Inverse Obstacle Scattering

Transcription

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

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

3 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

4 Incident plane wave

5 Scattered wave

6 Total wave

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

8 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

9 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

10 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

11 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

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

13 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

14 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 ( x + k 2 )Φ(x, y) = 0, x y ( + k 2 )Φ(, y) = δ y

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

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

17 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

18 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

19 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 Theorem (Holmgren) u Γ = ν u Γ = 0 for Γ D open u = 0 in D

20 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

21 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 + ū u dx D D

22 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

23 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.

24 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

25 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

26 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. Lemma (Rellich) { ( ) ( )} x 1 u + O, x x x u (ˆx) = 0 for all ˆx S 2 (or all ˆx Γ for an open Γ S 2 ) u(x) = 0 for all x IR 3 \ D

27 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

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

29 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

30 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

31 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

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

33 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

34 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 ( ) 1 [ ik u 2 ds + O = grad u s 2 k 2 u s 2] dx S 2 R D R

35 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 ( ) 1 [ ik u 2 ds + O = grad u s 2 k 2 u s 2] dx S 2 R D R u = 0 on S 2 u s = 0 in IR 3 \ D

36 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)

37 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

38 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 u = ϕ, ν u = iϕ on D

39 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 u = ϕ, ν u = iϕ on D i D ϕ 2 ds = D ū u ν ds = D { grad u 2 k 2 u 2} dx

40 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 u = ϕ, ν u = iϕ on D i D ϕ 2 ds = D ū u ν ds = D { grad u 2 k 2 u 2} dx ϕ = 0 Apply Riesz theory for compact operators

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

42 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

43 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

44 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.

45 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

46 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

47 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.

48 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

49 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

50 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 Question of uniqueness: Existence of additional closed surfaces with Bu = 0 u s

51 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!!

52 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???

53 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

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

55 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

56 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)

57 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

58 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

59 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

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

61 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

62 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

63 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!

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

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

66 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

67 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

68 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.

69 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 <

70 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.

71 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

72 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 D 0

73 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 u = 0 D 0

74 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 u = 0 D 0 General case: Use reflection principle to find a path to infinity

75 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

76 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

77 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)

78 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)

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

80 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

81 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

82 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

83 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

84 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

85 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) =,

86 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

87 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

88 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

89 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

90 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 Has been extended to a variety of other scattering problems.

91 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

92 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 Sampling method of Colton, Kirsch 1996 Use of w s (x, z) if x, z D D

93 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

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

95 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

96 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

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

98 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

99 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

100 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.

101 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

102 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

103 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

104 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

105 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

106 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

107 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

108 Newton iterations for boundary to far field map Numerical examples: Hohage, Hettlich, Kirsch, K., Murch et al, Roger, Rundell, Tobocman, , in 2D Farhat et al 2002, in 3D Harbrecht, Hohage 2005, in 3D

109 Newton iterations for boundary to far field map Numerical examples: Hohage, Hettlich, Kirsch, K., Murch et al, Roger, Rundell, Tobocman, , 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

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

111 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

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

113 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

114 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 Then setting ψ := J(p) u ν p S 2 e ik ˆx p(ŷ) ψ(ŷ) ds(ŷ), ˆx S 2 Data equation Field equation A (p, ψ) = u A(p, ψ) = u i p

115 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 ) )

116 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.

117 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)

118 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. Johansson, Sleeman 2007 A (p, ψ) = u A (p, ψ; q) = u A (p, ψ)

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

120 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)

121 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

122 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

123 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 K., Rundell 2005 Laplace equation Ivanyshyn, K. 2006,... Helmholtz equation

124 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).

125 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.

126 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.

127 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.

128 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. Slight modification leads to the hybrid decomposition method of K., Serranho 2003,...

129 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.

130 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. For example in first step represent u s (x) = 1 e ik x y 4π x y for some surface Γ D. Kirsch, K Γ ϕ(y) ds(y)

131 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)

132 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

133 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 u p + ((grad u) p) q = 0 for q. Equivalent to linearizing the field equation A(p, ψ) = u i p only with respect to the evaluation point. K., Serranho 2003,...

134 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

135 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

136 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

137 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 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.

138 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

139 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 and collocation.

140 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).

141 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.

142

143

144 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.

145 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

146 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

147 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

148 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 Consider ill-posed linear integral equation Fg(, z) = w ( i, z) for arbitrary source locations z

149 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

150 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

151 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

152 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

153 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

154 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 Good news: Method works well and can be justified by approximation arguments

155 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?

156 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)

157 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.

158 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)

159 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.

160 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

161 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

162 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

163 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

164 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

165 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 Good news: Method works well and can be justified by approximation arguments

166 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

167 References

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

169 References Serranho, P. A hybrid method for inverse scattering for sound-soft obstacles in IR 3. Inverse Problems and Imaging, 1, (2007). Ivanyshyn, O., Kress, R. and Serranho, P. Huygens principle and iterative methods in inverse obstacle scattering. Advances in Computational Mathematics 33, (2010). Ivanyshyn, O. and Kress, R. Identification of sound-soft 3D obstacles from phaseless data. Inverse Problems and Imaging 4, (2010). Colton, D. and Kress, R. Inverse scattering. In: Handbook of Mathematical Methods in Imaging (Scherzer, O., ed.) Springer-Verlag, pp (2011).