泊松白噪声驱动的一维随机波动方程

One-Dimensional Stochastic Wave Equations Driven by Poisson White Noise

摘要: 偏微分方程是一种确定性方程，虽然在多个领域具有广泛应用，但是不能很好的描述不确定的情况，因此探究随机噪声驱动的偏微分方程方程显得十分重要，所以本文主要利用了Picard迭代和Gronwall不等式证明了在一定条件下泊松白噪声驱动的一维随机波动方程的mild解的存在性和唯一性。

Abstract: Partial differential equation is a kind of deterministic equations, although with wide applications in many fields, but not a good description of uncertain situation, thus to explore the partial differential equation of random noise drive equation is very important, so this paper mainly use the Picard iteration and Gronwall inequality is proved under certain conditions poisson white noise driven one-dimensional random wave equation of mild existence and uniqueness of solution.

1. 引言

一维波动方程基本形式为：

$\{\begin{array}{l}\frac{{\partial }^{2}u}{\partial t^2}=a^2\frac{{\partial }^{2}u}{\partial x^2}\\ u\left(0,x\right)=\phi \left(x\right)\\ u_t\left(0,x\right)=\psi (\; x\; )\end{array}$

其中， $x\in \left(0,\infty \right)$，a是正常数， $\phi \left(x\right)$ 和 $\psi \left(x\right)$ 是定义在R上的已知函数。

一个微分方程中出现多元函数的偏导数，或者说如果未知函数和几个变量有关，而且方程中出现未知函数对应几个变量的导数，那么这种微分方程就是偏微分方程。在科学技术日新月异的发展过程中，人们研究的许多问题用一个自变量的函数来描述已经不够，故而常常用偏微分方程来描述世界现象的发展规律。但是，在现实生活中不确定性无处不在、不可避免，含有随机项的偏微分方程可以更加贴切地描述客观规律，在量子场论、统计力学、金融数学等领域有着广泛的应用。因此，研究含有随机项的偏微分方程有重大意义。

近些年来，Claudia Knoche [1] 证明了无限维波动方程解的存在性，唯一性和可微性及初始条件相关的规律性。Sergio Albeverio，Jiang-Lun Wu和Tu-Sheng Zhang [2] 建立了由纯泊松白噪声驱动的抛物型随机偏微分方程的解的存在唯一性。Caroline Cardon-Weber [3] 证明了受时空白噪声扰动的抛物线型微分方程的大偏差原理。Max-K. von Renesse和Michael Scheutzowr [4] 证明了随机泛函微分方程解的存在唯一性，基于此Michael Röckner和Tusheng Zhang [5] 提出了跳跃型随机偏微分方程的存在性，唯一性和大偏差原理。

对于白噪声驱动的随机微分方程解的相关问题，也有大量学者对其进行的研究阐述。陶文健 [6] 研究了具有乘性白噪声项的波动方程生成的一族半群以及其全局吸引子的存在性；朱星亮 [7] 研究了一类由Gaussian型Levy白噪声驱动的半线性随机波动方程Cauchy问题解的存在唯一性；Alebeverio，Wu和Zhang [2] 将泊松白噪声驱动的随机微分方程转化为带跳的随机积分方程，证明了希尔伯特空间中的解的存在唯一性。中国科学技术大学成丹 [8] 也曾研究过泊松白噪声驱动的一维随机方程解的存在唯一性，考虑了随机波动方程的mild解，利用Picard迭代的方法证明了一维随机波动方程解u的存在性，同时证明Picard迭代中的随机过程的un关于时间变量和空间变量具有正则性。

本文主要对泊松白噪声驱动的一维随机波动方程的解的存在唯一性进行证明。在文献 [8] 中给出的解的证明方法的基础上，假设方程(1)的解满足文中假设的条件下，证明方程(1)的解存在且唯一。

本文考虑如下泊松白噪声驱动的一维随机波动方程：

$\{\begin{array}{l}\frac{{\partial }^{2}u}{\partial t^2}\left(t,x,\omega \right)=\frac{{\partial }^{2}u}{\partial x^2}\left(t,x,\omega \right)+f\left(t,u\left(t,x\right)\right)+{\displaystyle {\int }_{u}g\left(t,x,u\left(t,x,\omega \right);y\right){\eta }_t\left(\text{d}y,\omega \right)}\\ u\left(0,x\right)=0\\ u_t\left(0,x\right)=0\end{array}$ (1)

其中， $t\in \left[0,\infty \right)$， $x\in R$， $\omega \in \Omega$。映射 $g:\left[0,\infty \right)\times R^2\times U\to R$，为 $B\left(\left[0,\infty \right)\right)\otimes B\left(R^2\right)\otimes B\left(U\right)/B\left(R\right)$ 可测函数， ${\eta }_t$ 是泊松白噪声，并称如下随机积分形式的解，为上述方程(1)的mild解：

$\begin{array}{c}u\left(t,x,\omega \right)={\displaystyle {\int }_0^{t+}{\displaystyle {\int }_u{\displaystyle {\int }_R\left(G_{t-s}\left(x-z\right)g\left(s,z,u\left(s,-z,\omega \right);y\right)\right)\text{d}zq\left(\text{d}s,\text{d}y,\omega \right)}}}\\ +{\displaystyle {\int }_0^t{\displaystyle {\int }_R\left(G_{t-s}\left(x-z\right)f\left(s,u\left(s,z\right)\right)\right)\text{d}s\text{d}z}}\end{array}$ (2)

这里 $t\in \left[0,\infty \right)$， $x\in R$。

2. 主要内容

2.1. 假设

在本文的这部分，我们给出本文的一些假设：

2.1.1. 假设1

随机过程 $u\left(t,x,\omega \right)\in \left[0,\infty \right)\times R\times \Omega$，是方程(1)的解，应当满足以下性质：

