This content has been downloaded from IOPscience. Please scroll down to see the full text.

Download details:

IP Address: 129.13.72.197

This content was downloaded on 07/05/2014 at 07:30

Please note that terms and conditions apply.

The factorization method for inverse obstacle scattering with conductive boundary condition

View the table of contents for this issue, or go to the journal homepage for more

2013 Inverse Problems 29 095021

(http://iopscience.iop.org/0266-5611/29/9/095021)

Home Search Collections Journals About Contact us My IOPscience

IOP PUBLISHING INVERSE PROBLEMS

Inverse Problems 29 (2013) 095021 (25pp) doi:10.1088/0266-5611/29/9/095021

The factorization method for inverse obstaclescattering with conductive boundary condition

Oleksandr Bondarenko1 and Xiaodong Liu2,3

1 Department of Mathematics, Karlsruhe Institute of Technology, D-76128 Karlsruhe, Germany2 Institute of Applied Mathematics, Chinese Academy of Sciences, 100190 Beijing,People’s Republic of China

E-mail: bondarenko@kit.edu and xdliu@amt.ac.cn

Received 18 June 2013, in final form 9 August 2013Published 3 September 2013Online at stacks.iop.org/IP/29/095021

AbstractThe inverse acoustic scattering by a penetrable obstacle with a generalconductive boundary condition is considered. Having established the wellposedness of the direct problem by a variational method, we study thefactorization method for recovering the location and the shape of the obstacle.One by-product of the method is an explicit proof of uniqueness of the inversescattering problem under certain assumptions. Some numerical experimentsare also presented to demonstrate the feasibility and effectiveness of thefactorization method.

(Some figures may appear in colour only in the online journal)

1. Introduction

In this paper, we are concerned with an inverse scattering problem which is derived as theTM-mode from the time-harmonic Maxwell system, where the scattering medium is coveredby a thin layer with very high conductivity. This model leads to the so-called conductivetransmission conditions, which can be regarded as a generalization of the well-knowntransmission boundary condition. The case of an impenetrable obstacle covered by a thinlayer leads to an impedance boundary condition. Recently, in [3], there has been publisheda result on inverse scattering for a generalized impedance boundary condition, where thesurface impedance involves a second-order surface operator, which provides a better modelfor complicated surface materials. The conductive boundary condition has been known for along time in the study of electromagnetic induction in the earth ( [30, 31] provide the physicalexplanation of the condition). We refer to [1] for the derivation of the conductive boundarycondition for the full Maxwell system, where an inhomogeneity is covered by an infinitelythin (the electric field would not penetrate into an ideal conductor of positive thickness) highlyconducting layer. The attention to such inverse problems is stipulated among others by the

3 Author to whom any correspondence should be addressed.

0266-5611/13/095021+25$33.00 © 2013 IOP Publishing Ltd Printed in the UK & the USA 1

Inverse Problems 29 (2013) 095021 O Bondarenko and X Liu

recent interest in using inflatable objects, covered, for example, by a thin aluminum sheet as adecoy. For the inhomogeneity represented by a constant refractive index, the direct and inversescattering problems have been studied in [2, 14, 15]. In this work, we assume the refractiveindex to be an L∞(D) function, where D represents the inhomogeneity.

For the TM-mode and an obstacle being an infinite cylinder, we arrive at the scalarHelmholtz equation with corresponding boundary conditions in R

2. For convenience, we willtreat the problem in R

3. However, all of the results hold also for the two-dimensional casewith possibly different constants, because we do not use any particular feature of R

3.Let ui = eikx·d be the incident plane wave which is described by the positive wave number

k = ω/c with frequency ω, sound speed c and incident direction d ∈ S2. Here and in thefollowing, S2 := {x ∈ R

3 : |x| = 1} denotes the unit sphere in R3. Then, the scattering

problem for an inhomogeneous medium with a thin layer of high conductivity is to find thetotal field u such that

�u + k2u = 0 in R3\D, (1.1)

�u + k2(1 + q)u = 0 in D, (1.2)

and the conductive boundary condition

u+ − u− = 0,∂u+∂ν

− λ∂u−∂ν

+ μu+ = 0 on ∂D, (1.3)

where ν is the unit outward normal to the boundary ∂D. Here, u± and ∂u±/∂ν denote thelimit of u and ∂u/∂ν from the exterior (+) and interior (−), respectively. The assumptions onD, q (or n), λ and μ will be given in the following sections. Furthermore, the scattered fieldus = u − ui satisfies the Sommerfeld radiation condition

limr→∞ r

(∂us

∂r− ikus

)= 0, r = |x|, (1.4)

which holds uniformly with respect to x = x/|x| ∈ S2.Roughly speaking, the Sommerfeld radiation condition (1.4) imposes that us behave as a

spherical wave propagating away from the obstacle D. More precisely, it is known (see [11])that us has the asymptotic behavior of an outgoing spherical wave

u(x) = eik|x|

4π |x|{

u∞(x) + O

(1

|x|)}

as |x| → ∞ (1.5)

uniformly in all directions x := x/|x|, where the function u∞(x) defined on the unit sphere S2

is known as the far-field pattern with x denoting the observation direction.Let u∞(x, d) be the far-field pattern corresponding to the observation direction x and the

incident direction d. The inverse problem we consider is to determine D from knowledge ofthe far-field patterns u∞(x, d) for all x, d ∈ S2.

In this paper, we will study the applicability of the factorization method (FM) for theproblem above. The FM has been introduced by Kirsch in [16] for scattering by impenetrablesound-soft or sound-hard obstacles. The essential idea of the FM is to decide for a givensampling point z ∈ R

3 whether or not the equation

Fg = e−ikz·x, x ∈ S2,

is solvable in L2(S2), which in turn is equivalent to whether or not the given point z belongsto D. Here, F is a self-adjoint operator which can be computed from the far-field operatorF : L2(S2) → L2(S2) given by

(Fg)(x) =∫

S2u∞(x, d)g(d) ds(d) for x ∈ S2. (1.6)

2

Inverse Problems 29 (2013) 095021 O Bondarenko and X Liu

The FM has been already employed for some special cases of (1.1)–(1.4). The reader canconsult the papers by Kirsch [17, 18] for λ = 1 and μ = 0, by Kirsch and Liu [24] for λ �= 1and μ = 0 and by Kirsch and Kleefeld [23] for q = 0, λ = 1 and μ = ic where c � 0. In thispaper, we will study the FM for the general case under assumption 2.1.

For the case of constant q ∈ C, it is known that the inverse problem admits at most onesolution [12, 13], i.e., the location and shape of the obstacle D can be uniquely recovered fromknowledge of the far-field pattern u∞(x, d) for all x, d ∈ S2. In fact, this also follows as acorollary of theorems 3.5 and 3.14 below, which thus provides a novel proof of this uniquenessresult.

This paper is organized as follows. In the next section, we will study the well posedness(existence, uniqueness and stability) of the direct problem by a variational method. Section 3is devoted to a study of the FM for recovering the shape and location of the penetrable obstacleD. Some numerical simulations in two dimensions will be presented in section 4 to justify thevalidity of our method.

2. The direct scattering problem

This section is devoted to the solution of the direct acoustic scattering problem (1.1)–(1.4).Let D denote a bounded Lipschitz domain and BR := {x ∈ R

3 : |x| < R}. Define the Sobolevspaces

H1(D) := {u : u ∈ L2(D), |∇u| ∈ L2(D)},H1

loc(R3\D) := {u : u ∈ H1(BR\D) for every R > 0, such that D ⊂ BR}.

Furthermore, H1/2(∂D) is the trace space of H1(D) and H−1/2(∂D) is the dual space ofH

12 (∂D).

Let n ∈ L∞(R3), n := 1 + q, denote the index of refraction. Here and throughout thispaper, we make the following general assumptions on D, q (or n), λ and μ.

Assumption 2.1.

(a) Let D ⊂ R3 be bounded domains, such that the exterior De := R

3\D of D is connected.Assume that the boundary ∂D is smooth enough, such that both the embedding of H1/2(∂D)

into H−1/2(∂D) and the embedding of H1(D) into L2(D) are compact.(b) The contrast function q ∈ L∞(D) satisfies �(1+q) > 0 and (q) � 0 almost everywhere

(a.e.) in D.(c) λ is a non-zero complex constant, such that

(1) there exists c > 0, such that �(λ) � c,(2) (λ) � 0,(3) (λn) � 0 a.e. in D.

(d) μ ∈ L∞(∂D) with �(μ) � 0 and (μ) � 0 a.e. on ∂D.We extend q by zero (or n by n = 1) outside of D.

Since the incident field ui satisfies the Helmholtz equation �ui +k2ui = 0 in all of R3, the

scattered field us solves the following problem with the boundary data f2 = (λ − 1) ∂ui

∂ν− μui

on ∂D and the source term f = k2qui in D:

�us + k2us = 0 in R3\D, (2.1)

�us + k2nus = − f in D, (2.2)

3

Inverse Problems 29 (2013) 095021 O Bondarenko and X Liu

us+ − us

− = 0,∂us

+∂ν

− λ∂us

−∂ν

+ μus+ = f2 on ∂D, (2.3)

limr→∞ r

(∂us

∂r− ikus

)= 0 uniformly in all directions x/r. (2.4)

For f ∈ L2(D) and f2 ∈ H−1/2(∂D), the solution us ∈ H1loc(R

3) of (2.1)–(2.3) has to beunderstood in the weak sense; that is,∫∫

R3\D[∇us · ∇ϕ − k2usϕ] dx + λ

∫∫D

[∇us · ∇ϕ − k2nusϕ] dx −∫

∂Dμusϕ ds

= λ

∫∫D

f ϕ dx −∫

∂Df2ϕ ds (2.5)

for any test function ϕ ∈ H1(R3) with compact support. A well-known regularity result forelliptic differential equations [28] yields that us is even analytic in De. In particular, radiationcondition (2.4) makes sense.

