) . sup |x1 − x2 |α |x1 −x2 |<x1 /2 Now, (2.61) results by (2.63) and (2.64).
(2.64)
2.2. Direct method of integral equations
129
Using the change of variables q = 1/p and z = 1/w, we write σ(q)K(z, q)dsq Γc± (x)
in the form
S c (w± )
σt (p)Kt (w, p)dsp ,
(2.65)
where S c (w± ) is the image of Γ± under the inversion: Kt (w, p) = K
1 1 , w p
and
σt = σ
1 1 ∈ Λ1−β (1/δ, ∞). p |p|2
Since ∂ ∂ |z| 1 1 = |w|2 = ±|w|2 Re (1 + iϕ± (u)) , log log ∂sz |z − q| ∂sw |w − p| p−w it follows that the norm of (2.65) in Λα −1−β (1/δ, ∞) does not exceed the Λ1−β (1/δ, ∞)-norm of
dp du
. (2.66) σt± (u)
du p − w c S (w± ) The function
1 dp , (p − w) du
considered as a function of u on S c (w± ), satisfies the conditions of Theorem 2.2.1 (see Appendix E(i)). Therefore, the norm of the integral (2.66) in Λ1−β (1/δ, ∞) does not exceed the norm of
dp dp −1
σt± (u)
du du in the same space and is compared with σ± Λα1+β (0,δ) . Finally, we obtain
∂ Vσ Λα1+β (Γ) c σ Λα1+β (Γ) . ∂s
Inequality sup |xβ (Vσ)± (x)| c σ Λαβ+1 (Γ)
x∈(0,δ)
is derived as follows. We start with the equality |z| dsq + (Jσ)(z) . Vσ(z) = σ(q) log + c c |z − q| Γ+ (x) Γ− (x)
(2.67)
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Chapter 2. Boundary Integral Equations in H¨ older Spaces
Note that |q|/|z| c |u|/|x| on Γ + (x) ∪ Γ − (x). Hence (1+ε)−1 x
q u
σ(q) log |1 − |dsq c |σ± (u)| log 1 + c du z x Γ± (x) 0 (1+ε)−1 x c |σ± (u)|udu c x−β σ Λαβ+1 (Γ) . x 0
To estimate the integral over Γr+ (x) ∪ Γr− (x), we observe that
z
= log 1 + q |q| c u c log
z−q z−q |z − q| |x − u| and u − x > εu. Then δ
q
σ(q) log |1 − |dsq c |σ± (u)|du c x−β σ Λαβ+1 (Γ) . (2.68)
z Γ± (x) (1−ε)−1 x By (2.67) and (2.68) sup |xβ (Jσ)± (x)| c σ Λαβ+1 (Γ) .
(2.69)
x∈(0,δ)
On Γc± (x) one has
z
|1 + iκ+ (x)/x| x
c .
cx
z−q |x − u| + |κ+ (x) − κ± (u)| |x − u| Hence
σ(q) log
Γc± (x)
(1−ε)−1 x |z| cx
dsq du |σ± (u)| log |z − q| |x − u| (1+ε)−1 x (1−ε)−1 x σ Λαβ+1 (Γ) u−β−1 log (1+ε)−1 x
= σ Λαβ+1 (Γ) x−β
(1−ε)−1
u−β−1 log
(1+ε)−1
= cx−β−1 σ Λαβ+1 (Γ) , which together with (2.69) proves (2.67). 2◦ The difference Vσ(z+ ) − Vσ(z− ) is equal to
1 − q/z
−
σ(q) log
dsq = I+ (x) + I− (x) . 1 − q/z + Γ+ ∪Γ− Let us show that I+ Λαβ−1 (0,δ) c σ Λα1+β (Γ) .
cx du |x − u| c du |1 − u|
2.2. Direct method of integral equations
131
We start by expressing I+ (x) as
1 − q/z
−
c r σ(q) log
+ + (x) + I+ (x) + I+ (x) .
dsq = I+ 1 − q/z+ Γ+ (x) Γc+ (x) Γr+ (x) −1 −1 − z− | c, it follows that Since |z+
1 − q/z
xu
−
.
1+c 1 − q/z+ |x − u|
r Hence, the sum I+ (x) + I+ (x) does not exceed
(1+ε)−1 x
c
|σ(u)|udu + c x
0
δ
(1−ε)−1 x
|σ(u)|du
and
r sup |xβ−1 I+ (x)| + sup |xβ−1 I+ (x)| c σ Λα1+β (0,δ) .
(0,δ)
(0,δ)
Now, suppose that x < x and |x − x | < (ε/2)x . We write I+ (x )− I+ (x ) in the form
1 − q/z 1 − q/z
− +
σ(q) log
1 − q/z dsq 1 − q/z Γ+ (x ) + − (2.70)
1 − q/z
−
+ σ(q) log
= J + J . 1 2 dsq 1 − q/z+ Γ+ (x )\Γ+ (x )
In order to estimate the first term, we write the product 1 − q/z+ 1 − q/z− 1 − q/z 1 − q/z+ −
as )(1 − q/z+ ) − (1 − q/z− )(1 − q/z+ ) (1 − q/z− (1 − q/z− )(1 − q/z+ ) q z1 − z1 + q z1 − z1 + q 2 z1z − z 1z + + − − − + − + . =1− )(1 − q/z ) (1 − q/z− +
1+
(2.71)
The difference (1/z+ ) − (1/z+ ) is equal to
(x − x ) + i(κ+ (x ) − κ+ (x )) x x iκ+ (x )/x x − x iκ+ (x )/x 1− = · 1− (1 + r1 (x )) , 1 + iκ+ (x )/x 1 + iκ+ (x )/x x x
(2.72)
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Chapter 2. Boundary Integral Equations in H¨ older Spaces
where |r1 (x )| c x . A similar representation holds for the difference 1 1 x − x − = − (1 + r2 (x )) , z− z− x x
(2.73)
z+ ) − 1/(z− z+ ) as where |r2 (x )| c x . Let us write 1/(z− z − z+ z− − z+ x − x + − = 2 (1 + r3 (x )) z− z+ z− z− z+ z+ (x ) x x − x x −x r5 (x ) , − 2 (1 + r4 (x )) = (x ) x x x
(2.74)
where |r3 (x )| c x , |r4 (x )| c x and |r5 (x )| c . Since |z | | |z− c x c x c and + c, − q| |x − u| |z+ − q| |x − u|
|z−
q ∈ Γ + (x ) ,
(2.75)
the quotient in (2.71) is estimated on Γ + (x ) by c (u/x ) |x − x |. Therefore,
1 − q/z
1 − q/z
u
−
−
− log
log 1 + c |x − x |
log
1 − q/z+ 1 − q/z+ x u |x − x | , q ∈ Γ + (x ) , x which implies the majorant for J1 : |J1 |
c σ Λα1+β (Γ) c σ Λα1+β (Γ)
(1+ε)−1 x
0
|x − x | x
u−β−1
u |x − x |du x
(1+ε)−1 x
u−β du
0
c σ Λα1+β (Γ) c(x )−α−β+1 |x − x |α . The second term in (2.70) is estimated as follows:
|J2 |
z
|σ+ (u)| log − dsq z+ −1 (1+ε) x (1+ε)−1 x
q − z
+
+c |σ+ (u)| log
dsq = J21 + J22 . q − z −1 (1+ε) x −
c
(1+ε)−1 x
Since z− /z+ = 1 + O(x ), we have
log z− c x z+
(2.76)
2.2. Direct method of integral equations
133
and
| c x |J21
(1+ε)−1 x
(1+ε)−1 x 1−β−α
c (x )
|σ+ (u)|du c (x )1−β σ Λα1+β (0,δ)
|x − x | x
|x − x |α σ Λα1+β (0,δ) .
In a similar way, the estimate for J22 results by
q − z |z+ − z− | c (x )2 (x )2
+
log 1 + c c x , 1 +
log
| q − z− q − z− |u − x | |u − x | where q ∈ Γ + (x )\Γ + (x ). Next, consider another difference r r I+ (x ) − I+ (x )
1 − q/z
1 − q/z
−
−
= σ(q) log
− σ(q) log
ds
q dsq r r 1 − q/z 1 − q/z Γ+ (x ) Γ+ (x ) + +
1 − q/z 1 − q/z
1 − q/z
− +
−
= σ(q) log
+ log
dsq dsq r r r 1 − q/z 1 − q/z 1 − q/z Γ+ (x ) Γ+ (x )\Γ+ (x ) + − +
= J1r + J2r . By (2.72)–(2.75) the absolute value of the right-hand side of (2.71) on Γr+ (x ) does not exceed |x − x | (x )2 c u2 c |x − x | . x x |x − u|2 Hence
1 − q/z 1 − q/z
− +
log 1 + c|x − x | c |x − x | .
log
1 − q/z+ 1 − q/z− Thus |J1r | c |x − x |
1/δ
(1−ε)−1 x
|σ+ (u)|du
c σ Λα1+β (Γ) |x − x |
1/δ
u−β−1 du
(1−ε)−1 x
c σ Λα1+β (Γ) |x − x |(x )−β = c σ Λα1+β (Γ) (x )−β+1 c σ Λα1+β (Γ) (x )1−β−α |x − x |α . The term J2r is estimated in the same way as J2 .
|x − x | x
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Chapter 2. Boundary Integral Equations in H¨ older Spaces
We make the change of variables p = 1/q and w = 1/z in the integral
1 − q/z
−
c I+ (x) = σ(q) log
dsq . c 1 − q/z + Γ+ (x)
Then
p − w
−
σt (p) log
dsp , p − w+ S c (w)
(2.77)
c where S c (w) = S c (w mapping w = 1/z and
Γ+ (x) under the−1 + ) is the image of 2
σt = σ(1/p) 1/|p| . The function log (p − w− )(p − w+ ) |, p ∈ S c (w), satisfies the conditions of Theorem 2.2.1 (see Appendix E(ii)), hence the norm of (2.77) considered as a function of y in Λα . 1−β (1/δ, ∞) does not exceed c σt+ Λα 1−β (1/δ,∞) c α Therefore, the norm of I+ (x) in Λα (0, δ) is dominated by c σ . + Λ1+β (0,δ) β−1 The integral I− (x) is estimated in a similar way. The theorem is proved.
2.3 Dirichlet and Neumann problems in a strip Let G ⊂ R2 be a domain with C ∞ -boundary such that the set {(τ, ν) ∈ G : τ 0} has a compact closure and {(τ, ν) ∈ G : τ > 0} = {(τ, ν) : τ > 0, |ν| < 1}. By C ,α (G), 0 < α < 1, we denote the completion of the set of functions which are smooth in G and have compact support in G in the norm |∇ ϕ(ζ1 ) − ∇ ϕ(ζ2 )| + sup |∇k ϕ(ζ)|, = 0, 1, 2 . α |ζ1 − ζ2 | ζ1 ,ζ2 ∈G ζ∈G
ϕC ,α (G) = sup
k=0
Also we introduce the space C −1,α (G) of distributions on G with the finite norm 3 ϕC −1,α (G) = inf ϕk C 0,α (G) , k=0
where the infimum is taken over all representations ϕ = ϕ0 + (∂/∂τ )ϕ1 + (∂/∂ν)ϕ2 with ϕj , j = 0, 1, 2, from C 0,α (G). Consider the space Cβ1,α (G) of functions ϕ such that exp(βτ )ϕ ∈ C 1,α (G) for any β ∈ R with the norm ϕC 1,α (G) = exp(βτ )ϕC 1,α (G) . β
The space Eβα (∂G), introduced for functions on ∂G, is endowed with norm ϕ Eβα (∂G) =
±
ϕ( · , ±1) Eβα (0,∞) + ϕ Λα (∂G∩{(τ,ν): τ <0}) .
2.3. Direct method of integral equations
135
By Eβ1,α (∂G) we mean the space of continuously differentiable functions on ∂G with the norm ϕ E 1,α (∂G) = ϕ( · , ±1) E 1,α (0,∞) + ϕ Λ1,α (∂G∩{(τ,ν): τ <0}) . β
β
±
In a similar way, one defines the space Eβα (∂Π) and Eβ1,α (∂Π). Here and henceforth Π stands for the strip Π = {(τ, ν) : τ ∈ R, |ν| < 1}. The following assertion is a particular case of the results obtained in [7]. Lemma 2.3.1. Let 0 < β < π/2. Given ψ ∈ Cβ−1,α (G), there exists one and only one solution ϕ ∈ Cβ1,α (G) of the problem Δϕ = ψ in G, ϕ = 0 on ∂G, satisfying ϕC 1,α (G) c ψC −1,α (G) . β
Moreover, if ψ ∈
Cβ0,α (G),
then ϕ ∈
β
Cβ2,α (G)
and
ϕC 2,α (G) c ψC 0,α (G) . β
β
Next we prove several assertions concerning properties of solutions to the Dirichlet and Neumann problems in a strip. Proposition 2.3.2. Let ϕ ∈ Eβα (∂G), 0 < α, β < 1. There exists a harmonic extension H of ϕ onto G, whose normal derivative on ∂G can be expressed as ∂H/∂n = ∂Q1 /∂s + Q2 + Q3 ,
(2.78)
where Q1 , Q2 , Q3 ∈ Eβα (∂G), supp Q1 and supp Q2 are subsets of {(τ, ±1) : τ > 0}, Q2 (τ, +1) = −Q2 (τ, −1) for τ > 0, and Q3 decreases at infinity faster than any power function. Moreover, Q1 Eβα (∂G) + Q2 Eβα (∂G) + Q3 Eβα (∂G) c ϕEβα (∂G) ,
(2.79)
where c is a constant independent of ϕ. Proof. (i) We start with the case when ϕ ∈ Eβα (∂G) vanishes outside {(τ, ±1) : τ > 1}. By ϕ± we denote the functions on R such that ϕ± (τ ) = ϕ(τ, ±1) for τ ≥ 0 and ϕ± (τ ) = 0 for τ < 0 . In general, ϕ± are not integrable and belong to the Schwartz class of slowly growing functions. Therefore, their Fourier transforms ϕ ± (ξ) are distributions of slow growth.
136
Chapter 2. Boundary Integral Equations in H¨ older Spaces
(+) Let us introduce a bounded harmonic function H (τ, ν) in Π, taking the values ϕ+ (τ ) + ϕ− (τ ) /2 at the points (τ, 1) and (τ, −1) on ∂Π. The Fourier transform of H (+) is a distribution of slow growth and hence it is a solution of the boundary value problem for the family of ordinary differential equations: ∂2 2 H (+) (ξ, ν) = 0 for |ν| < 1, − ξ ∂ν 2 (2.80) (+) (ξ, ±1) = ϕ H (ξ) + ϕ (ξ) /2 . + −
Solving this problem for any fixed ξ ∈ R, we find the Fourier transform in τ of H (+) (τ, ν): (+) (ξ, ν) = c cosh(ξν)(cosh ξ)−1 (ϕ H 1 + (ξ) + ϕ− (ξ)) . Here the equality is understood in the sense of distributions. The normal derivative (+) on ∂Π takes the form of H (+) −1 ∂H (ξ, ±1) = c1 ξ sinh(ξ) coshξ (ξ) ϕ + (ξ) + ϕ − (ξ) . ∂n Since the inverse Fourier transform of tanh ξ is sinh−1 ( π2 τ ) up to a constant factor, we find d ∂H (+) (τ, ±1) = ±c2 Q11 (τ, ±1) , ∂n dτ where
Q11 (τ, ±1) = ± R
−1 π ϕ+ (t) + ϕ− (t) sinh (τ − t) dt . 2
By Theorem 2.2.1, Q11 Eβα (∂Π) c ϕ+ + ϕ− Eβα (R) , where c is independent of ϕ. Consider function H (−) in Π, taking the values ϕ+ (τ )− a bounded harmonic ϕ− (τ ) /2 and ϕ− (τ ) − ϕ+ (τ ) /2 at (τ, +1) and (τ, −1) on ∂Π. The distributional Fourier transform of H (−) (τ, ν) with respect to τ is (−) (ξ, ν) =c sinh(ξν) sinh ξ −1 ϕ H 3 + (ξ) − ϕ − (ξ) −1 −1 ξ ϕ + 2 sinh(2ξ) = c3 sinh(ξν) tanh coshξ + (ξ) − ϕ − (ξ) . 2 The normal derivative of H (−) (τ, ν) obeys the equality (−) 1 ∂H ξ (ξ, ±1) = ±c3 ξ tanh + ϕ + (ξ) − ϕ − (ξ) . ∂n 2 sinh ξ
(2.81)
2.3. Direct method of integral equations
137
−2 The inverse Fourier transforms of ξ(sinh ξ)−1 and tanh(ξ/2) are cosh π2 τ and −1 up to a constant factor. We set sinh πτ −1 Q21 (τ, ±1) = ϕ+ (t) − ϕ− (t) sinhπ(τ − t) dt R
and
Q22 (τ, ±1) = ± R
−2 π ϕ+ (t) − ϕ− (t) cosh (τ − t) dt . 2
By Theorems 2.2.1 and 2.2.3, Q21 Eβα (∂Π) c ϕ+ − ϕ− Eβα (R) and Q22 Eβα (∂Π) c ϕ+ − ϕ− Eβα (R) , where c is independent of ϕ. Taking the inverse Fourier transforms in (2.81), we obtain ∂H (−) d (τ, ±1) = ±c4 Q21 (τ, ±1) + c5 Q22 (τ, ±1) . ∂n dτ Let χ ∈ C ∞ (R), and let χ be equal to 1 for τ > 2 and 0 for τ < 1. We define a function Ψ in G ∩ {(τ, ν) : τ ≥ 0} by ∂χ (−) ∂2χ ∂ (−) H − H Ψ = Δ χ(H (−) + H (+) ) = 2 + H (+) + H (+) ∂τ ∂τ ∂τ 2 and put Ψ = 0 in G ∩ {(τ, ν) : τ < 0}. Since the support of Ψ is a bounded set, it follows by (2.96) that Ψ C −1,α (G) c ϕ Eβα (∂G) . β
(2.82)
According to Lemma 2.3.1, the boundary value problem ΔZ = −Ψ in G, Z = 0 on ∂G has a solution satisfying the inequality ZCγ1,α (G) c ΨCγ−1,α (G)
(2.83)
with 0 < γ < π/2 and with a constant c depend on γ, but not on ϕ± . For such γ this implies |∇Z(τ, ν)| c exp(−γτ )ΨCγ−1,α (G) , (τ, ν) ∈ G .
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Chapter 2. Boundary Integral Equations in H¨ older Spaces
By (2.82) and (2.83), the normal derivative ∂Z/∂n satisfies ∂Z/∂nEβα(∂G) c ∇ZC 0,α (G) c ϕEβα (∂G) . β
The function H = Z + χ H (−) + H (+) is a harmonic function in G with the boundary value ϕ. Its normal derivative ∂H/∂n has the required representation with Q1 , Q2 and Q3 , given by χ (c1 Q11 + c4 Q21 ) on ∂G ∩ {(τ, ν) : τ 0} ; Q1 = 0 on ∂G ∩ {(τ, ν) : τ < 0} , c5 χQ22 on ∂G ∩ {(τ, ν) : τ 0} ; Q2 = 0 on ∂G ∩ {(τ, ν) : τ < 0} and
Q3 =
(∂Z/∂n) − (∂χ/∂s)(c1 Q11 + c2 Q21 ) ∂G ∩ {(τ, ν) : τ 0} ; ∂Z/∂n on ∂G ∩ {(τ, ν) : τ < 0} .
The inequality (2.79) results by the derived estimates for Q1 , Q2 and Q3 . (ii) Now, suppose that ϕ ∈ Eβα (∂G) is supported by ∂G outside {(τ, ν) : τ > 1}. Let θ denote a conformal mapping of the unit disk D on G, satisfying the condition θ(1) = ∞. We introduce the harmonic extension F of the continuous function ϕ ◦ θ on D. The normal derivative ∂F/∂n on ∂D can be written as ∂F 1 d (ζ) = ∂n 2π ds
π −π
t−s t − cot dt, ζ = eis . ϕ(θ(eit )) cot 2 2
By Theorem 1.3.3, ∂F/∂n is the derivative in variable s of a function in C 0,α (∂D). Hence, H = F ◦ θ−1 is a harmonic extension of ϕ onto G and ∂H/∂n = (∂/∂s)Q1 , where Q1 is defined by 1 Q1 ◦ θ(e ) = 2π
π
is
−π
t−s t − cot dt . ϕ(θ(eit )) cot 2 2
We prove the inequality Q1 Eβα (∂G) c ϕEβα (∂G) .
(2.84)
The function θ−1 is continuously differentiable at the boundary of G. Therefore, Q1 ∈ C 0,α (∂G ∩ {(τ, ν) : τ < 2}) and Q1 C 0,α (∂G∩{(τ ν) : τ <2}) c ϕEβα (∂G) .
(2.85)
Under the mapping (τ, ν) → ie−(π/2)(τ +iν) of G, the conformal mapping of the upper half-plane onto the image of G is analytically extended onto the lower halfplane. Hence, the restriction of this conformal mapping onto a neighborhood of
2.3. Direct method of integral equations
139
the origin is infinitely differentiable. This implies that the conformal mapping θ−1 of G onto D has the representation ζ = θ−1 (τ ± i) =
i − p(± exp(− π2 τ )) i + p(± exp(− π2 τ ))
on ∂G ∩ {(τ, ν) : τ > 1}), where p is a real-valued infinitely differentiable function and p(0) = 0. We have |Q1 (τ1 ± i) − Q1 (τ2 ± i)| |τ1 − τ2 |α |τ1 −τ2 |< 2 τ1
θ−1 (τ ± i) − θ−1 (τ ± i) α
1 2 c ϕ ◦ θC 0,α (∂D) max 1 τ1α+β
τ1 − τ2 |τ1 −τ2 |< 2 τ1
exp(− π τ ) − exp(− π τ ) α
2 1 2 2
c ϕEβα (∂G) max 1 τ1α+β
τ1 − τ2 |τ1 −τ2 |< 2 τ1 max 1
τ1α+β
c ϕEβα (∂G)
max
|τ1 −τ2 |< 12 τ1
(2.86)
τ1α+β exp(−τ1 απ/2))
c ϕEβα (∂G) . Since |Q1 ◦ θ(ζ)| c |ζ − 1| ϕEβα (∂G) ,
ζ = eis ,
it follows that
π
τ β |Q1 (τ ± i)| c ϕEβα (∂G) τ β p ± exp − τ
2 π c ϕEβα (∂G) τ β exp − τ c ϕEβα (∂G) . 2
(2.87)
The inequality (2.84) results from (2.85), (2.86) and (2.87). This and (i) imply (2.78). While proving Propositions 2.3.3–2.3.6, we omit the arguments already used in Proposition 2.3.2. Proposition 2.3.3. Let ϕ ∈ Eβ1,α (∂G) with 0 < α, β < 1. There exists a harmonic extension H of ϕ to G such that the normal derivative of H on ∂G can be represented as ∂H/∂n = Q1 + Q2 , (2.88) where Q1 ∈ Eβα (∂G), Q2 ∈ Eβ1,α (∂G), supp Q2 ⊂ {(τ, ±1) : τ > 0}, and Q2 (τ, +1) = −Q2 (τ, −1) for τ > 0. Moreover, the functions Q1 and Q2 satisfy the estimate Q1 Eβα (∂G) + Q2 E 1,α (∂G) c ϕE 1,α (∂G) , β
where c is a constant independent of ϕ.
β
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Chapter 2. Boundary Integral Equations in H¨ older Spaces
Proof. Assume that ϕ ∈ Eβ1,α (∂G) is a function vanishing outside {(τ, ν) : u > 1}. Let ϕ± be the function on R equal to ϕ(τ, ±1) for τ 0 and to 0 for τ < 0. Consider the bounded harmonic function H (+) in Π which takes the value (ϕ+ (τ ) + ϕ− (τ ))/2 at the points (τ, 1) and (τ, −1) on ∂Π. The Fourier transform of H (+) (τ, ν) with respect to τ is (+) (ξ, ν) = c cosh(ξν)(coshξ)−1 ϕ H 1 + (ξ) + ϕ − (ξ) , where ϕ ± stands for the Fourier transform of ϕ± . The normal derivatives satisfy (+) ∂H ) (ξ, ±1) = c1 sinh ξ(coshξ)−1 ϕ + (ξ) + ϕ− (ξ) . ∂n Making the inverse Fourier transform, we obtain ∂H (+) (τ, ±1) = c2 Q11 (τ, ±1) , ∂n where
Q11 (τ, ±1) =
R
−1 π ϕ+ (t) + ϕ− (τ ) sinh (τ − t) dt . 2
By Theorem 2.2.1, Q11 Eβα (∂G) c ϕ+ + ϕ− E 1,α (R) β
with c independent of ϕ. Now, let H (−) be the bounded harmonic function in Π which takes the values (ϕ+ (τ ) − ϕ− (τ ))/2 and (ϕ− (τ ) − ϕ+ (τ ))/2 at the points (τ, +1) and (τ, −1) on ∂Π, τ ∈ R. The Fourier transform of H (−) (τ, ν) with respect to τ is (−) (ξ, ν) = c sinh(ξν) sinhξ −1 ϕ H 3 + (ξ) − ϕ − (ξ) −1 −1 ξ ϕ + 2 sinh(2ξ) = c3 sinh(ξν) tanh coshξ + (ξ) − ϕ − (ξ) . 2 (−) /∂n)(ξ, ±1) in the form Let us write (∂ H ±c3 tanh
ξ (ξ) ± c (cosh ξ)−1 ϕ ϕ (ξ) − ϕ + (ξ) − ϕ − (ξ) . 3 − 2 +
We put Q21 (τ, ±1) = ± R
−1 ϕ+ (t) − ϕ− (t) sinhπ(τ − t) dt
2.3. Direct method of integral equations
and
Q22 (τ, ±1) = ±
R
141
−2 π ϕ+ (t) − ϕ− (t) cosh (τ − t) dt . 2
From Theorems 2.2.1 and 2.2.3 it follows that Q21 Eβα (∂G) c ϕ+ − ϕ− E 1,α (R) β
and Q22 E 1,α (∂G) c ϕ+ − ϕ− E 1,α (R) . β
β
Making the inverse Fourier transform, we find ∂H (−) (τ, ±1) = c4 Q21 (τ, ±1) + c5 Q22 (τ, ±1). ∂n Let χ ∈ C ∞ (R) be equal to 1 for τ > 2 and let it vanish for τ < 1 . Also let Ψ = Δ χ(H (−) + H (+) ) in G ∩ {(τ, ν) : τ ≥ 0} and let Ψ = 0 in G ∩ {(τ, ν) : τ < 0}. By (2.82), α Ψ C −1,α (G) c ϕ Eβ−1 (∂G) β
with c depending on β, but not on ϕ. According to Lemma 2.3.1, the boundary value problem ΔZ = −Ψ in G, Z = 0 on ∂G has a solution satisfying α ZC 1,α(G) c ΨC −1,α(G) c ϕ Eβ−1 (∂G) c ϕ E 1,α (∂G) β
β
β
with 0 < β < 1. Hence we have ∂Z/∂nEβα(∂G) c ∇ZC 0,α(G) c ΨC −1,α(G) c ϕE 1,α (∂G) . β
β
β
Thus H = Z + χ H (−) + H (+) is a harmonic function in G with the boundary value ϕ, and its normal derivative ∂H/∂n on ∂G has the required form with ∂Z/∂n + c2 χQ12 + c4 χQ21 on ∂G ∩ {(τ, ν) : τ 0}; Q1 = ∂Z/∂n on ∂G ∩ {(τ, ν) : τ < 0} and
Q2 =
c5 (χQ22 ) on ∂G ∩ {(τ, ν) : τ 0}; 0 on ∂G ∩ {(τ, ν) : τ < 0} .
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Chapter 2. Boundary Integral Equations in H¨ older Spaces
(ii) Now suppose that ϕ ∈ Eβα (∂G) is compactly supported. Let θ be a conformal mapping of the unit disk D to G such that θ(1) = ∞. Consider the harmonic extension F of the continuous function ϕ ◦ θ to the domain D. On ∂D, the normal derivative ∂F/∂n can be written as ∂F 1 (ζ) = ∂n 2π
π −π
d t−s ϕ(θ(eit )cot dt , ζ = eis . dt 2
This representation implies ∂F/∂n ∈ C 0,α (∂D) (cf. [5], p. 400). Hence, H = F ◦θ−1 is a harmonic extension of ϕ onto G, and ∂H/∂n has the required representation on ∂G with vanishing Q2 . As in (ii) of Proposition 2.3.3, we show that Q1 = ∂H/∂n satisfies Q1 Eβα (∂G) c ϕE 1,α (∂G) . β
Together with (i), this results in the required representation (2.88).
In the next two assertions we study the boundary behavior of solutions to the Neumann problem. Proposition 2.3.4. Let ϕ ∈ Eβα (∂G), and let 0 < α, β < 1. There exists a harmonic extension H of ϕ to the domain G such that on ∂G the conjugate function H admits representation in the form = Q1 + Q2 , H
(2.89)
where Q1 ∈ Eβα (∂G), Q2 ∈ Eβ1,α (∂G), supp Q2 ⊂ {(u, ±1) : u > 1}, and Q2 (u, +1) = Q2 (u, −1) for τ > 0. Moreover, Q1 Eβα (∂G) + Q2 E 1,α (∂G) c ϕEβα (∂G) β
with a constant c independent of ϕ. Proof. (i) Assume that ϕ belongs to Eβα (∂G) and that ϕ = 0 outside {(u, ±1) : u > 1}. As in Proposition 2.3.2, we introduce two functions ϕ± (τ ), τ ∈ R, equal (+) to ϕ(τ, ±1) for τ ≥ 1 and vanishing for τ < 0. Let be the bounded harmonic H function on Π taking the value ϕ+ (τ ) + ϕ− (τ ) /2 at the point (τ, +1) and (τ, −1) on ∂Π, τ ∈ R. The Fourier transform of H (+) (τ, ν) with respect to τ is (+) (ξ, ν) = c cosh(ξν)(coshξ)−1 ϕ H 1 + (ξ) + ϕ − (ξ) . By the Cauchy–Riemann conditions, the Fourier transform of one of the conjugate functions (let it be denoted by U (+) ) is (+) (ξ, ±1) = ±c i tanh(ξ) ϕ U + (ξ) + ϕ − (ξ) 1
2.3. Direct method of integral equations
143
on ∂Π. The inverse Fourier transform of tanhξ is sinh−1 ( π2 τ ) up to a constant factor. Hence, on ∂Π, −1 π (+) U (τ, ±1) = ±c2 ϕ+ (t) + ϕ− (t) sinh (τ − t) dt . 2 R
According to Theorem 2.2.1, U (+) Eβα (∂Π) c ϕ+ + ϕ− Eβα (R) .
(2.90)
(−) be a bounded harmonic function in the strip Π taking the opposite values Let H ϕ+ (τ ) − ϕ− (τ ) /2 and − ϕ+ (τ ) − ϕ− (τ ) /2 at the points (τ, 1) and (τ, −1). The Fourier transform of H (−) with respect to τ is (−) (ξ, ν) = c sinh(ξν)(sinhξ)−1 ϕ H 3 + (τ ) − ϕ − (τ ) .
The Fourier transform of ∂H (−) /∂n on ∂Π has the form ξ (−) ∂H ξ (ξ, ±1) = ±c3 + ξ tanh ϕ + (ξ) − ϕ − (ξ) . ∂n sinhξ 2 Let U (−) (τ, ν), (τ, ν) ∈ Π, stand for the function conjugate of H (−) . The Fourier transform of ∂U (−) /∂τ on ∂Π is (−) ∂U ξ ξ (ξ, ±1) = −c3 ξtanh + ϕ + (ξ) − ϕ − (ξ) . ∂τ 2 sinhξ The inverse Fourier transform of ξ/ sinh ξ is equal to cosh−2 ( π2 τ ) up to a constant (−) (−) factor. Therefore, ∂U (−) /∂τ = ∂U1 /∂τ + U2 on ∂Π, where −1 (−) U1 (τ, ±1) = c4 ϕ+ (t) − ϕ− (t) sinhπ(τ − t) dt R
and (−)
U2 (τ, ±1) = c5 R
−2 π ϕ+ (t) − ϕ− (t) cosh (τ − t) dt . 2
By Theorems 2.2.1 and 2.2.3, (−)
U1 Eβα (∂Π) c ϕ+ − ϕ− Eβα (R) , (−)
U2 Eβα (∂Π) c ϕ+ − ϕ− Eβα (R) .
(2.91)
∞ on R, Let χ be a C -function equal to 1 for τ > 2 and vanishing for τ < 1. We put Ψ = Δ χ H (+) + H (−) in G ∩ {(τ, ν) : τ ≥ 0} and Ψ = 0 in G ∩ {(τ, ν) : τ < 0}. It was proved in Proposition 2.3.2 that Ψ ∈ Cγ−1,α (G), γ ∈ R, and
ΨCγ−1,α (G) c ϕEβα (∂G) ,
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Chapter 2. Boundary Integral Equations in H¨ older Spaces
where c depends on γ and β, but not on ϕ. According to Lemma 2.3.1, the problem ΔZ = −Ψ in G, Z = 0 on ∂G has a solution satisfying ZCγ1,α (G) c ΨCγ−1,α (G) with γ ∈ (0, π/2) and with c depending on γ, but not on ϕ. For the conjugate vanishing at infinity we have function Z E α (∂G∩{(τ,ν):τ >1}) c Z 1,α Z E (∂G∩{(τ,ν):τ >1}) β 1+β
α α c ∂ Z/∂τ Eβ+1 (∂G∩{(τ,ν):τ >1}) = ∂Z/∂nEβ+1 (∂G∩{(τ,ν):τ >1})
(2.92)
c ZC 1,α (G) c ΨC −1,α(G) c ϕEβα (∂G) . β
The function
β
H = Z + χ H (−) + H (+)
is the harmonic extension of ϕ onto G. be the function conjugate of H and equal to Z + U (−) + U (+) in Let H G ∩ {(τ, ν) : τ > 1}. It follows from (2.90) and (2.91) that E α (∂G∩{(τ,ν) : τ >2}) c ϕE α (∂G) . H β β We introduce a conformal mapping θ of the unit disk D onto G such that θ(1) = ∞. ◦θ By Theorem 1.3.3, outside a small neighborhood of 1 + i0 on ∂D the function H satisfies the H¨older condition with exponent α and ◦ θC 0,α (∂D\B (1)) c ϕ ◦ θC 0,α (∂D\B (1)) H ε ε/2 for a small positive ε. Here Bε (1) is a disk of radius ε about z = 1 and c depends on ε, but not on ϕ. Hence, C 0,α (∂G∩{(τ,ν) : τ <3}) c ϕE α(∂G) . H β with This inequality and (2.90), (2.91) imply the required representation of H ⎧ τ (−) ⎨ c5 (χU2 )(t)dt on ∂G ∩ {(τ, ν) : τ ≥ 0}; Q2 (τ, ±1) = 0 ⎩ 0 on ∂G ∩ {(τ, ν) : τ < 0} , + U (+) + U (−) up to a constant factor on ∂G ∩ {(τ, ν) : and with Q1 equal to Z 1 τ > 2}. (ii) It remains to show that (2.89) holds for any ϕ from Eβα (∂G), vanishing for {(τ, ±1) : τ > 1}. Let, as before, θ be a conformal mapping of D onto G
2.3. Direct method of integral equations
145
satisfying the condition θ(1) = ∞. Let F denote the harmonic extension of ϕ ◦ θ onto D. Consider the conjugate harmonic function F (ζ) of the form 1 2π
π −π
t − s t − cot dt, ζ = eis . ϕ(θ(eit )) cot 2 2
By Theorem 1.3.3, F ∈ C 0,α (∂D). Then H = F ◦θ−1 is the harmonic extension of ϕ onto G. The function F ◦θ−1 is conjugate of H and has the required representation with Q1 = 0. Using the same argument as in Proposition 2.3.2, one shows that the Eβα (∂G)-norm of Q2 = F ◦ θ−1 does not exceed c ϕEβα (∂G) . Proposition 2.3.5. Let ϕ ∈ Eβα (∂G), 0 < α, β < 1 and ϕds = 0 .
