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

Authors: Fenglong Qu
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

This paper is concerned with the problem of scattering of time-harmonic electromagnetic
waves by a penetrable, inhomogeneous, Lipschitz obstacle covered with a thin layer of high
conductivity. The well posedness of the direct problem is established by the variational method. The
inverse problem is also considered in this paper. Under certain assumptions, a uniqueness result is
obtained for determining the shape and location of the obstacle and the corresponding surface parameterλ(x) from the knowledge of the near field data, assuming that the incident fields are electric
dipoles located on a large sphere with polarization p∈ℝ3. Our results extend those in the paper by F. Hettlich (1996) to the
case of inhomogeneous Lipschitz obstacles.


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## Body

## 1. 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∈C0,α(∂D) in [2]. Variational methods for the homogeneous isotropic medium problem were proposed in [1], under the assumption that the bounded domain D⊆ℝ3 with boundary ∂D in the class C2, and some additional conditions on k,nD,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 Section2, besides the formulation of the direct scattering problem in a penetrable, inhomogeneous, Lipschitz domain, we also provide 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

LetD⊂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 kD2=k2nD(x) and kb2=k2nb 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
(2.1)Ee(x,x0,p,kb)=ikbcurlxcurlxpeikb|x-x0|4π|x-x0|.
Denote by G(x,x0) the free space Green tensor of the background medium and define Ei(x)=Ei(x,x0,p)=pG(x,x0) which satisfies
(2.2)curlcurlEi(x)-kb2Ei(x)=pδ(x-x0)inℝ3,
where δ is the Dirac delta function. Note that Ei(x) can be written as
(2.3)Ei(x)=Ee(x,x0,p,kb)+Ebs(x),
where Ebs(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:(2.4)H(curl,D)={u∈(L2(D))3,curlu∈(L2(D))3},H(div,D)={u∈(L2(D))3,divu∈(L2(D))3},Lt2(∂D)={u∈(L2(∂D))3,ν·u=0on∂D},Hdiv-1/2(∂D)={u∈(H-1/2(∂D))3,ν·u=0,div∂Du∈H-1/2(∂D)},Hcurl-1/2(∂D)={u∈(H-1/2(∂D))3,ν·u=0,curl∂Du∈H-1/2(∂D)},H0(curl,D)={u∈(L2(D))3,curlu∈(L2(D))3,ν×u=0on∂D},
where ν denotes the exterior unit normal to ∂D. If D is unbounded, we denote by Hloc(curl,D) the space of functions u∈H(curl,K) for any compact set K  ⊂⊂  D. Introduce the space
(2.5)X~={u∈Hloc(curl,ℝ3),uT|Γ∈Lt2(Γ)},
where uT=ν×u×ν. Then the scattering problem can be formulated as follows. Given Ei, find the field V and the scattered field Es such that
(2.6)curlcurlV-kD2V=0,inD,(2.7)curlcurlEs-kb2Es=0,inDe,(2.8)ν×Es-ν×V=-ν×Ei,onΓ,(2.9)ν×curlEs-ν×curlV=ikλETs+ikλETi-ν×curlEi,onΓ,
and the scattered field Es=E-Ei is required to satisfy the Silver-Müller radiation condition
(2.10)limr→∞r(curlEs×x^-ikEs)=0
uniformly in x^=x/|x|, where r=|x|.We first have the following uniqueness result for the above scattering problem.Theorem 2.1.
The scattering problem (2.6)–(2.9) has at most one solution.Proof.
To prove the theorem, it is enough to consider the caseEi=0 whence E=Es. Taking the dot product of (2.6) with V over D and of (2.7) with E over DR=De∩BR with De=ℝ3∖D-, respectively, and integrating by parts, we obtain by using the conductive conditions (2.8) and (2.9) that
(2.11)∫D‍(|curlV|2-kD2|V|2)dx+∫DR‍(|curlE|2-kb2|E|2)dx-ik∫Γ‍λ|ET|2ds+ik∫SR‍ν×H·E-Tds=0,
where H=(1/ik)curlE is the corresponding scattered magnetic field. Taking the complex conjugate of both sides of (2.11) and using the fact that Im(nD),Im(nb), and Re(λ) are nonnegative gives
(2.12)Re∫SR‍ν×E·H-ds=-k-1∫D‍Im(nD)|V|2dx-k-1∫DR‍Im(nb)|E|2dx-∫Γ‍Re(λ)|ET|2ds≤0.
An application of the Rellich lemma yields that E=0 in ℝ3∖B-R (see [26, Theorem 6.10]). This, together with the unique continuation principle, implies that E=0 in ℝ3∖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[nD(x)]>0 for all x∈D, we have that V=0 in D.
