# Precisely, we can reconstruct the obstacle by the eigenvalues and eigenfunctions of the far-eld operator

## Full text

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ANISOTROPIC INHOMOGENEOUS BACKGROUND

PU-ZHAO KOW AND JENN-NAN WANG

Abstract. In this paper, we are interested in the problem of determining the shape of sound-soft or impedance obstacles in the acoustic wave scattering with an anisotropic inhomogeneous medium. The main theme is to extend the factorization method to this setting. Precisely, we can reconstruct the obstacle by the eigenvalues and eigenfunctions of the far-eld operator. We also provide some numerical simulations in the paper.

1. Introduction

Let an obstacle be embedded inside an anisotropic inhomogeneous background. This system causes the acoustic wave to scatter. In this work, we are interested in reconstructing the shape of the obstacle by the far-eld information assuming that the inhomogeneity is known. There are several existing methods for other types of inverse scattering problems. Here we will focus on the factorization method, which was rst introduced by Kirsch in [Ki98, Ki99] to treat the inverse scattering problems for the scalar Helmholtz equation. For the detailed development of the factorization method for the Helmholtz and time-harmonic Maxwell equations, we refer the reader to the monograph [KG08].

The factorization method shares the same spirit as the linear sampling method, where both methods try to reconstruct the shape of the scatterer (an obstacle or the support of the inhomogeneity) by determining whether a point is inside or outside the scatterer.

We now briey describe the problem considered in the paper. Let A(x) = (aij(x))be a real-symmetric matrix with C entries, which satises the following uniform elliptic condition: there exists a constant 0 < c < 1 such that

c|ξ|2 ≤X

ij

aij(x)ξiξj ≤ c−1|ξ|2

for all x ∈ Rd, ξ ∈ Rd, d = 2 or 3. Moreover, we assume that supp(A − I) is compact in Rd. Suppose that the acoustic refraction index n ∈ L(Rd) with n ≥ c, such that supp(n − 1) is compact in Rd. Let BR be the ball centered at origin with radius R > 0 which contains supp(A − I) and supp(n − 1). Let D be an open bounded domain where ∂D is Lipschitz, D ⊂ BR, and Rd\ D is connected. Let utoD(x, ˆz) = uincA,n(x, ˆz) + uscD(x, ˆz)with an appropriate incident eld uincA,n(x, ˆz)and the corresponding scattered eld uscD(x, ˆz), ˆz = z/|z|, satisfy the following acoustic equation





∇ · (A(x)∇utoD) + k2n(x)utoD = 0 in Rd\ D,

ButoD = 0 on ∂D,

uscD satises the Sommerfeld radiation condition at |x| → ∞,

(1.1)

1

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where ButoD is either Dirichlet or impedance boundary conditions. Let uD(ˆx, ˆz) be the far-eld pattern of uscD. The inverse obstacle problem is to determine D from the knowledge of uD(ˆx, ˆz)for all ˆx, ˆz ∈ Sd−1.

This type of inverse obstacle problem including theoretical and numerical investigations has been studied extensively. We will not try to exhaust all the related works here.

Instead, we refer to recent monographs [CCH16, KG08] and references therein for the detailed development of the problems and reconstruction methods. To put our problem in perspective, we mention several closely related results.

• Penetrable obstacle in an inhomogeneous background. In [KP98], the authors considered the acoustic scattering from a penetrable obstacle inside an inhomogeneous structure. In their work, they showed that, if the reference medium is known, then the penetrable obstacle can be uniquely determined by the far-eld data at a xed energy. The approach used in [KP98] was only for the uniqueness proof. Later, in [GMMR12], the authors designed a reconstruction algorithm in the spirit of the factorization method. For the case of anisotropic penetrable obstacle embedded inside of a homogeneous medium, a factorization method was developed in [KL13]. On the other hand, when there was an unknown cavity hidden inside of the penetrable obstacle, a reconstruction result based on the factorization method was investigated in [YZZ13].

• Impenetrable obstacle in a homogeneous background. A factorization method for reconstructing an impenetrable obstacle in a homogeneous medium (Helmholtz equation) using the spectral data of the far-eld operator was developed in [Ki98]

• Impenetrable obstacle in an inhomogeneous background. When the inhomogeneous background was described the Schrödinger equation with an inhomogeneous potential function, the theoretical study of the factorization method was carried out in [NPT07]. Moreover, the obstacle can be reconstructed without knowing the boundary condition on the boundary of the obstacle (either Dirichlet or Robin boundary conditions). Our work is closely related to this paper. In our paper, we consider an impenetrable obstacle in an anisotropic inhomogeneous background.

We would like to remark that, in [KG08, KP98], the far-eld operator is dened in terms of the far-eld pattern induced by the incident plane waves. In our case, we take a dierent incident eld. Roughly speaking, the incident eld used in our approach is the far-eld pattern of the outgoing fundamental solution for the background equation.

This idea is inspired by the work [GMMR12]. The main dierence between our work and [GMMR12] is that they consider penetrable obstacles (involving volume potentials);

while in our case, we consider impenetrable obstacles (involving surface potentials).

Indeed, in the case where the reference medium is homogeneous, the far-eld pattern of the standard outgoing fundamental solution is simply the plane waves. Since we assume that the background medium is known, the far-eld pattern of the outgoing fundamental solution for the background equation can be determined. However, from the numerics perspective, the computation of the far-eld of the outgoing fundamental solution directly is not ecient. Luckily, thanks to the mixed reciprocity relation (see Lemma 2.2), we can compute the far-eld of the outgoing fundamental solution by solving the total eld of the background equation with the incident plane waves.

