二维时间反向热传导问题的两种正则化方法及后验误差估计

Two Regularization Methods for the Two-Dimensional Time-Inverse Heat Conduction Problem and Its Posterior Error Estimation

摘要: 本文讨论了一类二维时间反向热传导问题，它从终值时刻 的温度分布来反演初始时刻的温度分布。该问题在图像处理方面有重要应用。这是一个严重不适定问题，它的解在一定条件下不连续依赖于数据。针对传统正则化方法的缺陷，本文采用拟逆正则化方法和分数次Tikhonov正则化方法，来恢复解对数据的依赖性。同时我们还给出了两种方法相应的后验参数选取规则及其正则解与精确解的误差估计。

Abstract: The time-inverse heat conduction problem was concerned in two-dimensional space, which retrieves the temperature distribution from the temperature distribution at the final moment. This problem has important application in image processing. This is a serious ill-posed problem, i.e. its solution is not continuously dependent on the data under certain conditions. For the defects of traditional regularization methods, the quasi-reversibility regularization method and the fractional Tikhonov regularization method are proposed to restore the dependence of the solution on the data. Meanwhile, the errors between the approximate solutions and the exact solution for the ill-posed problem are estimated, and the posteriori regularization parameter selection rules are given.

1. 引言

随着科学技术的发展，人们发现实际问题的建模中，不适定问题占有重要地位，于是不适定问题的研究得以发展。对于热传导问题各类典型的不适定问题，学者们对于该问题提出了一些正则化方法，如Tikhonov正则化方法 [1] [2]、拟逆正则化方法 [3] [4]、数值微分正则化方法 [5]、边界元方法 [6] [7] 等。同时也给出了相应近似解与精确解的误差估计。通过文献检索，我们发现对于一维空间中的时间反向热传导问题的研究相对较多，同时大多给出的误差估计为先验误差估计，而对二维空间的此类问题 [8] [9] [10] 研究相对较少。本文将对一类二维时间反向热传导问题进行讨论，并给出其后验误差估计。

本文考虑二维时间反向热传导问题 [11]

$\{\begin{array}{l}{u}_t\left(x\mathrm{,}y\mathrm{,}t\right)=\Delta u\left(x\mathrm{,}y\mathrm{,}t\right)\mathrm{,}\left(x\mathrm{,}y\right)\in R^2\mathrm{,0}\le t<T\mathrm{,}\\ u\left(x\mathrm{,}y\mathrm{,}T\right)=g\left(x\mathrm{,}y\right)\mathrm{,}\left(x\mathrm{,}y\right)\in R^2\mathrm{.}\end{array}$ (1)

其中 $\Delta =\frac{{\partial }^2}{\partial x^2}+\frac{{\partial }^2}{\partial y^2}$， $u\left(x\mathrm{,}y\mathrm{,}\cdot \right)$ 和 $g\left(x,y\right)$ 是定义在 $L^2\left(R^2\right)$ 上的函数。我们想由数据 $g\left(x\mathrm{,}y\right)$ 来确定

$u\left(x\mathrm{,}y\mathrm{,}t\right)$ 在 $0\le t<T$ 上的温度分布。而这里的输入数据 $g\left(x\mathrm{,}y\right)$ 往往是被测量出来的，记 $g^{\delta }\left(x\mathrm{,}y\right)\in L^2\left(R^2\right)$ 为带噪音的测量数据，且满足

$\Vert g-g^{\delta }\Vert \le \delta \mathrm{,}$ (2)

$\delta$ 表示输入数据的噪音水平且 $\delta >0$。 $\Vert \cdot \Vert$ 是 $L^2\left(R^2\right)$ 空间的范数。进一步，给出如下先验假设

$\Vert u\left(\cdot \mathrm{,}\cdot \mathrm{,0}\right)\Vert \le E\mathrm{,}$ (3)

其中E为大于0的有界常数。

2. 问题的解与不适定性分析

下面分析问题(1)在 $L^2\left(R^2\right)$ 空间中的不适定性。我们定义 $f\left(x\mathrm{,}y\right)\in L^2\left(R^2\right)$ 的Fourier变换 $\hat{f}\left(\xi \mathrm{,}\eta \right)$ 如下：

$\hat{f}\left(\xi \mathrm{,}\eta \right)\mathrm{<mo>\; :\; </mo>}=\frac{1}{2\pi }{\displaystyle {\int }_{R^2}}{\text{e}}^{-i\left(\xi x+\eta y\right)}f\left(x\mathrm{,}y\right)\text{d}x\text{d}y\mathrm{,}$

对问题(1)中的精确解 $u\left(x\mathrm{,}y\mathrm{,}t\right)$ 进行Fourier变换得到

$\hat{u}\left(\xi ,\eta ,t\right)={\text{e}}^{\left({\xi }^2+{\eta }^2\right)\left(T-t\right)}\hat{g}\left(\xi ,\eta \right),$ (4)

因此，作Fourier逆变换有

$u\left(x\mathrm{,}y\mathrm{,}t\right)=\frac{1}{2\pi }{\displaystyle {\int }_{R^2}}{\text{e}}^{i\left(\xi x+\eta y\right)}{\text{e}}^{\left({\xi }^2+{\eta }^2\right)\left(T-t\right)}\hat{g}\left(\xi \mathrm{,}\eta \right)\text{d}\xi \text{d}\eta \mathrm{,}$ (5)

当 $t=0$ 时，

$\hat{u}\left(\xi \mathrm{,}\eta \mathrm{,0}\right)={\text{e}}^{\left({\xi }^2+{\eta }^2\right)T}\hat{g}\left(\xi \mathrm{,}\eta \right)\mathrm{,}$ (6)

注意到， ${\text{e}}^{\left({\xi }^2+{\eta }^2\right)\left(T-t\right)}$ 为放大因子，即对固定的 $t\in \left(\mathrm{0,}T\right)$，它关于 $\xi \mathrm{,}\eta$ 是无界的。若假设精确解 $\hat{u}\left(\xi \mathrm{,}\eta \mathrm{,}t\right)={\text{e}}^{\left({\xi }^2+{\eta }^2\right)\left(T-t\right)}\hat{g}\left(\xi \mathrm{,}\eta \right)$ 在 $L^2\left(R^2\right)$ 中，则当 $\xi \mathrm{,}\eta \to \infty$ 时， $\hat{g}\left(\xi \mathrm{,}\eta \right)$ 必须为急降函数。否则当输入数据 $g\left(\xi \mathrm{,}\eta \right)$ 有微小扰动时，放大因子 ${\text{e}}^{\left({\xi }^2+{\eta }^2\right)\left(T-t\right)}$ 将解无限放大，导致解的爆破。我们只有带噪音的数据 ${\hat{g}}^{\delta }\left(\xi \mathrm{,}\eta \right)\in L^2\left(R^2\right)$，而 ${\hat{g}}^{\delta }\left(\xi \mathrm{,}\eta \right)$ 一般不是急降函数，故问题严重不适定。

