二维时间反向热传导问题的Fourier正则化

Fourier Regularization of the Two-Dimensional Time-Reverse Heat Conduction Problem

摘要: 本文用Fourier正则化的方法求解了二维时间反向热传导问题，即在无界区域中通过终值时刻的数据来确定温度的分布。该问题在图像处理方面有着广泛的应用。本文先用Fourier变换求出了问题的精确解，发现该反问题是严重不适定的。为了解决这一反问题，用Fourier正则化的方法构造出了问题的Fourier正则化解，并在先验条件的假设下给出了其精确解与正则化近似解之间的对数型误差估计。最后由偏差原理给出了近似解的后验误差估计。

Abstract: In this paper, the two-dimensional time reverse heat transfer problem is solved by the method of Fourier regularization, that is, the temperature distribution is determined by the data at the final value in the unbounded region. This problem has a wide range of applications in image processing. In this paper, the exact solution of the problem is first solved by using the Fourier transform, and it is found that the inverse problem is seriously ill-posed. In order to solve this inverse problem, the Fourier regular solution of the problem is constructed by the Fourier regularization method, and the logarithmic error estimation between the exact solution and the regularization approximate solution is given under the assumption of the prior conditions. Finally, the posterior error estimation of the approximate solution is given by the bias principle.

1. 引言

与热传导方程有关的反问题大多都是不适定的，也就是即使方程存在唯一的精确解，也不连续依赖给定的数据。近年来，该领域的大多数已发表文献都使用了正则化方法。如Fourier正则化方法 [1] [2] [3] ，Tikhonov正则化方法 [4] [5] [6] 、准边界法 [7] [8] 和拟逆正则化方法 [9] [10] 等。这些方法都被非常广泛的用于各种不适定问题的解决当中。而对于二维时间反向热传导问题，目前研究的还比较少 [11] [12] ，但二维时间反向热传导问题在数学建模以及图像处理等方面有着重要的应用。在文献中 [13] ，作者用拟逆正则化方法和分数次Tikhonov正则化方法求解了二维时间反向热传导问题。在文献 [14] 中，作者用Fourier正则化研究了一维的逆热传导问题。下面本文将用Fourier正则化的方法求解二维时间反向热传导方程，并分别给出了先验和后验参数选取下的误差估计。

考虑无界区域中的二维时间反向热传导方程

$\{\begin{array}{ll}{u}_t\left(x,y,t\right)=\Delta u\left(x,y,t\right),\hfill & -\infty <x,y<\infty ,0\le t<T,\hfill \\ u\left(x,y,T\right)=g_T\left(x,y\right),\hfill & -\infty <x,y<\infty .\hfill \end{array}$ (1)

下面，需通过数据 $g_T\left(x,y\right)$ 来确定温度分布 $u\left(x,y,t\right)\left(0\le t<T\right)$ 。

设 $H^s$ 是一个Sobolev空间，我们定义 $f\left(x,y\right)\in L\left(R\right)\times L\left(R\right)$ 的Fourier变换为 $\hat{f}\left(\xi ,\eta \right)$ ，由以下给出

$\hat{f}\left(\xi ,\eta \right)\text{:}=\frac{1}{2\pi }{\displaystyle {\iint }_{R^2}{\text{e}}^{-i\left(\xi x+\eta y\right)}f\left(x,y\right)\text{d}x\text{d}y}\text{,}-\infty <\xi ,\eta <\infty .$ (2)

函数 $\hat{f}\left(\xi ,\eta \right)$ 相应的Fourier逆变换为

$f\left(x,y\right):=\frac{1}{2\pi }{\displaystyle {\iint }_{R^2}{\text{e}}^{i\left(\xi x+\eta y\right)}\hat{f}\left(\xi ,\eta \right)\text{d}\xi \text{d}\eta ,-\infty <\xi ,\eta <\infty .}$ (3)

$f\left(x,y\right)$ 在Sobolev空间上的范数为

${\Vert f\left(x,y\right)\Vert }_{H^s}:={\left({\displaystyle {\iint }_{R^2}{\left|\hat{f}\left(\xi ,\eta \right)\right|}^2{\left(1+{\xi }^2+{\eta }^2\right)}^s\text{d}\xi \text{d}\eta }\right)}^{\frac{1}{2}}\text{.}$ (4)

当 $s=0$ 时， ${\Vert \left(\cdot ,\cdot \right)\Vert }_{H^0}=\Vert \left(\cdot ,\cdot \right)\Vert$ 为 $L_2$ 范数。

在本文的第二节中，用Fourier变换得到了问题的精确解并给出了反问题的不适定性分析；在第三节中，我们构造出了问题的Fourier正则化近似解，给出了先验参数选取下的对数型误差估计，并由偏差原理给出了后验参数选取下的Hölde型误差估计。最后，对本文进行了一个总结。

2. 问题的解及不适定性分析

首先对问题(1)关于变量 $x,y$ 作Fourier变换，可得

$\{\begin{array}{ll}{\hat{u}}_t\left(\xi ,\eta ,t\right)+\left({\xi }^2+{\eta }^2\right)\hat{u}\left(\xi ,\eta ,t\right)=0,\hfill & -\infty <\xi ,\eta <\infty ,0\le t<T,\hfill \\ \hat{u}\left(\xi ,\eta ,T\right)={\hat{g}}_T\left(\xi ,\eta \right),\hfill & -\infty <\xi ,\eta <\infty .\hfill \end{array}$ (5)

可以得到问题(5)的解为

$\hat{u}\left(\xi ,\eta ,t\right)={\text{e}}^{\left({\xi }^2\text{+}{\eta }^2\right)\left(T-t\right)}{\hat{g}}_T\left(\xi ,\eta \right).$ (6)

由Fourier逆变换，可以得到问题(1)的解为

$u\left(x,y,t\right):=\frac{1}{2\pi }{\displaystyle {\iint }_{R^2}{\text{e}}^{i\left(\xi x+\eta y\right)}{\text{e}}^{\left({\xi }^2+{\eta }^2\right)\left(T-t\right)}{\hat{g}}_T\left(\xi ,\eta \right)\text{d}\xi \text{d}\eta .}$ (7)

特别的，当 $t=0$ 时，由(6)式，有

$\hat{u}\left(\xi ,\eta ,0\right)={\text{e}}^{\left({\xi }^2\text{+}{\eta }^2\right)T}{\hat{g}}_T\left(\xi ,\eta \right).$ (8)

记 $g_0\left(x,y\right):=u\left(x,y,0\right)$ 考虑问题(1)在 $L^2\left(R\right)$ 中关于变量 $x,y$ 时，设 $g_0\left(x,y\right)$ 存在一个先验界：

$\Vert g_0\Vert :=\Vert u\left(\cdot ,\cdot ,0\right)\Vert \le E.$ (9)

