A space-time Trefftz method for the second order wave equation · Outline of the talk 1 Motivation...

A space-time Trefftz method for the second order waveequation

Lehel Banjai

The Maxwell Institute for Mathematical SciencesHeriot-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)

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Outline of the talk

1 Motivation

2 An interior-penalty space-time dG method

3 Damped wave equation

4 Numerical results

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Outline

1 Motivation



4 Numerical results

3 / 34

Acoustic wave equation

u −∇ ⋅ a∇u = 0 in Ω × [0,T ],

u = 0 on ∂Ω × [0,T ],

u(x ,0) = u0(x), u(x ,0) = v0(x), in Ω.

Set-up

Initial data u0 ∈ H10(Ω), v0 ∈ L2(Ω).

a(x) piecewise constant 0 < ca < a(x) < Ca.

Unique solution exists with

u ∈ L2([0,T ]; H1

0(Ω)), u ∈ L2([0,T ]; L2

(Ω)), u ∈ L2([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.

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Acoustic wave equation

u −∇ ⋅ a∇u = 0 in Ω × [0,T ],

u = 0 on ∂Ω × [0,T ],

u(x ,0) = u0(x), u(x ,0) = v0(x), in Ω.

Set-up

Initial data u0 ∈ H10(Ω), v0 ∈ L2(Ω).

a(x) piecewise constant 0 < ca < a(x) < Ca.

Unique solution exists with

u ∈ L2([0,T ]; H1

0(Ω)), u ∈ L2([0,T ]; L2

(Ω)), u ∈ L2([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

Frequency domain motivation

Frequency domain

Take cue from frequency domain

u(x) ≈k

∑j=1

fjeiωx⋅aj ,

where aj are directions, ∣aj ∣ = 1.

Motivation −∆u − ω2u = 0:

For large ω, minimize the number of degrees of freedom per wavelength.

Time-domain

Time-domain equivalent

u(x, t) ≈k

∑j=1

fj(t − x ⋅ aj)

≈k

∑j=1

p

∑`=0

αj ,`(t − x ⋅ aj)`.

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Frequency domain motivation

Frequency domain

Take cue from frequency domain

u(x, ω) ≈k

∑j=1

fj(ω)e iωx⋅aj ,

where aj are directions, ∣aj ∣ = 1.

Time-domain

Time-domain equivalent

u(x, t) ≈k

∑j=1

fj(t − x ⋅ aj)

≈k

∑j=1

p

∑`=0

αj ,`(t − x ⋅ aj)`.

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(A bit of) Literature on Trefftz methodsPlenty of literature in the frequency domain

O. Cessenat and B. Despres, Application of an ultra weak variationalformulation of elliptic PDEs to the two-dimensional Helmholtz equation,SIAM J. Numer. Anal., (1998)

R. Hiptmair, A. Moiola, and I. Perugia, Plane wave discontinuous Galerkinmethods 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 Galerkinmethod 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 analysisof space-time Trefftz discontinuous Galerkin methods for wave problems,IMA JNA, (2015).

LB, E. Georgoulis, O Lijoka, A Trefftz polynomial space-time discontinuousGalerkin method for the second order wave equation, accepted in SINUM(2016).

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Outline

1 Motivation



4 Numerical results

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DG setting

Time discretization 0 = t0 < t1 < ⋅ ⋅ ⋅ < tN = T , In = [tn, tn+1];τn = tn+1 − tn.

Spatial-mesh T n of Ω consisting of open simplices such thatΩ = ∪K∈TnK .

Space-time slabs Tn × In, h-space-time meshwidth.

The skeleton of the space mesh denoted Γn and Γn ∶= Γn−1 ∪ Γn.

Usual jump and average definitions (e = K+ ∩K− ∈ Γint)

u ∣e =1

2(u+ + u−), v ∣e =

1

2(v+ + v−),

[u] ∣e = u+n+ + u−n−, [v] ∣e = v+ ⋅ n+ + v− ⋅ n−,

and if e ∈ K+ ∩ ∂Ω,

v ∣e = v+, [u] ∣e = u+n+

Also⟦u(tn)⟧ = u(t+n ) − u(t−n ), ⟦u(t0)⟧ = u(t+0 ).