(2.93)
∂G
Then a solution H of the Neumann problem in G with the boundary value ϕ has the form ∂G, H = Q 1 + Q2 , (2.94) where Q1 and ∂Q2 /∂s belong to Eβ1,α (∂G), supp Q2 ⊂ {(u, ±1) : u > 0} and Q2 (u, +1) = Q2 (u, −1) for τ > 0. Moreover, Q1 E 1,α (∂G) + (∂/∂s)Q2 E 1,α (∂G) c ϕEβα (∂G) β
β
with a constant c independent of ϕ. Proof. (i) First, we treat the case of ϕ ∈ Eβα (∂G) such that ϕ = 0 on {(τ, ±1) : τ > 1} and (2.93) holds. Let θ be a conformal mapping of the unit disk D onto G satisfying θ(1) = ∞. Let F be a harmonic extension to D of a continuous function ψ on ∂D such that ψ(1 + i0) = 0 and (d/ds)ψ = −(ϕ ◦ θ)|θ |. We denote by F (ζ), ζ ∈ D, the conjugate function, which has the form 1 2π
π −π
t − s dt ψ(eit )cot 2
on ∂D. By Theorem 1.3.3, F ∈ Λα (∂D). We set H = F ◦ θ−1 . Since ∂ F ∂F ∂ψ 1 1 1 ∂H = ◦ θ−1 = − ◦ θ−1 = − ◦ θ−1 = ϕ, ∂n ∂n |θ | ∂s |θ | ∂s |θ | it follows that H is a solution to the Neumann problem with boundary data ϕ and H has a required representation with Q2 = 0. In the same way as in Proposition
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Chapter 2. Boundary Integral Equations in H¨ older Spaces
2.3.2 we show that the Eβα (∂G)-norm of (∂/∂s)H does not exceed c ϕEβα (∂G) . Hence, for Q1 = H we have Q1 E 1,α (∂G) (∂/∂s)HEβα (∂G) + HEβα(∂G) β
c (∂/∂s)HEβα (∂G) c ϕEβα(∂G) . (ii) Next, let ϕ ∈ Eβα (∂G) vanish outside {(u, ±1) : u > 1} and let (2.93) hold. For τ ∈ R we introduce ⎧ τ ⎨± ϕ(t, ±1)dt , τ 0, ψ± (τ ) = ⎩ 0 0, τ < 0. Since ϕ ∈ Eβα (∂G) we have ψ± ∈ Eβ1,α (∂Π). By U (+) we denote the harmonic function in Π, taking the equal values ψ+ (τ ) + ψ− (τ ) /2 at the points (τ, +1) and (τ, −1) and growing at infinity not faster than a power function. The Fourier transform of U (+) (τ, ν) with respect to τ is ) U + (ξ, ν) = c1 cosh(ξν)(coshξ)−1 ψ + (ξ) + ψ− (ξ) . Hence, on ∂Π we find ∂) U+ (ξ, ±1) = −c1 i tanh ξ ψ + (ξ) + ψ− (ξ) . ∂n Here ψ ± and ψ± stand for the Fourier transform of ψ± and ψ± respectively. Let (+) (+) be a function conjugate of U , whose Fourier transform on ∂Π is H ) + (ξ, ±1) = ±c i tanhξ ψ H + (ξ) + ψ− (ξ) . 2
Therefore, H
(+)
(τ, ±1) = ±c2 R
−1 π ψ+ (t) + ψ− (t) sinh (τ − t) dt . 2
By Theorem 2.2.1, H (+) E 1,α (∂Π) c ϕ+ + ϕ− Eβα (R) . β
(−) Let be a bounded function harmonic U in Π which takes the opposite values ψ+ (τ )−ψ− (τ ) /2 and − ψ+ (τ )−ψ− (τ ) /2 at the points (τ, 1) and (τ, −1). The Fourier transform of U (−) with respect to τ is
(−) (ξ, ν) = c sinh(ξν)(sinhξ)−1 ψ (τ ) − ψ (τ ) . U 3 + −
2.3. Direct method of integral equations
147
On ∂Π, the Fourier transform of the normal derivative ∂U (−) /∂n can be represented in the form ∂U (−) ξ ξ (ξ, ±1) = ±c3 ξ tanh + ψ+ (ξ) − ψ − (ξ) . ∂n 2 sinhξ Let H (−) (τ, ν) denote a function conjugate of U (−) (τ, ν) on Π. We take H (−) on (−) (−) ∂Π in the form H1 + H2 , where −1 (−) ψ+ (t) − ψ− (t) sinhπ(τ − t) dt H1 (τ, ±1) = c4 R (−)
and H2
is defined, up to a constant factor, by
(−)
∂H2 ∂τ
(τ, ±1) = c5 R
−2 π ψ+ (t) − ψ− (t) cosh (τ − t) dt . 2
(−)
(−)
By Theorems 2.2.1 and 2.2.3, the functions H1 (τ, ±1) and (∂H2 /∂τ )(τ, ±1) belong to Eβ1,α (∂Π) and (−)
(−)
H1 E 1,α (∂Π) + (∂/∂τ )H2 E 1,α (∂Π) c ϕ+ − ϕ− Eβα (R) . β
β
∞
Let χ be a C on R, equal to 1 for τ > 1 and vanishing for τ < 0. -function We set Ψ = Δ χ U (+) + U (−) in G ∩ {(τ, ν) : τ ≥ 0} and Ψ = 0 in G ∩ {(τ, ν) : τ < 0}. According to Proposition 2.3.2, α ΨC −1,α (G) cψEβ−1 (∂G) . β
In view of Lemma 2.3.1, the boundary value problem ΔZ = −Ψ in G, Z = 0 on ∂G has a solution satisfying the inequality 1 (G) c Ψ −1,α ZCα,γ Cγ (G)
with γ ∈ (0, π/2) and with a constant c depending on γ but not on ϕ. The function U = Z + χ U (−) + U (+) is a harmonic extension of ψ onto G. Let H be one of the functions conjugate of U . Then H is the required solution of the Neumann problem in G with the boundary data ϕ, because on ∂G one has ∂H = ϕ. ∂n
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Chapter 2. Boundary Integral Equations in H¨ older Spaces
We introduce the function Q2 : (−) on ∂G ∩ {(τ, ν) : τ 0}, χ H2 Q2 = 0 on ∂G ∩ {(τ, ν) : τ < 0} . Clearly, supp Q2 ⊂ {(τ, ±1), : τ > 0} and Q2 (τ, 1) = Q2 (τ, −1) for τ > 0 . Furthermore, (∂/∂s)Q2 ∈ Eβ1,α (∂G) and (∂/∂s)Q2 E 1,α (∂G) c ϕEβα (∂G) . β
The derivative (∂Q1 /∂s) of the function Q1 := H ∂G − Q2 obeys the relation ∂H ∂H ∂Q2 ∂U ∂Q2 ∂Z Q1 1 = − = − = +χ ∂s ∂s ∂s ∂n ∂s ∂n ∂s
(−)
−
∂H (+) ∂χ (−) − H . ∂s ∂s 2
The second and the third terms on the right-hand side belong to Eβα (∂G), while the first term ∂Z/∂n satisfies ∂Z/∂nEβα (∂G) ∇ZEβα (G) c ZE 1,α(G) ΨC −1,α (G) β
β
α c ψEβ−1 (∂G) c ∂ψ/∂sEβα (∂G) c ϕEβα (∂G) .
Thus the functions Q1 and Q2 have the required properties on ∂G. 1,α (∂G) E1−β
α E1−β (0, ∞),
and ϕ(τ, +1) − ϕ(τ, −1) ∈ 0< Proposition 2.3.6. Let ϕ ∈ α, β < 1. Then there exists a harmonic extension H of ϕ in G that has the normal α derivative (∂/∂n)H in E1−β (∂G). Moreover, α α (∂/∂n)HE1−β (2.95) (∂G) c ϕE 1,α (∂G) + ϕ(·, +1) − ϕ(·, −1)E1−β (0,∞) 1−β
with a constant c independent of ϕ. 1,α 1,α (∂G), ϕ = 0 outside {(τ, ν) : u > 1}, and let ϕ± ∈ E1−β (R) Proof. (i) Let ϕ ∈ E1−β be the functions on R equal to ϕ(τ, ±1) for τ ≥ 1 and ϕ± (τ ) = 0 for τ < 0. We introduce a bounded harmonic function H (+) in Π, having the value (ϕ+ (τ ) + ϕ− (τ ))/2 both at (τ, 1) and (τ, −1) on ∂Π. The Fourier transform of H (+) (τ, ν) with respect to τ is c1 cosh(ξν)(coshξ)−1 ϕ + (ξ) + ϕ − (ξ) ,
where ϕ ± stands for the Fourier transform of ϕ± . Hence, on ∂Π, −1 ∂H (+) π (τ, ±1) = ϕ+ (t) + ϕ− (t) sinh (τ − t) dt . ∂n 2 R
2.3. Direct method of integral equations
By Theorem 2.2.1,
149
∂H (+) c ϕE 1,α (R) . α 1−β ∂n E1−β (∂G)
Let H (−) be a bounded harmonic function in Π taking the opposite values (ϕ+ (τ ) − ϕ− (τ ))/2 and (ϕ− (τ ) − ϕ+ (τ ))/2 at the points (τ, +1) and (τ, −1) on ∂Π. The Fourier transform of H (−) (τ, ν) with respect to τ is −1 c3 sinh(ξν) sinhξ ϕ + (ξ) − ϕ − (ξ) −1 −1 ξ ϕ + tanh coshξ = c3 sinh(ξν) 2 sinh(2ξ) + (ξ) − ϕ − (ξ) . 2 As in Proposition 2.3.3, we set −1 Q1 (τ, ±1) = ± ϕ+ (t) − ϕ− (t) sinhπ(τ − t) dt R
and
Q2 (τ, ±1) = ±
−2 π ϕ+ (t) − ϕ− (t) cosh (τ − t) dt . 2
R
Then
∂H (−) (τ, ±1) = c4 Q21 (τ, ±1) + c5 Q22 (τ, ±1). ∂n According to Theorems 2.2.1 and 2.2.3, α Q1 E1−β (∂G) c ϕ+ − ϕ− E 1,α (R) 1−β
and α α Q2 E1−β (∂G) c ϕ+ − ϕ− E1−β (R) .
Let C ∞ (R), χ = 1 for τ > 2 and χ = 0 for τ < 1 . Further, let Ψ = χ ∈ (−) Δ χ(H + H (+) ) in G ∩ {(τ, ν) : τ ≥ 0} and Ψ = 0 in G ∩ {(τ, ν) : τ < 0}. Since (+) + H (−) is a harmonic extension of (∂/∂τ )ϕ± into Π, it follows by (∂/∂τ ) H Lemma 2.3.7 that α (∂/∂τ )(H (+) + H (−) )C 0,α (K) c(K)ϕ E1−β (∂G) c(K)ϕE 1,α (∂G) 1−β
for any subdomain K of Π. Here a constant c(K) does not depend on ϕ. This and Lemma 2.3.7 imply that the function Ψ=2
∂ (+) ∂χ ∂2χ (H − (H (+) + H (−) ) 2 + H (−) ) ∂τ ∂τ ∂τ
belongs to Cγ0,α (G), γ ∈ R, and ΨCγ0,α (G) c ϕE 1,α (∂G) , 1−β
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Chapter 2. Boundary Integral Equations in H¨ older Spaces
where a constant c depends on γ, but not on ϕ. In view of Lemma 2.3.1, the problem ΔZ = −Ψ in G, Z = 0 on ∂G has a solution satisfying ZCγ2,α(G) c ΨCγ0,α(G) with γ ∈ (0, π/2) and with a constant c depending on γ. Besides, α ∂Z/∂nE1−β (∂G) c ϕE 1,α (∂G) . 1−β
Therefore, H = Z + χ H (−) + H (+) is the harmonic function in G with the boundary data ϕ satisfying α α (∂/∂n)HE1−β (∂G) c ϕE 1,α (∂G) + ϕ+ − ϕ− E1−β (0,∞) . 1−β
(ii) Next, let ϕ ∈ ∂G and let ϕ have a compact support. By θ we denote a conformal mapping of D onto G with θ(1) = ∞. We introduce the harmonic extension F of the continuous function ϕ◦ θ onto D. On ∂D, the normal derivative of F has the form π ∂F d 1 t−s (ζ) = ϕ(θ(eit )cot dt , ζ = eis . ∂n 2π dt 2 −π
This representation implies ∂F/∂n ∈ Λα (∂D). Hence, H = F ◦θ−1 is the harmonic extension of ϕ into G. Using the same argument as in the proof (ii) of Proposition 2.3.2, we show that α ∂H/∂nE1−β (∂G) c ϕE 1,α (∂G) . 1−β
This inequality and part (i) imply the required estimate (2.95).
Next, we obtain an estimate for harmonic functions in the strip Π. Lemma 2.3.7. Let H be a harmonic extension of ϕ ∈ Eδα (∂Π) with 0 < α < 1 and δ ∈ R onto the strip Π. Then, for any subdomain K of Π, HC 0,α (K) c(K, δ) ϕEδα (∂Π) .
(2.96)
Proof. By μ we denote a function in C ∞ (R), which is equal to 1 for |τ | < 1 and to 0 for |τ | > 2. Then μH is harmonic in Π ∩ {(τ, ν) : |τ | < 1} and vanishes for |τ | > 2. Hence, Δ μH is infinitely differentiable in Π and vanishes for |τ | < 1 and |τ | > 2. Let G0 be a bounded subdomain of Π with smooth boundary, and let suppΨ0 ⊂ G0 and supp(μϕ) ⊂ ∂G0 ∩ ∂Π. Let U stand for a harmonic extension of μϕ on ∂G0 ∩ ∂Π , Φ= 0 on ∂G0 \∂Π
2.4. Integral equations of the Dirichlet and Neumann problems
151
onto G0 . Clearly, Φ ∈ C 0,α (∂G0 ). By Theorem 1.3.2, U ∈ C 0,α (G0 ) and U C 0,α (G0 ) c Φ C 0,α (∂G0 ) = c μϕ C 0,α (∂G0 ∩∂Π) c ϕ Eδα (∂Π)
(2.97)
with c depending on μ and a parameter δ from R. The function μH −U is harmonic in Π ∩ {(τ, ν) : |τ | < 1} = Π0 and vanishes on the boundary segments {(τ, ±1) : |τ | < 1}. The extension of the function in question, equal to μ(τ, ν)H(τ, ν) − U(τ, ν) for (τ, ν) ∈ Π0 , F (τ, ν) = −μ(τ, ν)H(τ, ν) + U(τ, ν) for (τ, ∓2 − ν) ∈ Π0 , is a harmonic function in {(τ, ν) : |τ | < 1, |ν| < 3}. By the interior estimate for a harmonic function one has max
{|τ |<1/2,|ν|<2}
|∇F | c max |F | max |H|+ U C 0,α (G0 ) . Π0
Π0
With the help of the integral representation of the harmonic function in the strip Π as well as the maximum modulus principle and (2.97), we find that max
{|τ |<1/2,|ν|<2}
|∇F | c ϕ Eδα (∂Π)
(2.98)
with c depending on δ. Using the shifts of μ, by (2.97) and (2.98) we arrive at (2.96).
2.4 Boundary integral equations of the Dirichlet and Neumann problems in domains with outward peak 2.4.1 Auxiliary assertions Recall that the functions v = ϕ+ (u) and v = ϕ− (u), u > 1/δ, whose graphs are the images of Γ+ and Γ− under the mapping w = 1/z, are twice continuously differentiable and obey the relations ϕ+ (u) > ϕ− (u) and
lim ϕ+ (u) = lim ϕ− (u) = 0
u→∞
u→∞
(cf. the proof of Theorem 2.2.7). We introduce the functions σ(u) = ϕ+ (u) − ϕ− (u),
and ϕ(u) =
1 (ϕ+ (u) + ϕ− (u)) , 2
and the domain D given by u > 1/δ,
ϕ− (u) < v < ϕ+ (u).
Let W (u + iv) stand for the conformal mapping of D onto G. The following assertion was proved by Warschawski [40].
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Chapter 2. Boundary Integral Equations in H¨ older Spaces
Theorem 2.4.1. Let ∞ 1/δ
(σ (s))2 ds < ∞ σ(s)
∞
and 1/δ
(ϕ (s))2 ds < ∞ . σ(s)
Then
u
W (u + iv) = 2 1/δ
v − ϕ(u) ds + 2i + λ + o(1) , σ(s) σ(u)
where λ is a real number. The functions ϕ± (u) admit the representations ϕ± (u) = −
κ± (x) κ± (x) u=− + O(x2 ) , x x2
where x =
1 1 +O 3 , u u
which are twice differentiable. Therefore, the finiteness of the integrals in the statement of the above theorem follows from 1 d 1 2 κ± u =O . du u u The function σ(u), equal to κ (x ) κ (x ) + + − − u, − x+ x− where x+ and x− are abscissas of images of u + iϕ− and u + iϕ+ , respectively, can be expressed in the form 1 κ+ (x+ ) − κ− (x+ ) u+O 2 . x+ u
σ(u) = The relation
κ+ (x+ ) − κ− (x+ ) u = (α+ − α− ) + o(1) x+ implies 1 1 = + o(1) . σ(u) α+ − α− Hence
u
1/δ
ds 1 = u + o(u). σ(s) α+ − α−
2.4. Integral equations of the Dirichlet and Neumann problems
153
As a result we obtain W (u + iv) =
α+ − α− 2 2 + λ + o(1) , v+ u + a(u) + i α+ − α− α+ − α− 2
where a(u) = o(u) and λ is a real number. Therefore, the conformal mapping ω(z) = W 1z of Ω+ onto G has the representation 2 2 ω(z) = x−1 + a(x−1 ) − i y x−2 α+ − α− α+ − α− (2.99) α+ + α− +i + λ + o(1) , z = x + iy . α+ − α− Let ρ stand for the inverse mapping ω −1 . We introduce the mapping ζ(z) of the exterior domain Ω− onto the upper half-plane R2+ with ζ(0) = 0. This mapping admits the representation ζ(z) = ±x1/2 ±c2 x3/2 log x±(c3 +ic± )x3/2 +O x3/2+ε , z = x+iy ∈ Γ± , (2.100) where ε ∈ (0, 1/2) (see Ch. 3, Lemma 3.1.2). Inverting the decomposition (2.100), we arrive at θ(τ + i0) = ζ −1 (τ + i0) = τ 2 + c4 τ 4 log |τ | + O(τ 4 ) as τ → 0 .
(2.101)
It follows from (2.99) and (2.100) that ω ◦ θ(τ ) =
2 τ −2 + b(τ ) + c5 log |τ | + O(1) , α+ − α−
(2.102)
where b(τ ) = o(τ −2 ). Therefore, the function d± (u) = ω ◦ θ(±u−1/2 ), u > 0, can be represented as d± (u) =
2 u + c(u) + c6 log u + O(1) , α+ − α−
(2.103)
where c(u) = o(u). The equation 1/2 , x ∈ (0, 1), ζ(x + iκ± (x)) = ± s± (x) defines the functions s± (x) subject to s± (x) = x + c7 x2 log(1/x) + O x2 .
(2.104)
All above decompositions are differentiable once and the derivatives of the mappings satisfy H¨older’s condition with an exponent less than 1.
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Chapter 2. Boundary Integral Equations in H¨ older Spaces
The next lemmas are frequently used in the sequel. Lemma 2.4.2. Let ϕ ∈ Λ1,α 2−β (0, 1) and let the functions X1 (x) and X2 (x) satisfy the relations X1 (x) = x + o(x), X2 (x) = x + o(x), Z(x) = X2 (x) − X1 (x) = O x2 , which are differentiable at least once. Then ϕ X1 (x) − ϕ X2 (x) ∈ Λα −β (0, 1).
(2.105)
Proof. We introduce the function L(x) = ϕ X2 (x) − ϕ X1 (x) , which admits the estimate X2 (x)
|ϕ (τ )|dτ c
|L(x)| X1 (x)
2 X1 (x)+O(x )
τ β−2 dτ c xβ .
(2.106)
X1 (x)
According to Lemma 2.1.1, we may assume that x < y and 2|x − y| < x. Let us write L(x) in the form L(x) = 0
1
d ϕ(X1 (x) + tZ(x))dt . dt
By this representation we deduce
−β+α
x L(x) − y −β+α L(y)
−β+α
x
1 Z(x)
ϕ X1 (x) + tZ(x) − ϕ X1 (y) + tZ(y) dt
0
+ x−β+α Z(x) − y −β+α Z(y)
1
ϕ X1 (y) + tZ(y) dt
0
1−α |x − y| |x − y|α c |x − y|α + x c |x − y|α . This estimate and (2.106) imply (2.105).
Lemma 2.4.3. Let w(ζ), ζ = ξ + iη, stand for a harmonic function in the semistrip G vanishing on ∂G\{O} and having at most power growth as ζ → ∞. Then w = 0 in G.
2.4. Integral equations of the Dirichlet and Neumann problems
155
Proof. Represent w(ζ) as the Fourier series w(ζ) =
∞
ck (ξ) sin πkη .
k=1
The coefficient ck (ξ) has the form αk eπkξ + βk e−πkξ . Since
1
−1
|w(ξ, η)|2 dη =
∞
|ck (ξ)|2 ,
k=1
where the left-hand side grows not faster than a power function as ξ → ∞, it follows that αk = 0, k = 1, 2, . . . . Hence w(ζ) is a bounded function in G, vanishing on ∂G. Thus w(ζ) = 0 in G. Lemma 2.4.4. Let w(ζ), ζ = ξ + iη, be a harmonic function in the semi-strip G with vanishing normal derivative on ∂G\{O} and growing not faster than a power function as ζ → ∞. Then w = const in G. The proof is similar to that of the previous lemma. Below we need two auxiliary assertions proved by integration by parts. They involve a function ϕ which is continuous on R, compactly supported and satisfies the inequality |ϕ(τ )| c τ γ with γ > 0. Given integer k ≥ 1 such that k − 1 < γ < k + 1, we introduce the function ν −k Φ(τ ) = τ k ϕ(ν)dν . τ −ν R
Lemma 2.4.5. Let k − 1 < γ < k. Then Φ(±u
−1/2
)= ∓u
−(k−1)/2
∞ 0
1 ∓ u−(k−1)/2 2
v (k−1)/2 ϕ(± v −1/2 )dv u−v
∞
v (k−2)/2 ϕ(± v −1/2 )dv u1/2 + v 1/2
0
(−1)k −(k−1)/2 ± u 2
∞ 0
v (k−2)/2 ϕ(∓ v −1/2 )dv . u1/2 + v 1/2
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Chapter 2. Boundary Integral Equations in H¨ older Spaces
Let k < γ < k + 1. Then Φ(±u
−1/2
)= ∓u
−(k−1)/2
∞ 0
1 ∓ u−k/2 2
∞
v (k−1)/2 ϕ(± v −1/2 )dv u−v
! v (k−2)/2 ϕ(± v −1/2 ) + (−1)k+1 ϕ(∓ v −1/2 ) dv
0
1 ± u−k/2 2
∞
v (k−1)/2 ϕ(± v −1/2 )dv + v 1/2
u1/2 0
k
(−1) −k/2 u ∓ 2
∞
v (k−1)/2 ϕ(∓ v −1/2 )dv . u1/2 + v 1/2
0
Suppose that ψ is continuously differentiable on R, compactly supported and
d
ψ(τ ) c τ γ
dν
with γ > 0. We set
Ψ(τ ) = τ k R
ν −k d ψ(ν)dν, τ − ν dν
where k ≥ 1 and k − 1 < γ < k + 1. Lemma 2.4.6. Let k − 1 < γ < k. Then * ∞ (k+2)/2 v d −(k+2)/2 Ψ(τ )dτ = ∓ u ψ(±v −1/2 )dv u − v dv 0
1 ∓ u−(k+2)/2 2
∞
v (k+1)/2 d ψ(±v −1/2 )dv u1/2 + v 1/2 dv
0
(−1)k −(k+2)/2 u ∓ 2
∞
+ v (k+1)/2 d ψ(∓v −1/2 )dv du . u1/2 + v 1/2 dv
0
Let k < γ < k + 1. Then * ∞ (k+2)/2 v d −(k+2)/2 ψ(±v −1/2 )dv Ψ(τ )dτ = ∓ u u − v dv 0
1 ∓ u−(k+3)/2 2
∞ v (k+1)/2 0
d d ψ(±v −1/2 ) + (−1)k ψ(∓v −1/2 ) dv dv dv
2.4. Integral equations of the Dirichlet and Neumann problems
1 ± u−(k+3)/2 2
∞
157
v (k+2)/2 d ψ(±v −1/2 )dv + v 1/2 dv
u1/2 0
(−1)k −(k+3)/2 u ± 2
∞
+ v (k+2)/2 d −1/2 ψ(∓v )dv du . u1/2 + v 1/2 dv
0
Here τ = ±u−1/2 .
2.4.2 Integral equation of the Dirichlet problem on a contour with outward peak In this section we obtain the solvability and describe zeros of the integral equation (2.4) on a contour with outward peak. %σ we denote the function Here and in what follows, by W %σ(z) = σ(q) ∂ log 1 dsq , z ∈ W / Γ. ∂sq |z − q| Γ
We start with a description of zeros of the operator πI − T given by (2.33). For this aim we need two auxiliary assertions. 1,α Lemma 2.4.7. Let Ω+ have an outward peak and let σ ∈ Lodd 2−β (Γ) ×R. Then
∂ 1
dsq c xβ−1 , z = x + iy ∈ Ω+ . log |W σ(z)| = σ(q) ∂nq |z − q| Γ Proof. Since ∂ ∂ 1 = log arg(q − z) ∂nq |z − q| ∂sq and the integral
∂
arg(q − z) dsq
∂s q Γ equals the increment of q → arg(q − z) as q changes along Γ, it follows that
1
∂ log
dsq c < ∞ .
|z − q| Γ ∂nq 1,α We represent σ ∈ Lodd 2−β (Γ) as the sum σ = σ1 + σ2 , where σ1 ∈ 1,α Λodd 2−β (Γ), σ1+ + σ1− = 0 on Γ+ ∪ Γ− and σ2 ∈ Λα −β (Γ). We have |σ2 (q)| c |q|β , β > 0.
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Chapter 2. Boundary Integral Equations in H¨ older Spaces
Hence
∂ 1
|σ2 (q)|
log
dsq ∂nq |z − q| Γ
1
∂ c log
dsq c < ∞ .
|z − q| Γ ∂nq
|W σ2 (z)|
(2.107)
The function σ1 obeys the inequality |σ1 (q)| |q|β−1 . Given z ∈ Ω+ with Re z = x δ, we introduce d = |y − κ± (x)| = |z − z± |, where z± = x + iκ± (x). It suffices to estimate ∂ 1 dsq σ1 (q) log ∂nq |z − q| Γ+ ∂ 1 dsq . σ1 (q) = + log ∂nq |z − q| Γ+ ∩{|q−z+ |
d/2} To find the majorant for the first integral, we use the equality ∂ 1 1 (1 + κ+ = Re (1 + iκ+ . (u)2 )1/2 log (u)) ∂nq |z − q| q−z We have
1
du
σ1+ (u)Re (1 + iκ+ (u)) q−z Γ+ ∩{|q−z+ |
Note that
∂ 1
log (u)2 )1/2
(1 + κ+ ∂nq |z − q| |κ+ (x) − κ+ (u) − κ+ (u)(x − u)| y − κ+ (x)| d + c+ . 2 2 |q − z| |q − z| |q − z|2 For the integral over Γ+ ∩ {|q − z+ | > d/2} with the kernel d/|z − q|2 we get du |σ+ (u)| d |q − z|2 Γ+ ∩{|q−z+ |>d/2} x−d/4 δ du du β−1 d u +d uβ−1 . 2 (x − u) (x − u)2 0 x+d/4
2.4. Integral equations of the Dirichlet and Neumann problems
159
The first integral on the right-hand side does not exceed d xβ−2
1−d/4x
0
uβ−1
du x c d xβ−1 = c xβ−1 (1 − u)2 d
and the second one is dominated by ∞ du x d xβ−2 uβ−1 c d xβ−1 = c xβ−1 . 2 (1 − u) d 1+d/4x Therefore,
1
du c xβ−1 . σ1+ (u)Re (1 + iκ+ (u)) q−z Γ+ ∩{|q−z+ |>d/2}
(2.109)
By (2.108) and (2.109) we conclude that |W σ1 (z)| c xβ−1 ,
which together with (2.107) completes the proof. Lemma 2.4.8. Let Ω+ have an outward peak and let 1,α σ ∈ Lodd 2−β (Γ) × R. Then
W σ(z) =
σ(q) ∂ log 1 dsq
c 1 , ∂nq |z − q| |z|2 Γ
z = x + iy ∈ Ω− .
Proof. It suffices to show that
1 ∂ 1
dsq c 2 , σ(q) log
∂n |z − q| |z| q Γ+ ∪Γ− assuming that z ∈ Ω− ∩ {|z| < δ} and σ is integrable on Γ. Let κ(x) = max{|κ+ (x)|, |κ− (x)|}. By D1 we denote the domain Ω− ∩ {z = x + iy : |y| < 2κ(x)} and let |z| < δ, x ∈ (0, δ). In the same way as in the preceding lemma, one shows that |W σ(z)| c |z|β−1 c |z|−1
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Chapter 2. Boundary Integral Equations in H¨ older Spaces
in D1 . For z ∈ D2 = Ω− ∩ {z : |z| < δ, Re z 0} we have |z − q| c|z|, Hence
|W σ(z)|
Γ+ ∩Γ−
|σ(q)|
q ∈ Γ± .
c 1 c dsq 2. |z − q| |z| |z|
(+) D 3
Let stand for the image of Ω− ∩ {z = x + iy : |z| < δ and y 2κ(x), x ∈ (0, δ)} under the mapping ξ = x, η = y − κ+ (x) . Then Γ+ is mapped onto the segment [0, δ] on the real line and the image of + : η = κ(ξ) − κ+ (ξ) = κ (ξ) with y = κ(x) with x ∈ [0, δ] is the curve Γ ξ ∈ [0, δ]. Note that κ (ξ) κ+ (ξ). Changing δ if necessary, we may assume that (+) of z ∈ D(+) and the |κ± (ξ)| < 1/2 for ξ ∈ [0, δ]. The image ζ = ξ + iη ∈ D 3 3 (+) of q ∈ D(+) obey the inequality image p = τ + iν ∈ D 3 3 |z − q|
1 |ζ − p| . 2
(+) we have Besides, for p ∈ [0, δ] and ζ = ξ + iη ∈ D 3 |p − ζ| |ξ − ζ| = |η| . Therefore,
∂ 1 dsq c
σ(q) log |σ(q)| |σ(q)|dsq . dsq c
∂nq |z − q| |z − q| η Γ+ Γ+ Γ+ In the same way we show that
c ∂ 1
dsq σ(q) log |σ(q)|dsq .
∂nq |z − q| η Γ− Γ− For ξ < |ζ|/2 we have η −1 |ζ|−1 , while for ξ |ζ|/2 we get η −1 c ξ −2 c |ζ|−2 . Hence
c ∂ 1 c
dsq 2 2 . σ(q) log
∂n |z − q| |ζ| |z| q Γ+ ∪Γ− A similar inequality holds for z ∈ D3 = Ω− ∩ {z : |z| < δ and y −2κ(x), x ∈ [0, δ]}. Since (+) (−) Ω− ∩ {z : |z| < δ} = D1 ∪ D2 ∪ D3 ∪ D3 , (−)
the result follows from the above estimates.
2.4. Integral equations of the Dirichlet and Neumann problems
161
Theorem 2.4.9. Let Ω+ have an outward peak and let α ˙ ˙ (Lodd )1,α 2−β (Γ)+R σ −→ πσ − T σ ∈ Λ−β (Γ)+R ,
as in (2.33). Then a) ker (πI − T ) = {0} for 0 < α < 1 and 1/2 < β < 1, b) dim ker (πI − T ) = 1 for 0 < α < 1 and 0 < β < 1/2. In the case b) the kernel ker (πI − T ) consists of the functions c Re (1/ζ0 ), where c ∈ R and ζ0 is a conformal mapping of Ω− onto R2+ , normalized by the conditions ζ0 (0) = 0, ζ0 (∞) = i. Proof. Let σ belong to ker (πI − T ). Since |W σ(z)| = O |z|−1 , z ∈ Ω+ , (cf. Lemma 2.4.7) and since the limit values of W σ(z) vanish on Γ\{O}, by Lemma %σ, being conjugate of W σ, is 2.4.3 we have W σ = 0 on Ω+ . Hence the function W + + % a constant in Ω . We put W σ = C in Ω and introduce the harmonic function %σ(z) − C), z ∈ Ω− . W (z) = (W σ)(z) + i(W − 2 Similarly to (2.99), let ξ + iη = ζ(z) be a conformal mapping of Ω onto R+ such that ζ(0) = 0. The function F (ξ + iη) = W ◦ θ (1/(ξ + iη)) is holomorphic on R2− = {ξ + iη, η < 0}, continuous up to the boundary, and Im F = 0 on the real axis. Since W σ grows not faster than a power function in Ω− (cf. Lemma 2.4.8), it follows that the holomorphic extension F ext of F onto C is an entire function with the real part satisfying the inequality
|Re F ext (ξ + iη)| c (ξ 2 + η 2 ) . By the Schwarz integral formula we conclude that F ext is a polynomial with real coefficients. Hence W σ(z) = F ext Re (1/ζ(z)) , z ∈ Ω− , and the jump formula for W σ implies 1 σ(z) = F ext Re (1/ζ(z)) , z ∈ Γ \ {0}. π 1,α ˙ and the fact that constants do not belong to Recalling that σ ∈ (Lodd )2−β (Γ)+R ker (πI − T ), we obtain σ = c Re (1/ζ). Therefore, by the integral representation for the harmonic function Re (1/ζ) in Ω− and by the limit relation for the double layer potential we conclude that 1 ∂ |z| 1 1 log dsq − Re Re π ∂nq ζ(q) |z − q| ζ(∞) Γ 1 1 ∂ 1 1 dsq − Re , z ∈ Γ \ {O} . = Re log π ζ(q) ∂nq |z − q| ζ(z) Γ
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Chapter 2. Boundary Integral Equations in H¨ older Spaces
Since (∂/∂n)Re (1/ζ) = 0 on Γ \ {O}, the zeros of πI − T have the form c Re (1/ζ), where c ∈ R, and ζ is an arbitrary conformal mapping of Ω− on R2+ , subject to the conditions ζ(0) = 0 and Re ζ(∞) = 0. Note that functions of the form Re (1/ζ), subject to ζ(0) = 0 and Re ζ(∞) = 0, differ by a positive constant factor, hence dim ker (πI − T ) does not exceed 1. The inclusion Re (1/ζ) ∈ (Lodd )1,α 2−β (Γ) holds only for β < 1/2, thus ker (πI − T ) is one-dimensional for 0 < β < 1/2 and trivial for 1/2 < β < 1. In our further notation for compositions of functions and conformal mappings we write conformal mappings as subscripts, for example ϕ ◦ ρ will be written as ϕρ and G ◦ ω ◦ θ will be Gω,θ . ˙ Our next assertion concerns the solvability in (Lodd )1,α 2−β (Γ)+R of the equation (2.4) with the right-hand side in Λα (Γ). −β Theorem 2.4.10. Let Ω+ have an outward peak and let 0 < β, α < 1, β = 1/2. Then the operator πI − T , defined by α ˙ ˙ (Lodd )1,α 2−β (Γ)+R σ −→ πσ − T σ ∈ Λ−β (Γ)+R ,
as in (2.33), is surjective. Proof. Suppose that ϕ ∈ Λα −β (Γ) and ϕ = 0 in some neighborhood of the peak. Let H stand for the harmonic extension of ϕρ ∈ Eβα (∂G) into G, constructed in Proposition 2.3.2. By he we denote the harmonic extension of ϕ onto exterior the domain Ω− , such that grad he = O |z|−1/2 . Since grad Hω = O |z|N with z ∈ Ω+ and integer N , we obtain ∂he 1 ∂Hω 1 − log dsq + he (∞), z ∈ Γ \ {O} . (2.110) ϕ(z) = 2π ∂n ∂n |z − q| Γ
The normal derivative ∂H/∂n can be written as ∂Q1 /∂s + Q2 + Q3 , where Q1 , Q2 and Q3 are described in Proposition 2.3.2. (i) As a solution (g1 )θ of the Neumann problem in R2+ with the boundary data (∂/∂s)(Q1 )ω,θ , consider the function ∂ 1 (Q1 )ω, θ (ν) dν . (g1 )θ (τ + iη) = − log |τ − ν + iη| − log |ν| π ∂s R
Then 1 (g1 )θ (τ + i0) = lim (g1 )θ (τ + i0) = − τ η→+0 π
R
ν −1 (Q1 )ω,θ (ν) dν , τ ∈ R. τ −ν
We apply Lemma 2.4.5 to the last integral. Since the functions Q1 (τ ± i), τ ≥ 1, belong to Eβα (1, ∞), it follows that Q1 ◦ d± also belong to Eβα (1, ∞). By Theorem
2.4. Integral equations of the Dirichlet and Neumann problems
163
2.2.1 and Lemma 2.2.4, the functions (g1 )θ (±u−1/2 ), u > 1, belong to Eβα (1, ∞) with 0 < β < 1/2 and their norms in Eβα (1, ∞) do not exceed c Q1 ◦ d± Eβα (1,∞) with c independent of ϕ. Hence (g1 )θ (±x1/2 ), x ∈ (0, 1), are elements of Λα −β (0, 1). Therefore, by Lemma 2.4.2 we conclude that the Λα −β (0, 1)-norms of the functions g1 (x + iκ± (x)) = (g1 )θ (±(s± (x))1/2 ), x ∈ (0, 1), are dominated by c ϕΛα−β (Γ) with c independent of ϕ. In the case 1/2 < β < 1, by Theorem 2.2.1 and Lemma 2.2.5 the functions (g1 )θ (±u−1/2 ) ∓ c1 (ϕ) u−1/2 , u > 1, belong to Eβα (1, ∞). Here c1 (ϕ) is a linear continuous functional in Λα −β (Γ). Hence 1/2 1/2 α (g1 )θ (±x ) ∓ c1 (ϕ) x , x ∈ (0, 1), are elements of the space Λ−β (0, 1). In view 1/2 of Lemma 2.4.2 the Λα is dominated −β (0, 1) norm of g1 (x + iκ± (x)) ∓ c1 (ϕ) x by c ϕΛα−β (Γ) with c independent of ϕ. , ( )
(ii) We introduce the function Q2
on ∂G, setting
⎧ 0, z ∈ ∂G ∩ {τ < 0}, ⎪ ⎪ ⎪ τ ⎪ ⎪ ⎨ , − ( ) Q2 (u, +1)du, z = τ + i, τ > 0, Q2 (z) = 0 ⎪ τ ⎪ ⎪ ⎪ ⎪ ⎩ Q2 (u, −1)du, z = τ − i, τ > 0 . 0 ,
( )
Clearly, (d/ds)Q2 (z) = Q2 (z), z ∈ ∂G, where ∂/∂s is the derivative with respect to the parameter s, equal to the directed arc length measured from some fixed point z ∈ ∂G ∩ {τ < 0} and varying from −∞ to ∞. As a solution (g2 )θ ,of the ( ) Neumann boundary value problem in R2+ with the boundary data (d/ds) Q2 ω,θ , consider the function τ ν d (, ) |τ − ν + iη| 1 + 2 Q log (ν) dν . (g2 )θ (τ + iη) = − π |τ + iη| τ + η 2 ds 2 ω,θ R
The boundary value of (∂/∂τ )(g2 )θ is ∂ 1 (g2 )θ (τ + i0) = − 2 ∂τ τ
R
ν 2 d (, ) Q (ν) dν , τ ∈ R. τ − ν dν 2 ω,θ
Let us apply Lemma 2.4.6 to the differential d(g2 )θ (τ ) = Φ± (u) du ,
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Chapter 2. Boundary Integral Equations in H¨ older Spaces
where τ = ±u−1/2 , u > 0. From the inclusion Q2 ∈ Eβα (∂G), it follows that the functions d (, ) −1/2 d (, ) d Q2 ω,θ ±v Q = (d± (v)) = ∓Q2 (d± (v)) d± (v) , v > 1, dv dv 2 dv belong to Eβα (1, ∞). By Theorem 2.2.1 and Lemmas 2.2.4, 2.2.5, the functions Φ± (u) belong to Eβα (1, ∞) with 0 < β < 1/2 and there exists a linear functional c2 (ϕ) continuous in Λα and such that Φ± (u) ∓ c2 (ϕ) u−1/2 ∈ Eβα (1, ∞) for −β (Γ) −1 −2 1/2 < β < 1. Hence, Φ± x x and Φ± x−1 x−2 ∓ c2 (ϕ) x−3/2 belong to Λα 2−β (0, 1) with 0 < β < 1/2 and with 1/2 < β < 1 respectively. Since g2 (x + iκ± (x)) = (g2 )θ ±(s± (x)1/2 , x ∈ (0, 1), it follows from Lemma 2.4.2 that the functions (d/dx)g2 (x + iκ± (x)) and
(d/dx) g2 (x + iκ± (x)) ± 2 c2 (ϕ) x−1/2
belong to Λα 2−β (0, 1), the first one with 0 < β < 1/2 and the second one with 1/2 < β < 1. Moreover, the Λ1,α 2−β (0, 1) norms of g2 (x + iκ± (x))
g2 (x + iκ± (x)) ± 2 c2 (ϕ) x−1/2
and
do not exceed c ϕΛα−β (Γ)) with c independent of ϕ for 0 < β < 1/2 and for 1/2 < β < 1, respectively. , ( )
(iii) Similarly to (ii), we introduce the function Q3
on ∂G such that
,
( )
(d/ds)Q3 (z) = Q3 (z). , ( ) In view of Q3 (τ, ν) = O exp(−γτ ) as τ → ∞ on ∂G, we may assume that Q3 vanishes at infinity. The solution (g3 )θ of the, Neumann problem in R2+ with the ( ) boundary condition (∂/∂n)(g3 )θ = (d/ds) Q3 ω,θ on ∂R2+ is 1 d (, ) Q log |τ − ν + iη| (ν)dν . (g3 )θ (τ + iη) = − π dν 3 ω,θ R
On R + i0 we have (g3 )θ (τ + i0) = lim (g3 )θ (τ + iη) = (g3 )θ (0) − η→+0
τ π
R
ν −1 (, ) Q3 ω,θ (ν) dν . τ −ν
Using the same argument as in (i), we prove that g3 (x + iκ± (x)) − g3 (0) belongs to Λα −β (0, 1) with 0 < β < 1/2 and there exists a linear functional c3 (ϕ) continuous in
2.4. Integral equations of the Dirichlet and Neumann problems
165
1/2 Λα belongs to Λα −β (Γ) and such that g3 (x+iκ± (x))−g3 (0)∓c3 (ϕ) x −β (0, 1) with 1/2 < β < 1. One can easily see that ϕ → g3 (0) is a linear functional, continuous in Λα −β (Γ).