由(8)，(9)式以及Parseval等式，有

${\Vert g_0\Vert }^2={\displaystyle {\iint }_{R^2}{\left|{\text{e}}^{\left({\xi }^2+{\eta }^2\right)T}{\hat{g}}_T\left(\xi ,\eta \right)\right|}^2\text{d}\xi \text{d}\eta <\infty }.$ (10)

由(6)式可知，当 $\left|\xi \right|\to \infty$ ， $\left|\eta \right|\to \infty$ 时， ${\text{e}}^{\left({\xi }^2+{\eta }^2\right)\left(T-t\right)}\to \infty$ ，因此 ${\text{e}}^{\left({\xi }^2+{\eta }^2\right)\left(T-t\right)}$ 是一个放大因子，当 $\left|\xi \right|\to \infty$ ， $\left|\eta \right|\to \infty$ 时， ${\hat{g}}_T\left(\xi ,\eta \right)$ 必须是急速衰减的，否则当 ${\hat{g}}_T\left(\xi ,\eta \right)$ 有微小的扰动时，放大因子 ${\text{e}}^{\left({\xi }^2+{\eta }^2\right)\left(T-t\right)}$ 将会把解无限的放大，最终导致解的爆破，但是在实际问题中T时刻的数据通常都是由测量

得到的，记为 $g_T^{\delta }\left(x,y\right)$ ，且 $g_T^{\delta }\left(x,y\right)$ 一般是不会满足急速衰减的，所以问题(1)是严重不适定的。

3. Fourier正则化及误差估计

由于 $g_T\left(x,y\right)$ 在实际中无法准确的知道，记 $g_T^{\delta }\left(x,y\right)$ 为T时刻的测量数据， $\delta >0$ 为噪音水平，满足：

$\Vert g_T^{\delta }-g_T\Vert \le \delta .$ (11)

对于 $s\ge 0$ ，假设一个如下的先验界：

${\Vert g_0\Vert }_{H^s}\le E.$ (12)

这里的E是一个正常数， ${\Vert \cdot \Vert }_{H^s}$ 表示的是Sobolev空间 $H^s$ 上的范数。

接下来，构造出问题(1)的含有噪音数据 $g_T^{\delta }\left(x,y\right)$ 的正则化近似解，称为问题(1)的Fourier正则化解，由以下给出

$u_{{\chi }_{\xi }{\chi }_{\eta }}^{\delta }\left(x,y,t\right):=\frac{1}{2\pi }{\displaystyle {\iint }_{R^2}{\text{e}}^{i\left(\xi x+\eta y\right)}{\text{e}}^{\left({\xi }^2+{\eta }^2\right)\left(T-t\right)}{\hat{g}}_T^{\delta }\left(\xi ,\eta \right){\chi }_{\xi }{\chi }_{\eta }\text{d}\xi \text{d}\eta }.$ (13)

等价于

${\hat{u}}_{{\chi }_{\xi }{\chi }_{\eta }}^{\delta }\left(\xi ,\eta ,t\right)={\text{e}}^{\left({\xi }^2+{\eta }^2\right)\left(T-t\right)}{\hat{g}}_T^{\delta }\left(\xi ,\eta \right){\chi }_{\xi }{\chi }_{\eta }.$ (14)

其中 ${\xi }_{\max },{\eta }_{\max }$ 是正常数， ${\chi }_{\xi },{\chi }_{\eta }$ 分别表示的是区间 $\left[-{\xi }_{\max },{\xi }_{\max }\right]$ 和 $\left[-{\eta }_{\max },{\eta }_{\max }\right]$ 上的特征函数， ${\xi }_{\max },{\eta }_{\max }$ 在这里扮演的是正则化参数的角色。

${\chi }_{\xi }=\{\begin{array}{l}1,\left|\xi \right|\le {\xi }_{\max },\\ 0,\left|\xi \right|>{\xi }_{\max }.\end{array}$

${\chi }_{\eta }=\{\begin{array}{l}1,\left|\eta \right|\le {\eta }_{\max },\\ 0,\left|\eta \right|>{\eta }_{\max }.\end{array}$

3.1. 先验参数选取下的误差估计

定理1设 $u\left(x,y,t\right)$ 和 ${\hat{u}}_{{\chi }_{\xi }{\chi }_{\eta }}^{\delta }\left(\xi ,\eta ,t\right)$ 分别为问题(1)的精确解和Fourier正则化解，假设先验条件(11)和(12)成立，如果选取正则化参数

${\xi }_{\max }={\eta }_{\max }={\left(\ln \left({\left(\frac{E}{\delta }\right)}^{\frac{1}{2T}}{\left(\ln \frac{E}{\delta }\right)}^{-\frac{s}{4T}}\right)\right)}^{\frac{1}{2}}.$ (15)

则有以下的估计

$\Vert u\left(\cdot ,\cdot ,t\right)-u_{{\chi }_{\xi }{\chi }_{\eta }}^{\delta }\left(\cdot ,\cdot ,t\right)\Vert \le E^{\left(1-\frac{t}{T}\right)}{\delta }^{\frac{t}{T}}{\left(\ln \frac{E}{\delta }\right)}^{-\frac{s\left(T-t\right)}{2T}}\left(3{\left(\frac{\ln \frac{E}{\delta }}{\frac{1}{T}\ln \frac{E}{\delta }+\ln {\left(\ln \frac{E}{\delta }\right)}^{-\frac{s}{2T}}}\right)}^{\frac{s}{2}}+1\right).$ (16)

证明：由Parseval等式和三角不等式，由(6)，(14)式有

$\begin{array}{c}\Vert u\left(\cdot ,\cdot ,t\right)-u_{{\chi }_{\xi }{\chi }_{\eta }}^{\delta }\left(\cdot ,\cdot ,t\right)\Vert =\Vert \hat{u}\left(\cdot ,\cdot ,t\right)-{\hat{u}}_{{\chi }_{\xi }{\chi }_{\eta }}^{\delta }\left(\cdot ,\cdot ,t\right)\Vert \\ =\Vert \hat{u}\left(\cdot ,\cdot ,t\right)-{\hat{u}}_{{\chi }_{\xi }{\chi }_{\eta }}\left(\cdot ,\cdot ,t\right)\Vert +\Vert {\hat{u}}_{{\chi }_{\xi }{\chi }_{\eta }}\left(\cdot ,\cdot ,t\right)-{\hat{u}}_{{\chi }_{\xi }{\chi }_{\eta }}^{\delta }\left(\cdot ,\cdot ,t\right)\Vert \\ =I_1+I_2\end{array}$