针对传统的正则化方法，如Fourier正则化方法和经典Tihonov正则化方法给出的近似解过度光滑，因此本文将用拟逆正则化方法和分数次Tihonov正则化方法来解决该不适定问题，并给出后验参数选取及相应的误差估计。

3. 预备定理

引理1. [12] 若常数 $\alpha >0$， $0<p<q$，且 $s\ge 0$，则有下述不等式成立

$\underset{s\ge 0}{\sup }\frac{{\text{e}}^{sp}}{1+\alpha {\text{e}}^{sq}}\le {\alpha }^{-\frac{p}{q}}.$

引理2. [13] 对于 $r\ge 0$，有下述不等式成立

$1-{\text{e}}^{-r}\le r.$

4. 分数次Tikhonov正则化方法后验参数选取及误差估计

由前述内容可知问题(1)的精确解为

$\hat{u}\left(\xi ,\eta ,t\right)={\text{e}}^{\left({\xi }^2+{\eta }^2\right)\left(T-t\right)}\hat{g}\left(\xi ,\eta \right),$

我们选取分数次Tihonov正则化解

${\hat{u}}_{\alpha ,\delta }\left(\xi ,\eta ,t\right)=\frac{{\text{e}}^{\left({\xi }^2+{\eta }^2\right)\left(T-t\right)}}{1+\alpha {\text{e}}^{\gamma \left({\xi }^2+{\eta }^2\right)T}}{\hat{g}}^{\delta }\left(\xi ,\eta \right).$ (7)

这里 $\alpha >0$ 为正则化参数， $1\le \gamma \le 2$ 为分数次参数。当 $\gamma =1$ 时为拟边界方法；当 $\gamma =2$ 时为经典Tihonov方法。记

${\hat{u}}_{\alpha }\left(\xi ,\eta ,t\right)=\frac{{\text{e}}^{\left({\xi }^2+{\eta }^2\right)\left(T-t\right)}}{1+\alpha {\text{e}}^{\gamma \left({\xi }^2+{\eta }^2\right)T}}\hat{g}\left(\xi ,\eta \right).$ (8)

下面我们考虑分数次Tihonov正则化方法的后验参数选取。当噪音水平 $\delta$ 已知时，一般采用Morozov’s偏差原理进行后验正则化参数选取，找到一个 $\alpha$ 满足方程

$\rho \left(\alpha \right)=\Vert \frac{1}{1+\alpha {\text{e}}^{\left({\xi }^2+{\eta }^2\right)T\gamma }}{\hat{g}}^{\delta }\left(\xi \mathrm{,}\eta \right)-{\hat{g}}^{\delta }\left(\xi \mathrm{,}\eta \right)\Vert =\tau \delta \mathrm{,}$ (9)

其中 $\tau >1$ 是一个常数， $\alpha >0$ 是正则化参数。

引理3. 设对固定的 $\delta >0$， $\rho \left(\alpha \right)$ 满足如下几条性质：

1) $\rho \left(\alpha \right)$ 是连续函数；

2) $\underset{\alpha \to 0}{\lim }\rho \left(\alpha \right)=0$ ；

3) $\underset{\alpha \to \infty }{\lim }\rho \left(\alpha \right)=\Vert g^{\delta }\Vert$ ；

4) $\rho \left(\alpha \right)$ 是严格单调增函数。

证明：由 $\rho \left(\alpha \right)$ 的表达式

$\begin{array}{c}\rho \left(\alpha \right)=\Vert \frac{1}{1+\alpha {\text{e}}^{\left({\xi }^2+{\eta }^2\right)T\gamma }}{\hat{g}}^{\delta }\left(\xi ,\eta \right)-{\hat{g}}^{\delta }\left(\xi ,\eta \right)\Vert \\ =\Vert \frac{\alpha {\text{e}}^{\left({\xi }^2+{\eta }^2\right)T\gamma }}{1+\alpha {\text{e}}^{\left({\xi }^2+{\eta }^2\right)T\gamma }}{\hat{g}}^{\delta }\left(\xi ,\eta \right)\Vert \end{array}$

容易验证上述结论成立。若选择常数 $\tau$ 满足 $0<\tau \delta <\Vert g^{\delta }\Vert$ 时，根据引理内容知(9)的解是存在且唯一的。

引理4. 假设噪音水平(2)和先验条件(3)成立，如果 $\alpha$ 是(9)的解，可得如下不等式

$\Vert \frac{1}{1+\alpha {\text{e}}^{\left({\xi }^2+{\eta }^2\right)T\gamma }}{\hat{g}}^{\delta }\left(\xi ,\eta \right)-\hat{g}\left(\xi ,\eta \right)\Vert \le \left(\tau +1\right)\delta .$ (10)

证明：由三角不等式，噪音水平(2)及(9)有

$\begin{array}{l}\Vert \frac{1}{1+\alpha {\text{e}}^{\left({\xi }^2+{\eta }^2\right)T\gamma }}{\hat{g}}^{\delta }\left(\xi ,\eta \right)-\hat{g}\left(\xi ,\eta \right)\Vert \\ =\Vert \frac{1}{1+\alpha {\text{e}}^{\left({\xi }^2+{\eta }^2\right)T\gamma }}{\hat{g}}^{\delta }\left(\xi ,\eta \right)-{\hat{g}}^{\delta }\left(\xi ,\eta \right)+{\hat{g}}^{\delta }\left(\xi ,\eta \right)-\hat{g}\left(\xi ,\eta \right)\Vert \\ \le \Vert \frac{1}{1+\alpha {\text{e}}^{\left({\xi }^2+{\eta }^2\right)T\gamma }}{\hat{g}}^{\delta }\left(\xi ,\eta \right)-{\hat{g}}^{\delta }\left(\xi ,\eta \right)\Vert +\Vert g^{\delta }\left(\xi ,\eta \right)-\hat{g}\left(\xi ,\eta \right)\Vert \\ \le \left(\tau +1\right)\delta .\end{array}$

引理5. 假设噪音水平(2)和先验界(3)成立。如果 $\alpha$ 是(9)的解，有不等式

${\alpha }^{-\frac{1}{\gamma }}\le \frac{E}{\left(\tau -1\right)\delta }\mathrm{.}$ (11)

