Direct and inverse acoustic scattering by a mixed-type scatterer

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2013 Inverse Problems 29 065005
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IOP PUBLISHING
INVERSE PROBLEMS
doi:10.1088/0266-5611/29/6/065005
Inverse Problems 29 (2013) 065005 (19pp)
Direct and inverse acoustic scattering by a mixed-type
scatterer
Andreas Kirsch 1 and Xiaodong Liu 1,2
1
Department of Mathematics, Karlsruhe Institute of Technology, D-76128 Karlsruhe, Germany
Institute of Applied Mathematics, Chinese Academy of Sciences, Beijing 100190, People’s
Republic of China
2
E-mail: [email protected] and [email protected]
Received 25 February 2013, in final form 28 March 2013
Published 26 April 2013
Online at stacks.iop.org/IP/29/065005
Abstract
The scattering of time-harmonic acoustic plane waves by a mixed-type scatterer
is considered. Such a scatterer is given as the union of a bounded impenetrable
obstacle and a penetrable inhomogeneous medium with compact support.
Having established the well-posedness of the direct problem by a variational
method, we study the factorization method for recovering the location and
shape of the mixed-type scatterer. Some numerical experiments are presented to
demonstrate its feasibility and effectiveness. One by-product of the factorization
method is that it provides an explicit proof of uniqueness of the inverse
scattering problem under certain assumptions.
(Some figures may appear in colour only in the online journal)
1. Introduction
The two basic problems in classical acoustic scattering theory are the scattering of timeharmonic waves by a bounded impenetrable obstacle and by a penetrable inhomogeneous
medium with compact support. Since the first uniqueness result in inverse acoustic scattering
by sound-soft obstacle (see [3, 12]), there has been an extensive study in these two problems and
rich results can be found in the literature. For a survey on the state of the art of the mathematical
theory and numerical approaches for solving these problems, we refer to the monographs of
Colton and Kress [3], Isakov [5] and Kirsch [10]. However, in practical applications such as
radar, remote sensing, geophysics and nondestructive testing, the scatterer may be given as
the union of a bounded impenetrable obstacle and a penetrable inhomogeneous medium with
compact support. In this paper, we have made the first step in this direction. For simplicity, we
will assume that the impenetrable obstacle is sound soft, i.e. the Dirichlet boundary condition
is imposed on the boundary of the obstacle. The case of impedance type (which includes the
Neumann type as a special case) can be treated analogously.
0266-5611/13/065005+19$33.00 © 2013 IOP Publishing Ltd
Printed in the UK & the USA
1
Inverse Problems 29 (2013) 065005
A Kirsch and X Liu
Let D ⊂ Rn (n = 2, 3) be an open and bounded domain such that the exterior
De := Rn \D of D is connected. The mixed-type scatterer D is given as the union of a
bounded impenetrable obstacle D1 and a penetrable inhomogeneous medium with compact
support D2 . The inhomogeneous medium is modeled by a ‘contrast function’ q ∈ L∞ (D2 ).
Furthermore, assume that D1 and D2 are disjoint, i.e. D1 ∩ D2 = ∅. Let k > 0 be the wave
number and
ui (x, d) = eikx·d ,
x ∈ Rn ,
be the incident plane wave with direction d ∈ Sn−1 . Here, Sn−1 = {x ∈ Rn : |x| = 1} denotes
the unit sphere in Rn . The scattering problem for the mixed-type scatterer D is to find the total
field u such that
u + k2 (1 + q)u = 0
u=u +u
i
s
u=0
in Rn \D1 ,
in R \D1 ,
n
(1.1)
(1.2)
on ∂D1 ,
(1.3)
∂us
(1.4)
lim r
− ikus = 0,
r→∞
∂r
where r = |x| and (1.4) is the Sommerfeld radiation condition which holds uniformly in all
directions x := x/|x| and physically implies that energy is transported to infinity. Note that we
have extended q by zero outside D1 in equation (1.1).
It is well known that us (x) has the following asymptotic behavior:
eikr
1
as |x| → ∞,
(1.5)
x) + O
us (x) = γn n−1 u∞ (
r
r 2
uniformly for all directions x := x/|x|, where
⎧
1
⎪
⎪
n = 3,
⎨ ,
4π
γn =
eiπ/4
⎪
⎪
,
n = 2,
⎩√
8kπ
depending on the dimension n and the function u∞ (
x ) defined on the unit sphere Sn−1 is known
x, d) be the far-field
as the far-field pattern with x denoting the observation direction. Let u∞ (
pattern corresponding to the observation direction x and the incident direction d. Then, the
inverse problem we consider in this paper is, given the wave number k and the far-field pattern
x, d) for all x, d ∈ Sn−1 , to determine the location and shape of the impenetrable obstacle
u∞ (
D1 and the contrast function q or at least its support D2 .
The factorization method has been introduced by Kirsch in [7] for the scattering by
impenetrable sound-soft or sound-hard obstacles and in [8, 9] for the scattering by an
inhomogeneous medium. One main character of the factorization method is that it provides
a criterion for sampling points z which is both necessary and sufficient. The factorization
method has been justified for a large number of inverse problems, see, e.g., the monograph
[11]. The factorization method has also been used in the problems of inverse scattering by
an impenetrable obstacle in an inhomogeneous background medium; we refer to the recent
papers [1, 15] and [16].
This paper is organized as follows. In the following section, we will study the well
posedness (existence, uniqueness and stability) of the direct problem by a variational method.
Section 3 is devoted to a study of the factorization method for recovering the shape of the
impenetrable obstacle D1 and the support D2 of the contrast function q. Some numerical
simulations in two dimensions will be presented in section 4 to justify the validity of our
method.
n−1
2
2
Inverse Problems 29 (2013) 065005
A Kirsch and X Liu
2. The direct scattering problem
This section is devoted to the solution of the direct acoustic scattering problem (1.1)–(1.4).
To this end, let G denote a bounded Lipschitz domain in Rn and BR := {x ∈ Rn : |x| < R}.
Define the Sobolev spaces
H 1 (G) := {u : u ∈ L2 (G), |∇u| ∈ L2 (G)},
1
Hloc
(Rn \G) := {u : u ∈ H 1 (BR \G) for every R > 0 such that BR \G = ∅}.
Recall that H 2 (∂G) is the trace space of H 1 (G) and H − 2 (∂G) is the dual space of H 2 (∂G).
For this and subsequent sections, we make the following general assumptions.
1
1
1
Assumption 2.1. Assume that the boundary ∂D1 of D1 is smooth enough such that the
imbedding of H 1/2 (∂D1 ) into H −1/2 (∂D1 ) is compact, the boundary ∂D2 of D2 is smooth
enough such that the imbedding of H 1 (D2 ) into L2 (D2 ) is compact and the contrast function
q ∈ L∞ (D2 ) satisfies the following.
