Reconstruction of penetrable obstacles in acoustic scattering

32  Download (0)

Full text

(1)

Reconstruction of penetrable obstacles in acoustic scattering

Sei Nagayasu

Gunther Uhlmann

Jenn-Nan Wang

Abstract

We develop a reconstruction algorithm to determine penetrable obstacles inside a domain in the plane from acoustic measurements made on the boundary. This algorithm uses complex geometrical op- tics solutions to the Helmholtz equation with polynomial-type phase functions.

1 Introduction

Let D be an unknown obstacle with an unknown index of refraction and a subset of a larger bounded domain Ω with a homogeneous index of refraction.

Assume that D is penetrable. Suppose that we are given all possible Cauchy data or the Dirichlet-to-Neumann map measured on ∂Ω. The inverse problem we consider in this paper is to determine the shape of D using such boundary measurements. This problem can be considered as the inverse scattering problem with near field measurements. In theory, it is known that knowing

Department of Mathematics, Hokkaido University, North 10, West 8, Kita-Ku, Sap- poro, Hokkaido 060-0810, Japan. Current address: Department of Mathematical Science, Graduate School of Material Science, University of Hyogo, 2167 Shosha, Himeji, Hyogo 671-2280, Japan. Email:sei@sci.u-hyogo.ac.jp

Department of Mathematics, University of Washington, Box 354305, Seattle, WA 98195-4350, USA and Department of Mathematics University of California, Irvine, CA 92697-3875, USA. Email:gunther@math.washington.edu

Department of Mathematics, Taida Institute of Mathematical Sci- ences, NCTS (Taipei), National Taiwan University, Taipei 106, Taiwan.

Email:jnwang@math.ntu.edu.tw

(2)

the far field data such as the scattering amplitude is equivalent to knowing the near field measurements (see, for example, [11, Chapter 6]).

In this paper, we consider this problem in the plane, that is, we assume D ⋐ Ω ⊂ R2. For simplicity, we suppose that both D and Ω have C2 boundaries. Let γD ∈ C2(D) satisfy γD ≥ cγ for some positive constant cγ

and denote eγ := 1 + γDχD, where χD denotes the characteristic function of D. Let k > 0 and consider the steady state acoustic wave equation in Ω with Dirichlet condition

(∇ · (eγ∇v) + k2v = 0 in Ω,

v = f on ∂Ω. (1.1)

In the case that eγ = 1, the problem (1.1) is the boundary value problem for the Helmholtz equation

(∆v0+ k2v0 = 0 in Ω,

v0 = f on ∂Ω. (1.2)

Throughout the paper, we assume that k2 is not a Dirichlet eigenvalue of the operator −∇ · (eγ∇•) and −∆ in Ω. It is known that for any f ∈ H1/2(∂Ω), there exists a unique solution v to (1.1). Thus, we can define the Dirichlet- to-Neumann map ΛD : H1/2(∂Ω)→ H−1/2(∂Ω) for (1.1) by

ΛDf := ∂v

∂ν

∂Ω

for f ∈ H1/2(∂Ω).

The inverse problem consists of determining D from ΛD. The domain D can also be treated as an inclusion embedded in Ω. The aim of this work is to give a reconstruction algorithm for this problem. Note that the information on the medium parameter γD inside D is not known a priori.

The main tool in our reconstruction method is the complex geometri- cal optics (CGO) solutions with polynomial-type phase functions for the Helmholtz equation. This type of CGO solutions has been introduced in [18]

for general second order elliptic equations or systems having the Laplacian as the principal part, which includes the Helmholtz equation. However, in order to obtain more explicit forms in the lower orders of the CGO solutions, we will not adopt the approach in [18]. Instead, we will take advantage of the transformation between the harmonic functions and the solutions to the

(3)

Helmholtz equation in two dimensions found by Vekua [20] (also see [8]) to construct the needed CGO solutions.

Having obtained the CGO solutions with polynomial phases for the Helmholtz equation, we apply them to determine the shape of D by ΛD. CGO solutions have been found to be useful in detecting some geometrical information of D in several inverse problems. For the inclusion problem in the static case, i.e., k = 0, there are several articles, and some of them include numerical results, dealing with either the conductivity equation (the first equation of (1.1) with k = 0) or the isotropic elasticity [1], [5], [6], [4], [17], [18], and [19].

This type of method was called the enclosure method by Ikehata. We refer to his survey paper [7] for some of the early developments.

For the reconstruction of penetrable obstacles or inclusions in acoustics by the enclosure type method, we mention the work [9] by Ikehata. In this paper he considers the reconstruction of a penetrable polygon having homoge- neous medium different from the background one by a single pair of Cauchy data in two dimensions. Using CGO solutions with linear phases, he showed that one can reconstruct the convex hull of the polygonal obstacle using a single measurement. In our paper, we consider a general penetrable obstacle and assume the medium inside of the penetrable obstacle is an unknown in- homogeneous function. Using CGO solutions with polynomial-type phases, we are able to reconstruct more information on the shape of the penetrable obstacle from the Dirichlet-to-Neumann map. Especially, in theory, we can completely reconstruct certain class of penetrable objects such as star-shaped obstacles. For other related results, we would like to mention that Nakamura and Yoshida [16] used CGO solutions with limiting Carleman weights intro- duced in [12] to reconstruct some non-convex sound-hard obstacles from the Dirichlet-to-Neumann map. The level set of the limiting Carleman weights are circles (in two dimensions) or spheres (in three dimensions). Also, we mention that the uniqueness of determining a penetrable obstacle by the scattering amplitude at a fixed energy was proven by Isakov [10] and Kirsch, Kress [13].