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Local Trefftz spacesThe space of piecewise polynomials on the time-space mesh denoted

Sh,pn ∶= u ∈ L2

(Ω × In) ∶ u∣K×In ∈ Pp(Rd+1

), K ∈ Tn ,

Let Sh,pn,Trefftz ⊂ Sh,p

n with

v(t, x) − a∆v(t, x) = 0, t ∈ In, x ∈ K , for any v ∈ Sh,pn,Trefftz.

The dimensions of the spaces Sh,pn and Sh,p

n,Trefftz for spatial dimension d are

1D 2D 3D

poly 12(p + 1)(p + 2) 1

6(p + 1)(p + 2)(p + 3) 2D × 14(p + 4)

Trefftz 2p + 1 (p + 1)2 16(p + 1)(p + 2)(2p + 3)

We expect the approximation properties of solutions of the waveequation to be the same for the two spaces of different dimension.

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Constructing the polynomial spaces

Choose directions ξj (see talk by Moiola):

(t −1

√a

x ⋅ ξj)αk

, αk - multi-index.

Alternatively propagate polynomial initial condition:

u(0) = xαk , u(0) = 0,

andu(0) = 0, u(0) = xβk ,

with ∣αk ∣ ≤ p and ∣βk ∣ ≤ p − 1.

An important obervation is that the trunctation of Taylor expansionof exact solution is a polynomial solution of the wave equation.

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The space on Ω × [0,T ] is then defined as

V h,pTrefftz = u ∈ L2

(Ω × [0,T ]) ∶ u∣Ω×In ∈ Sh,pn,Trefftz, n = 0,1 . . . ,N.

(Abuse of) Notation:

uh ∈ V h,pTrefftz-discrete function on Ω × [0,T ]

un ∈ Sh,pn,Trefftz, restriction of u on Ω × In.

uex– the exact solution.

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An interior penalty dG method

We start with

∫

tn+1

tn[∫

Ωuv +A∇u ⋅ ∇vdx − ∫

Γa∇u ⋅ [v]ds − ∫

Γ[u] ⋅ a∇vds

+ σ0∫Γ[u] ⋅ [v]ds ]dt = 0.

Testing with v = u gives

∫

tn+1

tn

d

dtE(t,u)dt = 0,

where the energy is given by

E(t,u) = 12∥u(t)∥2

Ω +12∥

√a∇u(t)∥2

Ω +12∥

√σ0 [u(t)] ∥2

Γ −∫Γa∇u ⋅ [u]ds.

Discrete inverse inequality in space and usual choice of penalty parametergives E(t,u) ≥ 0.

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Jumps in time

Summing over n gives

E(t−N) − E(t+0 ) −N−1

∑n=1

⟦E(tn)⟧ = 0.

To give a sign to the extra terms (Hughes, Hulbert ’88):

1

2⟦(u(tn), u(tn))L2(Ω)⟧ − (⟦u(tn)⟧, u(t+n ))L2(Ω) =

1

2(⟦u(tn)⟧, ⟦u(tn)⟧)L2(Ω).

Do this for all the terms, including the stabilization.

Obtain a dissipative method.

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Space-time dG formulation

a(u, v) ∶=N−1

∑n=0

(u, v)Ω×In + (⟦u(tn)⟧, v(t+n ))Ω

+ (a∇u,∇v)Ω×In + (⟦a∇u(tn)⟧,∇v(t+n ))Ω

− (a∇u , [v])Γn×In − (⟦a∇u(tn)⟧, [v(t+n )])Γn

− ([u] ,a∇v)Γn×In − (⟦[u(tn)]⟧,a∇v(t+n ))Γn

+ (σ0 [u] , [v])Γn×In + (σ0⟦[u(tn)]⟧, [v(t+n )])Γn

+ (σ1 [u] , [v])Γn×In + (σ2 [a∇u] , [a∇v])Γn×Inand

binit(v) ∶= (v0, v(t+0 ))Ω + (a∇u0,∇v(t+0 ))Ω − (a∇u0 , [v(t+0 )])Γ0

− ([u0] ,a∇v(t+0 ))Γ0+ (σ0 [u0] , [v(t+0 )])Γ0

.

