Multilayered Scattering Problem with Generalized Impedance Boundary Condition on the Core

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.


Introduction
This work is concerned with the scattering problem of timeharmonic 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 is  2 .
To be precise, let  2 ⊂  2 denote the impenetrable obstacle which is a bounded domain with a smooth boundary  1 (e.g.,  2 ).Assume that the unknown obstacle  2 is buried in a penetrable obstacle  with a closed  2 surface  0 such that  2 ⊂ .Denote by  1 =  \  2 a connected bounded domain filled with homogeneous medium and denote by  0 =  2 \  the unbounded connected domain occupied by another homogeneous medium.Let   =   /  > 0 be the wave number in terms of the frequency   and the sound speed   in the corresponding region   ( = 0, 1) (see Figure 1).
The scattering of time-harmonic acoustic plane waves by an obstacle with GIBC in a piecewise homogeneous medium in  2 can be modeled by the Helmholtz equation with boundary conditions on the boundary  1 and interface  0 : Here ] is the unit outward normal vector on the boundary  0 or  1 ;  + , ( + /])(V − , V − /]) denote the limit of , (/])(V, V/]) on the boundary  0 from the exterior (interior) of .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 total field  =   +   is decomposed into the given incident field   =   0 ⋅ ,  ∈ S 1 , (the unit sphere in  2 ) and the unknown scattered field   which is required to satisfy the Sommerfeld radiation condition [2] lim uniformly in x = /|| with  = ||.Further it is known that the scattered field   (, ) has the following asymptotic representation: uniformly for all directions x, where the function  ∞ ( x, ) defined on the unit sphere S 1 is known as the far field pattern with x and  denoting, respectively, the observation direction and the incident direction.
The direct problem is to seek functions  ∈  1 loc ( 0 ) and V ∈  1 ( 1 ) satisfying ( 1) and (2).In the next section, more general direct problem (4) will be considered.If the impenetrable obstacle  2 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 ∈  −1 ( 1 ) by the variational method; but for the case when ℎ belongs to  −3/2 ( 1 ), 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.
The inverse problem we consider in this paper is to determine the shape of the obstacle  2 and (, ) from the knowledge of the far field pattern  ∞ ( x, ) for all x,  ∈ S 1 with the given wave number   ( = 0, 1) and the positive constant  0 .
As usual in most of the inverse problems, the first issue is the uniqueness, that is, in what conditions, the shape of the obstacle  2 (or the parameters such as (, )) 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][7][8][9][10][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  2 and the impedances (, ) can be uniquely determined from far field measurements.In Section 4, a mathematical basis is given to reconstruct the shape of the obstacle  2 by using a modified linear sampling method.
Proof.Clearly, it is sufficient to show that  = 0 in  0 and V = 0 in  1 if  =  = 0 on  0 and ℎ = 0 on  1 .Denote by   a circle large enough with radius  such that  is contained in its interior.From Green's theorem, we obtain in the domain   \  and in the domain  1 .By the boundary conditions of (4), we conclude from the above two equations that Since  > 0,  0 > 0, Im() ≤ 0, and Im() ≥ 0, it follows that Im (∫ Rellich's lemma [2] shows that  = 0 in  2 \   and it follows by the unique continuation principle [2] that  = 0 in  0 .The transmission boundary conditions and Holmgren's uniqueness theorem [18] imply that V = 0 in  1 .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: where  ∈  −1/2 ( 0 ),  ∈  1/2 ( 0 ), and  ∈  −1/2 ( 1 ) are the unknown densities and = 0,1, is the fundamental solution of the Helmholtz equation in  2 .
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.
with , ,  = 0, 1. Referring to [20], we have mapping properties for , ,  = 0, 1 and −1 ≤  ≤ 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  0 ( + − V − )     0 = ( 0  00.0 +  00.1 )  + ( 00.0 +  00.1 )  −  10.1 , On the boundary  1 , we obtain that ) . ( Then the potential functions defined by ( 9) solve problem (4) provided the unknown densities , , and  solve the following boundary integral system: Defining the Sobolev spaces it is easy to see that the matrix operator  maps  continuously into .
Based on the following two lemmas, we show the solvability of ( 18) by using the Fredholm theory.
Lemma 5.The operator  given by ( 18) is Fredholm with index zero.
Proof.From [20], the operators  00.0 ,  00.1 , − 00.0 , and − 00.1 are positive and bounded up to a compact perturbation, respectively; we denote by  0 ,  1 ,  0 , and  1 the compact operators where ⟨ , ⟩ denotes the duality between  −1/2 ( 0 ) and  1/2 ( 0 ).Let  and   be the operators defined as  00.0 and   00.0 , respectively, with kernel Φ 0 (, ) replaced by Φ(, ) = −(1/2) ln | − |.Then   =  00. −  and    =   00. −   ( = 0, 1) are compact since they have continuous kernels.It is easy to show that  and   are adjoint since their kernels are real; that is, Now, we decompose  into two parts; that is, where   =  00. +   ,   =  00. +   for ( = 0, 1).Consider the following sesquilinear form for (, ) where  * is the dual space of  and (⋅, ⋅) denotes the scalar product on  2 ( 0 ).Due to the coercivity of   and −  , the adjoint between  and   , we obtain that the above sesquilinear form is coercive; that is, Re ⟨( Whence the operator is invertible.On the other hand, by our Assumption A, it can be seen that  11 :  −1/2 ( 1 ) →  −3/2 ( 1 ) is invertible (see Lemma 2.1 in [21]).So the operator  0 is invertible.The entries  10.1 ,   10.1 , , and  have continuous kernels, which means that they are compact operators.Due to the compact embedding theorem and the mapping properties of  11.1 and   11.1 , the entry  11 is compact.As stated above, the other entries are also compact.We conclude that   is compact.So we complete the proof of this lemma.Lemma 6.The operator  given by ( 18) has a trivial kernel.
Proof.Let  = (, , ) ⊤ ∈  satisfying  = 0. Define two potentials Using the jump relations of the single-and double-layer potentials across  0 , we have Since  = 0, it is easy to check that the potentials defined in ( 27) and (28) satisfy We can show that problem (30) has only trivial solution (see [2]).
Using the same (, , ) ⊤ , define two new potentials Then we can prove that  satisfies problem (4) with homogeneous boundary conditions.Lemma 3 shows that  = 0. Thus again by the jump relations of the single-and doublelayer potentials across  0 we have At this time, the potential given by (28) becomes and note that  = 0 on  1 because of the trivial solution of (30); we conclude that  satisfies the Helmholtz equation in  2 with homogeneous Dirichlet boundary condition if we let  ∈  2 .By our Assumption A,  2 1 is not a Dirichlet eigenvalue in  2 , which implies that  = 0 in  2 .Therefore, the jump relation across  1 shows that Then we complete the proof of this lemma.
By Fredholm theory, the above two lemmas show that the matrix operator  given by ( 18) has a bounded inverse; as a consequence, we have the following.Theorem 7.Under Assumption A, integral system (18) has a unique solution, and problem (4) has a unique solution given by ( 9) which satisfies

