数学物理学报  2017, Vol. 37 Issue (3): 457-468   PDF    
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本文作者相关文章
谢瓯
孟泽红
赵振宇
由雷
求解拉普拉斯方程柯西问题的截断赫尔米特展开方法
谢瓯1, 孟泽红2, 赵振宇1, 由雷1     
1. 广东海洋大学数学与计算机学院 广东湛江 524008;
2. 浙江财经大学数学与统计学院 杭州 310018
摘要:该文研究一类拉普拉斯方程的柯西问题.为了获得稳定的数值解,采用了基于赫尔米特函数展开的截断方法来克服问题的不适定性.通过偏差原理选取截断参数并建立了相应的误差估计.数值结果同样显示方法是有效的.
关键词不适定问题    拉普拉斯方程柯西问题    偏差原理    截断方法    
A Truncation Method Based on Hermite Functions Expansion for a Cauchy Problem of the Laplace Equation
Xie Ou1, Meng Zehong2, Zhao Zhenyu1, You Lei1     
1. Faculty of Mathematics and Computer Science, Guangdong Ocean University, Guangdong Zhanjiang 524088;
2. School of Mathematics and Statistics, Zhejiang University of Finance & Economics, Hangzhou 310018
Abstract: We investigate a Cauchy problem for the Laplace equation in this paper. To obtain a stable numerical solution for this ill posed problem, we present a truncation method based on Hermite functions expansion. Error estimate are obtained together with a discrepancy principle for the regularization parameter. Some numerical tests show that the method works effectively.
Key words: Ill-posed problem     Cauchy problem for Laplace equation     Regularization     Discrepancy principle     Truncation method    
1 引言

本文考虑如下拉普拉斯方程的柯西问题[1-5]

$\begin{equation}\label{cp} \begin{array}{lll} u_{xx}+u_{yy}=0, &-\infty<x<\infty, 0<y<1, \\ u(x, 0)=\phi(x), &-\infty<x<\infty, \\ u_y(x, 0)=0, &-\infty<x<\infty, \end{array} \end{equation}$ (1.1)

即通过数据$\phi(x)$确定$u(\cdot, y)$, $0<y\leq 1$.在实际过程中$y=0$处的数据一般是通过读取一些物理装置获得的, 所以它通常是不精确的, 只能给出它的一组扰动值$\phi^{\delta}(x)$.假设扰动数据与精确数据之间满足如下关系

$\begin{equation}\label{errorcondition} \|\phi-\phi^{\delta}\|\leq \delta, \end{equation}$ (1.2)

这里$\delta>0$表示测量误差的界, $\|\cdot\|$定义$L^2$ -范数.这个问题出现在很多实际领域当中, 比如等离子物理[6]、心脏病学[7]、生物相关领域[8]、无损探测[9]、地震学[3], 等等.问题的主要困难在于其不适定性:问题解(即使存在)不连续依赖于给定数据.因此, 要想获得稳定的数值解必须采用一些正则化手段.

目前, 针对此问题已有一些方法被提出:磨光方法[1-2]、小波正则化方法[3-4]、傅里叶正则化方法[5], 等等.在所有这些方法中, 正则化参数的选取都依赖于解的先验信息$\|u(\, \cdot\, , 1)\|_p\leq E$, 但在实际过程中$E$$p$都是难以估计的.

结合后验准则的正则化方法[10-13]可以较容易的拓展到此问题.本文中研究一种基于赫尔米特函数展开的方法来解决此问题.该方法已经被应用于数值微分问题[14]和数值解析延拓问题[15].这个方法中正则化参数的选取可以通过简单的偏差原理来实现, 无需解的先验信息.当精确解是解析的时候, 方法可以获得更好的收敛结果.

全文安排如下:第二节简单回顾一下所需的一些前期结果; 第三节介绍本文方法和相应的收敛结果; 最后给出一些数值实验来验证方法的有效性.

2 预备知识

$\Lambda=\{x|-\infty<x<\infty\}$, $L^2(\Lambda)$$\|\cdot\|$为一般意义下的勒贝格空间及相应范数. $l$次赫尔米特函数定义为

$\begin{equation} h_l(x)= (2^ll!\sqrt{\pi})^{-1/2}{\rm e}^{-\frac{1}{2}x^2}H_l(x), \end{equation}$ (2.1)

这里$H_l(x)$为通常(非标准)的赫尔米特多项式.赫尔米特函数为${L}^2(\Lambda)$ -正交系统[16], 即

$\begin{equation} \int_{\Lambda}h_l(x)h_m(x){\rm d}x=\delta_{l, m}. \end{equation}$ (2.2)

对于$\phi\in {L}^2(\Lambda)$空间函数, 其赫尔米特展开为

$\phi (x) = \sum\limits_{l = 0}^\infty {{\phi _l}} {h_l}(x),$

这里

$\begin{equation} {\phi}_l=\int_{\Lambda}\phi(x)h_l(x){\rm d}x,  l=0, 1, 2\cdots. \end{equation}$ (2.3)

