A space-time Trefftz method for the second order wave equation

Transcription

1 A space-time Trefftz method for the second order wave equation Lehel Banjai The Maxwell Institute for Mathematical Sciences Heriot-Watt University, Edinburgh & Department of Mathematics, University of Novi Sad RICAM, 9th Nov 2016 Joint work with: Emmanuil Georgoulis (U of Leicester), Oluwaseun F Lijoka (HW) 1 / 34

2 Outline of the talk 1 Motivation 2 An interior-penalty space-time dg method 3 Damped wave equation 4 Numerical results 2 / 34

3 Outline 1 Motivation 2 An interior-penalty space-time dg method 3 Damped wave equation 4 Numerical results 3 / 34

4 Acoustic wave equation ü a u = 0 in Ω [0, T ], u = 0 on Ω [0, T ], u(x, 0) = u 0 (x), u(x, 0) = v 0 (x), in Ω. Set-up Initial data u 0 H0 1(Ω), v 0 L 2 (Ω). a(x) piecewise constant 0 < c a < a(x) < C a. Unique solution exists with u L 2 ([0, T ]; H0 1 (Ω)), u L 2 ([0, T ]; L 2 (Ω)), ü L 2 ([0, T ]; H 1 (Ω)). 4 / 34

5 Acoustic wave equation Set-up ü a u = 0 in Ω [0, T ], u = 0 on Ω [0, T ], u(x, 0) = u 0 (x), u(x, 0) = v 0 (x), in Ω. Initial data u 0 H 1 0 (Ω), v 0 L 2 (Ω). a(x) piecewise constant 0 < c a < a(x) < C a. Unique solution exists with u L 2 ([0, T ]; H 1 0 (Ω)), u L 2 ([0, T ]; L 2 (Ω)), ü L 2 ([0, T ]; H 1 (Ω)). Aim Develop an efficient Trefftz type method: Approximate u in terms of local solutions of the wave equation. To do this develop and analyse a time-space dg method. 4 / 34

6 Frequency domain motivation Frequency domain Take cue from frequency domain û(x) where a j are directions, a j = 1. Motivation û ω 2 û = 0: k j=1 f j e iωx a j, For large ω, minimize the number of degrees of freedom per wavelength. 5 / 34

7 Frequency domain motivation Frequency domain Take cue from frequency domain û(x, ω) where a j are directions, a j = 1. k j=1 ˆf j (ω)e iωx a j, 5 / 34

8 Frequency domain motivation Frequency domain Take cue from frequency domain û(x, ω) where a j are directions, a j = 1. Time-domain Time-domain equivalent u(x, t) k j=1 k j=1 ˆf j (ω)e iωx a j, f j (t x a j ) 5 / 34

9 Frequency domain motivation Frequency domain Take cue from frequency domain û(x, ω) where a j are directions, a j = 1. Time-domain Time-domain equivalent u(x, t) k j=1 k k j=1 ˆf j (ω)e iωx a j, f j (t x a j ) p α j,l (t x a j ) l. j=1 l=0 5 / 34

10 (A bit of) Literature on Trefftz methods Plenty of literature in the frequency domain O. Cessenat and B. Després, Application of an ultra weak variational formulation of elliptic PDEs to the two-dimensional Helmholtz equation, SIAM J. Numer. Anal., (1998) R. Hiptmair, A. Moiola, and I. Perugia, Plane wave discontinuous Galerkin methods for the 2D Helmholtz equation: analysis of the p-version, SIAM J. Numer. Anal., (2011). 6 / 34

11 (A bit of) Literature on Trefftz methods Plenty of literature in the frequency domain O. Cessenat and B. Després, Application of an ultra weak variational formulation of elliptic PDEs to the two-dimensional Helmholtz equation, SIAM J. Numer. Anal., (1998) R. Hiptmair, A. Moiola, and I. Perugia, Plane wave discontinuous Galerkin methods for the 2D Helmholtz equation: analysis of the p-version, SIAM J. Numer. Anal., (2011). Fewer in time-domain S. Petersen, C. Farhat, and R. Tezaur, A space-time discontinuous Galerkin method for the solution of the wave equation in the time domain, Internat. J. Numer. Methods Engrg. (2009) F. Kretzschmar, A. Moiola, I. Perugia, S. M. Schnepp, A priori error analysis of space-time Trefftz discontinuous Galerkin methods for wave problems, IMA JNA, (2015). LB, E. Georgoulis, O Lijoka, A Trefftz polynomial space-time discontinuous Galerkin method for the second order wave equation, accepted in SINUM (2016). 6 / 34