首先对 $I_1$ 做估计，有

$\begin{array}{c}{I}_1=\Vert \hat{u}\left(\cdot ,\cdot ,t\right)-{\hat{u}}_{{\chi }_{\xi }{\chi }_{\eta }}\left(\cdot ,\cdot ,t\right)\Vert \\ =\Vert {\text{e}}^{\left({\xi }^2+{\eta }^2\right)\left(T-t\right)}{\hat{g}}_T\left(\xi ,\eta \right)-{\text{e}}^{\left({\xi }^2+{\eta }^2\right)\left(T-t\right)}{\hat{g}}_T\left(\xi ,\eta \right){\chi }_{\xi }{\chi }_{\eta }\Vert \\ =\Vert {\text{e}}^{-\left({\xi }^2+{\eta }^2\right)t}{\hat{g}}_0\left(\xi ,\eta \right)\left(1-{\chi }_{\xi }{\chi }_{\eta }\right)\Vert \\ ={\left({\displaystyle {\iint }_{{\Omega }_2}{\left|{\text{e}}^{-\left({\xi }^2+{\eta }^2\right)t}{\hat{g}}_0\left(\xi ,\eta \right)\right|}^2\text{d}\xi \text{d}\eta }\right)}^{\frac{1}{2}}+{\left({\displaystyle {\iint }_{{\Omega }_3}{\left|{\text{e}}^{-\left({\xi }^2+{\eta }^2\right)t}{\hat{g}}_0\left(\xi ,\eta \right)\right|}^2\text{d}\xi \text{d}\eta }\right)}^{\frac{1}{2}}\\ +{\left({\displaystyle {\iint }_{{\Omega }_4}{\left|{\text{e}}^{-\left({\xi }^2+{\eta }^2\right)t}{\hat{g}}_0\left(\xi ,\eta \right)\right|}^2\text{d}\xi \text{d}\eta }\right)}^{\frac{1}{2}}\\ =I_{11}+I_{12}+I_{13}\end{array}$

其中

${\Omega }_1=\left[-{\xi }_{\max },{\xi }_{\max }\right]\times \left[-{\eta }_{\max },{\eta }_{\max }\right]$ ;

${\Omega }_2=\left(\left(-\infty ,-{\xi }_{\max }\right]\cup \left[{\xi }_{\max },+\infty \right)\right)\times \left[-{\eta }_{\max },{\eta }_{\max }\right]$ ;

${\Omega }_3=\left[-{\xi }_{\max },{\xi }_{\max }\right]\times \left(\left(-\infty ,-{\eta }_{\max }\right]\cup \left[{\eta }_{\max },+\infty \right)\right)$ ;

${\Omega }_4=\left(\left(-\infty ,-{\xi }_{\max }\right]\cup \left[{\xi }_{\max },+\infty \right)\right)\times \left(\left(-\infty ,-{\eta }_{\max }\right]\cup \left[{\eta }_{\max },+\infty \right)\right)$ .

对 $I_{11}$ 作估计，有

$\begin{array}{c}{I}_{11}={\left({\displaystyle {\iint }_{{\Omega }_2}{\left|{\text{e}}^{-\left({\xi }^2+{\eta }^2\right)t}{\hat{g}}_0\left(\xi ,\eta \right)\right|}^2\text{d}\xi \text{d}\eta }\right)}^{\frac{1}{2}}\\ \le \frac{{\text{e}}^{-\left({\xi }_{\max }^2+{\eta }_{\max }^2\right)t}}{{\left(1+{\xi }_{\max }^2+{\eta }_{\max }^2\right)}^{\frac{s}{2}}}{\left({\displaystyle {\iint }_{{\Omega }_3}{\left(1+{\xi }^2+{\eta }^2\right)}^s{\left|{\hat{g}}_0\left(\xi ,\eta \right)\right|}^2\text{d}\xi \text{d}\eta }\right)}^{\frac{1}{2}}\\ \le \frac{{\text{e}}^{-\left({\xi }_{\max }^2+{\eta }_{\max }^2\right)t}}{{\left(1+{\xi }_{\max }^2+{\eta }_{\max }^2\right)}^{\frac{s}{2}}}E=\frac{{\text{e}}^{-2{\xi }_{\max }^{2}t}}{{\xi }_{\max }^s}E.\end{array}$

同理可以得到 $I_{12},I_{13}$ 的估计

$I_{12}={\left({\displaystyle {\iint }_{{\Omega }_3}{\left|{\text{e}}^{-\left({\xi }^2+{\eta }^2\right)t}{\hat{g}}_0\left(\xi ,\eta \right)\right|}^2\text{d}\xi \text{d}\eta }\right)}^{\frac{1}{2}}\le \frac{{\text{e}}^{-2{\xi }_{\max }^{2}t}}{{\xi }_{\max }^s}E;$

$I_{13}={\left({\displaystyle {\iint }_{{\Omega }_4}{\left|{\text{e}}^{-\left({\xi }^2+{\eta }^2\right)t}{\hat{g}}_0\left(\xi ,\eta \right)\right|}^2\text{d}\xi \text{d}\eta }\right)}^{\frac{1}{2}}\le \frac{{\text{e}}^{-2{\eta }_{\max }^{2}t}}{{\eta }_{\max }^s}E.$

对 $I_2$ 做估计，有

$\begin{array}{c}{I}_2=\Vert {\hat{u}}_{{\chi }_{\xi }{\chi }_{\eta }}\left(\cdot ,\cdot ,t\right)-{\hat{u}}_{{\chi }_{\xi }{\chi }_{\eta }}^{\delta }\left(\cdot ,\cdot ,t\right)\Vert \\ =\Vert {\text{e}}^{\left({\xi }^2+{\eta }^2\right)\left(T-t\right)}{\hat{g}}_T\left(\xi ,\eta \right){\chi }_{\xi }{\chi }_{\eta }-{\text{e}}^{\left({\xi }^2+{\eta }^2\right)\left(T-t\right)}{\hat{g}}_T^{\delta }\left(\xi ,\eta \right){\chi }_{\xi }{\chi }_{\eta }\Vert \\ =\Vert {\text{e}}^{\left({\xi }^2+{\eta }^2\right)\left(T-t\right)}\left({\hat{g}}_T\left(\xi ,\eta \right)-{\hat{g}}_T^{\delta }\left(\xi ,\eta \right)\right){\chi }_{\xi }{\chi }_{\eta }\Vert \\ ={\left({\displaystyle {\iint }_{{\Omega }_1}{\left|{\text{e}}^{\left({\xi }^2+{\eta }^2\right)\left(T-t\right)}\left({\hat{g}}_T^{\delta }\left(\xi ,\eta \right)-{\hat{g}}_T\left(\xi ,\eta \right)\right)\right|}^2\text{d}\xi \text{d}\eta }\right)}^{\frac{1}{2}}\\ \le {\text{e}}^{\left({\xi }_{\max }^2+{\eta }_{\max }^2\right)\left(T-t\right)}{\left({\displaystyle {\iint }_{{\Omega }_1}{\left|{\hat{g}}_T^{\delta }\left(\xi ,\eta \right)-{\hat{g}}_T\left(\xi ,\eta \right)\right|}^2\text{d}\xi \text{d}\eta }\right)}^{\frac{1}{2}}\\ \le {\text{e}}^{\left({\xi }_{\max }^2+{\eta }_{\max }^2\right)\left(T-t\right)}\delta ={\text{e}}^{2{\xi }_{\max }^2\left(T-t\right)}\delta .\end{array}$

