Boundary Value Problems
Volume 2009 (2009), Article ID 524846, 19 pages
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

Received 8 September 2009; Accepted 2 November 2009

Academic Editor: Salim Messaoudi

Copyright © 2009 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.

Abstract

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 𝑅 2 as cross section. We assume that the crack Γ 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.

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 in [1] by 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 ̈ o nch considered the same scattering problem for sound-hard crack [2], and in the same year, Alves and Ha Duong discussed the scattering problem but for flat cracks in [3]. Later in 2000, Kress’s work was continued by Kirsch and Ritter in [4] who used the factorization method to reconstruct the shape of the crack from the knowledge of the far-field pattern. In 2003, Cakoni and Colton in [5] considered the direct and inverse scattering problem for cracks which (possibly) coated on one side by a material with surface impedance 𝜆 . Later in 2008, Lee considered an inverse scattering problem from an impedance crack and tried to recover impedance function from the far field pattern in [6]. However, studying an inverse problem always requires a solid knowledge of the corresponding direct 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 in [1, 2, 4–12] and 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 crack Γ and a bounded domain 𝐷 in 𝑅 2 as cross section. We assume that the cylinder is (possibly) partially coated on one side by a material with surface impedance 𝜆 . 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 the 𝑇 𝑀 mode, this leads to a mixed boundary value problem for the Helmholtz equation defined in the exterior of a mixture in 𝑅 2 .

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 problems (see [13–17] and 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 operator 𝐴 which 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 crack Γ and a bounded domain 𝐷 in 𝑅 2 as cross section. For further considerations, we suppose that 𝐷 has smooth boundary 𝜕 𝐷 (e.g., 𝜕 𝐷 ∈ 𝐶 2 ), and the crack Γ (smooth) can be extended to an arbitrary smooth, simply connected, closed curve 𝜕 Ω enclosing a bounded domain Ω such that the normal vector 𝜈 on Γ coincides with the outward normal vector on 𝜕 Ω which we again denote by 𝜈 . The bounded domain 𝐷 is located inside the domain Ω , and ⋂ 𝜕 𝐷 𝜕 Ω = ∅ .

In the whole paper, we assume that 𝜕 𝐷 ∈ 𝐶 2 and 𝜕 Ω ∈ 𝐶 2 .

Suppose that

𝑧 𝑠 Γ = ( 𝑠 ) ∶ 𝑠 ∈ 0 , 𝑠 1 , ( 2 . 1 ) where 𝑧 ∶ [ 𝑠 0 , 𝑠 1 ] → 𝑅 2 is an injective piecewise 𝐶 1 function. We denote the outside of Γ with respect to the chosen orientation by Γ + and the inside by Γ − . Here we suppose that the Γ is divided into two parts Γ 1 and Γ 2 and consider the electromagnetic field E-polarized. Different boundary conditions on Γ ± 1 , Γ ± 2 , and 𝜕 𝐷 lead to the following problem:

Δ 𝑈 + 𝑘 2 𝑈 = 0 i n 𝑅 2 ⧵ , 𝑈 𝐷 ∪ Γ ± = 0 o n Γ ± 1 , 𝑈 − = 0 o n Γ − 2 , 𝜕 𝑈 + 𝜕 𝜈 + 𝑖 𝑘 𝜆 U + = 0 o n Γ + 2 , 𝑈 = 0 o n 𝜕 𝐷 , ( 2 . 2 ) where 𝑈 ± ( 𝑥 ) = l i m ℎ → 0 + 𝑈 ( 𝑥 ± ℎ 𝜈 ) for 𝑥 ∈ Γ and 𝜕 𝑈 ± / 𝜕 𝜈 = l i m ℎ → 0 + 𝜈 ⋅ ∇ 𝑈 ( 𝑥 ± ℎ 𝜈 ) for 𝑥 ∈ Γ . The total field 𝑈 is decomposed into the given incident field 𝑢 𝑖 ( 𝑥 ) = 𝑒 𝑖 𝑘 𝑥 ⋅ 𝑑 , | 𝑑 | = 1 , and the unknown scattered field 𝑢 which is required to satisfy the Sommerfeld radiation condition

l i m 𝜈 → ∞ √ 𝑟 𝜕 𝑢 𝜕 𝑟 − 𝑖 𝑘 𝑢 = 0 ( 2 . 3 ) uniformly in ̂ 𝑥 = 𝑥 / | 𝑥 | with 𝑟 = | 𝑥 | .

We recall some usual Sobolev spaces and some trace spaces on Γ in the following.