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 scatterer  2 and (, ) 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.
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.
Next, we consider the case  ∈  1 .From the boundary condition on  1 , we have that for the total fields V(⋅, ) and V(⋅, ) For the scattered fields   (⋅, ) and   (⋅, ) we still have equality (37), and for the incident plane wave   (⋅, ) and incident point source Φ 0 (⋅, ) we have By the well posedness of the direct problem and the interior elliptic regularity [22], V(⋅, x) ∈  ∞ ( 1 ) and   (⋅, ) ∈  as  → ∞.This together with the Cauchy-Schwarz inequality implies that the integral on (;   ) tends to 0 as  → ∞.By passing to the limit  → ∞ in (45) with  =   we have 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 It follows from ( 47) and ( 48) together with the boundary conditions on  0 and  1 and Green's second theorem that We conclude from (49) that Φ ∞ 0 (− x, ) =  0   (, x) + ( 0 − 1)  (, x), for  ∈  1 , x ∈ S 1 .Therefore the proof of this lemma is completed.Lemma 10.For the transmitted wave V of problem ( 1) and ( 2) associated with the incident plane wave   =   0 ⋅ , we have that Note that for the incident plane wave   , the regularity of elliptic equations shows that the solution of problem ( 1) and (2) belongs to  2 loc ( 0 ) ∩  2 ( 1 ).
Proof.Assume that  is a function in  −3/2 ( 1 ) such that for every  ∈ S 1 Consider the following problem: According to Theorem 7, this problem is well posedness and we have that the unique solution  ∈  1 loc ( 0 ) ∩  1 ( 1 ).Then we have Furthermore, from the transmission boundary condition, Green's second theorem, and the radiation condition for   and  we have Thus Rellich's lemma implies that  = 0 in  0 .Then the transmission boundary conditions on  0 and Holmgren's uniqueness theorem show that  = 0 in  1 ; hence  = 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].
For this purpose, let μ = − μ and λ = − λ, and denote by V the same total fields V and Ṽ.From the boundary conditions for the total field, we have that This equality should be understood in the weak sense.Thus for every  ∈  3/2 ( 1 ) we have With the help of Lemma 10, we obtain div  1 ( μ∇  1 ) + λ = 0, ∀ ∈  3/2 ( 1 ) .