Unlike the static case, for the enclosure type method in the Helmholtz equation, we need to analyze the effect coming from the term with k2. Roughly speaking, we have to compare the sizes ofR

|v−v0|2dx andR

D|∇v0|2dx (see estimates (3.16) and (3.19)). Note that in the static case, we only need to consider R

D|∇v0|2dx. The estimate (3.16) is somewhat straightforward.

Our main focus is on (3.19). In the impenetrable case, this can be achieved using elliptic regularity with smooth coefficient and the Sobolev embedding

(4)

theorem (see [16]). However, in the penetrable case, the coefficient is merely piecewise smooth. The Sobolev embedding theorem does not work because the solution is not smooth enough. One of the difficulties of this paper is to establish the estimate (3.19).

The rest of this paper is organized as follows. In section 2 we construct the CGO solutions to the Helmholtz equation with polynomial phase func- tions and their properties. In section 3 we establish some identities and the estimates we need. In section 4 we prove our main result on the determina- tion of D from ΛD under a “curvature assumption” on ∂D on the intersection of the level sets of the real part of the phases of the CGO with ∂D (see The- orem 4.1). In section 5 we show that the curvature assumption is satisfied for a large class of CGO solutions. In section 6 state the conclusion of this paper.

2 CGO solutions

In this section we want to construct CGO solutions with polynomial phases for the Helmholtz equation. We do this by combining the idea in [18] and the transform introduced by Vekua (see (13.9) on page 58 in [20]). Let us first introduce η(x) := c (x1− x∗,1) + i(x2− x∗,2)N

as the phase function, where c ∈ C satisfies |c| = 1, N is a positive integer, and x = (x∗,1, x∗,2)∈ R2\Ω.

Without loss of generality we may assume that x = 0 using an appropriate translation. We put ηR(x) := Re η(x). Note that

ηR(x) = rNcos N(θ− θ) for x = r(cos θ, sin θ) ∈ R2. We now define an open cone

Γ := n

r(cos θ, sin θ) : |θ − θ| < π 2N

o

with an opening angle π/N (see Figure 1). It is clear that ηR(x) > 0 for all x∈ Γ.

Given any h > 0, ˇVh(x) := exp η(x)/h

is a harmonic function. Following Vekua [20], we define a map Tk on any harmonic function ˇV (x) by

TkV (x) := ˇˇ V (x)− Z 1

0

V (tx)ˇ ∂

∂t

nJ0 k|x|√

1− to dt

= ˇV (x)− k|x|

Z 1 0

V (1ˇ − s2)x

J1 k|x|s ds

(5)

0 x2

x1

Γ

Figure 1: The definition of Γ

where Jm is the Bessel function of the first kind of order m. We now set Vh(x) := Tkh(x). Then Vh(x) satisfies the Helmholtz equation ∆Vh+k2Vh= 0 in R2. The function Vh is a CGO solution to the Helmholtz equation in Γ by the following lemma.

Lemma 2.1. We can write Vh(x) = exp

η(x) h



1 + R0(x)

in Γ, (2.1)

where R0(x) = R0(x; h) satisfies

|R0(x)| ≤ h k2|x|2R(x),

∂R0

∂xj

(x)

≤ Nk2|x|N +1

R(x) + h k2|xj|

R(x) in Γ.

Proof. Let x ∈ Γ. We note that

|R0(x)| ≤ k|x|

Z 1 0

exp

1 hRe

η (1− s2)x

− η(x)

J1 k|x|s ds

≤ k2|x|2 2

Z 1 0

exp

1 hRe

η (1− s2)x

− η(x)

s ds since

R0(x) =−k|x|

Z 1 0

exp

1 h



η (1− s2)x

− η(x)

J1 k|x|s

ds (2.2)

(6)

and |J1(t)| ≤ t/2 for t ≥ 0. On the other hand, we can see that Re

η (1− s2)x

− η(x)

= Re

−η(x) 1 − (1 − s2)N

=−ηR(x) 1− (1 − s2)N

≤ −ηR(x) s2 (2.3) for any 0 < s < 1 using the formula

1− s2− (1 − s2)N = (1− s2) 1− (1 − s2)N −1

≥ 0 for any s ∈ (0, 1).

Hence, since ηR(x) > 0 in Γ, we have

|R0(x)| ≤ k2|x|2 2

Z 1 0

exp

1 hRe

η (1− s2)x

− η(x)

s ds

≤ k2|x|2 2

Z 1 0

exp



−ηR(x) s2 h



s ds < h k2|x|2R(x).

In a similar fashion we can obtain the estimate for ∂R0/∂xj since we have

∇R0(x) =−k|x|

h Z 1

0

(1− s2)(∇η) (1 − s2)x

− ∇η(x)

× exp

1 h

η (1− s2)x

− η(x)

J1 k|x|s ds

− k2x Z 1

0

exp

1 h



η (1− s2)x

− η(x)

J0 k|x|s s ds and |J0(t)| ≤ 1 for any t ≥ 0.

