Uniqueness in Inverse Electromagnetic Conductive Scattering by Penetrable and Inhomogeneous Obstacles with a Lipschitz Boundary

and Applied Analysis 3 a proof of the well posedness for the direct problem by using a variational method. The uniqueness result for the inverse problem will be shown in Section 3. 2. The Direct Problem Let D ⊂ R3 be a bounded penetrable, inhomogeneous, isotropic domain with a Lipschitz boundary ∂D denoted by Γ and covered with a thin layer of high conductivity. Assume that the domain D is imbedded in a homogeneous background medium. Define k2 D k nD x and k2 b k nb with k > 0 being the wave number, where nD x and nb are the refractive index of the domain D and the background medium, respectively. Assume that nD ∈ C D with Re nD x > 0, Im nD x > 0 for all x ∈ D and nb is a complex constant with Im nb ≥ 0. Assume further that λ ∈ L∞ Γ with Re λ x ≥ 0 is a complex-valued function describing the surface impedance of the coating. The incident field is considered to be an electric dipole located at x0 on a large sphere SR0 {x ∈ R3 : |x| R0}with polarization p ∈ R3 given by Ee ( x, x0, p, kb ) i kb curlxcurlx p eb |x−x0| 4π |x − x0| . 2.1 Denote byG x, x0 the free space Green tensor of the backgroundmedium and define E x E x, x0, p pG x, x0 which satisfies curl curlE x − k2 bEi x pδ x − x0 in R3, 2.2 where δ is the Dirac delta function. Note that E x can be written as E x Ee ( x, x0, p, kb ) E b x , 2.3 where E b x is the scattered electric field due to the background medium and the electric dipole Ee x, x0, p, kb . In order to formulate precisely the scattering problem, recall the following Sobolev spaces: H curl, D { u ∈ ( L2 D )3 , curlu ∈ ( L2 D )3} , H div, D { u ∈ ( L2 D )3 , divu ∈ ( L2 D )3} , Lt ∂D { u ∈ ( L2 ∂D )3 , ν · u 0 on ∂D } , 4 Abstract and Applied Analysis H−1/2 div ∂D { u ∈ ( H−1/2 ∂D )3 , ν · u 0, div∂D u ∈ H−1/2 ∂D } , H−1/2 curl ∂D { u ∈ ( H−1/2 ∂D )3 , ν · u 0, curl∂D u ∈ H−1/2 ∂D } , H0 curl, D { u ∈ ( L2 D )3 , curlu ∈ ( L2 D )3 , ν × u 0 on ∂D } ,


