The Problem of Scattering by a Mixture of Cracks and Obstacles

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 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 . 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 and a bounded domain , 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 and a bounded domain in as cross section. We assume that the cylinder is (possibly) partially 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 mode, this leads to a mixed boundary value problem for the Helmholtz equation defined in the exterior of a mixture in .

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 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.

Consider the scattering of time-harmonic electromagnetic plane waves from an infinite cylinder with a mixture of an open crack and a bounded domain in as cross section. For further considerations, we suppose that has smooth boundary (e.g., ), and the crack (smooth) can be extended to an arbitrary smooth, simply connected, closed curve enclosing a bounded domain such that the normal vector on coincides with the outward normal vector on which we again denote by . The bounded domain is located inside the domain , and .

In the whole paper, we assume that and .

Suppose that

(2.1)

where is an injective piecewise function. We denote the outside of with respect to the chosen orientation by and the inside by . Here we suppose that the is divided into two parts and and consider the electromagnetic field E-polarized. Different boundary conditions on , and lead to the following problem:

(2.2)

where for and for . The total field is decomposed into the given incident field , and the unknown scattered field which is required to satisfy the Sommerfeld radiation condition

(2.3)

uniformly in with .

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

Let be a piece of the boundary. Use and to denote the usual Sobolev spaces, is the trace space, and we define

(2.4)

Just consider the scattered field , then (2.2) and (2.3) are a special case of the following problem.

Given , , and find such that

(2.5)

and is required to satisfy the Sommerfeld radiation condition (2.3). For simplicity, we assume that and .

Theorem 2.1.

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

Proof.

Let be a solution to the problem (2.5) with , we want to show that in .

Suppose that (with boundary ) is a sufficiently large ball which contains the domain . Obviously, to the Helmholtz equation in (2.5), the solution satisfies the following transmission boundary conditions on the complementary part of :

(2.6)

where " " denote the limit approaching from outside and inside , respectively. Applying Green's formula for and in and , we have

(2.7)

where is directed into the exterior of the corresponding domain.

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

(2.8)

Hence

(2.9)

So, from [13, Theorem ] and a unique continuation argument we obtain that in .

We use and to denote the jump of and across the crack , respectively. Then we have the following.

Lemma 2.2.

If is a solution of (2.5) and (2.3), then and .

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 , by Green representation formula

(2.10)

and for

(2.11)

where

(2.12)

is the fundamental solution to the Helmholtz equation in , and 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 (see [5, 11]) and approaching the boundary from inside , we obtain (for )

(2.13)

(2.14)

where , , , and are boundary integral operators:

(2.15)

defined by (for )

(2.16)

Similarly, approaching the boundary from inside we obtain (for )

(2.17)

(2.18)

From (2.13)–(2.18), we have

(2.19)

(2.20)

Restrict on , from (2.19) we have

(2.21)

where means a restriction to .

Define

(2.22)

Then zero extend , , and to the whole in the following:

(2.23)

By using the boundary conditions in (2.5), we rewrite (2.21) as

(2.24)

where

(2.25)

Furthermore, we modify (2.24) as

(2.26)

where the operator is the operator applied to a function with and evaluated on , with analogous definition for , , and . We have mapping properties (see [5, 11])

(2.27)

Again from (2.13)–(2.18), restricting to boundary we have

(2.28)

or

(2.29)

Like previous, define

(2.30)

Then we can rewrite (2.29) as

(2.31)

where

(2.32)

Similar to (2.26), we modify (2.31) as

(2.33)

and we have mapping properties:

(2.34)

Combining (2.13) and (2.14),

(2.35)

(2.36)

Using (2.17) and (2.18),

(2.37)

Then using (2.36),

(2.38)

From (2.29), we have

(2.39)

Restricting (2.38) to and using (2.39), we modify (2.38) as

(2.40)

where

(2.41)

for .

