Some Old and Some New Results in Inverse Obstacle Scattering Rainer - - PowerPoint PPT Presentation
Some Old and Some New Results in Inverse Obstacle Scattering Rainer Kress, Gttingen Analysis and Numerics of Acoustic and Electromagnetic Problems Linz, October 2016 Rainer Kress Some Old and Some New Results in Inverse Obstacle Scattering
Rainer Kress, Göttingen Analysis and Numerics
Linz, October 2016
Rainer Kress Some Old and Some New Results in Inverse Obstacle Scattering
ν
❳❳ ❳ ③ ✲ ✄ ✄✄ ✄
ui = eik x·d us, u = ui + us D ∆u + k2u = 0 in R3 \ ¯ D u = 0
∂us ∂r − ikus = o 1 r
r = |x| → ∞ us(x) = eik|x| |x|
x |x|
1 |x|
|x| → ∞ Given: Far field u∞ for one (or several) incident plane wave Find: Shape and location of scatterer D
Rainer Kress Some Old and Some New Results in Inverse Obstacle Scattering
1
Two old uniqueness proofs
2
Two new uniqueness results for the generalized impedance boundary condition
3
A recent iterative reconstruction algorithm for the generalized impedance boundary condition
4
Factorization method and transmission eigenvalues
Rainer Kress Some Old and Some New Results in Inverse Obstacle Scattering
Rellich’s Lemma: u∞ = 0
⇔ us = 0 in R3 \ D
Rainer Kress Some Old and Some New Results in Inverse Obstacle Scattering
Rellich’s Lemma: u∞ = 0
⇔ us = 0 in R3 \ D Far field u∞ uniquely determines total field u = ui + us D u = 0 ui us
❳❳ ❳ ③ ✲ ✄ ✄✄ ✄
Question of uniqueness: ⇔ Existence of additional closed surfaces with u = 0
Rainer Kress Some Old and Some New Results in Inverse Obstacle Scattering
Question of uniqueness: ⇔ Existence of additional closed surfaces on which u = 0 D u = 0 ui us
❳❳ ❳ ③ ✲ ✄ ✄✄ ✄
u = 0 No!!
Rainer Kress Some Old and Some New Results in Inverse Obstacle Scattering
Question of uniqueness: ⇔ Existence of additional closed surfaces on which u = 0 D u = 0 ui us
❳❳ ❳ ③ ✲ ✄ ✄✄ ✄
u = 0 Do not know???
Rainer Kress Some Old and Some New Results in Inverse Obstacle Scattering
Theorem (Schiffer ≈ 1960) For a sound-soft scatterer, assume that u∞,1(ˆ x, d) = u∞,2(ˆ x, d) for all observation directions ˆ x and all incident directions d. Then D1 = D2.
Rainer Kress Some Old and Some New Results in Inverse Obstacle Scattering
D1 D2 D∗ = unbounded component
us
1(· , d) = us 2(· , d)
in D∗
Rainer Kress Some Old and Some New Results in Inverse Obstacle Scattering
D1 D2 D∗ = unbounded component
us
1(· , d) = us 2(· , d)
in D∗ u1(x, d) = eik x·d + us
1(x, d)
In shaded domain: △u1 + k2u1 = 0 On boundary: u1 = 0 {u1(· , d) : d ∈ S2} linearly independent
Rainer Kress Some Old and Some New Results in Inverse Obstacle Scattering
D1 D2 D∗ = unbounded component
us
1(· , d) = us 2(· , d)
in D∗ u1(x, d) = eik x·d + us
1(x, d)
In shaded domain: △u1 + k2u1 = 0 On boundary: u1 = 0 {u1(· , d) : d ∈ S2} linearly independent Lax and Philipps 1967
Rainer Kress Some Old and Some New Results in Inverse Obstacle Scattering
Correct shaded domain: (R3 \ D∗) \ D1 D1 D2 D∗ Incorrect shaded domain: D2 \ (D1 ∩ D2) D1 D2
Rainer Kress Some Old and Some New Results in Inverse Obstacle Scattering
Schiffer’s proof does not work for other boundary conditions! D1 D2 Finite multiplicity of eigenvalues is based on the compact embedding of H1
0(Ω) into L2(Ω) for the shaded domain Ω.
However, in general, the embedding H1(Ω) into L2(Ω) is not compact.
Rainer Kress Some Old and Some New Results in Inverse Obstacle Scattering
u =
sound-soft ∂u ∂ν =
sound-hard ∂u ∂ν + ikλ =
impedance, ℜλ ≥ 0
Rainer Kress Some Old and Some New Results in Inverse Obstacle Scattering
Theorem (Kirsch, K. 1992) Assume that u∞,1(ˆ x, d) = u∞,2(ˆ x, d) for all observation directions ˆ x and all incident directions d. Then D1 = D2 and B1 = B2.
