# Multilayered Scattering Problem with Generalized Impedance Boundary Condition on the Core

Authors: Jun Guo; Guozheng Yan; Mingjian Cai
Journal: Journal of Applied Mathematics
 (2015)

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## Abstract

This paper is concerned with the scattering problem of time-harmonic acoustic plane waves by an impenetrable obstacle buried in a piecewise homogeneous medium. The so-called generalized impedance boundary condition is imposed on the boundary of the obstacle. Firstly, the well posedness of the solution to the direct scattering problem is established by using the boundary integral method. Then a uniqueness result for the inverse scattering problem is proved; that is, both of the obstacle’s shape and the impedances (μ, λ) can be uniquely determined from far field measurements. Furthermore, a mathematical basis is given to reconstruct the shape of the obstacle by using a modified linear sampling method.


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## Body

## 1. Introduction

This work is concerned with the scattering problem of time-harmonic acoustic plane waves by an impenetrable obstacle buried in a piecewise homogeneous medium. We set the generalized impedance boundary condition (GIBC) on the boundary of the obstacle and the transmission boundary conditions on the surface of the layered medium. The GIBC is commonly used to model thin coatings or gratings as well as more accurate models for imperfectly conducting obstacles. Addressing this problem is motivated by applications in nondestructive testing, medical imaging, remote sensing or radar, and so on; at the same time the background may be modeled as a layered medium. For simplicity, we just consider that the unknown obstacle is embedded in a two-layered medium, and the space isR2.To be precise, letD2⊂R2 denote the impenetrable obstacle which is a bounded domain with a smooth boundary S1 (e.g., C2). Assume that the unknown obstacle D2 is buried in a penetrable obstacle D with a closed C2 surface S0 such that D¯2⊂D. Denote by D1=D∖D¯2 a connected bounded domain filled with homogeneous medium and denote by D0=R2∖D¯ the unbounded connected domain occupied by another homogeneous medium. Let kl=ωl/cl>0 be the wave number in terms of the frequency ωl and the sound speed cl in the corresponding region Dl (l=0,1) (see Figure 1).Figure 1
Multilayered scattering problem.The scattering of time-harmonic acoustic plane waves by an obstacle with GIBC in a piecewise homogeneous medium inR2 can be modeled by the Helmholtz equation with boundary conditions on the boundary S1 and interface S0:(1)Δu+k02u=0,inD0,Δv+k12v=0,inD1,u+-v-=0,onS0,∂u+∂ν-λ0∂v-∂ν=0,onS0,∂v∂ν+divS1μ∇S1v+λv=0,onS1.Here ν is the unit outward normal vector on the boundary S0 or S1; u+, ∂u+/∂ν(v-,∂v-/∂ν) denote the limit of u, ∂u/∂ν(v,∂v/∂ν) on the boundary S0 from the exterior (interior) of D.Remark 1.
In the following discussion, we use “(·)±” or “(·)±” to denote the limit approaching the boundary from outside and inside to the corresponding domain, respectively.The constant surface impedance onS0 is supposed λ0>0 which is given by λ0=ρ0/ρ1 in terms of the density ρl in the corresponding region Dl (l=0,1). On the boundary S1, the impedances μ∈C1(S1) and λ∈C(S1) are complex-valued functions satisfying Im⁡(μ)≤0 and Im⁡(λ)≥0. The surface divergence divS1 and the surface gradient ∇S1 are precisely defined in Chapter 5 of [1]. In the two-dimensional case, the inhomogeneous Laplace-Beltrami differential operator becomes divS1(μ∇S1v)=d/dsμdv/ds, where d/ds is the tangential derivative and s is the arc length.The total fieldu=us+ui is decomposed into the given incident field ui=eik0x·d, d∈S1, (the unit sphere in R2) and the unknown scattered field us which is required to satisfy the Sommerfeld radiation condition [2](2)limr→∞⁡r∂us∂r-ik0us=0uniformly in x^=x/x with r=x. Further it is known that the scattered field us(x,d) has the following asymptotic representation:(3)usx,d=eik0xxu∞x^,d+O1xasx⟶∞uniformly for all directions x^, where the function u∞(x^,d) defined on the unit sphere S1 is known as the far field pattern with x^ and d denoting, respectively, the observation direction and the incident direction.Thedirect problem is to seek functions u∈Hloc1(D0) and v∈H1(D1) satisfying (1) and (2). In the next section, more general direct problem (4) will be considered. If the impenetrable obstacle D2 with GIBC is set in a homogeneous medium, it was shown in [3] that there exists a unique solution for the case when the data h∈H-1(S1) by the variational method; but for the case when h belongs to H-3/2(S1), this method is no longer valid and the difficulty has been resolved in [4] by the integral equation method with the help of the modified Green function technique in [5]. More related works can be found in [6, 7]. In this paper we will employ the integral equation method to solve direct problem (4) in some Sobolev spaces. The main challenge is to derive a suitable boundary integral system and show that the corresponding boundary integral operators are Fredholm of index zero.Theinverse problem we consider in this paper is to determine the shape of the obstacle D2 and (μ,λ) from the knowledge of the far field pattern u∞(x^,d) for all x^,d∈S1 with the given wave number kl(l=0,1) and the positive constant λ0.As usual in most of the inverse problems, the first issue is the uniqueness, that is, in what conditions, the shape of the obstacleD2 (or the parameters such as (μ,λ)) can be uniquely determined by the far field pattern. Through establishing a mixed reciprocity relation, we obtain a uniqueness result in Section 3 (see [6–11] and the references therein).We solve the above-mentioned inverse problem by using the linear sampling method which was discussed early in 1996 by Colton and Kirsch [12]. The linear sampling method has been developed greatly and applied to solve a variety of inverse problems; we can refer to [13, 14] and the references therein. Some other methods also can be used to reconstruct the buried obstacle, for example, the reciprocity gap functional method [15, 16] and the Newton iteration method [17].The remaining part of the paper is organized as follows. In the next section, we will use integral equation method to solve direct scattering problem (4) based on Fredholm theory. In Section 3, we give a uniqueness result, that is, both of the obstacle D2 and the impedances (μ,λ) can be uniquely determined from far field measurements. In Section 4, a mathematical basis is given to reconstruct the shape of the obstacle D2 by using a modified linear sampling method.