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Another focal point of this work is the numerical demonstration of increasing stability phenomenon with respect to the wave number k in the reconstruction of the obstacle. It is known that most of inverse problems are ill-posed including the problem we discuss here. However, in the case of identifying the potential in the Schrödinger equation by the Dirichlet-to-Neumann map, the stability increases as we increase the wave number (the logarithmic part decreases as the wave number increases) [INUW14]. A similar property was also established for the determination of the refraction index in the acoustic equation [NUW13]. We need to point out a major dierence between [INUW14]

and [NUW13]. In [INUW14], the constant appears in the Hölder part of the stability estimate depends polynomially in k. Hence, the stability of determining the potential in the Schrödinger equation tends a Hölder type as k increases. While, in [NUW13], the constant in the Hölder part of the stability estimate depends exponentially in k.

Consequently, the stability estimate in the acoustic case remains a logarithmic type when k is too large.

Considering the discussion above, even though the inverse obstacle problems for the Schrödinger equation and for the acoustic equation at a xed wave number are mathematically equivalent, it is necessary in practice to study the acoustic equation separately at least from the viewpoint of stability. We believe that it is an important and challenging problem to solve the inverse obstacle problem in the acoustic equation stably at high wave numbers.

The paper is organized as follows. In Section 2, we rst review some properties of the outgoing fundamental solution for the background equation and its far-eld pattern.

In Section 3, we prove our main reconstruction theorem for the sound-soft (Dirichlet) obstacle. The same ideas can be generalized to the obstacle with Neumann or impedance boundary conditions. Thus, in Section4, we state the reconstruction theorems without proofs corresponding to the cases of Neumann and impedance boundary conditions.

In Section 5, some numerical simulations are presented to show the eciency of our method. Finally, in Section 6, we compare the reconstruction results for the acoustic and the Schrödinger equations at dierent wave numbers to demonstrate the increasing stability phenomena.

2. Fundamental solution and Herglotz wave functions

In this section, we discuss some properties of the outgoing fundamental solution to the background equation (that is, the reference medium).

Denition 2.1. We say that a function u ∈ Hloc1 (Rd)satises the Sommerfeld radiation condition, if

ru(x) − iku(x) = o(|x|d−12 ) as |x| → ∞, where ∂r= ˆx · ∇.

For each x ∈ Rd, let ΦA,n(z, x) (for x 6= z) be the outgoing fundamental solution of the following anisotropic inhomogeneous acoustic equation

(∇z· (A(z)∇zΦA,n(z, x)) + k2n(z)ΦA,n(z, x) = δ(z − x) ∀ z ∈ Rd, ΦA,n(z, x) satises the Sommerfeld radiation condition at |z| → ∞,

see also [DHM18] where the existence of the fundamental solution, even for elliptic systems, was proved. Let uincref(x, ˆz) = eikx·ˆz be an incident plane wave. The presence

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of inhomogeneity (without the obstacle) in BR gives rise to a unique scattered eld uscref ∈ Hloc1 (Rd)which satises the Sommerfeld radiation condition. Then the total eld utoref = uscref + uincref satises

∇ · (A(x)∇utoref) + k2n(x)utoref = 0, ∀ x ∈ Rd. (2.1) We rst prove the following mixed reciprocity principle.

Lemma 2.2. Let ΦA,n(ˆz, x) be the far-eld pattern of ΦA,n(z, x). Then we have ΦA,n(ˆz, x) = utoref(x, −ˆz) ∀ x ∈ Rd, ˆz ∈ Sd−1,

where utoref(·, −ˆz) is given in (2.1).

Proof. Here we adopt the proof from [KG08]. Let Φ(z, x) = ΦI,1(z, x) be the outgoing fundamental solution of the standard Helmholtz equation. Dene

Ψ(z, x) := ΦA,n(z, x) − Φ(z, x).

Then Ψ(z, x) is a smooth at all z ∈ Rd and

zΨ(z, x) + k2Ψ(z, x)

= −(∇z· ((A(z) − I)∇zΦA,n(z, x)) + k2(n(z) − 1)ΦA,n(z, x)).

We can write Ψ(z, x)

= − Z

Rd

Φ(z, y)(∇y · ((A(y) − I)∇yΦA,n(y, x)) + k2(n(y) − 1)ΦA,n(y, x)) dy.

Since both A − I and n − 1 have compact supports, the integral equation above is well- dened. Note that the far-eld pattern of Φ(z, y) is Φ(ˆz, y) = e−ikˆz·y = uincref(y, −ˆz). We then see that the far-eld pattern of Ψ(z, x) is

Ψ(ˆz, x)

= − Z

Rd

e−ikˆz·y(∇y · ((A(y) − I)∇yΦA,n(y, x)) + k2(n(y) − 1)ΦA,n(y, x)) dy. (2.2) Since utoref(x, −ˆz) satises

x· (A(x)∇xutoref(x, −ˆz)) + k2n(x)utoref(x, −ˆz) = 0 for all x ∈ Rd, we have

x· (A(x)∇xuscref(x, −ˆz)) + k2n(x)uscref(x, −ˆz)

= − (∇x· (A(x)∇xuincref(x, −ˆz)) + k2n(x)uincref(x, −ˆz))

= − (∇x· ((A(x) − I)∇xuincref(x, −ˆz)) + k2(n(x) − 1)uincref(x, −ˆz)).

In view of the symmetry property ΦA,n(x, y) = ΦA,n(y, x), using integration by parts, we can derive

uscref(x, −ˆz)

= − Z

Rd

ΦA,n(y, x)(∇y· ((A(y) − I)∇yuincref(y, −ˆz)) + k2(n(y) − 1)uincref(y, −ˆz)) dy

= − Z

Rd

e−ikˆz·y(∇y · ((A(y) − I)∇yΦA,n(y, x)) + k2(n(y) − 1)ΦA,n(y, x)) dy. (2.3)

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Combining (2.2) and (2.3) implies that

Ψ(ˆz, x) = uscref(x, −ˆz), and hence

ΦA,n(ˆz, x) = Ψ(ˆz, x) + Φ(ˆz, x) = uscref(x, −ˆz) + uincref(x, −ˆz) = utoref(x, −ˆz),

which is our desired lemma. 