12 Outline 1 Motivation 2 An interior-penalty space-time dg method 3 Damped wave equation 4 Numerical results 7 / 34

13 DG setting Time discretization 0 = t 0 < t 1 < < t N = T, I n = [t n, t n+1 ]; τ n = t n+1 t n. Spatial-mesh T n of Ω consisting of open simplices such that Ω = K Tn K. Space-time slabs T n I n, h-space-time meshwidth. The skeleton of the space mesh denoted Γ n and ˆΓ n = Γ n 1 Γ n. Usual jump and average definitions (e = K + K Γ int ) and if e K + Ω, {u} e = 1 2 (u+ + u ), {v} e = 1 2 (v+ + v ), [u] e = u + n + + u n, [v] e = v + n + + v n, {v} e = v +, [u] e = u + n + Also u(t n ) = u(t + n ) u(t n ), u(t 0 ) = u(t + 0 ). 8 / 34

14 Local Trefftz spaces The space of piecewise polynomials on the time-space mesh denoted S h,p n = {u L 2 (Ω I n ) u K In P p (R d+1 ), K T n }, Let S h,p n,trefftz S n h,p with v(t, x) a v(t, x) = 0, The dimensions of the spaces S h,p n t I n, x K, for any v S h,p n,trefftz. and S h,p n,trefftz for spatial dimension d are 1D 2D 3D 1 poly 2 (p + 1)(p + 2) 1 6 (p + 1)(p + 2)(p + 3) 2D 1 4 (p + 4) Trefftz 2p + 1 (p + 1) (p + 1)(p + 2)(2p + 3) We expect the approximation properties of solutions of the wave equation to be the same for the two spaces of different dimension. 9 / 34

15 Constructing the polynomial spaces Choose directions ξ j (see talk by Moiola): (t 1 a x ξ j )α k, α k - multi-index. 10 / 34

16 Constructing the polynomial spaces Choose directions ξ j (see talk by Moiola): (t 1 a x ξ j )α k, α k - multi-index. Alternatively propagate polynomial initial condition: u(0) = x α k, u(0) = 0, and u(0) = 0, u(0) = x β k, with α k p and β k p / 34

17 Constructing the polynomial spaces Choose directions ξ j (see talk by Moiola): (t 1 a x ξ j )α k, α k - multi-index. Alternatively propagate polynomial initial condition: u(0) = x α k, u(0) = 0, and u(0) = 0, u(0) = x β k, with α k p and β k p 1. An important obervation is that the trunctation of Taylor expansion of exact solution is a polynomial solution of the wave equation. 10 / 34

18 The space on Ω [0, T ] is then defined as V h,p Trefftz = {u L2 (Ω [0, T ]) u Ω In S h,p n,trefftz, n = 0, 1..., N}. 11 / 34

19 The space on Ω [0, T ] is then defined as V h,p Trefftz = {u L2 (Ω [0, T ]) u Ω In S h,p n,trefftz, n = 0, 1..., N}. (Abuse of) Notation: u h V h,p Trefftz-discrete function on Ω [0, T ] u n S h,p n,trefftz, restriction of u on Ω I n. u ex the exact solution. 11 / 34

20 An interior penalty dg method We start with t n t n+1 [ Ω ü v + A u vdx Γ {a u} [ v] ds Γ [u] {a v} ds Testing with v = u gives where the energy is given by t n t n+1 + σ 0 Γ [u] [ v] ds ] dt = 0. d E(t, u)dt = 0, dt E(t, u) = 1 2 u(t) 2 Ω a u(t) 2 Ω σ 0 [u(t)] 2 Γ Γ {a u} [u] ds. Discrete inverse inequality in space and usual choice of penalty parameter gives E(t, u) / 34