Introduce the electric-to-magnetic Calderon operatorGe (see [27]), which maps the electric field boundary data φ on the surface of a large ball BR={x∈ℝ3:|x|<R} to the magnetic boundary data x^×H on SR=∂BR, where (E,H) satisfies
(2.13)curlE-ikH=0,inℝ3∖B-R,(2.14)curlH+ikE=0,inℝ3∖B-R,(2.15)x^×E=φ,onSR,(2.16)limr→∞(H×x-rE)=0.
Then the scattering problem (2.6)–(2.10) can be reformulated in the following mixed conductive boundary value problem (MOCKUP) over a bounded domain:
(2.17)curlcurlV-kD2V=0,inD,(2.18)curlcurlEs-kb2Es=0,inDR,(2.19)ν×Es-ν×V=-ν×Ei,onΓ,(2.20)ν×curlEs-ν×curlV=ikλETs+ikλETi-ν×curlEi,onΓ,(2.21)ν×1ikcurlEs=Ge(ν×Es),onSR,
where DR=De∩BR.
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 sphereSR are the vector spherical harmonics of order n given by
(2.22)Unm=1n(n+1)∇τ1Ynm,Vnm=x^×Unm
for n=1,2,… and m=-n,…,n. Here, as usual, ∇τ1 denotes the surface gradient on the surface of the unit sphere S1.
Forφ∈Hdiv-1/2(SR) given by φ=∑n=1∞‍∑m=-nn‍an,mUnm+bn,mVnm, the operator Ge can be defined by
(2.23)Geφ=∑n=1∞‍∑m=-nn‍{-ikRbn,mδnUnm+an,mδnikRVnm},
where
(2.24)δn=kR(hn(1))′(kR)hn(1)(kR)+1,
and hn(1)(kR) is the spherical Hankel function.
Ifk=i in (2.23), we will obtain another operator Ge~. Properties of Ge and Ge~ are collected in the following lemma (for a proof see [27]).Lemma 2.2.
The operatorGe~ is negative definite in the sense that
(2.25)〈Ge~φ,φ×x^〉SR<0
for any φ∈Hdiv-1/2(SR) with φ≠0. Furthermore,
(2.26)|〈Ge~φ,φ×x^〉SR|≥C∥φ∥Hdiv-1/2(SR)2,∀φ∈Hdiv-1/2(SR),Ge+ikGe~:Hdiv-1/2(div,SR)→Hdiv-1/2(SR)
is compact, where
(2.27)Hdiv-1/2(div,SR)={φ=∑n=1∞‍∑m=-nn‍bn,mVnm∣∑n=1∞‍∑m=-nn‍11+n(n+1)|bn,m|2<∞}.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
(2.28)X={u∈Hloc(curl,BR),uT|Γ∈Lt2(Γ)},
where D⊂BR. Then multiplying (2.17) and (2.18) by test function ϕ∈Hloc(curl,BR), 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
(2.29)∫D‍(curlw·curlϕ--kD2w·ϕ-)dx+∫DR‍(curlw·curlϕ--kb2w·ϕ-)dx-ik∫Γ‍λwT·ϕ-Tds+ik∫SR‍Ge(ν×w)·ϕ-Tds=ik∫SR‍[Ge(ν×Ei)-ν×Hi]·ϕ-Tds,
where Hi=(1/ikb)curlEi is the incident magnetic field and
(2.30)w={V,x∈D,E=Ei+Es,x∈ℝ3∖D-.We rewrite (2.29) as the problem of finding w∈X such that
(2.31)A(w,ϕ)=B(ϕ),
where the sesquilinear form A:X×X→C is defined by
(2.32)A(w,ϕ)=(curlw,curlϕ)D-(kD2w,ϕ)D+(curlw,curlϕ)DR-(kb2w,ϕ)DR-ik〈λwT,ϕT〉Γ+ik〈Ge(ν×w),ϕT〉SR,B(ϕ)=ik〈Ge(ν×Ei)-ν×Hi,ϕT〉SR.
Here (·,·)D denotes the (L2(D))3 scalar product, and 〈·,·〉∂D denotes the (L2(∂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(2.33)S={p∈H1(D)∩H1(DR)∣p=onΓ},
then we seek p∈S such that
(2.34)A(∇p,∇ξ)=B(∇ξ),∀ξ∈S.
