分段连续型随机微分方程平衡方法的均方收敛性与稳定性

The Mean Square Convergence and Stability of the Balanced Implicit Method of Stochastic Differential Equations with Piecewise Continuous Arguments

摘要: 针对一类分段连续型随机微分方程，证明了其采用全隐式平衡方法的情况下，该方法具有良好的均方收敛性和稳定性，并且证明了平衡方法的强收敛阶为1/2，同时通过数值分析实验验证了此类方程数学分析的正确以及合理性，证明了强平衡隐式方法以及弱平衡隐式方法都是均方稳定的。

Abstract: For a class of piecewise continuous stochastic differential equations, the mean square convergence and stability of the fully implicit method-balanced method are studied by using the balanced method. It is proved that the strong convergence order of the balanced method is 1/2, and the rationality and correctness of the analysis are verified by numerical examples. It is shown that both the strong balanced implicit method and the weak balanced implicit method are mean square stable.

1. 引言

随机微分方程[1]在经济学、控制理论、金融学、医学、生物学等各个领域[2] [3]都是很重要的数学工具并且都有较为广泛的应用，比如在金融数学里可以用来期权定价、风险度量；在物理学的统计力学方面有所体现，生物学中关于种群增长模型等，并且随机微分方程是一种考虑外部因素如白噪声对整个系统产生干扰的数学模型[4] [5]，其中本文研究的分段连续型随机微分方程，该方程具有的解是一个局部光滑且连续的函数，并在这些区间里满足该方程。该方程常用于神经网络模拟[6]、生物系统建模[7]等，目前对分段连续型随机微分方程已经有诸多成果[8]-[14]，关于该方程全隐式方法的结论还没有找到有关文献。本篇文章提出分段连续型随机微分方程在满足全局Lipschitz条件下，全隐式方法的数值方法，全隐式方法是对干扰项都进行全隐式处理，具有很好的稳定性，并且给出了方程数值解的均方收敛性和均方稳定性[15]。

考虑如下分段连续型随机微分方程

$\text{d}x\left(t\right)=f\left(x\left(t\right),x\left(\left[t\right]\right)\right)\text{d}t+g\left(x\left(t\right),x\left(\left[t\right]\right)\right)\text{d}W\left(t\right),t\in \left[0,T\right].$ (1.1)

其中初值为$x\left(0\right)=x_0\in {ℝ}^n$ ，漂移项系数$f:{ℝ}^d\times {ℝ}^d\to {ℝ}^d$ ，扩散项系数$g:{ℝ}^d\times {ℝ}^d\to {ℝ}^{d\times m}$ ， $W\left(t\right)={\left(W_1\left(t\right),W_2\left(t\right),\cdots ,W_d\left(t\right)\right)}^{\text{T}}$ 为$\left(\Omega ,ℱ,{\left\{{ℱ}_t\right\}}_{t\ge 0},P\right)$ 上的$d$ 维Brownian运动。

2. 平衡方法的收敛性

设$\left|x\right|$ 表示向量的Euclid范数，且$x\in {ℝ}^n$ 。用$\left[u\right]$ 表示$u$ 的整数部分。若$A$ 为一个向量或者矩阵，$A^{\text{T}}$ 称为转置，$\left|A\right|=\sqrt{tr\left(A^{\text{T}}A\right)}$ 为迹范数。

与(1.1)等价的积分形式如下：

$x\left(t\right)=x_0+{\displaystyle {\int }_0^{t}f\left(x\left(s\right),x\left(\left[s\right]\right)\right)\text{d}s}+{\displaystyle {\int }_0^{t}g\left(x\left(s\right),x\left(\left[s\right]\right)\right)\text{d}W\left(s\right)}$ (2.1)

下列两个方程(2.1)需要满足的条件，我们假设$f,g$ 满足如下假设：

(a) 全局Lipschitz条件：存在一个常数$K>0$ ，使得对$\forall x_i,y_i\in {ℝ}^d\left(i=1,2\right)$ 有

${\left|v\left(x_1,x_2\right)-v\left(y_1,y_2\right)\right|}^2\le K\left({\left|x_1-x_2\right|}^2+{\left|y_1-y_2\right|}^2\right)$ (2.2)

其中$v=f,g$ 。

(b) 线性增长条件：存在正常数$L$ 使得对所有$x\in {ℝ}^d$ 有

${\left|v\left(x_1,x_2\right)\right|}^2\le L\left(1+{\left|x_1\right|}^2+{\left|x_2\right|}^2\right)$ (2.3)

其中$v=f,g$ 。

2.1. 数值格式

用等步长$h$ 将区间$\left[0,T\right]$ 做均匀划分，其中$h=\frac{T}{J},t_n=nh,n=0,1,\cdots ,J$ ($J\in {\mathbb{Z}}^{+}$ ，${\mathbb{Z}}^{+}$ 为正整数)。将平衡方法应用到方程(2.1)得：

$Y_{n+1}=Y_n+f\left(Y_n,{\tilde{Y}}_n\right)h+g\left(Y_n,{\tilde{Y}}_n\right)\Delta W_n+C\left(Y_n,{\tilde{Y}}_n\right)\left(Y_n-Y_{n+1}\right)$ (2.4)

其中初值$Y_0=x\left(0\right)$ ，$Y_n$ 是解$x\left(t_n\right)$ 在分点$t_n$ 处的数值逼近值。$t_n=nh$ ，$\Delta W_n=W\left(t_{n+1}\right)-W\left(t_n\right)$ ，其中${\tilde{Y}}_n=Y_{\left[\frac{\left[t\right]}{h}\right]}$ 。

此处$d\times d$ 值矩阵函数$C\left(Y_n,{\tilde{Y}}_n\right)$ 用以下的形式表达：

$C\left(Y_n,{\tilde{Y}}_n\right)=C_0\left(Y_n,{\tilde{Y}}_n\right)h+C_1\left(Y_n,{\tilde{Y}}_n\right)\left|\Delta W_n\right|=C_{0n}h+C_{1n}\left|\Delta W_n\right|.$ (2.5)

此处$d\times d$ 值函数$C_0,C_1$ 我们视为控制函数，并且需要满足以下的假设：

若我们认为$C_0$ 和$C_1$ 具有一致有界性。之后假设对任意实数${\alpha }_0\in \left[0,\overline{\alpha }\right],{\alpha }_1\ge 0,\overline{\alpha }\ge h$ 以及$x\in ℝ$ ，矩阵函数$M\left(x\right)=I+{\alpha }_0{C}_0\left(x\right)+{\alpha }_1{C}_1\left(x\right)$ 是可逆的，且满足

$\left|{\left(M\left(x\right)\right)}^{-1}\right|\le H<\infty$ (2.6)

矩阵函数中的$I$ 为单位阵，$H$ 是正常数。

根据以上过程我们把方程(2.4)改写为下列形式

$Y_{n+1}=Y_n+{\left[I+C\left(Y_n,{\tilde{Y}}_n\right)\right]}^{-1}\left[f\left(Y_n,{\tilde{Y}}_n\right)h+g\left(Y_n,{\tilde{Y}}_n\right)\Delta W_n\right]$ (2.7)

我们给出如下阶梯型函数

$\overline{\Delta W}\left(t\right)={\displaystyle {\sum }_{k=0}^{\infty }\left|\Delta W_k\right|I_{\left[kh,\left(k+1\right)h\right)}}\left(t\right),\overline{h}\left(t\right)={\displaystyle {\sum }_{k=0}^{\infty }\left(t_{k+1}-t_k\right)I_{\left[kh,\left(k+1\right)h\right)}}\left(t\right),$

${\overline{Y}}_1\left(t\right)={\displaystyle {\sum }_{k=0}^{\infty }{Y}_k{I}_{\left[kh,\left(k+1\right)h\right)}}\left(t\right),{\overline{Y}}_2\left(t\right)={\displaystyle {\sum }_{k=0}^{\infty }{\tilde{Y}}_k{I}_{\left[kh,\left(k+1\right)h\right)}}\left(t\right),$

由此我们给出如下连续形式的数值解

$\begin{array}{c}{Y}^c\left(t\right)=Y\left(0\right)+{\displaystyle {\int }_0^t{\left(I+C_0\left({\overline{Y}}_1\left(s\right),{\overline{Y}}_2\left(s\right)\right)\overline{h}\left(s\right)+C_1\left({\overline{Y}}_1\left(s\right),{\overline{Y}}_2\left(s\right)\right)\overline{\Delta W}\left(s\right)\right)}^{-1}f\left({\overline{Y}}_1\left(s\right),{\overline{Y}}_2\left(s\right)\right)\text{d}s}\\ +{\displaystyle {\int }_0^t{\left(I+C_0\left({\overline{Y}}_1\left(s\right),{\overline{Y}}_2\left(s\right)\right)\overline{h}\left(s\right)+C_1\left({\overline{Y}}_1\left(s\right),{\overline{Y}}_2\left(s\right)\right)\overline{\Delta W}\left(s\right)\right)}^{-1}g\left({\overline{Y}}_1\left(s\right),{\overline{Y}}_2\left(s\right)\right)\text{d}W\left(s\right)}\end{array}$ (2.8)

