Error analysis of discontinuous Galerkin method for the time fractional KdV equation with weak singularity solution

Na An; Chaobao Huang; Xijun Yu

doi:10.3934/dcdsb.2019185

Article Contents

2020, Volume 25, Issue 1: 321-334. Doi: 10.3934/dcdsb.2019185

This issue Previous Article Pullback exponential attractors for differential equations with variable delays Next Article Multi-scale analysis for highly anisotropic parabolic problems

Error analysis of discontinuous Galerkin method for the time fractional KdV equation with weak singularity solution

1.
School of Mathematics and Statistics, Shandong Normal University, Jinan 250014, China
2.
School of Mathematic and Quantitative Economics, Shandong University of Finance and Economics, Jinan 250014, China
3.
Labroatory of Computational Physics, Institute of Applied Physics and Computational Mathematics, Beijing 100088, China

^* Corresponding author: huangcb@csrc.ac.cn
^* Corresponding author: huangcb@csrc.ac.cn

Received: December 31, 2017

Revised: February 28, 2019

Early access: September 2019

Published: January 2020

The authors are grateful to the National Natural Science Foundation of PR China (Grant Nos. 11801332, 11571002, and 11971276)

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Abstract

In this work, the time fractional KdV equation with Caputo time derivative of order $ \alpha \in (0,1) $ is considered. The solution of this problem has a weak singularity near the initial time $ t = 0 $. A fully discrete discontinuous Galerkin (DG) method combining the well-known L1 discretisation in time and DG method in space is proposed to approximate the time fractional KdV equation. The unconditional stability result and O$ (N^{-\min \{r\alpha,2-\alpha\}}+h^{k+1}) $ convergence result for $ P^k \; (k\geq 2) $ polynomials are obtained. Finally, numerical experiments are presented to illustrate the efficiency and the high order accuracy of the proposed scheme.

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Mathematics Subject Classification: Primary: 35R11, 65M60; Secondary: 65M12.

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Figure 1. The numerical solution for Example 4.1 with $ \alpha = 0.4 $

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Table 1. $ L^\infty(L^2) $ errors and orders of convergence on temporal direction for Example 4.1 with $ r = (2-\alpha)/\alpha $

	N = 32	N = 64	N = 128	N = 256	$ N = 512 $	N = 1024
$ \alpha = 0.4 $	3.0496E-2	1.1110E-2	3.9235E-3	1.3578E-3	4.6379E-4	1.5729E-4
		1.4567	1.5016	1.5307	1.5498	1.5600
$ \alpha = 0.6 $	3.8341E-2	1.5127E-2	5.8825E-3	2.2665E-3	8.6831E-4	3.3157E-4
		1.3417	1.3626	1.3759	1.3842	1.3888
$ \alpha = 0.8 $	5.9953E-2	2.6607E-2	1.1728E-2	5.1485E-3	2.2540E-3	9.8512E-4
		1.1720	1.1817	1.1878	1.1916	1.1941

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Table 2. Errors and orders of convergence on space direction for Example 4.1 with $ \alpha = 0.4 $

Polynomial	M	$ \\|u-u_h\\|_{L^2} $	Order	$ \\|u-u_h\\|_{L^{\infty}} $	Order
$ P^2 $	5	5.3831E-01	-	3.2328E-01	-
	10	7.8579E-02	2.7762	4.7729E-02	2.7598
	20	9.9319E-03	2.9840	6.2124E-03	2.9416
	40	1.1426E-04	3.1196	7.5845E-04	3.0340
$ P^3 $	5	1.7236E-02	-	1.3819E-02	-
	10	1.1399E-03	3.9184	8.7589E-04	3.9798
	15	2.2712E-04	3.9406	1.7695E-04	3.9667
	20	7.2979E-05	3.9418	6.1408E-04	3.9070

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Table 3. $ L^\infty(L^2) $ errors and orders of convergence on temporal direction for Example 4.2 with $ r = (2-\alpha)/\alpha $

	N = 32	N = 64	N = 128	N = 256	$ N = 512 $	N = 1024
$ \alpha = 0.4 $	2.6605E-2	9.8042E-3	3.4860E-3	1.2119E-3	4.1549E-4	1.4179E-4
		1.4402	1.4918	1.5243	1.5444	1.5510
$ \alpha = 0.6 $	3.0086E-2	1.2002E-2	4.6980E-3	1.8177E-3	6.9850E-4	2.6770E-4
		1.3258	1.3531	1.3699	1.3797	1.3836
$ \alpha = 0.8 $	4.1374E-2	1.8226E-2	7.9818E-3	3.4836E-3	1.5178E-3	6.6104E-4
		1.1827	1.1912	1.1961	1.1985	1.1992

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Table 4. Errors and orders of convergence on space direction for Example 4.2 with $ \alpha = 0.4 $

Polynomial	M	$ \\|u-u_h\\|_{L^2} $	Order	$ \\|u-u_h\\|_{L^{\infty}} $	Order
$ P^2 $	5	3.8931E-01	-	2.3483E-01	-
	10	5.6563E-02	2.7829	3.4418E-02	2.7704
	20	7.1139E-03	2.9911	4.4696E-03	2.9449
	40	7.7979E-04	3.1894	5.4210E-04	3.0435
$ P^3 $	5	1.2812E-02	-	1.0564E-02	-
	10	8.3809E-03	3.9342	6.7270E-04	3.9731
	15	1.6615E-04	3.9552	1.2940E-04	4.0071
	20	5.3162E-05	3.9564	4.3749E-05	3.9578

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Table 5. $ L^2 $ errors and orders of convergence on temporal direction for Example 4.3 with $ r = (2-\alpha)/\alpha $

	N = 64	N = 128	N = 256	N = 512	N = 1024
$ \alpha = 0.4 $	2.4959E-3	8.4989E-4	2.8741E-4	9.6635E-5	3.2367E-5
		1.5542	1.5641	1.5720	1.5784
$ \alpha = 0.6 $	6.0125E-3	2.3383E-3	8.9915E-4	3.4367E-4	1.3092E-4
		1.3624	1.3788	1.3875	1.3923
$ \alpha = 0.8 $	1.0359E-2	4.6808E-3	2.0899E-3	1.2645E-4	4.0879E-4
		1.1460	1.1633	1.1736	1.1803

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Table 6. $ L^2 $ errors and orders of convergence on temporal direction for Example 4.3 with $ r = 2(2-\alpha)/\alpha $

	N = 64	N = 128	N = 256	N = 512	N = 1024
$ \alpha = 0.4 $	6.0380E-3	2.2285E-3	7.2999E-4	2.4866E-4	8.4023E-5
		1.5110	1.5371	1.5536	1.5653
$ \alpha = 0.6 $	1.1214E-2	4.4769E-3	1.7480E-3	6.7438E-4	2.5839E-4
		1.3248	1.3567	1.3741	1.3839
$ \alpha = 0.8 $	1.5602E-2	7.0291E-3	3.1157E-3	1.3694E-3	5.9926E-4
		1.1503	1.1737	1.1859	1.1923

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