The variational problem (2.34) can be rewritten as
(2.35)A1(p,ξ)+A2(p,ξ)=B(∇ξ),∀ξ∈S,
where we define
(2.36)A1(p,ξ)=-(kD2∇p,∇ξ)D-(kb2∇p,∇ξ)DR+k2〈Ge~(ν×∇p),∇τξ〉SR,A2(p,ξ)=ik〈Ge+ikGe~(ν×∇p),∇τξ〉SR.
Here we have used ∇τξ to write the tangential component of the gradient of ξ in terms of the tangential gradient on the sphere SR. By Lemma 2.2, it follows that Ge~ is negative definite, then we obtain that A1(p,ξ) is a coercive sesquilinear form on S×S. Further by Lax-Milgram theorem, it is easy to see that A1(p,ξ) gives rise to a bijective operator. Since ν×∇p∈Hdiv-1/2(div,SR), still by Lemma 2.2, we know that A2(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.Lemma 2.3.
The variational problem (2.34) has at most one solution.Proof.
It suffices to consider the following equation:(2.37)-(kD2∇p,∇ξ)D-(kb2∇p,∇ξ)DR+ik〈Ge(ν×∇p),∇τξ〉SR=0,∀ξ∈S.
Choosing ξ=p, it is easy to see that
(2.38)ik〈Ge(ν×∇p),∇τp〉SR=(kD2∇p,∇p)D+(kb2∇p,∇p)DR.
By the definition of the operator Ge, if Es∈Hloc(curl,ℝ3∖B-) is the weak solution of the problem
(2.39)curlcurlEs-k2Es=0,inℝ3∖B-,ν×Es=ν×∇p,onSR,limr→∞(curlEs×x-ikrEs)=0,
then we have
(2.40)Ge(ν×∇p)=ν×Hs,onSR,
where
(2.41)Hs=1ikcurlEs.
Furthermore, we can compute that
(2.42)∫SR‍Hs·ν×Es¯ds=-〈ν×Hs,∇τp〉SR=-〈Ge(ν×∇p),∇τp〉SR=-1ik(kD2∇p,∇p)D-1ik(kb2∇p,∇p)DR,
which together with the fact ImnD≥0,Imnb≥0 implies
(2.43)Re∫SR‍ν×Es·Hs¯ds=Re∫SR‍Hs·ν×Es¯ds=-1k(ImkD2∇p,∇p)D-1k(Imkb2∇p,∇p)DR≤0.
Therefore the Rellich lemma ensures us that Es=0 in ℝ3∖B-. From (2.39), we see that ∇τp=0 on SR and then (kD2∇p,∇p)D+(kb2∇p,∇p)DR=0 which, together with the fact that p|Γ=0, implies p=0. This completes the proof of Lemma 2.3.Lemma2.3 together with the Fredholm alternative implies that there exits a unique solution p0∈S of the variational problem (2.34).Lemma 2.4.
The space(2.44)X0={u∈X∣A(u,∇ξ)=0,∀ξ∈S}
is compactly imbedded in (L2(B))3, where B is a ball with D⊂B⊂BR.Proof.
Consider a bounded set of functions{uj}j=1∞⊂X0. Each function uj∈X0 can be extended to all of ℝ3 by solving the exterior Maxwell equation
(2.45)∇×(∇×vj)-kb2vj=0,inℝ3∖B-,ν×vj=ν×uj,onSR,lim|x|→∞|x|((∇×vj)×ν-ikvj)=0.
Define
(2.46)uje={uj,ifx∈B,vj,ifx∈ℝ3∖B-.
Since the tangential components of uje are continuous across SR, it follows that uje∈Hloc(curl,ℝ3). By using the properties of the Calderon operator Ge and the conditions in X0, we see that the following equations hold true
(2.47)div(kD2u)=0,inD,div(kb2u)=0,inDR,kb2ν·u=-ik∇τ·Ge(ν×u),onSR.
Then, by the definition of Ge that Ge(ν×uj)=(1/ik)ν×curlvj and the relationship ∇τ·(ν×vj) = -ν·curlvj on SR, we immediately have
(2.48)kb2ν·uj=-ik∇τ·Ge(ν×uj)=-∇τ·(ν×curlvj)=ν·curlcurlvj=kb2ν·vj,onSR.
Thus, uje has a well-defined divergence and ∇·(k^2uje)=0 in ℝ3∖Γ, where
(2.49)k^2={kD2,x∈D,kb2,x∈ℝ3∖D-.
Now we choose a cut-off function χ∈C0∞(ℝ3) such that χ=1 in B- and χ is supported in a ball BR1⊃B-. Then one can use the general compactness theorem (Theorem 4.7 in [27]) to the sequence {χuje} and extract a subsequence converging strongly in (L2(B))3. This proves the lemma.From the above definitions ofS and X0, we have the following Helmholtz decomposition lemma.Lemma 2.5.
