A Space-Time Boundary Element Method for the Wave Equation

Transcription

1 W I S S E N T E C H N I K L E I D E N S C H A F T A Space-Time Boundary Element Method for the Wave Equation Marco Zank and Olaf Steinbach Institut für Numerische Mathematik Space-Time Methods for PDEs, Linz 10 November

2 Wave equation Consider u tt = u u(, 0) = 0 in R d \ Γ, u t (, 0) = 0 and Kirchhoff s formula with in (R d \ Γ) (0, ), in R d \ Γ u = Ṽ [[γ 1 u]] W [[γ 0 u]] [[γ 0 u]] := γ0 int u γext 0 u, [[γ 1 u]] := γ1 int u γext 1 u. 2

3 Table of contents 1. Laplace transform method 2. Space-time Sobolev spaces 3. Energy approach One-dimensional wave equation Two-dimensional wave equation 3

4 Laplace transform For causal function f : R X for ω = η + ισ C: F(ω) := L{f }(ω) := = f (t)e ιωt dt f (t)e σt e ιηt dt = F{f ( )e σ }(η) The set of Laplace transformable distributions with values in X: LT(X) := { f D + (X) e σ 0 f ( ) S +(X) } σ 0 R Holomorphic function: ω F(ω) = L{f }(ω) is holomorphic for Im ω > σ 0. 4

5 Laplace transform Convolution: If S LT(L(X, Y )), f LT(X), then S f LT(Y ) and L{S f }(ω) = L{S}(ω)L{f }(ω) Y. C σ0 := {ω C Im ω > σ 0 } Theorem (Paley-Wiener) Let σ 0 R be fixed. Let F : C σ0 X be a holomorphic function. Then the following conditions are equivalent: 1. There exists a distribution f LT(X) such that L{f }(ω) = F(ω). 2. There exists some σ 1 σ 0, some C > 0 and k 0 such that for all ω C σ0 with Im ω > σ 1. F (ω) X C(1 + ω ) k 5

6 Wave equation Let Ω int R d be a bounded Lipschitz domain with boundary Γ := Ω and Ω ext := R d \ Ω int. The wave equation u + u tt = 0 in (R d \ Γ) R +, u(, 0) = 0 in R d \ Γ, u t (, 0) = 0 in R d \ Γ, u Γ R + = g on Γ R + transforms for ω C 0 into the Helmholtz equation U(ω) H 1 (Ω int/ext ), U(ω) ω 2 U(ω) = 0 in R d \ Γ, γ int/ext 0 U(ω) = G(ω) on Γ. 6

7 Proposition The interior and exterior problem of the Helmholtz equation has a unique solution u H 1 (Ω int/ext ) for all ω C 0. The solution verifies U(ω) 2 H 1 (Ω int/ext ),ω C(σ 0) ω G(ω) 2 H 1/2 (Γ) for all ω {Im ω σ 0 > 0} with the norm v 2 H 1 (Ω int/ext ),ω := v 2 L 2 (Ω int/ext ) + ω 2 v 2 L 2 (Ω int/ext ). Single layer potential Ṽ (ω): H 1/2 (Γ) H 1 (Ω int/ext ) with U(ω) = Ṽ (ω)λ(ω) for ω C 0. 7

8 Single layer operator V (ω) = γ int/ext 0 Ṽ (ω): H 1/2 (Γ) H 1/2 (Γ) with V (ω) C(σ 0 ) ω and the coercivity estimate Re λ, ιωv (ω)λ C min{σ 0, 1} λ 2 H ω 1/2 (Γ) λ H 1/2 (Γ), inverse single layer operator V (ω) 1 : H 1/2 (Γ) H 1/2 (Γ) with for all ω {Im ω σ 0 > 0}. V (ω) 1 C(σ 0 ) ω 2 8

9 Hence, U(ω) = Ṽ (ω)λ(ω) with Λ(ω) H 1/2 (Γ) is the unique solution of: Find ϕ H 1/2 (Γ) such that Theorem ψ, ιωv (ω)ϕ = ψ, ιωg(ω) ψ H 1/2 (Γ). For all g LT(H 1/2 (Γ)) the interior and exterior problem of the wave equation has a unique solution u = Ṽ λ := L 1 (ω Ṽ (ω)λ(ω)) LT(H1 (Ω int/ext )) with the density λ LT(H 1/2 (Γ)). The integral representation for (x, t) (R d \ Γ) R + is t u(x, t) = Ṽ λ(x, t) = U ( x y, t τ)λ(y, τ)ds y dτ. 0 Γ 9

10 Standard space-time Sobolev spaces } Hσ(R, r H ν (Γ)) := {u LT(H ν (Γ)) u σ,r,h ν (Γ) < with σ > 0, r R, ν R and the norm +ισ u 2 σ,r,h ν (Γ) := ω 2r L{u}(ω) 2 H ν (Γ) dω. For r = k N it holds +ισ u 2 σ,k,h ν (Γ) R = 2π e 2σt d k u dt k (t) 2 H ν (Γ) dt = d k u dt k 2 σ,0,h ν (Γ). 10