$N$为任意正整数及

$\begin{equation} {\cal H}_N=\hbox{span}\{\, h_0(x), h_1(x), \cdots, h_N(x)\, \}. \end{equation}$ (2.4)

$L^2(\Lambda)$ -正交投影算子$P_N:{L}^2(\Lambda)\rightarrow{\cal H}_{{N}}$为如下映射:对任意$\phi\in {L}^2(\Lambda)$, 有

$\begin{equation} \langle\, \phi-P_N\phi\, , \, \varphi\, \rangle=0,  \forall \varphi\in\cal{H}_{{N}}, \end{equation}$ (2.5)

等价地

$\begin{equation} P_N\phi(x)=\sum\limits_{l=0}^N{\phi}_lh_l(x). \end{equation}$ (2.6)

$2l+1$的正平方根在全文中多次出现, 定义其为$\nu_l$.后面的行文中$c$定义为独立于任何函数及参数$N$的正常数.下列结果在后面的证明中需要用到.

引理2.1[17] 如果令

$\begin{equation} H(x, y;\lambda)=\sum\limits_{l=0}^{\infty}|h_l(x+{\rm i}y)|^2{\rm e}^{-2\lambda {\nu_l}}, \lambda>0, \end{equation}$ (2.7)

则如下不等式对任意$|y|\leq \lambda$成立

$\begin{equation} H(x, y;\lambda)<c|x|^{\frac{1}{2}}\exp[-2|x|(\lambda^2-y^2)^{1/2}]. \end{equation}$ (2.8)

引理2.2[17] 设$\phi(z)$为解析函数, 则傅里叶-赫尔米特级数

$\begin{equation} \sum\limits_{l=0}^{\infty}{\phi}_lh_l(z),  {\phi}_l=\int_{\Lambda}\phi(x)h_l(x){\rm d}x \end{equation}$ (2.9)

存在且在带状区域$S_{\eta}:-\eta<y<\eta$收敛于$\phi(z)$的充分必要条件为$\phi(z)$$S_{\eta}$中为全纯函数, 并且对于给定的$\beta$, $0\leq \beta<\eta$, 存在正常数$B(\beta)$满足

$\begin{equation}\label{vercon} |\phi(x+{\rm i}y)|\leq B(\beta)\exp[-|x|(\beta^2-y^2)^{1/2}] \end{equation}$ (2.10)

对任意$-\infty<x<\infty$, $-\beta\leq y\leq \beta$成立.此时${\phi}_l$对任意$\epsilon>0$满足不等式

$\begin{equation}\label{cofcondition} |{\phi}_l|\leq M(\epsilon){\rm e}^{-(\eta-\epsilon){\nu_l}}. \end{equation}$ (2.11)

引理2.3[18] 函数$f(\rho):(0, a]\rightarrow \Lambda$定义为

$\begin{equation}\label{taute1} f(\rho)=\rho^b\bigg(d\ln\frac{1}{\rho}\bigg)^{-c}, \end{equation}$ (2.12)

其中$c\in \Lambda$为常数, $a<1$, $b$$d$为正常数, 则其反函数$f^{-1}(\eta)$满足:对$\rho\rightarrow 0$, 有

$\begin{equation}\label{taute2} f^{-1}(\rho)=\rho^{\frac{1}{b}}\bigg(\frac{d}{b} \ln\frac{1}{\rho}\bigg)^{\frac{c}{b}}(1+o(1)) . \end{equation}$ (2.13)
3 截断方法及相应误差估计

函数$\phi\in L^2(\Lambda)$的傅里叶变换定义为

$\begin{equation} \hat{\phi}(\omega)={\cal F}[\phi(x)]=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\phi(x){\rm e}^{-{\rm i}\omega x}{\rm d}x. \end{equation}$ (3.1)

对问题(1.1) 关于变量$x$做傅里叶变换可得

$\begin{equation} \begin{array}{lll}\label{fcp} \hat{u}_{yy}(\omega, y)=\omega^2\hat{u}(\omega, y), &\omega\in \Lambda, y\in (0, 1), \\ \hat{u}(\omega, 0)=\hat{\phi}(\omega), &\omega\in \Lambda, \\ \hat{u}_y(\omega, 0)=0, &\omega\in \Lambda. \end{array} \end{equation}$ (3.2)

易得问题(3.2) 的解为

$\begin{equation} \hat{u}(\omega, y)=\hat{\phi}(\omega)\cosh(y\omega). \end{equation}$ (3.3)

方程(1.1) 可视为映函数$\phi(x)$$u(x, y)$的算子.我们定义$T_y$为此映射, 即

$\begin{equation}\label{Fp1} T_y\phi(x)=u(x, y)={\cal F}^{-1}\left[\hat{\phi}(\omega)\cosh(y\omega)\right]. \end{equation}$ (3.4)

显然, 精确解$\hat{\phi}(\omega)$$\omega\rightarrow \infty$时必须是快速衰减的.然而, 实际问题中$y=0$处所获取的数据含有误差.对应扰动数据的傅里叶变换一般无法满足这种衰减性质.下面, 我们提出一种基于赫尔米特展开的磨光方法来获得问题稳定的数值解.