Theorem 2.2. For any f ∈ L2(D) and f2 ∈ H−1/2(∂D), there exists at most one solutionv ∈ H1

loc(R3) of (2.1)–(2.4), or, equivalently (2.5), (2.4) has a unique solution in H1

loc(R3).

Proof. Let v be the difference of two solutions. Then v solves (2.5), (2.4) with f2 = 0 on ∂Dand f = 0 in D. To prove the uniqueness, we show that v vanishes in all of R

3.Choose a ball BR centered at the origin big enough such that D ⊂ BR. Let φ ∈ C∞(R3),

such that φ(x) = 1 for |x| < R and φ(x) = 0 for |x| � R + 1. Then, setting ϕ = φv, we obtainfor (2.5)∫∫

R�x�R+1[∇v · ∇ϕ − k2vϕ] dx +

∫∫BR\D

[|∇v|2 − k2|v|2] dx

+ λ

∫∫D

[|∇v|2 − k2n|v|2] dx −∫

∂Dμ|v|2ds = 0.

From interior regularity results (2.4), v is analytic outside of BR. Applying Green’s first theoremto the first integral yields

−∫

|x|=R

∂v

∂νv ds +

∫∫BR\D

[|∇v|2 − k2|v|2] dx + λ

∫∫D

[|∇v|2 − k2n|v|2] dx −∫

∂Dμ|v|2 ds = 0.

From this and the assumptions on μ, λ and n given in assumption 2.1, it follows that

∫

∂BR

v∂v

∂νds � 0.

Theorem 2.12 in [11] now shows v = 0 in R3\BR. By analytic continuation v = 0 in De and

thus the trace of v also vanishes on ∂D. Thus, v ∈ H1(R3) is a weak solution of �v+k2nv = 0in R

3 and is identically zero outside some ball. The unique continuation principle (see e.g.,theorem 6.4 in [21]) implies that v vanishes in all of R

3. This completes the proof. �Next, using variational techniques, we prove existence for (2.1)–(2.4) and show that the

solution depends continuously on the data f and f2. We write the problem (2.1)–(2.4) as anequivalent problem in a bounded domain. For this, we introduce the Dirichlet-to-Neumannoperator (DtN)

: v → ∂v

∂νon ∂BR,

mapping v to ∂v/∂ν, where v solves the exterior Dirichlet problem for the Helmholtz equation�v + k2v = 0 in R

3\BR with the Dirichlet boundary data v|∂BR = v. Again here, BR is a ballof radius R, such that D ⊂ BR. We will need the following important property of the DtNoperator (see e.g. [11, pp. 116–7] for details).

4

Inverse Problems 29 (2013) 095021 O Bondarenko and X Liu

Lemma 2.3. The DtN operator is a bounded linear operator from H1/2(∂BR) to H−1/2(∂BR).Furthermore, there exists a bounded operator 0 : H1/2(∂BR) → H−1/2(∂BR) satisfying that

−∫

∂BR

0ww ds � c‖w‖2H1/2(∂BR )

for some constant c > 0, such that − 0 : H1/2(∂BR) → H−1/2(∂BR) is compact.

We now reformulate problem (2.1)–(2.3) as follows: given f ∈ L2(D) and f2 ∈H−1/2(∂D), find v ∈ H1(BR) satisfying (2.1)–(2.3) and the boundary condition

∂v

∂ν= v on ∂BR. (2.6)

Again, problem (2.1)–(2.3), (2.6) has to be understood in the weak sense; that is, v ∈ H1(BR)

solves∫∫BR\D

[∇v · ∇ϕ − k2vϕ] dx −∫

∂BR

vϕ ds + λ

∫∫D

[∇v · ∇ϕ − k2nvϕ] dx

−∫

∂Dμvϕ ds = λ

∫∫D

f ϕ dx −∫

∂Df2ϕ ds. (2.7)

In exactly the same way as in the proof of lemma 5.22 in [4], one can show that a solutionv to problem (2.1)–(2.3) and (2.6) can be extended to a solution to the scattering problem(2.1)–(2.4) and conversely for a solution v to the scattering problem (2.1)–(2.4), v, restrictedto BR, solves problem (2.1)–(2.3) and (2.6). Therefore, by theorem 2.2, the problem (2.1)–(2.3)and (2.6) has at most one solution. We now have the following result on the well posedness ofthe problem (2.1)–(2.3) and (2.6).

Theorem 2.4. Let f ∈ L2(D) and f2 ∈ H−1/2(∂D). Then problem (2.7) has a unique solutionv ∈ H1(BR). Furthermore,

‖v‖H1(BR ) � C(‖ f2‖H−1/2(∂D) + ‖ f ‖L2(D)) (2.8)

with a positive constant C independent of f and f2.

Proof. We write (2.7) as

a(v, ϕ) = b(ϕ) ∀ϕ ∈ H1(BR), (2.9)

with

a(v, ϕ) =∫∫

BR\D[∇v · ∇ϕ − k2vϕ] dx −

∫∂BR

vϕ ds

+ λ

∫∫D

[∇v · ∇ϕ − k2nvϕ] dx −∫

∂Dμvϕ ds,

and

b(ϕ) = λ

∫∫D

f ϕ dx −∫

∂Df2ϕ ds.

We write a = a1 + a2, where

a1(v, ϕ) =∫∫

BR\D[∇v · ∇ϕ + vϕ] dx −

∫∂BR

0vϕ ds + λ

∫∫D

[∇v · ∇ϕ + vϕ] dx

−∫

∂Dμvϕ ds

and

a2(v, ϕ) = −∫∫

BR\D[1 + k2]vϕ dx −

∫∂BR

( − 0)uϕ ds − λ

∫∫BR\D

[1 + k2n]vϕ dx,

5

Inverse Problems 29 (2013) 095021 O Bondarenko and X Liu

where 0 is the operator defined in lemma 2.3. By the boundedness of 0 and the tracetheorem, a1 is bounded. By the Riesz representation theorem, there exists a bounded linearoperator A1 : H1(BR) → H1(BR), such that

a1(v, ϕ) = (A1v, ϕ) for all ϕ ∈ H1(BR).

On the other hand, by assumption 2.1 and lemma 2.3, for all v ∈ H1(BR),

�[a1(v, v)] = ‖v‖2H1(BR\D)

−∫

∂BR

0vϕ ds + �(λ)‖v‖2H1(D)

−∫

∂D�(μ)vϕ ds

� ‖v‖2H1(BR\D)

+ c‖v‖2

H12 (∂BR )

+ c‖v‖2H1(D)

� min{1, c}‖v‖2H1(BR )

,

that is, a1 is strictly coercive. The Lax–Milgram theorem (see theorem 13.26 in [26]) impliesthat the operator A1 : H1(BR) → H1(BR) has a bounded inverse. By the Riesz representationtheorem again, there exists a bounded linear operator A2 : H1(BR) → H1(BR), such that

a2(v, ϕ) = (A2v, ϕ) for all ϕ ∈ H1(BR).

By the compactness of − 0 and Rellich’s embedding theorem (that is, that the embeddingof H1(BR) into L2(BR)) is compact, it follows that A2 is compact. By the Riesz representationtheorem again, there is a function v ∈ H1(BR), such that

b(ϕ) = (v, ϕ) for all ϕ ∈ H1(BR).

Thus, the variational formulation (2.9) is equivalent to the problem

find v ∈ H1(BR) such that A1v + A2v = v, (2.10)

where A1 is bounded and strictly coercive and A2 is compact. The Riesz–Fredholm theory andthe uniqueness result (theorem 2.2) imply that problem (2.10) or, equivalently, problem (2.9)has a unique solution. Estimate (2.8) follows from the fact that ‖v‖H1(BR ) = ‖b‖ is boundedby ‖ f ‖L2(D) + ‖ f2‖H−1/2(∂D). �

3. Factorization of the far-field operator

In this section, we turn to the inverse problem of recovering the shape of D from knowledgeof the far-field patterns u∞(x, d) for all x, d ∈ S2.

To prove the applicability of the FM, we proceed in following steps:

• derive a factorization of the far-field operator of the form F = GT G∗,• characterize D by test functions,• link the test functions and the data operator F by the FM.

The main difficulty is to show the middle operator T satisfies the assumptions requiredby theorem 2.15 ([22]). For rigorous justification of the FM, we had to distinguish betweentwo cases, λ = 1 and λ �= 1 and derived different factorizations of F for each of them.

We first prove the FM for the case λ = 1. As in the case of scattering by an inhomogeneousmedium [22], we assume that the contrast q is locally bounded below.

Assumption 3.1. |q| is locally bounded below; that is, for every compact subset M ⊂ D, thereexists c > 0 (depending on M), such that

|q(x)| � c for almost all x ∈ M.

