Inverse problem for wave propagation in a perturbed layered half-space

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Mathematical and Computer Modelling 45 (2007) 21–33 www.elsevier.com/locate/mcm Inverse problem for wave propagation in a perturbed layered half-space Robert Gilbert a,* , Klaus Hackl b , Yongzhi Xu c a Department of Mathematics, University of Delaware, Newark, DE 19716, United States b Lehrstuhl f¨ ur Allgemeine Mechanik, Ruhr Universit¨ at Bochum, Bochum, Germany c Department of Mathematics, University of Louisville, Louisville, KY 40292, United States Received 3 March 2006; accepted 7 March 2006 Abstract This paper is concerned with the inverse medium scattering problem in a perturbed, layered, half-space, which is a problem related to the seismologial investigation of inclusions inside the earth’s crust. A wave penetrable object is located in a layer where the refraction index is different from the other part of the half-space. Wave propagation in such a layered half-space is different from that in a homogeneous half-space. In a layered half-space, a scattered wave consists of a free wave and a guided wave. In many cases, only the free-wave far-field or only the guided-wave far-field can be measured. We establish mathematical formulas for relations between the object, the incident wave and the scattered wave. In the ideal condition where exact data are given, we prove the uniqueness of the inverse problem. A numerical example is presented for the reconstruction of a penetrable object from simulated noise data. c 2006 Elsevier Ltd. All rights reserved. Keywords: Inverse problems 1. Introduction In this paper, we study the inverse medium scattering problem in a perturbed layered half-space. A wave penetrable object is located in a layer where the refraction index is different from the other part of the half-space. Let R 2 + ={x R 2 | x 2 0}, where x = (x 1 , x 2 ). n(x) is defined on R 2 + with the following properties: Let n 0 (x 2 ) = n 0 , for 0 < x 2 < h , 1, for h < x 2 < , (1.1) where n 0 > 1 is a constant. We assume that the inhomogeneity is contained in a bounded domain Ω R 2 h ={x R 2 | 0 < x 2 < h } with a C 2 boundary having an outward-pointing normal vector ν . n(x) = n 0 (x 2 ), for x Ω . (1.2) * Corresponding author. Tel.: +1 302 831 2315; fax: +1 302 831 4456. E-mail address: [email protected] (R. Gilbert). 0895-7177/$ - see front matter c 2006 Elsevier Ltd. All rights reserved. doi:10.1016/j.mcm.2006.03.020

Transcript of Inverse problem for wave propagation in a perturbed layered half-space

Page 1: Inverse problem for wave propagation in a perturbed layered half-space

Mathematical and Computer Modelling 45 (2007) 21–33www.elsevier.com/locate/mcm

Inverse problem for wave propagation in a perturbed layeredhalf-space

Robert Gilberta,∗, Klaus Hacklb, Yongzhi Xuc

a Department of Mathematics, University of Delaware, Newark, DE 19716, United Statesb Lehrstuhl fur Allgemeine Mechanik, Ruhr Universitat Bochum, Bochum, Germany

c Department of Mathematics, University of Louisville, Louisville, KY 40292, United States

Received 3 March 2006; accepted 7 March 2006

Abstract

This paper is concerned with the inverse medium scattering problem in a perturbed, layered, half-space, which is a problemrelated to the seismologial investigation of inclusions inside the earth’s crust. A wave penetrable object is located in a layer wherethe refraction index is different from the other part of the half-space. Wave propagation in such a layered half-space is differentfrom that in a homogeneous half-space. In a layered half-space, a scattered wave consists of a free wave and a guided wave. Inmany cases, only the free-wave far-field or only the guided-wave far-field can be measured.

We establish mathematical formulas for relations between the object, the incident wave and the scattered wave. In the idealcondition where exact data are given, we prove the uniqueness of the inverse problem. A numerical example is presented for thereconstruction of a penetrable object from simulated noise data.c© 2006 Elsevier Ltd. All rights reserved.

Keywords: Inverse problems

1. Introduction

In this paper, we study the inverse medium scattering problem in a perturbed layered half-space. A wave penetrableobject is located in a layer where the refraction index is different from the other part of the half-space.

Let R2+ = x ∈ R2

| x2 ≥ 0, where x = (x1, x2). n(x) is defined on R2+ with the following properties:

Let

n0(x2) =

n0, for 0 < x2 < h,

1, for h < x2 < ∞,(1.1)

where n0 > 1 is a constant. We assume that the inhomogeneity is contained in a bounded domain Ω ∈ R2h = x ∈

R2| 0 < x2 < h with a C2 boundary having an outward-pointing normal vector ν.

n(x) = n0(x2), for x 6∈ Ω . (1.2)

∗ Corresponding author. Tel.: +1 302 831 2315; fax: +1 302 831 4456.E-mail address: [email protected] (R. Gilbert).

