广州大学，数学与信息科学学院，广东 广州

收稿日期：2022年2月14日；录用日期：2022年3月8日；发布日期：2022年3月15日

摘要

通过谱分解的方法研究了高斯白噪声扰动下分数阶强阻尼波动方程的终边值问题，这类问题是不适定的，即解不连续依赖于终值条件，应用Fourier截断方法建立了问题的正则解，并给出正则解和精确解间的收敛性误差估计，这说明了正则化方法的有效性。

关键词

强阻尼波动方程，分数阶拉普拉斯算子，高斯白噪声，Fourier正则化

Fourier Regularization of Fractional Strongly Damped Wave Equations

Weijie Rong

School of Mathematics and Information Science, Guangzhou University, Guangzhou Guangdong

Received: Feb. 14^th, 2022; accepted: Mar. 8^th, 2022; published: Mar. 15^th, 2022

ABSTRACT

In this paper, by using spectral decomposition method, we study a final-boundary value problem of the fractional strongly damped wave equations with Gauss white noise. This problem is ill-posed in the sense of Hadamard, i.e. the solution discontinuity depends on the final condition. Hence, regularized solution is established by the Fourier truncation method. Moreover, we show the convergent error estimates between the regularized solution and the exact solution, which implies the regularized method is effective.

Keywords:Strongly Damped Wave Equation, Fractional Laplacian Operator, Gauss White Noise, Fourier Regularization

Copyright © 2022 by author(s) and Hans Publishers Inc.

This work is licensed under the Creative Commons Attribution International License (CC BY 4.0).

http://creativecommons.org/licenses/by/4.0/

1. 引言

强阻尼波动方程在模拟粘弹性材料振动中有广泛应用，它的终边值问题是一个不适定问题，即解不连续依赖于终值条件，这种不适定性不利于实际运用中数值模拟的实现，必须采用正则化方法进行处理，这些年来陆续出现各种关于强阻尼波动方程的正则化方法，包括Fourier截断正则化方法 [1] [2] [3]、拟边值方法 [4]、滤波方法 [5] 等等，其中Fourier截断正则化方法是一种高效且被广泛应用到各种偏微分方程的正则化方法，有许多相关的正则化结果 [6] [7] [8] [9] [10]。

近期随着分数阶理论的不断深入，谱分数阶拉普拉斯算子的引入有利于强阻尼波动方程搭建更多的数学模型及模拟更多的物理现象。相应地，涉及谱分数阶拉普拉斯算子的强阻尼波动方程正则化理论较为有限，本文应用Fourier截断正则化方法处理分数阶强阻尼波动方程，给出相关的误差估计，对这一块的正则化理论进行补充。

考虑如下分数阶强阻尼波动方程的终边值问题

$\{\begin{array}{ll}{u}_{tt}+\alpha {\left(-\Delta \right)}^{s_1}{u}_t+\beta {\left(-\Delta \right)}^{s_2}u=F\left(u,x,t\right),\hfill & \left(x,t\right)\in \Omega \times \left[0,T\right],\hfill \\ u\left(x,t\right)=0,\hfill & \left(x,t\right)\in \partial \Omega \times \left[0,T\right],\hfill \\ u\left(x,T\right)=f_0\left(x\right),\hfill & x\in \Omega ,\hfill \\ u_t\left(x,T\right)=f_1\left(x\right),\hfill & x\in \Omega .\hfill \end{array}$ (1)

其中 $\alpha >0,\beta >0,s_1>0,s_2>0,T>0$，$\Omega$ 为 ${ℝ}^n$ 上带有光滑边界 $\partial \Omega$ 的有界开区域。根据 $\left(-\Delta \right)$ 算子的谱分解定理，存在一族特征函数 ${\left\{{\phi }_k\right\}}_{k=1}^{\infty }$ 并且它们构成 $L^2\left(\Omega \right)$ 上的完备正交基使得

$\{\begin{array}{ll}\left(-\Delta \right){\phi }_k={\lambda }_k{\phi }_k,\hfill & x\in \Omega ,\hfill \\ {\phi }_k\left(x\right)=0,\hfill & x\in \partial \Omega .\hfill \end{array}$

这里 ${\lambda }_k$ 为对应 ${\phi }_k$ 的谱且有 $0<{\lambda }_1\le {\lambda }_2\le \cdots \le {\lambda }_k\le \cdots \nearrow \infty$。在实际测量中往往受到不可控外界因素的影响，如风、湿度等等。假设观察数据是在下述高斯白噪声模型 [10] 的驱动下获取的，以(2) (3)中的 ${\tilde{f}}_{0,N}\left(x\right)$ 和 ${\tilde{f}}_{1,N}\left(x\right)$ 分别作为 $f_0\left(x\right)$ 和 $f_1\left(x\right)$ 观测数据，其中

