The Problem of Scattering by a Mixture of Cracks and Obstacles

Boundary Value Problems

Volume 2009, Article ID 524846,19pages doi:10.1155/2009/524846

Research Article

The Problem of Scattering by a Mixture of

Cracks and Obstacles

Guozheng Yan

Department of Mathematics, Central China Normal University, Wuhan 430079, China

Correspondence should be addressed to Guozheng Yan,yan [email protected]

Received 8 September 2009; Accepted 2 November 2009

Recommended by Salim Messaoudi

Consider the scattering of an electromagnetic time-harmonic plane wave by an infinite cylinder having an open crack Γ and a bounded domain D in R2 _{as cross section. We assume that}

the crackΓ is divided into two parts, and one of the two parts is possibly coated on one side by a material with surface impedance λ. Different boundary conditions are given on Γ

and ∂D. Applying potential theory, the problem can be reformulated as a boundary integral system. We obtain the existence and uniqueness of a solution to the system by using Fredholm theory.

Copyrightq2009 Guozheng Yan. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

1. Introduction

Crack detection is a problem in nondestructive testing of materials which has been often addressed in literature and more recently in the context of inverse problems. Early works on the direct and inverse scattering problem for cracks date back to 1995 in1by Kress. In that paper, Kress considered the direct and inverse scattering problem for a perfectly conducting crack and used Newton’s method to reconstruct the shape of the crack from a knowledge of the far-field pattern. In 1997, M ¨onch considered the same scattering problem for sound-hard crack2, and in the same year, Alves and Ha Duong discussed the scattering problem but for flat cracks in3. Later in 2000, Kress’s work was continued by Kirsch and Ritter in4

problem. Therefore, in the following we just consider the direct scattering problem for a mixture of a crack Γ and a bounded domain D, and the corresponding inverse scattering problem can be considered by similar methods in1,2,4–12and the reference therein.

Briefly speaking, in this paper we consider the scattering of an electromagnetic time-harmonic plane wave by an infinite cylinder having an open crackΓand a bounded domain

DinR2_{as cross section. We assume that the cylinder ispossiblypartially coated on one side} by a material with surface impedanceλ. This corresponds to the situation when the boundary or more generally a portion of the boundary is coated with an unknown material in order to avoid detection. Assuming that the electric field is polarized in theTMmode, this leads to a mixed boundary value problem for the Helmholtz equation defined in the exterior of a mixture inR2.

Our aim is to establish the existence and uniqueness of a solution to this direct scattering problem. As is known, the method of boundary integral equations has widely applications to various direct and inverse scattering problemssee13–17and the reference therein. A few authors have applied such method to study the scattering problem with mixture of cracks and obstacles. In the following, we will use the method of boundary integral equations and Fredholm theory to obtain the existence and uniqueness of a solution. The difficult thing is to prove the corresponding boundary integral operatorAwhich is a Fredholm operator with index zero since the boundary is a mixture and we have complicated boundary conditions.

The outline of the paper is as follows. In Section 2, the direct scattering problem is considered, and we will establish uniqueness to the problem and reformulate the problem as a boundary integral system by using single- and double-layer potentials. The existence and uniqueness of a solution to the corresponding boundary integral system will be given in Section 3. The potential theory and Fredholm theory will be used to prove our main results.

2. Boundary Integral Equations of the Direct Scattering Problem

Consider the scattering of time-harmonic electromagnetic plane waves from an infinite cylinder with a mixture of an open crackΓand a bounded domainDinR2_{as cross section.} For further considerations, we suppose thatD has smooth boundary∂D e.g., ∂D ∈ C2_, and the crackΓ smoothcan be extended to an arbitrary smooth, simply connected, closed curve∂Ωenclosing a bounded domainΩsuch that the normal vectorνonΓcoincides with the outward normal vector on∂Ωwhich we again denote byν. The bounded domainDis located inside the domainΩ, and∂D∂Ω ∅.

