数学物理学报  2016, Vol. 36 Issue (5): 928-936   PDF    
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刘宏影
吕学琴
再生核结合配置法求解一类带有积分边值条件的四阶非线性微分方程
刘宏影, 吕学琴     
哈尔滨师范大学数学科学学院 哈尔滨 150025
摘要:该文主要研究了一类带有积分边值条件的四阶非线性微分方程的求解.利用再生核理论结合配置法来求解此类问题,并着重说明了用此种方法得到的解的一致收敛性.同时给出了算例说明了该方法的有效性.
关键词配置法     积分边值条件     四阶非线性微分方程     再生核空间    
Method for Solving a Class of Fourth Order Nonlinear Differential Equations with Integral Boundary Conditions by Combining Collocation Method and Reproducing Kernel Method
Liu Hongying, Lv Xueqin     
School of Mathematics and Sciences, Harbin Normal University, Harbin 150025
Abstract: In this paper, we study to solve a class of fourth order nonlinear differential equations with integral boundary conditions by combining collocation method and reproducing kernel method and especially illustrates the uniform convergence of the solution obtained by this method. Some numerical examples are displayed to demonstrate the validity and applicability of the present method.
Key words: Collocation method     Integral boundary conditions     Fourth order nonlinear differential equations     Reproducing kernel space    
1 引言

本文研究了一类带有非局部边界条件的微分方程数值求解,这类问题常应用于化学工程、热弹性学、周期轨道的提取、由一个无限域造成的不均匀性、非线性机械振荡器、疾病的预测、地下河流动和人口流动问题等,由于这些方面的重要性,国内外众多学者都在热衷于从事寻找这类微分方程数值求解的算法 (见文献 [1-5]). 因此,本文基于再生核空间数值分析理论,给出一类带有积分边值条件的微分方程的数值求解方法.文中主要研究了以下带有积分边值条件的四阶非线性微分方程的解

$\left\{ \begin{matrix} {{u}^{4}}(x)-\lambda f(x,u(x))=0,0<x<1,\\ u(0)=u(1)=\int_{0}^{1}{p}(x)u(x)dx,\\ {u}''(0)={u}''(1)=\int_{0}^{1}{q}(x){u}''(x)dx,\\ \end{matrix} \right.$ (1.1)

其中: $f:[0,1]\times [0,+ \infty)\rightarrow [0,+ \infty)$ 是连续函数,$\lambda $ 是非负实数,$p(x),q(x)\in C_{[0,1]}^{2}$.

方程 (1.1) 解的存在性在文献 [6-7] 中得到证明. 本文在求解的基础上,更加严密的证明了解的一致收敛性. 此类带有积分边界条件的微分方程求解问题,许多学者都进行了研究,如文献 [8]. 本文不但改进了再生核的求法,而且在证明解的一致收敛性的问题上,也有一定的突破. 许多学者在研究这类问题时都是在假设了解在范数有界的条件下,收敛性成立 (见文献 [8-9]). 本文完善了范数有界的这部分证明,从而让解的收敛性的证明更加严密.

2 预备知识
2.1 再生核空间$W_{2}^{5}[0,1]$
$\begin{align} & W_{2}^{5}[0,1]=\{u(x)|{{u}^{4}}(x)\text{ 是一个连续实值函数,}\ {{u}^{5}}(x)\in {{L}^{2}}[0,1],\\ & u(0)=u(1)=\int_{0}^{1}{p}(x)u(x)dx,{u}''(0)={u}''(1)=\int_{0}^{1}{q}(x){u}''(x)dx\}. \\ \end{align}$

$W_{2}^{5}[0,1]$的内积定义为

${{\langle u(x),v(x)\rangle }_{W_{2}^{5}}}=u(0)v(0)+{u}''(0){v}''(0)+{{u}^{(4)}}(0){{v}^{(4)}}(0)+\int_{0}^{1}{{{u}^{5}}}(x){{v}^{5}}(x)dx,$ (2.1)
$\|u{{\|}_{W_{2}^{5}}}=\sqrt{{{\langle u,u\rangle }_{W_{2}^{5}}}}.$