Rainer Kress Some Old and Some New Results in Inverse Obstacle Scattering
z∗ x∗ D wi(x, z) = eik|x−z| |x − z| = incident field, point source ws(x, z) = scattered field, w∞(ˆ x, z) = far field
Rainer Kress Some Old and Some New Results in Inverse Obstacle Scattering
z∗ x∗ D wi(x, z) = eik|x−z| |x − z| = incident field, point source ws(x, z) = scattered field, w∞(ˆ x, z) = far field reciprocity: u∞(ˆ x, d) = u∞(−d, −ˆ x), ws(x, z) = ws(z, x) mixed reciprocity: us(z, d) = w∞(−d, z)
Rainer Kress Some Old and Some New Results in Inverse Obstacle Scattering
mixed reciprocity: us(z, d) = w∞(−d, z) D1 D2 D∗ x⋆ ⋆z B1 B2
Rainer Kress Some Old and Some New Results in Inverse Obstacle Scattering
mixed reciprocity: us(z, d) = w∞(−d, z) D1 D2 D∗ x⋆ ⋆z B1 B2 u∞,1(ˆ x, d) = u∞,2(ˆ x, d) for |ˆ x| = |d| = 1 us
1(z, d)
= us
2(z, d)
for z ∈ D∗, |d| = 1 w∞,1(d, z) = w∞,2(d, z) for z ∈ D∗, |d| = 1 ws
1(x, z)
= ws
2(x, z)
for x, z ∈ D∗
Rainer Kress Some Old and Some New Results in Inverse Obstacle Scattering
D1 D2 D∗ x∗ ⋆ ⋆z B1 B2 ws
1(x, z) = ws 2(x, z)
for x, z ∈ D∗
Rainer Kress Some Old and Some New Results in Inverse Obstacle Scattering
D1 D2 D∗ x∗ ⋆ ⋆z B1 B2 ws
1(x, z) = ws 2(x, z)
for x, z ∈ D∗ lim
z→x∗ B1ws 1(x∗, z) = ∞,
lim
z→x∗ B1ws 2(x∗, z) = finite
⇒ D1 = D2 Holmgren’s theorem yields B1 = B2
Rainer Kress Some Old and Some New Results in Inverse Obstacle Scattering
D1 D2 D∗ x∗ ⋆ ⋆z B1 B2 ws
1(x, z) = ws 2(x, z)
for x, z ∈ D∗ lim
z→x∗ B1ws 1(x∗, z) = ∞,
lim
z→x∗ B1ws 2(x∗, z) = finite
⇒ D1 = D2 Holmgren’s theorem yields B1 = B2 Use of mixed reciprocity: Potthast 1999
Rainer Kress Some Old and Some New Results in Inverse Obstacle Scattering
∂u ∂ν + ik (λu − div∂D µ grad∂D u) = 0
ℜλ, ℜµ ≥ 0 Theorem (Bourgeois, Chaulet, Haddar 2012) Both the shape and the impedance functions of a scattering
determined by the far field patterns for an infinite number of incident waves with distinct incident directions and one fixed wave number.
Rainer Kress Some Old and Some New Results in Inverse Obstacle Scattering
∂u ∂ν + ik
ds µ du ds
Theorem (Cakoni, K. 2013) In two dimensions, for a given shape ∂D, three far field patterns corresponding to the scattering of three plane waves with different incident directions uniquely determine the impedance functions µ and λ.