## 2. The Direct Scattering Problem

In this section, we will establish the well posedness of the direct scattering problem by employing the integral equation method. Let us consider a more general direct scattering problem: Given the transmission boundary conditionsf∈H1/2(S0), g∈H-1/2(S0) and a general boundary data h∈H-3/2(S1), find u∈Hloc1(D0) and v∈H1(D1) such that(4)Δu+k02u=0,inD0,Δv+k12v=0,inD1,u+-v-=f,onS0,∂u+∂ν-λ0∂v-∂ν=g,onS0,∂v∂ν+divS1μ∇S1v+λv=h,onS1,limr→∞⁡r∂u∂r-ik0u=0,r=x.Remark 2.
Direct scattering problem (1) and (2) is a special case of problem (4) by taking u=us, f=-ui, g=-∂ui/∂ν, and h=0.Lemma 3.
Suppose thatk>0, λ0>0, Im⁡(μ)≤0, and Im⁡(λ)≥0; then problem (4) has at most one solution.Proof.
Clearly, it is sufficient to show thatu=0 in D0 and v=0 in D1 if f=g=0 on S0 and h=0 on S1. Denote by BR a circle large enough with radius R such that D is contained in its interior. From Green’s theorem, we obtain(5)∫∂BRu∂u¯∂νds=∫BR∖D¯∇u2-k2u2dx+∫S0u∂u¯∂νdsin the domain BR∖D¯ and(6)∫S0v∂v¯∂νds=∫D1∇v2-k2v2dx+∫S1v∂v¯∂νdsin the domain D1. By the boundary conditions of (4), we conclude from the above two equations that(7)∫∂BRu∂u¯∂νds=∫BR∖D¯∇u2-k2u2dx+λ0∫D1∇v2-k2v2dx+λ0∫S1μ¯dvds2ds-λ0∫S1λ¯v2ds.Since k>0, λ0>0, Im⁡(μ)≤0, and Im⁡(λ)≥0, it follows that(8)Im⁡∫∂BRu∂u¯∂νds=λ0Im⁡∫S1μ¯duds2ds-λ0Im⁡∫S1λ¯u2ds≥0.Rellich’s lemma [2] shows that u=0 in R2∖B¯R and it follows by the unique continuation principle [2] that u=0 in D0. The transmission boundary conditions and Holmgren’s uniqueness theorem [18] imply that v=0 in D1. Then we complete the proof of this lemma.In order to establish the existence of the solution to problem (4), we construct a solution to problem (4) in the form of combined single- and double-layer potentials as follows:(9)ux=λ0∫S0Φ0x,yφydsy+∫S0∂Φ0x,y∂νyψydsy,x∈D0,vx=∫S1Φ1x,yηydsy-∫S0Φ1x,yφydsy-∫S0∂Φ1x,y∂νyψydsy,x∈D1,where φ∈H-1/2(S0), ψ∈H1/2(S0), and η∈H-1/2(S1) are the unknown densities and(10)Φl≔i4H01klx-y,x≠y,l=0,1, is the fundamental solution of the Helmholtz equation in R2.Remark 4.
Based on the method proposed in [19] for the transmission problem and in [4] for the obstacle scattering with GIBC, we choose the solution as the form of (9). As the authors in [4] point out that the obtained integral equation fails to be uniquely solvable if the irregular frequencies occur. In order to exclude the irregular frequencies, we make the following assumption.
