A fast direct solver for high frequency scattering from a cavity in two dimensions

Transcription

1 1/31 A fast direct solver for high frequency scattering from a cavity in two dimensions Jun Lai 1 Joint work with: Leslie Greengard (CIMS) Sivaram Ambikasaran (CIMS) Workshop on fast direct solver, Dartmouth College, Courant Institute of Mathematical Sciences, New York University

2 2/31 Outline 1 Introduction 2 Integral formulation of the cavity problem 3 Discretization 4 Fast direct solver 5 Numerical experiment 6 Conclusion

3 3/31 Introduction Cavity structures exist in many electromagnetic applications Figure : F16 exhaust nozzle, from: f16model.blogspot.com

4 4/31 Geometry Consider the 2D wave scattering from a cavity embedded in the ground plane Γ 1 R 2 + \Ω 1 x2 Ω 1 Γ c P 1 B x1 B P2 Γ c Γ Ω 2 Γ 2 R 2 \Ω 2 Figure : Geometry of the problem

5 5/31 Related work Well-posedness(Ammari-Bao-Wood 02) Inverse scattering problem(bao-gao-li 11) Stability of the solution(bao-yun-zhou 12,Li-Ma-Sun 12) Numerical simulation: bounding and shooting rays(ling-chou-lee 89), Gaussian beam asymptotics(burkholder 91), finite difference(bao-sun 05), finite element(liu-jin 00), iterative physical optics(hemon-pouliguen 08), mode matching method(bao-gao-lin-zhang 12), Overfilled cavity(van-wood 04,Huang-Wood 05) Asymptotic of resonances(bonnetier-triki 14) Optimal design(bao-lai 14)

6 6/31 Formulation Assumption: permittivity ε 0, permeability µ 0 are constant everywhere Plane wave incidence: Total field u has the form: u inc = e i(k 0 cos θx k 0 sin θy) u(x, y) = u s (x, y) + u inc e i(k 0 cos θx+k 0 sin θy) Total field u satisfies the equation: u + k 2 u = 0 in Ω 1 R 2 + where k = ω ε 0 µ 0 is the wavenumber.

7 7/31 Helmholtz Equation The scattered wave satisfies the Helmholtz equation: { u s + k 2 u s = 0 in Ω 1 R 2 + u s = g on Γ Γ c (1) along with the Sommerfeld radiation condition (2), where g = (u i + u r ), along with the Sommerfeld radiation condition ( ) u s lim r iku s = 0 (2) r r Approach to find u s : Integral formulation.

8 8/31 Preliminary Let Φ(x, y) = i 4 H(1) 0 (k x y ) (3) Φ H (x, y) = i 4 H(1) 0 (k x y ) i 4 H(1) 0 (k x y ) (4) and S H and D H are the single layer and double layer operators in the half-space: SΓ H 1 σ = Φ H (x, y)σ(y)ds y (5) Γ 1 DΓ H Φ H (x, y) 1 µ = µ(y)ds y. (6) Γ 1 n(y)

9 x2 x1 9/31 Integral Formulation Assume Γ 1 is an artificial boundary that covers the cavity Ω 1. A natural formulation to represent the exterior field u s is: u s = S H Γ 1 σ + D H Γ 1 µ, for x R 2 +\Ω 1 (7) The scattered field u s in Ω 1 is represented by u s = S Γ1 σ + D Γ1 µ + D B µ + D Γ µ for x Ω 1 (8) Γ1 R 2 + \Ω1 Γ c P1 B Ω1 B P2 Γ c Γ Ω2 Γ2 R 2 \Ω2

10 Integral equation Combining the boundary condition and jump condition from layer potentials yields i.e. µ S Γ2 σ D Γ2 µ D B µ D Γ µ = 0, for x Γ 1 σ + N Γ2 σ + T Γ2 µ + T B µ + T Γ µ = 0, for x Γ µ + S Γ 1 σ + D Γ1 µ + D B µ + D Γ µ = g, for x Γ B A = where 1/2 D D S D 1/2 + D D S D D 1 D S T T T 1 + N N Γ σ = T Γ µ = Γ Γ µ B µ Γ µ Γ1 σ Γ1 (9) (10) Φ(x, y) n(x) σ(y)ds y (11) 2 Φ(x, y) n(x) n(y) µ(y)ds y (12) 10/31

