﻿ 高阶非线性微分方程非局部边值问题的解法 Solving Higher Order Nonlinear Differential Equation with Nonlocal Boundary Value Problem

Vol.06 No.08(2017), Article ID:22860,5 pages
10.12677/AAM.2017.68124

Solving Higher Order Nonlinear Differential Equation with Nonlocal Boundary Value Problem

Yongfang Zhou, Lijun Ma, Xiangmei Zhang, Dayong Jin, Guozhong Su

School of Science, Hebei University of Technology, Tianjin

Received: Nov. 13th, 2017; accepted: Nov. 23rd, 2017; published: Nov. 29th, 2017

ABSTRACT

This paper discusses the numerical method for the higher order nonlinear differential equation with nonlocal boundary value problem. By constructing the reproducing kernel space which satisfies the nonlocal boundary value conditions, the simple reproducing kernel numerical approximate method is established. Convergence of approximate solution and its derivatives is proved, respectively.

Keywords:Nonlocal Boundary Value Problems, Higher Order Nonlinear Differential Equation, Reproducing Kernel Space

1. 引言

$\left\{\begin{array}{l}{y}^{\left(n\right)}\left(x\right)=F\left(x,y\left(x\right),{y}^{\prime }\left(x\right),\cdots ,{y}^{\left(n-1\right)}\left(x\right)\right)\\ {y}^{\left(i-1\right)}\left({x}_{j}\right)={b}_{ij}\\ y\left({x}_{k+1}\right)-y\left({x}_{k+2}\right)={b}_{n}\end{array}$

$\left\{\begin{array}{l}{y}^{\left(n\right)}\left(x\right)=F\left(x,y\left(x\right),{y}^{\prime }\left(x\right),\cdots ,{y}^{\left(n-1\right)}\left(x\right)\right)\\ {y}^{\left(i-1\right)}\left({x}_{j}\right)=0\\ y\left({x}_{k+1}\right)-y\left({x}_{k+2}\right)=0\end{array}$ (1)

2. 再生核空间

${W}^{n}\left[a,b\right]$ 是再生核空间(证明参见文献 [4] )，对任意 $y\left(x\right),z\left(x\right)\in {W}^{n}\left[a,b\right]$ ，内积和范数分别为 ${〈y,z〉}_{{W}^{n}\left[a,b\right]}=\underset{i=0}{\overset{n-1}{\sum }}{y}^{\left(i\right)}\left(a\right){z}^{\left(i\right)}\left(a\right)+{\int }_{a}^{b}{y}^{\left(n\right)}\left(x\right){z}^{\left(n\right)}\left(x\right)\text{d}x$${‖y‖}_{{W}^{n}\left[a,b\right]}=\sqrt{{〈y,y〉}_{{W}^{n}\left[a,b\right]}}$

${W}^{n1}\left[a,b\right]$${W}^{n}\left[a,b\right]$ 的闭子空间(证明参见文献 [4] )， ${W}^{n1}\left[a,b\right]$ 是再生核空间。设 ${W}^{n1}\left[a,b\right]$ 的再生核函数为 $K\left(x,t\right)$ (具体表达式的确定参见文献 [4] )。

$W\left[a,b\right]$ 是再生核空间，对任意的 $y\left(x\right),z\left(x\right)\in W\left[a,b\right]$ ，内积和范数分别为

${〈y,z〉}_{W\left[a,b\right]}=y\left(a\right)z\left(b\right)+{\int }_{a}^{b}{y}^{\prime }\left(x\right){z}^{\prime }\left(x\right)\text{d}x$ , ${‖y‖}_{W\left[a,b\right]}=\sqrt{{〈y,y〉}_{W\left[a,b\right]}}$ .

$W\left[a,b\right]$ 的再生核函数为 $R\left(x,t\right)$ (具体表达式的确定参见文献 [4] )。

3. 近似解的构造

$Ty\left(x\right)=F\left(x,y\left(x\right),{y}^{\prime }\left(x\right),\cdots {y}^{\left(n-1\right)}\left(x\right)\right)$ , (2)

${\left\{{x}_{i}\right\}}_{i=1}^{\infty }$$\left[a,b\right]$ 上的稠密子集。

${\phi }_{i}\left(x\right)=R\left(x,{x}_{i}\right)$${\Psi }_{i}\left(x\right)={T}^{\ast }{\phi }_{i}\left(x\right)$ ，其中， ${T}^{\ast }$$T$ 的共轭算子。

$\begin{array}{c}{\Psi }_{i}\left(x\right)=\left({T}^{*}{\phi }_{i}\right)\left(x\right)={〈\left({T}^{*}{\phi }_{i}\right)\left(t\right),K\left(t,x\right)〉}_{{W}^{n1}\left[a,b\right]}\\ ={〈{\phi }_{i}\left(t\right),TK\left(t,x\right)〉}_{{W}^{n1}\left[a,b\right]}=TK\left(t,x\right)\end{array}$ ,

