﻿ 一类混合型积分微分方程的数值解法 A Numerical Solution of Mixed Integral Differential Equations

Vol.07 No.07(2018), Article ID:25820,9 pages
10.12677/AAM.2018.77090

A Numerical Solution of Mixed Integral Differential Equations

Yang Liu, Yulan Wang

Inner Mongolia University of Technology, Hohhot Inner Mongolia

Received: Jun. 16th, 2018; accepted: Jul. 4th, 2018; published: Jul. 11th, 2018

ABSTRACT

Polynomial approximation in the mathematical analysis and numerical approximation theory has an important position; it has been widely used in engineering calculation and the actual life. And the study of the numerical method for solving the integral differential equation is one of the important subjects exists in every field. This paper mainly based on Legendre polynomial rebuilding the reproducing kernel, through the “Gram-Schmidt”, the approximate solution of the equation is given. At the same time, it gives three numerical examples of Volterra-Fredholm integral differential equation. Compared with the traditional methods of reproducing kernel, we further verified that our method was effective and had high precision. All numerical calculations are given by the Mathematica 8.0 software.

Keywords:Legendre Polynomials, Reproducing Kernel, Numerical Solution, Volterra-Fredholm Integral Differential Equation

1. 引言

$\left\{\begin{array}{l}{q}_{1}\left(x\right){u}^{\left(n\right)}\left(x\right)+{q}_{2}\left(x\right){u}^{\left(n-1\right)}\left(x\right)+{q}_{3}\left(x\right)u\left(x\right)+{\int }_{0}^{x}{k}_{1}\left(t,x\right)u\left(t\right)\text{d}t+{\int }_{0}^{1}{k}_{2}\left(t,x\right)u\left(t\right)\text{d}t=f\left(x\right),0\le x\le 1\\ {u}^{\left(i\right)}\left(0\right)={\alpha }_{i},{u}^{\left(j\right)}\left(1\right)={\beta }_{j},0\le i,j\le n-1.\end{array}$ (1)

2. 勒让德多项式

${\int }_{-1}^{1}{P}_{n}\left(x\right){P}_{m}\left(x\right)\text{d}x=\frac{2}{2n+1}\text{\hspace{0.17em}},$ (2)

${\int }_{0}^{1}\sqrt{2n+1}\text{ }{L}_{n}\left(t\right)\sqrt{2n+1}\text{ }{L}_{m}\left(t\right)\text{d}t=1\text{\hspace{0.17em}}.$ (3)

${L}_{i}\left(t\right)$ 有如下的递推关系

$\begin{array}{l}{L}_{i+1}\left(t\right)=\frac{\left(2i+1\right)\left(2-1\right)}{i+1}{L}_{i}\left(t\right)-\frac{i}{i+1}{L}_{i-1}\left(t\right),i=1,2,\cdots \\ {L}_{0}\left(t\right)=1,{L}_{1}\left(t\right)=2t-1.\end{array}$ (4)

n阶移位的Legendre多项式的一般表达式为：

$\begin{array}{l}{L}_{n}\left(t\right)=\sum _{i=0}^{n}{\left(-1\right)}^{n+i}\frac{\left(n+i\right)!{t}^{i}}{\left(n-i\right)!\left(i\right){!}^{2}}\text{\hspace{0.17em}},\\ {L}_{n}\left(0\right)={\left(-1\right)}^{\left(n\right)},{L}_{n}\left(1\right)=1\text{\hspace{0.17em}}.\end{array}$ (5)

3. 造核

$〈{e}_{i}\left(t\right),{e}_{j}\left(t\right)〉={\delta }_{ij}=\left\{\begin{array}{l}0\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\left(i\ne j\right)\\ 1\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(i=j\right)\end{array},$ (6)