From the above lemma, we conclude that Vh is a CGO solution to the Helmholtz equation in Γ∩ Ω. We now extend it to the whole domain Ω by using an appropriate cut-off. Let ls :={x ∈ Γ : ηR(x) = 1/s} for s > 0. This is the level curve of ηR (see Figure 2). For ε > 0 small enough and t > 0 large enough, we define the function φt∈ C(R2) by

φt(x) =









1 for x∈ [

0<s<t+ε/2

ls, t∈ [0, t], 0 for x∈ R2\ [

0<s<t+ε

ls, t∈ [0, t]

(7)

0 x2

x1

ηR(x) > 1 s

ηR(x) < 1 s ηR(x) = 1

s

Figure 2: A level curve of ηR.

0 x2

x1

ηR(x) = 1 t + ε/2

ηR(x) = 1 t + ε φt(x) = 1

φt(x) = 0

Figure 3: The definition of φt

(8)

(see Figure 3) and

|∂xαφt(x)| ≤ Cφ for |α| ≤ 2, x ∈ Ω, t ∈ [0, t]

for some positive constant Cφ depending only on Ω, N, t and ε. Next we define the function Vt,h by

Vt,h(x) := φt(x) exp



− 1 ht



Vh(x) for x ∈ Ω.

Then we know by Lemma 2.1 that the dominant parts of Vt,h and its deriva- tives are as follows:

Vt,h(x) =

















0 for x∈ Ω \ [

0<s<t+ε

ls,

exp

1 h

−1

t + η(x)

φt(x) + S0(x)h

for x∈ Ω ∩ [

0<s<t+ε

ls,

(2.4)

∇Vt,h(x) =

















0 for x∈ Ω \ [

0<s<t+ε

ls, 1

hexp

1 h

−1

t + η(x)

φt(x)∇η(x) + S(x)h for x∈ Ω ∩ [

0<s<t+ε

ls

(2.5)

for t ∈ (0, t] and h ∈ (0, 1], where S0(x) = S0(x; t, h) and S(x) = S(x; t, h) satisfy

|S0(x)|, |S(x)| ≤ CV for any x∈ Ω ∩ [

0<s<t+ε

ls, t∈ (0, t], h∈ (0, 1]

with a positive constant CV depending only on Ω, N, t, ε and k. Unfor- tunately, the function Vt,h does not satisfy the Helmholtz equation in Ω.

However, if we let v0,t,h be the solution to the Helmholtz equation in Ω with boundary value ft,h := Vt,h|∂Ω, then the error between Vt,h and v0,t,h is expo- nentially small as shown in the following lemma.

(9)

Lemma 2.2. There exist constants C0, C0 > 0 and a > 0 such that kv0,t,h− Vt,hkH2(Ω) ≤ C0

h e−at/h≤ C0e−a/h

for any h ∈ (0, 1], where the constants C0 and C0 depend only on Ω, k, N, t and ε; the constant a depends only on t and ε; and we set at :=

1/t− 1/(t + ε/2).

This lemma can be proved in the same way as Lemma 4.1 in [18]. So we omit the details here.

For our inverse problem, the difference between the two Dirichlet-to- Neumann maps ΛD and Λ plays a crucial role. We define the functional E(t, h) by

E(t, h) :=

Z

∂Ω

D− Λ)ft,hft,hdσ. (2.6) Roughly speaking, for a fixed large N, we can show that if D∩ S

0<s<tls

=∅ then E(t, h) → 0 as h → +0; if D ∩ S

0<s<tls

 6= ∅ then E(t, h) → +∞ as h→ +0. We will state our main result more precisely in Theorem 4.1.

3 Some identities and estimates

In this section, we derive some identities and estimates for solutions to some Dirichlet problems which are needed later. We denote C > 0 a general constant in this section. The constant C depends only on Ω, D, γD and k.

When a constant depends on other data, we will denote the dependence by writing as a subscript, for example Cq the dependence of the constant on q. For a fixed f ∈ H1/2(∂Ω), let v0 and v be the solutions to the Dirichlet problems (1.2) and (1.1), respectively. As before, we put w = v − v0. Note that w satisfies the Dirichlet problem

(∇ · (eγ∇w) + k2w =−∇ · ((eγ − 1)∇v0) in Ω,

w = 0 on ∂Ω. (3.1)

We first show an estimate for w.

Lemma 3.1. For any 2 < q ≤ 4, we have

kwkL(Ω) ≤ Cqk∇v0kLq(D). (3.2)

(10)

Proof. Recall that k2 is not a Dirichlet eigenvalue of−∇ · (eγ∇•) in Ω. From the well-posedness of the boundary value problem (3.1) (see Corollary 8.7 in [3], for example), we have

kwkH1(Ω) ≤ C k(eγ − 1)∇v0kL2(Ω) ≤ C k∇v0kL2(D). On the other hand, since W := w satisfies the Dirichlet problem

(∇ · (eγ∇W ) = −k2w− ∇ · ((eγ − 1)∇v0) in Ω, W = 0 on ∂Ω,

we have the estimate

kwkL(Ω)≤ Cq kk2wkLq/2(Ω)+k(eγ − 1)∇v0kLq(Ω)



≤ Cq kwkL2(Ω)+k∇v0kLq(D)

by Theorem 8.16 in [3] and H¨older’s inequality. Hence we get that kwkL(Ω) ≤ Cq kwkL2(Ω)+k∇v0kLq(D)

≤ Cq kwkH1(Ω)+k∇v0kLq(D)



≤ Cq k∇v0kL2(D)+k∇v0kLq(D)

≤ Cqk∇v0kLq(D) by using H¨older’s inequality again.

We next prove some useful identities.