Find uh ∈ V h,pTrefftz(Ω × [0,T ]) such that

a(uh, v) = binit(v), ∀v ∈ V h,p

Trefftz(Ω × [0,T ]).14 / 34

Time-space dG as a time-stepping method

an(u, v) ∶= (u, 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×In − (a∇u(t+n ) , [v(t+n )])Γn

− ([u] ,a∇v)Γn×In − ([u(t+n )] ,a∇v(t+n ))Γn

+ (σ0 [u] , [v])Γn×In + (σ0 [u(t+n )] , [v(t+n )])Γn

+ (σ1 [u] , [v])Γn×In + (σ2 [a∇u] , [a∇v])Γn×In ,

bn(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+ (σ0 [u(t−n )] , [v(t+n )])Γn

,

Find un ∈ Sh,pn,Trefftz such that

an(un, v) = bn(un−1, v), ∀v ∈ Sh,pn,Trefftz.

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Consistency and stability

Theorem

The following statements hold:

1 The method is consistent.

2 There exists a choice of σ0 ∼ h−1, such that for any v ∈ Sh,pn,Trefftz and

t ∈ In the energy is bounded below as

E(t, v) ≥1

2∥v(t)∥2

L2(Ω) +1

4∥√

a∇v(t)∥2L2(Ω).

3 Let uh ∈ V h,pTrefftz discrete solution. Then

E(t−N ,uh) ≤ E(t−1 ,u

h).

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a(⋅, ⋅)1/2∶= ∣∣∣⋅∣∣∣ - a norm on V h,p

Trefftz

a(w ,w) = Eh(t−N ,w) + Eh(t+0 ,w) +N−1

∑n=1

(12∥⟦w(tn)⟧∥

2Ω + 1

2∥√

a⟦∇w(tn)⟧∥2Ω

− (⟦a∇w(tn)⟧, ⟦[w(tn)]⟧)Γn+ 1

2∥⟦√σ0 [w(tn)]⟧∥

2Γn)

+N−1

∑n=0

(∥√σ1 [w]∥

2Γn×In + ∥

√σ2 [a∇w]∥

2Γn×In).

Theorem

a(v , v)1/2 = ∣∣∣v ∣∣∣ = 0 Ô⇒ v = 0, for v ∈ V h,pTrefftz.

Hence, the time-space dG method

a(u, v) = binit(v), ∀v ∈ V h,p

Trefftz

has a unique solution in V h,pTrefftz.

The proof is by noticing that if ∣∣∣v ∣∣∣ = 0 then v is a smooth solution ofthe wave equation, uniquely determined by the initial condition.

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Convergence analysis

If we prove continuity of a(⋅, ⋅)

∣a(u, v)∣ ≤ C⋆∣∣∣u∣∣∣⋆∣∣∣v ∣∣∣, ∀u ∈ cont. sol. +V h,pTrefftz, v ∈ V h,p

Trefftz,

we can use Galerkin orthogonality to show, for any v ∈ V h,pTrefftz

∣∣∣uh− v ∣∣∣2 = a(uh

− v ,uh− v)

= a(uex − v ,uh− v)

≤ C⋆ ∣∣∣uex − v ∣∣∣⋆∣∣∣uh− v ∣∣∣

and hence we have quasi-optimality

∣∣∣uh− uex ∣∣∣ ≤ inf

v∈V h,pTrefftz

∣∣∣uh− v ∣∣∣ + ∣∣∣v − uex ∣∣∣

≤ infv∈V h,p

Trefftz

∣∣∣v − uex ∣∣∣ + C⋆∣∣∣v − uex ∣∣∣⋆

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Integrating by parts a few times (this is how to implement it)

a(w , v) =N−1

∑n=0

( (a∇w , [v])Γn×In − (σ0 [w] , [v])Γn×In − (w , [a∇v])Γintn ×In

+ (σ1 [w] , [v])Γn×In + (σ2 [a∇w] , [a∇v])Γn×In )

−N

∑n=1

( (w(t−n ), ⟦v(tn)⟧)Ω + (a∇w(t−n ), ⟦∇v(tn)⟧)Ω

− (a∇w(t−n ) , ⟦[v(tn)]⟧)Γn− ([w(t−n )] , ⟦a∇v(tn)⟧)Γn

+ (σ0 [w(t−n )] , ⟦[v(tn)]⟧)Γn).