In the whole paper, we assume that∂D∈C2_and∂Ω∈C2_. Suppose that

Γ {zs:s∈s0, s1}, 2.1

divided into two partsΓ1andΓ2and consider the electromagnetic field E-polarized. Different boundary conditions onΓ±₁,Γ±₂,and∂Dlead to the following problem:

ΔUk2_{U0 inR}2_\D∪Γ,

U± 0 onΓ±1,

U− 0 onΓ−2,

∂U

∂ν ikλU 0 onΓ

2,

U0 on∂D,

2.2

whereU_±x limh→0Ux±hνforx∈Γand∂U±/∂ν limh→0ν· ∇Ux±hνforx∈Γ. The total fieldUis decomposed into the given incident fieldui_{x e}ikx·d_{,|d| 1, and the}

unknown scattered fielduwhich is required to satisfy the Sommerfeld radiation condition

lim

ν→ ∞

√

r

∂u ∂r −iku

0 2.3

uniformly inxx/|x|withr|x|.

We recall some usual Sobolev spaces and some trace spaces onΓin the following. LetΓ⊆Γbe a piece of the boundary. UseH1_DandH1

locR2\Dto denote the usual

Sobolev spaces,H1/2_{Γis the trace space, and we define}

H1/2Γ u|_Γ:u∈H1/2Γ

,

H1/2Γ u∈H1/2Γ: suppu⊆Γ,

H−1/2ΓH1/2Γ the dual space ofH1/2Γ,

H−1/2ΓH1/2Γ the dual space ofH1/2Γ.

2.4

Just consider the scattered field u, then2.2 and 2.3 are a special case of the following problem.

Givenf∈H1/2_Γ

1,g∈H1/2Γ2,h∈H−1/2Γ2,andr ∈H1/2∂Dfindu∈H_loc1 R2\

D∪Γsuch that

Δuk2u0 in R2\D∪Γ,

u±f onΓ±1,

u−g onΓ−2,

∂u

∂ν ikλuh onΓ

2,

ur on∂D,

and u is required to satisfy the Sommerfeld radiation condition 2.3. For simplicity, we assume thatk >0 andλ >0.

Theorem 2.1. The problems2.5and2.3have at most one solution.

Proof. Letube a solution to the problem2.5withf g hr 0, we want to show that

u0 inR2_\D∪Γ.

Suppose that BR with boundary ∂BR is a sufficiently large ball which contains

the domain Ω. Obviously, to the Helmholtz equation in 2.5, the solution u ∈ H1_B

R \

ΩH1_{Ω\Dsatisfies the following transmission boundary conditions on the} complemen-tary part∂Ω\Γof∂Ω:

uu₋,

∂u

∂ν

∂u−

∂ν ,

2.6

where “±” denote the limit approaching ∂Ω from outside and inside Ω, respectively. Applying Green’s formula foruanduinΩ\DandBR\Ω, we have

Ω\DuΔu∇u· ∇udx

∂Ω\Γ

u−∂u_∂ν−ds

Γ1

u−∂u_∂ν−ds

Γ2

u−∂u_∂ν−ds,

BR\ΩuΔu∇u· ∇udx

∂BR

u∂u ∂νds

∂Ω\Γu

∂u

∂ν ds

Γ1

u∂u_∂νds

Γ2

u∂u_∂νds,

2.7

whereνis directed into the exterior of the corresponding domain.

Using boundary conditions onΓ1,Γ2and the above transmission boundary condition

2.6, we have

∂BR

u∂u ∂νds

BR\Ω

Ω\D

|∇u|2−k2|u|2dx

Γ2

ikλ|u|2ds. 2.8

Hence

Im

∂BR

u∂u ∂νds

≥0. 2.9

So, from13, Theorem 2.12and a unique continuation argument we obtain that u 0 in

R2_\D∪Γ.

We useu u−−uand∂u/∂ν ∂u−/∂ν−∂u/∂νto denote the jump ofuand

Lemma 2.2. Ifuis a solution of2.5and2.3, thenu∈H1/2_{Γand∂u/∂ν∈H}−1/2_Γ.

The proof of this lemma can be found in11.