其中$Y\left(0\right)=x_0$ 。可以容易得到$Y^c\left(t_n\right)=Y_n$ ，即$Y^c\left(t\right)$ 与$Y_n$ 的节点在整数的情况下是一致的。

在证明数值格式(2.8)的强收敛性用另一种数值格式需要连续性表达，其形式如下：

$\begin{array}{c}Y\left(t\right)=Y\left(0\right)+{\displaystyle {\int }_0^t{\left(I+D\left(s\right)\right)}^{-1}f\left({\overline{Y}}_1\left(s\right),{\overline{Y}}_2\left(s\right)\right)\text{d}s}\\ +{\displaystyle {\int }_0^t{\left(I+D\left(s\right)\right)}^{-1}g\left({\overline{Y}}_1\left(s\right),{\overline{Y}}_2\left(s\right)\right)\text{d}W\left(s\right)}\end{array}$ (2.9)

其中初值$Y\left(0\right)=x_0$ 。$D\left(t\right)=C_0\left(Y_n,{\tilde{Y}}_n\right)\left(t-t_n\right)+C_1\left(Y_n,{\tilde{Y}}_n\right)\left|W\left(t\right)-W\left(t_n\right)\right|$ ，$t\in \left[t_n,t_{n+1}\right)$ 。

2.2. 收敛性证明

引理1. 若全局Lipschitz条件(2.2)和线性增长条件(2.3)成立，则方程(2.1)的解满足：

$\mathbb{E}\left(\underset{0\le t\le T}{\text{sup}}{\left|x\left(t\right)\right|}^2\right)\le Q_1$

其中$Q_1$ 与h是无关常数。

证明. 由(2.1)式和(2.3)式，鞅的性质以及基本不等式${\left|a+b+c\right|}^2\le 3{\left|a\right|}^2+3{\left|b\right|}^2+3{\left|c\right|}^2$ 得

$\begin{array}{c}{\left|x\left(t\right)\right|}^2\le 3{\left|x\left(0\right)\right|}^2+3{\left|{\displaystyle {\int }_0^{t}f}\left(x\left(s\right),x\left(\left[s\right]\right)\right)\text{d}s\right|}^2+3{\left|{\displaystyle {\int }_0^{t}g}\left(x\left(s\right),x\left(\left[s\right]\right)\right)\text{d}W\left(s\right)\right|}^2\\ \le 3{\left|x\left(0\right)\right|}^2+3LT{\displaystyle {\int }_0^t\left(1+{\left|x\left(s\right)\right|}^2+{\left|x\left(\left[s\right]\right)\right|}^2\right)\text{d}s}+3{\left|{\displaystyle {\int }_0^{t}g\left(x\left(s\right),x\left(\left[s\right]\right)\right)\text{d}W\left(s\right)}\right|}^2\end{array}$

$\begin{array}{c}\mathbb{E}\left(\underset{0\le s\le t}{\sup }{\left|x\left(t\right)\right|}^2\right)\le 3\mathbb{E}{\left|x_0\right|}^2+3LT{\displaystyle {\int }_0^t\left(1+{\left|x\left(s\right)\right|}^2+{\left|x\left(\left[s\right]\right)\right|}^2\right)\text{d}s}\\ +3\mathbb{E}\underset{0\le s\le t}{\sup }{\left|{\displaystyle {\int }_0^{s}g\left(x\left(s_1\right),x\left(\left[s_1\right]\right)\right)\text{d}W\left(s_1\right)}\right|}^2\\ \le 3\mathbb{E}{\left|x_0\right|}^2+3L\left(T+4\right){\displaystyle {\int }_0^t\left(1+{\left|x\left(s\right)\right|}^2+{\left|x\left(\left[s\right]\right)\right|}^2\right)\text{d}s}\end{array}$

于是

$\begin{array}{c}1+\mathbb{E}\left(\underset{0\le s\le t}{\sup }{\left|x\left(s\right)\right|}^2\right)\le 1+3\mathbb{E}{\left|x_0\right|}^2+3L\left(T+4\right){\displaystyle {\int }_0^t\left(1+2\mathbb{E}{\left|x\left(s\right)\right|}^2\right)\text{d}s}\\ \le 1+3\mathbb{E}{\left|x_0\right|}^2+6L\left(T+4\right){\displaystyle {\int }_0^t\left(1+\mathbb{E}\left(\underset{0\le r\le s}{\sup }{\left|x\left(r\right)\right|}^2\right)\right)\text{d}s}\end{array}$

由Gronwall不等式得

$1+\mathbb{E}\left(\underset{0\le s\le t}{\sup }{\left|x\left(s\right)\right|}^2\right)\le \left(1+3\mathbb{E}{\left|x_0\right|}^2\right){\text{e}}^{6LT\left(T+4\right)}$

因此

$\mathbb{E}\left(\underset{0\le s\le t}{\sup }{\left|x\left(s\right)\right|}^2\right)\le \left(1+3\mathbb{E}{\left|x_0\right|}^2\right){\text{e}}^{6LT\left(T+4\right)}$

证毕。

引理2. 若条件(2.3)和条件(2.6)成立，则有

$\underset{0\le n\le J}{\max }\mathbb{E}{\left|Y_n\right|}^2\le Q_2$

证明. 根据$\mathbb{E}{\left|\Delta W_n\right|}^2=mh$ ，$\mathbb{E}\Delta W_n=0$ ，由(2.7)得

$\begin{array}{c}\mathbb{E}{\left|Y_{n+1}\right|}^2=\mathbb{E}{\left|Y_n\right|}^2+\mathbb{E}{\left|{\left[I+C\left(Y_n,{\tilde{Y}}_n\right)\right]}^{-1}\left[f\left(Y_n,{\tilde{Y}}_n\right)h+g\left(Y_n,{\tilde{Y}}_n\right)\Delta W_n\right]\right|}^2\\ +2h\mathbb{E}{Y}_n\mathbb{E}\left({\left[I+C\left(Y_n,{\tilde{Y}}_n\right)\right]}^{-1}f\left(Y_n,{\tilde{Y}}_n\right)h\right)\\ \le \mathbb{E}{\left|Y_n\right|}^2+2\left|{\left[I+C\left(Y_n,{\tilde{Y}}_n\right)\right]}^{-2}\right|\cdot h^2\cdot \mathbb{E}{\left|f\left(Y_n,{\tilde{Y}}_n\right)\right|}^2+2{\left[I+C\left(Y_n,{\tilde{Y}}_n\right)\right]}^{-2}\mathbb{E}{\left|g\left(Y_n,{\tilde{Y}}_n\right)\right|}^2\cdot {\left|\Delta W_n\right|}^2\\ +h\mathbb{E}\left({\left|Y_n\right|}^2+\left|{\left[I+C\left(Y_n,{\tilde{Y}}_n\right)\right]}^{-2}\right|\cdot {\left|f\left(Y_n,{\tilde{Y}}_n\right)\right|}^2\right)\\ \le \mathbb{E}{\left|Y_n\right|}^2+2H^2{h}^{2}L\mathbb{E}\left(1+{\left|Y_n\right|}^2+{\left|{\tilde{Y}}_n\right|}^2\right)+2H^{2}mhL\mathbb{E}\left(1+{\left|Y_n\right|}^2+{\left|{\tilde{Y}}_n\right|}^2\right)\\ +h\mathbb{E}{\left|Y_n\right|}^2+H^{2}hL\mathbb{E}\left(1+{\left|Y_n\right|}^2+{\left|{\tilde{Y}}_n\right|}^2\right)\\ \le \left(2H^2{h}^{2}L+2H^{2}mhL+H^{2}hL\right)+\left(1+h+4H^2{h}^{2}L+2H^{2}hL+4H^{2}mhL\right)\underset{0\le i\le J}{\max }\mathbb{E}{\left|Y_i\right|}^2\\ \le h\left(3H^{2}L+2H^{2}mL\right)+\left[1+h\left(1+6H^{2}L+4H^{2}mL\right)\right]\underset{0\le i\le J}{\max }\mathbb{E}{\left|Y_i\right|}^2\end{array}$

设$P=3H^{2}L+2H^{2}mL$ ，$Q=1+6H^{2}L+4H^{2}mL$ ，$S_n=\underset{0\le i\le J}{\max }\mathbb{E}{\left|Y_i\right|}^2$ ，上式可得：

$\begin{array}{c}{S}_{n+1}\le Ph+\left(1+Qh\right)S_n\\ =Ph+Ph\left(1+Qh\right)+Ph{\left(1+Qh\right)}^2+\cdots +{\left(1+Qh\right)}^{n+1}{S}_0\\ =\frac{P}{Q}\left[{\left(1+Qh\right)}^{n+1}-1\right]+{\left(1+Qh\right)}^{n+1}{S}_0\\ \le \frac{P}{Q}\left[{\left(1+Qh\right)}^{J+1}-1\right]+{\left(1+Qh\right)}^{J+1}{S}_0\\ =\frac{P}{Q}\left\{{\left[{\left(1+Qh\right)}^{\frac{1}{Qh}}\right]}^{Qh\left(J+1\right)}\right\}+{\left[{\left(1+Qh\right)}^{\frac{1}{Qh}}\right]}^{Qh\left(J+1\right)}\mathbb{E}{\left|Y_0\right|}^2\end{array}$