(iv) Thus we constructed the function g = g1 + g2 + g3 which is harmonic in Ω− and satisfies the boundary condition ∂g/∂n = ∂Hω /∂n. Using the estimate grad g = O |z|−1/2 , we see that the function w = g − he − g(∞) + he (∞) admits the representation ∂w ∂ 1 1 1 − dsq w(q) log (q) log w(z) = 2π ∂nq |z − q| ∂nq |z − q| Γ
in Ω− . By the limit relation for the single layer potential and (2.110) we obtain ∂ 1 1 dsq = 2(ϕ(z) + g(∞) − he (∞)) . w(q) log w(z) − π ∂nq |z − q| Γ
of the equation (2.4) Since T 1 = −π, we find that σ = (2π)−1 (g − ϕ) is a solution on Γ. Note that the functions of the form Re 1/ζ(z) on Γ \ {O}, where ζ is a conformal mapping of Ω− onto R2+ , subject to the conditions Re ζ(∞) = 0 and ζ(0) = 0, satisfies the homogeneous equation (2.4). By (i)–(iii) it follows that the solution σ + A Re 1/ζ with some real A can be written as σ(z) = σ1 (z) + σ2 (z), α where σ1 ∈ (Λodd )1,α 2−β (Γ), σ2 − σ2 (0) ∈ Λ−β (Γ). The functions σ1 and σ2 obey the inequality
σ1 (Λodd )1,α
2−β (Γ)
+ σ2 − σ2 (0)Λα−β (Γ) + |σ2 (0)| c ϕΛα−β (Γ) .
(2.111)
Given ϕ from Λα −β (Γ), there exists a sequence {ϕn }n≥1 of smooth functions on Γ \ {O}, vanishing near the peak, which converges in Λα −β (Γ) with α < α and β < β, and satisfies ϕn Λα−β (Γ) c ϕ Λα−β (Γ) . ˙ Let σn denote the solution of (2.4) in (Lodd )1,α 2−β (Γ)+R with the right-hand side ϕn , constructed as in (i)–(iii). By (2.111), σn (Lodd )1,α
˙
2−β (Γ)+R
c ϕ Λα−β (Γ) .
(2.112)
1,α α ˙ Since Λα −β (Γ) ⊂ Λ−β (Γ), the sequence {σn }n≥1 converges in (Lodd )2−β (Γ)+ R to a limit σ. In particular, {σn }n≥1 converges to σ pointwise on Γ \ {O}. Combined ˙ with (2.112), this implies σ ∈ (Lodd )1,α 2−β (Γ)+R. The operator
α ˙ ˙ (Lodd )1,α 2−β +R σ −→ πσ − T σ ∈ Λ−β (Γ)+R
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Chapter 2. Boundary Integral Equations in H¨ older Spaces
is continuous (cf. Theorem 2.2.7), hence, passing to the limit, we obtain −ϕ = (πI − T )σ. α ˙ ˙ Thus, the equation (2.4) is solvable in (Lodd )1,α 2−β (Γ)+R for ϕ in Λ−β (Γ)+R and
σ (Lodd )1,α
˙
2−β (Γ)+R
c ϕ Λα−β (Γ) .
The converse assertion is proved in Theorem 2.2.7.
2.4.3 Integral equation of the Neumann problem on a contour with outward peak We start with an estimate for the logarithmic potential near the origin. Lemma 2.4.11. The logarithmic potential
z
V σ (z) = σ(q) log
dsq , z−q Γ
z ∈ Ω + ∪ Ω−
with density σ such that |σ(q)| c|q|−α−1 ,
0 < α < 1,
admits the estimate |V σ(z)| c|z|−α as
z → 0.
Proof. It suffices to estimate the integral
z
σ(q) log
dsq z − q Γ+ ∪Γ− in a small neighborhood of z = 0. We put q = u + iκ± (u), u ∈ [0, δ], on Γ± . For 2|q| < |z| one has
q
|q|
log 1 −
c z |z| with some constant c. Hence
z
c
σ(q) log
|σ(q)| |q|dsq
dsq
z−q |z| Γ± ∩{|q|< |z| Γ± ∩{|q|< |z| 2 } 2 } |z|/2 c u−α du c |z|−α . |z| 0 Let 2|z| < |q|. Then
z
|z| u
+c + c,
log
log z−q |q| |z|
(2.113)
2.4. Integral equations of the Dirichlet and Neumann problems
and therefore,
∞
z
u
+ 1 du σ(q) log
u−α−1 log
dsq c z−q |z| Γ± ∩{|q|2|z|} |z| c|z|
−α
167
(2.114)
.
Now, suppose that |z|/2 < |q| < 2|z|. Then either |1 − q/z| < 1 and hence 1 1
log
log
1 − |q|/|z| , 1 − q/z|
or |1 − q/z| > 1 and
1
c .
log
1 − q/z
Thus, for |z|/2 < |q| < 2|z|,
z
1
+ c .
log
log
z−q 1 − |q|/|z|
We set
v = v± (u) = |q|u
1+
κ± (u) 2 u
.
Let δ > 0 be so small that κ± (u)
d κ± (u) 0 du u
for all u ∈ [0, δ]. The function v± (u) increases for 0 < u < δ and 0<
1 v± (u) c c
with some constant c. Introducing the notation ∗ σ± (u) = σ u + iκ± (u) , we have
z
σ(q) log
dsq
z−q Γ± ∩{|z|/2<|q|<2|z|} 2|z|
1
∗
+ 1 dv c |σ± (u)| log
1 − |u|/|z|
|z|/2 2
1
∗ c |z| |σ± (|z|u)| log
+ 1 du |1 − u| 1/2 2
1
c |z|−α
+ 1 du c |z|−α .
log |1 − u| 1/2
(2.115)
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Chapter 2. Boundary Integral Equations in H¨ older Spaces
Estimates (2.113), (2.114) and (2.115) imply the inequality
z
σ(q) log
dsq c |z|−α .
z − q Γ+ ∪Γ−
α Let (Λ0 )α −β (Γ) denote the subspace of functions ψ ∈ Λ−β (Γ), subject to
ψds = 0. Γ
Further, let D2 be the linear hull of the functions (∂/∂s)Re z and Re z −1/2 , z ∈ Γ\{O}. Recall that D1 is the linear hull of the function (∂/∂s)Re z, z ∈ Γ\{O}. The next assertion contains a description of the kernel of πI + S: α ˙ ˙ (Lodd )1,α (2.116) 2−β (Γ)+Dm σ −→ πσ + Sσ ∈ Λ0 −β (Γ)+D1 , where m is defined by (m − 1)/2 < β < m/2. Theorem 2.4.12. Let Ω+ have an outward peak and the operator πI + S be as in (2.116). Then a) ker (πI + S) = {0} for 0 < α < 1 and 1/2 < β < 1; b) dim ker (πI + S) = 1 for 0 < α < 1 and 0 < β < 1/2. In the case b) the kernel ker (πI + S) consists of the functions c (∂/∂n) log | hext (z) | , where c ∈ R and hext (z)is a conformal mapping of Ω− onto the exterior of the unit disk, such that hext (∞) = ∞. Proof. Let σ belong to ker(πI + S). Since (V σ)(z), z ∈ Ω+ , grows not faster than a power function as z → 0 (cf. Lemma 2.4.11) and since V σ is a solution of the Neumann boundary value problem with zero data on ∂Ω+ \{O}, it follows that V σ = const in Ω+ (cf. Lemma 2.4.4). We put V σ(z) = C, z ∈ Ω+ . By σ0 we denote the density of the equilibrium distribution on Γ (cf. Stoilov [39], Ch. V) and let −1 c0 = σ(q)dsq σ0 (q)dsq , z ∈ Ω+ . Γ
Γ
We introduce the holomorphic function σ(q) − c0 σ0 (q) log V(z) = −i Γ
1 dsq − C + γ0 , z ∈ Ω− . z−q
The function V is bounded at infinity and the constant γ0 is chosen in such a way that the imaginary part Im V vanishes on Γ \ {O}.
2.4. Integral equations of the Dirichlet and Neumann problems
169
Let ξ + iη = ζ(z) be a conformal mapping of Ω− onto R2+ , such that ζ(0) = 0. The function F (ξ+iη) = V ◦θ (1/(ξ+iη)) is holomorphic in R2− = {ξ+iη, η < 0}, continuous up to the boundary, and Im F = 0 on the real axis. The holomorphic extension of F ext of F to C is an entire function and Im F ext satisfies Im F ext (ξ + iη) = O (|ξ| + |η|)N , ξ + iη → 0, where N is an integer. By the Schwarz integral formula, F ext has the same order of growth at infinity. Hence F ext is a polynomial with real coefficients. Then Im (Vσ)(z) = F ext Im (1/ζ(z) , z ∈ Ω− , and the jump formula for the single layer potential implies σ(z) − c0 σ0 (z) =
1 1 ∂ ext F . z ∈ ∂Ω+ . Im 2π ∂n ζ(z)
(2.117)
˙ Since σ ∈ (Lodd )1,α 2−β(Γ) +Dm , where m satisfies the inequality (m − 1)/2 < β < −1/2 m/2, and σ0 (z) = O |z| , ∂Ω+ z → 0, it follows from (2.117) that σ−c0 σ0 = 0 on Γ \ {O}. The inclusion σ0 ∈ (Lodd )1,α 2−β (Γ) holds only with β < 1/2. Hence ker (πI + S) is one-dimensional for 0 < β < 1/2 and is trivial for 1/2 < β < 1. Now, we prove a solvability theorem for the equation (2.7). Theorem 2.4.13. Let Ω+ have an outward peak and let 0 < α, β < 1, β = 1/2. Then the operator πI + S defined by (2.116) is surjective. α Proof. Let ψ ∈ Λ0 −β (Γ) vanish in a neighborhood of the peak. By H we denote the harmonic function constructed in Proposition 2.3.5 and satisfying the boundα ary condition ∂H/∂n = ψρ |ρ | ∈ E2+β (∂G). On ∂G the function H has the form Q1 + Q2 with Q1 and Q2 specified in Proposition 2.3.5. (i) Let (f1 )θ stand for the Poisson integral of (Q1 )ω,θ in R2+ . The normal derivative (∂/∂n)(f1 )θ on R can be written as d ∂ dt (f1 )θ (τ + i0) = (Q1 )ω,θ (t) (2.118) ∂n dt τ −t R d d d dt dt t−2 2 =− (Q1 )ω,θ (t) − τ (Q1 )ω,θ (t) 2 + τ (Q1 )ω,θ (t) dt. dt t dt t dt τ −t R
R
R
The linear functionals d d dt dt c11 (ψ) = − (Q1 )ω,θ (t) , c12 (ψ) = − (Q1 )ω,θ (t) 2 dt t dt t R
R
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Chapter 2. Boundary Integral Equations in H¨ older Spaces
−1/2 are continuous in Λα , we can write the −β (Γ). Making the substitution τ = ±u expression ⎤ ⎡ −2 d t ⎣τ 2 (Q1 )ω,θ (t) dt⎦ dτ dt τ −t R
as F± (u)du with F± described in Lemma 2.4.6. Since (d/du)(Q1 ◦ d± ) ∈ α E2+β (1, ∞), it can be shown with the help of Theorem 2.2.1 and Lemmas 2.2.4 α and 2.2.5 that F± (u) ∈ E2+β (1, ∞) for 0 < β < 1/2 and that there exists a lin α α ear functional c13 (ψ) in Λ−β (Γ) such that F± (u) ∓ c13 (ψ) u−5/2 ∈ E2+β (1, ∞) for 1/2 < β < 1. Moreover, α F± E2+β (1,∞) c Q1 E 1,α (∂G) β
for 0 < β < 1/2
and α F± ∓ c13 (ψ)u−5/2 E2+β (1,∞) c Q1 E 1,α (∂G) β
for 1/2 < β < 1 .
Combining these estimates with (2.118), we find that the conjugate function f 1 satisfies the inequality m+1 c1k (ψ)(Re ζ)k (∂/∂s) f1 − k=1
Λα −β (Γ+ ∪Γ− )
c Q1 E 1,α (∂G) β
(2.119)
with functionals c1k (ψ), k = 1, . . . , m + 1, coinciding with c1k (ψ) up to a constant factor and with m subject to (m − 1)/2 < β < m/2. (ii) The normal derivative of the Poisson integral (f2 )θ of (Q2 )ω,θ has the form ∂ d dt (f2 )θ (τ + i0) = (Q2 )ω,θ (t) , τ ∈ R. ∂n dt τ −t R
Making the change of variable τ = ±u−1/2 and using Lemma 2.4.6, we write (∂/∂n)(f2 )θ (τ )dτ as 1 1 1 Φ± (u)du = − Φ± dx , u x x where x = u−1 and ∞ Φ± (u) = ∓ 0
1 ∓ 2
d 1 v (Q2 )ω,θ (±v −1/2 )dv u − v dv
∞ u1/2
1 d v 1/2 (Q2 )ω,θ (±v −1/2 )dv 1/2 dv +v
0
∓
1 2
∞ 0
1 1/2 d −1/2 (Q v ) (∓v )dv = (Ik )± (u) . 2 ω,θ dv u1/2 + v 1/2 3
k=1
2.4. Integral equations of the Dirichlet and Neumann problems
171
1,α Since (d/dv)Q2 belongs to Eβ+2 (∂G), we have 1,α (1, ∞). (d/dv)(Q2 )ω,θ (±v −1/2 ) ∈ Eβ+2
Hence, the functions
d d v (Q2 )ω,θ (±v −1/2 ) dv dv
α (∂G). Using the equality belong to Eβ+1
∞
d d v (Q2 )ω,θ (±v −1/2 ) dv = 0 , dv dv
0
we obtain d 1 (I1 )± (u) = ∓ du u
∞ 0
1 d d v v (Q2 )ω,θ (±v −1/2 ) dv. u − v dv dv
By Theorem 2.2.1 α (d/du)(I1 )± (u) ∈ Eβ+1 (1, ∞) α (1, ∞), and, in particular, for 0 < β < 1. Therefore, (1/u)(I1 )± (u) ∈ Eβ+1
1 d 1 (I1 )± ∈ Λα 2−β (0, 1) . dx x x In view of the equality 1 1 ∂ v 1/2 ∂ = , 1/2 1/2 1/2 1/2 ∂u u + v u ∂v u + v 1/2 we have d (I2 )± (u) = ±u−1 du
∞
d d 1 −1/2 v v ) (±v ) dv. (Q 2 ω,θ v 1/2 u1/2 + v 1/2 dv dv 1
0 α By Lemma 2.2.4 this implies that (d/du)(I2 )± (u) ∈ Eβ+1 (1, ∞) with 0 < β < 1/2. In a similar way, we show that
d α (I3 )± (u) ∈ Eβ+1 (1, ∞) du for 0 < β < 1/2.
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Chapter 2. Boundary Integral Equations in H¨ older Spaces
Let 1/2 < β < 1. As in Lemma 2.4.6, we write the sum (I2 )± (u) + (I3 )± (u) in the form 1 ∓ u−1/2 2
∞ v 1/2
! d (Q2 )ω,θ (±v −1/2 ) + (Q2 )ω,θ (∓v −1/2 ) dv dv
0
1 ± u−1/2 2
∞
1 d v (Q2 )ω,θ (±v −1/2 )dv u1/2 + v 1/2 dv
0
1 ± u−1/2 2
∞ u1/2
1 d v (Q2 )ω,θ (∓v −1/2 )dv 1/2 dv +v
0 (ψ) u−1/2 + (J2 )± (u) + (J3 )± (u), = ∓c21 (ψ), equal to where the functional c21
1 2
∞ v 1/2
! d (Q2 )ω,θ (±v −1/2 ) + (Q2 )ω,θ (∓v −1/2 ) dv , dv
0
is continuous in Λα −β (Γ). Taking into account that 1 d 1 v 1 d = − 3/2 , 1/2 1/2 1/2 1/2 du u u + v u dv u + v 1/2 we find d 1 (J2 )± (u) = ∓ 3/2 du u
∞
v d d −1/2 v (Q ) (±v ) dv. 2 ω,θ dv u1/2 + v 1/2 dv
0 α (1, ∞). In a similar By Lemma 2.2.5, the functions (d/du)(J2 )± (u) belong to Eβ+1 α way, we can show that (d/du)(J3 )± (u) ∈ Eβ+1 (1, ∞). As a result, we conclude that (d/dx)(x−1 (Jk )± (x−1 )) ∈ Λα 2−β (0, 1), k = 2, 3, with 1/2 < β < 1.
Thus, by Theorem 2.2.1 combined with Lemmas 2.2.4 and 2.2.5, we obtain that any function f 2 , conjugate of f2 satisfies (∂ 2 /∂s2 )f 2 Λα2−β (Γ+ ∪Γ− ) c (∂/∂s)Q2 E 1,α (∂G) β
for 0 < β < 1/2
and (∂ 2 /∂s2 ) f 2 − c21 (ψ) x1/2 Λα2−β (Γ+ ∪Γ− ) c (∂/∂s)Q2 E 1,α (∂G) β
for 1/2 < β < 1,
where the functional c21 (ψ) equals c21 (ψ) up to a constant factor.
2.4. Integral equations of the Dirichlet and Neumann problems
173
(iii) The function f = f 1 + f 2 is conjugate of the harmonic extension f of Hω into Ω− and can be written as k m+1 ζ(z) bk (ψ)Re + g1 (z) + g2 (z) , f(z) = 1 − iζ(z) k=1
1,α where (∂/∂s)g1 ∈ Λα −β (Γ), (∂/∂s)g2 ∈ (Lodd )2−β (Γ) and bk (ψ) are linear combinations of the functionals c21 (ψ) and c1 (ψ), = 1, . . . , m + 1 (see (2.119)). The restriction to Γ+ ∪ Γ− of the conformal mapping ζ(z) of Ω+ onto R2+ has the expansion Re ζ(z) = ±x1/2 ± c1 x3/2 log x ± c2 x3/2 + O x3/2+ε , 0 < ε < 1/2.
Since it is differentiable (cf. Sect. 2.4.1), we have k m+1 ∂ 1 ∂ ζ(z) bk (ψ)Re = b1 (ψ)Re z −1/2 + b2 (ψ) Re z + r(z), ∂s 1 − i ζ(z) 2 ∂s k=1
where r(z) ∈ (Lodd )1,α 2−β (Γ). Then ∂ ∂ f (z) = b2 (ψ) Re z + h1 (z) + h2 (z) ∂n ∂s for 0 < β < 1/2 and 1 ∂ ∂ f (z) = b1 (ψ)Re z −1/2 + b2 (ψ) Re z + h1 (z) + h2 (z) ∂n 2 ∂s for 1/2 < β < 1, where h1 and h2 satisfy the inequality h1 Λα−β (Γ) + h2 (Lodd )1,α
2−β (Γ)
c ϕΛα−β (Γ) .
We use grad Hω = O |z|N , z ∈ Ω+ , with an arbitrary integer N to get −1/2 , z ∈ Ω− . Hence grad f = O |z| 1 ∂f 1 (z) log dsq + f (∞), z ∈ Γ \ {O} . ψ(z) + Hω (z) = − 2π ∂n |z − q| Γ
Since Hω is a solution of the Neumann problem, it is defined up to a constant term. We choose Hω so that f (∞) = 0. Let us put σ(z) = −(2π)−1 ψ(z) + (∂/∂n)f (z) , z ∈ Γ \ {O} . Noting that V σ(z) grows not faster than a power function as z → 0, we conclude that the single layer potential V σ(z) coincides with Hω (z) on Ω+ . Hence σ(z) is ˙ a solution of (2.7) in the space (Lodd )1,α 2−β (Γ)+Dm and σ (Lodd )1,α
˙
2−β (Γ)+Dm
c ψ Λα−β (Γ) .
(2.120)
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Chapter 2. Boundary Integral Equations in H¨ older Spaces
α (iv) Given an arbitrary function ψ in Λ0 −β (Γ), there exists a sequence {ψn }n≥1 of smooth functions on Γ \ {O}, vanishing near the peak, converging in Λα −β (Γ) for α < α, β < β, and satisfying ψn Λα−β (Γ) c ψ Λα−β (Γ) . By σn we denote a solution of (2.7) with the right-hand side ˙ ψn ∈ (Lodd )1,α 2−β (Γ)+Dm , constructed as in (i)–(iii). By (2.120), σn (Lodd )1,α
2−β
˙ m (Γ)+D
c ψ Λα−β (Γ) .
(2.121)
α Since Λα −β (Γ) ⊂ Λ−β (Γ), it follows by (2.120) that the sequence {σn }n≥1 converges ˙ in (Lodd )1,α 2−β (Γ)+Dm to a limit σ. Note that {σn }n≥1 also converges pointwise to ˙ σ on Γ \ {O}. Combined with (2.121), this implies that σ ∈ (Lodd )1,α 2−β (Γ)+Dm . Using the continuity of the operator α ˙ ˙ (πI + S) : (Lodd )1,α 2−β (Γ)+Dm → Λ0 −β (Γ)+D1
and passing to the limit, we obtain −ψ = (πI + S)σ. Thus
1,α ˙ (Λ0 )α −β (Γ) ⊂ (πI + S) (Lodd )2−β (Γ)+Dm .
(2.122)
(v) Now we put ψ(z) = (∂/∂s)Re z, z ∈ Γ \ {O}. Clearly, the harmonic function h(z) = −Im z, z ∈ Ω+ , obeys the relation ∂h/∂n = ψ on Γ \ {O}. Since hρ ∈ Eβ1,α (∂G), it follows by (i) and (iii) that there exists a harmonic extension f ˙ of hρ into Ω− with ∂f /∂n in (Lodd )1,α 2−β (Γ)+Dm . Hence, the solution σ(z) = −(2π)−1 ψ(z) + (∂/∂n)f (z) , z ∈ Γ \ {O} , ˙ of the equation (2.7) belongs to (Lodd )1,α 2−β (Γ)+Dm . Therefore, by (2.122) 1,α ˙ ˙ (Λ0 )α −β (Γ)+D1 ⊂ (πI + S) (Lodd )2−β (Γ)+Dm . (vi) Using the integral representation for the harmonic function Im ζ(z) in Ω+ ∩ {|z| < 1/2} and the limit relation for the normal derivatives of a single layer potential, we find ∂ ∂ 1 ∂ dsq + π Im ζ(q) log Im ζ(z) ∂nq ∂nz |z − q| ∂nz Γ∩{|z|<1/2}
=
∂ h(z), z ∈ Γ ∩ {|z| < 1/2} , ∂nz
2.5. Integral equations of the Dirichlet and Neumann problems
175
where h(z) is a harmonic function in a neighborhood of the peak. Hence, (πI + ˙ S) (∂/∂s)Re ζ belongs to Λα −β (Γ)+D1 . In a similar way, we prove that (πI + ˙ S) (∂/∂s)Re ζ 2 ∈ Λα (Γ) +D . It follows from (2.100) and Lemma 3.1.2 that 1 −β (∂/∂s)Re ζ(z) − (1/2)Re z −1/2 ∈ Λα −β (Γ) and (∂/∂s)Re (ζ(z))2 − (∂/∂s)Re z ∈ Λα −β (Γ). Combined with Theorem 2.2.9, this proves the converse inclusion.
2.5 Boundary integral equations of the Dirichlet and Neumann problems in domains with inward peak In this section, unlike Section 2.4.1, by ω and ζ we denote the conformal mapping of Ω− and Ω+ onto G and R2+ , respectively. Also, we introduce the function I on Γ by putting I(z) = Im z 1/2 .
2.5.1 Integral equation of the Dirichlet problem on a contour with inward peak We start by showing the surjectivity of the operator Wβ introduced in the next theorem. Theorem 2.5.1. Let Ω+ have an inward peak and let 0 < α, β < 1, β = 1/2. Then the operator
Lodd
1,α 2−β
Wβ
˙ (Γ) σ − −→ πσ − T σ ∈ Λα −β (Γ)+R
(2.123)
with 0 < β < 1/2 and 1,α Wβ ˙ −→ πσ − T σ + tI ∈ Λα Lodd 2−β (Γ) × R (σ, t) − −β (Γ)+R
(2.124)
for 1/2 < β < 1 is surjective. Here I(z) = Im z 1/2 , z ∈ Γ \ {O}. Proof. (i) Let ϕ ∈ Λα −β (Γ) and let ϕ = 0 in a neighborhood of the peak. By hθ we denote the Poisson integral of ϕθ in R2+ . The conjugate function hθ has the restriction on R given by t−1 hθ (τ + i0) = τ ϕθ (t) dt . τ −t R
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Chapter 2. Boundary Integral Equations in H¨ older Spaces
We apply Lemma 2.4.5 to F± (u) = τ
R
ϕθ (t)
t−1 dt , τ −t
−1/2
where τ = ±u , u ∈ (1, ∞). Since ϕρ ∈ Eβα (∂G), we have ϕρ ◦ d± ∈ Eβα (1, ∞) (see (2.103)). By Theorem 2.2.1 combined with Lemmas 2.2.4 and 2.2.5, F± (u) ∈ Eβα (1, ∞) with 0 < β < 1/2 and there exists a linear functional c1 (ϕ), continuous in Λα −β (Γ), and such that F± (u) ∓ c1 (ϕ) u−1/2 ∈ Eβα (1, ∞) for 1/2 < β < 1 (cf. Lemma 2.4.5).
−1/2 The functions t± (u) determined by the equations ζ ◦ ρ(u) = ± t± (u) have the expansions t± (u) = u + O log u , which are at least once differentiable. Hence the function h(z), z ∈ Γ, conjugate of h, admits the representation h(z) = c1 (ϕ)Re z 1/2 + g(z), where g ∈ Λα −β (Γ). Moreover, |c1 (ϕ)| + gΛα−β (Γ) cϕΛα−β (Γ) with a constant c independent of ϕ. Here we assume that c1 (ϕ) = 0 for 0 < β < 1/2. We put t = c1 (ϕ) in (2.124). Consider the bounded harmonic function g = h(z)+ t Im z 1/2 , z ∈ Ω+ . Let he denote the harmonic of ϕ(z)+ t Im z 1/2 , −1/2extension − e − z ∈ Γ\{O}, into Ω , such that grad h (z) = O |z| , z ∈ Ω . Since grad g(z) = −1/2 + , z ∈ Ω , it follows that O |z| ∂he 1 ∂g 1 − log dsq + he (∞), z ∈ Γ \ {O} . g(z) = (2.125) 2π Γ ∂n ∂n |z − q| The function ( g )ρ belongs to Eβα (∂G). Hence, by Proposition 2.3.4, the harmonic extension of ( g )ρ into G has the conjugate function H, which admits the representation H = Q1 + Q2 with Q1 ∈ Eβα (∂G), ∂Q2 /∂ s ∈ Eβα (∂G), supp Q2 ⊂ {(u, ±1) : u > 1} and Q2 (u, 1) = Q2 (u, −1) for u > 1. Therefore, Hω is a so− lution in (Lodd )1,α with the boundary data 2−β (Γ) of the Neumann problem in Ω ∂Hω /∂n = −∂g/∂n. We may assume that H (∞) = 0. ω In view of the relation grad Hω = O |z|−3/2 , z ∈ Ω− , the function w = −Hω − he + he (∞) admits the representation in Ω− : ∂w ∂ 1 |z| 1 − d sq . w(q) log (q) log w(z) = 2π Γ ∂nq |z − q| ∂nq |z − q|
2.5. Integral equations of the Dirichlet and Neumann problems
177
By the limit relation for the double layer potential and by (2.125) we have ∂ 1 1 d sq = −2 g(z) − he (∞) , z ∈ Γ \ {O} . w(q) log w(z) − π ∂nq |z − q| Γ
Since T 1 = −π, the pair (σ, t) with σ(z) = −(2π)−1 Hω (z) + ϕ(z) + t Im z 1/2 is a solution of (2.5) and σ (Lodd )1,α
2−β (Γ)
+| t | c ϕ Λα−β (Γ) .
(2.126)
(ii) Now, let ϕ be an arbitrary function in Λα −β (Γ). There exists a sequence {ϕn }n≥1 of smooth functions on Γ \ {O}, vanishing near the peak and converging in Λα −β (Γ) with α < α and β < β, for which ϕn Λα−β (Γ) c ϕ Λα−β (Γ) . Let (σn , tn ) ∈ P1,α 2−β (Γ) × R denote the solution of (2.5) with the right-hand side ϕn constructed as in (i). By (2.126), σn (Lodd )1,α
2−β (Γ)
+| tn | c ϕ Λα−β (Γ)
(2.127)
α with c independent of n. Since Λα −β (Γ) ⊂ Λ−β (Γ), by (2.126) we obtain that the
sequence {σn }n≥1 converges in (Lodd )1,α 2−β (Γ) × R to a limit (σ, t). In particular, {σn }n≥1 converges pointwise to σ on Γ \ {O}. In view of (2.127) we find that (σ, t) belongs to (Lodd )1,α 2−β (Γ) × R. Using the continuity of the operator
α ˙ πI − T : (Lodd )1,α 2−β (Γ) → Λ−β (Γ)+R
(cf. Theorem 2.2.8) and passing to the limit, we conclude that −ϕ = (πI − T )σ + t I , where t = 0 for 0 < β < 1/2. This proves the inclusion 1,α ˙ Λα −β (Γ)+R ⊂ Wβ (Lodd )2−β (Γ) × R . Since Im z 1/2 ∈ Λα −β (Γ), the converse inclusion follows by Theorem 2.2.8.
Next, we prove that the kernel of the operator contains only zero. Theorem 2.5.2. Let Ω+ have an inward peak and let 0 < α, β < 1, β = 1/2. Then the operator Wβ defined by (2.123) for 0 < β < 1/2 and by (2.124) for 1/2 < β < 1 is injective.
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Chapter 2. Boundary Integral Equations in H¨ older Spaces
Proof. (i) Suppose that 0 < β < 1/2 and let σ ∈ (Lodd )1,α 2−β (Γ) be a solution of the homogeneous equation πσ − T σ = 0 . (2.128) Then W σ (z), z ∈ Ω+ , has zero limit values on Γ \ {O}. We introduce the holomorphic function W (z) = − W σ (z) + i W σ (z) z ∈ Ω+ . + 2 Let ζ(z) = ξ +iη be a conformal mapping of the domain Ω onto R+ satisfying the condition ζ(0) = 0. The function F (ξ + iη) = W ◦ θ (1/(ξ + iη)) is holomorphic in the lower half-plane R2− = {ξ + iη, η < 0}, continuous up to the boundary, and Im F = 0 on the real axis. Since W σ admits the estimate (W σ)(z) = O |z|−1 ,
the holomorphic extension F ext of F to C is an entire function satisfying the inequality |Re F ext (ξ + iη)| c |ξ + iη|2 . With the help of the integral Schwarz formula, it can be checked that the rate of growth of F ext is the same as that of Re F ext . Therefore, the function F ext is a polynomial with real coefficients, and %σ)(z) = Re F ext (1/ζ(z)), z ∈ Ω+ . −(W %σ)(x) = O(1) as R x → −0, the polynomial F ext is constant in C, and Since (W %σ is constant in Ω+ . Therefore, the function W %σ on Ω− has a constant hence W − %σ(z), z ∈ Ω , grows not faster than a power boundary value on Γ \ {O}. Since W % function as z → 0, we see that W σ = const in Ω− . Therefore, the conjugate function −W σ is constant in Ω− . The jump formula for the double layer potential W σ shows that σ = const on Γ \ {O}. Since no nonzero constant satisfies the homogeneous equation (2.128), the function σ is zero on Γ \ {O}. (ii) Now, let 1/2 < β < 1 and let the pair (σ, t) ∈ (Lodd )1,α 2−β (Γ) × R belong to ker πI − T + tI . As in (i), we show that W (z) = − W σ (z) + i W σ (z) + t z 1/2 is constant in Ω+ . By Theorem 2.5.1 and the statement proved in (i), the equation (2.4) with the right-hand side Im z 1/2 has a unique solution σ0 in (Lodd )1,α 2−β (Γ) 1,α with 0 < β < 1/2. Since (Lodd )1,α (Γ) ⊂ (L ) (Γ) and σ , being equivalent odd 2−β 0 2−β to c x−1/2 as z → 0, does not belong to (Lodd )1,α (Γ), we have t = 0. Applying 2−β the argument used in (i), we prove that σ = 0.
2.5. Integral equations of the Dirichlet and Neumann problems
179
2.5.2 Integral equation of the Neumann problem on contour with inward peak We start our study of the integral equation (2.8) on a contour with inward peak with a description of solutions to the corresponding homogeneous equation. The proof of the next assertion duplicates the proof of Lemma 2.4.11 with obvious changes. Lemma 2.5.3. The logarithmic potential
z
V σ (z) = σ(q) log
dsq , z − q Γ
z ∈ Ω+ ∪ Ω − ,
with density σ such that |σ(q)| c |q|−α ,
0 < α < 1,
obeys the inequality |V σ(z)| c log
1 |z|
for
z → 0.