定理 2.1$W_{2}^{5}[0,1]$ 是一个完备的再生核空间,即存在 ${{R}_{x}}(t)\in W_{2}^{5}[0,1]$ 对任意 $u(t)\in W_{2}^{5}[0,1]$和每一个固定的 $x\in [0,1]$,使得对任何 $t\in [0,1]$,有 ${{\langle u(t),{{R}_{x}}(t)\rangle }_{W_{2}^{5}}}=u(x)$. 再生核函数可以被写作

${{R}_{x}}(t)=\left\{ \begin{matrix} \sum\limits_{i=1}^{10}{{{a}_{i}}}(x){{t}^{i-1}}+\frac{{{c}_{1}}{{t}^{11}}}{11!}+{{c}_{2}}{{e}^{t}},t\le x,\\ \sum\limits_{i=1}^{10}{{{b}_{i}}}(x){{t}^{i-1}}+\frac{{{c}_{1}}{{t}^{11}}}{11!}+{{c}_{2}}{{e}^{t}},t>x. \\ \end{matrix} \right.$ (2.2)

(1) $W_{2}^{5}[0,1]$ 空间的完备性和再生性在参考文献 [7] 中已得到证明.

(2)再生核空间 $W_{2}^{5}[0,1]$ 中的再生核函数表达式 ${{\text{R}}_{X}}\text{(t)}$ 求解如下.

对 (2.1)式 进行分部积分,得到

$\begin{align} & {{\langle u(t),{{R}_{x}}(t)\rangle }_{W_{2}^{5}}}=u(0){{R}_{x}}(0)+{{u}^{''}}(0)R_{X}^{''}(0)+{{u}^{(4)}}(0)R_{x}^{(4)}(0) \\ & -\sum\limits_{i=0}^{4}{{{(-1)}^{4-i}}}{{u}^{(i)}}(0)R_{x}^{(9-i)}(0)+\sum\limits_{i=0}^{4}{{{(-1)}^{4-i}}}{{u}^{(i)}}(1)R_{x}^{(9-i)}(1) \\ & -\int_{0}^{1}{u(t)R_{x}^{(10)}(t)dt.} \\ \end{align}$

因为$u(x)\in W_{2}^{5}[0,1]$,则

$\begin{align} & u(0)=u(1)=\int_{0}^{1}{p(t)u(t)dt},{{u}^{''}}(0)={{u}^{''}}(1)=\int_{0}^{1}{q(t){{u}^{''}}(t)dt,} \\ & {{\langle u(t),{{R}_{x}}(t)\rangle }_{W_{2}^{5}}}=u(0){{R}_{x}}(0)+{{u}^{''}}(0){{R}_{{{x}''}}}(0)+{{u}^{(4)}}(0)R_{x}^{(4)}(0)-\sum\limits_{i=0}^{4}{{{(-1)}^{4-i}}}{{u}^{(i)}}(0)R_{x}^{(9-i)}(0) \\ & +\sum\limits_{i=0}^{4}{{{(-1)}^{4-i}}}{{u}^{(i)}}(1)R_{x}^{9-i}(1)-\int_{0}^{1}{u(t)R_{x}^{(10)}(t)dt} \\ & +{{c}_{1}}[u(0)-\int_{0}^{1}{p(t)u(t)dt}]+{{c}_{2}}[{{u}^{''}}(0)-\int_{0}^{1}{q(t){{u}^{''}}(t)dt}] \\ & =u(0){{R}_{x}}(0)+{{u}^{''}}(0){{R}_{{{x}''}}}(0)+{{u}^{(4)}}(0)R_{x}^{(4)}(0)-\sum\limits_{i=0}^{4}{{{(-1)}^{4-i}}}{{u}^{(i)}}(0)R_{x}^{(9-i)}(0) \\ & +\sum\limits_{i=0}^{4}{{{(-1)}^{4-i}}}{{u}^{(i)}}(1)R_{x}^{(9-i)}(1)-\int_{0}^{1}{u(t)R_{x}^{(10)}(t)dt} \\ & +{{c}_{1}}[u(0)-\int_{0}^{1}{p(t)u(t)dt}]+{{c}_{2}}[{{u}^{''}}(0)-{{u}^{'}}(1)q(1) \\ & +{u}'(0)q(0)+{q}'(1)u(1)-{q}'(0)u(0)-\int_{0}^{1}{{{q}^{''}}(t)u(t)dt}]. \\ \end{align}$

