带Poisson跳的随机时滞微分方程的矩有界性

Moment Boundedness of Stochastic Delay Differential Equations with Poisson Jumps

摘要: 研究一类带Poisson跳的随机时滞微分方程解的有界性。运用随机分析以及不等式技巧证明了该方程的解是p(p∈(0,1))-阶矩有界的。

Abstract: The boundedness of solutions for stochastic delay differential equations with Poisson jumps is in-vestigated in this paper. By using stochastic analysis and inequality techniques, it is proved that the p-th p(p∈(0,1)) moment of the solution of such equations is bounded.

1. 引言

随机微分方程广泛地应用于生物学、数学、经济、金融领域以及控制工程等诸多领域，见文献 [1] [2] [3] 及其中的参考文献。在许多实际工程系统中，时滞是这些实际系统经常发生振荡、不稳定性和性能差的原因。因此，很有必要研究带有时滞的随机微分方程，学者们在研究随机时滞微分方程方面做出了许多努力，参阅文献 [4] [5] [6] [7]。王等人在文献 [8] 和文献 [9] 中分别研究了带分布时滞的线性随机时滞微分方程的矩有界性和线性随机时滞微分方程的二阶矩有界性。文献 [10] 研究了脉冲随机时滞微分方程的有界性。

Brown运动是连续的随机过程，可以用来刻画连续的随机系统。然而许多实际的随机系统会遭遇跳跃形式的随机突发扰动。在这些情形下，不能用Brown运动来刻画这些随机系统，因此将带Poisson跳的过程引入到这些随机系统中来处理这些实际情形是合理的 [11] [12] [13]。本文研究了一类带Poisson跳的随机时滞微分方程的 $p\left(p\in \left(0,1\right)\right)$ -阶矩有界性。该文结构如下：第2节给出预备知识，第3节给出本文的主要结果。

2. 预备知识

记 $R^n$ 表示n-维的欧式空间。若 $x\in R^n$， $\left|x\right|$ 表示欧式范数。记 $R=\left(-\infty ,+\infty \right)$， $R^{+}=\left[0,+\infty \right)$。记一个完备的概率空间为 $\left(\Omega ,\mathcal{F},P\right)$，其滤子 ${\mathcal{F}}_t$ 满足通常的条件，即 ${\left\{{\mathcal{F}}_t\right\}}_{t\ge 0}$ 是右连续的且 ${\mathcal{F}}_0$ 包含所有的零测集。令 $\delta >0$， $C\left(\left[-\delta ,0\right];R^n\right)$ 表示所有定义在 $[-\delta ,0]$ 上的 $R^n$ -值连续函数 $\phi$。 $C_{F_0}^b\left(\Omega ;R^n\right)$ 表示所有 ${\mathcal{F}}_0$ -可测的且定义在 $C\left(\left[-\delta ,0\right];R^n\right)$ 上的有界函数全体，且其上的范数为 $\Vert \phi \Vert =\underset{-\delta \le \theta \le 0}{\sup }\left|\phi \left(\theta \right)\right|$。对 $\forall x,y\in R^n$， $\langle x,y\rangle$ 和 $x^{\text{T}}y$ 均表示内积。

${\left\{r\left(t\right)\right\}}_{t\ge 0}$ 表示的是定义在完备概率空间上的一个取值为 $S=\left\{1,2,\cdots ,N\right\}$ 的右连续时齐Markov链，其生成元(密度矩阵) $\Gamma ={\left({\gamma }_{ij}\right)}_{N\times N}$ 由转移概率矩阵确定，即

$P\left\{r\left(t+\Delta \right)=j|r\left(t\right)=i\right\}=\{\begin{array}{ll}{\gamma }_{ij}\Delta +o\left(\Delta \right),\hfill & 当i\ne j,\hfill \\ 1+{\gamma }_{ii}\Delta +o\left(\Delta \right),\hfill & 当i=j,\hfill \end{array}$

这里， $\Delta >0$， $\underset{\Delta \to 0}{\lim }o\left(\Delta \right)/\Delta =0$，对于 $i\ne j$， ${\gamma }_{ij}\ge 0$ 表示从状态i到状态j的转移速率且 ${\gamma }_{ii}=-{\displaystyle \underset{j\ne i}{\sum }{\gamma }_{ij}}$。作为一个事实， $r(\cdot )$ 的每个样本轨道是一个右连续的阶梯函数，且在 $R^{+}=\left[0,+\infty \right)$ 上的任何一个有限区间里至多存在有限个跳跃点(参阅 [14] )。

