带跳随机比例微分方程补偿分步θ方法的收敛性和稳定性

Convergence and Stability of Compenseted Split-Step Theta Methods for Stochastic Pantograph Differential Equations with Jumps

摘要: 我们探讨了带跳的随机比例微分方程的补偿分步θ方法。首先，证明了该数值方法收敛，并且收敛阶为12，其次证明了解析解的均方稳定性，并且证明了补偿分步θ方法能保持解析解的均方稳定性，最后给出数值算例验证理论结果的正确性。

Abstract: A semi-implicit compensated split-step θ method for stochastic pantograph differential equations with Poisson jumps is investigated in the paper. Firstly, the convergence of the numerical method is discussed, and it is proved that the convergence order is 1/2. Secondly, the mean-square stability of the analytical solution is proved, and it is found that the compensated split-step θ methods can maintain the mean square stability of the analytical solution under some conditions. Finally, two numerical examples are given to verify the correctness of the theoretical results.

1. 前言

随机比例微分方程(stochastic pantograph differential equations，简称SPDEs)是一类特殊的随机无界延迟微分方程，1971年，Ockendon和Tayler [1] 发现了电力机车的受电弓是如何收集电流的，因此提出了随机比例微分方程。随机比例微分方程的一般形式如下

$\text{d}x\left(t\right)=f\left(x\left(t\right),x\left(qt\right)\right)\text{d}t+g\left(x\left(t\right),x\left(qt\right)\right)\text{d}W\left(t\right)$ (1-1)

在许多领域，比如金融、控制和工程中都有广泛的应用。然而，一般来说，SPDEs没有显式解，因此对SPDEs数值解的研究就非常重要。在实践中，还广泛讨论了带泊松跳跃的随机微分方程和数值方法，带泊松跳的随机系统如今广泛应用于随机金融学、电子工程、生物学等领域，能够描述状态的意外、突然变化。将带跳的Poisson扰动项纳入随机比例延迟系统来建模突变，泊松跳跃在随机金融和模型种群动力学领域受到越来越多的关注。带跳的SPDE可视为随机比例微分方程的扩展。

2007年，Mao [2] 开始了对带跳的随机比例微分方程的研究，给出此方程的解存在且唯一的证明，在此基础上给出了Euler方法的数值解，证明了数值解在 $L^2$ 意义下收敛于精确解。2011年，Mao等人 [3] 研究了局部Lipschitz条件下带跳的随机比例微分方程半隐式Euler方法的数值解均方收敛于方程的精确解。2013年，Mao等人 [4] 研究了一类具有Markovian切换和Levy跳跃的随机比例微分方程，证明了在局部Lipschitz条件下Euler方法得到的数值解均方收敛且依概率收敛于方程的精确解。同年，Hu等人 [5] 建立了关于带泊松跳随机比例微分方程的平衡隐式方法，研究了该方法的均方收敛性和均方稳定性。证明了该方法强收敛阶为1/2，同时还证明了，对于线性标量方程，当步长充分小时，强平衡隐式方法和弱平衡隐式方法都是均方稳定的。2014年，Yang等人 [6] 研究了具有随机幅度的泊松驱动跳跃的随机比例微分方程，给出了数值方法收敛的强阶，并得到了数值方法的收敛性。但是，对于非线性的带泊松跳随机比例微分方程解析解的均方稳定性研究仍是一个空白，因此这成了本文研究的部分内容。

论文组织如下：在第2节中，我们将介绍带跳的SPDE，并定义补偿分步θ方法(2-7)，证明了补偿分步θ方法的均方收敛性；在第3节中，找到了方程解析解均方稳定的条件，证明了补偿分步θ方法的均方稳定性。

2. 补偿分步θ方法的收敛性

考虑n维带跳的随机比例微分方程

$\{\begin{array}{l}\text{d}x\left(t\right)=f\left(x\left(t\right),x\left(qt\right)\right)\text{d}t+g\left(x\left(t\right),x\left(qt\right)\right)\text{d}W\left(t\right)+h\left(x\left(t\right),x\left(qt\right)\right)\text{d}N\left(t\right),0\le t\le T,\\ x\left(0\right)=x_0\end{array}$ (2-1)

其中， $q\in \left(0,1\right)$ 是比例系数， $0<T<\infty$ ， $f,g,h:R^n\times R^n\to R^n$ 都是Borel可测函数。 $x\left(t\right)$ 是一个n维状态过程， $W\left(t\right)={\left(W_1\left(t\right),W_2\left(t\right),\cdots ,W_m\left(t\right)\right)}^{\text{T}}$ 是各分量相互独立的m维标准维纳过程， $N\left(t\right)$ 是参变量为 $\lambda \left(\lambda >0\right)$ 的标量泊松过程，初始条件 $x_0$ 是一个有界的 ${\mathcal{F}}_0$ 可测的随机向量，也即 $E{\left|x_0\right|}^2<\infty$ 成立。

方程(2-1)等价于

$x\left(t\right)=x\left(0\right)+{\displaystyle {\int }_0^{t}f\left(x\left(s\right),x\left(qs\right)\right)\text{d}s}+{\displaystyle {\int }_0^{t}g\left(x\left(s\right),x\left(qs\right)\right)\text{d}W\left(s\right)}+{\displaystyle {\int }_0^{t}h}\left(x\left(s\right),x\left(qs\right)\right)\text{d}N\left(s\right),t>0,$ (2-2)

为保证方程(2-1)解的存在唯一性，假设系数f，g，h满足如下条件：

(1) (全局Lipschitz条件)存在正常数K，使得对所有的 $x_1,x_2,y_1,y_2\in R^n$ ，有

${\left|u\left(x_1,y_1\right)-u\left(x_2,y_2\right)\right|}^2\le K\left({\left|x_1-x_2\right|}^2+{\left|y_1-y_2\right|}^2\right),u=f,g,h.$

(2) (线性增长条件)存在正常数L，使得对所有的 $x,y\in R^n$ ，有

${\left|u\left(x,y\right)\right|}^2\le L\left(1+{\left|x\right|}^2+{\left|y\right|}^2\right),u=f,g,h.$

由条件(1)可得，若 $u\left(0,0\right)=0$ ，则对条件(1)的正常数K，满足对任意 $x,y\in R^n$ ，有

${\left|u\left(x,y\right)\right|}^2\le K\left({\left|x\right|}^2+{\left|y\right|}^2\right),u=f,g,h.$ (2-3)

为了方便，本文假设 $u\left(0,0\right)=0$ ， $u=f,g,h$ ，易知方程(2-1)有解 $x\left(t\right)\equiv 0$ 。

利用补偿泊松过程 $\tilde{N}\left(t\right):=N\left(t\right)-\lambda t$ 是鞅，并且满足

$E\left[\tilde{N}\left(t+s\right)-\tilde{N}\left(t\right)\right]=0,E{\left[\tilde{N}\left(t+s\right)-\tilde{N}\left(t\right)\right]}^2=\lambda s,t,s\ge 0.$

于是方程(2-1)可写成如下等价的方程

$\{\begin{array}{l}\text{d}x\left(t\right)=f_{\lambda }\left(x\left(t\right),x\left(qt\right)\right)\text{d}t+g\left(x\left(t\right),x\left(qt\right)\right)\text{d}W\left(t\right)+h\left(x\left(t\right),x\left(qt\right)\right)\text{d}\tilde{N}\left(t\right),t\in \left[0,T\right]\\ x\left(0\right)=x_0,\end{array}$ (2-4)

其中 $f_{\lambda }\left(x,y\right)$ 定义如下：

$f_{\lambda }\left(x,y\right):=f\left(x,y\right)+\lambda h\left(x,y\right).$

方程(2-4)等价于如下积分形式：

$x\left(t\right)=x\left(0\right)+{\displaystyle {\int }_0^t{f}_{\lambda }}\left(x\left(s\right),x\left(qs\right)\right)\text{d}s+{\displaystyle {\int }_0^{t}g}\left(x\left(s\right),x\left(qs\right)\right)\text{d}W\left(s\right)+{\displaystyle {\int }_0^{t}h}\left(x\left(s\right),x\left(qs\right)\right)\text{d}\tilde{N}\left(s\right),t>0,$ (2-5)