Assumption  A. k12is not a Dirichlet eigenvalue of-Δoperator in the domainD2 which can guarantee the well posedness of the direct problem.For further consideration, we define the single- and double-layer operatorsSij.l and Kij.l, respectively, by(11)Sij.lφx≔∫SiφyΦlx,ydsy,x∈Sj,Kij.lφx≔∫SiφyΦlx,y∂νydsy,x∈Sjand the normal derivative operators Kij.l′ and Tij.l by(12)Kij.l′φx≔∫Siφy∂Φlx,y∂νxdsy,x∈Sj,Tij.lφx≔∂∂νx∫Siφy∂Φlx,y∂νydsy,x∈Sj,with i,j,l=0,1. Referring to [20], we have mapping properties(13)Sij.l:H-1/2+sSi⟶H1/2+sSj,Kij.l:H1/2+sSi⟶H1/2+sSj,Kij.l′:H-1/2+sSi⟶H-1/2+sSj,Tij.l:H1/2+sSi⟶H-1/2+sSj,for i,j,l=0,1 and -1≤s≤1.Now we try to establish an integral system by employing the boundary integral equation approach. According to the presentation of the solution in the form of (9) and by making use of the known jump relations of single- and double-layer potentials [19], we have that on the interface S0(14)u+-v-S0=λ0S00.0+S00.1φ+K00.0+K00.1ψ-S10.1η,∂u+∂ν-λ0∂v-∂νS0=λ0K00.0′+K00.1′φ+T00.0+λ0T00.1ψ-λ0K10.1′η.On the boundary S1, we obtain that(15)∂v∂ν+divS1μ∇S1v+λvS1=ddsμddsS11.1η-12η+K11.1′η+λS11.1η-ddsμddsS01.1φ-K01.1′φ-λS01.1φ-ddsμddsK01.1ψ-T01.1ψ-λK01.1ψ.Define bounded linear operatorsA11,B11:H-1/2(S1)→H-3/2(S1) by(16)A11.1η≔μd2ds2S11.1η+μ∫S1S11.1ηds,B11η≔dμdsddsS11.1η-12η+K11.1′η+λS11.1η-μ∫S1S11.1ηds.Let(17)M=ddsμddsS01.1φ+K01.1′φ+λS01.1φ,N=ddsμddsK01.1ψ+T01.1ψ+λK01.1ψ,ω=φψη,R=f-gλ0h,A=λ0S00.0+S00.1K00.0+K00.1-S10.1-K00.0′-K00.1′-1λ0T00.0-T00.1K10.1′-M-NA11+B11.Then the potential functions defined by (9) solve problem (4) provided the unknown densities φ, ψ, and η solve the following boundary integral system:(18)Aω=R.Defining the Sobolev spaces(19)H=H-1/2S0×H1/2S0×H-1/2S1,W=H1/2S0×H-1/2S0×H-3/2S1,it is easy to see that the matrix operator A maps H continuously into W.Based on the following two lemmas, we show the solvability of (18) by using the Fredholm theory.Lemma 5.
The operatorA given by (18) is Fredholm with index zero.Proof.
From [20], the operators S00.0, S00.1, -T00.0, and -T00.1 are positive and bounded up to a compact perturbation, respectively; we denote by L0, L1, J0, and J1 the compact operators(20)Ll:H-1/2S0⟶H1/2S0;Jl:H1/2S0⟶H-1/2S0,l=0,1such that(21)Re⁡S00.l+Llφ,φ¯≥CφH-1/2S02,forφ∈H-1/2S0,Re⁡-T00.l+Jlψ,ψ¯≥CψH1/2S02,forψ∈H1/2S0,where , denotes the duality between H-1/2(S0) and H1/2(S0).
LetK and K′ be the operators defined as K00.0 and K00.0′, respectively, with kernel Φ0(x,y) replaced by Φ(x,y)=-1/2πln⁡x-y. Then Klc=K00.l-K and Klc′=K00.l′-K′(l=0,1) are compact since they have continuous kernels. It is easy to show that K and K′ are adjoint since their kernels are real; that is,(22)Kψ,ϕ=ψ,K′ϕforψ∈H1/2S0,ϕ∈H-1/2S0.
Now, we decomposeA into two parts; that is,(23)A=λ0S0+S12K0-2K′-1λ0T0-T1000A11+-λ0L0-L1K0c+K1c-S10.1-K0c′-K1c′1λ0J0+J1K10.1′-M-NB11≔A0+Ac,where Sl=S00.l+Ll, Tl=T00.l+Jl for (l=0,1). Consider the following sesquilinear form for (φ,ψ)⊤∈U≔H-1/2(S0)×H1/2(S0):(24)λ0S0+S12K-2K′-1λ0T0-T1φψ,φ¯ψ¯U∗,U=λ0S0φ,φ+S1φ,φ+1λ0-T0ψ,ψ+-T1ψ,ψ+2Kψ,φ-2K′φ,ψ,where U∗ is the dual space of U and (·,·) denotes the scalar product on L2(S0). Due to the coercivity of Sl and -Tl, the adjoint between K and K′, we obtain that the above sesquilinear form is coercive; that is,(25)Re⁡λ0S0+S12K-2K′-1λ0T0-T1φψ,φ¯ψ¯U∗,U≥CφH-1/2S0+ψH1/2S0.Whence the operator(26)λ0S0+S12K-2K′-1λ0T0-T1:U⟶U∗is invertible. On the other hand, by ourAssumption  A, it can be seen that A11:H-1/2(S1)→H-3/2(S1) is invertible (see Lemma  2.1 in [21]). So the operator A0 is invertible.
The entriesS10.1, K10.1′, M, and N have continuous kernels, which means that they are compact operators. Due to the compact embedding theorem and the mapping properties of S11.1 and K11.1′, the entry B11 is compact. As stated above, the other entries are also compact. We conclude that Ac is compact. So we complete the proof of this lemma.Lemma 6.
The operatorA given by (18) has a trivial kernel.Proof.
Letξ=(α,β,δ)⊤∈H satisfying Aξ=0. Define two potentials(27)qx=λ0∫S0Φ0x,yαydsy+∫S0∂Φ0x,y∂νyβydsy,x∈D,(28)qx=∫S1Φ1x,yδydsy-∫S0Φ1x,yαydsy-∫S0∂Φ1x,y∂νyβydsy,x∈D0.Using the jump relations of the single- and double-layer potentials across S0, we have(29)q--q+S0=λ0S00.0+S00.1α+K00.0+K00.1β-S10.1η,∂q-∂ν-λ0∂q+∂νS0=λ0K00.0′+K00.1′α+T00.0+λ0T00.1β-λ0K10.1′δ.Since Aξ=0, it is easy to check that the potentials defined in (27) and (28) satisfy(30)Δq+k12q=0,inD0,Δq+k02q=0,inD,q+-q-=0,onS0,λ0∂q+∂ν-∂q-∂ν=0,onS0,limr→∞⁡r∂q∂r-ik1q=0,r=x.We can show that problem (30) has only trivial solution (see [2]).