21 Jumps in time Summing over n gives E(tN N 1 ) E(t+ 0 ) n=1 E(t n ) = 0. To give a sign to the extra terms (Hughes, Hulbert 88): 1 2 ( u(t n), u(t n )) L 2 (Ω) ( u(t n ), u(t + n )) L 2 (Ω) = 1 2 ( u(t n), u(t n ) ) L 2 (Ω). Do this for all the terms, including the stabilization. Obtain a dissipative method. 13 / 34

22 Space-time dg formulation N 1 a(u, v) = (ü, v) Ω In + ( u(t n ), v(t n + )) Ω n=0 + (a u, v) Ω In + ( a u(t n ), v(t n + )) Ω ({a u}, [ v]) Γn I n ( {a u(t n )}, [v(t + n )])ˆΓ n ([u], {a v}) Γn I n ( [u(t n )], {a v(t + n )})ˆΓn + (σ 0 [u], [ v]) Γn I n + (σ 0 [u(t n )], [v(t + n )])ˆΓ n and + (σ 1 [u], [v]) Γn I n + (σ 2 [a u], [a v]) Γn I n b init (v) = (v 0, v(t + 0 )) Ω + (a u 0, v(t + 0 )) Ω ({a u 0 }, [v(t + 0 )]) Γ0 ([u 0 ], {a v(t + 0 )}) Γ0 + (σ 0 [u 0 ], [v(t + 0 )]) Γ0. Find u h V h,p Trefftz (Ω [0, T ]) such that a(u h, v) = b init (v), v V h,p Trefftz (Ω [0, T ]). 14 / 34

23 Time-space dg as a time-stepping method a n (u, v) = (ü, v) Ω In + ( u(t + n ), v(t + n )) Ω + (a u, v) Ω In + (a u(t + n ), v(t + n )) Ω ({a u}, [ v]) Γn I n ({a u(t + n )}, [v(t + n )]) Γn ([u], {a v}) Γn I n ([u(t + n )], {a v(t + n )}) Γn + (σ 0 [u], [ v]) Γn I n + (σ 0 [u(t + n )], [v(t + n )]) Γn + (σ 1 [u], [v]) Γn I n + (σ 2 [a u], [a v]) Γn I n, b n (u, v) = ( u(t n ), v(t + n )) Ω + (a u(t n ), v(t + n )) Ω ({a u(t n )}, [v(t + n )]) Γn ([u(t n )], {a v(t + n )}) Γn 1 + (σ 0 [u(t n )], [v(t + n )])ˆΓ n, Find u n S h,p n,trefftz such that a n (u n, v) = b n (u n 1, v), v S h,p n,trefftz. 15 / 34

24 Consistency and stability Theorem The following statements hold: 1 The method is consistent. 2 There exists a choice of σ 0 h 1, such that for any v S h,p n,trefftz and t I n the energy is bounded below as E(t, v) 1 2 v(t) 2 L 2 (Ω) a v(t) 2 L 2 (Ω). 3 Let u h V h,p Trefftz discrete solution. Then E(t N, uh ) E(t 1, u h ). 16 / 34

25 a(, ) 1/2 = - a norm on V h,p Trefftz a(w, w) = E h (tn, w) + E h(t 0 + N 1, w) + ( 1 2 ẇ(t n) 2 Ω a w(t n ) 2 Ω n=1 ( {a w(t n )}, [w(t n )] )ˆΓ n σ 0 [w(t n )] 2ˆΓn ) N 1 + n=0 ( σ 1 [w] 2 Γ n I n + σ 2 [a w] 2 Γ n I n ). 17 / 34