根据再生性质,即 ${{\langle u(t),{{R}_{x}}(t)\rangle }_{W_{2}^{5}}}=u(x)$,容易知道${{\text{R}}_{x}}\text{(t) }$是以下微分方程的解

$\left\{ \begin{matrix} -(R_{x}^{(10)}(t)+{{c}_{1}}p(t)+{{c}_{2}}{{q}^{''}}(t))=\delta (t-x),\\ {{R}_{x}}(0)-R_{x}^{(9)}(0)-{{c}_{1}}+R_{x}^{(9)}(1)+{{c}_{2}}{{q}^{'}}(0)-{{c}_{2}}{{q}^{'}}(1)=0,\\ R_{x}^{(8)}(0)-{{c}_{2}}q(0)=0,\\ {{R}_{{{x}^{''}}}}(0)-R_{x}^{(7)}(0)-{{c}_{2}}+R_{x}^{(7)}(1)=0,\\ R_{x}^{(6)}(0)=0,\\ R_{x}^{(4)}(0)-R_{x}^{(5)}(0)=0,\\ R_{x}^{(8)}(1)-{{c}_{2}}q(1)=0,\\ R_{x}^{(6)}(1)=0,\\ R_{x}^{(5)}(1)=0. \\ \end{matrix} \right.$ (2.3)

当 $t\ne x$,容易证明 ${{\text{R}}_{x}}\text{(t) }$ 是下面十阶微分方程的解

$-[R_{x}^{(10)}(t)+{{c}_{1}}p(t)+{{c}_{2}}{{q}^{''}}(t)]=0,$ (2.4)

边界条件为

$\left\{ \begin{matrix} {{R}_{x}}(0)-R_{x}^{(9)}(0)-{{c}_{1}}+R_{x}^{(9)}(1)+{{c}_{2}}{{q}^{'}}(0)-{{c}_{2}}{{q}^{'}}(1)=0,\\ R_{x}^{(8)}(0)-{{c}_{2}}q(0)=0,\\ {{R}_{{{x}^{''}}}}(0)-R_{x}^{(7)}(0)-{{c}_{2}}+R_{x}^{(7)}(1)=0,\\ R_{x}^{(6)}(0)=0,\\ R_{x}^{(4)}(0)-R_{x}^{(5)}(0)=0,\\ R_{x}^{(8)}(1)-{{c}_{2}}q(1)=0,\\ R_{x}^{(6)}(1)=0,\\ R_{x}^{(5)}(1)=0. \\ \end{matrix} \right.$ (2.5)

容易知道 (2.4)式 的特征方程为 ${{\lambda }^{10}}=0$,具有 10 重特征根$\lambda =0$,所以 (2.4)式 的通解为

${{\text{R}}_{x}}\text{(t)}=\left\{ \begin{matrix} \sum\limits_{i=1}^{10}{{{a}_{i}}}{{t}^{i-1}}+\frac{{{c}_{1}}{{t}^{11}}}{11!}+{{c}_{2}}{{e}^{t}},& t\le x,\\ \sum\limits_{i=1}^{10}{{{b}_{i}}}{{t}^{i-1}}+\frac{{{c}_{1}}{{t}^{11}}}{11!}+{{c}_{2}}{{e}^{t}},& t>x. \\ \end{matrix} \right.$ (2.6)

因为

$-[R_{x}^{(10)}(t)+{{c}_{1}}p(t)+{{c}_{2}}{{q}^{''}}(t)]=\delta (t-x),$ (2.7)

所以

$\begin{align} & R_{x}^{(k)}(t+0)=R_{x}^{(k)}(t-0),k=0,1,2\cdots 8,\\ & R_{x}^{(9)}(t+0)-R_{x}^{(9)}(t-0)=1,\\ \end{align}$ (2.8)

