Inverse scattering problem from an impedance obstacle

Transcription

1 Inverse Inverse scattering problem from an impedance obstacle Department of Mathematics, NCKU 5 th Workshop on Boundary Element Methods, Integral Equations and Related Topics in Taiwan NSYSU, October 4, 2014

2 Outline Inverse 1 Direct 2 Inverse 3 Ellipse Peanut Bean

3 Scattering problem Inverse Direct Object : time harmonic acoustic scattering Modelling : Obstacle Exterior boundary value problem for the Helmholtz equation B) A typical boundary value problem: Point Source Approximation for Obstacle Problems p.5/47

4 Scattering problem Inverse Direct Object : time harmonic acoustic scattering Modelling : Obstacle Exterior boundary value problem for the Helmholtz equation B) A typical boundary value problem: Point Source Approximation for Obstacle Problems p.5/47

5 Scattering problem Inverse Direct Object : time harmonic acoustic scattering Modelling : Obstacle Exterior boundary value problem for the Helmholtz equation B) A typical boundary value problem: Point Source Approximation for Obstacle Problems p.5/47

6 Direct problem Inverse Direct Definition 1 Find: u s C 2 (R 2 \ D) C(R 2 \ D) satisfies 1 the Helmholtz equation u s + k 2 u s = 0, in R 2 \ D 2 the impedance boundary condition u ν for the total field u := u i + u s + λu = 0 on D (1) 3 the Sommerfeld radiation condition(src) ( ) lim r r u s ν ikus = 0, r := x, ˆx := x x

7 Inverse Green s representation formula Direct u s (x) = D u s Φ(x, y) (y) ν(y) us (y) Φ(x, y)ds(y), ν(y) x IR2 \ D (2)

8 Solution ansatz Inverse Direct u s (x) = D Φ(x, y) u(y) ν(y) +λ(y)u(y)φ(x, y)ds(y), x IR2 \D (3)

9 Integral operators Inverse Direct Sϕ(x) := 2 D K ϕ(x) := 2 D Φ(x, y)ϕ(y)ds(y) (4) Φ(x, y) ϕ(y)ds(y) (5) ν(y)

10 Inverse Well-posedness of DP Direct Theorem 1 The direct problem has a unique solution given by u s (x) = D Φ(x, y) u(y) + λ(y)u(y)φ(x, y)ds(y), x IR 2 \ ν(y) D (6) where (the total field) u is the (unique) solution to the following boundary integral equation u Ku S(λu) = 2u i, on D (7)

11 Inverse Direct Far field pattern u The far field pattern or the scattering amplitude is given by ( )} u s (x) = {u eik x 1 (ˆx) + O x x x uniformly for all directions ˆx Ω := {x R 2 x = 1}. In our case u (ˆx) = (c 1 < ν(y), ˆx > +c 2 λ(y)) e ik<ˆx,y> u(y)ds(y) (8) where c 1 = 1 i 4 D k π, c 2 = 1+i 4 kπ

12 Inverse Direct Far field pattern u The far field pattern or the scattering amplitude is given by ( )} u s (x) = {u eik x 1 (ˆx) + O x x x uniformly for all directions ˆx Ω := {x R 2 x = 1}. In our case u (ˆx) = (c 1 < ν(y), ˆx > +c 2 λ(y)) e ik<ˆx,y> u(y)ds(y) (8) where c 1 = 1 i 4 D k π, c 2 = 1+i 4 kπ

13 Inverse Summary : Direct Problem Direct The direct problem can be understood as the process of calculating the far-field pattern from an impedance obstacle. Mathematically, it is equivalent to the solving of the system: { u Ku S(λu) = 2u i, on D u (ˆx) = F( D, λ, u), ˆx Ω (9)

14 Inverse Summary : Direct Problem Direct The direct problem can be understood as the process of calculating the far-field pattern from an impedance obstacle. Mathematically, it is equivalent to the solving of the system: { u Ku S(λu) = 2u i, on D u (ˆx) = F( D, λ, u), ˆx Ω (9)

15 Inverse Problem Inverse Definition 2 (IP) Determine both the scatterer D and the impedance λ if the far field pattern u (, d) is known for one incident direction d and one wave number k > 0.