Proof. It suffices to estimate the integral
z
σ(q) log
dsq z−q Γ+ ∪Γ− in a small neighborhood of z = 0. For 2|q| < |z| one has
q
|q|
log 1 −
c z |z| with some constant c. Hence
z c
σ(q) log
|σ(q)| |q|dsq
dsq
|z| z − q |z| Γ± ∩{|q|< |z| Γ± ∩{|q|< 2 } 2 } |z|/2 c u−α+1 du c . |z| 0
(2.129)
In the case 2|z| < |q|,
z
u |z|
+cc +1 .
log
log z−q |q| |z| Hence
δ
z
u
+ 1 du σ(q) log
u−α log
dsq c z−q |z| Γ± ∩{|q|2|z|} |z| δ/|z| 1 . c|z|−α+1 u−α log u + 1 du c log |z| 2
(2.130)
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Chapter 2. Boundary Integral Equations in H¨ older Spaces
Next, suppose that |z|/2 < |q| < 2|z|. Then either |1 − q/z| < 1 and hence 1 1
log
log
1 − |q|/|z| , 1 − q/z|
or |1 − q/z| > 1 and hence
1
c .
log
1 − q/z
Thus, for |z|/2 < |q| < 2|z|, one has
z
1
+ c .
log
log
z−q 1 − |q|/|z|
We set v = v± (u) = |q|u 1 +
κ± (u) 2 u
and assume δ > 0 to be so small that κ± (u)
d κ± (u) 0 du u
for all u ∈ [0, δ]. Therefore, v± (u) increases for 0 < u < δ and 0<
1 v± (u) c . c
∗ (u) = σ u + iκ± (u) . We have Let σ±
z
σ(q) log
dsq
z − q Γ± ∩{|z|/2<|q|<2|z|} 2|z|
∗ c |σ± (u)| log
1
+ 1 dv 1 − |u|/|z|
|z|/2 2
1
∗ c|z| |σ± (|z|u)| log
+ 1 du |1 − u| 1/2 2
1
c |z|−α+1
+ 1 du c .
log |1 − u| 1/2
By (2.129), (2.130), and (2.131) we conclude that
z
1
. σ(q) log
dsq c log z − q |z| Γ+ ∪Γ−
(2.131)
2.5. Integral equations of the Dirichlet and Neumann problems
181
By B1 we denote the linear hull of the function Re z −1/2 , z ∈ Γ, and by B0 we mean the null-space consisting of the zero function only. As in (2.6), consider the function α+ − α− 3/2 α+ + α− 3/2 1/2 z (log z − πi) + i z , z ∈ Ω+ . R(z) = Re z + 4π 4 Theorem 2.5.4. Let Ω+ have an inward peak and let 0 < α, β < 1 and β = 1/2. Then the operator ∂ ˙ ˙ (Lodd )1,α R ∈ (Λ0 )α −β (Γ)+D1 2−β (Γ)+Bm−1 × R (σ, t) −→ πσ + Sσ + t ∂n (2.132) with m subject to m − 1 < 2β < m has a one-dimensional kernel. Elements of the kernel are pairs of the form ∂ c log |hext | , 0 , ∂n where c ∈ R and hext (z) is a conformal mapping of Ω− onto the exterior of the unit disk such that hext (∞) = ∞. Proof. Let the pair (σ, t) from the space ˙ (Lodd )1,α 2−β (Γ)+ Bm−1 × R belong to ker πI + S + t R . Then the harmonic function v(z) = V σ(z) + t R(z), z ∈ Ω+ , is a solution to the Neumann problem with zero boundary date on Γ \ {O}. By Lemma 2.5.3, |v(z)| c log(1/|z|) . (2.133) Therefore, the integral representation of the harmonic function v(z) and the limit relation for the double layer potential imply ∂ 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− . Since the double layer potential (W v)(z), z ∈ Ω− , grows not faster than a power function, and its values on Γ \ {O} are zero, we have (W v)(z) = 0 in Ω− . %v(z), z ∈ Ω− , is constant in Ω− . Therefore, the conjugate harmonic function W %v = C, we consider the holomorphic function Putting W %v − C), z ∈ Ω+ . W (z) = (W v)(z) + i(W
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Chapter 2. Boundary Integral Equations in H¨ older Spaces
Let ζ(z) = ξ + iη be a conformal mapping of Ω+ onto R2+ satisfying the condition ζ(0) = 0. The function F (ξ + iη) = (W ◦ θ)(1/(ξ + iη)) is holomorphic in the lower half-plane R2− , continuous up to the boundary, and Im F = 0 on R. The holomorphic extension F ext of F to C is an entire function with at most power growth as ξ + iη → ∞. Therefore,
(W v)(z) =
c(k) Re (1/γ(z))−k , z ∈ Ω+ ,
k=0
where c(k) , 0 k , are real coefficients and is a nonnegative integer. Using the jump formula for W v, we obtain v(z) = −(2π)−1
c(k) Re (1/γ(z))−k , z ∈ Γ \ {O}.
k=0
Then (2.133) implies that v(z) = c(0) on Γ \ {O}, whence (V σ)(z) + tR(z) = c(0) , z ∈ Ω− . Let Re denote a harmonic extension of the restriction of R to Γ \ {O} to Ω− , and let σ0 be the density of the equilibrium distribution on Γ. We put −1 c0 = σ(q)dsq σ0 (q)dsq . Γ
Γ
Since V σ − c0 V σ0 + tRe is a function bounded and harmonic on Ω− , and constant on Γ \ {O}, we have V σ(z) = c0 V σ0 (z) − tRe (z) + c(1) , z ∈ Ω+ . The jump formula for the normal derivative of a single layer potential yields 2πσ(z) = 2πc0 σ0 (z) − t
∂ ∂ R(z) + t Re (z) , z ∈ Γ \ {O} . ∂n ∂n
It can be easily seen that ∂ ∂ e R(z) ∼ aα± |z|1/2 , R (z) ∼ ±b |z|−3/2 , ∂n ∂n where a, b ∈ R, and that on Γ \ {O} the function σ0 satisfies the estimate |σ0 (z)| c exp(−γx), z → 0, ˙ where γ is a small positive number. Since σ ∈ (Lodd )1,α 2−β (Γ)+ Bm−1 , it follows that t = 0, and σ = c0 σ0 .
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183
Clearly, σ0 is a solution of the homogeneous equation πσ + Sσ = 0 , and σ0 ∈ (Lodd )1,α 2−β (Γ) for 0 < β < 1. Therefore, the kernel of the operator (2.132) is one-dimensional. As in Section 2.4.3, by (Λ0 )α −β (Γ) we denote the subspace of functions ψ ∈ subject to the orthogonality condition
Λα −β (Γ),
ψds = 0. Γ
Theorem 2.5.5. Let Ω+ have an inward peak and let 0 < α, β < 1, β = 1/2. Then the operator (2.132) is surjective. Proof. (i) Let ψ ∈ (Λ0 )α −β (Γ) be a function such that ψ = 0 near the peak. We , ( ) define a function ψ on Γ by the formula d (, ) ψ = ψ. ds Since θ(±x1/2 ) = x + O(x2 log x), we see that the functions , (d/dx) ψ ( ) θ (±x1/2 ) belong to Λα −β (0, 1). Therefore, , (d/du) ψ ( ) θ (±u−1/2 ) α are elements of Eβ+2 (1, ∞). We introduce the solution hθ of the Neumann bound , ary value problem with the boundary data (d/dτ ) ψ ( ) θ :
hθ (ζ) = −
log |ζ − ν| R
d (, ) ψ (ν)dν, ζ = τ + iη ∈ R2+ , θ dν
and represent ∂hθ /∂τ on R as follows: −τ 2 R
ν −2 d (, ) ψ (ν)dν + θ τ − ν dν
R
d (, ) dν ψ +τ (ν) θ dν ν
R
d (, ) dν ψ (ν) 2 . θ dν ν
It is clear that the linear functionals d (, ) d (, ) dν dν ψ and a ψ a1 (ψ) = (ν) (ψ) = (ν) 2 2 θ θ dν ν dν ν R
R
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Chapter 2. Boundary Integral Equations in H¨ older Spaces
are continuous on Λα −β (Γ). Now we apply Lemma 2.4.6 to the differentials
τ
2 R
ν −2 d (, ) ψ (ν)dν dτ = F± (u) d u , θ τ − ν dν
where τ and u are related by the formula τ = ±u−1/2 , u > 0. From Theorem 2.2.1 α and Lemmas 2.2.4, 2.2.5 it follows that F± (u) ∈ E2+β (1, ∞) for 0 < β < 1/2, and that there exists a continuous linear functional a3 (ψ) on the space Λα −β (Γ) such that α F± (u) ∓ a3 (ψ)u−5/2 ∈ E2+β (1, ∞) for 1/2 < β < 1 . 1/2 gives rise to functions t± admitting repreThe equation ζ ◦ ρ(u) = ± t± (u) sentation in the form t± (u) = u + O(log u), and these relations are differentiable at least once. Therefore, the function h(z), z ∈ Γ, satisfies h(z) − h(0) =
m+1
bk (ψ) Re
k=1
ζ(z) 1 − iζ(z)
k + g(z) ,
where g is subject to the inequality (∂/∂s)gΛα−β (Γ) cψΛα−β (Γ) , the functionals b1 (ψ), . . . , bm (ψ) are linear combinations of a1 (ψ), . . . , am (ψ), and the integer m is such that (m − 1)/2 < β < m/2. (ii) The harmonic extension H of gρ to G constructed in Proposition 2.3.3 α has the normal derivative: ∂H/∂n = Q1 + Q2 , where Q1 ∈ Eβ+2 (∂G), ∂Q2 /∂s ∈ α Eβ+2 (∂G), supp Q2 ⊂ {(u, ±1) : u > 0}, and Q2 (u, 1) = −Q2 (u, −1) for u > 1. Now it follows from Lemma 2.4.2 that ∂Hω /∂n can be represented in the α form q1 + q2 , where q1 ∈ Λα −β (Γ), (∂/∂s)q2 ∈ Λ2−β (Γ), and q2 (x + iκ+ (x)) = −q2 (x + iκ− (x)), x ∈ (0, 1]. According to Proposition 3.2.5, the bounded harmonic extension of m+1 k=1
bk (ψ) Re
ζ(z) 1 − iζ(z)
k − b1 (ψ)R(z), z ∈ Γ \ {O},
into Ω− has the normal derivative of the form q3 (z) − c(ψ)Re z −1/2 , where q3 ∈ Λα −β (Γ) and c(ψ) is a linear combination of the functionals b1 (ψ), . . . , bm+1 (ψ). Let f be a harmonic extension of the function h − b1 (ψ)R to Ω− . Since h is determined up to an additive constant, we may assume that f vanishes at infinity.
2.5. Integral equations of the Dirichlet and Neumann problems
185
Using grad h = O |z|−1/2 in Ω+ and grad f = O |z|−1/2 in Ω− , we obtain h(z) − b1 (ψ)R(z) ∂f 1 1 ∂R (q) − (q) log dsq , z ∈ Γ \ {O} . ψ(q) − b1 (ψ) = 2π ∂n ∂n |z − q| Γ
We put σ(z) =
3 1 ∂R ψ(z) − b1 (ψ) (z) − qk (z) + c(ψ)Re z −1/2 . 2π ∂n
(2.134)
k=1
Notice that σ(z) = O |z|−1/2 , z ∈ Γ, implies V σ(z) = O(1). Since h − b1 (ψ)R is bounded, the latter fact implies that V σ = h − b1 (ψ)R in Ω+ . Using the limit relation for the normal derivative of the potential V σ, we obtain ∂ ∂ V σ = −ψ + b1 (ψ) R on Γ \ {O} . πσ + ∂n ∂n Therefore, the pair (σ, t), where t = −b1 (ψ) and σ is given by (2.134), is a solution of the equation (2.8). Since ∂R/∂n ∈ Λα −β (Γ), we see that the density σ can be represented in the form σ(z) = σ (0) (z) + with σ(0) (z) =
1 c(ψ)Re z −1/2 , z ∈ Γ \ {O}. 2π
(2.135)
3 1 ∂R ψ(z) − b1 (ψ) (z) − qk (z) . 2π ∂n k=1
Moreover, σ (0) (Lodd )1,α
2−β (Γ)
+ |c(ψ)| c ψΛα−β (Γ) .
(2.136)
α (iii) Now let ψ be an arbitrary function in Λ0 −β (Γ). There exists a sequence {ψn }n≥1 of smooth functions on Γ \ {O} which are equal to zero near the peak, converge in Λα −β (Γ) for α < α and β < β, and satisfy the estimate ψn Λα−β (Γ) c ψ Λα−β (Γ) . ˙ Let (σn , tn ) stand for the solution of (2.8) in (Lodd )1,α 2−β (Γ) × Bm−1 +R with the right-hand side ψn , constructed as in (i)–(ii). Here the coefficient tn is equal to b1 (ψn ) and 1 c(ψn )Re z −1/2 , z ∈ Γ \ {O} σn (z) = σn(0) (z) + 2π
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Chapter 2. Boundary Integral Equations in H¨ older Spaces
α (cf. (2.135)). In view of the embedding Λα −β (Γ) ⊂ Λ−β (Γ), by (2.136) the se ˙ quence {(σn , tn )}n≥1 converges to a limit (σ, t) in (Lodd )1,α 2−β (Γ) × Bm−1 +R. In particular, {σn }n≥1 converges to σ pointwise on Γ \ {O}. Owing to the estimate
σn (Lodd )1,α
2−β (Γ)×Bm−1
+ | tn | c ψΛα−β (Γ) ,
˙ the pair (σ, t) belongs to (Lodd )1,α 2−β (Γ) × Bm−1 +R and σ(Lodd )1,α
2−β (Γ)×Bm−1
+ | t | c ψΛα−β (Γ) .
Using the continuity of the operator (2.59) (cf. Theorem 2.2.11) and passing to the limit in the equality πσn + Sσn + tn
∂ R = ψn on Γ \ {O} , ∂n
we conclude that (σ, t) satisfies (2.8). Now, let ψ(z) = (∂/∂s)Re z, z ∈ Γ. Clearly, the harmonic function h(z) = −Im z, z ∈ Ω+ , satisfies the equality ∂h/∂n = ψ. Since ∂h/∂s ∈ Λα −β (Γ), it follows from Proposition 2.3.3 that there exists a harmonic extension f of h into the domain Ω− with the normal derivative in (Lodd )1,α 2−β (Γ). In the same way as in (ii) we show that the pair (σ, 0) with σ(z) =
1 ∂ ψ− f 2π ∂n
satisfies (2.8) with ψ as the right-hand side. Thus ∂ ˙ ˙ R (Lodd )1,α × R . (Λ0 )α (Γ) +B m−1 −β (Γ)+D1 ⊂ πI + S + t 2−β ∂n ∂ R ∈ Λα −β (Γ), by Theorem 2.2.11 we have ∂n ∂ ˙ (Γ) × R ⊂ (Λ0 )α πI + S + t R (Lodd )1,α −β (Γ)+D1 . 2−β ∂n
(iv) Since
(2.137)
We choose ψ ∈ (Λ0 )α −β (Γ) in such a way that the coefficient c(ψ) in (2.135) does not vanish (cf. Theorem 2 [20]). Note that σ (0) in (2.135) belongs to (Lodd )1,α 2−β (Γ), ˙ therefore, by Theorem 2.2.11, (πI + S)σ0 ∈ (Λ0 )α (Γ) +D . This proves the em1 −β ˙ bedding (πI + S) B1 ⊂ (Λ0 )α (Γ) +D . Combining this with (2.137), we arrive 1 −β at the converse embedding.
2.6. Integral equation of the first kind
187
2.6 Integral equation of the first kind on a contour with peak In this section we establish the solvability and describe the kernel for the integral equation of the first kind (2.3) on a contour with peak. We start with three easily proved auxiliary assertions. 1,α 1 1 Lemma 2.6.1. Suppose that f (u) ∈ Λ1,α 1+β (0, d). Then g(u) = f ( u ) ∈ Λ1−β ( d , ∞).
Lemma 2.6.2. Suppose that ψ ∈ Λ1,α 1+β (0, d). Then (ψθ ) (±u1/2 ) Λαβ+1/2 (0,d) ψ Λαβ+1 (0,d) . Lemma 2.6.3. Suppose that ϕ ∈ Λβ+γ (R). Then uγ ϕ Λαβ (R) ϕ Λαβ+γ (R) . Theorem 2.6.4. Let Ω+ have an outward or inward peak and let 0 < α, β < 1, β = 1/2. Then the operator Λα −→ Vσ + c ∈ Nβ1,α (Γ) 1+β (Γ) × R {σ, c} −
(2.138)
is surjective. Proof. To be definite, suppose that Ω+ has an outward peak. Let ϕ ∈ Nβ1,α (Γ) and ϕ = 0 in a neighborhood of the peak. We show that the integral equation (2.3) is solvable in Λα 1+β (Γ) × R. We introduce the function ϕθ = ϕ ◦ θ on R and consider the harmonic extension of ϕθ into the upper half-plane: d 1 1 τ ϕθ (τ )Im log 1 + dτ for 0 < β < Gθ (ξ, η) = − π R dτ ζ −τ 2 and d τ 1 1 τ ϕθ (τ )Im log 1 + − dτ for < β < 1. Gθ (ξ, η) = − π R dτ ζ −τ ζ −τ 2 The normal derivative of Gθ on R + i0 has the form ∂ d 1 1 τ dτ Gθ (ξ, 0) = − ϕθ (τ ) for 0 < β < ∂n πξ R dτ ξ−τ 2 and ∂ d 1 1 τ 2 dτ Gθ (ξ, 0) = − 2 ϕθ (τ ) for < β < 1. ∂n πξ R dτ ξ−τ 2 Using Lemma 2.4.6 for 0 < β < 1/2 (the case k = −1), we write the differential Φ(ξ)dξ = (∂G/∂n)(ξ, 0)dξ
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Chapter 2. Boundary Integral Equations in H¨ older Spaces
in the form Φ(ξ)dξ = ∓ u−1/2
∞
v 1/2 d ϕθ (±v −1/2 )dv u − v dv 0 1 d 1 −1/2 ∞ √ √ ϕθ (±v −1/2 )dv ∓ u 2 dv u + v 0 3 1 d 1 −1/2 ∞ −1/2 √ √ ϕθ (∓v )dv du = u−1/2 Ik (u)du ± u 2 dv u + v 0 1
and for 1/2 < β < 1 (the case k = −2) in the form
∞
Φ(ξ)dξ = ∓ 0
1 ∓ 2 ∓
1 2
1 d ϕθ (±v −1/2 )dv u − v dv ∞
v −1/2 d √ √ ϕθ (±v −1/2 )dv u + v dv
∞
v −1/2 d √ √ ϕθ (∓v −1/2 )dv du = u−1/2 Ik (u)du , u + v dv 1
0
0
3
where ξ = ±u−1/2 . In view of Lemmas 2.6.2, 2.6.2, and 2.2.5 we have d [ϕθ (±v −1/2 )] Λα1−β (R) dv d ϕθ (±v −1/2 )v −3/2 Λα1−β (R) =c dτ d ϕθ (±v 1/2 )v 3/2 Λαβ−1 (R) =c dτ d ϕθ (±v 1/2 ) Λαβ+1/2 (0,d) c ϕ Λαβ+1 (0,d) . c dτ
u−1/2 I2 (u) Λα1−β (R) c
In a similar way, for I3 we obtain u−1/2 I3 (u) Λα1−β (R) c ϕ Λαβ+1 (0,d) . By Theorem 2.2.1 and Lemma 2.6.2, u−1/2 I1 (u) Λα1−β (R) c I1 (u) Λα1/2−β (R) d c v 1/2 [ϕθ (v −1/2 )] Λα1/2−β (R) dv d d ϕθ (v −1/2 ) Λα1/2−β (R) c v ϕθ (v 1/2 ) Λαβ−1/2 (0,d) c v −1 dτ dτ d 1/2 ϕθ (v ) Λαβ+1/2 (0,d) c ϕ Λαβ+1 (0,d) . c dτ
2.6. Integral equation of the first kind
189
Thus, for 0 < β < 1/2, we conclude that Φ Λα1−β (R) =
∂ G(ξ, 0) Λα1−β (R) c ϕ Λαβ+1 (0,d) . ∂η
For 1/2 < β < 1 we use a similar argument. Indeed, by Lemma 2.2.4 we have d ϕθ (±v −1/2 ) Λα1−β (R) dv c ϕ Λ1,α (0,d) .
max{ I2 Λα1−β (R) , I3 Λα1−β (R) } c
1+β
Hence by Theorem 2.2.1 I1 Λα1−β (R) c
d ϕθ (±v −1/2 ) Λα1−β (R) c ϕ Λ1,α (0,d) . 1+β dv
Note that ϕθ (±u−1/2 ) = ϕρ,ω,θ (±u−1/2 ) = ϕρ,d± (u) (see (2.103)). Since the first derivatives of d± belong to Λα (1/d, ∞), we see that ϕρ,d± and ϕρ both belong to Λ1,α 1−β (1/d, ∞). Therefore, by Lemma 2.6.1 ϕθ (±u−1/2 ) and ϕρ (u) are in Λ1,α (1/d, ∞). 1−β We have Φ(ξ)dξ = I(u)du, where ξ = ±u−1/2 and I(u) ∈ Λα 1−β (1/d, ∞). The function G, equal to Gθ,ζ , is a harmonic extension of ϕ onto Ω− . Let F denote the conjugate harmonic function. Then
∂ ∂
Gθ (ξ, η)
Fθ (ξ, η)
= . ∂η ∂ξ η=+i0 η=+i0 Since ∂ ∂ Gθ (±u−1/2 )u−3/2 = ∓2 Fθ (±u−1/2 ) ∈ Λα 1−β (1/d, ∞) , ∂η ∂u we have Fθ (±u−1/2 ) ∈ Λ1,α 1−β (1/d, ∞). Let us write Fθ as Fρ,ω,θ (u) = (Fρ )ω◦θ (u). Then (Fρ )ω,θ (±u−1/2 ) = Fρ,d± (u), that is Fθ (±u−1/2 ) = Fρ,d± (u). The functions 1,α Fρ and Fρ,d± belong to Λ1,α 1−β (1/d, ∞). Therefore, F is an element of Λ1+β (0, d). α Hence the function ∂G/∂n, equal to ∂F /∂s on Γ± , belongs to Λ1+β (0, d). From the integral representation for Gθ it follows that −1 ∂ G(z) = O |z| for 0 < β < 1/2 ∂n and
∂ G(z) = O |z|−3/2 for ∂n
1/2 < β < 1,
Ω− z → 0.
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Chapter 2. Boundary Integral Equations in H¨ older Spaces
1 Consider a harmonic extension H of the function ϕρ from Λ1,α 1−β ( d , ∞) satisfying 1 α the condition ϕρ+ − ϕρ− ∈ Λ1−β ( d , ∞). Such an extension was constructed in Proposition 2.3.6. Note that for any integer N ,
N ∂ Hω (z) = O |z| , ∂n
z ∈ Ω+ .
Hence 1 ϕ(z) = 2π
∂G |z| ∂Hω − log dsq + G(∞), ∂n ∂n |z − q| Γ
z ∈ Γ\ {O} .
Thus the pair (σ, t) with σ=
∂G 1 ∂Hω − 2π ∂n ∂n
and t = G(∞)
obeys (2.3) and σ Λα1+β (Γ) +|t| c ϕ N 1,α (Γ) , β
where the constant c is independent of ϕ. Given any ϕ ∈ Nβ1,α (Γ), there exists a sequence {ϕn }n1 of functions which are smooth on Γ\{O}, vanish near the peak, converge to ϕ in Nβ1,α (Γ) with α < α and β > β, and satisfy the estimate ϕn Λα1+β (Γ) c ϕ N 1,α (Γ) β
(2.139)
with c independent of n. Let (σn , tn ) ∈ Λα 1+β (Γ) × R be the above-constructed solution of (2.3) with the right-hand side ϕn . By (2.139) we have σn Λα1+β (Γ) +|tn | c ϕ N 1,α (Γ) . β
(2.140)
(Γ), the sequence {(σn , tn )}n1 converges to (σ, t) in Since Nβ1,α (Γ) ⊂ Nβ1,α α Λ1+β (Γ). In particular, {σn }n1 converges to σ pointwise on Γ\{O}. It follows from (2.140) that the pair (σ, t) belongs to Λα 1+β (Γ) × R. Using the continuity of
−→ Nβ1,α (Γ) and passing to the limit in the equality the operator V : Λα 1+β (Γ) − V σn + tn = ϕn , we find that V σ + t = ϕ. This proves the inclusion ˙ Nβ1,α (Γ) ⊂ V Λα 1+β (Γ) +R . By Theorem 2.2.12 the converse embedding holds, ˙ Nβ1,α (Γ) ⊃ V Λα 1+β (Γ) +R . The proof is complete.
2.6. Integral equation of the first kind
191
Theorem 2.6.5. Let Ω+ have either outward or inward peak and let V be the operator of the integral equation (2.3). Then a) ker V = {0} for 0 < β < 1/2, b) dim ker V = 1 for 1/2 < β < 1. Proof. Let Ω+ have an outward peak and let {σ, C} belong to ker V. The potential V σ equals −C on Γ\{0} and admits the estimate |V σ(z)| c |z|−β , z = 0 . Therefore, V σ = −C on 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−q
The branch of arg is chosen so that arg 1 = 0. Let ζ = γ(z) be a conformal mapping of Ω− on R2+ , γ(0) = 0. The function W (ζ) = (V ◦ γ −1 )(1/ζ) is holomorphic on the lower half-plane, continuous up to the boundary and Im W = 0 on the real axis. The holomorphic extension W ext of W onto the upper half-plane is an entire function with imaginary part satisfying |Im W ext (ζ)| c |ζ|2β . From the Schwarz integral formula it follows that W ext has at most the same order of growth at infinity. Since β < 1, we see that W ext is a linear function of ζ. Therefore, V σ = c0 + c1 Im (1/γ) . From the jump formula for (∂/∂n)V σ we have σ(z) = −
c1 ∂ 1 Im ∼ c x−3/2 as z → 0, z ∈ Γ\{O}. 2π ∂n γ(z)
By the integral representation for the harmonic function Im (1/γ) on Ω− and a limit relation for the double layer potential we obtain 1 ∂ |z| 1 1 log dsq − Im Im π Γ ∂nq γ(q) |z − q| γ(∞) (2.141) 1 ∂ 1 1 1 dsq − Im Im log = π Γ γ(q) ∂nq |z − q| γ(z) for z ∈ Γ\{O}. Since Im (1/γ(z)) = 0 on Γ\{O}, it follows that zeros of V have the form 1 ∂ 1 1 c Im , c Im , c ∈ R. π ∂n γ γ(∞)
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Chapter 2. Boundary Integral Equations in H¨ older Spaces
We note that Im (1/γ) is defined modulo a constant positive factor so that dim ker V is at most 1. Since (∂/∂n)Im (1/γ) ∈ Λα β+1 (Γ) only for β > 1/2, we obtain that ker V is one-dimensional for 1/2 < β < 1 and zero for 0 < β < 1/2. The result for the case of inward peak can be obtained similarly using the identity ∂ |z| 1 1 log dsq Im π Γ ∂nq γ(q) |z − q| 1 1 ∂ 1 1 dsq + Im , z ∈ Γ\{O}, = Im log π Γ γ(q) ∂nq |z − q| γ(z) instead of (2.141).
In passing, we have proved the following characterization of the kernel of V in the case 1/2 < β < 1. Proposition 2.6.6. a) Let Ω+ have an outward peak and let 1/2 < β < 1. Then t ∂ 1 1 ker V = Im (out) , t Im (out) , 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 < 1. Then 1 t ∂ ker V = Im (in) , 0 , π ∂n γ where t ∈ R and γ (in) is a conformal mapping of Ω+ onto R2+ normalized by the conditions γ (in) (0) = 0 and γ (in) (z0 ) = i with a fixed point z0 ∈ Ω+ .
2.7. Appendices
193
2.7 Appendices Appendix A: Proof of Theorem 2.2.1 We introduce an even C ∞ -function χ, putting 1, t ∈ [−3/4, 3/4], χ(t) = 0, t ∈ / [−1, 1] . Further, let χr (t) = χ r−1 (t − (x + y)/2) , ηr (t) = 1 − χr (t) , where r = |x − y|. Following [30], we represent the difference T f (x) − T f (y) in the form T {(f − f (x))χr }(x) − T {(f − f (y))χr }(y) + [f (x) − f (y)]T χr (y) + f (x)[T χr (x) − T χr (y)] + [K(x, t) − K(y, t)]f (t)ηr (t)dt R
= T {(f − f (x))χr }(x) − T {(f − f (y))χr }(y) + [K(x, t) − K(y, t)][f (t) − f (x)]ηr (t)tdt
(2.142)
R
+ [f (x) − f (y)]T χr (y) + f (x)[T 1 (x) − T 1 (y)] =
5
Ik ,
k=1
where Ik correspond to the terms on the left-hand side. The continuity of the operator T in E0α (R) was proved in [30] under more general conditions on the kernel K(x, y). Hence it suffices to prove the required estimate outside a neighborhood of the origin for min(|x|, |y|) 1. In view of Lemma 2.1.2 we assume that 2|x − y| min(|x|, |y|). Further, let f Eβα (R) = 1. We estimate each term in (2.142). (i) Consider the term I1 in (2.142). For |t − (x + y)/2| < r we have |t − x| < (3/2)|y − x|. Therefore,
T {χr (f − f (x))}(x) =
K(x, t)χr (t)(f (t) − f (x))dt
R 1 |x|−α−β |x − t|α dt |K(x, t)||f (t) − f (x)|dt c x+y x+y |x − t| |t− 2 |
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Chapter 2. Boundary Integral Equations in H¨ older Spaces
(ii) The term I2 is estimated in a similar way. Thus,
T {χr (f −f (y))}(y) c |y|−β−α |x − y|α c |x|−β−α |x − y|α , because x and y are equivalent and x/2 < y < 2x. (iii) For |t − (x + y)/2| > r we have min(|t − x|, |t − y|)
1 |x − y|. 4
Therefore, properties (a) and (b) of K(x, t) (cf. (2.26)) imply |K(x, t) − K(y, t)| c
|x − y| . |x − t|2 (1 + |x − t|J )
Note that for 2|x − y| min(|x|, |y|) the product xy is negative. We have
I3 = [K(x, t) − K(y, t)]ηr (t)[f (t) − f (x)]dt
R (2.143) |K(x, t) − K(y, t)||f (t) − f (x)|dt . + t>x+ 14 |x−y|
t
We estimate |f (t) − f (x)|: α+β α+β |f (t) − f (x)| c x−α−β |(1 + x2 ) 2 f (x) − (1 + t2 ) 2 f (t)| + x−α−β |f (t)||(1 + t2 )
α+β 2
− (1 + x2 )
α+β 2
| .
(2.144)
Since f ∈ Eβα (R), the first term on the right-hand side of (2.144) does not exceed c x−α−β |x − y|α . For |t| > 2x we have (1 + x2 ) x2 < 1. < 2 t (1 + t2 ) Hence the second term on the right-hand side of (2.144) for such t is dominated by α+β (1 + t2 ) α+β 2 − (1 + x2 ) 2 | |x − t|α α+β 1 c x−α−β |x − t|α . c x−α−β (1 + t2 )−β/2 |x − t|α (1 + t2 ) 2 t−α x α 1 − |t|
x−α−β (1 + t2 )−β/2 |x − t|α
For |t| x/2 we find x−α−β |f (t)||(1 + t2 )
α+β 2
− (1 + x2 )
c x−α−β (1 + t2 )−β/2 (1 + x2 )
α+β 2
α+β 2
| 1 α 1 − | xt |
|x − t|α x−α
c x−α−β |x − t|α (1 + t2 )−β/2 (1 + x2 )β/2 .
2.7. Appendices
195
Finally, for x/2 < |t| < 2x we obtain |(1 + t ) 2
α+β 2
− (1 + x ) 2
α+β 2
x
α+β−1
| = (α + β) (1 + u2 ) 2 du
|t|
c (1 + x ) 2
α+β−1 2
|x − t| .
Therefore, x−α−β |f (t)||(1 + t2 )
α+β 2
− (1 + x2 )
cx−α−β (1 + t2 )−β/2 (1 + x2 )| c x−α−β (1 + x2 )α/2 Thus
|f (t) − f (x)|
α+β 2
|
α+β−1 2
|x − t| c x−α−β (1 + x2 )α/2
|x − t| x
|x − t|α c x−α−β |x − t|α . xα
c x−α−β |x − t|α for |t| > x/2 , c x−α−β |x − t|α (1 + x2 )β/2 (1 + t2 )−β/2 for |t| < x/2 .
(2.145)
Hence on the set {t : t > x + 14 |x − y|} we have |f (t) − f (x)| c x−β−α |x − t|α . Now we see that the first integral on the right-hand side of (2.143) does not exceed |x − t|α−2 dt c x−α−β |x − y| t−x> 14 |x−y| ∞
= c x−α−β |x − y|
1 4 |x−y|
uα−2 du = c x−α−β |x − y|α
and the second integral on the right-hand side of (2.143) is dominated by |y − t|α−2 dt c x−α−β |x − y| y−t> 14 |x−y|
+ (1 + x2 )β/2
|t|< x 2
(1 + t2 )−β/2 |x − t|α−J−2 dt = K1 + K2 .
(2.146)
The term K1 in (2.146) is majorized by c x−α−β |x−y|α , and K2 obeys the estimate −α−β 2 β/2 α−J−2 |x − y|(1 + x ) x (1 + t2 )−β/2 dt K2 c x |t| x
2 ⎧ 2 β−1/2 ⎪ (1 + x ) (1 + x−β+1 ) , 0 < β < 1, ⎪ ⎨ |x − y| α −α−β x x c (1 + x2 )−J/2 log x , β = 1, ⎪ x ⎪ ⎩ (1 + x2 )β−J−1/2 , β > 1,
c x−α−β |x − y|α .
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Chapter 2. Boundary Integral Equations in H¨ older Spaces
Thus we conclude that |I3 | c |x − y|α x−α−β . (iv) Now we show that |T χr (x)| c < ∞. Note that T χr (x) = K(x, t)χr (t)dt R = lim − K(x, t)ηr (t)dt + ε→0 a→∞
ε<|x−t|
= −T ηr (x) + lim
ε→0 a→∞
= −T ηr (x) +
1 2
R
x+a
K(x, t)dt
ε<|x−t|
[K(x, t) + K(x, 2x − t)]dt
x+ε
[K(x, x + t) + K(x, x − t)]dt = −T ηr (x) + O(1),
because of the property (c) of the kernel K(x, t). Let us represent T ηr (x) in the form [K(x, t) − K( x+y , t)]η (t)dt + K( x+y (2.147) r 2 2 , t)ηr (t)dt = J1 + J2 . R
R
We have
|J2 | =
1 2
∞ x+y
2 R
!
x+y K( x+y 2 , t) + K( 2 , x + y − t) ηr (t)dt
x+y x+y
x+y x+y
K(
2 , 2 + t) + K( 2 , 2 − t) dt = O(1)
and
|J1 | c |x − y|
dt
1 |t− x+y 2 |> 2 |x−y|
|t −
x+y 2 2 |
= c |x − y|
1 2 |x−y|
du = O(1) u2
by the condition b) of the theorem. Thus |I4 | = |f (x) − f (y)||T χr (x)| c |x − y|α |x|−α−β . (v) By the property (d) of the kernel K(x, t), |I5 | = |f (x)[T 1(x) − T 1(y)]| c x−β
|x − y| α c x−α−β |x − y| , |x|
because 2|x − y| < min{|x|, |y|}. In view of (i)–(v) and (2.142), |T f (x) − T f (y)| c x−α−β |x − y| . α
2.7. Appendices
197
(vi) Next, we show that |T f (x)| c |x|−β . Consider
∞
T f (x) =
K(x, t)f (t)dt =
− 23 x
+
−∞
0
We have
23 x
K(x, t)f (t)dt c
− 23 x
2 3x
−2x
2 3x
+
− 23 x
4 3x
∞
+
2 3x
K(x, t)f (t)dt .
4 3x
(1 + t2 )−β/2 dt |x − t|(1 + |x − t|J )
3 ⎧ 2/3 −β ⎪ t ⎪ ⎪ c x−β dt c x−β , 0 < β < 1 , ⎪ ⎪ 1 − t ⎪ −2/3 ⎪ ⎪ (2.148) ⎪ 2/3 ⎨ 1 1 2 −β/2 −β c cx , β = 1, (1 + t ) dt c ⎪ x(1 + xJ ) −2/3 x(1 + xJ ) ⎪ ⎪ ⎪ 2/3 ⎪ ⎪ ⎪ 1 ⎪c ⎪ (1 + t2 )−β/2 dt c x−J−1 c x−β , β > 1 , ⎩ x(1 + xJ ) −2/3
as well as
∞ 4 3x
K(x, t)f (t)dt c
∞
x−β
4 3x
t−β dt dt = c x−β |1 − t|
(2.149)
and
− 23 x
−∞
K(x, t)f (t)dt c
∞
(1 + t2 )−β/2 dt x+t 2/3 ∞ −β t = c x−β dt = cx−β . 2/3 1 + t
(2.150)
Besides,
4 3x 2 3x
K(x, t)f (t)dt
=
4 3x
x
K(x, 2x − t)f (2x − t)dt
x
4 3x
4 3x
K(x, t)f (t)dt +
|K(x, t) + K(x, 2x − t)||f (t)|dt
x
+
4 3x
x
c x−β + 0
|K(x, 2x − t)||f (2x − t) − f (t)|dt
1 3x
R
|K(x, t) + K(x, 2x − t)|dt
|K(x, x − t)||f (x − t) − f (x + t)|dt.
(2.151)
198
Chapter 2. Boundary Integral Equations in H¨ older Spaces
Here we used the estimate (2.145): |f (x + t) − f (x − t)| ctα |x − t|−α−β , proved above for 0 < t < 13 x. Then, by (2.151),
4x
3
c x−β K(x, t)f (t)dt |K(x, x + t) + K(x, x − t)|dt
2
R
3x
+
1 3x
(x − t)−α−β tα−1 dt c x−β + c x−β
1 3
0
0
(1 − t)−α−β dt = c x−β . t1−α
Combining this with (2.148)–(2.150), we obtain |T f (x)| c x−β . (vii) For 1 < y < x and |x − y| > y/2 we have |x − y| > x/3. Therefore, |xα−β T f (x) − |y α−β T f (y)| |x|α−β |T f (x)| |y|α−β |T f (y)| + α |x − y| |x − y|α |x − y|α |x|α−β |T f (x)| |y|α−β |T f (y)| 1 α 1 α + 3x 2y −β −β c |x| |T f (x)| + |y| |T f (y) | c
in view of (vi). The proof of Theorem 2.2.1 is complete.