0895-7177/$ - see front matter c© 2006 Elsevier Ltd. All rights reserved.doi:10.1016/j.mcm.2006.03.020

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22 R. Gilbert et al. / Mathematical and Computer Modelling 45 (2007) 21–33

We consider the following problem: given a point source or a plane, incident wave ui satisfying

4ui+ k2n2

0(x2)ui= f (x), in R2

+, (1.3)

find the total field u = us+ ui such that

4us+ k2n2(x)us

= 0, in R2+, (1.4)

u(x) = us(x) + ui (x) = 0, when x2 = 0. (1.5)

Let ∂u−

∂ν(x1, x2) and u− (x1, x2) be the limits of ∂u

∂ν(x1, x2) and u (x1, x2) as (x1, x2) approaches ∂Ω from the interior

of Ω , and let ∂u+

∂ν(x1, x2) and u+ (x1, x2) be the limits of ∂u

∂ν(x1, x2) and u (x1, x2) as (x1, x2) approaches ∂Ω from

the exterior of Ω , respectively. We assume, for some constants ρ0 and ρ, that u satisfies interface conditions

ρ0u+(x) = ρu−(x),∂u+(x)

∂ν=

∂u−(x)

∂ν, on ∂Ω . (1.6)

Moreover, us also satisfies the out-going radiation condition; that is, no wave is coming in from infinity except theprescribed incident wave plane wave.

Here, 4 =∂2

∂x21

+∂2

∂x22

, ui , us and u are the incident, scattered and total field of the time harmonic waves.

This setting has considerable significance in the seismological investigation of geological objects that are locatedin the earth’s crust, particularly volcanic magma chambers. In this case, the refraction index is normalized to 1 withinthe earth’s mantle, while it is n0 within the crust, where it is known to be significantly different, mainly due to thelower density of the crust material. The refraction index of the mantle material inside the magma chamber is close tobut different from that of the mantle itself.

Until today, only the depth of the volcanic magma chambers is roughly known, but no information is availableconcerning the size, shape and exact location. Estimates range from between 1 and 2 km and about 30 km for the depthand between 0.1 and 0.2 km and about 5 km for the diameter. There is even the possiblity of multiple interconnectedchambers. For detailed information, see [1,3–6].

On the other hand, there is a rich natural source of incoming waves due to micro-earthquakes occuring permanentlybeneath active volcanoes and in the vicinity of magma chambers. The strength and location of those earthquakes canbe determined quite accurately, which defines a dense field of point sources surrounding the chamber at variousdistances.

The methodology developed in this paper applied to the seismological data available should provide a verysignificant increase in information on volcanic magma chambers compared with what is known at present. It mighteven be possible to monitor processes occuring within the chamber in this manner, such as the intrution of freshmagma rising from the mantle. Knowledge of this kind may then contribute to more accurate predicitons of volcaniceruptions.

2. Green’s function and its far-field behavior

In this section, we outline the construction of the Green’s function for the time-harmonic waves in a layered half-space and obtain its free-wave and guided-wave far-field behavior.

A function G( · ; x01 , x0

2) is called the outgoing Green’s function from the source at x0= (x0

1 , x02) for the time-

harmonic wave in a layered half-space, if G(x1, x2; x01 , x0

2) satisfies

4G + k2n20(x2)G = −δ(|x − x0

|), in R2+ (2.1)

in the generalized function sense,

G(x1, 0; x01 , x0

2) = 0, (2.2)

and satisfies the out-going radiation condition.

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R. Gilbert et al. / Mathematical and Computer Modelling 45 (2007) 21–33 23

Using the Fourier transform, and considering the outgoing property in the horizontal direction, we obtain arepresentation for G(x1, x2; x0

1 , x02) as

G(x1, x2; x01 , x0

2) =1

∫∞

−∞

G(ξ, x2; x01 , x0

2)eiξ |x1−x01 |dξ, (2.3)

where G(ξ, x2; x01 , x0

2) satisfies

G ′′(x2) + [k2n20(x2) − ξ2

]G(x2) = −δ(x2 − x02), 0 < x2 < ∞, (2.4)

G(ξ, 0; x01 , x0

2) = 0, (2.5)

and

G(ξ, x2; x01 , x0

2) = Aei√

k2n20−ξ2x2

+ O

(1

|x2|

), as x2 → ∞, (2.6)

where A is a constant.