6

Inverse Problems 29 (2013) 095021 O Bondarenko and X Liu

Furthermore, let � ⊂ D be relatively open, such that μ �= 0 on � and μ = 0 on ∂D\�.We rewrite problem (2.1)–(2.4) as follows. Let f ∈ L2(D) and f2 ∈ H−1/2(∂D) be given. Findu ∈ H1

loc(R3), such that

�u + k2(1 + q)u = −k2 q√|q| f in R3\∂D, (3.1)

∂u+∂ν

− ∂u−∂ν

+ μu = − f2 on �,∂u+∂ν

− ∂u−∂ν

= 0 on ∂D\�, (3.2)

u+ = u− on ∂D, (3.3)

limr→∞ r

(∂u

∂r− iku

)= 0 uniformly in all directions x/r. (3.4)

Define the data-to-pattern operator G : L2(D) × H−1/2(�) → L2(S2) by

G :

(ff2

)→ v∞,

where v∞ is the far-field pattern of the solution to (3.1)–(3.4), and the operator H : L2(S2) →L2(D)×H−1/2(�), Hg =

(H1gH2g

), where H1 : L2(S2) → L2(D) and H2 : L2(S2) → H−1/2(�)

are given by

(H1ψ)(x) =√

|q(x)|∫

S2ψ(θ ) eikx·θ ds(θ ), x ∈ D, (3.5)

and

(H2ϕ)(x) = μ(x)

∫S2

ϕ(θ ) eikx·θ ds(θ ), x ∈ �. (3.6)

Then, by the superposition principle, F = GH.

The adjoint of H, H∗ : L2(D) × H1/2(�) → L2(S2) is given by

H∗(

ϕ1

ϕ2

)(x) =

∫∫D

ϕ1(y)√

|q(y)| e−ikx·y dy +∫

�

ϕ2(y)μ(y) e−ikx·y ds(y), x ∈ S2.

From the asymptotic behavior of the fundamental solution, it follows that H∗(ϕ1, ϕ2)� is the

far field w∞ of the function w, which is the sum of the volume and the single-layer potentials

w(x) =∫∫

Dϕ1(y)

√|q(y)|�(x, y) dy +

∫�

ϕ2(y)μ(y)�(x, y) dy, x ∈ R3\∂D.

By properties of the volume [20] and the single-layer potentials [28], w ∈ H1loc(R

3) is radiatingand satisfies

�w + k2w = −ϕ1

√|q| in R

3\∂D, (3.7)

w+ = w− on ∂D, (3.8)

∂w+∂ν

− ∂w−∂ν

= −ϕ2μ on �,∂w+∂ν

− ∂w−∂ν

= 0 on ∂D\� (3.9)

in the weak sense.We also can write (3.7)–(3.9) as

�w + k2(1 + q)w = −k2 q√|q|(

q

k2|q|ϕ1 −√

|q|w)

in R3\∂D, (3.10)

w+ = w− on ∂D, (3.11)

7

Inverse Problems 29 (2013) 095021 O Bondarenko and X Liu

∂w+∂ν

− ∂w−∂ν

+ μw = −(ϕ2μ − μw) on �,∂w+∂ν

− ∂w−∂ν

= 0 on ∂D\�. (3.12)

Since, (3.7)–(3.9) is uniquely solvable, H∗(ϕ1, ϕ2)� = w∞ = G(

qϕ1

k2|q| − √|q|w, ϕ2μ − μw)�

for all (ϕ1, ϕ2)� ∈ L2(D) × H1/2(�). We define an operator T : L2(D) × H1/2(�) →

L2(D) × H−1/2(�) by

T

(ϕ1

ϕ2

)=

( qϕ1

k2|q| − √|q|wϕ2μ − μw

).

Then H∗ = GT or H = T ∗G∗. Thus, the far-field operator F can be represented as F = GT ∗G∗.The next theorem provides a link between D and the range of the data-to-pattern

operator G.

Theorem 3.2. For any z ∈ R3, define φz by

φz(x) := e−ikx·z, x ∈ S2. (3.13)

Then,

z ∈ D ⇐⇒ φz ∈ R(G). (3.14)

Proof. First we assume that z ∈ D. Let B[z, e] ⊂ D be some closed ball centered at z withradius ε. Choose a cut-off function ψ ∈ C∞(R) with ψ(t) = 1 for |t| > ε and ψ(t) = 0 for|t| � ε/2. Define v ∈ C∞(R3) by

v(x) := ψ(|x − z|) eik|x−z|

4π |x − z| , x ∈ R3.

The far-field pattern of v is given by v∞ = φz. Also, v solves (3.1)–(3.4) with f =−�v + k2(1 + q)v and f2 = −μv. Therefore, G( f , f2)

� = φz.Let now z /∈ D and assume on the contrary that there exists ( f , f2)

T ∈ H1/2(∂D) ×H−1/2(∂D), such that G( f , f2)

T = φz. Let u be the solution of (3.1)–(3.4) determined by fand f1, and u∞ = G( f1, f2)

T be its far-field pattern. Since, φz is the far-field pattern of �(·, z),by Rellich’s lemma and analytic continuation u(x) = �(x, z) for all x ∈ De\{z}. If z ∈ De,this contradicts the fact that u is analytic in De while �(·, z) is singular at x = z. If z ∈ ∂D,we have that u+ ∈ H1/2(∂D) by trace theorem. However, �(x, z) cannot be in H1/2(∂D). Weobserve that ∇�(x, z) = O(1/|x − z|2) as x → z, and thus �(·, z) is neither in H1(D) nor inH1

loc(De). The proof is finished. �

Let 〈·, ·〉 denote the dual form⟨(ψ1

ψ2

),

(ϕ1

ϕ2

)⟩= (ψ1, ϕ1)L2(D) +

∫�

ψ2ϕ2 ds

for all(

ψ1

ψ2

) ∈ L2(D) × H−1/2(�) and(

ϕ1

ϕ2

) ∈ L2(D) × H1/2(�); the integral∫�

ψ2ϕ2 ds has tobe understood as the dual form in the dual system 〈H−1/2(�), H1/2(�)〉.

In following, we collect properties of G and T in order to apply the functional analyticresult first stated by Kirsch [16] and further refined by Lechleiter [27]. This result will enableus to express the range of the unknown operator G by the range of the far-field operator F . Wemake the following additional assumption on the contrast q.

Assumption 3.3. There exist t ∈ [0, π ] and c0 > 0, such that

�[eitq(x)] � c0|q(x)|a.e. in D.

8

Inverse Problems 29 (2013) 095021 O Bondarenko and X Liu

For a real-valued q, this condition is satisfied (with c0 = 1 and t = 0 if q is positive ort = π if q is negative). If q is complex valued and �q � 0, we can set t = π/4. For the case�q � 0, lemma 4.11 in [22] provides a sufficient condition for existence of t ∈ [0, π ].

Theorem 3.4.

(a) G is compact with dense range.(b) Let t and c0 be such that the assumption 3.3 is satisfied. Then, the middle operator T

has a decomposition of the form T = C + K, where K is compact and the real part�[eitC] : L2(D) × H1/2(�) → L2(D) × H−1/2(�) is coercive, in particular,

�[

eit

⟨C

(ϕ1

ϕ2

),

(ϕ1

ϕ2

)⟩]� 1

2min

{c0

k2, 1

} ∥∥∥∥(ϕ1

ϕ2

)∥∥∥∥2

L2(D)×H1/2(�)

(3.15)

for all

(ϕ1

ϕ2

)∈ L2(D) × H1/2(�). (3.16)

(c) For T holds ⟨T

(ϕ1

ϕ2

),

(ϕ1

ϕ2

)⟩� 0 for all

(ϕ1

ϕ2

)∈ L2(D) × H1/2(�).

(d) T is one-to-one.

Proof.

(a) The proof for the compactness of G is analogous to the arguments provided in the proofof lemma 1.13 in [22] and is omitted for brevity.

We consider the L2−adjoint G∗ of G and show that it is injective, which implies thedenseness of the range of G. Recall that u is a solution of (3.1)–(3.4) with f ∈ L2(D) andf2 ∈ H−1/2(�). Let v be a solution of the boundary value problem defined in (3.20)–(3.23)with λ = 1 and boundary data (vg, ∂vg/∂ν + μvg), where vg is the trace of the Herglotzwave function; that is, a function given by

vg(y) =∫

S2eikx·yg(x) ds(x), y ∈ ∂D.