11 Energy space-time Sobolev spaces } H r;ν σ,γ {u := LT(H ν (Γ)) u σ,γ;r;ν < with σ > 0, r R, ν R and the norm +ισ u 2 σ,γ;r;ν := ω 2r L{u}(ω) 2 ν,ω,γ dω. +ισ The ω-dependent norm ν,ω,γ is defined analogously by v 2 ν,ω,r d := ( ω 2 + ξ 2 ) ν F{v}(ξ) 2 dξ R d for v H ν (R d ). It holds H r+ν σ (R, H ν (Γ)) H r;ν σ,γ H r σ(r, H ν (Γ)), H r;ν σ,γ = Hσ(R, r H ν (Γ)) Hσ r+ν (R, L 2 (Γ)). 11

12 Boundedness of the single layer operator It holds for σ σ 0 > 0 and C = C(Γ, σ 0 ) V λ σ,0,h 1/2 (Γ) C λ σ,1,h 1/2 (Γ), V λ σ,γ;0;1/2 C λ σ,γ;1; 1/2, V 1 g C g σ,0,h 1/2 σ,2,h (Γ) 1/2 (Γ), V 1 g C g σ,γ;1;1/2. σ,γ;0; 1/2 Loss of regularity in time V 1 V : H 3 σ(r, H 1/2 (Γ)) H 0 σ(r, H 1/2 (Γ)), V 1 V : H 2; 1/2 σ,γ H 0; 1/2 σ,γ. 12

13 Coercivity in the time-domain Bilinear form in the frequency domain for ω {Im ω σ 0 > 0} : ψ, ιωv (ω)ϕ ϕ, ψ H 1/2 (Γ). Define bilinear form for σ σ 0 > 0 ( a σ (λ, τ) := e τ(t), 2σt V dλ ) (t) dt. 0 dt Γ For C = C(Γ, σ 0 ) there are the coercivity estimates a σ (λ, λ) C λ 2 σ, 1/2,H 1/2 (Γ), a σ (λ, λ) C λ 2 σ,γ;0; 1/2, λ H 1/2 σ λ H 0; 1/2 σ,γ and the boundedness estimates a σ (λ, τ) C λ σ,1,h 1/2 (Γ) τ σ,1,h 1/2 (Γ), a σ (λ, τ) C λ σ,γ;1; 1/2 τ σ,γ;1; 1/2. (R, H 1/2 (Γ)), 13

14 Variational formulation Let g Hσ 3/2 (R, H 1/2 (Γ)) be the given Dirichlet data. Find λ Hσ(R, 1 H 1/2 (Γ)) with V λ Hσ(R, 1 H 1/2 (Γ)) such that dg a σ (λ, τ) = e 2σt (t), τ(t) dt 0 dt Γ for all τ H 1 σ(r, H 1/2 (Γ)). Find λ h, t V m 1,m 2 h, t := V m 1 h V m 2 t such that a σ (λ h, t, τ h, t ) = for all τ h, t V m 1,m 2 h, t. 0 e 2σt dg dt (t), τ h, t(t) dt Γ 14

15 Error estimates For sufficiently regular λ it holds for m 1 = 0, m 2 = 1 λ λh, t σ, 1/2,H 1/2 (Γ) =O(h1/2 + h 1/2 t 1/2 ), λ λ h, t σ,0,h 0 (Γ) =O(h 1 + t 1/2 ) and λ λ h, t σ,γ;0; 1/2 =O(max{h 1/2, t 1/2 }(h 2/3 + t 2 )), λ λ h, t σ,γ;0;0 =O(max{h 1, t 1 }(h 2/3 + t 2 )). For h t in d = 2 with σ = 0 numerical experiments [Gläfke] show λ λ L h, t 2(0,T ;L2(Γ)) = O(h + t). 15

16 An energy approach [Aimi et al. 2008] For any solution u of the wave equation tt u(x, t) u(x, t) = 0 for x R d \ Γ, t (0, T ), t u(x, 0) = u(x, 0) = 0 for x R d \ Γ, u(x, t) = g(x, t) for (x, t) Σ := Γ [0, T ] it holds 0 = ( tt u u) t u ( 1 = t 2 ( tu) ) 2 u 2 ( t u u). 16

17 With Green s formula it follows 0 = 1 { ( t u(x, T )) 2 + u(x, T ) 2} dx 2 R d \Γ T Define the energy E(T ) := t γ int 0 u(, t), [[γ 1u]](, t) Γ dt. R d \Γ { ( t u(x, T )) 2 + u(x, T ) 2} dx and represent the solution by u = Ṽ λ. Hence, E(T ) = T Define the bilinear form 0 t (V λ)(t), λ(t) Γ dt = t (V λ), λ Σ. a E (λ, ϕ) := t (V λ), ϕ Σ. 17

18 One-dimensional wave equation Q := (0, L) (0, T ) and Σ := {0, L} [0, T ]. Single layer potential: Ṽ w(x, t) = 1 2 t x w 0 (s)ds for t R, x R \ {0, L} and a density w = t x L ( w0 w L ). w L (s)ds 18