1) $u\left(t,x,\omega \right)$ 是 $\left\{F_t\right\}$ -适应过程；

2) $u\left(t,x,\omega \right)$ 是 $L^2\left(\Omega \right)$ 值的随机过程，关于变量 $t\in \left[0,\infty \right)$ 是右连左极限存在的；

3) 固定的t，x，方程成立。

2.1.2. 假设2

给定T>0若存在实数 $\epsilon$， $L_T>0$，正的实函数 $K_T\left(x\right)\in L^2\left(R\right)\cap L^{2+\epsilon }\left(R\right)$，对于 $\left(t,x,z\right)\in \left[0,T\right]\times R\times R$ 满足：

(c1)： ${\displaystyle {\int }_U{\left[g\left(t,x,z;y\right)\right]}^{2}v\left(\text{d}y\right)}\le K_T\left(x\right)\left|z\right|$

(c2)： ${\left|f\left(t,z\right)\right|}^2\le a\left(t\right)+k\left(t\right)\left|z\right|$

(c3)： ${\displaystyle {\int }_U{\left[g\left(t,x,z_1;y\right)-g\left(t,x,z_2;y\right)\right]}^{2}v\left(\text{d}y\right)}\le K_T\left(x\right){\left|z_1-z_2\right|}^2$

(c4)： ${\left|f\left(t,z_1\right)-f\left(t,z_2\right)\right|}^2\le B_T\left|z_1-z_2\right|$

对于 $\left(t,x,z\right)\in \left(0,T\right]\times R\times R$ 成立，那么方程存在唯一的解。

2.2. 推导过程

在本文的这部分，主要给出方程的解存在唯一性的证明过程，首先通过Picard迭代的方法来证明解的存在性，之后，再利用Grownwall不等式进行唯一性的证明。

为方便证明，我们把解分为两个部分：

设 $u\left(t,x,\omega \right)=I_1+I_2$，其中：

$I_1:{\displaystyle {\int }_0^{t+}{\displaystyle {\int }_u{\displaystyle {\int }_R\left(G_{t-s}\left(x-z\right)g\left(s,z,u\left(s,-z,\omega \right);y\right)\right)\text{d}zq\left(\text{d}s,\text{d}y,\omega \right)}}}$

$I_2:{\displaystyle {\int }_0^t{\displaystyle {\int }_R{G}_{t-s}\left(x-z\right)f\left(s,u\left(s,z\right)\right)\text{d}s\text{d}z}}$

首先，对于 $\left(t,x,\omega \right)$ 和 $n\in N$，令 $u_1\left(t,x,\omega \right)=0$，由Picard迭代定义：

$I_1:u_{n+1}\left(t,x,\omega \right)={\displaystyle {\int }_0^{t+}{\displaystyle {\int }_u{\displaystyle {\int }_R\left(G_{t-s}\left(x-z\right)g\left(s,z,u\left(s,-z,\omega \right);y\right)\right)\text{d}zq\left(\text{d}s,\text{d}y,\omega \right)}}}=I_1\left(t,x,\omega \right)$ (3)

$I_2:u_{n+1}\left(t,x\right)={\displaystyle {\int }_0^t{\displaystyle {\int }_R\left(G_{t-s}\left(x-z\right)f\left(s,u_n\left(s,z\right)\right)\right)\text{d}z\text{d}s}}=I_1\left(t,x\right)$ (4)

我们必须证明对于任意的 $n\in N$，公式(3)和(4)是有意义的，具体为：

$I_1:$

引理1若 $u_n\in N$ 则对于任意固定的 $t\in \left[0,T\right]$， $x\in R$，

$h_{t,x}\left(s,y,\omega \right):={\displaystyle {\int }_R\left(G_{t-s}\left(x-z\right)g\left(s,z,u_n\left(s,z,\omega \right);y\right)\right)\text{d}z}\in H$

证明：

对于 $\left(s,z,\omega \right)\in \left[0,T\right]\times R\times \Omega$，定义：

${\mu }_n^{\left(m\right)}\left(s,z,\omega \right):={\mu }_n\left(0,z,\omega \right)+\underset{k=0}{\overset{2^m-1}{{\displaystyle \sum }}}{\mu }_n\left(\frac{kt}{2^m},z,\omega \right)1_{\left(\frac{kt}{2^m},\frac{\left(k+1\right)t}{2^m}\right)}\left(s\right),m\in N$

则 ${\mu }_n^{\left(m\right)}\left(s,z,\omega \right)$ 是 $\left\{F_s\right\}$ -可料的。

对于所有的 $\left(s,z,\omega \right)\in \left[0,T\right]\times U\times \Omega$ 令：

$h_{t,x}^m\left(s,y,\omega \right):={\displaystyle {\int }_R{G}_{t-s}}\left(x-z\right)g\left(s,z,u_n^{\left(m\right)}\left(s,z,\omega \right);y\right)\text{d}z,m\in N.$

我们首先证明： $h_{t,x}^m\left(s,y,\omega \right)\in H_p^2$

1) 证明 $h_{t,x}^m\left(s,y,\omega \right)$ 是 $\left\{F_s\right\}$ -可料的。

2) 证明 $h_{t,x}^m\left(s,y,\omega \right)$ 是 $B\left[0,T\right]\otimes R\otimes F/B\left(R\right)$ -可测的，由Fubini定理，只需证明：

${\displaystyle {\int }_R{G}_{t-s}\left(x-z\right)\left|g\left(s,z,u_n^{\left(m\right)}\left(s,z,\omega \right);y\right)\right|\text{d}z}<\infty ,$ (5)

