Error analysis and fast solvers for high-frequency scattering problems
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1 Error analysis and fast solvers for high-frequency scattering problems I.G. Graham (University of Bath) Woudschoten October 2014
2 High freq. problem for the Helmholtz equation Given an object Ω R d, with boundary Γ and exterior Ω, Incident plane wave, e.g. : u I (x) = exp(ikx â) Ω Ω â Γ Total wave u = u I + u S, where Scattered wave u S satisfies: plus boundary condition u S + k 2 u S = 0 in Ω (Mostly u I + u S = 0 on Γ) and radiation condition: us r ikus = o(r (d 1)/2 ) as r
3 Numerical-asymptotic methods Constant wavenumber k - asymptotic information (mostly BEM) Computing in time independent of frequency. Links to Daan and Simon s talks Geometry dependent methods
4 Numerical-asymptotic methods Constant wavenumber k - asymptotic information (mostly BEM) Computing in time independent of frequency. Links to Daan and Simon s talks Geometry dependent methods
5 Numerical-asymptotic methods Still a role for conventional BEM
6 Second Talk: Truncated problems u + k 2 u = 0 in Ω B R u = u I on Γ u n iku = 0 on B R B R Ω Γ Ω for large R Model cavity problem u + k 2 u = f in bounded domain Ω u iku n = g on Γ := Ω
7 Heterogeneity Seismic inversion problem: ( ) ωl 2 u u = f, c(x) ω = frequency solve for u with approximate c. Second talk: Conventional discretisation and fast solvers Key reference: Erlangga, Osterlee, Vuik, Link to Martin s talks and Domain Decomposition
8 Outline of my talks: Two problems on conventional methods. 1. When is the error in the h version BEM bounded independently of k? 2. Give an analysis of preconditioning methods for standard h version FEM Both have solutions which use high-frequency analysis.
9 References for the talks: I.G. Graham, M. Loehndorf, J.M. Melenk and E.A. Spence, When is the error in the h-bem for solving the Helmholtz equation bounded independently of k? To appear in BIT, M. J. Gander, I. G. Graham and E. A. Spence, How should one choose the shift for the shifted Laplacian to be a good preconditioner for the Helmholtz equation?, submitted I.G. Graham, E.A. Spence and E. Vainikko, Convergence of additive Schwarz methods for the Helmholtz equation with and without absorption, in preparation. S.N. Chandler-Wilde, I.G. Graham, S. Langdon and E.A. Spence, Boundary Integral Equation Methods for High Frequency Scattering Problems Acta Numerica 2012.
10 References for the talks: I.G. Graham, M. Loehndorf, J.M. Melenk and E.A. Spence, When is the error in the h-bem for solving the Helmholtz equation bounded independently of k? To appear in BIT, M. J. Gander, I. G. Graham and E. A. Spence, How should one choose the shift for the shifted Laplacian to be a good preconditioner for the Helmholtz equation?, submitted I.G. Graham, E.A. Spence and E. Vainikko, Convergence of additive Schwarz methods for the Helmholtz equation with and without absorption, in preparation. S.N. Chandler-Wilde, I.G. Graham, S. Langdon and E.A. Spence, Boundary Integral Equation Methods for High Frequency Scattering Problems Acta Numerica 2012.
11 First problem When is the error in the h version boundary element method bounded independently of k?
12 Fundamental solution for the Helmholtz equation u + k 2 u = 0 G k (x, y) = i 4 H(1) 0 (k x y ) 2D exp(ik x y ) 4π x y 3D Phase: k x y single layer potential : (S k φ)(x) = Γ G k(x, y)φ(y)ds(y), double layer: (D k φ)(x) = Γ [ n(y)g k (x, y)]φ(y)ds(y), adjoint double layer: D k (switch roles of x and y).