Denition 2.3. For each g ∈ L2(Sd−1), we dene the modied Herglotz wave function as

uA,n,g(x) :=

Z

Sd−1

ΦA,n(ˆz, x)g(ˆz) ds(ˆz) for all x ∈ Rd. It is well-known that

g 7→

Z

Sd−1

eikx·ˆzg(ˆz) ds(ˆz) = Z

Sd−1

uincref(x, ˆz)g(ˆz) ds(ˆz)

is injective. Since each uincref(x, ˆz) leads a unique scattered eld uscref(x, ˆz), and thus induces a unique total eld utoref(x, ˆz). Combining these discussions and by the superposition property, we can prove the following lemma.

Lemma 2.4. For coecients A and refraction index n prescribed above, the mapping g 7→ uA,n,g is injective.

3. Reconstruction of sound-soft obstacle

By Lemma2.2, we can choose the incident eld uincA,n(x, ˆz) = ΦA,n(ˆz, x)instead of the usual plane wave uincref(x, ˆz) = eikx·ˆz, also see [GMMR12] in which the similar incident

eld was used. In practice, this can be achieved by using a two-stage detection as in Algorithm 1. By choosing this incident eld, no scattering occurs if there is no obstacle. Now let uscDir(x, ˆz) be the corresponding scattered eld, then the total eld utoDir(x, ˆz) = uincA,n(x, ˆz) + uscDir(x, ˆz)satises





∇ · (A(x)∇utoDir) + k2n(x)utoDir = 0 in Rd\ D,

utoDir = 0 on ∂D,

uscDir satises Sommerfeld radiation condition at |x| → ∞,

(3.1)

that is,





∇ · (A(x)∇uscDir) + k2n(x)uscDir = 0 in Rd\ D,

uscDir = −uincA,n on ∂D,

uscDir satises Sommerfeld radiation condition at |x| → ∞.

(3.2)

Denote uDir be the far-eld pattern of uscDir, which satises uscDir(x) = γd eik|x|

|x|d−12 uDir(ˆx) + O(|x|d+12 ) as |x| → ∞ uniformly for all directions ˆx = x/|x|, where

γd :=

(eiπ/4/√

8πk if d = 2,

1/4π if d = 3, (3.3)

see e.g. [GMMR12].

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The well-posedness of uscD is guaranteed by the following well-known fact:

Lemma 3.1. Given any f ∈ H1/2(∂D), there exists a unique v ∈ Hloc1 (Rd\ D) such

that 





∇ · (A(x)∇v) + k2n(x)v = 0 in Rd\ D,

v = f on ∂D,

v satises Sommerfeld radiation condition at |x| → ∞.

(3.4)

3.1. The factorization of the far-eld operator. For each g ∈ L2(Sd−1), recall the modied Herglotz wave function is dened by

uA,n,g(x) :=

Z

Sd−1

ΦA,n(ˆz, x)g(ˆz) ds(ˆz)

= Z

Sd−1

uincA,n(x, ˆz)g(ˆz) ds(ˆz), (3.5) which is simply the superposition of the incident elds. So, the far-eld of the scattered

eld is the superposition of uDir. Thus, we can dene the far-eld operator as follows:

Denition 3.2. The far-eld operator FDir: L2(Sd−1) → L2(Sd−1) is given by FDir(g)(ˆx) =

Z

Sd−1

uDir(ˆx, ˆz)g(ˆz) ds(ˆz) for all ˆx ∈ Sd−1.

With Lemma 3.1, we also dene an operator closely related to the far-eld operator:

Denition 3.3. The data-to-pattern operator GDir : H1/2(∂D) → L2(Sd−1) is dened by GDirf = v, where v is the far-eld of v ∈ Hloc1 (Rd\ D) which satises (3.4).

To discuss the relation between the far-eld operator and the data-to-pattern operator, we introduce the single-layer potential operator S dened by

(Sφ)(x) :=

Z

∂D

ΦA,n(x, z)φ(z) ds(z) for all x ∈ ∂D.

We rst establish the following mapping property of S.

Lemma 3.4. The potential S : H−1/2(∂D) → H1/2(∂D) is continuous.

Proof. Let ΦA,1(x, z) be the outgoing fundamental solution of

(∇x· (A(x)∇xΦA,1(x, z)) + k2ΦA,1(x, z) = δ(x − z) ∀ x ∈ Rd, ΦA,1(x, z) satises the Sommerfeld radiation condition at |x| → ∞.

Since the coecient A is smooth, ΦA,1(x, z) is C for x 6= z. By the techniques of pseudodierential operators (see [McL00, Theorem 6.11]), let

( ˜Sφ)(x) :=

Z

∂D

ΦA,1(x, z)φ(z) ds(z) for all x ∈ ∂D, then ˜S : H−1/2(∂D) → H1/2(∂D) is continuous. Note that

x· (A(x)∇xA,n− ΦA,1)) + k2n(x)(ΦA,n− ΦA,1) = k2(1 − n)ΦA,1.

Observe that k2(1−n)ΦA,1 ∈ L2(Rd)(uniformly in z). By the elliptic estimate theorem, we have that

A,n− ΦA,1)(·, z) ∈ H2,2(D).

