Runge-Kutta convolution quadrature for acoustic scatteringhiptmair/org/Oberwolfach/Slides10/...Lehel...

Runge-Kutta convolution quadrature for acousticscattering

Lehel Banjai

Max-Planck Research GroupMPI for Mathematics in the Sciences, Leipzig

Oberwolfach, 15. 02. 2010

L. Banjai (MPI Leipzig) RK boundary integral methods Oberwolfach, 15. 02. 2010 1 / 23

Outline of the talk

1 Convolution quadrature: Introduction and convergence analysis

2 Easy (, robust, and efficient) implementation

3 Acoustic scatteringConvergence orderQualitative convergence propertiesNumerical experiments

4 Conclusions


Time convolution

Non-sectorial operators

LetK (s) be analytic and |K (s)| ≤ C (σ)|s|µ, Res > σ > 0.

We will consider time convolutions,

Operational notation

if K (s) = Lk(s) =∫∞

0 e−stk(t)dt and g(t) ≡ 0, for t < 0,

K (∂t)g :=

∫ t

0k(t − τ)g(τ)dτ :=

∫σ±i∞

estK (s)Lg(s)ds,

Motivation for the notation

For K (s) = s and g(0) = 0

K (∂t)g = ∂tg .


Convolution quadrature, Ch. LubichApproximation in the form of a discrete convolution:∫ tn

0k(tn − τ)g(τ)dτ ≈

n∑j=0

ω∆tn−j(K )g(tj), tj = j∆t

=n∑

j=0

ω∆tj (K )g(tn − tj),

g(t) ≡ 0, t < 0

Taking the Laplace transform, s = σ + iω, σ > 0,

K (s)Lg(s) ≈

∞∑j=0

ω∆tj (K )e−sj∆t

Lg(s).

So we need a function of ∆t, e−s∆t ,K that approximates K (s)

e.g. K (s) ≈ K(F (e−s∆t ,∆t)

)with F (e−s∆t ,∆t) ≈ s.

NB: Require ReF (e−s∆t ,∆t) > 0 for Res > 0.


Convolution quadrature, Ch. LubichApproximation in the form of a discrete convolution:∫ tn

0k(tn − τ)g(τ)dτ ≈

n∑j=0


=∞∑j=0

ω∆tj (K )g(tn − tj), g(t) ≡ 0, t < 0


K (s)Lg(s) ≈

∞∑j=0


Lg(s).


e.g. K (s) ≈ K(F (e−s∆t ,∆t)




Convolution quadrature, Ch. LubichApproximation in the form of a discrete convolution:∫ t

0k(t − τ)g(τ)dτ ≈

n∑j=0


=∞∑j=0

ω∆tj (K )g(t − tj), g(t) ≡ 0, t < 0


K (s)Lg(s) ≈

∞∑j=0


Lg(s).


e.g. K (s) ≈ K(F (e−s∆t ,∆t)




Convolution quadrature, Ch. LubichApproximation in the form of a discrete convolution:∫ t

0k(t − τ)g(τ)dτ ≈

n∑j=0


=∞∑j=0

ω∆tj (K )g(t − tj), g(t) ≡ 0, t < 0


K (s)Lg(s) ≈

∞∑j=0


Lg(s).


e.g. K (s) ≈ K(F (e−s∆t ,∆t)


NB: Require ReF (e−s∆t ,∆t) > 0 for Res > 0.L. Banjai (MPI Leipzig) RK boundary integral methods Oberwolfach, 15. 02. 2010 4 / 23

Linear Multistep formulas

BDF11− e−s∆t

∆t= s+sO(s∆t) Trapez.

2

∆t

1− e−s∆t

1 + e−s∆t= s+sO((s∆t)2).

i.e., take the choice F (ζ,∆t) = δ(ζ)∆t - δ- generating fn of LMS.

Re δ(ζ) > 0 for |ζ| < 1 ⇐⇒ A-stability =⇒ Order ≤ 2.