假设$\phi^{\delta}$满足

$\left\| {{\phi ^\delta }} \right\| > \tau \delta .$ (3.5)

首先由扰动数据$\phi^{\delta}$重构出函数$\varphi^{\delta}$

$\begin{equation}\label{prosolution} {\phi}^{\delta}\rightarrow \varphi^{n, \delta}=P_n\phi^{\delta}=\sum\limits_{l=0}^n{\mathbf \phi}_l^{\delta} h_l(x). \end{equation}$ (3.6)

正则化参数$n$由如下偏差原理获得

$\begin{equation}\label{discre1} \|\, (I-P_n)\phi^{\delta}\, \|\leq \tau\delta<\|\, (I-P_{n-1})\phi^{\delta}\, \|. \end{equation}$ (3.7)

然后取

$\begin{equation} u^{n, \delta}(x, y)=T_y\varphi^{n, \delta} \end{equation}$ (3.8)

作为$u(x, y)$的近似解.我们知道[19]:对于$S_{\eta}$内解析的函数$\psi(x)$, 有

$\begin{equation} \psi(z)=\psi(x+{\rm i}y)=\sum\limits_{n=0}^{\infty}\frac{\psi^{(n)}(x)}{n!}({\rm i}y)^n=\sum\limits_{n=0}^{\infty}\frac{({\rm i}y)^n}{n!}D^n\psi(x), |y|< \eta, \end{equation}$ (3.9)

这里$D^n=\frac{\partial^n}{\partial x^n}$.关于变量$x$做傅里叶变换可得

$\begin{equation}\label{Ft} \widehat{\psi(\cdot+{\rm i}y)}(\omega, y)={\rm e}^{-y\omega}\hat{\psi}(\omega). \end{equation}$ (3.10)

$\begin{equation} {\cal H}_l(x, y)=h_l(x+{\rm i}y)+h_l(x-{\rm i}y), \end{equation}$ (3.11)

$\begin{eqnarray}\label{uxyde} u^{n, \delta}(x, y)&=&T_y\varphi^{n, \delta} =\displaystyle{{\cal F}^{-1}\left[\widehat{\varphi^{n, \delta}}(\omega)\cosh(y\omega)\right] ={\cal F}^{-1}\left[\sum\limits_{l=0}^n\phi^{\delta}_l\hat{h}_l(\omega)\cosh(y\omega)\right]} \\ &=&\displaystyle{\frac{1}{2}\sum\limits_{l=0}^n{\mathbf \phi}_l^{\delta} {\left[h_l(x+{\rm i}y)+h_l(x-{\rm i}y)\right]}}=\frac{1}{2}\sum\limits_{l=0}^n{\mathbf \phi}_l^{\delta} {\cal H}_l(x, y). \end{eqnarray}$ (3.12)

下面我们导出方法的误差估计, 首先给出相关的一些辅助结果.

引理3.1 设$\psi\in{\cal H}_N$, 则如下不等式对于任意$|y|<\lambda$成立,

$\begin{equation}\label{ampl} \|T_y\psi\|\leq c {\rm e}^{\lambda \nu_N}\|\psi\|. \end{equation}$ (3.13)

 由Hölder不等式, 有

$\begin{eqnarray} \|T_y\psi\|^2&=&\displaystyle{\left\|\frac{1}{2}\sum\limits_{l=0}^N {\mathbf \psi}_l {\cal H}_l(x, y)\right\|^2}=\displaystyle{\frac{1}{4}\int_{-\infty}^{+\infty}\left|\sum\limits_{l=0}^N {\mathbf \psi}_l {\cal H}_l(x, y)\right|^2{\rm d}x} \\&=&\displaystyle{\frac{1}{4}\int_{-\infty}^{+\infty}\left|\sum\limits_{l=0}^N {\mathbf \psi}_l {\rm e}^{\lambda\nu_l} {\cal H}_l(x, y){\rm e}^{-\lambda\nu_l}\right|^2{\rm d}x} \\ &\leq& \displaystyle{\frac{1}{4}\bigg(\sum\limits_{l=0}^N {\mathbf \psi}^2_l {\rm e}^{2\lambda\nu_l}\bigg)\int_{-\infty}^{+\infty} \bigg(\sum\limits_{l=0}^N| {\cal H}_l(x, y)|^2{\rm e}^{-2\lambda\nu_l}\bigg){\rm d}x}. \end{eqnarray}$ (3.14)