当$h\to 0$ 时，收敛于$\frac{P}{Q}\left({\text{e}}^{2QT}-1\right)+{\text{e}}^{2QT}{x}_0^2$ 。

对$k=0,1,2,\cdots ,n$ ，$S_k\le \frac{P}{Q}\left({\text{e}}^{2QT}-1\right)+{\text{e}}^{2QT}{x}_0^2$ 。

当$k=0,-1,-2,\cdots$ ，$Y_k=x_0^2$ 。

取$Q_2=\frac{P}{Q}\left({\text{e}}^{2QT}-1\right)+{\text{e}}^{2QT}{x}_0^2$ ，故$\mathbb{E}{\left|Y_n\right|}^2\le Q_2$ 。

引理3. 若条件(2.3)和条件(2.6)成立，则有

$\mathbb{E}\underset{0\le t\le T}{\text{sup}}{\left|Y\left(t\right)-{\overline{Y}}_1\left(t\right)\right|}^2\le Q_{3}h$

其中$Q_3$ 与h是无关常数。

证明. 设$n_t=\left[t/h\right]\le J\in Z^{+}$ ，由${\left|a+b+c\right|}^2\le 3{\left|a\right|}^2+3{\left|b\right|}^2+3{\left|c\right|}^2$ 得

$\begin{array}{l}\mathbb{E}\underset{0\le t\le T}{\text{sup}}{\left|Y\left(t\right)-{\overline{Y}}_1\left(t\right)\right|}^2\\ \le 3\mathbb{E}\underset{0\le t\le T}{\text{sup}}{\left|{\displaystyle \sum _{i=0}^{n_t-1}{\displaystyle {\int }_{t_i}^{t_{i+1}}{\left(I+D\left(s\right)\right)}^{-1}f\left({\overline{Y}}_1\left(s\right),{\overline{Y}}_2\left(s\right)\right)\text{d}s}}-{\displaystyle \sum _{i=0}^{n_t-1}{\left(I+C\left(Y_i,{\tilde{Y}}_i\right)\right)}^{-1}f\left(Y_i,{\tilde{Y}}_i\right)h}\right|}^2\\ +3\mathbb{E}\underset{0\le t\le T}{\text{sup}}{\left|{\displaystyle \sum _{i=0}^{n_t-1}{\displaystyle {\int }_{t_i}^{t_{i+1}}{\left(I+D\left(s\right)\right)}^{-1}g\left({\overline{Y}}_1\left(s\right),{\overline{Y}}_2\left(s\right)\right)\text{d}W\left(s\right)}}-{\displaystyle \sum _{i=0}^{n_t-1}{\left(I+C\left(Y_i,{\tilde{Y}}_i\right)\right)}^{-1}g\left(Y_i,{\tilde{Y}}_i\right)\Delta W_i}\right|}^2\\ +3\mathbb{E}\underset{0\le t\le T}{\text{sup}}{\left|{\displaystyle {\int }_{t_{n_t}}^t{\left(I+D\left(s\right)\right)}^{-1}f\left({\overline{Y}}_1\left(s\right),{\overline{Y}}_2\left(s\right)\right)\text{d}s}+{\displaystyle {\int }_{t_{n_t}}^t{\left(I+D\left(s\right)\right)}^{-1}g\left({\overline{Y}}_1\left(s\right),{\overline{Y}}_2\left(s\right)\right)\text{d}W\left(s\right)}\right|}^2\end{array}$ (2.10)

注意到$C_0\left(Y_i,{\tilde{Y}}_i\right),C_1\left(Y_i,{\tilde{Y}}_i\right)$ 一致有界，$\exists Q_0$ 正常数使得$\left|C_j\left(Y_i,{\tilde{Y}}_i\right)\right|\le Q_0\left(j=1,2,\cdots ;i=0,1,2,\cdots \right)$ 。由基本不等式${\displaystyle {\sum }_{i=0}^k{\left|a_i\right|}^2}\le \left(k+1\right){\displaystyle {\sum }_{i=0}^k{\left|a_i\right|}^2}$ ，(2.3)，(2.6)，引理2和$\mathbb{E}{\left|\Delta W_n\right|}^2=mh$ ，我们可以推出

(2.11)

同理可得

$\begin{array}{l}\mathbb{E}\underset{0\le t\le T}{\text{sup}}{\left|{\displaystyle \sum _{i=0}^{n_t-1}{\displaystyle {\int }_{t_i}^{t_{i+1}}{\left(I+D\left(s\right)\right)}^{-1}g\left({\overline{Y}}_1\left(s\right),{\overline{Y}}_2\left(s\right)\right)\text{d}W\left(s\right)}}-{\displaystyle \sum _{i=0}^{n_t-1}{\left(I+C\left(Y_i,{\tilde{Y}}_i\right)\right)}^{-1}g\left(Y_i,{\tilde{Y}}_i\right)\Delta W_i}\right|}^2\\ \le 2\mathbb{E}\underset{0\le t\le T}{\text{sup}}{\left|{\displaystyle \sum _{i=0}^{n_t-1}\left[{\displaystyle {\int }_{t_i}^{t_{i+1}}\left(-D\left(s\right)\right){\left(I+D\left(s\right)\right)}^{-1}g\left({\overline{Y}}_1\left(s\right),{\overline{Y}}_2\left(s\right)\right)\text{d}W\left(s\right)}\right]}\right|}^2\\ +2\mathbb{E}\underset{0\le t\le T}{\text{sup}}{\left|{\displaystyle \sum _{i=0}^{n_t-1}\left[{\displaystyle {\int }_{t_i}^{t_{i+1}}g\left({\overline{Y}}_1\left(s\right),{\overline{Y}}_2\left(s\right)\right)\text{d}W\left(s\right)}-{\left(I+C\left(Y_i,{\tilde{Y}}_i\right)\right)}^{-1}g\left(Y_i,{\tilde{Y}}_i\right)\Delta W_i\right]}\right|}^2\end{array}$ (2.12)

注意到对任意的$J\in {\mathbb{Z}}^{+}$ ，离散列

${\zeta }_n=\left\{{\displaystyle \sum _{i=0}^{n-1}\left[{\displaystyle {\int }_{t_i}^{t_{i+1}}\left(-D\left(s\right)\right){\left(I+D\left(s\right)\right)}^{-1}g\left({\overline{Y}}_1\left(s\right),{\overline{Y}}_2\left(s\right)\right)\text{d}W\left(s\right)}\right]}\right\}$

是$\left\{{ℱ}_{t_n}:0\le n\le J\right\}$ 可测的。由Doob鞅不等式，$W\left(t\right)$ 的性质，$\mathbb{E}{\left|\Delta W_n\right|}^2=mh$ ，$\mathbb{E}{\left|\Delta W_n\right|}^4=3m^2{h}^2$ ，条件(2.3)和条件(2.6)，可得

$\begin{array}{l}\mathbb{E}\underset{0\le t\le T}{\text{sup}}{\left|{\displaystyle \sum _{i=0}^{n_t-1}\left[{\displaystyle {\int }_{t_i}^{t_{i+1}}\left(-D\left(s\right)\right)}{\left(I+D\left(s\right)\right)}^{-1}g\left({\overline{Y}}_1\left(s\right),{\overline{Y}}_2\left(s\right)\right)\text{d}W\left(s\right)\right]}\right|}^2\\ \le 4\mathbb{E}{\left|{\displaystyle \sum _{i=0}^{J-1}\left[{\displaystyle {\int }_{t_i}^{t_{i+1}}\left(-D\left(s\right)\right){\left(I+D\left(s\right)\right)}^{-1}g\left({\overline{Y}}_1\left(s\right),{\overline{Y}}_2\left(s\right)\right)\text{d}W\left(s\right)}\right]}\right|}^2\\ =4{\displaystyle \sum _{i=0}^{J-1}\left[{\displaystyle {\int }_{t_i}^{t_{i+1}}\mathbb{E}}{\left|\left(-D\left(s\right)\right){\left(I+D\left(s\right)\right)}^{-1}g\left({\overline{Y}}_1\left(s\right),{\overline{Y}}_2\left(s\right)\right)\text{d}W\left(s\right)\right|}^2\text{d}s\right]}\\ \le 8H^2{Q}_0^{2}LT\left(T+m\right)\left(1+2Q_2\right)h\end{array}$ (2.13)

利用$\mathbb{E}\left[{\displaystyle {\int }_{t_i}^{t_{i+1}}g\left(s\right)\text{d}W\left(s\right)}{\displaystyle {\int }_{t_j}^{t_{j+1}}g\left(s\right)\text{d}W\left(s\right)}\right]=0$ ，$\mathbb{E}\left[g\left(Y_i\right)\Delta W_{i}g\left(Y_j\right)\Delta W_j\right]=0\left(i\ne j\right)$ ，可导出