${\tilde{f}}_{0,N}\left(x\right)={\displaystyle \underset{k=1}{\overset{N}{\sum }}\langle f_0,{\phi }_k\rangle {\phi }_k}+{\epsilon }_0\langle {\xi }_0,{\phi }_k\rangle {\phi }_k,k=1,2,\cdots ,N\left({\epsilon }_0,{\epsilon }_1\right),$ (2)

${\tilde{f}}_{1,N}\left(x\right)={\displaystyle \underset{k=1}{\overset{N}{\sum }}\langle f_1,{\phi }_k\rangle {\phi }_k}+{\epsilon }_1\langle {\xi }_1,{\phi }_k\rangle {\phi }_k,k=1,2,\cdots ,N\left({\epsilon }_0,{\epsilon }_1\right).$ (3)

其中N称为离散观测步数， ${\epsilon }_0>0,{\epsilon }_1>0$ 统称为噪声幅值， $\langle {\xi }_0,{\phi }_k\rangle$，$\langle {\xi }_1,{\phi }_k\rangle$ 均独立同分布于标准正态分布。

本文对问题(1)的讨论总是基于如下假设进行，记 ${\Lambda }_k={\alpha }^2{\lambda }_k^{2s_1}-4\beta {\lambda }_k^{s_2}$。

(A1) 源项 $F\left(u,x,t\right)$ 满足全局Lipschitz条件，即

${\Vert F\left(u_1,x,t\right)-F\left(u_2,x,t\right)\Vert }_2\le L{\Vert u_1-u_2\Vert }_2,\left(x,t\right)\in \Omega \times \left[0,T\right].$

(A2) 当 $2s_1>s_2$ 时，约定

${\lambda }_1^{2s_1-s_2}>\frac{2\beta s_2}{{\alpha }^2{s}_1},{\alpha }^2{\lambda }_1^{2s_1}-4\beta {\lambda }_1^{s_2}>0.$

(A3) 当 $2s_1<s_2$ 时，约定

${\lambda }_1^{2s_1-s_2}<\frac{2\beta s_2}{{\alpha }^2{s}_1},{\alpha }^2{\lambda }_1^{2s_1}-4\beta {\lambda }_1^{s_2}<0.$

(A2)保证了此时每一 ${\Lambda }_k$ 为正数且单调递增，(A3)保证了此时每一 ${\Lambda }_k$ 为负数且单调递减。

2. 预备知识

如果一个函数 $\varphi \in L^2\left(\Omega \right)$ 表示成

$\varphi ={\displaystyle \underset{k=1}{\overset{\infty }{\sum }}\langle \varphi ,{\phi }_k\rangle {\phi }_k}$

的形式，那么它的 $L^2\left(\Omega \right)$ 范数可以写成

${\Vert \varphi \Vert }_{L^2\left(\Omega \right)}=\sqrt{{\displaystyle \underset{k=1}{\overset{\infty }{\sum }}{\langle \varphi ,{\phi }_k\rangle }^2}}:={\Vert \varphi \Vert }_2.$

Hilbert空间 $H^{\mu }\left(\Omega \right)\left(\mu >0\right)$

$H^{\mu }\left(\Omega \right):=\left\{\varphi \in L^2\left(\Omega \right)|{\displaystyle \underset{k=1}{\overset{\infty }{\sum }}{\lambda }_k^{\mu }{\langle \varphi ,{\phi }_k\rangle }^2<\infty }\right\},$

并配备了范数

${\Vert \varphi \Vert }_{H^{\mu }\left(\Omega \right)}=\sqrt{{\displaystyle \underset{k=1}{\overset{\infty }{\sum }}{\lambda }_k^{\mu }{\langle \varphi ,{\phi }_k\rangle }^2}}:={\Vert \varphi \Vert }_{H^{\mu }}.$

定义在 $H^{2\mu }\left(\Omega \right)$ 上的谱分数阶拉普拉斯算子 ${\left(-\Delta \right)}^{\mu }$

${\left(-\Delta \right)}^{\mu }\varphi ={\displaystyle \underset{k=1}{\overset{\infty }{\sum }}{\lambda }_k^{\mu }\langle \varphi ,{\phi }_k\rangle {\phi }_k}.$

给定一个完备的概率测度空间 $\tilde{\Omega }$，空间 $L^2\left(\tilde{\Omega },L^2\left(\Omega \right)\right)$ 为 $\varphi :\tilde{\Omega }\to L^2\left(\Omega \right)$ 的所有可测函数，并配备了范数 $\sqrt{{\rm E}{\Vert \cdot \Vert }_2^2}$，其中 ${\rm E}$ 为数学期望。