Introduction
In this paper we are interested in determining the shape and location of a penetrable, inhomogeneous, isotropic, Lipschitz obstacle surrounded by a piecewise homogeneous, isotropic medium. The obstacle is covered with a thin layer of high conductivity. Such penetrable obstacles lead to conductive boundary conditions; for the precise mathematical description, the reader is referred to 1-3 . In this paper, it is shown that the shape and location of the obstacle and the corresponding surface parameter are uniquely determined from a knowledge of the near field data of the scattered electromagnetic wave at a fixed frequency. To this end, we need a well posedness result for the direct problem.
The well posedness of the Helmholtz equation for a penetrable, inhomogeneous, anisotropic medium has been studied recently in 4 . In 5 , the authors provided a proof for the well posedness of the scattering problem for a dielectric that is partially coated by a highly conductive layer in the TM case in 2007.
In the case of exterior Maxwell problem for the partially coated Lipschitz domains, the authors in 6 have established the well posedness of a unique solution by variational methods in 2004. For the homogeneous isotropic medium problem, by means of an integral equation method, Angell and Kirsch proved the existence and uniqueness of the classical solution for Maxwell's equations with conductive boundary conditions assuming λ ∈ C 0,α ∂D in 2 . Variational methods for the homogeneous isotropic medium problem were proposed in 1 , under the assumption that the bounded domain D ⊆ R 3 with boundary ∂D in the class C 2 , and some additional conditions on k, n D , μ, λ. It is also shown that the obstacle is uniquely determined by the far field patterns of all incident waves with a fixed wave number. For the inhomogeneous anisotropic media, the well posedness of the direct problem was proved in 7 .
The uniqueness result for the inverse medium scattering problem was first provided by Isakov see 8, 9 , in which it is shown that the shape of a penetrable, inhomogeneous, isotropic medium is uniquely determined by its far field pattern of all incident plane waves. The idea is to construct singular solutions of the boundary value problem with respect to two different scattering obstacles with identical far field patterns. Our uniqueness proof is based on this idea. The idea of Isakov was modified by Kirsh and Kress 10 using potential theory for the impenetrable obstacle case with Neumann boundary conditions. By the same technique, the authors in 11 proved the case of a penetrable obstacle with constant index of refraction. The use of potential theory will require strong smoothness assumptions on the scattering object. Then D. Mitrea and M. Mitrea 12 improved the previous results to the case of Lipschitz domains. In 13 , they extended Isakov's approach to the case of a penetrable obstacle for Hemholtz equations.The uniqueness theorem of Helmholtz equations for partially coated buried obstacle problem was shown in 14, 15 , assuming that the scattering fields were known with point sources as incident fields.
Recently, uniqueness for the inverse scattering problem in a layered medium has attracted intensive studies. For the sound-soft or sound-hard obstacle case, based on Schiffer's idea, 16 proved a uniqueness result. But their method can not be extended to other boundary conditions. In recent years, by employing the generalized mixed reciprocity relation, it was proved in 17, 18 that both the obstacle and its physical property can be uniquely determined for different boundary conditions. For the inverse acoustic scattering by an impenetrable obstacle in a two-layered medium case, it is shown in 19 that interface is uniquely determined from the far field pattern. Unfortunately, this method can not be extended to the electromagnetic case, but using ideas in 20 , a different method was used in 21 to establish such a uniqueness result for the electromagnetic case.
There are also some uniqueness results for partial differential equations with constant coefficients by integral equation methods. see 22, 23 . However, integral equation methods are not well tailored for partial differential equations having inhomogeneous coefficients of the highest derivatives. Consequently, in 24 , the author brought together the variational approach and the idea from 8, 9 to provide a uniqueness proof of Helmholtz equations with inhomogeneous coefficients for a penetrable, anisotropic obstacle. Their method depends on a regularity theorem for the direct problem and the well posedness of the interior transmission problem related to the direct problem. This idea has been extended to the case of electromagnetic scattering problem for anisotropic media in 25 .
The outline of this paper is as follows. In Section 2, besides the formulation of the direct scattering problem in a penetrable, inhomogeneous, Lipschitz domain, we also provide where H 1/ik curl E is the corresponding scattered magnetic field. Taking the complex conjugate of both sides of 2.11 and using the fact that Im n D , Im n b , and Re λ are nonnegative gives

2.12
An application of the Rellich lemma yields that E 0 in R 3 \ B R see 26, Theorem 6.10 . This, together with the unique continuation principle, implies that E 0 in R 3 \ D. From the trace theorem, it follows that ν × E 0 on Γ. Thus, taking the imaginary part of 2.11 and using the assumption that Im n D x > 0 for all x ∈ D, we have that V 0 in D.
Introduce the electric-to-magnetic Calderon operator G e see 27 , which maps the electric field boundary data ϕ on the surface of a large ball B R {x ∈ R 3 : |x| < R} to the magnetic boundary data Then the scattering problem 2.6 -2.10 can be reformulated in the following mixed conductive boundary value problem MOCKUP over a bounded domain: In the following, we introduce some properties of the Calderon operator that will be frequently used in the rest of this section. The basis functions for tangential fields on a sphere S R are the vector spherical harmonics of order n given by If k i in 2.23 , we will obtain another operator G e . Properties of Ge and G e are collected in the following lemma for a proof see 27 .