考虑如下形式的带Poisson跳的随机时滞微分方程：

$\begin{array}{l}\text{d}x\left(t\right)=\left[f\left(x\left(t\right),t,r\left(t\right)\right)+u\left(x\left(t-\delta \right),t,r\left(t\right)\right)\right]\text{d}t+g\left(x\left(t\right),t,r\left(t\right)\right)\text{d}\omega \left(t\right)\\ +h\left(x\left(t\right),t,r\left(t\right)\right)\text{d}N\left(t\right),t\ge t_0,\end{array}$ (1)

初始值 $x_{t_0}=\phi \left(\theta \right)=\left\{x\left(t_0+\theta \right):-\delta \le \theta \le 0\right\}\in C_{F_0}^b\left(\Omega ;R^n\right)$， $r\left(0\right)=i_0\in S$，其中 $f,g,h,u:R^n\times R^{+}\times S\to R^n$， $\omega \left(t\right)$ 和 $N\left(t\right)$ 分别是定义在该概率空间上的1-维Brown运动和强度为 $\lambda$ 的Poisson过程，这里假设Markov链 $r\left(t\right)$ 与Brown运动 $B\left(t\right)$ 和Poisson过程 $N\left(t\right)$ 是相互独立的。 $\tilde{N}\left(t\right)=N\left(t\right)-\lambda t$ 表示 $N\left(t\right)$ 的补偿Poisson过程。本文假设对于任意的 $x_1,x_2\in R^n,t\in R^{+}$ 和 $i\in S$，函数 $f,g,h,u$ 满足如下条件：

$\begin{array}{l}\left|f\left(x_1,t,i\right)-f\left(x_2,t,i\right)\right|\le L_1\left|x_1-x_2\right|,\\ \left|g\left(x_1,t,i\right)-g\left(x_2,t,i\right)\right|\le L_2\left|x_1-x_2\right|,\\ \left|h\left(x_1,t,i\right)-h\left(x_2,t,i\right)\right|\le L_3\left|x_1-x_2\right|,\\ \left|u\left(x_1,t,i\right)-u\left(x_2,t,i\right)\right|\le K\left|x_1-x_2\right|.\end{array}$ (2)

且对 $\forall t\ge 0$ 和 $i\in S$，假设 $f\left(0,t,i\right)=g\left(0,t,i\right)=h\left(0,t,i\right)=0$。因此，当方程(1)的初始值 $x_0=0$ 时，该方程存在平凡解 $x\left(t\right)\equiv 0$。进一步，经计算对 $\forall \left(x,i,t\right)\in R^n\times S\times R^{+}$ 有

$\left|f\left(x,t,i\right)\right|\le L_1\left|x\right|,\left|g\left(x,t,i\right)\right|\le L_2\left|x\right|,\left|h\left(x,t,i\right)\right|\le L_3\left|x\right|,\left|u\left(x,t,i\right)\right|\le K\left|x\right|.$

注在假设条件(2)下，方程(1)存在唯一的全局解且 $E{\left|x\left(t\right)\right|}^2<\infty$ ( $\forall t\ge 0$ )。

令 $C^{2,1}\left(R^n\times R^{+}\times S;R^{+}\right)$ 表示的是对 $\forall i\in S$ 关于变量x二阶连续可导且关于变量t一阶连续可导的全体非负函数 $V\left(x,t,i\right)$ 的集合。给定任意的 $V\left(x,t,i\right)\in C^{2,1}\left(R^n\times R^{+}\times S;R^{+}\right)$，定义算子

$\begin{array}{c}LV\left(x,t,i\right)=V_t\left(x,t,i\right)+V_x\left(x,t,i\right)\left[f\left(x,t,i\right)+u\left(x,t,i\right)\right]\\ +\frac{1}{2}trace\left[g^{\text{T}}\left(x,t,i\right)V_{xx}\left(x,t,i\right)g\left(x,t,i\right)\right]\\ +\lambda \left[V\left(x+h\left(x,t,i\right),t,i\right)-V\left(x,t,i\right)\right]\\ +{\displaystyle {\sum }_{j=1}^N{\gamma }_{ij}V}\left(x,t,j\right),\end{array}$

其中， $V_t\left(x,t,i\right)=\frac{\partial v\left(x,t,i\right)}{\partial t}$， $V_{xx}\left(x,t,i\right)={\left(\frac{{\partial }^{2}v\left(x,t,i\right)}{\partial x_i\partial x_j}\right)}_{n\times n}$，