Here and in the following, z denotes the complex conjugate of z ∈ C. Then, we obtainthat the adjoint operator G∗ : L2(S2) → L2(D) × H1/2(∂D) is given by

G∗g =(

v

v−

). (3.17)

Let, f ∈ L2(D), f2 ∈ H−1/2(�) and g ∈ L2(S2). Then,

(G( f1, f2)T , g)L2(S2 ) =

∫S2

u∞(x)g(x) ds(x)

=∫

S2

(∫∂D

[u+(y)

∂e−ikx·y

∂ν(y)− e−ikx·y ∂u+(y)

∂ν

]ds(y)

)g(x) ds(d)

=∫

∂D

[u+(y)

∂vg(y)

∂ν− vg(y)

∂u+(y)

∂ν

]ds(y)

=∫

∂D

[u+(y)

(∂v−(y)

∂ν− ∂v+(y)

∂ν− μ(y)v−(y)

)−(v−(y) − v+(y))

∂u+(y)

∂ν

]ds(y)

9

Inverse Problems 29 (2013) 095021 O Bondarenko and X Liu

=∫

∂D

[u+(y)

(∂v−(y)

∂ν− μ(y)v−(y)

)− v−(y)

∂u+(y)

∂ν

]ds(y)

=∫

∂D

[u+(y)

(∂v−(y)

∂ν− μ(y)v−(y)

)−v−(y)

(∂u−(y)

∂ν− μ(y)u+(y)

)]ds(y)+

∫�

v−(y) f2(y) ds(y)

=∫

∂D

[u+(y)

∂v−(y)

∂ν− v−(y)

∂u−(y)

∂ν

]ds(y)+

∫�

v−(y) f2(y) ds(y)

=∫∫

Du(−k2(1 + q(x))v(x)) + v(x)(k2(1 + q(x))u(x))

+ v(x) f (x) dx +∫

�

v−(y) f2(y) ds(y)

=∫∫

Dv(x) f (x) dx +

∫�

v−(y) f2(y) ds(y),

where we have used the conductive boundary conditions (3.22) and (3.2) in the fourth andthe sixth equality for v and u, respectively. The fifth equality holds because both u andv are radiating solutions in De. In the eighth equality, we have applied Green’s theorem,(3.1) and (3.21). Thus, G∗g = (v, v−)� for all g ∈ L2(S2).

We proceed by showing that the adjoint operator G∗ is injective. Let, g ∈ L2(S2)

be such that G∗g = 0, i.e., (v|D, v−)� = (0, 0)�. Thus, v− = ∂v−/∂ν = 0 on ∂D andby boundary conditions (3.22), v+ = −vg and ∂v+/∂ν = −∂vg/∂ν on ∂D. We define afunction w by

w ={−vg in D,

v in De.(3.18)

Then, w is an entire solution of the Helmholtz equation satisfying the radiation condition.From this we conclude that w = 0 in R

3 and thus vg = 0 in D. By theorem 3.15 in [11],g = 0, which implies that G∗ is injective.

(b) We decompose T into the sum T = C + K with C(

ϕ1

ϕ2

) = ( qk2 |q| ϕ1

ϕ2

)and K

(ϕ1

ϕ2

) =−( √|q|w

(1−μ)ϕ2+μw

). It has been shown in the previous section that the mapping ϕ1 → w|D

from L2(D) into H1(D) is bounded. The operator K is compact because of the compactembeddings of H1(D) into L2(D) and H1/2(�) into H−1/2(�). For �[eitC], we have

�[

eit

⟨C

(ϕ1

ϕ2

),

(ϕ1

ϕ2

)⟩]= 1

k2

∫∫D

�[

eit q

|q|]|ϕ1|2 dx +

∫�

|ϕ2|2ds

� c0

k2‖ϕ1‖2

L2(D)+ ‖ϕ2‖2

H1/2(�)

� 1

2min

{c0

k2, 1

}∥∥∥∥(ϕ1

ϕ2

)∥∥∥∥2

L2(D)×H1/2(�)

.

(c) Let

(ϕ1

ϕ2

)∈ L2(D) × H1/2(�). Then,⟨

T

(ϕ1

ϕ2

),

(ϕ1

ϕ2

)⟩=

∫∫D

(1

k2

q

|q|ϕ1 −√

|q|w)

ϕ1 dx +∫

�

(μϕ2 − μw)ϕ2 ds

=∫∫

D

1

k2

q

|q| |ϕ1|2 dx +∫

�

μ|ϕ2|2 ds −∫∫

D

√|q|wϕ1 dx −

∫�

μwϕ2 ds.

10

Inverse Problems 29 (2013) 095021 O Bondarenko and X Liu

Application of Green’s theorem in D and in {x /∈ D, |x| < R} yields⟨T

(ϕ1

ϕ2

),

(ϕ1

ϕ2

)⟩=

∫∫D

1

k2

q

|q| |ϕ1|2 dx +∫

�

μ|ϕ2|2 ds +∫∫

D[−|∇w|2 + k2|w|2] dx

+∫

�

∂w−∂ν

w ds −∫

�

μwϕ2 ds(y)

=∫∫

D

1

k2

q

|q| |ϕ1|2 dx +∫

�

μ|ϕ2|2 ds

+∫∫

D[−|∇w|2 + k2|w|2] dx +

∫�

∂w+∂ν

w ds

=∫∫

D

1

k2

q

|q| |ϕ1|2 dx +∫

�

μ|ϕ2|2 ds

+∫∫

BR

[−|∇w|2 + k2|w|2] dx +∫

|x|=R

∂w

∂νw ds.

Finally, by the radiation condition, the last term converges to −ik/(4π)2∫

S2 |w∞|2 ds.Therefore, since by assumption 2.1 q � 0 and μ � 0, we have

⟨T

(ϕ1

ϕ2

),

(ϕ1

ϕ2

)⟩=

∫∫D

1

k2 q

|q| |ϕ1|2 dx +∫

�

μ|ϕ2|2 ds − k2

(4π)2

∫S2

|w∞|2 � 0.

(d) Let(

ϕ1

ϕ2

) ∈ L2(D) × H1/2(�), such that T(

ϕ1

ϕ2

) = ( 00

). Then, (3.10)–(3.12) becomes

�w + k2(1 + q)w = 0 in R3\∂D,

w+ = w− on ∂D,

∂w+∂ν

− ∂w−∂ν

+ μw = 0 on ∂D.

From the uniqueness of the solution to (2.1)–(2.4), it follows that w vanishes in all of R3.

Thus, k2q|q| ϕ1 = 0 in D and μϕ2 = 0 on �. Since |q| is locally bounded below and μ �= 0

on �, (ϕ1, ϕ2)� = (0, 0). �

Now we can state the first main result of this section.

Theorem 3.5. Let the assumptions 2.1, 3.1 and 3.3 hold. For z ∈ R3, define φz ∈ L2(S2) by

(3.13). Then,

z ∈ D ⇐⇒ φz ∈ R(F1/2

�

),

and consequently

z ∈ D ⇐⇒∞∑j=1

|〈φz, ψ j〉L2(S2 )|2|λ j| < ∞, (3.19)

where F� = |�[eitF]| + |F| and (λ j, ψ j) is its eigensystem. In other words, the sign of thefunction

W (z) =⎡⎣ ∞∑

j=1

|〈φz, ψ j〉L2(S2 )|2|λ j|

⎤⎦−1

is the characteristic function of D.

11

Inverse Problems 29 (2013) 095021 O Bondarenko and X Liu

We proceed with the case when λ is complex valued and λ �= 1. Since ui solves theHelmholtz equation in all of R

3, a function u := (ui|D + us|D, us|De ) solves the followingboundary value problem with f1 = ui and f2 = ∂ui

∂ν+ μui. Given f1 ∈ H1/2(∂D) and

f2 ∈ H−1/2(∂D) find u|D ∈ H1(D), u|De ∈ H1loc(D

e), such that

�u + k2u = 0 in De, (3.20)

�u + k2nu = 0 in D, (3.21)

u+ − u− = − f1,∂u+∂ν

− λ∂u−∂ν

+ μu+ = − f2 on ∂D, (3.22)

limr→∞ r

(∂u

∂r− iku

)= 0 r = |x|. (3.23)

The well posedness of (3.20)–(3.23) can be established following the same steps as in theproof of theorem 2.2 and theorem 2.4.

Now we derive a factorization of the far-field operator F . Define the data-to-patternoperator G : H1/2(∂D) × H−1/2(∂D) → L2(S2) by

G

(f1

f2

)= v∞,

where v∞ is the far-field pattern of the solution to (3.20)–(3.23). Besides, we also define anauxiliary operator H : L2(S2) → H1/2(∂D) × H−1/2(∂D) by

(Hg)(x) =( ∫

S2 eikx·dg(d) ds(d)∫S2

(∂eikx·d∂ν(x)

+ μeikx·d)

g(d) ds(d)

), x ∈ ∂D. (3.24)

From the superposition principle, we observe that

F = GH. (3.25)

For the subsequent analysis, we need the following two assumptions regarding the wavenumber k.

Assumption 3.6.

• k2 is not a Dirichlet eigenvalue of the following boundary value problem:

�u + k2nu = 0 in D, (3.26)u = f on ∂D, (3.27)

i.e., the only solution u ∈ H1(D) of (3.26),(3.27) for f = 0 is the trivial one.• k2 is not a Neumann eigenvalue of the following boundary value problem:

�u + k2nu = 0 in D, (3.28)

λ∂u

∂ν= g on ∂D, (3.29)

i.e., the only solution u ∈ H1(D) of (3.28),(3.29) for g = 0 is the trivial one.

We make yet another assumption on the wave number k in order to ensure the injectivityof the far-field operator.

Theorem 3.7. Assume that k2 is not an eigenvalue of the following interior transmissionproblem in D, i.e., the only solution (u, v) ∈ H1(D) × H1(D) of

�u + k2nu = 0, �v + k2v = 0 in D, (3.30)

u = v, λ∂u

∂ν= ∂v

∂ν+ μv on ∂D (3.31)

is the trivial one u = v = 0. Then, F is injective and its range R(F ) is dense in L2(S2).

12

Inverse Problems 29 (2013) 095021 O Bondarenko and X Liu

Proof. The proof follows from the same arguments as in theorem 4.4 of [22] and is omittedhere for the sake of brevity. �

Remark 3.8. With the help of Green’s theorem and the unique continuation principle, one canshow that there are no transmission eigenvalues if (λ) < 0 or (λn) > 0 on some subsetD0 ⊂ D. We refer to Paivarinta and Sylvester [29], Kirsch [20], Cakoni and her collaborators[5–9] for further existence results.

