Inverse Obstacle Scattering - PDF Free Download

, Göttingen AIP 2011, Pre-Conference Workshop Texas A&M University, May 2011

Scattering theory Scattering theory is concerned with the effects that obstacles and inhomogenities have on the propagation of waves

Scattering theory Scattering theory is concerned with the effects that obstacles and inhomogenities have on the propagation of waves Restrict to time-harmonic waves and obstacles

Incident plane wave

Scattered wave

Total wave

Direct and inverse Scattering Direct scattering problem Given: Incident field and scatterer Find: Scattered field

Direct and inverse Scattering Direct scattering problem Given: Incident field and scatterer Find: Scattered field Inverse scattering problem Given: Incident and scattered field Find: Position and shape of scatterer

Outline 1 The Helmholtz equation 2 The direct scattering problem b) Uniqueness b) Existence c) Numerical solution 3 The inverse scattering problem b) Uniqueness b) Iterative solution methods c) Decomposition methods d) Sampling and probe methods

Time-harmonic waves Wave equation: U = 1 c 2 2 U t 2 U = velocity potential, electric field c = speed of sound, speed of light

Time-harmonic waves Wave equation: U = 1 c 2 2 U t 2 U = velocity potential, electric field c = speed of sound, speed of light U(x, t) = R { u(x)e iωt} Helmholtz equation: u + k 2 u = 0 ω = frequency, k = ω/c = wave number

Helmholtz equation Two main tools: Fundamental solution and Green s integral theorems

Helmholtz equation Two main tools: Fundamental solution and Green s integral theorems Φ(x, y) := 1 4π e ik x y x y, x y, in IR3 Φ(x, y) := i 4 H(1) 0 (k x y ), x y, in IR2 H (1) 0 = J 0 + iy 0 Hankel function

Fundamental solution in three dimensions R eir r = cos r r I eir r = sin r r

Fundamental solution in two dimensions RiH (1) 0 (r) = Y 0(r) IiH (1) 0 (r) = J 0(r)

Green s integral formula D D C 2 ν u C 2 (D) C 1 ( D D) u + k 2 u = 0 in D u(x) = D { } u Φ(x, y) (y) Φ(x, y) u(y) ds(y), ν ν(y) x D

Green s integral formula D D C 2 ν u C 2 (D) C 1 ( D D) u + k 2 u = 0 in D u(x) = D { } u Φ(x, y) (y) Φ(x, y) u(y) ds(y), ν ν(y) x D Solutions to Helmholtz equation inherit properties of fundamental solution. Solutions to Helmholtz equation are analytic

Laplace versus Helmholtz equation Helmholtz equation shares many properties with Laplace equation. i 4 H(1) 0 1 4π e ik x y x y = 1 4π 1 x y + ik + O( x y ) 4π 1 (k x y ) = 2π ln 1 x y + i 4 1 2π ln k 2 C +O( x y ) 2π However no maximum-minimum principle and no coercitivity ū u { ν ds = grad u 2 k 2 u 2} dx D D In particular, there exist Dirichlet and Neumann eigenvalues, that is, wave numbers k and solutions u 0 to u + k 2 u = 0 in D with homogeneous boundary data u = 0 or ν u = on D, respectively.

Sommerfeld radiation condition Consider Helmholtz equation in IR 3 \ D. Sommerfeld radiation condition requires ( ) u 1 r iku = o, r = x r uniformly for all directions. Radiating solutions

Sommerfeld radiation condition Consider Helmholtz equation in IR 3 \ D. Sommerfeld radiation condition requires ( ) u 1 r iku = o, r = x r uniformly for all directions. Radiating solutions Equivalent to u(x) = eik x x u is called the far field pattern and defined on the unit sphere S 2. { ( ) ( )} x 1 u + O, x x x

Direct obstacle scattering U(x, t) = R { u(x)e iωt} 3 D u i u s u i : incident field, plane wave, u i (x, d) = e ik x d, d = 1 u s : scattered field u = u i + u s : total field

Direct obstacle scattering ν 3 D u i u s, u = u i + u s u + k 2 u = 0 in IR 3 \ D

Direct obstacle scattering ν 3 D u i u s, u = u i + u s u + k 2 u = 0 in IR 3 \ D Bu = 0 on D

Direct obstacle scattering ν 3 D u i u s, u = u i + u s u + k 2 u = 0 in IR 3 \ D Bu = 0 u s r on D iku s = o ( 1 r ), r = x

Direct obstacle scattering ν 3 D u i u s, u = u i + u s u + k 2 u = 0 in IR 3 \ D Bu = 0 u s r Boundary condition: Bu = u Bu = u ν on D iku s = o ( 1 r ), r = x sound-soft + ikλu impedance, λ 0

Existence and uniqueness Theorem The direct scattering problem for a sound-soft obstacle has a unique solution u H 1 loc (IR3 \ D).