因此，有

$\begin{array}{c}\Vert u\left(\cdot ,\cdot ,t\right)-u_{{\chi }_{\xi }{\chi }_{\eta }}^{\delta }\left(\cdot ,\cdot ,t\right)\Vert \le 3\frac{{\text{e}}^{-2{\xi }_{\max }^{2}t}}{{\xi }_{\max }^s}E+{\text{e}}^{2{\xi }_{\max }^2\left(T-t\right)}\delta \\ =3\frac{{\text{e}}^{-2t\text{l}\ln \left({\left(\frac{E}{\delta }\right)}^{\frac{1}{2T}}{\left(\ln \frac{E}{\delta }\right)}^{-\frac{s}{4T}}\right)}}{{\left(\ln \left({\left(\frac{E}{\delta }\right)}^{\frac{1}{2T}}{\left(\ln \frac{E}{\delta }\right)}^{-\frac{s}{4T}}\right)\right)}^{\frac{s}{2}}}E+{\text{e}}^{2\left(T-t\right)\ln \left({\left(\frac{E}{\delta }\right)}^{\frac{1}{2T}}{\left(\ln \frac{E}{\delta }\right)}^{{}^{-\frac{s}{4T}}}\right)}\delta \\ =3{\left(\frac{E}{\delta }\right)}^{-\frac{t}{T}}{\left(\ln \frac{E}{\delta }\right)}^{\frac{st}{2T}}{\left(\frac{\ln \frac{E}{\delta }}{\frac{1}{T}\ln \frac{E}{\delta }+\ln {\left(\ln \frac{E}{\delta }\right)}^{-\frac{s}{2T}}}\right)}^{\frac{s}{2}}{\left(\ln \frac{E}{\delta }\right)}^{-\frac{s}{2}}E+E^{\left(1-\frac{t}{T}\right)}{\delta }^{\frac{t}{T}}{\left(\ln \frac{E}{\delta }\right)}^{-\frac{s\left(T-t\right)}{2T}}\\ =E^{\left(1-\frac{t}{T}\right)}{\delta }^{\frac{t}{T}}{\left(\ln \frac{E}{\delta }\right)}^{-\frac{s\left(T-t\right)}{2T}}\left(3{\left(\frac{\ln \frac{E}{\delta }}{\frac{1}{T}\ln \frac{E}{\delta }+\ln {\left(\ln \frac{E}{\delta }\right)}^{-\frac{s}{2T}}}\right)}^{\frac{s}{2}}+1\right).\end{array}$

所以，定理1得证。

注1：当 $s=0$ 时，估计(16)式为

$\Vert u\left(\cdot ,\cdot ,t\right)-u_{{\chi }_{\xi }{\chi }_{\eta }}^{\delta }\left(\cdot ,\cdot ,t\right)\Vert \le 4E^{\left(1-\frac{t}{T}\right)}{\delta }^{\frac{t}{T}}.$

此时为 $L_2\left(R\right)$ 中的Hölde型最优稳定性估计。当 $t\to 0^{+}$ 时正则化解的精度逐渐降低，在 $t=0$ 处误差为4E。

注2：在 $t=0$ 处，当 $s>0$ ， $\delta \to 0^{\text{+}}$ 时，估计(16)式为

$\Vert u\left(\cdot ,\cdot ,t\right)-u_{{\chi }_{\xi }{\chi }_{\eta }}^{\delta }\left(\cdot ,\cdot ,t\right)\Vert \le E{\left(\ln \frac{E}{\delta }\right)}^{-\frac{s}{2}}\left(3{\left(\frac{\ln \frac{E}{\delta }}{\frac{1}{T}\ln \frac{E}{\delta }+\ln {\left(\ln \frac{E}{\delta }\right)}^{-\frac{s}{2T}}}\right)}^{\frac{s}{2}}+1\right)\to 0.$

此时(16)式仍为最优估计。

3.2. 后验参数选取下的误差估计

当噪音水平 $\delta$ 已知时，一般采用偏差原理进行后验正则化参数的选取，定义新的正则化算子 ${\rho }_{\xi },{\rho }_{\eta }$ 。找到 ${\xi }_{\max },{\eta }_{\max }$ 满足下面的方程

$\Vert \left(1-{\rho }_{\xi }{\rho }_{\eta }\right){\hat{g}}_T^{\delta }\left(\xi ,\eta \right)\Vert =r\delta .$ (17)

其中 $r>1$ 是常数， ${\xi }_{\max },{\eta }_{\max }$ 均为正常数在这里扮演的是正则化参数的角色。

${\rho }_{\xi }=\{\begin{array}{l}1,\left|\xi \right|\le {\xi }_{\max },\\ {\text{e}}^{-\left(T-t\right)\left({\xi }^2-{\xi }_{\max }^2\right)},\left|\xi \right|>{\xi }_{\max }.\end{array}$

${\rho }_{\eta }=\{\begin{array}{l}1,\left|\eta \right|\le {\eta }_{\max },\\ {\text{e}}^{-\left(T-t\right)\left({\eta }^2-{\eta }_{\max }^2\right)},\left|\eta \right|>{\eta }_{\max }.\end{array}$

构造出问题(1)的Fourier正则化解

$u_{{\rho }_{\xi }{\rho }_{\eta }}^{\delta }\left(x,y,t\right):=\frac{1}{2\pi }{\displaystyle {\iint }_{R^2}{\text{e}}^{i\left(\xi x+\eta y\right)}{\text{e}}^{\left({\xi }^2+{\eta }^2\right)\left(T-t\right)}{\hat{g}}_T^{\delta }\left(\xi ,\eta \right){\rho }_{\xi }{\rho }_{\eta }\text{d}\xi \text{d}\eta }.$

本节考虑 $s=0$ 时，即 $L_2\left(R\right)$ 中后验参数的选取，假设一个新的先验界：

$\Vert g_0\Vert \le M.$ (18)

这里的M是一个正常数， $\Vert \cdot \Vert$ 表示的是 $L^2\left(R^2\right)$ 空间上的范数。

引理1设 $P\left({\xi }_{\max },{\eta }_{\max }\right)=\Vert \left(1-{\rho }_{\xi }{\rho }_{\eta }\right){\hat{g}}_T^{\delta }\left(\xi ,\eta \right)\Vert$ ，如果当 $0<\delta <\Vert {\hat{g}}_T^{\delta }\left(\xi ,\eta \right)\Vert$ ，则 $P\left({\xi }_{\max },{\eta }_{\max }\right)$ 满足以下性质