Let τ =

√k2 − ξ2 and τ0 =

√k2n2

0 − ξ2. Define

φ+(ξ, x2) =

eiτ x2 , h < x2 < ∞

B1eiτ0x2 + B2e−iτ0x2 , 0 < x2 < h,(2.7)

φ−(ξ, x2) =

sin(τ0x2), 0 < x2 < hC1 cos(τ x2) + C2 sin(τ x2), h < x2 < ∞,

(2.8)

where

B1 =12

(1 +

τ

τ0

)ei(τ−τ0)h,

B2 =12

(1 −

τ

τ0

)ei(τ+τ0)h,

C1 = cos(τh) sin(τ0h) −τ0

τsin(τh) cos(τ0h),

C2 = sin(τh) sin(τ0h) +τ0

τcos(τh) cos(τ0h).

Then G(ξ, x2; x01 , x0

2) has the representation

G(ξ, x2; x01 , x0

2) =φ+(ξ, x2>)φ−(ξ, x2<)

W (φ+, φ−), (2.9)

where

x2> = maxx2, x02, x2< = minx2, x0

2,

W (φ+, φ−) is the Wronskian:

W (φ+, φ−) = φ+d

dx2φ−

− φ−d

dx2φ+

= ei(τh) [τ0 cos(τ0h) − iτ sin(τ0h)] . (2.10)

We obtain, from (2.3) and (2.9),

G(x1, x2; x01 , x0

2) =1

∫∞

−∞

φ+(ξ, x2>)φ−(ξ, x2<)

W (p+, p−)eiξ |x1−x0

1 |dξ. (2.11)

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24 R. Gilbert et al. / Mathematical and Computer Modelling 45 (2007) 21–33

Fig. 1. Contour and branch cuts.

Setting the Wronskian to 0, we have

tan(τ0h) = iτ

τ0. (2.12)

Eq. (2.12) has at most a finite number of solusions for ξ2 > k2. The outgoing radiation condition as |x1| → ∞ impliesthat the positive zeros should be used. Denote them by k < ξ1 < ξ2 < · · · < ξN . These are poles of the integrand.There are branch points at ±k and ±kn0. Choosing the branch cuts C and C0 shown in Fig. 1 for τ =

√k2 − ξ2 and

τ0 =

√k2n2

0 − ξ2, and denoting τn =√

k2 − ξ2n and τ0n =

√k2n2

0 − ξ2n , we can represent the Green’s function as:

G(x1, x2; x01 , x0

2) = G f (x1, x2; x01 , x0

2) + Ge(x1, x2; x01 , x0

2) + Gg(x1, x2; x01 , x0

2)

=1

∫Γ

φ+(ξ, x2>)φ−(ξ, x2<)eiξ |x1−x01 |

eiτh[τ0 cos(τ0h) − iτ sin(τ0h)]dξ +

12π

∫Γ0

φ+(ξ, x2>)φ−(ξ, x2<)eiξ |x1−x01 |

eiτh[τ0 cos(τ0h) − iτ sin(τ0h)]dξ

+

N∑n=1

φ+(ξn, x2>)φ−(ξn, x2<)eiξn |x1−x01 |

ddξ

W (φ+, φ−)

∣∣∣ξ=ξn

, (2.13)

where Γ = AO ∪ O B ∪ BO ′ ∪ O ′ A′, and Γ0 = DC ∪ C D′.A straightforward computation shows that

ddξ

W (φ+, φ−)

∣∣∣∣ξ=ξn

=ξ(i + hτn)eiτnh

τ 20nτn

(k2n20 − k2) sin(τ0nh).

Since all ξn > k, τn = i |τn|. For x2 > h,

Gg(x1, x2; x01 , x0

2) =

N∑n=1

e−|τn |x2 sin(ξn, x02)eiξn |x1−x0

1 |

ξ(i+hτn)eiτn h

τ 20nτn

(k2n20 − k2) sin(τ0nh)

= O(

e−|τ1|x2)

. (2.14)

Gg(x1, x2; x01 , x0

2) corresponds to the guided wave that is trapped in the layer 0 < x2 < h.