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From the trace theorem and the symmetry of the fundamental solutions, it follows that

|∇xA,n− ΦA,1)(x, z)| = |∇zA,n− ΦA,1)(x, z)| ∈ H1/2(∂D) uniformly for x ∈ ∂D. Therefore, we can show that the operator

S0 : φ 7→

Z

∂D

A,n− ΦA,1)(x, z)φ(z)ds(z) (3.6) maps H−1/2(∂D)to H1(∂D), in particular, S0 : H−1/2(∂D) → H1/2(∂D)is continuous.

Since S = ˜S + S0, Lemma 3.4 follows.

 Following the ideas in [KG08, Theorem 1.15], we can obtain the following proposition, which contains the main idea of this work.

Proposition 3.5. Let GDir : L2(Sd−1) → H−1/2(∂D) be the adjoint of GDir and S : H−1/2(∂D) → H1/2(∂D) be the adjoint of S. Then

FDir = −GDirSGDir. (3.7) Proof. We consider an auxiliary operator HDir : L2(Sd−1) → H1/2(∂D) dened as the restriction of the modied Herglotz wave function on ∂D, that is,

(HDirg)(x) :=

Z

Sd−1

ΦA,n(ˆz, x)g(ˆz) ds(ˆz) for all x ∈ ∂D.

From (3.2), we know that

FDir = −GDirHDir. (3.8)

Also, it is not dicult to check that the adjoint HDir : H−1/2(∂D) → L2(Sd−1) of HDir

is given by

(HDir φ)(ˆz) = Z

∂D

ΦA,n(ˆz, x)φ(x) dx

for all φ ∈ H−1/2(∂D). Observe that (HDir φ)(ˆz) is nothing but the far-eld pattern of v(z) =

Z

∂D

ΦA,n(z, x)φ(x) dx, which shows that HDir = GDirS, that is,

HDir = SGDir. (3.9)

Putting together (3.8) and (3.9) implies (3.7). 

Similar to [KG08, Theorem 1.12], we can show that

Proposition 3.6. ΦA,n(·, z) ∈ range(GDir) if and only if z ∈ D.

Proof. First of all, we assume that ΦA,n(·, z) ∈ range(GDir), i.e., there exists f ∈ H1/2(∂D) such that GDirf = ΦA,n(·, z). Let w ∈ Hloc1 (Rd\ D)be the unique solution to (3.4). Since both

wand ΦA,n(·, z) have the same far-eld pattern ΦA,n(·, z), by Rellich's lemma, there exists r0 > 0 such that

w(x) = ΦA,n(x, z) for |x| > r0.

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Since Rd\ D is connected, and both w and ΦA,n satisfy the anisotropic inhomogeneous acoustic equation (2.1) in Rd\ D, by the unique continuation property of this equation, we have that

w(x) = ΦA,n(x, z) for all x ∈ Rd\ D.

Note that w ∈ Hloc1 (Rd\ D). In other words, w has no singularity in Rd\ D and thus z ∈ D.

Conversely, let z ∈ D. Dene w = ΦA,n(·, z) in Rd\ D and f = w|∂D, we can easily see that GDirf = ΦA,n(·, z), that is, ΦA,n(·, z) ∈ range(GDir).  With Proposition 3.5 and 3.6 at hand, it is helpful to study the operators S, GDir, and FDir.

3.2. The properties of the single-layer potential S. In this section, we shall study the properties of the far-eld operator. With the help of [McL00, Theorem 7.5], we can further improve Lemma3.4.

Lemma 3.7. If k2 is not an eigenvalue of

(∇ · (A(x)∇u) + k2n(x)u = 0 in D,

u = 0 on ∂D, (3.10)

then S : H−1/2(∂D) → H1/2(∂D) is an isomorphism.

Next, by mimicking the proofs of [KG08, Lemma 1.14(b),(c),(d)], we can obtain the following lemmas. Since we are considering a dierent equation, we include the proofs here for the sake of completeness.

Lemma 3.8. If k2 is not an eigenvalue of (3.10), then =hφ, Sφi ≤ 0 for all φ ∈ H1/2(∂D), where h·, ·i is the H−1/2(∂D) × H1/2(∂D) duality pair and = denotes the imaginary part of a complex number. Moreover,

=hφ, Sφi = 0 if and only if φ ≡ 0.

Proof. For each φ ∈ H−1/2(∂D), dene v(x) = (SL φ)(x) =

Z

∂D

ΦA,n(x, y)φ(y) ds(y) for x ∈ Rd\ ∂D.

Then, by [McL00, Theorem 6.11], we have v ∈ H1(D) ∩ Hloc1 (Rd\ D) and the traces on

∂D with the outer normal normal ν(x), v±(x) = lim

h→0+

v(x ± hν(x)), ν(x) · A(x)∇v±(x) = lim

h→0+

ν(x) · A(x ± hν(x))∇v(x ± hν(x)), exist and satisfy the jump condition

v = v± = Sφ for all φ ∈ H−1/2(∂D).

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Choose φ = ν · A∇v− ν · A∇v+, then hφ, Sφi = hν · A∇v− ν · A∇v+, vi

= Z

∂D

(ν · A∇v)v ds − Z

∂D

(ν · A∇v+)v ds

= Z

∂D

(ν · A∇v)v ds + Z

∂(BR\D)

(ν · A∇v+)v ds − Z

|x|=R

(ν · ∇v)v ds

= Z

D∪(BR\D)

(∇ · (A∇v))v dx + Z

D∪(BR\D)

∇¯v · A∇v dx − Z

|x|=R

(∂rv)v ds.

