Solving the wave equation with multifrequency oscillations

Marissa Condon; Arieh Iserles; Karolina Kropielnicka; Pranav Singh

doi:10.3934/jcd.2019012

Article Contents

2019, Volume 6, Issue 2: 239-249. Doi: 10.3934/jcd.2019012

This issue Previous Article A new class of integrable Lotka–Volterra systems Next Article Principal symmetric space analysis

Solving the wave equation with multifrequency oscillations

1.
School of Electronic Engineering, Dublin City University, DCU Glasnevin Campus, Dublin 9, Ireland
2.
DAMTP, Centre for Mathematical Sciences, University of Cambridge, Cambridge CB3 0WA, United Kingdom
3.
Institute of Mathematics, University of Gdańsk, ul. Wit Stwosz 57, 80-308, Gdańsk, Poland
4.
Department of Mathematical Sciences, University of Bath, Bath BA2 7AY, United Kingdom

Received: March 2019

Revised: October 2019

Published: November 2019

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Abstract

We explore a new asymptotic-numerical solver for the time-dependent wave equation with an interaction term that is oscillating in time with a very high frequency. The method involves representing the solution as an asymptotic series in inverse powers of the oscillation frequency. Using the new scheme, high accuracy is achieved at a low computational cost. Salient features of the new approach are highlighted by a numerical example.

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Mathematics Subject Classification: Primary: 65M70; Secondary: 41A60, 35L05, 65L05.

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Table 1. [Example 1] Error and cost for the proposed asymptotic method $ \mathcal{A} $ compared to the Lanczos solver $ \mathcal{S} $. Note that each step of $ \mathcal{A} $ uses $ \mathcal{S} $ twice. $ N $ is the number of time steps ($ h = T/N $, $ T = 10 $)

	$ N $	1	2	4	8	16	64	256	1024
	$ h $	10	5	2.5	1.25	0.625	0.156	0.039	0.010
Calls	$ \mathcal{S} $	1	2	4	8	16	64	256	1024
to $ \mathcal{S} $	$ \mathcal{A} $	2	4	8	16	32	128	512	2048
Error	$ \mathcal{S} $	4147	1.1e5	152.3	44.3	79.4	0.163	0.030	0.007
$ \omega = 25 $	$ \mathcal{A} $	0.593	0.254	0.251	0.206	0.252	0.249	0.248	0.248
Error	$ \mathcal{S} $	4147	6.8e4	9794	960.2	41.8	0.176	0.037	0.009
$ \omega = 50 $	$ \mathcal{A} $	0.140	0.062	0.081	0.047	0.033	0.032	0.032	0.032
Error	$ \mathcal{S} $	4147	1.6e5	7914	67.95	80.2	2.081	0.039	0.009
$ \omega = 100 $	$ \mathcal{A} $	0.077	0.015	0.021	0.011	0.001	0.001	0.001	0.001

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Table 2. [Example 2] Error and cost for the proposed asymptotic method $ \mathcal{A} $ compared to the Lanczos solver $ \mathcal{S} $. Note that each step of $ \mathcal{A} $ uses $ \mathcal{S} $ twice. $ N $ is the number of time steps ($ h = T/N $, $ T = 5 $)

$ N $		1	2	4	8	16	32	64	128	256
$ h $		5	2.5	1.25	0.625	0.313	0.156	0.078	0.039	0.02
Calls	$ \mathcal{S} $	1	2	4	8	16	32	64	128	256
to $ \mathcal{S} $	$ \mathcal{A} $	2	4	8	16	32	64	128	256	512
Error	$ \mathcal{S} $	4.237	88.3	96.3	98.1	1.18	0.447	0.189	0.092	0.047
$ \omega = 25 $	$ \mathcal{A} $	2.14	1.82	1.022	0.409	0.391	0.389	0.388	0.388	0.388
Error	$ \mathcal{S} $	2.55	88.7	82.9	8.845	13.1	0.262	0.128	0.058	0.027
$ \omega = 50 $	$ \mathcal{A} $	0.452	0.380	0.198	0.045	0.038	0.039	0.039	0.039	0.039
Error	$ \mathcal{S} $	2.39	32.5	20.1	18.1	2.5	2.9	0.106	0.045	0.020
$ \omega = 100 $	$ \mathcal{A} $	0.117	0.100	0.053	0.008	0.004	0.003	0.003	0.003	0.003

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