Using the same(α,β,δ)⊤, define two new potentials(31)px=λ0∫S0Φ0x,yαydsy+∫S0∂Φ0x,y∂νyβydsy,x∈D0,px=∫S1Φ1x,yδydsy-∫S0Φ1x,yαydsy-∫S0∂Φ1x,y∂νyβydsy,x∈D1.Then we can prove that p satisfies problem (4) with homogeneous boundary conditions. Lemma 3 shows that p=0. Thus again by the jump relations of the single- and double-layer potentials across S0 we have(32)β=p+-q-S0=0,λ0α=∂q-∂ν-∂p+∂νS0=0.
At this time, the potential given by (28) becomes(33)qx=∫S1Φ1x,yδydsy,and note that q=0 on S1 because of the trivial solution of (30); we conclude that q satisfies the Helmholtz equation in D2 with homogeneous Dirichlet boundary condition if we let x∈D2. By ourAssumption  A, k12 is not a Dirichlet eigenvalue in D2, which implies that q=0 in D2. Therefore, the jump relation across S1 shows that(34)δ=∂q-∂ν-∂p+∂νS1=0.Then we complete the proof of this lemma.By Fredholm theory, the above two lemmas show that the matrix operatorA given by (18) has a bounded inverse; as a consequence, we have the following.Theorem 7.
UnderAssumption  A, integral system (18) has a unique solution, and problem (4) has a unique solution given by (9) which satisfies(35)uHloc1D0+vH1D1≤CfH1/2S0+gH-1/2S0+hH-3/2S1.


## 3. A Uniqueness Result of the Inverse Problem

As usual in most of the inverse problems, the first question to ask is the identifiability, that is, whether the scattererD2 and (μ,λ) can be identified from a knowledge of the far field pattern. Mathematically, the identifiability is the uniqueness issue which is of theoretical interest and is required in order to proceed to efficient numerical methods of solutions.Let us go back to scattering problem (1) and (2). The incident wave has two choices: the incident plane wave ui(·,d) (i.e., ui(x,d)=eikx·d) and the incident point source Φ(·,z).(1)
To the incident plane waveui(·,d), we use us(·,d) and u∞(·,d) to denote the scattered field and the corresponding far field pattern, respectively.(2)
To the incident point sourceΦ(·,z), we use us(·,z) and Φ∞(·,z) to denote the scattered field and the corresponding far field pattern, respectively.The uniqueness result is based on the following mixed reciprocity relation.Lemma 8.
For the scattering of plane wavesui(·,d) with d∈S1 and point sources Φ0(·,z), we have(36)Φ0∞x^,z=γusz,-x^z∈D0,x^∈S1,λ0γusz,-x^+λ0-1γuiz,-x^z∈D1,x^∈S1,where γ=eiπ/4/8k0π.Remark 9.
The mixed reciprocity relation has been established in the case of obstacle scattering problem [7, 11, 18]; here we extend the result to the scattering problem by an obstacle with GIBC buried in a piecewise homogeneous medium.Proof.
We consider the casez∈D0 firstly. By Green’s second theorem and the Sommerfeld radiation condition we have that(37)∫S0u+sy,z∂u+sy,d∂νy-∂u+sy,z∂νyu+sy,ddsy=0for z∈D0, d∈S1. By the boundary conditions and Green’s second theorem, the total fields u(·,d)=us(·,d)+ui(·,d) and u(·,z)=us(·,z)+Φ(·,z) satisfy(38)∫S0u+y,z∂u+y,d∂νy-∂u+y,z∂νyu+y,ddsy=λ0∫S0v-y,z∂v-y,d∂νy-∂v-y,z∂νyv-y,ddsy=λ0∫S1μ∇S1v·,z·∇S1v·,d-λv·,zv·,dds-λ0∫S1μ∇S1v·,d·∇S1v·,z+λv·,dv·,zds=0.Since the incident plane wave ui(·,d) and incident point source Φ0(·,z) solve the Helmholtz equation inside D, we obtain that from the above two equalities(39)∫S0u+sy,z∂uiy,d∂νy-∂u+sy,z∂νyuiy,ddsy=∫S0u+sy,d∂Φ0y,z∂νy-∂u+sy,d∂νyΦ0y,zdsy.Using Green’s representation formula for us(·,x^), we obtain the representation(40)usz,x^=∫S0u+sy,x^∂Φ0y,z∂νy-∂u+sy,x^∂νyΦ0y,zdsyfor x^∈S1. The far field pattern Φ0∞(x^,z) has the following integral representation:(41)Φ0∞-x^,z=γ∫S0u+sy,z∂uiy,x^∂νy-∂u+sy,z∂νyuiy,x^dsy.Thus we conclude that Φ0∞(-x^,z)=γus(z,x^) for z∈D0, x^∈S1 from (39), (40), and (41) with d replaced by x^.
Next, we consider the casez∈D1. From the boundary condition on S1, we have that for the total fields v(·,d) and v(·,z)(42)∫S1vy,z∂vy,d∂νy-∂vy,z∂νyvy,ddsy=0.For the scattered fields us(·,d) and us(·,z) we still have equality (37), and for the incident plane wave ui(·,d) and incident point source Φ0(·,z) we have(43)∫S1uiy,d∂Φ0y,z∂νy-∂uiy,d∂νyΦ0y,zdsy=0by Green’s second theorem.