Rainer Kress Some Old and Some New Results in Inverse Obstacle Scattering
∂u ∂ν + ik
ds µ du ds
Theorem (Cakoni, K. 2013) In two dimensions, for a given shape ∂D, three far field patterns corresponding to the scattering of three plane waves with different incident directions uniquely determine the impedance functions µ and λ. ik d ds µ
du2 ds − u2 du1 ds
= u1 ∂u2 ∂ν − u2 ∂u1 ∂ν
⇒ α3 W(u1, u2) = α1 W(u2, u3) = α2 W(u3, u1)
Rainer Kress Some Old and Some New Results in Inverse Obstacle Scattering
Iterative methods: Reformulate inverse problem as nonlinear ill-posed operator equation. Solve by iteration methods such as regularized Newton methods, Landweber iterations or conjugate gradient methods
Rainer Kress Some Old and Some New Results in Inverse Obstacle Scattering
Iterative methods: Reformulate inverse problem as nonlinear ill-posed operator equation. Solve by iteration methods such as regularized Newton methods, Landweber iterations or conjugate gradient methods Qualitative methods: Develop criterium in terms of behaviour
whether a point lies inside or outside the scatterer. Linear sampling, factorization, probe methods, etc
Rainer Kress Some Old and Some New Results in Inverse Obstacle Scattering
Assume that k2 is not a Dirichlet eigenvalue for −∆ in D us(x) = 1 4π
eik|x−y| |x − y| ϕ(y) ds(y), x ∈ R3 \ D
Rainer Kress Some Old and Some New Results in Inverse Obstacle Scattering
Assume that k2 is not a Dirichlet eigenvalue for −∆ in D us(x) = 1 4π
eik|x−y| |x − y| ϕ(y) ds(y), x ∈ R3 \ D Data equation u∞(ˆ x) = 1 4π
e−ik ˆ
x·y ϕ(y) ds(y),
ˆ x ∈ S2 Field equation ui(x) = − 1 4π
eik|x−y| |x − y| ϕ(y) ds(y), x ∈ ∂D Inverse problem is equivalent to solving the two equations for the two unknowns ∂D and ϕ
Rainer Kress Some Old and Some New Results in Inverse Obstacle Scattering
∂D = {p(ˆ x) : ˆ x ∈ S2}, set ψ = ϕ ◦ p Define S, S∞ : C2(S2, R3) × L2(S2, C) → L2(S2, C) by S(p, ψ)(ˆ x) := 1 4π
eik|p(ˆ
x)−p(ˆ y)|
|p(ˆ x) − p(ˆ y)| J(ˆ y) ψ(ˆ y) ds(ˆ y), ˆ x ∈ S2 S∞(p, ψ)(ˆ x) := 1 4π
x·p(ˆ y)ψ(ˆ
y) J(ˆ y) ds(ˆ y), ˆ x ∈ S2
Rainer Kress Some Old and Some New Results in Inverse Obstacle Scattering
∂D = {p(ˆ x) : ˆ x ∈ S2}, set ψ = ϕ ◦ p Define S, S∞ : C2(S2, R3) × L2(S2, C) → L2(S2, C) by S(p, ψ)(ˆ x) := 1 4π
eik|p(ˆ
x)−p(ˆ y)|
|p(ˆ x) − p(ˆ y)| J(ˆ y) ψ(ˆ y) ds(ˆ y), ˆ x ∈ S2 S∞(p, ψ)(ˆ x) := 1 4π
x·p(ˆ y)ψ(ˆ
y) J(ˆ y) ds(ˆ y), ˆ x ∈ S2 Data equation S∞(p, ψ) = u∞ Field equation S(p, ψ) = −ui ◦ p Linearize both equations with respect to p and ψ (and iterate). Ivanyshyn, K., Rundell, Serranho 2005 . . .
Rainer Kress Some Old and Some New Results in Inverse Obstacle Scattering
One incident wave with k = 1 Reconstruction of a cushion for 5% noise
Rainer Kress Some Old and Some New Results in Inverse Obstacle Scattering
Step 1. Given an approximation for p solve the field equation S(p, ψ) = −ui ◦ p for the density ψ. Step 2. Keeping ψ fixed, linearize the data equation S∞(p, ψ) = u∞ to obtain S′
∞(p, ψ; q) = u∞ − S∞(p, ψ)
and solve for q to update p into p + q. Johansson, Sleeman 2007
Rainer Kress Some Old and Some New Results in Inverse Obstacle Scattering
∂D = {z(t) : 0 ≤ t ≤ 2π} (Sψ)(t) := i 4 2π H(1)
0 (k|z(t)−z(τ)|) |z′(τ)| ψ(τ) dτ,
t ∈ [0, 2π] (S∞ψ)(ˆ x) := ei π
4
√ 8πk 2π e−ik ˆ
x·yψ(τ) dτ,
ˆ x ∈ S1
Rainer Kress Some Old and Some New Results in Inverse Obstacle Scattering
∂D = {z(t) : 0 ≤ t ≤ 2π} (Sψ)(t) := i 4 2π H(1)
0 (k|z(t)−z(τ)|) |z′(τ)| ψ(τ) dτ,
t ∈ [0, 2π] (S∞ψ)(ˆ x) := ei π
4
√ 8πk 2π e−ik ˆ
x·yψ(τ) dτ,
ˆ x ∈ S1 Data equation S∞ψ = u∞ Field equation Sψ = −ui ◦ z
Rainer Kress Some Old and Some New Results in Inverse Obstacle Scattering
(Nψ)(t) := ik 4 2π ν(z(t)) · [z(τ) − z(t)] |z(t) − z(τ)| H(1)