26 a(, ) 1/2 = - a norm on V h,p Trefftz a(w, w) = E h (tn, w) + E h(t 0 + N 1, w) + ( 1 2 ẇ(t n) 2 Ω a w(t n ) 2 Ω n=1 Theorem ( {a w(t n )}, [w(t n )] )ˆΓ n σ 0 [w(t n )] 2ˆΓn ) N 1 + n=0 ( σ 1 [w] 2 Γ n I n + σ 2 [a w] 2 Γ n I n ). a(v, v) 1/2 = v = 0 v = 0, for v V h,p Trefftz. Hence, the time-space dg method a(u, v) = b init (v), v V h,p Trefftz has a unique solution in V h,p Trefftz. The proof is by noticing that if v = 0 then v is a smooth solution of the wave equation, uniquely determined by the initial condition. 17 / 34

27 Convergence analysis If we prove continuity of a(, ) a(u, v) C u v, u cont. sol. + V h,p Trefftz, v V h,p Trefftz, 18 / 34

28 Convergence analysis If we prove continuity of a(, ) a(u, v) C u v, u cont. sol. + V h,p Trefftz, v V h,p Trefftz, we can use Galerkin orthogonality to show, for any v V h,p Trefftz u h v 2 = a(u h v, u h v) and hence we have quasi-optimality = a(u ex v, u h v) C u ex v u h v u h u ex inf u h v + v u ex v V h,p Trefftz inf v V h,p Trefftz v u ex + C v u ex 18 / 34

29 Integrating by parts a few times (this is how to implement it) a(w, v) = Recall N 1 n=0 ( ({a ẇ}, [v]) Γn In (σ 0 [ẇ], [v]) Γn In ({ẇ}, [a v]) Γ int n I n + (σ 1 [w], [v]) Γn I n + (σ 2 [a w], [a v]) Γn I n ) N ( (ẇ(tn ), v(t n ) ) Ω + (a w(tn ), v(t n ) ) Ω n=1 ({a w(t n )}, [v(t n )] ) Γn ([w(t n )], {a v(t n )} ) Γn + (σ 0 [w(t n )], [v(t n )] ) Γn ). v 2 = E(t N, v) + E(t+ 0, v) + N 1 ( 1 2 v(t n) 2 Ω v(t n) 2 Ω n=1 + ( { v(t n )}, [v(t n )] )ˆΓn σ 0 [v(t n )] 2ˆΓ n ) N 1 + n=0 ( σ 1 [v] 2 Γ I n + σ 2 [ v] 2 Γ I n ), 19 / 34

30 Integrating by parts a few times (this is how to implement it) a(w, v) = N 1 n=0 ( ({a ẇ}, [v]) Γn In (σ 0 [ẇ], [v]) Γn In ({ẇ}, [a v]) Γ int n I n + (σ 1 [w], [v]) Γn I n + (σ 2 [a w], [a v]) Γn I n ) N ( (ẇ(tn ), v(t n ) ) Ω + (a w(tn ), v(t n ) ) Ω n=1 Hence define, w 2 = 1 2 ({a w(t n )}, [v(t n )] ) Γn ([w(t n )], {a v(t n )} ) Γn + (σ 0 [w(t n )], [v(t n )] ) Γn ). N n=1 N 1 + n=0 ( ẇ(t n ) 2 Ω + a w(t n ) 2 Ω + σ 0 [w(t n )] 2 Γ n + σ 1/2 0 {a w(t ( σ 1 [w] 2 Γ n I n + σ 2 [a w] 2 Γ n I n + σ 1/2 2 {ẇ} 2 Γ int + σ 1/2 1 {a ẇ} 2 Γ n I n + σ 0 σ 1/2 1 [ẇ] 2 Γ n I n ). n I n 19 / 34

31 The choice of stabilization parameters and convergence Let τ n = t n+1 t n, h = diam(k), (x, t) K (t n, t n+1 ), K T n. diam(k)/ρ K c T, K T n, n = 0, 1,..., N 1, where ρ K is the radius of the inscribed circle of K. Assume space-time elements star-shaped with respect to a ball. Choice of parameters σ 0 = p 2 c T Ca 2 C inv (c a h) 1, σ 1 = C a p 3 (hτ n ) 1 σ2 = h(c a τ n ) 1. Theorem For sufficiently smooth solution and h τ u h u ex = O(h p 1/2 ). Proof uses truncated Taylor expansion. 20 / 34