Recall

∣∣∣v ∣∣∣2 = E(t−N , v) + E(t+0 , v) +N−1

∑n=1

(12∥⟦v(tn)⟧∥

2Ω + 1

2∥⟦∇v(tn)⟧∥2Ω

+ (⟦∇v(tn)⟧, ⟦[v(tn)]⟧)Γn+ 1

2∥⟦√σ0 [v(tn)]⟧∥

2Γn)

+N−1

∑n=0

(∥√σ1 [v]∥2

Γ×In + ∥√σ2 [∇v]∥2

Γ×In),

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Integrating by parts a few times (this is how to implement it)

a(w , v) =N−1

∑n=0

( (a∇w , [v])Γn×In − (σ0 [w] , [v])Γn×In − (w , [a∇v])Γintn ×In

+ (σ1 [w] , [v])Γn×In + (σ2 [a∇w] , [a∇v])Γn×In )

−N

∑n=1

( (w(t−n ), ⟦v(tn)⟧)Ω + (a∇w(t−n ), ⟦∇v(tn)⟧)Ω

− (a∇w(t−n ) , ⟦[v(tn)]⟧)Γn− ([w(t−n )] , ⟦a∇v(tn)⟧)Γn

+ (σ0 [w(t−n )] , ⟦[v(tn)]⟧)Γn).

Hence define,

∣∣∣w ∣∣∣2⋆ =

12

N

∑n=1

(∥w(t−n )∥2Ω + ∥

√a∇w(t−n )∥

2Ω + ∥

√σ0 [w(t−n )] ∥

2Γn+ ∥σ

−1/20 a∇w(t−n ) ∥

2Γn)

+N−1

∑n=0

(∥√σ1 [w]∥

2Γn×In + ∥

√σ2 [a∇w]∥

2Γn×In + ∥σ

−1/22 w∥

2Γintn ×In

+ ∥σ−1/21 a∇w∥

2Γn×In + ∥σ0σ

−1/21 [w]∥

2Γn×In) .

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The choice of stabilization parameters and convergence

Let τn = tn+1 − tn, h = diam(K), (x , t) ∈ K × (tn, tn+1),K ∈ Tn.

diam(K)/ρK ≤ cT , ∀K ∈ Tn, n = 0,1, . . . ,N − 1, where ρK is theradius of the inscribed circle of K .

Assume space-time elements star-shaped with respect to a ball.

Choice of parameters σ0 = p2cT C 2

a Cinv(cah)−1, σ1 = Cap3(hτn)

−1

σ2 = h(Caτn)−1.

Theorem

For sufficiently smooth solution and h ∼ τ

∣∣∣uh− uex∣∣∣ = O(hp−1/2

).

Proof uses truncated Taylor expansion.

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Error estimate in mesh independent norm

Using a Gronwall argument we can show for v ∈ Sh,pn,Trefftz

∥v∥2Ω×In + ∥

√a∇v∥2

Ω×In ≤ τneC(tn+1−tn)/hn (∥v(t−n+1)∥2Ω + ∥

√a∇v(t−n+1)∥

2Ω) ,

hn ∶= minx∈Ω h(x , t), t ∈ In. Let τ = max τn and h = min hn. Then

∥V ∥2Ω×(0,T) + ∥

√a∇V ∥

2Ω×(0,T) ≤ CτeCτ/h∣∣∣V ∣∣∣

2⋆, ∀V ∈ V h,p

Trefftz.

Proposition

∥uh− uex∥

2Ω×[0,T ] + ∥∇uh

−∇uex∥2Ω×[0,T ]

≤ C infV ∈V h,p

Trefftz

(τeCτ/h∣∣∣V − uex∣∣∣2⋆

+ ∥V − uex∥2Ω×[0,T ] + ∥

√a∇(V − uex)∥

2Ω×[0,T ]).

