# 具有Markov切换Poisson跳的随机微分方程的均方指数稳定性Mean Square Exponential Stability of Stochastic Differential Equations with Markovian Switching and Poisson Jumps

Abstract: This paper investigates the mean square exponential stability of stochastic differential equations with Markovian switching and Poisson jumps. By using Lyapunov stability theory, stochastic anal-ysis and inequality techniques, some sufficient conditions are derived to obtain the mean square exponential stability of the trivial solution. At last, an example is presented to illustrate the ob-tained results.

1. 引言

Brown运动是连续的随机过程，然而许多实际系统会遭受跳跃形式的随机突发扰动，如随机故障、地震、海啸等。在这些情形下，不能用Brown运动来刻画这些系统，因此将带跳过程引入到系统中来处理这些实际情形是合理的，见文献 [9] [10] [11]。近年来，关于带跳的随机微分方程的稳定性研究也获得了相应的成果，参阅文献 [12] [13] 及其中的参考文献。

2. 预备知识

$P\left\{r\left(t+\Delta \right)=j|r\left(t\right)=i\right\}=\left\{\begin{array}{ll}{\gamma }_{ij}\Delta +o\left(\Delta \right),\hfill & 当\text{\hspace{0.17em}}\text{\hspace{0.17em}}i\ne j,\hfill \\ 1+{\gamma }_{ii}\Delta +o\left(\Delta \right),\hfill & 当\text{\hspace{0.17em}}\text{\hspace{0.17em}}i=j,\hfill \end{array}$

$\begin{array}{l}\text{d}x\left(t\right)=f\left(x\left(t\right),t,r\left(t\right)\right)\text{d}t+g\left(x\left(t\right),t,r\left(t\right)\right)\text{d}B\left(t\right)\\ \text{}+h\left(x\left(t\right),t,r\left(t\right)\right)\text{d}N\left(t\right),\text{}t\ge 0,\end{array}$ (1)

$E{|x\left(t\right)|}^{2}\le K{\text{e}}^{-\gamma t}E{|{x}_{0}|}^{2},\text{}t\ge 0$

${C}^{2,1}\left({R}^{m}×{R}^{+};{R}^{+}\right)$ 表示关于变量 $x$ 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 $V\left(x,t\right)$ 的集合。给定任意的 $V\left(x,t\right)\in {C}^{2,1}\left({R}^{m}×{R}^{+};{R}^{+}\right)$，定义算子

$\begin{array}{c}LV\left(x,t,i\right)={V}_{t}\left(x,t\right)+{V}_{x}\left(x,t\right)f\left(x,t,i\right)\\ +\frac{1}{2}\text{trace}\left[{g}^{\text{T}}\left(x,t,i\right){V}_{xx}\left(x,t\right)g\left(x,t,i\right)\right]\\ +\lambda \left[V\left(x+h\left(x,t,i\right),t\right)-V\left(x,t\right)\right],\end{array}$

3. 主要结果

${x}^{\text{T}}\left(t\right)f\left(x\left(t\right),t,i\right)+\frac{1}{2}{|g\left(x\left(t\right),t,i\right)|}^{2}\le {\alpha }_{i1}{|x\left(t\right)|}^{2},$ (2)

${x}^{\text{T}}\left(t\right)h\left(x\left(t\right),t,i\right)\le {\alpha }_{i2}{|x\left(t\right)|}^{2},$ (3)

${|h\left(x\left(t\right),t,i\right)|}^{2}\le {\alpha }_{i3}{|x\left(t\right)|}^{2}.$ (4)

${\alpha }_{0}+2{\alpha }_{i1}+\lambda \left(2{a}_{i2}+{\alpha }_{i3}\right)<0$ (5)