又因为 ${{R}_{x}}(y)\in W_{2}^{5}[0,1]$,故

$\left\{ \begin{array}{*{35}{l}} {{R}_{x}}(0)={{R}_{x}}(1),\\ {{R}_{x}}(0)=\int_{0}^{1}{p(t){{\text{R}}_{x}}\text{(t) }dt,} \\ {{R}_{{{x}^{''}}}}(0)={{R}_{{{x}^{''}}}}(1),\\ {{R}_{{{x}^{''}}}}(0)=\int_{0}^{1}{q(t){{R}_{{{x}^{''}}}}(t)dt}. \\ \end{array} \right.$ (2.9)

不难看出,(2.5),(2.7),(2.8)和 (2.9)式 提供了 22 个方程,这样容易计算出 ${{a}_{i}},{{b}_{i}}(i=1,2,\cdots )$ 和 c1,c2,这样便得到了再生核函数表达式 ${{\text{R}}_{x}}\text{(t) }$.

2.2 再生核空间 $W_{2}^{1}[0,1]$

再生核空间 $W_{2}^{1}[0,1]$ 的构造在文献 [10] 中给出.且其再生核函数 Qx(y) 的表达式为

${{Q}_{x}}(y)=\left\{ \begin{array}{*{35}{l}} 1+x,& x\le y,\\ 1+y,& x>y. \\ \end{array} \right.$ (2.10)
3 对方程 (1.1) 的分析

引进一个线性算子$L:W_{2}^{5}[0,1]\to W_{2}^{1}[0,1]$,令 $L(u(x))={{u}^{(4)}}(x)$,则很明显的知道 L 是一个有界线性算子,则方程 (1.1) 等价转化为

$\left\{ \begin{array}{*{35}{l}} Lu=\lambda f(x,u(x)),0<x<1,\\ u(0)=u(1)=\int_{0}^{1}{p}(x)u(x)dx,\\ {{u}^{''}}(0)={{u}^{''}}(1)=\int_{0}^{1}{q}(x){{u}^{''}}(x)dx,\\ \end{array} \right.$ (3.1)

其中 $u(x)\in W_{2}^{5}[0,1],\lambda f(x,u(x))\in W_{2}^{1}[0,1]$.

令 $\{{{x}_{i}}\}_{i=1}^{\infty }$ 是 [0,1] 上的稠密点集,且 ${{\varphi }_{i}}(y)={{Q}_{{{x}_{i}}}}(y)$. 再令 ${{\psi }_{i}}(x)={{L}^{*}}{{\varphi }_{i}}(x)$ 其中 L* 是 L 的共轭算子. 则 $\{{{\psi }_{i}}(x)\}_{i=1}^{\infty }$ 是$W_{2}^{5}[0,1]$中的完全系.

对完全系 $\{{{\psi }_{i}}(x)\}_{i=1}^{\infty }$ 进行 Gram-Schmidt 正交化,得到再生核空间$W_{2}^{5}[0,1]$的一组标准正交基

$\overline{{{\psi }_{i}}}(x)=\sum\limits_{k=1}^{i}{{{\beta }_{ik}}}{{\psi }_{k}}(x),$ (3.2)

其中 ${{\beta }_{ik}}$ 是正交化系数,${{\beta }_{ii}}>0,i=1,2\cdots $.

利用配置法可以将非线性方程 (3.1) 的解设为

${{u}_{n}}(x)=\sum\limits_{i=1}^{n}{{{a}_{i}}}\overline{{{\psi }_{i}}}(x),$ (3.3)

其中 ${{a}_{i}}(i=1,2\cdots )$ 为待定的系数.