易知 $f_{\lambda }\left(x,y\right)$ 仍满足全局Lipschitz条件和线性增长条件，只需将相应的Lipschitz常数和线性增长系数改为更大的常数：

$K_{\lambda }=2\left({\lambda }^2+1\right)K,L_{\lambda }=2\left({\lambda }^2+1\right)L.$ (2-6)

定理2.1若条件(1)和(2)成立，则方程(2-4)存在唯一解 $x\left(t\right)$ ，且

$E\left(\underset{0\le t\le T}{\sup }{\left|x\left(t\right)\right|}^2\right)<\infty .$

证明 解的存在唯一性参考文献 [2] ，下证 $E\left(\underset{0\le t\le T}{\sup }{\left|x\left(t\right)\right|}^2\right)<\infty$ 。

由基本不等式 ${\left|a+b+c+d\right|}^2\le 4{\left|a\right|}^2+4{\left|b\right|}^2+4{\left|c\right|}^2+4{\left|d\right|}^2$ 及方程(2-5)可得

${\left|x\left(t\right)\right|}^2\le 4{\left|x_0\right|}^2+4{\left|{\displaystyle {\int }_0^t{f}_{\lambda }}\left(x\left(s\right),x\left(qs\right)\right)\text{d}s\right|}^2+4{\left|{\displaystyle {\int }_0^{t}g}\left(x\left(s\right),x\left(qs\right)\right)\text{d}W\left(s\right)\right|}^2+4{\left|{\displaystyle {\int }_0^{t}h}\left(x\left(s\right),x\left(qs\right)\right)\text{d}\tilde{N}\left(s\right)\right|}^2$

对上式两边取上界，再取期望并由Doob鞅不等式、鞅等距映射性质及条件(2)可得

$\begin{array}{l}E\left(\underset{0\le s\le t}{\sup }{\left|x\left(s\right)\right|}^2\right)\\ \le 4E{\left|x_0\right|}^2+4E\left(\underset{0\le s\le t}{\sup }{\left|{\displaystyle {\int }_0^s{f}_{\lambda }}\left(x\left(v\right),x\left(qv\right)\right)\text{d}v\right|}^2\right)+4E\left(\underset{0\le s\le t}{\sup }{\left|{\displaystyle {\int }_0^{s}g}\left(x\left(v\right),x\left(qv\right)\right)\text{d}W\left(v\right)\right|}^2\right)\\ +4E\left(\underset{0\le s\le t}{\sup }{\left|{\displaystyle {\int }_0^{s}h}\left(x\left(v\right),x\left(qv\right)\right)\text{d}\tilde{N}\left(v\right)\right|}^2\right)\\ \le 4E{\left|x_0\right|}^2+4TE{\left|{\displaystyle {\int }_0^t{f}_{\lambda }}\left(x\left(s\right),x\left(qs\right)\right)\right|}^2\text{d}s+16E{\left|{\displaystyle {\int }_0^{t}g}\left(x\left(s\right),x\left(qs\right)\right)\text{d}W\left(s\right)\right|}^2\\ +16E{\left|{\displaystyle {\int }_0^{t}h}\left(x\left(s\right),x\left(qs\right)\right)\text{d}\tilde{N}\left(s\right)\right|}^2\end{array}$

$\begin{array}{l}\le 4E{\left|x_0\right|}^2+4TE{\left|{\displaystyle {\int }_0^t{f}_{\lambda }}\left(x\left(s\right),x\left(qs\right)\right)\right|}^2\text{d}s+16E{\left|{\displaystyle {\int }_0^{t}g}\left(x\left(s\right),x\left(qs\right)\right)\right|}^2\text{d}s\\ +16\lambda E{\left|{\displaystyle {\int }_0^{t}h}\left(x\left(s\right),x\left(qs\right)\right)\right|}^2\text{d}s\\ \le 4E{\left|x_0\right|}^2+4L_{\lambda }\left(T+4+4\lambda \right){\displaystyle {\int }_0^t\left(1+E{\left|x\left(s\right)\right|}^2+E{\left|x\left(qs\right)\right|}^2\right)\text{d}s}\\ \le 4E{\left|x_0\right|}^2+4L_{\lambda }T\left(T+4+4\lambda \right)+8L_{\lambda }\left(T+4+4\lambda \right){\displaystyle {\int }_0^{t}E}\left(\underset{0\le s\le t}{\sup }{\left|x\left(s\right)\right|}^2\right)\text{d}s.\end{array}$

由Gronwall不等式可得

$E\left(\underset{0\le s\le t}{\sup }{\left|x\left(s\right)\right|}^2\right)\le \left(4E{\left|x_0\right|}^2+4L_{\lambda }T\left(T+4+4\lambda \right)\right)\exp \left(8L_{\lambda }\left(T+4+4\lambda \right)T\right).$

故

$E\left(\underset{0\le s\le t}{\sup }{\left|x\left(s\right)\right|}^2\right)\le \left(4E{\left|x_0\right|}^2+4L_{\lambda }T\left(T+4+4\lambda \right)\right)\exp \left(8L_{\lambda }\left(T+4+4\lambda \right)T\right)<\infty .$ □

分步θ方法 [7] (Split-step θ method，简称SS θ method)应用于方程(2-4)得到如下的补偿分步θ方法(Compensated split-step θ method，简称CSS θ method)

$\begin{array}{l}{y}_n^{\ast }=y_n+\theta \Delta t{f}_{\lambda }\left(y_n^{\ast },y_{\left[qn\right]}\right),\\ y_{n+1}=y_n+\Delta t{f}_{\lambda }\left(y_n^{\ast },y_{\left[qn\right]}\right)+g\left(y_n^{\ast },y_{\left[qn\right]}\right)\Delta W_n+h\left(y_n^{\ast },y_{\left[qn\right]}\right)\Delta {\tilde{N}}_n.\end{array}$ (2-7)

其中初值 $y_0=x\left(0\right)$ ，步长 $\Delta t=\frac{T}{M}$ ，M为正整数，参数 $\theta \in \left[0,1\right]$ ， $t_n=n\Delta t$ ， $y_n$ 是 $x\left(t_n\right)$ 的近似数值解，即 $y_n\approx x\left(t_n\right)$ ， $y_{\left[qn\right]}\approx x\left(q{t}_n\right)$ 。 $\left[a\right]$ 是指a的整数部分， $\Delta W_n=W\left(t_{n+1}\right)-W\left(t_n\right)$ 是维纳增量， $\Delta {\tilde{N}}_n=\tilde{N}\left(t_{n+1}\right)-\tilde{N}\left(t_n\right)$ 是补偿泊松增量， $n=0,1,\cdots ,M-1$ 。

注意到对每个 $n\left(=0,1,\cdots ,M-1\right)$ ， $y_n^{\ast },\Delta W_n$ 和 $\Delta {\tilde{N}}_n$ 相互独立，且 $y_n^{\ast }$ 是 ${\mathcal{F}}_{t_n}$ -可测的。当 $\theta =1$ ，补偿分步θ方法就成了补偿分步向后欧拉法(Compensated split-step back ward Euler Method，简称CSSBE method)；当 $\theta =0$ ，CSS θ方法就成了显式Euler方法。

由(2-7)式可得

$y_{n+1}=y_n+\Delta t{f}_{\lambda }\left(y_n^{\ast },y_{\left[qn\right]}\right)+g\left(y_n^{\ast },y_{\left[qn\right]}\right)\Delta W_n+h\left(y_n^{\ast },y_{\left[qn\right]}\right)\Delta {\tilde{N}}_n.$

对任意 $t\in \left[t_{n+1},t_n\right)$ ， $n=0,1,\cdots ,M-1$ ，定义

$y\left(t\right)=y_n+\left(t-t_n\right)f_{\lambda }\left(y_n^{\ast },y_{\left[qn\right]}\right)+g\left(y_n^{\ast },y_{\left[qn\right]}\right)\left(W\left(t_{n+1}\right)-W\left(t_n\right)\right)+h\left(y_n^{\ast },y_{\left[qn\right]}\right)\left(\tilde{N}\left(t_{n+1}\right)-\tilde{N}\left(t_n\right)\right).$ (2-8)

易知 $y\left(t_n\right)=y_n$ ，定义分段函数

$Z_1\left(t\right)={\displaystyle \underset{n=0}{\overset{\infty }{\sum }}{y}_n^{\ast }{\mathcal{X}}_{\left[t_n,t_{n+1}\right)}}\left(t\right),$

