#### 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 (, ) 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.

#### 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 is .

To be precise, let denote the impenetrable obstacle which is a bounded domain with a smooth boundary (e.g., ). Assume that the unknown obstacle is buried in a penetrable obstacle with a closed surface such that . Denote by a connected bounded domain filled with homogeneous medium and denote by the unbounded connected domain occupied by another homogeneous medium. Let be the wave number in terms of the frequency and the sound speed in the corresponding region () (see Figure 1).

The scattering of time-harmonic acoustic plane waves by an obstacle with GIBC in a piecewise homogeneous medium in can be modeled by the Helmholtz equation with boundary conditions on the boundary and interface :Here is the unit outward normal vector on the boundary or ; , denote the limit of , on the boundary 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 constant surface impedance on is supposed which is given by in terms of the density in the corresponding region (). On the boundary , the impedances and are complex-valued functions satisfying and . The surface divergence and the surface gradient are precisely defined in Chapter 5 of [1]. In the two-dimensional case, the inhomogeneous Laplace-Beltrami differential operator becomes , where is the tangential derivative and is the arc length.

The total field is decomposed into the given incident field , , (the unit sphere in ) and the unknown scattered field which is required to satisfy the Sommerfeld radiation condition [2]uniformly in with . Further it is known that the scattered field has the following asymptotic representation:uniformly for all directions , where the function defined on the unit sphere is known as the far field pattern with and denoting, respectively, the observation direction and the incident direction.

The* direct problem* is to seek functions and satisfying (1) and (2). In the next section, more general direct problem (4) will be considered. If the impenetrable obstacle 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 by the variational method; but for the case when belongs to , 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 and from the knowledge of the far field pattern for all with the given wave number and the positive constant .

As usual in most of the inverse problems, the first issue is the uniqueness, that is, in what conditions, the shape of the obstacle (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 [6–11] 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 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 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 conditions , and a general boundary data , find and such that

*Remark 2. *Direct scattering problem (1) and (2) is a special case of problem (4) by taking , , , and .

Lemma 3. *Suppose that , , , and ; then problem (4) has at most one solution.*

*Proof. *Clearly, it is sufficient to show that in and in if on and on . Denote by a circle large enough with radius such that is contained in its interior. From Green’s theorem, we obtainin the domain andin the domain . By the boundary conditions of (4), we conclude from the above two equations thatSince , , , and , it follows thatRellich’s lemma [2] shows that in and it follows by the unique continuation principle [2] that in . The transmission boundary conditions and Holmgren’s uniqueness theorem [18] imply that in . 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 , , and are the unknown densities and, is the fundamental solution of the Helmholtz equation in .

*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*. * is not a Dirichlet eigenvalue of ** operator in the domain * which can guarantee the well posedness of the direct problem.

For further consideration, we define the single- and double-layer operators and , respectively, byand the normal derivative operators and bywith . Referring to [20], we have mapping propertiesfor and .

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 On the boundary , we obtain that

Define bounded linear operators byLetThen the potential functions defined by (9) solve problem (4) provided the unknown densities , , and solve the following boundary integral system:Defining the Sobolev spacesit 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 , , , and are positive and bounded up to a compact perturbation, respectively; we denote by , , , and the compact operatorssuch thatwhere denotes the duality between and .

Let and be the operators defined as and , respectively, with kernel replaced by . Then and 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 , for . Consider the following sesquilinear form for :where is the dual space of and denotes the scalar product on . Due to the coercivity of and , the adjoint between and , we obtain that the above sesquilinear form is coercive; that is,Whence the operatoris invertible. On the other hand, by our* Assumption A*, it can be seen that is invertible (see Lemma 2.1 in [21]). So the operator is invertible.

The entries , , , and have continuous kernels, which means that they are compact operators. Due to the compact embedding theorem and the mapping properties of and , the entry 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 . Define two potentialsUsing the jump relations of the single- and double-layer potentials across , we haveSince , it is easy to check that the potentials defined in (27) and (28) satisfyWe can show that problem (30) has only trivial solution (see [2]).