另外

$\begin{equation} \sum\limits_{l=0}^N {\mathbf \psi}^2_l {\rm e}^{2\lambda\nu_l}\leq {\rm e}^{2\lambda\nu_N}\sum\limits_{l=0}^N {\mathbf \psi}^2_l= {\rm e}^{2\lambda\nu_N}\|\psi\|. \end{equation}$ (3.15)

则由引理2.1, 积分

$\begin{eqnarray} \displaystyle{\int_{-\infty}^{+\infty} \bigg(\sum\limits_{l=0}^N|{\cal H}_l(x, y)|^2{\rm e}^{-2\lambda\nu_l}\bigg){\rm d}x} &<& \displaystyle{\int_{-\infty}^{+\infty} \bigg(2\sum\limits_{l=0}^N{\left[|h_l(x +{\rm i}y)|^2+|h_l(x-{\rm i}y)|^2\right]}{\rm e}^{-2\lambda\nu_l}\bigg) {\rm d}x}\\ &<&\displaystyle{2\int_{-\infty}^{+\infty}{H(x, y;\lambda) +H(x, -y;\lambda)}{\rm d}x}\\ & \leq & c \int_{-\infty}^{+\infty}|x|^{\frac{1}{2}} \exp[-2|x|(\lambda^2-y^2)^{1/2}]{\rm d}x \end{eqnarray}$ (3.16)

对于任意$|y|<\lambda$有界, 因此结论成立.

引理3.2 如果条件(2.11) 成立, 则

$\begin{equation}\label{ine} \|\phi-P_N\phi\|\leq cM(\epsilon)N^{\frac{1}{4}}{\rm e}^{-(\eta-\epsilon){\nu_N}}. \end{equation}$ (3.17)

 由条件(2.11), 有

$\begin{equation} \|\phi-P_N\phi\|^2=\sum\limits_{l={N+1}}^{\infty}{\mathbf \phi}_l^2\leq M^2(\epsilon)\sum\limits_{l=N+1}^{\infty}{\rm e}^{-2(\eta-\epsilon){\nu_l}}. \end{equation}$ (3.18)

进而, 注意到和式

$\begin{equation} Q\equiv\sum\limits_{l=N+1}^{\infty}{\rm e}^{-2(\eta-\epsilon){\nu_l}} \end{equation}$ (3.19)

为如下积分的复合矩形公式近似

$\begin{equation} I=\int_{N}^{\infty}{\rm e}^{-2(\eta-\epsilon)\sqrt{2x+1}}{\rm d}x, \end{equation}$ (3.20)

而且矩形始终处于被积函数的下方, 显然

$\begin{equation} Q\leq I. \end{equation}$ (3.21)

由分部积分可得

$\begin{eqnarray} I&=&{\displaystyle\int_{N}^{\infty}{\rm e}^{-2(\eta-\epsilon)\sqrt{2x+1}} {\rm d}x\leq\int_{N}^{\infty}{\rm e}^{-2(\eta-\epsilon)\sqrt{2x}}{\rm d}x }\\ &=&{\displaystyle\frac{1}{\sqrt{2}(\eta-\epsilon)}\sqrt{N}{\rm e}^{-2(\eta-\epsilon) \sqrt{2N}}+\frac{1}{4(\eta-\epsilon)^2}{\rm e}^{-2(\eta-\epsilon)\sqrt{2N}}} \\ &\leq& c\sqrt{N}{\rm e}^{-2(\eta-\epsilon)\nu_N}. \end{eqnarray}$ (3.22)

因此结论成立.

引理3.3 对于$0<y<\mu$, 有

$\begin{equation}\label{ssss} \cosh(y\omega)^{\mu/y}\leq \cosh(\mu\omega). \end{equation}$ (3.23)

 对于任意固定的$\omega$, 容易验证$f(y)=\cosh(y|\omega|)^{\mu/y}$$0<y<\mu$单调递增, 因此(3.23) 式成立.

引理3.4 如果$\|T_\mu \phi\|\in L^2(\Lambda)$, 则对于$0<y<\mu$, 有

$\begin{equation}\label{hol1} \|T_y\phi\|\leq \|\phi\|^{\frac{\mu-y}{\mu}}\|T_{\mu}\phi\|^{\frac{y}{\mu}}. \end{equation}$ (3.24)