对 (3.3) 式左右两边同时用 $\overline{{{\psi }_{i}}}(x)$ 做内积

$\begin{align} & \langle {{u}_{n}}(x),\overline{{{\psi }_{i}}}(x)\rangle =\langle {{u}_{n}}(x),\sum\limits_{k=1}^{i}{{{\beta }_{ik}}}{{\psi }_{k}}(x)\rangle \\ & =\sum\limits_{k=1}^{i}{{{\beta }_{ik}}}{{\langle {{u}_{n}}(x),{{L}^{*}}{{\varphi }_{k}}(x)\rangle }_{W_{2}^{5}}} \\ & =\sum\limits_{k=1}^{i}{{{\beta }_{ik}}}{{\langle L{{u}_{n}}(x),{{\varphi }_{k}}(x)\rangle }_{W_{2}^{1}}} \\ & =\sum\limits_{k=1}^{i}{{{\beta }_{ik}}}{{\langle \lambda f(x,{{u}_{n}}(x)),{{\varphi }_{k}}(x)\rangle }_{W_{2}^{1}}} \\ & =\sum\limits_{k=1}^{i}{{{\beta }_{ik}}}\lambda f({{x}_{k}},{{u}_{n}}({{x}_{k}})). \\ & \langle \sum\limits_{i=1}^{n}{{{a}_{i}}}\overline{{{\psi }_{i}}}(x),\overline{{{\psi }_{i}}}(x)\rangle =\langle {{a}_{1}}\overline{{{\psi }_{1}}}(x)+{{a}_{2}}\overline{{{\psi }_{2}}}(x)+\cdots +{{a}_{n}}\overline{{{\psi }_{n}}}(x),\overline{{{\psi }_{i}}}(x)\rangle ={{a}_{i}}. \\ \end{align}$

进而

${{a}_{i}}=\sum\limits_{k=1}^{i}{{{\beta }_{ik}}}\lambda f({{x}_{k}},{{u}_{n}}({{x}_{k}})).$

所以

${{u}_{n}}(x)=\sum\limits_{i=1}^{n}{\sum\limits_{k=1}^{i}{{{\beta }_{ik}}}}\lambda f({{x}_{k}},{{u}_{n}}({{x}_{k}}))\overline{{{\psi }_{i}}}(x).$ (3.4)

下面构造一个迭代格式

$\left\{ \begin{array}{*{35}{l}} {{u}_{n}}(x)=\sum\limits_{i=1}^{n}{{{A}_{i}}}\overline{{{\psi }_{i}}}(x),\\ {{A}_{i}}=\sum\limits_{k=1}^{i}{{{\beta }_{ik}}}\lambda f({{x}_{k}},{{u}_{n-1}}({{x}_{k}}))(i=1,2,\cdots ,n),\\ \end{array} \right.$ (3.5)

其中 u0(x)=0.

引理 3.1 (3.2) 式中的正交化系数 ${{\beta }_{ik}}$ 满足 $\sum\limits_{k=1}^{\infty }{\sum\limits_{i=k}^{\infty }{\beta _{ik}^{2}}}<+\infty $ 成立.

设 $\hat{u}(x)\in W_{2}^{5}[0,1]$ ,且 $\hat{u}(x)$ 是关于 $x\in [0,1]$ 的连续函数. 在 $W_{2}^{5}[0,1]$ 中解微分方程

$\left\{ \begin{array}{*{35}{l}} L\hat{u}(x)=1,\\ \hat{u}(0)=\hat{u}(1),\\ {\hat{u}}''(0)={\hat{u}}''(1),\\ \end{array} \right.$

$\begin{align} & \hat{u}(x)=\sum\limits_{i=1}^{\infty }{\langle \hat{u}(}x),\overline{{{\psi }_{i}}}(x)\rangle \overline{{{\psi }_{i}}}(x)=\sum\limits_{i=1}^{\infty }{\sum\limits_{k=1}^{i}{{{\beta }_{ik}}}}\langle \hat{u}(x),{{L}^{*}}{{\varphi }_{k}}(x)\rangle \overline{{{\psi }_{i}}}(x) \\ & =\sum\limits_{i=1}^{\infty }{\sum\limits_{k=1}^{i}{{{\beta }_{ik}}}}\langle L\hat{u}(x),{{\varphi }_{k}}(x)\rangle \overline{{{\psi }_{i}}}(x)=\sum\limits_{i=1}^{\infty }{\sum\limits_{k=1}^{i}{{{\beta }_{ik}}}}L\hat{u}({{x}_{k}})\overline{{{\psi }_{i}}}(x)= \\ & \sum\limits_{i=1}^{\infty }{\sum\limits_{k=1}^{i}{{{\beta }_{ik}}}}\overline{{{\psi }_{i}}}(x)=\sum\limits_{k=1}^{\infty }{\sum\limits_{i=k}^{\infty }{{{\beta }_{ik}}}}\overline{{{\psi }_{i}}}(x),\\ \end{align}$