(1) q 0 in D2 .
(2) There exists c0 ∈ (0, 1), such that 1 + q c0 in D2 .
(3) |q| is locally bounded below, i.e. for every compact subset M ⊂ D2 , there exists
c1 ∈ (0, 1 − c0 ) (depending on M), such that |q| c1 in M. Note that this is satisfied for
a continuous contrast q which vanishes at most on ∂D2 .
We extend q by zero in Rn \D2 .
Since the incident field ui satisfies the Helmholtz equation ui + k2 ui = 0 in all of Rn ,
the scattered field √
us solves the following problem with boundary data f1 = ui on ∂D1 and the
source term f2 = |q|ui in D2 .
1
(Rn \D1 ), such that
Given f1 ∈ H 1/2 (∂D1 ) and f2 ∈ L2 (D2 ), find v ∈ Hloc
q
(2.1)
v + k2 (1 + q)v = −k2 √ f2 in Rn \D1 ,
|q|
v = − f1 on ∂D1 ,
(2.2)
∂v
n−1
lim r 2
− ikv = 0 uniformly in all directions x/r.
(2.3)
r→∞
∂r
Here, and in the following, we do not distinguish between a function defined in D2 and its
extension by zero to all of Rn \D1 . Equation (2.1) has to be understood in the variational sense,
1
(Rn \D1 ) has to satisfy
i.e. v ∈ Hloc
q
[∇v · ∇ϕ − k2 (1 + q)vϕ] dx =
k2 √ f2 ϕ dx,
(2.4)
|q|
Rn \D1
D2
for any test function ϕ with compact support in Rn \D1 , the boundary condition (2.2) is
understood in the sense of the trace operator. A well-known regularity result for elliptic
differential equations (see e.g. [6]) yields that v is analytic in De . In particular, the radiation
condition (2.3) makes sense.
Theorem 2.2. For any f1 ∈ H 1/2 (∂D1 ) and f2 ∈ L2 (D2 ), there exists at most one solution
1
(Rn \D1 ) of (2.1)–(2.3).
v ∈ Hloc
Proof. Clearly, it is sufficient to show that v = 0 in Rn \D1 if f1 = 0 on ∂D1 and f2 = 0 in D2 .
Choose a large ball BR centered at the origin such that D ⊂ BR . Denote by ν the exterior unit
normal to ∂BR . From Green’s theorem and equation (2.1), we obtain
∂v
v
[|∇v|2 − k2 (1 + q)|v|2 ] dx.
ds =
∂BR ∂ν
BR \D1
3
Inverse Problems 29 (2013) 065005
A Kirsch and X Liu
From this, since k > 0 and q 0, it follows that
∂v
ds = k2
v
q|v|2 dx 0.
∂ν
∂BR
BR \D1
Theorem 2.12 in [3] now shows that v = 0 in Rn \BR and it follows by the unique continuation
principle (see e.g. theorem 6.4 in [10]) that v = 0 in Rn \D1 .
The boundary value problems arising in scattering theory are formulated in unbounded
domains. In order to solve such problems by the variational method, we need to write it as an
equivalent problem in a bounded domain. Recall that BR is chosen to be a ball centered at the
origin such that the domain D is contained in the ball and define the Dirichlet-to-Neumann
operator
∂
w
on ∂BR ,
:w→
∂ν
w
, where w
solves the exterior Dirichlet problem for the Helmholtz equation
which maps w to ∂
∂ν
2
w+k w
= 0 in Rn \BR with the Dirichlet boundary data w
|∂BR = w. Since BR is a ball,
then, by separating variables, we can find a solution to the exterior Dirichlet problem outside
BR in the form of a series expansion involving Hankel functions. Based on this result, the
following important properties of the Dirichlet-to-Neumann operator can be established (see
[3, pp 116–7] or [2, theorem 5.20] for details).
Lemma 2.3. The Dirichlet-to-Neumann operator is a bounded linear operator from
1
1
1
H 2 (∂BR ) to H − 2 (∂BR ). Furthermore, there exists a bounded operator 0 : H 2 (∂BR ) →
1
H − 2 (∂BR ) satisfying that
0 ww ds cw2 1
−
H 2 (∂BR )
∂BR
for some constant c > 0, such that − 0 : H 2 (∂BR ) → H − 2 (∂BR ) is compact.
1
1
We now reformulate the problem (2.1)–(2.3) as follows: given f1 ∈ H 1/2 (∂D1 ) and
f2 ∈ L2 (D2 ), find v ∈ H 1 (BR \D1 ) satisfying (2.1)–(2.2) and the boundary condition
∂v
= v
on ∂BR .
(2.5)
∂ν
In exactly the same way as in the proof of lemma 5.22 in [2], one can show that a solution
v to the problem (2.1)–(2.2) and (2.5) can be extended to a solution to the scattering problem
(2.1)–(2.3) and conversely, for a solution v to the scattering problem (2.1)–(2.3), v, restricted
to BR \D1 , solves the problems (2.1)–(2.2) and (2.5). Therefore, by theorem 2.2, the problems
(2.1)–(2.2) and (2.5) have at most one solution. We now have the following result on the well
posedness of the problems (2.1)–(2.2) and (2.5).
Theorem 2.4. Let f1 ∈ H 1/2 (∂D1 ) and f2 ∈ L2 (D2 ). Then the problems (2.1)–(2.2) and (2.5)
have a unique solution v ∈ H 1 (BR \D1 ) satisfying
vH 1 (BR \D1 ) C( f1 1
H 2 (∂D1 )
+ f2 L2 (D2 ) )
(2.6)
with a positive constant C independent of f1 and f2 .
Proof. Let v0 ∈ H 1 (BR \D1 ) be such that v0 = − f1 on ∂D1 , v0 = 0 on ∂BR and
(this is possible by e.g. theorem 3.37 in [14]). Then, for
v0 H 1 (BR \2 ) C f1 12
H (∂D1 )
every solution v to the problems (2.1)–(2.2) and (2.5), w = v − v0 is in the Sobolev space
H01 (BR \D1 ) which is defined by
H01 (BR \D1 ) := {w ∈ H 1 (BR \D1 ) : w = 0 on ∂D1 }.