$\begin{array}{l}\mathbb{E}\underset{0\le t\le T}{\text{sup}}{\left|{\displaystyle \sum _{i=0}^{n_t-1}\left[{\displaystyle {\int }_{t_i}^{t_{i+1}}g\left({\overline{Y}}_1\left(s\right),{\overline{Y}}_2\left(s\right)\right)\text{d}W\left(s\right)}-{\left(I+C\left(Y_i,{\tilde{Y}}_i\right)\right)}^{-1}g\left(Y_i,{\tilde{Y}}_i\right)\Delta W_i\right]}\right|}^2\\ \le {\displaystyle \sum _{i=0}^{J-1}\mathbb{E}{\left|\left(I-{\left(I+C\left(Y_i,{\tilde{Y}}_i\right)\right)}^{-1}\right)g\left(Y_i,{\tilde{Y}}_i\right)\Delta W_i\right|}^2}\\ ={\displaystyle \sum _{i=0}^{J-1}\mathbb{E}{\left|C\left(Y_i,{\tilde{Y}}_i\right){\left(I+C\left(Y_i,{\tilde{Y}}_i\right)\right)}^{-1}g\left(Y_i,{\tilde{Y}}_i\right)\Delta W_i\right|}^2}\\ \le 2H^2{Q}_0^{2}LT\left(mT+3m^2\right)\left(1+2Q_2\right)h\end{array}$ (2.14)

将(2.13)和(2.14)代入(2.12)得

$\begin{array}{l}\mathbb{E}\underset{0\le t\le T}{\text{sup}}{\left|{\displaystyle \sum _{i=0}^{n_t-1}{\displaystyle {\int }_{t_i}^{t_{i+1}}{\left(I+D\left(s\right)\right)}^{-1}g\left({\overline{Y}}_1\left(s\right),{\overline{Y}}_2\left(s\right)\right)\text{d}W\left(s\right)}}-{\displaystyle \sum _{i=0}^{n_t-1}{\left(I+C\left(Y_i,{\tilde{Y}}_i\right)\right)}^{-1}g\left(Y_i,{\tilde{Y}}_i\right)\Delta W_i}\right|}^2\\ \le H^2{Q}_0^{2}LT\left[8\left(T+m\right)+2\left(mT+3m^2\right)\right]\left(1+2Q_2\right)h\end{array}$ (2.15)

不难得到

$\begin{array}{l}\mathbb{E}\underset{0\le t\le T}{\text{sup}}{\left|{\displaystyle {\int }_{t_{n_t}}^t{\left(I+D\left(s\right)\right)}^{-1}f\left({\overline{Y}}_1\left(s\right),{\overline{Y}}_2\left(s\right)\right)\text{d}s}+{\displaystyle {\int }_{t_{n_t}}^t{\left(I+D\left(s\right)\right)}^{-1}g\left({\overline{Y}}_1\left(s\right),{\overline{Y}}_2\left(s\right)\right)\text{d}W\left(s\right)}\right|}^2\\ =\underset{0\le i\le J-1}{\max }\mathbb{E}\underset{t_i\le t\le t_{i+1}}{\text{sup}}{\left|{\displaystyle {\int }_{t_i}^t{\left(I+D\left(s\right)\right)}^{-1}f\left({\overline{Y}}_1\left(s\right),{\overline{Y}}_2\left(s\right)\right)\text{d}s}+{\displaystyle {\int }_{t_{n_t}}^t{\left(I+D\left(s\right)\right)}^{-1}g\left({\overline{Y}}_1\left(s\right),{\overline{Y}}_2\left(s\right)\right)\text{d}W\left(s\right)}\right|}^2\\ \le 2\underset{0\le i\le J-1}{\max }{H}^{2}h{\displaystyle {\int }_{t_i}^{t_{i+1}}\mathbb{E}{\left|f\left({\overline{Y}}_1\left(s\right),{\overline{Y}}_2\left(s\right)\right)\right|}^2\text{d}s}+8\underset{0\le i\le J-1}{\max }{H}^2\mathbb{E}{\left|g\left({\overline{Y}}_1\left(s\right),{\overline{Y}}_2\left(s\right)\right)\right|}^2\text{d}s\\ \le 2H^{2}L\left(1+2Q_2\right)\left(T+4\right)h\end{array}$ (2.16)

将(2.11)，(2.15)，(2.16)带代入(2.10)可得

$\begin{array}{l}\mathbb{E}\underset{0\le t\le T}{\text{sup}}{\left|Y\left(t\right)-{\overline{Y}}_1\left(t\right)\right|}^2\\ \le 6H^{2}L\left(1+2Q_2\right)\left\{\left(T+4\right)+4T^2{Q}_0^2\left(T+m\right)+2T^2{Q}_0^2\left[4\left(T+m\right)+\left(mT+3m^2\right)\right]\right\}\\ =Q_{3}h\end{array}$

引理4. 若线性增长条件(2.3)和条件(2.6)成立，则有

$\mathbb{E}\underset{0\le t\le T}{\text{sup}}{\left|Y\left(\left[t\right]\right)-{\overline{Y}}_2\left(t\right)\right|}^2\le Q_{4}h$

其中$Q_4$ 与h是无关常数。

证明. 对任给$t\in \left[0,T\right]$ ，存在某个$k$ ，使得$t\in \left[kh,\left(k+1\right)h\right)$ ，有

$Y\left(\left[t\right]\right)-{\overline{Y}}_2\left(t\right)=Y\left(\left[t\right]\right)-{\tilde{Y}}_k=Y\left(\left[t\right]\right)-Y_{\left[\frac{\left[t\right]}{h}\right]}=Y\left(\left[t\right]\right)-Y^C{}_{\left(\left[\frac{\left[t\right]}{h}\right]\cdot h\right)}$

其中$t-1\le \left[t\right]\le t,t-1-h\le \left(\frac{\left[t\right]}{h}-1\right)h\le \left[\frac{\left[t\right]}{h}\right]\cdot h\le t$ 。

这里只有一种情况，就是当$\left[t\right]\ge \left[\frac{\left[t\right]}{h}\right]\cdot h\ge 0$ ，

$\left[t\right]-\left[\frac{\left[t\right]}{h}\right]\cdot h\le t-t+1+h=1+h\le \left(\gamma +1\right)h$

这里的$\gamma$ 是大于等于1的常数，且与$Q_4$ 与h无关。

记$n_t=\left[t/h\right]\le J$

$\begin{array}{l}\mathbb{E}\underset{0\le t\le T}{\text{sup}}{\left|Y\left(\left[t\right]\right)-{\overline{Y}}_2\left(t\right)\right|}^2\\ =\mathbb{E}\underset{0\le t\le T}{\text{sup}}{\left|Y\left(\left[t\right]\right)-Y\left(n_t\cdot h\right)+Y\left(n_t\cdot h\right)-Y^C\left(n_t\cdot h\right)\right|}^2\\ \le 2\mathbb{E}\underset{0\le t\le T}{\text{sup}}{\left|Y\left(\left[t\right]\right)-Y\left(n_t\cdot h\right)\right|}^2+2\mathbb{E}\underset{0\le t\le T}{\text{sup}}{\left|Y\left(n_t\cdot h\right)-Y^C\left(n_t\cdot h\right)\right|}^2\\ \le 2\mathbb{E}\underset{0\le t\le T}{\text{sup}}{\left|{\displaystyle {\int }_{n_t\cdot h}^{\left[t\right]}{\left(I+D\left(s\right)\right)}^{-1}f\left({\overline{Y}}_1\left(s\right),{\overline{Y}}_2\left(s\right)\right)\text{d}s}+{\displaystyle {\int }_{n_t\cdot h}^{\left[t\right]}{\left(I+D\left(s\right)\right)}^{-1}g\left({\overline{Y}}_1\left(s\right),{\overline{Y}}_2\left(s\right)\right)\text{d}W\left(s\right)}\right|}^2\\ +4\mathbb{E}\underset{0\le t\le T}{\text{sup}}{\left|{\displaystyle \sum _{i=0}^{n_t-1}{\displaystyle {\int }_{t_i}^{t_{i+1}}{\left(I+D\left(s\right)\right)}^{-1}f\left({\overline{Y}}_1\left(s\right),{\overline{Y}}_2\left(s\right)\right)\text{d}s}}-{\displaystyle \sum _{i=0}^{n_t-1}{\left(I+C\left(Y_i,{\tilde{Y}}_i\right)\right)}^{-1}f\left(Y_i,{\tilde{Y}}_i\right)h}\right|}^2\\ +4\mathbb{E}\underset{0\le t\le T}{\text{sup}}{\left|{\displaystyle \sum _{i=0}^{n_t-1}{\displaystyle {\int }_{t_i}^{t_{i+1}}{\left(I+D\left(s\right)\right)}^{-1}g\left({\overline{Y}}_1\left(s\right),{\overline{Y}}_2\left(s\right)\right)\text{d}W\left(s\right)}}-{\displaystyle \sum _{i=0}^{n_t-1}{\left(I+C\left(Y_i,{\tilde{Y}}_i\right)\right)}^{-1}g\left(Y_i,{\tilde{Y}}_i\right)\Delta W_i}\right|}^2\end{array}$ (2.17)