19 Single layer operator V : L 2 (Σ) H 1 {0} (Σ) ( V w(t) = 1 t 0 w 0(s)ds + ) t L 0 w L (s)ds 2 t L 0 w 0 (s)ds + t 0 w L(s)ds for t [0, T ] with L 2 (Σ) := H 1 {0} (Σ) := {v = ( v0 { v = ( v0 v L ) } : v 0, v L L 2 (0, T ), ) } : v v 0, v L H 1 (0, T ), γv 0 (0) = γv L (0) = 0. L 19

20 Theorem (Aimi et al. 2008) The bilinear form a E : L 2 (Σ) L 2 (Σ) R, a E (w, v) := t (V w), v L2 (Σ), is bounded and coercive: a E (ϕ, ϕ) C(T ) ϕ 2 L 2 (Σ) ϕ L 2 (Σ) Find w L 2 (Σ) for given g H{0} 1 (Σ) such that a E (w, v) = 1 2 tg, v L 2 (Σ) + t(k g), v L 2 (Σ) where the double layer operator is given by ( ) g0 K g(t) = K (t) = 1 ( ) gl (t L) g L 2 g 0 (t L) v L 2 (Σ), for t [0, T ]. 20

21 Decomposition Σ = N 0 +N L i=1 τ i. A conforming ansatz space for L 2 (Σ) is S t 0 (Σ) :=S0 t 0 (0, T ) S t 0 L (0, T ) {( )} ϕ 0 N0 {( =span 0,i 0 span 0 i=1 { } =span ˆϕ 0 N0 +N L i. The resulting linear system is i=1 V t w = g ϕ 0 L,j )} NL with V t R (N 0+N L ) (N 0 +N L ), g R N 0+N L and w R N 0+N L. j=1 21

22 A posteriori error estimator [Steinbach, Schulz 2000] η i := γ int 1 u w t Approximate solution in Q : L 2 (τ i ) ũ t := Ṽ w t W g Neumann trace of the approximate solution: Local error estimator: for i = 1,..., N 0 + N L w t := γ int 1 ũ t = 1 2 w t + K w t + D g η i := w t w t L 2 (τ i ) for i = 1,..., N 0 + N L For a parameter θ [0, 1] refine all elements τ i where η i θ max j η j. 22

23 Numerical examples with L = 3 and T = 6 23

24 Numerical examples with L = 3 and T = 6 23

25 Numerical examples with L = 3, T = 6 L γ int L N 0 + N L 1 u w t 2 (Σ) eoc u ũ t L 2 (Q) eoc e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e

26 Numerical examples with L = 3, T = 6 1 γ 1 int u w t L 2 (Σ) uniform θ = 0.05 N ,000 10,000 degrees of freedom N : =N 0 + N L 25

27 Numerical examples with L = 3 and T = 6 26

28 Numerical examples with L = 3 and T = 6 26

29 Numerical examples with L = 3, T = 6 L γ int L N 0 + N L 1 u w t 2 (Σ) eoc u ũ t L 2 (Q) eoc e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e

30 Numerical examples with L = 3, T = 6 10 γ 1 int u w t L 2 (Σ) uniform θ = 0.05 θ = 0.5 N 1 N 1/ ,000 10,000 degrees of freedom N : =N 0 + N L 28

31 Numerical examples with L = 3, T = 6, θ = t 2 1 x

32 Numerical examples with L = 3, T = 6, θ = t 2 1 x

33 Numerical examples with L = 3, T = 6, θ = t 2 1 x

34 Numerical examples with L = 3, T = 6, θ = t 2 1 x

35 Numerical examples with L = 3, T = 6, θ = t 2 1 x

36 Two-dimensional wave equation [Aimi et al. 2009] Let Γ = {(x, 0) R 2 : x [0, L]} be an open arc and consider tt u(x, y, t) u(x, y, t) = 0 for (x, y) R 2 \ Γ, t (0, T ), t u(x, y, 0) = u(x, y, 0) = 0 for (x, y) R 2 \ Γ, u(x, y, t) = g(x, y, t) for (x, y, t) Σ. Reinterpret as jump problem. Apply Fourier transform with respect to x. Klein-Gordon equation with mass ξ 2. Single layer potential for the Klein-Gordon equation. Limit y 0. Apply Fourier transform with respect to t. 30

37 Fourier representation For ϕ C0 (Σ) it follows a E (ϕ, ϕ) = t (V ϕ), ϕ L 2 (Σ) = 1 2 R ω > ξ ω ϕ(ξ, ω) 2 ω2 ξ dωdξ. 2 It holds for all ϕ, ψ L 2 (0, T ; H 1/4 (Γ)) a E (ϕ, ψ) C (1 + T ) ϕ L 2 (0,T ;H 1/4 (Γ)) ψ L 2 (0,T ;H 1/4 (Γ)). For any s R there exists a sequence (ϕ n ) n of non-vanishing functions in C0 (Σ) such that a E (ϕ n, ϕ n ) lim n ϕ n 2 = 0. H s (R 2 ) 31

38 Summary and outlook Laplace transform method Loss of regularity in time Energy approach One-dimensional wave equation A posteriori error estimator Error analysis for 1D? Bilinear form for 2D, 3D? 32

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