Lemma 2.2. The operator G e is negative definite in the sense that
for any ϕ ∈ H −1/2 div S R with ϕ / 0. Furthermore, In the remainder of this paper we will refer to 2.17 -2.21 as CBP . Here we will adapt the variational approach used in 6, 27 to prove the existence of a unique solution to our CBP . Define where D ⊂ B R . Then multiplying 2.17 and 2.18 by test function φ ∈ H loc curl, B R , using formally integration by parts and using the conductive boundary conditions on Γ, we can derive the following equivalent variational formulation for CBP . Find w ∈ X such that We rewrite 2.29 as the problem of finding w ∈ X such that

2.32
Here ·, · D denotes the L 2 D 3 scalar product, and ·, · ∂D denotes the L 2 ∂D 3 scalar product. We will use a Helmholtz decomposition to factor out the nullspace of the curl operator and then to prove the existence of a unique solution to CBP . Define The variational problem 2.34 can be rewritten as Abstract and Applied Analysis where we define

2.36
Here we have used ∇ τ ξ to write the tangential component of the gradient of ξ in terms of the tangential gradient on the sphere S R . By Lemma 2.2, it follows that G e is negative definite, then we obtain that A 1 p, ξ is a coercive sesquilinear form on S × S. Further by Lax-Milgram theorem, it is easy to see that A 1 p, ξ gives rise to a bijective operator. Since ν × ∇p ∈ H −1/2 div div, S R , still by Lemma 2.2, we know that A 2 p, ξ gives rise to a compact operator. In order to apply the Fredholm alternative to the variational problem 2.34 , we need to prove the following uniqueness lemma. Proof. It suffices to consider the following equation: Choosing ξ p, it is easy to see that

2.38
By the definition of the operator G e , if E s ∈ H loc curl, R 3 \ B is the weak solution of the problem which together with the fact Im n D ≥ 0, Im n b ≥ 0 implies

2.43
Therefore the Rellich lemma ensures us that Proof. Consider a bounded set of functions {u j } ∞ j 1 ⊂ X 0 . Each function u j ∈ X 0 can be extended to all of R 3 by solving the exterior Maxwell equation

2.46
Since the tangential components of u e j are continuous across S R , it follows that u e j ∈ H loc curl, R 3 . By using the properties of the Calderon operator G e and the conditions in X 0 , we see that the following equations hold true

2.47
Then, by the definition of G e that Ge ν × u j

2.48
Thus, u e j has a well-defined divergence and

2.49
Now we choose a cut-off function χ ∈ C ∞ 0 R 3 such that χ 1 in B and χ is supported in a ball B R 1 ⊃ B. Then one can use the general compactness theorem Theorem 4.7 in 27 to the sequence {χu e j } and extract a subsequence converging strongly in L 2 B 3 . This proves the lemma.
From the above definitions of S and X 0 , we have the following Helmholtz decomposition lemma.
Lemma 2.5. The spaces ∇S and X 0 are closed subspaces of X. The space X is the direct sum of the spaces ∇S and X 0 , that is, The proof of this Helmholtz decomposition Lemma is entirely classical see 27, 28 . We now look for a solution of the variational problem 2.31 in the form w w 0 ∇p 0 , where w 0 ∈ X 0 and p 0 ∈ S is the unique solution of 2.34 . We observe that A w 0 , ∇ξ 0 for all ξ ∈ S by the definition of X 0 . Hence the problem of determining w ∈ X is equivalent to the problem of determining w 0 ∈ X 0 such that where the operator G 1e is a compact operator from X 0 into H −1/2 div S R and the operator G 2e satisfies ik G 2e ν × ϕ , ϕ T S R ≥ 0. We now split the sesquilinear form A ·, · into A a b with The sesquilinear form a ·, · is obviously bounded and a direct computation verifies that with some constant α > 0. Hence by Lax-Milgram theorem, a ·, · gives rise to a bijective operator and by the compact embedding of X 0 in L 2 B 3 and the fact that G 1e is a compact operator from X 0 into H −1/2 div S R , the second term b ·, · gives rise to a compact operator. Then a standard argument implies that the Fredholm alternative can be applied. Finally, the uniqueness theorem yields the existence result. We summarize the above analysis in the following theorem. Theorem 2.6. For any incident field E i , there exists a unique solution w ∈ X of (CBP) which depends continuously on the incident field E i .

