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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 eigen- functions of the far-field 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 re- constructing the shape of the obstacle by the far-field 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 first 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 sam- pling 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 briefly describe the problem considered in the paper. Let A(x) = (aij(x)) be a real-symmetric matrix with C entries, which satisfies 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 field uincA,n(x, ˆz) and the corresponding scattered field 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 satisfies 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-field 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 investiga- tions 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 inhomo- geneous structure. In their work, they showed that, if the reference medium is known, then the penetrable obstacle can be uniquely determined by the far-field data at a fixed 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 ob- stacle 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-field operator was developed in [Ki98]

• Impenetrable obstacle in an inhomogeneous background. When the inhomoge- neous background was described the Schrödinger equation with an inhomoge- neous 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 Dirich- let or Robin boundary conditions). Our work is closely related to this paper.

In our paper, we consider an impenetrable obstacle in an anisotropic inhomoge- neous background.

We would like to remark that, in [GMMR12, KG08, KP98], the far-field operator is defined in terms of the far-field pattern induced by the incident plane waves. In our case, we take a different incident field. Roughly speaking, the incident field used in our approach is the far-field pattern of the outgoing fundamental solution for the background equation. Indeed, in the case where the reference medium is homogeneous, the far-field pattern of the standard outgoing fundamental solution is simply the plane waves. Since we assume that the background medium is known, the far-field pattern of the outgoing fundamental solution for the background equation can be determined.

However, from the numerics perspective, the computation of the far-field of the outgoing fundamental solution directly is not efficient. Luckily, thanks to the mixed reciprocity relation (see Lemma 2.2), we can compute the far-field of the outgoing fundamental solution by solving the total field of the background equation with the incident plane waves.

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.

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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 acous- tic equation [NUW13]. We need to point out a major difference between [INUW14] and [NUW13]. In [INUW14], the constant appears in the Hölder part of the stability es- timate 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 fixed 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 first review some properties of the outgoing fundamental solution for the background equation and its far-field 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 efficiency of our method. Finally, in Section 6, we compare the reconstruction results for the acoustic and the Schrödinger equations at different 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).

Definition 2.1. We say that a function u ∈ Hloc1 (Rd) satisfies 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) satisfies the Sommerfeld radiation condition at |z| → ∞.

Let uincref(x, ˆz) = eikx·ˆz be an incident plane wave. The presence of inhomogeneity (with- out the obstacle) in BR gives rise to a unique scattered field uscref ∈ Hloc1 (Rd) which

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satisfies the Sommerfeld radiation condition. Then the total field utoref = uscref + uincref satisfies

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

Lemma 2.2. Let ΦA,n(ˆz, x) be the far-field 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. Define

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- defined. Note that the far-field pattern of Φ(z, y) is Φ(ˆz, y) = e−ikˆz·y = uincref(y, −ˆz).

We then see that the far-field 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) satisfies

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. 

Definition 2.3. For each g ∈ L2(Sd−1), we define the modified 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 field uscref(x, ˆz), and thus induces a unique total field utoref(x, ˆz). Combining these discussions and by the super- position property, we can prove the following lemma.

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

3. Reconstruction of sound-soft obstacle

By Lemma 2.2, we can choose the incident field uincA,n(x, ˆz) = ΦA,n(ˆz, x) instead of the usual plane wave uincref(x, ˆz) = eikx·ˆz. By choosing this incident field, no scattering occurs if there is no obstacle. Now let uscDir(x, ˆz) be the corresponding scattered field, then the total field utoDir(x, ˆz) = uincA,n(x, ˆz) + uscDir(x, ˆz) satisfies





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

utoDir = 0 on ∂D,

uscDir satisfies 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 satisfies Sommerfeld radiation condition at |x| → ∞.

(3.2)

Denote uDir be the far-field pattern of uscDir, which satisfies 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].

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

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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 satisfies Sommerfeld radiation condition at |x| → ∞.