Let Γ ⊆ Γ be a piece of the boundary. Use 𝐻 1 ( 𝐷 ) and 𝐻 1 𝑙 𝑜 𝑐 ( 𝑅 2 ⧵ 𝐷 ) to denote the usual Sobolev spaces, 𝐻 1 / 2 ( Γ ) is the trace space, and we define

𝐻 1 / 2 Γ = Γ 𝑢 | ∶ 𝑢 ∈ 𝐻 1 / 2 , 𝐻 ( Γ ) 1 / 2 Γ = 𝑢 ∈ 𝐻 1 / 2 Γ , 𝐻 ( Γ ) ∶ s u p p 𝑢 ⊆ − 1 / 2 Γ = 𝐻 1 / 2 Γ 𝐻 t h e d u a l s p a c e o f 1 / 2 Γ , 𝐻 − 1 / 2 Γ = 𝐻 1 / 2 ( Γ ) t h e d u a l s p a c e o f 𝐻 1 / 2 Γ . ( 2 . 4 ) Just consider the scattered field 𝑢 , then (2.2) and (2.3) are a special case of the following problem.

Given 𝑓 ∈ 𝐻 1 / 2 ( Γ 1 ) , 𝑔 ∈ 𝐻 1 / 2 ( Γ 2 ) , ℎ ∈ 𝐻 − 1 / 2 ( Γ 2 ) , and 𝑟 ∈ 𝐻 1 / 2 ( 𝜕 𝐷 ) find 𝑢 ∈ 𝐻 1 𝑙 𝑜 𝑐 ( 𝑅 2 ⧵ ( 𝐷 ∪ Γ ) ) such that

Δ 𝑢 + 𝑘 2 𝑢 = 0 i n 𝑅 2 ⧵ , 𝑢 𝐷 ∪ Γ ± = 𝑓 o n Γ ± 1 , 𝑢 − = 𝑔 o n Γ − 2 , 𝜕 𝑢 + 𝜕 𝜈 + 𝑖 𝑘 𝜆 𝑢 + = ℎ o n Γ + 2 , 𝑢 = 𝑟 o n 𝜕 𝐷 , ( 2 . 5 ) and 𝑢 is required to satisfy the Sommerfeld radiation condition (2.3). For simplicity, we assume that 𝑘 > 0 and 𝜆 > 0 .

Theorem 2.1. The problems (2.5) and (2.3) have at most one solution.

Proof. Let 𝑢 be a solution to the problem (2.5) with 𝑓 = 𝑔 = ℎ = 𝑟 = 0 , we want to show that 𝑢 = 0 in 𝑅 2 ⧵ ( 𝐷 ∪ Γ ) .

Suppose that 𝐵 𝑅 (with boundary 𝜕 𝐵 𝑅 ) is a sufficiently large ball which contains the domain Ω . Obviously, to the Helmholtz equation in (2.5), the solution 𝑢 ∈ 𝐻 1 ( 𝐵 𝑅 ⧵ ⋃ 𝐻 Ω ) 1 ( Ω ⧵ 𝐷 ) satisfies the following transmission boundary conditions on the complementary part 𝜕 Ω ⧵ Γ of 𝜕 Ω :

𝑢 + = 𝑢 − , 𝜕 𝑢 + = 𝜕 𝜈 𝜕 𝑢 − , 𝜕 𝜈 ( 2 . 6 ) where “ ± ” denote the limit approaching 𝜕 Ω from outside and inside Ω , respectively. Applying Green’s formula for 𝑢 and 𝑢 in Ω ⧵ 𝐷 and 𝐵 𝑅 ⧵ Ω , we have

Ω ⧵ 𝐷 𝑢 Δ 𝑢 + ∇ 𝑢 ⋅ ∇ 𝑢 𝑑 𝑥 = 𝜕 Ω ⧵ Γ 𝑢 − 𝜕 𝑢 − 𝜕 𝜈 𝑑 s + Γ 1 𝑢 − 𝜕 𝑢 − 𝜕 𝜈 𝑑 𝑠 + Γ 2 𝑢 − 𝜕 𝑢 − 𝜕 𝜈 𝑑 𝑠 , 𝐵 𝑅 ⧵ Ω 𝑢 Δ 𝑢 + ∇ 𝑢 ⋅ ∇ 𝑢 = 𝑑 𝑥 𝜕 𝐵 𝑅 𝑢 𝜕 𝑢 𝜕 𝜈 𝑑 𝑠 + 𝜕 Ω ⧵ Γ 𝑢 + 𝜕 𝑢 + 𝜕 𝜈 𝑑 𝑠 + Γ 1 𝑢 + 𝜕 𝑢 + 𝜕 𝜈 𝑑 𝑠 + Γ 2 𝑢 + 𝜕 𝑢 + 𝜕 𝜈 𝑑 𝑠 , ( 2 . 7 ) where 𝜈 is directed into the exterior of the corresponding domain.