Uniqueness for the Inverse Problem
In this section we will show that the scattering obstacle D and the corresponding parameter λ are uniquely determined from the knowledge of the scattered fields E s 1,x 0 x, x 0 | x∈∂Ω for all x 0 ∈ S R 0 , where S R 0 is the surface of a large ball B R 0 with D ⊂ Ω ⊂ B R 0 . By some properties of the scattered fields, we can derive a relationship between them, then constructing special singular solutions which satisfy the relationship. Finally, we can obtain the uniqueness result by using the singularities of the singular solutions that we constructed.

Abstract and Applied Analysis
Proof. For simplicity, we only prove statement ii . Case i can be proved similarly. Let g ∈ L 2 t ∂Ω 0 be such that Then it follows that By 3.2 , it is easy to see that for arbitrary polarization p x 0 in the tangential plane to S R 0 at x 0 , we have From the definition of 3.3 , we immediately have Due to the symmetry of the background Green function, F * 2 x 0 as a function of x 0 solves curl Hence, F * x 0 satisfies the Maxwell's equation in R 3 \ ∂Ω 0 . By 3.6 and the fact that p x 0 is an arbitrary polarization in the tangential plane to S R 0 at x 0 , we immediately have that ν × F * x 0 | S R 0 0. The uniqueness of the exterior problem implies that F * x 0 0 in R 3 \ B R 0 . Thus, the unique continuation principle ensures us that F * x 0 0 in Ω e 0 R 3 \Ω 0 . By trace theorem, it follows that ν×F * x 0 0 and ν×curl x 0 F * x 0 Abstract and Applied Analysis 13 0 on ∂Ω 0 . By the definition of F * x 0 and the jump relations of the vector potential across ∂Ω 0 , it can be checked that F * x 0 satisfies the following equations:

3.9
Therefore, the uniqueness theorem of the interior problem for Maxwell's equations implies that F * x 0 0 in Ω 0 . Finally, from the jump relations of the vector potential across ∂Ω 0 , we have which completes the proof.
We now consider two obstacles D 1 and D 2 with the refractive index n D j and the surface impedance λ j , j 1, 2. Let U denote the unbounded part of R 3 \ D 1 ∪ D 2 and D 0 R 3 \ U its open complement. From the proof of Theorem 2.6, it follows that the total field w j , j 1, 2 satisfies for any large ball B R with D ⊂ B R and all test function φ ∈ X, where It is convenient to introduce the following space: where D j ⊂ B R . The relationship derived in the following lemma plays a central role in the proof of the main result in this section.

Lemma 3.2. Assume that k 2 b is not an eigenvalue of Maxwell equation in
1,x 0 x, x 0 and E s 2,x 0 x, x 0 be the scattered fields with respect to D 1 and D 2 , respectively, produced by the same incident field pG ·, x 0 . Assume that ν × E s 1,x 0 | ∂Ω ν × E s 2,x 0 | ∂Ω for all x 0 ∈ S R 0 with the radius R 0 > R for a fixed wave number k. Then we have 3.14 14

Abstract and Applied Analysis
Here V j ∈ X 1 ∩ X 2 satisfies the following variational problem: Proof. i We first prove that for any fixed z ∈ U, the scattered fields E s 1,z x, z E s 2,z x, z , where V j,z , E s j,z , j 1, 2 is the solution of the following problem: By Lemma 3.1 and the fact that k 2 b is not an eigenvalue of Maxwell equation in Ω, it follows that there exists a sequence a n ∈ R and x n 0 ∈ S R 0 such that ν × pG ·, z − n a n ν × pG ·, x Let E x pG ·, z − n a n pG ·, ν × E f, on ∂Ω

and 3.17 imply that
This, together with the fact that div E 0, implies see 28 Abstract and Applied Analysis 15 Then by 3.19 , 3.20 , and the trace theorem, it can be proved that