Since ∇ · (A∇v) + k2n(x)v = 0 in Rd\ ∂D, together with the Sommerfeld radiation condition, then

hφ, Sφi = Z

D∪(BR\D)

(∇¯v · A∇v − k2n(x)|v|2) dx − Z

|x|=R

(∂rv)v ds

= Z

D∪(BR\D)

(∇¯v · A∇v − k2n(x)|v|2) dx − ik Z

|x|=R

|v|2ds + o(Rd−12 )

(3.11)

as R → ∞. Then

=hφ, Sφi = −k Z

|x|=R

|v|2ds + o(Rd−12 ), and thus

=hφ, Sφi = −k lim

R→∞

Z

|x|=R

|v|2ds = −k lim

R→∞

Z

Sd−1

|R(d−1)/2v|2ds(ˆx)

= −|γd|2k Z

Sd−1

|v|2ds(ˆx) ≤ 0,

(3.12)

where γd is given in (3.3).

Finally, we want to show that =hφ, Sφi = 0 leads to φ ≡ 0. If =hφ, Sφi = 0, then v ≡ 0 from (3.12). Since Rd \ D is connected, from Rellich's lemma and unique continuation property, we know that v = 0 in Rd\ D. Therefore, by the trace theorem, we have Sφ = 0. By Lemma3.7, we know that S is an isomorphism and thus φ ≡ 0.  Lemma 3.9. Let Si be the single-layer operator corresponding to the wave number k = i = √

−1. Then Si : H−1/2(∂D) → H1/2(∂D) is self-adjoint and coercive. Moreover, S − Si : H−1/2(∂D) → H1/2(∂D) is compact.

Proof. Substituting k = i into (3.11) gives hφ, Siφi =

Z

D∪(BR\D)

(∇¯v · A∇v + n(x)|v|2) dx + Z

|x|=R

|v|2ds + o(Rd−12 ) as R → ∞. By taking R → ∞, we reach

hφ, Siφi = Z

Rd\∂D

(∇¯v · A∇v + n(x)|v|2) dx, which shows that Si is self-adjoint.

Since A is uniform elliptic and n(x) ≥ c, then

hφ, Siφi ≥ ckvk2H1(Rd\∂D).

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By trace theorem, together with Lemma3.7, there exist constants c1, c2 with 0 < c2 <

c1 < c such that

hφ, Siφi ≥ c1kvk2H1/2(∂D) = c1kSiφk2H1/2(∂D) ≥ c2kφk2H−1/2(∂D),

that is, Si is coercive. Note that the operator S − Si has the same mapping property as S0 in (3.6). We thus immediately conclude that

S − Si : H−1/2(∂D) → H1/2(∂D)

is compact since the embedding H1(∂D) into H1/2(∂D)is compact.  3.3. The properties of the data-to-pattern operator GDir. We can prove following two lemmas about GDir.

Lemma 3.10. GDir : H1/2(∂D) → L2(Sd−1) is injective.

Proof. This is a direct consequence of Rellich's lemma and unique continuation property.

 Lemma 3.11. GDir : H1/2(∂D) → L2(Sd−1) is compact.

Proof. This lemma can be proved following the line of [KG08, Lemma 1.13]. Let R > 1 be a large number. For each f ∈ H1/2(∂D), let v ∈ Hloc1 (Rd\ D) satisfy (3.4). By the representation formula

v(ˆx) = Z

∂D



(ν · A(z)∇ΦA,n(ˆx, z))v(z) − ΦA,n(ˆx, z)(ν · A(z)∇v(z))



ds(z) (3.13) for ˆx ∈ Sd−1, we can decompose GDir into GDir = G2◦ G1, where

G1 : H1/2(∂D) → C(∂BR) × C(∂BR) G2 : C(∂BR) × C(∂BR) → L2(Sd−1) are dened by

G1f =



v|∂BR, ν · A∇v|∂BR

 , G2(g, h) =

Z

∂BR



(ν · A∇ΦA,n(ˆx, z))g(z) − ΦA,n(ˆx, z)h(z)

 ds(z).

By interior regularity results for elliptic equation, we know that G1 : H1/2(∂D) → C(∂BR) × C(∂BR) is bounded. Moreover, from the analyticity of ΦA,n(ˆx, z) in ˆx, it immediately follows that G2 : C(∂BR) × C(∂BR) → L2(Sd−1) is compact and the proof

is completed. 

Modify the ideas in [AC06, Theorem 3.1], we can establish the denseness of range(GDir). Lemma 3.12. ran(GDir) is dense in L2(Sd−1).

Proof. Let the orthogonal complement of range(GDir)

ran(GDir) := g ∈ L2(Sd−1) (g, ψ)L2(Sd−1) = 0 for all ψ ∈ ran(GDir) . Our aim is to show that ran(GDir) = 0. Recall that

ran(GDir) = ker(GDir).

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Given any f ∈ H1/2(∂D), let v ∈ Hloc1 (Rd\ D) be the unique solution to (3.4) and GDirf = v be its far-eld pattern. Again, using the representation formula (3.13), we have

(g, GDirf )L2(Sd−1)= (g, v)L2(Sd−1)

= Z

Sd−1

g(ˆx)

Z

∂D

(ν · A(z)∇ΦA,n(ˆx, z))v(z) ds(z)

 ds(ˆx)

− Z

Sd−1

g(ˆx)

Z

∂D

ΦA,n(ˆx, z)(ν · A(z)∇v(z)) ds(z)

 ds(ˆx)

= Z

∂D

ν · A(z)∇

 Z

Sd−1

ΦA,n(ˆx, z)g(ˆx) ds(ˆx)



v(z) ds(z)

− Z

∂D

 Z

Sd−1

ΦA,n(ˆx, z)g(ˆx) ds(ˆx)



ν · A∇v(z) ds(z)

= Z

∂D



(ν · A(z)∇uA,n,g)f − uA,n,g(ν · A(z)∇v)



ds. (3.14) Let u be the unique solution to (3.4) with boundary data u = uA,n,g on ∂D, that is,





∇ · (A(x)∇u) + k2n(x)u = 0 in Rd\ D,

u = uA,n,g on ∂D,

u satises Sommerfeld radiation condition at |x| → ∞.