所以有

$\begin{align} & {{\left\| {\hat{u}} \right\|}^{2}}=\langle \hat{u}(x),\hat{u}(x)\rangle =\langle \sum\limits_{k=1}^{\infty }{\sum\limits_{i=k}^{\infty }{{{\beta }_{ik}}}}\overline{{{\psi }_{i}}}(x),\sum\limits_{k=1}^{\infty }{\sum\limits_{i=k}^{\infty }{{{\beta }_{ik}}}}\overline{{{\psi }_{i}}}(x)\rangle \\ & =\sum\limits_{k,{k}'=1}^{\infty }{\langle }\sum\limits_{i=k}^{\infty }{{{\beta }_{ik}}}\overline{{{\psi }_{i}}}(x),\sum\limits_{i={k}'}^{\infty }{{{\beta }_{i{k}'}}}\overline{{{\psi }_{i}}}(x)\rangle ,\\ \end{align}$

$\begin{align} & \sum\limits_{k,{k}'=1}^{\infty }{\langle }\sum\limits_{i=k}^{\infty }{{{\beta }_{ik}}}\overline{{{\psi }_{i}}}(x),\sum\limits_{i={k}'}^{\infty }{{{\beta }_{i{k}'}}}\overline{{{\psi }_{i}}}(x)\rangle \\ & \ge \sum\limits_{k,{k}'=1,k\le {k}'}^{\infty }{\langle }\sum\limits_{i=k}^{\infty }{{{\beta }_{ik}}}\overline{{{\psi }_{i}}}(x),\sum\limits_{i={k}'}^{\infty }{{{\beta }_{i{k}'}}}\overline{{{\psi }_{i}}}(x)\rangle \\ & =\sum\limits_{k=1}^{\infty }{\langle }\sum\limits_{i=k}^{\infty }{{{\beta }_{ik}}}\overline{{{\psi }_{i}}}(x),\sum\limits_{i=k}^{\infty }{{{\beta }_{ik}}}\overline{{{\psi }_{i}}}(x)\rangle \\ & =\sum\limits_{k=1}^{\infty }{\sum\limits_{i,{i}'=k}^{\infty }{\langle {{\beta }_{ik}}\overline{{{\psi }_{i}}}(x),{{\beta }_{{i}'k}}\overline{{{\psi }_{{{i}'}}}}(x)\rangle }}=\sum\limits_{k=1}^{\infty }{\sum\limits_{i=k}^{\infty }{\beta _{ik}^{2}}},\\ \end{align}$

因此有

$\sum\limits_{k=1}^{\infty }{\sum\limits_{i=k}^{\infty }{\beta _{ik}^{2}}}\le \|\hat{u}{{\|}^{2}}$. 即 $\sum\limits_{k=1}^{\infty }{\sum\limits_{i=k}^{\infty }{\beta _{ik}^{2}}}<+\infty $.

引理 3.2 设 $\lambda f(x,u(x))$ 是有界函数,也是关于 $x\in [0,1]$ 的连续函数,则 $\left\| {{u}_{n}} \right\|\le M$.

因为

$\begin{align} & {{u}_{n}}(x)=\sum\limits_{i=1}^{n}{\sum\limits_{k=1}^{i}{{{\beta }_{ik}}}}\lambda f({{x}_{k}},{{u}_{n-1}}({{x}_{k}}))\overline{{{\psi }_{i}}}(x) \\ & =\sum\limits_{k=1}^{n}{\sum\limits_{i=k}^{n}{{{\beta }_{ik}}}}\overline{{{\psi }_{i}}}(x)\lambda f({{x}_{k}},{{u}_{n-1}}({{x}_{k}})),\\ \end{align}$