(可以由下面的证明得到)。

3) $E{\displaystyle {\int }_0^t{\displaystyle {\int }_u{\left[h_{t,x}^m\left(s,y,\omega \right)\right]}^{2}v\left(\text{d}y\right)\text{d}s}}<\infty$

首先，由Schwarz不等式得：

$\begin{array}{l}{\displaystyle {\int }_R{G}_{t-s}\left(x-z\right)\left|g\left(s,z,u_n^{\left(m\right)}\left(s,z,\omega \right);y\right)\right|\text{d}z}\\ \le {\left\{{\displaystyle {\int }_R{G}_{t-s}\left(x-z\right)\text{d}z}\right\}}^{\frac{1}{2}}{\left\{{\displaystyle {\int }_R{G_{t-s}\left(x-z\right)\left(g\left(s,z,\omega \right);y\right)]}^2\text{d}z}\right\}}^{\frac{1}{2}}<\infty \end{array}$

那么，由假设(c1)， $K_T\left(z\right)\in L^2\left(R\right)$ 和 ${\left[G_{t-s}\left(x-z\right)\right]}^2=G_{t-s}\left(x-z\right)$ 得：

$\begin{array}{l}{\displaystyle {\int }_u{\displaystyle {\int }_R{G}_{t-s}}}\left(x-z\right){\left[g\left(s,z,U_n^{\left(m\right)}\left(s,z,\omega \right);y\right)\right]}^2\text{d}zv\left(\text{d}y\right)\\ \le T{\displaystyle {\int }_R{\displaystyle {\int }_u{G}_{t-s}}}\left(x-z\right){\left[g\left(s,z,U_n^{\left(m\right)}\left(s,z,\omega \right);y\right)\right]}^2\text{d}zv\left(\text{d}y\right)\\ =T{\displaystyle {\int }_R{G}_{t-s}}\left(x-z\right){\left[g\left(s,z,U_n^{\left(m\right)}\left(s,z,\omega \right);y\right)\right]}^2\text{d}zv\left(\text{d}y\right)\\ \le T{\displaystyle {\int }_R{G}_{t-s}}\left(x-z\right)K_T\left(z\right)\left|U_n^{\left(m\right)}\left(s,z,\omega \right)\right|\text{d}z\\ \le T{\left\{{\displaystyle {\int }_R{G}_{t-s}\left(x-z\right)K_T{\left(z\right)}^2\text{d}z}\right\}}^{\frac{1}{2}}{\left\{2{\displaystyle {\int }_R{G}_{t-s}}\left(x-z\right){\left(\left|U_n^{\left(m\right)}\left(s,z,\omega \right)\right|\right)}^2\text{d}z\right\}}^{\frac{1}{2}}\\ \le \sqrt{2}T{\left\{{\displaystyle {\int }_R{K}_T{\left(Z\right)}^2\text{d}z}\right\}}^{\frac{1}{2}}{\left\{{\displaystyle {\int }_R{\left(\left|U_N^{\left(m\right)}\left(s,z,\omega \right)\right|\right)}^2\text{d}z}\right\}}^{\frac{1}{2}}<\infty \end{array}$ (6)

由以上公式(6)，可知公式(5)成立，而且有：

$E\left\{{\displaystyle {\int }_0^T{\displaystyle {\int }_U{\left[h_{t,x}^m\left(s,y,\omega \right)\right]}^{2}v\left(\text{d}y\right)\text{d}s}}\right\}\le TE\left\{{\displaystyle {\int }_0^T{\displaystyle {\int }_U{\displaystyle {\int }_R{G}_{t-s}}}}\left(x-z\right){\left[g\left(s,z,u_n^{\left(m\right)}\left(s,z,\omega \right);y\right)\right]}^2\text{d}zv\left(\text{d}y\right)\text{d}s\right\}<\infty$

综上所述，我们可以得到 $h_{t,x}^m\left(s,y,\omega \right)\in H_p^2$，通过文献中利用假设(c1)的证明，最终得到 $h_{t,x}\left(s,y,\omega \right)\in H$。

引理2如果 $u_n\in U_T$，则对于任意固定的 $\left(t,x\right)\in \left[0,T\right]\times R$， $I_1\left(t,x,\cdot \right)$ 是可测的，即 $I_1$ 关于 $\left\{F_t\right\}$ -适应，并且存在 $I_1$ 关于 $t\in \left[0,T\right]$ 右连续左极限存在的修正，关于 $x\in R$ 有连续的修正。

(通过文献 [8] 可得引理2的证明过程，在此省略)

引理3令 $\left\{X\left(x\right)\right\}$ 是一个实值的随机过程，如果对任意的L > 0存在正实数 $d_L,{\beta }_L$ 和 ${\epsilon }_L$ (只依赖于L)满足 $E\left\{\left|X\left(x_1\right)-X\left(x_2\right)\right|\right\}\le {\beta }_L{\left|x_1-x_2\right|}^{1+\epsilon L}$，则 $\left\{X\left(x\right),x\in R\right\}$ 有连续版本。

下面来进行证明：

(这部分与文献 [8] 中证明类似，但为了保持文章完整性，在此列出具体过程)