4
Inverse Problems 29 (2013) 065005
A Kirsch and X Liu
Multiplying equation (2.1) by a test function ϕ ∈ H01 (BR \D1 ), integrating by parts and
using the boundary conditions on ∂D1 , ∂BR , we obtain the following variational formulation
for the problems (2.1)–(2.2) and (2.5): find w ∈ H01 (BR \D1 ) such that
a(w, ϕ) = −a(v0 , ϕ) + b(ϕ)
∀ϕ ∈ H01 (BR \D1 ),
(2.7)
where, for u, ϕ ∈ H01 (BR \D1 ),
2
[∇u · ∇ϕ − k (1 + q)uϕ] dx −
uϕ ds,
a(u, ϕ) =
BR \D1
∂BR
q f2 ϕ dx.
b(ϕ) = k2
D2
We write a = a1 + a2 with
a1 (u, ϕ) =
BR \D1
and
[∇u · ∇ϕ + uϕ] dx −
∂BR
a2 (u, ϕ) = −
0 uϕ ds
[1 + k (1 + q)]uϕ dx −
2
BR \D1
∂BR
( − 0 )uϕ ds,
where 0 is the operator defined in lemma 2.3. By the boundedness of 0 and the trace
theorem, a1 is bounded. On the other hand, for all w ∈ H01 (BR \D1 ),
a1 (w, w) = w2H 1 (BR \D1 ) −
0 ww ds
∂BR
0
w2H 1 (BR \D1 )
0
+ cw2
1
H 2 (∂BR )
w2H 1 (BR \D1 ) ,
0
that is, a1 is strictly coercive. The Lax–Milgram theorem implies that there exists a bijective
bounded linear operator A1 : H01 (BR \D1 ) → H01 (BR \D1 ) satisfying that
a1 (w, ϕ) = (A1 w, ϕ)
for all ϕ ∈ H01 (BR \D1 ).
By the Riesz representation theorem, there exists a bounded linear operator A2 : H01 (BR \D1 ) →
H01 (BR \D1 ), such that
a2 (w, ϕ) = (A2 w, ϕ)
for all ϕ ∈ H01 (BR \D1 ).
By the compactness of − 0 and the Rellich selection theorem (that is, the compact
imbedding of H01 (BR \D1 ) into L2 (BR \D1 )), it follows that A2 is compact. By the Riesz
representation theorem again, one can find a function v ∈ H01 (BR \D1 ), such that
F (ϕ) := −a(v0 , ϕ) + b(ϕ) = (
v, ϕ)
for all ϕ ∈ H01 (BR \D1 ).
Thus, the variational formulation (2.7) is equivalent to the problem
v,
find w ∈ H01 (BR \D1 ) such that A1 w + A2 w = (2.8)
where A1 is bounded and strictly coercive and A2 is compact. The Riesz–Fredholm theory and
the uniqueness result (theorem 2.2) imply that the problem (2.8) or equivalently the problem
(2.7) has a unique solution. The estimate (2.6) follows from the fact that, by a duality argument,
vH 1 (BR \D1 ) = F is bounded by f2 L2 (D2 ) and v0 H 1 (BR \2 ) which in turn is bounded by
0
.
f1 12
H (∂D )
1
5
Inverse Problems 29 (2013) 065005
A Kirsch and X Liu
3. The factorization method for the inverse scattering problem
We begin with the following functional analysis result behind the factorization method. The
proof is completely analogous to theorem 2.15 in [11] and theorem 2.1 in [13]. We omit the
proof here.
Theorem 3.1. Let X ⊂ U ⊂ X ∗ be a Gelfand triple with a Hilbert space U and a reflexive
Banach space X such that the imbedding is dense. Furthermore, let Y be a second Hilbert
space and let A : Y → Y, B : Y → X and C : X → X ∗ be linear bounded operators such that
A = B∗CB.
We make the following assumptions.
(1) B is compact with a dense range.
+ K with some
(2) There exists t ∈ [0, 2π ] such that [eit C] has the form [eit C] = C
compact operator K and some self-adjoint and coercive operator C : X → X ∗ .
(3) C is non-negative or non-positive on X, i.e. Cφ, φ 0 or Cφ, φ 0 for all
φ ∈ X.
(4) (a) C is injective or
(b) C is positive or negative on the closure R(B) of R(B).
Then, the operator A = |[eit A]| + |A| is positive, and the ranges of B : Y → X and
A1/2
: Y → Y coincide.
We want to remark that, in this paper, the real and the imaginary parts of an operator T
are self-adjoint operators given by
T + T∗
T − T∗
(T ) =
and (T ) =
.
2
2i
x, d), x, d ∈ Sn−1 , define the far-field operator F : L2 (Sn−1 ) →
The far-field patterns u∞ (
L2 (Sn−1 ) by
(Fg)(
x) =
u∞ (
x, d)g(d) ds(d) for x ∈ Sn−1 .
(3.1)
Sn−1
From the definition of the far-field operator, we note that Fg is the far-field pattern
corresponding to the incident field given as a Herglotz wavefunction vg, i.e. a function of
the form
eikx·d g(d) ds(d), x ∈ Rn ,
(3.2)
vg (x) =
Sn−1
with density g ∈ L2 (Sn−1 ).
The essential idea of the factorization method is to decide for any z ∈ Rn whether or not
the equation
= φz
Fg
is a self-adjoint operator which can be explicitly computed
is solvable in L2 (Sn−1 ). Here, F
2 n−1
from F, and φz ∈ L (S ) is given by the exponential function
φz (x)
ˆ = e−ikz·xˆ ,
xˆ ∈ Sn−1 .
(3.3)
In what follows, we will consider the following two cases of mixed problem.
• First case: their exists c0 > 0 with
q
−c0
|q|
6
almost everywhere in D2 .
(3.4)
Inverse Problems 29 (2013) 065005
A Kirsch and X Liu
• Second case: their exists t ∈ (π , 2π ) and c > 0 such that
it e q
c
almost everywhere in D2 .
|q|
(3.5)
Note that (3.5) is√
always satisfied for every q, such that (q) does not change sign. Indeed,
(3.5) holds with c = 2/2 and
7π /4, if (q) 0;
t=
5π /4, if (q) < 0,
since |(q)| + (q) |q|. Note that we have assumed that (q) 0 in assumption 2.1.
In the first case, the original factorization method can be justified rigorously. We will
do this in subsection 3.1. In the case q does not satisfy (3.4), we will assume that q satisfies
(3.5). Then, this case requires—at least to our knowledge—a modification of the factorization
method based on some a priori information on the location of the components of D.
3.1. The factorization method for q satisfies (3.4)
We begin our theory analysis with the following factorization result.
Theorem 3.2. Let assumption 2.1 hold. Then we have the following.
(a) The far-field operator F : L2 (Sn−1 ) → L2 (Sn−1 ) can be factorized as
F = GM ∗ G∗ .
(3.6)
1
2
Here, the data-to-pattern operator G : H (∂D1 ) × L (D2 ) → L (S
f
G 1 = v∞,
f2
2
2
n−1
) is given by
(3.7)
where v ∞ is the far-field pattern of the solution to (2.1)–(2.3) with f1 ∈ H 1/2 (∂D1 ) and
1
1
f2 ∈ L2 (D2 ). The middle operator M : H − 2 (∂D1 ) × L2 (D2 ) → H 2 (∂D1 ) × L2 (D2 ) has
the form
−Si 0
+ M1
M = M0 + M1 :=
(3.8)
|q|
0
k2 q
with some compact operator M1 .