可以推出

$\begin{array}{l}\mathbb{E}\underset{0\le t\le T}{\text{sup}}{\left|{\displaystyle {\int }_{n_t\cdot h}^{\left[t\right]}{\left(I+D\left(s\right)\right)}^{-1}f\left({\overline{Y}}_1\left(s\right),{\overline{Y}}_2\left(s\right)\right)\text{d}s}+{\displaystyle {\int }_{n_t\cdot h}^{\left[t\right]}{\left(I+D\left(s\right)\right)}^{-1}g\left({\overline{Y}}_1\left(s\right),{\overline{Y}}_2\left(s\right)\right)\text{d}W\left(s\right)}\right|}^2\\ =\underset{0\le i\le J-1}{\max }\mathbb{E}\underset{t-1\le \left[t\right]\le t}{\text{sup}}{\left|{\displaystyle {\int }_{n_t\cdot h}^{\left[t\right]}{\left(I+D\left(s\right)\right)}^{-1}f\left({\overline{Y}}_1\left(s\right),{\overline{Y}}_2\left(s\right)\right)\text{d}s}+{\displaystyle {\int }_{n_t\cdot h}^{\left[t\right]}{\left(I+D\left(s\right)\right)}^{-1}g\left({\overline{Y}}_1\left(s\right),{\overline{Y}}_2\left(s\right)\right)\text{d}W\left(s\right)}\right|}^2\\ \le 2H^2\left(\gamma +1\right)h{\displaystyle {\int }_{n_t\cdot h}^t\mathbb{E}{\left|f\left({\overline{Y}}_1\left(s\right),{\overline{Y}}_2\left(s\right)\right)\right|}^2\text{d}s}+8H^2\left(\gamma +1\right)h{\displaystyle {\int }_{n_t\cdot h}^t\mathbb{E}{\left|g\left({\overline{Y}}_1\left(s\right),{\overline{Y}}_2\left(s\right)\right)\right|}^2\text{d}s}\\ \le 2H^2\left(\gamma +1\right)hL{\displaystyle {\int }_{n_t\cdot h}^t\left(1+\mathbb{E}{\left|{\overline{Y}}_1\left(s\right)\right|}^2+\mathbb{E}{\left|{\overline{Y}}_2\left(s\right)\right|}^2\right)\text{d}s}+8H^{2}L{\displaystyle {\int }_{n_t\cdot h}^t\left(1+\mathbb{E}{\left|{\overline{Y}}_1\left(s\right)\right|}^2+\mathbb{E}{\left|{\overline{Y}}_2\left(s\right)\right|}^2\right)\text{d}s}\\ \le 2H^2\left(\gamma +1\right)hL\left(1+2Q_2\right)\left[\left(\gamma +1\right)h+4\right]h\end{array}$ (2.18)

由(2.11)得

$\begin{array}{l}\mathbb{E}\underset{0\le t\le T}{\text{sup}}{\left|{\displaystyle \sum _{i=0}^{n_t-1}\left[{\displaystyle {\int }_{t_i}^{t_{i+1}}{\left(I+D\left(s\right)\right)}^{-1}f\left({\overline{Y}}_1\left(s\right),{\overline{Y}}_2\left(s\right)\right)\text{d}s}-{\left(I+C\left(Y_i,{\tilde{Y}}_i\right)\right)}^{-1}f\left(Y_i,{\tilde{Y}}_i\right)h\right]}\right|}^2\\ \le 8H^2{Q}_0^{2}L{T}^2\left(T+m\right)\left(1+2Q_2\right)h\end{array}$ (2.19)

由(2.15)得

$\begin{array}{l}\mathbb{E}\underset{0\le t\le T}{\text{sup}}{\left|{\displaystyle \sum _{i=0}^{n_t-1}{\displaystyle {\int }_{t_i}^{t_{i+1}}{\left(I+D\left(s\right)\right)}^{-1}g\left({\overline{Y}}_1\left(s\right),{\overline{Y}}_2\left(s\right)\right)\text{d}W\left(s\right)}}-{\displaystyle \sum _{i=0}^{n_t-1}{\left(I+C\left(Y_i,{\tilde{Y}}_i\right)\right)}^{-1}g\left(Y_i,{\tilde{Y}}_i\right)\Delta W_i}\right|}^2\\ \le H^2{Q}_0^{2}LT\left[8\left(T+m\right)+2\left(mT+3m^2\right)\right]\left(1+2Q_2\right)h\end{array}$ (2.20)

将(2.18)，(2.19)，(2.20)代入(2.17)可得

$\mathbb{E}\underset{0\le t\le T}{\text{sup}}{\left|Y\left(\left[t\right]\right)-{\overline{Y}}_2\left(t\right)\right|}^2\le Q_{4}h$

引理5. 若条件(2.2)~(2.6)成立，则数值解(2.9)满足

$\mathbb{E}\underset{0\le t\le T}{\text{sup}}{\left|Y\left(t\right)-x\left(t\right)\right|}^2\le Q_{5}h$ (2.21)

其中$Q_5$ 与h是无关常数。

证明. 对于任意的$0\le t_1\le T$ ，有

$\begin{array}{l}\mathbb{E}\underset{0\le t\le t_1}{\text{sup}}{\left|Y\left(t\right)-x\left(t\right)\right|}^2\\ \le 4\mathbb{E}\underset{0\le t\le t_1}{\text{sup}}{\left|{\displaystyle {\int }_0^t{\left(I+D\left(s\right)\right)}^{-1}\left[f\left({\overline{Y}}_1\left(s\right),{\overline{Y}}_2\left(s\right)\right)-f\left(x\left(s\right),x\left(\left[s\right]\right)\right)\right]\text{d}s}\right|}^2\\ +4\mathbb{E}\underset{0\le t\le t_1}{\text{sup}}{\left|{\displaystyle {\int }_0^t{\left(I+D\left(s\right)\right)}^{-1}\left[g\left({\overline{Y}}_1\left(s\right),{\overline{Y}}_2\left(s\right)\right)-g\left(x\left(s\right),x\left(\left[s\right]\right)\right)\right]\text{d}W\left(s\right)}\right|}^2\\ +4\mathbb{E}\underset{0\le t\le t_1}{\text{sup}}{\left|{\displaystyle {\int }_0^t\left[{\left(I+D\left(s\right)\right)}^{-1}-I\right]f\left(x\left(s\right),x\left(\left[s\right]\right)\right)\text{d}s}\right|}^2\\ +4\mathbb{E}\underset{0\le t\le t_1}{\text{sup}}{\left|{\displaystyle {\int }_0^t\left[{\left(I+D\left(s\right)\right)}^{-1}-I\right]g\left(x\left(s\right),x\left(\left[s\right]\right)\right)\text{d}W\left(s\right)}\right|}^2\end{array}$ (2.22)

第一项：

$\begin{array}{l}\mathbb{E}\underset{0\le t\le t_1}{\text{sup}}{\left|{\displaystyle {\int }_0^t{\left(I+D\left(s\right)\right)}^{-1}\left[f\left({\overline{Y}}_1\left(s\right),{\overline{Y}}_2\left(s\right)\right)-f\left(x\left(s\right),x\left(\left[s\right]\right)\right)\right]\text{d}s}\right|}^2\\ \le H^{2}T\mathbb{E}{\displaystyle {\int }_0^{t_1}{\left|f\left({\overline{Y}}_1\left(s\right),{\overline{Y}}_2\left(s\right)\right)-f\left(x\left(s\right),x\left(\left[s\right]\right)\right)\right|}^2\text{d}s}\\ \le H^{2}TK\mathbb{E}{\displaystyle {\int }_0^{t_1}\left({\left|{\overline{Y}}_1\left(s\right)-x\left(s\right)\right|}^2+{\left|{\overline{Y}}_2\left(s\right)-x\left(\left[s\right]\right)\right|}^2\right)\text{d}s}\\ =H^{2}TK{\displaystyle {\int }_0^{t_1}\mathbb{E}}{\left|{\overline{Y}}_1\left(s\right)-Y\left(s\right)+Y\left(s\right)-x\left(s\right)\right|}^2+{\left|{\overline{Y}}_2\left(s\right)-Y\left(\left[s\right]\right)+Y\left(\left[s\right]\right)-x\left(\left[s\right]\right)\right|}^2\text{d}s\\ \le 2H^{2}TK{\displaystyle {\int }_0^{t_1}\mathbb{E}}{\left|{\overline{Y}}_1\left(s\right)-Y\left(s\right)\right|}^2+\mathbb{E}{\left|Y\left(s\right)-x\left(s\right)\right|}^2+\mathbb{E}{\left|{\overline{Y}}_2\left(s\right)-Y\left(\left[s\right]\right)\right|}^2+\mathbb{E}{\left|Y\left(\left[s\right]\right)-x\left(\left[s\right]\right)\right|}^2\text{d}s\\ \le 2H^2{T}^{2}K\left(Q_3+Q_4\right)h+4H^{2}TK\mathbb{E}{\displaystyle {\int }_0^{t_1}\underset{0\le \delta \le s}{\text{sup}}{\left|Y\left(\delta \right)-x\left(\delta \right)\right|}^2\text{d}s}\end{array}$ (2.23)