证明：由噪音水平(2)和(9)，可得

$\begin{array}{c}\tau \delta =\Vert \frac{1}{1+\alpha {\text{e}}^{\left({\xi }^2+{\eta }^2\right)T\gamma }}{\hat{g}}^{\delta }\left(\xi \mathrm{,}\eta \right)-{\hat{g}}^{\delta }\left(\xi \mathrm{,}\eta \right)\Vert \\ =\Vert \frac{\alpha {\text{e}}^{\left({\xi }^2+{\eta }^2\right)T\gamma }}{1+\alpha {\text{e}}^{\left({\xi }^2+{\eta }^2\right)T\gamma }}{\hat{g}}^{\delta }\left(\xi \mathrm{,}\eta \right)\Vert \\ =\Vert \frac{\alpha {\text{e}}^{\left({\xi }^2+{\eta }^2\right)T\gamma }}{1+\alpha {\text{e}}^{\left({\xi }^2+{\eta }^2\right)T\gamma }}\left({\hat{g}}^{\delta }\left(\xi \mathrm{,}\eta \right)-\hat{g}\left(\xi \mathrm{,}\eta \right)+\hat{g}\left(\xi \mathrm{,}\eta \right)\right)\Vert \\ \le \Vert \frac{\alpha {\text{e}}^{\left({\xi }^2+{\eta }^2\right)T\gamma }}{1+\alpha {\text{e}}^{\left({\xi }^2+{\eta }^2\right)T\gamma }}\left({\hat{g}}^{\delta }\left(\xi \mathrm{,}\eta \right)-\hat{g}\left(\xi \mathrm{,}\eta \right)\right)\Vert +\Vert \frac{\alpha {\text{e}}^{\left({\xi }^2+{\eta }^2\right)T\left(\gamma -1\right)}}{1+\alpha {\text{e}}^{\left({\xi }^2+{\eta }^2\right)T\gamma }}{\text{e}}^{\left({\xi }^2+{\eta }^2\right)T}\hat{g}\left(\xi \mathrm{,}\eta \right)\Vert \\ \le \delta +\alpha E\underset{\xi \mathrm{,}\eta \in R}{\sup }\left(\frac{{\text{e}}^{\left({\xi }^2+{\eta }^2\right)T\left(\gamma -1\right)}}{1+\alpha {\text{e}}^{\left({\xi }^2+{\eta }^2\right)T\gamma }}\right)\\ \le \delta +E{\alpha }^{\frac{1}{\gamma }}\mathrm{,}\end{array}$

由此，有

${\alpha }^{-\frac{1}{\gamma }}\le \frac{E}{\left(\tau -1\right)\delta }\mathrm{.}$

定理1：假设噪音水平(2)及先验条件(3)成立。正则化参数 $\alpha$ 是由Morozov’s偏差原理选取的，则有如下误差估计

$\Vert u_{\alpha \mathrm{,}\delta }\left(\cdot \mathrm{,}\cdot \mathrm{,}t\right)-u\left(\cdot \mathrm{,}\cdot \mathrm{,}t\right)\Vert \le {\left(\frac{\tau }{\tau -1}\right)}^{1-\frac{t}{T}}{\left(\tau +1\right)}^{\frac{t}{T}}{E}^{1-\frac{t}{T}}{\delta }^{\frac{t}{T}}\mathrm{.}$ (12)

证明：设 $J=\Vert u\left(\cdot \mathrm{,}\cdot \mathrm{,}t\right)-u_{\alpha \mathrm{,}\delta }\left(\cdot \mathrm{,}\cdot \mathrm{,}t\right)\Vert$。由Parseval’s等式，(4)及(7)有

$\begin{array}{c}{J}^2={\Vert u\left(\cdot \mathrm{,}\cdot \mathrm{,}t\right)-u_{\alpha \mathrm{,}\delta }\left(\cdot \mathrm{,}\cdot \mathrm{,}t\right)\Vert }^2\\ ={\Vert \hat{u}\left(\cdot \mathrm{,}\cdot \mathrm{,}t\right)-{\hat{u}}_{\alpha \mathrm{,}\delta }\left(\cdot \mathrm{,}\cdot \mathrm{,}t\right)\Vert }^2\\ ={\Vert {\text{e}}^{\left({\xi }^2+{\eta }^2\right)\left(T-t\right)}\hat{g}\left(\xi \mathrm{,}\eta \right)-\frac{{\text{e}}^{\left({\xi }^2+{\eta }^2\right)\left(T-t\right)}}{1+\alpha {\text{e}}^{\left({\xi }^2+{\eta }^2\right)T\gamma }}{\hat{g}}^{\delta }\left(\xi \mathrm{,}\eta \right)\Vert }^2\\ ={\Vert {\text{e}}^{\left({\xi }^2+{\eta }^2\right)\left(T-t\right)}\left(\hat{g}\left(\xi \mathrm{,}\eta \right)-\frac{1}{1+\alpha {\text{e}}^{\left({\xi }^2+{\eta }^2\right)T\gamma }}{\hat{g}}^{\delta }\left(\xi \mathrm{,}\eta \right)\right)\Vert }^2\mathrm{,}\end{array}$

由Hölder不等式，得

$\begin{array}{c}{J}^2={\displaystyle {\int }_{R^2}}{\text{e}}^{2\left({\xi }^2+{\eta }^2\right)\left(T-t\right)}{\left|\hat{g}\left(\xi \mathrm{,}\eta \right)-\frac{1}{1+\alpha {\text{e}}^{\left({\xi }^2+{\eta }^2\right)T\gamma }}{\hat{g}}^{\delta }\left(\xi \mathrm{,}\eta \right)\right|}^2\text{d}\xi \text{d}\eta \\ ={\displaystyle {\int }_{R^2}}{\text{e}}^{2\left({\xi }^2+{\eta }^2\right)\left(T-t\right)}{\left|\hat{g}\left(\xi \mathrm{,}\eta \right)-\frac{1}{1+\alpha {\text{e}}^{\left({\xi }^2+{\eta }^2\right)T\gamma }}{\hat{g}}^{\delta }\left(\xi \mathrm{,}\eta \right)\right|}^{2\left(1-\frac{t}{T}\right)}\\ \mathrm{<mtext>\; \hspace{0.17em}\; </mtext>\; <mtext>\; \hspace{0.17em}\; </mtext>\; <mtext>\; \hspace{0.17em}\; </mtext>}\cdot {\left|\hat{g}\left(\xi \mathrm{,}\eta \right)-\frac{1}{1+\alpha {\text{e}}^{\left({\xi }^2+{\eta }^2\right)T\gamma }}{\hat{g}}^{\delta }\left(\xi \mathrm{,}\eta \right)\right|}^{\frac{2t}{T}}\text{d}\xi \text{d}\eta \end{array}$