 由Parseval公式, (3.4) 式, 引理3.3及Hölder不等式, 有

$\begin{eqnarray} \|T_y\phi\|^2&=&\displaystyle{\int_{-\infty}^{\infty}|\hat{\phi}(\omega)\cosh(y\omega)|^2{\rm d}\omega =\int_{-\infty}^{\infty}|\hat{\phi}(\omega)|^{2(1-\frac{y}{\mu})}|\hat{\phi}(\omega)|^{\frac{2y}{\mu}}\cosh^2(y\omega){\rm d}\omega}\\ & \leq & \displaystyle{\left[\int_{-\infty}^{\infty}\bigg(|\hat{\phi}(\omega)|^{2(1-\frac{y}{\mu})}\bigg)^{\frac{\mu}{\mu-y}}{\rm d}\omega\right]^{1-\frac{y}{\mu}} \left[\int_{-\infty}^{\infty}\bigg(|\hat{\phi}(\omega)|^{\frac{2y}{\mu}}\cosh^2(y\omega)\bigg)^{\frac{\mu}{y}}{\rm d}\omega\right]^{\frac{y}{\mu}}}\\ &=&\displaystyle{\|\phi\|^{2(1-\frac{y}{\mu})}\|T_\mu \phi\|^{\frac{2y}{\mu}}}. \end{eqnarray}$ (3.25)

证毕.

定理3.1 如果$\phi(z)$$S_{\eta}$上为全纯函数以及对于任意给定的$\beta$, $0\leq \beta<\eta$, 存在正常数$B(\beta)$满足

$|\phi (x + {\text{i}}y)| \leqslant B(\beta )\exp [ - |x|{({\beta ^2} - {y^2})^{1/2}}],$ (3.26)

$-\infty<x<\infty$, $-\beta\leq y\leq \beta$成立. $\varphi_{n, \delta}$由(3.6) 和(3.7) 式定义, 其中$\tau>1$.则对于任意$0<y< \eta$, $0<\epsilon<\eta-y$, 我们有

$\begin{equation}\label{convergence1} \|u_{n, \delta}(\, \cdot\, , \, y\, )-u(\, \cdot\, , \, y)\|=O\bigg(\delta^{\frac{\eta_{\epsilon}-y}{\eta_{\epsilon}}}\bigg(\ln\frac{{C_{\tau, \epsilon}}}{{\delta}}\bigg)^{\frac{y}{2\eta_{\epsilon}}}\bigg), \end{equation}$ (3.27)

其中$\eta_{\epsilon}=\eta-\epsilon$, $C_{\tau, \epsilon}$为常数.

 由三角不等式可得

$\begin{eqnarray}\label{m1} \|u_{n, \delta}(\, \cdot\, , \, y\, )-u(\, \cdot\, , \, y)\|&=&\|T_{y}P_n\phi^{\delta}-T_{y}P_n\phi+T_{y}P_n\phi-T_{y}\phi\| \\ &\leq&\|T_{y}P_n(\phi^{\delta}-\phi)\|+\|T_{y}(I-P_n)\phi\|\\ & =: & I_1+I_2. \end{eqnarray}$ (3.28)

现在分别估计$I_1$$I_2$.对于$I_1$, 按照(3.5) 式和不等式(3.13), 可得

$\begin{equation}\label{I1} I_1=\|T_{y}P_n(\phi^{\delta}-\phi)\|\leq c {\rm e}^{(y+\frac{\epsilon}{2})\nu_n}\|P_n(\phi^{\delta}-\phi)\|\leq c {\rm e}^{(y+\frac{\epsilon}{2})\nu_n}\delta. \end{equation}$ (3.29)

另外

$\begin{eqnarray}\label{x1} \|P_{n-1}\phi-\phi\|&=&\|(P_{n-1}\phi^{\delta}-\phi^{\delta})-(I-P_{n-1})(\phi-\phi^{\delta})\| \\ & \geq & \|(I-P_{n-1})\phi^{\delta}\|-\|(I-P_{n-1})(\phi-\phi^{\delta})\|. \end{eqnarray}$ (3.30)

以及

$\begin{equation}\label{x2} \|(I-P_{n-1})(\phi-\phi^{\delta})\|\leq\|\phi-\phi^{\delta}\|\leq \delta. \end{equation}$ (3.31)

由(3.30), (3.31), (3.7) 和(3.17) 式可得

$\begin{equation} (\tau-1)\delta\leq\|P_{n-1}\phi-\phi\|\leq cM\bigg(\frac{\epsilon}{2}\bigg)(n-1)^{\frac{1}{4}}{\rm e}^{-(\eta-\frac{\epsilon}{2}){\nu_{n-1}}}\leq \bar{c}M\bigg(\frac{\epsilon}{2}\bigg)n^{\frac{1}{4}}{\rm e}^{-(\eta-\frac{\epsilon}{2}){\nu_n}}. \end{equation}$ (3.32)

$\begin{equation}\label{e37} n^{\frac{1}{4}}{\rm e}^{-(\eta-\frac{\epsilon}{2}){\nu_n}}\geq \frac{(\tau-1)}{\bar{c}M\big(\frac{\epsilon}{2}\big)}\delta. \end{equation}$ (3.33)