1 (k|z(t)−z(τ)|) |z′(τ)| ψ(τ) dτ
Rainer Kress Some Old and Some New Results in Inverse Obstacle Scattering
(Nψ)(t) := ik 4 2π ν(z(t)) · [z(τ) − z(t)] |z(t) − z(τ)| H(1)
1 (k|z(t)−z(τ)|) |z′(τ)| ψ(τ) dτ
Data equation S∞ψ = u∞ Field equation 1 2 ψ − Nψ − ik
|z′| d dt µ |z′| d dt Sψ
where g := ∂ui ∂ν
+ ik
ds µ d ds
Inverse problem is equivalent to solving the two equations for the two unknowns z and ϕ
Rainer Kress Some Old and Some New Results in Inverse Obstacle Scattering
(Sψ)(t) := i 4 2π H(1)
0 (k|z(t) − z(τ)|) |z′(τ)| ψ(τ) dτ
(Nψ)(t) := ik 4 2π ν(z(t)) · [z(τ) − z(t)] |z(t) − z(τ)| H(1)
1 (k|z(t)−z(τ)|) |z′(τ)| ψ(τ) dτ
Field equation 1 2 ψ − Nψ − ik
|z′| d dt µ |z′| d dt Sψ
= g ◦ z
Rainer Kress Some Old and Some New Results in Inverse Obstacle Scattering
(Sψ)(t) := i 4 2π H(1)
0 (k|z(t) − z(τ)|) |z′(τ)| ψ(τ) dτ
(Nψ)(t) := ik 4 2π ν(z(t)) · [z(τ) − z(t)] |z(t) − z(τ)| H(1)
1 (k|z(t)−z(τ)|) |z′(τ)| ψ(τ) dτ
Field equation 1 2 ψ − Nψ − ik
|z′| d dt µ |z′| d dt Sψ
= g ◦ z A : H1/2[0, 2π] → H−1/2[0, 2π] is Fredholm with index zero (provided |µ| > 0). Solve via trigonometric quadrature and differentiation with spectral convergence Cakoni, K. 2013 for Laplace, K. 2016 for Helmholtz
Rainer Kress Some Old and Some New Results in Inverse Obstacle Scattering
The impedance functions
1 2 3 4 5 6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 7 6 Rainer Kress Some Old and Some New Results in Inverse Obstacle Scattering
Two incident waves with k = 2
0.5 1
0.2 0.4 0.6 0.8 1 reconstructed exact initial guess
0.5 1
0.2 0.4 0.6 0.8 1 reconstructed exact initial guess
Reconstruction of an apple for exact data and for 5% noise
Rainer Kress Some Old and Some New Results in Inverse Obstacle Scattering
Two incident waves with k = 2
0.5 1
0.2 0.4 0.6 0.8 1 reconstructed exact initial guess
0.5 1
0.2 0.4 0.6 0.8 1 reconstructed exact initial guess
Reconstruction of a peanut for exact data and for 5% noise
Rainer Kress Some Old and Some New Results in Inverse Obstacle Scattering
Assume that k2 is not a Dirichlet eigenvalue of −∆ for D. Define far field operator F : L2(S2) → L2(S2) by Fg(ˆ x) :=
x, d)g(d) ds(d), ˆ x ∈ S2 and recall the far field of a point source w∞(ˆ x, z) = e−ik ˆ
x·z,
ˆ x ∈ S2, z ∈ R3
Rainer Kress Some Old and Some New Results in Inverse Obstacle Scattering
Assume that k2 is not a Dirichlet eigenvalue of −∆ for D. Define far field operator F : L2(S2) → L2(S2) by Fg(ˆ x) :=
x, d)g(d) ds(d), ˆ x ∈ S2 and recall the far field of a point source w∞(ˆ x, z) = e−ik ˆ
x·z,
ˆ x ∈ S2, z ∈ R3 Theorem (Kirsch 1998) The equation (F ∗F)1/4g(· , z) = w∞(· , z) is solvable in L2(S2) if and only if z ∈ D. Use Picard criterium for numerical implementation
Rainer Kress Some Old and Some New Results in Inverse Obstacle Scattering
∆u + k2nu = 0 in R3 u = ui + us ∂us ∂r − ikus = o 1 r
r = |x| → ∞ where n > 0 is the refractive index with supp(n − 1) = D Theorem (Kirsch 1999) The equation (F ∗F)1/4g(· , z) = w∞(· , z) is solvable in L2(S2) if and only if z ∈ D. Here k is required not to be an interior transmission eigenvalue
Rainer Kress Some Old and Some New Results in Inverse Obstacle Scattering
k is called a transmission eigenvalue it there exist nontrivial functions v, w ∈ H2(D) such that ∆v + k2v = 0, ∆w + k2nv = 0 in D and v = w, ∂v ∂ν = ∂w ∂ν
Rainer Kress Some Old and Some New Results in Inverse Obstacle Scattering
k is called a transmission eigenvalue it there exist nontrivial functions v, w ∈ H2(D) such that ∆v + k2v = 0, ∆w + k2nv = 0 in D and v = w, ∂v ∂ν = ∂w ∂ν
Can boundary integral equations be used to analyze and compute transmission eigenvalues in the case n = const in D?