Using boundary conditions on Γ 1 , Γ 2 and the above transmission boundary condition (2.6), we have

𝜕 𝐵 𝑅 𝑢 𝜕 𝑢 𝜕 𝜈 𝑑 𝑠 = 𝐵 𝑅 ⧵ Ω + Ω ⧵ 𝐷 | | | | ∇ 𝑢 2 − 𝑘 2 | 𝑢 | 2 𝑑 𝑥 + Γ 2 | | 𝑢 𝑖 𝑘 𝜆 + | | 2 d 𝑠 . ( 2 . 8 ) Hence

I m 𝜕 𝐵 𝑅 𝑢 𝜕 𝑢 𝜕 𝜈 𝑑 𝑠 ≥ 0 . ( 2 . 9 ) So, from [13, Theorem 2 . 1 2 ] and a unique continuation argument we obtain that 𝑢 = 0 in 𝑅 2 ⧵ ( 𝐷 ∪ Γ ) .

We use [ 𝑢 ] = 𝑢 − − 𝑢 + and [ 𝜕 𝑢 / 𝜕 𝜈 ] = ( 𝜕 𝑢 − / 𝜕 𝜈 ) − ( 𝜕 𝑢 + / 𝜕 𝜈 ) to denote the jump of 𝑢 and 𝜕 𝑢 / 𝜕 𝜈 across the crack Γ , respectively. Then we have the following.

Lemma 2.2. If 𝑢 is a solution of (2.5) and (2.3), then 𝐻 [ 𝑢 ] ∈ 1 / 2 ( Γ ) and 𝐻 [ 𝜕 𝑢 / 𝜕 𝜈 ] ∈ − 1 / 2 ( Γ ) .

The proof of this lemma can be found in [11].

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

𝑢 ( 𝑥 ) = 𝜕 Ω 𝜕 𝑢 𝜕 𝜈 Φ ( 𝑥 , 𝑦 ) − 𝑢 𝜕 Φ ( 𝑥 , 𝑦 ) 𝜕 𝜈 𝑑 𝑠 𝑦 + 𝜕 𝐷 𝜕 𝑢 𝜕 𝜈 Φ ( 𝑥 , 𝑦 ) − 𝑢 𝜕 Φ ( 𝑥 , 𝑦 ) 𝜕 𝜈 𝑑 𝑠 𝑦 ( 2 . 1 0 ) and for 𝑥 ∈ R 2 ⧵ Ω

𝑢 ( 𝑥 ) = 𝜕 Ω 𝑢 𝜕 Φ ( 𝑥 , 𝑦 ) − 𝜕 𝜈 𝜕 𝑢 𝜕 𝜈 Φ ( 𝑥 , 𝑦 ) 𝑑 𝑠 𝑦 , ( 2 . 1 1 ) where

𝑖 Φ ( 𝑥 , 𝑦 ) = 4 𝐻 0 ( 1 ) 𝑘 | | | | 𝑥 − 𝑦 ( 2 . 1 2 ) is the fundamental solution to the Helmholtz equation in 𝑅 2 , and 𝐻 0 ( 1 ) 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 (see [5, 11]) and approaching the boundary 𝜕 Ω from inside Ω ⧵ 𝐷 , we obtain (for 𝑥 ∈ 𝜕 Ω )

𝑢 − ( 𝑥 ) = 𝑆 Ω Ω 𝜕 𝑢 − 𝜕 𝜈 − 𝐾 Ω Ω 𝑢 − + 2 𝜕 𝐷 𝜕 𝑢 ( 𝑦 ) 𝜕 𝜈 Φ ( 𝑥 , 𝑦 ) − 𝑢 ( 𝑦 ) 𝜕 Φ ( 𝑥 , 𝑦 ) 𝜕 𝜈 𝑑 𝑠 𝑦 , ( 2 . 1 3 ) 𝜕 𝑢 − ( 𝑥 ) 𝜕 𝜈 = 𝐾 Ω Ω 𝜕 𝑢 − 𝜕 𝜈 − 𝑇 Ω Ω 𝑢 − 𝜕 + 2 𝜕 𝜈 ( 𝑥 ) 𝜕 𝐷 𝜕 𝑢 ( 𝑦 ) 𝜕 𝜈 Φ ( 𝑥 , 𝑦 ) − 𝑢 ( 𝑦 ) 𝜕 Φ ( 𝑥 , 𝑦 ) 𝜕 𝜈 𝑑 𝑠 𝑦 , ( 2 . 1 4 ) where 𝑆 Ω Ω , 𝐾 Ω Ω , 𝐾 Ω Ω , and 𝑇 Ω Ω are boundary integral operators:

𝑆 Ω Ω ∶ 𝐻 − 1 / 2 ( 𝜕 Ω ) ⟶ 𝐻 1 / 2 ( 𝜕 Ω ) , 𝐾 Ω Ω ∶ 𝐻 1 / 2 ( 𝜕 Ω ) ⟶ 𝐻 1 / 2 𝐾 ( 𝜕 Ω ) Ω Ω ∶ 𝐻 − 1 / 2 ( 𝜕 Ω ) ⟶ 𝐻 − 1 / 2 ( 𝜕 Ω ) , 𝑇 Ω Ω ∶ 𝐻 1 / 2 ( 𝜕 Ω ) ⟶ 𝐻 − 1 / 2 ( 𝜕 Ω ) , ( 2 . 1 5 ) defined by (for 𝑥 ∈ 𝜕 Ω )