Existence and uniqueness Theorem The direct scattering problem for a sound-soft obstacle has a unique solution u H 1 loc (IR3 \ D). Uniqueness: D R := { x IR 3 \ D : x R } x =R ū s us ν ds ū s us [ D ν ds = grad u s 2 + ū s u s] dx D R

Existence and uniqueness Combined double- and single-layer potential { } Φ(x, y) u s (x) = iφ(x, y) ϕ(y) ds(y), ν(y) D x IR 3 \ D solves sound-soft scattering problem if ϕ H 1/2 ( D) satisfies { } 1 Φ(x, y) 2 ϕ(x)+ iφ(x, y) ϕ(y) ds(y) = u i (x), x D D ν(y) Let ϕ solve the homogeneous equation and set u := u s u + = 0 on D u = 0 in IR 3 \ D

Numerical solution { } 1 Φ(x, y) 2 ϕ(x)+ iφ(x, y) ϕ(y) ds(y) = u i (x), D ν(y) x D

Numerical solution { } 1 Φ(x, y) 2 ϕ(x)+ iφ(x, y) ϕ(y) ds(y) = u i (x), D ν(y) x D Advantages of boundary integral equation method versus variational approach 1. Radiation condition automatically satisfied 2. Reduce dimension by one

Numerical solution { } 1 Φ(x, y) 2 ϕ(x)+ iφ(x, y) ϕ(y) ds(y) = u i (x), D ν(y) x D Solve numerically via collocation method plus quadrature. 1. Approximate ϕ by low order spline functions 2. Map boundary D on circle or sphere and approximate ϕ by trigonometric polynomials or spherical harmonics, respectively K.E. Atkinson 1997 The most efficient numerical methods for solving boundary integral equations on smooth planar boundaries are those based on trigonometric polynomial approximations. When calculations using piecewise polynomial approximations are compared with those using trigonometric polynomial approximations, the latter are almost always the more efficient.

Inverse obstacle scattering ν 3 D u i u s, u = u i + u s u + k 2 u = 0 in IR 3 \ D Bu = 0 on D u s ( ) 1 iku s = o, r = x r r u s (x) = eik x x { u ( x x ) + O ( )} 1, x x

Inverse obstacle scattering ν 3 D u i u s, u = u i + u s u + k 2 u = 0 in IR 3 \ D Bu = 0 on D u s ( ) 1 iku s = o, r = x r r u s (x) = eik x x { u ( x x ) + O Given: Far field u for one incident plane wave Find: Shape and location of scatterer D ( )} 1, x x

Inverse obstacle scattering ν 3 D u i u s, u = u i + u s u + k 2 u = 0 in IR 3 \ D Bu = 0 on D u s ( ) 1 iku s = o, r = x r r u s (x) = eik x x { u ( x x ) + O ( )} 1, x x Given: Far field u for one incident plane wave Find: Shape and location of scatterer D Nonlinear and ill-posed Model problem: nondestructive evaluation, radar, sonar etc.

Uniqueness, i.e., identifiability Recall Rellich s Lemma: u (ˆx) = 0 for all directions ˆx (or all ˆx from a limited aperture) u s (x) = 0 for all x IR 3 \ D

Uniqueness, i.e., identifiability Recall Rellich s Lemma: u (ˆx) = 0 for all directions ˆx (or all ˆx from a limited aperture) u s (x) = 0 for all x IR 3 \ D Far field u uniquely determines total field u = u i + u s Bu = 0 3 D u i u s

Uniqueness, i.e., identifiability Question of uniqueness: Existence of additional closed surfaces on which Bu = 0 Bu = 0 3 D u i u s Bu = 0 No!!