1
(b) (M) is non-negative on H − 2 (∂D1 ) × L2 (D2 ), i.e. ϕ, Mϕ 0 for all ϕ ∈
1
H − 2 (∂D1 ) × L2 (D2 ).
(c) G is compact with a dense range in L2 (Sn−1 ). For any z ∈ Rn , define φz ∈ L2 (Sn−1 ) as in
(3.3), then z ∈ D if and only if φz belongs to the range R(G) of G.
Proof.
1
(a) We define the operators H1 : L2 (Sn−1 ) → H 2 (∂D1 ) and H2 : L2 (Sn−1 ) → L2 (D2 ),
respectively, by
(H2 g)(x) = |q(x)|vg (x), x ∈ D2 ,
(H1 g)(x) = vg (x), x ∈ ∂D1 ,
where vg is the Herglotz wavefunction given in (3.2). As an auxiliary operator, we define
1
H : L2 (Sn−1 ) → H 2 (∂D1 ) × L2 (D2 ) by
H1 g
.
(3.9)
Hg =
H2 g
7
Inverse Problems 29 (2013) 065005
A Kirsch and X Liu
Its adjoint H ∗ : H − 2 (∂D1 ) × L2 (D2 ) → L2 (Sn−1 ) is given by
φ
H∗
(
x) =
φ(y) e−ikx·y ds(y) +
ψ (y) e−ikx·y |q(y)| dy,
ψ
∂D1
D2
1
x ∈ Sn−1 .
(3.10)
It follows from the definition (3.1) that Fg is the far-field pattern v ∞ of the solution to
(2.1)–(2.3) corresponding to the data
f1 (x) = vg (x), x ∈ ∂D1 ,
f2 (y) = |q(y)| vg (y), y ∈ D2 .
Thus, from the definitions (3.7) of the data-to-pattern operator G and (3.9) of the auxiliary
operator H, we have
F = GH.
(3.11)
φ
We note that by the asymptotic behavior of the fundamental solution H ∗
is just the
ψ
far-field pattern of the combination of a volume potential and a single-layer potential
φ(y)(x, y) ds(y) +
ψ (y)(x, y) |q(y)| dy, x ∈ Rn .
w(x) =
∂D1
D2
Recall that the fundamental solution of the Helmholtz equation u + k2 u = 0 is
given by
⎧ ik|x−y|
e
⎪
⎨
for x, y ∈ R3 , x = y,
(x, y) = 4π |x − y|
(3.12)
⎪
⎩ i H (1) (k|x − y|)
2
for x, y ∈ R , x = y,
4 0
where H0(1) is the Hankel function of the first kind of order zero. The single-layer boundary
operator S : H −1/2 (∂D1 ) → H 1/2 (∂D1 ) is given by
(Sφ)(x) =
φ(y)(x, y) ds(y), x ∈ ∂D1 .
∂D1
Denote by Si the single-layer operator corresponding to the wave number k = i.
It is easy to check that w is a radiating solution (in the weak sense) of
w + k2 w = − |q|ψ
in Rn \D1 ,
w=
D2
ψ (y)(x, y) |q(y)| dy∂D1 + Sφ
on ∂D1 .
(3.14)
Rewrite (3.13) in the form
q
|q|
w + k2 (1 + q)w = −k2 √
− |q|w + 2 ψ
k q
|q|
We set
f1 := −w|∂D1 = −
D2
and
8
in Rn \D1 .
ψ (y)(x, y) |q(y)| dy∂D1 − Sφ
|q|
f2 := − |q|w + 2 ψ.
k q
(3.13)
Inverse Problems 29 (2013) 065005
A Kirsch and X Liu
Then, G( f1 , f2 ) = w ∞ = H ∗ (φ, ψ ). Define M : H − 2 (∂D1 ) × L2 (D2 ) → H 2 (∂D1 ) ×
L2 (D2 ) by
φ
f1
M
:=
f2
ψ
√
−Si 0
−(S − Si )φ − D2 ψ (y)(x, y) |q(y)| dy
φ
=
+
∂D1
|q|
√
0
ψ
k2 q
− |q|w
φ
φ
=: M0
+ M1
,
ψ
ψ
where w is a radiating solution of (3.13)–(3.14). Then,
1
H ∗ = GM.
1
(3.15)
Substituting this into (3.11) yields the factorization (3.6). The compactness property of
the operator M1 follows from the well posedness of the direct problem, the compactness of
H 1 (D2 ) to L2 (D2 ), the mapping property (see theorem 8.2 in [3]) for the volume potential
and the compactness of the difference S − Si (see lemma 1.14 in [11]).
1
(Rn \D1 ),
(b) From the theory of potentials it is known (see [3, 14]) that w ∈ H 1 (D1 ) ∩ Hloc
√
∂w±
2
w + k w = − |q|ψ in D2 , the traces w± and ∂ν exist on ∂D1 in the variational sense
√
−
+
with w± = D2 ψ (y)(x, y) |q(y)| dy|∂D1 + Sψ and φ = ∂w
− ∂w
. Here, ν denotes
∂ν
∂ν
∂w±
the exterior unit normal vector on ∂D1 and w± , ∂ν denote the limit from the exterior
(+) and interior (−), respectively. Choose a large ball BR centered at the origin such that
D ⊂ BR . Therefore, using Green’s theorem in D and in BR \D we conclude that
0
φ
φ
φ
−φ
w±
√
,M
=
, |q|
+
,
ψ
ψ
ψ
ψ
|q|w
−ψ
k2 q
∂w+
∂w
q
w
− ∂ν−
2
∂ν
= 2 ψL2 (D2 ) +
,
2
w
w + k w
k |q|
∂w
q
2
2
2
2
= 2 ψL2 (D2 ) +
(−|∇w| + k |w| ) dx +
w ds
k |q|
∂BR ∂ν
BR
q
= 2 ψ2L2 (D2 ) +
(−|∇w|2 + k2 |w|2 ) dx
k |q|
BR
1
|w|2 ds + O
+ ik
R
∂BR
as R tends to infinity. Taking the imaginary part yields
q
φ
φ
2
|w|2 ds
,M
= 2 ψL2 (D2 ) + k lim
ψ
ψ
R→∞ ∂B
k |q|
R
q
|w ∞ |2 ds
= 2 ψ2L2 (D2 ) + k|γn |2
n−1
k |q|
S
0,
since q 0.
(c) The compactness of G and the denseness of R(G) follow from similar arguments as in
the proof of lemma 1.13 in [11]. Here, we just prove that, for any z ∈ Rn ,
z ∈ D ⇐⇒ φz ∈ R(G).