$\begin{array}{c}LV\left(x\left(t\right),t,i\right)={\alpha }_{0}{\text{e}}^{{\alpha }_{0}t}{|x\left(t\right)|}^{2}+2{\text{e}}^{{\alpha }_{0}t}{x}^{\text{T}}\left(t\right)f\left(x\left(t\right),t,i\right)+{\text{e}}^{{\alpha }_{0}t}{|g\left(x\left(t\right),t,i\right)|}^{2}\\ +{\text{e}}^{{\alpha }_{0}t}\lambda \left(2{x}^{\text{T}}\left(t\right)h\left(x\left(t\right),t,i\right)+{|h\left(x\left(t\right),t,i\right)|}^{2}\right)\\ ={\text{e}}^{{\alpha }_{0}t}\left[{\alpha }_{0}{|x\left(t\right)|}^{2}+2{x}^{\text{T}}\left(t\right)f\left(x\left(t\right),t,i\right)+{|g\left(x\left(t\right),t,i\right)|}^{2}\\ +\lambda \left(2{x}^{\text{T}}\left(t\right)h\left(x\left(t\right),t,i\right)+{|h\left(x\left(t\right),t,i\right)|}^{2}\right)\right].\end{array}$

$LV\left(x\left(t\right),t,i\right)\le {\text{e}}^{{\alpha }_{0}t}\left[{\alpha }_{0}+2{\alpha }_{i1}+\lambda \left(2{a}_{i2}+{\alpha }_{i3}\right)\right]{|x\left(t\right)|}^{2}.$

$ELV\left(x\left(t\right),t,i\right)\le 0.$

$EV\left(x\left(t\right),t\right)-EV\left(x\left(0\right),0\right)\le {\int }_{0}^{t}ELV\left(x\left(s\right),s,r\left(s\right)\right)\text{d}s\le 0,$

$EV\left(x\left(t\right),t\right)\le EV\left(x\left(0\right),0\right).$

${\text{e}}^{{\alpha }_{0}t}E{|x\left(t\right)|}^{2}\le E{|{x}_{0}|}^{2},$

$E{|x\left(t\right)|}^{2}\le {\text{e}}^{-{\alpha }_{0}t}E{|{x}_{0}|}^{2}.$

4. 例子

$\text{d}x\left(t\right)=f\left(x\left(t\right),t,r\left(t\right)\right)\text{d}t+g\left(x\left(t\right),t,r\left(t\right)\right)\text{d}B\left(t\right)+h\left(x\left(t\right),t,r\left(t\right)\right)\text{d}N\left(t\right),$ (6)

$\text{Γ}=\left(\begin{array}{cc}-1& 1\\ 2& -2\end{array}\right).$

$i=1$ 时， $f\left(x,t,1\right)=-2x,\text{}g\left(x,t,1\right)=x,\text{}h\left(x,t,1\right)=x$

$i=2$ 时， $f\left(x,t,1\right)=-3x,\text{}g\left(x,t,1\right)=2x,\text{}h\left(x,t,1\right)=-\frac{1}{2}x$

$\forall i\in S$，可知 $f\left(0,t,i\right)=g\left(0,t,i\right)=h\left(0,t,i\right)=0$。经计算得，当 $i=1$ 时，

${x}^{\text{T}}\left(t\right)f\left(x\left(t\right),t,1\right)+\frac{1}{2}{|g\left(x\left(t\right),t,1\right)|}^{2}=-\frac{3}{2}{|x\left(t\right)|}^{2},$

${x}^{\text{T}}\left(t\right)h\left(x\left(t\right),t,1\right)={|x\left(t\right)|}^{2},$

${|h\left(x\left(t\right),t,1\right)|}^{2}={|x\left(t\right)|}^{2}.$

$i=2$ 时，

${x}^{\text{T}}\left(t\right)f\left(x\left(t\right),t,2\right)+\frac{1}{2}{|g\left(x\left(t\right),t,2\right)|}^{2}=-{|x\left(t\right)|}^{2},$

${x}^{\text{T}}\left(t\right)h\left(x\left(t\right),t,2\right)=-\frac{1}{2}{|x\left(t\right)|}^{2},$

${|h\left(x\left(t\right),t,1\right)|}^{2}=\frac{1}{4}{|x\left(t\right)|}^{2}.$

${\alpha }_{0}=1$，由此可知对 $\forall i\in S$，有 ${\alpha }_{0}+2{\alpha }_{i1}+\lambda \left(2{a}_{i2}+{\alpha }_{i3}\right)<0$ 成立。根据第三节中的定理可得方程(6)的平凡解是均方指数稳定的。

NOTES

*通讯作者。

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