We are now ready to prove the existence of a solution to the above scattering problem by using an integral equation approaching. Forx∈Ω\D, by Green representation formula

ux

∂Ω

∂u ∂νΦ

x, y−u∂Φ

x, y ∂ν

dsy

∂D

∂u ∂νΦ

x, y−u∂Φ

x, y ∂ν

dsy 2.10

and forx∈R2\Ω

ux

∂Ω

u∂Φ

x, y

∂ν −

∂u ∂νΦ

x, y

dsy, 2.11

where

Φx, y i

4H

1 0

kx−y 2.12

is the fundamental solution to the Helmholtz equation inR2_{, andH}1

0 is a Hankel function of the first kind of order zero.

By making use of the known jump relationships of the single- and double-layer potentials across the boundary∂Ω see5,11and approaching the boundary∂Ωfrom inside

Ω\D, we obtainforx∈∂Ω

u−x SΩΩ∂u_∂ν− −KΩΩu−2

∂D

∂uy

∂ν Φ

x, y−uy∂Φ

x, y ∂ν

dsy, 2.13

∂u−x

∂ν K

ΩΩ

∂u−

∂ν −TΩΩu−2 ∂ ∂νx

∂D

∂uy

∂ν Φ

x, y−uy∂Φ

x, y ∂ν

dsy, 2.14

whereSΩΩ,KΩΩ,K_ΩΩ , andTΩΩare boundary integral operators:

SΩΩ:H−1/2∂Ω−→H1/2∂Ω, KΩΩ:H1/2∂Ω−→H1/2∂Ω

K_ΩΩ :H−1/2∂Ω−→H−1/2∂Ω, TΩΩ:H1/2∂Ω−→H−1/2∂Ω,

2.15

defined byforx∈∂Ω

S_ΩΩϕx 2

∂Ω

ϕyΦx, ydsy, KΩΩϕx 2

∂Ω

ϕyΦ

x, y ∂νy

dsy,

K_ΩΩϕx 2

∂Ωϕ

y∂Φ

x, y ∂νx

dsy, TΩΩϕx 2

∂ ∂νx

∂Ωϕ

y∂Φ

x, y ∂νy

dsy.

Similarly, approaching the boundary∂Ωfrom insideR2_{\Ωwe obtainforx∈∂Ω}

ux −S_ΩΩ∂u

∂ν KΩΩu, 2.17

∂ux

∂ν −K

ΩΩ

∂u

∂ν TΩΩu. 2.18

From2.13–2.18, we have

u₋uS_ΩΩ ∂u₋ ∂ν − ∂u ∂ν

−K_ΩΩu₋−u

2

∂D

∂uy

∂ν Φ

x, y−uy∂Φ

x, y ∂ν

dsy,

2.19

∂u₋

∂ν

∂u

∂ν K

ΩΩ ∂u₋ ∂ν − ∂u ∂ν

−T_ΩΩu₋−u

2 ∂

∂νx

∂D

∂uy

∂ν Φ

x, y−uy∂Φ

x, y ∂ν

dsy.

2.20

RestrictuonΓ±₁, from2.19we have

2fx SΩΩ

∂u− ∂ν − ∂u ∂ν Γ1

−KΩΩu−−u|Γ1

2

∂D

∂uy

∂ν Φx, ydsy Γ1 −2 ∂D

uy∂Φx, y

∂ν dsy

Γ1

2.21

where·|_Γ1means a restriction toΓ1. Define

SΩΓ1ϕx 2

∂ΩϕyΦx, ydsy

Γ1 ,

KΩΓ1ϕx 2

∂Ω

∂Φx, y

∂ν ϕydsy

Γ1 ,

SDΓ1ϕx 2

∂D

ϕyΦx, ydsy

Γ1 , ∂u ∂ν ∂D a, ∂u ∂ν Γ1 ∂u− ∂ν − ∂u ∂ν Γ1 b, ∂u ∂ν Γ2 ∂u− ∂ν − ∂u ∂ν Γ2

c, u|_Γ2 u−−u|Γ2 d.