The spaces∇S and X0 are closed subspaces of X. The space X is the direct sum of the spaces ∇S and X0, that is,
(2.50)X=X0⊕∇S.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=w0+∇p0, where w0∈X0 and p0∈S is the unique solution of (2.34). We observe that A(w0,∇ξ)=0 for all ξ∈S by the definition of X0. Hence the problem of determining w∈X is equivalent to the problem of determining w0∈X0 such that
(2.51)A(w0,ϕ0)=B(ϕ0)-A(∇p0,ϕ0),∀ϕ0∈X0.
From Chapter 10.3.2 in [27] we know that for φ∈Hdiv-1/2(SR)(2.52)Geφ=G1eφ+G2eφ,
where the operator G1e is a compact operator from X0 into Hdiv-1/2(SR) and the operator G2e satisfies ik〈G2e(ν×φ),φT〉SR≥0. We now split the sesquilinear form A(·,·) into A=a+b with
(2.53)a(w0,ϕ0)=(curlw0,curlϕ0)D+(w0,ϕ0)D+(curlw0,curlϕ0)DR+(w0,ϕ0)DR-ik〈λ(w0)T,(ϕ0)T〉Γ+ik〈G2e(ν×w0),(ϕ0)T〉SR,(2.54)b(w0,ϕ0)=-((kD2+1)w0,ϕ0)D-((kb2+1)w0,ϕ0)DR+ik〈G1e(ν×w0),(ϕ0)T〉SR.
The sesquilinear form a(·,·) is obviously bounded and a direct computation verifies that
(2.55)|a(w0,w0)|⩾α∥w0∥X2,∀w0∈X0,
with some constant α>0.Hence by Lax-Milgram theorem,a(·,·) gives rise to a bijective operator and by the compact embedding of X0 in (L2(B))3 and the fact that G1e is a compact operator from X0 into Hdiv-1/2(SR), 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 fieldEi, there exists a unique solution w∈X of (CBP) which depends continuously on the incident field Ei.


## 3. Uniqueness for the Inverse Problem

In this section we will show that the scattering obstacleD and the corresponding parameter λ are uniquely determined from the knowledge of the scattered fields E1,x0s(x,x0)∣x∈∂Ω for all x0∈SR0, where SR0 is the surface of a large ball BR0 with D-⊂Ω-⊂BR0. 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.Lemma 3.1.
Assume thatkb2 is not an eigenvalue of Maxwell equation for the domain Ω0. Then we have(i)
the restriction to∂Ω0 of {ν×p(x0)G(x,x0):x0∈SR0} is complete in Hdiv-1/2(∂Ω0);(ii)
the restriction to∂Ω0 of {ν×curlxp(x0)G(x,x0)-ikλp(x0)G(x,x0)T:x0∈SR0} is complete in Lt2(∂Ω0).Proof.
For simplicity, we only prove statement (ii). Case (i) can be proved similarly.
Letg∈Lt2(∂Ω0) be such that
(3.1)∫∂Ω0‍[ν(x)×curlxp(x0)G(x,x0)-ikλp(x0)G(x,x0)T]·g-(x)ds(x)=0.
Then it follows that
(3.2)∫∂Ω0‍[curlx0(g-(x)×ν(x))G(x,x0)-ikλg-(x)G(x,x0)]ds(x)·p(x0)=0.
Define
(3.3)F1*(x0)=∫∂Ω0‍curlx0(g-(x)×ν(x))G(x,x0)ds(x),(3.4)F2*(x0)=-∫∂Ω0‍ikλg-(x)G(x,x0)ds(x),(3.5)F*(x0)=F1*(x0)+F2*(x0).
By (3.2), it is easy to see that for arbitrary polarization p(x0) in the tangential plane to SR0 at x0, we have
(3.6)F*(x0)·p(x0)=0,∀x0∈SR0.
From the definition of (3.3), we immediately have
(3.7)curlx0curlx0F1*(x0)-kb2F1*(x0)=0,∀x0∈ℝ3∖∂Ω0.
Due to the symmetry of the background Green function, F2*(x0) as a function of x0 solves curlx0curlx0F2*(x0)-kb2F2*(x0)=0,∀x0∈ℝ3∖∂Ω0. Hence, F*(x0) satisfies the Maxwell's equation in ℝ3∖∂Ω0. By (3.6) and the fact that p(x0) is an arbitrary polarization in the tangential plane to SR0 at x0, we immediately have that ν×F*(x0)|SR0=0.