(3.15)

Since both v and u satises ∇·(A(x)∇u)+k2n(x)u = 0in Rd\Dand satisfy Sommerfeld radiation condition, then

0 = Z

∂D



(ν · (A∇u))f − uA,n,g(ν · (A∇v))



ds (3.16)

Subtracting (3.14) by (3.16) yields (g, GDirf )L2(Sd−1)=

Z

∂D

(ν · (A∇(uA,n,g − u)))f ds, that is,

GDirg = ν · A∇(uA,n,g− u).

If g ∈ ker(GDir), then ν · A∇uA,n,g = ν · A∇u. Combining this with (3.15), we can extend u in D by dening u = uA,n,g, and then u satises

∇ · (A∇u) + k2n(x)u = 0 in Rd.

Since u satises Sommerfeld radiation condition at |x| → ∞, by uniqueness result, we see that u ≡ 0, which gives uA,n,g = 0 in D. By the unique continuation property, we hence have uA,n,g ≡ 0. Finally, Lemma2.4 implies g ≡ 0, that is, ran(GDir) = 0.  3.4. The properties of the far-eld operator FDir. In view of the argument in [KG08, Theorem 1.8], we can prove an important property of FDir.

Lemma 3.13. The far-eld operator FDir : L2(Sd−1) → L2(Sd−1) satises

FDir− FDir = i2kγ2FDir FDir, (3.17)

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where FDir : L2(Sd−1) → L2(Sd−1) is the adjoint operator of FDir, and the constant γ is given in (3.3).

Proof. Given any g, h ∈ L2(Sd−1), choose vinc = uA,n,g and winc = uA,n,h, which are modied Herglotz wave functions given in (3.5). Let v and w be the solutions of the scattering problem (3.1), with corresponding scattered elds vsc = v − vin and wsc= w − win, which are solution to (3.2), and the far-eld patterns of vsc and wsc are v and w, respectively. By denition of FDir, we have

v= FDirg and w= FDirh.

Note that 0 =

Z

BR\D

(v∇ · (A∇w) − w∇ · (A∇v)) dx = Z

|x|=R



v(∂rw) − w(∂rv)



ds. (3.18) Clearly, for each R > 0,

Z

|x|=R



vinc(∂rwinc) − winc(∂rvinc)



ds = 0. (3.19)

By the Sommerfeld radiation condition and the asymptotic expansion of the scattered

eld, we can compute

vsc(x)(∂rwsc(x)) − wsc(x)(∂rvsc(x))

= −ikvsc(x)wsc(x) − ikwsc(x)vsc(x) + o(|x|d−12 )

= −ik

 γ eik|x|

|x|d−12 v(ˆx) + O(|x|d+12 )



γe−ik|x|

|x|d−12 w(ˆx) + O(|x|d+12 )



− ik



γe−ik|x|

|x|d−12 w(ˆx) + O(|x|d+12 )



γ eik|x|

|x|d−12 v(ˆx) + O(|x|d+12 )



+ o(|x|d−12 )

= −i2kγ2

|x|d−1v(ˆx)w(ˆx) + O(|x|−d) uniformly for all ˆx = x/|x|. Hence we have

R→∞lim Z

|x|=R



vsc(∂rwsc) − wsc(∂rvsc)

 ds

= −i lim

R→∞

Z

|x|=R

2kγ2

|x|d−1v(ˆx)w(ˆx) ds

= −i lim

R→∞

2kγ2 Rd−1

Z

|x|=R

v(ˆx)w(ˆx) ds

= −i lim

R→∞

2kγ2 Rd−1

Z

Sd−1

v(ˆx)w(ˆx)Rd−1ds

= −i2kγ2 Z

Sd−1

v(ˆx)w(ˆx) ds

= −i2kγ2(FDirg, FDirh)L2(Sd−1). (3.20)

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Next, for each R > 1, by Fubini's theorem and the representation formula of the far-eld pattern, we can obtain that

Z

|x|=R



vinc(∂rwsc) − wsc(∂rvinc)

 ds(x)

= Z

Sd−1

g(ˆz)

 Z

|x|=R

ΦA,n(ˆz, x)(∂rwsc) − wsc(∂rΦA,n(ˆz, x)) ds(x)

 ds(ˆz)

= − Z

Sd−1

g(ˆz)w(ˆz) ds(ˆz)

= −(g, FDirh)L2(Sd−1), (3.21)

and similarly,

Z

|x|=R



vsc(∂rwinc) − winc(∂rvsc)

 ds(x)

= (FDirg, h)L2(Sd−1). (3.22)

Combining (3.18), (3.19), (3.20), (3.21), and (3.22) gives

−i2kγ2(FDirg, FDirh)L2(Sd−1)− (g, FDirh)L2(Sd−1)+ (FDirg, h)L2(Sd−1) = 0,

which is our desired result. 

With the help of (3.17), we can investigate the range property of the far-eld operator FDir.

Lemma 3.14. Let SDir : L2(Sd−1) → L2(Sd−1) be the scattering operator dened by SDir = I + i2kγ2FDir,

then SDir is unitary, i.e., SDir SDir = SDirSDir = I. Moreover, FDir is normal.

Proof. It is straightforward to see that SDir SDir =



I − i2kγ2FDir



I + i2kγ2FDir



= I + i2kγ2(FDir− FDir ) + 4k2γ4FDir FDir.