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Outline

1 Motivation



4 Numerical results

22 / 34

Wave equation with damping

Damped wave equation:u + αu −∆u = 0.

The extra term decreases the energy:

d

dtE(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−αtp1(x , t) + p2(x , t)

with p1 and p2 polynomial in x and t.

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Solution formula in 1D

Instead, use basis functions obtained by propagating polynomial initialdata:

u(x ,0) = u0(x) = xαj , u(x ,0) = v0(x) = 0,

andu0(x) = 0, v0(x) = xβj

with∣αj ∣ ≤ p ∣βj ∣ ≤ p − 1.

In 1D solution is then given by the d’Alambert-like formula

u(x , t) =1

2[u0(x − t) + u0(x + t)] e−αt/2

+α

4e−αt/2

∫

x+t

x−tu0(s)I0 (ρ(s)α2 ) +

t

ρ(s)I1 (ρ(s)α2 )ds

+1

2e−αt/2

∫

x+t

x−tv0(s)I0 (ρ(s)α2 )ds, ρ(s; x , t) =

√t2 − (x − s)2.

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Solution formula ctd.

Rearranging (the last term)

1

2te−αt/2

∫

1

0[v0(x + st) + v0(x − st)] I0 (αt2

√1 − s2)ds.

For example for v0(x) = x2

x2te−αt/2∫

1

0I0 (αt2

√1 − s2)ds + t3e−αt/2

∫

1

0s2I0 (αt2

√1 − s2)ds.

Corresponding term in 3D

te−αt/2

1

∫

0

−∫∂B(0,1)

[v0(x + tsz) +Dv0(x + tsz)] ⋅zdSz I0 (αt2

√1 − s2)ds

Hence need efficient representation of functions of the type

%j(t) = ∫1

0s j I0 (αt2

√1 − s2)ds

In Matlab: Chebfun is ideal.

25 / 34

Solution formula ctd.

Rearranging (the last term)

1

2te−αt/2

∫

1

0[v0(x + st) + v0(x − st)] I0 (αt2

√1 − s2)ds.

For example for v0(x) = x2

x2te−αt/2∫

1

0I0 (αt2

√1 − s2)ds + t3e−αt/2

∫

1

0s2I0 (αt2

√1 − s2)ds.

Corresponding term in 3D

te−αt/2

1

∫

0

−∫∂B(0,1)

[v0(x + tsz) +Dv0(x + tsz)] ⋅zdSz I0 (αt2

√1 − s2)ds

Hence need efficient representation of functions of the type

%j(t) = ∫1

0s j I0 (αt2

√1 − s2)ds

In Matlab: Chebfun is ideal.25 / 34

Changes to the analysis

Polynomial in space Ô⇒ the same discrete inverse inequalities used+ the extra term decreases energy Ô⇒ 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− topolynomials and propagate.

At boundary, extend exact solution anti-symmetrically and againproject and propagate.

26 / 34

Outline

1 Motivation



4 Numerical results

27 / 34

One dimensional settingSimple 1D setting:

Ω = (0,1), a ≡ 1.

Initial data

u0 = e−( x−5/8

δ)2

, v0 = 0, δ ≤ δ0 = 7.5 × 10−2.

Interested in having few degrees of freedom for decreasing δ ≤ δ0.

0 0.2 0.4 0.6 0.8 1

-0.5

-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5

t = 1.5, δ = δ0

0 0.2 0.4 0.6 0.8 1

-0.5

-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5

t = 1.5, δ = δ0/4

Energy of exact solution

exact energy = 12∥ux(x ,0)∥2

Ω ≈ 2δ−1∫

∞

−∞y 2e−2y2

dy = δ−1

√π

2√

2.

We compare with full polynomial space.Note for polynomial order p

2p + 1 Trefftz 12(p + 1)(p + 2)full polynomial space.

In all 1D experiments square space-time elements.

28 / 34

One dimensional settingSimple 1D setting:

Ω = (0,1), a ≡ 1.

Initial data

u0 = e−( x−5/8

δ)2

, v0 = 0, δ ≤ δ0 = 7.5 × 10−2.