Then zero extendb,c, anddto the whole∂Ωin the following:

b

⎧ ⎨ ⎩

0, on∂Ω\Γ1,

b, onΓ1,

c

⎧ ⎨ ⎩

0, on∂Ω\Γ2,

c, onΓ2,

d

⎧ ⎨ ⎩

0, on∂Ω\Γ2,

d, onΓ2.

2.23

By using the boundary conditions in2.5, we rewrite2.21as

SDΓ1aSΩΓ1

bc−KΩΓ1dp1x, 2.24

where

p1x 2fx 2

∂D

∂Φx, y

∂νy rydsy

Γ1

. 2.25

Furthermore, we modify2.24as

SDΓ1aSΓ1Γ1bSΓ2Γ1c−KΓ2Γ1dp1x, 2.26

where the operatorSΓ2Γ1 is the operator applied to a function with supp⊆Γ2and evaluated onΓ1, with analogous definition forSDΓ1,SΓ1Γ1, andKΓ2Γ1. We have mapping propertiessee

5,11

SDΓ1 :H

−1/2_∂D−→H1/2_Γ 1,

SΓ1Γ1 :H− 1/2_Γ

1−→H1/2Γ1,

SΓ2Γ1 :H− 1/2_Γ

2−→H1/2Γ1,

K_Γ2Γ1:H1/2Γ2−→H1/2Γ1.

2.27

Again from2.13–2.18, restrictinguto boundaryΓ−₂ we have

2gx S_ΩΩ

∂u−

∂ν − ∂u

∂ν

Γ2

−K_ΩΩu₋−u|_Γ2 u₋−u|_Γ2

2

∂D

∂uy

∂νyΦx, ydsy

Γ2

−2

∂D

∂Φx, y

∂νy rydsy

Γ2

or

2

∂D

∂uy

∂νyΦx, ydsy

Γ2

SΩΩ

∂u−

∂ν − ∂u

∂ν

Γ2

−KΩΩu−−u|Γ2 u−−u|Γ2

2gx 2

∂D

∂Φx, y

∂νy rydsy

Γ2 .

2.29

Like previous, define

SΩΓ2ϕx 2

∂ΩϕyΦx, ydsy

Γ2 ,

K_ΩΓ2ϕx 2

∂Ω

∂Φx, y

∂ν ϕydsy

Γ2 ,

SDΓ2ϕx 2

∂D

ϕyΦx, ydsy

Γ2 .

2.30

Then we can rewrite2.29as

SDΓ2aSΩΓ2

bc−K_ΩΓ2ddp2x, x∈Γ−2, 2.31

where

p2x 2gx 2

∂D

∂Φx, y

∂νy rydsy

Γ−

2

. 2.32

Similar to2.26, we modify2.31as

SDΓ2aSΓ1Γ2bSΓ2Γ2c I−KΓ2Γ2dp2x 2.33

and we have mapping properties:

SDΓ2 :H−

1/2_∂D−→H1/2_Γ 2,

SΓ2Γ2 :H− 1/2_Γ

2−→H1/2Γ2,

S_Γ1Γ2 :H−1/2Γ1−→H1/2Γ2,

KΓ2Γ2:H 1/2_Γ

2−→H1/2Γ2.

Combining2.13and2.14,

−ikλ

S_ΩΩ∂u−

∂ν −KΩΩu−

−ikλ

u−−2

∂D

∂uy

∂ν Φ

x, y−u∂Φ

x, y ∂νy

dsy

−ikλu−2ikλ

∂D

∂uy

∂ν Φ

x, ydsy−2ikλ

∂D

ry∂Φ

x, y ∂νy dsy,

−K_ΩΩ∂u−

∂ν TΩΩu−

−∂u−

∂ν 2

∂ ∂νx

∂D

∂uy

∂ν Φ

x, ydsy−2

∂ ∂νx

∂D

ry∂Φ

x, y ∂ν dsy,

2.35

−ikλu₋−∂u− ∂ν

−ikλu−−u−

∂u− ∂ν − ∂u ∂ν

−ikλu−∂u_∂ν.