$\begin{align} & {{\left\| {{u}_{n}} \right\|}^{2}}=\langle {{u}_{n}}(x),{{u}_{n}}(x)\rangle \\ & =\langle \sum\limits_{k=1}^{n}{\sum\limits_{i=k}^{n}{{{\beta }_{ik}}}}\overline{{{\psi }_{i}}}(x)\lambda f({{x}_{k}},{{u}_{n-1}}({{x}_{k}})),\sum\limits_{k=1}^{n}{\sum\limits_{i=k}^{n}{{{\beta }_{ik}}}}\overline{{{\psi }_{i}}}(x)\lambda f({{x}_{k}},{{u}_{n-1}}({{x}_{k}}))\rangle \\ & =\sum\limits_{k,{k}'=1}^{n}{\langle }\sum\limits_{i=k}^{n}{{{\beta }_{ik}}}\overline{{{\psi }_{i}}}(x),\sum\limits_{i={k}'}^{n}{{{\beta }_{i{k}'}}}\overline{{{\psi }_{i}}}(x)\rangle \lambda f({{x}_{k}},{{u}_{n-1}}({{x}_{k}}))\lambda f({{x}_{{{k}'}}},{{u}_{n-1}}({{x}_{{{k}'}}})) \\ & =\sum\limits_{k=1}^{n}{\langle }\sum\limits_{i=k}^{n}{{{\beta }_{ik}}}\overline{{{\psi }_{i}}}(x),\sum\limits_{i=k}^{n}{{{\beta }_{ik}}}\overline{{{\psi }_{i}}}(x)\rangle {{[\lambda f({{x}_{k}},{{u}_{n-1}}({{x}_{k}}))]}^{2}} \\ & =\sum\limits_{k=1}^{n}{\langle }\sum\limits_{i,{i}'=k}^{n}{{{\beta }_{ik}}}\overline{{{\psi }_{i}}}(x),{{\beta }_{{i}'k}}\overline{{{\psi }_{i}}}(x)\rangle {{[\lambda f({{x}_{k}},{{u}_{n-1}}({{x}_{k}}))]}^{2}} \\ & =\sum\limits_{k=1}^{n}{\sum\limits_{i=k}^{n}{\beta _{ik}^{2}}}{{[\lambda f({{x}_{k}},{{u}_{n-1}}({{x}_{k}}))]}^{2}}. \\ \end{align}$

所以由引理 3.1 我们有 ${{\left\| {{u}_{n}} \right\|}^{2}}\le M$,即 $\left\| {{u}_{n}} \right\|\le \sqrt{M}$.

定理 3.1 (收敛性) 设 $\{{{x}_{i}}\}_{i=1}^{\infty }$是 [0,1] 上的密集点集,且 $f(x,u(x))\in W_{2}^{1}[0,1],u(x)\in W_{2}^{5}[0,1]$,则由 (3.4)式得到的 un(x) 一致收敛于方程 (3.1) 的真解 u(x),且 $u(x)=\sum\limits_{i=1}^{n}{{{A}_{i}}}\overline{{{\psi }_{i}}}(x)$.

首先,我们将证明 un(x) 的收敛性.由 (3.5) 式有 ${{u}_{n}}(x)={{u}_{n-1}}(x)+{{A}_{n}}\overline{{{\psi }_{n}}}(x)$,因为 $\{\bar{\psi }(x)\}_{i=1}^{\infty }$ 是一组标准正交基,则${{\left\| {{u}_{n}} \right\|}^{2}}={{\left\| {{u}_{n-1}} \right\|}^{2}}+{{({{A}_{n}})}^{2}}={{\left\| {{u}_{n-2}} \right\|}^{2}}+{{({{A}_{n-1}})}^{2}}+{{({{A}_{n}})}^{2}}=\cdots ={{\left\| {{u}_{0}} \right\|}^{2}}+\sum\limits_{i=1}^{n}{{{({{A}_{i}})}^{2}}}$. 由 $\left\| {{u}_{n}} \right\|$ 的有界性,有 $\sum\limits_{i=1}^{\infty }{{{({{A}_{i}})}^{2}}}<\infty $,即 $\{{{A}_{i}}\}\in {{l}^{2}}(i=1,2\cdots )$. 不妨令 m>n,则 $({{u}_{m}}-{{u}_{m-1}})\bot ({{u}_{m-1}}-{{u}_{m-2}})\bot \cdots \bot ({{u}_{n}}-{{u}_{n-1}})$,故