Uniqueness, i.e., identifiability Question of uniqueness: Existence of additional closed surfaces on which Bu = 0 Bu = 0 3 D u i Bu = 0 u s Do not know???

Schiffer s theorem Theorem (Schiffer 1960) For a sound-soft scatterer, assume that u,1 (ˆx, d) = u,2 (ˆx, d) for all observation directions ˆx and all incident directions d. Then D 1 = D 2. D 1 D 2

Idea of Schiffer s proof D = unbounded component of IR 3 \ (D 1 D 2 ) D 1 D 2

Idea of Schiffer s proof D = unbounded component of IR 3 \ (D 1 D 2 ) D 1 D 2 u1 s(, d) = us 2 (, d) in D

Idea of Schiffer s proof D = unbounded component of IR 3 \ (D 1 D 2 ) D 1 D 2 u1 s(, d) = us 2 (, d) in D u 1 (x, d) = e ik x d + u1 s (x, d)

Idea of Schiffer s proof D = unbounded component of IR 3 \ (D 1 D 2 ) D 1 D 2 u1 s(, d) = us 2 (, d) in D u 1 (x, d) = e ik x d + u1 s (x, d) In shaded domain: u 1 + k 2 u 1 = 0

Idea of Schiffer s proof D = unbounded component of IR 3 \ (D 1 D 2 ) D 1 D 2 u1 s(, d) = us 2 (, d) in D u 1 (x, d) = e ik x d + u1 s (x, d) In shaded domain: u 1 + k 2 u 1 = 0 On boundary: u 1 = 0

Idea of Schiffer s proof D = unbounded component of IR 3 \ (D 1 D 2 ) D 1 D 2 u1 s(, d) = us 2 (, d) in D u 1 (x, d) = e ik x d + u1 s (x, d) In shaded domain: u 1 + k 2 u 1 = 0 On boundary: u 1 = 0 {u 1 (, d) : d S 2 } linearly independent

Idea of Schiffer s proof Correct shaded domain: (IR 3 \ D ) \ D 1 D 1 D 2 D

Idea of Schiffer s proof Correct shaded domain: (IR 3 \ D ) \ D 1 D 1 D 2 D Incorrect shaded domain: D 2 \ (D 1 D 2 ) D 1 D 2 D

Idea of Schiffer s proof Correct shaded domain: (IR 3 \ D ) \ D 1 D 1 D 2 D Incorrect shaded domain: D 2 \ (D 1 D 2 ) D 1 D 2 D Lax and Philipps 1967

Idea of Schiffer s proof Correct shaded domain: (IR 3 \ D ) \ D 1 D 1 D 2 D Incorrect shaded domain: D 2 \ (D 1 D 2 ) D 1 D 2 D Lax and Philipps 1967 This proof does not work for other boundary conditions!

Uniqueness for one incident wave Quasi counter example for D = ball of radius R

Uniqueness for one incident wave Quasi counter example for D = ball of radius R u i (x) = sin k x x

Uniqueness for one incident wave Quasi counter example for D = ball of radius R u i (x) = sin k x x u s sin kr (x) = e ikr e ik x x

Uniqueness for one incident wave Quasi counter example for D = ball of radius R u i (x) = sin k x x u s sin kr (x) = e ikr e ik x x u(x) = sin k( x R) e ikr x u = 0 on spheres x = R + nπ, n = 0, 1, 2... k

Uniqueness for one incident wave Theorem (Colton, Sleeman 1983) Under the a priori assumption k diamd < 2π a sound-soft obstacle D is uniquely determined by the far field for one incident plane wave. Schiffer s proof plus monotonicity of eigenvalues with respect to the domain. Gintides 2005 k diamd < 8.99...

Uniqueness for one incident wave Theorem (Liu 1997) A sound-soft ball is uniquely determined by the far field pattern for one incident plane wave.