The uniqueness of the exterior problem(3.8)curlx0curlx0F*(x0)-kb2F*(x0)=0,inℝ3∖B-R0,ν×F*(x0)=0,onSR0,lim|x0|→∞|x0|(curlx0F*(x0)×ν(x0)-ikF*(x0))=0
implies that F*(x0)=0 in ℝ3∖B-R0. Thus, the unique continuation principle ensures us that F*(x0)=0 in Ω0e=ℝ3∖Ω-0. By trace theorem, it follows that ν×F*(x0)=0 and ν×curlx0F*(x0)=0 on ∂Ω0. By the definition of F*(x0) and the jump relations of the vector potential across ∂Ω0, it can be checked that F*(x0) satisfies the following equations:
(3.9)curlx0curlx0F*(x0)-kb2F*(x0)=0,inΩ0,ν×curlx0F*(x0)-ikλν×F*(x0)=0,on∂Ω0.
Therefore, the uniqueness theorem of the interior problem for Maxwell's equations implies that F*(x0)=0 in Ω0. Finally, from the jump relations of the vector potential across ∂Ω0, we have
(3.10)0=[ν×F*(x0)×ν|∂Ω0=-g-(x0),
which completes the proof.We now consider two obstaclesD1 and D2 with the refractive index nDj and the surface impedance λj,j=1,2. Let U denote the unbounded part of ℝ3∖(D-1∪D-2) and D0=ℝ3∖U- its open complement. From the proof of Theorem 2.6, it follows that the total field wj,j=1,2 satisfies
(3.11)(curlwj,curlϕ)BR-(kj*2wj,ϕ)BR-ik〈λjwjT,ϕT〉Γj=-〈ν×curlwj,ϕ〉SR,
for any large ball BR with D⊂BR and all test function ϕ∈X~, where
(3.12)kj*2={kDj2,x∈Dj;kb2,x∈ℝ3∖D-j.
It is convenient to introduce the following space:
(3.13)Xj={E∈H(curl,BR),ET|Γj∈Lt2(Γj)},j=1,2,
where Dj⊂BR. 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 thatkb2 is not an eigenvalue of Maxwell equation in Ω. Let BR⊂ℝ3 be a ball with D-0⊂Ω-⊂BR. Let E1,x0s(x,x0) and E2,x0s(x,x0) be the scattered fields with respect to D1 and D2, respectively, produced by the same incident field pG(·,x0). Assume that ν×E1,x0s|∂Ω=ν×E2,x0s|∂Ω for all x0∈SR0 with the radius R0>R for a fixed wave number k. Then we have
(3.14)k2((nD1-nb)V1,V-2)D1+ik〈λ1(V1)T,(V-2)T〉Γ1=k2((nD2-nb)V1,V-2)D2+ik〈λ2(V1)T,(V-2)T〉Γ2.
Here Vj∈X1∩X2 satisfies the following variational problem:
(3.15)(ajcurlVj,curlV)BR-(κjVj,V)BR-ik〈λ~j(Vj)T,VT〉Γj=0,
for all V∈H0(curl,BR), where the coefficients aj,κj∈L∞(BR) and λ~j∈L∞(Γj) satisfy that aj|D0=1,κj|D0=(kj*2)|D0 and λ~j|Γj∩D0=λj|Γj,j=1,2.Proof.
(i) We first prove that for any fixedz∈U, the scattered fields E1,zs(x,z)=E2,zs(x,z), where (Vj,z,Ej,zs),j=1,2 is the solution of the following problem:
(3.16)curlcurlV-kDj2V=0,inDj,curlcurlEs-kb2Es=0,inℝ3∖D-j,ν×(V-Es)=f,onΓj,ν×curl(V-Es)=-ikλjETs+r+h,onΓj,
with the incident field Ei=pG(·,z) and f=ν×Ei, h=ν×curlEi, r=-ikλjETi. By Lemma 3.1 and the fact that kb2 is not an eigenvalue of Maxwell equation in Ω, it follows that there exists a sequence an∈ℝ and x0(n)∈SR0 such that
(3.17)∥ν×pG(·,z)-∑n‍anν×pG(·,x0(n))∥Hdiv-1/2(∂Ω)<ε,∀ε>0.
Let E(x)=pG(·,z)-∑n‍anpG(·,x0(n)), then it satisfies the Maxwell equation curlcurlE-kb2E=0 in Ω∖{z}. Let f=(ν×E)|∂Ω, then the well posedness of the problem
(3.18)curlcurlE-kb2E=0,inΩ∖{z},ν×E=f,on∂Ω
and (3.17) imply that
(3.19)∥E∥H(curl,Ω)≤C1∥ν×E∥Hdiv-1/2(∂Ω)<C1ε,∀ε>0.