$\begin{array}{c}〈f\left(t\right),{K}_{s}\left(t\right)〉=〈\sum _{i=1}^{n}{c}_{i}{e}_{i}\left(t\right),\sum _{j=1}^{n}{e}_{i}\left(t\right),{e}_{i}\left(s\right)〉\\ =\sum _{i=1}^{n}{c}_{i}〈{e}_{i}\left(t\right),\sum _{j=1}^{n}{e}_{j}\left(t\right){e}_{j}\left(s\right)〉\\ =\sum _{i=1}^{n}{c}_{i}{e}_{i}\left(s\right)\\ =f\left(s\right).\end{array}$ (7)

3.1. 构造再生核空间(RKM)

$K\left(x,y\right)=1+3\left(2x-1\right)\left(2y-1\right)+5\frac{3{\left(2x-1\right)}^{2}}{2}\frac{3{\left(2y-1\right)}^{2}}{2}+\cdots +\left(2n+1\right){Q}_{n}\left(x\right){Q}_{m}\left(y\right).$ (8)

3.2. 近似解的表示

$v\left(x\right)=u\left(x\right)+{a}_{1}+{a}_{2}x+{a}_{3}{x}^{2}+\cdots +{a}_{n}{x}^{n-1}$ ，则方程(1)转化为：

$\left\{\begin{array}{l}{q}_{1}\left(x\right){v}^{\left(n\right)}\left(x\right)+{q}_{2}\left(x\right){v}^{\left(n-1\right)}\left(x\right)+{q}_{3}\left(x\right)v\left(x\right)+{\int }_{0}^{x}{k}_{1}\left(t,x\right)v\left(t\right)\text{d}t+{\int }_{0}^{1}{k}_{2}\left(t,x\right)v\left(t\right)\text{d}t=f\left(x\right),0\le x\le 1\\ {v}^{\left(i\right)}\left(0\right)=0,{v}^{\left(j\right)}\left(1\right)=0,0\le i,j\le n-1\text{\hspace{0.17em}}.\end{array}$ (9)

$Av={q}_{1}\left(x\right){v}^{\left(n\right)}\left(x\right)+{q}_{2}\left(x\right){v}^{\left(n-1\right)}\left(x\right)+{q}_{3}\left(x\right)v\left(x\right)+{\int }_{0}^{x}{k}_{1}\left(t,x\right)v\left(t\right)\text{d}t+{\int }_{0}^{1}{k}_{2}\left(t,x\right)v\left(t\right)\text{d}t\text{\hspace{0.17em}},$ (10)

$\left(Av\right)\left(x\right)=f\left(x\right).$ (11)

${\psi }_{i}\left(x\right)={{A}_{y}K\left(x,y\right)|}_{y={x}_{i}},i=0,1,2,\cdots$$〈{\psi }_{j}\left(x\right),{\psi }_{i}\left(x\right)〉={{A}_{x}{A}_{y}K\left(x,y\right)|}_{x={x}_{j}},i,j=1,2,\cdots$ 此时， ${\left\{{\psi }_{i}\left(x\right)\right\}}_{i=1}^{\infty }$ 是完全函数系。具体证明见文献 [12] 。对此完全系做Gram-Schimdt，我们可以得到标准正交基 ${\left\{{\stackrel{˜}{\psi }}_{i}\left(x\right)\right\}}_{i=1}^{\infty }$${\stackrel{˜}{\psi }}_{i}\left(x\right)=\sum _{k=1}^{i}{\beta }_{ik}{\psi }_{i}\left(x\right),i=1,2,\cdots$ 其中 ${\beta }_{ik}$ 为正交化系数。