$Z_2\left(t\right)={\displaystyle \underset{n=0}{\overset{\infty }{\sum }}{y}_{\left[qn\right]}{\mathcal{X}}_{\left[t_n,t_{n+1}\right)}}\left(t\right).$

其中 ${\mathcal{X}}_F\left(t\right)=\{\begin{array}{l}1,t\in F\\ 0,t\notin F\end{array}$ ，又 $y\left(0\right)=y_0$ ，于是(2-8)式可以表示为

$y\left(t\right)=y\left(0\right)+{\displaystyle {\int }_0^t{f}_{\lambda }}\left(Z_1\left(s\right),Z_2\left(s\right)\right)\text{d}s+{\displaystyle {\int }_0^{t}g}\left(Z_1\left(s\right),Z_2\left(s\right)\right)\text{d}W\left(s\right)+{\displaystyle {\int }_0^{t}h}\left(Z_1\left(s\right),Z_2\left(s\right)\right)\text{d}\tilde{N}\left(s\right).$ (2-9)

引理2.2设条件(2)成立， $0\le \theta \le 1$ ， $0<\Delta t<\min \left\{1,\frac{1}{4\theta L_{\lambda }}\right\}$ ，那么存在正常数 $A=4\left(1+L_{\lambda }\right)$ ， $B=4L_{\lambda }$ ，使得 $E{\left|y_n^{\ast }\right|}^2\le A\underset{0\le k\le n}{\max }E{\left|y_k\right|}^2+B$ ，其中 $y_n$ 与 $y_n^{\ast }$ 由(2-7)式确定。

证明 由(2-7)式，基本不等式 $2ab\le a^2+b^2$ ，条件(2)及(2-6)式可得，

$\begin{array}{c}{\left|y_n^{\ast }\right|}^2={\left|y_n+\theta \Delta t{f}_{\lambda }\left(y_n^{\ast },y_{\left[qn\right]}\right)\right|}^2\\ ={\left|y_n\right|}^2+{\left|\theta \Delta t{f}_{\lambda }\left(y_n^{\ast },y_{\left[qn\right]}\right)\right|}^2+2\theta \Delta t{f}_{\lambda }\left(y_n^{\ast },y_{\left[qn\right]}\right)y_n\\ \le {\left|y_n\right|}^2+{\theta }^2{\left(\Delta t\right)}^2{\left|f_{\lambda }\left(y_n^{\ast },y_{\left[qn\right]}\right)\right|}^2+\theta \Delta t\left({\left|f_{\lambda }\left(y_n^{\ast },y_{\left[qn\right]}\right)\right|}^2+{\left|y_n\right|}^2\right)\\ \le {\left|y_n\right|}^2+2\theta \Delta t{\left|f_{\lambda }\left(y_n^{\ast },y_{\left[qn\right]}\right)\right|}^2+\theta \Delta t{\left|y_n\right|}^2\\ \le \left(1+\theta \Delta t\right){\left|y_n\right|}^2+2\theta \Delta t{L}_{\lambda }\left(1+{\left|y_n^{\ast }\right|}^2+{\left|y_{\left[qn\right]}\right|}^2\right).\end{array}$

上式两边取期望，得

$\begin{array}{c}E{\left|y_n^{\ast }\right|}^2\le \left(1+\theta \Delta t\right)E{\left|y_n\right|}^2+2\theta \Delta t{L}_{\lambda }\left(1+E{\left|y_n^{\ast }\right|}^2+E{\left|y_{\left[qn\right]}\right|}^2\right)\\ \le \left(1+\theta \Delta t+2\theta \Delta t{L}_{\lambda }\right)\underset{0\le k\le n}{\max }E{\left|y_k\right|}^2+2\theta \Delta t{L}_{\lambda }\left[1+E{\left|y_n^{\ast }\right|}^2\right].\end{array}$

由 $0\le \theta \le 1$ ， $\exists \Delta t$ 充分小，使得 $2\theta \Delta t{L}_{\lambda }\le \frac{1}{2}$ ，所以 $\frac{1}{1-2\theta \Delta t{L}_{\lambda }}\le 2$ 。

$\begin{array}{l}\therefore E{\left|y_n^{\ast }\right|}^2\le 2\left(1+\theta \Delta t+2\theta \Delta t{L}_{\lambda }\right)\underset{0\le k\le n}{\max }E{\left|y_k\right|}^2+4\theta \Delta t{L}_{\lambda }\\ \le 2\left(1+1+2L_{\lambda }\right)\underset{0\le k\le n}{\max }E{\left|y_k\right|}^2+4L_{\lambda }\\ =4\left(1+L_{\lambda }\right)\underset{0\le k\le n}{\max }E{\left|y_k\right|}^2+4L_{\lambda }\\ =A\underset{0\le k\le n}{\max }E{\left|y_k\right|}^2+B,\end{array}$

其中 $A=4\left(1+L_{\lambda }\right)$ ， $B=4L_{\lambda }$ 。□

引理2.3设条件(1)和(2)成立， $0\le \theta \le 1$ ， $0<\Delta t<\min \left\{1,\frac{1}{4\theta L_{\lambda }}\right\}$ ，那么由(2-7)式得到的离散数值解满足

$\underset{0\le k\le n}{\max }E{\left|y_k\right|}^2\le C_1,\underset{0\le k\le n}{\max }E{\left|y_k^{\ast }\right|}^2\le C_2.$

证明 由(2-7)式、(2-9)式及 $y\left(t_{n+1}\right)=y_{n+1}$ 可得

$y_{n+1}=y\left(0\right)+{\displaystyle {\int }_0^{t_{n+1}}{f}_{\lambda }}\left(Z_1\left(s\right),Z_2\left(s\right)\right)\text{d}s+{\displaystyle {\int }_0^{t_{n+1}}g}\left(Z_1\left(s\right),Z_2\left(s\right)\right)\text{d}W\left(s\right)+{\displaystyle {\int }_0^{t_{n+1}}h}\left(Z_1\left(s\right),Z_2\left(s\right)\right)\text{d}\tilde{N}\left(s\right),$

其中 $n=0,1,\cdots ,M-1$ ，对上式两边取平方再取期望并由基本不等式 ${\left|a+b+c+d\right|}^2\le 4{\left|a\right|}^2+4{\left|b\right|}^2+4{\left|c\right|}^2+4{\left|d\right|}^2$ ，得

$\begin{array}{c}E{\left|y_{n+1}\right|}^2\le 4E{\left|y_0\right|}^2+4E{\left|{\displaystyle {\int }_0^{t_{n+1}}{f}_{\lambda }}\left(Z_1\left(s\right),Z_2\left(s\right)\right)\text{d}s\right|}^2+4E{\left|{\displaystyle {\int }_0^{t_{n+1}}g}\left(Z_1\left(s\right),Z_2\left(s\right)\right)\text{d}W\left(s\right)\right|}^2\\ +4E{\left|{\displaystyle {\int }_0^{t_{n+1}}h}\left(Z_1\left(s\right),Z_2\left(s\right)\right)\text{d}\tilde{N}\left(s\right)\right|}^2.\end{array}$

由Cauchy-Schwarz不等式，条件(2)，(2-6)式，鞅等距映射性质及Fubini定理可得

$\begin{array}{l}\underset{1\le k\le n+1}{\max }E{\left|y_k\right|}^2\\ \le 4E{\left|y_0\right|}^2+4TE{\displaystyle {\int }_0^{t_{n+1}}{\left|f_{\lambda }\left(Z_1\left(s\right),Z_2\left(s\right)\right)\right|}^2\text{d}s}+4E{\displaystyle {\int }_0^{t_{n+1}}{\left|g\left(Z_1\left(s\right),Z_2\left(s\right)\right)\right|}^2\text{d}s}\\ +4\lambda E{\displaystyle {\int }_0^{t_{n+1}}{\left|h\left(Z_1\left(s\right),Z_2\left(s\right)\right)\right|}^2\text{d}s}\\ \le 4E{\left|y_0\right|}^2+4L_{\lambda }\left(T+1+\lambda \right){\displaystyle {\int }_0^{t_{n+1}}\left(1+E{\left|Z_1\left(s\right)\right|}^2+E{\left|Z_2\left(s\right)\right|}^2\right)\text{d}s}\\ \le 4E{\left|y_0\right|}^2+4L_{\lambda }T\left(T+1+\lambda \right)+4L_{\lambda }\left(T+1+\lambda \right)\Delta t\left(\underset{i=0}{\overset{n}{{\displaystyle \sum }}}E{\left|y_i\right|}^2+\underset{i=0}{\overset{n}{{\displaystyle \sum }}}E{\left|y_i^{\ast }\right|}^2\right)\end{array}$