This, together with the fact that divE=0, implies (see [28])
(3.20)∥E∥H1(Ω)≤C2(∥E∥H(curl,Ω)+∥E∥H(div,Ω)+∥ν×E∥Hdiv-1/2(∂Ω))≤C2ε,∀ε>0.
Then by (3.19), (3.20), and the trace theorem, it can be proved that
(3.21)∥ν×E∥Hdiv-1/2(Γj)+∥ν×curlE∥Lt2(Γj)+∥ν×curlE-ikλjET∥Lt2(Γj)≤C3(∥E∥H(curl,Dj)+∥ET∥Lt2(Γj))≤C4(∥E∥H(curl,Dj)+∥E∥H1(Dj))≤C5(∥E∥H(curl,Ω)+∥E∥H1(Ω))≤C5ε,j=1,2,∀ε>0.
Denote by E~1s and E~2s the scattered fields with respect to D1 and D2 produced by the same incident field ∑n‍anpG(·,x0(n)). By the assumption ν×E1,x0s|∂Ω=ν×E2,x0s|∂Ω for all x0∈SR0, it is easy to see that ν×E~1s|∂Ω=ν×E~2s|∂Ω. Then by the uniqueness theorem of the exterior scattering problem, it follows that E~1s=E~2s in ℝ3∖Ω-, which together with the unique continuation principle ensures that E~1s=E~2s 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)∥E1,zs(x,z)-E~1s(x)∥H(curl,K)<ε,∀ε>0,∥E1,zs(x,z)-E~2s(x)∥H(curl,K)<ε,∀ε>0,
for any fixed z∈K.
Therefore, the factE~1s=E~2s in U ensures us that
(3.23)∥E1,zs(x,z)-E2,zs(x,z)∥H(curl,K)<ε,∀ε>0.
The arbitrarity of ε implies that E1,zs(x,z)=E2,zs(x,z) for any fixed z∈K.
(ii) Next we will show that the identity (3.14) holds. Set w*=w2-w1, then it follows from (3.11) that
(3.24)(curlw*,curlV)BR-(k2*2w*,V)BR-ik〈λ2wT*,VT〉Γ2=-((k1*2-k2*2)w1,V)BR-ik〈λ1(w1)T,VT〉Γ1+ik〈λ2(w1)T,VT〉Γ2
for all V∈X*={V∈X1∩X2,VT|SR=0}. Choose two domains Ω1,Ω2⊂BR with D-0⊂Ω1⊂Ω2 and define a smooth function ψ∈C∞(ℝ3) with ψ=1 in Ω1 and ψ=0 in ℝ3∖Ω2. Let a2=ψ, κ2=ψk2*2, λ~2=λ2, it is easy to see that a2,κ2,λ~2 satisfy the assumptions of the lemma. We further assume V2 satisfies (3.15) with respect to a2=ψ, κ2=ψk2*2, λ~2=λ2, so that substituting V=ψV2¯ into the left hand of (3.24) and noting that w*=0 in U yield that
(3.25)(curlw*,curl(ψV2¯))BR-(k2*2w*,ψV2¯)BR-ik〈λ2wT*,ψ(V2)T¯〉Γ2=-(a2curlV2,curlw-*)BR-(κ2V2,w-*)BR-ik〈λ~2V2T,w-T*〉Γ2=0.
Hence substituting V=ψV2¯ into (3.24), it follows from the right hand of (3.24) that
(3.26)-((k1*2-k2*2)w1,V-2)BR-ik〈λ1(w1)T,(V-2)T〉Γ1+ik〈λ2(w1)T,(V-2)T〉Γ2=0.
We define f∈X1 by
(3.27)(f,ϕ)X1=-((k1*2-k2*2)V2,ϕ)BR-ik〈λ1(V2)T,ϕT〉Γ1+ik〈λ2(V2)T,ϕT〉Γ2,
for all ϕ∈X1. By Theorem 2.6, it follows that there exists a unique solution w0∈X1 of the problem
(3.28)(curlw0,curlϕ)BR-(k1*2w0,ϕ)BR-ik〈λ1(w0)T,ϕT〉Γ1+ik〈Ge(ν×w0),ϕT〉SR=(f,ϕ)X1,
for all ϕ∈X1. Choose two domains Ωe,Ωi⊆BR with Ω-⊆Ωe,D-0⊂Ωi⊂Ω and define smooth functions ψe,ψi∈C∞(ℝ3) with ψe=1,x∈Ω,ψe=0,x∈ℝ3∖Ωe and ψi=1,x∈Ωi,ψi=0,x∈ℝ3∖Ω. Take j=1,ϕ=ψew0¯ in (3.11), it is seen that
(3.29)(curlw1,curl(ψew0¯))BR-(k1*2w1,ψew0¯)BR-ik〈λ1(w1)T,(w-0)T〉Γ1=0.