32 Error estimate in mesh independent norm Using a Gronwall argument we can show for v S h,p n,trefftz v 2 Ω I n + a v 2 Ω I n τ n e C(t n+1 t n)/h n ( v(t n+1 ) 2 Ω + a v(t n+1) 2 Ω ), h n = min x Ω h(x, t), t I n. Let τ = max τ n and h = min h n. Then V 2 Ω (0,T ) + a V 2 Ω (0,T ) Cτe Cτ/ h V 2, V V h,p Trefftz. Proposition u h u ex 2 Ω [0,T ] + uh u ex 2 Ω [0,T ] C inf (τe Cτ/ h V u ex 2 V V h,p Trefftz + V u ex 2 Ω [0,T ] + a (V u ex ) 2 Ω [0,T ] ). 21 / 34

33 Outline 1 Motivation 2 An interior-penalty space-time dg method 3 Damped wave equation 4 Numerical results 22 / 34

34 Wave equation with damping Damped wave equation: ü + α u u = 0. The extra term decreases the energy: d dt E(t) = α u 2, hence only a minor modification to the DG formulation needed. 23 / 34

35 Wave equation with damping Damped wave equation: ü + α u u = 0. The extra term decreases the energy: d dt E(t) = α u 2, hence only a minor modification to the DG formulation needed. However: truncations of the Taylor expansion are no longer solutions! 23 / 34

36 Wave equation with damping Damped wave equation: ü + α u u = 0. The extra term decreases the energy: d dt E(t) = α u 2, hence only a minor modification to the DG formulation needed. However: truncations of the Taylor expansion are no longer solutions! Grysa, Maciag, Adamczyk-Krasa 14 consider solutions of the form e αt p 1 (x, t) + p 2 (x, t) with p 1 and p 2 polynomial in x and t. 23 / 34

37 Solution formula in 1D Instead, use basis functions obtained by propagating polynomial initial data: u(x, 0) = u 0 (x) = x α j, u(x, 0) = v 0 (x) = 0, and with u 0 (x) = 0, v 0 (x) = x β j α j p β j p / 34

38 Solution formula in 1D Instead, use basis functions obtained by propagating polynomial initial data: u(x, 0) = u 0 (x) = x α j, u(x, 0) = v 0 (x) = 0, and with u 0 (x) = 0, v 0 (x) = x β j α j p β j p 1. In 1D solution is then given by the d Alambert-like formula u(x, t) = 1 2 [u 0(x t) + u 0 (x + t)] e αt/2 + α 4 e αt/ e αt/2 x+t x t x+t x t u 0 (s) {I 0 (ρ(s) α 2 ) + t ρ(s) I 1 (ρ(s) α )} 2 ds v 0 (s)i 0 (ρ(s) α 2 ) ds, ρ(s; x, t) = t 2 (x s) / 34

39 Solution formula ctd. Rearranging (the last term) te αt/2 [v 0(x + st) + v 0 (x st)] I 0 ( αt s 2 ) ds. For example for v 0 (x) = x 2 1 x 2 te αt/2 I 0 ( αt s 2 ) ds + t 3 e αt/2 0 s2 I 0 ( αt 2 1 s 2 ) ds / 34

40 Solution formula ctd. Rearranging (the last term) te αt/2 [v 0(x + st) + v 0 (x st)] I 0 ( αt s 2 ) ds. For example for v 0 (x) = x 2 1 x 2 te αt/2 I 0 ( αt s 2 ) ds + t 3 e αt/2 0 s2 I 0 ( αt 2 1 s 2 ) ds. Corresponding term in 3D 1 te αt/2 0 [v 0 (x + tsz) + Dv 0 (x + tsz)] zds z I 0 ( αt B(0,1) 2 1 s 2 ) ds 1 25 / 34

41 Solution formula ctd. Rearranging (the last term) te αt/2 [v 0(x + st) + v 0 (x st)] I 0 ( αt s 2 ) ds. For example for v 0 (x) = x 2 1 x 2 te αt/2 I 0 ( αt s 2 ) ds + t 3 e αt/2 0 s2 I 0 ( αt 2 1 s 2 ) ds. Corresponding term in 3D 1 te αt/2 0 [v 0 (x + tsz) + Dv 0 (x + tsz)] zds z I 0 ( αt B(0,1) 2 1 s 2 ) ds 1 Hence need efficient representation of functions of the type ϱ j (t) = 0 sj I 0 ( αt 2 1 s 2 ) ds In Matlab: Chebfun is ideal / 34