$\begin{array}{c}\le {\left({\displaystyle {\int }_{R^2}}{\left({\text{e}}^{2\left({\xi }^2+{\eta }^2\right)\left(T-t\right)}{\left|\hat{g}\left(\xi \mathrm{,}\eta \right)-\frac{1}{1+\alpha {\text{e}}^{\left({\xi }^2+{\eta }^2\right)T\gamma }}{\hat{g}}^{\delta }\left(\xi \mathrm{,}\eta \right)\right|}^{2\left(1-\frac{t}{T}\right)}\right)}^{\frac{T}{T-t}}\text{d}\xi \text{d}\eta \right)}^{1-\frac{t}{T}}\\ \mathrm{<mtext>\; \hspace{0.17em}\; </mtext>\; <mtext>\; \hspace{0.17em}\; </mtext>\; <mtext>\; \hspace{0.17em}\; </mtext>}\cdot {\left({\displaystyle {\int }_{R^2}}{\left({\left|\hat{g}\left(\xi \mathrm{,}\eta \right)-\frac{1}{1+\alpha {\text{e}}^{\left({\xi }^2+{\eta }^2\right)T\gamma }}{\hat{g}}^{\delta }\left(\xi \mathrm{,}\eta \right)\right|}^{\frac{2t}{T}}\right)}^{\frac{T}{t}}\text{d}\xi \text{d}\eta \right)}^{\frac{t}{T}}\\ \le {\left({\displaystyle {\int }_{R^2}}{\text{e}}^{2\left({\xi }^2+{\eta }^2\right)T}{\left|\hat{g}\left(\xi \mathrm{,}\eta \right)-\frac{1}{1+\alpha {\text{e}}^{\left({\xi }^2+{\eta }^2\right)T\gamma }}{\hat{g}}^{\delta }\left(\xi \mathrm{,}\eta \right)\right|}^2\text{d}\xi \text{d}\eta \right)}^{1-\frac{t}{T}}\\ \mathrm{<mtext>\; \hspace{0.17em}\; </mtext>\; <mtext>\; \hspace{0.17em}\; </mtext>\; <mtext>\; \hspace{0.17em}\; </mtext>}\cdot {\left({\displaystyle {\int }_{R^2}}{\left|\hat{g}\left(\xi \mathrm{,}\eta \right)-\frac{1}{1+\alpha {\text{e}}^{\left({\xi }^2+{\eta }^2\right)T\gamma }}{\hat{g}}^{\delta }\left(\xi \mathrm{,}\eta \right)\right|}^2\text{d}\xi \text{d}\eta \right)}^{\frac{t}{T}}\mathrm{,}\end{array}$

因此有

$\begin{array}{c}{J}^2\le {\Vert {\text{e}}^{\left({\xi }^2+{\eta }^2\right)T}\left(\hat{g}\left(\xi \mathrm{,}\eta \right)-\frac{1}{1+\alpha {\text{e}}^{\left({\xi }^2+{\eta }^2\right)T\gamma }}{\hat{g}}^{\delta }\left(\xi \mathrm{,}\eta \right)\right)\Vert }^{2\left(1-\frac{t}{T}\right)}{\Vert \hat{g}\left(\xi \mathrm{,}\eta \right)-\frac{1}{1+\alpha {\text{e}}^{\left({\xi }^2+{\eta }^2\right)T\gamma }}{\hat{g}}^{\delta }\left(\xi \mathrm{,}\eta \right)\Vert }^{\frac{2t}{T}}\\ ={\Vert {\text{e}}^{\left({\xi }^2+{\eta }^2\right)T}\hat{g}\left(\xi \mathrm{,}\eta \right)-\frac{{\text{e}}^{\left({\xi }^2+{\eta }^2\right)T}}{1+\alpha {\text{e}}^{\left({\xi }^2+{\eta }^2\right)T\gamma }}{\hat{g}}^{\delta }\left(\xi \mathrm{,}\eta \right)\Vert }^{2\left(1-\frac{t}{T}\right)}{\Vert \hat{g}\left(\xi \mathrm{,}\eta \right)-\frac{1}{1+\alpha {\text{e}}^{\left({\xi }^2+{\eta }^2\right)T\gamma }}{\hat{g}}^{\delta }\left(\xi \mathrm{,}\eta \right)\Vert }^{\frac{2t}{T}}\\ \le {\left(\Vert {\text{e}}^{\left({\xi }^2+{\eta }^2\right)T}\hat{g}\left(\xi \mathrm{,}\eta \right)\left(1-\frac{1}{1+\alpha {\text{e}}^{\left({\xi }^2+{\eta }^2\right)T\gamma }}\right)\Vert +\Vert \frac{{\text{e}}^{\left({\xi }^2+{\eta }^2\right)T}}{1+\alpha {\text{e}}^{\left({\xi }^2+{\eta }^2\right)T\gamma }}\left(\hat{g}\left(\xi \mathrm{,}\eta \right)-{\hat{g}}^{\delta }\left(\xi \mathrm{,}\eta \right)\right)\Vert \right)}^{2\left(1-\frac{t}{T}\right)}\\ \mathrm{<mtext>\; \hspace{0.17em}\; </mtext>\; <mtext>\; \hspace{0.17em}\; </mtext>\; <mtext>\; \hspace{0.17em}\; </mtext>}\cdot {\Vert \hat{g}\left(\xi \mathrm{,}\eta \right)-\frac{1}{1+\alpha {\text{e}}^{\left({\xi }^2+{\eta }^2\right)T\gamma }}{\hat{g}}^{\delta }\left(\xi \mathrm{,}\eta \right)\Vert }^{\frac{2t}{T}}\mathrm{,}\end{array}$

根据引理1、引理4、引理5，易得

$\begin{array}{c}{J}^2\le {\left(E+\delta \underset{\xi \mathrm{,}\eta \in R}{\sup }\left(\frac{{\text{e}}^{\left({\xi }^2+{\eta }^2\right)T}}{1+\alpha {\text{e}}^{\left({\xi }^2+{\eta }^2\right)T\gamma }}\right)\right)}^{2\left(1-\frac{t}{T}\right)}{\left(\left(\tau +1\right)\delta \right)}^{\frac{2t}{T}}\\ \le E^{2\left(1-\frac{t}{T}\right)}{\delta }^{\frac{2t}{T}}{\left(\frac{\tau }{\tau -1}\right)}^{2\left(1-\frac{t}{T}\right)}{\left(\tau +1\right)}^{\frac{2t}{T}}\mathrm{,}\end{array}$

即

${\Vert u_{\alpha \mathrm{,}\delta }\left(\cdot \mathrm{,}\cdot \mathrm{,}t\right)-u\left(\cdot \mathrm{,}\cdot \mathrm{,}t\right)\Vert }^2\le {\left(\frac{\tau }{\tau -1}\right)}^{2\left(1-\frac{t}{T}\right)}{\left(\tau +1\right)}^{\frac{2t}{T}}{E}^{2\left(1-\frac{t}{T}\right)}{\delta }^{\frac{2t}{T}}\mathrm{,}$

故

$\Vert u_{\alpha \mathrm{,}\delta }\left(\cdot \mathrm{,}\cdot \mathrm{,}t\right)-u\left(\cdot \mathrm{,}\cdot \mathrm{,}t\right)\Vert \le {\left(\frac{\tau }{\tau -1}\right)}^{1-\frac{t}{T}}{\left(\tau +1\right)}^{\frac{t}{T}}{E}^{1-\frac{t}{T}}{\delta }^{\frac{t}{T}}\mathrm{.}$