Appendix B: Proof of Corollary 2.2.2 We need to prove only parts (iii) and (vi). Let α+ β > 0. For |t| < x/2 the second term on the right-hand side of (2.144) is dominated by c x−α−β
1 + t2 −β/2 1 + x2
|x − t|α α c x−α−β |x−t|α . (2.152) 1 − |t|/x
(1+x2 )(α+β)/2 x−α α+β
α+β
The expressions (1 + t2 ) 2 and (1 + x2 ) 2 are equivalent for x/2 |t| 2x. Hence the second term on the right-hand side of (2.144) does not exceed
x
α+β−1
c x−α−β (1 + t2 )−β/2
(1 + u2 ) 2 du
|t|
1 + t2 −β/2
|x − t| (1 + x2 )α/2 1 + x2 x −α−β 2 α/2 −α α cx (1 + x ) x |x − t| c x−α−β |x − t|α .
c x−α−β
(2.153)
2.7. Appendices
199
For |t| > 2x the second term on the right-hand side of (2.144) is majorized by
α+β
(1 + t2 ) α+β 2 − (1 + x2 ) 2
(1 + t ) |x − t| cx |x − t|α −α α+β c x−α−β (1 + t2 )−β/2 |x − t|α (1 + t2 ) 2 t−α 1 − x/|t| −α−β
2 −β/2
α
(2.154)
c x−α−β |x − t|α . Now, let α + β < 0. Then the estimates (2.152) and (2.154) should be replaced by c x−α−β (1 + t2 )−β/2 (1 + t2 )
α+β 2
1 α 1 − |t|/x
|x − t|α x−α
(2.155)
c x−α−β |x − t|α , and
α+β
(1 + t2 ) α+β 2 − (1 + x2 ) 2
(1 + t ) |x − t| cx |x − t|α α+β 1 α c x−α−β (1 + t2 )−β/2 |x − t|α (1 + x2 ) 2 t−α 1 − x/|t| 1 + x2 β/2 c x−α−β |x − t|α . 1 + t2 −α−β
2 −β/2
α
(2.156)
Since f ∈ Eβα (R), the first term on the right-hand side of (2.144) does not exceed c x−α−β |x − t|α . Thus |f (t)−f (x)| ⎧ −α−β cx |x − t|α for α + β > 0, ⎪ ⎪ ⎪ ⎨ −α−β |x − t|α , |t| < 2x , for α + β < 0, cx ⎪ ⎪ 1 + x2 β/2 ⎪ −α−β ⎩ cx , |t| > 2x , |x − t|α 1 + 1 + t2
for α + β < 0 .
The first integral on the right-hand side of (2.143) is dominated by c x−α−β |x − y|
∞
uα−2 du + c x−α−β
1 4 |x−t|
3 2x 3 2y
|t − x|α−1 dt c x−α−β |x − y|α .
The second integral on the right-hand side of (2.143) is majorized by −α−β
cx
|x − y|
∞
u
α−2
−α−β
x/2
du+x
1 4 |x−y|
(y − t)α−1 dt
y/2 −α−β
cx
|x − y| . α
(2.157)
200
Chapter 2. Boundary Integral Equations in H¨ older Spaces
Thus the term I3 in (2.142) satisfies the estimate |I3 | c |x − y|α x−α−β . Under the additional assumptions (2.27) on the kernel K(x, t) the function T f (x) admits the estimate
|(T f )(x)| =
=
3 2x 1 2x 3 2x
K(x, t)f (t)dt
K(x, t)f (t)dt +
x
K(x, 2x − t)f (2x − t)dt
x
3 2x
3 2x
|K(x, t) + K(x, 2x − t)||f (t)|dt
x
(2.158)
3 2x
+ x
c x−β
∞
|K(x, t) + K(x, 2x − t)|dt
−∞
1 2x
+
|K(x, 2x − t)||f (2x − t) − f (t)|
|K(x, x − t)||f (x − t) − f (x + t)|dt .
0
In view of the properties (c) and (a) of the kernel K(x, t) by (2.158) we have | T f (x)| c x−β
R
|K(x, x + t) + K(x, x − t)|dt +
c x−β + x−β
0
1 2x
(x − t)−α−β tα−1 dt
0 1 2
(1 − t)−α−β dt = c x−β . t1−α
The other parts of the proof remain unchanged.
Appendix C: To proof of Theorem 2.2.7 (i) With the notation of Theorem 2.2.7, we show that arg(p − w) with p ∈ S c (w− ) and w ∈ S+ obey (2.26). Let the images of Γ± be denoted by S± . The curves S± are the graphs of the functions η = ϕ± (ξ) with asymptotics ϕ± = −α∓ + o(1), ϕ± (ξ) = O(1/ξ) and ϕ± (ξ) = O(1/ξ 2 ) .
(2.159)
Let us introduce the notation K(ξ, u) = arg(p−w), where p = u+iv = u+iϕ+ (u), w = ξ + iτ = ξ + iϕ− (ξ). The property (a) results from
c Im (p − w)
, | arg(p − w)| = arctan
Re (p − w) |Re (p − w)|
p ∈ S c (w− ) .
2.7. Appendices
201
Let w1 and w2 ∈ S+ , and let |w1 − w2 |
ε min{|w1 |, |w2 |} , 2
where ε is introduced in Section 2.2.2. In order to prove the relation (b) in (2.26) we start with the estimate
Im (p − w ) Im (p − w )
1 2
− | arg(p − w1 ) − arg(p − w2 )|
Re (p − w1 ) Re (p − w2 )
1 1
− |Im (p − w1 )|
Re (p − w1 ) Re (p − w2 ) 1 |Im (p − w1 ) − Im (p − w2 )| + |Re (p − w1 )| |Im (w1 − w2 )| |Re (w1 − w2 )| + . |Re (p − w1 )Re (p − w2 )| Re (p − w2 ) Since ϕ± (ξ) = O 1/ξ , it follows from the mean value theorem that |Im (w1 − w2 )|
|Re (w1 − w2 )| . |Re (p − w1 )|
Therefore, for w1 and w2 under consideration we have |K(ξ1 , u) − K(ξ2 , u)| = | arg(p − w1 ) − arg(p − w2 )|
|Re (w1 − w2 )| . |Re (p − w1 )|2
We introduce more notation. Let γ1 (ξ) and γ2 (ξ) stand for the projections onto x-axes of the end points w and wr of the arc S c (w− ) with representations (1 − ε)ξ + O(1/ξ) and (1 + ε)ξ + O(1/ξ). Also let γ1 (ξ) − ξ and γ2 (ξ) − ξ be denoted by δ1 (ξ) and δ2 (ξ) respectively. Since |K(ξ, ξ + τ ) + K(ξ, ξ − τ )| |(ϕ+ (ξ + τ ) − ϕ+ (ξ − τ ))/τ | , we obtain |K(ξ, ξ + τ ) + K(ξ, ξ − τ )|dτ R
δ2 (ξ)
δ1 (ξ)
|ϕ+ (ξ + τ ) − ϕ+ (ξ − τ )|
dτ , (2.160) |τ |
where δ1 (ξ) and δ2 (ξ) are equivalent to −εξ and εξ. The numerator of the integrand admits the estimate 1
1 d
dt
ϕ+ (ξ + τ t)dt c |τ | |ϕ+ (ξ + τ ) − ϕ+ (ξ − τ )| =
−1 dt −1 (ξ + τ t)
τ d(ξ + t)
ξ + τ
= c
= c log
. ξ−τ −τ (ξ + t)
202
Chapter 2. Boundary Integral Equations in H¨ older Spaces
Hence by (2.160), |K(ξ, ξ + τ ) + K(ξ, ξ − τ )|dτ c R
δ2 (ξ)
δ1 (ξ)
ξ + τ
dτ
c,
log ξ − τ |τ |
which proves the inequality (c) in (2.26). Next, we prove (d). Let us start with the equality (1+ε)ξ+O(1/ξ) ! ϕ+ (ξ) − ϕ− (τ ) + π dτ . arctan K(ξ, τ )dτ = − ξ − τ R (1−ε)ξ+O(1/ξ)
(2.161)
The derivative in ξ of the integral on the left-hand side of (2.161) is equal to the sum of the derivative of the integrand and two more terms, whose sum can be written as −π + O 1/ξ , that is (1+ε)ξ+O(1/ξ) d d ξ−τ arctan dτ − π + O 1/ξ . K(ξ, τ )dτ = dξ R dξ ϕ (ξ) − ϕ (τ ) + − (1−ε)ξ+O(1/ξ) We have ϕ+ (ξ) − ϕ+ (u) + (u − ξ)ϕ+ (ξ) d ξ−τ arctan = dξ ϕ+ (ξ) − ϕ− (τ ) (ξ − u)2 + ϕ+ (ξ − ϕ− (u))2 ϕ+ (u) − ϕ− (u) + = J1 (ξ, u) + J2 (ξ, u) . (ξ − u)2 + (ϕ+ (ξ) − ϕ− (u))2 The term J1 (ξ, u) is estimated by the Taylor formula: |ϕ+ (ξ) − ϕ+ (u) + (u − ξ)ϕ+ (ξ)| 1 (ξ − u)2 c c 2. 2 2 (ξ − u) + (ϕ+ (ξ) − ϕ− (u)) ξ2 1 + (ξ − u)2 ξ Hence
(1+ε)ξ+O(1/ξ)
(1−ε)ξ+O(1/ξ)
c
J1 (ξ, u)du 2 2εξ + O 1/ξ = O 1/ξ . ξ
We may assume that |ϕ+ (ξ) − ϕ− (u)| (α+ − α− )/2 on S c (w− ), w ∈ S+ . For J2 (ξ, u) we can write (α+ − α− )/2 + O 1/ξ J2 (ξ, u) = 2 (ξ − u)2 + (α+ − α− )/2 + O 1/ξ (α+ − α− )/2 1 = . 2 + O 1/ξ 2 1 + (ξ − u)2 (ξ − u) + (α+ − α− )/2 Therefore, (1+ε)ξ+O(1/ξ)
(1+ε)ξ+O(1/ξ)
J2 (ξ, u)du = (1−ε)ξ+O(1/ξ)
(1−ε)ξ+O(1/ξ)
(α+ − α− )/2 2 du (ξ − u)2 + (α+ − α− )/2
(1+ε)ξ+O(1/ξ)
+ O(1/ξ) (1−ε)ξ+O(1/ξ)
du = π + O(1/ξ) . 1 + (ξ − u)2
2.7. Appendices
203
Thus we arrive at the required estimate d K(ξ, τ )dτ = O(1/ξ) as dξ R
ξ → ∞.
(ii) We show that the function (∂/∂np ) log |p − w|, where p = u + iv = u + iϕ− (u), w = ξ + iη = ξ + iϕ+ (ξ) and |u − ξ| < εξ obeys the relations (2.28). Let us start with the estimate (a). We have
(∂/∂np ) log |p − w| = (∂/∂sp ) arg(p − w) c (∂/∂u) arg(p − w)
|ϕ+ (ξ)(ξ − u) − ϕ+ (ξ) + ϕ+ (u)| |ϕ+ (u) − ϕ− (u)| + 1 + (ξ − u)2 1 + (ξ − u)2 1 1 , c c 1 + (ξ − u)2 1 − |ξ − u|
because |ϕ+ (ξ)(ξ − u) − ϕ+ (ξ) + ϕ+ (u)| c
(ξ − u)2 c and |ϕ+ (u) − ϕ− (u)| c. ξ2
Thus, the estimate (a) is proved. Let w1 and w2 belong to S+ and let |w1 − w2 | (ε/2) min{|w1 |, |w2 |}. In order to estimate the difference (∂/∂sp )[arg(p − w1 ) − arg(p − w2 )], we use the equality ϕ (u)(u − ξ) − ϕ− (u) + ϕ+ (ξ) ∂ ∂ Im (p − w) arg(p − w) = arctan = − . ∂u ∂u Re (p − w) |p − w|2 We have
∂
(arg(p − w1 ) − arg(p − w2 ))
∂sp
1 1
c
−
|ϕ− (u)(u − ξ1 ) − ϕ− (u) + ϕ+ (ξ1 )| |p − w1 |2 |p − w2 |2 1 (|ϕ (u)| |ξ2 − ξ1 | + |ϕ+ (ξ2 ) − ϕ+ (ξ1 )|) = J1 + J2 . +c |p − w2 |2 − Since there exists a constant c satisfying 1/c |p − w1 |/|p − w2 | c, it follows that
1 |w1 − w2 | 1
− .
c 2 2 |p − w1 | |p − w2 | |p − w2 |2
204
Chapter 2. Boundary Integral Equations in H¨ older Spaces
Using |ϕ+ (ξ2 ) − ϕ− (u) + ϕ− (u)(u − ξ2 )| + |ϕ+ (ξ2 ) − ϕ− (ξ2 )| |ξ − u|2 2 c + 1 c, ξ22 we obtain |J1 | c
|ξ1 − ξ2 | . 1 + |ξ2 − u|2
(2.162)
Applying the equivalence of ξ1 and ξ2 , by the mean value theorem and the estimate (2.159) for ϕ+ we find |ϕ− (u)| |ξ2 − ξ1 | + |ϕ+ (ξ2 ) − ϕ+ (ξ1 )| c Hence we conclude that |J2 | c
|ξ2 − ξ1 | c |ξ2 − ξ1 | . u
|ξ1 − ξ2 | . 1 + |ξ2 − u|2
(2.163)
The inequalities (2.162) and (2.163) imply the property (b). To show that (c) holds, we use the equality
γ2 (ξ)
γ1 (ξ)
∂ arg(p − w)du ∂u = arctan
ϕ− (γ1 (ξ)) − ϕ+ (ξ) ϕ− (γ2 (ξ)) − ϕ+ (ξ) + arctan , γ2 (ξ) − ξ γ1 (ξ) − ξ
where γ1 (ξ) = Re w = (1 − ε)ξ + O(1/ξ) and γ2 (ξ) = Re wr = (1 + ε)ξ + O(1/ξ). We estimate the expression
∂ ϕ− (γi (ξ)) − ϕ+ (ξ)
arctan
∂ξ γi (ξ) − ξ
∂ ϕ (γ (ξ)) (γi (ξ) − ξ)2 ∂ ϕ+ (ξ)
− i =
−
, (γi (ξ) − ξ)2 + (ϕ− (γi (ξ)) − ϕ+ (ξ))2 ∂ξ γi (ξ) − ξ ∂ξ γi (ξ) − ξ where i = 1, 2. Since ϕ± (ξ) = O(1) and ϕ± (ξ) = O(1/ξ), we see that ϕ (γi (ξ))γi (ξ) ϕ− (γi (ξ)) ∂ ϕ− (γi (ξ)) = − − (γi (ξ) − ξ) = O(1/ξ 2 ) . ∂ξ γi (ξ) − ξ γi (ξ) − ξ (γi (ξ) − ξ)2 The derivative (∂/∂ξ) ϕ+ (ξ)/(γi (ξ) − ξ) is estimated in a similar way. Therefore,
∂
∂ ξ2 c
log |p − w|dsp c O(1/ξ 2 ) ,
∂ξ S c (w− ) ∂np 1 + ξ2 1 + ξ2
ξ → ∞.
2.7. Appendices
205
Appendix D: To proof of Theorem 2.2.9 Making the change of variables p = 1/q, w = 1/z, p = u + iv, w = ξ + iη in the integral 1 dsq , σ(q)Re z−q Γc− (z) we find that it is equal to σ(1/p)Re S(w− )
1 w dsp , p−w p
p ∈ S c (w− ), w ∈ S+ .
Let us show that the function Re
1 w p−w p
obeys the conditions (a), (c), (d) in (2.26) and (b ) in (2.27). Clearly, (a) holds. Let us estimate the difference 1 w 1 w 1 2 − Re Re p − w1 p p − w2 p 1 w −w w2 w1 − w2 1 2 + Re = Re (w1 − p)(w2 − p) p p − w1 p & on S c (w1− ) S c (w2− ). Since u |ξ1 − u| on S c (w− ), we have
Re and
Re
|ξ1 − ξ2 | |ξ1 − ξ2 | 1 w1 − w2
|w1 − w2 | 1 c
p − w1 p |w1 − p| |p| |ξ1 − u|u |ξ1 − u|2
|ξ1 − ξ2 | |ξ1 − ξ2 | w1 − w2
c
c (w1 − p)(w2 − p) |(ξ1 − u)(ξ2 − u)| |ξ1 − u|2
under the condition that |ξ1 − ξ2 | (1/2) min{|ξ1 − u|, |ξ2 − u|}. Thus, (b ) holds as well. Next, we check the property (d). Since the end points w and wr of S c (w− ) are equivalent to (1 − ε)ξ and (1 + ε)ξ, and their representations can be differentiated, we see that the terms outside the integral in 1 w d 1 w d dsp = Re Re dsp dξ S c (w− ) p−w p dξ S c (w− ) p − w p are O 1/ξ . The derivative in ξ of the integrand is p dw d 1 w 1 . = 2 dξ p − w p (p − w) p dξ
206
Chapter 2. Boundary Integral Equations in H¨ older Spaces
Hence,
Re
d 1 w
dsp
S c (w− ) dξ p − w p
dw 1 p
dp
dp −1
= Re dp .
dξ S c (w− ) (p − w)2 p du du
(2.164)
Note, that p = u + iϕ− (u) on S c (w− ) and dp/du = 1 + iϕ− (u) = 1 + O(1/u). Therefore,
dp dp −1 1 1 p p−p
and
. =1+ =1+O =1+O p p u du du u Thus the right-hand side of (2.164) does not exceed
dw
dw 1
1 1
dp
dp + O
Re
Re 2 2 dξ S c (w− ) (p − w) dξ S c (w− ) (p − w) u γ2 (ξ)
dw 1
1 1 du 1
− c . Re
+c 2+1 u dξ wr − w w − w (u − ξ) ξ γ1 (ξ) Here γ1 (ξ) and γ2 (ξ) are projections of wr and w onto the x-axis, equivalent to (1 − ε)ξ and (1 + ε)ξ. We introduce the notation pt = (ξ+t)+iϕ− (ξ+t) and estimate the expression w 1 1 w + pt − w pt p−t − w p−t w w 1 1 1 w + = Re = J1 (t) + J2 (t) . − − pt − w p−t − w pt p−t − w pt p−t
Re
For J2 we have
w |w| |pt − p−t | w
1
−
Re p−t − w pt p−t |pt p−t | |p−t − w| t |ξ| |ξ| . c 2 c 2 2 2 1/2 |ξ − t | (t + 1) |ξ − t2 | Hence
δ2 (ξ)
δ1 (ξ)
|J2 (t)|dt
δ2 (ξ)
δ1 (ξ)
|ξ| dt = |ξ 2 − t2 |
δ2 (ξ)/ξ
δ1 (ξ)/ξ
dt c < ∞, 1 − t2
where δ1 (ξ) and δ2 (ξ) are equivalent to −εξ and εξ. We represent Re
1 1 + pt − w p−t − w
2.7. Appendices
207
in the form pt + p−t − 2w w ϕ− (ξ + t) + ϕ− (ξ − t) − 2ϕ− (ξ) w =i . (pt − w)(p−t − w) pt (pt − w)(p−t − w) pt This implies the estimate
|J1 (t)| |ϕ− (ξ + t) + ϕ− (ξ − t) − 2ϕ− (ξ)| Im
w
1
(pt − w)(p−t − w) pt w
t2 1
Re + Im w , c 2 ξ |(pt − w)(p−t − w)| pt pt
because |ϕ− (ξ + t) + ϕ− (ξ − t) − 2ϕ− (ξ)| c t2 /ξ 2 . In view of
Re w + Im w c pt pt and
1 c w
,
(pt − w)(p−t − w) pt 1 + t2 we obtain 1 t2 c 2. |J1 (t)| c 2 2 1+t ξ ξ Thus we arrive at
εξ+O(1/ξ)
−εξ+O(1/ξ)
|J1 (t)|dt c ,
ξ → ∞,
and conclude that (c) holds as well.
Appendix E: To proof of Theorem 2.2.12 (i) We show that the function 1 dp on S c (w± ) p − w du considered as a function of the variable u, satisfies the conditions in Corollary 2.2.2 of Theorem 2.2.1. The property (a) holds, because |u − ξ| |p − w|. For w1 = ξ1 + iϕ± (ξ1 ) and w2 = ξ2 + iϕ± (ξ2 ) we have
1
w1 − w2 1
−
=
p − w1 p − w2 (p − w1 )(p − w2 ) |ξ1 − ξ2 | |ξ1 − ξ2 | c on S c (w1± ) ∩ S c (w2± ) , |ξ1 − u| |ξ2 − u| |ξ1 − u|2 provided that |ξ1 − ξ2 | (1/2) min{|ξ1 − u|, |ξ2 − u|}. Hence the property (b ) holds.
208
Chapter 2. Boundary Integral Equations in H¨ older Spaces
Next, we check (c). We put pt = ξ + t + iϕ± (ξ + t). Representing the sum 1 1 + pt − w p−t − w in the form pt + p−t − 2w ϕ± (ξ + t) + ϕ± (ξ − t) − 2ϕ± (ξ) =i , (pt − w)(p−t − w) (pt − w)(p−t − w) we find
1
1 1
+
ϕ± (ξ + t) + ϕ± (ξ − t) − 2ϕ± (ξ)
pt − w p−t − w |pt − w| |p−t − w| 2 1 c t 2, c 2 ξ |pt − w| |p−t − w| ξ because |p±t − w| |t|. Thus εξ+O(1/ξ)
1
1 +
du c,
p−t − w −εξ+O(1/ξ) pt − w
ξ → ∞.
Let us check the property (d). We have γ2± (ξ) 1 + iϕ± (u) wr± − w dp du = = log , p − w p − w w ± − w c S (w± ) γ1± (ξ) where wr± and w ± are end points of the arcs S c (w± ) . Hence d wr± − w d γ2± (ξ) 1 + iϕ± (u) du = log dξ γ1± (ξ) p−w dξ w ± − w c d d 1 1 (wr± − w) − (w ± − w) , = wr± − w dξ w ± − w dξ ξ because wr± −w = εξ +O(1/ξ), w ± −w = −εξ +O(1/ξ), and since these equalities can be differentiated. (ii) Let us show that the function
p − w
−
K(y, v) = log
, p ∈ S c (w) , p − w+ satisfies the conditions of Theorem 2.2.1. Noting that
p − w
c |w+ − w− | |w+ − w− |
−
,
log 1 +
log
p − w+ |p − w+ | |p − w+ | |v − y| we conclude that (a) holds.
2.7. Appendices
209
In order to check that (b) holds, we estimate the product p − w+ p − w− p − w p − w+ −
assuming that Re w+ = y < y = Re w+
and |y − y | (1/2)|y | .
With this aim in view, we represent that product as (p − w− )(p − w+ ) − (p − w+ )(p − w− ) )(p − w ) (p − w+ − p(w+ − w+ ) + p(w− − w− ) + (w− w+ − w− w+ ) . =1+ (p − w+ )(p − w− )
1+
(2.165)
Consider each difference in the numerator of the right-hand side. By (2.72) the difference (1/z+ ) − (1/z+ ) can be written as 1 1 x − x − = (1 + r1 (x )) , z+ z+ x x where |r1 (x )| c x . Since x = 1/y + O(1/y 3 ), we have − w− = w+
where
1 1 − z = (y − y )(1 + r+ (y )) , z+ − c y
|r+ (y )|
as y → ∞ .
Using (2.73), we show in a similar way that − w− = (y − y )(1 + r− (y )) , w−
where |r− (y )|
c , y
as y → ∞ .
In view of (2.74) we have 1 z z− +
−
1 z z− +
=
x − x r(x ) , x x
where r(x ) is bounded as x → 0. Hence
w w − w w =
− + − +
1 z z− +
−
1 z z− +
c |y − y | .
210
Chapter 2. Boundary Integral Equations in H¨ older Spaces
Thus the numerator on the right-hand side of (2.165) admits the estimate
p(w+ − w+ ) + p(w− − w− ) + (w− w+ − w− w+ ) c |y − y | on S c (w ) ∩ S c (w ). Besides, on the intersection of arcs S c (w ) ∩ S c (w ) we have
1
1
c c
and
|y − v| + 1 . p − w+ |y − v| p − w+ Therefore,
p − w p − w
c |y − y |
− +
. 1 +
p − w p − w+ |y − v|(|y − v| + 1) − Finally,
p − w p − w
|y − y |
− +
log 1 + c
log
p − w p − w+ |y − v| |y − v| − on S c (w ) ∩ S c (w ). The last estimate implies
p − w p − w
|y − y |
− +
,
c
log
p − w+ p − w− |y − v|2 if
p ∈ S c (w ) ∩ S c (w ) ,
|y − y | (1/2) min{|y − v|, |y − v|} .
Thus K(y, v) obeys the condition (b). Next, we show that K(y, v) satisfies the condition (c). Putting pt = (y + t) + iϕ+ (y + t), we write the product pt − w− p−t − w− pt − w+ p−t − w+ in the form 1 + (w+ − w− )
(w+ − w− )2 pt + p−t − 2w+ + . (pt − w+ )(p−t − w− ) (pt − w+ )(p−t − w− )
Clearly, the last term has absolute value not exceeding c/t2 . Hence we only need to estimate the second term. Since ϕ+ (y) = −iα+ + o(1), we find that |pt + p−t − 2w+ | c on S c (w). Therefore,
p + p − 2w
c
t −t +
2 as t → ∞ . (pt − w+ )(p−t − w− ) t As a result we obtain
p − w p − w
c
t − −t −
1+ 2 .
pt − w+ p−t − w+ t
2.7. Appendices
211
This proves the estimate |t|>1
|K(y, y + t) + K(y, y − t)|dt c .
Hence, for |t| 1,
w − w
p − w
c
+
±t −
−
. log
log 1 +
log 1 + p±t − w+ p±t − w+ |t| Since the last logarithm is an integrable function on the interval (−1, 1), we have |K(y, y + t) + K(y, y − t)|dt c . |t|1
Thus the condition (c) is satisfied. Next, we check that the condition (d) holds for the function log
p − w+ dp , p − w− dv
p ∈ S c (w),
w ∈ S+ .
Using the equality log(p − w+ )dp S c (w)
= (wr − w+ ) log(wr − w+ ) − 1 − (w − w+ ) log(w − w+ ) − 1 , we obtain d dy
S c (w)
log(p − w+ )dp
(2.166) d d (wr − w+ ) log(wr − w+ ) − (w − w+ ) log(w − w+ ) . dy dy Since p = v + iϕ+ (v) on S c (w), where v ∈ γ1 (y), γ2 (y) , it follows that the right-hand side of (2.166) is equal to =
wr − w+ d ϕ+ (γ2 (y)) − ϕ+ (y) log(wr − w+ ) +i w − w+ dy d d ϕ+ (γ1 (y)) − ϕ+ (y) log(w − w+ ) + (wr − w ) −i dy dy d wr − w+ ϕ+ (γ2 (y)) − ϕ+ (γ1 (y)) log(wr − w+ ) +i = log w − w+ dy d wr − w+ d ϕ+ (γ1 (y)) − ϕ+ (y) log (wr − w ) . +i + dy w − w+ dy
log
(2.167)
212
Chapter 2. Boundary Integral Equations in H¨ older Spaces
The derivative of the integral
S c (w)
log(p − w− )dp
obeys a similar formula: d log(p − w− )dp dy S c (w) d wr − w− ϕ+ (γ2 (y)) − ϕ+ (γ1 (y)) log(wr − w− ) +i = log w − w− dy d wr − w− d ϕ+ (γ1 (y)) − ϕ− (y) log (wr − w ) . +i + dy w − w− dy
(2.168)
Taking the difference of the equalities (2.167) and (2.168), we have w − w w − w d p − w+ r +
− log dp = log dy S c (w) p − w− w − w+ wr − w− wr − w+ d ϕ+ (γ2 (y)) − ϕ+ (γ1 (y)) log +i dy wr − w− (2.169) d wr − w+ ϕ+ (γ1 (y)) − ϕ+ (y) log +i dy w − w+ d wr − w− ϕ+ (γ1 (y)) − ϕ− (y) log −i . dy w − w− Using the formula log
w+ − w− wr − w+ =1+ = 1 + O 1/y , wr − w− w− − wr
we find that the first term on the right-hand side of (2.169) has the representation w − w w − w w − w w − w r +
− r +
− = log = O 1/y . log w − w+ wr − w− wr − w− w − w+ Since wr − w− = O(1) and w − w− we finally conclude that
d ϕ+ (γ1 (y)) − ϕ− (y) = O 1/y , dy
d p − w+
c
log dp .
dy S c (w) p − w− y
Next, in order to show that the condition (d) holds for the kernel K(y, v), it suffices to check that the second term on the right-hand side in
p − w
p − w p − w
+ +
+ Re log dp = log
ϕ+ (v) dv , p ∈ S c (w) ,
− arg p − w− p − w− p − w−
2.7. Appendices
213
obeys the condition (d). Note that the terms outside the integral sign in d dy
γ2 (y)
γ1 (y)
arg(p − w± )ϕ+ (v)dv
have the order O 1/y , because arg is a bounded function and ϕ+ (v) = O 1/v . For the derivative of the integrand we have
d
1 1
, p ∈ S c (w) ,
arg(p − w± ) c 2 + dy y (y − v)2 + 1 since ϕ+ (y) = O 1/y 2 . Taking into account that the primitive function of −1 is bounded and that γ1 (y) and γ2 (y) are equivalent to y, we (y − v)2 + 1 obtain
d γ2 (y)
p − w+ c
as y → ∞. arg ϕ+ (v)dv
dy γ1 (y) p − w− y The proof is complete.
Chapter 3
Asymptotic Formulae for Solutions of Boundary Integral Equations Near Peaks A classical method for solving Dirichlet and Neumann boundary value problems for the Laplace equation is the representation of their solutions in the form of double layer potentials W σ and single layer potentials V τ . For the internal Dirichlet problem and for the external Neumann problem the densities of the corresponding potentials can be found from the boundary integral equations −πσ + T σ = g,
(3.1)
where T σ is the value of the potential W σ at a boundary point, and −πτ + Sτ = h,
(3.2)
where Sτ is the value of the normal derivative of the single layer potential V τ . In this chapter we develop an asymptotic theory of the equations (3.1) and (3.2) on contours with several peaks. We give a short qualitative description of our results concerning equations (3.1) and (3.2). We show that the number of linearly independent solutions of the homogeneous equation (3.1) in a certain class M is equal to the number of outward peaks. We prove that (3.1) is solvable in M provided that the right-hand side belongs to a certain class N of continuous functions with prescribed asymptotics near peaks. We give an example of the equation (3.1) with a continuous right-hand side on the contour with outward peak, which is unsolvable in the class M. In the presence of outward peaks we achieve the unique solvability of (3.1) reducing both classes of solutions and right-hand sides. These new smaller classes will be denoted by Mext and Next respectively.
216
Chapter 3. Asymptotic Formulae for Solutions of Integral Equations
We turn to the equation (3.2). It appears that the presence of peaks does not violate the uniqueness in the class M. If the contour has no outward peaks, the equation (3.2) is solvable in M for an arbitrary h ∈ N with zero mean value. If the contour has outward peaks, the solvability in M holds under the orthogonality of h to zeros of (3.1) from the class M. Therefore, for the contour which contains outward peaks, it is preferable to express a solution of the outward Neumann problem as the sum of V τ and a linear combination of explicitly written functions. The resulting integral equation proves to be solvable in M. We introduce some basic notation. Consider a plane simply connected domain Ω+ with compact closure bounded by a piecewise C ∞ -smooth contour Γ. Let Γ have the outward peaks en , 1 n N , and the inward peaks im , 1 m M . The set of all peaks will be denoted by Y . With each peak z0 we associate a Cartesian coordinate system, in which either Ω+ or its complementary domain Ω− are given by the inequalities κ− (x) < y < κ+ (x), 0 < x < δ, where κ± are C ∞ -functions on [0, δ] satisfying the conditions: κ± (0) = κ± (0) = 0, κ+ (0) > κ− (0). Thus we restrict ourselves to contours with peaks of the first-order tangency. This requirement is unimportant for the method, but facilitates calculations. The arcs {(x, κ ± (x)) : x ∈ [0, δ]} will be denoted by Γ± (z0 ). The above-mentioned classes M, Mext and N, Next of solutions and righthand sides respectively are defined as follows. By M we denote the class of C ∞ -functions σ on Γ\Y such that σ(z) = O (z − z0 )β(z0 ) , β(z0 ) > −1, for each peak z0 . The subset Mext of M is defined by the additional condition β(ep ) > −1/2 for outward peaks ep , p = 1, . . . , N . Let N denote the class of functions ϕ on Γ admitting the representation ϕ± (x) = xν(z0 ) ψ± (x)
on Γ± (z0 )
for each peak z0 . Here ψ± are C ∞ -functions on [0, δ], |ψ+ (0)| + |ψ− (0)| = 0, and ν(z0 ) > 0. A subclass of N with ν(ep ) > 1/2, p = 1, . . . , N will be denoted by Next . We briefly describe our approach of studying boundary integral equations under the assumption that Ω+ is bounded by a smooth curve Γ with a single singular point O. Under certain general conditions on the function ϕ there exist harmonic extensions u(i) onto Ω+ and u(e) onto Ω− of ϕ satisfying ϕ(z) =
1 lim 2π ε→0
{Γ:|ζ|>ε}
∂u(i) ∂nζ
(ζ) −
∂u(e) 1 dsζ + u(e) (∞) (ζ) log ∂nζ |z − ζ|
(3.3)
Chapter 3. Asymptotic Formulae for Solutions of Integral Equations
217
on Γ \ {O}. Let v (e) denote a solution of the Neumann problem in Ω− , vanishing at infinity, with the function ∂u(i) /∂n on Γ \ {O} in the boundary condition. We can choose v (e) so that for w = v (e) − u(e) + u(e) (∞) on z ∈ Γ \ {O} the equality ∂ 1 dsζ = −2π(ϕ(z) − u(e) (∞)) πw(z) − lim w(ζ) log (3.4) ε→0 ∂nζ |z − ζ| {Γ:|ζ|>ε}
holds. Solutions of the equations (3.1) and (3.2) are constructed by means of (3.3) and (3.4). Thus the function σ=
1 (e) (v − ϕ) 2π
(3.5)
is a solution of (3.1). A solution of (3.2) can be obtained as follows. Let us introduce the solution v (e) of the Neumann problem in Ω− with the boundary data ψ, vanishing at infinity, and the harmonic extension u(i) onto Ω+ of the restriction of v (e) to Γ \ {O}. Under sufficiently general assumptions on the function ψ, we can select v (e) and u(i) so that the density 1 ∂u(i) −ψ (3.6) τ= 2π ∂n satisfies (3.2). Now we are in a position to give a more precise account of our results. Let u(i) denote the solution of the internal Dirichlet problem D(i) Δu(i) = 0
in Ω+ , u(i) = g
on Γ, g ∈ N.
If we are seeking u(i) in the form of a double layer potential W σ(z), then the density σ is found from the integral equation (3.1) valid on Γ\Y . The kernel of the operator πI − W in the space M consists of linear combinations of the functions Re (1/ζk ), where ζk is the conformal mapping of Ω− onto the upper half-plane subject to the conditions ζk (ek ) = 0, Re ζk (∞) = 0 and Re (1/ζk )(z) = ± x−1/2 + O(1) on Γ± (ek ) in the local coordinate system (see Sect. 3.3). The solvability of (3.1) with a right-hand side g ∈ N in the class M is proved in the same section. There we also study the asymptotic behavior of the solution near the peak. We also deal with the equation (3.2) and the external Neumann problem N (e) , Δv (e) = 0 in Ω+ , (∂/∂n)v (e) = h on Γ, v(z) = O(1 + |z|−1 ), |z| → ∞, where h is a function from N. We are looking for the solution v (e) in the form v
(e)
(z) = V τ (z) +
N n=1
tn n (z),
(3.7)
218
Chapter 3. Asymptotic Formulae for Solutions of Integral Equations
where
n (z) = Re
(z − en )(z0 − en ) z0 − z
1/2 ,
z0 is a fixed point in Ω+ , and the branch of the square root has its positive real part in the upper half-plane. The boundary equation corresponding to the representation (3.7) takes the form −πτ + Sτ +
N k=1
tk
∂ k = h. ∂n
(3.8)
Its solution is a pair (τ, t), where t = (t1 , . . . , tN ). A uniqueness theorem for the equation (3.8) in the class of pairs (τ, t) with τ ∈ M and t ∈ RN and a theorem on solvability of (3.8) are proved in Section 3.4, where also asymptotic formulae for solutions near peaks are given. At the end of this section, we obtain the abovementioned information on the equation (3.2) from our previous results on (3.8). We prove that (3.2) is uniquely solvable in M for every h ∈ N subjected to the orthogonality conditions hds = 0 and h Re (1/ζk )ds = 0, k = 1, . . . , N. Γ
Γ
The integral equation for the outward Dirichlet problem D(e) and that for the inward Neumann problem N (i) are briefly discussed in Section 3.3.3 and 3.4.3. We shall use the notion of the finite part of a divergent integral introduced by Hadamard [1]. Let us define it for integrals of the form
h
I(ε) = ε
f (x) dx xr+1
where r is positive noninteger. If the function f has a sufficient number of derivatives at the point O, then there exist coefficients ck (0 k < r, k is an integer) such that the difference I(ε) −
[r]
ck εk−r
k=0
has a finite limit as ε → +0. This limit is what Hadamard called the finite part of the integral I(0) and denoted by the symbol
0
h
f (x) dx . xr+1
3.1. Preliminary facts
219
3.1 Preliminary facts 3.1.1 Asymptotics of a conformal mapping of a domain with outward peak onto a strip Let L+ and L− be two curves in the plane {w ∈ C : w = u + iv} given by the equations v = g± (u), u u0 , g+ (u) > g− (u) . Further, let L1 be a Jordan curve lying in the half-plane u u0 and connecting the initial points of the curves L+ and L− (the curve L1 may consist of one point). The contour L, composed of L+ , L− and L1 , divides the plane into two parts. The one that contains the domain {w : g− (u) < v < g+ (u), u > u0 } will be denoted by D and called the curvilinear semi-infinite strip. First, we look for an asymptotic representation at ∞ of the imaginary part of the mapping ζ(u, v) of D onto the strip Π = {ζ = τ +iν : |ν| < π/2} satisfying the condition Re ζ(∞) = ∞. The function ν = Im ζ is a bounded harmonic function equal to π/2 on an arc containing the curve L+ and to −π/2 on the complementary arc of the boundary curve containing L− . Suppose that the functions g+ (u), g− (u) have the asymptotic representations (k) g± (u) = α± u−k , u → ∞. (3.9) k0
We look for an asymptotic expansion of ν as a series in nonnegative powers of 1/u: ν(u, v) = pk (v)u−k . (3.10) k0
In the kth step we construct the polynomial pk (v) in such a way that the coefficient of u−k is compensated in the preceding discrepancy. As a result, the polynomials pk (v), k = 0, 1, . . . , are uniquely defined by the relations p0 (g± (u)) p0 (g± (u)) + p1 (g± (u))u−1 p0 (g± (u)) + p1 (g± (u))u−1 + p1 (g± (u))u−2 . . . . . . . . . . . . . . . . . . . . . . . (u
−k
pk (v))vv
+
(u
pk (v) pk−2 (v))vv
−k+2
= =
0, 0,
= = = . . k k
±π/2 + O(u−1 ), ±π/2 + O(u−2 ), ±π/2 + O(u−3 ), . . . . . . . . = =
0, 1, 2, 3, 4, . . . .