$\begin{array}{c}v\left(x\right)=\sum _{i=1}^{\infty }〈v,{\stackrel{˜}{\psi }}_{i}〉{\stackrel{˜}{\psi }}_{i}\left(x\right)\\ =\sum _{i=1}^{\infty }〈v,\sum _{k=1}^{i}{\beta }_{ik}{\psi }_{k}〉{\stackrel{˜}{\psi }}_{i}\left(x\right)\\ =\sum _{i=1}^{\infty }\sum _{k=1}^{i}{\stackrel{˜}{\beta }}_{ik}〈v,{A}_{y}K\left(x,y\right)〉{\stackrel{˜}{\psi }}_{i}\left(x\right)\\ =\sum _{i=1}^{\infty }\sum _{k=1}^{i}{\stackrel{˜}{\beta }}_{ik}Av\left({x}_{k}\right){\stackrel{˜}{\psi }}_{i}\left(x\right)\\ =\sum _{i=1}^{\infty }\sum _{k=1}^{i}{\stackrel{˜}{\beta }}_{ik}f\left({x}_{k}\right){\stackrel{˜}{\psi }}_{i}\left(x\right)\text{\hspace{0.17em}}.\end{array}$ (12)

${v}_{n}\left(x\right)=\sum _{i=1}^{n}\sum _{k=1}^{i}{\stackrel{˜}{\beta }}_{ik}f\left({x}_{k}\right){\stackrel{˜}{\psi }}_{i}\left(x\right)\text{\hspace{0.17em}}.$ (13)

${u}_{n}\left(x\right)=\sum _{i=1}^{n}\sum _{k=1}^{i}{\stackrel{˜}{\beta }}_{ik}f\left({x}_{k}\right){\stackrel{˜}{\psi }}_{i}\left(x\right)-{a}_{1}-{a}_{2}x-{a}_{3}{x}^{2}-\cdots -{a}_{n}{x}^{n-1}\text{\hspace{0.17em}}.$ (14)

4. 数值算例

$\left\{\begin{array}{l}x{u}^{″}\left(x\right)-x{u}^{\prime }\left(x\right)+2u\left(x\right)-{\int }_{0}^{x}\left(x-t\right)u\left(t\right)\text{d}t-{\int }_{0}^{1}\left(x+t\right)u\left(t\right)\text{d}t=\frac{1}{12}{x}^{4}-\frac{1}{6}{x}^{3}-\frac{1}{2}{x}^{2}-\frac{13}{6}x+\frac{17}{12},\\ 0

Figure 1. Absolute error of the improved RKM of Example 1

Figure 2. Absolute error of the traditional RKM of Example 1

$\left\{\begin{array}{l}{x}^{2}{u}^{\left(4\right)}\left(x\right)-{u}^{″}\left(x\right)+{u}^{\prime }\left(x\right)-{\int }_{0}^{x}\left(x-t\right)u\left(t\right)\text{d}t+2{\int }_{0}^{1}\left(x+t\right)u\left(t\right)\text{d}t=-\frac{1}{12}{x}^{4}-\frac{1}{2}{x}^{2}+\frac{14}{3}x-\frac{1}{2},\\ 0

Table 1. Absolute error of two regenerated kernel methods of Example 1

Figure 3. Absolute error of two RKM of Example 2

$\left\{\begin{array}{l}{u}^{″}\left(x\right)+\frac{1}{\sqrt{x}}{u}^{\prime }\left(x\right)+\frac{1}{x}u\left(x\right)+{\int }_{0}^{x}\left(t+x\right)u\left(t\right)\text{d}t+{\int }_{0}^{1}txu\left(t\right)\text{d}t-\frac{1}{1+\mathrm{sin}\left({u}^{2}\left(x\right)\right)}-{\text{e}}^{{u}^{9}\left(x\right)}+{u}^{11}\left(x\right)=f\left(x\right),\\ 0

Figure 4. Absolute error of the improved RKM of Example 3

Figure 5. Absolute error of the traditional RKM of Example 3

Table 2. Absolute error of two regenerated kernel methods of Example 3

5. 结论

A Numerical Solution of Mixed Integral Differential Equations[J]. 应用数学进展, 2018, 07(07): 749-757. https://doi.org/10.12677/AAM.2018.77090

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