$\begin{array}{l}\le 4E{\left|y_0\right|}^2+4L_{\lambda }T\left(T+1+\lambda \right)+4L_{\lambda }\left(T+1+\lambda \right)\Delta t\left(\underset{i=0}{\overset{n}{{\displaystyle \sum }}}E{\left|y_i\right|}^2+\underset{i=0}{\overset{n}{{\displaystyle \sum }}}\left(A\underset{0\le k\le i}{\max }E{\left|y_k\right|}^2+B\right)\right)\\ \le 4E{\left|y_0\right|}^2+4L_{\lambda }T\left(T+1+\lambda \right)+8L_{\lambda }T\left(T+1+\lambda \right)B+4L_{\lambda }\left(1+A\right)\left(T+1+\lambda \right)\Delta t\underset{i=0}{\overset{n}{{\displaystyle \sum }}}\left(\underset{0\le k\le i}{\max }E{\left|y_k\right|}^2\right)\\ =k_1+k_2\Delta t\underset{i=0}{\overset{n}{{\displaystyle \sum }}}\left(\underset{0\le k\le i}{\max }E{\left|y_k\right|}^2\right),\end{array}$

其中 $k_1=4E{\left|y_0\right|}^2+4L_{\lambda }T\left(T+1+\lambda \right)+8L_{\lambda }T\left(T+1+\lambda \right)B$ ， $k_2=4L_{\lambda }\left(1+A\right)\left(T+1+\lambda \right)$ ， $k_1,k_2$ 均与∆t无关。

由离散的Gronwall不等式，我们可得

$\underset{0\le k\le n}{\max }E{\left|y_k\right|}^2\le k_1{\text{e}}^{k_{2}T}:=C_1.$

由引理2.2，则

$E{\left|y_n^{\ast }\right|}^2\le A\underset{0\le k\le n}{\max }E{\left|y_k\right|}^2+B.$

即

$\underset{0\le k\le n}{\max }E{\left|y_k^{\ast }\right|}^2\le A\underset{0\le k\le n}{\max }E{\left|y_k\right|}^2+B:=C_2.$ □

为了证明主要结论，给出如下引理2.4：

引理2.4设条件(1)和(2)成立， $0\le \theta \le 1$ ， $0<\Delta t<\min \left\{1,\frac{1}{4\theta L_{\lambda }}\right\}$ ， $y\left(t\right)$ 由(2-8)式给出，则存在与∆t无关的正常数 $C_3$ 和 $C_4$ 使得

$E{\left|y\left(t\right)-Z_1\left(t\right)\right|}^2\le C_3\Delta t,$

$E{\left|y\left(qt\right)-Z_2\left(t\right)\right|}^2\le C_4\Delta t\text{.}$

证明 由(2-7)式及(2-9)式可得

$\begin{array}{c}y\left(t\right)-y_n^{\ast }={\displaystyle {\int }_{t_n}^t{f}_{\lambda }}\left(Z_1\left(s\right),Z_2\left(s\right)\right)\text{d}s+{\displaystyle {\int }_{t_n}^{t}g}\left(Z_1\left(s\right),Z_2\left(s\right)\right)\text{d}W\left(s\right)\\ +{\displaystyle {\int }_{t_n}^{t}h}\left(Z_1\left(s\right),Z_2\left(s\right)\right)\text{d}\tilde{N}\left(s\right)-\theta \Delta t{f}_{\lambda }\left(y_n^{\ast },y_{\left[qn\right]}\right).\end{array}$

由不等式 ${\left|a+b+c\right|}^2\le 3{\left|a\right|}^2+3{\left|b\right|}^2+3{\left|c\right|}^2$ ，条件(2)和(2-6)式可知，对 $\forall t\in \left[t_n,t_{n+1}\right)$ ，

$\begin{array}{l}E{\left|y\left(t\right)-Z_1\left(t\right)\right|}^2\\ \le 3E{\left|{\displaystyle {\int }_{t_n}^t{f}_{\lambda }}\left(Z_1\left(s\right),Z_2\left(s\right)\right)\text{d}s-\theta \Delta t{f}_{\lambda }\left(y_n^{\ast },y_{\left[qn\right]}\right)\right|}^2+3E{\left|{\displaystyle {\int }_{t_n}^{t}g}\left(Z_1\left(s\right),Z_2\left(s\right)\right)\text{d}W\left(s\right)\right|}^2\\ +3E{\left|{\displaystyle {\int }_{t_n}^{t}h}\left(Z_1\left(s\right),Z_2\left(s\right)\right)\text{d}\tilde{N}\left(s\right)\right|}^2\\ \le 6E{\left|{\displaystyle {\int }_{t_n}^t{f}_{\lambda }}\left(Z_1\left(s\right),Z_2\left(s\right)\right)ds\right|}^2+6{\theta }^2{\left(\Delta t\right)}^{2}E{\left|f_{\lambda }\left(y_n^{\ast },y_{\left[qn\right]}\right)\right|}^2\\ +3E{\left|{\displaystyle {\int }_{t_n}^{t}g}\left(Z_1\left(s\right),Z_2\left(s\right)\right)\text{d}W\left(s\right)\right|}^2+3E{\left|{\displaystyle {\int }_{t_n}^{t}h}\left(Z_1\left(s\right),Z_2\left(s\right)\right)\text{d}\tilde{N}\left(s\right)\right|}^2\end{array}$

$\begin{array}{l}\le 6T{\displaystyle {\int }_{t_n}^t{\left|f_{\lambda }\left(Z_1\left(s\right),Z_2\left(s\right)\right)\right|}^2\text{d}s}+6{\theta }^2{\left(\Delta t\right)}^{2}E{\left|f_{\lambda }\left(y_n^{\ast },y_{\left[qn\right]}\right)\right|}^2\\ +3E{\displaystyle {\int }_{t_n}^t{\left|g\left(Z_1\left(s\right),Z_2\left(s\right)\right)\right|}^2\text{d}s}+3\lambda E{\displaystyle {\int }_{t_n}^t{\left|h\left(Z_1\left(s\right),Z_2\left(s\right)\right)\right|}^2\text{d}s}\\ \le 3\left[2T+1+\lambda \right]L_{\lambda }{\displaystyle {\int }_{t_n}^{t_{n+1}}\left(1+E{\left|Z_1\left(s\right)\right|}^2+E{\left|Z_2\left(s\right)\right|}^2\right)\text{d}s}\\ +6{\theta }^2{\left(\Delta t\right)}^2{L}_{\lambda }\left(1+E{\left|y_n^{\ast }\right|}^2+E{\left|y_{\left[qn\right]}\right|}^2\right).\end{array}$

由引理2.3知 $\underset{0\le k\le n}{\max }E{\left|y_k\right|}^2\le C_1$ ， $\underset{0\le k\le n}{\max }E{\left|y_k^{\ast }\right|}^2\le C_2$ ，

$\therefore {\left|Z_1\left(s\right)\right|}^2\le C_2,{\left|Z_2\left(s\right)\right|}^2\le C_1.$

再由 $0\le \theta \le 1$ ， $0<\Delta t<\min \left\{1,\frac{1}{4\theta L_{\lambda }}\right\}$ 可得

$\begin{array}{l}E{\left|y\left(t\right)-Z_1\left(t\right)\right|}^2\\ \le 3\left[2T+1+\lambda \right]L_{\lambda }{\displaystyle {\int }_{t_n}^{t_{n+1}}\left(1+C_1+C_2\right)\text{d}s}+6{\theta }^2{\left(\text{Δ}t\right)}^2{L}_{\lambda }\left(1+C_1+C_2\right)\\ \le 3L_{\lambda }\left[2T+1+\lambda \right]\left(1+C_1+C_2\right)\text{Δ}t+6L_{\lambda }\left(1+C_1+C_2\right)\text{Δ}t\\ :=C_3\text{Δ}t,\forall t\in \left[0,T\right].\end{array}$

