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The Reciprocity Gap Method An Analytic Case Numerical Examples

Numerical studies of the reciprocity gap functional method in inverse scattering

Jiguang Sun

Delaware State University AIP2009, Vienna, Austria

The Reciprocity Gap Method An Analytic Case Numerical Examples

Outline

1

The Reciprocity Gap Method

2

An Analytic Case

3

Numerical Examples

The Reciprocity Gap Method An Analytic Case Numerical Examples

Introduction

Γ D Ω x0 B

The unknown scattering obstacle D is inside some domain B. Ω ⊂ B and the boundary of Ω is given by Γ. D ⊂ Ω. Point sources located at x0 lie on a smooth curve ∂B that is homotopic to Γ.

The Reciprocity Gap Method An Analytic Case Numerical Examples

The direct scattering problem △u + k2n(x)u = 0 in R\ D ∪ {x0}

• ,

(1) u = 0

• n

∂D, (2) u = us + ui, (3) lim

r→∞

√ r ∂us ∂r − ikus

• = 0,

(4) where k is the wave number and r = |x|. Note that when n(x) is constant, ui is given by Φ(x) defined as Φ(x, x0) = i 4H(1)

0 (k

√ n|x − x0|) (5) where H(1) is the Hankel function of first kind and order zero.

The Reciprocity Gap Method An Analytic Case Numerical Examples

The inhomogeneous background Green’s function When the background is inhomogeneous, ui can be written as ui(x, x0) = G(x, x0) (6) where G(x, x0) is the radiating solution of △G(x, x0) + k2n(x)G(x, x0) = −δ(x − x0) in R2 (7) with δ denoting the Dirac delta function. The inverse scattering problem we are interested in is to determine the support of the scattering obstacle D from the knowledge of the Cauchy data of u on ∂Ω.

The Reciprocity Gap Method An Analytic Case Numerical Examples

The Reciprocity Gap Functional H(Ω) := {v ∈ H1(Ω) : ∆v + k2n(x)v = 0 in Ω} U is the set of solutions to (1-4) for all x0 ∈ ∂B. For v ∈ H(Ω) and u ∈ U we define the reciprocity gap functional by R(ux0, v) =

• ∂Ω
• ux0

∂v ∂ν − v ∂ux0 ∂ν

• ds.

(8) Define R : H(Ω) → L2(∂B) defined by R(v)(x0) = R(u, v).

The Reciprocity Gap Method An Analytic Case Numerical Examples

Herglotz wave functions H(Ω) contains the set of Herglotz wave functions defined by vg,k1(x) =

• S

eik1ˆ

d·xg(ˆ

d)ds(ˆ d), (9) where g ∈ L2(S), with S = {ˆ d ∈ R2 : |ˆ d| = 1} and k2

1 = k2n0

for some constant n0. Let Φk1,z = i

4H(1) 0 (k1|x − z|). For z ∈ Y, a sampling domain

contained in Ω and containing D inside, we look for a solution g ∈ L2(S1) to R(u, vg) = R(u, Φk1,z) for all u ∈ U, (10) where vg is a Herglotz wave function defined by (9).

The Reciprocity Gap Method An Analytic Case Numerical Examples

Characterization of the target Theorem Assume that k2 is not a Dirichlet eigenvalue for D. (a) If z ∈ D then there exist a sequence {gn}, such that lim

n→∞ R(u, vgn) = R(u, Φz)

for all u ∈ U. Furthermore, vgn converges in H1(D) and vgn → Φz in H1/2(∂D). (b) If z ∈ Ω \ D then for every sequence {vgn}, such that lim

n→∞ R(u, vgn) = R(u, Φz)

for all u ∈ U. we have lim

n→∞ vgnH1(D) = ∞.

The Reciprocity Gap Method An Analytic Case Numerical Examples

Now equation (10) can be written in the form Ag(·, z) = φ(·, z) (11) where A : L2(S) → L2(C) is the integral operator with kernel K : S × C ∈ C defined by K(ˆ d, x0) = R(u(·, x0), v(·, ˆ d)) with v(x, ˆ d) = exp(ik ˆ d · x) and where φ(x0, z) = R(u(·, x0), Φz). We expect that for g(·, z) being the solution to (11) where A is replaced by a regularized operator, g(·, z)L2(S) should be large for z ∈ Ω \ D and bounded for z ∈ D.

The Reciprocity Gap Method An Analytic Case Numerical Examples

Regularization Scheme To obtain regularized solution, Tikhonov regularization can be

• used. This means solving

(α + A∗A)gα(·, z) = A∗φ(·, z) where α is the regularization parameter and A∗ is the adjoint of

• A. If A is a compact operator with singular system (σn; vn, un),

we have gα =

∞

• n=1

σn σ2

n + α < φ, un > vn.

