在再生核空间中,解决如下分片光滑边值问题

(1.1) $ \begin{eqnarray} \left\{ \begin{array}{l} u''(x)+a_1(x)u'(x)+a_0(x)u(x) = f(x), {\quad} x\in[0, 1], \\ u(0) = r_0, u(1) = r_1, \end{array} \right. \end{eqnarray} $

其中$ a_1(x), a_0(x), f(x) $在区间$ [0, 1] $上点$ x_1, x_2, \cdots, x_n $不连续.

微分方程边值问题在物理、生物、控制等科学及工程领域有着极其广泛的应用^[1-3].对此类问题的研究也备受关注,近年来,很多方法被提出用于解决此类问题^[4-8].尤其,利用再生核方法(RKM)求解微分方程初始边值问题的研究越来越多^{[9-13, 15-18]}.再生核方法的优点是理论严谨,易于实现,且在建立再生核空间后,可将边界条件吸收到空间中,这使得在后面的算法中不必再考虑边界条件,从而简化了这类边值问题的求解.文献[11, 19, 21]讨论了该算法的一致收敛性.文献[20]将三次样条理论与再生核理论相结合,利用再生核函数巧妙地构造了三次样条函数空间的一组基底,基于三次样条插值的高收敛的特点,得到了微分方程边值问题近似解的一种新的求解方法. Zhao等^[21]提出了再生核方法求得的近似解具有不低于二阶的收敛性.Xu等^[22]建立了衔接再生核空间,并将其应用于求解一维椭圆边界问题.下面,将构造再生核空间来解决这类具有分段光滑或不连续系数的微分方程边值问题.

文章的结构安排如下:第2节回顾了关于RKM的一些重要定义,第3节引入RKM方法求解分片光滑微分方程边值问题并对收敛性进行了分析,在第4节中将此方法应用于一些数值算例,最后是本文的结论及致谢.

首先,提出后面算法中要将要用到的一些定义和定理.

定义2.1 再生核空间$ W_{2}^{3}[0, 1] $

$ W_{2}^{3}[0, 1] = \{u(t)|u''(t)\ \hbox{在$[0, 1]$上绝对连续, }\ u'''(t)\in L^2[0, 1] \}, $

其内积定义为

$ \left\{ \begin{array}{ll} { } \langle u, v\rangle_{_{W_{2}^{3}[0, 1]}} = \sum\limits_{i = 0}^{2}u^{(i)}(0)v^{(i)}(0)+\int_{0}^{1}u^{'''}v^{'''}{\rm d}t, u, v\in W_{2}^{3}[0, 1], \\ \left\| u \right\|_{W_{2}^{3}[0, 1]} = \sqrt{\langle u, u\rangle_{_{W_{2}^{3}}}}. \end{array} \right. $

记$ R_{x}(y) $为空间$ W_{2}^{3}[0, 1] $的再生核函数.由再生核函数的定义,可得$ R_{x}(y) $表达式^[23]

$ R_{x}(y) = \left\{ \begin{array}{ll} { } 1+\frac{x^5}{120}+(x-\frac{x^4}{24})y+\frac{1}{2}(\frac{x^2}{2}+\frac{x^3}{6})y^2, &y\geq x, \\ { } 1+\frac{y^5}{120}+(y-\frac{y^4}{24})x+\frac{1}{2}(\frac{y^2}{2}+\frac{y^3}{6})x^2, &y<x. \end{array} \right. $

定义线性算子$ {\cal L}: W_{2}^{3}[0, 1]\rightarrow L^2[0, 1] $

$ {\cal L}u(x) = u''(x)+a_1(x)u'(x)+a_0(x)u(x). $

相应地,方程(1.1)等价为

(2.1) $ \begin{eqnarray} \left\{ \begin{array}{ll} {\cal L}u(x) = f(x), x\in[0, 1], \\ u(0) = r_0, u(1) = r_1. \end{array} \right. \end{eqnarray} $

注2.1  算子$ {\cal L} $的像空间不再需要满足连续的条件.而在以前的研究中,这个条件是必需的.