Changing variable ζ =

√k2 − ξ2, we have τ0 =

√ζ 2 + k2(n2

0 − 1), ξ =

√k2 − ζ 2, and the contour Γ becomes

−∞ < ζ < ∞:

G f (x1, x2; x01 , x0

2) =1

∫Γ

eiξ |x1−x01 |+iτ x2 sin(τ0x0

2)

eiτh[τ0 cos(τ0h) − iτ sin(τ0h)]dξ

=1

∫∞

−∞

ei√

k2−ζ 2|x1−x01 |+i

√ζ 2+k2(n2

0−1)x2Φ f (ζ )dζ

=1

∫∞

−∞

eirΘ f (ζ )Φ f (ζ )dζ (2.15)

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R. Gilbert et al. / Mathematical and Computer Modelling 45 (2007) 21–33 25

where (|x1 − x01 |, x2) = r(| cos θ |, sin θ),

Θ f (ζ ) =

√k2 − ζ 2| cos θ | +

√ζ 2 + k2(n2

0 − 1) sin θ,

and

Φ f (ζ ) =

−ζ sin(√

ζ 2 + k2(n20 − 1)x0

2

)eiζh

√k2 − ζ 2

/ [√ζ 2 + k2(n2

0 − 1) cos(√

ζ 2 + k2(n20 − 1)h

)−iζ sin

(√ζ 2 + k2(n2

0 − 1)h

)].

The phase function Θ f (ζ ) has stationary points:

ζ = 0, and, ζ± = ±ζ0 := ±k√

1 − n20 cos2 θ.

Using the Method of Stationary Phase, we obtain an approximation of G f (x1, x2; x01 , x0

2) for x2 > h and r = |x|

large:

G f (x1, x2; x01 , x0

2) =eikr

√2π ir

[G∞

f (θ; x01 , x0

2)]

+ O

(1

r3/2

)

:=eikr

√2π ir

[Φ f (ζ+)√Θ ′′(ζ+)

+Φ f (0)

√Θ ′′(0)

+Φ f (ζ−)√Θ ′′(ζ−)

]+ O

(1

r3/2

). (2.16)

G f (x1, x2; x01 , x0

2) corresponds to the free wave that propagates to the half-plane. G∞

f (θ; x01 , x0

2) is the free-wavefar-field pattern.

To estimate Ge, we change variable ζ = −i√

k2n20 − ξ2, where we have τ = i

√ζ 2 + k2(n2

0 − 1), ξ =√k2n2

0 + ζ 2, and the contour Γ0 becomes −∞ < ζ < ∞:

Ge(x1, x2; x01 , x0

2) =1

∫Γ0

eiξ |x1−x01 |+iτ x2 sin(τ0x0

2)

eiτh[τ0 cos(τ0h) − iτ sin(τ0h)]dξ

=1

∫∞

−∞

ei√

ζ 2+k2n20|x1−x0

1 |−

√ζ 2+k2(n2

0−1)x2Φe(ζ )dζ

=1

∫∞

−∞

eirΘe(ζ )Φe(ζ )dζ (2.17)

where (|x1 − x01 |, x2) = r(| cos θ |, sin θ),

Θe(ζ ) =

√ζ 2 + k2n2

0| cos θ |,

and

Φe(ζ ) =ζ sin(ζ x0

2)e−

√ζ 2+k2(n2

0−1)(x2−h)√ζ 2 + k2n2

0

/ [ζ cos(ζh) +

√ζ 2 + k2(n2

0 − 1) sinh(ζh)

].

The phase function Θe(ζ ) has stationary point ζ = 0. But

Φe(0) = limζ→0

Φe(ζ ) = 0.

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26 R. Gilbert et al. / Mathematical and Computer Modelling 45 (2007) 21–33

Using the Method of Stationary Phase, we obtain an approximation of Ge(x1, x2; x01 , x0

2) for x2 > h and r = |x|

large:

Ge(x1, x2; x01 , x0

2) = O

(1

r3/2

). (2.18)

Ge(x1, x2; x01 , x0

2) corresponds to the evanescent wave that does not propagate in any direction.For a related discussion concerning free waves and guided waves in stratified media, see [14–19]. For inverse

problems in a homogeneous half-space, see, for example, [2,9] and the papers cited therein.