Lemma 3.2. We have Z

∂Ω

D− Λ)f f dσ = Re Z

(eγ − 1)∇v · ∇v0dx. (3.3) Proof. It is clear that

Z

∂Ω

∂v

∂ν ϕ dσ = Z

∂Ωeγ∂v

∂νϕ dσ = Z

∇ · eγ ϕ ∇v dx

= Z

∇ · eγ∇v

ϕ dx + Z

eγ∇v · ∇ϕ dx

=−k2 Z

v ϕ dx + Z

eγ ∇v · ∇ϕ dx

(11)

for any ϕ∈ H1(Ω). Since v = v0 = f on ∂Ω, the left-hand side of the identity above has the same value whether we take ϕ = v or ϕ = v0 and it is equal to R

∂ΩΛDf f dσ. Thus we have Z

∂Ω

ΛDf f dσ =−k2 Z

v v0dx + Z

eγ ∇v · ∇v0dx

=−k2 Z

|v|2dx + Z

eγ |∇v|2dx∈ R.

The right-hand side of the identity above is real. Hence, by taking the real part, we obtain

Z

∂Ω

ΛDf f dσ =−k2Re Z

v v0dx + Re Z

eγ ∇v · ∇v0dx. (3.4) In the same way, we show that

Z

∂Ω

Λf f dσ =−k2Re Z

v v0dx + Re Z

∇v · ∇v0dx (3.5) since

Z

∂Ω

∂v0

∂ν ϕ dσ = Z

∆v0ϕ dx + Z

∇v0· ∇ϕ dx

=−k2 Z

v0ϕ dx + Z

∇v0· ∇ϕ dx for any ϕ ∈ H1(Ω). Now (3.3) follows easily from (3.4) and (3.5).

The estimates (3.8) and (3.9) in the following lemma play an essential role in our reconstruction algorithm.

Lemma 3.3.

Z

∂Ω

D − Λ)f f dσ

=− Z

eγ|∇w|2dx + k2 Z

|w|2dx + Z

eγ − 1

|∇v0|2dx, (3.6) Z

∂Ω

D − Λ)f f dσ

= Z

|∇w|2dx− k2 Z

|w|2dx + Z

eγ − 1

|∇v|2dx. (3.7)

(12)

In particular, we have Z

∂Ω

D− Λ)f f dσ≤ k2 Z

|w|2dx + Z

D

γD|∇v0|2dx, (3.8) Z

∂Ω

D− Λ)f f dσ≥ Z

D

γD

1 + γD|∇v0|2dx− k2 Z

|w|2dx. (3.9) Proof. By multiplying the identity

0 =∇ · (eγ∇w) + ∇ · ((eγ − 1)∇v0) + k2w by w and integrating the result over Ω, we get

0 = Z

∇ · eγ∇w

w dx + Z

∇ · (eγ − 1)∇v0

w dx + Z

k2w w dx

=− Z

eγ |∇w|2dx + Z

∂Ωeγ∂w

∂ν w dσ

− Z

(eγ − 1)∇v0 · ∇w dx + Z

∂Ω(eγ − 1)∂v0

∂ν w dσ + k2 Z

|w|2dx

=− Z

eγ |∇w|2dx− Z

(eγ − 1)∇v0· ∇w dx + k2 Z

|w|2dx

=− Z

eγ |∇w|2dx− Z

(eγ − 1)∇v0· ∇v dx +

Z

(eγ − 1)|∇v0|2dx + k2 Z

|w|2dx.

Taking the real part of this identity and substituting the identity (3.3) im- mediately leads to (3.6). The identity (3.8) is an easy consequence of (3.6).

Similarly, by multiplying the identity

0 =∇ · ((eγ − 1) ∇v) + ∆w + k2w by w and integrating the result over Ω, we obtain

0 = Z

∇ · (eγ − 1)∇v

w dx + Z

∆w w dx + Z

k2w w dx

=− Z

(eγ − 1)∇v · ∇w dx − Z

|∇w|2dx + Z

k2|w|2dx

(13)

=− Z

(eγ − 1)|∇v|2dx + Z

(eγ − 1)∇v · ∇v0dx

− Z

|∇w|2dx + k2 Z

|w|2dx,

which implies (3.7). Finally, (3.9) follows from (3.7) and the formula

|∇w|2+ eγ − 1

|∇v|2 = eγ |∇v|2− 2 Re ∇v · ∇v0+|∇v0|2

= eγ ∇v −

1 eγ∇v0

2

+

 1− 1



|∇v0|2.

In view of (3.8) and (3.9), we need to estimate kwkL2(Ω). To begin with, we consider the boundary value problem

(∇ · (eγ∇p) + k2p = w in Ω,

p = 0 on ∂Ω. (3.10)

Note that there exists a unique solution p∈ H1(Ω) to (3.10). We can derive the following estimates for p.

Lemma 3.4. Let p be the solution to (3.10), then

kpkH1(Ω) ≤ CkwkL2(Ω), (3.11) kpkL(Ω) ≤ CkwkL2(Ω). (3.12) Furthermore, for any 2 < q ≤ 4 and any 0 < α < 1, we have

kpkCα(Ω) ≤ Cq,α

kwkL2(Ω)+k∇v0kLq(D)

. (3.13)

Proof. The estimate (3.11) follows directly from the well-posedness of the boundary value problem (3.10). On the other hand, we have

kpkL(Ω) ≤ Ck−k2p + wkL2(Ω) (3.14) by Theorem 8.16 in [3] since P := p satisfies the Dirichlet problem

(∇ · (eγ∇P ) = −k2p + w in Ω,

P = 0 on ∂Ω. (3.15)

(14)

Combining (3.14) and (3.11), we obtain (3.12). Finally, by virtue of Corol- lary 7.3 in [14], we have

kpkCα(Ω) ≤ Cαk−k2p + wkL(Ω)

Then (3.13) follows from by this estimate, (3.12) and (3.2).