$V_x\left(x,t,i\right)=\left(\frac{\partial v\left(x,t,i\right)}{\partial x_1},\frac{\partial v\left(x,t,i\right)}{\partial x_2},\cdots ,\frac{\partial v\left(x,t,i\right)}{\partial x_n}\right).$

3. 主要结果

令 $p\in \left(0,1\right)$。本节将证明方程(1)的解是p-阶矩有界的。

定理 若条件(1)成立，则对所有的 $t_0\ge 0$ 和 $T\ge 0$，

$\underset{t_0\le t\le t_0+T+\delta }{\sup }E{\left|x\left(t\right)\right|}^p\le M_1^{\frac{p}{2}}E{\Vert \phi \Vert }^2,$ (3)

$E\left(\underset{t_0\le t\le t_0+T+\delta }{\sup }{\left|x\left(t\right)\right|}^p\right)\le M_2^{\frac{p}{2}}E{\Vert \phi \Vert }^2,$ (4)

其中， $M_1=M_1\left(p,\delta ,T\right)=\left(1+k\delta \right){\text{e}}^{\left[2\left(L_1+\lambda L_3\right)+L_2^2+\lambda L_3^2+2K\right]\left(T+\delta \right)}$， $M_2=M_2\left(p,\delta ,T\right)=5\left[1+L_1^2\left(T+\delta \right)+K^2\left(T+\delta \right)+4L_2^2+2{\lambda }^2{L}_3^2\left(T+\delta \right)+8\lambda L_3^2\right]\left(T+\delta \right)M_1$。

证明(1) 对 ${\left|x\left(t\right)\right|}^2$ 应用Itô公式，由方程(1)可得

$\begin{array}{c}\text{d}{\left|x\left(t\right)\right|}^2=2\langle x\left(t\right),f\left(x\left(t\right),t,r\left(t\right)\right)+u\left(x\left(t-\delta \right),t,r\left(t\right)\right)\rangle \text{d}t\\ +2\langle x\left(t\right),g\left(x\left(t\right),t,r\left(t\right)\right)\rangle \text{d}\omega \left(t\right)+{\left|g\left(x\left(t\right),t,r\left(t\right)\right)\right|}^2\text{d}t\\ +\left[{\left|x\left(t\right)+h\left(x\left(t\right),t,r\left(t\right)\right)\right|}^2-x^2\right]\text{d}N(\; t\; )\end{array}$

$\begin{array}{l}=2\langle x\left(t\right),f\left(x\left(t\right),t,r\left(t\right)\right)+u\left(x\left(t-\delta \right),t,r\left(t\right)\right)\rangle \text{d}t\\ +2\langle x\left(t\right),g\left(x\left(t\right),t,r\left(t\right)\right)\rangle \text{d}\omega \left(t\right)+{\left|g\left(x\left(t\right),t,r\left(t\right)\right)\right|}^2\\ +\lambda \left[h^2\left(x\left(t\right),t,r\left(t\right)\right)+2x^{\text{T}}\left(t\right)h\left(x\left(t\right),t,r\left(t\right)\right)\right]\text{d}t\\ +\left[h^2\left(x\left(t\right),t,r\left(t\right)\right)+2x^{\text{T}}\left(t\right)h\left(x\left(t\right),t,r\left(t\right)\right)\right]\text{d}\tilde{N}\left(t\right)\end{array}$ (5)

对(3)式两边同时取期望，计算得

$\begin{array}{c}E{\left|x\left(t\right)\right|}^2=E{\left|x\left(t_0\right)\right|}^2+2E{\displaystyle {\int }_{t_0}^t{x}^{\text{T}}}\left(s\right)f\left(x\left(s\right),s,r\left(s\right)\right)\text{d}s\\ +2E{\displaystyle {\int }_{t_0}^t{x}^{\text{T}}}\left(s\right)u\left(x\left(s-\delta \right),s,r\left(s\right)\right)\text{d}s\\ +2E{\displaystyle {\int }_{t_0}^t{\left|g\left(x\left(s\right),s,r\left(s\right)\right)\right|}^2\text{d}s}\\ +\lambda E{\displaystyle {\int }_{t_0}^t\left[h^2\left(x\left(s\right),s,r\left(s\right)\right)+2x^{\text{T}}\left(s\right)h\left(x\left(s\right),s,r\left(s\right)\right)\right]\text{d}s}.\end{array}$ (6)