其中 $C_3=3L_{\lambda }\left[2T+3+\lambda \right]\left(1+C_1+C_2\right)$ 。

同样地，

$\begin{array}{c}y\left(qt\right)-Z_2\left(t\right)=y\left(qt\right)-y_{\left[qn\right]}=y\left(qt\right)-y\left(\left[qn\right]\Delta t\right)\\ ={\displaystyle {\int }_{\left[qn\right]\Delta t}^{qt}{f}_{\lambda }}\left(Z_1\left(s\right),Z_2\left(s\right)\right)\text{d}s+{\displaystyle {\int }_{\left[qn\right]\Delta t}^{qt}g}\left(Z_1\left(s\right),Z_2\left(s\right)\right)\text{d}W\left(s\right)\\ +{\displaystyle {\int }_{\left[qn\right]\Delta t}^{qt}h}\left(Z_1\left(s\right),Z_2\left(s\right)\right)\text{d}\tilde{N}(\; s\; )\end{array}$

由基本不等式 ${\left|a+b+c\right|}^2\le 3{\left|a\right|}^2+3{\left|b\right|}^2+3{\left|c\right|}^2$ ，条件(2)及(2-6)式可知，对 $\forall t\in \left[t_n,t_{n+1}\right)$ ，

$\begin{array}{l}E{\left|y\left(qt\right)-Z_2\left(t\right)\right|}^2\\ \le 3E{\left|{\displaystyle {\int }_{\left[qn\right]\Delta t}^{qt}{f}_{\lambda }}\left(Z_1\left(s\right),Z_2\left(s\right)\right)\text{d}s\right|}^2+3E{\left|{\displaystyle {\int }_{\left[qn\right]\Delta t}^{qt}g}\left(Z_1\left(s\right),Z_2\left(s\right)\right)\text{d}W\left(s\right)\right|}^2\\ +3E{\left|{\displaystyle {\int }_{\left[qn\right]\Delta t}^{qt}h}\left(Z_1\left(s\right),Z_2\left(s\right)\right)\text{d}\tilde{N}\left(s\right)\right|}^2\\ \le 3TE{\displaystyle {\int }_{\left[qn\right]\Delta t}^{qt}{\left|f_{\lambda }\left(Z_1\left(s\right),Z_2\left(s\right)\right)\right|}^2\text{d}s}+3E{\displaystyle {\int }_{\left[qn\right]\Delta t}^{qt}{\left|g\left(Z_1\left(s\right),Z_2\left(s\right)\right)\right|}^2\text{d}s}\\ +3\lambda E{\displaystyle {\int }_{\left[qn\right]\Delta t}^{qt}{\left|h\left(Z_1\left(s\right),Z_2\left(s\right)\right)\right|}^2\text{d}s}\end{array}$

$\begin{array}{l}\le 3\left(T+1+\lambda \right)L_{\lambda }{\displaystyle {\int }_{\left[qn\right]\Delta t}^{qt}\left(1+E{\left|Z_1\left(s\right)\right|}^2+E{\left|Z_2\left(s\right)\right|}^2\right)\text{d}s}\\ \le 3\left(T+1+\lambda \right)L_{\lambda }{\displaystyle {\int }_{\left[qn\right]\Delta t}^{qt}\left(1+C_1+C_2\right)\text{d}s}\\ \le 6L_{\lambda }\left(T+1+\lambda \right)\left(1+C_1+C_2\right)\text{Δ}t\\ :=C_4\text{Δ}t,\forall t\in \left[0,T\right],\end{array}$

其中 $C_4=6L_{\lambda }\left(T+1+\lambda \right)\left(1+C_1+C_2\right)$ 。□

下面给出了CSS θ方法的收敛性。

定理2.5 设条件(1)和(2)成立， $0\le \theta \le 1$ ， $0<\Delta t<\min \left\{1,\frac{1}{4\theta L_{\lambda }}\right\}$ ，由(2-9)式给出的连续时间数值解 $y\left(t\right)$ 将均方收敛于方程(2-4)的真解 $x\left(t\right)$ ，即

$E\left[\underset{0\le t\le T}{\sup }{\left|y\left(t\right)-x\left(t\right)\right|}^2\right]\le C_5\text{Δ}t,$

其中 $C_5$ 是独立于 $\Delta t$ 的常数。

证明 令 $f_{\lambda }{}_1\left(Z_1\left(s\right),Z_2\left(s\right)\right)=f_{\lambda }\left(x\left(s\right),x\left(qs\right)\right)-f_{\lambda }\left(Z_1\left(s\right),Z_2\left(s\right)\right)$ ，

$g_1\left(Z_1\left(s\right),Z_2\left(s\right)\right)=g\left(x\left(s\right),x\left(qs\right)\right)-g\left(Z_1\left(s\right),Z_2\left(s\right)\right),$

$h_1\left(Z_1\left(s\right),Z_2\left(s\right)\right)=h\left(x\left(s\right),x\left(qs\right)\right)-h\left(Z_1\left(s\right),Z_2\left(s\right)\right).$

对任意的 $0\le r\le T$ ，由基本不等式 ${\left|a+b+c\right|}^2\le 3{\left|a\right|}^2+3{\left|b\right|}^2+3{\left|c\right|}^2$ ，Cauchy-Schwarz不等式，Doob鞅不等式，鞅等距映射性质，条件(1)，(2-6)式及引理2.4可得

$\begin{array}{l}E\left[\underset{0\le t\le T}{\sup }{\left|x\left(t\right)-y\left(t\right)\right|}^2\right]\\ \le 3E\left[\underset{0\le t\le T}{\sup }{\left|{\displaystyle {\int }_0^t{f}_{\lambda }{}_1\text{d}s}\right|}^2\right]+3E\left[\underset{0\le t\le T}{\sup }{\left|{\displaystyle {\int }_0^t{g}_1\text{d}W\left(s\right)}\right|}^2\right]+3E\left[\underset{0\le t\le T}{\sup }{\left|{\displaystyle {\int }_0^t{h}_1\text{d}\tilde{N}}\left(s\right)\right|}^2\right]\\ \le 3TE{\displaystyle {\int }_0^r{\left|f_{\lambda }{}_1\right|}^2\text{d}s}+12E{\displaystyle {\int }_0^r{\left|g_1\right|}^2\text{d}s}+12\lambda E{\displaystyle {\int }_0^r{\left|h_1\right|}^2\text{d}s}\\ \le 3\left(T+4+4\lambda \right)E{\displaystyle {\int }_0^r{K}_{\lambda }}\left({\left|x\left(s\right)-Z_1\left(s\right)\right|}^2+{\left|x\left(qs\right)-Z_2\left(s\right)\right|}^2\right)\text{d}s\end{array}$

$\begin{array}{l}\le 6\left(T+4+4\lambda \right)K_{\lambda }E{\displaystyle {\int }_0^r\left({\left|x\left(s\right)-y\left(s\right)\right|}^2+{\left|y\left(s\right)-Z_1\left(s\right)\right|}^2+{\left|x\left(qs\right)-y\left(qs\right)\right|}^2+{\left|y\left(qs\right)-Z_2\left(s\right)\right|}^2\right)\text{d}s}\\ \le 6\left(T+4+4\lambda \right)K_{\lambda }\left(C_3+C_4\right)T\Delta t+12\left(T+4+4\lambda \right)K_{\lambda }{\displaystyle {\int }_0^{r}E}\left[\underset{0\le w\le s}{\sup }{\left|x\left(w\right)-y\left(w\right)\right|}^2\right]\text{d}s\\ =P_1\Delta t+P_2{\displaystyle {\int }_0^{r}E}\left[\underset{0\le w\le s}{\sup }{\left|x\left(w\right)-y\left(w\right)\right|}^2\right]\text{d}s.\end{array}$

由Gronwall不等式及 $r\in \left[0,T\right]$ 的任意性可得

$E\left[\underset{0\le t\le T}{\sup }{\left|x\left(t\right)-y\left(t\right)\right|}^2\right]\le C_5\text{Δ}t,$