𝑆 Ω Ω 𝜑 ( 𝑥 ) = 2 𝜕 Ω 𝜑 ( 𝑦 ) Φ ( 𝑥 , 𝑦 ) 𝑑 𝑠 𝑦 , 𝐾 Ω Ω 𝜑 ( 𝑥 ) = 2 𝜕 Ω 𝜑 ( 𝑦 ) Φ ( 𝑥 , 𝑦 ) 𝜕 𝜈 𝑦 𝑑 𝑠 𝑦 , 𝐾 Ω Ω 𝜑 ( 𝑥 ) = 2 𝜕 Ω 𝜑 ( 𝑦 ) 𝜕 Φ ( 𝑥 , 𝑦 ) 𝜕 𝜈 𝑥 𝑑 𝑠 𝑦 , 𝑇 Ω Ω 𝜕 𝜑 ( 𝑥 ) = 2 𝜕 𝜈 𝑥 𝜕 Ω 𝜑 ( 𝑦 ) 𝜕 Φ ( 𝑥 , 𝑦 ) 𝜕 𝜈 𝑦 𝑑 𝑠 𝑦 . ( 2 . 1 6 ) Similarly, approaching the boundary 𝜕 Ω from inside 𝑅 2 ⧵ Ω we obtain (for 𝑥 ∈ 𝜕 Ω )

𝑢 + ( 𝑥 ) = − 𝑆 Ω Ω 𝜕 𝑢 + 𝜕 𝜈 + 𝐾 Ω Ω 𝑢 + , ( 2 . 1 7 ) 𝜕 𝑢 + ( 𝑥 ) 𝜕 𝜈 = − 𝐾 Ω Ω 𝜕 𝑢 + 𝜕 𝜈 + 𝑇 Ω Ω 𝑢 + . ( 2 . 1 8 ) From (2.13)–(2.18), we have

𝑢 − + 𝑢 + = 𝑆 Ω Ω 𝜕 𝑢 − − 𝜕 𝜈 𝜕 𝑢 + 𝜕 𝜈 − 𝐾 Ω Ω 𝑢 − − 𝑢 + + 2 𝜕 𝐷 𝜕 𝑢 ( 𝑦 ) 𝜕 𝜈 Φ ( 𝑥 , 𝑦 ) − 𝑢 ( 𝑦 ) 𝜕 Φ ( 𝑥 , 𝑦 ) 𝜕 𝜈 𝑑 𝑠 𝑦 , ( 2 . 1 9 ) 𝜕 𝑢 − + 𝜕 𝜈 𝜕 𝑢 + 𝜕 𝜈 = 𝐾 Ω Ω 𝜕 𝑢 − − 𝜕 𝜈 𝜕 𝑢 + 𝜕 𝜈 − 𝑇 Ω Ω 𝑢 − − 𝑢 + 𝜕 + 2 𝜕 𝜈 ( 𝑥 ) 𝜕 𝐷 𝜕 𝑢 ( 𝑦 ) 𝜕 𝜈 Φ ( 𝑥 , 𝑦 ) − 𝑢 ( 𝑦 ) 𝜕 Φ ( 𝑥 , 𝑦 ) 𝜕 𝜈 𝑑 𝑠 𝑦 . ( 2 . 2 0 ) Restrict 𝑢 on Γ ± 1 , from (2.19) we have

2 𝑓 ( 𝑥 ) = 𝑆 Ω Ω 𝜕 𝑢 − − 𝜕 𝜈 𝜕 𝑢 + | | | | 𝜕 𝜈 Γ 1 − 𝐾 Ω Ω ( 𝑢 − − 𝑢 + ) | | Γ 1 + 2 𝜕 𝐷 𝜕 𝑢 ( 𝑦 ) 𝜕 𝜈 Φ ( 𝑥 , 𝑦 ) 𝑑 𝑠 𝑦 | | | | Γ 1 − 2 𝜕 𝐷 𝑢 ( 𝑦 ) 𝜕 Φ ( 𝑥 , 𝑦 ) 𝜕 𝜈 𝑑 𝑠 𝑦 | | | | Γ 1 ( 2 . 2 1 ) where ( ⋅ ) | Γ 1 means a restriction to Γ 1 .