定理2.1 如果$ a_0(x), a_1(x) $在$ [0, 1] $上有界,则$ {\cal L} $是线性有界算子.

证  显然, $ {\cal L} $是线性算子.

对任意$ u\in W_{2}^{3}[0, 1] $,由$ W_{2}^{3}[0, 1] $、$ L^2[0, 1] $内积的定义及$ a_0(x), a_1(x) $的有界性,可得

$ \begin{eqnarray*} \left\| {\cal L}u \right\|_{L^2}^{2}& = &\int_{0}^{1}({\cal L}u)^{2} {\rm d}x\\ & = & \int_{0}^{1}\left[ u''+a_1(x)u'+a_0(x)u\right]^{2} {\rm d}x\\ &\leq& \int_{0}^{1}\left[|u''|^2+M_1|u'|^2+M_2|u|^2+M_3|u'||u''|+M_4|u||u''|+M_5|u||u'|\right]{\rm d}x. \end{eqnarray*} $

其中$ M_i(1\leq i\leq 5) $是常数.由于

$ \int_{0}^{1}|u^{(i)}||u^{(j)}|{\rm d}x\leq \sqrt{\int_{0}^{1}(|u^{(i)}|)^2{\rm d}x\int_{0}^{1}(|u^{(j)}|)^2{\rm d}x}{\quad} (i, j = 0, 1, 2). $

$ \int_{0}^{1}|u^{(i)}||u^{(j)}|{\rm d}x\leq C_{ij} \int_{0}^{1}\sum\limits_{i = 0}^{2}(u^{(i)})^2{\rm d}x, $

由再生核的再生性,可得

$ u(x) = \langle u, R_{x}\rangle_{W^{3}_{2}}, \quad u^{(i)}(x) = \langle u, {\partial} ^{i}_{x}R_{x}\rangle_{W^{3}_{2}}, \quad 0\leq i\leq 2, $

$ |u^{(i)}(x)| = |\langle u, {\partial}^{i}_{x}R_{x}\rangle_{W^{3}_{2}}|\leq \|u\|_{W^{3}_{2}} \| {\partial} ^{i}_{x}R_{x}\|_{W^{3}_{2}}\leq D_{i}\|u\|_{W^{3}_{2}} , $

从而

$ \int_0^1 \sum\limits_{i = 0}^2 (u^{(i)})^2 {\rm d}x\leq \int_0^1 \sum\limits_{i = 0}^{2}D_{i}^2 {\|u\|}_{W^{3}_{2}}^2 {\rm d}x+\int_0^1 (u''')^2{\rm d}x\leq C{\|u\|}_{W^{3}_{2}}^2. $

因此,有如下不等式

$ \left\| {\cal L}u \right\|_{L^2}\leq M\left\| u \right\|_{W_2^{3}}, $

其中$ C_{ij}, D_{i}, M $是常数.即证得$ {\cal L} $是线性有界算子.

取$ [0, 1] $上的稠密点集$ \{x_{i}\}_{1}^{\infty} $,令

$ \psi_{i}(x) = {\cal L}R_{x}(x_{i}), \varphi_{0}(x) = R_{x}(0), \varphi_{1}(x) = R_{x}(1). $

由再生核$ R_{x}(y) $的再生性,可得如下定理.

定理2.2 对于任意$ i $, $ \psi_{i}(x) = {\cal L}R_{x}(x_{i})\in W_{2}^{3}[0, 1] $.即: $ {\psi}''_{i}(x) $在$ [0, 1] $上绝对连续, $ {\psi}'''_{i}(x)\in L^2[0, 1] $.