2.36

Using2.17and2.18,

∂u

∂ν ikλu−K

ΩΩ

∂u

∂ν TΩΩuikλ

K_ΩΩu−S_ΩΩ∂u ∂ν

K_ΩΩ ∂u₋ ∂ν − ∂u ∂ν

−TΩΩu−−u ikλSΩΩ

∂u₋ ∂ν − ∂u ∂ν

−ikλKΩΩu−−u−ikλ

SΩΩ∂u_∂ν− −KΩΩu−

−K_ΩΩ∂u−

∂ν TΩΩu−

K_ΩΩ ∂u− ∂ν − ∂u ∂ν

−TΩΩu−−u ikλSΩ

∂u− ∂ν − ∂u ∂ν

−ikλKΩΩu−−u−ikλu−−∂u_∂ν−

2ikλ

∂D

∂uy

∂ν Φ

x, ydsy2

∂ ∂νx

∂D

∂uy

∂ν Φ

x, ydsy

−2ikλ

∂D

ry∂Φ

x, y ∂νy dsy−2

∂ ∂νx

∂D

ry∂Φ

x, y ∂νy dsy.

Then using2.36,

2

∂u

∂ν ikλu

K_ΩΩ ∂u− ∂ν − ∂u ∂ν

−TΩΩu−−u ikλSΩΩ

∂u− ∂ν − ∂u ∂ν

−ikλKΩΩu−−u−ikλu−−u−

∂u− ∂ν − ∂u ∂ν 2ikλ ∂D

∂uy

∂ν Φ

x, ydsy2

∂ ∂νx

∂D

∂uy

∂ν Φ

x, ydsy

−2ikλ

∂D

ry∂Φ

x, y ∂νy dsy−2

∂ ∂νx

∂D

ry∂Φ

x, y ∂νy dsy.

2.38

From2.29, we have

2ikλgx ikλ S_ΩΩ ∂u₋ ∂ν − ∂u ∂ν _Γ 2

− K_ΩΩu₋−u|_Γ2 u₋−u|_Γ2

2

∂D

∂uy

∂ν Φx, ydsy Γ2 − 2 ∂D

ry∂Φx, y ∂νy dsy

Γ2 . 2.39

Restricting2.38toΓ₂ and using2.39, we modify2.38as

2 ∂

∂νx

∂D

∂uy

∂ν Φx, ydsy Γ 2 K_ΩΩ ∂u− ∂ν − ∂u ∂ν Γ 2

−TΩΩu−−u_Γ 2− ∂u− ∂ν − ∂u ∂ν Γ 2

−2ikλu−−u|Γ

2 p3x,

2.40

where

p3x 2hx−2ikλgx

∂D

ry∂Φx, y ∂νy dsy

Γ 2

2.41

forx∈Γ. Define

K_DΓ

2ϕx 2 ∂ ∂νx

∂D

ϕyΦx, ydsy

Γ 2

, 2.42

and using the notation in previous, we can rewrite2.40as

K_DΓ 2aK

ΩΓ2

or

K_DΓ 2aK

Γ1Γ2b

K_Γ 2Γ2−I

c−TΓ2Γ22ikλIdp3x, 2.44

where the operatorsK_Γ 1Γ2,K

Γ2Γ2, andTΓ2Γ2are restriction operatorssee2.29. As before, we have mapping properties:

K_DΓ 2:H

−1/2_∂D−→H−1/2_Γ 2,

K_Γ 1Γ2:H

−1/2_Γ

1−→H−1/2Γ2,

K_Γ 2Γ2:H

−1/2_Γ

2−→H−1/2Γ2,

T_Γ2Γ2:H1/2Γ2−→H−1/2Γ2.

2.45

By using Green formula and approaching the boundary∂Dfrom insideΩ\Dwe obtainfor

x∈∂D

ux 2

∂D

∂uy

∂ν Φ

x, ydsy2

∂D

∂Φx, y

∂ν u

ydsy

2

∂Ω

∂u−y

∂ν Φ

x, y−u₋y∂Φ

x, y ∂ν

dsy.