42 Changes to the analysis Polynomial in space the same discrete inverse inequalities used + the extra term decreases energy choice of parameters, stability and quasi-optimality proof identical. Approximation properties of the discrete space (in 1D): Away from the boundary, in each space-time element K (t, t + ) project solution to K and neighbouring elements at time t to polynomials and propagate. At boundary, extend exact solution anti-symmetrically and again project and propagate. 26 / 34

43 Outline 1 Motivation 2 An interior-penalty space-time dg method 3 Damped wave equation 4 Numerical results 27 / 34

44 One dimensional setting Simple 1D setting: Ω = (0, 1), a 1. Initial data u 0 = e x 5/8 ( ) 2 δ, v 0 = 0, δ δ 0 = Interested in having few degrees of freedom for decreasing δ δ t = 1.5, δ = δ t = 1.5, δ = δ 0 / / 34

45 One dimensional setting Simple 1D setting: Ω = (0, 1), a 1. Initial data u 0 = e x 5/8 ( ) 2 δ, v 0 = 0, δ δ 0 = Energy of exact solution exact energy = 1 2 u x(x, 0) 2 Ω 2δ 1 y 2 e 2y 2 π dy = δ We compare with full polynomial space. Note for polynomial order p 2p + 1 Trefftz 1 2 (p + 1)(p + 2)full polynomial space. In all 1D experiments square space-time elements. 28 / 34

46 Error in dg-norm, δ = δ 0, T = 1/4: Error Error Error Trefftz poly Full poly p = 1 p = 2 p = 3 p = 4 p = h Trefftz p-convergence (fixed h in space and time): p = 1 p = 2 p = 3 p = 4 p = h p 29 / 34

47 Energy Energy Energy conservation For δ = δ 0 /4: Trefftz poly Full poly Energy, p=1 Energy, p=2 5 Energy, p=3 Energy, p=4 Exact Energy Time 10 Energy, p=1 Energy, p=2 5 Energy, p=3 Energy, p=4 Exact Energy Time 30 / 34

48 Error Relative error for decreasing δ error δ = ( δ 2 u(, T ) u h(, T ) 2 Ω + δ 2 u(, T ) u h(, T ) 2 Ω ) 1/ p=2, / = / 0 p=3, / = / 0 p=4,/ = / 0 p=5, / = / 0 p=2, / = / 0=2 p=3, / = / 0=2 p=4, / = / 0=2 p=5, / = / 0=2 p=2, / = / 0=4 p=3, / = / 0=4 p=4, / = / 0=4 p=5, / = / 0= h=/ 31 / 34

49 Error Error 2D experiment On square [0, 1] 2 with exact solution Energy of error at final time: u(x, y, t) = cos( 2πt) sin πx sin πy. error = ( 1 2 u(, T ) u h(, T ) 2 Ω u(, T ) u h(, T ) 2 Ω )1/2. Trefftz poly Full poly p = 1 p = 2 p = 3 p = Mesh-size p = 1 p = 2 p = 3 p = Mesh-size 32 / 34

50 Error Error Damped wave equation in 1D Error in dg norm. Exact solution on Ω = (0, 1): u(x, t) = e ( αt/2) sin(πx) cos π 2 α2 4 t + α 2 π 2 α 2 /4 sin π 2 α2 4 t. Trefftz poly Full poly p = 1 p = 2 p = 3 p = p = 1 p = 2 p = 3 p = Number of DOF Number of DOF 33 / 34

51 Conclusions A space-time interior penalty dg method for the acoustic wave equation in second order form: To do: Allows Trefftz basis functions, polynomial in space. Stability and convergence analysis available. Also practical for damped wave equation. A posteriori error analysis Adaptivity in h, p, wave directions. 34 / 34