$\begin{align} & \left\| {{u}_{m}}-{{u}_{n}} \right\|_{w_{2}^{5}}^{2}=\left\| {{u}_{m}}-{{u}_{m-1}}+{{u}_{m-1}}-{{u}_{m-2}}+\cdots +{{u}_{n}}-{{u}_{n-1}} \right\|_{w_{2}^{5}}^{2} \\ & \le {{\left\| {{u}_{m}}-{{u}_{m-1}} \right\|}^{2}}+\cdots +{{\left\| {{u}_{n}}-{{u}_{n-1}} \right\|}^{2}}=\sum\limits_{i=n+1}^{m}{{{({{A}_{i}})}^{2}}}\to 0. \\ \end{align}$

这样由空间 $W_{2}^{5}[0,1]$ 的完备性,存在$u(x)\in W_{2}^{5}[0,1]$,下式成立

${{u}_{n}}(x)\xrightarrow{\|\cdot {{\|}_{W_{2}^{5}}}}u(x),n\to \infty .$

其次证明 u(x) 即为 方程(3.1) 的解. 此部分定理证明见文献 [11].

定理 3.2假设 u(x) 是方程 (3.1) 的解,令 ${{\varepsilon }_{n}}$ 是近似解 un(x) 和真解 u(x) 间的误差,则 ${{\varepsilon }_{n}}$ 在 $\|.{{\|}_{W_{2}^{5}}}$ 下单调递减.

由 (3.4)式,得

$\begin{align} & {{\left\| {{\varepsilon }_{n}} \right\|}_{W_{2}^{5}}}={{\left\| u(x)-{{u}_{n}}(x) \right\|}_{W_{2}^{5}}}={{\left\| \sum\limits_{i=n+1}^{\infty }{\sum\limits_{k=1}^{i}{{{\beta }_{ik}}}}\lambda f({{x}_{k}},{{u}_{n}}({{x}_{k}}))\overline{{{\psi }_{i}}}(x) \right\|}_{W_{2}^{1}}} \\ & =\sum\limits_{i=n+1}^{\infty }{[}\sum\limits_{k=1}^{i}{{{\beta }_{ik}}}\lambda f({{x}_{k}},{{u}_{n}}({{x}_{k}})){{]}^{2}}\le \sum\limits_{i=n}^{\infty }{[}\sum\limits_{k=1}^{i}{{{\beta }_{ik}}}\lambda f({{x}_{k}},{{u}_{n}}({{x}_{k}})){{]}^{2}} \\ & ={{\left\| {{\varepsilon }_{n-1}} \right\|}_{W_{2}^{5}}},\\ \end{align}$

这表明误差 ${{\varepsilon }_{n}}$ 在 $\|.{{\|}_{W_{2}^{5}}}$ 下单调递减.

4 数值算例

在这部分内容中,所有数值结果均通过 Mathematica 5.0 计算所得.

例1 考虑下面的带有积分边值条件的四阶非线性微分方程问题

$\left\{ \begin{array}{*{35}{l}} {{u}^{(4)}}(x)={{u}^{2}}(x)+f(x),0<x<1,\\ u(0)=u(1)=\int_{0}^{1}{xu(x)dx},\\ {u}''(0)={u}''(1)=\int_{0}^{1}{{{e}^{x}}{{u}^{''}}(x)dx},\\ \end{array} \right.$

其精确解为

$\begin{align} & u(x)=-\frac{17-6e}{5(-2+e)}-\frac{(20-7e)x}{-2+e}-\frac{6(-3+e){{x}^{2}}}{-2+e}-2{{x}^{3}}+{{x}^{4}}, \\ & f(x)=24-{{(-\frac{17-6e}{5(-2+e)}-\frac{(20-7e)x}{-2+e}-\frac{6(-3+e){{x}^{2}}}{-2+e}-2{{x}^{3}}+{{x}^{4}})}^{2}}. \\ \end{align}$

数值计算结果见表 4.1和 表 4.2.

表 4.1 例 1 精确解和近似解的数值计算结果

表 4.2 例 1 各阶导数均方差
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