With the two assumptions, we can define the interior DtN operator T : H1/2(∂D) →H−1/2(∂D) by

T f = λ∂u

∂νon ∂D, (3.32)

where u solves the interior Dirichlet boundary value problem (3.26)–(3.27) with boundary dataf ∈ H1/2(∂D) and the interior Neumann-to-Dirichlet operator T −1 : H−1/2(∂D) → H1/2(∂D)

by

T −1g = u on ∂D, (3.33)

where u solves the interior Neumann boundary value problem (3.28)–(3.29) with boundarydata g ∈ H−1/2(∂D). We note that T and T −1 are bounded linear operators.

Properties of G are collected in the following lemma.

Theorem 3.9. Assume that k2 is not a Dirchlet eigenvalue of the boundary value problem(3.26)–(3.27).

(a) The data-to-pattern operator G : H1/2(∂D) × H−1/2(∂D) → L2(S2) is compact withdense range in L2(S2).

(b) The kernel space of G is given by N (G) = {( f , T f )T : f ∈ H1/2(∂D)}.(c) For any z ∈ R

3, define φz by

φz(x) := e−ikx·z, x ∈ S2. (3.34)

Then,

z ∈ D ⇐⇒ φz ∈ R(G). (3.35)

Proof.

(a) This follows by the similar arguments as in the case of λ = 1 (see theorem 3.4(a)).(b) We show first that ( f , T f ) ⊆ N(G), i.e., that the far field of u|D ∈ H1(D), u|De ∈ H1

loc(De),

satisfying (3.20)–(3.23) with f1 = f and f2 = T f is identically zero.Define u|D ∈ H1(D), u|De ∈ H1

loc(De) by

u ={v in D,

0 in De,(3.36)

with v ∈ H1(D) being the solution of

�v + k2nv = 0 in D,

v = f on ∂D.

Then, λ∂ u−/∂ν = T f and, since (3.20)–(3.23) is uniquely solvable, u = u is the solutionof (3.20)–(3.23) with f1 = f and f2 = T f . Thus, G( f , T f ) = u∞ = 0.

It remains to prove that the set {( f , T f )T : f ∈ H1/2(∂D)} forms the entire kernelof G. Let ( f , g)T ∈ H1/2(∂D) × H−1/2(∂D) be from N (G). By linearity, we have

13

Inverse Problems 29 (2013) 095021 O Bondarenko and X Liu

that G(0, g − T f )T = 0. This means that the far-field patterns of the solution u of theboundary value problem (3.20)–(3.23) with boundary data ( f1, f2) = (0, g− T f ) vanish,and therefore u = 0 in De by an application of Rellich’s lemma and unique continuation.From this we conclude that u− = 0 and λ∂u−/∂ν = g − T f on ∂D. Assumption 3.6implies that g − T f = 0.

(c) Let first z ∈ D and define

v(x) := �(x, z) = eik|x−z|

4π |x − z| , x ∈ De.

The far-field pattern of v is given by v∞ = φz. Define u|D ∈ H1(D), u|De in H1loc(D

e) by

u ={

0 in D,

v in De.(3.37)

Then, u solves (3.20)–(3.23) with f1 = −v+ and f2 = −∂v+/∂ν − μv+. Thus, since uand v coincide in the exterior of D, we have that u∞ = v∞ = φz. Therefore, for chosenf1 and f2 holds G( f1, f2) = φz.The rest of the proof follows by the same arguments used in theorem 3.2. �Define, for any ϕ ∈ H−1/2(∂D), the single-layer potential SLϕ by

(SLϕ)(x) :=∫

∂Dϕ(y)�(x, y) ds(y), x ∈ R

3\∂D, (3.38)

and for any ψ ∈ H1/2(∂D), the double-layer potential KLϕ by

(KLψ)(x) :=∫

∂Dψ(y)

∂�(x, y)

∂ν(y)ds(y), x ∈ R

3\∂D, (3.39)

where � denotes the fundamental solution of the Helmholtz equation in R3 and is given by

�(x, y) = eik|x−y|

4π |x − y| , x, y ∈ R3, x �= y.

We return now to the factorization of the far-field operator F . The adjoint operatorH∗ : H−1/2(∂D) × H1/2(∂D) → L2(S2) of H (see 3.24) is given by(

H∗(

ϕ

ψ

))(x) =

∫∂D

(e−ikx·yϕ(y) +

(∂e−ikx·y

∂ν(y)+ μ(y) e−ikx·y

)ψ(y)

)dy, x ∈ S2. (3.40)

By the asymptotic behavior of the fundamental solution, H∗(ϕ, ψ)T is just the far-field patternof

u(x) = (SLϕ)(x) + [(KL + SLμ)ψ](x), x ∈ De.

The theory of potentials yields (see e.g. [28]) that u ∈ H1loc(D

e) is a radiating solution of theHelmholtz equation (3.20) with traces

u+ = Sϕ + Kψ + Sμψ + 1

2ψ,

∂u+∂ν

= K′ϕ + Nψ + K′μψ − 1

2ϕ − μ

2ψ on ∂D. (3.41)

where for any ϕ ∈ H−1/2(∂D), ψ ∈ H1/2(∂D), the boundary integral operators S :H−1/2(∂D) → H1/2(∂D), K : H1/2(∂D) → H1/2(∂D), K′ : H−1/2(∂D) → H−1/2(∂D)

and N : H1/2(∂D) → H−1/2(∂D) on ∂D are given by

(Sϕ)(x) =∫

∂Dϕ(y)�(x, y) ds(y), x ∈ ∂D,

(Kψ)(x) =∫

∂Dψ(y)

∂�(x, y)

∂ν(y)ds(y), x ∈ ∂D,

14

Inverse Problems 29 (2013) 095021 O Bondarenko and X Liu

(K′ϕ)(x) =∫

∂Dψ(y)

∂�(x, y)

∂ν(x)ds(y), x ∈ ∂D,

(Nψ)(x) = ∂

∂ν(x)

∫∂D

ψ(y)∂�(x, y)

∂ν(y)ds(y), x ∈ ∂D.

We extend u by a solution of the equation (3.21) in D such that u|D ∈ H1(D) andu|De ∈ H1

loc(De) satisfies (3.20)–(3.23) with boundary data ( f1, f2) = (u− − u+, λ∂u−/∂ν −

∂u+/∂ν − μu+) on ∂D. From the definition of the data-to-pattern operator G, we obtainG(u− − u+, λ∂u−/∂ν − ∂u+/∂ν − μu+)T = H∗(ϕ, ψ)T . Since (u−, λ∂u−/∂ν)T ∈ N (G),we obtain

H∗ = G

( −S −K − Sμ − 12 I

−K′ + 12 I − μS −N − K′μ + μ

2 I − μK − μSμ − μ

2 I

). (3.42)

Combining (3.42) with the earlier established decomposition (3.25), we arrive at the followingfactorization of the far-field operator:

F = G

( −S −K − Sμ − 12 I

−K′ + 12 I − μS −N − K′μ + μ

2 I − μK − μSμ − μ

2 I

)∗G∗. (3.43)

For the FM, we need to show that the real part of the middle operator can be represented asa sum of a compact and a coercive operator. Similar to the case of the mixed boundary valueproblem ([22]), due to the fact that the real part of S is the sum of a positively coercive anda compact part, while the real part of N is the sum of a negatively coercive and a compactpart, the coercivity condition for the middle operator cannot be established. We follow theidea in Charalambopoulus et al’s paper [10] and decompose the operator G, such that only itsinjective part is retained:

G = G

(I 00 I

)= 1

2G

(I T −1

T I

)+ 1

2G

(I −T −1

−T I

)= 1

2G

(IT

) (I T −1

) + 1

2G

(T −1

−I

) (T −I

)= G1

(T −I

). (3.44)

Here, we have used the fact that the range of the operator (I, T )T : H1/2(∂D) →H1/2(∂D) × H−1/2(∂D) belongs to the kernel N (G) by theorem 3.9. The operator G1 :H−1/2(∂D) → L2(S2) is given by

G1 = 1

2G

(T −1

−I

). (3.45)

From the derivation of (3.44) and the definition of G1, we see that G1 inherits the range of G.Substituting decomposition (3.44) of the operator G into factorization (3.43) of the operatorF , we obtain a new factorization for F as follows:

F = G1(T −I

) ( −S −K − Sμ − 12 I

−K′ + 12 I − μS −N − K′μ + μ

2 I − μK − μSμ − μ

2 I

)∗ (T ∗

−I

)G∗

1

= G1M∗G∗1, (3.46)

where the operator M∗ is the adjoint of the operator M : H1/2(∂D) → H−1/2(∂D) defined by

M = −T ST ∗ + T K + T Sμ + T

2+ K′T ∗ − N − K′μ − T ∗

2+ μ

2I

+μST ∗ − μK − μSμ − μ

2I. (3.47)

We summarize these results in the following theorem.