We now prove the first upper bound on the L2 norm of w.

Lemma 3.5. We have Z

|w|2dx≤ C Z

D|∇v0|2dx. (3.16) Proof. By the first equation of (3.10), we see that

Z

|w|2dx = Z

w

∇ · eγ∇p

+ k2p dx

=− Z

∇w · eγ∇p dx + k2 Z

w p dx

= Z

∇ · eγ∇w

+ k2w

p dx =− Z

∇ ·

eγ − 1

∇v0 p dx

= Z

eγ− 1

∇v0· ∇p dx = Z

D

γD∇v0· ∇p dx. (3.17) Hence we get (3.16) by the Cauchy-Schwarz inequality and (3.11).

From (3.8), (3.9) and (3.16), we immediately obtain Corollary 3.6.

Z

∂Ω

D − Λ)f f dσ ≤ C

Z

D|∇v0|2dx. (3.18) Now we prove another bound on the L2 norm of w. We first define

Ix0:=

Z

∂D

∂v0

∂ν (x)

|x − x0|αdσ(x)

for any x0 ∈ Ω and 0 < α < 1.

(15)

Lemma 3.7. For any x0 ∈ Ω, 0 < α < 1 and 2 < q ≤ 4, we have Z

|w|2dx≤ Cq,α

Ix20+ Ix0k∇v0kLq(D)+kv0k2L2(D)

. (3.19)

Proof. By (3.17), we get Z

|w|2dx = Z

D

γD∇v0· ∇p dx

=− Z

D∇ · γD∇v0

p dx + Z

∂D

γD

∂v0

∂ν p dσ

=− Z

D

∇γD · ∇v0p dx− Z

D

γD∆v0p dx + Z

∂D

γD

∂v0

∂ν p dσ

=− Z

D∇γD · ∇v0p dx + k2 Z

D

γDv0p dx + Z

∂D

γD

∂v0

∂ν p dσ. (3.20) Hence it is enough to estimate each term of the right-hand side of (3.20).

We first show that we can estimate the first term as

Z

D∇γD · ∇v0p dx

≤ Ckv0kL2(D)kwkL2(Ω). (3.21) Note that

Z

D∇γD · ∇v0p dx =

Z

D

v0∇ · (p∇γD) dx + Z

∂D

v0 ∂γD

∂ν p dσ

≤ C

kv0kL2(D)kpkH1(D)+kv0kH−1/2(∂D)kpkH1/2(∂D)



≤ C

kv0kL2(D)+kv0kH−1/2(∂D)

kpkH1(D).

Therefore, using (3.11), we can obtain (3.21) if we show that

kv0kH−1/2(∂D) ≤ Ckv0kL2(D). (3.22) To derive (3.22), we remark that

kψkH−1/2(∂D) ≤ CkψkH(D)

holds for ψ ∈ D(D) satisfying ψ ∈ L2(D) and ∆ψ ∈ L2(D), wherekψkH(D) is defined bykψk2H(D):=kψk2L2(D)+k∆ψk2L2(D). (see, for example, Lemma 1.1 in [2]). Using this fact, we immediately obtain that

kv0kH−1/2(∂D) ≤ C

kv0kL2(D)+k∆v0kL2(D)

= C

kv0kL2(D)+kk2v0kL2(D)

,

(16)

which is (3.22). To estimate the second term, we simply use (3.11) and obtain

Z

D

γDv0p dx

≤ Ckv0kL2(D)kpkL2(D)≤ Ckv0kL2(D)kwkL2(Ω). (3.23) We now estimate the last term of the right-hand side of (3.20). We have

Z

∂D

γD

∂v0

∂ν p dσ

≤ Z

∂D

γD(x)∂v0

∂ν(x) p(x)− p(x0) dσ(x)

+

p(x0)

Z

∂D

γD

∂v0

∂ν dσ

≤ Cq,α

kwkL2(Ω)+k∇v0kLq(D)



Ix0+ CkwkL2(Ω)

Z

∂D

γD

∂v0

∂ν dσ by (3.13) and (3.12). On the other hand, using (3.22) and H¨older’s inequality, we can estimate

Z

∂D

γD

∂v0

∂ν dσ =

Z

∂D

∂γD

∂ν v0dσ + Z

D

γD∆v0dx− Z

D

∆γDv0dx

= Z

∂D

∂γD

∂ν v0dσ− k2 Z

D

γDv0dx− Z

D

∆γDv0dx

≤ C

kv0kH−1/2(∂D)+kv0kL1(D)

≤ Ckv0kL2(D).

Consequently, we get

Z

∂D

γD

∂v0

∂ν p dσ

≤ Cq,α

kwkL2(Ω)+k∇v0kLq(D)

Ix0+ CkwkL2(Ω)kv0kL2(D). (3.24)

Combining (3.20), (3.21), (3.23) and (3.24), we then have Z

|w|2dx =− Z

D∇γD· ∇v0p dx + k2 Z

D

γDv0p dx + Z

∂D

γD

∂v0

∂ν p dσ

≤ Ckv0kL2(D)kwkL2(Ω)+ Cq,α

kwkL2(Ω)+k∇v0kLq(D) Ix0

≤ eεkwk2L2(Ω)+ C

eεkv0k2L2(D)

+ eεkwk2L2(Ω)+ Cq,α

e

ε Ix20+ Cq,αk∇v0kLq(D)Ix0. The lemma follows by taking eε > 0 small enough.