3.21
Denote by E s 1 and E s 2 the scattered fields with respect to D 1 and D 2 produced by the same incident field n a n pG ·, x n 0 . By the assumption ν × E s 1, Then by the uniqueness theorem of the exterior scattering problem, it follows that E s 1 E s 2 in R 3 \ Ω, which together with the unique continuation principle ensures that E s 1 E s 2 in U. Now, by 3.21 and the well posedness of the direct problem 2.18 , it can be checked that for any compact set K ⊂⊂ U, we have

3.22
for any fixed z ∈ K. Therefore, the fact E s The arbitrarity of ε implies that E s 1,z x, z E s 2,z x, z for any fixed z ∈ K. ii Next we will show that the identity 3.14 holds. Set w * w 2 − w 1 , then it follows from 3.11 that and define a smooth function ψ ∈ C ∞ R 3 with ψ 1 in Ω 1 and ψ 0 in R 3 \ Ω 2 . Let a 2 ψ, κ 2 ψk 2 2 * , λ 2 λ 2 , it is easy to see that a 2 , κ 2 , λ 2 satisfy the assumptions of the lemma. We further assume V 2 satisfies 3.15 with respect to a 2 ψ, κ 2 ψk 2 2 * , λ 2 λ 2 , so that substituting V ψV 2 into the left hand of 3.24 and noting that w * 0 in U yield that curl w * , curl ψV 2 3.25 16 Abstract and Applied Analysis Hence substituting V ψV 2 into 3.24 , it follows from the right hand of 3.24 that We define f ∈ X 1 by for all φ ∈ X 1 . By Theorem 2.6, it follows that there exists a unique solution w 0 ∈ X 1 of the problem Equation 3.28 with φ replaced by ψ i w 1 yields By 3.26 and 3.27 , it can be shown that

3.31
Taking the difference of 3.29 and 3.30 , we have that curl ψ e w 0 · curl w 1 − k 2 b ψ e w 0 · w 1 dx.

3.32
By 3.28 , we can deduce that w 0 is a radiating solution of the corresponding Maxwell's equations in B R , then it can be extended to all of R 3 denoted by w e 0 by solving the exterior Maxwell's equation in R 3 \ B R with ν × w e 0 ν × w 0 on S R , which also satisfies the Silver-Müler radiation condition at infinity. By applying the vector Green formula to 3.32 , it can be proved that In view of the fact

3.34
Application of the vector Green formula again and noting that both E s 1 and the extended function w e 0 satisfy the Silver-Müler radiation condition, it follows that

3.35
Hence the Stratton-Chu formula combines with 3.34 implies that

3.36
Since p x 0 is an arbitrary polarization in the tangential plane to S R 0 at x 0 , we obtain that ν × w e 0 x 0 | S R 0 0. By the fact that w e 0 is a radiating solution of Maxwell's equation in R 3 \ B R 0 , it follows that w e 0 0 in R 3 \ B R 0 . Hence the unique continuation principle implies that w e 0 0 in R 3 \ U. Therefore, w e 0 can be used as a test function for V 1 , which satisfies 3.15 with a 1 ψ i , κ 1 ψ i k 2 1 * , λ 1 λ 1 . So that from the left hand of 3.30 , we deduce that

3.37
Thus, it follows from the right hand of 3.30 that f, ψ i V 1 X 1 0. Furthermore, from 3.27 with φ replaced by ψ i V 1 , it can be shown that From the definitions of k 2 j * , we observe that 3.39 which combines 3.38 , the definition of the scalar product ·, · D , and the fact that implies that 3.14 holds. This ends the proof of this lemma.
The main result of this section is contained in the following theorem.