Uniqueness for one incident wave Theorem (Liu 1997) A sound-soft ball is uniquely determined by the far field pattern for one incident plane wave. Theorem ( Cheng, Yamamoto 2003, Alessandrini, Rondi 2005, Elschner,Yamamoto 2006, Liu, Zou 2006) A polyhydral sound-soft obstacle is uniquely determined by the far field pattern for one incident plane wave. D

Uniqueness of obstacle plus boundary condition Theorem (Kirsch, K. 1992) Assume that u,1 (ˆx, d) = u,2 (ˆx, d) for all observation directions ˆx and all incident directions d. Then D 1 = D 2 and B 1 = B 2. B 1 D 1 D 2 B 2

Idea of proof x z D w i (x, z) = eik x z x z w s (x, z) = scattered field, = incident field, point source w (ˆx, z) = far field

Idea of proof x z D w i (x, z) = eik x z x z w s (x, z) = scattered field, = incident field, point source w (ˆx, z) = far field reciprocity: u (ˆx, d) = u ( d, ˆx), w s (x, z) = w s (z, x)

Idea of proof x z D w i (x, z) = eik x z x z w s (x, z) = scattered field, = incident field, point source w (ˆx, z) = far field reciprocity: u (ˆx, d) = u ( d, ˆx), w s (x, z) = w s (z, x) mixed reciprocity: u s (z, d) = w ( d, z)

Idea of proof mixed reciprocity: u s (z, d) = w ( d, z) z x B 1 D D 1 D 2 B 2

Idea of proof mixed reciprocity: u s (z, d) = w ( d, z) z x B 1 D D 1 D 2 B 2 u,1 (ˆx, d) = u,2 (ˆx, d) for ˆx = d = 1

Idea of proof mixed reciprocity: u s (z, d) = w ( d, z) z x B 1 D D 1 D 2 B 2 u,1 (ˆx, d) = u,2 (ˆx, d) for ˆx = d = 1 u s 1 (z, d) = us 2 (z, d) for z D, d = 1

Idea of proof mixed reciprocity: u s (z, d) = w ( d, z) z x B 1 D D 1 D 2 B 2 u,1 (ˆx, d) = u,2 (ˆx, d) for ˆx = d = 1 u s 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

Idea of proof mixed reciprocity: u s (z, d) = w ( d, z) z x B 1 D D 1 D 2 B 2 u,1 (ˆx, d) = u,2 (ˆx, d) for ˆx = d = 1 u s 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 w1 s(x, z) = w 2 s (x, z) for x, z D

Idea of proof z B 1 x D D 1 D 2 B 2 w1 s (x, z) = w 2 s (x, z) for x, z D

Idea of proof z B 1 x D D 1 D 2 B 2 w1 s (x, z) = w 2 s (x, z) for x, z D lim B 1w s z x 1 (x, z) =,

Idea of proof z B 1 x D D 1 D 2 B 2 w1 s (x, z) = w 2 s (x, z) for x, z D lim B 1w s z x 1 (x, z) =, lim B z x 1w2 s (x, z) = finite

Idea of proof z B 1 x D D 1 D 2 B 2 w1 s (x, z) = w 2 s (x, z) for x, z D lim B 1w s z x 1 (x, z) =, lim B z x 1w2 s (x, z) = finite D 1 = D 2

Idea of proof z B 1 x D D 1 D 2 B 2 w1 s (x, z) = w 2 s (x, z) for x, z D lim B 1w s z x 1 (x, z) =, lim B z x 1w2 s (x, z) = finite D 1 = D 2 Holmgren s theorem yields B 1 = B 2

Idea of proof z B 1 x D D 1 D 2 B 2 w1 s (x, z) = w 2 s (x, z) for x, z D lim B 1w s z x 1 (x, z) =, lim B z x 1w2 s (x, z) = finite D 1 = D 2 Holmgren s theorem yields B 1 = B 2 Use of mixed reciprocity: Potthast 1999

Reconstruction methods Reconstruction methods connected to the uniqueness proof of Kirsch, K.: Singular source method of Potthast 2001 Needle method of Ikehata 2000 Use of w s (x, z), if x, z IR 3 \ D D z D x

Inverse obstacle scattering ν 3 D u i u s, u = u i + u s u + k 2 u = 0 in IR 3 \ D u = 0 u s r on D iku s = o ( ) 1, r = x r u s (x) = eik x x { ( ) ( )} x 1 u + O, x x x Given: Far field u for one incident plane wave Find: Shape and location of scatterer D Nonlinear and ill-posed

Example for nonlinearity A priori information: D is ball of radius R centered at origin Incident field: u i (x) = sin k x x