16 Unique solvability Inverse? Uniqueness Not available Existence Not available

17 Unique solvability Inverse? Uniqueness Not available Existence Not available

18 Unique solvability Inverse? Uniqueness Not available Existence Not available

19 Unique solvability Inverse? Uniqueness Not available Existence Not available

20 Unique solvability Inverse? Uniqueness Not available Existence Not available

21 Inverse Comments on Existence Solving the inverse problem means to solve the far field equation F( D, λ, u) = u (10) However (10) is an equation of the first kind The operator F is compact F has no bounded inverse in general This means that equation (10) cannot be resonably solved!

22 Inverse Comments on Existence Solving the inverse problem means to solve the far field equation F( D, λ, u) = u (10) However (10) is an equation of the first kind The operator F is compact F has no bounded inverse in general This means that equation (10) cannot be resonably solved!

23 Inverse Comments on Existence Solving the inverse problem means to solve the far field equation F( D, λ, u) = u (10) However (10) is an equation of the first kind The operator F is compact F has no bounded inverse in general This means that equation (10) cannot be resonably solved!

24 Inverse Fredholm integral Equations 2. Kind ϕ(x) (x+1)e xy ϕ(y)dy = e x e (x+1), 0 x 1 Trapzoidal rule n x = 0 x = 0.5 x = Simpson s rule n x = 0 x = 0.5 x =

25 Inverse Fredholm integral Equations 1. Kind 1 0 (x + 1)e xy ϕ(y)dy = 1 e (x+1), 0 x 1 Trapzoidal rule n x = 0 x = 0.5 x = Simpson s rule n x = 0 x = 0.5 x =

26 Inverse Ill-Posed Problems : Definition 3 () Assume X, Y are normed spaces. Let the operator A : X Y be linear, bounded and injective. A family of bounded linear operators R α : Y X, α > 0 is called a regularization scheme for Aϕ = f, if it satisfies the following pointwise convergence lim R αaϕ = ϕ, for all ϕ X α 0 In this case, the parameter α is called the regularization parameter.

27 Inverse Ill-Posed Problems : Definition 3 () Assume X, Y are normed spaces. Let the operator A : X Y be linear, bounded and injective. A family of bounded linear operators R α : Y X, α > 0 is called a regularization scheme for Aϕ = f, if it satisfies the following pointwise convergence lim R αaϕ = ϕ, for all ϕ X α 0 In this case, the parameter α is called the regularization parameter.

28 Inverse : Error Find a stable approximation to the equation Aϕ = f The regularized approximation ϕ δ α := R α f δ The total approximation error ϕ δ α ϕ = R α f δ R α f + R α Aϕ ϕ We have ϕ δ α ϕ δ R α + R α Aϕ ϕ

29 Inverse : Error Find a stable approximation to the equation Aϕ = f The regularized approximation ϕ δ α := R α f δ The total approximation error ϕ δ α ϕ = R α f δ R α f + R α Aϕ ϕ We have ϕ δ α ϕ δ R α + R α Aϕ ϕ

30 Inverse : Parameter How to choose the regularization parameter α? 1 a priori choice based on some information of the solution. In general not available 2 a posteriori choice based on the data error level δ Discrepancy Principle of Morozov : AR α f δ f δ = γδ, γ 1

31 Inverse : Parameter How to choose the regularization parameter α? 1 a priori choice based on some information of the solution. In general not available 2 a posteriori choice based on the data error level δ Discrepancy Principle of Morozov : AR α f δ f δ = γδ, γ 1

32 Inverse : Parameter How to choose the regularization parameter α? 1 a priori choice based on some information of the solution. In general not available 2 a posteriori choice based on the data error level δ Discrepancy Principle of Morozov : AR α f δ f δ = γδ, γ 1

33 Inverse : Parameter How to choose the regularization parameter α? 1 a priori choice based on some information of the solution. In general not available 2 a posteriori choice based on the data error level δ Discrepancy Principle of Morozov : AR α f δ f δ = γδ, γ 1

34 Inverse : Example X, Y Hilbert spaces. Theorem 2 Assume A : X Y compact and linear. Then for every α > 0, the operator αi + A A : X X is bijective and has a bounded inverse. Furthermore, if the operator A is injective, then R α := (αi + A A) 1 A, α > 0 describes a regularization scheme with R α 1 2 α.