其中 $C_5=P_1{\text{e}}^{P_{2}T}$ ， $P_1=6\left(T+4+4\lambda \right)K_{\lambda }\left(C_3+C_4\right)T$ ， $P_2=12\left(T+4+4\lambda \right)K_{\lambda }$ 。□

3. 补偿分步θ方法的均方稳定性

考虑CSS θ方法求解非线性带跳的随机比例微分方程(2-1)的均方稳定性。为了讨论稳定性，我们再假设系数 $f_{\lambda }$ 满足如下条件 [8] ：

(3) 对 $\forall t\in \left[0,T\right]$ ， $x,y\in R^n$ ，存在正常数 ${\mu }_1,{\mu }_2$ 使得

$2x{f}_{\lambda }\left(x,y\right)\le -{\mu }_1{\left|x\right|}^2+{\mu }_2{\left|y\right|}^2.$

定理3.1设条件(1)、(2)和(3)成立，当下列不等式成立：

${\mu }_1>{\mu }_2+2K_{\lambda }\left(2+\lambda \right),$ (3-1)

则对任意步长，方程(2-1)是均方稳定的，也即

$\underset{t\to \infty }{\lim }E{\left|x\left(t\right)\right|}^2=0.$

证明由方程(2-1)和Itȏ公式，我们有

$\begin{array}{c}\text{d}{\left|x\left(t\right)\right|}^2=\left[2x\left(t\right)f\left(x\left(t\right),x\left(qt\right)\right)+{\left|g\left(x\left(t\right),x\left(qt\right)\right)\right|}^2\right]\text{d}t+2x\left(t\right)g\left(x\left(t\right),x\left(qt\right)\right)\text{d}W\left(t\right)\\ +\left({\left|x\left(t\right)+h\left(x\left(t\right),x\left(qt\right)\right)\right|}^2-{\left|x\left(t\right)\right|}^2\right)\text{d}N\left(t\right).\end{array}$

由 $\tilde{N}\left(t\right)=N\left(t\right)-\lambda t$ ， $f_{\lambda }=f+\lambda h$ 我们可得

$\begin{array}{c}\text{d}{\left|x\left(t\right)\right|}^2=\left[2x\left(t\right)f\left(x\left(t\right),x\left(qt\right)\right)+{\left|g\left(x\left(t\right),x\left(qt\right)\right)\right|}^2\right]\text{d}t+2x\left(t\right)g\left(x\left(t\right),x\left(qt\right)\right)\text{d}W\left(t\right)\\ +\left({\left|x\left(t\right)+h\left(x\left(t\right),x\left(qt\right)\right)\right|}^2-{\left|x\left(t\right)\right|}^2\right)\text{d}\tilde{N}\left(t\right)+\lambda \left({\left|x\left(t\right)+h\left(x\left(t\right),x\left(qt\right)\right)\right|}^2-{\left|x\left(t\right)\right|}^2\right)\text{d}t\\ =\left[2x\left(t\right)f_{\lambda }\left(x\left(t\right),x\left(qt\right)\right)+{\left|g\left(x\left(t\right),x\left(qt\right)\right)\right|}^2+\lambda {\left|h\left(x\left(t\right),x\left(qt\right)\right)\right|}^2\right]\text{d}t\\ +2x\left(t\right)g\left(x\left(t\right),x\left(qt\right)\right)\text{d}W\left(t\right)+\left({\left|x\left(t\right)+h\left(x\left(t\right),x\left(qt\right)\right)\right|}^2-{\left|x\left(t\right)\right|}^2\right)\text{d}\tilde{N}\left(t\right).\end{array}$

定义 $\phi \left(t\right)=E{\left|x\left(t\right)\right|}^2$ ，我们有

$\begin{array}{l}\frac{\phi \left(t+\Delta t\right)-\phi \left(t\right)}{\Delta t}=E\frac{{\left|x\left(t+\Delta t\right)\right|}^2-{\left|x\left(t\right)\right|}^2}{\Delta t}\\ =E\frac{1}{\Delta t}\{{\displaystyle {\int }_t^{t+\Delta t}\left[2x\left(s\right)f_{\lambda }\left(x\left(s\right),x\left(qs\right)\right)+{\left|g\left(x\left(s\right),x\left(qs\right)\right)\right|}^2+\lambda {\left|h\left(x\left(s\right),x\left(qs\right)\right)\right|}^2\right]\text{d}s}\\ +{\displaystyle {\int }_t^{t+\Delta t}2x\left(s\right)g\left(x\left(s\right),x\left(qs\right)\right)\text{d}W\left(s\right)}+{\displaystyle {\int }_t^{t+\Delta t}\left({\left|x\left(s\right)+h\left(x\left(s\right),x\left(qs\right)\right)\right|}^2-{\left|x\left(s\right)\right|}^2\right)\text{d}\tilde{N}\left(s\right)}\}\\ =E\frac{1}{\Delta t}{\displaystyle {\int }_t^{t+\Delta t}\left[2x\left(s\right)f_{\lambda }\left(x\left(s\right),x\left(qs\right)\right)+{\left|g\left(x\left(s\right),x\left(qs\right)\right)\right|}^2+\lambda {\left|h\left(x\left(s\right),x\left(qs\right)\right)\right|}^2\right]\text{d}s}.\end{array}$

由控制收敛定理可得

$\begin{array}{l}\underset{\Delta t\to 0}{\lim }\frac{\phi \left(t+\Delta t\right)-\phi \left(t\right)}{\Delta t}\\ =\underset{\Delta t\to 0}{\lim }E\left[\frac{1}{\Delta t}{\displaystyle {\int }_t^{t+\Delta t}\left(2x\left(s\right)f_{\lambda }\left(x\left(s\right),x\left(qs\right)\right)+{\left|g\left(x\left(s\right),x\left(qs\right)\right)\right|}^2+\lambda {\left|h\left(x\left(s\right),x\left(qs\right)\right)\right|}^2\right)\text{d}s}\right]\\ =E\left[\underset{\Delta t\to 0}{\lim }\frac{1}{\Delta t}{\displaystyle {\int }_t^{t+\Delta t}\left(2x\left(s\right)f_{\lambda }\left(x\left(s\right),x\left(qs\right)\right)+{\left|g\left(x\left(s\right),x\left(qs\right)\right)\right|}^2+\lambda {\left|h\left(x\left(s\right),x\left(qs\right)\right)\right|}^2\right)\text{d}s}\right]\\ =E\left[2x\left(t\right)f_{\lambda }\left(x\left(t\right),x\left(qt\right)\right)+{\left|g\left(x\left(t\right),x\left(qt\right)\right)\right|}^2+\lambda {\left|h\left(x\left(t\right),x\left(qt\right)\right)\right|}^2\right].\end{array}$

因此

${\phi }^{\prime }\left(t\right)=E\left[2x\left(t\right)f_{\lambda }\left(x\left(t\right),x\left(qt\right)\right)+{\left|g\left(x\left(t\right),x\left(qt\right)\right)\right|}^2+\lambda {\left|h\left(x\left(t\right),x\left(qt\right)\right)\right|}^2\right].$

由条件(3)和(2-3)式，可得

$\begin{array}{l}E\left[2x\left(t\right)f_{\lambda }\left(x\left(t\right),x\left(qt\right)\right)+{\left|g\left(x\left(t\right),x\left(qt\right)\right)\right|}^2+\lambda {\left|h\left(x\left(t\right),x\left(qt\right)\right)\right|}^2\right]\\ \le \left(-{\mu }_{1}E{\left|x\left(t\right)\right|}^2+{\mu }_{2}E{\left|x\left(qt\right)\right|}^2\right)+K\left(1+\lambda \right)\left(E{\left|x\left(t\right)\right|}^2+E{\left|x\left(qt\right)\right|}^2\right)\\ \le \left(-{\mu }_{1}E{\left|x\left(t\right)\right|}^2+{\mu }_{2}E{\left|x\left(qt\right)\right|}^2\right)+K_{\lambda }\left(2+\lambda \right)\left(E{\left|x\left(t\right)\right|}^2+E{\left|x\left(qt\right)\right|}^2\right)\\ \le \left[-{\mu }_1+{\mu }_2+2K_{\lambda }\left(2+\lambda \right)\right]E\left(\underset{0\le s\le t+\Delta t}{\sup }{\left|x\left(s\right)\right|}^2\right).\end{array}$