11 11/31 Numerical result The formulation is not Fredholm equation of second kind. Numerically it is difficult to discretize T B µ to high order. Table : Results for pot shaped cavity at different wavenumber k Error

12 x2 x1 12/31 Conclusion New formulation We therefore propose the following formulation: { u s = S H Γ 1 σ + D H Γ 1 µ + D B µ, for x R 2 +\Ω 1 u s = S H Γ 1 σ + D H Γ 1 µ + D B µ + D Γ µ, for x Ω 1 (13) Compared to the old formulation: u s = { S H Γ1 σ + D H Γ 1 µ, for x R 2 +\Ω 1 S Γ1 σ + D Γ1 µ + D B µ + D Γ µ for x Ω 1 (14) we add some non-physical terms in the new formulation. Γ1 R 2 + \Ω1 Ω1 Γ c P1 B B P2 Γ c Γ Ω2 Γ2 R 2 \Ω2

13 x2 x1 13/31 Conclusion Integral equation The new formulation leads to the following integral equation: µ D Γ µ = 0, for x Γ 1 σ + T Γ µ = 0, for x Γ µ + SH Γ 1 σ + DΓ H 1 µ + D B µ + D Γ µ = g, for x B Γ Advantage: the system is still second kind. More importantly, there is no hypersingular term anymore. One can even eliminate µ and σ on Γ 1 by hand and reduce the unknown to µ on B Γ only. Γ1 R 2 + \Ω1 Γ c P1 B Ω1 B P2 Γ c Γ Ω2 Γ2 R 2 \Ω2

14 14/31 Conclusion Well-posedness We have the following uniqueness result: Theorem Given k > 0, the following system µ D Γ µ = 0, for x Γ 1 σ + T Γ µ = 0, for x Γ µ + SH Γ 1 σ + DΓ H 1 µ + D B µ + D Γ µ = 0, for x B Γ only has a zero solution for µ C(B) C(Γ) C(Γ 1 ) and σ C(Γ 1 ). from which the existence of the solution follows.

15 15/31 Discretization In our numerical simulation, we apply Nyström discetization to the integral equation. In other words, we replace the integral K (x, y)σ(y)ds y (15) by the quadrature N i=1 j=1 Γ p K(x l,m, y i,j )σ(y i,j )w l,m,i,j (16) where x l,m is the m-th Gauss-Legendre node on panel l, y i,j is the j-th Gauss-Legendre node on panel i, w l,m,i,j is the quadrature weight and K is the "quadrature kernel".

16 16/31 Dyadic refinement In the existence of corners, we apply the dyadic refinement at the end of each segment Figure : Graded mesh on a smooth segment Error analysis: The formal error analysis depends on the regularity of σ and µ. Qualitatively, if ɛ denotes the length of the finest panel, then the error is proportional to O(e p + ɛ).

17 17/31 Fast direct solver A large dense linear system will be generated from the cavity problem, especially when the frequency is high. Conventional linear solver, either direct or iterative, requires an enormous amount of time. We therefore choose fast direct solver to solve the problem. The advantages of fast direct solver are: The direct solver is insensitive to multiple reflections inside the cavity, especially at high frequencies and hence is much faster than iterative solver. The solver scales exceptionally well with multiple right hand sides.

18 18/31 The linear matrix to be solved The following is the system that we need to solve: A = 1/2 D 0 0 D 1/2 + D D H S H 0 D T 0 1 µ B µ Γ µ Γ1 σ Γ1 (17)

19 19/31 Low rank interaction for far field Low Rank Interaction Figure : Low Rank interaction between two different segments The rank of the far field interaction is roughly on the order of log(k), which implies the matrix, A R N N, has a hierarchical off-diagonal low-rank (HODLR) structure.