Method of proof (Lubich ’94):

K (∂t)g − K (∂∆tt )g =

∫σ+iR

est

K (s)− K

(δ(e−s∆t)

∆t

)Lg(s)ds

In the case |s∆t| ≤ 1 use approximation property and for |s| > 1/∆t use∫|s|>1/∆t

∣∣∣∣K (s)− K

(δ(e−s∆t)

∆t

)∣∣∣∣ |s|−r |Lg (r)(s)||ds|.


Runge-Kutta methodsEvaluate g at stages tj + cl∆t, c = (c1, . . . , cm)T

L (g(tj + cl∆t))ml=1 = ecs∆tLg(s).

K (s)ecs∆t ≈ K

(δ(e−s∆t)

∆t

)ecs∆t .

For RKc A

bTand R(∞) = 0: δ(ζ) = A−1 − ζA−1

1bTA−1.

For method with stage order q we have

δ(e−s∆t)

∆tecs∆t = secs∆t + sO((s∆t)q).

Further reduction of order for µ > 1 due to:

‖δ(e−s∆t)/∆t‖ ≤ C max(|s|,∆t−1).


Convergence order

For sufficiently smooth and compatible data convergence of O(∆tp)for A-stable linear multistep methods; (Lubich 1994)

For sufficiently smooth and compatible data convergence ofO(∆tq+1−µ) for A-stable RK methods (Banjai, Lubich 2010).

Numerical example: Kµ(s) = sµ

1−e−s . Applying the 3-stage Radau IIA

method (stage order 3) for g(t) = e−0.4t sin6 t.

N µ = −1 µ = −1/2 µ = 1/2 µ = 1

4 0.04 0.04 0.18 0.688 0.05 0.06 0.15 0.75

16 0.05 0.04 0.19 0.8032 0.04 0.03 0.20 0.8164 0.04 0.02 0.20 0.81

128 0.04 0.03 0.20 0.81

Relative `2 error divided by (∆t)4−µ.


Convergence order

For sufficiently smooth and compatible data convergence of O(∆tp)for A-stable linear multistep methods; (Lubich 1994)

For sufficiently smooth and compatible data convergence ofO(∆tq+1−µ) for A-stable RK methods (Banjai, Lubich 2010).

Numerical example: Kµ(s) = sµ

1−e−s . Applying the 3-stage Radau IIA

method (stage order 3) for g(t) = e−0.4t sin6 t.

N µ = −1 µ = −1/2 µ = 1/2 µ = 1

4 0.04 0.04 0.18 0.688 0.05 0.06 0.15 0.75

16 0.05 0.04 0.19 0.8032 0.04 0.03 0.20 0.8164 0.04 0.02 0.20 0.81

128 0.04 0.03 0.20 0.81

Relative `2 error divided by (∆t)4−µ.L. Banjai (MPI Leipzig) RK boundary integral methods Oberwolfach, 15. 02. 2010 7 / 23

Solving the triangular Toeplitz system

0BBBB@ω0 0 · · · 0

ω1 ω0

. . ....

.... . . 0

ωN · · · ω1 ω0

1CCCCA0BBB@

φ0

φ1

...φN

1CCCA =

0BBB@g0

g1

...gN

1CCCAProblem: O(N2) to solve, ω∆t

j implicitly definedN-number of time steps.

Solution [Hairer et. al ’86]: O(N log2 N)- solve the same small system T many times.- apply the Toeplitz matrix using FFT.

- The system T can also be solved by FFT[Banjai, Sauter ’09]


Approximation by Helmholtz operators [Banjai ’09]

Toeplitz matrix =

λ−n−1Λ−1

λn+1

ωn+1

λn

ωn · · ·

λ1

ω1

λn+2

ωn+2. . .

. . .

λ2

ω2...

. . .. . .

...