15

Inverse Problems 29 (2013) 095021 O Bondarenko and X Liu

Theorem 3.10. Assume that k2 is neither a Dirchlet eigenvalue of the boundary value problem(3.26)–(3.27) nor a Neumann eigenvalue of the boundary value problem (3.28)–(3.29). Then,the far-field operator F : L2(S2) → L2(S2) from (1.6) has a factorization of the form

F = G1M∗G∗1,

where the operator G1 : H−1/2(∂D) → L2(S2) is given by (3.45) and the operatorM : H1/2(∂D) → H−1/2(∂D) is defined by (3.47). G1 is compact, injective with denserange in L2(S2). For any z ∈ R

3, let φz be given by (3.13). Then,

z ∈ D ⇐⇒ φz ∈ R(G1). (3.48)

We write M = M0 + M1 with

M0 = −T ST ∗ + T K + T

2+ K′T ∗ − N − T ∗

2, (3.49)

M1 = T Sμ − K′μ + μ

2I + μST ∗ − μK − μSμ − μ

2I. (3.50)

M1 : H1/2(∂D) → H−1/2(∂D) is compact because the operators S, K and K′ are compact fromH1/2(∂D) into H−1/2(∂D) and the embedding from H1/2(∂D) into H−1/2(∂D) is compact.For the convenience of the subsequent analysis, we will indicate the dependence of M0 on thewave number k by writing M0(k), similarly for the operators S, K, K′, N and T . The differencesS(k) − S(i), K(k) − K(i), K′(k) − K′(i), N(k) − N(i) are compact since the kernels of theseoperators are sufficiently smooth. We note that T (k) f − T (i) f = λ∂ u/∂ν, where u = uk − ui

solves

�u + k2nu = −(1 + k2)nui in D, u = 0 on ∂D

in the weak sense, with uk and ui being solutions of the boundary value problem (3.26)–(3.27),corresponding to wave numbers k and i, respectively. The boundedness of f → ui and ui → ufrom H1/2(∂D) into H1(D) and from L2(D) into H1(D), respectively, and the compactness ofthe embedding H1(D) into L2(D) implies that T (k) − T (i) is compact from H1/2(∂D) intoH−1/2(∂D). These compactness results imply that the difference of M0(k)− M0(i) is compactfrom H1/2(∂D) into H−1/2(∂D). So far we have shown M0(k) = M0(i) + (M0(k) − M0(i)),where M0(k) − M0(i) is compact.

For the FM, we need a further property of the operator M0(i) and introduce three more DtNoperators +(i), −(i) and Tn0 . Precisely, the exterior DtN operator +(i) : H1/2(∂D) →H−1/2(∂D) is defined by +(i) f = ∂u+/∂ν on ∂D, where u solves the exterior Dirichletboundary value problem (1.1) and (1.4) with k = i and boundary data f ∈ H1/2(∂D), whilethe interior DtN operator −(i) : H1/2(∂D) → H−1/2(∂D) is defined by −(i) f = ∂v−/∂ν

on ∂D, where v solves the equation �v − v = 0 in D with boundary data f ∈ H1/2(∂D).Tn0 : H1/2(∂D) → H−1/2(∂D) is defined by Tn0 f = λ∂w−/∂ν on ∂D, where w solves thefollowing boundary value problem

�w − n0w = 0 in D, w = f on ∂D (3.51)

with some chosen constant n0 > 0. We note that ±(i) : H1/2(∂D) → H−1/2(∂D) areself-adjoint operators.

In the subsequent analysis, we shall use c to denote a generic positive constant whosevalues may change in different inequalities but remain always bounded away from infinity.

Lemma 3.11. The operator M0(i) has a new representation

M0(i) = −BS(i)B∗ + B + R, (3.52)

16

Inverse Problems 29 (2013) 095021 O Bondarenko and X Liu

where B = Tn0 −+(i) and R = −(T (i)−Tn0 )S(i)(T (i)−+(i))∗−(Tn0 −+(i))S(i)(T (i)−Tn0 )

∗ + (T (i) − Tn0 ) have the property that �B is coercive and R is compact from H1/2(∂D)

into H−1/2(∂D).

Proof. Using the operator identities (see lemma 2.6 in [24])

S(i)[−(i) − +(i)] = I = [−(i) − +(i)]S(i), (3.53)

+(i)S(i) = − 12 I + K′(i), (3.54)

S(i)−(i) = 12 I + K(i), (3.55)

−(i)S(i)+(i) = +(i)S(i)−(i) = N(i), (3.56)

one can derive new representation (3.52).To prove the coercivity property of the operator B : H1/2(∂D) → H−1/2(∂D), let v be the

radiating solution of �v − v = 0 in De with boundary data v+ = f ∈ H1/2(∂D) on ∂D and w

the solution of (3.51). Then, using Green’s theorem in D and in DR := {x ∈ De : |x| < R}, weobtain

〈B f , f 〉 = 〈Tn0 f , f 〉 − 〈+(i) f , f 〉=

⟨λ

∂w−∂ν

,w−

⟩−

⟨∂v+∂ν

, v+

⟩= λ

∫∫D(n0|w|2 + |∇w|2) dx +

∫∫DR

(|v|2 + |∇v|2) dx −∫

|x|=R

∂v

∂νv ds

= λ

∫∫D(n0|w|2 + |∇w|2) dx +

∫∫DR

(|v|2 + |∇v|2) dx +∫

|x|=R|v|2 ds + O

(1

R

),

and thus as R → ∞ (note that v decays exponentially)

〈B f , f 〉 = λ

∫∫D(n0|w|2 + |∇w|2) dx +

∫∫De

(|v|2 + |∇v|2) dx. (3.57)

Since, n0 > 0 and �(λ) � c � 0, the trace theorem now yields the existence of c > 0, suchthat

�〈B f , f 〉 � ‖v‖2H1(De)

� c‖v‖2H1/2(∂D)

= c‖ f ‖2H1/2(∂D)

.

Analogously to the proof of the compactness of the operator T (k) − T (i), we obtain thecompactness result for T (i) − Tn0 . Thus, the operator R is also compact from H1/2(∂D) intoH−1/2(∂D). �

Lemma 3.12. Denote by M0(i) := −BS(i)B∗ + B. Then, M0(i) has the following properties.

(a) If �(λ) − 1 < 0, then �M0(i) is positively coercive, i.e., there exists c > 0 with

�〈M0(i) f , f 〉 � c‖ f ‖2H1/2(∂D)

for all f ∈ H1/2(∂D).

(b) If �(λ) − 1 > 0, then �M0(i) is negatively coercive, i.e, there exists c > 0 with

−�〈M0(i) f , f 〉 � c‖ f ‖2H1/2(∂D)

for all f ∈ H1/2(∂D).

(c) If �(λ) − 1 = 0 and (λ) �= 0, then �M0(i) is negatively coercive.

17

Inverse Problems 29 (2013) 095021 O Bondarenko and X Liu

Proof. Define, for any f ∈ H1/2(∂D), the single-layer potentials u and v by u(x) = SL(i)B∗ fin R

3\∂D and v(x) = SL(i)S−1(i) f in De, respectively. Then, u|D ∈ H1(D) and u|De ∈ H1(De)

is a solution of the equation �u − u = 0 in R3\∂D and, in terms of traces, u± and ∂u±/∂ν are

given by

u± = S(i)B∗ f and B∗ f = ∂u−∂ν

− ∂u+∂ν

on ∂D, (3.58)

and v ∈ H1(De) is a solution of �v − v = 0 in De with

v = f on ∂D. (3.59)

Using Green’s theorem in D, we see that T ∗n0

: H1/2(∂D) → H−1/2(∂D) is defined byT ∗

n0f = λ∂w/∂ν, where w satisfies

�w − n0w = 0 in D,

w = f on ∂D. (3.60)

From the definition of the operator B, we conclude that

B∗ f = λ∂w

∂ν− ∂v

∂νon ∂D, (3.61)

where we have used the fact that the operator +(i) is self-adjoint.Using the traces introduced in (3.58)–(3.61), we can obtain the following three

representations:

〈M0(i) f , f 〉 = −⟨u, λ

∂w

∂ν− ∂v

∂ν

⟩+

⟨f ,

∂u−∂ν

− ∂u+∂ν

⟩= −

⟨u, λ

∂w

∂ν

⟩+

⟨u,

∂v

∂ν

⟩+

⟨w,

∂u−∂ν

⟩−

⟨v,

∂u+∂ν

⟩= −

⟨u, λ

∂w

∂ν

⟩+

⟨u,

∂v

∂ν

⟩+

⟨w,

∂u−∂ν

⟩−

⟨v,

∂u+∂ν

⟩= −

⟨u, λ

∂w

∂ν

⟩+

⟨w,

∂u−∂ν

⟩, (3.62)

〈M0(i) f , f 〉 = −⟨u,

∂u−∂ν

− ∂u+∂ν

⟩+

⟨f ,

∂u−∂ν

− ∂u+∂ν

⟩= −

⟨u,

∂u−∂ν

⟩+

⟨u,

∂u+∂ν

⟩+

⟨w,

∂u−∂ν

⟩−

⟨v,

∂u+∂ν

⟩, (3.63)

〈M0(i) f , f 〉 = −⟨u, λ

∂w

∂ν− ∂v

∂ν

⟩+

⟨f , λ

∂w

∂ν− ∂v

∂ν

⟩= −

⟨u, λ

∂w

∂ν

⟩+

⟨u,

∂v

∂ν

⟩+

⟨w, λ

∂w

∂ν

⟩−

⟨v,

∂v

∂ν

⟩. (3.64)