5. 拟逆正则化方法后验参数选取及误差估计

对于问题(1)给出拟逆正则化方法近似如下：

$\{\begin{array}{l}{u}_t\left(x\mathrm{,}y\mathrm{,}t\right)-\alpha {\left(\Delta u\right)}_t\left(x\mathrm{,}y\mathrm{,}t\right)-\Delta u\left(x\mathrm{,}y\mathrm{,}t\right)=\mathrm{0,}\left(x\mathrm{,}y\right)\in R^2\mathrm{,}t\in (0\mathrm{,}T)\mathrm{,}\\ u\left(x\mathrm{,}y\mathrm{,}T\right)=g^{\delta }\left(x\mathrm{,}y\right)\mathrm{,}\left(x\mathrm{,}y\right)\in R^2\mathrm{,}\end{array}$ (13)

这里的 $\alpha >0$ 为正则化参数。对(13)作Fourier变换有

$\{\begin{array}{l}{\hat{u}}_t\left(\xi \mathrm{,}\eta \mathrm{,}t\right)+\alpha \left({\xi }^2+{\eta }^2\right){\hat{u}}_t\left(\xi \mathrm{,}\eta \mathrm{,}t\right)+\left({\xi }^2+{\eta }^2\right)\hat{u}\left(\xi \mathrm{,}\eta \mathrm{,}t\right)=\mathrm{0,}\left(\xi \mathrm{,}\eta \right)\in R^2\mathrm{,}t\in (0\mathrm{,}T)\mathrm{,}\\ \hat{u}\left(\xi \mathrm{,}\eta \mathrm{,}T\right)={\hat{g}}^{\delta }\left(\xi \mathrm{,}\eta \right)\mathrm{,}\left(\xi \mathrm{,}\eta \right)\in R^2\mathrm{.}\end{array}$ (14)

其频域上相应的正则解为

${\hat{u}}_{\alpha \mathrm{,}\delta }\left(\xi \mathrm{,}\eta \mathrm{,}t\right)={\text{e}}^{\frac{\left({\xi }^2+{\eta }^2\right)\left(T-t\right)}{1+\alpha \left({\xi }^2+{\eta }^2\right)}}{\hat{g}}^{\delta }\left(\xi \mathrm{,}\eta \right)\mathrm{.}$

记

${\hat{u}}_{\alpha }\left(\xi \mathrm{,}\eta \mathrm{,}t\right)={\text{e}}^{\frac{\left({\xi }^2+{\eta }^2\right)\left(T-t\right)}{1+\alpha \left({\xi }^2+{\eta }^2\right)}}\hat{g}\left(\xi \mathrm{,}\eta \right)\mathrm{.}$ (15)

下面我们考虑拟逆正则化方法的后验参数选取以及相应的误差估计。当噪音水平 $\delta$ 已知时，一般采用Morozov’s偏差原理进行后验正则化参数选取，找到一个 $\alpha$ 满足方程

$\begin{array}{c}\rho \left(\alpha \right)=\Vert {\text{e}}^{-\frac{\alpha {\left({\xi }^2+{\eta }^2\right)}^2\left(T-t\right)}{1+\alpha \left({\xi }^2+{\eta }^2\right)}}{\hat{g}}^{\delta }\left(\xi \mathrm{,}\eta \right)-{\hat{g}}^{\delta }\left(\xi \mathrm{,}\eta \right)\Vert \\ =\delta +\tau {\left(\log \log \left(\frac{1}{\delta }\right)\right)}^{-1}\mathrm{.}\end{array}$ (16)

其中 $\tau >0$ 是一个常数， $\alpha >0$ 是正则化参数。

引理6. 设对固定的 $\delta >0$， $\rho \left(\alpha \right)$ 满足以下几条性质：

1) $\rho \left(\alpha \right)$ 是连续函数；

2) $\underset{\alpha \to 0}{\lim }\rho \left(\alpha \right)=0$ ；

3) $\underset{\alpha \to \infty }{\lim }\rho \left(\alpha \right)=\Vert g^{\delta }\Vert$ ；

4) $\rho \left(\alpha \right)$ 是严格单调增函数。

证明：由 $\rho \left(\alpha \right)$ 的表达式

$\rho \left(\alpha \right)=\Vert {\text{e}}^{-\frac{\alpha {\left({\xi }^2+{\eta }^2\right)}^2\left(T-t\right)}{1+\alpha \left({\xi }^2+{\eta }^2\right)}}{\hat{g}}^{\delta }\left(\xi \mathrm{,}\eta \right)-{\hat{g}}^{\delta }\left(\xi \mathrm{,}\eta \right)\Vert$

容易验证上述结论成立。若选择常数 $\tau$ 满足 $0<\delta +\tau {\left(\log \log \left(\frac{1}{\delta }\right)\right)}^{-1}<\Vert {\hat{g}}^{\delta }\Vert$ 时，根据引理内容知(16)的解是存在且唯一的。

引理7. 如果 $\alpha$ 是(16)的解，则有下述不等式成立

$\Vert \hat{g}\left(\xi \mathrm{,}\eta \right)-{\text{e}}^{-\frac{\alpha {\left({\xi }^2+{\eta }^2\right)}^2\left(T-t\right)}{1+\alpha \left({\xi }^2+{\eta }^2\right)}}{\hat{g}}^{\delta }\left(\xi \mathrm{,}\eta \right)\Vert \le 2\delta +\tau {\left(\log \log \left(\frac{1}{\delta }\right)\right)}^{-1}\mathrm{.}$ (17)

证明：由三角不等式和噪音水平(2)可得

$\begin{array}{l}\Vert \hat{g}\left(\xi ,\eta \right)-{\text{e}}^{-\frac{\alpha {\left({\xi }^2+{\eta }^2\right)}^2\left(T-t\right)}{1+\alpha \left({\xi }^2+{\eta }^2\right)}}{\hat{g}}^{\delta }\left(\xi ,\eta \right)\Vert \\ =\Vert {\text{e}}^{-\frac{\alpha {\left({\xi }^2+{\eta }^2\right)}^2\left(T-t\right)}{1+\alpha \left({\xi }^2+{\eta }^2\right)}}{\hat{g}}^{\delta }\left(\xi ,\eta \right)-{\hat{g}}^{\delta }\left(\xi ,\eta \right)+{\hat{g}}^{\delta }\left(\xi ,\eta \right)-\hat{g}\left(\xi ,\eta \right)\Vert \\ \le \Vert {\text{e}}^{-\frac{\alpha {\left({\xi }^2+{\eta }^2\right)}^2\left(T-t\right)}{1+\alpha \left({\xi }^2+{\eta }^2\right)}}{\hat{g}}^{\delta }\left(\xi ,\eta \right)-{\hat{g}}^{\delta }\left(\xi ,\eta \right)\Vert +\Vert {\hat{g}}^{\delta }\left(\xi ,\eta \right)-\hat{g}\left(\xi ,\eta \right)\Vert \\ \le 2\delta +\tau {\left(\log \log \left(\frac{1}{\delta }\right)\right)}^{-1}.\end{array}$