2.46

The last term in2.46can be reformulated as

2

∂Ω

∂u−y

∂ν Φ

x, y−u−y∂Φ

x, y

∂ν u

y

dsy

2

∂Ω

∂u−y

∂ν −

∂uy

∂ν

Φx, y−u−y−uy∂Φ

x, y ∂ν

dsy

2

∂Ω

∂uy

∂ν Φ

x, y−uy∂Φ

x, y ∂ν

dsy.

2.47

Sincex∈∂Dandy∈∂Ωin2.47, we have the following resultsee13.

Lemma 2.3. By using Green formula and the Sommerfeld radiation condition2.3, one obtains

∂Ω

∂uy

∂ν Φ

x, y−uy∂Φ

x, y ∂ν

dsy 0. 2.48

Proof. Denote by BR a sufficiently large ball with radius R containing Ω and use Green

formula insideBR\Ω. Furthermore noticingx∈∂D,y∈∂Ω, and the Sommerfeld radiation

Combining2.46,2.47, andLemma 2.3and restrictingxto∂Dwe have

2

∂D

∂uy

∂ν Φx, ydsy

∂D

2

∂Ω

∂u−y

∂ν −

∂uy

∂ν

Φx, ydsy

∂D

−2

∂Ω

u−y−uy∂Φ_∂νx, ydsy

∂D

p0x,

2.49

where

p0x rx−2

∂D

ry∂Φx, y

∂ν dsy

∂D

. 2.50

Define

SDDϕx 2

∂D

ϕyΦx, ydsy

∂D

, 2.51

and then we can rewrite2.49as

SDDaSΓ1DbSΓ2Dc−KΓ2Ddp0x. 2.52

Similarly,S_ΓDandKΓDare restriction operators as before, and we have mapping properties:

SDD :H−1/2∂D−→H1/2∂D,

SΓD:H−1/2Γ−→H1/2∂D,

K_ΓD:H1/2Γ−→H1/2∂D.

2.53

Combining2.52,2.26,2.33, and2.44, we have

SDDaSΓ1DbSΓ2Dc−KΓ2Ddp0x,

SDΓ1aSΓ1Γ1bSΓ2Γ1c−KΓ2Γ1dp1x,

SDΓ2aSΓ1Γ2bSΓ2Γ2c I−KΓ2Γ2dp2x,

K_DΓ 2aK

Γ1Γ2b

K_Γ 2Γ2−I

c−T_Γ2Γ22ikλIdp3x.

If we define

A

⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝

SDD SΓ1D SΓ2D −KΓ2D SDΓ1 SΓ1Γ1 SΓ2Γ1 −KΓ2Γ1 SDΓ2 SΓ1Γ2 SΓ2Γ2 I−KΓ2Γ2 K_DΓ

2 K

Γ1Γ2 K

Γ2Γ2−I −TΓ2Γ22ikλI ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠,

− →_p

⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝

p0x

p1x

p2x

p3x ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠,

2.55

then2.54can be rewritten as a boundary integral system:

A ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ a

b

c

d ⎞ ⎟ ⎟ ⎟ ⎟ ⎟

⎠

−

→_{p . 2.56}

Remark 2.4. If the above system2.56has a unique solution, our problem2.5with2.3will have a unique solutionsee13,14.

3. Existence and Uniqueness

Based on the Fredholm theory, we show the existence and uniqueness of a solution to the integral system2.56.

Define

HH−1/2∂D×H−1/2Γ1×H−1/2Γ2×H1/2Γ2 3.1

and its dual space

H∗H1/2∂D×H1/2Γ1×H1/2Γ2×H−1/2Γ2. 3.2

Theorem 3.1. The operatorAmapsHcontinuously intoH∗and is Fredholm with index zero.