Example for nonlinearity A priori information: D is ball of radius R centered at origin Incident field: u i (x) = sin k x x Scattered field: u s sin kr (x) = e ikr e ik x x

Example for nonlinearity A priori information: D is ball of radius R centered at origin Incident field: u i (x) = sin k x x Scattered field: u s sin kr (x) = e ikr sin kr Far-field pattern: u (ˆx) = e ikr e ik x x Nonlinear equation for the unknown radius R

Example for ill-posedness sin kr Perturbed data: u (ˆx) = e ikr + εy n(ˆx)

Example for ill-posedness sin kr Perturbed data: u (ˆx) = e ikr + εy n(ˆx) Total field: sin k( x R) u(x) = e ikr x ( ) + ε k i n+1 h (1) x n (k x )Y n x

Example for ill-posedness sin kr Perturbed data: u (ˆx) = e ikr + εy n(ˆx) Total field: sin k( x R) u(x) = e ikr x ( ) + ε k i n+1 h (1) x n (k x )Y n x ( ) u(x) = ε k i n+1 h (1) x n (kr)y n, x = R x

Example for ill-posedness sin kr Perturbed data: u (ˆx) = e ikr + εy n(ˆx) Total field: sin k( x R) u(x) = e ikr x ( ) + ε k i n+1 h (1) x n (k x )Y n x ( ) u(x) = ε k i n+1 h (1) x n (kr)y n, x = R x u(x) ε k ( ) 2n n ( ) x Y n, x = R ekr x Small errors in data u can cause large errors in solution, or solution may not exist anymore.

Existence??? ν 3 D u i u s, u, u = u i + u s Wrong question to ask: Would need to characterize far-field patterns for which the corresponding total field vanishes on a closed surface. Main Task: Assuming correct data or perturbed correct data, design methods for a stable approximate solution

Iterative methods versus qualitative methods 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 of certain ill-posed linear integral equations that decide on whether a point lies inside or outside the scatterer. Linear sampling, factorization, probe methods, etc

Iterative methods for boundary to far field map 3 D u i u Interpret inverse problem as operator equation F( D) = u For simplicity: D = {p(ˆx) : ˆx S 2 }, p : S 2 IR 3 Then F : C 2 (S 2, IR 3 ) L 2 (S 2, IC), F : p u Inverse problem: Solve F(p) = u Linearize: F(p + q) = F(p) + F (p; q) + o(q) and, given an approximation p, solve F (p) + F (p; q) = u for q to update p into p + q. Regularization required

Fréchet derivative Theorem Fréchet derivative is given by F (p; ) : q v q, D p ν p + q q p D p+q where v q, is the far field of radiating solution to v q + k 2 v q = 0 v q = ν q u ν Proof by hand waving in IR 3 \ D p on D p F(p + q) = F(p) + F (p; q) + o(q) 0 = u( D p+q ) Dq+q u( D p ) Dp + [u ( D)q] Dp + grad u( D p ) Dp q }{{}}{{}}{{} = 0 = v q = ν u ν

Fréchet derivative Theorem Fréchet derivative is given by F (p; ) : q v q, D p ν p + q q p D p+q where v q, is the far field of radiating solution to v q + k 2 v q = 0 v q = ν q u ν in IR 3 \ D p on D p F(p + q) = F(p) + F (p; q) + o(q) Roger 1981, hand waving Kirsch, K. 1991, Hilbert space methods, domain derivative Potthast 1992, boundary integral equations

Newton iterations for boundary to far field map Numerical examples: Hohage, Hettlich, Kirsch, K., Murch et al, Roger, Rundell, Tobocman,... 1991-..., in 2D Farhat et al 2002, in 3D Harbrecht, Hohage 2005, in 3D Pros: Conceptually simple Very good reconstructions Contras: Need efficient forward solver and good a priori information Convergence not completely settled

Huygens principle u s (x) = 1 4π D e ik x y x y u (y) ds(y), ν x IR3 \ D

Huygens principle u s (x) = 1 4π Data equation u (ˆx) = 1 4π Field equation u i (x) = 1 4π D D e ik x y x y D u (y) ds(y), ν x ik ˆx y u e (y) ds(y), ν e ik x y x y u (y) ds(y), ν x Two integral equations for two unknowns IR3 \ D ˆx S2 D