Equation (3.28) with ϕ replaced by ψiw1¯ yields
(3.30)(curlw0,curl(ψiw1¯))BR-(k1*2w0,ψiw1¯)BR-ik〈λ1(w0)T,(w-1)T〉Γ1=(f,ψiw1¯)X1.
By (3.26) and (3.27), it can be shown that
(3.31)(f,ψiw1¯)X1=-((k1*2-k2*2)V2,w-1)BR-ik〈λ1(V2)T,(w-1)T〉Γ1+ik〈λ2(V2)T,(w-1)T〉Γ2=0.
Taking the difference of (3.29) and (3.30), we have that
(3.32)0=∫Ω∖Ωi‍curlw0·curl(ψiw1-w1)-kb2w0·(ψiw1-w1)dx-∫Ωe∖Ω‍curl(ψew0)·curlw1-kb2(ψew0)·w1dx.
By (3.28), we can deduce that w0 is a radiating solution of the corresponding Maxwell's equations in BR, then it can be extended to all of ℝ3 denoted by w0e by solving the exterior Maxwell's equation in ℝ3∖B-R with ν×w0e=ν×w0 on SR, which also satisfies the Silver-Müler radiation condition at infinity. By applying the vector Green formula to (3.32), it can be proved that
(3.33)0=∫∂Ω‍[w1·(ν×curlw0e)-w0e·(ν×curlw1)]ds.
In view of the fact w1=E1=E1s+Ei and Ei=pG(·,x0) in ℝ3∖D-1, we immediately have
(3.34)0=∫∂Ω‍E1s·(ν×curlw0e)-w0e·(ν×curlE1s)ds+∫∂Ω‍[(ν×curlw0e)·pG(·,x0)-w0e·(ν×curl(pG(·,x0)))]ds.
Application of the vector Green formula again and noting that both E1s and the extended function w0e satisfy the Silver-Müler radiation condition, it follows that
(3.35)∫∂Ω‍E1s·(ν×curlw0e)-w0e·(ν×curlE1s)ds=∫SR‍E1s·(ν×curlw0e)-w0e·(ν×curlE1s)ds=0.
Hence the Stratton-Chu formula combines with (3.34) implies that
(3.36)0=p(x0)·w0e(x0),∀x0∈SR0,∀p(x0)∈ℝ3.
Since p(x0) is an arbitrary polarization in the tangential plane to SR0 at x0, we obtain that ν×w0e(x0)|SR0=0. By the fact that w0e is a radiating solution of Maxwell's equation in ℝ3∖B-R0, it follows that w0e=0 in ℝ3∖B-R0. Hence the unique continuation principle implies that w0e=0 in ℝ3∖U-. Therefore, w0e can be used as a test function for V1, which satisfies (3.15) with a1=ψi,κ1=ψik1*2,λ~1=λ1. So that from the left hand of (3.30), we deduce that
(3.37)(curlw0e,curl(ψiV1¯))BR-(k1*2w0e,ψiV1¯)BR-ik〈λ1(w0e)T,(V-1)T〉Γ1=(a1curlV1,curlw-0e)BR-(κ1V1,w-0e)BR-ik〈λ~1(V1)T,(w-0e)T〉Γ1=0.
Thus, it follows from the right hand of (3.30) that (f,ψiV1¯)X1=0. Furthermore, from (3.27) with ϕ replaced by ψiV1¯, it can be shown that
(3.38)-((k1*2-k2*2)V2,V1)BR-ik〈〈λ1(V1)T,(V2)T〉Γ1+ik〈λ2(V2)T,(V1)T〉Γ2=0.
From the definitions of kj*2, we observe that
(3.39)k1*2-k2*2={k2nD1-k2nb,x∈D1,k2nD2-k2nb,x∈D2,0,x∈BR∖(D-1∪D-2),
which combines (3.38), the definition of the scalar product (·,·)D, and the fact that ((k1*2-k2*2)V2,V1)BR=k2((nD1-nb)V2,V1)D1-k2((nD2-nb)V2,V1)D2 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.