Hence it is very helpful to study the singular values of A in different cases.

The Reciprocity Gap Method An Analytic Case Numerical Examples

Setting of the Problem Now we consider the simplified case when ∂D, ∂Ω and ∂B are all circles whose centers are the origin of the coordinate

• system. Assume that the incident field ui is due to a point

source at z = (rs, φ) and consider the scattering of ui by an infinite cylinder whose cross section is a circle with radius rD. The forward problem is to find a complex valued radiating function us(x) such that △us + k2us = 0 in R2 \ ¯ D us(r, θ) = −Φ(k|x − z|)

• n

∂D where Φ(k|x − z|) = i 4H(1)

0 (k|x − z|).

The Reciprocity Gap Method An Analytic Case Numerical Examples

Expansion of the solution Suppose D is a circle centered at the origin with radius rD. Due to symmetry, u(r, θ) =

∞

• n=0

αnH(1)

n (kr) cos nθ,

r > rD. On ∂D where r = rD, we have u(rD, θ) =

∞

• n=0

αnH(1)

n (krD) cos nθ.

(12) On the other hand, let x = (ρ, ϕ), we have |z − x| =

• r2

s − 2rsρ cos θ + ρ2 where θ = φ − ϕ and

H1

0 (|z − x|) = J0(ρ)H(1) 0 (rs) + 2 ∞

• n=1

Jn(ρ)H(1)

n (rs) cos nθ

(13) for rs > ρ.

The Reciprocity Gap Method An Analytic Case Numerical Examples

Expansion coefficients Function Jn is the Bessel’s function of the first kind of order n. Using (13), (12) and the boundary condition, we obtain α0 = − i 4 J0(krD)H(1)

0 (krs)

H(1)

0 (krD)

, αn = − i 4 2Jn(krD)H(1)

n (krs)

H(1)

n (krD)

n ≥ 1. d dz H(1)

n (z) = H(1) n−1(z)− n

z H(1)

n (z) = −H(1) n+1(z)+ n

z H(1)

n (z). (14)

The Reciprocity Gap Method An Analytic Case Numerical Examples

Hence u(r, θ) = − i 4 J0(krD)H(1)

0 (krs)

H(1)

0 (krD)

H1

0(kr)

− i 2

∞

• n=1

Jn(krD)H(1)

n (krs)

H(1)

n (krD)

H1

n(kr) cos(nθ),

∂u(r, θ) ∂r = −ik 4 J0(krD)H(1)

0 (krs)

H(1)

0 (kr0)

{−H1

1(kr)}

−ik 2

∞

• n=1

Jn(krD)H(1)

n (krs)

H(1)

n (krD)

{H1

n−1(kr) − n

kr H1

n(kr) cos(nθ)}.

The Reciprocity Gap Method An Analytic Case Numerical Examples

In particular, if the point source is at z = (rs, φ) on a circle with radius rs, then the scattered field can be obtained easily from the above, u(r, θ) = − i 4 J0(krD)H(1)

0 (krs)

H(1)

0 (krD)

H1

0(kr)

− i 2

∞

• n=1

Jn(krD)H(1)

n (krs)

H(1)

n (krD)

H1

n(kr) cos(n(θ − φ)),

∂u(r, θ) ∂r = −ik 4 J0(krD)H(1)

0 (krs)

H(1)

0 (krD)

{−H1

1(kr)}

−ik 2

∞

• n=1

Jn(krD)H(1)

n (krs)

H(1)

n (krD)

{H1

n−1(kr) − n

kr H1

n(kr)} cos(n(θ − φ)).

The Reciprocity Gap Method An Analytic Case Numerical Examples

Now we consider the case when Γ = ∂Ω := {x, |x| = rΓ > rD}. The solutions of the Helmholtz equation in Ω can be written as vn = Jn(krΓ) cos(nθ) or wn = Jn(krΓ) sin(nθ). (15) ∂vn ∂r = k

• Jn−1(kr) − n

kr Jn(kr)

• cos(nθ),

∂wn ∂r = k

• Jn−1(kr) − n

kr Jn(kr)

• sin(nθ).