所有 $L^2\left(\Omega \right)$ 可测过程组成的函数空间 $L^{\infty }\left(\left[0,T\right],L^2\left(\tilde{\Omega },L^2\left(\Omega \right)\right)\right)$，$\varphi :\left[0,T\right]\to L^2\left(\tilde{\Omega },L^2\left(\Omega \right)\right)$，配备了范数 $\underset{t\in \left[0,T\right]}{\sup }\sqrt{{\rm E}{\Vert \cdot \Vert }_2^2}$，记为 ${\Vert \cdot \Vert }_{L_2^{\infty }}$。

引理1 [10] 若 $f_0\left(x\right),f_1\left(x\right)\in H^{\mu }\left(\Omega \right)$，则

${\rm E}{\Vert {\tilde{f}}_{0,N}\left(x\right)-f_0\left(x\right)\Vert }_2^2\le {\epsilon }_0^{2}N+{\lambda }_N^{-\mu }{\Vert f_0\left(x\right)\Vert }_{H^{\mu }}^2,{\rm E}{\Vert {\tilde{f}}_{1,N}\left(x\right)-f_1\left(x\right)\Vert }_2^2\le {\epsilon }_1^{2}N+{\lambda }_N^{-\mu }{\Vert f_1\left(x\right)\Vert }_{H^{\mu }}^2.$

3. 温和解推导和不适定性分析

为了不适定性分析和误差估计，需要导出问题(1)的温和解。根据特征函数分解定理，假设问题(1)有

形如 $u={\displaystyle \underset{k=1}{\overset{\infty }{\sum }}\langle u,{\phi }_k\rangle {\phi }_k}:={\displaystyle \underset{k=1}{\overset{\infty }{\sum }}{u}_k\left(t\right){\phi }_k}$ 的解，记含参数算子 $S_{ij}\left(t\right)\varphi :L^2\left(\Omega \right)\to L^2\left(\Omega \right),i=1,2,3;j=1,2$ ：

$\begin{array}{l}\begin{array}{l}{S}_{11}\left(t\right)\varphi ={\displaystyle \underset{k=1}{\overset{\infty }{\sum }}\exp \left\{\frac{-\alpha {\lambda }_k^{s_1}t}{2}\right\}\left[\frac{-\alpha {\lambda }_k^{s_1}}{\sqrt{{\Lambda }_k}}\sinh \left(\frac{\sqrt{{\Lambda }_k}t}{2}\right)+\cosh \left(\frac{\sqrt{{\Lambda }_k}t}{2}\right)\right]\langle \varphi ,{\phi }_k\rangle {\phi }_k,}\hfill \\ S_{12}\left(t\right)\varphi ={\displaystyle \underset{k=1}{\overset{\infty }{\sum }}\frac{2}{\sqrt{{\Lambda }_k}}\exp \left\{\frac{-\alpha {\lambda }_k^{s_1}t}{2}\right\}\sinh \left(\frac{\sqrt{{\Lambda }_k}t}{2}\right)\langle \varphi ,{\phi }_k\rangle {\phi }_k,}\hfill \\ S_{21}\left(t\right)\varphi ={\displaystyle \underset{k=1}{\overset{\infty }{\sum }}\left(1+\frac{\alpha {\lambda }_k^{s_1}t}{2}\right)}\exp \left\{\frac{-\alpha {\lambda }_k^{s_1}t}{2}\right\}\langle \varphi ,{\phi }_k\rangle {\phi }_k,\hfill \end{array}\hfill \\ S_{22}\left(t\right)\varphi ={\displaystyle \underset{k=1}{\overset{\infty }{\sum }}t}\exp \left\{\frac{-\alpha {\lambda }_k^{s_1}t}{2}\right\}\langle \varphi ,{\phi }_k\rangle {\phi }_k,\hfill \\ S_{31}\left(t\right)\varphi ={\displaystyle \underset{k=1}{\overset{\infty }{\sum }}\exp \left\{\frac{-\alpha {\lambda }_k^{s_1}t}{2}\right\}\left[\frac{-\alpha {\lambda }_k^{s_1}}{\sqrt{{\Lambda }_k}}\sin \left(\frac{\sqrt{-{\Lambda }_k}t}{2}\right)+\cos \left(\frac{\sqrt{-{\Lambda }_k}t}{2}\right)\right]\langle \varphi ,{\phi }_k\rangle {\phi }_k,}\hfill \\ S_{32}\left(t\right)\varphi ={\displaystyle \underset{k=1}{\overset{\infty }{\sum }}\frac{2}{\sqrt{-{\Lambda }_k}}\exp \left\{\frac{-\alpha {\lambda }_k^{s_1}t}{2}\right\}\sin \left(\frac{\sqrt{-{\Lambda }_k}t}{2}\right)\langle \varphi ,{\phi }_k\rangle {\phi }_k.}\hfill \end{array}$