35 Inverse : Example X, Y Hilbert spaces. Theorem 2 Assume A : X Y compact and linear. Then for every α > 0, the operator αi + A A : X X is bijective and has a bounded inverse. Furthermore, if the operator A is injective, then R α := (αi + A A) 1 A, α > 0 describes a regularization scheme with R α 1 2 α.

36 Inverse Tikhonov Theorem 3 Let A : X Y be a linear and bounded operator. Assme α > 0. Then for each f Y there exists a unique ϕ α X such that { Aϕ α f + α ϕ α = inf ϕ X Aϕ f 2 + α ϕ 2} The minimizer ϕ α is given by the unique solution of the equation αϕ α + A Aϕ α = A f and depends continuously on f.

37 Inverse : Newton s method F( D, λ, u) = u 3 F Θ = u F (11) where αi βi γi + A A Θ = A (u F) (12) F D 0 0 A = F 0 λ 0 F 0 0 u

38 Inverse : Newton s method F( D, λ, u) = u 3 F Θ = u F (11) where αi βi γi + A A Θ = A (u F) (12) F D 0 0 A = F 0 λ 0 F 0 0 u

39 Inverse Modified Newton s Method Recall the solution of direct problem (9) { u Ku S(λu) = 2u i, on D u (ˆx) = F (γ, λ, u), ˆx Ω Split the inverse problem into two parts: 1 B(γ, λ)u = 2u i (13) 2 F (u)(γ, λ) = u (14) (13) is solved as a (well-posed) direct problem. (14) is solved as an ill-posed problem with two regularization parameters.

40 Inverse Modified Newton s Method Recall the solution of direct problem (9) { u Ku S(λu) = 2u i, on D u (ˆx) = F (γ, λ, u), ˆx Ω Split the inverse problem into two parts: 1 B(γ, λ)u = 2u i (13) 2 F (u)(γ, λ) = u (14) (13) is solved as a (well-posed) direct problem. (14) is solved as an ill-posed problem with two regularization parameters.

41 Inverse Modified Newton s Method Recall the solution of direct problem (9) { u Ku S(λu) = 2u i, on D u (ˆx) = F (γ, λ, u), ˆx Ω Split the inverse problem into two parts: 1 B(γ, λ)u = 2u i (13) 2 F (u)(γ, λ) = u (14) (13) is solved as a (well-posed) direct problem. (14) is solved as an ill-posed problem with two regularization parameters.

42 Inverse Modified Newton s Method Recall the solution of direct problem (9) { u Ku S(λu) = 2u i, on D u (ˆx) = F (γ, λ, u), ˆx Ω Split the inverse problem into two parts: 1 B(γ, λ)u = 2u i (13) 2 F (u)(γ, λ) = u (14) (13) is solved as a (well-posed) direct problem. (14) is solved as an ill-posed problem with two regularization parameters.

43 Inverse Modified Newton s Method Recall the solution of direct problem (9) { u Ku S(λu) = 2u i, on D u (ˆx) = F (γ, λ, u), ˆx Ω Split the inverse problem into two parts: 1 B(γ, λ)u = 2u i (13) 2 F (u)(γ, λ) = u (14) (13) is solved as a (well-posed) direct problem. (14) is solved as an ill-posed problem with two regularization parameters.

44 Inverse Modified Newton s Method Recall the solution of direct problem (9) { u Ku S(λu) = 2u i, on D u (ˆx) = F (γ, λ, u), ˆx Ω Split the inverse problem into two parts: 1 B(γ, λ)u = 2u i (13) 2 F (u)(γ, λ) = u (14) (13) is solved as a (well-posed) direct problem. (14) is solved as an ill-posed problem with two regularization parameters.

45 Inverse Modified Newton s Method : iterative scheme 1 Given a pair of initial guesses γ 0, λ 0. 2 Solve (13) for u. 3 Solve the regularized version of (14) for updates of γ, λ : ([ ] ) [ ] αi 0 + A χ A = A (u 0 βi q F) (15) where A = [ F γ 0 0 F λ ] 4 Set γ 0 = γ 0 + χ, λ 0 = λ 0 + q and repeat steps 2,3 until some criterion is fullfilled.

46 Inverse Modified Newton s Method : iterative scheme 1 Given a pair of initial guesses γ 0, λ 0. 2 Solve (13) for u. 3 Solve the regularized version of (14) for updates of γ, λ : ([ ] ) [ ] αi 0 + A χ A = A (u 0 βi q F) (15) where A = [ F γ 0 0 F λ ] 4 Set γ 0 = γ 0 + χ, λ 0 = λ 0 + q and repeat steps 2,3 until some criterion is fullfilled.