第二项：类似(2.22)得

$\begin{array}{l}\mathbb{E}\underset{0\le t\le t_1}{\text{sup}}{\left|{\displaystyle {\int }_0^t{\left(I+D\left(s\right)\right)}^{-1}\left[g\left({\overline{Y}}_1\left(s\right),{\overline{Y}}_2\left(s\right)\right)-g\left(x\left(s\right),x\left(\left[s\right]\right)\right)\right]\text{d}W\left(s\right)}\right|}^2\\ \le 8H^{2}TK\left(Q_3+Q_4\right)h+16H^{2}TK\mathbb{E}{\displaystyle {\int }_0^{t_1}\underset{0\le \delta \le s}{\text{sup}}{\left|Y\left(\delta \right)-x\left(\delta \right)\right|}^2\text{d}s}\end{array}$ (2.24)

第三项：

$\begin{array}{l}\mathbb{E}\underset{0\le t\le t_1}{\text{sup}}{\left|{\displaystyle {\int }_0^t\left[{\left(I+D\left(s\right)\right)}^{-1}-I\right]f\left(x\left(s\right),x\left(\left[s\right]\right)\right)\text{d}s}\right|}^2\\ \le 2H^{2}T{Q}_0^2\left(T+m\right)h\mathbb{E}{\displaystyle {\int }_0^{t_1}{\left|f\left(x\left(s\right),x\left(\left[s\right]\right)\right)\right|}^2\text{d}s}\\ \le 2H^{2}T{Q}_0^2\left(T+m\right)h\mathbb{E}{\displaystyle {\int }_0^{t_1}\left(1+{\left|x\left(s\right)\right|}^2+{\left|x\left(\left[s\right]\right)\right|}^2\right)\text{d}s}\\ \le 2H^2{T}^{2}L{Q}_0^2\left(T+m\right)\left(1+2Q_1\right)h\end{array}$ (2.25)

第四项：类似于(2.24)得

$\begin{array}{l}\mathbb{E}\underset{0\le t\le t_1}{\text{sup}}{\left|{\displaystyle {\int }_0^t\left[{\left(I+D\left(s\right)\right)}^{-1}-I\right]g\left(x\left(s\right),x\left(\left[s\right]\right)\right)\text{d}W\left(s\right)}\right|}^2\\ \le 8H^2{T}^{2}L{Q}_0^2\left(T+m\right)\left(1+2Q_1\right)h\end{array}$ (2.26)

把(2.23)，(2.24)，(2.25)和(2.26)代入(2.22)，由Gronwall不等式得到

$\mathbb{E}\underset{0\le t\le t_1}{\text{sup}}{\left|Y\left(t\right)-x\left(t\right)\right|}^2\le Q_{5}h$

下面给出本章节主要定理：

定理1. 若条件(2.2)~(2.6)成立，则数值解(2.9)满足

$\mathbb{E}\underset{0\le t\le T}{\text{sup}}{\left|Y^C\left(t\right)-x\left(t\right)\right|}^2\le Q_{6}h$ (2.27)

其中$Q_6$ 与h是无关常数。

证明. 由(2.3)，(2.6)，(2.9)以及引理2和引理3得

$\begin{array}{l}\mathbb{E}\underset{0\le t\le T}{\text{sup}}{\left|Y\left(t\right)-Y^C\left(t\right)\right|}^2\\ =\mathbb{E}\underset{0\le t\le T}{\text{sup}}{\left|Y\left(t\right)-{\overline{Y}}_1\left(t\right)+{\overline{Y}}_1\left(t\right)-Y^C\left(t\right)\right|}^2\\ \le 2\mathbb{E}\underset{0\le t\le T}{\text{sup}}{\left|Y\left(t\right)-{\overline{Y}}_1\left(t\right)\right|}^2\\ +2\mathbb{E}\underset{0\le t\le T}{\text{sup}}|{\left[I+C_0\left(Y_{n_t},{\tilde{Y}}_{n_t}\right)h+C_1\left(Y_{n_t},{\tilde{Y}}_{n_t}\right)\left|\Delta W_{n_t}\right|\right]}^{-1}f\left(Y_{n_t},{\tilde{Y}}_{n_t}\right)\left(t-t_{n_t}\right)\\ +{{\left[I+C_0\left(Y_{n_t},{\tilde{Y}}_{n_t}\right)h+C_1\left(Y_{n_t},{\tilde{Y}}_{n_t}\right)\left|\Delta W_{n_t}\right|\right]}^{-1}g\left(Y_{n_t},{\tilde{Y}}_{n_t}\right)\left(W\left(t\right)-W\left(t_{n_t}\right)\right)|}^2\\ \le 2Q_{3}h+4H^2\mathbb{E}\underset{0\le t\le T}{\text{sup}}{\left|\left(t-t_{n_t}\right)f\left(Y_{n_t},{\tilde{Y}}_{n_t}\right)\right|}^2+4H^2\mathbb{E}\underset{0\le t\le T}{\text{sup}}{\left|\left(W\left(t\right)-W\left(t_{n_t}\right)\right)g\left(Y_{n_t},{\tilde{Y}}_{n_t}\right)\right|}^2\\ \le 2Q_{3}h+4H^2{h}^2\mathbb{E}\underset{0\le t\le T}{\text{sup}}{\left|f\left(Y_{n_t},{\tilde{Y}}_{n_t}\right)\right|}^2+4H^{2}mh\mathbb{E}\underset{0\le t\le T}{\text{sup}}{\left|g\left(Y_{n_t},{\tilde{Y}}_{n_t}\right)\right|}^2\\ \le 2Q_{3}h+4H^2{h}^{2}L\left(1+2Q_2\right)+4H^{2}mhL\left(1+2Q_2\right)\\ \le \left[2Q_3+4H^{2}TL\left(1+2Q_2\right)+4H^{2}mL\left(1+2Q_2\right)\right]h\end{array}$

所以

$\begin{array}{l}\mathbb{E}\underset{0\le t\le T}{\text{sup}}{\left|Y^C\left(t\right)-x\left(t\right)\right|}^2\\ =\mathbb{E}\underset{0\le t\le T}{\text{sup}}{\left|Y^C\left(t\right)-Y\left(t\right)+Y\left(t\right)-x\left(t\right)\right|}^2\\ \le 2\mathbb{E}\underset{0\le t\le T}{\text{sup}}{\left|Y^C\left(t\right)-Y\left(t\right)\right|}^2+2\mathbb{E}\underset{0\le t\le T}{\text{sup}}{\left|Y\left(t\right)-x\left(t\right)\right|}^2\\ \le 2\left[2Q_3+4H^{2}TL\left(1+2Q_2\right)+4H^{2}mL\left(1+2Q_2\right)\right]h+2Q_{5}h\\ \le Q_{6}h\end{array}$

证毕。

3. 平衡方法的均方稳定性

对于方程(1.1)，我们考虑其线性标量的形式：

$\{\begin{array}{l}\text{d}x\left(t\right)=\left[a_{1}x\left(t\right)+a_{2}x\left(\left[t\right]\right)\right]\text{d}t+\left[b_{1}x\left(t\right)+b_{2}x\left(\left[t\right]\right)\right]\text{d}W\left(t\right)\\ x\left(0\right)=x_0\end{array}$ (3.1)

引理6. [16]若方程(3.1)满足如下条件：

$2a_1+b_1^2+b_2^2+2\left|a_2+b_1{b}_2\right|<0$ (3.2)

成立，则方程(3.1)的解析解为均方稳定的。

3.1. 强平衡隐式方法的均方稳定性

定义1. 在给定的序列$\left\{a_1,a_2,b_1,b_2\right\}$ 以及步长h，对于任意$Y_0$ ，将平衡隐式方法应用于方程(3.1)得到的数值解$Y_n$ 满足

$\underset{n\to \infty }{\lim }\mathbb{E}{\left(Y_n\right)}^2=0$ ,

则称平衡隐式方法为均方稳定。

定理2. 若条件(2.6)和(3.2)成立，则数值方法是均方稳定的。

证明. (2.4)应用于(3.1)得

$Y_{n+1}=Y_n+\left(a_1{Y}_n+a_2{\tilde{Y}}_n\right)h+\left(b_1{Y}_n+b_2{\tilde{Y}}_n\right)\Delta W_n+C_n\left(Y_n-Y_{n+1}\right)$ (3.3)