利用主方程(1)左端的半线性结构，两边同时关于 ${\phi }_k$ 作 $L^2\left(\Omega \right)$ 内积，根据特征函数的正交性质和分部积分法可以反解出

$u\left(x,t\right)=\{\begin{array}{c}{S}_{11}\left(t-T\right)f_0+S_{12}\left(t-T\right)f_1-{\displaystyle \underset{t}{\overset{T}{\int }}{S}_{12}\left(t-\tau \right)F\left(u\right)\text{d}\tau ,{\left\{{\Lambda }_k\right\}}_{k=1}^{\infty }>0,}\\ S_{21}\left(t-T\right)f_0+S_{22}\left(t-T\right)f_1-{\displaystyle \underset{t}{\overset{T}{\int }}{S}_{22}\left(t-\tau \right)F\left(u\right)\text{d}\tau ,{\left\{{\Lambda }_k\right\}}_{k=1}^{\infty }=0,}\\ S_{31}\left(t-T\right)f_0+S_{32}\left(t-T\right)f_1-{\displaystyle \underset{t}{\overset{T}{\int }}{S}_{32}\left(t-\tau \right)F\left(u\right)\text{d}\tau ,{\left\{{\Lambda }_k\right\}}_{k=1}^{\infty }<0.}\end{array}$ (4)

在(4)中，由指数项变化推知 $S_{ij}\left(t-T\right),i=1,2,3;j=1,2$ 为无界算子，这就说明了问题(1)是不适定的。

4. Fourier截断正则化理论

Fourier截断正则化方法的思想就是通过截断高频分量的方式将上述的无界算子变成有界算子，关于问题(1)是通过以下积分方程给出相应地正则解：

${\tilde{u}}_J=\{\begin{array}{c}{S}_{11}^J\left(t-T\right){\tilde{f}}_{0,N}+S_{12}^J\left(t-T\right){\tilde{f}}_{1,N}-{\displaystyle \underset{t}{\overset{T}{\int }}{S}_{12}^J\left(t-\tau \right)F\left({\tilde{u}}_J\right)\text{d}\tau ,{\left\{{\Lambda }_k\right\}}_{k=1}^{\infty }>0,}\\ S_{21}^J\left(t-T\right){\tilde{f}}_{0,N}+S_{22}^J\left(t-T\right){\tilde{f}}_{1,N}-{\displaystyle \underset{t}{\overset{T}{\int }}{S}_{22}^J\left(t-\tau \right)F\left({\tilde{u}}_J\right)\text{d}\tau ,{\left\{{\Lambda }_k\right\}}_{k=1}^{\infty }=0,}\\ S_{31}^J\left(t-T\right){\tilde{f}}_{0,N}+S_{32}^J\left(t-T\right){\tilde{f}}_{1,N}-{\displaystyle \underset{t}{\overset{T}{\int }}{S}_{32}^J\left(t-\tau \right)F\left({\tilde{u}}_J\right)\text{d}\tau ,{\left\{{\Lambda }_k\right\}}_{k=1}^{\infty }<0.}\end{array}$ (5)

其中含参数算子 $S_{ij}^J\left(t\right)\varphi ={\displaystyle \underset{k=1}{\overset{J}{\sum }}\langle S_{ij}\left(t\right)\varphi ,{\phi }_k\rangle {\phi }_k},i=1,2,3;j=1,2;1\le J<N$。

根据表达式(4) (5)可得到本文的主要结果，即在精确解先验界下正则解和精确解间的收敛性误差估

计。定理1若终值条件 $f_0\left(x\right),f_{\text{1}}\left(x\right)\in H^{\mu }\left(\Omega \right)\left(\mu >0\right)$，(A1)~(A3)成立，则(5)中的每一积分方程均在 $L_2^{\infty }$ 有唯一解。如果假设问题(1)的精确解 $u\in C\left(\left[0,T\right],H^{2s_1}\left(\Omega \right)\cap H^{2s_2}\left(\Omega \right)\right)$，并且

1) 当 $2s_1>s_2$ 或 $2s_1=s_2,{\alpha }^2-4\beta >0$ 时，若

$\underset{t\in \left[0,T\right]}{\sup }{\displaystyle \underset{k=1}{\overset{\infty }{\sum }}\exp \left\{\left(\alpha {\lambda }_k^{s_1}+\sqrt{{\Lambda }_k}\right)t\right\}}{\langle u,{\phi }_k\rangle }^2\le M_1,$