利用Hölder不等式可得

$\begin{array}{l}2KE{\displaystyle {\int }_{t_0}^t\left|x\left(s\right)\right|\left|x\left(s-\delta \right)\right|\text{d}s}\\ \le KE{\displaystyle {\int }_{t_0}^t{\left|x\left(s\right)\right|}^2\text{d}s}+KE{\displaystyle {\int }_{t_0}^t{\left|x\left(s-\delta \right)\right|}^2\text{d}s}\\ =KE{\displaystyle {\int }_{t_0}^t{\left|x\left(s\right)\right|}^2\text{d}s}+KE{\displaystyle {\int }_{t_0-\delta }^{t-\delta }{\left|x\left(\nu \right)\right|}^2\text{d}\nu }\\ =KE{\displaystyle {\int }_{t_0}^t{\left|x\left(s\right)\right|}^2\text{d}s}+KE{\displaystyle {\int }_{t_0-\delta }^{t_0}{\left|x\left(\nu \right)\right|}^2\text{d}\nu }+KE{\displaystyle {\int }_{t_0}^{t-\delta }{\left|x\left(\nu \right)\right|}^2\text{d}\nu }\\ \le 2KE{\displaystyle {\int }_{t_0}^t{\left|x\left(s\right)\right|}^2\text{d}s}+K\delta E{\Vert \phi \Vert }^2.\end{array}$ (7)

将(7)代入到(6)中可得

$\begin{array}{c}E{\left|x\left(t\right)\right|}^2\le E{\left|x\left(t_0\right)\right|}^2+\left[2\left(L_1+\lambda L_3\right)+L_2^2+\lambda L_3^2+2K\right]{\displaystyle {\int }_{t_0}^{t}E}{\left|x\left(s\right)\right|}^2\text{d}s+K\delta E{\Vert \phi \Vert }^2\\ \le \left(1+K\delta \right)E{\Vert \phi \Vert }^2+\left[2\left(L_1+\lambda L_3\right)+L_2^2+\lambda L_3^2+2K\right]{\displaystyle {\int }_{t_0}^t\left(\underset{t_0\le \nu \le s}{\sup }E{\left|x\left(\nu \right)\right|}^2\right)\text{d}s}.\end{array}$ (8)

从而

$\underset{t_0\le \nu \le t_0+T+\delta }{\sup }E{\left|x\left(\nu \right)\right|}^2\le \left(1+K\delta \right)E{\Vert \phi \Vert }^2+\left[2\left(L_1+\lambda L_3\right)+L_2^2+\lambda L_3^2+2K\right]{\displaystyle {\int }_{t_0}^{t_0+T+\delta }\left(\underset{t_0\le \nu \le s}{\sup }E{\left|x\left(\nu \right)\right|}^2\right)\text{d}s}.$ (9)

进一步利用Gronwall不等式得

$\underset{t_0\le \nu \le t_0+T+\delta }{\sup }E{\left|x\left(\nu \right)\right|}^2\le \left(1+K\delta \right){\text{e}}^{\left[2\left(L_1+\lambda L_3\right)+L_2^2+\lambda L_3^2+2K\right]\left(T+\delta \right)}E{\Vert \phi \Vert }^2=M_{1}E{\Vert \phi \Vert }^2.$

最后运用Hölder不等式，进一步得

$\underset{t_0\le t\le t_0+T+\delta }{\sup }E{\left|x\left(t\right)\right|}^p\le M_1^{\frac{p}{2}}E{\Vert \phi \Vert }^p.$

(2)由方程(1)可得

$\begin{array}{c}x\left(t\right)=x\left(t_0\right)+{\displaystyle {\int }_{t_0}^t\left[f\left(x\left(s-\delta \right),s,r\left(s\right)\right)+u\left(x\left(s-\delta \right),s,r\left(s\right)\right)\right]\text{d}s}\\ +{\displaystyle {\int }_{t_0}^{t}g\left(x\left(s\right),s,r\left(s\right)\right)\text{d}\omega \left(s\right)}+{\displaystyle {\int }_{t_0}^{t}h\left(x\left(s\right),s,r\left(s\right)\right)\text{d}N}\left(s\right).\end{array}$ (10)