Here, the functions p0 , p1 and p2 are linear. In what follows we shall use the following explicit expression for p0 : p0 (v) = a + bv = −
π α+ + α− π + v. 2 α+ − α− α+ − α−
We set p1 (v) = α + βv and p2 (v) = ω + γv.
220
Chapter 3. Asymptotic Formulae for Solutions of Integral Equations
Lemma 3.1.1. Suppose that the asymptotic equality (3.9) can be differentiated three times. Then ν(u, v) has the representation (3.11) ν(u, v) = p0 (v) + p1 (v)u−1 + p2 (v)u−2 + O u−3 as u → ∞ , which can be differentiated once, i.e., νu (u, v) = −p1 (v)u−2 − 2p2 (v)u−3 + O u−4 , νv (u, v) = p0 (v) + p1 (v)u−1 + p2 (v)u−2 + O u−3 . Proof. According to (3.10), 5 Δ ν(u, v) − pk (v)u−k = O u−6 , k=0
ν(u, g± (u)) −
5
pk (g± (u))u−k = O u−6 .
k=0
We set W (u, ν) = u
4
ν(u, v) −
5
pk (v)u−k
(3.12)
k=0
and note that u4 Δ u−4 W = ΔW − 8u−1 ∂W /∂u + 20u−2 W . Let η ∈ C ∞ be a cut-off function vanishing for u < 1 and equal to 1 for u > 2. We set ηR (u) = η(u/R). By Λ0 we denote the restriction of the Laplace operator to the space W21 (D) and by ΛR we mean the operator defined on W21 (D) by −8u−1∂/∂u + 20u−2 . 1 ˚ (D) → W −1 (D) is invertible, the operator Since the operator Λ0 W 2 2 1 ˚ (D) → W −1 (D) is also invertible for large R. Hence for such Λ0 + ΛR W 2 2 R, the boundary value problem, "1 = F in D , W "1 = 0 on ∂D, Λ 0 + ΛR W (3.13) ˚ 1 (D) for any F in L2 (D). is uniquely solvable in W 2 Let χ be a cut-off function from C0∞ − 1, 1 , equal to 1 for −1/2 < u < 1/2 and let χτ (u) = χ(u − τ ), u ∈ R. From the well-known local estimate for solutions of elliptic boundary value problems and from the Sobolev embedding theorem (see [17], Theorem 1.4.5(e)), "1 are bounded in D. "1 and its gradient ∇W it follows that the function W
3.1. Preliminary facts
221
"2 we denote We set Q(u, v) = u4 Δ u−4 W (u, v) and q± = W (u, g± (u)). By W the function ηR (u) q− (u)(g+ (u) − v) + q+ (u)(v − g− (u)) /(g+ (u) − g− (u)), which belongs to W23 (D) and is bounded on the strip D together with its gradient. "2 belonging to W 1 (D) as F (u, v) in We take the function Q(u, v) − Λ0 + ΛR W 2 "=W "1 + W "2 is a solution of the boundary (3.13). Then, for sufficiently large R, W value problem 8 ∂ " 20 " = Q in D, W + ηR (u) 2 W u ∂u u " = 0 on L1 , " = η q± on L± , W W R
" − η (u) ΔW R
in W21 (D). This and (3.12) imply that the harmonic function " (u, v) + ν1 (u, v) = u−4 W
5
pk (v)u−k
k=0
satisfies the relation ν1 (g± (u)) = ±π/2 for sufficiently large u. We show that ν2 = ν −ν1 decays exponentially in D as u → ∞. Let W = ω(ζ) denote the conformal mapping of the right half-plane {ζ = η + iξ, η > 0} onto the curvilinear semi-infinite strip D, normalized by the condition ω(0) = ∞. The function ν2 ◦ ω(ζ) is harmonic and bounded in the right half-plane, and it vanishes on some interval of the imaginary axis containing the origin, say at {ξ : |ξ| < ξ0 }. Representing this function as the Poisson integral, we find that |ν2 ◦ ω(ζ)| c |ζ| . According to the Warschawski theorem on the asymptotics of the conformal mapping of the curvilinear semi-infinite strip (see Theorem 2.4.1), the mapping ζ = ω −1 (w), inverse to w = ω(ζ), has the representation log ω −1 (w) = −bw − β log w + O(1) for u = Re w → ∞. This implies |ν2 (w)| c exp(−bu/2) .
(3.14)
Thus ν(u, v) = ν1 (u, v) + O exp(−bu/2) . Hence ν(u, v) = p0 (v) + p1 (v)u−1 + p2 (v)u−2 + O u−3 ,
u → ∞.
222
Chapter 3. Asymptotic Formulae for Solutions of Integral Equations
We now show that the gradient of ν2 decays exponentially as u → ∞. The func˚ 1 (D) if R is large enough. After integration by parts and tion ηR ν2 belongs to W 2 application of the Schwarz inequality, we obtain c 2 2 2 |∇(ηR ν2 )| dudv |ν2 | dudv + |∇(ηR ν2 )| dudv . R R
(3.15)
Consider the function χτ ν2 which satisfies the boundary value problem Δu = 2∇χτ · ∇ν2 + ∇χτ ν2 in D, u = 0 on ∂D , for large positive τ . A local a priori estimate and the inequalities (3.14), (3.15) yield |ν2 |2 dudv c exp(−bτ /2) . χτ ν2 2 3 c W2 (D)
uτ /2
This and the Sobolev embedding theorem (see [17], Theorem 1.4.5(e))) imply max
|u−τ |<1/2
|∇ν2 (u, v)| exp(−bu/4)
for sufficiently large u, which proves the possibility of differentiation of the asymptotic decomposition of ν. The lemma is proved. Under the assumption that (3.9) can be differentiated four times, one can prove that the relation (3.11) can be differentiated twice. The real part τ (u, v) can be restored from the imaginary part of the conformal mapping ζ(u + iv) of the curvilinear semi-infinite strip D on to the strip Π. We have τ (u, v) = λ + bu = β log u − γu−1 + O u−2 , (3.16) where b, β and γ are coefficients of the polynomials p0 , p1 and p2 , and λ is an arbitrary constant. We set λ = β log b. Unifying (3.11) and (3.16), we find ζ(w) = bw + β log w + (β log b + ia) − (γ − iα)w−1 + O w−2 . The inversion of this expansion gives the conformal mapping of the strip Π onto the semi-infinite strip D: w = w(ζ) = b−1 ζ − β log ζ − ia + β 2 ζ −1 log ζ ! + (γb + i(aβ − αb))ζ −1 + o ζ −1 ,
Re ζ → ∞ .
(3.17)
3.1. Preliminary facts
223
The inversion u + iv = (x + iy)−1 maps the domain Ω+ with an outward cusp described in the introduction to this chapter onto the curvilinear semi-infinite strip D, bounded for large u by the curves L+ and L− defined by the functions (1) g± (u) = α± + α± u−1 + · · · , where (0)/2, α± = −κ∓
α+ > α− ,
and α± = −κ∓ (0)/6 . (1)
Using the expansion (3.17) and making the inversion, we find the following representation of the conformal mapping w(ζ) of the strip Π onto Ω+ : w(ζ) = b ζ −1 + bβζ −2 log ζ + iabζ −2 + βbζ −3 (log ζ)2 − βb(β − 2ia)ζ −3 log ζ − b[(a2 + bγ) + i(βa − αb)]ζ −3 + O(ζ −3 ) .
(3.18)
Separating the real and imaginary parts in (3.18), we find x = Re w(τ ± iπ/2) = b[τ −1 + βτ −2 log τ + β 2 τ −3 (log τ )2 − β 2 τ −3 log τ − b(γ − bα2∓ )τ −3 + o(τ −3 )], y = Im w(τ ± iπ/2)
(3.19)
= −b2 [α± τ −2 + 2βα± τ −3 log τ + bα± τ −3 + o(τ −3 )] . (1)
Taking the inverse of (3.19), we obtain τ = τ± (x) = bx−1 + β log(x−1 ) + β log b − (γ − bα2± )x + o(x) .
(3.20)
Finally, straightforward calculations give the representation |w (τ ± iπ/2)| = bτ −2 + 2bβτ −3 log τ − bβτ −3 + 3bβ 2 τ −4 (log τ )2 − 5bβ 2 τ −4 log τ − b(3bσ + b2 α2± − β 2 )τ −4 + o(τ −4 )
(3.21)
as τ → ∞.
3.1.2 Asymptotics of a conformal mapping of a domain with inward peak onto the upper half-plane Let Ω+ be the domain with inward peak described in the introduction to this chapter. By the transformation u + iv = (x + iy)1/2 , we map the plane, with the cut along the positive part of the real axis, onto the upper half-plane. The image "+ bounded in a neighborhood of the origin by the graph of of Ω+ is the domain Ω the function v = v(u) with the expansion (1)
(2)
v(u) = β± u3 + β± u4 + β± u5 + o(u5 ), (0)/4, β± = −κ± (0)/12. where β± = κ± (1)
u → ±0 ,
(3.22)
224
Chapter 3. Asymptotic Formulae for Solutions of Integral Equations
We denote by θ(ζ) the conformal mapping of the upper half-plane {ζ = "+ normalized by the condition θ(0) = 0. From the Kellogg η + iξ : ξ > 0} onto Ω + i0) has theorem (see Theorem 1.3.2) it follows that the function u(η) = Re θ(η the representation u(η) = aθ η + bθ η 2 + O |η|2+ε , 0 < ε < 1 , (3.23) as η → 0. Substituting (3.23) into (3.22), we obtain + i0) v(η) = Im θ(η
(1) = a3θ β± η 3 + (3a2θ bθ β± + a4θ β± )η 4 + O η|4+ε ,
(3.24) 0 < ε < 1.
The derivative θ (ζ) belongs to the Hardy space H1 in the upper half-plane. Therefore, u = −H(v ) + c, where H is the Hilbert transform. Lemma 3.1.2. Suppose that ϕ has the expansion ϕ(t) = a± t3 + b± t4 + O |t|4+ε , 0 < ε < 1,
t → ±0 ,
which can be differentiated once, and let ϕ (t) be integrable on the axis (−∞, ∞). Then the function H(ϕ ) admits the representation η ∞ dt 3(a+ − a− ) 2 1 ∞ dt η log |η| ϕ (t) − ϕ (t) 2 + H(ϕ )(η) = − π −∞ t π −∞ t π
η 2
∞ dt 4 η 4
∞ dt 3 −
ϕ (t) 3 + (b+ − b− )η log |η| −
ϕ (t) 4 + O |η 3+ε | . π −∞ t π π −∞ t Proof. For positive η we have 1 ∞ dt H(ϕ )(η) = ϕ (t) π −∞ η−t ! dt 1 2η dt 1 η + ϕ (η − t) − ϕ (η + t) ϕ (−t) = π 0 t π η+t ∞ 0 tdt 2 dt 2 ∞ ϕ o (t) 2 ϕ e (t) 2 + η , + π 2η η − t2 π 2η η − t2
(3.25)
where ϕ o (t) and ϕ e (t) are odd and even components of ϕ , respectively. Straightforward calculations yield ! dt 1 η 12 80 = − a+ η 2 − b+ η 3 + O η 3+ε , ϕ (η − t) − ϕ (η + t) π 0 t π 3π 2η (3.26) 3+ε 1 3 8 dt 4 2 3 = a− η log 3 − b− − log 3 η + O η . ϕ (−t) π 0 η+t π π 3
3.1. Preliminary facts
225
Further, using the equality (1 − t2 )−1 = 1 + t2 +
t4 , 1 − t2
we obtain 2 π
∞
(ϕ )o (t)
2η
∞
tdt − t2
η2
dt 3(a+ − a− ) 2 + η log η t π −∞ * +
∞
3 dt 2 1
6 + log 3 (a+ − a− ) −
ϕ (t) 3 η + π 2 t −∞ 1 56 − 2 log 3 (b+ + b− )η 3 + O η 3+ε . + π 3
=−
1 π
ϕ (t)
(3.27)
Finally, 2 η π
∞
(ϕ )e (t)
2η
η2
dt − t2
3 dt 1 6 − log 3 (a+ + a− )η 2 ϕ (t) 2 + t π 2 −∞ 4 1 8 + 2 log 3)(b+ − b− )η 3 + (b+ − b− )η 3 log η + π π
1 3
∞ dt − η
ϕ (t) 4 + O η 3+ε . π t
η =− π
∞
(3.28)
−∞
Substituting (3.26)–(3.28) into (3.25) and taking into account that dt dt 1 ∞ 1 ∞ ϕ (t) ϕ (−t) =− π −∞ η−t π −∞ |η| − t for η < 0, we obtain the required asymptotic expansion of H(ϕ ). The lemma is proved. We set ϕ = v in the last lemma. By (3.23) and (3.24) we have + i0) = aθ η + bθ η 2 − a3 (β+ − β− )π −1 η 3 log |η| + (d + ia3 β± )η 3 θ(η θ θ 4+ε 4 4 (1) 4 . − f η log |η| + (h + iaθ β± )η + O |η| Let us make the coefficient bθ vanish by a conformal mapping of the upper halfplane onto itself. After standard calculations we obtain the following expansion of
226
Chapter 3. Asymptotic Formulae for Solutions of Integral Equations
the conformal mapping of the upper half-plane onto Ω+ : θ(η + i0) = a2θ η 2 − 2π −1 a4θ (β+ − β− )η 4 log |η| + 2aθ (d + ia3θ β± )η 4 (1) −2aθ f η 5 log |η| + 2aθ (h + ia4θ β± )η 5 + O |η|5+ε , which can be differentiated twice. In particular, x = Re θ(η + i0) = a2θ η 2 − 2π −1 a4θ (β+ − β− )η 4 log |η| + 2aθ dη 4 − 2aθ f η 5 log |η| + 2aθ hη 5 + O |η|5+ε , 0 < ε < 1 .
(3.29)
Taking the inverse of (3.29), we find η = η± (x) 1/2 3/2 3/2 = ± a−1 − π −1 a−1 log(aθ x−1/2 ) − da−4 θ x θ (β+ − β− )x θ x 2+ε 2 −1/2 2 ∓ f a−5 , 0 < ε < 1/2 . ) ∓ ha−5 θ x log(aθ x θ x +O x
(3.30)
Finally, by straightforward calculation, we find the asymptotics of the modulus of the derivative of the conformal mapping |θ (η + i0)| = 2a2θ |η| − 8π −1 a4θ (β+ − β− )|η|3 log |η| + 8aθ d − 2π −1 a4θ (β+ − β− ) |η|3 + O |η|3+ε , 0 < ε < 1 .
(3.31)
3.2 The Dirichlet and Neumann problems in domains with peaks In the sequel we need several auxiliary facts concerning the solvability of the boundary value problems as well as asymptotic formulae for their solutions near peaks. For proofs of Lemmas 3.2.1–3.2.4 below see Appendices. Lemma 3.2.1. Let Ω+ have one outward peak at z = 0 and let g be infinitely differentiable on Γ\{z = 0}. Suppose that g has the asymptotic decomposition (±)
q0 xν +
n+1
(±) qk xk+ν + O xn+ν+2
k=1
for ν ≥ 0, ν = l/2, l ∈ Z, and
n+1 (±) (±) (±) q0,0 + q0,1 log x xν + Pk+2 (log x) xk+ν + O xn+2+ν− k=1 (±)
for ν ≥ 0, ν = l/2, where ε is a small positive number and Pj of degree j.
is a polynomial
3.2. The Dirichlet and Neumann problems for domains with peaks
227
We assume that these decompositions are at least three times differentiable. Then the problem D(i) has a solution with the following asymptotic properties. In a neighborhood of z = 0, (3.32) (z) + O |z|n+[ν] , u(z) = Re ϕ(ext) n where ϕ(ext) (z) = n
n
βk z k+ν−1
for ν = l/2,
(3.33)
k=0
or (z) = ϕ(ext) n
2
β0,r (log z)r z ν−1 +
r=0
n
Qk+1 (log z) z k+ν−1
for
ν = l/2. (3.34)
k=1
Here Qj is a polynomial of degree j and the coefficient β0 is β0 = −i
(+)
(−)
q0 − q0 2 . ν − 1 κ+ (0) − κ− (0)
The coefficients β0,r are expressed differently for ν = 1 and ν = 1. Namely, (+)
(−)
q0,1 − q0,1 2 , β0,2 = 0, ν − 1 κ+ (0) − κ− (0) (+) (−) q0,0 − q0,0 1 2 + Im β0,1 = −i ν − 1 κ+ (0) − κ− (0) 2
β0,1 = −i β0,0 for ν = 1, and
(+)
β0,0 = 0, β0,1 = −2i
(−)
q0,0 − q0,0
κ+ (0) − κ− (0)
(+)
, β0,2 = −i
(−)
q0,1 − q0,1
κ+ (0) − κ− (0)
for ν = 1. The relation (3.32) is differentiable at least once. Lemma 3.2.2. Let Ω+ have one inward peak at the point z = 0 and let g be a C ∞ function on Γ\{z = 0}. Suppose that g admits one of the following asymptotic representations near z = 0: (±)
q0 xν +
n+1
(±) qk xk+ν + O xn+ν+2
k=1
for ν > 0, and
n+1 (±) (±) (±) qk xk + O xn+2 q0,0 + q0,1 log x + k=1
228
Chapter 3. Asymptotic Formulae for Solutions of Integral Equations
for ν = 0. Assume that both representations can be differentiated at least three times. Then the problem D(i) has a solution with the properties: In a neighborhood of z = 0 the solution u can be written as , , ( ) ( ) (3.35) u(z) = Re ϕn (z) + ψn (z) + O |z|n+[ν] , where either
, ( )
ϕn (z) =
n
αk z k+ν
for
ν = l/2, l ∈ Z,
(3.36)
k=0
or , ( ) ϕn (z)
=
2
r
α0,r (log z) z ν
r=0 n
+
(3.37) Qk+1 (log z) z
k+ν
for ν = l/2,
k=1
and , ( ) ψn (z)
2(n+[ν])
= r0 z
1/2
+
Rk (log z) z k/2 .
k=2
Here Rk is a polynomial of degree [(k − 1)/2] and Qj is a polynomial of degree j. The coefficient α0 is given by q0 e−2πνi − q0 . sin 2πν are expressed differently for ν = 0 and for ν = 0. Namely, (+)
(−)
α0 = i
The coefficients α0,r
(+)
(+)
Re α0,0 = q0 , α0,1 = i
q0
(−)
− (−1)2ν q0 2π
, α0,2 = 0
for ν = 0, and (+)
Re α0,0 =
(+) q0,0 ,
α0,1 =
(+) q0,1
(−)
(+)
(−)
q0,0 − q0,0 q0,1 − q0,1 , α0,2 = i +i 2π 4π
for ν = 0. Besides,
1 1
∂ ds for 0 ν < 1/2 , u(i) (z) Im
πaθ Γ ∂n ζ(z) 1 ∂ 1 Im r0 = u(i) (z) Im ds for ν > 1/2 , πaθ Γ ∂n ζ(z)
Im r0 =
and Im r0 =
1 q + + q0− 1
∂ ds − 0 , u(i) (z) Im
πaθ Γ ∂n ζ(z) π
The relation (3.35) can be differentiated at least once.
for
ν = 1/2 .
3.2. The Dirichlet and Neumann problems for domains with peaks
229
Lemma 3.2.3. Let Ω+ have one outward peak at z = 0 and let h be a C ∞ -function on Γ\{z = 0} with the following asymptotic representation on the arcs Γ± (0): (±)
h0 xν +
n+1
(±)
ν > −2 .
hk xk+ν + O(xn+ν+2 ),
k=1
Suppose that this representation can be differentiated at least once and assume that V.P. hds = 0. Γ
has a solution v with the following properties: Then the problem N In a neighborhood of z = 0, (z) + ψn(ext) (z) + O |z|n+[ν] , v(z) − v(0) = Re ϕ(ext) n (e)
where either (z) = ϕ(ext) n
n
βk z k+ν
ν = l/2, l ∈ Z,
for
(3.38)
(3.39)
k=1
or (z) = ϕ(ext) n
2
β1,r (log z)r z ν+1
r=0 n
+
(3.40) Qk+1 (log z) z
k+ν
for
ν = l/2,
k=2 (ext)
and ψn
is given by ψn(ext) (z) = r0 z 1/2 +
n0
Rk (log z) z k/2 .
k=2
Here Rk is a polynomial of degree [(k − 1)/2], Qj is a polynomial of degree j and z0 is a fixed point in Ω+ . The coefficient β1 is given by (−)
β1 =
h0
(+)
(+)
h + h0 cos 2πν −i 0 . (ν + 1) sin 2πν ν +1
The coefficients β1,r are calculated differently for ν = −1 and ν = −1. They are: Im β1,0 = −
(+)
(+)
(−)
+ (−1)2ν h0 h0 h , β1,1 = 0 ν +1 2π(ν + 1)
, β1,2 = 0
for ν = −1 and (+)
Im β1,0 = 0, β1,1 = −ih0 , β1,2 =
(+)
h0
(−)
+ h0 4π
for ν = −1. The relation (3.38) can be differentiated once.
(3.41)
230
Chapter 3. Asymptotic Formulae for Solutions of Integral Equations
Lemma 3.2.4. Suppose that Ω+ has one inward peak. Let h be an infinitely differentiable function on Γ\{z = 0} admitting one of the following expansions on the arcs Γ± (0): n+1
(±)
hk xk+ν +
k=0
n0
(±)
Tk
(log x)x−1+k/2 + O(xγ+1 )
k=1
for ν > −1, ν = l/2, l ∈ Z, and
n+1 (±) (±) (±) h0,0 + h0,1 log x xν + Pk+1 (log x)xk+ν k=1
+
n0
(±)
Tk
(log x)x−1+k/2 + O(xγ+1 )
k=1 (±)
(±)
for ν > −1, ν = l/2. Here Tk are polynomials of degree [(k − 1)/2], Pj are polynomials of degree j, and n + ν < γ < n0 . Suppose that these expansions can be differentiated at least twice and that hds = 0. Γ
Then the problem N (e) has a solution admitting the following representation in a neighborhood of z = 0: , ( ) (3.42) v(z) − v(0) = Re ϕn (z) + O |z|n+[ν] , where
, ( ) ϕn (z)
=
n
αk z
k+ν
+
k=0
n0
Uk (log z) z −1+k/2
(3.43)
k=1
for ν = l/2, and , ( )
ϕn (z) =
2
r
α0,r (log z) z ν +
r=0
+
n k=1 n0
Qk+1 (log z) z k+ν (3.44) Uk (log z) z
−1+k/2
k=1
for ν = l/2. Here Uk are polynomials of degree [(k − 1)/2] and Qj are polynomials of degree j. The relation (3.42) can be differentiated at least once. The next assertions are corollaries of Lemmas 3.2.1–3.2.4.
3.2. The Dirichlet and Neumann problems for domains with peaks
231
Proposition 3.2.5. Let g be infinitely differentiable on Γ\Y . Suppose that g vanishes outside a small neighborhood of the peak ep and admits one of the following asymptotic representations near ep : (±)
q0 xν +
n+1
(±) qk xk+ν + O xn+ν+2
k=1
for ν ≥ 0, ν = l/2, l ∈ Z, and
n+1 (±) (±) (±) q0,0 + q0,1 log x xν + Pk+2 (log x) xk+ν + O xn+2+ν− k=1 (±)
for ν ≥ 0, ν = l/2, where ε is a small positive number and Pj are polynomials of degree j. Suppose that these representations are differentiable at least three times. Then the problem D(i) has a solution with the asymptotic properties: a) In a neighborhood of ep the following holds: n+[ν] , u(z) = Re ϕ(ext) p,n (z) + O |z − ep |
(3.45)
where either ϕ(ext) p,n (z) =
n
βk (z − ep )k+ν−1
for ν = l/2,
(3.46)
k=0
or ϕ(ext) p,n (z) +
n
=
2
β0,r (log(z − ep ))r (z − ep )ν−1
r=0
(3.47)
Qk+1 (log(z − ep )) (z − ep )
k+ν−1
for ν = l/2
k=1
with Qj being polynomials of degree j. b) In a neighborhood of er , r = p, u(z) = O |z − er |n+[ν] .
(3.48)
c) In a neighborhood of im we have
, ( ) u(z) = Re ψm,n (z) + O |z − im |n+[ν] ,
(3.49)
, ( )
where ψm,n is given by , ( ) ψm,n (z)
2(n+[ν])
=
Rk (log(z − im )) (z − im )k/2 .
k=1
Here Rk (t) are polynomials of degree [(k − 1)/2]. The equalities (3.45), (3.48), and (3.49) can be differentiated at least once.
232
Chapter 3. Asymptotic Formulae for Solutions of Integral Equations
Proposition 3.2.6. Let g be a C ∞ -function on Γ\Y . Suppose that g vanishes outside a small neighborhood of the peak ip and admits one of the following two asymptotic representations near ip : (±)
q0 xν +
n+1
(±) qk xk+ν + O xn+ν+2
k=1
for ν > 0, and
n+1 (±) (±) (±) q0,0 + q0,1 log x + qk xk + O xn+2 k=1
for ν = 0. We assume that both representations are differentiable at least three times. Then the problem D(i) has a solution with the following properties: a) In a neighborhood of er
u(z) = O |z − er |n+[ν] .
(3.50)
b) In a neighborhood of im , m = p,
, ( ) u(z) = Re ψm,n (z) + O |z − im |n+[ν] .
c) In a neighborhood of ip , , , ( ) ( ) u(z) = Re ϕp,n (z) + ψp,n (z) + O |z − ip |n+[ν] ,
(3.51)
(3.52)
where either , ( )
ϕp,n (z) =
n
αk (z − ip )k+ν
ν = l/2, l ∈ Z
for
(3.53)
k=0
or , ( )
ϕp,n (z) =
2
r
α0,r (log(z − ip )) (z − ip )ν
r=0
+
n
(3.54) Qk+1 (log(z − ip )) (z − ip )
k+ν
for ν = l/2
k=1
and , ( ) ψr,n (z)
2(n+[ν])
=
Rk (log(z − ir )) (z − ir )k/2 .
k=1
Here Rk are polynomials of degree [(k − 1)/2]. The equalities (3.50), (3.51), and (3.52) can be differentiated at least once.
3.2. The Dirichlet and Neumann problems for domains with peaks
233
Proposition 3.2.7. Let h be a C ∞ -function on Γ\Y , vanishing outside a small neighborhood of the peak ep , and having the following asymptotic representation in local coordinates on the arcs Γ± (ep ): (±)
h0 xν +
n+1
(±)
ν > −2 .
hk xk+ν + O(xn+ν+2 ),
k=1
Suppose that this decomposition can be differentiated at least once. Assume also that V.P. hds = 0. Γ
Then the problem N (e) has a solution v with the following asymptotic properties: a) In a neighborhood of ep (ext) n+[ν] , v(z) − v(ep ) = Re ϕ(ext) (z) + ψ (z) + O |z − e | p p,n p,n
(3.55)
where either ϕ(ext) p,n (z) =
n
k+ν
βk (z − ep )
for
ν = l/2, l ∈ Z,
(3.56)
k=1
or ϕ(ext) p,n (z) =
2
r
ν+1
β1,r (log(z − ep )) (z − ep )
r=0
+
n
(3.57)
Qk+1 (log(z − ep )) (z − ep )
k+ν
for
ν = l/2 .
k=2
Here Qj are polynomials of degree j. b) In a neighborhood of er , r = p,
(ext) (z) + O |z − er |n+[ν] . v(z) − v(er ) = Re ψr,n
(3.58)
c) In a neighborhood of im
v(z) − v(im ) = O |z − im |n+[ν] .
(ext)
Here ψr,n
(3.59)
is given by
(ext) (z) = ψr,n
n0
Rk (log(z − er )) (z − er )k/2 ,
r = 1, . . . , N ,
k=1
and Rk are polynomials of degree [(k − 1)/2]. The equalities (3.55), (3.58), and (3.59) can be differentiated at least once.
234
Chapter 3. Asymptotic Formulae for Solutions of Integral Equations
Proposition 3.2.8. Let h be infinitely differentiable on Γ\Y , vanishing outside a small neighborhood of the peak ip , and having one of the following asymptotic representations in local coordinates on the arcs Γ± (ip ): n+1
(±)
hk xk+ν +
k=0
n0
(±)
Tk
(log x)x−1+k/2 + O(xγ+1 )
k=1
for ν > −1, ν = l/2, l ∈ Z, and
n+1 (±) (±) (±) h0,0 + h0,1 log x xν + Pk+1 (log x)xk+ν k=1
+
n0
(±)
Tk
(log x)x−1+k/2 + O(xγ+1 )
k=1 (±)
(±)
for ν > −1, ν = l/2. Here Tk are polynomials of degree [(k − 1)/2], Pj are polynomials of degree j, and n+ν < γ < n0 . We assume that these representations can be differentiated at least once and that hds = 0. Γ
Then the problem N
(e)
has a solution v which can be represented as follows:
a) In a neighborhood of em
(ext) v(z) − v(em ) = Re ψm,n (z) + O |z − em |n+[ν] .
(3.60)
(ext)
The function ψm,n in (3.60) is defined by n0
(ext) ψm,n =
Rk (log(z − em )) (z − em )k/2 ,
k=1
where Rk are polynomials of degree [(k − 1)/2]. b) In a neighborhood of ir , r = p, v(z) − v(ir ) = O |z − ir |n+[ν] .
(3.61)
c) In a neighborhood of ip
, ( ) v(z) − v(ip ) = Re ϕp,n (z) + O |z − ip |n+[ν] ,
where we use the notation , ( )
ϕp,n (z) =
n k=0
αk (z − ip )k+ν +
n0 k=1
Uk (log(z − ip )) (z − ip )−1+k/2
(3.62)
3.3. Integral equations of the Dirichlet problem
235
for ν = l/2, and , ( ) ϕp,n (z)
=
2
α0,r (log(z − ip ))r (z − ip )ν
r=0
+
n
Qk+1 (log(z − ip )) (z − ip )k+ν +
k=1
n0
Uk (log(z − ip )) (z − ip )−1+k/2
k=1
for ν = l/2. Here Uk are polynomials of degree [(k − 1)/2] and Qj are polynomials of degree j. The equalities (3.60), (3.61), (3.62) can be differentiated once.
3.3 Integral equations of the Dirichlet problem 3.3.1 Homogeneous integral equation of the problem D (i) In the next two lemmas we suppose that a domain Ω+ is bounded by a closed, piecewise smooth curve Γ with a single peak at the origin. We suppose that near the point O either the domain Ω+ or the domain Ω− , complementary to it, is given in the Cartesian coordinates x, y by the inequalities κ− (x) < y < κ+ (x) and 0 < x < δ, where κ± are C ∞ functions on [0, δ] satisfying the conditions κ± (0) = κ± (0) = 0 and κ+ (0) > κ− (0). In the first case we speak of an inward peak and in the second of an outward peak. Lemma 3.3.1. Suppose that Ω+ is a domain with an outward peak. Then, for any function σ of the class M, the representation ∂ 1 dsq = −πσ± (x) + O(1) , z = x + iy ∈ Ω+ , σ(q) log ∂n |q − z| q Γ± holds, where σ± (x) = σ(x + iκ± (x)). Proof. For simplicity we assume that the domain Ω+ in a neighborhood of the origin is bounded by the curves y − α± x2 = 0, and α+ > α− . Suppose that the point p0 (x0 , y0 ) lies on Γ+ , and construct a new coordinate system with the origin at this point. We direct the axis Oξ along the tangent to Γ+ and the axis Oη along the outward normal to Γ+ . Let Γ+ (x0 ) denote the arc of the curve Γ+ which admits a Cartesian representation on the segment {|ξ| < x0 /2}. If q is a point of the arc Γ+ (x0 ) with coordinates (x, y) and (ξ, η), then, as x0 → 0, x = x0 + γ(ξ) = x0 + ξ(1 + 4α2+ x20 )−1/2 + O ξ 2 , η = η(ξ) = α+ ξ 2 (1 + 4α2+ x20 )−3/2 + O ξ 3 . We suppose that the point p ∈ Ω+ lies on the axis Oη, ν = |p − p0 |, r = |p − q|. Let nq be the vector of the outward normal to Γ+ at the point
236
Chapter 3. Asymptotic Formulae for Solutions of Integral Equations
q = (ξ, η) and let rpq be the vector with the starting point p and the endpoint q. Below cos(n q , 0ξ), cos(n q , 0η) and cos(rpq , 0ξ), cos(r pq , 0η) are unit vectors nq and rpq /r in the local coordinate system. We have cos(r pq , nq ) = cos(r pq , 0η) cos(n q , 0η) + cos(rpq , 0ξ) cos(nq , 0ξ) . 2 −1/2 Since cos(r , it may be assumed pq , 0η) = (ν +η)/r and cos(n q , 0η) = (1+(ηξ ) ) −1 cos(r that cos(n q , 0η) 1/2, and cos(nq , 0ξ) = O(|ξ|). Further, r pq , nq ) = O(1) for |x − x0 | > x0 /2 (x0 → +0), and it may be assumed that r2 > (|ξ|2 + ν 2 )/2. Then cos(r cos(r pq , nq ) pq , nq ) dsq = dsq σ(q) σ(q) + r r Γ± Γ+ (x0 ) Γ+ \Γ+ (x0 ) cos(r pq , nq ) | cos(n , 0η)|−1 dξ + O(1) , = σ ˜ (ξ) r |ξ|<x0 /2
where σ ˜ (ξ) = σ(x0 + γ(ξ)). Since cos(r pq , nq ) cos(nq , 0ξ) = ν + O(|ξ|), = cos(r pq , 0η) + cos(rpq , 0ξ) r cos(n cos(n q , 0η) q , 0η) it follows that
cos(r pq , nq ) dsq = ν σ(q) r Γ±
σ ˜ (ξ)r−2 dξ + O(1) .
|ξ|<x0 /2
We write the integral on the right-hand side in the form ! σ ˜ (ξ) − σ ˜ (0) r−2 dξ σ ˜ (ξ)r−2 dξ = |ξ|<x0 /2 |ξ|<x0 /2 r−2 dξ + σ(x) + σ(x0 ) − σ(x) |ξ|<x0 /2
(3.63) r−2 dξ . |ξ|<x0 /2
For |ξ| < x0 /2 we have
γ(ξ)
σ (x0 + τ )dτ
|˜ σ (ξ) − σ ˜ (0)| =
0 |ξ|(1+o(1)) c (x0 + τ )β−1 dτ c |ξ|xβ−1 , 0 0
where c does not depend on x0 . This implies
! −2
σ ˜ (ξ) − σ ˜ (0) r dξ c ν xβ−1 |ξ|(ξ 2 + ν 2 )−1 dξ
ν 0
|ξ|<x0 /2
|ξ|<x0 /2 log(x0 ν −1 ) = O(1) c ν xβ−1 0
3.3. Integral equations of the Dirichlet problem
237
as x0 → +0. We consider the last term on the right-hand side of (3.63). Since η = O ξ 2 and ν = O x20 , it follows that 2 2 −1 −2 2 −1 ξ + (ν + η(ξ)) ξ + ν2 r dξ = ν dξ = ν dξ ν |ξ|<x0 /2
|ξ|<x0 /2
|ξ|<x0 /2
2 −1 −1 ! ξ + (ν + η(ξ))2 dξ = π + O(x0 ) , − (ξ 2 + ν 2
−ν |ξ|<x0 /2
because
1 η(ξ) |ν + η(ξ)| + η(ξ)2 1
c −
ξ 2 + (ν + η(ξ))2 ξ2 + ν 2 (ξ 2 + ν 2 )(ξ 2 + (ν + η(ξ))2 ) |ξ| + 1 . c 2 ξ + ν2
Noting that |σ(x) − σ(x0 )| =
O(x30 )
0
we obtain
σ(q)
∂ 1 log dsq = −πσ+ (x) + O(1) . ∂nq |z − q|
σ(q)
∂ 1 dsq = −πσ− (x) + O(1) . log ∂nq |z − q|
Γ+
Similarly,
Γ−
= O x0 , |σ (x0 + τ )|dτ = O xβ+2 0
Lemma 3.3.2. Suppose that Ω+ is a domain with an inward peak and α = max |κ+ (0)|, |κ− (0)| . Then 1) for any function σ of the class M the asymptotic expansion ∂ 1 dsq = ∓πσ± (x) + O(1) σ(q) log ∂n |z − q| q Γ± holds for κ+ (x) < y < αx2 , and ∂ 1 dsq = ±πσ± (x) + O(1) σ(q) log ∂n |z − q| q Γ± holds for −αx2 < y < κ− (x); 2) for any function σ of the class M the representation δ σ± (τ ) ∂ 1 dsq = ∓ dτ + O(1) σ(q) log ∂sq |z − q| Γ± 0 x−τ holds for −αx2 < y < αx2 .