Proof. As is known, the operator SDD is positive and bounded below up to a compact

perturbationsee18; that is, there exists a compact operator

such that

Re$SDDLDψ, ψ

%

≥C&&ψ&&2_H−1/2_∂D, forψ∈H−1/2∂D 3.4

where,denote the duality betweenH−1/2_∂DandH1/2_∂D. For convenience, in the following discussion we define

S0SDDLD. 3.5

Similarly, the operators SΩΩ and −TΩΩ are positive and bounded below up to compact

perturbationssee18, that is, there exist compact operators

L_Ω :H−1/2_∂Ω−→H1/2_∂Ω

LT :H1/2∂Ω−→H−1/2∂Ω

3.6

such that

Re$SΩΩLΩψ, ψ%≥C&&ψ&&2_H−1/2_∂D, forψ∈H−1/2∂Ω,

Re$−TΩΩLTϕ, ϕ

%

≥C&&ϕ&&2_H1/2_∂D, forϕ∈H1/2∂Ω.

3.7

DefineS1 SΩΩLΩ and T1 −TΩΩLT, thenS1 and T1 are bounded below up and positive.

Takea∈H−1/2_{∂D, and letb∈H}−1/2_∂Ω,c∈H−1/2_{∂Ω, andd∈H}1/2_{∂Ωbe the} extension by zero to∂Ωofb∈H−1/2Γ1,c∈H−1/2Γ2, andd∈H1/2Γ2,respectively.

Denote→−ξ a, b, c, dT.

It is easy to check that the operatorsSΓ1D,SΓ2D,SDΓ1,SDΓ2,KΓ2D, andKDΓ2are compact operators, and then we can rewriteA→−ξ as the following:

A→−ξ A ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ a

b

c

d ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠A0

⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ a

b

c

d ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠AC

⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ a

b

c

d ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠A0

− →

with AC − → ξ ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝

−LDa|∂DSΓ1DbSΓ2Dc−KΓ2Dd

SDΓ1a−

L_ΩbL_Ωc

Γ1 SDΓ2a−

L_ΩbL_Ωc

Γ2 K_DΓ

2LT d Γ2 ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ , A0 − → ξ ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝

S0a

S1bS1c

Γ1

−KΓ2Γ1d

S1bS1c

Γ2

I−K_Γ2Γ2d

K_Γ 1Γ2b

K_Γ

2Γ2−I

c T1d

Γ2

−2ikλd ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ , 3.9

whereAC:H → H∗is compact andA0 :H → H∗defines a sesquilinear form, that is,

' A0→−ξ ,→−ξ

(

H,H∗ S0a, a

S1bS1c, b

−KΓ2Γ1d, b|Γ1

S1bS1c, c

d, c−K_Γ2Γ2d, c|_Γ 2

K_Γ1Γ2b, d

Γ2

K_Γ2Γ2c, d

Γ2

−c, d T1d,d

−2ikλd, d.

3.10

Hereu, vdenotes the scalar product onL2_∂DorL2_{∂Ωdefined by})

∂Duvdsor

)

∂Ωuvds,

andu, v|_Γi is the scalar product onL2_Γ

i i1,2.

By properties of the operatorsS0,S1, andT1, we have

ReS0a, a≥Ca2_H−1/2_∂D,

Re*S1bS1c,b

S1bS1c, c

T1d,d +

Re*S1

bc,bcT1d,d +

≥C&&&bc&&&

H−1/2_∂Ω && & d&&&

H1/2_∂Ω

Cb2_H−1/2_Γ 1c

2

H−1/2_Γ2d 2

H1/2_Γ 2

Similarly,

Re

K_Γ 1Γ2b, d

Γ2

−K_Γ2Γ1d, b|_Γ 1

Re*b, K_Γ2Γ1d|_Γ

1−KΓ2Γ1d, b|Γ1 +

Re

KΓ2Γ1d, b

Γ1−

KΓ2Γ1d, b|Γ1

0,

Re

K_Γ2Γ2c, d

Γ2

− KΓ2Γ2d, c|Γ2

0,

Red, c−c, d−2idλd, d Re*d, c−d, c−2ikλd, d+

0.

3.12

So the operatorA0is coercive, that is,

Re,A−Acξ, ξ

-H,H∗

≥Cξ2_H forξ∈H, 3.13

whence the operatorAis Fredholm with index zero.

Theorem 3.2. The operatorAhas a trivial kernel if−k2 _{is not Dirichlet eigenvalue of the Laplace}

operator inD.