From (37) and (41) we get(44)Φ0∞-x^,z=γ∫S0u+sy,z∂u+y,x^∂νy-u+y,x^∂u+sy,z∂νydsy.Let S(z;ϵ)≔{y∈R2:y-z=ϵ} be a sphere contained in D1. Applying Green’s second theorem in the domain B(z;ϵ)≔{y∈D1:y-z>ϵ} and taking into account the boundary condition on S0 we further have(45)Φ0∞-x^,z=γλ0∫S1usy,z∂vy,x^∂νy-vy,x^∂usy,z∂νydsy+γλ0∫Sz;ϵusy,z∂vy,x^∂νy-vy,x^∂usy,z∂νydsy+γλ0∫Bz;ϵk12-k02Φ0y,zvy,x^dy+1-λ0γ∫S0v-y,x^∂Φ0y,z∂νydsy.By the well posedness of the direct problem and the interior elliptic regularity [22], v(·,x^)∈C∞(D1) and us(·,z)∈H2(V) for any compact subset V of D1. Therefore, there is a sequence ϵj such that ϵj→0 and(46)∫Sz;ϵjusy,z2+∇usy,z2dsy⟶0as j→∞. This together with the Cauchy-Schwarz inequality implies that the integral on S(z;ϵj) tends to 0 as j→∞. By passing to the limit j→∞ in (45) with ϵ=ϵj we have(47)Φ0∞-x^,z=γλ0∫S1usy,z∂vy,x^∂νy-vy,x^∂usy,z∂νydsy+γλ0∫D1k12-k02Φ0y,zvy,x^dy+1-λ0γ∫S0v-y,x^∂Φ0y,z∂νydsy.The volume integral exists as an improper integral since its integrand is weakly singular.
On the other hand, by Green’s representation formula and Green’s second theorem, we have that(48)γλ0vz,x^=γλ0∫S0Φ0z,y∂v-y,x^∂νy-v-y,x^∂Φ0z,y∂νydsy-γλ0∫S1Φ0z,y∂vy,x^∂νy-vy,x^∂Φ0z,y∂νydsy+γλ0∫D1k12-k02Φ0z,yvy,x^dy.It follows from (47) and (48) together with the boundary conditions on S0 and S1 and Green’s second theorem that(49)Φ0∞-x^,z-γλ0vz,x^=1-λ0γ∫S0v-y,x^∂Φ0y,z∂νydsy-γλ0∫S0Φ0z,y∂v-y,x^∂νy-v-y,x^∂Φ0z,y∂νydsy=γ∫S0u+y,x^∂Φ0z,y∂νy-Φ0z,y∂u+y,x^∂νydsy=γ∫S0uiy,x^∂Φ0z,y∂νy-Φ0z,y∂uiy,x^∂νydsy+γ∫S0u+sy,x^∂Φ0z,y∂νy-Φ0z,y∂u+sy,x^∂νydsy=-γuiz,x^.We conclude from (49) that Φ0∞(-x^,z)=λ0γus(z,x^)+(λ0-1)γui(z,x^), for z∈D1, x^∈S1. Therefore the proof of this lemma is completed.Lemma 10.
For the transmitted wavev of problem (1) and (2) associated with the incident plane wave ui=eik0x·d, we have that V={v·,dS1,d∈S1} is complete in H3/2(S1).Note that for the incident plane waveui, the regularity of elliptic equations shows that the solution of problem (1) and (2) belongs to Hloc2(D0)∩H2(D1).Proof.
Assume thatϕ is a function in H-3/2(S1) such that for every d∈S1(50)∫S1vϕds=0.Consider the following problem:(51)Δω+k02ω=0,inD0,Δω+k12ω=0,inD1,ω+-ω-=0,onS0,∂ω+∂ν-λ0∂ω-∂ν=0,onS0,∂ω∂ν+divS1μ∇S1ω+λω=ϕ,onS1,limr→∞⁡r∂ω∂r-ik0ω=0,r=x.According to Theorem 7, this problem is well posedness and we have that the unique solution ω∈Hloc1(D0)∩H1(D1). Then we have(52)0=∫S1vϕds=∫S1v∂ω∂ν+divS1μ∇S1ω+λωds=∫S1v∂ω∂νds+∫S1ωdivS1μ∇S1v+λvds=∫S1v∂ω∂ν-ω∂v∂νds.Furthermore, from the transmission boundary condition, Green’s second theorem, and the radiation condition for us and ω we have(53)0=1λ0∫S0u+∂ω+∂ν-ω+∂u+∂νds=1λ0∫S0u+s∂ω+∂ν-ω+∂u+s∂νds+1λ0∫S0ui∂ω+∂ν-ω+∂ui∂νds=-1γλ0ω∞-d.Thus Rellich’s lemma implies that ω=0 in D0. Then the transmission boundary conditions on S0 and Holmgren’s uniqueness theorem show that ω=0 in D1; hence ϕ=0 from the trace theorem. We complete the proof of this lemma.We are now in the position to present the uniqueness result based on the idea in [11].Theorem 11.
Given the interfaceS0, the positive constant λ0, and the wave numbers k0 and k1, due to incident plane waves ui=eik0x·d, assume that the two far field patterns u∞(x^,d) and u~∞(x^,d) corresponding to the two scattered fields us(x,d) and u~s(x,d) which are arisen by two obstacles D2 with impedances (μ,λ) and D~2 with impedances (μ~,λ~), respectively, coincide at a fixed frequency for all x^∈S1 and d∈S1; then D2=D~2 and (μ,λ)=(μ~,λ~).Proof.