因此，由 $E\left(\underset{0\le s\le T}{\sup }{\left|x\left(s\right)\right|}^2\right)<\infty$ ，只要 $-{\mu }_1+{\mu }_2+2K_{\lambda }\left(2+\lambda \right)<0$ ，也即 ${\mu }_1>{\mu }_2+2K_{\lambda }\left(2+\lambda \right)$ ，即可得

$\underset{t\to \infty }{\lim }E{\left|x\left(t\right)\right|}^2=0.$ □

定理3.2设条件(1)、(2)、(3)和(3-1)式成立，总存在一个正数 $h_0$ ，使得当 $\Delta t\in \left(0,h_0\right)$ 时，CSS θ方法(2-6)是均方稳定的，也即

$\underset{n\to \infty }{\lim }E{\left|y_n\right|}^2=0.$

证明 由(2-7)式得

$y_{n+1}=y_n^{\ast }+\left(1-\theta \right)\Delta t{f}_{\lambda }\left(y_n^{\ast },y_{\left[qn\right]}\right)+g\left(y_n^{\ast },y_{\left[qn\right]}\right)\Delta W_n+h\left(y_n^{\ast },y_{\left[qn\right]}\right)\Delta {\tilde{N}}_n.$

上式两边平方，得

$\begin{array}{c}{\left|y_{n+1}\right|}^2={\left|y_n^{\ast }\right|}^2+{\left(1-\theta \right)}^2{\left(\Delta t\right)}^2{\left|f_{\lambda }\left(y_n^{\ast },y_{\left[qn\right]}\right)\right|}^2+{\left|g\left(y_n^{\ast },y_{\left[qn\right]}\right)\Delta W_n\right|}^2+{\left|h\left(y_n^{\ast },y_{\left[qn\right]}\right)\Delta {\tilde{N}}_n\right|}^2\\ +2\left(1-\theta \right)\Delta t{y}_n^{\ast }{f}_{\lambda }\left(y_n^{\ast },y_{\left[qn\right]}\right)+2y_n^{\ast }\left(g\left(y_n^{\ast },y_{\left[qn\right]}\right)\Delta W_n\right)+2y_n^{\ast }\left(h\left(y_n^{\ast },y_{\left[qn\right]}\right)\Delta {\tilde{N}}_n\right)\\ +2\left(1-\theta \right)\Delta t{f}_{\lambda }\left(y_n^{\ast },y_{\left[qn\right]}\right)\left(g\left(y_n^{\ast },y_{\left[qn\right]}\right)\Delta W_n\right)+2\left(1-\theta \right)\Delta t{f}_{\lambda }\left(y_n^{\ast },y_{\left[qn\right]}\right)\left(h\left(y_n^{\ast },y_{\left[qn\right]}\right)\Delta {\tilde{N}}_n\right)\\ +2\left(g\left(y_n^{\ast },y_{\left[qn\right]}\right)\Delta W_n\right)\left(h\left(y_n^{\ast },y_{\left[qn\right]}\right)\Delta {\tilde{N}}_n\right).\end{array}$ (3-2)

由 $E\left(\Delta W_n\right)=0$ ， $E{\left(\Delta W_n\right)}^2=\Delta t$ ， $E\left(\Delta {\tilde{N}}_n\right)=0$ ， $E{\left(\Delta {\tilde{N}}_n\right)}^2=\lambda \Delta t$ ， $y_n^{\ast }$ 与 $y_{\left[qn\right]}$ 都是 ${\mathcal{F}}_{t_n}$ -可测以及 $\Delta W_n$ 与 $\Delta {\tilde{N}}_n$ 相互独立可知

$E{\left|g\left(y_n^{\ast },y_{\left[qn\right]}\right)\Delta W_n\right|}^2=\Delta tE{\left|g\left(y_n^{\ast },y_{\left[qn\right]}\right)\right|}^2,$

$E{\left|h\left(y_n^{\ast },y_{\left[qn\right]}\right)\Delta {\tilde{N}}_n\right|}^2=\lambda \Delta tE{\left|h\left(y_n^{\ast },y_{\left[qn\right]}\right)\right|}^2.$

令

$\begin{array}{c}{Q}_n=2y_n^{\ast }\left(g\left(y_n^{\ast },y_{\left[qn\right]}\right)\Delta W_n\right)+2y_n^{\ast }\left(h\left(y_n^{\ast },y_{\left[qn\right]}\right)\Delta {\tilde{N}}_n\right)+2\left(1-\theta \right)\Delta t{f}_{\lambda }\left(y_n^{\ast },y_{\left[qn\right]}\right)\left(g\left(y_n^{\ast },y_{\left[qn\right]}\right)\Delta W_n\right)\\ +2\left(1-\theta \right)\Delta t{f}_{\lambda }\left(y_n^{\ast },y_{\left[qn\right]}\right)\left(h\left(y_n{}^{*},y_{\left[qn\right]}\right)\Delta {\tilde{N}}_n\right)+2\left(g\left(y_n^{\ast },y_{\left[qn\right]}\right)\Delta W_n\right)\left(h\left(y_n^{\ast },y_{\left[qn\right]}\right)\Delta {\tilde{N}}_n\right),\end{array}$

则 $E\left(Q_n\right)=0$ 。

对(3-2)式取期望，并由条件(1)，条件(3)，(2-3)式，(2-6)式， $0\le K\le K_{\lambda }$ 及 $0<\Delta t<1$ ， $0<\theta \le 1$ 可得，

$\begin{array}{c}E{\left|y_{n+1}\right|}^2=E{\left|y_n^{\ast }\right|}^2+{\left(1-\theta \right)}^2{\left(\Delta t\right)}^{2}E{\left|f_{\lambda }\left(y_n^{\ast },y_{\left[qn\right]}\right)\right|}^2+\Delta tE{\left|g\left(y_n^{\ast },y_{\left[qn\right]}\right)\right|}^2\\ +\lambda \text{Δ}tE{\left|h\left(y_n^{\ast },y_{\left[qn\right]}\right)\right|}^2+2\left(1-\theta \right)\Delta tE\left(y_n^{\ast }{f}_{\lambda }\left(y_n^{\ast },y_{\left[qn\right]}\right)\right)\\ \le E{\left|y_n^{\ast }\right|}^2+{\left(1-\theta \right)}^2{\left(\Delta t\right)}^2{K}_{\lambda }\left(E{\left|y_n^{\ast }\right|}^2+E{\left|y_{\left[qn\right]}\right|}^2\right)+\Delta tK\left(E{\left|y_n^{\ast }\right|}^2+E{\left|y_{\left[qn\right]}\right|}^2\right)\\ +\lambda \text{Δ}tK\left(E{\left|y_n^{\ast }\right|}^2+E{\left|y_{\left[qn\right]}\right|}^2\right)+\left(1-\theta \right)\Delta t\left(-{\mu }_{1}E{\left|y_n^{\ast }\right|}^2+{\mu }_{2}E{\left|y_{\left[qn\right]}\right|}^2\right)\end{array}$

$\begin{array}{c}\le \left(1+{\left(1-\theta \right)}^2{\left(\Delta t\right)}^2{K}_{\lambda }+\Delta t{K}_{\lambda }+\lambda \text{Δ}t{K}_{\lambda }-\left(1-\theta \right)\Delta t{\mu }_1\right)E{\left|y_n^{\ast }\right|}^2\\ +\left({\left(1-\theta \right)}^2{\left(\Delta t\right)}^2{K}_{\lambda }+\Delta t{K}_{\lambda }+\lambda \text{Δ}t{K}_{\lambda }+\left(1-\theta \right)\Delta t{\mu }_2\right)E{\left|y_{\left[qn\right]}\right|}^2\\ \le \left(1+\Delta t{K}_{\lambda }\left(2+\lambda \right)-\left(1-\theta \right)\Delta t{\mu }_1\right)E{\left|y_n^{\ast }\right|}^2+\left(\Delta t{K}_{\lambda }\left(2+\lambda \right)+\left(1-\theta \right)\Delta t{\mu }_2\right)E{\left|y_{\left[qn\right]}\right|}^2\\ =K_{1}E{\left|y_n^{\ast }\right|}^2+K_{2}E{\left|y_{\left[qn\right]}\right|}^2,\end{array}$ (3-3)