那么有

$\begin{array}{c}{\rm E}{\Vert {\tilde{u}}_J-u\Vert }_2^2\le 12\exp \left\{\left(\alpha {\lambda }_J^{s_1}+\sqrt{{\Lambda }_J}+12L^{2}T{\Lambda }_1^{-1}\right)\left(T-t\right)\right\}[\left(1+{\alpha }^2{\lambda }_J^{2s_1}{\Lambda }_1^{-1}\right)\left({\epsilon }_0^{2}N+{\lambda }_N^{-\mu }{\Vert f_0\Vert }_{H^{\mu }}^2\right)\\ +2{\Lambda }_1^{-1}\left({\epsilon }_1^{2}N+{\lambda }_N^{-\mu }{\Vert f_1\Vert }_{H^{\mu }}^2\right)]+2M_1\exp \left\{-\left(\alpha {\lambda }_J^{s_1}+\sqrt{{\Lambda }_J}\right)t+4L^{2}T\left(T-t\right){\Lambda }_1^{-1}\right\}.\end{array}$ (6)

2) 当 $2s_1=s_2,{\alpha }^2-4\beta =0$ 时，若

$\underset{t\in \left[0,T\right]}{\sup }{\displaystyle \underset{k=1}{\overset{\infty }{\sum }}\exp \left\{\alpha {\lambda }_k^{s_1}t\right\}}{\langle u,{\phi }_k\rangle }^2\le M_2,$

那么有

$\begin{array}{c}{\rm E}{\Vert {\tilde{u}}_J-u\Vert }_2^2\le \left[\left(12+3{\alpha }^2{\lambda }_J^{2s_1}{T}^2\right)\left({\epsilon }_0^{2}N+{\lambda }_N^{-\mu }{\Vert f_0\Vert }_{H^{\mu }}^2\right)+6T^2\left({\epsilon }_1^{2}N+{\lambda }_N^{-\mu }{\Vert f_1\Vert }_{H^{\mu }}^2\right)\right]\\ \times \exp \left\{\left(3L^2{T}^3-\alpha {\lambda }_J^{s_1}\right)t\right\}+2M_2\exp \left\{L^2{T}^3\left(T-t\right)-\alpha {\lambda }_J^{s_1}t\right\}.\end{array}$ (7)

3) 当 $2s_1<s_2$ 或 $2s_1=s_2,{\alpha }^2-4\beta <0$ 时，若

$\underset{t\in \left[0,T\right]}{\sup }{\displaystyle \underset{k=1}{\overset{\infty }{\sum }}\exp \left\{\alpha {\lambda }_k^{s_1}t\right\}}{\langle u,{\phi }_k\rangle }^2\le M_3,$

那么有

$\begin{array}{c}{\rm E}{\Vert {\tilde{u}}_J-u\Vert }_2^2\le 12\left[\left(1-{\alpha }^2{\lambda }_J^{2s_1}{\Lambda }_1^{-1}\right)\left({\epsilon }_0^{2}N+{\lambda }_N^{-\mu }{\Vert f_0\Vert }_{H^{\mu }}^2\right)-2{\Lambda }_1^{-1}\left({\epsilon }_1^{2}N+{\lambda }_N^{-\mu }{\Vert f_1\Vert }_{H^{\mu }}^2\right)\right]\\ \times \exp \left\{\left(\alpha {\lambda }_J^{s_1}-12L^{2}T{\Lambda }_1^{-1}\right)\left(T-t\right)\right\}+2M_3\exp \left\{-\left(\alpha {\lambda }_J^{s_1}t+4L^{2}T\left(T-t\right){\Lambda }_1^{-1}\right)\right\}.\end{array}$ (8)

证明 本定理的证明以(6)为例，将证明过程分为两部分，(7) (8)的对应部分证明是类似的，这里略去。

第一部分，存在唯一性的证明。考虑积分方程

${\tilde{u}}_J=S_{11}^J\left(t-T\right){\tilde{f}}_{0,N}+S_{12}^J\left(t-T\right){\tilde{f}}_{1,N}-{\displaystyle \underset{t}{\overset{T}{\int }}{S}_{12}^J\left(t-\tau \right)F\left({\tilde{u}}_J\right)\text{d}\tau }.$ (9)

定义算子 ${\rm A}\left(u\right):L_2^{\infty }\to L_2^{\infty }$

${\rm A}\left(u\right)=S_{11}^J\left(t-T\right){\tilde{f}}_{0,N}+S_{12}^J\left(t-T\right){\tilde{f}}_{1,N}-{\displaystyle \underset{t}{\overset{T}{\int }}{S}_{12}^J\left(t-\tau \right)F\left(u\right)\text{d}\tau }.$