证

$ \psi_{i}(x) = {\cal L}R_{x}(x_{i}) = \frac{\partial^{2}}{\partial x_{i}^{2}}R_{x}(x_i)+a_1(x_i)\frac{\partial}{\partial x_{i}}R_{x}(x_i)+a_0(x_i)R_{x}(x_i). $

对每个固定的$ i $和变量$ x $,由算子$ {\cal L} $的连续性,可得

$ {\psi}''_{i}(x) = {\cal L}\frac{\partial^{2}}{\partial x^{2}}R_{x}(x_{i}) = \frac{\partial^{4}}{\partial x_{i}^{2}\partial x^{2}}R_{x}(x_i)+a_1(x_i)\frac{\partial^{3}}{\partial x_{i}\partial x^{2}}R_{x}(x_i)+a_0(x_i)\frac{\partial^{2}}{\partial x^{2}}R_{x}(x_{i}). $

$ {\psi}'''_{i}(x) = {\cal L}\frac{\partial^{3}}{\partial x^{3}}R_{x}(x_{i}) = \frac{\partial^{5}}{\partial x_{i}^{2}\partial x^{3}}R_{x}(x_i)+a_1(x_i)\frac{\partial^{4}}{\partial x_{i}\partial x^{3}}R_{x}(x_i)+a_0(x_i)\frac{\partial^{3}}{\partial x^{3}}R_{x}(x_{i}). $

由再生核$ R_{x}(y) $的性质^[23],可得出:当$ 0\leq j+k \leq4 $时, $ \frac{\partial^{j+k}}{\partial x^{j}\partial y^{k}}R_{x}(y) $是关于$ x $和$ y $的绝对连续函数;当$ j+k = 5 $时, $ \frac{\partial^{j+k}}{\partial x^{j}\partial y^{k}}R_{x}(y)\in L^2[0, 1] $.因此,对任意$ i $, $ {\psi}''_{i}(x) $是$ [0, 1] $上的绝对连续函数, $ {\psi}'''_{i}(x)\in L^2[0, 1] $.

注2.2  $ [0, 1] $上的稠密子集$ \{x_{i}\}_{1}^{\infty} $可任意选择,不管是否包含不连续点.

假设方程(1.1)的解存在且唯一,在此条件下,有如下定理成立.

定理3.1  $ \{\psi_{i}\}_{1}^{\infty} $是$ W_2^{3}[0, 1] $中的完全系.

证 对任意$ u\in W_2^{3}[0, 1] $,如果$ \langle u(x), \psi_{i}(x)\rangle_{W_2^3} = 0 $,需证$ u(x) = 0 $.

$ 0 = \langle u(x), \psi_{i}(x)\rangle_{W_2^3} = \langle u(x), {\cal L}R_{x}(x_i)\rangle_{W_2^3} = {{\cal L}}_{x_i}\langle u(x), R_{x}(x_i)\rangle_{W_2^3} = {\cal L}u(x_i). $

考虑到$ {\cal L}u $分片光滑及$ \{x_{i}\}_{1}^{\infty} $的稠密性,可得

$ {\cal L}u = 0. $

由于$ {\cal L} $是可逆算子,从而有$ u(x) = 0 $.

令

$ S_{n+2} = {\rm span}\{\{\psi_{i}\}_{1}^{n}\bigcup \{\varphi_{0}, \varphi_{1}\}\}\in W_2^{3}[0, 1]. $

记$ {{\cal P}}_{n} $为从$ W_2^3 $到$ S_{n+2} $的映射算子.

定理3.2 对每个固定的$ n $, $ \{\psi_{i}\}_{1}^{n}\bigcup \{\varphi_{0}, \varphi_{1}\} $在$ {W_2^3}[0, 1] $中是线性无关的.

定理3.3 如果$ u(x) $是方程(2.1)的解,那么$ u_n\triangleq{{\cal P}}_{n}u $满足如下方程

(3.1) $ \begin{eqnarray} \left\{ \begin{array}{ll} \langle v, \psi_{i}\rangle = f(x_{i}), i = 1, 2, \cdots , n, \\ \langle v, \varphi_{0}\rangle = r_{0}, \langle v, \varphi_{1}\rangle = r_{1}. \end{array} \right. \end{eqnarray} $