利用Hölder不等式和条件(2)可得

$\begin{array}{c}{\left|x\left(t\right)\right|}^2\le 5{\left|x\left(t_0\right)\right|}^2+5{\left|{\displaystyle {\int }_{t_0}^{t}f}\left(x\left(s\right),s,r\left(s\right)\right)\text{d}s\right|}^2+5{\left|{\displaystyle {\int }_{t_0}^{t}u}\left(x\left(s-\delta \right),s,r\left(s\right)\right)\text{d}s\right|}^2\\ +5{\left|{\displaystyle {\int }_{t_0}^{t}g}\left(x\left(s\right),s,r\left(s\right)\right)\text{d}\omega \left(s\right)\right|}^2+5{\left|{\displaystyle {\int }_{t_0}^{t}h}\left(x\left(s\right),s,r\left(s\right)\right)\text{d}N\left(s\right)\right|}^2\\ \le 5{\left|x\left(t_0\right)\right|}^2+5\left(t-t_0\right){\displaystyle {\int }_{t_0}^t{f}^2}\left(x\left(s\right),s,r\left(s\right)\right)\text{d}s+5\left(t-t_0\right){\displaystyle {\int }_{t_0}^t{u}^2}\left(x\left(s-\delta \right),s,r\left(s\right)\right)\text{d}s\\ +5{\left|{\displaystyle {\int }_{t_0}^{t}g}\left(x\left(s\right),s,r\left(s\right)\right)\text{d}\omega \left(s\right)\right|}^2+10{\lambda }^2\left(t-t_0\right){\displaystyle {\int }_{t_0}^t{h}^2}\left(x\left(s\right),s,r\left(s\right)\right)\text{d}s\\ +10{\left|{\displaystyle {\int }_{t_0}^{t}h}\left(x\left(s\right),s,r\left(s\right)\right)\text{d}\tilde{N}\left(s\right)\right|}^2.\end{array}$

利用Doob鞅不等式可得

$\begin{array}{l}E\left(\underset{t_0\le t\le t_0+T+\delta }{\sup }{\left|x\left(t\right)\right|}^2\right)\\ \le 5E{\left|x\left(t_0\right)\right|}^2+5\left(T+\delta \right)E{\displaystyle {\int }_{t_0}^{t_0+T+\delta }{f}^2}\left(x\left(s\right),s,r\left(s\right)\right)\text{d}s\\ +5\left(T+\delta \right)E{\displaystyle {\int }_{t_0}^{t_0+T+\delta }{u}^2}\left(x\left(s-\delta \right),s,r\left(s\right)\right)\text{d}s+20E{\displaystyle {\int }_{t_0}^{t_0+T+\delta }{g}^2}\left(x\left(s\right),s,r\left(s\right)\right)\text{d}s\\ +10{\lambda }^2\left(T+\delta \right)E{\displaystyle {\int }_{t_0}^{t_0+T+\delta }{h}^2}\left(x\left(s\right),s,r\left(s\right)\right)\text{d}s+40\lambda E{\displaystyle {\int }_{t_0}^{t_0+T+\delta }{h}^2}\left(x\left(s\right),s,r\left(s\right)\right)\text{d}s\\ \le 5E{\Vert \phi \Vert }^2+5L_1^2\left(T+\delta \right){\displaystyle {\int }_{t_0}^{t_0+T+\delta }E}{\left|x\left(s\right)\right|}^2\text{d}s+5K^2\left(T+\delta \right){\displaystyle {\int }_{t_0}^{t_0+T+\delta }E}{\left|x\left(s-\delta \right)\right|}^2\text{d}s\\ +20L_2^2{\displaystyle {\int }_{t_0}^{t_0+T+\delta }E}{\left|x\left(s\right)\right|}^2\text{d}s+40\lambda L_3^{2}E{\displaystyle {\int }_{t_0}^{t_0+T+\delta }E}{\left|x\left(s\right)\right|}^2\text{d}s.\end{array}$

利用(3)，进一步可得

$\begin{array}{l}E\left(\underset{t_0\le t\le t_0+T+\delta }{\sup }{\left|x\left(t\right)\right|}^2\right)\\ \le 5\left[1+L_1^2\left(T+\delta \right)+K^2\left(T+\delta \right)+4L_2^2+2{\lambda }^2{L}_3^2\left(T+\delta \right)+8\lambda L_3^2\right]\times \left(T+\delta \right)M_{1}E{\Vert \phi \Vert }^2\\ =\text{:}{M}_{2}E{\Vert \phi \Vert }^2.\end{array}$

再次运用 不等式可得

$E\left(\underset{t_0\le t\le t_0+T+\delta }{\sup }{\left|x\left(t\right)\right|}^2\right)\le M_3^{\frac{p}{2}}E{\Vert \phi \Vert }^2.$

证毕。

基金项目

国家自然科学基金项目(No. 11901398)。

NOTES

^*通讯作者。