分片光滑边值问题的再生核方法

数学物理学报 ›› 2020, Vol. 40 ›› Issue (5): 1333-1340.

分片光滑边值问题的再生核方法

赵志红*(),林迎珍

^{北京理工大学珠海学院广东珠海 519088}

收稿日期:2018-11-21 出版日期:2020-10-26 发布日期:2020-11-04
通讯作者: 赵志红 E-mail:zhaozh39@163.com
基金资助:
广东省普通高校青年创新人才项目(2018KQNCX338);广东省普通高校特色创新项目(2019KTSCX217)

Reproducing Kernel Method for Piecewise Smooth Boundary Value Problems

Zhihong Zhao*(),Yingzhen Lin

^{School of Science, Zhuhai Campus, Beijing Institute of Technology, Guangdong Zhuhai 519088}

Received:2018-11-21 Online:2020-10-26 Published:2020-11-04
Contact: Zhihong Zhao E-mail:zhaozh39@163.com
Supported by:
the Young Innovative Talents Program in Universities and Colleges of Guangdong Province(2018KQNCX338);the Characteristic Innovative Scientific Research Project of Guangdong Province(2019KTSCX217)

摘要/Abstract

摘要：

一种高效的再生核算法（RKM）被提出用于解决分片光滑边值问题（BVP）.通过定义算子 ${\cal L}：{W_{2}^{3}[0, 1]}\rightarrow{L_{2}[0, 1]}$ ，应用再生核算法，解决了较为复杂的分片光滑边值问题.对定理的证明保证了算法理论的正确性，进而得到近似解 $u_{n}（x）$ 以 $O（h^2）$ 收敛于精确解.即：在范数 $\left\|\displaystyle.\right\|_{W_2^{3}}$ 意义下 $u_{n}（x）$ 有不低于二阶的收敛性.数值算例表明算法正确、简便、有效.

关键词: 再生核方法, 分片光滑, 收敛阶

Abstract:

In this paper, an efficient reproducing kernel method(RKM) is proposed for solving the piecewise smooth BVP. By defining an operator ${\cal L}:{W_{2}^{3}[0, 1]}\rightarrow {L_{2}[0, 1]}$ and applying the reproducing kernel property, we have skillfully solved the complex piecewise smooth BVP. The theorems show the accuracy of the algorithm. Furthermore, we get that the approximate solution $u_{n}(x)$ convergence to the exact one with order $O(h^2)$ . That is, $u_{n}(x)$ has second order convergence in the sense of norm $\left\|\displaystyle. \right\|_{W_2^{3}}$ . Finally, some numerical experiments are given to illustrate the algorithm is accuracy, simple and efficient.

Key words: Reproducing kernel method, Piecewise smooth, Convergence order

中图分类号:

O241

赵志红,林迎珍. 分片光滑边值问题的再生核方法[J]. 数学物理学报, 2020, 40(5): 1333-1340.

Zhihong Zhao,Yingzhen Lin. Reproducing Kernel Method for Piecewise Smooth Boundary Value Problems[J]. Acta mathematica scientia,Series A, 2020, 40(5): 1333-1340.

图/表 3

表 1

算例4.1的最大绝对误差及收敛阶"

$n$	$\|u_{n}(x)-u(x)\|_{\max}^{[14]}$	$\|u_{n}(x)-u(x)\|_{\max}$	$\|u'_{n}(x)-u'(x)\|_{\max}$	$\|u''_{n}(x)-u''(x)\|_{\max}$
$10$	$4.0\times10^{-4}$	$5.877\times10^{-5}$	$2.867\times10^{-4}$	$1.263\times10^{-5}$
$100$	$4.0\times10^{-6}$	$5.894\times10^{-7}$	$2.868\times10^{-6}$	$1.260\times10^{-7}$
$1000$	$/$	$5.893\times10^{-9}$	$2.867\times10^{-8}$	$1.263\times10^{-9}$
阶	$/$	$2.000$	$2.000$	$2.000$

表 1

表 2

算例4.2的最大绝对误差和收敛阶"

$n$	$\|u_{n}(x)-u(x)\|_{\max}$	$\|u'_{n}(x)-u'(x)\|_{\max}$	$\|u''_{n}(x)-u''(x)\|_{\max}$
$100$	$1.868\times10^{-6}$	$1.051\times10^{-5}$	$1.839\times10^{-5}$
$1000$	$1.868\times10^{-8}$	$1.051\times10^{-7}$	$1.839\times10^{-7}$
阶	$2.000$	$2.000$	$2.000$

表 2

图 1

左侧曲线是算例4.2 $f(x)$ 的图像;中间曲线是算例4.2 $u_{1000}(x)$ 与 $u(x)$ 的之间的误差;右侧曲线是算例4.2 $u'_{1000}(x)$ 与 $u'(x)$ 之间的误差"

图 1

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