Parameterized equations Recall D = {p(ˆx) : ˆx S 2 }

Parameterized equations Recall D = {p(ˆx) : ˆx S 2 } Define A, A : C 2 (S 2, IR 3 ) L 2 (S 2, IC) L 2 (S 2, IC) by A(p, ψ)(ˆx) := 1 4π S2 e ik p(ˆx) p(ŷ) p(ˆx) p(ŷ) ψ(ŷ) ds(ŷ), ˆx S2 and A (p, ψ)(ˆx) := 1 4π S 2 e ik ˆx p(ŷ) ψ(ŷ) ds(ŷ), ˆx S 2

Derivatives of operators A(p, ψ)(ˆx) := 1 4π A (p, ψ; q)(ˆx) = 1 4π S2 e ik p(ˆx) p(ŷ) ψ(ŷ) ds(ŷ) p(ˆx) p(ŷ) S2 grad eik p(ˆx) p(ŷ) ψ(ŷ) ds(ŷ) p(ˆx) p(ŷ) [q(ˆx) q(ŷ)] and A (p, ψ)(ˆx) := 1 e ik ˆx p(ŷ) ψ(ŷ) ds(ŷ) 4π S 2 A (p, ψ; q)(ˆx) = ik e ik ˆx p(ŷ) ˆx q(ŷ) ψ(ŷ) ds(ŷ) 4π S 2 Linearizations in the sense A(p + q, ψ) A(p, ψ) A (p, ψ; q) L 2 (S 2 ) = o( q C 2 (S 2 ) )

Linearization of the data equation If k 2 is not a Dirichlet eigenvalue of the negative Laplacian in D, then A(p, ) : H 1/2 (S 2 ) H 1/2 (S 2 ) is a homeomorphism.

Linearization of the data equation If k 2 is not a Dirichlet eigenvalue of the negative Laplacian in D, then A(p, ) : H 1/2 (S 2 ) H 1/2 (S 2 ) is a homeomorphism. Given an approximation for p solve the field equation A(p, ψ) = u i p for the density ψ, that is, ψ = [A(p, )] 1 (u i p) Keeping ψ fixed, linearize the data equation to obtain for q to update p into p + q. A (p, ψ) = u A (p, ψ; q) = u A (p, ψ)

Linearization of the data equation Recall boundary to far field operator F : p u.

Linearization of the data equation Recall boundary to far field operator F : p u. Can represent with derivative F(p) = A (p, [A(p, )] 1 (u i p)) F (p; q) = A (p, [A(p, )] 1 (u i p); q) +A (p, [A(p, )] 1 A (p, [A(p, )] 1 (u i p); q)) A (p, [A(p, )] 1 ((grad u i ) p) q) Linearization of the data equation corresponds to Newton iteration for F(p) = u with the derivate of F approximated through the first term

Simultaneous linearization of both equations Given approximations p and ψ linearize both equations to obtain and A (p, ψ; q) + A (p, χ) = A (p, ψ) + u A (p, ψ; q) + ((grad u i ) p) q + A(p, χ) = A(p, ψ) u i p to be solved for q and χ to update p and ψ into p + q and ψ + χ

Simultaneous linearization of both equations Theorem (Ivanyshyn, Kress 2008) Assume that k 2 is not a Dirichlet eigenvalue of the negative Laplacian in D and set ψ := [A(p, )] 1 (u i p).

Simultaneous linearization of both equations Theorem (Ivanyshyn, Kress 2008) Assume that k 2 is not a Dirichlet eigenvalue of the negative Laplacian in D and set ψ := [A(p, )] 1 (u i p). Provided q satisfies linearized boundary to far field equation F (p; q) = u F(p) then q and χ := [A(p, )] 1 (A (p, ψ; q) + ((grad u i ) p) q) satisfy linearized integral equations and A (p, ψ; q) + A (p, χ) = A (p, ψ) + u A (p, ψ; q) + ((grad u i ) p) q + A(p, χ) = A(p, ψ) u i p and vice versa.

Linearization of the field equation If k 2 is not a Dirichlet eigenvalue of the negative Laplacian in D, then A (p, ) : L 2 (S 2 ) L 2 (S 2 ) is injective and has dense range.