(3.64)
238
Chapter 3. Asymptotic Formulae for Solutions of Integral Equations
Proof. The assertion 1) is proved in the same way as Lemma 3.3.1. It is enough to prove 2) under the assumption that α− (x) < y < α+ (x). Suppose that the point p0 (x0 , y0 ) lies on the curve Γ+ . In a new coordinate system with the origin at (x0 , y0 ), we direct the axis Oξ along the tangent to Γ+ , and the axis Oη along the outward normal to Γ+ . If q is a point at Γ+ with coordinates (x, y) and (ξ, η), we have, as x0 → 0, x = x0 + γ(ξ) = x0 + ξ(1 + 4α2+ x20 )−1/2 + O(ξ 2 ), η = η(ξ) = α+ ξ 2 (1 + 4α2+ x20 )−3/2 + O(ξ 3 ) . Suppose that a point p in Ω+ belongs to the axis Oη. Let nq and τq stand for the outward normal vector and the tangent vector to Γ+ at the point q. We set r = |p − q|, rpq is the vector with the starting point p and the endpoint q, rpq /r = cos(r pq , 0ξ), cos(r pq , 0η) and τq = cos(τq , 0ξ), cos(τ q , 0η) . We have cos(r pq , 0τ ) = cos(r pq , 0η) cos(τ q , 0η) + cos(rpq , 0ξ) cos(τq , 0ξ) . Since cos(τ q , 0η) = − cos(nq , 0ξ) and cos(τq , 0ξ) = cos(n q , 0η), we see that cos(r ∂ 1 cos(r pq , 0η) pq , 0ξ) cos(n cos(n log = − q , 0ξ) + q , 0η), ∂sq r r r
−
where cos(n q , 0ξ) = O(|ξ|) and cos(n q , 0η) = O(1). Hence cos(n cos(n q , 0ξ) q , 0ξ) = O(1) and dsq = O(1) . cos(r σ(q) cos(r pq , 0η) pq , 0η) r r Γ+
Therefore, −
σ(q)
Γ+
1 ∂ log dsq = ∂s r
σ(q)
cos(r pq , 0ξ) cos(n q , 0η)dsq + O(1) r
Γ+
δ =
σ+ (x)
cos(r pq , 0ξ) dx + O(1) . r
0
Omitting details, we write the integral on the right-hand side as δ
cos(r pq , 0ξ) dx = σ+ (x) r
0
δ σ+ (x) 0
δ = 0
cos(r p0 q , 0ξ) dx + O(1) |rp0 q |
δ cos(r 1 dx p0 q , 0ξ) − dx + σ+ (x) σ+ (x) + O(1) , |rp0 q | x − x0 x − x0 0
where rp0 q = q − p0 . Since the first term on the right-hand side is a bounded function, we obtain (3.64).
3.3. Integral equations of the Dirichlet problem
239
Further we assume that Γ is the same as in introduction to this chapter. Lemma 3.3.3. Let g coincide with the restriction of a C ∞ -function defined on R2 to Γ. Then the integral equation of the problem D(i) , −πσ + W σ = g, has a solution in the class M which is bounded in a neighborhood of each point ek , k = 1, . . . , N . Proof. Suppose that ek coincides with the origin. Then g has the representation (2)
g(z) = c(0) + c(1) x + c± x2 + · · · on the arcs Γ± (0) = Γ± . By Proposition 3.2.5, the complex potential of the problem D(i) can be written in the form ϕ(z) = α + βz + γz 2 + · · · , where (2)
α = c(0) , Im β = −2
(2)
c + − c− , Re β = c(1) , κ+ (0) − κ− (0) (2)
Re γ = 2
(2)
α+ c− − α− c+ . κ+ (0) − κ− (0)
The normal derivative ∂(Re ϕ)/∂n admits the asymptotic decomposition ∂Re ϕ ∂Im ϕ = = ∓Im β ± d± x + · · · . ∂n ∂s Therefore, the boundary function h of the problem N (e) has the representation h(z) = ±Im β ± d± x + · · · on the arcs Γ± . By Proposition 3.2.7, the holomorphic function ϕ(z), whose real part is a solution of the problem N (e) , has the form ϕ(z) = c0 + c1 z 1/2 + (β1,1 log z + β1,0 )z + · · · . From the expression (3.41) for β1,1 it is clear that β1,1 = 0, i.e., ϕ(z) = c0 + c1 z 1/2 + β1,0 z + · · · . Hence the function σ = (Re ϕ − g)/2π admits the estimate σ(z) = O(1)
240
Chapter 3. Asymptotic Formulae for Solutions of Integral Equations
on the arcs Γ± (en ). By Propositions 3.2.6 and 3.2.8 we have σ(z) = δm x−1/2 + O(1) on the arcs Γ± (im ), m = 1, . . . , M . It follows from (3.5) that σ is a solution of the integral equation (3.1) of the problem D(i) with required properties. Let ζ(z) be the conformal mapping of the domain Ω− onto the upper halfplane normalized by the conditions Re ζ(∞) = 0 and ζ(0) = 0. Theorem 3.3.4. Let σ ∈ M be a solution of the homogeneous integral equation of the problem D(i) , −πσ + W σ = 0. Then σ=
N
ck Re (1/ζk ).
k=1
Proof. (i) Let σ be a solution of the homogeneous equation in the class M. Consider C ∞ -functions χk , k = 1, . . . , N , such that χk = 1 in a small neighborhood of ek with supp χk ∩ supp χn = ∅, k = n, and im ∈ supp χk , m = 1, . . . , M . Since (πI − W )(σχj ) = −(πI − W )(1 − χj )σ , it follows by Lemma 3.3.3 that the equation (πI − W )σj = −(πI − W )(σχj ) has a solution σj belonging to M and bounded in a neighborhood of each point ek , k = 1, . . . , N . The function θj = σj +σχj is a solution of the homogeneous integral equation, bounded in a neighborhood of each point ek , k = 1, . . . , N, k = j. By W± θj we denote the double layer potential in domains Ω+ and Ω− respectively. According to the statement a) of Lemma 3.3.1, we have W+ θj (z) = −π (θj )+ + (θj )− (x) + O(1), z = x + iy → ej , in an appropriate coordinate system. Since the boundary values of W+ θj vanish on Γ\Y , we see that (θj )+ + (θj )− (x) = O(1) as z → ej . This implies that W+ θj is bounded in a neighborhood of each ek , k = 1, . . . , N . (ii) We fix an integer m ≥ 1 not exceeding M and choose a coordinate system so that, in a neighborhood of im , the domain Ω− is defined by {κ− (x) < y < κ+ (x), 0 < x < δ}
3.3. Integral equations of the Dirichlet problem
241
with functions κ± described above. By the statement 1) of Lemma 3.3.2, the potential W+ θj (z) is equal (up to a sign) to π[(θj )+ − (θj )− ](x) + O(1) on the set # $ Ω+ ∩ z = x + iy : either κ+ (x) < y < αx2 or − αx2 < y < κ− (x) ,
where α = max (|κ+ (0)|, |κ− (0)|). This implies that W+ θj is bounded on the set Ω+ ∩ {z : −αx2 < y < αx2 }. Further, let χ be a cut-off function on the plane with small support, and let χ = 1 near the origin. We introduce the function σe on the arcs Γ± (im ) as σe (x) = [(θj )+ (x) + (θj )− (x)]/2. ˜ + (χσe ) are bounded on By Lemma 3.3.1, W+ (χσe ) and its harmonic conjugate W + 2 2 the set Ω ∩ {z : −αx < y < αx }. Besides, using explicit expressions for these functions, one can show that ˜ + (χσe )(z) = O(|z|−2 ) W+ (χσe )(z) = O(|z|−2 ), W in Ω+ as |z| → 0. By the Phragm´en-Lindel¨ of principle (cf. [39], p. 262), the holomorphic func˜ + (χσe ) is bounded near the origin. Since tion W+ (χσe ) + iW (θj )+ (x) − (θj )− (x) = O(1) as x → 0, it follows that (W+ θj )(z) is also bounded near the origin. Thus the potential is bounded in a neighborhood of each point im , m = 1, . . . , M , and hence it is bounded in Ω+ . Noting that W+ θj is a harmonic function vanishing on Γ\Y , we ˜ + θj is a constant. see that W+ θj = 0 in Ω+ . Therefore, its conjugate W For each n, n M , by Lemma 3.3.2,
δ
[((θj )+ − (θj )− ) (x + τ ) − ((θj )+ − (θj )− ) (x − τ )]
0
dτ = O(1). τ
Since in Ω− , ˜ − θj )(z) = − (W 0
δ
[((θj )+ − (θj )− ) (x + τ ) − ((θj )+ − (θj )− ) (x − τ )]
dτ + O(1), τ
˜ − θj is a bounded function in a neighborhood of as x → +0, we conclude that W each point im , m = 1, . . . , M .
242
Chapter 3. Asymptotic Formulae for Solutions of Integral Equations
(iii) In a neighborhood of each ek , k = 1, . . . , N, k = j, the function ˜ + θj = c. Therefore, by Lemma (W− θj )(z) is bounded. Furthermore, we have W ˜ − θj )(z) is bounded on the set 3.3.1, the function (W $ # Ω− ∩ z : −αx2 < y < αx2 in an appropriate coordinate system, where α = max(|κ+ (0)|, |κ− (0)|). Since ˜ − θj )(z) = O(|z|−2 ), the Phragm´en-Lindel¨ (W of principle, applied to the function ˜ W− θj + iW− θj , implies the boundedness of the potential in each neighborhood in ˜ − θj is bounded inside a small neighborhood of ek . question. Thus W ˜ + θj be equal to a constant C in Ω+ . Consider the function (iv) Let W ˜ − θj (z) − C . W (z) = W− θj (z) + i W It is holomorphic in Ω− and its imaginary part vanishes on Γ\Y . half-plane Let z = λj (ζ), ζ = η + iξ, be the conformal mapping of the upper onto Ω− , inverse of ζ = ζj (z). The imaginary part v(ζ) of W λj (ζ) = u(ζ)+ iv(ζ) vanishes at the boundary of the upper half-plane except for the points ζj (T ). By v˜ we denote the odd extension of the function v. Clearly, v˜ is harmonic on the punctured complex plane C\{0} and bounded at infinity. Therefore, v˜ can be expanded as v˜(reiγ ) = dk r−k sin kγ. k≥1
The conjugate function u˜ has the expansion dk r−k cos kγ. u ˜(reiγ ) = k≥0
Using the relation
W λj (ζ) = O(|ζ|−4 ) as ζ → 0,
we see that the function u ˜(reiγ ) + i˜ v (reiγ ), which is holomorphic in C\{0}, has a pole at the origin. Note that u ˜ equals −2π(θj ◦ λj )(η + i0), η ∈ R, at the boundary of the upper half-plane. Hence u ˜ = O(|η|2β ) as η + i0 → 0, where β > −1. Then u ˜(ζ) = d0 + d1 r−1 cos γ = d0 + d1 Re (1/ζ) and, consequently, θj (z) = −(2π)−1 d0 + d1 Re 1/ζj (z) , z ∈ Γ. Since a non-zero constant does not satisfy the homogeneous integral equation of the problem D(i) , it follows that θj (z) = −(2π)−1 d1 (Re 1/ζj )(z) = cj (Re 1/ζj )(z).
3.3. Integral equations of the Dirichlet problem
243
(v) We set σ ˜=σ−
N
σχj
1
and obtain σ=σ ˜+ The density σ ˜−
N
1N
σχj = σ ˜−
N
1
j=1
σj +
N
1
cj (Re 1/ζj ).
1
σj is a solution of the homogeneous equation
N N σ− σj = (πI − W )˜ (πI − W )σj (πI − W ) σ ˜− 1
1
= (πI − W )˜ σ+
N
(πI − W )(σχj )
1 N = (πI − W ) σ ˜+ σχj = (πI − W )σ = 0. 1
The function σ ˜−
1N
σj is bounded in a neighborhood of each ek , k = 1, . . . , N . 1N Repeating the arguments used in (ii) and (iii) for the function σ ˜ − j=1 σj , we prove that j=1
N W+ σ ˜− σj = 0 in Ω+ j=1
and N W− σ ˜− σj = 0 in Ω− . j=1
The limit relations for the double layer potential imply σ ˜−
N
σj = 0.
j=1
Thus σ=
N
cj Re 1/ζj .
1
The theorem is proved.
The next result follows directly from Theorem 3.3.4. Corollary 3.3.5. The homogeneous integral equation of the problem D(i) has only the trivial solution in the class Mext .
244
Chapter 3. Asymptotic Formulae for Solutions of Integral Equations
3.3.2 Solvability of the integral equation of the problem D (i) Theorem 3.3.6. Let g be a function from N. Then the boundary equation −πσ(p) + W σ(p) = g(p), p ∈ Γ\Y,
(3.65)
has a solution in the class M with the following representation in local coordinates: a) on the arcs Γ± (en ), σ(z) = ±(β0 + β1 log x)x−1+ν(en ) + O(1);
(3.66)
b) on the arcs Γ± (im ), σ(z) = (α0 + α1 log x)−1+ν(im ) + δx−1/2 + O(1).
(3.67)
Moreover, the space of solutions of the homogeneous equation is N -dimensional. Proof. Consider the single layer potential V σ(z) = V.P. σ(q) log Γ
1 dsq . |z − q|
Let u(i) and u(e) be solutions of the problems D(i) and D(e) . By Proposition 3.2.5, the holomorphic function ϕ(i) , whose real part is a harmonic extension of u(i) , admits the following decomposition in local variables: ϕ(i) (z) = β0,0 + β0,1 log z z ν−1 + β1 z ν + O(z) in a neighborhood of en . Here β0,1 = 0 only for ν = 1. Hence (±)
(∂u(i) /∂n)(z) = ±β0 xν−2 + β1 xν−1 + O(1) on the arcs Γ± (en ). In a neighborhood of im one has ϕ(i) (z) = α0,0 + α0,1 log z)z ν + δ0 z 1/2 + O(z) in a local coordinate system. Here α0,1 = 0 only for ν = l/2, l ∈ Z (cf. Proposition 3.2.6). This implies the decomposition (∂u(i) /∂n)(z) = (α0,0 + α0,1 log x)xν−1 + δ0 x−1/2 + O(1) (±)
(±)
(±)
on the arcs Γ± (im ). Here α0,1 = 0 only for ν = l/2. Then g(p) =
1 V 2π
∂u(i) ∂u(e) − ∂n ∂n
(p) + u(e) (∞), p ∈ Γ\Y.
3.3. Integral equations of the Dirichlet problem
245
Let v denote a solution of the problem N (e) : Δv = 0 in Ω− , ∂v/∂n = ∂u(i) /∂n on Γ\Y and let v vanish at infinity. In a neighborhood of en the holomorphic function ϕ, whose real part is the solution v, has the form ϕ(z) = β1
(z − en )(z0 − en ) z0 − z
ν−1
+ δ0
(z − en )(z0 − en ) z0 − z
−1/2 + O(1)
for ν = l/2 and ν−1 (z − en )(z0 − en ) (z − en )(z0 − en ) z0 − z z0 − z −1/2 (z − en )(z0 − en ) + γ0 + O(1) z0 − z
ϕ(z) =
β1,0 + β1,1 log
for ν = l/2. Therefore, by Proposition 3.2.7, v(z) = ± β0 xν−1 ± γ0 x−1/2 + O(1) for ν = l/2 and v(z) = ± (β0 + β1 log x)xν−1 ± γ0 x−1/2 + O(1) for ν = l/2 in a local coordinate system on the arcs Γ± (en ). By Proposition 3.2.8, ν−1 (z − im )(z0 − im ) (z − im )(z0 − im ) ϕ(z) = α0,0 + α0,1 log z0 − z z0 − z −1/2 (z − im )(z0 − im ) + δ0 + O(1) z0 − z in a neighborhood of im . Hence v(z) = (α0 + α1 log x)xν−1 + δx−1/2 + O(1) in a local coordinate system on the arcs Γ± (im ). It follows from the integral representation for the harmonic function w = v − u(e) + u(e) (∞) in Ω− and from the formula for limit values of the double layer potential that i ∂u ∂u(e) πw − W w = −V − = −2π(g − u(e) (∞)) ∂n ∂n
246
Chapter 3. Asymptotic Formulae for Solutions of Integral Equations
on Γ\Y . Consequently, the density σ = (2π)−1 (w − u(e) (∞)) = (2π)−1 (v − g) is a solution of the equation (3.65). By Theorem 3.3.4, the functions of the form N σ = (2π)−1 v − g + ck Re 1/ζk 1
are solutions of (3.65) in the class M.
Explicit formulae for the coefficients α0 , α1 , β0 , β1 , and δ in (3.66), (3.67) were given in [19]. We can summarize Theorem 3.3.4 and Theorem 3.3.6 in the following statement. Theorem 3.3.7. Let g ∈ Next . Then the equation (3.65) is uniquely solvable in Mext . Remark 3.3.8. The above considerations allow one to conclude that the boundary equation (3.65) is applicable to solving the problem D(i) in domains with peaks and with boundary data in the class N. We look for the solution u(i) of the problem D(i) as the sum of an explicitly written harmonic function u0 and of the double layer potential W σ with unknown density σ. By Propositions 3.2.5, the solution of the problem D(i) admits the representation: u(i) (z) = Re ϕ(ext) (z) + O |z − ep |1+[ν(ep )] p in a neighborhood of ep , where (z) = (β0,0 + β0,1 log(z − ep )) (z − ep )ν(ep )−1 + β1 (z − ep )ν(ep ) . ϕ(ext) p
We set u0 (z) = Re
N
' (ext) (ext) pk (z) ϕk (z)
,
k=1 (ext)
where pk
are interpolation polynomials such that (ext)
pk
(ext)
(ek ) = 1, Dzn pk
(ext)
Dzn pk
(ek ) = 0, n = 1, . . . , 1 + [ν(ek )],
(er ) = 0, n = 0, . . . , 1 + [ν(er )], r = k,
(see [36], Ch. I). Then
(u(i) − u0 )(z) = O |z − z0 |1+[ν(z0 )]
in a neighborhood of each outward peak z0 . By Theorem 3.3.7, the boundary integral equation (3.65) with the right-hand side g − u0 is uniquely solvable in Mext .
3.3. Integral equations of the Dirichlet problem
247
3.3.3 Integral equation of the problem D (e) Let g be a function from N. We look for the solution u of the external Dirichlet problem D(e) in the form of the potential ∂ 1 σ(q) log + 1 dsq , z ∈ Ω− , r = |z − q|, u(z) = ∂nq r Γ with the density σ satisfying the equation ∂ 1 log + 1 dsq = g(p). πσ(p) + σ(q) ∂nq r Γ
(3.68)
We refer to u(i) , u(e) as harmonic extensions of g to Ω+ and Ω− . By Propositions 3.2.5 and 3.2.6, we have the decompositions (±)
(∂u(e) /∂n)(z) = ±β0 xν−2 + β1 xν−1 + O(1) on the arcs Γ± (im ) and (∂u(e) /∂n)(z) = (α0,0 + α0,1 log x)xν−1 + δ0 x−1/2 + O(1) (±)
(±)
on the arcs Γ± (en ). Let v be a solution of the Neumann problem Δv = 0 in Ω+ , ∂v/∂n = ∂u(e) /∂n on Γ, normalized by the condition
gds − 2u(e) (∞),
vds = Γ
Γ
where u(e) (∞) is the limit value of u(e) at infinity. By Propositions 3.2.7 and 3.2.8 we have v(z) = ± β0 xν−1 ± γ0 x−1/2 + O(1) for ν = l/2, v(z) = ± (β0 + β1 log x)xν−1 ± γ0 x−1/2 + O(1) for ν = l/2 on the arcs Γ± (im ), and v(z) = (α0 + α1 log x)xν−1 + δx−1/2 + O(1)
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Chapter 3. Asymptotic Formulae for Solutions of Integral Equations
on the arcs Γ± (en ). Then σ = (2π)−1 (v − g) is a solution of the equation (3.68) in the class M, and furthermore, σ has the representations: a)
σ(z) = ±(α0 + α1 log x)x−1+ν(im ) + O(1)
on the arcs Γ± (im ), b)
σ(z) = (α0 + α1 log x)−1+ν(en ) + δx−1/2 + O(1)
on the arcs Γ± (en ). As in Theorem 3.3.4, solutions of the homogeneous equation (3.68) in M are functions of the form M
(i)
ck Re (1/ζk ),
k=1 (i)
where ck ∈ R and ζk is the conformal mapping of Ω+ onto the upper half-plane, normalized by the conditions (i) (i) ζk (ik ) = 0, Re (1/ζk )ds = 0, k = 1, . . . , M. Γ
3.4 Integral equations of the Neumann problem 3.4.1 Homogeneous integral equation of the problem N (e) Theorem 3.4.1. The homogeneous boundary equation of the problem N (e) ∂ ∂ V τ (p) + tk k (p) = 0 ∂np ∂np N
−πτ (p) +
k=1
considered on the set of pairs {τ, t}, where τ ∈ M and t ∈ RN , has only the trivial solution. Proof. The functions V τ (z) and n (z), n = 1, . . . , N , z ∈ Ω− , are harmonic, tend to zero at infinity and are bounded. Hence V τ (z) −
N
tn n (z)
n=1
is a solution of the Neumann problem with zero boundary condition. By the uniqueness theorem for the problem N (e) , V τ (z) −
N n=1
tn n (z) = c.
3.4. Integral equations of the Neumann problem
249
Since the solution vanishes at infinity, we conclude that V τ (z) =
N
tn n (z)
n=1
in Ω− . (i) Let n (z) denote the bounded harmonic extension of n (p), p ∈ Γ, to Ω+ , and let z = ω(ζ) be a conformal mapping of the strip Π = {(η, ξ) : 0 < ξ < 1} onto Ω+ with Re ω −1 (en ) = ∞. Then the function 0 (z) = V τ (z) −
N
tn (i) n (z)
n=1
is bounded in Ω+ and vanishes on Γ. Therefore, the Fourier decomposition of w(ζ) = 0 (ω(ζ)) has the form w(z) =
∞
ck (η) sin πkξ.
k=1
The coefficients ck (η) are equal to αk eπkη + βk e−πkη . Since 1
∞
|w(η, ξ)|2 dξ =
1 |ck (η)|2 2 k=1
0
and since the left-hand side of this equality is bounded, it follows that αk = 0, k = 1, 2, . . .. Hence 0 (z) and the gradient ∇0 (z) exponentially decay as z → en . We have N V τ (z) = tn (i) n (z) + 0 (z) n=1
and therefore, τ (p) =
N n=1
tn
N (i) ∂0 ∂n ∂n (p) − (p) + (p), p ∈ Γ, tn ∂n ∂n ∂n n=1
(i)
where the functions (∂n /∂n)(p) and (∂n /∂n)(p) have different orders of singularities, and (∂0 /∂n)(p) exponentially decays as p → en . Since τ ∈ M, the coefficients tn , n = 1, . . . , N , vanish. Consequently, V τ (z) = 0 in Ω− . The potential V τ (z) is continuous in Ω+ \Y and bounded in Ω+ , therefore V τ = 0 in Ω+ . The jump formula for the single layer potential implies τ = 0.
250
Chapter 3. Asymptotic Formulae for Solutions of Integral Equations
3.4.2 Solvability of the integral equation of the problem N (e) Theorem 3.4.2. Let h be a function from N and let hds = 0. Γ
Then the boundary equation ∂ ∂ V τ (p) + k (p) = h(p), tk ∂n ∂n N
−πτ (p) +
(3.69)
k=1
considered on the set of pairs {τ, t} with τ ∈ M and t ∈ RN , is uniquely solvable. The density τ has the following representations in local coordinates: a) on the arcs Γ± (en ), τ (z) = ± (β1,0 + β1,1 log x)xν−1 ± (μ1,0 + μ1,1 log x)x−1/2 + O(log x);
(3.70)
b) on the arcs Γ± (im ), τ (z) = (α0,0 + α0,1 log x)xν−1 + μ0 x−1/2 + O(1).
(3.71)
Proof. By Proposition 3.2.7, the holomorphic function, whose real part is a solution of the problem N (e) with the boundary data h, has the decomposition 1/2 (z − en )(z0 − en ) (z − en )(z0 − en ) + γ2,0 ϕ(z) = γ0 + γ1,0 z0 − z z0 − z 3/2 (z − en )(z0 − en ) (z − en )(z0 − en ) + γ3,0 + γ3,1 log z0 − z z0 − z 1+ν (z − en )(z0 − en ) (z − en )(z0 − en ) + β1,0 + β1,1 log z0 − z z0 − z 2 + O |z − en | log |z − en | in a neighborhood of en . Here β1,1 = 0 only for ν = l/2, l ∈ Z. We choose the parameter tn to equal Re γ1,0 . Then the problem N (e) with the boundary function N ∂ h− n tn ∂n n=1 has a solution v with the following representation in local coordinates on the arcs Γ± (en ): (±) (±) v(z) = γ0 + γ1 x + γ2,0 + γ2,1 log x x3/2 (±) (±) + β0,0 + β0,1 log x x1+ν + O(x2 log x).
3.4. Integral equations of the Neumann problem
251
On the arcs Γ± (im ) we have (±)
v(z) = α0 + α1 xν + α2
xν+1 + O(x2 )
in corresponding coordinate systems (cf. Proposition 3.2.8). Moreover, it follows from Proposition 3.2.5 that the complex potential of the problem D(i) with the boundary function v(p), p ∈ Γ\Y , has the representation ϕ(i) (z) = β0 + (β1,0 + β1,1 log(z − en )) (z − en )ν
+ (γ1,0 + γ1,1 log(z − en )) (z − en )1/2 + O |z − en | log |z − en |
in a neighborhood of en . The coefficient β1,1 does not vanish only for ν = l/2. Therefore, (∂u(i) /∂n)(z) = ± (β1,0 + β1,1 log x) xν−1 ± (γ1,0 + γ1,1 log x) x−1/2 + O(log x), on the arcs Γ± (en ), where γ1,1 = 0 only for ν = l/2. In a neighborhood of im we obtain ϕ(i) (z) = α + (α0,0 + α0,1 log(z − im ))(z − im )ν + δ0 (z − im )1/2 + O(|z − im |) (cf. Proposition 3.2.6) with α0,1 = 0 only for ν = l/2. This implies the decomposition (∂u(i) /∂n)(z) = (α0,0 + α0,1 log x) xν−1 + δ0 x−1/2 + O(1) on the arcs Γ± (im ). Here α0,1 = 0 only for ν = l/2. On Γ we have N 1 (i) V ∂u /∂n − h + tk ∂k /∂n . v= 2π 1 Consider the difference 1 V v0 (z) = v(z) − 2π
∂u(i) /∂n − h +
N
tk ∂k /∂n (z), z ∈ Ω− .
1
Let z = ω(ζ), ζ = η + iξ, be a conformal mapping of the strip Π = {(η, ξ) : 0 < ξ < 1} onto Ω− with Re ω −1 (im ) = ∞. The function w(ζ) = v0 (ω(ζ)) is a solution of the Laplace equation in the strip Π, it grows at infinity not faster than a power function and vanishes on ∂Π except for a finite set of points.
252
Chapter 3. Asymptotic Formulae for Solutions of Integral Equations
We take the Fourier decomposition of w(ζ), w(ζ) =
∞
ck (η) sin πkξ.
k=1
The coefficients ck (η) have the form αk eπkη + βk e−πkη . Since
0
1
∞ 1 |w(η, ξ)| dξ = |ck (η)|2 2 2
k=1
and since the left-hand side of this equality does not grow faster than a power function as η tends to infinity, it follows that αk = 0, k = 1, 2, . . . . Hence v0 (z) is bounded in a neighborhood of im . Thus v(z) is a bounded harmonic function in Ω− vanishing on Γ\Y , and therefore, v vanishes in Ω− . Hence we conclude that the density N τ = (2π)−1 ∂u(i) /∂n − h + tk ∂k /∂n 1
belongs to the class M, satisfies the boundary equation of the problem N (e) and has the required asymptotic representation. Theorem 3.4.1 implies that the solution of the equation (3.69), just constructed, is unique. Explicit formulas for the coefficients β1,0 , β1,1 , μ1,0 , μ1,1 , α0,0 , α0,1 and μ0 in (3.70), (3.71) were given in [20]. We apply Theorems 3.4.1 and 3.4.2 to obtain the following result. Theorem 3.4.3. Let the function h belong to N. Suppose that hds = 0 and h Re (1/ζk )ds = 0, k = 1, . . . , N. Γ
Γ
Then the equation (3.2) is uniquely solvable in M. Proof. We denote the unique solution of (3.8) by {τ, t}, where τ ∈ M and t ∈ RN . Applying the Green formula to the solution v (e) of the problem N (e) and to the function Re (1/ζk ) in Ω− \{|z − ek | < ε} and then passing to the limit as ε → 0, we obtain tk = (1/π) h Re (1/ζk )ds = 0, k = 1, . . . , N. Γ
Therefore τ is a unique solution of (3.2).
3.4. Integral equations of the Neumann problem
253
3.4.3 Integral equation of the problem N (i) Given h ∈ N, one seeks a solution of the internal Neumann problem N (i) as the sum of a simple layer potential and of a linear combination of the functions δn , n = 1, . . . , M , with unknown real coefficients v (i) (z) = V τ (z) +
M
tn δn (z),
n=1
where δn (z) = Re (z − in )1/2 , z ∈ Ω+ . The density τ and the vector t = (t1 , . . . , tM ) satisfy the equation
∂ 1 log dsq + tk δk (p) = h(p). ∂np r M
τ (q)
πτ (p) + Γ
(3.72)
k=1
Using the same arguments as in Theorems 3.4.1 and 3.4.2, we prove that (3.72) is uniquely solvable on the set of pairs {τ, t}, τ ∈ M, t ∈ RM . The only difference is that the harmonic function v (i) , satisfying the boundary condition ∂v (i) /∂n = h −
M
tk δ k ,
k=1
should be normalized so that the harmonic extension u(e) of v (i) from Γ to Ω− vanishes at infinity. The density τ has the representations: a)
τ (z) = ± (β1,0 + β1,1 log x)xν(im )−1 ± (μ1,0 + μ1,1 log x)x−1/2 + O(log x)
on the arcs Γ± (im ), b)
τ (z) = (α0,0 + α0,1 log x)xν(en )−1 + μ0 x−1/2 + O(1)
on the arcs Γ± (en ).
254
Chapter 3. Asymptotic Formulae for Solutions of Integral Equations
3.5 Appendices Appendix A: Counterexample Here we present an equation (3.1) having no integrable solution, although its righthand side is continuous. Consider the function on the contour Γ with an outward peak at the origin defined by cos log (1/x) ω(x) = ± , 0 < α < 1, (log (1/x))α on the arcs Γ± in a certain coordinate system. The normal derivative of the solution of the problem D(i) has the decomposition cos log (1/x) ∂u 1 (x, y) = ± 2 on Γ± (O). +O ∂n x (log (1/x))α (log (1/x))α The solution v of the problem N (e) with the boundary data ∂u/∂n can be represented in the form v(x, y) = ∓
sin log (1/x) − cos log (1/x) 1 tanh(π) · + (an integrable function). 2 x(log (1/x))α
Then the density σ = 2π(v − ω) is a nonintegrable solution of the integral equation of the problem D(i) with the right-hand side ω (the integral in the double layer potential is understood in the sense of the principal value ∂ 1 lim dsq , σ(q) log ε→0 ∂nq |z − q| {Γ:|q|>ε}
where z ∈ Γ\{O}). Among solutions of the homogeneous equation (3.1) there are no densities represented in the form ±c
sin log(1/x) − cos log(1/x) + (an integrable function ). x(log(1/x))α
Indeed, the potential W σ has a power growth as z → 0, and it vanishes on Γ\{O}. Therefore, W σ is equal to zero in Ω+ . Let u(e) denote the solution of the problem D(e) with a boundary function σ. We have ∂u(e) V − W u(e) = 2πu(e) (∞), z ∈ Ω+ . ∂n
3.5. Appendices
255
Since W u(e) vanishes in Ω+ , we have V
∂u(e) (z) = u(e) (∞), z ∈ Ω+ . ∂n
It follows from the limit relation for the single layer potential that ∂ ∂u(e) V + πu(e) (z) = 0, z ∈ Γ\{O}. ∂n ∂n Thus we have V
∂u(e) (z) = −2πu(e) (z) + u0 (z), z ∈ Ω− , ∂n
(3.73)
where u0 is a solution of the problem Δu0 = 0 in Ω− ,
∂u0 = 0 on Γ\{O}. ∂n
We substitute the integral representation of u(e) into (3.73). Then W u(e) (z) = u0 (z) − 2πu(e) (∞), z ∈ Ω− . Since the potential W u(e) vanishes at infinity, we obtain u0 (∞) = 2πu(e) (∞). The formulae for limit values of the double layer potential imply σ(z) = u(e) (z) = (2π)−1 (u0 (z) − u0 (∞)). The functions u(e) and W u(e) have a power growth as z → 0. Hence the function u0 does not grow faster than a power function. Since σ(z) = O
1 , z ∈ Γ\{O}, x(log x)α
the function u0 (z) coincides with Re (1/ζ(z)), where ζ(z) is the conformal mapping of Ω− onto the upper half-plane subject to the conditions ζ(0) = 0 and Re ζ(∞) = 0. Therefore, the equation (3.1) with the right-hand side g equal to ω on Γ± is unsolvable in L(Γ).
256
Chapter 3. Asymptotic Formulae for Solutions of Integral Equations
Appendix B: Proof of Lemma 3.2.1 (ext) un (z) = Re χ(z) ϕn+1 (z) , z ∈ Ω+ .
Let (ext)
Here ϕn+1 is given by (3.33) for ν = l/2 and by (3.34) for ν = l/2, and χ is a cutoff C ∞ -function supported in a small neighborhood of the point z = 0. Coefficients (ext) of ϕn+1 are chosen to satisfy (g − un )(z) = O |z|n+1+ν , (3.74) ∇k (g − un )(z) = O |z|n+1+ν−k , k = 1, 2, 3, z ∈ Γ± (0). We are looking for the coefficients β0 and Re β1 in (3.33). We choose them in such a way that the restriction of ϕ(z) = β0 z ν−1 + β1 z ν on Γ± coincides with (±) the sum of q0 xν and the terms of the form c± xi (log x)j admitting the estimate ν+1 O(x ). We decompose the function ϕ(z) = β0 z ν−1 + β1 z ν on Γ± in powers of x and identify the coefficients of xν−1 and xν with coefficients of g(x). Thus we arrive at the algebraic system ⎧ Re β0 = 0, ⎪ ⎪ ⎪ ⎨ (+) (−) (0) − κ− (0)) = q0 − q0 , −2Im β0 (ν − 1))(κ+ ⎪ ⎪ 1 ⎪ (+) (−) ⎩ − Im β0 (ν − 1)(κ+ (0) + κ− (0)) + Re β1 = q0 + q0 2 with the solutions β0 = −
(+)
κ+ (0)q0 − κ− (0)q0 (0) − κ (0) κ+ − (−)
(−)
q0 − q0 2i (0) − κ (0) , ν − 1 κ+ −
Re β1 =
(+)
.
Next we are looking for the coefficients β0,r , r = 0, 1, 2, in the decomposition (3.34). Consider two cases: ν = 1 and ν = 1. For ν = /2 and ν = 1 we decompose the restriction of ϕ(z) = (β0,0 + β0,1 log z)ν−1 + (β1,0 + β1,1 log z)ν on Γ± in powers of x and identify the coefficients of xν and xν log x with the corresponding coefficients in the decomposition of g(z). Hence we obtain ⎧ ⎪ ⎪ Re β0,0 = 0, Re β0,1 = 0, ⎪ ⎪ ⎨ 1 (±) (0) + Reβ1,1 = q0,1 , − Im β0,1 (ν − 1)κ± 2 ⎪ ⎪ ⎪ 1 ⎪ 1 (±) ⎩ − Im β0,0 (ν − 1)κ± (0) − Im β0,1 κ± (0) + Re β0,1 = q0,0 . 2 2
3.5. Appendices
257
Solutions of this system are q0,1 − q0,1 κ− (0)q0,1 − κ+ (0)q0,1 2i , Re β , = 1,1 ν − 1 κ+ (0) − κ− (0) κ+ (0) − κ− (0) (+) (−) 1 2i q0,0 − q0,0 β0,0 = − + Im β , 0,1 (0) − κ (0) ν − 1 κ+ 2 − 1 (+) 1 (−) = (q0,0 + q0,0 ) + Im β0,0 (ν − 1)(κ+ (0) + κ− (0)) 2 2 1 (0) + κ− (0)) . + Im β0,1 (κ+ 2 (+)
(−)
(+)
(−)
β0,1 = −
Re β1,0
Finally, for ν = 1, we expand the function ϕ(z) = (β0,1 log z + β0,2 log2 z) + (β1,0 + β1,1 log z)z into a power series in x on Γ± and identify it with the series for g(z). Then we find ⎧ Re β0,1 = 0, Re β0,2 = 0, ⎪ ⎪ ⎪ ⎨ 1 (±) (0) + Re β1,0 = q0,0 , − Im β0,1 κ± 2 ⎪ ⎪ ⎪ ⎩ (±) (0) + Re β1,1 = q0,1 . −Im β0,2 κ± Hence
⎧ (+) (−) (+) (−) q0,0 − q0,0 q0,1 − q0,1 ⎪ ⎪ ⎪ β0,1 = −2i β0,2 = −i ⎪ (0) , (0) , ⎪ κ+ (0) − κ− κ+ (0) − κ− ⎪ ⎨ 1 (+) 1 (−) Re β (q Im β = + q ) + (κ (0) + κ (0)) , 1,0 0,1 ⎪ + − 0,0 0,0 ⎪ 2 2 ⎪ ⎪ ⎪ ⎪ ⎩ Re β = 1 (q (+) + q (−) ) + Im β (κ (0) + κ (0)) . 1,1 0,2 + − 0,1 2 0,1
To conclude the calculation of coefficients in the representation of the function (ext) ϕp,n+1 satisfying (3.74), it suffices to show the existence of numbers α and β such that the expansion of the restriction of ϕ(z) = αz ν−1 (log z)m + βz ν (log z)m ν on Γ± is equal to x)m plus the terms of the form c± xi (log x)j admitting νA± x (log m−1 the estimate O x (log x) . Here ν = 1, ν > 0, m = 0, 1, 2 . . . , and the value of Re α is known. Next we take a power expansion of ϕ in x on Γ± and compare the coefficients of xν (log x)m with the given A± :
1 − Im α(ν − 1)κ± (0) + Re β = A± . 2
258
Chapter 3. Asymptotic Formulae for Solutions of Integral Equations
This implies Im α = −
A+ − A− 2 (0) − κ (0) , ν − 1 κ+ −
A+ κ− (0) − A− κ+ (0) . κ+ (0) − κ− (0)
Re β =
(ext)
Repeating this procedure, we find all coefficients of ϕp,n+1 (z) in (3.33) for ν = /2 and in (3.34) for ν = /2. u
(1)
We represent the harmonic extension of g − un as the sum u(1) + u(2) , where = g − un on Γ\{0} and ∇k u(1) (z) = O(|z − ep |n+1+ν−k ),
k = 0, 1, 2, 3.