Assume first that z ∈ D. Let be the distance between z and the boundary ∂D of D.
Clearly, > 0. Choose a function χ ∈ C∞ (R) with χ (t ) = 1 for |t| /2 and χ (t ) = 0
for |t| /3 and define v ∈ C∞ (Rn ) by v(x) = χ (|x − z|)(x, z) in Rn . Define f1 := −v
9
Inverse Problems 29 (2013) 065005
A Kirsch and X Liu
√
√
√
|q|
|q|
2
2
on ∂D1 and f2 := − k2|q|
[v
+
k
(1
+
q)v]
=
−
[v
+
k
v]
−
v in D2 . Then,
2
q
k q
k2
f1 ∈ H 1/2 (∂D1 ) and f2 ∈ L2 (D2 ) by assumption 2.1 (3) and the fact that v + k2 v = 0
near the boundary ∂D2 . Noting that v(x) = (x, z) for x ∈ Rn \D and therefore the
far-field pattern of v is given by
ˆ
v ∞ (x)
ˆ = e−ikx·z
,
xˆ ∈ Sn−1 ,
f
which coincides with φz . This implies G 1 = φz , i.e. φz ∈ R(G).
f2
Letnow
z
∈
/
D
and
assume
in
contrast
that
their exists f1 ∈ H 1/2 (∂D1 ) and f2 ∈ L2 (D2 )
f
with G 1 = φz . Let v be the solution of the exterior problem (2.1)–(2.3) with f1 and f2
f2 f
and v ∞ = G 1 = φz be its far-field pattern. By the Rellich lemma and unique continuation
f2
/ D, this contradicts the
principle, we find that v(x) = (x, z) for x ∈ Rn \(D ∪ {z}). If z ∈
fact that v is analytic in Rn \D and (x, z) has a singularity at x = z. If z ∈ ∂D, we have that
(x, z) = v(x) for x ∈ ∂D, x = z. This implies (x, z) ∈ H 1/2 (∂D). However, this contradicts
1
(Rn \D), since ∇(x, z) = O(1/|x−z|n−1 )
the fact that (·, z) is certainly not in H 1 (D) or Hloc
as x → z.
Recall that q satisfies (3.4), that is, their exists c0 > 0 with
q
−c0
|q|
almost everywhere in D2 . Then, for all ψ ∈ L2 (D2 ), we have
q
c0
|q|
− ψ, 2 ψ = − 2 ψ2L2 (D2 ) 2 ψ2L2 (D2 ) .
k q
k |q|
k
It is known that the operator Si is self-adjoint and positive coercive, i.e. there exists c > 0 with
φ, Si φ cφ2H −1/2 (∂D1 )
(3.16)
−1/2
for all φ ∈ H
(∂D1 ) (see lemma 1.14 in [11]) and this implies that, for those contrasts q
c > 0 with
satisfying (3.4), the operator M0 is negative coercive, i.e. there exists φ
φ
−
c φ2H −1/2 (∂D1 ) + ψ2L2 (D2 )
, M0
ψ
ψ
for all φ ∈ H −1/2 (∂D1 ) and ψ ∈ L2 (D2 ). The middle operator M is injective under the
condition that k2 is not a Dirichlet eigenvalue of − in D1 . Indeed, assume that
φ
0
M
=
ψ
0
for some φ ∈ H −1/2 (∂D1 ) and some ψ ∈ L2 (D2 ) which means
ψ (y)(x, y) |q(y)| dy − Sφ = 0 on ∂D1 ,
f1 = −
D2
f2 =
|q|
ψ − |q|v = 0
2
k q
∂D1
in D2 .
(3.17)
(3.18)
The uniqueness result of the direct problem (see theorem 2.2) implies that v = 0 in Rn \D1
and therefore ψ = 0 in D2 and Sφ = 0 on ∂D1 . The assumption that k2 is not a Dirichlet
eigenvalue of − in D1 implies S is an isomorphism (see lemma 1.14 in [11]) and thus φ = 0
on ∂D1 .
We summarize these results in the following theorem.
10
Inverse Problems 29 (2013) 065005
A Kirsch and X Liu
Theorem 3.3. Let M : H − 2 (∂D1 ) × L2 (D2 ) → H 2 (∂D1 ) × L2 (D2 ) be defined as in the
previous theorem.
1
1
(a) In addition to assumption 2.1, let q ∈ L∞ satisfies (3.4). Then −(M0 ) is positive coercive.
(b) Assume that k2 is not a Dirichlet eigenvalue of − in D1 . Then M is injective.
For the further investigation, we recall the auxiliary self-adjoint positive operator
F : L2 (Sn−1 ) → L2 (Sn−1 ) which is given by
F := |(F )| + |(F )|.
(3.19)
Combining theorems 3.2 and 3.3, with the help of the well-known range identity theorem
(see e.g. theorem 2.15 in [11] and theorem 2.1 in [13]), we obtain the first main result of this
paper.
Theorem 3.4. In addition to assumption 2.1, we further assume that q satisfies (3.4) and k2 is
not a Dirichlet eigenvalue of − in D1 . For any z ∈ Rn , define again φz ∈ L2 (Sn−1 ) by (3.3),
i.e.
ˆ = e−ikz·xˆ ,
φz (x)
Then,
xˆ ∈ Sn−1 .
z ∈ D ⇐⇒ φz ∈ R F1/2
and consequently
z ∈ D ⇐⇒
∞
|φz , ψ j L2 (Sn−1 ) |2
j=1
|λ j |
< ∞,
(3.20)
where (λ j , ψ j ) is the eigensystem of the operator F : L2 (Sn−1 ) → L2 (Sn−1 ) given by (3.19).
In other words, the sign of the function
⎡
⎤−1
∞
|φz , ψ j L2 (Sn−1 ) |2
⎦
(3.21)
W (z) = ⎣
|λ j |
j=1
is just the characteristic function of D.
3.2. A modified factorization method for q satisfies (3.5)
From the decomposition M = M0 + M1 and the facts that M0 is a bounded and invertible
operator under the condition that k2 is not a Dirichlet eigenvalue of − in D1 and M1 is
compact, we conclude that M is a Fredholm operator of index zero. Thus the injective property
of M implies that it is an isomorphism. Eliminating G in (3.11) by using (3.15) yields the
following factorization:
F = H ∗ M −1 H.
The middle operator M
M
−1
−1
1
2
: H (∂D1 ) × L (D2 ) → H
f , f > 0,
−1
(3.22)
2
− 12
(∂D1 ) × L (D2 ) satisfies
2
for any f = ( f1 , f2 ) ∈ R(H ) with f = 0.
T
(3.23)
Indeed, let now M f , f = 0 for some f = ( f1 , f2 ) ∈ R(H ). We will show that f = 0.