Define

(2.42)

and using the notation in previous, we can rewrite (2.40) as

(2.43)

or

(2.44)

where the operators , , and are restriction operators (see (2.29)). As before, we have mapping properties:

(2.45)

By using Green formula and approaching the boundary from inside we obtain (for )

(2.46)

The last term in (2.46) can be reformulated as

(2.47)

Since and in (2.47), we have the following result (see [13]).

Lemma 2.3.

By using Green formula and the Sommerfeld radiation condition (2.3), one obtains

(2.48)

Proof.

Denote by a sufficiently large ball with radius containing and use Green formula inside . Furthermore noticing , , and the Sommerfeld radiation condition (2.3), we can prove this lemma.

Combining (2.46), (2.47), and Lemma 2.3 and restricting to we have

(2.49)

where

(2.50)

Define

(2.51)

and then we can rewrite (2.49) as

(2.52)

Similarly, and are restriction operators as before, and we have mapping properties:

(2.53)

Combining (2.52), (2.26), (2.33), and (2.44), we have

(2.54)

If we define

(2.55)

then (2.54) can be rewritten as a boundary integral system:

(2.56)

Remark 2.4.

If the above system (2.56) has a unique solution, our problem (2.5) with (2.3) will have a unique solution (see [13, 14]).

Based on the Fredholm theory, we show the existence and uniqueness of a solution to the integral system (2.56).

Define

(3.1)

and its dual space

(3.2)

Theorem 3.1.

The operator maps continuously into and is Fredholm with index zero.

Proof.

As is known, the operator is positive and bounded below up to a compact perturbation (see [18]); that is, there exists a compact operator

(3.3)

such that

(3.4)

where denote the duality between and .

For convenience, in the following discussion we define

(3.5)

Similarly, the operators and are positive and bounded below up to compact perturbations (see [18]), that is, there exist compact operators

(3.6)

such that

(3.7)

Define and , then and are bounded below up and positive.

Take , and let , , and be the extension by zero to of , , and respectively.

Denote .

It is easy to check that the operators , , , , , and are compact operators, and then we can rewrite as the following:

(3.8)

with

(3.9)

where is compact and defines a sesquilinear form, that is,

(3.10)

Here denotes the scalar product on or defined by or , and is the scalar product on ( ).

By properties of the operators , , and , we have

(3.11)

Similarly,

(3.12)

So the operator is coercive, that is,

(3.13)

whence the operator is Fredholm with index zero.

Theorem 3.2.

The operator has a trivial kernel if is not Dirichlet eigenvalue of the Laplace operator in .

Proof.

In this part, we show that . To this end let be a solution of the homogeneous system , and we want to prove that .

However, means that

(3.14)

Define a potential

(3.15)

where , and have the same meaning as before and

(3.16)

This potential satisfies Helmholtz equation in and the Sommerfeld radiation condition (see [13, 14]).

Considering the potential inside and approaching the boundary ( ), we have

(3.17)

and (3.14) implies that

(3.18)

Similarly, considering the potential inside and approaching the boundary ( ), then restricting to the partial boundary :

(3.19)

and restricting to the partial boundary , we have

(3.20)

Now, we consider the potential in the region and approach the boundary ( ), and then restricting to the partial boundary , similar to (3.19), we have

(3.21)

Refering to (3.20),

(3.22)

Combining (3.22), from (3.14) we have

(3.23)

From (3.18)–(3.23), the potential satisfies the following boundary value problem:

(3.24)

and the Sommerfeld radiation condition:

(3.25)

uniformly in with .

The uniqueness result Theorem 2.1 in Section 2 implies that

(3.26)

Notice that is not Dirichlet eigenvalue of the Laplace operator in , and so

(3.27)

Therefore, the well-known jump relationships (see [13, 14]) imply that

(3.28)

So we complete the proof of the theorem.

Combining Theorems 3.1 and 3.2, we have the following

Theorem 3.3.

The boundary integral system (2.56) has a unique solution.

Remark 3.4.

If we remove the condition that " is not Dirichlet eigenvalue of the Laplace operator in ," instead of it by the assumption that Im  , then Theorem 2.1 in Section 2 and Theorem 3.3 in Section 3 are also true.