其中$C_n=C_{0n}h+C_{1n}\left|\Delta W_n\right|$ 。我们只考虑$C_{0n},C_{1n}\in ℝ\left(n=0,1,\cdots \right)$ ，将(3.3)式改写为以下形式：

$Y_{n+1}=\left[1+\left(a_{1}h+b_1\Delta W_n\right){\left(1+C_n\right)}^{-1}\right]Y_n+\left[\left(a_{2}h+b_2\Delta W_n\right){\left(1+C_n\right)}^{-1}\right]{\tilde{Y}}_n$ (3.4)

对(3.4)式两边平方并取期望得：

$\begin{array}{c}\mathbb{E}{Y}_{n+1}^2=\mathbb{E}{Y}_n^2\mathbb{E}{\left[1+\left(a_{1}h+b_1\Delta W_n\right){\left(1+C_n\right)}^{-1}\right]}^2+\mathbb{E}{\tilde{Y}}_n^2\mathbb{E}{\left[\left(a_{2}h+b_2\Delta W_n\right){\left(1+C_n\right)}^{-1}\right]}^2\\ +2\mathbb{E}{Y}_n\mathbb{E}{\tilde{Y}}_n\mathbb{E}\left\{\left[1+\left(a_{1}h+b_1\Delta W_n\right){\left(1+C_n\right)}^{-1}\right]\cdot \left[\left(a_{2}h+b_2\Delta W_n\right){\left(1+C_n\right)}^{-1}\right]\right\}\\ =\mathbb{E}{Y}_n^2\mathbb{E}{\alpha }_1^2+\mathbb{E}{\tilde{Y}}_n^2\mathbb{E}{\alpha }_2^2+2\mathbb{E}{Y}_n{\tilde{Y}}_n\mathbb{E}\left({\alpha }_1{\alpha }_2\right)\\ \le \left(\mathbb{E}{\alpha }_1^2+\mathbb{E}{\alpha }_2^2+2\left|\mathbb{E}\left({\alpha }_1{\alpha }_2\right)\right|\right){\max }_{0\le i\le n}\mathbb{E}{Y}_i^2\end{array}$ (3.5)

其中${\alpha }_1=1+\left(a_{1}h+b_1\Delta W_n\right){\left(1+C_n\right)}^{-1}$ ， ${\alpha }_2=\left(a_{2}h+b_2\Delta W_n\right){\left(1+C_n\right)}^{-1}$ 于是

$\begin{array}{l}\mathbb{E}{\alpha }_1^2+\mathbb{E}{\alpha }_2^2+2\left|\mathbb{E}\left({\alpha }_1{\alpha }_2\right)\right|\\ =1+2a_{1}h\mathbb{E}{\left(1+C_n\right)}^{-1}+2b_2\mathbb{E}\left[\Delta W_n{\left(1+C_n\right)}^{-1}\right]+\left(a_1^2{h}^2+a_2^2{h}^2\right)\mathbb{E}{\left(1+C_n\right)}^{-2}\\ +\left(b_1^2+b_2^2\right)\mathbb{E}\left[\Delta W_n^2{\left(1+C_n\right)}^{-2}\right]+\left(2a_1{b}_{1}h+2a_2{b}_{2}h\right)\mathbb{E}\left[\Delta W_n{\left(1+C_n\right)}^{-2}\right]\\ +2|a_{2}h\mathbb{E}{\left(1+C_n\right)}^{-1}+b_2\mathbb{E}\left[\Delta W_n{\left(1+C_n\right)}^{-1}\right]+a_1{a}_2{h}^2\mathbb{E}{\left(1+C_n\right)}^{-2}\\ +\left(a_1{b}_{2}h+a_2{b}_{1}h\right)\mathbb{E}\left[\Delta W_n{\left(1+C_n\right)}^{-2}\right]+b_1{b}_2+\mathbb{E}\left[\Delta W_n^2{\left(1+C_n\right)}^{-2}\right]|\end{array}$ (3.6)

设$\xi$ 为服从标准正态分布随机变量，则

$\begin{array}{l}\mathbb{E}\left[\Delta W_n{\left(1+C_n\right)}^{-1}\right]\\ =\mathbb{E}\left(\sqrt{h}\xi {\left(1+C_{0n}h+C_{1n}\sqrt{h}\left|\xi \right|\right)}^{-1}\right)\\ =\frac{\sqrt{h}}{\sqrt{2\pi }}{\displaystyle {\int }_{-\infty }^{+\infty }{\text{e}}^{-\frac{x^2}{2}}x{\left(1+C_{0n}h+C_{1n}\sqrt{h}\left|x\right|\right)}^{-1}\text{d}x}\\ =0\end{array}$ (3.7)

类似可得

$E\left[\Delta W_n{\left(1+C_n\right)}^{-2}\right]=0$ (3.8)

而

$\begin{array}{l}\left(b_1^2+b_2^2\right)\mathbb{E}\left[\Delta W_n^2{\left(1+C_n\right)}^{-2}\right]\\ =\left(b_1^2+b_2^2\right)\mathbb{E}\left[\Delta W_n^2\left(1-\frac{2C_n}{1+C_n}+\frac{C_n^2}{{\left(1+C_n\right)}^2}\right)\right]\\ =\left(b_1^2+b_2^2\right)h-2\left(b_1^2+b_2^2\right)\mathbb{E}\left[\Delta W_n^2\frac{C_n}{1+C_n}\right]+\left(b_1^2+b_2^2\right)\mathbb{E}\left[\Delta W_n^2\frac{C_n^2}{{\left(1+C_n\right)}^2}\right].\end{array}$ (3.9)

其中

$\begin{array}{l}\left|-2\left(b_1^2+b_2^2\right)\mathbb{E}\left[\Delta W_n^2\frac{C_n}{1+C_n}\right]\right|\\ \le 2\left(b_1^2+b_2^2\right)\mathbb{E}\left|\Delta W_n^2\frac{C_n}{1+C_n}\right|\\ \le 2\left(b_1^2+b_2^2\right)H\mathbb{E}\left[\Delta W_n^2\left|C_{0n}h+C_{1n}\left|\Delta W_n\right|\right|\right]\\ =O\left(h^{\frac{3}{2}}\right),\end{array}$ (3.10)

$\begin{array}{l}\left(b_1^2+b_2^2\right)\mathbb{E}\left[\Delta W_n^2\frac{C_n^2}{{\left(1+C_n\right)}^2}\right]\\ \le \left(b_1^2+b_2^2\right)H^2\mathbb{E}\left[\Delta W_n^2{\left|C_{0n}h+C_{1n}\left|\Delta W_n\right|\right|}^2\right]\\ =O\left(h^2\right)\end{array}$ (3.11)

将(3.10)，(3.11)代入(3.9)得

$\left(b_1^2+b_2^2\right)\mathbb{E}\left[\Delta W_n^2{\left(1+C_n\right)}^{-2}\right]=\left(b_1^2+b_2^2\right)h+O\left(h\right)$ (3.12)

同理可以得到

$\mathbb{E}\left[{\left(1+C_n\right)}^{-2}\right]=\mathbb{E}\left(1-\frac{2C_n}{1+C_n}+\frac{C_n^2}{{\left(1+C_n\right)}^2}\right)=1+O\left(h^{\frac{1}{2}}\right)$ (3.13)

$\mathbb{E}\left[{\left(1+C_n\right)}^{-1}\right]=\mathbb{E}\left[1-\frac{C_n}{1+C_n}\right]=1+O\left(h^{\frac{1}{2}}\right)$ (3.14)

现在将(3.7)，(3.8)，(3.12)，(3.13)，(3.14)代入(3.6)得

$\begin{array}{l}\mathbb{E}{\alpha }_1^2+\mathbb{E}{\alpha }_2^2+2\left|\mathbb{E}\left({\alpha }_1{\alpha }_2\right)\right|\\ =1+2a_{1}h+\left(a_1^2{h}^2+a_2^2{h}^2\right)+\left(b_1^2+b_2^2\right)h+2\left|a_{2}h+a_1{a}_2{h}^2+b_1{b}_{2}h\right|+o\left(h\right)\\ \le 1+\left[2a_1+b_1^2+b_2^2+2\left|a_2+b_1{b}_2\right|\right]h+o\left(h\right)\end{array}$ (3.15)

当$h\to 0$ 时，可得结论强平衡方法均方稳定。

3.2. 弱平衡隐式方法的均方稳定性

考虑弱平衡方法

$Y_{n+1}=Y_n+\left(a_1{Y}_n+a_2{\tilde{Y}}_n\right)h+\left(b_1{Y}_n+b_2{\tilde{Y}}_n\right)\overline{\Delta W_n}+\overline{C_n}\left(Y_n-Y_{n+1}\right)$ (3.16)

其中$Y_0=x\left(0\right)$ ，$\overline{C_n}=C_{0n}h+C_{1n}\left|\overline{\Delta W_n}\right|$ 。$P\left(\overline{\Delta W_n}=\sqrt{h}\right)=P\left(\overline{\Delta W_n}-\sqrt{h}\right)=\frac{1}{2}$ ，不难证明$\mathbb{E}\left(\overline{\Delta W_n}\right)=0$ ，$\mathbb{E}\left({\overline{\Delta W_n}}^2\right)=h$ 。