$\sigma$满足

$\begin{equation}\label{ph1} \sigma^{\frac{1}{2}}{\rm e}^{-(\eta-\frac{\epsilon}{2})\sigma} =\frac{2^{\frac{1}{4}}(\tau-1)}{\bar{c}M\big(\frac{\epsilon}{2}\big)}\delta=\frac{\delta}{C_{\tau, \epsilon}}. \end{equation}$ (3.34)

可得存在$\delta_0>0$满足

$\begin{equation}\label{est1} \nu_n\leq \sigma, \forall \delta\in (0, \delta_0]. \end{equation}$ (3.35)

定义

$\begin{equation}\label{invfu} {\rm e}^{-(\eta-\frac{\epsilon}{2})\sigma}:=\rho\in(0, 1), \end{equation}$ (3.36)

$\begin{equation} \sigma=\frac{1}{\eta-\frac{\epsilon}{2}}\ln\frac{1}{\rho}. \end{equation}$ (3.37)

(3.34) 式变为

$\begin{equation} \rho\bigg(\frac{1}{\eta-\frac{\epsilon}{2}}\ln\frac{1}{\rho}\bigg)^{\frac{1}{2}}=\frac{\delta}{C_{\tau, \epsilon}}, \end{equation}$ (3.38)

相当于在(2.12) 式取$b=1$, $d=\frac{1}{\eta-\frac{\epsilon}{2}}$$c=-\frac{1}{2}$.由(2.13) 式可得

$\begin{equation}\label{fe} \rho=\frac{\delta}{C_{\tau, \epsilon}}\bigg(\frac{1}{\eta-\frac{\epsilon}{2}}\ln\frac{{C_{\tau, \epsilon}}}{{\delta}}\bigg)^{-\frac{1}{2}}(1+o(1)). \end{equation}$ (3.39)

取上式中$\rho$的主部, 利用(3.36) 式可得

$\begin{equation} {\rm e}^{-(\eta-\frac{\epsilon}{2})\sigma}=\frac{\delta}{C_{\tau, \epsilon}}\bigg(\frac{1}{\eta-\frac{\epsilon}{2}}\ln\frac{{C_{\tau, \epsilon}}}{{\delta}}\bigg)^{-\frac{1}{2}} \end{equation}$ (3.40)

$\begin{equation}\label{est2} \sigma=\frac{1}{\eta-\frac{\epsilon}{2}}\ln\left[\frac{{C_{\tau, \epsilon}}}{{\delta}}\bigg(\frac{1}{\eta-\frac{\epsilon}{2}}\ln\frac{{C_{\tau, \epsilon}}}{{\delta}}\bigg)^{\frac{1}{2}}\right]. \end{equation}$ (3.41)

结合(3.35) 式, (3.41) 及(3.29) 式, 可得

$\begin{eqnarray}\label{I1est} I_1 & \leq & \displaystyle{{\rm e}^{(y+\frac{\epsilon}{2})\nu_n}\delta\leq {\rm e}^{(y+\frac{\epsilon}{2})\sigma}\delta}\\ &=&\displaystyle{\left[\frac{\delta}{C_{\tau, \epsilon}} \bigg(\frac{1}{\eta-\frac{\epsilon}{2}}\ln\frac{{C_{\tau, \epsilon}}} {{\delta}}\bigg)^{-\frac{1}{2}}\right]^{-\frac{y+\frac{\epsilon}{2}} {\eta-\frac{\epsilon}{2}}}\delta} \leq \displaystyle{ \left[\frac{\delta}{C_{\tau, \epsilon}} \bigg(\frac{1}{\eta-\frac{\epsilon}{2}}\ln\frac{{C_{\tau, \epsilon}}} {{\delta}}\bigg)^{-\frac{1}{2}}\right]^{-\frac{y}{\eta_\epsilon}}\delta } \\ &=&\displaystyle{\bigg(\eta-\frac{\epsilon}{2}\bigg)}^{-\frac{y}{2\eta_\epsilon}} C_{\tau, \epsilon}^{\frac{y}{\eta_\epsilon}}\delta^{\frac{ \eta_\epsilon-y}{\eta_\epsilon}}\bigg(\ln\frac{{C_{\tau, \epsilon}}} {{\delta}}\bigg)^{\frac{y}{2\eta_\epsilon}}. \end{eqnarray}$ (3.42)

下面我们估计$I_2$.由(3.7) 式可得

$\begin{eqnarray}\label{t21} \|(I-P_n)\phi\|&=&\|(I-P_n)\phi^{\delta}+(I-P_n)(\phi-\phi^{\delta})\|\\ & \leq & \|(I-P_n)\phi^{\delta}\|+\|(I-P_n)(\phi-\phi^{\delta})\| \leq (\tau+1)\delta. \end{eqnarray}$ (3.43)

