An efficient finite element method and error analysis for fourth order problems in a spherical domain

Na Peng; Jiayu Han; Jing An

doi:10.3934/dcdsb.2022021

Article Contents

2022, Volume 27, Issue 11: 6807-6821. Doi: 10.3934/dcdsb.2022021

This issue Previous Article On the quasineutral limit for the compressible Euler-Poisson equations Next Article Local and parallel finite element algorithms for the incompressible Navier-Stokes equations with damping

An efficient finite element method and error analysis for fourth order problems in a spherical domain

School of Mathematical Sciences, Guizhou Normal University, Guiyang 550025, China

^*Corresponding author: Jing An
^*Corresponding author: Jing An

Received: November 2020

Revised: November 2021

Early access: February 2022

Published: November 2022

This work is supported by National Natural Science Foundation of China (Grant No. 12061023), Guizhou Provincial Education Department Foundation (Qianjiaohe No. KY[2018]041), and Guizhou Provincical Science and Technology Foundation (ZK[2021]012).

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Abstract

An efficient finite element method for the fourth order problems in a spherical domain will be solved in this paper. Initially, we derive the necessary pole conditions with the intention of overcoming the difficulty of singularity introduced by spherical coordinate transformation. Then the original problem is transformed into a series of equivalent one-dimensional problems by utilizing spherical harmonic functions expansion. Secondly, we introduce some appropriate weighted Sobolev spaces and derive weak form and corresponding discrete form for each one-dimensional fourth order problem based on these pole conditions. In addition, we illustrate the error estimate of the approximate solutions by employing Lax-milgram lemma and approximation property of the cubic Hermite interpolation operator. Eventually, we present the algorithm in detail and show its efficiency through some numerical examples.

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Mathematics Subject Classification: 65N25, 65N30.

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Figure 1. The figures of the numerical solution (left) and the exact solution (right) for $ N = 20 $, $ M = 12 $

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Figure 2. The error figures between numerical solution and exact solution with $ N = 10 $, $ M = 6 $(left) and $ N = 25 $, $ M = 10 $(right), respectively

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Figure 3. The figures of the numerical solution $ u_{0h}^{0} $ and the exact solution $ u_{0}^{0} $ for $ N = 10 $(left) and $ N = 25 $(right) with $ m = 0 $, respectively

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Figure 4. The figures of the numerical solution (left) and the exact solution (right) for $ N = 20 $, $ M = 12 $

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Figure 5. The error $ e(u_{Mh}(x, y, z), u(x, y, z)) $ figures between numerical solution and exact solution with $ N = 10 $, $ M = 6 $(left) and $ N = 25 $, $ M = 10 $(right), respectively

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Table 1.1. The error $e(u_{Mh}(x, y, z), u(x, y, z))$ for different $M$ and $N$

N	$M = 2$	$M = 4$	$M = 6$	$M = 8$
10	5.9016e-06	5.9016e-06	5.9041e-06	8.1147e-06
15	1.1920e-06	1.1920e-06	1.1920e-06	1.1920e-06
20	3.8196e-07	3.8196e-07	3.8196e-07	3.8196e-07
25	1.5769e-07	1.5769e-07	1.5769e-07	1.5769e-07

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Table 1.2. The convergence $r_h(\widetilde e)$ with $m = 0$

N	$N = {4}$	$N = {8}$	$N = {16}$	$N = {32}$	$N = {64}$
$M = 0$	1.9279	1.9910	2.0000	1.9880	1.9997

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Table 1.3. The error $e(u_{Mh}(x, y, z), u(x, y, z))$ between numerical solutions and exact solution for different $M$ and $N$

N	$M = 2$	$M = 4$	$M = 6$	$M = 8$
10	0.0163	4.3402e-04	8.9630e-04	0.0117
15	0.0164	3.7736e-04	1.6157e-05	1.4130e-05
20	0.0164	3.7154e-04	7.0821e-06	4.2228e-06
25	0.0164	3.6885e-04	4.7106e-06	1.7312e-06

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