其中 $K_1=1+\Delta t{K}_{\lambda }\left(2+\lambda \right)-\left(1-\theta \right)\Delta t{\mu }_1$ ， $K_2=\Delta t{K}_{\lambda }\left(2+\lambda \right)+\left(1-\theta \right)\Delta t{\mu }_2$ 。

接下来估计 $E{\left|y_n^{\ast }\right|}^2$ ，由(2-7)式得

$y_n^{\ast }-\theta \Delta t{f}_{\lambda }\left(y_n^{\ast },y_{\left[qn\right]}\right)=y_n.$

两边平方后取期望，由条件(3)得

$\begin{array}{c}E{\left|y_n^{\ast }\right|}^2\le E{\left|y_n\right|}^2+2\theta \Delta tE\left(y_n^{\ast }{f}_{\lambda }\left(y_n^{\ast },y_{\left[qn\right]}\right)\right)\\ \le E{\left|y_n\right|}^2+\theta \Delta t\left(-{\mu }_{1}E{\left|y_n^{\ast }\right|}^2+{\mu }_{2}E{\left|y_{\left[qn\right]}\right|}^2\right).\end{array}$

整理得

$E{\left|y_n^{\ast }\right|}^2\le \frac{1}{1+\theta \Delta t{\mu }_1}E{\left|y_n\right|}^2+\frac{\theta \Delta t{\mu }_2}{1+\theta \Delta t{\mu }_1}E{\left|y_{\left[qn\right]}\right|}^2.$

代入(3-3)式得

$\begin{array}{c}E{\left|y_{n+1}\right|}^2\le \frac{K_1}{1+\theta \Delta t{\mu }_1}E{\left|y_n\right|}^2+\left(\frac{\theta \Delta t{\mu }_2{K}_1}{1+\theta \Delta t{\mu }_1}+K_2\right)E{\left|y_{\left[qn\right]}\right|}^2\\ \le \left(\frac{K_1+\theta \Delta t{\mu }_2{K}_1}{1+\theta \Delta t{\mu }_1}+K_2\right)\max \left\{E{\left|y_n\right|}^2,E{\left|y_{\left[qn\right]}\right|}^2\right\}.\end{array}$

要使 $E{\left|y_{n+1}\right|}^2\to 0\left(n\to \infty \right)$ ，则要满足

$\frac{K_1+\theta \Delta t{\mu }_2{K}_1}{1+\theta \Delta t{\mu }_1}+K_2<1.$

上式等价于

$L_1{\left(\Delta t\right)}^2+L_2\Delta t<0,$ (3-4)

其中

$L_1=\theta K_{\lambda }\left(2+\lambda \right)\left({\mu }_1+{\mu }_2\right),$

$L_2=2K_{\lambda }\left(2+\lambda \right)+\left({\mu }_2-{\mu }_1\right).$

由 $\theta ,K_{\lambda },\lambda ,{\mu }_1,{\mu }_2$ 均为正数，可知 $L_1$ 恒大于0，由抛物线对称轴的公式及 $\Delta t>0$ 可知，需满足

$-\frac{2K_{\lambda }\left(2+\lambda \right)+\left({\mu }_2-{\mu }_1\right)}{2\theta K_{\lambda }\left(2+\lambda \right)\left({\mu }_1+{\mu }_2\right)}>0.$

从而需满足

$2K_{\lambda }\left(2+\lambda \right)+\left({\mu }_2-{\mu }_1\right)<0.$

故当 ${\mu }_1>{\mu }_2+2K_{\lambda }\left(2+\lambda \right)$ 时，总存在一个正数 ${\tilde{h}}_0$ ，使得当 $\Delta t\in \left(0,{\tilde{h}}_0\right)$ 时，(3-4)式成立。由 $L_1{\left(\Delta t\right)}^2+L_2\Delta t<0$ ，得

$\Delta t<-\frac{L_2}{L_1}={\tilde{h}}_0.$

因此，只要取 $h_0=\min \left(1,{\tilde{h}}_0\right)$ ，便能保证(3-4)式成立。□

4. 数值算例

考虑如下非线性带跳随机比例微分方程：

$\{\begin{array}{l}\text{d}x\left(t\right)=ax\left(t\right)\text{d}t+b\sin x\left(qt\right)\text{d}W\left(t\right)+cx\left(t\right)\text{d}N\left(t\right),0<t\le T\\ x\left(0\right)=1\end{array}$ (4-1)

选取如下两组系数：

算例1： $a=-3;b=0.3;c=4;q=0.7;\lambda =0.7$ 。

算例2： $a=-2;b=0.2;c=1.7;q=0.8;\lambda =1$ 。

首先验证补偿分步θ方法(2-7)的收敛性。在收敛性试验中，我们用1000个样本轨道平均来模拟数值解的期望。因方程(4-1)没有显式的解析解，我们用步长为 $\Delta t=2^{-10}$ 得到的数值解来近似地替代解析解。选择四种步长，分别为2∆t，4∆t，8∆t，16∆t， $\Delta t=2^{-10}$ ，均方误差 ${\epsilon }^r={\epsilon }_{2^{1+r}\Delta t}$ ， $r=0,1,2,3$ ，给出了在终点时刻 $T=1$ 处的误差，选取的1000个样本点记为 ${\omega }_i\left(1\le i\le 1000\right)$ ，其均方误差由以下方式估计：

${\epsilon }^r\cong \frac{1}{1000}\underset{i=1}{\overset{1000}{{\displaystyle \sum }}}{\left|x_{r,T}\left({\omega }_i\right)-Y_{r,n}\left({\omega }_i\right)\right|}^2$

在图1中我们以对数为底的坐标系中画出 $\sqrt{{\epsilon }^r}$ ，并以斜率为1/2的红色虚线为参照，从图1中可以看出蓝色曲线基本平行于这条参照线。这清楚表明步长分步θ方法(2-7)是均方收敛的。

(a) (b)

Figure 1. Convergence rate of the CSS θ method (left: example 1, right: example 2)

图1. 补偿分步θ方法的收敛性(左：算例1，右：算例2)

其次验证补偿分步θ方法的稳定性。同样选取上述两组算例，对于算例1，易知 $f_{\lambda }=f+\lambda h$ 以常数 $K_{\lambda }=0.04$ 满足全局Lipschitz条件，以常数 ${\mu }_1=0.4,{\mu }_2=0$ 满足条件(3)，从而满足(3-1)式，因而方程(4-1)的解析解是均方稳定的。同样地，对于算例2， $K_{\lambda }=0.09$ ， ${\mu }_1=0.6,{\mu }_2=0$ 也满足(3-1)式，方程的解析解也均方稳定。由定理3.2知， ${\tilde{h}}_0=-\frac{L_2}{L_1}$ ，对算例1， ${\tilde{h}}_0=8.\stackrel{\dot{}}{5}\stackrel{\dot{}}{1}\stackrel{\dot{}}{8}$ ，从而 $h_0=\min \left(1,{\tilde{h}}_0\right)=1$ ，因此保持数值方法稳定的步长范围是开区间 $\left(0,1\right)$ ；而对算例2， ${\tilde{h}}_0=0.\stackrel{\dot{}}{7}\stackrel{\dot{}}{4}\stackrel{\dot{}}{0}$ ，从而 $h_0=\min \left(1,{\tilde{h}}_0\right)=0.74$ ，因此保持数值方法稳定的步长范围是开区间 $\left(0,0.74\right)$ 。

(a) (b)

Figure 2. Mean-square stability of the CSS θ method (left: example 1, right: example 2; Δt = 0.125)

图2. 补偿分步θ方法的均方稳定性(左：算例1，右：算例2；Δt = 0.125)

(a) (b)

Figure 3. Mean-square stability of the CSS θ method (left: example 1, right: example 2; Δt = 0.25)

图3. 补偿分步θ方法的均方稳定性(左：算例1，右：算例2；Δt = 0.25)

(a) (b)

Figure 4. Mean-square stability of the CSS θ method (left: example 1, right: example 2; Δt = 0.5)

图4. 补偿分步θ方法的均方稳定性(左：算例1，右：算例2；Δt = 0.5)

图2~4表明了当步长 $\Delta t=0.5,0.25,0.125$ 时，补偿分步θ方法都保持稳定，这与上述分析一致。