LetE1s and E2s be the scattered fields with respect to D1 and D2, respectively, and λ1,λ2 the corresponding impedances. Suppose that the assumptions in Lemma 3.2 hold true and Γj∩(Di∖Dj) is not empty for i,j=1,2,i≠j. If one of the following assumptions holds, then we have D1=D2. Consider(i)
Reλj≥δ>0;(ii)
Imλj≥δ>0orImλj≤-δ<0.Proof.
Let us assume thatD-1 is not included in D-2. Since D2e=ℝ3∖D-2 is connected, we can find a point z∈Γ1∖D2 and a sufficiently small ε>0 with the following properties:(i)
B2ε(z)∩D-2=∅;(ii)
the pointszn=z+(ε/n)ν(z) lie in B2ε(z) for all n∈N, where ν(z) is the unit normal to Γ1 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)curlcurlV-kD2V=0,inD,curlcurlEs-kb2Es=0,inℝ3∖D-,ν×(Es-V)=-ν×Em(·,zn),on∂D,ν×curl(Es-V)=ikλ1(ETs+Em(·,zn))-ν×curlEm(·,zn),onΓ1∩∂D.
Here Es satisfies the Silver-Müler radiation condition at infinity, and Em denotes the magnetic dipole defined by
(3.41)Em(x,zn)=curlx(ν(z)eikb|x-zn||x-zn|).
Define
(3.42)F1(x,zn)={V(x),ifx∈D,Es(x)+Em(x,zn),ifx∈ℝ3∖(D-∪zn).
It can be proved that F1(x,zn) is a solution of Maxwell's equations with homogeneous conductive boundary value conditions on ∂D in any domain Ω⊂ℝ3 with D-0⊂Ω and zn⊈Ω.
Define(3.43)κ1(x)={k1*2(x)ifx∈ℝ3∖(D1∖D¯),kb2ifx∈D1∖D,
and λ~1(x)=λ1.
In view of the above definitions ofκ1(x) and λ~1(x), it follows that F1(x,zn) 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 zn is bounded away from ∂D imply that the solution (V,Es) of (3.40) is uniformly bounded in X~. We now define another singular solution with respect to D2 by
(3.44)F2(x,zn)={V~(x),ifx∈D2,E~s(x)+Em(x,zn),ifx∈ℝ3∖(D-2∪zn),
where Em is a magnetic dipole defined in (3.41), and (V~,E~s) is a solution of the problem
(3.45)curlcurlV~-kD22V~=0,inD2,curlcurlE~s-kb2E~s=0,inℝ3∖(D-2∪zn),ν×(E~s-V~)=-ν×Em(·,zn),on∂D,ν×curl(E~s-V~)=ikλ2(E~Ts+(Em(·,zn))T)-ν×curlEm(·,zn),onΓ2.
Here E~s satisfies the Silver-Müler radiation condition at infinity. Noting that F2(x,zn) satisfies the variational equation (3.15) in Lemma 3.2 with κ2(x)=k2*2 and λ~2=λ2, it follows that both F1(x,zn) and F2(x,zn) satisfy the relationship (3.14), then we obtain
(3.46)k2((nD-nb)F1,F-2)D+ik〈λ1(F1)T,(F-2)T〉∂D=k2((nD2-nb)F1,F-2)D2+ik〈λ2(F2)T,(F-1)T〉Γ2.
For case (i), by the fact that z∈Γ1∖D2 and the singularities of the magnetic dipole Em defined in (3.41), it can be proved that |k2((nD-nb)F1,F-2)D+ik〈λ1(F1)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 D1⊂D2. By choosing z∈Γ2∖D1 and using the similar analysis as in the proof above, one can prove that D2⊂D1. Finally, we obtain that D1=D2. For other cases, due to the singularities of Fj(j=1,2), a contradiction also arises in (3.46) as n→∞. This proves the theorem.Theorem 3.4.
AssumeD1=D2 with parameters λj∈C(∂D-j) and the scattered fields Ej,x0s,(j=1,2) satisfy ν×E1,x0s|∂Ω=ν×E2,x0s|∂Ω for all x0∈SR0, then we have λ1=λ2 on ∂D.Proof.
From the proof of Theorem3.3, it follows that there exists two singular solutions F1,F2 of the conductive boundary problem with respect to the obstacle D for some z∈∂D. By Lemma 3.2 and the identity D1=D2, it can be checked that
(3.47)∫D‍(k1*2-k2*2)F1(x,zn)·F2(x,zn)dx+ik∫∂D‍(λ1-λ2)F1(x,zn)·F2(x,zn)ds(x)=0.
The singularities of F1,F2 ensure that λ1=λ2. This completes the proof of the theorem.



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*Source: 306272-2012-12-31.xml*