Energy of exact solution

exact energy = 12∥ux(x ,0)∥2

Ω ≈ 2δ−1∫

∞

−∞y 2e−2y2

dy = δ−1

√π

2√

2.

We compare with full polynomial space.

Note for polynomial order p

2p + 1 Trefftz 12(p + 1)(p + 2)full polynomial space.

In all 1D experiments square space-time elements.28 / 34

Error in dG-norm, δ = δ0, T = 1/4:

Trefftz poly Full poly

10 -3 10 -2 10 -1

h

10 -8

10 -6

10 -4

10 -2

10 0

10 2

Err

or

p = 1p = 2p = 3p = 4p = 5

10 -3 10 -2 10 -1

h

10 -8

10 -6

10 -4

10 -2

10 0

10 2

Err

or

p = 1p = 2p = 3p = 4p = 5

Trefftz p-convergence (fixed h in space and time):

1 2 3 4 5 6 7 8 9 10p

10 -8

10 -6

10 -4

10 -2

10 0

10 2

Err

or

29 / 34

Energy conservation

For δ = δ0/4:


10 -3 10 -2 10 -1 10 0 10 1 10 2 10 3

Time

0

5

10

15

20

25

30

35

Ener

gy

Energy, p=1Energy, p=2Energy, p=3Energy, p=4Exact Energy

10 -3 10 -2 10 -1 10 0 10 1 10 2 10 3

Time

0

5

10

15

20

25

30

35

Ener

gyEnergy, p=1Energy, p=2Energy, p=3Energy, p=4Exact Energy

30 / 34

Relative error for decreasing δ

errorδ = (δ

2∥u(⋅,T ) − uh(⋅,T

−)∥

2Ω +

δ

2∥∇u(⋅,T ) −∇uh(⋅,T

−)∥

2Ω)

1/2

10 -2 10 -1 10 0 10 1

h=/

10 -10

10 -8

10 -6

10 -4

10 -2

10 0

Err

or

p=2, / = /0p=3, / = /0p=4,/ = /0p=5, / = /0p=2, / = /0=2p=3, / = /0=2p=4, / = /0=2p=5, / = /0=2p=2, / = /0=4p=3, / = /0=4p=4, / = /0=4p=5, / = /0=4

31 / 34

2D experimentOn square [0,1]2 with exact solution

u(x , y , t) = cos(√

2πt) sinπx sinπy .

Energy of error at final time:

error = (12∥u(⋅,T ) − uh(⋅,T

−)∥

2Ω + 1

2∥∇u(⋅,T ) −∇uh(⋅,T−)∥

2Ω)

1/2.


0.02 0.03 0.04 0.05 0.06 0.07 0.080.090.1Mesh-size

10 -8

10 -6

10 -4

10 -2

10 0

10 2

Err

or

p = 1p = 2p = 3p = 4

0.02 0.03 0.04 0.05 0.06 0.07 0.080.090.1Mesh-size

10 -8

10 -6

10 -4

10 -2

10 0

10 2

Err

or

p = 1p = 2p = 3p = 4

32 / 34

Damped wave equation in 1D

Error in dG norm. Exact solution on Ω = (0,1):

u(x , t) = e(−αt/2) sin(πx)

⎡⎢⎢⎢⎢⎣

cos

√

π2 −α2

4t +

α

2√π2 − α2/4

sin

√

π2 −α2

4t

⎤⎥⎥⎥⎥⎦

.


10 1 10 2 10 3 10 4

Number of DOF

10 -8

10 -6

10 -4

10 -2

10 0

10 2

Err

or

p = 1p = 2p = 3p = 4

10 1 10 2 10 3 10 4

Number of DOF

10 -8

10 -6

10 -4

10 -2

10 0

10 2

Err

or

p = 1p = 2p = 3p = 4

33 / 34

Conclusions

A space-time interior penalty dG method for the acoustic wave equation insecond order form:

Allows Trefftz basis functions, polynomial in space.

Stability and convergence analysis available.

Also practical for damped wave equation.

To do:

A posteriori error analysis

Adaptivity in h, p, wave directions.

34 / 34

A space-time Trefftz method for the second order wave equation · Outline of the talk 1 Motivation...

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