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

A space-time Trefftz method for the second order wave equation Lehel Banjai The Maxwell Institute for Mathematical Sciences Heriot-Watt University, Edinburgh Rome, 10th Apr 2017 Joint work with: Emmanuil Georgoulis (Leicester & Athens), Oluwaseun F Lijoka (HW) 1 / 36

Outline of the talk 1 Motivation 2 An interior-penalty space-time dg method 3 Damped wave equation 4 Numerical results 2 / 36

Outline 1 Motivation 2 An interior-penalty space-time dg method 3 Damped wave equation 4 Numerical results 3 / 36

Acoustic wave equation Find u(t) H0 1 (Ω), t [0, T ], s.t. (ü, v) L 2 (Ω) + (a u, v) L 2 (Ω) = 0 u(x, 0) = u 0 (x), u(x, 0) = v 0 (x), in Ω. for all v H 1 0 (Ω), 4 / 36

Acoustic wave equation Find u(t) H0 1 (Ω), t [0, T ], s.t. (ü, v) L 2 (Ω) + (a u, v) L 2 (Ω) = 0 u(x, 0) = u 0 (x), u(x, 0) = v 0 (x), in Ω. for all v H 1 0 (Ω), 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)) u C([0, T ]; H 1 0 (Ω)), u C([0, T ]; L 2 (Ω)). For smooth enough solution we have the transmission conditions u j = u k, a j n u j = a k n u k, on Ω j Ω k, u j = u Ωj, u k = u Ωk, where Ω j and Ω k are subsets of Ω with a a k in Ω k and a a j in Ω j. 4 / 36

How to discretize the wave equation? The usual approach: Construct a spatial mesh and a corresponding spatially discrete space: locally polynomial, continuous or discontinuous across the boundaries of the spatial elements (the spatial skeleton). Finite difference approximation in time. Solution computed by time-stepping. In this talk (Trefftz): Construct a space-time mesh and corresponding fully discrete space: In each space time element exact (polynomial or non-polynomial) exact solution of the wave equation. Discontinous across the space-time skeleton. Solve either by time-stepping or as a large system. 5 / 36

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. 6 / 36

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, 6 / 36

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 ) 6 / 36

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 6 / 36

(A bit of) Literature on Trefftz methods for waves 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, I. Perugia, A survey of Trefftz methods for the Helmholtz equation. Springer Lect. Notes Comput. Sci. Eng., 2016, pp. 237-278. 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 J. Numer. Anal., 36(4) 2016, pp. 1599-1635. L. Banjai, E. Georgoulis, O Lijoka, A Trefftz polynomial space-time discontinuous Galerkin method for the second order wave equation, SIAM J. Numer. Anal. 55-1 (2017), pp. 63 86. 7 / 36

Outline 1 Motivation 2 An interior-penalty space-time dg method 3 Damped wave equation 4 Numerical results 8 / 36

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. In each simplex K, a(x) is constant. 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 ). 9 / 36

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) 2 1 6 (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. 10 / 36

Constructing the polynomial spaces Choose directions ξ j : (t 1 l x ξ j ), l = 0, 1,..., p. a 11 / 36

Constructing the polynomial spaces Choose directions ξ j : (t 1 l x ξ j ), l = 0, 1,..., p. a 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, α k, β k multi-indices. Important observation: trunctation of the Taylor expansion of exact solution is a polynomial solution of the wave equation. 11 / 36

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}. 12 / 36

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. 12 / 36

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 + 1 2 a u(t) 2 Ω + 1 2 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) 0. 13 / 36

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. 14 / 36

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 ]). 15 / 36

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. 16 / 36

Consistency and stability Theorem The following statements hold: 1 The method is consistent for a sufficiently smooth solution u. 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) + 1 4 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 ). 17 / 36

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 + 1 2 a w(t n ) 2 Ω n=1 ( {a w(t n )}, [w(t n )] )ˆΓ n + 1 2 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 ). 18 / 36

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 + 1 2 a w(t n ) 2 Ω n=1 Theorem ( {a w(t n )}, [w(t n )] )ˆΓ n + 1 2 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. 18 / 36

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, 19 / 36

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 19 / 36

Integrating by parts a few times (this is how to implement the method) 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 + 1 2 v(t n) 2 Ω n=1 + ( { v(t n )}, [v(t n )] )ˆΓn + 1 2 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 ), 20 / 36

Integrating by parts a few times (this is how to implement the method) 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 20 / 36

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. 21 / 36

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 ] ). Hence in mesh independent energy norm we expect error O(h p ). 22 / 36

Outline 1 Motivation 2 An interior-penalty space-time dg method 3 Damped wave equation 4 Numerical results 23 / 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. 24 / 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. 24 / 36