进而, 利用Hölder不等式可得

$\begin{eqnarray} \|T_{\eta_{\epsilon}}(I-P_n)\phi\|^2&=&\displaystyle{\left\| \frac{1}{2}\sum\limits_{l=n+1}^{\infty}{\mathbf \phi}_l{\cal H}_l (x, \eta_{\epsilon})\right\|^2 }\\ & = & \displaystyle{\frac{1}{4}\int_{-\infty}^{+\infty}\left| \sum\limits_{l=n+1}^{\infty}{\mathbf \phi}_l{\rm e}^{(\eta- \frac{\epsilon}{2})\nu_l}{\cal H}_l(x, \eta_{\epsilon}) {\rm e}^{-(\eta-\frac{\epsilon}{2})\nu_l}\right|^2{\rm d}x}\\ &\leq&\displaystyle{\frac{1}{4}\bigg(\sum\limits_{l=n+1}^{\infty} {\mathbf \phi}^2_l{\rm e}^{2(\eta-\frac{\epsilon}{2})\nu_l}\bigg) \int_{-\infty}^{+\infty}\frac{H(x, \eta_{\epsilon}; \eta-\frac{\epsilon}{2})+ H(x, -\eta_{\epsilon};\eta-\frac{\epsilon}{2})}{2}{\rm d}x}. \qquad \end{eqnarray}$ (3.44)

由(2.11) 式及引理3.2的证明过程, 可得

$\begin{equation} \sum\limits_{l=n+1}^{\infty}{\mathbf \phi}^2_l{\rm e}^{2(\eta-\frac{\epsilon}{2})\nu_l}\leq M\sum\limits_{l=n+1}^{\infty}{\rm e}^{-\epsilon\nu_l}\leq c_1n^{\frac{1}{2}}{\rm e}^{-\epsilon\nu_l}. \end{equation}$ (3.45)

注意到积分

$\begin{equation} \int_{\Lambda}{H(x, \eta_{\epsilon};\eta-\frac{\epsilon}{2})+ H(x, -\eta_{\epsilon};\eta-\frac{\epsilon}{2})}{\rm d}x \end{equation}$ (3.46)

有界.所以

$\begin{equation}\label{t22} \|T_{\eta_{\epsilon}}(I-P_n)\phi\|^2\leq c_2n^{\frac{1}{2}}{\rm e}^{-\epsilon\nu_l}. \end{equation}$ (3.47)

这表明$\|T_{\eta_{\epsilon}}(I-P_n)\phi\|$有界, 所以由引理3.4, 有

$\begin{equation}\label{t23} I_2=\|T_y(I-P_n)\phi\|\leq \|(I-P_n)\phi\|^{\frac{\eta_{\epsilon}-y}{\eta_{\epsilon}}}\|T_{\eta_{\epsilon}}(I-P_n)\phi\|^{\frac{y}{\eta_{\epsilon}}}. \end{equation}$ (3.48)

带入(3.43) 和(3.47) 式到(3.48) 式, 并且令$\epsilon\rightarrow0$, 可得

$\begin{equation}\label{t24} I_2\leq c_3(\tau+1)^{\frac{\eta-y}{\eta}}\delta^{\frac{\eta-y}{\eta}}n^{\frac{y}{4\eta}}. \end{equation}$ (3.49)

由(3.35) 式, 有

$\begin{eqnarray}\label{I2est1} I_2 & \leq & c_3(\tau+1)^{\frac{\eta-y}{\eta}}\delta^{\frac{\eta-y}{\eta}} \bigg(\frac{v_n^2-1}{2}\bigg)^{\frac{y}{4\eta}}\\ &\leq& c_3(\tau+1)^{\frac{\eta-y}{\eta}}\delta^{\frac{\eta-y}{\eta}}\bigg(\frac{\sigma^2-1}{2}\bigg)^{\frac{y}{4\eta}} \\ & \leq & c_4(\tau+1)^{\frac{\eta-y}{\eta}}\delta^{\frac{\eta-y}{\eta}}\sigma^{\frac{y}{2\eta}}. \end{eqnarray}$ (3.50)

将(3.41) 式带入(3.50) 式, 得到对$ \delta\rightarrow 0$, 有

$\begin{eqnarray} \label{I2est} I_2 & \leq & c_4(\tau+1)^{\frac{\eta-y}{\eta}}\delta^{\frac{\eta-y}{\eta}} \left\{\frac{1}{\eta} \ln\left[\frac{{C_{\tau}}}{{\delta}} \bigg(\frac{1}{\eta}\ln\frac{{C_{\tau}}}{{\delta}}\bigg)^{\frac{1}{2}} \right]\right\}^{\frac{y}{2\eta}} \\ &=&c_4(\tau+1)^{\frac{\eta-y}{\eta}} \eta^{-\frac{y}{2\eta}}\delta^{\frac{\eta-y}{\eta}} \bigg(\ln\frac{{C_{\tau}}}{{\delta}}\bigg)^{\frac{y}{2\eta}}(1+o(1)) . \end{eqnarray}$ (3.51)

由(3.28), (3.42) 和(3.51) 式可得结论成立.