1) $P\left({\xi }_{\max },{\eta }_{\max }\right)$ 为连续函数；

2) $P\left({\xi }_{\max },{\eta }_{\max }\right)$ 为单调增函数；

3) $P_{{\rho }_{\xi }\to 0+,{\rho }_{\eta }\to 0+}\left({\xi }_{\max },{\eta }_{\max }\right)=0$ ；

4) $P_{{\rho }_{\xi }\to +\infty ,{\rho }_{\eta }\to +\infty }\left({\xi }_{\max },{\eta }_{\max }\right)=\Vert g_T^{\delta }\Vert$ 。

证明：由于

$P\left({\xi }_{\max },{\eta }_{\max }\right)=\{\begin{array}{l}\Vert \left(1-1\right){\hat{g}}_T^{\delta }\left(\xi ,\eta \right)\Vert ,\left|\xi \right|\le {\xi }_{\max },\left|\eta \right|\le {\eta }_{\max },\\ \Vert \left(1-{\text{e}}^{-\left(T-t\right)\left({\xi }^2-{\xi }_{\max }^2\right)}\right){\hat{g}}_T^{\delta }\left(\xi ,\eta \right)\Vert ,\left|\xi \right|>{\xi }_{\max },\left|\eta \right|\le {\eta }_{\max },\\ \Vert \left(1-{\text{e}}^{-\left(T-t\right)\left({\eta }^2-{\eta }_{\max }^2\right)}\right){\hat{g}}_T^{\delta }\left(\xi ,\eta \right)\Vert ,\left|\xi \right|\le {\xi }_{\max },\left|\eta \right|>{\eta }_{\max },\\ \Vert \left(1-{\text{e}}^{-\left(T-t\right)\left({\xi }^2-{\xi }_{\max }^2\right)-\left(T-t\right)\left({\eta }^2-{\eta }_{\max }^2\right)}\right){\hat{g}}_T^{\delta }\left(\xi ,\eta \right)\Vert ,\left|\xi \right|>{\xi }_{\max },\left|\eta \right|>{\eta }_{\max }.\end{array}$

由以上易知 $P\left({\xi }_{\max },{\eta }_{\max }\right)$ 为连续函数，因此1)成立。

又因为 ${\text{e}}^{-\left(T-t\right)\left({\xi }^2-{\xi }_{\max }^2\right)}$ ， ${\text{e}}^{-\left(T-t\right)\left({\eta }^2-{\eta }_{\max }^2\right)}$ ， ${\text{e}}^{-\left(T-t\right)\left({\xi }^2-{\xi }_{\max }^2\right)-\left(T-t\right)\left({\eta }^2-{\eta }_{\max }^2\right)}$ 均为单调减函数，所以 $P\left({\xi }_{\max },{\eta }_{\max }\right)$ 为单调增函数，因此2) 得证；

$\begin{array}{c}而P_{{\rho }_{\xi }\to 0+,{\rho }_{\eta }\to 0+}\left({\xi }_{\max },{\eta }_{\max }\right)=\underset{{\rho }_{\xi }\to 0+,{\rho }_{\eta }\to 0+}{\lim }\Vert \left(1-{\rho }_{\xi }{\rho }_{\eta }\right){\hat{g}}_T^{\delta }\left(\xi ,\eta \right)\Vert \\ =\Vert \left(1-1\right){\hat{g}}_T^{\delta }\left(\xi ,\eta \right)\Vert =0,\end{array}$

$\begin{array}{c}{P}_{{\rho }_{\xi }\to +\infty ,{\rho }_{\eta }\to +\infty }\left({\xi }_{\max },{\eta }_{\max }\right)=\underset{{\rho }_{\xi }\to +\infty ,{\rho }_{\eta }\to +\infty }{\lim }\Vert \left(1-{\rho }_{\xi }{\rho }_{\eta }\right){\hat{g}}_T^{\delta }\left(\xi ,\eta \right)\Vert \\ =\Vert \left(1-0\right){\hat{g}}_T^{\delta }\left(\xi ,\eta \right)\Vert =\Vert {\hat{g}}_T^{\delta }\left(\xi ,\eta \right)\Vert ,\end{array}$

可得性质3)，4)。

引理2假设条件(11)和(18)成立，且 ${\xi }_{\max },{\eta }_{\max }$ 是(17)式的解，则有以下不等式成立

${\text{e}}^{{\eta }_{\max }^2}={\text{e}}^{{\xi }_{\max }^2}\le {\left(\frac{3M}{\left(r-1\right)\delta }\right)}^{\frac{1}{2T}}.$ (19)

证明：由(8)式，有

$\begin{array}{l}\Vert \left(1-{\rho }_{\xi }{\rho }_{\eta }\right){\hat{g}}_T^{\delta }\left(\xi ,\eta \right)\Vert \\ =\Vert \left(1-{\rho }_{\xi }{\rho }_{\eta }\right)\left({\hat{g}}_T^{\delta }\left(\xi ,\eta \right)-{\hat{g}}_T\left(\xi ,\eta \right)+{\hat{g}}_T\left(\xi ,\eta \right)\right)\Vert \\ \le \Vert \left(1-{\rho }_{\xi }{\rho }_{\eta }\right)\left({\hat{g}}_T^{\delta }\left(\xi ,\eta \right)-{\hat{g}}_T\left(\xi ,\eta \right)\right)\Vert +\Vert \left(1-{\rho }_{\xi }{\rho }_{\eta }\right){\hat{g}}_T\left(\xi ,\eta \right)\Vert \\ \le \delta +\Vert \left(1-{\rho }_{\xi }{\rho }_{\eta }\right){\hat{g}}_T\left(\xi ,\eta \right)\Vert \\ \le \delta +\Vert \left(1-{\rho }_{\xi }{\rho }_{\eta }\right){\text{e}}^{-\left({\xi }^2+{\eta }^2\right)T}{\hat{g}}_0\left(\xi ,\eta \right)\Vert \end{array}$