The Reciprocity Gap Method An Analytic Case Numerical Examples

Now the reciprocity gap functional can be seen as an operator R : L2[0, 2π] → L2[0, 2π] (from L2(∂Ω) → L2(∂B)) defined by R(v)(φ) = R(u, v). Thus combining the above equations we obtain, for n = 0, R(v0)(φ) =

• Γ

u ∂v0 ∂r − v0 ∂u ∂r ds = −iπ 2 J0(krD)H(1)

0 (krs)

H(1)

0 (krD)

H1

0(krΓ)(−kJ1(krΓ))

+ikπ 2 J0(krD)H(1)

0 (krs)

H(1)

0 (krD)

{−H1

1(krΓ)}J0(krΓ)

= λ0v0

The Reciprocity Gap Method An Analytic Case Numerical Examples

Case of n > 1 R(vn)(φ) =

• Γ

u ∂vn ∂r − vn ∂u ∂r ds = −ikπ 2

• Jn(krD)H(1)

n (krs)

H(1)

n (krD)

H1

n(krΓ)

• Jn−1(krΓ) − n

krΓ Jn(krΓ)

• −Jn(krD)H(1)

n (krs)

H(1)

n (krD)

{H1

n−1(krΓ) − n

krΓ H1

n(krΓ)}Jn(krΓ)

• cos(nφ)

= λnvn

The Reciprocity Gap Method An Analytic Case Numerical Examples

Using the recurrence relation of Bessel’s functions, we obtain λn = −ikπc 2 Jn(krD)H(1)

n (krs)

H(1)

n (krD)

, n ≥ 1 (16) where c = H1

1(krΓ)J0(krΓ) − H1 0(krΓ)J1(krΓ)

(17) Similarly, we have R(wn)(φ) = λnwn.

The Reciprocity Gap Method An Analytic Case Numerical Examples

Right hand side Next, we consider the right hand side of the integral equation. Let x = (r, θ) and z = (ρ, φ). We have Φ(k|x−z|) = i 4J0(kρ)H(1)

0 (kr)+ i

2

∞

• n=1

Jn(kρ)H(1)

n (kr) cos n(θ−φ)

(18) and ∂Φ(k|x − z|) ∂r = −ik 4 J0(kρ)H(1)

1 (kr)

+ik 2

∞

• n=1

Jn(kρ)(H1

n−1(kr) − n

kr H1

n(kr)) cos n(θ − φ).

The Reciprocity Gap Method An Analytic Case Numerical Examples

Integrating the gap reciprocity functional R(vn, Φz) (similarly for wn), we obtain R(v0, Φz) = −ikπc 2 J0(kρ) := µ0v0 and R(vn, Φz) = ikπ 2 Jn(kρ)

• Jn(krΓ)H1

n−1(krΓ) − Jn−1(krΓ)H1 n(krΓ)

• cos(nφ)

= −ikπc 2 Jn(kρ) cos(nφ) := µn cos(nφ).

The Reciprocity Gap Method An Analytic Case Numerical Examples

Let ϕ = 0 and ρ = 0.1, 0.3, 0.5, 0.7. The first two points are inside D. The third is on ∂D and the last point is outside D.

1 2 3 4 5 6 7 8 9 −25 −20 −15 −10 −5 |λ| ρ=0.1 ρ=0.3 ρ=0.5 ρ=0.7

The Reciprocity Gap Method An Analytic Case Numerical Examples

Verification of the theorem

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 200 400 600 800 1000 1200 1400 1600 1800 2000

ρ ||gz||l

2

rD=0.5

k=1 k=3 k=6 k=9 k=12

The norm of the regularized solution remains small as ρ < rD = 0.5, the radius of the circular target, and starts to become larger when ρ ≥ rD. α = 10−6.

The Reciprocity Gap Method An Analytic Case Numerical Examples

Various targets

−2 −1.5 −1 −0.5 0.5 1 1.5 2 −2 −1.5 −1 −0.5 0.5 1 1.5 2 −2 −1.5 −1 −0.5 0.5 1 1.5 2 −2 −1.5 −1 −0.5 0.5 1 1.5 2 −2 −1.5 −1 −0.5 0.5 1 1.5 2 −2 −1.5 −1 −0.5 0.5 1 1.5 2 −2 −1.5 −1 −0.5 0.5 1 1.5 2 −2 −1.5 −1 −0.5 0.5 1 1.5 2 −2 −1.5 −1 −0.5 0.5 1 1.5 2 −2 −1.5 −1 −0.5 0.5 1 1.5 2

The Reciprocity Gap Method An Analytic Case Numerical Examples

Singular values for different k

10 20 30 40 50 60 70 −16 −14 −12 −10 −8 −6 −4 −2 2

log(svd(A)) n

k=1 k=3 k=6 k=9

Computed singular values of A of a triangular obstacle for different wave numbers (k = 1, 3, 6, 9).