Rainer Kress Some Old and Some New Results in Inverse Obstacle Scattering
k is called a transmission eigenvalue it there exist nontrivial functions v, w ∈ H2(D) such that ∆v + k2v = 0, ∆w + k2nv = 0 in D and v = w, ∂v ∂ν = ∂w ∂ν
Can boundary integral equations be used to analyze and compute transmission eigenvalues in the case n = const in D? Cossonnière, Haddar 2013 System of two integral equations for v and ∂νv from Green’s representation formula
Rainer Kress Some Old and Some New Results in Inverse Obstacle Scattering
Alternative approach using the Dirichlet-to-Neumann operator DtNk,n : ϕ → ∂u ∂ν , where u is the unique solution to ∆u + k2nu = in D u = ϕ
assuming that k2 is not an eigenvalue for this problem. If we further assume that k2 is not an eigenvalue for the case when n = 1, then k is a transmission eigenvalue if and only if the kernel of the operator M(k) := DtNk,n − DtNk,1 is nontrivial.
Rainer Kress Some Old and Some New Results in Inverse Obstacle Scattering
Theorem (Cakoni, K. 2016) Let κ > 0 and κn = κ√n. Then (κ2 − κ2
n)M(iκ) : H− 1
2 (∂D) → H 1 2 (∂D)
is coercive. Further let kn = k√n. Then the operator M(k) + k2 − k2
n
|k|2 − |kn|2 M(i|k|) : H− 1
2 (∂D) → H 1 2 (∂D)
is compact. Idea of proof: Write Mk = (I + Nk)S−1
k
− (I + Nkn)S−1
kn and use
Green’s integral theorem and mapping properties for difference
Rainer Kress Some Old and Some New Results in Inverse Obstacle Scattering
Corollary M(k) : H− 1
2 (∂D) → H 1 2 (∂D) is a Fredholm operator with index
zero and analytic in C \ E E is the set of all positive k such that k2 or k2
n is a Dirichlet
eigenvalue for −∆ in D. Can use the effective numerical method for non-linear eigenvalue problems proposed by Beyn 2012 to compute transmission eigenvalues. The drawback that Dirichlet eigenvalues have to be excluded can be remedied by an analogous (but more involved) analysis using a Robin-to-Dirichlet map.
Rainer Kress Some Old and Some New Results in Inverse Obstacle Scattering
Transmission eigenvalues for ellipse with major axis a = 1
n = 2 b = 1 b = 0.8 b = 0.5 b = 0.3 7.37512 7.63521 8.99951 12.32237 7.39666 8.13114 9.08377 12.34655 7.39666 8.43084 10.97143 16.48683 7.98435 8.51387 11.03990 16.49174 7.98435 8.95319 12.30361 17.01368 8.02926 9.00645 12.31572 17.14467 8.02926 9.24582 13.40175 19.44708 8.21647 9.34204 13.41828 19.44871 8.21647 9.47369 14.21896 19.92760 8.67540 9.76915 14.47108 19.94432
Rainer Kress Some Old and Some New Results in Inverse Obstacle Scattering
Cakoni, F . and Kress, R.: Integral equation methods for the inverse obstacle problem with generalized impedance boundary condition. Inverse Problems 29, 015005 (2013). Cakoni, F ., Hu, Y. and Kress, R.: Simultaneous reconstruction of shape and generalized impedance functions in electrostatic imaging. Inverse Problems 30, 105009 (2014). Cakoni, F . and Kress, R.: A boundary integral equation method for the transmission eigenvalue problem. Applicable Analysis (2016). Kress, R.: Integral equation methods in inverse obstacle scattering with a generalized impedance boundary condition. Submitted
Rainer Kress Some Old and Some New Results in Inverse Obstacle Scattering
Rainer Kress Some Old and Some New Results in Inverse Obstacle Scattering