Linearization of the field equation If k 2 is not a Dirichlet eigenvalue of the negative Laplacian in D, then A (p, ) : L 2 (S 2 ) L 2 (S 2 ) is injective and has dense range. Given an approximation for p, find a regularized solution ψ of the data equation A (p, ψ) = u. Keeping ψ fixed, linearize the field equation to obtain A(p, ψ) = u i p A (p, ψ; q) + ((grad u i ) p) q = A(p, ψ) u i p for q to update p into p + q.

Decomposition methods Decompose the inverse problem into a linear ill-posed part and a nonlinear part. Step 1. Reconstruct scattered field u s from the given far field pattern u. Step 2. Find unknown boundary D as location where the boundary condition u i + u s = 0 for the total field is satisfied.

Decomposition methods Given an approximation for p, find a regularized solution ψ of the data equation A (p, ψ) = u and define the single-layer potential u(x) = u i (x) + 1 4π S2 e ik x p(ŷ) ψ(ŷ) ds(ŷ). x p(ŷ) Find an update p + q by linearizing the boundary condition u (p + q) = 0, that is, by solving the linear equation for q. u p + ((grad u) p) q = 0

Implementation: Hybrid method 1. Given an approximation p, solve ill-posed integral equation A (p, ψ) = u, that is, S 2 e ik ˆx p(ŷ) ψ(ŷ) ds(ŷ) = u (ˆx), ˆx S 2

Implementation: Hybrid method 1. Given an approximation p, solve ill-posed integral equation A (p, ψ) = u, that is, S 2 e ik ˆx p(ŷ) ψ(ŷ) ds(ŷ) = u (ˆx), ˆx S 2 Smooth integrand, for example, Gauss-trapezoidal rule Severe ill-posedness requires regularization, for example, via Tikhonov regularization

Implementation: Hybrid method 2. Approximate u(x) = u i (x) + 1 4π S2 e ik x p(ŷ) ψ(ŷ) ds(ŷ) x p(ŷ) and evaluate boundary values and normal derivatives on D = {p(ˆx) : ˆx S 2 } by jump relations. Use spectral quadrature rules of Wienert 1990, Ganesh, Graham, Sloan 2002 2004 Find update q by solving Insert q(ˆx) = r(ˆx)ˆx with u p + ((grad u) p) q = 0 r = M m m=0 n= m a mn Y mn, collocate at L points on D and solve L (M + 1) 2 linear system for a mn by penalized least squares.

Implementation: Linearize data equation 1. Given an approximation p, solve well-posed integral equation that is, A(p, ψ) = u, 1 4π S2 e ik p(ˆx) p(ŷ) p(ˆx) p(ŷ) ψ(ŷ) ds(ŷ) = ui (p(ˆx)), ˆx S 2 Use spectral quadrature rules of Wienert 1990, Ganesh, Graham, Sloan 2002 2004 and collocation.

Implementation: Linearize data equation 2. Find update q by solving ill-posed linearized data equation, that is, A (p, ψ; q) = u A (p, ψ) by Tikhonov regularization (smooth integrands again).

Implementation: Linearize data equation 2. Find update q by solving ill-posed linearized data equation, that is, A (p, ψ; q) = u A (p, ψ) by Tikhonov regularization (smooth integrands again). Insert q(ˆx) = r(ˆx)ˆx with r = M m m=0 n= m a mn Y mn, collocate at L points on D and solve L (M + 1) 2 linear system for a mn by Tikhonov regularization.

Sampling and probe methods Develop criterium in terms of behaviour of certain ill-posed linear integral equations that decide on whether a point z lies inside or outside the scatterer D.