$\begin{array}{l}E{\left[J_t\left(r,x_1,.\right)-J_t\left(r,x_2,.\right)\right]}^2\\ =E\left\{{{\displaystyle {\int }_0^{r+}\left[{\displaystyle {\int }_R\left|G_{t-s}\left(x_1-z\right)-G_{t-s}\left(x_2-z\right)\right|g\left(s,z,u_n\left(s,z\right);y\right)\text{d}z}\right]}}^{2}v\left(\text{d}y\right)\text{d}s\right\}\\ \le E\left\{{\displaystyle {\int }_0^{r+}{\displaystyle {\int }_R\left|G_{t-s}\left(x_1-z\right)-G_{t-s}\left(x_2-z\right)\right|\text{d}z}}\times {\displaystyle {\int }_R\left|G_{t-s}\left(x_1-z\right)-G_{t-s}\left(x_2-z\right)\right|}{\displaystyle {\int }_u{\left\{g\left(s,z,u_n\left(s,z,.\right);y\right)\right\}}^2}\right\}v\left(\text{d}y\right)\text{d}z\text{d}s\\ \le E\left\{{\displaystyle {\int }_0^{r+}{\displaystyle {\int }_R\left|G_{t-s}\left(x_1-z\right)-G_{t-s}\left(x_2-z\right)\right|\text{d}z}}\times {\displaystyle {\int }_R\left\{\left|G_{t-s}\left(x_1-z\right)-G_{t-s}\left(x_2-z\right)\right|K_T\left(Z\right)\left|u_n\left(s,z,.\right)\right|\text{d}z\text{d}s\right\}}\right\}\end{array}$

首先计算下列式子，不妨假设 $x_1\le x_2$

$\begin{array}{l}{\displaystyle {\int }_R\left|G_{t-s}\left(x_1-z\right)-G_{t-s}\left(x_2-z\right)\right|\text{d}z}\\ ={\displaystyle {\int }_R\left|1_{\left\{x_1-\left(t-s\right)\le z\le x_1+\left(t-s\right)\right\}}-1_{\left\{x_2-\left(t-s\right)\le z\le x_2+\left(t-s\right)\right\}}\right|\text{d}z}\\ =4\left(t-s\right)1_{\left\{s\ge t-\frac{\left(x_2-x_1\right)}{2}\right\}}+2\left(x_2-x_1\right)1_{\left\{s<t-\frac{\left(x_2-x_1\right)}{2}\right\}}\end{array}$

那么，我们可以得到对于任意的 $x_1\in R$ 和 $x_2\in R$ ：

${\displaystyle {\int }_R\left|G_{t-s}\left(x_1-z\right)-G_{t-s}\left(x_2-z\right)\right|\text{d}z}=4\left(t-s\right)1_{\left\{s\ge t-\frac{\left(x_2-x_1\right)}{2}\right\}}+2\left(x_2-x_1\right)1_{\left\{s<t-\frac{\left(x_2-x_1\right)}{2}\right\}}$

由Hölder不等式，我们可以得到：

$\begin{array}{l}E{\left[I_1\left(t,x_1,.\right)-I_1\left(t,x_2,.\right)\right]}^2\\ =E{\left[J_t\left(t,x_1,.\right)-J_t\left(t,x_2,.\right)\right]}^2\\ \le E\left\{{\displaystyle {\int }_0^t{\displaystyle {\int }_R\left|G_{t-s}\left(x_1-z\right)-G_{t-s}\left(x_2-z\right)\right|\text{d}z}}\right\}\times \left[{\displaystyle {\int }_R\left|G_{t-s}\left(x_1-z\right)-G_{t-s}\left(x_2-z\right)\right|K_T\left(Z\right)\left|u_n\left(s,z,.\right)\right|\text{d}z}\right]\text{d}s\\ \le E\{{\displaystyle {\int }_0^t{\displaystyle {\int }_R\left|G_{t-s}\left(x_1-z\right)-G_{t-s}\left(x_2-z\right)\right|\text{d}z}}\times {\left[{\displaystyle {\int }_R{\left|G_{t-s}\left(x_1-z\right)-G_{t-s}\left(x_2-z\right)\right|}^{\frac{4+2\epsilon }{\epsilon }}\text{d}z}\right]}^{\frac{4+2\epsilon }{\epsilon }}\\ {\left[{\displaystyle {\int }_R{\left(K_T\left(Z\right)\right)}^{2+\epsilon }\text{d}z}\right]}^{\frac{1}{2+\epsilon }}\times {\left[{\displaystyle {\int }_R{\{u_n\left(s,z,.\right)]}^2\text{d}z}\right]}^{\frac{1}{2}}\}\text{d}s\end{array}$

令： $h\left(\epsilon \right)=\left(4+3\epsilon \right)/\left(4+2\epsilon \right)$ ：

上式 $\le E{\displaystyle {\int }_0^t{\left[{\displaystyle {\int }_R\left|G_{t-s}\left(x_1-z\right)-G_{t-s}\left(x_2-z\right)\right|\text{d}z}\right]}^{h\left(\epsilon \right)}}\times {\left\{{\displaystyle {\int }_R{\left(K_T\left(z\right)\right)}^{2+\epsilon }\text{d}z}\right\}}^{\frac{1}{2+\epsilon }}{\left[{\displaystyle {\int }_R{\left[u_n\left(s,z,.\right)\right]}^2\text{d}z}\right]}^{\frac{1}{2}}$

定义

$c\left(s\right)={\left\{{\displaystyle {\int }_R{\left(K_T\left(z\right)\right)}^{2+\epsilon }\text{d}z}\right\}}^{\frac{1}{2+\epsilon }}E{\left[{\displaystyle {\int }_R{\left[u_n\left(s,z,\cdot \right)\right]}^2\text{d}z}\right]}^{\frac{1}{2}}$

因为 $E{J}_R{\left[u\left(t,s,w\right)\right]}^2\text{d}x<\infty$，对于所有的 $t\in \left[0,T\right]$，并且 $u\left(t,x,w\right)$ 关于是右连续左极限存在的，所以