From the proof of theorem 3.2, we conclude that w ∞ = 0. Then Rellich’s lemma and the
unique continuation principle imply that w = 0 outside D. Therefore, f1 = 0 on ∂D1 by the
trace theorem. Since f = ( f1 , f2 )T ∈ R(H ), there exists g j ∈ L2 (Sn−1 ), such that Hg j → f
1
1
in H 2 (∂D1 ) × L2 (D2 ). Thus, f1 = 0 implies H1 g j → 0 in H 2 (∂D1 ). Note that k2 is not a
T
11
Inverse Problems 29 (2013) 065005
A Kirsch and X Liu
Dirichlet eigenvalue of − in D1 ; we conclude that g j → 0 in L2 (Sn−1 ) and therefore f2 = 0
in D2 .
Difficulties arise immediately if q does not satisfy (3.4). Indeed, the real part M0 of M0
fails to be coercive for such a case. Numerical tests indicate that the factorization method still
works for this case, but a rigorous proof is not known. However, motivated by the ideas from
[4], we will modify the operator F appropriately to treat the case of scatterers with a priori
separated impenetrable and penetrable parts. For the further analysis, we make the following
assumptions on D and q.
Assumption 3.5. We assume that we a priori know the open and bounded domains 1 and 2
with the Lipschitz boundaries 1 and 2 , respectively, such that
D1 ⊂ 1
D2 ⊂ 2 .
and
Here, the closures of the domains 1 and 2 are disjoint and their complements are connected.
2 is also chosen appropriately such that k2 is not a Dirichlet eigenvalue of − in 2 .
Furthermore, assume that q satisfies (3.5), that is, their exists t ∈ (π , 2π ) and c > 0, such
that
it e q
c
|q|
almost everywhere in D2 .
2 : L2 (Sn−1 ) → H 2 (2 ) which denotes
For further investigation, we define the operators H
the Herglotz operator on 2 , i.e.
2 g)(x) =
(H
eikx·d g(d) ds(d), x ∈ 2 ,
1
Sn−1
1
1
and S2 : H − 2 (2 ) → H 2 (2 ) which denotes the single-layer boundary operator on 2 , i.e.
(S2 ψ )(x) =
ψ (y)(x, y) ds(y), x ∈ 2 ,
2
with kernel (·, y) being the fundamental solution (3.12). Furthermore, we define
1 :L2 (Sn−1 ) → L2 (1 ) by
H
(H1 g)(x) =
eikx·d g(d) ds(d), x ∈ 1
Sn−1
1
and T1 : L2 (1 ) → L2 (1 ) by T1 f = k2 i( f + v|1 ), where v ∈ Hloc
(Rn ) is the radiating
solution of
v + k2 (1 + q1 )v = −k2 q1 f
in Rn
(3.24)
with
i, in 1 ;
q1 =
0, in Rn \1 .
For any positive parameter ρ, we introduce two modified operators Fo and Fm by
2 ,
1 .
2 ∗ S2 −1 H
1 ∗ T1 H
Fo = F − ρ H
Fm = F + ρ H
∗
−1
(3.25)
(3.26)
2 S2 H
2 is the factorization of the far-field operator corresponding
We want to remark that −H
to the inverse scattering of plane wave by a sound-soft obstacle 2 (see remark in page 19 of
1 is the factorization of the far-field operator corresponding to the inverse
1 ∗ T1 H
[11]) and H
scattering of plane wave by an inhomogeneous medium with contrast function q1 defined by
(3.25) (see theorem 4.5 in [11]).
1
1
Let us first collect some properties of the operators S2 : H − 2 (2 ) → H 2 (2 ) and
T1 : L2 (1 ) → L2 (1 ) which will be used in further arguments.
12
Inverse Problems 29 (2013) 065005
A Kirsch and X Liu
Lemma 3.6. Assume that k2 is not a Dirichlet eigenvalue of − in 2 . Then the following
results hold.
1
1
(a) S2 is an isomorphism from H − 2 (2 ) onto H 2 (2 ).
1
(b) For all φ ∈ H − 2 (2 ) with φ = 0, we have
(3.27)
φ, S2 φ < 0.
(c) Denote by Si2 the single-layer boundary operator corresponding to the wave number
1
1
k = i. Then it is a self-adjoint and coercive operator from H − 2 (2 ) onto H 2 (2 ), i.e.
their exists c2 > 0 with
1
2
for all ψ ∈ H − 2 (2 ).
(3.28)
ψ, S
1
i2 ψ c2 ψ
H − 2 (2 )
−
(d) The difference S2 − S
i2 is compact from H 2 (2 ) into H 2 (2 ).
2
(e) For all f ∈ L (1 ), with f = 0,
(T1 f , f )L2 ( ) > 0.
1
1
(3.29)
1
2
2
(f) T1 can be written in the form of T1 = T
10 + T11 , where T10 = k i f for f ∈ L (1 ) and
2
2
it T11 : L (1 ) → L (1 ) is compact. The operator [e T10 ] is coercive with t ∈ (π , 2π )
chosen such that (3.5) holds.
Proof. The results in (a)–(c) are exactly from lemma 1.14 in [11]. Here, we just prove (e) and
(f).
(e) By literally the same proof as in theorem 4.8 from [11], one shows
2
2
2
| f + v| dx + k|γn |
|v ∞ |2 ds 0,
(T1 f , f )L2 (1 ) = k
1
Sn−1
1
where v ∈ Hloc
(Rn ) is the radiating solution of (3.24). Let now (T1 f , f )L2 (1 ) = 0 for some
2
f ∈ L (1 ). Then, v ∞ = 0 and f = −v in 1 . Rellich’s lemma and the unique continuation
principle imply that v = 0 in Rn \1 and thus v = ∂v/∂ν = 0 on ∂1 . The fact f = −v in
1 implies that v is the solution of the Helmholtz equation in 1 . Therefore, by Holmgren’s
uniqueness theorem, we conclude that v = 0 in 1 and thus f = 0.
(f) The operator [eit T
10 ] is clearly positive coercive because t ∈ (π , 2π ) and
2
2
for all f ∈ L2 (1 ).
(3.30)
[eit (T
10 f , f )L2 ( ) ] = − sin(t )k f 2
L (1 )
1
2
2
1
The compactness of T
11 : L (1 ) → L (1 ) follows from the compact imbedding of H (1 )
2
into L (1 ).
Then we have the following result.
Theorem 3.7. In addition to assumptions 2.1 and 3.5, we assume that k2 is not a Dirichlet
eigenvalue of − in D1 .