13 Combined potential boundary integral formulations Exterior scattering problem with incident field u I : Green s identity for u S in Ω : S k ( n u S + n u I ) D k (u S + u I ) = ( u S + 0) in Ω (1) Limit to boundary Γ: Equation for unknown v := n u but with spurious frequencies. Take normal derivative in (1) and combine with (1): combined potential formulation ( ) 1 R k v := 2 I + D k v iks k v = n u I iku I, or k η Alternative indirect method: ( ) 1 R k φ := 2 I + D k φ iks k φ = u I,
14 Combined potential boundary integral formulations Exterior scattering problem with incident field u I : Green s identity for u I in Ω: S k ( n u S + n u I ) D k (u S + u I ) = ( u S + 0) in Ω (1) Limit to boundary Γ: Equation for unknown v := n u but with spurious frequencies. Take normal derivative in (1) and combine with (1): combined potential formulation ( ) 1 R k v := 2 I + D k v iks k v = n u I iku I, or k η Alternative indirect method: R k φ := ( ) 1 2 I + D k φ iks k φ = u I,
15 Combined potential boundary integral formulations Exterior scattering problem with incident field u I : Green s identity for u I in Ω: S k ( n u S + n u }{{} I ) D k (u S + u I ) = ( u }{{} S + 0) in Ω }{{} (1) nu =0 u I Limit to boundary Γ: Equation for unknown v := n u but with spurious frequencies. Take normal derivative in (1) and combine with rik (1): combined potential formulation ( ) 1 R k v := 2 I + D k v iks k v = n u I iku I, or k η Alternative indirect method: R k φ := ( ) 1 2 I + D k φ iks k φ = u I,
16 Combined potential boundary integral formulations Exterior scattering problem with incident field u I : Green s identity for u I in Ω: S k ( n u S + n u }{{} I ) D k (u S + u I ) = ( u }{{} S + 0) in Ω }{{} (1) nu =0 u I Limit to boundary Γ: Equation for unknown v := n u but with spurious frequencies. Take normal derivative in (1) and combine with ik (1): combined potential formulation ( ) 1 R k v := 2 I + D k v iks k v = n u I iku I, or k η Alternative indirect method: R k φ := ( ) 1 2 I + D k φ iks k φ = u I,
17 Combined potential boundary integral formulations Exterior scattering problem with incident field u I : Green s identity for u I in Ω: S k ( n u S + n u }{{} I ) D k (u S + u I ) = ( u }{{} S + 0) in Ω }{{} (1) nu =0 u I Limit to boundary Γ: Equation for unknown v := n u but with spurious frequencies. Take normal derivative in (1) and combine with ik (1): direct combined potential formulation ( ) 1 R k v := 2 I + D k v iks k v = n u I iku I, or k η Alternative indirect method: R k φ := ( ) 1 2 I + D k φ iks k φ = u I,
18 Combined potential boundary integral formulations Exterior scattering problem with incident field u I : Green s identity for u I in Ω: S k ( n u S + n u }{{} I ) D k (u S + u I ) = ( u }{{} S + 0) in Ω }{{} (1) nu =0 u I Limit to boundary Γ: Equation for unknown v := n u but with spurious frequencies. Take normal derivative in (1) and combine with ik (1): direct combined potential formulation ( ) 1 R k v := 2 I + D k v iks k v = n u I iku I, or k η Alternative indirect method: R k φ := ( ) 1 2 I + D k φ iks k φ = u I,
19 BEM analysis - Classical setting Fredholm integral equations of the Second kind R k v = (λi + L k )v = f k R k φ = (λi + L k )φ = g k (λ = 1/2) Galerkin method in approximating space V N (or V h ). e.g. piecewise polynomials of fixed degree p. Solution v N or φ N, e.g. (λi + P N L k )v N = P N f k
20 BEM analysis - Classical setting Fredholm integral equations of the second kind R k v = (λi + L k )v = f k R k φ = (λi + L k )φ = g k (λ = 1/2) Galerkin method in approximating space V N (or V h ). e.g. piecewise polynomials of fixed degree p. Solution v N or φ N, e.g. (λi + P N L k )v N = P N f k v v N = λ (λi P N L k }{{ ) 1 (v P } N v) }{{} stability best approx
21 Question 1 (best approximation error) When are inf wn V N v w N L 2 (Γ) v L 2 (Γ) and inf wn V N φ w N L 2 (Γ) bounded independently of k? φ L 2 (Γ)
22 Question 2 (quasioptimality) When are v v N L 2 (Γ) inf wn V N v w N L 2 (Γ) and φ φ N L 2 (Γ) bounded independently of k? inf wn V N φ w N L 2 (Γ)
23 If both hold...(bound on relative errors) v v N L 2 (Γ) v L 2 (Γ) and φ φ N L 2 (Γ) φ L 2 (Γ) bounded indpendently of k.
24 Answers: Question 1 ( direct version v = n u) When is inf wn V N v w N L 2 (Γ) bounded independently of k? v L 2 (Γ) Theorem If Ω is C and convex then for h BEM, inf w h V h v w h L 2 (Γ) (hk)p v L 2 (Γ) so hk 1 is sufficient for Question 1.
25 Proof uses v(x) := u/ n(x) = kv (x, k) exp(ikx â), x Γ, Theorem Dominguez, IGG, Smyshlyaev, 2007 { Cn, n = 0, 1, D n V (x, k) C n k 1 (k 1/3 + dist(x, SB)) (n+2) n 2, where SB = {x Γ : n(x).â = 0} shadow boundary. SB â Proves, e.g. v H 1 (Γ) k v L 2 (Γ)
26 Answers: Question 1 ( direct version v = n u) When is inf wn V N v w N L 2 (Γ) bounded independently of k? v L 2 (Γ) Theorem If Ω is a convex polygon then there is a mesh with O(N) points so that, inf w h V h v w h L 2 (Γ) k N v L 2 (Γ) so k/n 1 is sufficient for Question 1. (Requires sup x Ω u(x).)