${\Psi }_{i}\left(x\right)\in {W}^{n1}\left[a,b\right]$

${〈y\left(x\right),{\Psi }_{i}\left(x\right)〉}_{{W}^{n1}\left[a,b\right]}=0,\text{\hspace{0.17em}}i=1,2,\cdots$ ，其中 $y\left(x\right)\in {W}^{n1}\left[a,b\right]$ ，即得

${〈y\left(x\right),{T}^{*}{\phi }_{i}\left(x\right)〉}_{{W}^{n1}\left[a,b\right]}={〈Ty\left(x\right),{\phi }_{i}\left(x\right)〉}_{{W}^{n1}\left[a,b\right]}=Ty\left({x}_{i}\right)=0,\text{\hspace{0.17em}}i=1,2,\cdots$ .

${\left\{{x}_{i}\right\}}_{i=1}^{\infty }$$\left[a,b\right]$ 上稠密，故 $Ty\left(x\right)=0$ 。由 ${T}^{-1}$ 的存在性可知 $y\left(x\right)\equiv 0$ ，定理得证。

$y\left(x\right)=\underset{i=1}{\overset{\infty }{\sum }}\underset{k=1}{\overset{i}{\sum }}{\beta }_{ik}{\alpha }_{k}{\stackrel{¯}{\Psi }}_{k}\left(x\right)$ , (3)

$\begin{array}{c}y\left(x\right)=\underset{i=1}{\overset{\infty }{\sum }}\underset{k=1}{\overset{i}{\sum }}{〈y\left(x\right),{\stackrel{¯}{\Psi }}_{k}\left(x\right)〉}_{{W}^{n1}\left[a,b\right]}{\stackrel{¯}{\Psi }}_{k}\left(x\right)=\underset{i=1}{\overset{\infty }{\sum }}\underset{k=1}{\overset{i}{\sum }}{\beta }_{ik}{〈y\left(x\right),{\Psi }_{k}\left(x\right)〉}_{{W}^{n1}\left[a,b\right]}{\stackrel{¯}{\Psi }}_{k}\left(x\right)\\ =\underset{i=1}{\overset{\infty }{\sum }}\underset{k=1}{\overset{i}{\sum }}{\beta }_{ik}{〈y\left(x\right),{T}^{*}{\phi }_{k}\left(x\right)〉}_{{W}^{n1}\left[a,b\right]}{\stackrel{¯}{\Psi }}_{k}\left(x\right)=\underset{i=1}{\overset{\infty }{\sum }}\underset{k=1}{\overset{i}{\sum }}{\beta }_{ik}{〈Ty\left(x\right),{\phi }_{k}\left(x\right)〉}_{{W}^{n1}\left[a,b\right]}{\stackrel{¯}{\Psi }}_{k}\left(x\right)\\ =\underset{i=1}{\overset{\infty }{\sum }}\underset{k=1}{\overset{i}{\sum }}{\beta }_{ik}{〈Ty\left(x\right),R\left(x,{x}_{k}\right)〉}_{{W}^{n1}\left[a,b\right]}{\stackrel{¯}{\Psi }}_{k}\left(x\right)=\underset{i=1}{\overset{\infty }{\sum }}\underset{k=1}{\overset{i}{\sum }}{\beta }_{ik}Ty\left({x}_{k}\right){\stackrel{¯}{\Psi }}_{k}\left(x\right)\\ =\underset{i=1}{\overset{\infty }{\sum }}\underset{k=1}{\overset{i}{\sum }}{\beta }_{ik}F\left({x}_{k},y\left({x}_{k}\right),{y}^{\prime }\left({x}_{k}\right),\cdots ,{y}^{\left(n-1\right)}\left({x}_{k}\right)\right){\stackrel{¯}{\Psi }}_{k}\left(x\right)=\underset{i=1}{\overset{\infty }{\sum }}\underset{k=1}{\overset{i}{\sum }}{\beta }_{ik}{\alpha }_{k}{\stackrel{¯}{\Psi }}_{k}\left(x\right)\end{array}$ .

${y}_{n}\left(x\right)=\underset{i=1}{\overset{n}{\sum }}\underset{k=1}{\overset{i}{\sum }}{\beta }_{ik}{\alpha }_{k}{\stackrel{¯}{\Psi }}_{k}\left(x\right)$ ,(4)

${‖y\left(x\right)-{y}_{n}\left(x\right)‖}_{C}\to 0,\text{\hspace{0.17em}}{‖{y}^{\left(i\right)}\left(x\right)-{y}_{n}^{\left(i\right)}\left(x\right)‖}_{C}\to 0,\text{\hspace{0.17em}}n\to \infty ,\text{\hspace{0.17em}}i=1,2,\cdots ,n$ .