(3.4)

3.1. The factorization of the far-field operator. For each g ∈ L2(Sd−1), recall the modified Herglotz wave function is defined 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 fields. So, the far-field of the scattered field is the superposition of uDir. Thus, we can define the far-field operator as follows:

Definition 3.2. The far-field operator FDir: L2(S2) → L2(S2) 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 define an operator closely related to the far-field operator:

Definition 3.3. The data-to-pattern operator GDir : H1/2(∂D) → L2(Sd−1) is defined by GDirf = v, where v is the far-field of v ∈ Hloc1 (Rd\ D) which satisfies (3.4).

To discuss the relation between the far-field operator and the data-to-pattern oper- ator, we introduce the single-layer potential operator S defined by

(Sφ)(x) :=

Z

∂D

ΦA,n(x, z)φ(z) ds(z) for all x ∈ ∂D.

We first 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) satisfies the Sommerfeld radiation condition at |x| → ∞.

Since the coefficient A is smooth, ΦA,1(x, z) is C for x 6= z. By the techniques of pseudodifferential 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) defined as the restriction of the modified 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 difficult 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-field 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 solu- tion to (3.4). Since both

w and ΦA,n(·, z) have the same far-field 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. Define 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-field 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 different 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), define v(x) = (Sφ)(x) =

Z

∂D

ΦA,n(x, y)φ(y) ds(y) for x ∈ R3\ ∂D.

Then, by [McL00, Theorem 6.11], we have v ∈ H1(D) ∩ Hloc1 (R3\ 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 R3 \ ∂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 R3 \ D is connected, from Rellich’s lemma and unique continuation property, we know that v = 0 in R3\ 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

R3\∂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(R3\∂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 defined 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-field 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 satisfies Sommerfeld radiation condition at |x| → ∞.

(3.15)

Since both v and u satisfies ∇·(A(x)∇u)+k2n(x)u = 0 in Rd\D and 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 defining u = uA,n,g, and then u satisfies

∇ · (A∇u) + k2n(x)u = 0 in R3.

Since u satisfies 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, Lemma 2.4 implies g ≡ 0, that is, ran(GDir) = 0.  3.4. The properties of the far-field 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-field operator FDir : L2(Sd−1) → L2(Sd−1) satisfies

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 modified Herglotz wave functions given in (3.5). Let v and w be the solutions of the scattering problem (3.1), with corresponding scattered fields vsc = v − vin and wsc= w − win, which are solution to (3.2), and the far-field patterns of vsc and wsc are v and w, respectively. By definition 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 field, 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-field 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-field operator FDir.

Lemma 3.14. Let SDir : L2(Sd−1) → L2(Sd−1) be the scattering operator defined 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 first establish the injectivity. Given g ∈ ker(FDir). Recall that FDirg is the far-field pattern corresponding to the incident field

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

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

Since k2 is not an eigenvalue for (3.10), then vinc = 0 in D. By the unique continuation property for the acoustic equation, we then have vinc = uA,n,g ≡ 0 in 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 suffices 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-field 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 reflexive 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 field, 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 first discuss the Neumann case (sound-hard). The (FF )1/4-method can be easily modified to treat this case as in [KG08, Chapter 1]. Let FNeu be the far-field operator corresponding to the sound-hard obstacles, which is defined 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 FImp be the corre- sponding far-field 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 overcomed by the F]-method, which can be found in [KG08, Chapter 2]. The self-adjoint operator F] is defined 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)|

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 efficiency 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 first need to compute the incident field uincA,n(x, ˆz) = ΦA,n(ˆz, x) = utoref(x, −ˆz).

By the mixed reciprocity relation described above, it suffices to compute the total field for the acoustic equation (without obstacle) with plane incident fields. After generating the required incident field, we then simulate the scattered field and compute the far- field pattern for the acoustic equation with a sound-soft obstacle. The simulation is explained in the following algorithm.

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

1: Choose an incident direction ˆz ∈ S1;

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

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

4: Compute the scattered field uscDir(x, ˆz) for the acoustic equation (3.2) with incident field uincA,n(x, ˆz) = utoref(x, −ˆz);

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

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

( 1 + e

1

1+x2+y2 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 infinity. 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 approxi- mated total field ˜utoref = ˜uscref + uincref and the needed incident field

˜

uincI,n(x, ˆz) = ˜utoref(x, −ˆz).

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