Abstract

We consider the acoustic scattering problem from a crack which has Dirichlet boundary condition on one side and impedance boundary condition on the other side. The inverse scattering problem in this paper tries to determine the shape of the crack and the surface impedance coefficient from the near-field measurements of the scattered waves, while the source point is placed on a closed curve. We firstly establish a near-field operator and focus on the operator’s mathematical analysis. Secondly, we obtain a uniqueness theorem for the shape and surface impedance. Finally, by using the operator’s properties and modified linear sampling method, we reconstruct the shape and surface impedance.

1. Introduction

In this paper, we consider the scattering of an electromagnetic time-harmonic plane wave by an infinite cylinder having an open crack as cross section in . We assume that the cylinder is coated on one side by a material with surface impedance . 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 TM mode (see [1–3]), this leads to a mixed boundary value problem for the Helmholtz equation defined in the exterior of an open arc in .

Briefly speaking, let be an oriented piecewise smooth nonintersecting crack without cups; that is, , where is an injective piecewise function and the crack is contained in a closed curve . Then the mixed boundary value problem for the Helmholtz equation in can be formulated as follows: where is the wave number and is the surface impedance. is the total wave of the scattered wave and the incident wave ; that is, , and is the fundamental solution to the Helmholtz equation defined by with being a Hankel function of the first kind of order zero. The scattered field is required to satisfy the Sommerfeld radiation condition uniformly in with .

Remark 1. for , and for (for the details, see Section 2). In the following discussion, means the limit approaching the boundary from outside and inside the domain.

The inverse scattering problem in this paper is trying to determine the shape of the arc (or crack) and the surface impedance coefficient from the near-field measurements of the scattered waves, while the source point is placed on a closed curve, the results are as follows.

Inverse Problem (Ip). In this paper, the inverse problem we are concerned about is to determine the crack and the surface impedance from these measurements for .

In 1995, Kress considered the inverse scattering problem for cracks with sound-soft boundary condition in [4]. The case of a sound-hard crack was considered by Monch in 1997 in [5]. Both of the authors used Newton’s method to reconstruct the shape of the crack from a knowledge of the far-field pattern. In 2003, Cakoni and Colton considered an inverse scattering problem by cracks, and they reconstructed the cracks by using the linear sampling method in [1]. In 2005, Colton and Haddar discussed similar inverse scattering problem by cracks, and they recovered the cracks by using the reciprocity gap functional method in [6]. Zeev and Cakoni considered the inverse scattering problem for a crack embedded in a known inhomogeneous background and recovered the crack (with a point source as incident field) in 2009 in [3]. More related research works can be found in [2, 3, 7] and the references therein.

This paper is arranged as follows. In the next section, we formulate the scattering problem mathematically and prove that the associated near-field operator is injective with dense range under appropriate assumptions. In Section 3, we show that the crack and the surface impedance are uniquely determined from the near-field measurements of the scattered waves, while the source point is placed on a closed curve. We modify the linear sampling method and reconstruct the shape of the crack (or the surface impedance coefficient) in Section 4.

2. The Formulation of the Problem

We suppose that the crack can be extended to an arbitrary piecewise smooth, simply connected, and closed curve enclosing a bounded domain such that the normal vector on coincides with outward normal vector on which we again denote by . is a closed curve; we denote by the domain surrounded by . We suppose that is completely contained in , and we assume the normal vector on and is mapped to the exteriors of the domain and the domain , respectively.

In order to formulate our scattering problem more precisely, we need to properly define the trace space on and . Let be a bounded domain and let be an open subset of the boundary . If , , and denote the usual Sobolev spaces, we define the following spaces [8]: and we have the chain

Then problem (1) can be rewritten as and is required to satisfy Sommerfeld radiation condition (3).

Remark 2. By using similar method in [1], we can obtain the existence and uniqueness of solution to the direct problem (6). Here we use the point source as incident wave, while the plane wave was used as the incident in [1].

We define the near-field operator by where the function is the solution of problem (6). According to Green’s representation formula, we have On the boundary , we have Then, by substituting (9) into (8), we have By changing the order of and , we have

Then we have the following result.

Lemma 3. For the problem (6), one has for .

Proof. Applying Green’s second theorem and (3), we obtain Substituting (13) into (11) and (14) into (12), respectively, we have where . By using the boundary conditions in (6), we have This implies that . So, we complete the proof of this lemma.