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/2 + 1 2 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) 2. 25 / 36

Solution formula ctd. Rearranging (the last term) 1 1 2 te αt/2 [v 0(x + st) + v 0 (x st)] I 0 ( αt 0 2 1 s 2 ) ds. For example for v 0 (x) = x 2 1 x 2 te αt/2 I 0 ( αt 0 2 1 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 26 / 36

Damped wave equation: Efficient implementation Need efficient representation of functions of the type 1 ϱ j (t) = 0 sj I 0 ( αt 2 1 s 2 ) ds In Matlab: Chebfun works well. Note computations are done on a small space-time element, i.e. t (0, h) for a small h > 0. Can use truncated Taylor expansions of analytic functions I 0 and I 1 with efficiency increasing for decreasing h. We expect similar effect for any lower order terms. 27 / 36

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. 28 / 36

Outline 1 Motivation 2 An interior-penalty space-time dg method 3 Damped wave equation 4 Numerical results 29 / 36

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 = 7.5 10 2. Interested in having few degrees of freedom for decreasing δ δ 0. 0.5 t = 1.5, δ = δ 0 0.5 t = 1.5, δ = δ 0 /4 0.4 0.4 0.3 0.3 0.2 0.2 0.1 0.1 0 0-0.1-0.1-0.2-0.2-0.3-0.3-0.4-0.4-0.5 0 0.2 0.4 0.6 0.8 1-0.5 0 0.2 0.4 0.6 0.8 1 30 / 36

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 = 7.5 10 2. Energy of exact solution exact energy = 1 2 u x(x, 0) 2 Ω 2δ 1 y 2 e 2y 2 π dy = δ 1 2 2. 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. 30 / 36

Error in dg-norm, δ = δ 0, T = 1/4: Error Error Error Trefftz poly Full poly 10 2 10 2 10 0 10 0 10-2 10-2 10-4 10-4 10-6 10-8 p = 1 p = 2 p = 3 p = 4 p = 5 10-3 10-2 10-1 h Trefftz p-convergence (fixed h in space and time): 10 2 10-6 10-8 p = 1 p = 2 p = 3 p = 4 p = 5 10-3 10-2 10-1 h 10 0 10-2 10-4 10-6 10-8 1 2 3 4 5 6 7 8 9 10 p 31 / 36

Energy Energy Energy conservation For δ = δ 0 /4: Trefftz poly Full poly 35 35 30 30 25 25 20 20 15 15 10 Energy, p=1 Energy, p=2 5 Energy, p=3 Energy, p=4 Exact Energy 0 10-3 10-2 10-1 10 0 10 1 10 2 10 3 Time 10 Energy, p=1 Energy, p=2 5 Energy, p=3 Energy, p=4 Exact Energy 0 10-3 10-2 10-1 10 0 10 1 10 2 10 3 Time 32 / 36

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 ) 1/2 10 0 10-2 10-4 10-6 10-8 10-10 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=4 10-2 10-1 10 0 10 1 h=/ 33 / 36

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 + 1 2 u(, T ) u h(, T ) 2 Ω )1/2. Trefftz poly Full poly 10 2 10 2 10 0 10 0 10-2 10-2 10-4 10-4 10-6 10-8 p = 1 p = 2 p = 3 p = 4 0.02 0.03 0.04 0.05 0.06 0.070.080.090.1 Mesh-size 10-6 10-8 p = 1 p = 2 p = 3 p = 4 0.02 0.03 0.04 0.05 0.06 0.070.080.090.1 Mesh-size 34 / 36

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 10 2 10 0 p = 1 p = 2 p = 3 p = 4 10 2 10 0 p = 1 p = 2 p = 3 p = 4 10-2 10-2 10-4 10-4 10-6 10-6 10-8 10 1 10 2 10 3 10 4 Number of DOF 10-8 10 1 10 2 10 3 10 4 Number of DOF 35 / 36

Conclusions A space-time interior penalty dg method for the acoustic wave equation in second order form: Can be considered either as implicit time-stepping method or a large space-time system. Allows Trefftz basis functions, polynomial in space. Fewer degrees of freedom than full polynomial method and requires integration only over the space-time skeleton. Stability and convergence analysis available. Also practical for damped wave equation. 36 / 36

Conclusions A space-time interior penalty dg method for the acoustic wave equation in second order form: To do: Can be considered either as implicit time-stepping method or a large space-time system. Allows Trefftz basis functions, polynomial in space. Fewer degrees of freedom than full polynomial method and requires integration only over the space-time skeleton. Stability and convergence analysis available. Also practical for damped wave equation. A posteriori error analysis Adaptivity in h, p, wave directions. Tent-pitching meshes allow quasi explicit time-stepping. p-analysis in higher dimension. Applications. 36 / 36