下面应用数学归纳法证明， $\forall u_1,u_2\in L_2^{\infty }$，估计

${\rm E}{\Vert {{\rm A}}^m\left(u_1\right)-{{\rm A}}^m\left(u_2\right)\Vert }_2^2\le \frac{{\left[C\left(T-t\right)\right]}^m}{m!}{\Vert u_1-u_2\Vert }_{L_2^{\infty }}^2$ (10)

关于 $m\in {\mathbb{N}}^{+}$ 成立，其中 $C=4L^{2}T\exp \left\{\left(\alpha {\lambda }_J^{s_1}+\sqrt{{\Lambda }_J}\right)T\right\}{\Lambda }_1^{-1}$。

当 $m=1$ 时，由Hölder不等式及 ${\sinh }^2\left(a\right)\le \exp \left\{2a\right\},a>0$ 可推知

$\begin{array}{c}{\rm E}{\Vert {\rm A}\left(u_1\right)-{\rm A}\left(u_2\right)\Vert }_2^2={\rm E}{\Vert {\displaystyle \underset{t}{\overset{T}{\int }}{S}_{12}^J\left(t-\tau \right)\left(F\left(u_1\right)-F\left(u_2\right)\right)\text{d}\tau }\Vert }_2^2\\ \le T{\rm E}\left({\displaystyle \underset{t}{\overset{T}{\int }}{\displaystyle \underset{k=1}{\overset{J}{\sum }}4{\Lambda }_k^{-1}}\exp \left\{\alpha {\lambda }_k^{s_1}\left(\tau -t\right)\right\}{\sinh }^2\left(\frac{\sqrt{{\Lambda }_k}\left(\tau -t\right)}{2}\right){\langle F\left(u_1\right)-F\left(u_2\right),{\phi }_k\rangle }^2\text{d}\tau }\right)\\ \le 4T\exp \left\{\left(\alpha {\lambda }_J^{s_1}+\sqrt{{\Lambda }_J}\right)T\right\}{\Lambda }_1^{-1}{\displaystyle \underset{t}{\overset{T}{\int }}{\rm E}{\Vert F\left(u_1\right)-F\left(u_2\right)\Vert }_2^2\text{d}\tau }\\ \le C{\displaystyle \underset{t}{\overset{T}{\int }}{\rm E}{\Vert u_1-u_2\Vert }_2^2\text{d}\tau }\\ \le C\left(T-t\right){\Vert u_1-u_2\Vert }_{L_2^{\infty }}^2.\end{array}$

这证明了估计(10)关于 $m=1$ 成立，假设(10)关于 $m=m_0$ 成立，当 $m=m_0+1$ 时有

$\begin{array}{c}{\rm E}{\Vert {{\rm A}}^{m_0+1}\left(u_1\right)-{{\rm A}}^{m_0+1}\left(u_2\right)\Vert }_2^2\le C{\displaystyle \underset{t}{\overset{T}{\int }}{\rm E}{\Vert A^{m_0}\left(u_1\right)-A^{m_0}\left(u_2\right)\Vert }_2^2\text{d}\tau }\\ \le \frac{C^{m_0+1}}{m_0!}{\displaystyle \underset{t}{\overset{T}{\int }}{\left(T-\tau \right)}^{m_0}\text{d}\tau }{\Vert u_1-u_2\Vert }_{L_2^{\infty }}^2\\ \le \frac{{\left[C\left(T-t\right)\right]}^{m_0+1}}{\left(m_0+1\right)!}{\Vert u_1-u_2\Vert }_{L_2^{\infty }}^2.\end{array}$

这说明估计(10)关于 $m=m_0+1$ 成立，意味着估计(10)成立。对(10)式两边关于时间 $t\in \left[0,T\right]$ 取上确界得到

${\Vert {{\rm A}}^m\left(u_1\right)-{{\rm A}}^m\left(u_2\right)\Vert }_{L_2^{\infty }}\le \sqrt{\frac{{\left[CT\right]}^m}{m!}}{\Vert u_1-u_2\Vert }_{L_2^{\infty }},\underset{m\to \infty }{\lim }\sqrt{\frac{{\left[CT\right]}^m}{m!}}=0.$

根据Banach不动点定理可知，积分方程(9)在 $L_2^{\infty }$ 上有唯一的不动点 ${\tilde{u}}_J$，这就完成了存在唯一性的证明。

第二部分，误差估计和收敛分析。

建立如下积分方程

$u_J=S_{11}^J\left(t-T\right)f_0+S_{12}^J\left(t-T\right)f_1-{\displaystyle \underset{t}{\overset{T}{\int }}{S}_{12}^J\left(t-\tau \right)F\left(u_J\right)\text{d}\tau },$