证 由算子$ {\cal L} $的定义及$ R_{x}(y) $的再生性,可得

$ \begin{eqnarray*} \langle u_{n}(x), \psi_{i}(x)\rangle_{W_2^3}& = &\langle {{\cal P}}_{n}u(x), \psi_{i}(x)\rangle_{W_2^3} = \langle u(x), {{\cal P}}_{n}\psi_{i}(x)\rangle_{W_2^3}\\ & = &\langle u(x), \psi_{i}(x)\rangle_{W_2^3} = \langle u(x), {\cal L}R_{x}(x_i)\rangle_{W_2^3}\\ & = &{\cal L}_{x_i}\langle u(x), R_{x}(x_i)\rangle_{W_2^3} = {\cal L}_{x_i}u(x_i)\\ & = & f(x_i), {\quad} i = 1, 2, \cdots , n. \end{eqnarray*} $

且

$ \begin{eqnarray*} \langle u_{n}(x), \varphi_{i}(x)\rangle_{W_2^3}& = &\langle u_{n}(x), R_{x}(x_i)\rangle_{W_2^3} = \langle {{\cal P}}_{n}u(x), R_{x}(x_i)\rangle_{W_2^3}\\ & = &\langle u(x), R_{x}(x_i)\rangle_{W_2^3} = u(x_i) = r_i, {\quad} i = 0, 1. \end{eqnarray*} $

定理3.3证毕.

进而,有下面的定理.

定理3.4  在区间$ [0, 1] $上, $ u_{n}(x) $在范数$ \left\|. \right\|_{W_2^{3}} $意义下一致收敛于$ u(x) $.

证  对任意$ x\in[0, 1] $,有

$ \begin{eqnarray*} |u_{n}(x)-u(x)| = |\langle u_{n}(x)-u(x), R_{x}\rangle| \leq\| u_{n}(x) -u(x)\|\|R_{x}(x)\|\longrightarrow0. \end{eqnarray*} $

定理3.4证毕.

事实上, $ u_n(x) $是方程(1.2)精确解$ u(x) $的近似解.基于以上研究可知,必有$ \alpha_{i} $和$ \beta_{k} $满足

(3.2) $ \begin{eqnarray} u_{n}(x) = \sum\limits_{i = 1}^{n}\alpha_{i}\psi_{i}(x)+\sum\limits_{k = 0}^{1}\beta_{k}\varphi_{k}(x). \end{eqnarray} $

为了求得近似解$ u_n(x) $,只需求得$ \alpha_{i} $和$ \beta_{k} $.将(3.2)式代入方程(3.1)中,可得

(3.3) $ \begin{eqnarray} \left\{ \begin{array}{ll} { } \sum\limits_{i = 1}^{n}\alpha_{i}\langle \psi_{i}, \psi_{j}\rangle+\beta_{0}\langle r_0, \psi_{j}\rangle+\beta_{1}\langle r_1, \psi_{j}\rangle = f(x_{j}), j = 1, 2, \cdots , n, \\ { } \sum\limits_{i = 1}^{n}\alpha_{i}\langle \psi_{i}, r_k\rangle+\beta_{0}\langle \varphi_{0}, \varphi_{k}\rangle+\beta_{1}\langle \varphi_{1}, \varphi_{k}\rangle = r_k, k = 0, 1. \end{array} \right. \end{eqnarray} $

由于$ \psi_{i}, \varphi_{0}, \varphi_{1} $是线性无关的,上述线性方程组(3.3)有唯一解.在下一节中,将给出算例来呈现以上算法的有效性和正确性.

注3.1 不失一般性,这类边值问题可以一般化.

基于定理3.4的一致收敛性,可得如下结果.

定理3.5 如果在$ W_{2}^{3}[0, 1] $中$ {u_n}(x)\longrightarrow u(x) $,那么,在$ [0, 1] $上$ u_n^{(i)} $一致收敛于$ u^{(i)} $$ (i = 1, 2) $.