(a) Then the operators Fo and Fm from (3.26) have the following factorizations:
∗
∗
1
1
H1
H1
H
H
Fo = Mo ,
Fm =
Mm
H2
H2
H2
H2
(3.31)
with some Fredholm operators Mo : H 2 (∂D1 ) × H 2 (2 ) → H − 2 (∂D1 ) × H − 2 (2 ) and
Mm : L2 (1 ) × L2 (D2 ) → L2 (1 ) × L2 (D2 ). Here, Mo and Mm have the following forms:
ρk2 i 0
−Si−1
0
Mm =
(3.32)
Mo =
−1 + M1o,
k2 q + M1m
0
0
−ρ S
i2
|q|
1
1
1
1
with some compact operators M1o and M1m .
13
Inverse Problems 29 (2013) 065005
A Kirsch and X Liu
(b) The operators (Mo ) and (Mm ) are non-negative, i.e. Moϕ, ϕ 0 for all
1
1
ϕ ∈ H 2 (∂D1 ) × H 2 (2 ) and Mm χ , χ 0 for all χ ∈ L2 (1 ) × L2 (D2 ).
2 )T ) of R((H1 , H
2 )T ), i.e. Moϕ, ϕ >
(c) (Mo ) is strictly positive on the closure R((H1 , H
T
0 for all ϕ ∈ R((H1 , H2 ) ) with ϕ = 0. (Mm ) is strictly positive on the closure
1 , H2 )T ) of R((H
1 , H2 )T ), i.e. Mm χ , χ > 0 for all χ ∈ R((H
1 , H2 )T ) with
R((H
χ = 0.
(d) The real parts (Mo ) and (eit Mm ) have decompositions in the form
(Mo ) = Co + Ko
and
(eit Mm ) = Cm + Km
with self-adjoint and coercive operators Co and Cm and compact operators Ko and Km . Here,
t is the parameter of assumption 3.5.
Proof.
2 , where R2 : H 1/2 (2 ) → L2 (D2 ) is defined by
(a) We first note that H2 = R2 H
(R2 f )(x) = |q(x)|v(x), x ∈ D2 ,
(3.33)
where v ∈ H (2 ) solves the interior Dirichlet boundary value problem
1
v + k2 v = 0 in 2 ,
v = f on 2 .
The operator R2 is compact because H (D2 ) is compactly imbedded in L2 (D2 ).
We write the factorization (3.22) in the form
∗ H1
H1
I 0
0
−1 I
M
F= 2
0 R∗2
0 R2
H2
H
1
and thus
∗ 0
0
H1
H
I 0
0
−1 I
M
−
Fo = 1
−1
2 .
0 R∗2
0 R2
H2
H
0 ρ S2
Using (3.8), the operator M −1 can be written as
−Si−1 0
−Si−1
−1
−1
2
M =
M
−
M
1
k q
0
0
|q|
Substituting this into (3.34) yields
∗ −1
−Si
H
Fo = 1
H2
0
0
k2 q
|q|
(3.34)
.
0
H1
−1 + M1o
2 ,
H
−ρ S
i2
(3.35)
(3.36)
where we have collected all compact parts in the operator M1o. This proves the factorization
1
1
1
1
(3.26) for Fo with the middle operator Mo : H 2 (∂D1 )×H 2 (2 ) → H − 2 (∂D1 )×H − 2 (2 )
in the form
0
0
0
I 0
−1 I
M
−
(3.37)
Mo =
−1 .
0 R2
0 R∗2
0 ρ S2
For Fm , we proceed completely analogously. It is H1 = R1 H1 , where R1 : L2 (1 ) →
1
(Rn ) is the radiating solution of
H (∂D1 ) is given by R1 f = −v|∂D1 , where v ∈ Hloc
(3.24). Also, the operator R1 is compact because of interior regularity results. Then we
follow the lines of the proof for Fo. We write the factorization (3.22) in the form
∗ ∗
1
1
R1 0
R1 0
H
H
M −1
F=
0 I
0 I
H2
H2
1/2
14
Inverse Problems 29 (2013) 065005
and thus
∗ ∗
1
R1
H
Fm =
0
H2
0
R1
M −1
I
0
Substituting (3.35) into (3.38) yields
∗ 2
1
ρk i
H
Fm =
H2
0
A Kirsch and X Liu
0
ρ T1
+
I
0
0
k2 q
|q|
1
0
H
.
H2
0
(3.38)
1
H
,
+ M1m
H2
(3.39)
where we have collected all compact parts in the operator M1m . This proves the
factorization (3.26) for Fm with the middle operator Mm : L2 (1 ) × L2 (D2 ) →
L2 (1 ) × L2 (D2 ) in the form
∗
R1 0
R1 0
ρ T1 0
M −1
+
Mm =
.
(3.40)
0 I
0 I
0 0
1
1
) is non-negative. For any ϕ = (ϕ1 , ϕ2 )T ∈ H 2 (∂D1 ) × H 2 (2 )
(b) We first prove that
(Mo
I 0
ϕ. Then,
define ϕ
= M −1
0 R2
"
! −1
Moϕ, ϕ = ϕ , M
ϕ − ρ S2 ϕ2 , ϕ2 0,
where we have used (3.27) and the fact that (M) is non-negativeby theorem
3.2.
0
T
2
2
−1 R1
For any χ = (χ1 , χ2 ) ∈ L (1 ) × L (D2 ) define χ
= M
χ . Then,
0 I
Mm χ , χ = χ , M
χ + ρT1 χ1 , χ1 0,
where we have used (3.29) and the fact that (M) is non-negative by theorem 3.2.
(c) We only prove the property of (Mo ). For (Mm ), this can be proven analogously. From
1
1
the arguments in (b) we know that (Mo ) is non-negative on H 2 (∂D1 ) × H 2 (2 ). Let
T
2 )T ). We will show that ϕ = 0.
now Moϕ, ϕ = 0 for
someϕ = (ϕ1 , ϕ2 ) ∈ R((H1 , H
I 0
2 = R2 H2 .
ϕ, then ϕ ∗ ∈ R(H ) because of the relation H
Indeed, define ϕ ∗ =
0 R2
Again from the arguments in (b), we conclude that
"
! −1
M −1 ϕ ∗ , ϕ ∗ = S2 ϕ2 , ϕ2 = 0.
From (3.23) and (3.27), we find that ϕ2 = 0, ϕ ∗ = 0 and therefore ϕ1 = 0.
(d) This follows from the assumption (3.5) on q, the decomposition (3.32), the coercive
it properties (3.16) of Si , (3.28) of S
i2 and (3.30) of [e T0 ].
Now, we proceed to characterize the parts of the scatterer by the corresponding data-topattern operators Fo and Fm in (3.26).
Lemma 3.8. In addition to assumptions 2.1 and 3.5, we assume that k2 is not a Dirichlet
eigenvalue of − in D1 . For any z ∈ Rn , define again φz ∈ L2 (Sn−1 ) by (3.3).