Define

𝑆 Ω Γ 1 𝜑 ( 𝑥 ) = 2 𝜕 Ω 𝜑 ( 𝑦 ) Φ ( 𝑥 , 𝑦 ) 𝑑 𝑠 𝑦 | | | | Γ 1 , 𝐾 Ω Γ 1 𝜑 ( 𝑥 ) = 2 𝜕 Ω 𝜕 Φ ( 𝑥 , 𝑦 ) 𝜕 𝜈 𝜑 ( 𝑦 ) 𝑑 𝑠 𝑦 | | | | Γ 1 , 𝑆 𝐷 Γ 1 𝜑 ( 𝑥 ) = 2 𝜕 𝐷 𝜑 ( 𝑦 ) Φ ( 𝑥 , 𝑦 ) 𝑑 𝑠 𝑦 | | | | Γ 1 , 𝜕 𝑢 | | | 𝜕 𝜈 𝜕 𝐷 = 𝑎 , 𝜕 𝑢 | | | 𝜕 𝜈 Γ 1 = 𝜕 𝑢 − − 𝜕 𝜈 𝜕 𝑢 + | | | | 𝜕 𝜈 Γ 1 = 𝑏 , 𝜕 𝑢 | | | 𝜕 𝜈 Γ 2 = 𝜕 𝑢 − − 𝜕 𝜈 𝜕 𝑢 + | | | | 𝜕 𝜈 Γ 2 [ 𝑢 ] | | = 𝑐 , Γ 2 = 𝑢 − − 𝑢 + | | Γ 2 = 𝑑 . ( 2 . 2 2 ) Then zero extend 𝑏 , 𝑐 , and 𝑑 to the whole 𝜕 Ω in the following:

̃ 𝑏 = 0 , o n 𝜕 Ω ⧵ Γ 1 , 𝑏 , o n Γ 1 , ̃ 𝑐 = 0 , o n 𝜕 Ω ⧵ Γ 2 , 𝑐 , o n Γ 2 , 𝑑 = 0 , o n 𝜕 Ω ⧵ Γ 2 , 𝑑 , o n Γ 2 . ( 2 . 2 3 ) By using the boundary conditions in (2.5), we rewrite (2.21) as

𝑆 𝐷 Γ 1 𝑎 + 𝑆 Ω Γ 1 ̃ 𝑏 + ̃ 𝑐 − 𝐾 Ω Γ 1 𝑑 = 𝑝 1 ( 𝑥 ) , ( 2 . 2 4 ) where

𝑝 1 ( 𝑥 ) = 2 𝑓 ( 𝑥 ) + 2 𝜕 𝐷 𝜕 Φ ( 𝑥 , 𝑦 ) 𝜕 𝜈 ( 𝑦 ) 𝑟 ( 𝑦 ) 𝑑 𝑠 𝑦 | | | | Γ 1 . ( 2 . 2 5 ) Furthermore, we modify (2.24) as

𝑆 𝐷 Γ 1 𝑎 + 𝑆 Γ 1 Γ 1 𝑏 + 𝑆 Γ 2 Γ 1 𝑐 − 𝐾 Γ 2 Γ 1 𝑑 = 𝑝 1 ( 𝑥 ) , ( 2 . 2 6 ) where the operator 𝑆 Γ 2 Γ 1 is the operator applied to a function with s u p p ⊆ Γ 2 and evaluated on Γ 1 , with analogous definition for 𝑆 𝐷 Γ 1 , 𝑆 Γ 1 Γ 1 , and 𝐾 Γ 2 Γ 1 . We have mapping properties (see [5, 11])

𝑆 𝐷 Γ 1 ∶ 𝐻 − 1 / 2 ( 𝜕 𝐷 ) ⟶ 𝐻 1 / 2 Γ 1 , 𝑆 Γ 1 Γ 1 ∶ 𝐻 − 1 / 2 Γ 1 ⟶ 𝐻 1 / 2 Γ 1 , 𝑆 Γ 2 Γ 1 ∶ 𝐻 − 1 / 2 Γ 2 ⟶ 𝐻 1 / 2 Γ 1 , 𝐾 Γ 2 Γ 1 ∶ 𝐻 1 / 2 Γ 2 ⟶ 𝐻 1 / 2 Γ 1 . ( 2 . 2 7 ) Again from (2.13)–(2.18), restricting 𝑢 to boundary Γ − 2 we have

2 𝑔 ( 𝑥 ) = 𝑆 Ω Ω 𝜕 𝑢 − − 𝜕 𝜈 𝜕 𝑢 + | | | | 𝜕 𝜈 Γ 2 − 𝐾 Ω Ω ( 𝑢 − − 𝑢 + ) | | Γ 2 + ( 𝑢 − − 𝑢 + ) | | Γ 2 + 2 𝜕 𝐷 𝜕 𝑢 ( 𝑦 ) 𝜕 𝜈 ( 𝑦 ) Φ ( 𝑥 , 𝑦 ) 𝑑 𝑠 𝑦 | | | | Γ 2 − 2 𝜕 𝐷 𝜕 Φ ( 𝑥 , 𝑦 ) 𝜕 𝜈 ( 𝑦 ) 𝑟 ( 𝑦 ) 𝑑 𝑠 𝑦 | | | | Γ 2 ( 2 . 2 8 ) or