${\sup }_{s\in \{0,T]}C\left(s\right)<\infty$

因此

$\begin{array}{l}E{\left[I_1\left(t,x_1,\cdot \right)-I_1\left(t,x_2,\cdot \right)\right]}^2\\ \le {\sup }_{s\in \{0,T]}C\left(s\right)\left[2^{h\left(\epsilon \right)-1}\frac{1}{h\left(\epsilon \right)+1}{\left|x_2-x_1\right|}^{h\left(\epsilon \right)+1}+2^{h\left(\epsilon \right)}{\left|x_2-x_1\right|}^{h\left(\epsilon \right)}t-2^{h\left(\epsilon \right)-1}{\left|x_2-x_1\right|}^{h\left(\epsilon \right)+1}\right]\\ \le {\sup }_{s\in \{0,T]}C\left(s\right){\displaystyle {\int }_0^t{\left[{\displaystyle {\int }_R\left|G_{t-s}\left(x_1-z\right)-G_{t-s}\left(x_2-z\right)\right|\text{d}z}\right]}^{h\left(\epsilon \right)}\text{d}s}\\ \le {\sup }_{s\in \{0,T]}C\left(s\right)\left[{\displaystyle {\int }_{t-\frac{\left|x_2-x_1\right|}{2}}^t{4}^{h\left(\epsilon \right)}{\left(t-s\right)}^{h\left(\epsilon \right)}\text{d}s}+{\displaystyle {\int }_0^{t-\frac{\left|x_2-x_1\right|}{2}}{2}^{h\left(\epsilon \right)}{\left(x_2-x_1\right)}^{h\left(\epsilon \right)}\text{d}s}\right]\\ \le {\sup }_{s\in \{0,T]}C\left(s\right)\left[2^{h\left(\epsilon \right)-1}\frac{1}{h\left(\epsilon \right)+1}{\left|x_2-x_1\right|}^{h\left(\epsilon \right)+1}+2^{h\left(\epsilon \right)}{\left|x_2-x_1\right|}^{h\left(\epsilon \right)}t-2^{h\left(\epsilon \right)-1}{\left|x_2-x_1\right|}^{h\left(\epsilon \right)+1}\right]\end{array}$

令

$C_T=\max \left\{{\sup }_{s\in \{0,T]}c\left(s\right),2^{h\left(\epsilon \right)-1}\frac{1}{h\left(\epsilon \right)+1},2^{h\left(\epsilon \right)}T\right\}$

这是个只依赖T的常数，因此，根据Schwarz不等式，Fubini定理和假设(c3)，我们可以得到：

$\sup E{\left[I_1\left(t,x_1,\cdot \right)-I_1\left(t,x_2,\cdot \right)\right]}^2\le C_T{\left|x_2-x_1\right|}^{h\left(\epsilon \right)}t\in \{0,T].$

因为 $h\left(\epsilon \right)>1$，则引理3得证。

引理4对任意的 $n\in R$， $u_n\in U_T$。

证明：我们只需证明 $u_n\in U_T$，则 $u_{n+1}\in U_T$，由数学归纳法，即可证明引理4。

由面前的结果可知，我们只需要证明 $E{\displaystyle {\int }_R{\left[u_{n+1}\left(t,x,\cdot \right)\right]}^2\text{d}x}<\infty ,t\in \left[0,T\right]$。由Schwarz不等式，Fubini定理和假设(c3)和(c4)，我们可以得到：

$I_1:$

$\begin{array}{l}E{\displaystyle {\int }_R{\left[U_{n+1}\left(t,x,.\right)\right]}^2\text{d}x}\\ =E{\displaystyle {\int }_R{\left[I_1\left(t,x_1,\cdot \right)\right]}^2\text{d}s}=E_K{\left\{{\displaystyle {\int }_t\left(t,s,.\right)}\right\}}^2\text{d}x\\ ={\displaystyle {\int }_{R}E}{\left[{\displaystyle {\int }_0^t{\displaystyle {\int }_R{G}_{t-s}}}\left(x-z\right)g\left(s,z,u_z\left(s,z,\omega \right);y\right)\right]}^2\text{d}zv\left(\text{d}y\right)\text{d}s\}\text{d}x\\ \le T{\displaystyle {\int }_{R}E\left\{{\displaystyle {\int }_0^t{\displaystyle {\int }_u{\displaystyle {\int }_R{G}_{t-s}}}}\left(x-z\right){\left\{g\left(s,z,u_z\left(s,z,\omega \right);y\right)\right\}}^2\right\}}\\ \le T{\displaystyle {\int }_{R}E\{{\displaystyle {\int }_0^t{\displaystyle {\int }_R\left[{\displaystyle {\int }_R{G}_{t-s}\left(x-z\right)\text{d}z}\right]K_T\left(z\right)\left|u_n\left(s,z,\cdot \right)\right|\text{d}z\text{d}s}}}\end{array}$

$\begin{array}{l}\le T^{2}E\left\{{\displaystyle {\int }_0^t{\displaystyle {\int }_R{K}_T\left(z\right)\left|u_n\left(s,z,\cdot \right)\right|\text{d}z\text{d}s}}\right\}\\ \le T^{2}E\{{\displaystyle {\int }_0^t{\left\{{\displaystyle {\int }_R{\left[K_T\left(Z\right)\right]}^2\text{d}z}\right\}}^{\frac{1}{2}}}{\displaystyle {\int }_0^t{\left\{{\displaystyle {\int }_R{\left|u_n\left(s,z,\cdot \right)\right|}^2\text{d}z\text{d}s}\right\}}^{\frac{1}{2}}}\\ \le T^2{\displaystyle {\int }_0^t{\left\{{\displaystyle {\int }_R{\left[K_T\left(z\right)\right]}^2\text{d}z}\right\}}^{\frac{1}{2}}E{\left[{\displaystyle {\int }_R{\left|u_n\left(s,z,\cdot \right)\right|}^2\text{d}z}\right]}^{\frac{1}{2}}\text{d}s}<\infty \end{array}$