(a) A point z ∈ Rn \2 belongs to D1 if and only if the function φz belongs to the range of the
2 ∗ ) : H − 12 (∂D1 ) × H − 12 (2 ) → L2 (Sn−1 ).
operator (H1∗ , H
(b) A point z ∈ Rn \1 belongs to D2 if and only if the function φz belongs to the range of the
1 ∗ , H ∗ ) : L2 (1 ) × L2 (D2 ) → L2 (Sn−1 ).
operator (H
2
15
Inverse Problems 29 (2013) 065005
A Kirsch and X Liu
Proof.
(a) Let first z ∈ D1 . From theorem 1.12 in [11] and the invertibility of the single-layer
1
boundary operator S, we conclude that φz ∈ R(H1∗ ), i.e. there exists φ1 ∈ H − 2 (∂D1 )
∗
2 φ2 , i.e. φz belongs to
such that φz = H1∗ φ1 . Setting φ2 = 0 yields that φz = H1∗ φ1 + H
∗ ∗
the range of the operator (H1 , H2 ).
2 ∗ φ2 for some
Second, let z ∈
/ D1 and assume in contrast that φz = H1∗ φ1 + H
− 12
− 12
φ1 ∈ H (∂D1 ) and φ2 ∈ H (2 ). Noting that both sides are far-field patterns, then by
Rellich’s lemma and the unique continuation, we have
(x, y)φ1 (y) ds(y) +
(x, y)φ2 (y) ds(y), x ∈ Rn \{D1 ∪ 2 ∪ {z}}.
(x, z) =
∂D1
2
This leads to a contradiction because the left-hand side has a singular at x = z, while the
right-hand side is continuous in Rn .
(b) This can be proved analogously with the help of theorem 4.6 (instead of theorem 1.12) in
[11].
Application of theorem 3.1 and combination with theorems 3.7 and 3.8 yield the second
main result of this paper.
Theorem 3.9. In addition to assumptions 2.1 and 3.5, we assume that k2 is not a Dirichlet
eigenvalue of − in D1 . For any z ∈ Rn , define again φz ∈ L2 (Sn−1 ) by (3.3), i.e.
ˆ = e−ikz·xˆ ,
φz (x)
xˆ ∈ Sn−1 .
Then, the following characterization holds.
(a) For any point z ∈
/ 2 , we have
z ∈ D1 ⇐⇒ φz ∈ R Fo1/2 ,
where Fo = |(Fo )| + |(Fo )| and Fo is given in (3.26) and consequently
z ∈ D1 ⇐⇒
∞
|φz , ψ j L2 (Sn−1 ) |2
j=1
|λ j |
< ∞,
(3.41)
where (λ j , ψ j ) is the eigensystem of the operator Fo : L2 (Sn−1 ) → L2 (Sn−1 ).
(b) For any point z ∈
/ 1 , we have
1/2 ,
z ∈ D2 ⇐⇒ φz ∈ R Fm
where Fm = |(eit Fm )| + |(Fm )| and Fm is given in (3.26) and consequently
z ∈ D2 ⇐⇒
∞
|φz , ψ j L2 (Sn−1 ) |2
j=1
|λ j |
< ∞,
(3.42)
where (λ j , ψ j ) is the eigensystem of the operator Fm : L2 (Sn−1 ) → L2 (Sn−1 ).
Finally, we want to remark that theorems 3.4 and 3.9 also provide an explicit and
constructive proof of uniqueness of the inverse scattering problem under certain conditions.
16
Inverse Problems 29 (2013) 065005
A Kirsch and X Liu
8
6
4
2
0
−2
−4
−6
−8
−8
−6
−4
−2
0
2
4
6
8
Figure 1. The original domain.
4. Numerical tests
In this section, we present some numerical examples in two dimensions to illustrate the
applicability and effectiveness of the factorization method. The scatterer under concern in
system (1.1)–(1.4) is given as the union of two disjoint components. The plot of the region
is shown in figure 1. The one in the southwest corner is a sound-soft unit circle with center
(−4, −4), and the other one is a penetrable kite-shaped medium with the boundary described
by the parametric representation
x(t ) = (3.5 + cos(t ) + 0.65 cos(2t ) − 0.65, 3.5 + 1.5 sin(t )),
t ∈ [0, 2π ].
The sampling region G is deliberately chosen as [−8, 8] and a 321 × 321 uniform mesh is
used to cover the sampling region G.
To generate the measured far-field data F and the artificial far-field data Fo and Fm , we used
the boundary integral equation method. For the numerical treatment of the integral equations,
the Nystr¨om method, with 64 quadrature points on each boundary and each interface, has been
applied.
The following pictures show the values of the function:
⎡
⎤−1
64
|φz , ψ j L2 (Sn−1 ) |2
⎦ .
(4.1)
W (z) = ⎣
|λ
|
j
j=1
2
In the first example, q = −0.2 + i in D2 which satisfies relation (3.4). Figure 2 shows the
reconstruction where the function W is computed by using the eigensystem of the operator
F : L2 (Sn−1 ) → L2 (Sn−1 ).
In the second example, q = 0.2 + i in D2 which satisfies relation (3.5) with t = 7π
. We
4
assume that the unit disc is contained in a disc 1 of radius 2.5 with center (−4, −4), while
the ‘kite’ is contained in a disc 2 of radius 2.5 with center (3.5, 3.5). Because we are just
interested in the sampling points located in 1 ∪2 , the values of the function W are set to be 0
17
Inverse Problems 29 (2013) 065005
A Kirsch and X Liu
Figure 2. Reconstruction for the wave number k = 1, contrast function q = −0.2 + i, no noise
(left) and 2% noise (right).
Figure 3. Reconstruction for ρ = 0 (left), ρ = 0.1 (middle), ρ = 1 (right) and no noise.
Figure 4. Reconstruction for ρ = 0 (left), ρ = 0.1 (middle), ρ = 1 (right) and 2% noise.
deliberately for those points lie outside of 1 ∪ 2 . For the sampling points located in 1 , the
function W is computed by using the eigensystem of the operator Fo : L2 (Sn−1 ) → L2 (Sn−1 ),
while for the sampling points located in 2 , it is computed by the eigensystem of the operator
Fm : L2 (Sn−1 ) → L2 (Sn−1 ). Figures 3 and 4 show the reconstruction without noise and with
2% extra noise added to the far-field data F, respectively.
We observe that the reconstruction is independent of the choice of ρ. We also note that
the factorization method works for ρ = 0. However, a rigorous proof for this case is still not
established.
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Inverse Problems 29 (2013) 065005
A Kirsch and X Liu
Acknowledgments
The research of XL was supported by the NNSF of China under grants 11101412 and 11071244
and the Alexander von Humboldt Foundation. The authors would like to thank the referees for
their invaluable comments which helped improve the presentation of the paper.
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