λ2n+1

ω2n+1 · · ·

λn+1

ωn+1

Λ

Where 0 < λ < 1 and Λ = diag(1, λ, . . . , λn).

(−1)l2n+1∑j=0

λjωjζ−lj2n+2 = (−1)lK () +O(λ2n+2)

where ζ2n+2 = e2πi/(2n+2).

Careful: restricted accuracy epsλ−n + λ2n+2 ∼ eps2/3.



Toeplitz matrix =

λ−n−1Λ−1

λn+1ωn+1 λnωn · · · λ1ω1

λn+2ωn+2. . .

. . . λ2ω2...

. . .. . .

...λ2n+1ω2n+1 · · · λn+1ωn+1

Λ


(−1)l2n+1∑j=0

λjωjζ−lj2n+2 = (−1)lK

(δ(λζ−l

2n+2)/∆t)

+O(λ2n+2)





Toeplitz matrix =

λ−n−1Λ−1

λn+1ωn+1 λnωn · · · λ1ω1

λn+2ωn+2. . .

. . . λ2ω2...

. . .. . .

...λ2n+1ω2n+1 · · · λn+1ωn+1

Λ


(−1)l2n+1∑j=0

λjωjζ−lj2n+2 = (−1)lK (sl) +O(λ2n+2)




Remarks on the algorithm

A block time stepping method: at each (block) time step a smalllower triangular Toeplitz system (T ) needs to be solved (directly orvia the decoupling [Banjai, Sauter ’09]).

Update of the right-hand side done in parallel at each block time step.

For the latter step we need to compute

K (sl), with sl = δ(λζl)/∆t, ζNl = 1.

Due to A-stability eigenvalues of δ(λζl) in the right half-plane.

Is δ(ζ) diagonalizable for |ζ| < 1?

For 2 step Radau IIA δ(ζ) diagonalizable for ζ 6= 3√

3− 5For 3 step Radau IIA δ(ζ) diagonalizable for

ζ, ζ 6= 0.004598175 · · ·+ i0.069213506 · · ·

Costs O(mN log2 N) + O(Nm3), with diagonalization cost O(Nm3)essentially negligible if discretization in space necessary.


Acoustic scattering application

Let u satisfy, for bounded Ω ⊂ R3 with Γ = ∂Ω

∂2t u −∆u = 0 in Ωc × [0,T ]

u(x , 0) = ∂tu(x , 0) = 0 in Ωc

u(x , t) = g(x , t), on Γ× [0,T ]

Then u can be written as

u(x , t) =

∫ t

0

∫Γ

δ(t − τ − |x − y |)4π|x − y |

φ(y , τ)dΓydτ.

Find φ such that

g(x , t) =

∫ t

0

∫Γ

δ(t − τ − |x − y |)4π|x − y |

φ(y , τ)dΓydτ, (x , t) ∈ Γ× [0,T ].


Since Lδ(t − |x − y |) = e−s|x−y | we need to solve (compute)

g = V (∂t)φ ⇐⇒ φ = V−1(∂t)g

where

V (s)ϕ :=

∫Γ

e−s|x−y |

4π|x − y |ϕ(y)dΓy .

Convolution weights ωj(V ) : H−1/2(Γ)→ H1/2(Γ)

ω∆tj (V )ϕ =

∫Γ

ωj(|x − y |/∆t)

4π|x − y |ϕ(y)dΓy ,

that isωj : R≥0 → Rm×m, ω0(d) = e−δ(0)d .


Convolution weights: kernel functions

BDF2 Trapezoid

For both methods ωj ≈ 0 for d > Cj∆t.

For BDF2 ωj(d) ∼ dbj/2c for d → 0 and ωj ≈ 0 for j∆t > Cjdiam(Γ).[Hackbusch, Kress, Sauter ’06]


Convolution weights: kernel functions

Radau IIA, order 3: (ωj(d))11 (ωj(d))22

For m-stage Runge-Kutta method ωj : R→ Rm×m

Again ωj(d) ∼ d j for d → 0; ω0(d) = e−A−1d .