$\begin{array}{c}|y\left(x\right)-{y}_{n}\left(x\right)|=|{〈y\left({x}^{\prime }\right)-{y}_{n}\left({x}^{\prime }\right),K\left(x,{x}^{\prime }\right)〉}_{{W}^{n1}\left[a,b\right]}|\\ \le {‖y\left({x}^{\prime }\right)-{y}_{n}\left({x}^{\prime }\right)‖}_{{W}^{n1}\left[a,b\right]}{‖K\left(x,{x}^{\prime }\right)‖}_{{W}^{n1}\left[a,b\right]}\\ \le {\stackrel{˜}{C}}_{1}{‖y\left({x}^{\prime }\right)-{y}_{n}\left({x}^{\prime }\right)‖}_{{W}^{n1}\left[a,b\right]}\to 0\end{array}$ ,

$\begin{array}{c}|{y}^{\left(i\right)}\left(x\right)-{y}_{n}^{\left(i\right)}\left(x\right)|=|\frac{{\text{d}}^{i}}{\text{d}{x}^{i}}\left(y\left({x}^{\prime }\right)-{y}_{n}\left({x}^{\prime }\right)\right)|\\ =|{〈\frac{{\text{d}}^{i}}{\text{d}{x}^{i}}\left(y\left({x}^{\prime }\right)-{y}_{n}\left({x}^{\prime }\right)\right),K\left(x,{x}^{\prime }\right)〉}_{{W}^{n1}\left[a,b\right]}|\\ =|{〈y\left({x}^{\prime }\right)-{y}_{n}\left({x}^{\prime }\right),\frac{{\text{d}}^{i}}{\text{d}{x}^{i}}K\left(x,{x}^{\prime }\right)〉}_{{W}^{n1}\left[a,b\right]}|\\ \le {‖y\left({x}^{\prime }\right)-{y}_{n}\left({x}^{\prime }\right)‖}_{{W}^{n1}\left[a,b\right]}{‖\frac{{\text{d}}^{i}}{\text{d}{x}^{i}}K\left(x,{x}^{\prime }\right)‖}_{{W}^{n1}\left[a,b\right]}\\ \le {\stackrel{˜}{C}}_{2}{‖y\left({x}^{\prime }\right)-{y}_{n}\left({x}^{\prime }\right)‖}_{{W}^{n1}\left[a,b\right]}\to 0\end{array}$

4. 数值算法

$f\left({\alpha }_{1},{\alpha }_{2},\cdots ,{\alpha }_{n}\right)=\underset{i=1}{\overset{n}{\sum }}{\left[F\left({x}_{k},y\left({x}_{k}\right),{y}^{\prime }\left({x}_{k}\right),\cdots {y}^{\left(n-1\right)}\left({x}_{k}\right)\right)-{\alpha }_{k}\right]}^{2}$ ,

Step1：取初值 ${\alpha }_{k0}\left(k=1,2,\cdots ,n\right)$

Step2：计算 ${y}_{n0}\left(x\right)=\underset{i=1}{\overset{n}{\sum }}\underset{k=1}{\overset{i}{\sum }}{\beta }_{ik}{\alpha }_{n0}{\stackrel{¯}{\Psi }}_{k}\left(x\right)$

Step3：计算 $f\left({\alpha }_{10},{\alpha }_{20},\cdots ,{\alpha }_{n0}\right)$

Step4：如果 $f\left({\alpha }_{10},{\alpha }_{20},\cdots ,{\alpha }_{n0}\right)<\epsilon$ ( $\epsilon$ 是充分小的正数)，则计算终止；否则，计算

${\alpha }_{k1}=F\left({x}_{k},{y}_{n1}\left({x}_{k}\right),{{y}^{\prime }}_{n1}\left({x}_{k}\right),\cdots ,{y}_{n1}^{\left(n-1\right)}\left({x}_{k}\right)\right),\text{\hspace{0.17em}}k=1,2,\cdots ,n$ ;

Step5：计算 $f\left({\alpha }_{11},{\alpha }_{21},\cdots ,{\alpha }_{n1}\right)$

Step6：如果 $f\left({\alpha }_{11},{\alpha }_{21},\cdots ,{\alpha }_{n1}\right) ，则用 ${\alpha }_{k1}$ 代替 ${\alpha }_{k0}$ ，计算

${y}_{n1}\left(x\right)=\underset{i=1}{\overset{n}{\sum }}\underset{k=1}{\overset{i}{\sum }}{\beta }_{ik}{\alpha }_{n1}{\stackrel{¯}{\Psi }}_{k}\left(x\right)$ ,

${y}_{np}\left(x\right)=\underset{i=1}{\overset{n}{\sum }}\underset{k=1}{\overset{i}{\sum }}{\beta }_{ik}{\alpha }_{np}{\stackrel{¯}{\Psi }}_{k}\left(x\right)$ .(5)

5. 结论

Solving Higher Order Nonlinear Differential Equation with Nonlocal Boundary Value Problem[J]. 应用数学进展, 2017, 06(08): 1034-1038. http://dx.doi.org/10.12677/AAM.2017.68124

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