注3.1 实际计算中$\eta$的值是无需知道的, 我们只需要确认$\eta>1$.

注3.2 定理条件要强于文献[1-5], 但当相应条件成立时, 本文方法可以获得更好结果.

4 数值检验

本节给出一些数值结果来验证方法的有效性.离散节点为$x_i=-B+ih, i=0, 1, \cdots, m; h=2B/m$, 其中$m=256$, $B$为正实数, 满足当$|x|>B$$\phi(x)$接近于0.扰动数据通过如下方式获取

$\begin{equation} \phi^{\delta}=\phi+\hbox{randn}(\hbox{size}(\phi))\times\delta_1, \end{equation}$ (4.1)

其中"randn($\cdot$)"为Matlab中生成正态分布随机数的函数.

计算中需要计算如下积分

$\begin{equation} \langle \phi^{\delta}, h_j\rangle =\int_{\Lambda} \phi^{\delta}(x)h_j(x){\rm d}x. \end{equation}$ (4.2)

对此我们先通过扰动数据$\phi^{\delta}(x_i)$, 构造

$\begin{equation} \phi^{\delta}(x)=\left\{\begin{array}{lll} s^{\delta}(x) & , & x\in[-B, B], \\ 0 & , & \hbox{其它, }\\ \end{array}\right. \end{equation}$ (4.3)

其中$s^{\delta}$为三次样条差值.鉴于误差生成的方式我们在实际计算中取

$\delta=\frac{\sqrt{3}}{6}B\delta_1. $

下面我们将给出几个数值例子来验证方法的有效性, 并把本文方法(M1) 和文献[5]中方法(M2) 进行对比, 所有例子中取$\tau=1.01$$B=10$.

例1 易得

$\begin{equation} u(x, y)={\rm e}^{y^2-x^2}\cos(2xy) \end{equation}$ (4.4)

为问题(1.1) 在如下初始数据中的精确解

$\begin{equation} u(x, 0)={\rm e}^{-x^2}=\phi(x), \qquad \frac{\partial}{\partial x}u(x, 0)=0. \end{equation}$ (4.5)

表 1中给出了两种方法在$y=0, 0.2, \cdots, 1$时的相对误差

表 1 例1数值结果
$\begin{equation} {\cal E}_{y}=\frac{\displaystyle{ \sqrt{\frac{1}{m}\sum\limits_{i=1}^{m}\left|{\bf u}^{\delta}(x_i, y)-u(x_i, y)\right|^2}}}{\displaystyle{\sqrt{\frac{1}{m}\sum\limits_{i=1}^{m}\left|u(x_i, y)\right|^2}}}, \end{equation}$ (4.6)

其中${\bf u}^{\delta}$为近似解.

表 1可以看出当误差水平$\delta_1$$0.1$降到$0.0001$, 数值解的相对误差相应减小.结果显示了方法的有效性, 并且可以看出M1的结果优于M2.

图 1给出了数值解与精确解在不同误差水平及$y$不同取值时的对比.其中, 实线表示精确解虚线表示近似解.可以看出随着$y$的增大逼近效果变差, 这和理论分析相吻合.

图 1 精确解及其近似

例2 一般来讲问题(1.1) 的解析解是不易获得的, 下面例子采用如下方式:给定$\psi(x)\in L^2(\Lambda)$进而通过解适定问题

$\begin{equation}\label{cp1} \begin{array}{lll} u_{xx}+u_{yy}=0, & x\in \Lambda, y \in (0, 1), \\ u(x, 1)=\psi(x), & x\in \Lambda, \\ u_y(x, 0)=0, & x\in\Lambda, \end{array} \end{equation}$ (4.7)

获得$\phi(x)$.然后在$\phi(x)$施加扰动获得$\phi^{\delta}$.本例中取$\psi$

$\begin{equation} \psi(x)=\left\{\begin{array}{lll} x+4, &-4\leq x<0;\\ -x+4, & 0\leq x\leq 4. \end{array}\right. \end{equation}$ (4.8)

表 2图 2给出了本例的相关结果, 可以看出此时方法依然是有效的.此时方法M1依然优于M2, 尤其是对较大的$y$和较小的$\delta_1$.

表 2 例2数值结果

图 2 精确解及其近似
5 结论

本文采用了一种截断方法来求解拉普拉斯方程柯西问题, 方法中正则化参数由偏差原理确定.相应的理论结果和数值结果都表明了所提方法的有效性.新方法的收敛条件要强于以往文献[1-5], 但在相应条件下可以获得更优的结果.

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