(17)

4 The main theorem and its proof

In this section, we prove our main theorem, Theorem 4.1, which is the key to our reconstruction method. Here we denote the general constants by C, c > 0.

The constants C and c depend only on Ω, D, γD, k, N, c, t and ε. As in Section 3, when a constant depends on other data, we denote the dependence by subscript.

To begin, substituting v0 = v0,t,h and f = ft,h (= v0,t,h|∂Ω) to (3.18) yields

|E(t, h)| ≤ C Z

D|∇v0,t,h|2dx. (4.1)

On the other hand, applying estimate (3.19) to (3.9) for f = ft,h, we have E(t, h) ≥ c

Z

D|∇v0,t,h|2dx (4.2)

− Cq,α

It,h,x2 0+ It,h,x0k∇v0,t,hkLq(D)+kv0,t,hk2L2(D)



for x0 ∈ Ω, 2 < q ≤ 4 and 0 < α < 1, where It,h,x0 :=

Z

∂D

∂v0,t,h

∂ν (x)

|x − x0|αdσ(x).

Next we introduce a notion of relative curvature with respect to the level curve of ηR. Assume D∩ Γ 6= ∅ and put ΘD := supx∈D∩ΓηR(x). Let x0 ∈ {x ∈ Γ : ηR(x) = ΘD} ∩ ∂D. By simple translation and rotation T , i.e. the change of variables z = T (x−x0), we can take the unit outer normal vector of T (D−x0) at z = 0 as the vector (0, 1). We now put H(z) := ηR(T−1z +x0)− ΘD, and apply the coordinates transformation (ξ1, ξ2) = Ξ(z) := (z1, H(z)) to a neighborhood of z = 0. The transformation Ξ maps ηR(x) = ΘD near x0 to the line ξ2 = 0, and D near x0 to a subdomain of the half space {ξ ∈ R2 : ξ2 ≤ 0}. Let Σ be the image of ∂D near x0 by this transformation (see Figure 4). We then call the curvature of Σ at ξ = 0 the relative curvature to ηR(x) = ΘD of ∂D at x0. We now can state our main theorem.

Theorem 4.1. Assume D∩Γ 6= ∅. Suppose that {x ∈ Γ : ηR(x) = ΘD}∩∂D consists only of one point x0 and the relative curvature to ηR(x) = ΘD of ∂D at x0 is not zero. Then there exist positive constants C1, c1 and h1 such that for any 0 < t ≤ t and 0 < h≤ h1 the following holds:

(18)

0 x2

x1

x0

D

ηR(x) = ΘD

z = T (x− x0) -

z2

z1

0 H(z) = 0

ξ = Ξ(z)-

0 ξ2

ξ1

Σ

Figure 4: The coordinates transformation in defining a relative curvature.

(I) if 1/t > ΘD then

|E(t, h)| ≤







 C1

h2 exp

2 h



−1

t + 1 t + ε/2



if ΘD ≤ 1 t + ε/2, C1

h2 exp

2 h



−1 t + ΘD



if 1

t + ε/2 < ΘD < 1 t. (II) if 1/t≤ ΘD then

E(t, h)≥ c1exp

2 h

−1

t + ΘD

h−1/2. Remark 4.2. If D∩ Γ = ∅ then we can prove

|E(t, h)| ≤ C1

h2 exp

2 h

−1

t + 1 t + ε/2



in the same way as the proof of Theorem 4.1 (I) since ∇Vt,h ≡ 0 in D by (2.5).

Remark 4.3. In the main theorem, Theorem 4.1, we impose some restriction on the curvature of ∂D at x0. However, in Section 5, we will show that the curvature assumption is always satisfied as long as N is large enough.

Proof of Theorem 4.1. (I) By estimate (4.1), Lemma 2.2, and formula (2.5), it is easy to see that

|E(t, h)| ≤ Ck∇Vt,hk2L2(D)+ C

h2e−2at/h

≤ C h2exp

2 h

−1 t + ΘD



+ C

h2e−2at/h for 0 < h≤ 1.

(19)

Thus the estimates of E(t, h) in (I) is obvious.

(II) In view of (4.2), it suffices to estimateR

D|∇v0,t,h|2dx from below and other remaining terms in the right side of (4.2) from above. Using Lemma 2.2 and (2.5), we can get that

k∇v0,t,hk2L2(D)

≥ 1

2k∇Vt,hk2L2(D)− k∇(Vt,h − v0,t,h)k2L2(D)

≥ 1 2

1 hexp

1 h

−1

t + η(x)

φt(x)∇η(x) + S(x)h

2 L2(D)

− C02e−2a/h

≥ 1 4

1 hexp

1 h

−1

t + η(x)

φt(x)∇η(x)

2

L2(D)

−1 2

exp

1 h

−1

t + η(x)

S(x)

2 L2(D)

− C02e−2a/h

≥ c 1 h2

exp

1 h

−1

t + ηR(x)

2

L2(D′′)

− C exp

1 h

−1

t + ηR(x)

2 L2(D)

− C02e−2a/h, (4.3)

where we set D := D ∩S

0<s<t+εls and D′′ := D ∩S

0<s<t+ε/2ls. On the other hand, by (2.4), (2.5), Lemma 2.2 and the Sobolev embedding theorem (for two dimensions), we have