47 Inverse Modified Newton s Method : iterative scheme 1 Given a pair of initial guesses γ 0, λ 0. 2 Solve (13) for u. 3 Solve the regularized version of (14) for updates of γ, λ : ([ ] ) [ ] αi 0 + A χ A = A (u 0 βi q F) (15) where A = [ F γ 0 0 F λ ] 4 Set γ 0 = γ 0 + χ, λ 0 = λ 0 + q and repeat steps 2,3 until some criterion is fullfilled.

48 Inverse Modified Newton s Method : iterative scheme 1 Given a pair of initial guesses γ 0, λ 0. 2 Solve (13) for u. 3 Solve the regularized version of (14) for updates of γ, λ : ([ ] ) [ ] αi 0 + A χ A = A (u 0 βi q F) (15) where A = [ F γ 0 0 F λ ] 4 Set γ 0 = γ 0 + χ, λ 0 = λ 0 + q and repeat steps 2,3 until some criterion is fullfilled.

49 Inverse Modified Newton s Method : iterative scheme 1 Given a pair of initial guesses γ 0, λ 0. 2 Solve (13) for u. 3 Solve the regularized version of (14) for updates of γ, λ : ([ ] ) [ ] αi 0 + A χ A = A (u 0 βi q F) (15) where A = [ F γ 0 0 F λ ] 4 Set γ 0 = γ 0 + χ, λ 0 = λ 0 + q and repeat steps 2,3 until some criterion is fullfilled.

50 Major Advantages Inverse No extra equations Two smaller systems Fréchet derivatives are easily obtained The unique solvability of the numerical scheme followed straight forward

51 Inverse Numerical Settings: Solution spaces V m := span{1, cos t, cos 2t,..., cos mt; sin t,..., sin(m 1)t} Γ := (γ 1 (t), γ 2 (t)) V m V m λ V k Stopping criterion for the Newton s method: Discrepancy Principle: u,k u ɛ

52 Inverse Numerical Settings: Solution spaces V m := span{1, cos t, cos 2t,..., cos mt; sin t,..., sin(m 1)t} Γ := (γ 1 (t), γ 2 (t)) V m V m λ V k Stopping criterion for the Newton s method: Discrepancy Principle: u,k u ɛ

53 Inverse Numerical Settings: Solution spaces V m := span{1, cos t, cos 2t,..., cos mt; sin t,..., sin(m 1)t} Γ := (γ 1 (t), γ 2 (t)) V m V m λ V k Stopping criterion for the Newton s method: Discrepancy Principle: u,k u ɛ

54 Inverse Ellipse with exact data Ellipse Peanut Bean Γ = (0.4 cos t, 0.3 sin t)

55 Inverse Ellipse with exact data Ellipse Peanut Bean λ = cos t

56 Inverse Ellipse with 3% noise Ellipse Peanut Bean Γ = (0.4 cos t, 0.3 sin t)

57 Inverse Ellipse with 3% noise Ellipse Peanut Bean λ = cos t

58 Peanut A Inverse Ellipse Peanut Bean Γ = γ(t)(cos t, sin t), γ(t) = cos 2 t sin 2 t

59 Peanut A Inverse Ellipse Peanut Bean λ = sin t

60 Peanut B Inverse Ellipse Peanut Bean Γ = γ(t)(cos t, sin t), γ(t) = cos 2 t sin 2 t

61 Peanut B Inverse Ellipse Peanut Bean λ = 0.1 sin 2t cos t

62 Peanut C Inverse Ellipse Peanut Bean Γ = γ(t)(cos t, sin t), γ(t) = cos 2 t sin 2 t

63 Peanut C Inverse Ellipse Peanut Bean λ = 0.3e 0.25 cos3 t

64 Inverse Bean with exact data Ellipse Peanut Bean Γ = γ(t)(cos t, sin t), γ(t) = cos t sin(2t) cos t

65 Inverse Bean with exact data Ellipse Peanut Bean λ = sin t

66 Inverse Bean with 3% noise Ellipse Peanut Bean Γ = γ(t)(cos t, sin t), γ(t) = cos t sin(2t) cos t

67 Inverse Bean with 3% noise Ellipse Peanut Bean λ = sin t