Using the same , define two new potentialsThen we can prove that satisfies problem (4) with homogeneous boundary conditions. Lemma 3 shows that . Thus again by the jump relations of the single- and double-layer potentials across we haveAt this time, the potential given by (28) becomesand note that on because of the trivial solution of (30); we conclude that satisfies the Helmholtz equation in with homogeneous Dirichlet boundary condition if we let . By our* Assumption A*, is not a Dirichlet eigenvalue in , which implies that in . Therefore, the jump relation across shows thatThen 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*

#### 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 scatterer 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.

Let us go back to scattering problem (1) and (2). The incident wave has two choices: the incident plane wave (i.e., ) and the incident point source .(1)To the incident plane wave , we use and to denote the scattered field and the corresponding far field pattern, respectively.(2)To the incident point source , we use and 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 waves with and point sources , we havewhere .*

*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 case firstly. By Green’s second theorem and the Sommerfeld radiation condition we have thatfor , . By the boundary conditions and Green’s second theorem, the total fields and satisfySince the incident plane wave and incident point source solve the Helmholtz equation inside , we obtain that from the above two equalitiesUsing Green’s representation formula for , we obtain the representationfor . The far field pattern has the following integral representation:Thus we conclude that for , from (39), (40), and (41) with replaced by .

Next, we consider the case . From the boundary condition on , we have that for the total fields and For the scattered fields and we still have equality (37), and for the incident plane wave and incident point source we haveby Green’s second theorem.

From (37) and (41) we getLet be a sphere contained in . Applying Green’s second theorem in the domain and taking into account the boundary condition on we further haveBy the well posedness of the direct problem and the interior elliptic regularity [22], and for any compact subset of . Therefore, there is a sequence such that andas . This together with the Cauchy-Schwarz inequality implies that the integral on tends to as . By passing to the limit in (45) with we haveThe 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 thatIt follows from (47) and (48) together with the boundary conditions on and and Green’s second theorem thatWe conclude from (49) that , for , . Therefore the proof of this lemma is completed.

Lemma 10. *For the transmitted wave of problem (1) and (2) associated with the incident plane wave , we have that is complete in .*

Note that for the incident plane wave , the regularity of elliptic equations shows that the solution of problem (1) and (2) belongs to .

*Proof. *Assume that is a function in such that for every Consider the following problem:According to Theorem 7, this problem is well posedness and we have that the unique solution . Then we haveFurthermore, from the transmission boundary condition, Green’s second theorem, and the radiation condition for and we haveThus Rellich’s lemma implies that in . Then the transmission boundary conditions on and Holmgren’s uniqueness theorem show that in ; hence 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 interface , the positive constant , and the wave numbers and , due to incident plane waves , assume that the two far field patterns and corresponding to the two scattered fields and which are arisen by two obstacles with impedances and with impedances , respectively, coincide at a fixed frequency for all and ; then and .*

*Proof. *If the obstacles are not the same, that is, , let be the unbounded component of . By Rellich’s lemma the scattered fields and corresponding to the incident plane wave coincide in the unbounded domain . Without loss of generality, we may assume that there exist and open set such that and . We can choose such that the sequenceis contained in , where is the normal to at .

Consider the solution to problem (1) and (2) due to the incident point source . By Lemma 8, the far fields and coincide in . Then Rellich’s lemma implies that for . From this we have by denoting Considering as the scattered field corresponding to , using the boundary condition on for , we see thatFrom the well posedness of problem (1) and (2) and the regularity of elliptic equations we obtainin . On the other hand, as the same argument in Theorem 3.1 of [11], does not belong to . This is a contradiction, which implies that .

Next, we show that . 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, let and , and denote by the same total fields and . From the boundary conditions for the total field, we have thatThis equality should be understood in the weak sense. Thus for every we haveWith the help of Lemma 10, we obtainChoosing in the above equation leads to . The above equation also implies thatAssume that for some ; then, for example, without loss of generality. Since is continuous there exists such that for all . Let us choose as a smooth and compactly supported function in ; we obtain thatand then on ; that is, is a constant on , which is a contradiction. We hence have on 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 obstacle by using the modified linear sampling method (see [23]).

We do some preparation firstly. Consider the total wave () such thatRecalling that and are the solution to scattering problem (1) and (2) for incident plane wave with the direction , it is easy to verify that the fields , and , solve the following boundary value problem:where . 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 operator bywhere is the far field pattern of the wave field of problem (64).

The auxiliary operator bywhere is the solution of (63) corresponding to the incident wave .

The far field operator bywhere is the far field pattern of the scattered wave