It,h,x0≤ C 1 h

Z

∂D

exp

1 h

−1

t + ηR(x)

|x − x0|αdσ(x) + Ce−a/h, (4.4) k∇v0,t,hkLq(D)≤ C 1

h exp

1 h

−1

t + ηR(x)

Lq(D)

+ Cqe−a/h, (4.5)

kv0,t,hkL2(D)≤ C exp

1 h



−1

t + ηR(x)

L2(D)

+ C0e−a/h. (4.6)

Therefore, our task now is to estimate

exp

1 h

−1

t + ηR(x)

2 L2(D′′)

(4.7)

(20)

from below and exp

1 h

−1

t + ηR(x)

Lq(D)

, (4.8)

Z

∂D

exp

1 h

−1

t + ηR(x)

|x − x0|αdσ(x) (4.9) from above, where the index q in (4.8) is q = 2 (for (4.3) and (4.6)) or 2 < q ≤ 4 (for (4.5)).

We first look at (4.7). By translation and rotation with the orthogonal matrix T , we can assume that x0 = 0 and the unit outer normal vector of

∂D at x0 = 0 is (0, 1). Then we can see that (4.7) =

ZZ

T (D′′−x0)

exp

2 h



H(z)− 1 t + ΘD



dz

= exp

2 h

−1 t + ΘD

 Z Z

T (D′′−x0)

e2H(z)/hdz

where we have used H(z) = ηR(T−1z + x0)− ΘD. We now make the change of variables (ξ1, ξ2) = Ξ(z) := (z1, H(z)). Notice that there exists a neigh- borhood U0 of z = 0 such that the map ξ = Ξ(z) is injective from U0 to Ξ(U0) since we have det(∂Ξ/∂z)(0) = N|x0|N −1 6= 0. In particular, there exist positive constants a and a such that

a ≤ det∂Ξ

∂z ≤ a in U0, i.e. 1

a ≤ det∂z

∂ξ ≤ 1 a

in Ξ(U0).

Consequently, we have ZZ

T (D′′−x0)

e2H(z)/hdz ≥ ZZ

T (D′′−x0)∩U0

e2H(z)/hdz ≥ 1 a

ZZ

Ue0

e2/hdξ,

where eU0 := Ξ(T (D′′−x0)∩U0). We now parameterize the boundary ∂ eU0near 0. We remark that the boundary ∂ eU0 near 0 is the image of ∂D near x0under the coordinates transform given above. Therefore we can parameterize the boundary ∂ eU0near 0 by ξ2= l(ξ1) (we may choose a smaller neighborhood U0

if needed), and express eU0 near 0 as ξ2 ≤ l(ξ1). Moreover, by the assumption on the curvature of ∂D, there exist positive constants K and K such that

Kξ2 ≤ −l(ξ1)≤ Kξ2 for (ξ1, l(ξ1))∈ Ξ(U0).

(21)

0 ξ2

ξ1

Ue0

ξ2 = l(ξ1)

0 ξ2

ξ1

δ1

−δ1

ξ2 =−Kξ12 ξ2 =−Kξ12 Figure 5: The choice of δ1 > 0.

Then for δ1 > 0 small enough (see Figure 5), we can estimate Z Z

Ue0

e2/hdξ≥ ZZ

1|≤δ1

−Kδ12≤ξ2≤l(ξ1)

e2/h12

≥ ZZ

1|≤δ1

−Kδ12≤ξ2≤−Kξ12

e2/h12

= 2

√K Z 0

−Kδ21

e2/hp

−ξ22

= h3/2 (2K)1/2

Z 2Kδ12/h 0

e−ττ1/2dτ ≥ c h3/2 for any 0 < h ≪ 1. Summing up, we obtain

exp

1 h

−1

t + ηR(x)

2 L2(D′′)

≥ c h3/2exp

2 h

−1 t + ΘD



(4.10) for any 0 < h ≪ 1.

Next, we estimate (4.8). It is enough to estimate the integral on some neighborhood of x0. Indeed, when Ux0 is a neighborhood of x0, there exists

(22)

0 ξ2

ξ1 ξ2 = l(ξ1)

Ue

0 ξ2

ξ1

ξ2 =−Kξ12 ξ2 =−Kξ12 δ3

−δ3

Figure 6: The choice of δ3 > 0.

δ2 > 0 such that D\ Ux0 ⊂ {x ∈ Γ : ηR(x)≤ ΘD− δ2}. Then we obtain

exp

1 h

−1

t + ηR(x)

Lq(D\Ux0)

≤ exp

1 h

−1 t + ΘD

 e−δ2/h

Lq(D\Ux0)

≤ C exp

1 h

−1

t + ΘD − δ2



. (4.11)

Here we use the same notations as in estimating (4.7) (Denote U0 = T (Ux0− x0)). Using the similar arguments as above, we can derive

exp

1 h

−1

t + ηR(x)

q

Lq(D∩Ux0)

= exp

q h

−1 t + ΘD

 ZZ

T((D∩Ux0)−x0)eqH(z)/hdz

≤ 1 a

exp

q h

−1 t + ΘD

 ZZ

Ue

e2/hdξ (4.12)

where eU := Ξ (T ((D∩ Ux0)− x0)). Now we can take δ3 > 0 such that Ue ⊂ {ξ ∈ R2 :|ξ1| ≤ δ3, −Kδ32 ≤ ξ2 ≤ l(ξ1)} by choosing the neighborhood Ux0 of x0 small enough (see Figure 6). Then we have

(23)

ZZ

Ue

e2/hdξ ≤ Z Z

1|≤δ3

−Kδ32≤ξ2≤l(ξ1)

e2/h

≤ Z Z

1|≤δ3

−Kδ32≤ξ2≤−Kξ12

e2/hdξ ≤ Cqh3/2 (4.13)

for any 0 < h ≪ 1. Combining (4.11), (4.12), (4.13) yields

exp

1 h

−1

t + ηR(x)

Lq(D)

≤ Cqh3/(2q)exp

1 h

−1 t + ΘD

 (4.14)

for any 0 < h ≪ 1.