Again ωj ≈ 0 for j∆t > Cjdiam(Γ).


Convergence order

V (s) is analytic for Re s > σ0 > 0 and

‖V−1(s)‖H−1/2(Γ)←H1/2(Γ) ≤ C (σ0)1

Res|s|2, Bamberger, Ha Duong 1986.

For all the other operators, transmission problem, FEM-BEM coupling. . . see Sayas, Laliena 2009.

But, for circle numerical experiments suggest

‖V−1‖ = supn

(1 + n2)−1/2|λn(s)|−1 . |λ0(s)|−1 =

∣∣∣∣ 2s

1− e−2s

∣∣∣∣ .


A non-convex domain

Discretized by 14440 triangles.

g(x , t) = cos(ωπ(t − α.x))e−( t−α.x−Aσ )

2


101

10−2

10−1

100

N

error

3-stage Radau IIAO( 1

N3 )

We plot the error for a fixed x ∈ Γ

‖ϕN(x , ·)− ϕ2N(x , ·)‖`2 =

∆tN∑

j=0

(ϕN(x , tj)− ϕ2N(x , tj))2

1/2

.


Qualitative convergence properties

101

102

103

10−2

10−1

100

101

N (mN for the Radau IIA)

erro

r

BDF2Trapezoid2 stage Radau IIA3 stage Radau IIA

Relative `2 error plotted for V−1(∂∆tt )g with

g(x , t) = e−0.4t cos(8πt) sin6(t)(Y l0 + Y l

20), t ∈ [0, 6].

NB: We are solving both the interior and exterior problem!


A non-convex domain

Discretized by 14440 triangles.

g(x , t) = cos(0.5π(t − α.x))e−( t−α.x−40.5 )

2


Total field inside the cavity

2 3 4 5 6 7 8 9 10−0.4

−0.3

−0.2

−0.1

0

0.1

0.2

0.3

t

uto

t(0

,t)

Radau IIA, N = 60BDF2, N = 180Radau IIA, N = 80BDF2, N = 240

Computations done with on a large parallel cluster.


Numerical results II: Long time stability

0 10 20 30 40 50 60 70 80−8

−6

−4

−2

0

2

4

6

8

t

A(x

,t)

50 55 60 65 70 75 80−8

−6

−4

−2

0

2

4

6

8

t

A(x

,t)

Unit sphere discretized by 8192 triangles, BDF2 with N = 960, 3-stageRadau IIA N = 320, ωj ≈ 0 for large enough j .

g(x , t) =

0, if t − α.x − 1 < 0,

e−1

t−α.x−1 sinπ(t − α.x − 1), if t − α.x − 1 ≥ 0.


Conclusions

Analysis and construction of a multistep-multistage convolutionquadrature.

Efficient, robust, easily parallelizable algorithm for the solution of theresulting system.

Very good performance of Radau IIA methods even at moderateaccuracies

Very good long time stability properties.

One of the main advantages: a time-domain method that usesexclusively the Laplace transform of the distributional fundamentalsolution.


Issues open or undiscussed

Efficient space computationI H-matrices for inversion of operators on the diagonalI FMM with diagonal expansions for V (sl) with Imsl 1I Making use of sparsity of Galerkin discretization of operators ωj .I Long time computations when strong Huygens’ principle is missing

(Volker Gruhne).

Search for optimal multistep-multistage method

How does the shape of the domain effect µ in ‖V−1(s)‖ ≤ C |s|µ andhence convergence rates of the RK-CQ? Is there a domain for whichµ = 2 is reached?

How about more general problems? (see [Laliena, Sayas ’09]).

Can we give a rigorous answer to the question: How do I choose ∆t?

Non-smooth, non-compatible data, i.e. slow decay of Lg(s)? (seework of M. Schanz and his group)


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