The Modified Linear Sampling Method
In this part, we give a mathematical basis to reconstruct the shape of the obstacle  2 by using the modified linear sampling method (see [23]).We do some preparation firstly.Consider the total wave  0 =   +   0 (  (, ) =   0 ⋅ ) such that Recalling that (, ) and V(, ) are the solution to scattering problem (1) and ( 2) for incident plane wave   (, ) =   0 ⋅ with the direction  ∈ S 1 , it is easy to verify that the fields () :=  −  0 ,  ∈  0 and () := V −  0 ,  ∈  1 solve the following boundary value problem: where  = −( 0 /]) − div  1 (∇  1  0 ) −  0 .The well posedness of this boundary value problem has been established in Section 1.
We define four operators in the following.
The far field operator  :  2 (S 1 ) →  2 (S 1 ) by where  ∞ is the far field pattern of the scattered wave   of problem ( 1) and ( 2).The far field operator  0 :  2 (S 1 ) →  2 (S 1 ) by where  ∞ 0 is the far field pattern of the scattered wave   0 of problem (63).
Note that From the boundary conditions on  1 for V(, ) and (, ), we can factorize the operator  −  0 as  −  0 = −. (71) Let  0 (⋅, ),  ∈  2 be the Green function for problem (63) of scattering by the background medium.We now define the modified far field equation where  ∞ 0 ( x, ) ∈  2 (S 1 ) is the far field pattern of the Green function  0 (, ).We will characterize the obstacle  2 by the behavior of an approximate solution   of far field equation (72).
To prove the existence of an approximate solution of (72), we firstly explore the related properties of the operators  and .
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 for  in  1 and  in  0 , we can decompose the operator  as  =  1  2 , where  2 : where  =  /4 /√8 0 .The interior regularity of the solution to problem (64) implies that the operator  2 is bounded.So the operator  is compact since the operator  1 is compact.
To show denseness of the range of  we just need to prove that the adjoint operator  * is injective.To this end, let  and  be the solution of problem (1) and (2) with incident plane wave We remind the reader that  ∈  2 loc ( 0 ) and  ∈  2 ( 1 ) by the regularity of elliptic equations.
For any  ∈  −3/2 ( 1 ), let  and  be the solution to problem (64) with the boundary data .Then one can derive that by Green's second theorem and the radiation condition of scattered field Therefore, we conclude that  0 S 0  * () is just the far field pattern of the potential for  ∈  2 \  1 .
Due to the fact that S 0 is unitary (see [24]), we just need to show that  = 0 under the assumption  ∞ ( x) = 0. Next we will prove this assertion.
In fact, if  ∞ ( x) = 0, then Rellich's lemma implies that () = 0 for  ∈  2 \  2 .Then by using the jump relations of single-and double-potentials, we get Thus  satisfies the Helmholtz equation in  2 with GIBC: By assumption on , there exists  0 ∈  1 such that Im(()) ̸ = 0 in a small neighborhood Λ( 0 ) ⊂  0 .Green's theorem in  2 and the divergence theorem on  1 imply By taking the imaginary part of the above equation we can obtain that  = 0 on Λ( 0 ); then the boundary condition shows that /] = 0 on Λ( 0 ).Thus Holmgren's uniqueness theorem implies that  = 0 in  2 ; we then obtain  = 0 from the jump relations.This lemma is then proved.
Finally, we give the main result in this paper, that is, recovering the obstacles  2 by a modified linear sampling method.

Remark 18.
(1) From Theorem 17, we have to obtain the far field pattern of Green function  0 (, ) 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  0 instead of  0 .
(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.
2 1 is not a Dirichlet eigenvalue of −Δ operator in the domain  2 which can guarantee the well posedness of the direct problem.For further consideration, we define the single-and double-layer operators  . and  ., respectively, by Since the incident plane wave   (⋅, ) and incident point source Φ 0 (⋅, ) solve the Helmholtz equation inside , we obtain that from the above two equalities (, x))  () .