定理3. 条件(2.6)和(3.2)成立，则弱平衡隐式方法(3.16)是均方稳定的。

证明. 类似于定理2，可得

$\mathbb{E}{Y}_{n+1}^2\le \left(\mathbb{E}{\overline{{\alpha }_1}}^2+\mathbb{E}{\overline{{\alpha }_2}}^2+2\left|\mathbb{E}\left(\overline{{\alpha }_1}\overline{{\alpha }_2}\right)\right|\right){\max }_{0\le i\le n}\mathbb{E}{Y}_i^2$

其中$\overline{{\alpha }_1}=1+\left(a_{1}h+b_1\overline{\Delta W_n}\right){\left(1+\overline{C_n}\right)}^{-1}$ ，$\overline{{\alpha }_2}=\left(a_{2}h+b_2\overline{\Delta W_n}\right){\left(1+\overline{C_n}\right)}^{-1}$ 。

$\begin{array}{l}\mathbb{E}{\overline{{\alpha }_1}}^2+\mathbb{E}{\overline{{\alpha }_2}}^2+2\left|\mathbb{E}\left(\overline{{\alpha }_1}\overline{{\alpha }_2}\right)\right|\\ =1+2a_{1}h\mathbb{E}{\left(1+\overline{C_n}\right)}^{-1}+2b_2\mathbb{E}\left[\overline{\Delta W_n}{\left(1+\overline{C_n}\right)}^{-1}\right]+\left(a_1^2{h}^2+a_2^2{h}^2\right)\mathbb{E}{\left(1+\overline{C_n}\right)}^{-2}\\ +\left(b_1^2+b_2^2\right)\mathbb{E}\left[{\overline{\Delta W_n}}^2{\left(1+\overline{C_n}\right)}^{-2}\right]+\left(2a_1{b}_{1}h+2a_2{b}_{2}h\right)\mathbb{E}\left[\overline{\Delta W_n}{\left(1+\overline{C_n}\right)}^{-2}\right]\\ +2|a_{2}h\mathbb{E}{\left(1+\overline{C_n}\right)}^{-1}+b_2\mathbb{E}\left[\overline{\Delta W_n}{\left(1+\overline{C_n}\right)}^{-1}\right]+a_1{a}_2{h}^2\mathbb{E}{\left(1+\overline{C_n}\right)}^{-2}\\ +\left(a_1{b}_{2}h+a_2{b}_{1}h\right)\mathbb{E}\left[\overline{\Delta W_n}{\left(1+\overline{C_n}\right)}^{-2}\right]+b_1{b}_2+\mathbb{E}\left[{\overline{\Delta W_n}}^2{\left(1+\overline{C_n}\right)}^{-2}\right]|\end{array}$ (3.17)

因$P\left(\overline{\Delta W_n}=\sqrt{h}\right)=P\left(\overline{\Delta W_n}-\sqrt{h}\right)=\frac{1}{2}$ ，可以推出

$\begin{array}{l}\mathbb{E}\left[\overline{\Delta W_n}{\left(1+\overline{C_n}\right)}^{-1}\right]\\ =\mathbb{E}\left(\overline{\Delta W_n}{\left(1+C_{0n}h+C_{1n}\left|\overline{\Delta W_n}\right|\right)}^{-1}\right)\\ =\frac{1}{2}{\left[\sqrt{h}\left(1+C_{0n}h+C_{1n}\sqrt{h}\right)\right]}^{-1}+\frac{1}{2}{\left[\sqrt{h}\left(1+C_{0n}h+C_{1n}\sqrt{h}\right)\right]}^{-1}\\ =0\end{array}$ (3.18)

类似的有

$\mathbb{E}\left[\overline{\Delta W_n}{\left(1+\overline{C_n}\right)}^{-2}\right]=0$ (3.19)

注意到

$\begin{array}{l}\mathbb{E}\left[{\overline{\Delta W_n}}^2{\left(1+\overline{C_n}\right)}^{-2}\right]\\ =\mathbb{E}{\left({\overline{\Delta W_n}}^2\left(1+C_{0n}h+C_{1n}\left|\overline{\Delta W_n}\right|\right)\right)}^{-2}\\ =h{\left(1+C_{0n}h+C_{1n}\sqrt{h}\right)}^{-2}\end{array}$

那么一定存在$\widehat{h^{*}}=\min \left\{1/\left({\left|C_0\right|}^2+{\left|C_1\right|}^2\right),1\right\}$ 。对任意的$h\in \left(0,\widehat{h^{*}}\right)$ ，使得$\left|C_{0n}h+C_{1n}\sqrt{h}\right|<1$ 。由泰勒展开得

${\left(1+C_{0n}h+C_{1n}\sqrt{h}\right)}^{-2}=1-2C_{1n}\sqrt{h}+o\left(\sqrt{h}\right).$

因此可以得到

$\mathbb{E}\left[{\overline{\Delta W_n}}^2{\left(1+\overline{C_n}\right)}^{-2}\right]=h\left(1-2C_{1n}\sqrt{h}+o\left(\sqrt{h}\right)\right)=h+o\left(h\right).$ (3.20)

类似可以得到

$\mathbb{E}{\left(1+\overline{C_n}\right)}^{-2}=\mathbb{E}{\left(1+\overline{C_n}\right)}^{-1}=1+O\left(\sqrt{h}\right).$ (3.21)

将(3.18)，(3.19)，(3.20)和(3.21)代入(3.17)得

$\begin{array}{l}\mathbb{E}{\overline{{\alpha }_1}}^2+\mathbb{E}{\overline{{\alpha }_2}}^2+2\left|\mathbb{E}\left(\overline{{\alpha }_1}\overline{{\alpha }_2}\right)\right|\\ =1+2a_{1}h+\left(a_1^2{h}^2+a_2^2{h}^2\right)+\left(b_1^2+b_2^2\right)h+2\left|a_{2}h+a_1{a}_2{h}^2+b_1{b}_{2}h\right|+o\left(h\right)\end{array}$

同理证得弱平衡方法均方稳定。

4. 数值算例

首先，验证平衡方法的收敛性，我们考虑其线性标量的形式：

$\text{d}x\left(t\right)=\left[a_{1}x\left(t\right)+a_{2}x\left(\left[t\right]\right)\right]\text{d}t+\left[b_{1}x\left(t\right)+b_{2}x\left(\left[t\right]\right)\right]\text{d}W\left(t\right)$ (4.1)

其中初值$x\left(0\right)=1$ 。

选取如下三组系数：

算例1. $a_1=1;a_2=0.1;b_1=0.3;b_1=0.2$ 。

算例2. $a_1=-4;a_2=0.3;b_1=0.3;b_2=0.2$ 。

算例3. $a_1=-2;a_2=0.1;b_1=0.4;b_2=0.1$ 。

选择算例1、2、3作为方程(4.1)的系数，同时为了方便，令$C_{0j}=C_{1j}=1$ $\left(j=0,1,\cdots \right)$ ，并选择步长$h=8\Delta t,16\Delta t,32\Delta t,64\Delta t$ 。$\Delta t=h$ ，均方误差${\epsilon }^r={\epsilon }_{2^{3+r}\Delta t},r=0,1,2,3$ 均在终点时刻$T=1$ 处的由以下方式估计，选取2000个样本来模拟。

${\epsilon }^r\cong \frac{1}{2000}{\displaystyle {\sum }_{j=1}^{2000}{\left|x_{r,T}\left({\omega }_j\right)-Y_{r,i}\left({\omega }_j\right)\right|}^2}.$

在图1以对数为底的坐标系中画出$\sqrt{{\epsilon }^r}$ ，表明平衡方法是均方收敛的。

(a) 算例1 (b) 算例2

(c) 算例3

Figure 1. The convergence of the strong balance method

图1. 强平衡方法的收敛性

其次验证平衡方法的均方稳定性，这里取$a_1=-2;a_2=1;b_1=0.1;b_2=1$ ，

Figure 2. Mean-square stability of balanced implicit methods

图2. 平衡方法的均方稳定性

如图2，当步长$h=0.125,0.25,0.5$ 时，所有的曲线趋势都趋于零，因此平衡方法是均方稳定的。

5. 结论

本文针对满足全局Lipschitz条件的分段连续型随机微分方程，研究了平衡方法的均方收敛性，在此基础上，文章还构建了线性标量形式的随机微分方程，得到了强平衡方法有均方稳定性，弱平衡方法也具有均方稳定性。最终通过一系列数值实验分析证明了理论结果的可行合理性。此外，若本文放宽对满足全局Lipschitz条件的要求，满足关于其中一个变量的单边Lipschitz条件也是一个研究方向，讨论精确解是否有界，并研究其收敛性和稳定性。

基金项目

国家自然科学基金资助项目(11801238)；国家自然科学基金资助项目(11561028)。

NOTES

^*通讯作者。