2 𝜕 𝐷 𝜕 𝑢 ( 𝑦 ) 𝜕 𝜈 ( 𝑦 ) Φ ( 𝑥 , 𝑦 ) 𝑑 𝑠 𝑦 | | | | Γ 2 + 𝑆 Ω Ω 𝜕 𝑢 − − 𝜕 𝜈 𝜕 𝑢 + | | | | 𝜕 𝜈 Γ 2 − 𝐾 Ω Ω ( 𝑢 − − 𝑢 + ) | | Γ 2 + ( 𝑢 − − 𝑢 + ) | | Γ 2 = 2 𝑔 ( 𝑥 ) + 2 𝜕 𝐷 𝜕 Φ ( 𝑥 , 𝑦 ) 𝜕 𝜈 ( 𝑦 ) 𝑟 ( 𝑦 ) 𝑑 𝑠 𝑦 | | | | Γ 2 . ( 2 . 2 9 ) Like previous, define

𝑆 Ω Γ 2 𝜑 ( 𝑥 ) = 2 𝜕 Ω 𝜑 ( 𝑦 ) Φ ( 𝑥 , 𝑦 ) 𝑑 𝑠 𝑦 | | | | Γ 2 , 𝐾 Ω Γ 2 𝜑 ( 𝑥 ) = 2 𝜕 Ω 𝜕 Φ ( 𝑥 , 𝑦 ) 𝜕 𝜈 𝜑 ( 𝑦 ) 𝑑 𝑠 𝑦 | | | | Γ 2 , 𝑆 𝐷 Γ 2 𝜑 ( 𝑥 ) = 2 𝜕 𝐷 𝜑 ( 𝑦 ) Φ ( 𝑥 , 𝑦 ) 𝑑 𝑠 𝑦 | | | | Γ 2 . ( 2 . 3 0 ) Then we can rewrite (2.29) as

𝑆 𝐷 Γ 2 𝑎 + 𝑆 Ω Γ 2 ̃ 𝑏 + ̃ 𝑐 − 𝐾 Ω Γ 2 𝑑 + 𝑑 = 𝑝 2 ( 𝑥 ) , 𝑥 ∈ Γ − 2 , ( 2 . 3 1 ) where

𝑝 2 ( 𝑥 ) = 2 𝑔 ( 𝑥 ) + 2 𝜕 𝐷 𝜕 Φ ( 𝑥 , 𝑦 ) 𝜕 𝜈 ( 𝑦 ) 𝑟 ( 𝑦 ) 𝑑 𝑠 𝑦 | | | | Γ − 2 . ( 2 . 3 2 ) Similar to (2.26), we modify (2.31) as

𝑆 𝐷 Γ 2 𝑎 + 𝑆 Γ 1 Γ 2 𝑏 + 𝑆 Γ 2 Γ 2 𝑐 + 𝐼 − 𝐾 Γ 2 Γ 2 𝑑 = 𝑝 2 ( 𝑥 ) ( 2 . 3 3 ) and we have mapping properties:

𝑆 𝐷 Γ 2 ∶ 𝐻 − 1 / 2 ( 𝜕 𝐷 ) ⟶ 𝐻 1 / 2 Γ 2 , 𝑆 Γ 2 Γ 2 ∶ 𝐻 − 1 / 2 Γ 2 ⟶ 𝐻 1 / 2 Γ 2 , 𝑆 Γ 1 Γ 2 ∶ 𝐻 − 1 / 2 Γ 1 ⟶ 𝐻 1 / 2 Γ 2 , 𝐾 Γ 2 Γ 2 ∶ 𝐻 1 / 2 Γ 2 ⟶ 𝐻 1 / 2 Γ 2 . ( 2 . 3 4 ) Combining (2.13) and (2.14),

𝑆 − 𝑖 𝑘 𝜆 Ω Ω 𝜕 𝑢 − 𝜕 𝜈 − 𝐾 Ω Ω 𝑢 − 𝑢 = − 𝑖 𝑘 𝜆 − − 2 𝜕 𝐷 𝜕 𝑢 ( 𝑦 ) 𝜕 𝜈 Φ ( 𝑥 , 𝑦 ) − 𝑢 𝜕 Φ ( 𝑥 , 𝑦 ) 𝜕 𝜈 ( 𝑦 ) 𝑑 𝑠 𝑦 = − 𝑖 𝑘 𝜆 𝑢 − + 2 𝑖 𝑘 𝜆 𝜕 𝐷 𝜕 𝑢 ( 𝑦 ) 𝜕 𝜈 Φ ( 𝑥 , 𝑦 ) 𝑑 𝑠 𝑦 − 2 𝑖 𝑘 𝜆 𝜕 𝐷 𝑟 ( 𝑦 ) 𝜕 Φ ( 𝑥 , 𝑦 ) 𝜕 𝜈 ( 𝑦 ) 𝑑 𝑠 𝑦 , − 𝐾 Ω Ω 𝜕 𝑢 − 𝜕 𝜈 + 𝑇 Ω Ω 𝑢 − = − 𝜕 𝑢 − 𝜕 𝜕 𝜈 + 2 𝜕 𝜈 ( 𝑥 ) 𝜕 𝐷 𝜕 𝑢 ( 𝑦 ) 𝜕 𝜈 Φ ( 𝑥 , 𝑦 ) 𝑑 𝑠 𝑦 𝜕 − 2 𝜕 𝜈 ( 𝑥 ) 𝜕 𝐷 𝑟 ( 𝑦 ) 𝜕 Φ ( 𝑥 , 𝑦 ) 𝜕 𝜈 𝑑 𝑠 𝑦 , ( 2 . 3 5 ) − 𝑖 𝑘 𝜆 𝑢 − − 𝜕 𝑢 − 𝑢 𝜕 𝜈 = − 𝑖 𝑘 𝜆 − − 𝑢 + − 𝜕 𝑢 − − 𝜕 𝜈 𝜕 𝑢 + 𝜕 𝜈 − 𝑖 𝑘 𝜆 𝑢 + − 𝜕 𝑢 + . 𝜕 𝜈 ( 2 . 3 6 ) Using (2.17) and (2.18),