The function u(2) is found from the boundary value problem Δu(2) = −Δu(1) in Ω+ ,
u(2) = 0 on Γ.
(3.75)
Making the change of variable z = ζ −1 (ζ = ξ + iη), we rewrite the equation (3.75) for U (2) (ζ) = u(2) (1/ζ) in the form ΔU (2) = F (1)
in Λ,
where the curvilinear half-strip Λ is the image of Ω+ and F (1) satisfies ∇k F (1) (ζ) = O |ζ|−n−1−ν−k , k = 0, 1. Let η(ξ) be a function in the class C ∞ , equal to zero for ξ < 1 and to 1 for ξ > 2. We set ηr (ξ) = η(ξ/r). Clearly, nξ −1 ξ n¯ Δξ −¯n U (2) = ΔU (2) − 2¯
∂ (2) U +n ¯ (¯ n + 1)ξ −2 U (2) , ∂ξ
where n ¯ = n + [ν]. Let Lr stand for the operator defined on the space W21 (Λ) and given by −2¯ nηr (ξ)ξ −1 ∂/∂ξ + n ¯ (¯ n + 1)ηr (ξ)ξ −1 . The boundary value problem ˜ (2) = F (2) ˜ (2) + Lr U ΔU
in Λ,
˜ (2) = 0 on ∂Λ , and U
with F (2) (ξ, η) = ξ n¯ F (1) (ξ, η) and ∇k F (2) (ξ, η) = O(ξ −1−k ), k = 0, 1, is uniquely ˚21 (Λ) for large r. It follows from the local estimate solvable in W ˜ (2) L (Λ) , ˜ (2) W 3 (Λ∩{ −1<ξ< +1}) c χF (2) W 1 (Λ) + χU U 2 2 2 where χ belongs to C0∞ ( − 2, + 2) and equals 1 in ( − 1, + 1), and from the ˜ (2) and its first Sobolev embedding theorem (see [17], Theorem 1.4.5(e)) that U ˜ (2) (ξ, η). Then derivatives are bounded as ξ → ∞. We put U (3) (ξ, η) = ξ −¯n U ∇k U (3) (ξ, η) = O(ξ −¯n ),
k = 0, 1.
3.5. Appendices
259
˚21 (Λ) and satisfies the equation Clearly, U (3) belongs to W ΔU (3) = F (1)
for
ξ > 2r .
Using a partition of unity and the above estimate, we show that 1 ˚ ∩ W 3 (Λ2r ), U (2) − U (3) ∈ W 2 2 where Λ2r = Λ ∩ {ξ > 2r}. Let D(∂ξ , ∂η ) stand for the operator Δ in the strip (0)/2 < η < κ− (0)/2}, Π = {(ξ, η) : −κ+ 1 ˚ ∩ W 3 (Π) in W 1 (Π) for performing a continuous mapping of W 2,β 2,β 2,β (0) − κ− (0))−1 , β = 2πj(κ+
j = ±1, ±2, . . . .
Since D(∂ξ , ∂η ) is the “limit” operator for Δ + Lr (see [11]), we have U (2) − U (3) ∈ 1 ˚ ∩ W 3 (Λ). Noticing that U (2) = U (3) + (U (2) − U (3) ), we conclude by the W 2,β 2,β Sobolev embedding theorem that ∇k U (2) (ξ, η) = O |ξ|−¯n = O |ξ|−n−[ν] for k = 0, 1 . Thus u = un + u(1) + u(2) is a solution of the problem D(i) with the required representation.
Appendix C: Proof of Lemma 3.2.2 , ( ) un (z) = Re χ(z − ip ) ϕn+3 (z) , z ∈ Ω+ .
Let , ( )
Here ϕp,n+3 is given by (3.36) for ν = /2 and by (3.37) for ν = /2, and χ is a cut-off C ∞ function supported by a small neighborhood of the point ip . The , ( )
coefficients of ϕp,n+3 are chosen to satisfy (g − un )(z) = O |z − ip |n+3+ν , ∇k (g − un )(z) = O |z − ip |n+3+ν−k ,
z ∈ Γ± (ip ),
k = 1, 2, 3 .
We plug the power expansion in x of ϕ = α0 z ν , ν = /2, ∈ Z, along Γ± (0) into (±)
Re ϕ(z)|Γ± (0) = q0 xν . The coefficient α0 obeys the system (+) Re α0 = q0 , (−) cos 2πνRe α0 − sin 2πνIm α0 = q0 ,
260
Chapter 3. Asymptotic Formulae for Solutions of Integral Equations
which gives q0 e−2πiν − q0 . sin 2πν In the case ν = /2, /2 ∈ Z, = 0, let us look for the coefficients in (3.37) comparing the expansion of ϕ(z) = (α0,0 + α0,1 log z)z ν in variable x along Γ± (0) (±) with q0 xν . Then we obtain (+) Re α0,0 = q0 , Re α0,1 = 0, (+)
(−)
α0 = i
(−)
Re α0,0 − 2πIm α0,1 = (−1)2ν q0
.
Solving this system, we find (+)
Re α0,0 = q0 ,
(+)
Re α0,1 = 0
and Im α0,1 =
q0
(−)
− q0 (−1)2ν . 2π
Consider the case ν = 0. We plug the expansion of ϕ(z) = α0,0 + α0,1 log z + α0,3 log2 z in the variable x along Γ± (0) into the equation (±)
(±)
Re ϕ(z)|Γ± (0) = q0,0 + q0,1 log x . Equalizing the coefficients of logm z, m = 0, 1, 2, we arrive at the algebraic system ⎧ (+) Re α0,2 = 0, Re α0,0 = q0,0 , ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ Re α0,0 − 2πIm α0,1 = q (−) , 0,0 (−) ⎪ ⎪ Re α0,1 − 4πIm α0,2 = q0,1 , ⎪ ⎪ ⎪ ⎩ (+) Re α0,1 = q0,1 . Hence (+)
Re α0,0 = q0,0 , (+)
Im α0,1
(+)
Re α0,1 = q0,1 , (−)
q0,0 − q0,0 , = 2π
(+)
α0,2
(−)
q0,1 − q0,1 . =i 4π
It remains to consider the case ν > 0, ν = /2, ∈ Z, with the right-hand side g(x) = qk,m xk+ν logm x,
m 1.
It suffices to show that there exist a function ϕ(z) whose expansion on Γ± (0) is the sum of g(x) and the terms c± xi logj x, admitting the estimate O(xk+ν (log x)m−1 ).
3.5. Appendices
261
We expand the function ϕ(z) = αk,m+1 z k+ν (log z)m+1 + αk,m z k+ν (log z)m + αk,m−1 z k+ν (log z)m−1 in variable x on Γ± (0) and compare with q(x). As a result we obtain ⎧ ⎨ Re αk,m+1 = 0, Re αm,k = q (+) , k,m
⎩ −2π(m + 1)(−1)2ν Im αk,m+1 + (−1)2ν Re αk,m = q (−) . k,m Hence
(−)
(+)
qk,m − (−1)2ν qk,m
(m + 1) . 2π At the next step we find Re αk,m−1 and Im αk,m . We continue until the degree m of log z becomes zero. Then we proceed in the same way as in the case ν = /2, ∈ Z. This step allows us to increase the degree k of z in the discrepancy. Im αk,m+1 =
Let u(1) be a harmonic function equal to g − un on Γ\{0}, subject to u(1) (z) = O |z|n+ν+3 and ∇k u(1) (z) = O |z|n+ν+3−k , k = 1, 2, 3, near z = 0. By u(2) we denote a solution of the Dirichlet problem Δu(2) = −Δu(1)
in Ω+ ,
˚21 (Ω+ ) . u(2) ∈ W
Let Λ stand for the image of Ω+ under the mapping (r, θ) = (t, θ), where (r, θ) are polar coordinates of (x, y) and t = log 1r . Then the problem takes the form ˚21 (Λ) , (∂t2 + ∂θ2 )U (2) (t, θ) = F (1) (t, θ) ∈ W where U (2) (t, θ) = u(2) (x, y), F (1) (t, θ) = − exp(2t)Δu(1) (x, y). The local estimate U (2) W23 (Λ∩{ −1<ξ< +1}) c χF (1) W21 (Λ) + χU (2) L2 (Λ) , χ belongs to C0∞ ( − 2, + 2) and equals 1 on ( − 1, + 1), implies U (2) ∈ where 3 1 ˚ (Λ). By D(∂t , ∂θ ) we denote the operator Δ in the strip Π = {(t, θ) : W2 ∩ W 2 0 < θ < 2π, t ∈ R}. The eigenvalues of the operator pencil D(ik, ∂θ ) arek = /2, ˚ 1 ∩ W 3 (Π) into = ±1, ±2, . . . . The operator D(∂t , ∂θ ) continuously maps W 2,β 2,β 1 W2,β (Π) for β = /2, = ±1, ±2, . . . (see [11]). The dimension of the eigenspace corresponding to each eigenvalue is 1. Hence the strip 0 < Im z < β with β ∈ (n+[ν]+1, n+[ν]+3/2) contains p = 2(n+[ν]+1) eigenvalues of the operator pencil 1 D(ik, ∂θ ). Since F (1) ∈ W2,β (Λ), it follows that U (2) admits the representation U (2) =
p
V (k) + W (1) ,
k=1
where V (k) are linearly independent functions satisfying ΔV (k) = 0 in Λr = 3 3 Λ ∩ {t > r} and vanishing on ∂Λ ∩ {t > r}, V (k) ∈ / W2,β (Λr ) and W (1) ∈ W2,β (Λr ).
262
Chapter 3. Asymptotic Formulae for Solutions of Integral Equations
Making the reverse change of variable, we obtain u
(2)
(r, θ) =
p
v (k) (r, θ) + w(1) (r, θ) ,
k=1
where v (k) (r, θ) = V (k) (log 1r , θ) and ∇ w(1) (r, θ) = O rn+[ν]+1− for = 0, 1. Let us show that for k = 2, 3, . . . and m 1 there exist C and D depending on k and m, such that ϕ(z) = Cz k/2 logm z + Dz k/2 logm+1 obeys the relations
(±) Re ϕ(z) = A± xk/2 logm x + the terms of the form xk/2+r Qm−1 (log x) , Γ±
(±)
where r is a nonnegative integer, Qm−1 are polynomials of degree at most m − 1 and the coefficients are real numbers. With this aim in view, we use the expansions k (±) z k/2 = ± xk/2 + i α± xk/2+1 + xk/2+2 P[n−1−k/2] (x) + O(xn+1 ) , 2 (±)
(±)
logm z = logm x + ma± logm−1 x+Q1,m−1 (log x) + xQ2,m−1 (log x) + · · · (±) +xn Qn+1,m−1 (log x) + O xn+1 logm−1 x , (±)
(±)
where a+ = 0, a− = 2π, Pk and Q ,k are polynomials of degree k. We make a power expansion of ϕ(z)|Γ and equalize the coefficients of xk/2 log x and A± . We ±
find Re C and Im D from the system Re C = A+ , Re C + 2π(m + 1)Im D = A− . Since Re iz
m/2
|Γ
±
n+[ν] m m/2+1 = α± x + Ak± xm/2+k + O(xn+[ν]+1 ) , 2 k=2
we can find polynomials Pm/2,k of degree k such that Re
n+[ν] k=1
Pm/2,k (log z)z m/2+k
Γ±
= O(xn+[ν]+1 ) ,
3.5. Appendices
263
using the same procedure as above. The restriction of
s(z) = Re
n+[ν]
Pm/2,k (log z)z m/2+k χ(z) onto Γ±
k=1 5/2
belongs to V2,γ (Γ), where γ ∈ (−n − [ν], −n − [ν] + 1/2). Then the problem D(i) 3 with the boundary data s on Γ has a unique solution wm ∈ V2,γ (Ω+ ), m = 1, . . . , p,
n+[ν]−
and ∇ wm (z) = O(|z| ), = 0, 1. The functions v m (z) = Re
n+[ν]
Pm/2,k (log z)z m/2+k − wm (z),
m = 1, . . . , p,
k=1
are harmonic in Ω+ and vanish on Γ± . Since these functions are linearly independent, there exist αm such that p k=1
v (k) (r, θ) =
p
αm v m (r, θ) .
m=1
Thus u = un + u(1) + u(2) is a solution of the problem D(i) and it has a required representation. Let ζ stand for a conformal mapping of Ω+ onto the upper half-plane. By Kellogg’s theorem (see Theorem 1.3.2), 1+ε √ ζ(z) = a−1 z + O |z| 2 , θ
0 < ε < 1,
and this representation can be differentiated once. Then −1+ε aθ 1 = √ + O |z| 2 ζ(z) z and therefore, Im
−1+ε ϕ 1 = aθ r−1/2 sin + O r 2 , ζ(z) 2
where r = |z|, ϕ = arg z ∈ (0, 2π). To find r0 we apply the Green formula to the solution u(i) of the problem 1 (i) D and to the function Im ζ(z) in Ω+ \{|z| < ε}. First, let ν > 0, ν = /2. In polar coordinates the solution u(i) admits the representation u(i) (z) = Re α0 rν cos νϕ−Im α0 rν sin νϕ + Re r0 r1/2 cos ϕ/2 −Im r0 r1/2 sin ϕ/2 + O(|z|) .
264
Chapter 3. Asymptotic Formulae for Solutions of Integral Equations
Let Γε denote the arcs Γ∩{|z| < ε}. The Green formula for the harmonic functions u(i) and Im 1/ζ implies ∂ 1 1 ∂ (i) u u(i) Im − Im ds ∂n ζ ζ ∂n Γ ε ∂ 1 1 ∂ (i) + u u(i) Im − Im ds = 0 . ∂n ζ ζ ∂n {r=ε}∩Ω+ We find representations for the integrals ∂ 1 u(i) Im ds and ∂n ζ {r=ε}∩Ω+
Im {r=ε}∩Ω+
1 ∂ (i) u ds ζ ∂n
as functions of ε as ε → 0. It is easily seen that ∂ 1 1 π u(i) Im ds ∼ Aεν−1/2 − aθ Im r0 + o(1) , ∂n ζ 2 2 + {r=ε}∩Ω and
Im {r=ε}∩Ω+
1 ∂ (i) π u ds ∼ νAεν−1/2 + aθ Im r0 + o(1) , ζ ∂n 2
where A can be expressed by α0 . Note also that Im 1/ζ vanishes on Γε . Hence
1 1
∂ ds for 0 < ν < 1/2 Im r0 = u(i) (z) Im πaθ Γ ∂n ζ(z) and Im r0 =
1 πaθ
u(i) (z) Γ
1 ∂ Im ds ∂n ζ(z)
for
ν > 1/2
as ε → 0. We apply a similar argument for ν = 0. In the case ν = 1/2 we take Im α0,0 = 0. We have u(i) (z) = Re α0,0 r1/2 cos ϕ/2 + Re α0,1 r1/2 log r cos ϕ/2 − Im α0,1 r1/2 log r sin ϕ/2 − Re α0,1 r1/2 ϕ sin ϕ/2 − Im α0,1 r1/2 ϕ cos ϕ/2 + Re r0 r1/2 cos ϕ/2 − Im r0 r1/2 sin ϕ/2 + O(r) . Performing calculations analogous to those above, we find
1 1
∂ 2Im α0,1 + Im r0 = ds , u(i) (z) Im πaθ Γ ∂n ζ(z) that is
∂ 1 1 q0+ + q0− 1
∂ (i) (z) Im ds − Im α0,1 = Im ds − . u ∂n ζ(z) πaθ Γ ∂n ζ(z) π
3.5. Appendices
265
Appendix D: Proof of Lemma 3.2.3 We introduce the function vn :
(ext) vn (z) = Re χ(z)ϕn+1 (z) ,
z ∈ Ω− ,
(ext)
where ϕn+1 is defined in (3.39) for ν = l/2 and in (3.40) for ν = l/2, and χ is a smooth cut-off function supported near z = 0. (ext) We choose ϕn+1 in such a way that hn (z) = (h − ∂vn /∂n)(z) = O(|z|n+ν+1 ). To this end, we introduce the function
z
H(z) =
h(s)ds, z (0)
where z and z (0) belong to Γ. For ν = −1 this function has the power expansion near z = 0 on Γ: (±) h h0 k xν+1 ∓ xν+k+1 + O(xn+ν+3 ), ν +1 ν +k+1 n+1
H(z) = ∓
(±)
z ∈ Γ± .
k=1
In the case ν = −1 the power expansion of H(z) can be written as H(z) =
(±) ∓h0 log z
∓
n+1 k=1
(ext)
(±)
hk xk + O(xn+2 ) . ν+k+1
(ext)
Let ϕ˜n+2 = −iϕn+1 (z). We see that in a small neighborhood of zero, the equality holds (ext)
(ext)
hn (z) = h(z) − (∂/∂n) Re iϕ n+2 (z) = (∂/∂s)H(z) + (∂/∂n)Im ϕ n+2 (z) (ext)
= (∂/∂s)(H(z) − Re ϕ n+2 (z)) . (ext)
The function ϕ n+2 (z) is to be constructed in the same way as in Lemma 3.2.2. It has the form n+2 (ext) ϕ n+2 (z) = αk z k+ν (3.76) k=0
for ν = /2 and (ext)
ϕ n+2 (z) =
2 r=0
α0,r (log z)r z ν +
n+2
Qk+1 (log z)z k+ν
k=1
(3.77)
266
Chapter 3. Asymptotic Formulae for Solutions of Integral Equations
for ν = /2, ∈ Z, where Qj are polynomials of degree j. The coefficients α0 and α0,r with r = 0, 1, 2 in (3.76) and (3.77) are: α0 = −
h0 e−2πiν + h0 (ν + 1) sin 2πν (+)
(−)
for ν = /2
and Re α0,0 = −
(+)
(+)
(−)
+ h0 (−1)2ν h0 h , α0,1 = −i 0 , ν +1 (ν + 1)2π α0,2 = 0 for ν = /2 and ν = −1, (+)
Re α0,0 = 0,
(+)
α0,1 = −h0 ,
(−)
α0,2 = −i
h0,1 + h0,1 4π
hn (s)ds
on Γ± (O)
for ν = −1 .
Let v˜(1) (z) be such that
z
v˜(1) (z) = ∓ 0
and ∇k v˜(1) (z) = O(|z|n+ν+1−k ), k = 0, 1, 2, in a vicinity of the point O in Ω− . We denote by v˜(2) a solution of the problem v (1) , Δ˜ v (2) = −Δ˜
˚21 (Ω− ). v˜(2) ∈ W
(3.78)
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). After this change of variables, the problem (3.78) takes the form (∂t2 + ∂θ2 )V (2) (t, θ) = F (1) (t, θ) where
V (2) (t, θ) = v˜(2) (x, y),
in the strip Λ
F (1) (t, θ) = −(1/r2 )Δ˜ v (1) (x, y) .
Using the same argument as in Lemma 3.2.2, we show that V (2) admits the representation p " (1) . (k) + W V (2) = U k=1
(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˜ = v˜n + 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
Chapter 3. Asymptotic Formulae for Solutions of Integral Equations
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
Chapter 3. Asymptotic Formulae for Solutions of Integral Equations
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 Lam´e 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)
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Chapter 4. Integral Equations of Plane Elasticity in Domains with Peak
Inserting these expressions into (4.1), one arrives at the homogeneous Lam´e system μΔu + (λ + μ)∇div u = 0 ,
(4.3)
where λ and μ are physical constants related to the media. Called the Lam´e 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 Lam´e 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
Chapter 4. Integral Equations of Plane Elasticity in Domains with Peak
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 Lam´e 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 Lam´e 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
Chapter 4. Integral Equations of Plane Elasticity in Domains with Peak
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
Chapter 4. Integral Equations of Plane Elasticity in Domains with Peak
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
Chapter 4. Integral Equations of Plane Elasticity in Domains with Peak
˚ (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)
4.2. Boundary value problems of elasticity
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
Chapter 4. Integral Equations of Plane Elasticity in Domains with Peak
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.
4.2. Boundary value problems of elasticity
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
Chapter 4. Integral Equations of Plane Elasticity in Domains with Peak
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
4.2. Boundary value problems of elasticity
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<ξ< +1}) 2 2
where χ belongs to C0∞ (−2, +2) and equals 1 in (−1, +1), and from the Sobolev (2) and its derivatives up embedding theorem, that the vector-valued function U to order n are bounded as ξ → ∞. We set (2) (ξ, η) and ∇k U (3) (ξ, η) = O(ξ −n ), k = 0, . . . , n. U (3) (ξ, η) = ξ −n U ˚ 1 (Λ) and satisfies Clearly, U (3) belongs to the space W 2 L(∂ξ , ∂η )U (3) = F (1) for ξ > 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
Chapter 4. Integral Equations of Plane Elasticity in Domains with Peak
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
4.2. Boundary value problems of elasticity
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
Chapter 4. Integral Equations of Plane Elasticity in Domains with Peak
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
4.2. Boundary value problems of elasticity
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<ξ< +1}) c χF (1) W2n (Λ) + χU (2) L2 (Λ) , 2
(4.44)
∞ where χ belongs to C 0 ( − 2, + 2) and equals 1 in ( − 1, + 1), it follows that ˚ 1 (Λ). U (2) ∈ W2n+2 ∩ W 2 1 ˚ ∩ By D(∂t , ∂θ ) we denote the operator Δ∗ + K continuously mapping W 2,β n+2 n W2,β (Π) into W2,β (Π), where Π = {(t, θ) : 0 < θ < 2π, t ∈ R}.
Eigenvalues of the operator pencil D(ik, ∂θ ) are the numbers k = i/2, where ∈ Z, = 0. The multiplicity of each eigenvalue is equal to 2 and the maximum length of the Jordan chain for each eigenvector is equal to 1. Let us check these assertions. Linearly independent solutions of the equation D(ik, ∂θ )V = 0 are
(i−k)(θ−π) e , ie(i−k)(θ−π) , −(i−k)(θ−π) e , −ie−(i−k)(θ−π) , (k + iκ)e−(i+k)(θ−π) , (ik + κ)e−(i+k)(θ−π) , − (k + iκ)e(i+k)(θ−π) , (ik + κ)e(i+k)(θ−π) .
(4.45)
The boundary value problem D(ik, ∂θ )V = 0 for
0 < θ < 2π
and V |θ=0 = V |θ=2π = 0
has a nontrivial solution only if the determinant Det of the system ⎧ α1 e−kπ + α2 ekπ + α3 (k + iκ)e−kπ − α4 (k + iκ)ekπ = 0, ⎪ ⎪ ⎪ ⎪ ⎨ α1 ekπ + α2 e−π + α3 (k + iκ)ekπ − α4 (k + iκ)e−kπ = 0, ⎪ iα1 e−kπ − iα2 ekπ + α3 (ik + κ)e−kπ − α4 (ik + κ)ekπ = 0, ⎪ ⎪ ⎪ ⎩ iα1 ekπ − iα2 e−kπ + α3 (ik + κ)ekπ − α4 (ik + κ)e−kπ = 0
(4.46)
292
Chapter 4. Integral Equations of Plane Elasticity in Domains with Peak
vanishes. One checks directly that Det = −16κ2 (sinh 2kπ)2 . Hence, Det vanishes only for k = i/2, ∈ Z, and the multiplicity of every root is 2. Let us show that k = 0 is not an eigenvalue of the operator pencil. Linearly independent solutions of the equation D(i0, ∂θ )V = 0 are the vector-valued functions (a1 + b1 (θ − π))ei(θ−π) , (−b1 + iκa1 + iκb1 (θ − π))ei(θ−π) and
(a2 + b2 (θ − π))e−i(θ−π) , (−b2 − iκa2 − iκb2 (θ − π))e−i(θ−π) .
We plug the sum of these functions into the boundary condition of the problem (4.46). One easily checks that the system of linear equations in a1 , a2 , b1 , b2 has only the trivial solution. Next we show that there are no generalized eigenvectors for each eigenvector of the pencil D(ik, ∂θ ). We are looking for a generalized eigenvector V as a solution of the problem D(ik, ∂θ )V = −∂k D(ik, ∂θ )V ,
θ ∈ (0, 2π) ,
V |θ=0 = V |θ=2π = 0 ,
(4.47)
where V = (V1 , V2 ) is an eigenvector. Nonsolvability of this problem follows from ∂k D(ik, ∂θ )V, V = 2k (λ + 2μ) V1 2L2 (0,2π) +μ V2 2L2 (0,2π) > 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) ,
4.2. Boundary value problems of elasticity
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
Chapter 4. Integral Equations of Plane Elasticity in Domains with Peak
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)
4.2. Boundary value problems of elasticity
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± .
4.2. Boundary value problems of elasticity
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
Chapter 4. Integral Equations of Plane Elasticity in Domains with Peak
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)
4.2. Boundary value problems of elasticity
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
Chapter 4. Integral Equations of Plane Elasticity in Domains with Peak
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<ξ< +1}) c χHn W2n (Λ) + χUn L2 (Λ) , 2
(4.55)
where χ belongs to C0∞ ( − 2, + 2) and is equal to 1 on the interval ( − 1, + 1), together with the Sobolev embedding theorem imply that the vector-valued functions Un and all their derivatives up to order n are bounded as ξ → ∞. Setting Un(1) (ξ, η) = ξ −n Un (ξ, η) , we obtain
∇k Un(1) (ξ, η) = O(ξ −n ), k = 0, . . . , n. (1)
Clearly, Un
˚ 1 (Λ) and satisfies belongs to W 2 Δ2 Un(1) = H
as ξ > 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.
4.2. Boundary value problems of elasticity
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
Chapter 4. Integral Equations of Plane Elasticity in Domains with Peak
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.
4.2. Boundary value problems of elasticity
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
Chapter 4. Integral Equations of Plane Elasticity in Domains with Peak
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
Chapter 4. Integral Equations of Plane Elasticity in Domains with Peak
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}.
4.2. Boundary value problems of elasticity
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
Chapter 4. Integral Equations of Plane Elasticity in Domains with Peak
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) = −
4.2. Boundary value problems of elasticity
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 Lam´e 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)
4.2. Boundary value problems of elasticity
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
Chapter 4. Integral Equations of Plane Elasticity in Domains with Peak
4.2.4 Uniqueness theorems Here we collect classical uniqueness theorems for the Dirichlet and Neumann problems for the Lam´e 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 Lam´e 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 Lam´e 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
Chapter 4. Integral Equations of Plane Elasticity in Domains with Peak
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.
4.3. Integral equations on a contour with inward peak
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
Chapter 4. Integral Equations of Plane Elasticity in Domains with Peak
(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−
4.3. Integral equations on a contour with inward peak
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)
We represent the displacement vector u0 by the Airy function U in the domain Ωδ = Ω− ∩ {|z| < δ}. Since ∂U ∂U +i = −i T u0 ds + c , (4.76) ∂x ∂y (0z) where (0z) stands for the arc of Γ connecting 0 and z, and since (∂U/∂x) + i(∂U/∂y) is defined up to a constant term, we assume that ∂U ∂U +i = 0 on Γ± . ∂x ∂y Therefore, U is equal to constants c± on Γ± . We use the Kelvin transform V(ζ) = |ζ|2 U (1/ζ), ζ ∈ Λδ , where Λδ is the image of Ωδ under the mapping ζ = 1/z. The expressions V(ζ) and ∇k V(ζ), k = 1, 2, grow not faster than a power function. Therefore, V belongs 2 to W2,β (Λδ ) with β < 0, V is biharmonic in Λδ and satisfies the conditions V(ζ) = c± |ζ|2 ,
(∂V/∂n) = c± (∂|ζ|2 /∂n) on ∂Λδ ∩ {|ζ| > δ −1 }.
From the local estimate VW23 (Λ1 ∩{ −1
δ
2
δ
(4.77)
318
Chapter 4. Integral Equations of Plane Elasticity in Domains with Peak
where χ is a cut-off C ∞ -function which equals 1 on ( − 1, + 1) and vanishes 3 outside ( − 2, + 2), it follows that V belongs to W2,β (Λδ ). Noting that derivatives of the second order of w0 are represented as linear combinations of derivatives of V up to the third order with coefficients growing 2 not faster than a power function, we have w0 ∈ W2,β (Λδ ). Let N (∂ξ , ∂η ) denote the operator of the boundary value problem L(∂ξ , ∂η ) = Δ∗ u = f # $ in Π = (ξ, η) : ξ ∈ R, −κ+ (0)/2 < η < −κ− (0)/2 , # $ T (∂ξ , ∂η )u = h on (ξ, η) : ξ ∈ R, η = −κ± (0)/2 ,
(4.78)
1/2
2 (Π) into L2,β (Π) × W2,β (∂Π). continuously mapping W2,β Consider the auxiliary problem with a complex parameter k corresponding to (4.78):
L(ik, ∂η )U (η) = 0, η ∈ (−κ+ (0)/2, −κ−(0)/2, T (ik, ∂η )U (η) = 0, η = −
κ (0) , 2
(4.79)
where U (η) takes values in R2 . The eigenvalues of (4.79) obey the equation k 2 α2 = (sinh k α)
2
with α = κ+ (0) − κ− (0) /2. One can check that the dimension of the null space of the operator in (4.79) (0) equals 2. The vector-valued function U1 (η) = (0, 1) is the basis element of mul(0) tiplicity 4. The second linear independent eigenvector U2 (η) = (1, 0) has multi(0) (0) plicity 2. The generalized eigenvectors corresponding to u1 , u2 are 1 (1) U1 (η) = −iη − i (α+ + α− ), 0 , 2 λ 1 1 λ (2) η2 − (α+ + α− )η , U1 (η) = 0, − 2 λ + 2μ 2 λ + 2μ i 3λ + 4μ i 3λ + 4μ 2 (3) (4.80) η 3 − (α+ + α− ) η U1 (η) = − 6 λ + 2μ 4 λ + 2μ λ+μ α+ α− η, 0 , − 2i λ + 2μ λ (1) η , U2 (η) = 0, −i λ + 2μ where α± = κ± (0)/2. The operator pencil (4.79) has no other eigenvalues on R.
4.3. Integral equations on a contour with inward peak
319
Let us show that the second generalized eigenvector does not exist. For this aim we need to prove that the problem 1 2 (0) (1) (2) ∂ N (i0, ∂η )U2 + ∂k N (i0, ∂η )U2 + N (i0, ∂η )U2 = 0 2 k with N (ik, ∂η ) = L(ik, ∂η ), T (ik, ∂η ) has no solutions. In fact, the first compo(2) (2) nent U21 (η) of U2 (η) should be a solution of the problem κ− (0) κ+ (0) ∂ 2 (2) 3λ + 4μ = 0, η ∈ − ,− , U − ∂η 2 21 λ + 2μ 2 2 κ (0) ∂ (2) λ U21 = − η, η = − ± . ∂η λ + 2μ 2 The general solution (2)
U21 (η) =
1 3λ + 4μ 2 η + αη + β , 2 λ + 2μ
found from the first equation, does not satisfy the boundary condition for any α and β. (0) (1) (2) (3) Let us check that the Jordan chain U1 , U1 , U1 , U1 cannot be extended. The system 1 2 (2) (3) (4) ∂ N (i0, ∂η )U1 + ∂k N (i0, ∂η )U1 + N (i0, ∂η )U1 = 0 2 k (4)
(4)
contains the following equation for the second component U12 of U1 : κ (0) κ (0) ∂ 2 (4) ∂ (3) (2) U , U (η) = 2μU (η) − i(λ + μ) (η), η ∈ − − ,− + 12 ∂η 2 12 ∂η 11 2 2 κ (0) ∂ (4) (3) (λ + 2μ) U12 (η) + iλU11 (η) = 0, η = − ∓ . ∂η 2 Let us integrate the first equation over the interval − κ− /2, η . Then plug the resulting relation into the left-hand side of the boundary condition and write the boundary data in the form −α− (3) (3) (2) i U11 (−α− ) − U11 (−α+ ) − 2 U12 (η)dη . (λ + 2μ)
−α+
(3)
(2)
Take the expressions for U11 (η) and U11 (η) from (4.80). Simple calculations imply (5λ + 4μ)(α2+ + α2− ) − 2λα+ α− = 0 . Now the result follows since the above quadratic form in α+ and α− is positive definite.
320
Chapter 4. Integral Equations of Plane Elasticity in Domains with Peak
Therefore, near the origin u0 (z) is represented in the form u0 (z) =
6
dk Zk (z) + u1 (z)
k=0
(cf. [18]). Here Zk are linear independent solutions of the equation Δ∗ u = 0 in a vicinity of the peak in Ω− satisfying the boundary condition T u = 0 near O on Γ. The last term u1 and its gradient vanish exponentially as z → 0. Three of the vector-valued functions Zk , 1 k 6, say Z1 , Z2 , Z3 , form rigid displacements. The others are represented in the form Z4 (z) = x−1 + O(1), Z5 (z) = ix−3 + O(x−2 ), Z6 (z) = ix−4 + O(x−3 ). Since the functions w1− , w2− , w3− in (4.75) and Z4 , Z5 , Z6 have different orders of singularities and since σ ∈ M, we have c1 = c2 = c3 = d4 = d5 = d6 = 0. In particular, u0 (z) = d1 Z1 (z) + d2 Z2 (z) + d3 Z3 (z) + u1 (z). Hence it follows that u0 and its gradient ∇u0 are bounded as z → 0. Besides, according to (4.74) we have u0 (z) − u0 (∞) = O(|z|−1 ),
∇u0 (z) = O(|z|−2 ) as z → ∞.
By the classical uniqueness theorem (see Theorem 4.2.17) it follows that u0 (z) = u0 (∞), z ∈ Ω− . Thus, σ(z) = −u0 (z) + u− (∞) = c on Γ\{O}. Noting that a non-zero constant does not satisfy the homogeneous equation (4.70), we conclude that σ = 0. Consider the equation 2−1 σ − W σ − c1 A1 − c2 A2 − c3 A3 = −g on Γ
(4.81)
with respect to a density σ and a vector (c1 , c2 , c3 ). The functions A1 , A2 , A3 have been defined by (4.21). Next we show that the equation (4.81) is solvable and we find the main term in the asymptotics of its solution. Theorem 4.3.3. Let Ω+ have an inward peak and let g belong to the class Nν , ν > 3. Then the equation (4.81) has a unique solution {σ, c1 , c2 , c3 } in M × R3 , and the density σ can be represented in the form σ(z) = α(log x)2 + β log x + γ x−1/2 + O(x−ε ) on Γ± with a small positive ε.
4.3. Integral equations on a contour with inward peak
321
Proof. Let Um,k , m = 1, . . . , 4, and k = 1, 2, denote solutions of the problem Δ∗ u = 0 in Ω+ ∩ Br ,
u = 0 on Br ∩ Γ\{O}
corresponding to the eigenvalues k = i/2, ∈ Z, = 0 of the operator pencil introduced in Theorem 4.2.5. Here Br = {|z| < r} with a small positive r. We choose the functions Um,k in the following way: i 2κ Im z 1/2 − z −1/2 Im 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 + O(z 5/2 (log z)2 ), 2μ 1 2κ Im z 1/2 + z −1/2 Im z U12 (z) = − 2μ (κ + 1)(α+ − α− ) 2κ Im (z 3/2 log z) + 3z 1/2 log z Im z − 8πκμ (κ + 1)α + 3/2 + 2z 1/2 Im z + z + O(z 5/2 (log z)2 ), 2μ U11 (z) =
U21 (z) = i
(κ − 1)(α+ − α− ) κ−1 Im z + i 2κ Im (z 2 log z) μ 4πμκ (κ − 1)α+ 2 − 4z log z Im z − 2z Im z − i z + O(z 3 log2 z), μ
(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 + O(z 3 (log z)2 ), μ i 2κ Im z 3/2 − 3z 1/2 Im z + O(z 5/2 log z), U31 (z) = 2μ 1 2κ Im z 3/2 + 3z 1/2 Im z + O(z 5/2 log z), U32 (z) = − 2μ U22 (z) = −
U41 (z) =
! i κ Im z 2 − 2z Im z + O(z 3 log z), μ
U42 (z) = −
! 1 κ Im z 2 + 2z Im z + O(z 3 log z), μ
where α± = κ± (0)/2.
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Chapter 4. Integral Equations of Plane Elasticity in Domains with Peak
According to Theorem 4.2.5, the solution of the problem D(i) with the boundary data g admits the representation +
u (z) =
2 4 m=1
dmk Umk (z) + W (z),
k=1
where the coefficients dm,k are defined uniquely and W (z) = O(|z|2+ε ), ε > 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}.
4.3. Integral equations on a contour with inward peak
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
Chapter 4. Integral Equations of Plane Elasticity in Domains with Peak
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+
4.3. Integral equations on a contour with inward peak
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
Chapter 4. Integral Equations of Plane Elasticity in Domains with Peak
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)
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Chapter 4. Integral Equations of Plane Elasticity in Domains with Peak
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=
4.4. Integral equations on a contour with outward peak
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
Chapter 4. Integral Equations of Plane Elasticity in Domains with Peak
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.
4.4. Integral equations on a contour with outward peak
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
Chapter 4. Integral Equations of Plane Elasticity in Domains with Peak
We extend the functions k (z), z ∈ Γ, k = 1, 2, onto Ω+ as solutions of the + problem D(i) for the Lam´e 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,
4.4. Integral equations on a contour with outward peak
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
Chapter 4. Integral Equations of Plane Elasticity in Domains with Peak
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
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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 Lam´e 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 Poincar´e 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