Inverse scattering problem from an impedance obstacle

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

Outline Inverse 1 Direct 2 Inverse 3 Ellipse Peanut Bean

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

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

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)

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)

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)

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)

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

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)

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.

Unique solvability Inverse? Uniqueness Not available Existence Not available

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!

Inverse Fredholm integral Equations 2. Kind ϕ(x) 1 2 1 0 (x+1)e xy ϕ(y)dy = e x 1 2 +1 2 e (x+1), 0 x 1 Trapzoidal rule n x = 0 x = 0.5 x = 1 4-0.007146-0.010816-0.015479 8-0.001788-0.002711-0.003882 16-0.000447-0.000678-0.000971 32-0.000112-0.000170-0.000243 Simpson s rule n x = 0 x = 0.5 x = 1 4-0.00006652-0.00010905-0.00021416 8-0.00000422-0.00000692-0.00001366 16-0.00000026-0.00000043-0.00000086 32-0.00000002-0.00000003-0.00000005

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 = 1 4 0.4057 0.3705 0.1704 8-4.5989 14.6094-4.4770 16-8.5957 2.2626-153.4805 32 3.8965-32.2907 22.5570 64-88.6474-6.4484-182.6745 Simpson s rule n x = 0 x = 0.5 x = 1 4 0.0997 0.2176 0.0566 8-0.5463 6.0868-1.7274 16-15.4796 50.5015-53.8837 32 24.5929-24.1767 67.9655 64 23.7868-17.5992 419.4284

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.

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

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

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.

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.

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

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.

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.

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

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

Inverse Ellipse with exact data Ellipse Peanut Bean 0.8 0.6 0.4 0.2 0 0.2 0.4 0.6 0.8 0.8 0.6 0.4 0.2 0 0.2 0.4 0.6 0.8 Γ = (0.4 cos t, 0.3 sin t)

Inverse Ellipse with exact data Ellipse Peanut Bean 0.8 0.6 0.4 0.2 0 0.2 0 1 2 3 4 5 6 7 λ = 0.5 + 0.2 cos t

Inverse Ellipse with 3% noise Ellipse Peanut Bean 0.5 0.4 0.3 0.2 0.1 0 0.1 0.2 0.3 0.4 0.5 0.5 0.4 0.3 0.2 0.1 0 0.1 0.2 0.3 0.4 0.5 Γ = (0.4 cos t, 0.3 sin t)

Inverse Ellipse with 3% noise Ellipse Peanut Bean 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0.1 0.2 0 1 2 3 4 5 6 7 λ = 0.5 + 0.2 cos t

Peanut A Inverse Ellipse Peanut Bean 1.5 1 0.5 0 0.5 1 1.5 1.5 1 0.5 0 0.5 1 1.5 Γ = γ(t)(cos t, sin t), γ(t) = cos 2 t + 0.25 sin 2 t

Peanut A Inverse Ellipse Peanut Bean 0.8 0.6 0.4 0.2 0 0.2 0 1 2 3 4 5 6 7 λ = 0.3 + 0.1 sin t

Peanut B Inverse Ellipse Peanut Bean 1.5 1 0.5 0 0.5 1 1.5 1.5 1 0.5 0 0.5 1 1.5 Γ = γ(t)(cos t, sin t), γ(t) = cos 2 t + 0.25 sin 2 t

Peanut B Inverse Ellipse Peanut Bean 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0 1 2 3 4 5 6 7 λ = 0.1 sin 2t + 0.3 0.2 cos t

Peanut C Inverse Ellipse Peanut Bean 1.5 1 0.5 0 0.5 1 1.5 1.5 1 0.5 0 0.5 1 1.5 Γ = γ(t)(cos t, sin t), γ(t) = cos 2 t + 0.25 sin 2 t

Peanut C Inverse Ellipse Peanut Bean 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0 1 2 3 4 5 6 7 λ = 0.3e 0.25 cos3 t

Inverse Bean with exact data Ellipse Peanut Bean 1.5 1 0.5 0 0.5 1 1.5 1.5 1 0.5 0 0.5 1 1.5 Γ = γ(t)(cos t, sin t), γ(t) = 1 + 0.9 cos t + 0.1 sin(2t) 1 + 0.75 cos t

Inverse Bean with exact data Ellipse Peanut Bean 0.5 0.4 0.3 0.2 0.1 0 0.1 0.2 0 1 2 3 4 5 6 7 λ = 0.3 + 0.1 sin t

Inverse Bean with 3% noise Ellipse Peanut Bean 1.5 1 0.5 0 0.5 1 1.5 1.5 1 0.5 0 0.5 1 1.5 Γ = γ(t)(cos t, sin t), γ(t) = 1 + 0.9 cos t + 0.1 sin(2t) 1 + 0.75 cos t

Inverse Bean with 3% noise Ellipse Peanut Bean 0.5 0.4 0.3 0.2 0.1 0 0.1 0.2 0 1 2 3 4 5 6 7 λ = 0.3 + 0.1 sin t