3. Scattered wave in perturbed layered half-space

Now we consider the case where there exists a given inhomogeneity in the layer 0 < x2 < h. We assume that theincident wave is from a point source located at (x0

1 , x02) above the layer; i.e., x0

2 > h.Denote

m(x) = k2[n(x) − n0(x2)], for x ∈ R2

+. (3.1)

We rewrite the Eqs. (1.3) and (1.4) in the form

4u + k2n20u = −m(x)u − δ(x1 − x0

1)δ(x2 − x02), a.e. in R2

+. (3.2)

Let G(x1, x2; x01 , x0

2) be the Green’s function for the layered half-space with a source at(x0

1 , x02

). Multiplying both

sides of Eq. (3.2) by G(ξ1, ξ2; x1, x2) and integrating over Ω , we have∫Ω

G(ξ1, ξ2; x1, x2)[4u(ξ1, ξ2) + (k2n20)

2u(ξ1, ξ2)] dξ1 dξ2

=

∫Ω

G(ξ1, ξ2; x1, x2)[−m(ξ1, ξ2)u(ξ, ζ ) − δ(ξ1 − x01) δ(ξ2 − x0

2)] dξ1 dξ2. (3.3)

It follows by Green’s theorem that

u(x1x2) = G(

x01 , x0

2 ; x1, x2

)−

∫Ω

G(ξ1, ξ2; x1, x2)m(ξ1, ξ2)u(ξ1, ξ2)dξ1 dξ2

+

∫∂Ω

(u+ − u−)∂G(ξ1, ξ2; x1, x2)

∂νds, (x1, x2) ∈ R+

2 . (3.4)

Now we deduce the integral equations that determine u(x1, x2). Let

φ(ξ1, ξ2) = u+(ξ1, ξ2) − u−(ξ1, ξ2). (3.5)

From Eq. (3.4), it follows that

u(x1, x2) +

∫Ω

G(ξ1, ξ2; x1, x2)m(ξ1, ξ2)u(ξ1, ξ2)dξ1 dξ2 −

∫∂Ω

φ(ξ1, ξ2)∂G

∂ν(ξ1, ξ2; x1, x2)ds

= G(x01 , x0

2 ; x1, x2), (x1, x2) ∈ Ω . (3.6)

Using the interface conditions (1.6), we deduce that

φ(x1, x2) +2 (ρ0 − ρ)

ρ0 + ρ

∫∂Ω

φ(ξ1, ξ2)∂G

∂ν(ξ1, ξ2; x1, x2)ds

−2(ρ0 − ρ)

ρ0 + ρ

∫Ω

G(ξ1, ξ2; x1, x2)m(ξ1, ξ2)u(ξ1, ξ2)dξ1 dξ2

= −2(ρ0 − ρ)

ρ0 + ρG(x0

1 , x02 ; x1, x2), (x1, x2) ∈ ∂Ω . (3.7)

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R. Gilbert et al. / Mathematical and Computer Modelling 45 (2007) 21–33 27

Here we have used the facts that

G(x01 , x0

2 ; x1, x2) −

∫Ω

G(ξ1, ξ2; x1, x2)m(ξ1, ξ2)u(ξ1, ξ2)dξ1 dξ2

is continuous across ∂Ω , and that

v(x1, x2) =

∫∂Ω

φ(ξ1, ξ2)∂G

∂ν(ξ1, ξ2; x1, x2)ds

satisfies the jump conditions

v+(x1, x2) =12φ(x1, x2) +

∫∂Ω

φ(ξ1, ξ2)∂G

∂ν(ξ1, ξ2; x1, x2) dξ1 dξ2,

v−(x1, x2) = −12φ(x1, x2) +

∫∂Ω

φ (ξ1, ξ2)∂G

∂ν(ξ1, ξ2; x1, x2) dξ1 dξ2,

v+(x1, x2) − v−(x1, x2) = φ(x1, x2), (x1, x2) ∈ ∂Ω .

Using the above analysis, we have the following theorem. (See [20] for a discussion of a similar theorem.)

Theorem 3.1. If (u, φ) satisfies the direct scattering problem (1.4)–(1.6), then (u, φ) satisfies the integral equations(3.6) and (3.7).

Conversely, if (u, φ) ∈ C(Ω) × C(∂Ω) is a solution of the integral equations (3.6) and (3.7), then (u, φ) is asolution of the direct scattering problem.

Theorem 3.2. If M := maxm(ξ1, ξ2) and |ρ0 − ρ| are small enough, then the system of integral equations (3.6)and (3.7) has a unique solution.

Proof. Rewrite (3.6) in operator form as

u + Tu = f, (3.8)

where

Tu(x1, x2) :=

∫Ω

G(ξ1, ξ2; x1, x2)m(ξ1, ξ2)u(ξ1, ξ2)dξ1 dξ2

and

f (x1, x2) :=

∫∂Ω

φ(ξ1, ξ2)∂G

∂ν(ξ1, ξ2; x1, x2)ds + G

(x0

1 , x02 ; x1, x2

).