Theorem 3.3. Let E s
1 and E s 2 be the scattered fields with respect to D 1 and D 2 , respectively, and λ 1 , λ 2 the corresponding impedances. Suppose that the assumptions in Lemma 3.2 hold true and Γ j ∩ D i \D j is not empty for i, j 1, 2, i / j. If one of the following assumptions holds, then we have D 1 D 2 . Consider i Re λ j ≥ δ > 0; ii Im λ j ≥ δ > 0 or Im λ j ≤ −δ < 0.
Proof. Let us assume that D 1 is not included in D 2 . Since D e 2 R 3 \ D 2 is connected, we can find a point z ∈ Γ 1 \ D 2 and a sufficiently small ε > 0 with the following properties: ii the points z n z ε/n ν z lie in B 2ε z for all n ∈ N, where ν z is the unit normal to Γ 1 at z.
Denote D D 1 \ D 2 o , the inner part of the domain D 1 \ D 2 . We consider the unique solution of the following problem:

3.40
Here E s satisfies the Silver-Müler radiation condition at infinity, and E m denotes the magnetic dipole defined by E m x, z n curl x ν z e ik b |x−z n | |x − z n | .

3.41
Define It can be proved that F 1 x, z n is a solution of Maxwell's equations with homogeneous conductive boundary value conditions on ∂D in any domain Ω ⊂ R 3 with D 0 ⊂ Ω and z n / ⊆ Ω. Define and λ 1 x λ 1 . In view of the above definitions of κ 1 x and λ 1 x , it follows that F 1 x, z n satisfies the variational equation 3.15 in Lemma 3.2 for the obstacle D. The well posedness of the direct problem for CBP and the fact that z n is bounded away from ∂D imply that the solution V, E s of 3.40 is uniformly bounded in X. We now define another singular solution with respect to D 2 by where E m is a magnetic dipole defined in 3.41 , and V , E s is a solution of the problem curl curl E s − k 2 b E s 0, in R 3 \ D 2 ∪ z n , ν × E s − V −ν × E m ·, z n , on ∂D, ν × curl E s − V ikλ 2 E s T E m ·, z n T − ν × curl E m ·, z n , on Γ 2 .

3.45
Here E s satisfies the Silver-Müler radiation condition at infinity. Noting that F 2 x, z n satisfies the variational equation 3.15 in Lemma 3.2 with κ 2 x k 2 2 * and λ 2 λ 2 , it follows that both F 1 x, z n and F 2 x, z n satisfy the relationship 3.14 , then we obtain

3.46
For case i , by the fact that z ∈ Γ 1 \D 2 and the singularities of the magnetic dipole E m defined in 3.41 , it can be proved that |k 2 n D − n b F 1 , F 2 D ik λ 1 F 1 T , F 2 T ∂D | → ∞ as n → ∞, this, together with the fact that the other terms in the right hand of 3.46 are bounded, leads to a contradiction. Hence we have D 1 ⊂ D 2 . By choosing z ∈ Γ 2 \ D 1 and using the similar analysis as in the proof above, one can prove that D 2 ⊂ D 1 . Finally, we obtain that D 1 D 2 . For other cases, due to the singularities of F j j 1, 2 , a contradiction also arises in 3.46 as n → ∞. This proves the theorem. Theorem 3.4. Assume D 1 D 2 with parameters λ j ∈ C ∂D j and the scattered fields E s j,x 0 , j 1, 2 satisfy ν × E s 1,x 0 | ∂Ω ν × E s 2,x 0 | ∂Ω for all x 0 ∈ S R 0 , then we have λ 1 λ 2 on ∂D.
Proof. From the proof of Theorem 3.3, it follows that there exists two singular solutions F 1 , F 2 of the conductive boundary problem with respect to the obstacle D for some z ∈ ∂D. By Lemma 3.2 and the identity D 1 D 2 , it can be checked that D k 2 1 * − k 2 2 * F 1 x, z n · F 2 x, z n dx ik ∂D λ 1 − λ 2 F 1 x, z n · F 2 x, z n ds x 0.

3.47
The singularities of F 1 , F 2 ensure that λ 1 λ 2 . This completes the proof of the theorem.