Theorem 4. The near-field operator defined by (7) is injective and has dense range.

Proof. From , the adjoint of is given by Then we have , where . Thus, operator is injective if and only if is injective. Since in a Hilbert space, our proof will be finished by only showing that operator is injective.
Let ; we need to show that . Define It is easy to verify that satisfies the exterior problem This exterior Dirichlet problem has only zero solution (see [9]). Then the unique continuation principle now yields in . Therefore, Now define By the boundary conditions (6) and (20) together with the jump relationship of the single-layer potential on the boundary , we conclude that satisfies the following problem: On the boundary , due to the jump relationship of the single-layer potential, we have and . So, we get and by the boundary condition of (22) on . Then Holmgren’s principle and uniqueness continuation principle imply that on . From this, we obtain that .
Therefore, in the domain , we have the following problem for : The problem has only zero solution which implies that ; thus . Hence, is injective.

3. Uniqueness for the Inverse Problem

Based on the idea of [10, 11], we firstly conclude that is uniquely determined from without knowing a priori. Secondly, we show that the surface impedance can be uniquely determined by (see [12]).

Theorem 5. Assume that and are two cracks with corresponding impedance and impedance such that for a fixed wave number the corresponding scattering fields and coincide on for all point sources . Then and .

Proof. We consider problem (6) with replaced by . By using the same method in Lemma 3 in Section 2, we have where .
In the domain , let . Then satisfies problem (19) replacing the corresponding domain with ; that is, where we used the condition on for all point sources .
The only zero solution of this problem and the unique continuation principle imply that in ; that is, for all and .
By reciprocity (24) we have that for all and . Then again arguing as above we have that for all , .
Now we assume that . Without loss of generality, there exists but . Choose such that the sequence , , is contained in , where the unit normal vector to the boundary is directed into the exterior of . Considering as the solution of problem (6) with replaced by corresponding to and , then for . For simplicity, we use to denote the solution in ; that is, Consider the crack ; then , and, on the two sides of , we know that and on are uniformly bounded.
The well-posedness of the solution to the corresponding problem implies that the limit is bounded as , so by the trace theorem is uniformly bounded with respect to , where is a small neighborhood centered at not intersecting .
On the other hand, consider the solution of (6) with respect to the crack ; in this case . From the boundary condition on the crack , we have that as since as . This is a contradiction. Therefore .
Now let and assume that for . Then, from relation (24) and the unique continuation principle, we know that in . Then and on ; that is, and on . Let ; from the boundary condition (6), we have and thus Hence on since . Notice that on ; then we have This problem has only zero solution; that is, in .
Now choose sufficiently small such that ; that is, , where , , and is the unit outward normal to . Let ; then the limit of is bounded because that problem (6) is well posed, but is unbounded which leads to a contraction. So, we complete the proof of the theorem.

4. The Linear Sampling Method

The inverse scattering problem in this paper is trying to determine the shape of the crack and the surface impedance coefficient from the near-field measurements of the scattered waves, while the source point is placed on a closed curve. In this part, we provide the mathematical basis to reconstruct the crack from the knowledge of for by using the linear sampling method; that is, we want to determine from a knowledge of for , where is a circle centered at the origin; that is, . Based on the ideas of [2, 3], we introduce the near-field equation that is, where and is smooth nonintersecting arc and and .

We want to characterize the crack by using the behavior of an approximate solution of the near-field equation (30).

Now consider the following problem: for and . From [1], we know that this problem has a unique solution such that , where is a constant and does not depend on and .

To understand the near-field equation better, we define an operator which maps the boundary data to the solution on . We have the following conclusions about this operator .

Theorem 6. Operator is injective and compact and has dense range in .

Proof. Let ; that is, on ; we want to show that and . From on , we know that satisfies problem (19) which has only zero solution on . By the unique continuation principle, we have that in and thus and . So, from the boundary conditions in (33), we can get and , which implies that operator is injective.
Define where are polar coordinates of , , and . Clearly satisfies (33) with and . Since , the completeness of the trigonometric sequence in shows that operator has dense range.
We now show that operator is compact. Choose a disk such that . Using Green’s representation formula for , we can decompose operator as , where is defined by and is defined by The regularity of the solution to problem (33) implies that operator is bounded. So, operator is compact since operator is compact.