(3.23)

Combining this and (3.17), we obtain SDir SDir = I. Since FDir : L2(Sd−1) → L2(Sd−1) is a compact operator, then SDir is a compact perturbation of the identity, and so SDir

is bijective. Hence, SDir−1 = SDir , which shows that SDir is unitary. Note that SDirSDir =



I + i2kγ2FDir



I − i2kγ2FDir



= I + i2kγ2(FDir− FDir ) + 4k2γ4FDirFDir .

(3.24)

By comparing (3.23) and (3.24), we conclude that FDir is normal.  Lemma 3.15. If k2 is not an eigenvalue of (3.10), then FDir : L2(Sd−1) → L2(Sd−1) is injective and ran(FDir) is dense in L2(Sd−1).

Proof. We rst establish the injectivity. Given g ∈ ker(FDir). Recall that FDirg is the far-eld pattern corresponding to the incident eld

vinc(x) = uA,n,g(x).

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Let vscbe the associated scattered eld. As above, applying Rellich's lemma and unique continuation property, we obtain vsc = 0 in Rd. This shows that vinc satises (3.10).

Since k2 is not an eigenvalue for (3.10), then vinc = 0in D. By the unique continuation property for the acoustic equation, we then have vinc = uA,n,g ≡ 0in Rd. The injectivity of g → uA,n,g in Lemma 2.4 implies g ≡ 0.

Next, to prove that ran(FDir) is dense in L2(Sd−1), it suces to show that FDir : L2(Sd−1) → L2(Sd−1) is injective. For each g, h ∈ L2(Sd−1), the reciprocity relation for the far-eld pattern gives

(g, FDirh)L2(Sd−1)= Z

Sd−1

g(ˆx)

Z

Sd−1

vDir(ˆx, ˆz)h(ˆz) ds(ˆz)

 ds(ˆx)

= Z

Sd−1

g(ˆx)

Z

Sd−1

vDir(−ˆz, −ˆx)h(ˆz) ds(ˆz)

 ds(ˆx)

= Z

Sd−1

 Z

Sd−1

vDir(−ˆz, −ˆx)g(ˆx) ds(ˆx)



h(ˆz) ds(ˆz)

=

 Z

Sd−1

vDir(−ˆz, −ˆx)g(ˆx) ds(ˆx), h



L2(Sd−1)

, that is,

(FDir g)(ˆz) = Z

Sd−1

vDir(−ˆz, −ˆx)g(ˆx) ds(ˆx).

Therefore, if g ∈ ker(FDir ), then by the injectivity of F , we conclude that g ≡ 0.  3.5. The (FF )1/4-method. Now we are going to derive our main reconstruction method. The result is based on the following theorem in [KG08].

Theorem 3.16. [KG08, Theorem 1.23] Let H be a Hilbert space and X be a reexive Banach space. Assume that the compact operator F : H → H have a factorization of the form

F = B ˜AB with operators B : X → H and ˜A : X → X such that

(1) =hϕ, ˜Aϕi 6= 0 for all 0 6= ϕ ∈ ran(B).

(2) ˜A = ˜A0 + C for some compact operator C and some self-adjoint operator ˜A0 which is coercive on ran(B).

If F is injective and I + irF is unitary for some r > 0, then ran(B) = ran((FF )1/4). Putting together Proposition3.5, Lemma3.83.11,3.14,3.15, and applying the above theorem, we can immediately obtain the following lemma.

Lemma 3.17. If k2 is not an eigenvalue of (3.10), then ran(GDir) = ran((FDir FDir)1/4). Finally, we want to characterize ran((FDir FDir)1/4) as in [KG08] for the purpose of numerical simulations. Since FDir : L2(Sd−1) → L2(Sd−1) is compact and normal, there exists a set of (complex) eigenvalues {λjDir}j∈N with corresponding eigenfunctions {φjDir}j∈N in L2(Sd−1). Furthermore, the set of eigenfunctions forms an orthonormal

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basis of L2(Sd−1), see e.g. [Zim90, Corollary 3.2.9]. For g ∈ L2(Sd−1), we can write (FDir FDir)1/4g =X

j∈N

q

jDir|(g, φjDir)L2(Sd−1)φjDir, ΦA,n(·, z) =X

j∈N

A,n(·, z), φjDir)L2(Sd−1)φjDir.

Therefore, if ΦA,n(·, z) ∈ ran((FDir FDir)1/4), then there exists a g ∈ L2(Sd−1) such that X

j∈N

q

jDir|(g, φjDir)L2(Sd−1)φjDir=X

j∈N

A,n(·, z), φjDir)L2(Sd−1)φjDir. Equivalently,

q

jDir|(g, φjDir)L2(Sd−1) = (ΦA,n(·, z), φjDir)L2(Sd−1) for all j ∈ N, or

(g, φjDir)L2(Sd−1) = (ΦA,n(·, z), φjDir)L2(Sd−1) q

jDir|

for all j ∈ N.

Hence, ΦA,n(·, z) ∈ ran((FDir FDir)1/4) if and only if X

j∈N

|(ΦA,n(·, z), φjDir)L2(Sd−1)|2

jDir| < ∞.

Combining this with Lemma3.17and Proposition3.6, we get the following key theorem.

Theorem 3.18. If k2 is not an eigenvalue of (3.10), then the following are equivalent:

(1) z ∈ D;

(2) ΦA,n(·, z) ∈ ran((FDir FDir)1/4); (3) WDir(z) :=

 X

j∈N

|(ΦA,n(·, z), φjDir)L2(Sd−1)|2

jDir|

−1

> 0.

In other words, the characteristic function of the sound-soft obstacle D is given by χD(z) = sign(WDir(z)).