那么有

$E{\Vert {\tilde{u}}_J-u\Vert }_2^2\le 2E{\Vert {\tilde{u}}_J-u_J\Vert }_2^2+2E{\Vert u_J-u\Vert }_2^2.$ (11)

应用 ${\left(a+b+c\right)}^2\le 3a^2+3b^2+3c^2$ 和Hölder不等式对(12)右端的第一项进行放缩

$\begin{array}{c}E{\Vert {\tilde{u}}_J-u_J\Vert }_2^2=E{\Vert S_{11}^J\left(t-T\right)\left({\tilde{f}}_{0,N}-f_0\right)+S_{12}^J\left(t-T\right)\left({\tilde{f}}_{1,N}-f_1\right)-{\displaystyle \underset{t}{\overset{T}{\int }}{S}_{12}^J\left(t-\tau \right)\left(F\left({\tilde{u}}_J\right)-F\left(u_J\right)\right)\text{d}\tau }\Vert }_2^2\\ \le 3\left(E{\Vert S_{11}^J\left(t-T\right)\left({\tilde{f}}_{0,N}-f_0\right)\Vert }_2^2+E{\Vert S_{12}^J\left(t-T\right)\left({\tilde{f}}_{1,N}-f_1\right)\Vert }_2^2+E{\Vert {\displaystyle \underset{t}{\overset{T}{\int }}{S}_{12}^J\left(t-\tau \right)\left(F\left({\tilde{u}}_J\right)-F\left(u_J\right)\right)\text{d}\tau }\Vert }_2^2\right)\\ \le 6\exp \left\{\left(\alpha {\lambda }_J^{s_1}+\sqrt{{\Lambda }_J}\right)\left(T-t\right)\right\}\left[\left(1+{\alpha }^2{\lambda }_J^{2s_1}{\Lambda }_1^{-1}\right)E{\Vert {\tilde{f}}_{0,N}-f_0\Vert }_2^2+2{\Lambda }_1^{-1}E{\Vert {\tilde{f}}_{1,N}-f_1\Vert }_2^2\right]\\ +12L^{2}T{\Lambda }_1^{-1}{\displaystyle \underset{t}{\overset{T}{\int }}\exp \left\{\left(\alpha {\lambda }_J^{s_1}+\sqrt{{\Lambda }_J}\right)\left(\tau -t\right)\right\}E{\Vert {\tilde{u}}_J-u_J\Vert }_2^2\text{d}\tau }.\end{array}$

两边同时乘以 $\exp \left\{\left(\alpha {\lambda }_J^{s_1}+\sqrt{{\Lambda }_J}\right)t\right\}$ 得到

$\begin{array}{l}\exp \left\{\left(\alpha {\lambda }_J^{s_1}+\sqrt{{\Lambda }_J}\right)t\right\}E{\Vert {\tilde{u}}_J-u_J\Vert }_2^2\\ \le 6\exp \left\{\left(\alpha {\lambda }_J^{s_1}+\sqrt{{\Lambda }_J}\right)T\right\}\left[\left(1+{\alpha }^2{\lambda }_J^{2s_1}{\Lambda }_1^{-1}\right)E{\Vert {\tilde{f}}_{0,N}-f_0\Vert }_2^2+2{\Lambda }_1^{-1}E{\Vert {\tilde{f}}_{1,N}-f_1\Vert }_2^2\right]\\ +12L^{2}T{\Lambda }_1^{-1}{\displaystyle \underset{t}{\overset{T}{\int }}\exp \left\{\left(\alpha {\lambda }_J^{s_1}+\sqrt{{\Lambda }_J}\right)\tau \right\}E}{\Vert {\tilde{u}}_J-u_J\Vert }_2^2\text{d}\tau .\end{array}$

依Gronwall不等式知

$E{\Vert {\tilde{u}}_J-u_J\Vert }_2^2\le 6\exp \left\{\left(\alpha {\lambda }_J^{s_1}+\sqrt{{\Lambda }_J}+12L^{2}T{\Lambda }_1^{-1}\right)\left(T-t\right)\right\}\left[\left(1+{\alpha }^2{\lambda }_J^{2s_1}{\Lambda }_1^{-1}\right)E{\Vert {\tilde{f}}_{0,N}-f_0\Vert }_2^2+2{\Lambda }_1^{-1}E{\Vert {\tilde{f}}_{1,N}-f_1\Vert }_2^2\right].$ (12)