引理8. 假设噪音水平(2)成立，正则化参数 $\alpha$ 是由Morozov’s偏差原理(16)选取，则有下述不等式成立：

$\frac{1}{\alpha }\le \frac{4\left(T-t\right)E}{\tau {\text{e}}^2{T}^2}\log \log \left(\frac{1}{\delta }\right).$ (18)

证明：由噪音水平(2)和先验界(3)以及(16)，得

$\begin{array}{c}\delta +\tau {\left(\log \log \left(\frac{1}{\delta }\right)\right)}^{-1}=\Vert {\text{e}}^{-\frac{\alpha {\left({\xi }^2+{\eta }^2\right)}^2\left(T-t\right)}{1+\alpha \left({\xi }^2+{\eta }^2\right)}}{\hat{g}}^{\delta }\left(\xi ,\eta \right)-{\hat{g}}^{\delta }\left(\xi ,\eta \right)\Vert \\ =\Vert \left(1-{\text{e}}^{-\frac{\alpha {\left({\xi }^2+{\eta }^2\right)}^2\left(T-t\right)}{1+\alpha \left({\xi }^2+{\eta }^2\right)}}\right){\hat{g}}^{\delta }\left(\xi ,\eta \right)\Vert \\ \le \Vert \left(1-{\text{e}}^{-\frac{\alpha {\left({\xi }^2+{\eta }^2\right)}^2\left(T-t\right)}{1+\alpha \left({\xi }^2+{\eta }^2\right)}}\right)\left({\hat{g}}^{\delta }\left(\xi ,\eta \right)-\hat{g}\left(\xi ,\eta \right)\right)\Vert +\Vert \left(1-{\text{e}}^{-\frac{\alpha {\left({\xi }^2+{\eta }^2\right)}^2\left(T-t\right)}{1+\alpha \left({\xi }^2+{\eta }^2\right)}}\right)\hat{g}\left(\xi ,\eta \right)\Vert \\ \le \delta +\Vert \left(1-{\text{e}}^{-\frac{\alpha {\left({\xi }^2+{\eta }^2\right)}^2\left(T-t\right)}{1+\alpha \left({\xi }^2+{\eta }^2\right)}}\right){\text{e}}^{-\left({\xi }^2+{\eta }^2\right)T}{\text{e}}^{\left({\xi }^2+{\eta }^2\right)T}\hat{g}\left(\xi ,\eta \right)\Vert \\ \le \delta +E\sup \left(\left(1-{\text{e}}^{-\frac{\alpha {\left({\xi }^2+{\eta }^2\right)}^2\left(T-t\right)}{1+\alpha \left({\xi }^2+{\eta }^2\right)}}\right){\text{e}}^{-\left({\xi }^2+{\eta }^2\right)T}\right)\\ \le \delta +EG\left(\xi ,\eta \right),\end{array}$

令

$G\left(\xi \mathrm{,}\eta \right)\mathrm{<mo>\; :\; </mo>}=\underset{\xi \mathrm{,}\eta \in R}{\sup }\left(\left(1-{\text{e}}^{-\frac{\alpha {\left({\xi }^2+{\eta }^2\right)}^2\left(T-t\right)}{1+\alpha \left({\xi }^2+{\eta }^2\right)}}\right){\text{e}}^{-\left({\xi }^2+{\eta }^2\right)T}\right),$

由于

$1-{\text{e}}^{-\frac{\alpha {\left({\xi }^2+{\eta }^2\right)}^2\left(T-t\right)}{1+\alpha \left({\xi }^2+{\eta }^2\right)}}\le \frac{\alpha {\left({\xi }^2+{\eta }^2\right)}^2\left(T-t\right)}{1+\alpha \left({\xi }^2+{\eta }^2\right)}\le \alpha {\left({\xi }^2+{\eta }^2\right)}^2\left(T-t\right),$

故有

$G\left(\xi \mathrm{,}\eta \right)\le \underset{\xi \mathrm{,}\eta \in R}{\sup }\left(\alpha {\left({\xi }^2+{\eta }^2\right)}^2\left(T-t\right){\text{e}}^{-\left({\xi }^2+{\eta }^2\right)T}\right)\mathrm{,}$

令 ${\xi }^2+{\eta }^2=m$，则对m求导，得到极大值点 $m=\frac{2}{T}$，则

$\underset{\xi \mathrm{,}\eta \in R}{\sup }G\left(\xi \mathrm{,}\eta \right)\le \underset{\xi \mathrm{,}\eta \in R}{\sup }\left(\alpha {\left({\xi }^2+{\eta }^2\right)}^2\left(T-t\right){\text{e}}^{-\left({\xi }^2+{\eta }^2\right)T}\right)\le \frac{4\alpha \left(T-t\right)}{{\text{e}}^2{T}^2}\mathrm{,}$

于是有

$\tau {\left(\log \log \left(\frac{1}{\delta }\right)\right)}^{-1}\le \frac{4\alpha \left(T-t\right)}{{\text{e}}^2{T}^2}E\mathrm{,}$

从而

$\frac{1}{\alpha }\le \frac{4\left(T-t\right)E}{\tau {\text{e}}^2{T}^2}\log \log \left(\frac{1}{\delta }\right)\mathrm{.}$

定理2. 假设噪音水平(2)和先验条件(3)成立，并存在 $\tau >0$，使得 $0<\delta +\tau {\left(\log \log \left(\frac{1}{\delta }\right)\right)}^{-1}<\Vert g^{\delta }\Vert$，对于 $0\le t<T$，正则化参数 $\alpha$ 是由后验正则化参数选取得到的，则有如下估计：

$\Vert {\hat{u}}_{\alpha \mathrm{,}\delta }\left(\cdot \mathrm{,}\cdot \mathrm{,}t\right)-\hat{u}\left(\cdot \mathrm{,}\cdot \mathrm{,}t\right)\Vert \le \delta \cdot {\left(\frac{4\left(T-t\right)E}{\tau e^2{T}^2}\log \log \left(\frac{1}{\delta }\right)\right)}^{T-t}+E^{1-\frac{t}{T}}{\left(2\delta +\tau {\left(\log \log \left(\frac{1}{\delta }\right)\right)}^{-1}\right)}^{\frac{t}{T}}\mathrm{.}$

证明：令 $I=\Vert u_{\alpha \mathrm{,}\delta }\left(\cdot \mathrm{,}\cdot \mathrm{,}t\right)-u\left(\cdot \mathrm{,}\cdot \mathrm{,}t\right)\Vert$，由Parseval’s等式及三角不等式得

$\begin{array}{c}I=\Vert u_{\alpha ,\delta }\left(\cdot ,\cdot ,t\right)-u\left(\cdot ,\cdot ,t\right)\Vert \\ =\Vert {\hat{u}}_{\alpha ,\delta }\left(\cdot ,\cdot ,t\right)-\hat{u}\left(\cdot ,\cdot ,t\right)\Vert \\ \le \Vert {\hat{u}}_{\alpha ,\delta }\left(\cdot ,\cdot ,t\right)-{\hat{u}}_{\alpha }\left(\cdot ,\cdot ,t\right)\Vert +\Vert {\hat{u}}_{\alpha }\left(\cdot ,\cdot ,t\right)-\hat{u}\left(\cdot ,\cdot ,t\right)\Vert \\ =A_1+A_2.\end{array}$