4. The determination of the obstacle with other boundary conditions We can extend the factorization developed above to the obstacle with other types of boundary conditions on ∂D, i.e., Neumann or impedance boundary conditions. In the case of Helmholtz equation (homogeneous medium) with the obstacle having these two types of boundary conditions, the factorization method has been discussed in detail in [KG08, Chapter 1,2]. For the acoustic equation considered here, once we use ΦA,n(ˆz, x) as the incident eld, the factorization method for obstacles with Neumann or impedance boundary conditions can be established following exactly the arguments in [KG08, Chapter 1,2]. Since we have presented the detailed proof for the Dirichlet case in Section3, we will not repeat the arguments here. Instead, we state the main theorems of the reconstruction method without proofs.

We rst discuss the Neumann case (sound-hard). The (FF )1/4-method can be easily modied to treat this case as in [KG08, Chapter 1]. Let FNeu be the far-eld operator corresponding to the sound-hard obstacles, which is dened similarly as FDir.

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Theorem 4.1. If k2 is not an eigenvalue of

(∇ · (A(x)∇u) + k2n(x)u = 0 in D,

ν · (A(x)∇u) = 0 on ∂D, (4.1)

then the following are equivalent:

(1) z ∈ D;

(2) ΦA,n(·, z) ∈ ran((FNeu FNeu)1/4); (3) WNeu(z) :=

 X

j∈N

|(ΦA,n(·, z), φjNeu)L2(Sd−1)|2

jNeu|

−1

> 0, where {λjNeu, φjNeu} is an eigen-system of the operator FNeu.

In other words, the characteristic function of the sound-hard obstacle D is given by χD(z) = sign(WNeu(z)).

Now we consider the obstacle with impedance condition. Let FImpbe the corresponding far-eld operator. In this case, FImp fails to be normal in general, unless λ is real-valued (i.e. Robin boundary condition). However, the problem can be overcome by the F]- method, which can be found in [KG08, Chapter 2]. The self-adjoint operator F] is dened as follows:

F]= |<FImp| + |=FImp|, where

<FImp := 1

2(FImp+ FImp ) and =FImp = 1

2i(FImp− FImp ).

In this case, the result reads:

Theorem 4.2. Let λ ∈ L(∂D) be a complex-valued function with non-negative the imaginary part on ∂D. If k2 is not an eigenvalue of

(∇ · (A(x)∇u) + k2n(x)u = 0 in D,

ν · (A(x)∇u) + λ(x)u = 0 on ∂D, (4.2)

then the following are equivalent:

(1) z ∈ D;

(2) ΦA,n(·, z) ∈ ran(F]1/2); (3) WImp(z) :=

 X

j∈N

|(ΦA,n(·, z), φjImp)L2(Sd−1)|2

jImp|

−1

> 0, where {λjImp, φjImp} is an eigen-system of the operator F].

5. Numerical results

In this section, we present some numerical simulation results to show the eciency of the method developed in Section 3. For simplicity, we treat the acoustic equation in two dimensions. The simulations are obtained using MATLAB 2020a (with PDE Toolbox). We consider four shapes of sound-soft obstacles whose parametric equations are listed in Table1.

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Obstacle shape Parametrization (anti-clockwise oriented) circle x(t) = 0.5 cos t

0 ≤ t ≤ 2π y(t) = 0.5 sin t

peanut [YZZ13] x(t) = cos t ·√

cos2t + 0.25 sin2t

0 ≤ t ≤ 2π y(t) = sin t ·√

cos2t + 0.25 sin2t kite [YZZ13] x(t) = 0.5 cos t + 0.325 cos 2t − 0.325

0 ≤ t ≤ 2π y(t) = 0.75 sin t

heart x(t) = −0.5 sin3t

0 ≤ t ≤ 2π (Wolfram Mathworld) y(t) = 1332cos t −325 cos 2t −161 cos 3t − 321 cos 4t

Table 1. Parametrization of the obstacles

To carry out the simulations, we rst need to compute the incident eld uincA,n(x, ˆz) = ΦA,n(ˆz, x) = utoref(x, −ˆz).

By the mixed reciprocity relation described above, it suces to compute the total eld for the acoustic equation (without obstacle) with plane incident elds. After generating the required incident eld, we then simulate the scattered eld and compute the far-

eld pattern for the acoustic equation with a sound-soft obstacle. The simulation is explained in the following algorithm.

Algorithm 1 Simulation of the scattered eld uscDir and the far-eld pattern uDir

1: Choose an incident direction ˆz ∈ S1;

2: Compute the scattered eld uscref(x, −ˆz) for the acoustic equation (1.1) without obstacle by the incident eld uincref(x, −ˆz) = eikx·(−ˆz);

3: Compute the total eld utoref(x, −ˆz) = uscref(x, −ˆz) + uincref(x, −ˆz);

4: Compute the scattered eld uscDir(x, ˆz) for the acoustic equation (3.2) with incident

eld uincA,n(x, ˆz) = utoref(x, −ˆz);

5: Compute the far-eld of uDir and generate the far-eld operator FDir.

We now discuss the algorithm in more detail. In our simulation, we take A ≡ I and n(x, y) =

(

1 + e1+x2+y21 for x2+ y2 < 1

1 otherwise

which has jump discontinuities at x2+ y2 = 1. The original problem was formulated in R2, together with the Sommerfeld radiation condition at innity. Here, we restrict the computational domain in the ball x2 + y2 < 4 and approximate the Sommerfeld radiation condition by the impedance condition:

(∆˜uscref + k2n˜uscref = k2(1 − n)uincref in x2+ y2 < 4,

rscref − ik ˜uscref = 0 on x2+ y2 = 4.

We solve this boundary valued problem FEM with mesh size ≤ 0.1. Then the approximated total eld ˜utoref = ˜uscref + uincref and the needed incident eld

˜

uincI,n(x, ˆz) = ˜utoref(x, −ˆz).

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