𝜕 𝑢 + 𝜕 𝜈 + 𝑖 𝑘 𝜆 𝑢 + = − 𝐾 Ω Ω 𝜕 𝑢 + 𝜕 𝜈 + 𝑇 Ω Ω 𝑢 + 𝐾 + 𝑖 𝑘 𝜆 Ω Ω 𝑢 + − 𝑆 Ω Ω 𝜕 𝑢 + 𝜕 𝜈 = 𝐾 Ω Ω 𝜕 𝑢 − − 𝜕 𝜈 𝜕 𝑢 + 𝜕 𝜈 − 𝑇 Ω Ω 𝑢 − − 𝑢 + + 𝑖 𝑘 𝜆 𝑆 Ω Ω 𝜕 𝑢 − − 𝜕 𝜈 𝜕 𝑢 + 𝜕 𝜈 − 𝑖 𝑘 𝜆 𝐾 Ω Ω 𝑢 − − 𝑢 + 𝑆 − 𝑖 𝑘 𝜆 Ω Ω 𝜕 𝑢 − 𝜕 𝜈 − 𝐾 Ω Ω 𝑢 − − 𝐾 Ω Ω 𝜕 𝑢 − 𝜕 𝜈 + 𝑇 Ω Ω 𝑢 − = 𝐾 Ω Ω 𝜕 𝑢 − − 𝜕 𝜈 𝜕 𝑢 + 𝜕 𝜈 − 𝑇 Ω Ω 𝑢 − − 𝑢 + + 𝑖 𝑘 𝜆 𝑆 Ω 𝜕 𝑢 − − 𝜕 𝜈 𝜕 𝑢 + 𝜕 𝜈 − 𝑖 𝑘 𝜆 𝐾 Ω Ω 𝑢 − − 𝑢 + − 𝑖 𝑘 𝜆 𝑢 − − 𝜕 𝑢 − 𝜕 𝜈 + 2 𝑖 𝑘 𝜆 𝜕 𝐷 𝜕 𝑢 ( 𝑦 ) 𝜕 𝜈 Φ ( 𝑥 , 𝑦 ) 𝑑 𝑠 𝑦 𝜕 + 2 𝜕 𝜈 ( 𝑥 ) 𝜕 𝐷 𝜕 𝑢 ( 𝑦 ) 𝜕 𝜈 Φ ( 𝑥 , 𝑦 ) 𝑑 𝑠 𝑦 − 2 𝑖 𝑘 𝜆 𝜕 𝐷 𝑟 ( 𝑦 ) 𝜕 Φ ( 𝑥 , 𝑦 ) 𝜕 𝜈 ( 𝑦 ) 𝑑 𝑠 𝑦 𝜕 − 2 𝜕 𝜈 ( 𝑥 ) 𝜕 𝐷 𝑟 ( 𝑦 ) 𝜕 Φ ( 𝑥 , 𝑦 ) 𝜕 𝜈 ( 𝑦 ) 𝑑 𝑠 𝑦 . ( 2 . 3 7 ) Then using (2.36),