依精确解的先验界对(11)右端的第二项放缩

$\begin{array}{c}E{\Vert u_J-u\Vert }_2^2=E{\Vert {\displaystyle \underset{t}{\overset{T}{\int }}{S}_{12}^J\left(t-\tau \right)\left(F\left(u_J\right)-F\left(u\right)\right)\text{d}\tau }\Vert }_2^2\\ +{\displaystyle \underset{k=J+1}{\overset{\infty }{\sum }}\exp \left\{\left(\alpha {\lambda }_k^{s_1}+\sqrt{{\Lambda }_k}\right)t\right\}}\exp \left\{-\left(\alpha {\lambda }_k^{s_1}+\sqrt{{\Lambda }_k}\right)t\right\}{\langle u,{\phi }_k\rangle }^2\\ \le 4L^{2}T{\Lambda }_1^{-1}{\displaystyle \underset{t}{\overset{T}{\int }}\exp \left\{\left(\alpha {\lambda }_J^{s_1}+\sqrt{{\Lambda }_J}\right)\left(\tau -t\right)\right\}E{\Vert u_J-u\Vert }_2^{2}d\tau }\\ +{\displaystyle \underset{k=J+1}{\overset{\infty }{\sum }}\exp \left\{\left(\alpha {\lambda }_k^{s_1}+\sqrt{{\Lambda }_k}\right)t\right\}}\exp \left\{-\left(\alpha {\lambda }_J^{s_1}+\sqrt{{\Lambda }_J}\right)t\right\}{\langle u,{\phi }_k\rangle }^2.\end{array}$

两边同时乘以 $\exp \left\{\left(\alpha {\lambda }_J^{s_1}+\sqrt{{\Lambda }_J}\right)t\right\}$ 得到

$\exp \left\{\left(\alpha {\lambda }_J^{s_1}+\sqrt{{\Lambda }_J}\right)t\right\}E{\Vert u_J-u\Vert }_2^2\le 4L^{2}T{\Lambda }_1^{-1}{\displaystyle \underset{t}{\overset{T}{\int }}\exp \left\{\left(\alpha {\lambda }_J^{s_1}+\sqrt{{\Lambda }_J}\right)\tau \right\}E{\Vert u_J-u\Vert }_2^2\text{d}\tau }+M_1,$

依Gronwall不等式知

$E{\Vert u_J-u\Vert }_2^2\le M_1\exp \left\{-\left(\alpha {\lambda }_J^{s_1}+\sqrt{{\Lambda }_J}\right)t+4L^{2}T\left(T-t\right){\Lambda }_1^{-1}\right\}.$ (13)

联立(11) (12) (13)和引理1，就完成了误差估计(6)的证明。

这里(6)的收敛阶为

$\max \left\{{\epsilon }_0^{2}N\exp \left\{\left(\alpha {\lambda }_J^{s_1}+\sqrt{{\Lambda }_J}\right)T\right\},{\epsilon }_1^{2}N\exp \left\{\left(\alpha {\lambda }_J^{s_1}+\sqrt{{\Lambda }_J}\right)T\right\},{\lambda }_N^{-\mu }\exp \left\{\left(\alpha {\lambda }_J^{s_1}+\sqrt{{\Lambda }_J}\right)T\right\}\right\}.$

只要取 $\epsilon =\max \left\{{\epsilon }_0,{\epsilon }_1\right\},N={\epsilon }^{\nu }\left(-2<\nu <0\right)$，并令正则化参数J随 ${\epsilon }_0,{\epsilon }_1$ 趋于0满足

$\exp \left\{\left(\alpha {\lambda }_J^{s_1}+\sqrt{{\Lambda }_J}\right)T\right\}~N^{\theta }\left(\theta >0\right)$

就能保证(6)的收敛，其中

${\lambda }_N~N^{2n^{-1}}\left(N\to \infty \right),0<\theta <2\mu n^{-1},-2{\left(\theta +1\right)}^{-1}<\nu <0.$

这样就说明了Fourier截断正则化方法对问题(1)的有效性。

5. 结论

本文利用Fourier截断正则化方法研究了高斯白噪声扰动下分数阶强阻尼波动方程的正则化理论，通过温和解的表达式阐明了问题的不适定性，在精确解先验界下得到了正则解和精确解间的收敛性误差估计式，这一结果说明了正则化方法的有效性，并为分数阶强阻尼波动方程正则化研究提供了研究启示。

文章引用

容伟杰. 分数阶强阻尼波动方程的Fourier正则化
Fourier Regularization of Fractional Strongly Damped Wave Equations[J]. 应用数学进展, 2022, 11(03): 1013-1020. https://doi.org/10.12677/AAM.2022.113109

参考文献