If M is small enough, then the operator I + T has a bounded inverse operator (I + T)−1, where I denotes the identityoperator in L2(Ω). Substituting u = (I + T)−1 f into (3.7), we obtain

φ(x1, x2) +2(ρ0 − ρ)

ρ0 + ρSφ(x1, x2) −

2(ρ0 − ρ)

ρ0 + ρT (I + T)−1 f (x1, x2)

= −2(ρ0 − ρ)

ρ0 + ρG(x0

1 , x02 ; x1, x2), (3.9)

where

Sφ(x1, x2) =

∫∂Ω

φ(ξ1, ξ2)∂G

∂ν(ξ1, ξ2; x1, x2)ds.

The operators S and T (I + T)−1 are bounded in L2(∂Ω) and L2(Ω), respectively, as are their their composites.If |ρ0 − ρ| is small enough, then (3.9) has a unique solution φ ∈ L2(∂Ω) and, by substituting φ into (3.8), we candetermine u uniquely.

For a given Ω , m(ξ1, ξ2), ρ0, ρ and source location (x01 , x0

2), we can determine u(x1, x2) in Ω and φ = u+− u−

on ∂Ω by integral equations (3.6) and (3.7). Hence we can compute the wave field using (3.4).

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28 R. Gilbert et al. / Mathematical and Computer Modelling 45 (2007) 21–33

Fig. 2. Point-source waves: source in the layer.

From (3.4), (2.14), (2.16) and (2.18), us(x1x2) = u(x1x2) − G(x01 , x0

2 ; x1, x2) has asymptotic representation

us(x1, x2; x01 , x0

2) = usf (x1, x2; x0

1 , x02) + us

e(x1, x2; x01 , x0

2) + usg(x1, x2; x0

1 , x02).

Here

usg(x1, x2; x0

1 , x02) =

N∑n=1

ugn(x0

1 , x02)e−|τn |x2eiξn |x1−x0

1 |= O

(e−|τ1|x2

), (3.10)

where usg(x1, x2; x0

1 , x02) corresponds to the guided wave for the inhomogeneous waveguide,

usf (x1, x2; x0

1 , x02) =

eikr

√2π ir

[u∞

f (θ, x01 , x0

2)]

+ O

(1

r3/2

). (3.11)

usf (x1, x2; x0

1 , x02) corresponds to the free wave that propagates in the half-plane.

use(x1, x2; x0

1 , x02) = O

(1

r3/2

). (3.12)

use(x1, x2; x0

1 , x02) corresponds to the evanescent waves that do not propagate in any direction.

In the case that ρ0 = ρ, (3.9) becomes

φ(x1, x2) = 0, (x1, x2) ∈ ∂Ω , (3.13)

and the system of integral equations reduces to a single integral equation:

u(x1, x2) +

∫Ω

G(ξ1, ξ2; x1, x2)m(ξ1, ξ2)u(ξ1, ξ2)dξ1 dξ2 = G(x01 , x0

2 ; x1, x2). (3.14)

If M is small, we can approximate u(x1, x2) using the algorithm

un+1(x1, x2) = G(x01 , x0

2 ; x1, x2) −

∫Ω

G(ξ1, ξ2; x1, x2)m(ξ1, ξ2)un(ξ1, ξ2; x01 , x0

2)dξ1 dξ2, (3.15)

for n = 0, 1, 2, . . ., where u0 = 0.Figures showing the acoustic field in the wave guide for sources in and exterior to the layer are shown in Figs. 2–4.

4. Inverse scattering problem and a uniqueness theorem

In this section, we assume that ρ = ρ0. Let

Γ = (x1, x2) ∈ R2+ | x2 = x0

2 = constant, (4.1)

and

Γs = (x s1, x s

2) ∈ R2h | x s

2 = x s02 = constant. (4.2)

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R. Gilbert et al. / Mathematical and Computer Modelling 45 (2007) 21–33 29

Fig. 3. Point-source waves: source out of the layer.

We assume that both Γ and Γs are ‘above’ the layer R2h , i.e., minx2 | x ∈ Ω > h. The inverse problem that we

consider is the following:Given us(x, x s) for x ∈ Γ and x s

∈ Γs , determine n(x).We first prove the following uniqueness result.

Theorem 4.1. Assume that ρ = ρ0. Let n1, n2 be two indices of refraction with n1(x) = n2(x) = n0(x2) for all x 6∈ Ωwhere Ω ⊂ R2

h , such that n1 − n0, n2 − n0 ∈ C2(R2h). Let u1(x, xs), u2(x, xs) be the corresponding scattered waves

from an acoustic source xs . If u1(x, xs) = u2(x, xs) for all x ∈ Γ and xs ∈ Γs , then n1 = n2.