$I_2:$

${\displaystyle {\int }_R\left|G_{t-s}\left(x-z\right)f\left(s,u_n\left(s,z\right)\right)\right|\text{d}z}\le {\left\{{\displaystyle {\int }_R{G}_{t-s}\left(x-z\right)\text{d}z}\right\}}^{\frac{1}{2}}{\left\{{\displaystyle {\int }_R{G}_{t-s}}\left(x-z\right){\left[{{\displaystyle \int }}^{\text{}}\left(s,u_n\left(s,z\right)\right)\right]}^2\text{d}z\right\}}^{\frac{1}{2}}$

而

$\begin{array}{l}{\left\{{\displaystyle {\int }_R{G}_{t-s}\left(x-z\right){\left[f\left(s,u_n\left(s,z\right)\right)\right]}^2\text{d}z}\right\}}^{\frac{1}{2}}\\ \le 2\left({\displaystyle {\int }_R\left(a\left(s\right)+k\left(s\right)\left|u_n\left(s,z\right)\right|\right)\text{d}z}+1\right)\\ \le {\Vert a\Vert }_1+{\Vert k\Vert }_2{\Vert U_n\left(s\right)\Vert }_2+1<\infty \end{array}$

2.3. 存在性证明

$I_1:$

令 $F_n\left(t\right):=E\left\{{\displaystyle {\int }_R{\left[u_{n+1}\left(s,z,\cdot \right)-u_n\left(s,z,\cdot \right)\right]}^2\text{d}x}\right\}$ 其中 $t\in \left[0,T\right]$， $n\in N$，由公式(3)和假设(c3)，我们可以得到：

$\begin{array}{l}E{\displaystyle {\int }_R{\left\{{\displaystyle {\int }_0^{t+}{\displaystyle {\int }_U{\displaystyle {\int }_R{G}_{t-s}}}}\left(x-z\right)\times \left[g\left(s,z,u_n\left(s,z,\omega \right);y\right)-g\left(s,z,u_{n-1}\left(s,z,\omega \right);y\right)\right]\text{d}zq\left(\text{d}s,\text{d}y,\cdot \right)\right\}}^2\text{d}x}\\ =E{{\displaystyle {\int }_R{\displaystyle {\int }_0^t{\displaystyle {\int }_U\left[{\displaystyle {\int }_R{G}_{t-s}\left(x-z\right)\times \left[g\left(s,z,u_n\left(s,z,\omega \right);y\right)-g\left(s,z,u_{n-1}\left(s,z,\omega \right);y\right)\right]\text{d}z}\right]}}}}^{2}v\left(\text{d}y\right)\text{d}s\text{d}x\\ \le L_{T}E{\displaystyle {\int }_R{\displaystyle {\int }_0^t{\displaystyle {\int }_R{G}_{t-s}}}}\left(x-z\right){\left[u_n\left(s,z,\omega \right)-u_{n-1}\left(s,z,\omega \right)\right]}^2\text{d}z\text{d}s\text{d}x\\ =L_{T}E{\displaystyle {\int }_R{\displaystyle {\int }_0^t\left\{{\displaystyle {\int }_R{G}_{t-s}\left(x-z\right)\text{d}x}\right\}{\left[u_n\left(s,z,\omega \right)-u_{n-1}\left(s,z,\omega \right)\right]}^2\text{d}z\text{d}s}}\\ \le T{L}_T{\displaystyle {\int }_0^{t}E}{\displaystyle {\int }_R{\left[u_n\left(s,z,\omega \right)-u_{n-1}\left(s,z,\omega \right)\right]}^2\text{d}z\text{d}s}\\ =C_T{L}_T{\displaystyle {\int }_0^t{F}_{n-1}\left(s\right)\text{d}s}\end{array}$

其中， $C_T=T$。

因此，通过递推，我们可以得到：

$F_n\left(t\right)\le {\left[G_T{L}_T\right]}^{n-1}{\displaystyle {\int }_0^t\text{d}{t}_1}{\displaystyle {\int }_0^{t_1}\text{d}{t}_2}\cdots {\displaystyle {\int }_0^{t_{n-2}}{F}_1\left(t_{n-1}\right)\text{d}{t}_{n-1}}$

另外，通过公式(3)和假设(c1)，我们有：

$\begin{array}{c}{F}_1\left(t\right)=E\left\{{\displaystyle {\int }_R{\left[u_2\left(t,x,\cdot \right)-u_1\left(t,x,\cdot \right)\right]}^2\text{d}x}\right\}\\ \le C_T{\displaystyle {\int }_R{\displaystyle {\int }_0^t{\displaystyle {\int }_R{G}_{t-s}}}}\left(x-z\right)K_T\left(z\right)\left|u_1\left(s,z,\cdot \right)\right|\text{d}z\text{d}s\\ =C_T{\displaystyle {\int }_R{\displaystyle {\int }_0^t\left\{{\displaystyle {\int }_R{G}_{t-s}\left(x-z\right)\text{d}x}\right\}{K}_T\left(z\right)\text{d}z\text{d}s}}\\ \le T{C}_T{\displaystyle {\int }_0^t{\displaystyle {\int }_R{K}_T\left(z\right)\text{d}z\text{d}s}}\end{array}$