则先对 $A_1$ 做估计，根据引理1有

$\begin{array}{c}{A}_1=\Vert u_{\alpha ,\delta }\left(\cdot ,\cdot ,t\right)-u_{\alpha }\left(\cdot ,\cdot ,t\right)\Vert \\ =\Vert {\hat{u}}_{\alpha ,\delta }\left(\cdot ,\cdot ,t\right)-{\hat{u}}_{\alpha }\left(\cdot ,\cdot ,t\right)\Vert \\ =\Vert {\text{e}}^{\frac{\left({\xi }^2+{\eta }^2\right)\left(T-t\right)}{1+\alpha \left({\xi }^2+{\eta }^2\right)}}{\hat{g}}^{\delta }\left(\xi ,\eta \right)-{\text{e}}^{\frac{\left({\xi }^2+{\eta }^2\right)\left(T-t\right)}{1+\alpha \left({\xi }^2+{\eta }^2\right)}}\hat{g}\left(\xi ,\eta \right)\Vert \\ =\Vert {\text{e}}^{\frac{\left({\xi }^2+{\eta }^2\right)\left(T-t\right)}{1+\alpha \left({\xi }^2+{\eta }^2\right)}}\left({\hat{g}}^{\delta }\left(\xi ,\eta \right)-\hat{g}\left(\xi ,\eta \right)\right)\Vert \\ \le \delta \cdot {\text{e}}^{\frac{T-t}{\alpha }}.\end{array}$

由引理8，则

$A_1\le \delta \cdot {\text{e}}^{\frac{4{\left(T-t\right)}^{2}E}{\tau {\text{e}}^2{T}^2}\log \log \left(\frac{1}{\delta }\right)}\mathrm{.}$

再对 $A_2$ 做估计，由Hölder不等式有

$\begin{array}{c}{A}_2^2={\Vert u_{\alpha }\left(\cdot \mathrm{,}\cdot \mathrm{,}t\right)-u\left(\cdot \mathrm{,}\cdot \mathrm{,}t\right)\Vert }^2\\ ={\Vert {\hat{u}}_{\alpha }\left(\cdot \mathrm{,}\cdot \mathrm{,}t\right)-\hat{u}\left(\cdot \mathrm{,}\cdot \mathrm{,}t\right)\Vert }^2\\ =\Vert {\text{e}}^{\frac{\left({\xi }^2+{\eta }^2\right)\left(T-t\right)}{1+\alpha \left({\xi }^2+{\eta }^2\right)}}\hat{g}\left(\xi \mathrm{,}\eta \right)-{\text{e}}^{\left({\xi }^2+{\eta }^2\right)\left(T-t\right)}\hat{g}\left(\xi \mathrm{,}\eta \right)\Vert \\ ={\Vert {\text{e}}^{\left({\xi }^2+{\eta }^2\right)\left(T-t\right)}\left(\hat{g}\left(\xi \mathrm{,}\eta \right)-{\text{e}}^{-\frac{\alpha {\left({\xi }^2+{\eta }^2\right)}^2\left(T-t\right)}{1+\alpha \left({\xi }^2+{\eta }^2\right)}}\hat{g}\left(\xi \mathrm{,}\eta \right)\right)\Vert }^2\\ ={\displaystyle {\iint }_{R^2}}{\text{e}}^{2\left({\xi }^2+{\eta }^2\right)\left(T-t\right)}{\left|\hat{g}\left(\xi \mathrm{,}\eta \right)-{\text{e}}^{-\frac{\alpha {\left({\xi }^2+{\eta }^2\right)}^2\left(T-t\right)}{1+\alpha \left({\xi }^2+{\eta }^2\right)}}\hat{g}\left(\xi \mathrm{,}\eta \right)\right|}^{2\left(1-\frac{t}{T}\right)}\end{array}$

$\begin{array}{c}\mathrm{<mtext>\; \hspace{0.17em}\; </mtext>\; <mtext>\; \hspace{0.17em}\; </mtext>}\cdot {\left|\hat{g}\left(\xi \mathrm{,}\eta \right)-{\text{e}}^{-\frac{\alpha {\left({\xi }^2+{\eta }^2\right)}^2\left(T-t\right)}{1+\alpha \left({\xi }^2+{\eta }^2\right)}}\hat{g}\left(\xi \mathrm{,}\eta \right)\right|}^{\frac{2t}{T}}\text{d}\xi \text{d}\eta \\ \le {\left({\displaystyle {\iint }_{R^2}}{\left({\text{e}}^{2\left({\xi }^2+{\eta }^2\right)\left(T-t\right)}{\left|\hat{g}\left(\xi \mathrm{,}\eta \right)-{\text{e}}^{-\frac{\alpha {\left({\xi }^2+{\eta }^2\right)}^2\left(T-t\right)}{1+\alpha \left({\xi }^2+{\eta }^2\right)}}\hat{g}\left(\xi \mathrm{,}\eta \right)\right|}^{2\left(1-\frac{t}{T}\right)}\right)}^{\frac{T}{T-t}}\text{d}\xi \text{d}\eta \right)}^{\frac{T-t}{T}}\\ \mathrm{<mtext>\; \hspace{0.17em}\; </mtext>\; <mtext>\; \hspace{0.17em}\; </mtext>}\cdot {\left({\displaystyle {\iint }_{R^2}}{\left({\left|\hat{g}\left(\xi \mathrm{,}\eta \right)-{\text{e}}^{-\frac{\alpha {\left({\xi }^2+{\eta }^2\right)}^2\left(T-t\right)}{1+\alpha \left({\xi }^2+{\eta }^2\right)}}\hat{g}\left(\xi \mathrm{,}\eta \right)\right|}^{\frac{2t}{T}}\right)}^{\frac{T}{t}}\text{d}\xi \text{d}\eta \right)}^{\frac{t}{T}}\\ \le {\left({\displaystyle {\iint }_{R^2}}{\text{e}}^{2\left({\xi }^2+{\eta }^2\right)\left(T-t\right)}{\left|\hat{g}\left(\xi \mathrm{,}\eta \right)-{\text{e}}^{-\frac{\alpha {\left({\xi }^2+{\eta }^2\right)}^2\left(T-t\right)}{1+\alpha \left({\xi }^2+{\eta }^2\right)}}\hat{g}\left(\xi \mathrm{,}\eta \right)\right|}^2\text{d}\xi \text{d}\eta \right)}^{\frac{T-t}{T}}\\ \mathrm{<mtext>\; \hspace{0.17em}\; </mtext>\; <mtext>\; \hspace{0.17em}\; </mtext>}\cdot {\left({\displaystyle {\iint }_{R^2}}{\left|\hat{g}\left(\xi \mathrm{,}\eta \right)-{\text{e}}^{-\frac{\alpha {\left({\xi }^2+{\eta }^2\right)}^2\left(T-t\right)}{1+\alpha \left({\xi }^2+{\eta }^2\right)}}\hat{g}\left(\xi \mathrm{,}\eta \right)\right|}^2\text{d}\xi \text{d}\eta \right)}^{\frac{t}{T}}\mathrm{,}\end{array}$