Our uniqueness theorem depends on the following three Lemmas.

Lemma 4.2. Let n ∈ C2(R2h) with n(x) = n0 for x 6∈ Ω . Let u(·, xs) be the total field corresponding to the incident

field G(·, xs), xs ∈ Γs = x2 = x2s | (x1, x2) ∈ R2+. Define the space

H := v ∈ C2(B) | 4v + mv = 0 in B

where Ω ⊂ B ⊂ R2h is a simple region with C2 boundary. Then Spanu (·, xs)|Ω : xs ∈ Γs is dense in H |Ω with

respect to the norm in L2(Ω).

Proof. Let v ∈ H with

(v, u(·, xs))L2(Ω) :=

∫Ω

v(x)u(x, xs)dx = 0, ∀xs ∈ Γs .

Using

u(x1, x2) = G (x1s, x2s; x1, x2) −

∫Ω

G(ξ1, ξ2; x1, x2)(1 − n(ξ1, ξ2))u(ξ1, ξ2)dξ1 dξ2 (4.3)

and M small, it follows that

u = (I + T )−1G, (4.4)

where

T u(x) :=

∫Ω

G(ξ1, ξ2; x1, x2)(m(ξ1, ξ2))u(ξ1, ξ2)dξ1 dξ2, x ∈ Ω .

It follows that

0 = (v, (I + T )−1G(·, (x1s, x2s)))L2(Ω) = ((I + T ∗)−1v, G(·, (xs, zs)))L2(Ω), ∀x1s, x2s ∈ Γs . (4.5)

Here,

T ∗v(x1, x2) = m(x1, x2) ×

∫Ω

G (x1, x2; ξ1, ξ2) v (ξ1, ξ2) dξ1 dξ2, (x1, x2) ∈ Ω .

Define w := (I + T ∗)−1v. Then w ∈ L2(Ω) and satisfies

v(x1, x2) = w(x1, x2) + m(x1, x2)

∫Ω

G(ξ1, ξ2; x1, x2)w(ξ1, ξ2)dξ1 dξ2, (x1, x2) ∈ Ω . (4.6)

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30 R. Gilbert et al. / Mathematical and Computer Modelling 45 (2007) 21–33

Now set

w(x1, x2) :=

∫Ω

w(ξ1, ξ2)G (ξ1, ξ2; x1, x2) dξ1 dξ2, (x1, x2) ∈ R2+.

Then w satisfies

4w + k2n20(x2)w = 0, for x 6∈ Ω .

It follows from (4.5) that

w|Γs =

∫Ω

w(ξ1, ξ2)G(ξ1, ξ2; x1, x2)

∣∣∣∣(x1,x2)∈Γs

dξ1 dξ2

= ((I + T ∗)−1v, G(·, (x1, x2))|(x1,x2)∈Γs )L2(Ω)= 0. (4.7)

Also, as w satisfies the outgoing radiation condition, it follows that w(x1, x2) = 0 for all (x1, x2) 6∈ Ω .Now let v j ∈ H with v j → v in L2(Ω). Then∫

Ωvv j dx =

∫Ω

wv j dx +

∫Ω

mwv j dx =

∫Ω

wv j dx +

∫Ω

w[4v j + k2n20v j ]dx

=

∫Ω

w

[v j +

∫Ω

G(ξ1, ξ2; x1, x2)[4v j + k2n20v j ]dξ1 dξ2

]dx

=

∫Ω

w

∫∂Ω

[G (ξ1, ξ2; x1, x2)

∂v j

∂νy− v j

∂G(ξ1, ξ2; x1, x2)

∂νy

]dsydx.

Since v j satisfies the Helmholtz equation

4v j + k2n20v j = 0 for x ∈ B \ Ω ,

for any smooth contour C ⊂ B \ Ω , it follows from Green’s theorem that∫Ω

vv j dx =

∫Ω

w

∫C

[G(ξ1, ξ2; x1, x2)

∂v j

∂νy− v j

∂G(ξ1, ξ2; x1, x2)

∂νy

]dsydx

=

∫C

[w

∂v j

∂νy− v j

∂w

∂νy

]dsy = 0.

The last equality is due to the fact that w = 0 for x 6∈ Ω . Let j → ∞. then v j → v and v = 0 in Ω .