Note that the last equality in (3.62) holds since both u and v decay exponentially in De.Subtracting (3.63) from (3.64), we have that

0 = −⟨v − u,

∂(v − u+)

∂ν

⟩+

⟨u,

∂u−∂ν

⟩−

⟨w,

∂u−∂ν

⟩−

⟨u, λ

∂w

∂ν

⟩+

⟨w, λ

∂w

∂ν

⟩. (3.65)

18

Inverse Problems 29 (2013) 095021 O Bondarenko and X Liu

(a) We first consider the case when �(λ) − 1 < 0. Furthermore, we choose n0 < 1. Adding(3.62) to (3.65), using Green’s theorem and the fact that both u and v decay exponentiallyas |x| → ∞, we see that

�〈M0(i) f , f 〉 =∫∫

De

(|u − v|2 + |∇(u − v)|2) dx + �(λ)

×∫∫

D[n0|u − w|2 + |∇(u − w)|2] dx

+∫∫

D[(1 − �(λ)n0)|u|2 + (1 − �(λ))|∇u|2] dx

� min{1 − �(λ)n0, 1 − �(λ)}‖u‖2H1(D)

� c‖S(i)B∗ f ‖2H1/2(∂D)

� c‖B∗ f ‖2H−1/2(∂D)

� c|〈B∗ f , f 〉|2‖ f ‖2

H1/2(∂D)

� c|�〈B f , f 〉|2‖ f ‖2

H1/2(∂D)

� c‖ f ‖2H1/2(∂D)

,

where we have used the trace theorem in the second inequality, the fact that the operatorS(i) is continuously invertible in the third inequality, the Cauchy–Schwarz inequality inthe fourth inequality and the coercivity property of the operator �(B) (see lemma 3.11)in the last inequality.

(b) Considering now the case of �(λ) − 1 > 0. We choose n0 > 1. Subtracting (3.62) from(3.65), again using Green’s theorem and the fact that u and v decay exponentially as|x| → ∞, one obtains

− �〈M0(i) f , f 〉 =∫∫

De

(|u − v|2 + |∇(u − v)|2) dx +∫∫

D(|u − w|2 + |∇(u − w)|2) dx

+∫∫

D[(�(λ)n0 − 1)|w|2 + (�(λ) − 1)|∇w|2] dx

� min{�(λ)n0 − 1,�(λ) − 1}‖w‖2H1(D)

� c‖ f ‖2H1/2(∂D)

, (3.66)

where the last inequality is justified by the trace theorem.(c) Let �(λ) − 1 = 0 but (λ) �= 0. From (3.66), we observe that

−�〈M0(i) f , f 〉 � (n0 − 1)‖w‖2L2(D)

,

which implies that −�[M(i)] is non-negative and injective for some chosen n0 > 1. DefineB := Tn0 − −(i). One readily sees that B is compact from H1/2(∂D) into H−1/2(∂D).With the help of operator identity (3.53), from the definition of B = Tn0 − +(i), wederive that B = B − S(i)−1 and from this

M0(i) = −BS(i)B∗ − B∗.

Then,

−�[M0(i)] = �(B)S(i)�(B) + �(B) + (B)S(i)(B).

19

Inverse Problems 29 (2013) 095021 O Bondarenko and X Liu

From (3.57), we derive that

−〈B f , f 〉 = − (λ)

∫∫D(n0|w|2 + |∇w|2) dx

� − (λ) min{n0, 1}‖w‖2H1(D)

� c‖ f ‖2H1/2(∂D)

,

where we again have used the trace theorem. By the Lax–Milgram theorem, we concludethat (B) = (B) is bijective. This, together with the fact that S(i) is bijective (e.g., seelemma 1.14 in [22]), imply that (B)S(i)(B) is bijective. Thus, the operator −�[M0(i)]is an isomorphism from H1/2(∂D) into H−1/2(∂D). Noting that −�[M0(i)] is a boundednon-negative operator, we have (see inequality (4.5) in [19])

− �〈M0(i) f , f 〉 �‖ − �[M0(i)] f ‖2

H−1/2(∂D)

‖�[M0(i)]‖� c‖ f ‖2

H1/2(∂D). (3.67)

�Lemma 3.13. Assume that k2 is neither a Dirichlet eigenvalue of boundary value problem(3.26)–(3.27) nor an eigenvalue of the interior transmission problem (3.30)–(3.31). Then〈M f , f 〉 < 0 for all f ∈ H1/2(∂D) with f �= 0.

Proof. Define, for any f ∈ H1/2(∂D), the combination v of the single-layer and double-layerpotentials by

v = (SLT ∗ − KL − SLμ) f in R3\∂D.

Then, v ∈ H1(D) and v|De ∈ H1loc(D

e) is a radiating solution of the Helmholtz equation

�v + k2v = 0 in R3\∂D,

and

v± =(

ST ∗ − K − Sμ ∓ 1

2I

)f and

∂v±∂ν

=(

K′T ∗ − N − K′μ ∓ 1

2T ∗ ± μ

2I

)f on ∂D

(3.68)

in terms of traces. Consequently, we have that

v− − v+ = f and∂v−∂ν

− ∂v+∂ν

= (T ∗ − μ) f on ∂D. (3.69)

From the definition (3.47) of the operator M, we may rewrite

M = −(T − μ)

[ST ∗ − K − Sμ − 1

2I

]+

[K′T ∗ − N − K′μ − 1

2T ∗ + μ

2I

].

Using the relations given in (3.68) and (3.69), we have

〈M f , f 〉 = −⟨(T − μ)v+ − ∂v+

∂ν, f

⟩= −

⟨v+,

∂v−∂ν

− ∂v+∂ν

⟩+

⟨∂v+∂ν

, v− − v+

⟩= −

⟨v+,

∂v−∂ν

− ∂v+∂ν

⟩−

⟨∂v+∂ν

, v+

⟩+

⟨∂v+∂ν

, v−

⟩= −

⟨v+,

∂v−∂ν

− ∂v+∂ν

⟩−

⟨∂v+∂ν

, v+

⟩+

⟨∂v−∂ν

− T ∗ f + μ f , v−

⟩= −

⟨v+,

∂v−∂ν

− ∂v+∂ν

⟩−

⟨∂v+∂ν

, v+

⟩+

⟨∂v−∂ν

, v−

⟩− ⟨

T ∗ f − μ f , f + v+⟩

= − 2�[⟨

v+,∂v−∂ν

− ∂v+∂ν

⟩]−

⟨∂v+∂ν

, v+

⟩+

⟨∂v−∂ν

, v−

⟩− 〈 f , T f 〉 + μ〈 f , f 〉. (3.70)

20

Inverse Problems 29 (2013) 095021 O Bondarenko and X Liu

We now look at the imaginary part for each term on the right-hand side (rhs) of (3.70). The firstterm on the rhs of (3.70) is real valued. Using Green’s theorem in DR := {x ∈ De : |x| < R},we obtain

−⟨∂v+∂ν

, v+

⟩=

∫∫DR

[v�v + |∇v|2] dx −∫

|x|=R

∂v

∂νv ds

=∫∫

DR

[−k2|v|2 + |∇v|2] dx − ik∫

|x|=R|v|2 ds + O

(1

R

)as R tends to infinity. Taking the imaginary part yields

−⟨∂v+∂ν

, v+

⟩= −k lim

R→∞

∫|x|=R

|v|2 ds = − k

(4π)2

∫S2

|v∞|2 ds � 0.

Using the Green’s theorem in D again, we find⟨v−,

∂v−∂ν

⟩=

∫∫D

[v�v + |∇v|2] dx =∫∫

D[−k2|v|2 + |∇v|2] dx,

and therefore the third term on the rhs of (3.70) is real valued. Recall that the DtN operatorT : H1/2(∂D) → H−1/2(∂D) is defined by T f = λ∂u/∂ν, where u solves the interior Dirichletboundary value problem (3.26)–(3.27) with boundary data f ∈ H1/2(∂D). Using Green’stheorem in D, we have

〈 f , T f 〉 = λ

∫∂D

u∂u

∂νds = λ

∫∫D

[−k2n|u|2 + |∇u|2] dx.

Taking the imaginary part yields

−〈 f , T f 〉 =∫∫

D[−k2(λn)|u|2 + (λ)|∇u|2] dx,

and, since by assumption, (λ) � 0 and (λn) � 0, the imaginary part of the fourth term onthe rhs of (3.70) is non-positive. The assumption (μ) � 0 implies that the imaginary part ofthe fifth term on the rhs of (3.70) is non-positive.

Together with the above analysis, we have in fact proved that 〈M f , f 〉 � 0 for allf ∈ H1/2(∂D). Let us now assume that 〈M f , f 〉 = 0 for some f ∈ H1/2(∂D). Then v∞ = 0.Theorem 2.13 in [11] now shows that v = 0 in De. Then, from the boundary condition (3.69),we see that

v− = f and∂v−∂ν

= T ∗ f − μ f on ∂D.

By Green’s theorem, we have that the adjoint operator T ∗ : H1/2(∂D) → H−1/2(∂D) isgiven by

T ∗ f = λ∂u

∂νon ∂D,

where u solves the boundary value problem

�u + k2nu = 0 in D, u = f on ∂D.