20 20/31 Inverse of a matrix with low rank perturbation One can quickly find the inverse of a matrix with low rank perturbation if the inverse of the matrix is known through Sherman-Morrison formula: If A R n n is invertible, u R n and v R n, then (A + uv T ) 1 = A 1 A 1 uv T A v T A 1 u (18) Generalization to rank k perturbation to A is the Sherman-Morrison-Woodbury formula: (A + UV T ) 1 = A 1 A 1 U(I + V T A 1 U) 1 V T A 1 (19)

21 21/31 HODLR matrix A 2-level HODLR matrix can be written in the form shown in equation (21). A = [ = A (1) 1 U (1) 1 K (1) 1,2 V ] (1)T 2 U (1) 2 K (1) 2,1 V (1)T 1 A (1) 2 [ (2) A 1 U (2) 1 K (2) 1,2 V (2)T 2 U (2) 2 K (2) 2,1 V (2)T 1 A (2) 2 U (1) 2 K (1) 2,1 V (1)T 1 ] U (1) 1 K (1) 1,2 V (1)T 2 [ (2) A 3 U (2) 3 K (2) 3,4 V ] (2)T 4 U (2) 4 K (2) 4,3 V (2)T 3 A (2) 4 The low-rank decomposition of the off-diagonal blocks is obtained using the adaptive cross approximation (ACA) algorithm, which is a minor modification of the partially pivoted LU algorithm. (20) (21)

22 22/31 Multiple level Full rank; Low rank; Figure : A hierarchical off-diagonal low-rank matrix at different levels.

23 23/31 factorization The matrix A is factored as shown in Equation (22). A = A κ A κ 1 A 1 A 0 (22) where A i s are block diagonal matrices with 2 i diagonal blocks and each block is a low-rank perturbation to identity matrix. = A A 3 A 2 A 1 A 0 Full rank; Low-rank; Identity matrix; Zero matrix; Figure : Factorization of a HODLR matrix at level 3.

24 24/31 Inversion To invert the matrix after the factorization, one simply needs to apply the Sherman-Morrison-Woodbury formula recursively. The whole cost is on O(N). For multiple right hand sides, the additional cost is negligible.

25 25/31 Numerical experiment We are trying to mimic the following field in Ω: u(x) = i 4 H(1) 0 (k (x x 0) ) + i 4 H(1) 0 (k (x x 0 ) ), (23) by solving the following equation µ + D Γ µ = u(x), for x Γ 1 σ + T Γ µ = u(x) n, for x Γ µ + SH Γ 1 σ + DΓ H 1 µ + D B µ + D Γ µ = u(x), for x B Γ (24) where x 0 = (5, 12), x 0 = (5, 12) and the center of the cavity is at (0.5, 0).

26 Pot shaped cavity (a) Figure : Example 1. (a)real part of the scattered field for a pot shaped cavity from the normal incidence of a plane wave with wavenumber k=160. (b)the backscatter RCS in db for the pot shaped cavity at k= (b) /31

27 27/31 Pot shaped cavity Table : Results for pot shaped cavity at different wavenumber k N mid N over N tot T factor (s) T solve (s) E error

28 Engine shaped cavity (a) (b) Figure : Example 2. (a)real part of the scattered field for a engine shaped cavity from the 45 o incidence of a plane wave with wavenumber k=160. (b)the backscatter RCS in db for the engine shaped cavity at k= /31

29 Rough bottom cavity (a) (b) Figure : Example 3. (a)real part of the scattered field for a rough bottom cavity from the normal incidence of a plane wave with wavenumber k=160. (b)the backscatter RCS in db for the rough bottom cavity at k= /31

30 30/31 Conclusion Propose an integral formulation for the cavity problem that leads to a high order discretization Apply the fast direct solver for the resulted linear system The extension to TE case is straightforward Future work includes the impedance boundary problem, optimal design problem

31 31/31 Thank you