27 Proof uses: Theorem Chandler-Wilde and Langdon (2007) γ u (s) = 2 ui n n (s) + eiks v + (s) + e iks v (s) where s is distance along γ, and k n v (n) + (s) { Cn (ks) 1/2 n, ks 1, C n (ks) α n, 0 < ks 1, where α < 1/2 depends on the corner angle.
28 Answers: Question 1: Indirect method λφ = L k φ = iks k φ + D k φ To estimate the derivatives of φ: S k H 1 L 2 k (d 1)/2 (Γ Lipschitz) D k H 1 L 2 k (d+1)/2 (Γ smooth enough) These imply φ H 1 (Γ) k (d+1)/2 φ L 2 (Γ) And so hk (d+1)/2 1 is sufficient for Question 1.
29 Answers: Question 2 (classical approach) R k v := (λi + L k )v = f k compact perturbation (λi + P h L k )v h = P h f k Galerkin method Lemma [Atkinson, Anselone, 1960 s...] If (I P h )L k (λi + L k ) 1 << 1, then v v h (λi + L k ) 1 inf w h V h v w h Application: and in addition: (I P h )L k h L k L 2 H1 hk(d+1)/2 (λi + L k ) 1 1 [Chandler-Wilde & Monk, 2008] Lipschitz star-shaped Theorem Hence quasioptimality if hk (d+1)/2 C
30 Tools We used in this talk k explicit bounds on norms of L k, L k (where R k = 1 2 I + L k ), etc. needed smooth enough domains k explicit bounds on inverses (R k ) 1, (R k ) 1 needed Lipschitz star-shaped We will need in the next talk Bound on the solution operator for the Helmhotz BP PDE itself. connection between the two illustrates the role of star-shaped.
31 The Subtlety of Behaviour of L k and R 1 k Equivalently L k and (R k ) 1 L k, R 1 k k 1/3, 1 k 1/2, 1 k 1/2 m, k 7/5 m k 1/2 m, e γkm Chandler-Wilde, IGG et al (2009), Betcke, Chandler-Wilde, IGG et al (2011).
32 Numerical Experiments: domain [0, 0.5] [0, 5] rectangle 1 + γ 2 p v v N L 2 (Γ) inf wn V N v w N L 2 (Γ) norm estimate 10 1 A A 1 O(k 1/2 ) degree p = 0, wavenumber k N k γ 0 γ hk 1
33 Numerical Experiments: trapping domain C shaped domain r r r 3 r r 3 norm estimate 10 1 A A 1 O(k 1/2 ) O(k 0.9 ) wavenumber k m k N γ 0 γ
34 Tools Open question: Prove hk 1 sufficient for quasioptimality We used in this talk k explicit bounds on norms of L k, L k (where R k = 1 2 I + L k ), etc. needed smooth enough domains k explicit bounds on inverses (R k ) 1, (R k ) 1 needed Lipschitz star-shaped We will need in the next talk Bound on the solution operator for the Helmholtz BVP PDE itself. connection between the two illustrates the role of star-shaped.
35 Tools Open question: Prove hk 1 sufficient for quasioptimality We used in this talk k explicit bounds on norms of L k, L k (where R k = 1 2 I + L k ), etc. needed smooth enough domains k explicit bounds on inverses (R k ) 1, (R k ) 1 needed Lipschitz star-shaped We will need in the next talk Bound on the solution operator for the Helmholtz BVP PDE itself. connection between the two illustrates the role of star-shaped.
36 Trapping domains Can things get bad in the non-star-shaped case? Theorem If the exterior domain Ω contains a square Q of side length a and the boundary Γ coincides with two parallel sides of Q, then if 2ak = mπ for any positive integer m, R 1 k (ak)9/10. Ω Q Ω
37 Quasimodes (family of) sources g and solutions v of Helmholtz problem v + k 2 v = g in Ω v = 0 on Γ + Sommerfeld condition, where v L 2 (Ω ) M k g L 2 (Ω ), with M k large. This could contradict the bound v L 2 (Ω ) 1 k f L 2 (Ω ) which holds in star-shaped case (see next lecture). In fact R k ( nv) = ( n v I ikv I ) where v I is the Newtonian potential generated by g implies growth of (R k ) 1 More generally...
38 (R k ) 1 k (d 2) M k O(k (d 1)/2 ) With elliptic cavity M k can increase exponentially. [Betcke, Chandler-Wilde, IGG, Langdon, Lindner, 2011]
39 Final theorem for today... Model interior impedance problem: u k 2 u = f in bounded domain Ω u iku n = g on Γ := Ω...Also truncated sound-soft scattering problems in Ω Theorem (Stability) Assume Ω is Lipschitz and star-shaped. Then, u 2 L 2 (Ω) + k2 u 2 L 2 (Ω) }{{} =: u 2 1,k f 2 L 2 (Ω) + g 2 L 2 (Γ), k [Melenk 95, Cummings & Feng 06] Central result in next lecture
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