If the obstacles are not the same, that is,D2≠D~2, let G be the unbounded component of R2∖(D¯2∪D~¯2). By Rellich’s lemma the scattered fields us(·,d) and u~s(x,d) corresponding to the incident plane wave ui(·,d) coincide in the unbounded domain G. Without loss of generality, we may assume that there exist z0 and open set Γ such that z0∈Γ⊂∂D2 and Γ∩D~¯2=∅. We can choose h>0 such that the sequence(54)zj=z0+hjνz0,j=1,2,…is contained in G, where ν(z0) is the normal to ∂D2 at z0.
Consider the solutionus(·,zj) to problem (1) and (2) due to the incident point source Φ0(·,zj). By Lemma 8, the far fields Φ0∞(·,zj) and Φ~0∞(·,zj) coincide in G. Then Rellich’s lemma implies that us(x,zj)=u~s(x,zj) for x∈G. From this we have by denoting Pv=∂v/∂ν+divS1(μ∇S1v)+λv(55)Pus·,zj=Pu~s·,zjonΓ.
Consideringu~s(x,zj) as the scattered field corresponding to D~2, using the boundary condition on S1 for us(x,zj), we see that(56)Pu~s·,zj=-PΦ·,zjonΓ.From the well posedness of problem (1) and (2) and the regularity of elliptic equations we obtain(57)limj→∞⁡Pu~s·,zj=∂u~s·,z0∂ν+divS1μ∇S1u~s·,z0+λu~s·,z0in L2(Γ). On the other hand, as the same argument in Theorem  3.1 of [11], PΦ(·,z0) does not belong to L2(Γ). This is a contradiction, which implies that D2=D~2.
Next, we show that(μ,λ)=(μ~,λ~). The proof is based on Theorem  3.1 in [11]; for the reader’s convenience, we give its proof but make some slight modifications.
For this purpose, letμ^=μ-μ~ and λ^=λ-λ~, and denote by v^ the same total fields v and v~. From the boundary conditions for the total field, we have that(58)divS1μ^∇S1v^+λ^v^=0onS1.This equality should be understood in the weak sense. Thus for every θ∈H3/2(S1) we have(59)v^,divS1μ^∇S1θ+λ^θH3/2S1,H-3/2S1=0.With the help of Lemma 10, we obtain(60)divS1μ^∇S1θ+λ^θ=0,∀θ∈H3/2S1.Choosing θ=1 in the above equation leads to λ^=0. The above equation also implies that(61)∫S1μ^∇S1θ2ds=0,∀θ∈H3/2S1.Assume that μ^(x0)≠0 for some x0∈S1; then, for example, Re⁡μ^(x0)>0 without loss of generality. Since μ^ is continuous there exists ɛ>0 such that Re⁡μ^(x)>0 for all x∈S1∩B(x0;ɛ). Let us choose θ as a smooth and compactly supported function in S1∩B(x0;ɛ); we obtain that(62)∫S1∩Bx0;ɛRe⁡μ^∇S1θ2ds=0,and then ∇S1θ=0 on S1∩B(x0;ɛ); that is, θ is a constant on S1∩B(x0;ɛ), which is a contradiction. We hence have μ^=0 on S1 and the proof is completed.


## 4. The Modified Linear Sampling Method

In this part, we give a mathematical basis to reconstruct the shape of the obstacleD2 by using the modified linear sampling method (see [23]).We do some preparation firstly. Consider the total waveu0=ui+u0s (ui(x,d)=eik0x·d) such that(63)Δu0+k02u0=0,inD0,Δu0+k12u0=0,inD,u0+-u0-=0,onS0,∂u0+∂ν-λ0∂u0-∂ν=0,onS0,limr→∞⁡r∂u0s∂r-ik0u0s=0,r=x.Recalling that u(x,d) and v(x,d) are the solution to scattering problem (1) and (2) for incident plane wave ui(x,d)=eik0x·d with the direction d∈S1, it is easy to verify that the fields w(x)≔u-u0, x∈D0 and z(x)≔v-u0, x∈D1 solve the following boundary value problem:(64)Δw+k02w=0,inD0,Δz+k12z=0,inD1,w+-z-=0,onS0,∂w+∂ν-λ0∂z-∂ν=0,onS0,∂z∂ν+divS1μ∇S1z+λz=ϕ,onS1,limr→∞⁡r∂w∂r-ik0w=0,r=x,where ϕ=-∂u0/∂ν-divS1(μ∇S1u0)-λu0. The well posedness of this boundary value problem has been established in Section 1.We define four operators in the following.The data-to-pattern operatorG:H-3/2(S1)→L2(S1) by(65)Gϕx^=w∞,where w∞ is the far field pattern of the wave field w of problem (64).The auxiliary operatorH:L2(S1)→H-3/2(S1) by(66)Hgx=∫S1∂u0x,d∂νx+divS1μ∇S1u0+iλu0x,dgddsd,where u0 is the solution of (63) corresponding to the incident wave ui(·,d).The far field operatorF:L2(S1)→L2(S1) by(67)Fgx^=∫S1u∞x^,dgddsd,where u∞ is the far field pattern of the scattered wave us of problem (1) and (2).The far field operatorF0:L2(S1)→L2(S1) by(68)F0gx^=∫S1u0∞x^,dgddsd,where u0∞ is the far field pattern of the scattered wave u0s of problem (63).Note that(69)F-F0gx^=∫S1u∞x^,d-u0∞x^,dgddsdis just the far field pattern of the radiating function(70)∫S1ux,d-uix,d-u0x,d-uix,dgddsd=∫S1ux,d-u0x,dgddsd.From the boundary conditions onS1 for v(x,d) and z(x,d), we can factorize the operator F-F0 as(71)F-F0=-GH.LetG0(·,m), m∈R2 be the Green function for problem (63) of scattering by the background medium. We now define the modified far field equation(72)F-F0gmx^=G0∞x^,mforg∈L2S1,x^∈S1,where G0∞(x^,m)∈L2(S1) is the far field pattern of the Green function G0(x,m). We will characterize the obstacle D2 by the behavior of an approximate solution gm of far field equation (72).To prove the existence of an approximate solution of (72), we firstly explore the related properties of the operators G and H.Lemma 12.