Proof. In this part, we show that KernA {0}. To this end let→−ψ a, b, c, dT ∈ H be a solution of the homogeneous systemA→−ψ →−0 , and we want to prove that→−ψ ≡→−0 .

However,A→−ψ →−0 means that

SDDaSΓ1DbSΓ2Dc−KΓ2Dd0,

SDΓ1aSΓ1Γ1bSΓ2Γ1c−KΓ2Γ1d0,

SDΓ2aSΓ1Γ2bSΓ2Γ2c I−KΓ2Γ2d0,

K_DΓ 2aK

Γ1Γ2b

K_Γ 2Γ2−I

c−T_Γ2Γ22ikλId0.

3.14

Define a potential

whereb,canddhave the same meaning as before and

SDϕx

∂D

ϕyΦx, ydsy,

S_Ωϕx

∂Ωϕ

yΦx, ydsy,

KΩϕx

∂Ωϕ

y∂Φ

x, y ∂νy dsy.

3.16

This potentialvxsatisfies Helmholtz equation inR2_{\D∪Γand the Sommerfeld radiation} conditionsee13,14.

Considering the potentialvxinsideΩ\Dand approaching the boundary∂Dx → ∂D, we have

vx 1

2{SDDaSΓ1DbSΓ2Dc−KΓ2Dd} 3.17

and3.14implies that

vx|_∂D0. 3.18

Similarly, considering the potential vxinside Ω\D and approaching the boundary∂Ω x → ∂Ω, then restrictingvxto the partial boundaryΓ−₁:

vx|_Γ−

1 1

2{SDΓ1aSΓ1Γ1bSΓ2Γ1c−KΓ2Γ1d}0, x∈Γ−1, 3.19

and restrictingvxto the partial boundaryΓ−₂, we have

vx|_Γ−

2 1

2{SDΓ2aSΓ1Γ2bSΓ2Γ2cd−KΓ2Γ2d}0, x∈Γ−2. 3.20

Now, we consider the potential vx in the regionR2 _{\Ωand approach the boundary∂Ω}

x → ∂Ω, and then restrictingvxto the partial boundaryΓ₁, similar to3.19, we have

vx|_Γ

1 0, x∈Γ

−

1. 3.21

Refering to3.20,

vx|_Γ

2 −v|Γ2 0, x∈Γ

2,

∂vx ∂ν

Γ 2

1

2 K

DΓ2aK

Γ1Γ2b

K_Γ 2Γ2−I

c−TΓ2Γ2d

Combining3.22, from3.14we have

2∂vx

∂ν ikλvx

Γ 2

K_DΓ2aK_Γ1Γ2bK_Γ2Γ2−Ic−TΓ2Γ2d−2ikλd

0.

3.23

From3.18–3.23, the potentialvxsatisfies the following boundary value problem:

Δvk2v0 inR2\D∪Γ,

v± 0 onΓ±1,

v₋ 0 onΓ−₂,

∂v

∂ν ikλv 0 onΓ

2,

v0 on∂D,

3.24

and the Sommerfeld radiation condition:

lim

ν→ ∞

√

r

∂v ∂r −ikv

0 3.25

uniformly inxx/|x|withr|x|.

The uniqueness resultTheorem 2.1inSection 2implies that

vx 0, x∈R2\D∪Γ. 3.26

Notice that−k2_{is not Dirichlet eigenvalue of the Laplace operator inD, and so}

vx 0, x∈D. 3.27

Therefore, the well-known jump relationshipssee13,14imply that

−

→_{ψ a, b, c, d}T _0,0,0,0T_{. 3.28}

So we complete the proof of the theorem.

Combining Theorems3.1and3.2, we have the following

Theorem 3.3. The boundary integral system2.56has a unique solution.

Remark 3.4. If we remove the condition that “−k2 _{is not Dirichlet eigenvalue of the Laplace} operator inD,” instead of it by the assumption that Imk > 0, thenTheorem 2.1inSection 2

Acknowledgment

This research is supported by NSFC Grant no. 10871080, Laboratory of Nonlinear Analysis of CCNU, COCDM of CCNU.

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