The Reciprocity Gap Method An Analytic Case Numerical Examples

Reconstruction of the triangle

−1 −0.5 0.5 1 −1 −0.5 0.5 1 −1 −0.5 0.5 1 −1 −0.5 0.5 1 −1 −0.5 0.5 1 −1 −0.5 0.5 1 −1 −0.5 0.5 1 −1 −0.5 0.5 1

Contour plots of 1/gz for the follower using different wave numbers (Up left: k = 1, Up right: k = 3, Bottom left: k = 6, Bottom right: k = 9).

The Reciprocity Gap Method An Analytic Case Numerical Examples

Singular values of A for different targets (k = 3).

10 20 30 40 50 60 70 −18 −16 −14 −12 −10 −8 −6 −4 −2 2

log10(svd(A)) n

Circle Ellipse Kite Flower

The Reciprocity Gap Method An Analytic Case Numerical Examples

Different Source Geometries (∂B)

−8 −6 −4 −2 2 4 6 8 −8 −6 −4 −2 2 4 6 8 10 20 30 40 50 60 70 −18 −16 −14 −12 −10 −8 −6 −4 −2 2 log10(svd(A)) n Circle Square Triangle

Singular values for various geometries of points sources (k = 3) including a triangle, a square and a circle. Note that the dashed line is Γ, the collection geometries given by a circle with radius 3.

The Reciprocity Gap Method An Analytic Case Numerical Examples

Different Collection Geometries (Γ)

−6 −4 −2 2 4 6 −6 −4 −2 2 4 6

10 20 30 40 50 60 70 −18 −16 −14 −12 −10 −8 −6 −4 −2 2

log10(svd(A)) n

Circle Square Triangle

Singular values for various Γ’s (k = 3): a circle, a square and a triangle.

The Reciprocity Gap Method An Analytic Case Numerical Examples

Reconstruction for different Γ

−1.5 −1 −0.5 0.5 1 1.5 −1.5 −1 −0.5 0.5 1 1.5 −1.5 −1 −0.5 0.5 1 1.5 −1.5 −1 −0.5 0.5 1 1.5 −1.5 −1 −0.5 0.5 1 1.5 −1.5 −1 −0.5 0.5 1 1.5

Contour plots for different collection geometries Γ. From left to right: a circle, a square and a triangle.

The Reciprocity Gap Method An Analytic Case Numerical Examples

Limit aperture data (sources)

5 10 15 20 25 30 35 −18 −16 −14 −12 −10 −8 −6 −4 −2 2

log(svd(A))

π/2 π 3π/2 2π

Singular values of A for different source apertures (k = 3).

The Reciprocity Gap Method An Analytic Case Numerical Examples

Limit aperture data (receivers)

10 20 30 40 50 60 70 −18 −16 −14 −12 −10 −8 −6 −4 −2 2

log10(svd(A)) n

π/2 π 3π/2 2π

Singular values of A for different collection apertures (k = 3).

The Reciprocity Gap Method An Analytic Case Numerical Examples

Reconstruction by limited aperture data (A)

−5 −4 −3 −2 −1 1 2 3 4 5 −5 −4 −3 −2 −1 1 2 3 4 5 −5 −4 −3 −2 −1 1 2 3 4 5 −5 −4 −3 −2 −1 1 2 3 4 5 −5 −4 −3 −2 −1 1 2 3 4 5 −5 −4 −3 −2 −1 1 2 3 4 5 −1.5 −1 −0.5 0.5 1 1.5 −1.5 −1 −0.5 0.5 1 1.5 −1.5 −1 −0.5 0.5 1 1.5 −1.5 −1 −0.5 0.5 1 1.5 −1.5 −1 −0.5 0.5 1 1.5 −1.5 −1 −0.5 0.5 1 1.5

The Reciprocity Gap Method An Analytic Case Numerical Examples

Reconstruction by limited aperture data (B)

−5 −4 −3 −2 −1 1 2 3 4 5 −5 −4 −3 −2 −1 1 2 3 4 5 −5 −4 −3 −2 −1 1 2 3 4 5 −5 −4 −3 −2 −1 1 2 3 4 5 −5 −4 −3 −2 −1 1 2 3 4 5 −5 −4 −3 −2 −1 1 2 3 4 5 −1.5 −1 −0.5 0.5 1 1.5 −1.5 −1 −0.5 0.5 1 1.5 −1.5 −1 −0.5 0.5 1 1.5 −1.5 −1 −0.5 0.5 1 1.5 −1.5 −1 −0.5 0.5 1 1.5 −1.5 −1 −0.5 0.5 1 1.5

The Reciprocity Gap Method An Analytic Case Numerical Examples

Future works How to improve reconstruction using limited aperture data? What is the optimal aperture?

The Reciprocity Gap Method An Analytic Case Numerical Examples

Thank you!