The data-to-pattern operatorG:H-3/2(S1)→L2(S1) is injective and compact and has dense range in L2(S1).Proof.
First, injectivity is a direct consequence of Rellich’s lemma and analytic continuation of the solution to (64).
To prove compactness, using Green’s representation formula forz in D1 and w in D0, we can decompose the operator G as G=G1G2, where G2:H-3/2(S1)→H1/2(S0)×H-1/2(S0) is defined by G2(ϕ)=(z-S0,λ0∂z-/∂νS0) and G1:H1/2(S0)×H-1/2(S0)→L2(S1) is defined by(73)G1w+,∂w+∂νx^=γ∫S0w+y∂e-ik0x^·y∂νy-∂w+y∂νye-ik0x^·ydsy,x^∈S1,where γ=eiπ/4/8k0π. The interior regularity of the solution to problem (64) implies that the operator G2 is bounded. So the operator G is compact since the operator G1 is compact.
To show denseness of the range ofG we just need to prove that the adjoint operator G∗ is injective. To this end, let U and V be the solution of problem (1) and (2) with incident plane wave(74)Ui=∫S1e-ik0x^·ygx^¯dsx^,g∈L2S1.We remind the reader that U∈Hloc2(D0) and V∈H2(D1) by the regularity of elliptic equations.
For anyϕ∈H-3/2(S1), let w and z be the solution to problem (64) with the boundary data ϕ. Then one can derive that by Green’s second theorem and the radiation condition of scattered field(75)ϕ,G∗gH-3/2S1,H3/2S1=Gϕ,gL2S1,L2S1=∫S1w∞x^gx^¯dsx^=γ∫S1∫S0w+y∂e-ik0x^·y∂νy-∂w+y∂νye-ik0x^·ydsygx^¯dsx^=γ∫S0w+y∂Uiy∂νy-∂w+y∂νyUiydsy=γ∫S0w+y∂U+y∂νy-∂w+y∂νyU+ydsy=γλ0∫S1zy∂Vy∂νy-∂zy∂νyVydsy=-γλ0∫S1zydivS1μ∇S1Vy+λVy+∂zy∂νyVydsy=-γλ0∫S1∂zy∂νy+divS1μ∇S1zy+λzyVydsy=-γλ0∫S1ϕyVydsy.Thus we obtain that(76)G∗g=-λ0γ¯V¯S1.
LettingG∗g=0, we have VS1=0 and then ∂V/∂νS1=0 by the GIBC. Holmgren’s uniqueness theorem implies that V=0 in D1 and then U=0 in D0; this leads to Ui=0 since Ui does not satisfy the radiation condition; however Us does. Then we can conclude that g=0 which means that G∗ is injective. We complete the proof of this theorem.Lemma 13.
LetG be defined by (65), for any m∈R2; then G0∞(x^,m)∈L2(S1) belongs to the range of R(G0) if and only if m∈D2.Proof.
Firstly, letm∈D2; then G0(x,m) satisfies problem (64) with(77)ϕ=∂G0∂ν+divS1μ∇S1G0+λG0S1.By the definition of the operator G, we get Gϕ coinciding with G0∞.
Now letm∉D2 and assume on the contrary that there exists ϕ∈H-3/2(S1) such that Gϕ=G0∞. Let w, z be the solution of problem (64) with boundary data ϕ; then w∞=G0∞. By Rellich’s lemma we conclude that w(x)=G0(x,z) in D0∖{m} and then as a direct derivation z=G0(x,z) in D1∖{m}.
Ifm∈D1, this contradicts the fact that z is analytic in D0 but G0(x,m) is singular at x=m. If m∈S1, by the well posedness of the direct problem zH1/2(S1) is bounded which contradicts that G0H1/2(S1)=∞. For the case m∈D0 or m∈S0 we also can infer a contradiction, so we complete the proof of this lemma.Now, we turn our attention to the operatorH. In order to obtain the required properties of the operator H, we need the following two results in [23].Lemma 14 (mixed reciprocity relation).
Forx^∈S1, we have(78)G0∞x^,m=u0m,-x^,m∈D0,x^∈S1,λ0u0m,-x^,m∈D,x^∈S1,where G0∞(x^,m) is the far field pattern of Green function G0(x,m) and u0(m,-x^) is the solution of (63) with incident plane wave ui(m,-x^).Lemma 15.
For allm∈D, x^∈S1, we have(79)u0m,-x^=S0u0m,·¯x^,where S0=I+2ik0|γ|2F0 introduced in [23] is the scattering operator.Lemma 16.
IfIm⁡(λ)≢0 on S1, then the operator H defined by (66) has dense range.Proof.
The adjoint operator ofH is H∗:H3/2(S1)→L2(S1) with(80)H∗ξx^=∫S1∂u0y,x^∂νy¯+divS1μ¯∇S1u0y,x^¯+λ¯u0y,x^¯ξydsy.