$\begin{array}{l}=\delta +{\left({\displaystyle {\iint }_{{\Omega }_2}{\left|\left(1-{\text{e}}^{-\left(T-t\right)\left({\xi }^2-{\xi }_{\max }^2\right)}\right){\hat{g}}_0\left(\xi ,\eta \right){\text{e}}^{-\left({\xi }^2+{\eta }^2\right)T}\right|}^2\text{d}\xi \text{d}\eta }\right)}^{\frac{1}{2}}\\ +{\left({\displaystyle {\iint }_{{\Omega }_3}{\left|\left(1-{\text{e}}^{-\left(T-t\right)\left({\eta }^2-{\eta }_{\max }^2\right)}\right){\hat{g}}_0\left(\xi ,\eta \right){\text{e}}^{-\left({\xi }^2+{\eta }^2\right)T}\right|}^2\text{d}\xi \text{d}\eta }\right)}^{\frac{1}{2}}\\ +{\left({\displaystyle {\iint }_{{\Omega }_4}{\left|\left(1-{\text{e}}^{-\left(T-t\right)\left({\xi }^2-{\xi }_{\max }^2\right)-\left(T-t\right)\left({\eta }^2-{\eta }_{\max }^2\right)}\right){\hat{g}}_0\left(\xi ,\eta \right){\text{e}}^{-\left({\xi }^2+{\eta }^2\right)T}\right|}^2\text{d}\xi \text{d}\eta }\right)}^{\frac{1}{2}}\\ \le \delta +{\left({\displaystyle {\iint }_{{\Omega }_2}{\left|{\hat{g}}_0\left(\xi ,\eta \right){\text{e}}^{-\left({\xi }^2+{\eta }^2\right)T}\right|}^2\text{d}\xi \text{d}\eta }\right)}^{\frac{1}{2}}+{\left({\displaystyle {\iint }_{{\Omega }_3}{\left|{\hat{g}}_0\left(\xi ,\eta \right){\text{e}}^{-\left({\xi }^2+{\eta }^2\right)T}\right|}^2\text{d}\xi \text{d}\eta }\right)}^{\frac{1}{2}}\\ +{\left({\displaystyle {\iint }_{{\Omega }_4}{\left|{\hat{g}}_0\left(\xi ,\eta \right){\text{e}}^{-\left({\xi }^2+{\eta }^2\right)T}\right|}^2\text{d}\xi \text{d}\eta }\right)}^{\frac{1}{2}}\end{array}$

$\begin{array}{l}=\delta +{\text{e}}^{-\left({\xi }_{\max }^2+{\eta }_{\max }^2\right)T}M+{\text{e}}^{-\left({\xi }_{\max }^2+{\eta }_{\max }^2\right)T}M+{\text{e}}^{-\left({\xi }_{\max }^2+{\eta }_{\max }^2\right)T}M\\ =\delta +{\text{e}}^{-2{\xi }_{\max }^{2}T}M+{\text{e}}^{-2{\eta }_{\max }^{2}T}M+{\text{e}}^{-2{\xi }_{\max }^{2}T}M\\ =\delta +3{\text{e}}^{-2{\xi }_{\max }^{2}T}M\end{array}$

则有

$r\delta =\Vert \left(1-{\rho }_{\xi }{\rho }_{\eta }\right){\hat{g}}_T^{\delta }\left(\xi ,\eta \right)\Vert \le \delta +3{\text{e}}^{-2{\xi }_{\max }^{2}T}M$

$\left(r-1\right)\delta \le 3{\text{e}}^{-2{\xi }_{\max }^{2}T}M$

${\text{e}}^{{\xi }_{\max }^2}\le {\left(\frac{3M}{\left(r-1\right)\delta }\right)}^{\frac{1}{2T}}$

定理2设 $u\left(x,y,t\right)$ 和 ${\hat{u}}_{{\rho }_{\xi }{\rho }_{\eta }}^{\delta }\left(\xi ,\eta ,t\right)$ 分别为问题(1)的精确解和Fourier正则化解，假设条件(11)和(18)成立，且 ${\xi }_{\max },{\eta }_{\max }$ 是方程(17)的解，则有以下估计

$\Vert u_{{\rho }_{\xi }{\rho }_{\eta }}^{\delta }\left(\cdot ,\cdot ,t\right)-u\left(\cdot ,\cdot ,t\right)\Vert \le M^{1-\frac{t}{T}}{\delta }^{\frac{t}{T}}\left(4{\left(\frac{3}{r-1}\right)}^{1-\frac{t}{T}}+3{\left(r+1\right)}^{\frac{t}{T}}\right).$ (20)

证明：由Parseval等式和三角不等式，有

$\begin{array}{c}\Vert u_{{\rho }_{\xi }{\rho }_{\eta }}^{\delta }\left(\cdot ,\cdot ,t\right)-u\left(\cdot ,\cdot ,t\right)\Vert =\Vert {\hat{u}}_{{\rho }_{\xi }{\rho }_{\eta }}^{\delta }\left(\cdot ,\cdot ,t\right)-\hat{u}\left(\cdot ,\cdot ,t\right)\Vert \\ =\Vert {\hat{u}}_{{\rho }_{\xi }{\rho }_{\eta }}^{\delta }\left(\cdot ,\cdot ,t\right)-{\hat{u}}_{{\rho }_{\xi }{\rho }_{\eta }}\left(\cdot ,\cdot ,t\right)\Vert +\Vert {\hat{u}}_{{\rho }_{\xi }{\rho }_{\eta }}\left(\cdot ,\cdot ,t\right)-\hat{u}\left(\cdot ,\cdot ,t\right)\Vert \\ =S_1+S_2\end{array}$

下面我们对 $S_1,S_2$ 分别作估计，对 $S_1$ 做估计有

$\begin{array}{c}{S}_1=\Vert {\hat{u}}_{{\rho }_{\xi }{\rho }_{\eta }}^{\delta }\left(\cdot ,\cdot ,t\right)-{\hat{u}}_{{\rho }_{\xi }{\rho }_{\eta }}\left(\cdot ,\cdot ,t\right)\Vert \underset{{}_{}}{}\\ =\Vert {\text{e}}^{\left({\xi }^2+{\eta }^2\right)\left(T-t\right)}{\hat{g}}_T^{\delta }\left(\xi ,\eta \right){\rho }_{\xi }{\rho }_{\eta }-{\text{e}}^{\left({\xi }^2+{\eta }^2\right)\left(T-t\right)}{\hat{g}}_T\left(\xi ,\eta \right){\rho }_{\xi }{\rho }_{\eta }\Vert \\ =\Vert {\text{e}}^{\left({\xi }^2+{\eta }^2\right)\left(T-t\right)}{\rho }_{\xi }{\rho }_{\eta }\left({\hat{g}}_T^{\delta }\left(\xi ,\eta \right)-{\hat{g}}_T\left(\xi ,\eta \right)\right)\Vert \\ ={\left({\displaystyle {\iint }_{{\Omega }_1}{\left|{\text{e}}^{\left({\xi }^2+{\eta }^2\right)\left(T-t\right)}\left({\hat{g}}_T^{\delta }\left(\xi ,\eta \right)-{\hat{g}}_T\left(\xi ,\eta \right)\right)\right|}^2\text{d}\xi \text{d}\eta }\right)}^{\frac{1}{2}}\\ +{\left({\displaystyle {\iint }_{{\Omega }_2}{\left|{\text{e}}^{\left({\xi }^2+{\eta }^2\right)\left(T-t\right)}{\text{e}}^{-\left(T-t\right)\left({\xi }^2-{\xi }_{\max }^2\right)}\left({\hat{g}}_T^{\delta }\left(\xi ,\eta \right)-{\hat{g}}_T\left(\xi ,\eta \right)\right)\right|}^2\text{d}\xi \text{d}\eta }\right)}^{\frac{1}{2}}\end{array}$