则由(16)式有

$\begin{array}{c}{A}_2^2={\Vert u_{\alpha }\left(\cdot ,\cdot ,t\right)-u\left(\cdot ,\cdot ,t\right)\Vert }^2\\ ={\Vert {\text{e}}^{\left({\xi }^2+{\eta }^2\right)\left(T-t\right)}\left(\hat{g}\left(\xi ,\eta \right)-{\text{e}}^{-\frac{\alpha {\left({\xi }^2+{\eta }^2\right)}^2\left(T-t\right)}{1+\alpha \left({\xi }^2+{\eta }^2\right)}}\hat{g}\left(\xi ,\eta \right)\right)\Vert }^{2\left(1-\frac{t}{T}\right)}{\Vert \hat{g}\left(\xi ,\eta \right)-{\text{e}}^{-\frac{\alpha {\left({\xi }^2+{\eta }^2\right)}^2\left(T-t\right)}{1+\alpha \left({\xi }^2+{\eta }^2\right)}}\hat{g}\left(\xi ,\eta \right)\Vert }^{\frac{2t}{T}}\\ ={\Vert {\text{e}}^{\left({\xi }^2+{\eta }^2\right)\left(T-t\right)}\hat{g}\left(\xi ,\eta \right)\left(1-{\text{e}}^{-\frac{\alpha {\left({\xi }^2+{\eta }^2\right)}^2\left(T-t\right)}{1+\alpha \left({\xi }^2+{\eta }^2\right)}}\right)\Vert }^{2\left(1-\frac{t}{T}\right)}{\Vert \hat{g}\left(\xi ,\eta \right)\left(1-{\text{e}}^{-\frac{\alpha {\left({\xi }^2+{\eta }^2\right)}^2\left(T-t\right)}{1+\alpha \left({\xi }^2+{\eta }^2\right)}}\right)\Vert }^{\frac{2t}{T}}\\ \le E^{2\left(1-\frac{t}{T}\right)}\cdot {\Vert \left(\hat{g}\left(\xi ,\mathrm{<mtext>\; \hspace{0.17em}\; </mtext>}\eta \right)-{\hat{g}}^{\delta }\left(\xi ,\mathrm{<mtext>\; \hspace{0.17em}\; </mtext>}\eta \right)+{\hat{g}}^{\delta }\left(\xi ,\mathrm{<mtext>\; \hspace{0.17em}\; </mtext>}\eta \right)\right)\left(1-{\text{e}}^{-\frac{\alpha {\left({\xi }^2+{\eta }^2\right)}^2\left(T-t\right)}{1+\alpha \left({\xi }^2+{\eta }^2\right)}}\right)\Vert }^{\frac{2t}{T}}\end{array}$

$\begin{array}{c}\le E^{2\left(1-\frac{t}{T}\right)}\cdot {\left(\Vert \left(\hat{g}\left(\xi ,\mathrm{<mtext>\; \hspace{0.17em}\; </mtext>}\eta \right)-{\hat{g}}^{\delta }\left(\xi ,\mathrm{<mtext>\; \hspace{0.17em}\; </mtext>}\eta \right)\right)\left(1-{\text{e}}^{-\frac{\alpha {\left({\xi }^2+{\eta }^2\right)}^2\left(T-t\right)}{1+\alpha \left({\xi }^2+{\eta }^2\right)}}\right)\Vert +\Vert {\hat{g}}^{\delta }\left(\xi ,\mathrm{<mtext>\; \hspace{0.17em}\; </mtext>}\eta \right)\left(1-{\text{e}}^{-\frac{\alpha {\left({\xi }^2+{\eta }^2\right)}^2\left(T-t\right)}{1+\alpha \left({\xi }^2+{\eta }^2\right)}}\right)\Vert \right)}^{\frac{2t}{T}}\\ \le E^{2\left(1-\frac{t}{T}\right)}\cdot {\left(\Vert \hat{g}\left(\xi ,\mathrm{<mtext>\; \hspace{0.17em}\; </mtext>}\eta \right)-{\hat{g}}^{\delta }\left(\xi ,\mathrm{<mtext>\; \hspace{0.17em}\; </mtext>}\eta \right)\Vert +\Vert {\text{e}}^{-\frac{\alpha {\left({\xi }^2+{\eta }^2\right)}^2\left(T-t\right)}{1+\alpha \left({\xi }^2+{\eta }^2\right)}}{\hat{g}}^{\delta }\left(\xi ,\mathrm{<mtext>\; \hspace{0.17em}\; </mtext>}\eta \right)-{\hat{g}}^{\delta }\left(\xi ,\mathrm{<mtext>\; \hspace{0.17em}\; </mtext>}\eta \right)\Vert \right)}^{\frac{2t}{T}}\\ \le E^{2\left(1-\frac{t}{T}\right)}\cdot {\left(2\delta +\tau \log \log \left(\frac{1}{\delta }\right)\right)}^{\frac{2t}{T}}.\end{array}$

于是

$A_2\le E^{1-\frac{t}{T}}\cdot {\left(2\delta +\tau \log \log \left(\frac{1}{\delta }\right)\right)}^{\frac{t}{T}},$

从而

$\begin{array}{l}\Vert u_{\alpha ,\delta }\left(\cdot ,\cdot ,t\right)-u\left(\cdot ,\cdot ,t\right)\Vert \\ =\Vert {\hat{u}}_{\alpha ,\delta }\left(\cdot ,\cdot ,t\right)-\hat{u}\left(\cdot ,\cdot ,t\right)\Vert \\ \le \Vert {\hat{u}}_{\alpha ,\delta }\left(\cdot ,\cdot ,t\right)-{\hat{u}}_{\alpha }\left(\cdot ,\cdot ,t\right)\Vert +\Vert {\hat{u}}_{\alpha }\left(\cdot ,\cdot ,t\right)-\hat{u}\left(\cdot ,\cdot ,t\right)\Vert \\ \le \delta \cdot {\text{e}}^{\frac{4{\left(T-t\right)}^{2}E}{\tau {\text{e}}^2{T}^2}\log \log \left(\frac{1}{\delta }\right)}+E^{1-\frac{t}{T}}\cdot {\left(2\delta +\tau \log \log \left(\frac{1}{\delta }\right)\right)}^{\frac{t}{T}}.\end{array}$

6. 结束语

本文针对二维空间的时间反向热传导问题采用分数次Tihonov正则化方法和拟逆正则化方法进行了正则化，并得到了正则解。同时给出了后验参数选取规则以及正则解与精确解之间稳定的误差估计。

参考文献