Thus (u, v) solves the interior conductive problem (3.30)–(3.31) and consequently u = v = 0by taking into account the assumption on k2. From this we conclude that f = 0. �

Combining theorem 3.10, lemma 3.12 and lemma 3.13, with the help of the well-knownrange identity theorem (see e.g., theorem 2.15 in [22]), we obtain the following main result ofthis paper.

21

Inverse Problems 29 (2013) 095021 O Bondarenko and X Liu

Theorem 3.14. In addition to assumption 3.6, we further assume that k2 is not an eigenvalueof the interior transmission problem (3.30)–(3.31). For z ∈ R

3, define φz ∈ L2(S2) by (3.13).Then

z ∈ D ⇐⇒ φz ∈ R(F1/2

�

),

and consequently

z ∈ D ⇐⇒∞∑j=1

|〈φz, ψ j〉L2(S2 )|2|λ j| < ∞. (3.71)

where (λ j, ψ j) is an eigensystem of the operator F� : L2(S2) → L2(S2) given by

F� = |�F| + |F|. (3.72)

So again, the sign of the function

W (z) =⎡⎣ ∞∑

j=1

|〈φz, ψ j〉L2(S2 )|2|λ j|

⎤⎦−1

is just the characteristic function of D.

4. Numerical results

In this section, we study the applicability of our method through some numerical simulations inR

2. In each example, the forward data was generated for a kite-shaped object by coupling of thefinite element and boundary integral equation method as suggested in [25]. For the numericaltreatment of the integral equations, we applied the Nystrom method with 128 quadraturepoints, for the finite element method, we used the MATLAB PDE toolbox.

In all of the examples, the computed data set is represented by a C64×64 matrix F ,

where each entry is the far-field pattern u∞(θ j, θl ), j, l ∈ {1, . . . 64}, with θ j = 2π j/64and θl = 2π l/64 denoting the corresponding incident direction of the plane wave and theobservation point, respectively. Furthermore, we compute the matrix

F� = |�[eitF]| + |F|,which represents a discretized version of the operator F� defined in (3.19) (t is chosen tomatch the assumption 3.3) or in (3.72) (here, t = 0). The real and imaginary part of a matrixA ∈ C

N×N is given by

�(A) = A + A∗

2and (A) = A − A∗

2i,

respectively. We define the absolute value of a matrix A ∈ CN×N with a singular value

decomposition A = UV ∗ as

|A| = U ||V ∗,

with || = diag|λ j|, j = 1, . . . N. For our reconstructions, we used a grid G of 200 × 200equally spaced sampling points on the rectangle [−4, 4] × [−4, 4]. Let {(σn, ψn) : n =1, . . . , 64} represent the eigensystem of the matrix F�. Then, the analogous W of the indicatorfunction in (3.73) is given by

W (z) :=[

64∑j=1

|φ∗z ψn|2|σn|

]−1

, z ∈ G,

22

Inverse Problems 29 (2013) 095021 O Bondarenko and X Liu

Figure 1. Reconstruction of the inhomogeneity with n(x, y) = 1.1 + i(x4 + y4), k = 4,μ(x, y) = i − 2x2 and λ = 1.

Figure 2. Reconstruction of the inhomogeneity with n(x, y) = 1 + x2 + i(x2 + y2 + 2), k = 3,μ(x, y) = −2 + i(x2 + y2) and λ = 1 − i.

where φz = (e−ikθ1·z, e−ikθ2·z, . . . , e−ikθ64·z)� ∈ C64. Although, the sum is finite, we expect the

value of W (z) to be much larger for the points belonging to D than for those lying outside ofthe domain.

Figure 1 represents a reconstruction for the kite-shaped inhomogeneity D with n(x, y) =1.1 + i(x4 + y4), (x, y) ∈ D, the wave number k = 4, μ(x, y) = i − 2x2, (x, y) ∈ ∂D andλ = 1. For the reconstruction, we choose t = π/4.

In figure 2, k = 3, n(x, y) = 1+x2+i(x2+y2+2), λ = 1−i and μ(x, y) = −2+i(x2+y2).Figure 3 corresponds to k = 4, n(x, y) = 1 + 10i| sin(y)| + x2 + y2, λ = 1 and

μ(x, y) = i − 2x2. In this last example, the condition (3.1) on q is violated. Although we

23

Inverse Problems 29 (2013) 095021 O Bondarenko and X Liu

Figure 3. Reconstruction of the inhomogeneity with n(x, y) = 1 + 10i| sin(y)| + x2 + y2, k = 4,μ(x, y) = i − 2x2 and λ = 1.

do not have a theoretical justification for this case, figure 3 demonstrates that F� still enablesus to reconstruct the inhomogeneity.

Acknowledgments

The research of XL was supported by the NNSF of China under grants 11101412 and 11071244and the Alexander von Humboldt Foundation. The authors would like to thank the referees fortheir invaluable comments which helped improve the presentation of the paper.

References

[1] Angell T S and Kirsch A 1992 The conductive boundary condition for Maxwell’s equations SIAM J. Appl.Math. 52 1597–610

[2] Angell T S, Kleinman R E and Hettlich F 1990 The resistive and conductive problems for the exterior Helmholtzequation SIAM J. Appl. Math. 50 1607–22

[3] Bourgeois L, Chaulet N and Haddar H 2012 On simultaneous identification of a scatterer and its generalizedimpedance boundary condition SIAM J. Sci. Comput. 34 A1824–48

[4] Cakoni F and Colton D 2006 Qualitative Methods in Inverse Scattering Theory: An Introduction (Berlin:Springer)

[5] Cakoni F, Colton D and Haddar H 2010 The interior transmission problem for regions with cavities SIAM J.Math. Anal. 42 145–62

[6] Cakoni F and Gintides D 2010 New results on transmission eigenvalues Inverse Problems Imaging 4 39–48[7] Cakoni F, Gintides D and Haddar H 2010 The existence of an infinite discrete set of transmission eigenvalues

SIAM J. Math Anal. 42 237–55[8] Cakoni F and Haddar H 2009 On the existence of transmission eigenvalues in an inhomogeneous medium Appl.

Anal. 88 475–93[9] Cakoni F and Kirsch A 2010 On the interior transmission eigenvalue problem Int. J. Comput. Sci. Math.

3 142–67[10] Charalambopoulus A, Kirsch A, Anagnostopoulus K A, Gintides D and Kiriaki K 2007 The factorization method

in inverse elastic scattering from penetrable bodies Inverse Problems 23 27–51[11] Colton D and Kress R 1998 Inverse Acoustic and Electromagnetic Scattering Theory (Applied Mathematical

Sciences vol 93) 2nd edn (Berlin: Springer)

24

Inverse Problems 29 (2013) 095021 O Bondarenko and X Liu

[12] Gerlach T and Kress R 1996 Uniqueness in inverse obstacle scattering with conductive boundary conditionInverse Problems 12 619–25 (Corrigendum)

Gerlach T and Kress R 1996 Inverse Problems 12 1075[13] Hettlich F 1994 On the uniqueness of the inverse conductive scattering problem for the Helmholtz equation

Inverse Problems 10 129–44[14] Hettlich F 1994 Two methods for solving an inverse conductive scattering problem Inverse Problems 10 375–85[15] Hettlich F 1998 The Landweber iteration applied to inverse conductive scattering problems Inverse

Problems 14 931–47[16] Kirsch A 1998 Charaterization of the shape of a scattering obstacle using the spectral data of the far field

operator Inverse Problems 14 1489–512[17] Kirsch A 1999 Factorization of the far field operator for the inhomogeneous medium case and an application in

inverse scattering theory Inverse Problems 15 413–29[18] Kirsch A 2002 The MUSIC-algorithm and the factorization method in inverse scattering theory for

inhomogeneous media Inverse Problems 18 1025–40[19] Kirsch A 2005 The factorization method for a class of inverse elliptic problems Math. Nachr. 278 258–77[20] Kirsch A 2009 On the existence of transmission eigenvalues Inverse Problems Imaging 3 155–72[21] Kirsch A 2011 An Introduction to the Mathematical Theory of Inverse Problems 2nd edn (Berlin: Springer)[22] Kirsch A and Grinberg N 2008 The Factorization Method for Inverse Problems (Oxford: Oxford University

Press)[23] Kirsch A and Kleefeld A 2012 The factorization method for a conductive boundary condition J. Integral Eqns.

Appl. 24 575–601[24] Kirsch A and Liu X 2013 The factorization method for inverse acoustic scattering by a penetrable anisotropic

obstacle Math. Methods Appl. Sci. DOI: 10.1002/mma.2877[25] Kirsch A and Monk P 1994 An analysis of the coupling of finite-element and Nystrom methods in acoustic

scattering IMA J. Numer. Anal. 14 523–44[26] Kress R 1999 Linear Integral Equation 2nd edn (Berlin: Springer)[27] Lechleiter A 2009 The factorization method is independent of transmission eigenvalues Inverse Problems

Imaging 3 123–38[28] Mclean W 2000 Strongly Elliptic Systems and Boundary Integral Equation (Cambridge: Cambridge University

Press)[29] Paivarinta L and Sylvester J 2008 Transmission eigenvalues SIAM J. Math. Anal. 40 738–53[30] Schmucker U 2008 Interpretation of induction anomalies above nonuniform surface layers

Geophysics 36 156–65[31] Vasseur G and Weidelt P 1977 Bimodal electromagnetic induction in nonuniform thin sheets with an application

to the northern Pyrenean induction anomaly Geophys. J. R. Astron. Soc. 51 669–90

25