$\begin{array}{c}+{\left({\displaystyle {\iint }_{{\Omega }_3}{\left|{\text{e}}^{\left({\xi }^2+{\eta }^2\right)\left(T-t\right)}{\text{e}}^{-\left(T-t\right)\left({\eta }^2-{\eta }_{\max }^2\right)}\left({\hat{g}}_T^{\delta }\left(\xi ,\eta \right)-{\hat{g}}_T\left(\xi ,\eta \right)\right)\right|}^2\text{d}\xi \text{d}\eta }\right)}^{\frac{1}{2}}\\ +{\left({\displaystyle {\iint }_{{\Omega }_4}{\left|{\text{e}}^{\left({\xi }^2+{\eta }^2\right)\left(T-t\right)}{\text{e}}^{-\left(T-t\right)\left({\xi }^2-{\xi }_{\max }^2\right)-\left(T-t\right)\left({\eta }^2-{\eta }_{\max }^2\right)}\left({\hat{g}}_T^{\delta }\left(\xi ,\eta \right)-{\hat{g}}_T\left(\xi ,\eta \right)\right)\right|}^2\text{d}\xi \text{d}\eta }\right)}^{\frac{1}{2}}\\ =I_1+{\left({\displaystyle {\iint }_{{\Omega }_2}{\left|{\text{e}}^{\left({\xi }_{\max }^2+{\eta }^2\right)\left(T-t\right)}\left({\hat{g}}_T^{\delta }\left(\xi ,\eta \right)-{\hat{g}}_T\left(\xi ,\eta \right)\right)\right|}^2\text{d}\xi \text{d}\eta }\right)}^{\frac{1}{2}}\\ +{\left({\displaystyle {\iint }_{{\Omega }_3}{\left|{\text{e}}^{\left({\xi }^2+{\eta }_{\max }^2\right)\left(T-t\right)}\left({\hat{g}}_T^{\delta }\left(\xi ,\eta \right)-{\hat{g}}_T\left(\xi ,\eta \right)\right)\right|}^2\text{d}\xi \text{d}\eta }\right)}^{\frac{1}{2}}\\ +{\left({\displaystyle {\iint }_{{\Omega }_4}{\left|{\text{e}}^{\left({\xi }_{\max }^2+{\eta }_{\max }^2\right)\left(T-t\right)}\left({\hat{g}}_T^{\delta }\left(\xi ,\eta \right)-{\hat{g}}_T\left(\xi ,\eta \right)\right)\right|}^2\text{d}\xi \text{d}\eta }\right)}^{\frac{1}{2}}\end{array}$

$\begin{array}{c}\le {\text{e}}^{2{\xi }_{\max }^2\left(T-t\right)}\delta +{\text{e}}^{\left({\xi }_{\max }^2+{\eta }_{\max }^2\right)\left(T-t\right)}\delta +{\text{e}}^{\left({\xi }_{\max }^2+{\eta }_{\max }^2\right)\left(T-t\right)}\delta +{\text{e}}^{\left({\xi }_{\max }^2+{\eta }_{\max }^2\right)\left(T-t\right)}\delta \stackrel{\begin{array}{l}\\ \end{array}}{{}_{\begin{array}{l}\\ \end{array}}}\\ \le 4{\text{e}}^{2{\xi }_{\max }^2\left(T-t\right)}\delta \le 4\delta {\left({\left(\frac{3M}{\left(r-1\right)\delta }\right)}^{\frac{1}{2T}}\right)}^{2\left(T-t\right)}\\ =4\delta {\left(\frac{3M}{\left(r-1\right)\delta }\right)}^{1-\frac{t}{T}}=4M^{1-\frac{t}{T}}{\delta }^{\frac{t}{T}}{\left(\frac{3}{r-1}\right)}^{1-\frac{t}{T}}.\end{array}$

另一方面，对 $S_2$ 作估计

$\begin{array}{c}{S}_2=\Vert \hat{u}\left(\cdot ,\cdot ,t\right)-{\hat{u}}_{{\rho }_{\xi }{\rho }_{\eta }}\left(\cdot ,\cdot ,t\right)\Vert =\Vert {\text{e}}^{\left({\xi }^2+{\eta }^2\right)\left(T-t\right)}{\hat{g}}_T\left(\xi ,\eta \right)-{\text{e}}^{\left({\xi }^2+{\eta }^2\right)\left(T-t\right)}{\hat{g}}_T\left(\xi ,\eta \right){\rho }_{\xi }{\rho }_{\eta }\Vert \\ =\Vert {\text{e}}^{\left({\xi }^2+{\eta }^2\right)\left(T-t\right)}{\hat{g}}_T\left(\xi ,\eta \right)\left(1-{\rho }_{\xi }{\rho }_{\eta }\right)\Vert =\Vert {\text{e}}^{-\left({\xi }^2+{\eta }^2\right)t}{\hat{g}}_0\left(\xi ,\eta \right)\left(1-{\rho }_{\xi }{\rho }_{\eta }\right)\Vert \\ ={\left({\displaystyle {\iint }_{{\Omega }_2}{\left|\left(1-{\text{e}}^{-\left(T-t\right)\left({\xi }^2-{\xi }_{\max }^2\right)}\right){\hat{g}}_0\left(\xi ,\eta \right){\text{e}}^{-\left({\xi }^2+{\eta }^2\right)t}\right|}^2\text{d}\xi \text{d}\eta }\right)}^{\frac{1}{2}}\\ +{\left({\displaystyle {\iint }_{{\Omega }_3}{\left|\left(1-{\text{e}}^{-\left(T-t\right)\left({\eta }^2-{\eta }_{\max }^2\right)}\right){\hat{g}}_0\left(\xi ,\eta \right){\text{e}}^{-\left({\xi }^2+{\eta }^2\right)t}\right|}^2\text{d}\xi \text{d}\eta }\right)}^{\frac{1}{2}}\\ +{\left({\displaystyle {\iint }_{{\Omega }_4}{\left|\left(1-{\text{e}}^{-\left(T-t\right)\left({\xi }^2-{\xi }_{\max }^2\right)-\left(T-t\right)\left({\eta }^2-{\eta }_{\max }^2\right)}\right){\hat{g}}_0\left(\xi ,\eta \right){\text{e}}^{-\left({\xi }^2+{\eta }^2\right)t}\right|}^2\text{d}\xi \text{d}\eta }\right)}^{\frac{1}{2}}\\ =S_{21}+S_{22}+S_{23}\end{array}$