We just need to show that the operatorH∗ is injective. Using Lemmas 14 and 15, we have for every ξ∈H3/2(S1)(81)λ0S0H∗ξx^=λ0S0∫S1∂u0y,·∂νy¯+divS1μ¯∇S1u0y,·¯+λ¯u0y,·¯ξydsy=λ0∫S1∂u0y,-x^∂νy+divS1μ¯∇S1u0y,-x^+λ¯u0y,-x^ξydsy=∫S1∂G0∞x^,y∂νy+divS1μ¯∇S1G0∞x^,y+λ¯G0∞x^,yξydsy.Therefore, we conclude that λ0S0H∗(ξ) is just the far field pattern of the potential(82)Wx=∫S1∂G0x,y∂νy+divS1μ¯∇S1G0x,y+λ¯G0x,yξydsy=∫S1∂G0x,y∂νyξy+G0x,yλ¯ξy+divS1μ¯∇S1ξydsyfor x∈R2∖S1.
Due to the fact thatS0 is unitary (see [24]), we just need to show that ξ=0 under the assumption W∞(x^)=0. Next we will prove this assertion.
In fact, ifW∞(x^)=0, then Rellich’s lemma implies that W(x)=0 for x∈R2∖D¯2. Then by using the jump relations of single- and double-potentials, we get(83)W-S1=W-S1-W+S1=-ξ,∂W-∂νS1=∂W-∂νS1-∂W+∂νS1=λ¯ξ+divS1μ¯∇S1ξ.Thus W¯ satisfies the Helmholtz equation in D2 with GIBC:(84)ΔW¯+k12W¯=0,inD2,∂W¯∂ν+divS1μ∇S1W¯+λW¯=0,onS1.By the assumption on λ, there exists x0∈S1 such that Im⁡(λ(x))≠0 in a small neighborhood Λ(x0)⊂S0. Green’s theorem in D2 and the divergence theorem on S1 imply(85)0=∫D2ΔW+k12WW¯dx=-∫D2∇W2-k12W2dx+∫S1μ¯dWds2ds-∫S1λ¯W2ds.By taking the imaginary part of the above equation we can obtain that W=0 on Λ(x0); then the boundary condition shows that ∂W/∂ν=0 on Λ(x0). Thus Holmgren’s uniqueness theorem implies that W=0 in D2; we then obtain ξ=0 from the jump relations. This lemma is then proved.Finally, we give the main result in this paper, that is, recovering the obstaclesD2 by a modified linear sampling method.Theorem 17.
UnderAssumption  A, then we have the following results:(1)
Ifm∈D2, then for every ϵ>0 there exists a solution gmϵ∈L2(S1) to the far field equation (72) such that(86)F-F0gmϵ+G0∞L2S1<ϵ.(2)
Ifm∉D2, then for every ϵ>0 and δ>0 there exists a function gmϵ,δ∈L2(S1) such that(87)F-F0gmϵ,δ+G0∞L2S1<ϵ+δ,limδ→0⁡gmϵ,δL2S1=∞.Proof.
Ifm∈D2, by using Lemma 13 there exists ϕ∈H-3/2(S1) such that Gϕ=G0∞. From Lemma 16, for every ϵ0>0 there exists a function gmϵ0∈L2(S1) such that(88)Hgmϵ0-ϕH-1/2S1<ϵ0.The operator G is bounded from Lemma 12; then we have(89)GHgmϵ0-GϕL2S1<c1ϵ0,where c1 is a constant; that is,(90)F-F0gmϵ+G0∞L2S1<ϵ,where ϵ=c1ϵ0.
Next, we assume thatm∉D2. In this case, by Lemma 13  G0∞ is not in the range of G. But from Lemma 12 we know that the operator G is compact and injective with dense range in L2(S1). Hence for every δ>0 we can construct a unique Tikhonov regularized solution ϕρ∈H-3/2(S1) of equation Gϕρ=G0∞, such that(91)Gϕρ-G0∞L2S1<δ,where ρ is the regularization parameter (chosen by a regular regularization strategy, e.g., the Morozov discrepancy principle [25]). Then we have ϕρH-3/2(S1)→∞ as ρ→0. By Lemma 16  H has dense range; hence for ϵ>0 sufficiently small there exists gmϵ,ρ such that(92)Hgmϵ,ρ-ϕρH-3/2S1<ϵc1.Combining (91) and (92) we obtain that for every ϵ>0 and δ>0 there exists gmϵ,ρ∈L2(S1) such that(93)F-F0gmϵ,ρ+G0∞L2S1=GHgmϵ,ρ-G0∞L2S1≤GHgmϵ,ρ-GϕρL2S1+Gϕρ-G0∞L2S1<ϵ+δ.Since limδ→0ρ(δ)=0 we have that limδ→0ϕρH-3/2(S1)→∞. From (92) we have that limδ→0Hgmϵ,δH-3/2(S1)→∞. By the definition of the operator H given by (66) we obtain that limδ→0gmϵ,δL2(S1)→∞. Then we complete the proof of this theorem.Remark 18.
(1) From Theorem17, we have to obtain the far field pattern of Green function G0(x,m) which is defined in the layered background medium. Typically, this is a quite difficult task; however, with the help of Lemma 14 we only need to solve transmission problem (63) to get u0 instead of G0.
(2) In this paper, we just consider the case of two-layered background medium; in fact, our result can be extended to the case of multilayered piecewise homogeneous medium.



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