证 注意到

$ u_{n}^{(i)}(x) = \langle u_{n}, \partial_{x}^{(i)}R_{x}\rangle_{W_{2}^{3}}, {\quad} u^{(i)}(x) = \langle u, \partial_{x}^{(i)}R_{x}\rangle_{W_{2}^{3}}. $

由Cauchy-Schwarz不等式及$ \partial_{x}^{(i)}R_{x}(x) $的连续性,可得

$ \begin{eqnarray*} |u_{n}^{(i)}(x)-u^{(i)}(x)|& = &|\langle u_{n}-u, \partial_{x}^{(i)}R_{x}\rangle| \leq\| u_{n} -u\|_{W_{2}^{3}} \| \partial_{x}^{(i)}R_{x}(x)\|_{W_{2}^{3}}\\ &\leq& M_{i}\| u_{n} -u\|_{W_{2}^{3}}\longrightarrow0. \end{eqnarray*} $

因而, $ u_n^{(i)} $一致收敛于$ u^{(i)} (i = 1, 2) $.

从作者的另一篇关于再生核算法的收敛阶研究论文^[21],可得近似解$ u_{n}(x) $有如下不低于二阶的收敛性

$ |u_{n}(x)-u(x)|\leq C_{1}h^2, {\quad} |u'_{n}(x)-u'(x)|\leq C_{2}h^2, {\quad} |u''_{n}(x)-u''(x)|\leq C_{3}h^2. \label{formulall} $

其中$ h = \frac{1}{n} $.

上面研究了再生核算法的基本理论,下面通过具体数值算例实现算法的效果.下面两个算例中,均取$ x_{i} = \frac{i}{n}, i = 1, 2, \cdots, n-1 $.

算例4.1

考虑如下微分方程^[14]

$ u''(x)-|x-\frac{1}{2}|u(x) = f(x), {\quad} x\in[0, 1], $

其边值条件

$ u(0) = -\frac{1}{2}, {\quad} u(1) = \sin\frac{1}{2}, $

其中

$ f(x) = \left\{ \begin{array}{ll} { } (x-\frac{1}{2})^2, &{ } 0 \leq x \leq {\frac{1}{2}}, \\ { } -(\frac{1}{2}+x)\sin(x-\frac{1}{2}), &{ } {\frac{1}{2}}\leq x \leq 1. \end{array} \right. $

其精确解为

$ u(x) = \left\{ \begin{array}{ll} { } x-\frac{1}{2}, &{ } 0 \leq x \leq {\frac{1}{2}}, \\ { } \sin(x-\frac{1}{2}), &{ } {\frac{1}{2}}\leq x \leq 1. \end{array} \right. $

表 1还给出了与文献[14]对比的绝对误差.

算例4.2 考虑如下微分方程

$ u''(x)+a_1(x)u'(x)+a_0(x)u(x) = f(x), {\quad} x\in[0, 1], $

其边值条件

$ u(0) = -\sin(0.75) , {\quad} u(1) = \frac{95}{384}, $

其中

$ a_1(x) = \left\{ \begin{array}{ll} 1.75- x, &0 \leq x<0.75, \\ 2 e^{(0.75 - x) }, &0.75< x\leq1, \end{array} \right. $

$ a_0(x) = \left\{ \begin{array}{ll} -\cos(x-0.75), &0 \leq x<0.75, \\ 2 , &0.75<x\leq1, \end{array} \right. $

$ f(x) = \left\{ \begin{array}{ll} { } (1.75- x) \cos(0.75 - x) + \sin(0.75 - x) + \frac{1}{2} \sin2(0.75 - x), &0 \leq x<0.75, \\ { } 2 e^{(0.75 - x) }[1 - \frac{1}{2}(x-0.75 )^2] + (x-0.75) - \frac{1}{3}(x-0.75 )^3 , &0.75< x\leq1. \end{array} \right. $

其精确解为

$ u(x) = \left\{ \begin{array}{ll} \sin(x-0.75), &0\leq x\leq0.75, \\ { } (x-0.75)-\frac{(x-0.75)^3}{6}, &0.75\leq x\leq1. \end{array} \right. $

基于再生核空间的高效数值算法被提出用于解决分片光滑边值问题.通过与数值算例中方程的精确解相比较,结果表明算法正确且高效.