对 $S_{21}$ 作估计，有

$\begin{array}{c}{S}_{21}={\left({\displaystyle {\iint }_{{\Omega }_2}{\left|\left(1-{\text{e}}^{-\left(T-t\right)\left({\xi }^2-{\xi }_{\max }^2\right)}\right){\hat{g}}_0\left(\xi ,\eta \right){\text{e}}^{-\left({\xi }^2+{\eta }^2\right)t}\right|}^2\text{d}\xi \text{d}\eta }\right)}^{\frac{1}{2}}\\ ={\left({\displaystyle {\iint }_{{\Omega }_2}{\left|\left(1-{\text{e}}^{-\left(T-t\right)\left({\xi }^2-{\xi }_{\max }^2\right)}\right){\hat{g}}_0\left(\xi ,\eta \right){\text{e}}^{-\left({\xi }^2+{\eta }^2\right)t}\right|}^2\text{d}\xi \text{d}\eta }\right)}^{\frac{1}{2}}\\ \le {\left({\displaystyle {\iint }_{{\Omega }_2}{\left|{\hat{g}}_0\left(\xi ,\eta \right)\right|}^2\text{d}\xi \text{d}\eta }\right)}^{\frac{1}{2}\left(1-\frac{t}{T}\right)}{\left({\displaystyle {\iint }_{{\Omega }_2}{\left|{\left({\hat{g}}_0\left(\xi ,\eta \right)\right)}^{\frac{t}{T}}{\text{e}}^{-\left({\xi }^2+{\eta }^2\right)t}\left(1-{\text{e}}^{-\left(T-t\right)\left({\xi }^2-{\xi }_{\max }^2\right)}\right)\right|}^2\text{d}\xi \text{d}\eta }\right)}^{\frac{1}{2}}\\ \le M^{1-\frac{t}{T}}{\left({\displaystyle {\iint }_{{\Omega }_2}{\left|{\left({\hat{g}}_0\left(\xi ,\eta \right)\right)}^{\frac{t}{T}}{\text{e}}^{-\left({\xi }^2+{\eta }^2\right)t}\left(1-{\text{e}}^{-\left(T-t\right)\left({\xi }^2-{\xi }_{\max }^2\right)}\right)\right|}^2\text{d}\xi \text{d}\eta }\right)}^{\frac{1}{2}}\\ \le M^{1-\frac{t}{T}}{\left({\displaystyle {\iint }_{{\Omega }_2}{\left|{\left({\hat{g}}_T\left(\xi ,\eta \right){\text{e}}^{\left({\xi }^2+{\eta }^2\right)T}\right)}^{\frac{t}{T}}{\text{e}}^{-\left({\xi }^2+{\eta }^2\right)t}\left(1-{\text{e}}^{-\left(T-t\right)\left({\xi }^2-{\xi }_{\max }^2\right)}\right)\right|}^2\text{d}\xi \text{d}\eta }\right)}^{\frac{1}{2}}\end{array}$

$\begin{array}{c}\le M^{1-\frac{t}{T}}{\left({\displaystyle {\iint }_{{\Omega }_2}{\left|{\left({\hat{g}}_T\left(\xi ,\eta \right)\right)}^{\frac{t}{T}}\left(1-{\text{e}}^{-\left(T-t\right)\left({\xi }^2-{\xi }_{\max }^2\right)}\right)\right|}^2\text{d}\xi \text{d}\eta }\right)}^{\frac{1}{2}}\\ \le M^{1-\frac{t}{T}}{\left({\displaystyle {\iint }_{{\Omega }_2}{\left|{\hat{g}}_T\left(\xi ,\eta \right)\left(1-{\text{e}}^{-\left(T-t\right)\left({\xi }^2-{\xi }_{\max }^2\right)}\right)\right|}^2\text{d}\xi \text{d}\eta }\right)}^{\frac{t}{2T}}{\left({\displaystyle {\iint }_{{\Omega }_2}{\left|1-{\text{e}}^{-\left(T-t\right)\left({\xi }^2-{\xi }_{\max }^2\right)}\right|}^2\text{d}\xi \text{d}\eta }\right)}^{2\left(1-\frac{t}{T}\right)}\\ \le M^{1-\frac{t}{T}}{\left({\displaystyle {\iint }_{{\Omega }_2}{\left|{\hat{g}}_T\left(\xi ,\eta \right)\left(1-{\rho }_{\xi }{\rho }_{\eta }\right)\right|}^2\text{d}\xi \text{d}\eta }\right)}^{\frac{t}{2T}}\underset{\xi ,\eta \in {\Omega }_2}{\sup }{\left(1-{\text{e}}^{-\left(T-t\right)\left({\xi }^2-{\xi }_{\max }^2\right)}\right)}^{1-\frac{t}{T}}\\ \le M^{1-\frac{t}{T}}{\left({\displaystyle {\iint }_{{\Omega }_2}{\left|\left({\hat{g}}_T\left(\xi ,\eta \right)-{\hat{g}}_T^{\delta }\left(\xi ,\eta \right)+{\hat{g}}_T^{\delta }\left(\xi ,\eta \right)\right)\left(1-{\rho }_{\xi }{\rho }_{\eta }\right)\right|}^2\text{d}\xi \text{d}\eta }\right)}^{\frac{t}{2T}}\\ \le M^{1-\frac{t}{T}}{\left(\left(r+1\right)\delta \right)}^{\frac{t}{T}}\end{array}$

同理，可以得到

$S_{22}\le M^{1-\frac{t}{T}}{\left(\left(r+1\right)\delta \right)}^{\frac{t}{T}};S_{23}\le M^{1-\frac{t}{T}}{\left(\left(r+1\right)\delta \right)}^{\frac{t}{T}}.$

则，有

$S_2\le 3M^{1-\frac{t}{T}}{\left(\left(r+1\right)\delta \right)}^{\frac{t}{T}}.$

所以

$\begin{array}{c}\Vert u_{{\rho }_{\xi }{\rho }_{\eta }}^{\delta }\left(\cdot ,\cdot ,t\right)-u\left(\cdot ,\cdot ,t\right)\Vert \le 4M^{1-\frac{t}{T}}{\delta }^{\frac{t}{T}}{\left(\frac{3}{r-1}\right)}^{1-\frac{t}{T}}+3M^{1-\frac{t}{T}}{\left(\left(r+1\right)\delta \right)}^{\frac{t}{T}}\\ =M^{1-\frac{t}{T}}{\delta }^{\frac{t}{T}}\left(4{\left(\frac{3}{r-1}\right)}^{1-\frac{t}{T}}+3{\left(r+1\right)}^{\frac{t}{T}}\right)\end{array}$

所以，定理2得证。

4. 结论

本文解决了一个不适定问题，即二维时间反向热传导方程在无界区域中通过终值时刻的数据来确定温度的分布。本文先通过Fourier变换推导出了问题的精确解，然后构造出了问题的Fourier正则化解。最后分别给出了先验和后验参数选取下的误差估计。