Sampling and probe methods Develop criterium in terms of behaviour of certain ill-posed linear integral equations that decide on whether a point z lies inside or outside the scatterer D. D Evaluate the criterium numerically for a grid of points Need full data, i.e, u (ˆx, d) for all ˆx, d S 2

Linear sampling method Define far field operator F : L 2 (S 2 ) L 2 (S 2 ) by Fg(ˆx) := u (ˆx, d)g(d) ds(d), ˆx S 2 S 2

Linear sampling method Define far field operator F : L 2 (S 2 ) L 2 (S 2 ) by Fg(ˆx) := u (ˆx, d)g(d) ds(d), ˆx S 2 S 2 Recall point source w i (x, z) = eik x z = incident field x z w s (x, z) = scattered field, w (ˆx, z) = far field w (ˆx, i z) = e ik z ˆx = far field of incident field

Linear sampling method Let z D and g be a solution of S 2 u (ˆx, d)g(d, z) ds(d) = e ik z ˆx, ˆx S 2

Linear sampling method Let z D and g be a solution of u (ˆx, d)g(d, z) ds(d) = e ik z ˆx, S 2 ˆx S 2 u s (x, d)g(d, z) ds(d) = w i (x, z), S 2 x IR 3 \ D

Linear sampling method Let z D and g be a solution of u (ˆx, d)g(d, z) ds(d) = e ik z ˆx, S 2 ˆx S 2 u s (x, d)g(d, z) ds(d) = w i (x, z), S 2 x IR 3 \ D S 2 e ik x d g(d, z) ds(d) }{{} Hg(, z)(x) = eik x z x z, x D g(, z) L 2 (S 2 ), Hg(, z) L 2 ( D), z D Bad news: Integral equation, in general, not solvable

Linear sampling method Theorem (Colton, Kirsch 1996) For every ε > 0 and z D there exists g(, z) L 2 (S 2 ) such that Fg(, z) w i (, z) L 2 (S 2 ) ε and g(, z) L 2 (S 2 ), Hg(, z) L 2 ( D), z D Arens 2003 Why linear sampling works?

Factorization method Recall far field operator F : L 2 (S 2 ) L 2 (S 2 ) with Fg(ˆx) := u (ˆx, d)g(d) ds(d), ˆx S 2 S 2 and consider ill-posed linear integral equation (F F) 1/4 g(, z) = w ( i, z)

Factorization method Recall far field operator F : L 2 (S 2 ) L 2 (S 2 ) with Fg(ˆx) := u (ˆx, d)g(d) ds(d), ˆx S 2 S 2 and consider ill-posed linear integral equation (F F) 1/4 g(, z) = w ( i, z) Theorem (Kirsch 1998) The (F F) 1/4 equation is solvable if and only if z D.

Singular source method Exploit the uniqueness proof and characterize the boundary of the scatterer D by the points z where becomes large. Potthast 2001, pointwise Ikehata 2000, in energy norm w s (z, z)

Singular source method Exploit the uniqueness proof and characterize the boundary of the scatterer D by the points z where w s (z, z) Λ becomes large. Potthast 2001, pointwise Ikehata 2000, in energy norm D z Approximate incident point source field w i (, z) by linear combination of plane waves.

Singular source method Let z IR 3 \ D and g be a solution of S2 e ik x d g(d, z) ds(d) = eik x z x z, x Λ

Singular source method Let z IR 3 \ D and g be a solution of S2 e ik x d g(d, z) ds(d) = eik x z x z, x Λ u s (x, d)g(d, z) ds(d) = w s (x, z), x IR 3 \ D S 2

Singular source method Let z IR 3 \ D and g be a solution of S2 e ik x d g(d, z) ds(d) = eik x z x z, x Λ u s (x, d)g(d, z) ds(d) = w s (x, z), x IR 3 \ D S 2 u (ˆx, d)g(d, z) ds(d) = w (ˆx, z) S 2 = u s (z, ˆx), ˆx S 2 w s (z, z) = u ( ˆx, d) g(ˆx, z) g(d, z) ds(ˆx) ds(d) S 2 Bad news: Integral equation, in general, not solvable

Sampling and probe methods Pros: Nice mathematics Simple implementation No a priori information needed Contras: Need a lot of data No sharp boundaries ( =?) Sensitive to noise

References

References Serranho, P. A hybrid method for inverse scattering for sound-soft obstacles in IR 3. Inverse Problems and Imaging, 1, 691 712 (2007). Ivanyshyn, O., Kress, R. and Serranho, P. Huygens principle and iterative methods in inverse obstacle scattering. Advances in Computational Mathematics 33, 413 429 (2010). Ivanyshyn, O. and Kress, R. Identification of sound-soft 3D obstacles from phaseless data. Inverse Problems and Imaging 4, 111 130 (2010).