由公式(4)和假设(c2)，我们可以得到：

$\begin{array}{l}E{\displaystyle {\int }_R{\left\{{\displaystyle {\int }_0^{t+}{\displaystyle {\int }_R{G}_{t-s}}}\left(x-z\right)\times \left\{f\left(s,u_n\left(s,z\right)\right)-f\left(s,u_{n-1}\left(s,z\right)\right)\right\}\text{d}z\text{d}s\right\}}^2\text{d}x}\\ \le B_{T}E{\displaystyle {\int }_R{\displaystyle {\int }_0^t{\displaystyle {\int }_R{G}_{t-s}}}}\left(x-z\right)\times {\left[u_n\left(s,z\right)-u_{n-1}\left(s,z\right)\right]}^2\text{d}z\text{d}s\text{d}x\\ =B_{T}E{\displaystyle {\int }_R{\displaystyle {\int }_0^t\left\{{\displaystyle {\int }_R{G}_{t-s}\left(x-z\right)\text{d}x}\right\}}}{\left[u_n\left(s,z\right)-u_{n-1}\left(s,z\right)\right]}^2\text{d}z\text{d}s\\ \le T{B}_T{\displaystyle {\int }_0^t{F}_{n-1}\left(s\right)\text{d}s}\end{array}$

通过递推得到：

$F_n\left(t\right)\le {\left[C_T{B}_T\right]}^{n-1}{\displaystyle {\int }_0^t\text{d}{t}_1}{\displaystyle {\int }_0^{t_1}\text{d}{t}_2}\cdots {\displaystyle {\int }_0^{t_{n-2}}{F}_1\left(t_{n-1}\right)\text{d}{t}_{n-1}}$

之后，通过公式(4)和假设(c2)，我们有：

$\begin{array}{c}{F}_1\left(t\right)=E\left\{{\displaystyle {\int }_R{\left[u_2\left(t,x\right)-u_1\left(t,x\right)\right]}^2\text{d}x}\right\}\\ \le C_T{\displaystyle {\int }_R{\displaystyle {\int }_0^t{\displaystyle {\int }_R{G}_{t-s}\left(x-z\right){\left[a\left(t\right)+k\left(t\right)\left|u_1\left(s,z\right)\right|\right]}^2\text{d}z\text{d}s\text{d}x}}}\\ \le T{C}_T{\displaystyle {\int }_0^t{\displaystyle {\int }_R\left\{a\left(t\right)+k\left(t\right)\left|u_1\left(s,z\right)\right|\right\}\text{d}z\text{d}s\text{d}x}}\end{array}$

上式的最后一项也是一个依赖于T的常数，记为 $C_1$，因此：

$0\le F_n\left(t\right)\le \frac{C_1{\left[C_T{B}_{T}T\right]}^{n-1}}{\left(n-1\right)!},n\in N$

所以序列 ${\displaystyle {\sum }_{n\in N}{F}_n\left(t\right)}\left(\le C_0{\text{e}}^{T\left(C_T{L}_T\right)}\right)$ 和 ${\displaystyle {\sum }_{n\in N}{F}_n\left(t\right)}\left(\le C_1{\text{e}}^{T\left(C_T{B}_T\right)}\right)$ 在 $\left[0,t\right]$ 上一致收敛。

因此对于序列 ${\left\{U_n\left(t,\cdot ,\omega \right):\left(t,\omega \right)\in \{0,T]\times \Omega \right\}}_{n\in N}$ 在 $\left[0,T\right]$ 上一致收敛。

定义容易证明，对 $\forall t\in \left[0,t\right],u\left(t,\cdot ,\omega \right)\in L^2\left(R\right)$，并且是 $\left\{F_t\right\}$ -适应过程。

下面继续证明u是满足方程(1)：

$I_1:$

$u\left(t,x,\omega \right)={\displaystyle {\int }_0^{t+}{\displaystyle {\int }_u{\displaystyle {\int }_R(G_{t-s}\left(x-z\right)g\left(s,z,u\left(s-,z,\omega \right);y\right)\text{d}zq\left(\text{d}s,\text{d}y,\omega \right)}}}$ (7)

$\begin{array}{l}E\left\{{\displaystyle {\int }_R\left[u_n\left(t,x,\omega \right)\right]}-{\displaystyle {\int }_0^{t+}{\displaystyle {\int }_u{\displaystyle {\int }_R{G}_{t-s}}}}\left(x-z\right)g{\left(s,z,u\left(s-,z,\omega \right);y\right)}^2\text{d}x\right\}\\ \le C_T{L}_T{\displaystyle {\int }_0^{t}E}\left\{{\displaystyle {\int }_R{\left[u_{n-1}\left(s,z,\cdot \right)-u\left(s,z,\cdot \right)\right]}^2\text{d}x}\right\}\text{d}s\end{array}$ (8)

${\left\{u_n\left(t,\cdot ,\omega \right):\left(t,\omega \right)\in \left[0,T\right]\times \Omega \right\}}_{n\in N}$

$t\in \left[0,T\right]\times L^2\times (\; R\; )$

$I_2:$

$\left(t,x\right)={\displaystyle {\int }_0^t{\displaystyle {\int }_R{G}_{t-s}\left(x-z\right)f\left(s,u\left(s,z\right)\right)\text{d}s\text{d}z}}$ (9)

$\begin{array}{l}E\left\{{\displaystyle {\int }_R\left[u_n\left(t,x\right)\right]}-{\displaystyle {\int }_0^{t+}{\displaystyle {\int }_R{G}_{t-s}}}\left(x-z\right)f{\left(s,u\left(s-,z\right)\right)}^2\text{d}x\right\}\\ \le C_T{B}_T{\displaystyle {\int }_0^{t}E}\left\{{\displaystyle {\int }_R{\left[u_{n-1}\left(s,z\right)-u\left(s,z\right)\right]}^2\text{d}x}\right\}\text{d}s\end{array}$ (10)

并且序列 ${\left\{u_n\left(t,x\right)\right\}}_{n\in N}$ 在 $t\in \left[0,T\right]\times L^2$ 一致收敛。