Lastly, we turn to (4.9). As before, it suffices to estimate the integral on some neighborhood of x0. Thus we compute

Z

∂D∩Ux0

exp

1 h

−1

t + ηR(x)

|x − x0|αdσ(x)

= exp

1 h

−1 t + ΘD

 Z

T((∂D∩Ux0)−x0)eH(z)/h|z|αdσ(z).

Then by choosing a sufficiently small neighborhood Ux0 of x0, we have Z

T((∂D∩Ux0)−x0)eH(z)/h|z|αdσ(z)

≤ C Z δ4

−δ4

e−Kz21/h|z1|αdz1

= h(α+1)/2 C K(α+1)/2

Z Kδ24/h 0

e−ττ(α−1)/2dτ ≤ Cαh(α+1)/2 for any 0 < h ≪ 1 since

Kz12 ≤ −H(z) ≤ Kz12 for z ∈ T (∂D − x0) close to 0.

Therefore we obtain Z

∂D

exp

1 h

−1

t + ηR(x)

|x − x0|αdσ(x)

≤ Cαh(α+1)/2exp

1 h

−1 t + ΘD



(4.15)

(24)

for any 0 < h ≪ 1.

Now by (4.2), (4.3), (4.4), (4.5), (4.6), (4.10), (4.14) and (4.15), we con- clude that

E(t, h)≥

ch12 − Cq,α



hα−1+ hα2+23q32 + h32

× exp

2 h

−1 t + ΘD

 (4.16)

for any 0 < h ≪ 1. It is easy to see that we can choose 0 < α0 < 1 and 2 < q0 ≤ 4 such that

β := max

0

2 + 3 2q0 − 3

2, α0− 1



>−1 2 Then (4.16) implies

E(t, h)≥ ch−1/2− Cq00hβ exp

2 h

−1 t + ΘD



≥ ch−1/2exp

2 h



−1 t + ΘD



for any 0 < h ≪ 1.

In view of Theorem 4.1, we are able to reconstruct some part of ∂D using boundary measurements on ∂Ω by looking into the asymptotic behavior of E(t, h) for various t’s. More precisely, let

tD := supn

t∈ (0, t) : lim

h→0E(t, h) = 0o .

It should be noted that E(t, h) depends on, besides h and t, Ω, k, D, γD (see (1.1)), c, x, N (appear in the phase function η(x)), ε, t (appear in the cut-off function φt(x)). Thus tD depends on Ω, k, D, γD , c, x, N, ε and t. If tD = t, then D∩S

0<s<tls = ∅. On the other hand, if tD < t, then there exists a point xD ∈ ltD ∩ ∂D.

By taking N arbitrarily large (the opening angle of Γ(N, θ) becomes arbitrarily small), we can reconstruct even more information of ∂D. A point x0 on ∂D is said to be detectable if there exists a semi-straight line L starting from x0 such that L does not intersect ∂D except x0. For example, if D is star-shaped, every point of ∂D is detectable. We can prove the following corollary similarly to Corollary 5.4 in [18].

Corollary 4.4. Every detectable point of ∂D can be reconstructed from ΛD.

(25)

0 x2

x1

x0

ηR(x; c, N) = 1/t0

ηR(x; c, N) = 1/t

Figure 7: Intersection of two level curves.

5 Remarks on the curvature assumption

In this section we would like to show that the curvature condition assumed in Theorem 4.1 always holds provided the degree N of η is sufficiently large.

To this end, in order to indicate the dependence on c, θ, and N, we write η(x; c, N) = η(x) = c(x1 + ix2)N and similarly write ηR(x) = ηR(x; c, N).

Also, we denote

Γ(N, θ) :=n

r(cos θ, sin θ) : |θ − θ| < π 2N

o .

Recall that ηR(x) = rNcos N(θ−θ) for x = r(cos θ, sin θ)∈ R2 and ηR(x) >

0 when x ∈ Γ(N, θ).

Lemma 5.1. Let c ∈ C satisfy |c| = 1, N be a positive integer and t0 > 0.

Assume that x0 is on the level curve ηR(x; c, N) = 1/t0. Then there exist c ∈ C with |c| = 1, θ ∈ R, a positive integer N > N and a positive number t such that ηR(x; c, N) = rNcos N(θ− θ) and the following holds:

(i) The point x0 is on the level curve ηR(x; c, N) = 1/t.

(ii) If ηR(x; c, N) ≥ 1/t and x ∈ Γ(N, θ)\ {x0} then ηR(x; c, N) >

1/t0.

Proof. Without loss of generality, we can assume that c = 1. Let x0 = r0(cos θ0, sin θ0). By the assumption, we have rN0 cos Nθ0 = 1/t0 and |θ0| <

π/(2N). Now we choose a positive integer N > N such that |(N− N)θ0

Figure

Updating...

References

Related subjects :