Lemma 4.3. Let n1, n2 be two indices of refraction with n1(x) = n2(x) = n0(x2) for all x 6∈ Ω . Letu1(x, xs), u2(x, xs) be the corresponding outgoing scattered waves from the source xs . If u1(x, xs) = u2(x, xs)

for all x ∈ Γ and xs ∈ Γs , then∫Ω

v1(x)v2(x) [n1(x) − n2(x)] dx = 0

for all solutions v j ∈ C2(B) of

4v j + k21n jv j = 0, j = 1, 2

in B, where the C2 simple region B ⊃ Ω .

Proof. Let v1 be any fixed solution of

4v1 + k2n21v1 = 0 in Ω .

For any xs ∈ Γs , define u(·, xs) = u1(·, xs) − u2(·, xs). Then u(·, xs) satisfies

4u + k2n21u = k2(n2

2 − n21)u2

k2∫Ω

(n22 − n2

1)u2(·, xs)v1dx =

∫Ω

[v1 4 u(·, xs) − u(·, xs) 4 v1] dx

=

∫∂Ω

(v1

∂u(·, xs)

∂ν− u(·, xs)

∂v1

∂ν

)dsx.

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R. Gilbert et al. / Mathematical and Computer Modelling 45 (2007) 21–33 31

Fig. 4. Reconstruction of a wave penetrable object.

Recall that

u|Γ = u1|Γ − u2|Γ = 0.

Moreover,

4u + k2n20u = 0 for x 6∈ Ω , u = 0 for x2 = 0,

and u is an outgoing solution. It implies that u = 0 for x 6∈ Ω . The boundary integral vanishes and

k2∫Ω

(n22 − n2

1)u2(·, xs)v1dx = 0

for all xs ∈ Γ . By Theorem 3.2, u2(·, xs) is complete in H |Ω . So

k2∫Ω

(n22 − n2

1)v2v1dx = 0.

Lemma 4.4. Let Ω ⊂ R2h be a bounded domain and let n1, n2 ∈ C2(Ω) such that n1 − n0 and n2 − n0 have

compact support in Ω . Then the set of products u1u2, where u j ∈ C2(Ω), j = 1, 2 solves 4v + k2n2v = 0 forn = n j , j = 1, 2, respectively, is dense in L2(Ω).

This is a well-known result first proved by Sylvester and Uhlmann [13] for Ω ⊂ R3.Now Theorem 4.1 can be proved as follows.

Proof of Theorem 4.1. By Lemma 4.2, we know that∫Ω

v1(x)v2(x) [n1(x) − n2(x)] dx = 0

for all solutions v1, v2 satisfying in B

4v j + k2n2jv j = 0, j = 1, 2.

Hence, n21 − n2

2 = 0 by Lemma 4.4.

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32 R. Gilbert et al. / Mathematical and Computer Modelling 45 (2007) 21–33

Now we present a numerical example for the inverse problem. We use a regularized Born approximation methodto reconstruct the unknown inhomogeniety. Assuming that M is small, we have u(x, z) ' G(x1s, x2s; x1, x2) for(x1, x2) ∈ Ω . Therefore, the scattered field operator

F(mG)(x, z) :=

∫Ω

G (ξ1, ξ2; x1, x2) k 2G (ξ1, ξ2; x1s, x2s) dξ1 dξ2, for (x1, x2) ∈ Γ (4.8)

is the approximation of F(mu)(x1, x2) and

F(mG)(x1, x2) ' G (x1s, x2s; x1, x2) − u(x1, x2) =: us∗(x1, x2), for (x1, x2) ∈ Γ . (4.9)

Note that F is a linear operator of m. Discretizing (4.9), we obtain an ill-conditioned linear system Fm = us∗. The

regularized Born approximation gives the system

(ε I + F∗F)m = F∗u∗.

The following is a numerical example. The distributed inhomogeneity is contained in a rectangle (x1, x2) | 50 <

x1 < 70, 10 < x2 < 25. The measured data are from Γd = (−140 + 0.5m, 80) | m = 0, 1, 2, . . . , 200. The inputdata for the inverse problem are obtained by adding 3% white noise to the accurate solution of the direct problem.(We use iteration (3.15) until |un+1 − un| < 10−12.) The minimization problem is solved by the regularized Bornapproximation.

Example 1. The inhomogeniety is

n2(x1, x2) =

0.2

(52− (z − 82.5)2)(6.62

− (x − 60)2)

33.32

(x, z) ∈ [53.3, 66.6] × [77.5, 87.5]

0 otherwise.

The first oneis the original, and the reconstruction is shown in the second one.

For other results in inverse scattering problems, see, for example, [7,8,10–12], etc.

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