2 𝜕 𝑢 + 𝜕 𝜈 + 𝑖 𝑘 𝜆 𝑢 + = 𝐾 Ω Ω 𝜕 𝑢 − − 𝜕 𝜈 𝜕 𝑢 + 𝜕 𝜈 − 𝑇 Ω Ω 𝑢 − − 𝑢 + + 𝑖 𝑘 𝜆 𝑆 Ω Ω 𝜕 𝑢 − − 𝜕 𝜈 𝜕 𝑢 + 𝜕 𝜈 − 𝑖 𝑘 𝜆 𝐾 Ω Ω 𝑢 − − 𝑢 + 𝑢 − 𝑖 𝑘 𝜆 − − 𝑢 + − 𝜕 𝑢 − − 𝜕 𝜈 𝜕 𝑢 + 𝜕 𝜈 + 2 𝑖 𝑘 𝜆 𝜕 𝐷 𝜕 𝑢 ( 𝑦 ) 𝜕 𝜈 Φ ( 𝑥 , 𝑦 ) 𝑑 𝑠 𝑦 𝜕 + 2 𝜕 𝜈 ( 𝑥 ) 𝜕 𝐷 𝜕 𝑢 ( 𝑦 ) 𝜕 𝜈 Φ ( 𝑥 , 𝑦 ) 𝑑 𝑠 𝑦 − 2 𝑖 𝑘 𝜆 𝜕 𝐷 𝑟 ( 𝑦 ) 𝜕 Φ ( 𝑥 , 𝑦 ) 𝜕 𝜈 ( 𝑦 ) 𝑑 𝑠 𝑦 𝜕 − 2 𝜕 𝜈 ( 𝑥 ) 𝜕 𝐷 𝑟 ( 𝑦 ) 𝜕 Φ ( 𝑥 , 𝑦 ) 𝜕 𝜈 ( 𝑦 ) 𝑑 𝑠 𝑦 . ( 2 . 3 8 ) From (2.29), we have

𝑆 2 𝑖 𝑘 𝜆 𝑔 ( 𝑥 ) = 𝑖 𝑘 𝜆 Ω Ω 𝜕 𝑢 − − 𝜕 𝜈 𝜕 𝑢 + | | | | 𝜕 𝜈 Γ 2 − 𝐾 Ω Ω 𝑢 − − 𝑢 + | | Γ 2 + 𝑢 − − 𝑢 + | | Γ 2 + 2 𝜕 𝐷 𝜕 𝑢 ( 𝑦 ) 𝜕 𝜈 Φ ( 𝑥 , 𝑦 ) 𝑑 𝑠 𝑦 | | | | Γ 2 − 2 𝜕 𝐷 𝑟 ( 𝑦 ) 𝜕 Φ ( 𝑥 , 𝑦 ) 𝜕 𝜈 ( 𝑦 ) 𝑑 s 𝑦 | | | | Γ 2 . ( 2 . 3 9 ) Restricting (2.38) to Γ + 2 and using (2.39), we modify (2.38) as

2 𝜕 𝜕 𝜈 ( 𝑥 ) 𝜕 𝐷 𝜕 𝑢 ( 𝑦 ) 𝜕 𝜈 Φ ( 𝑥 , 𝑦 ) 𝑑 𝑠 𝑦 | | | | Γ + 2 + 𝐾 Ω Ω 𝜕 𝑢 − − 𝜕 𝜈 𝜕 𝑢 + | | | | 𝜕 𝜈 Γ + 2 − 𝑇 Ω Ω ( 𝑢 − − 𝑢 + ) | | Γ + 2 − 𝜕 𝑢 − − 𝜕 𝜈 𝜕 𝑢 + | | | | 𝜕 𝜈 Γ + 2 − 2 𝑖 𝑘 𝜆 ( 𝑢 − − 𝑢 + ) | | Γ + 2 = 𝑝 3 ( 𝑥 ) , ( 2 . 4 0 ) where

𝑝 3 ( 𝑥 ) = 2 ℎ ( 𝑥 ) − 2 𝑖 𝑘 𝜆 𝑔 ( 𝑥 ) + 𝜕 𝐷 𝑟 ( 𝑦 ) 𝜕 Φ ( 𝑥 , 𝑦 ) 𝜕 𝜈 ( 𝑦 ) 𝑑 𝑠 𝑦 | | | | Γ + 2 ( 2 . 4 1 ) for 𝑥 ∈ Γ + .

Define

𝐾 𝐷 Γ 2 𝜑 𝜕 ( 𝑥 ) = 2 𝜕 𝜈 ( 𝑥 ) 𝜕 𝐷 𝜑 ( 𝑦 ) Φ ( 𝑥 , 𝑦 ) 𝑑 𝑠 𝑦 | | | | Γ + 2 , ( 2 . 4 2 ) and using the notation in previous, we can rewrite (2.40) as

𝐾 𝐷 Γ 2 𝑎 + 𝐾 Ω Γ 2 ̃ 𝑏 + ̃ 𝑐 − 𝑇 Ω Γ 2 𝑑 − 𝑐 − 2 𝑖 𝑘 𝜆 𝑑 = 𝑝 3 ( 𝑥 ) ( 2 . 4 3 ) or

𝐾 𝐷 Γ 2 𝑎 + 𝐾 Γ 1 Γ 2 𝐾 𝑏 + Γ 2 Γ 2 𝑇 − 𝐼 𝑐 − Γ 2 Γ 2 + 2 𝑖 𝑘 𝜆 𝐼 𝑑 = 𝑝 3 ( 𝑥 ) , ( 2 . 4 4 ) where the operators 𝐾 Γ 1 Γ 2 ,