则 ${\sup }_{s\in \left[0,T\right]}E\left\{{\displaystyle {\int }_R{\left[u_{n-1}\left(s,z,\cdot \right)-u\left(s,z,\cdot \right)\right]}^2\text{d}x}\right\}$ 和 ${\sup }_{s\in \left[0,T\right]}E\left\{{\displaystyle {\int }_R{\left[u_{n-1}\left(s,z\right)-u\left(s,z\right)\right]}^2\text{d}x}\right\}$ 是有界的，由收敛控制定理可以得到上述公式(8)和公式(10)的最后一项收敛到0。进而公式(8)和(10)成立。从以上论述可以得到是可以关于t左极右连，且关于x连续。因此，u是方程(1)的解，进而解的存在性得证。

下面再进行解的唯一性证明。

2.4. 唯一性证明

对唯一性的证明采用反证法，假设方程(1)存在两个不同的解： $u_1$ 和 $u_2$。则 $u_1$ 和 $u_2$ 属于 $u_t$，因此 $u_1$ 和 $u_2$ 满足方程(7)和(9)。

$I_1:$

$H\left(t\right):=E\left\{{\displaystyle {\int }_R{\left[u_2\left(t,x,\cdot \right)-u_1\left(t,x,\cdot \right)\right]}^2\text{d}x}\right\},t\in \left[0,T\right]$

$\begin{array}{l}E\left\{{\displaystyle {\int }_R{\left[u_2\left(t,x,\cdot \right)-u_1\left(t,x,\cdot \right)\right]}^2\text{d}x}\right\}\\ =E{\displaystyle {\int }_R{\left\{{\displaystyle {\int }_0^t{\displaystyle {\int }_u{\displaystyle {\int }_R{G}_{t-s}}}}\left(x-z\right)\left[g\left(s,z,u_2\left(s-,z,\omega \right);y\right)-g\left(s,z,u_1\left(s-,z,\omega \right);y\right)\right]\text{d}zq\left(\text{d}s,\text{d}y,\cdot \right)\right\}}^2\text{d}x}\\ =E{\displaystyle {\int }_R{\displaystyle {\int }_0^t{\displaystyle {\int }_u{\left\{{\displaystyle {\int }_R{G}_{t-s}}\left(x-z\right)\times \left[g\left(s,z,u_2\left(s,z,\omega \right);y\right)-g\left(s,z,u_1\left(s,z,\omega \right);y\right)\right]\text{d}z\right\}}^{2}v\left(\text{d}y\right)\text{d}s\text{d}x}}}\\ \le L_T{\displaystyle {\int }_R{G}_{t-s}\left(x-z\right){\left[u_2\left(s,z,\omega \right)-u_1\left(s,z,\omega \right)\right]}^2\text{d}s\text{d}x}\\ \le T{L}_T{\displaystyle {\int }_0^t\text{d}s}{\displaystyle {\int }_R{\left[u_2\left(s,z,\omega \right)-u_1\left(s,z,\omega \right)\right]}^2\text{d}z\text{d}s}\\ =C_T{L}_T{\displaystyle {\int }_0^{t}H\left(s\right)\text{d}s}\end{array}$

$I_2:$

$H\left(t\right):=E\left\{{\displaystyle {\int }_R{\left[u_2\left(t,x\right)-u_1\left(t,x\right)\right]}^2\text{d}x}\right\},t\in \left[0,T\right]$

$\begin{array}{l}E\left\{{\displaystyle {\int }_R{\left[u_2\left(t,x\right)-u_1\left(t,x\right)\right]}^2\text{d}x}\right\}\\ =E{\displaystyle {\int }_R{\left\{{\displaystyle {\int }_0^t{\displaystyle {\int }_u{\displaystyle {\int }_R{G}_{t-s}}}}\left(x-z\right)\left[f\left(s,u_2\left(s-,z\right)\right)-f\left(s,u_1\left(s-,z\right)\right)\right]\text{d}z\text{d}s\right\}}^2\text{d}x}\\ =E{\displaystyle {\int }_R{\displaystyle {\int }_0^t{\displaystyle {\int }_u{\left\{{\displaystyle {\int }_R{G}_{t-s}}\left(x-z\right)\left[f\left(s,u_2\left(s-,z\right)\right)-f\left(s,u_1\left(s-,z\right)\right)\right]\text{d}z\text{d}s\right\}}^2\text{d}s\text{d}x}}}\\ \le B_T{\displaystyle {\int }_R{\displaystyle {\int }_0^t{G}_{t-s}}}\left(x-z\right){\left[\left(u_2\left(s-,z\right)\right)-\left(u_1\left(s-,z\right)\right)\right]}^2\text{d}z\text{d}s\\ \le T{B}_T{\displaystyle {\int }_0^t\text{d}s}{\displaystyle {\int }_R{\left[\left(u_2\left(s-,z\right)\right)-\left(u_1\left(s-,z\right)\right)\right]}^2\text{d}z}\\ =C_T{B}_T{\displaystyle {\int }_0^{t}H\left(s\right)\text{d}s}\end{array}$

$H\left(t\right)$ 关于变量t是左极右连的。因此 ${\sup }_{t\in \left[0,t\right]}H\left(t\right)<\infty$，通过解的存在性证明，我们得到 $H\left(t\right)\le C_T{L}_T{\displaystyle {\int }_0^{t}H\left(s\right)\text{d}s}$。由Gronwall不等式，我们可以得到 ${\sup }_{t\in \left[0,t\right]}H\left(t\right)=0$，即对于所有的 $t\in \left[0,T\right]$ 和所有的 $x\in R$ 得到 $u_2\left(t,x,\cdot \right)=u_1\left(t,x,\cdot \right)$ 和 $u_2\left(t,x\right)=u_1\left(t,x\right)$，即方程不存在两个不同的解，与原假设矛盾，唯一性得证。

基金项目

天津市教委科研计划项目(No. 2019KJ131)。

参考文献