随着微分方程在生物学、金融学、工程学和环境科学等领域的广泛应用, 如何有效刻画具有正性约束的随机系统已成为当前研究的重点问题. 这类系统在描述种群演化、资产定价、化学反应等实际过程时, 其数学模型的解必须严格保持正值才具有物理意义.

近年来, 众多学者（见文献[1-8]）致力于发展具有保正性、精度与稳定性的数值方法. Neuenkirch与Szpruch在文献[9]中针对取值于特定区域的SDE, 提出了漂移隐式Euler-Maruyama(EM)方法. 然而值得注意的是, 尽管该方法在理论上具有重要意义, 但其隐式特性导致计算成本较高. 此后, Mao^[10]开创性地提出截断EM方法, 这是一种用于研究高度非线性随机微分方程数值解的显式方法. Mao和Hu^{[11, 12]}通过改进步长约束条件, 进一步提升了该方法的适用性. 然而, 上述方法在处理解必须保持正性的SDEs时存在固有缺陷. 难以直接应用于正约束随机模型. 针对这一挑战, 众多学者相继提出了显式保正EM格式的创新解法. 其中, 对数截断EM方法（见文献[13, 14]）因其计算效率与保正特性的良好平衡而备受关注. 最近, 文献[15]通过弱化约束条件, 建立了标量SDEs对数截断EM方法的理论框架. 但需要指出的是, 其框架无法直接推广至多维情形. 因此, 需对假设条件进行适应性修正以拓展至多维SDEs. 文献[16-18]研究了具体的一类多维的随机微分方程的保正数值方法, 但该方法限制于标量的布朗运动.

本文核心目标是进一步研究对数截断EM方法, 将对数截断EM方法拓展到多维情况, 并将布朗运动拓广到$ m $维. 对原SDE施加假设, 结合对数变换相关的性质, 进而使得原随机微分方程数值解保持正性. 在此框架下, 我们通过构造新型截断函数, 证明了数值解的$ L^p $收敛性, 其收敛速率在适当条件下可达经典的$ 1/2 $阶. 同时, 该方法可有效应用于含负幂项系数的一类多维SDEs, 使得该方法可处理更广泛的非线性项.

全文结构如下: 第2节介绍符号体系与基本假设, 建立关键引理; 第3节构建多维SDEs的对数截断EM方法; 第4节分析其收敛速率; 第5节通过一个典型算例验证该方法在一般参数下对多维SDEs的有效性; 第6节基于算例进行数值模拟以佐证理论结论.

在全文论述中, 若无特殊说明, 我们设定以下基础框架: 设$ (\Omega, \mathcal{F}, \mathcal{P}) $为完备概率空间, 其配备满足常规条件的滤子$ {\left\{\mathcal{F}_{t}\right\}}_{t\geqslant0} $ (即右连续递增且$ \mathcal{F}_0 $包含所有$ \mathcal{P} $-零测集). 记$ \mathbb{E} $为关于测度$ \mathbb{P} $的数学期望算子. 在此空间上定义$ m $维$ \mathcal{F}_t $-适应布朗运动$ B(t)=(B_1(t), B_2(t), \ldots, B_m(t))^T $. 其中各分量$ B_i(t) $ ($ i = 1, \ldots, m $) 是相互独立、标准的一维布朗运动. 由布朗运动的性质可知$ B_i(t)-B_i(s) $服从以$ 0 $为均值、以$ t-s $为方差的正态分布.

关于符号体系作如下约定：

● 向量或矩阵$ A $的转置记为$ A^T $;

● 对$ x = (x_1, \ldots, x_d)^T \in \mathbb{R}_+^d \triangleq (0, \infty)^d $, 其欧几里得范数为$ \Vert x\Vert $;

● 若$ A $表示矩阵, 用$ \Vert A \Vert = \sqrt{\text{trace}(A^{T}A}) $表示其迹范数；

● 向量$ x $的逐元倒数记为$ (x^{-1}) $或$ \left(\frac{1}{x}\right) $, 即$ (x_1^{-1}, \ldots, x_d^{-1})^T $;

● 若$ A $表示一个矩阵或向量, 则用$ A_i $表示$ A $的第$ i $行;

● 对任意实数$ a, b $, 定义$ a \vee b = \max(a, b) $, $ a \wedge b = \min(a, b) $;

● 集合$ G $的示性函数记为$ I_G $, 即$ I_G(x) = 1 $当且仅当$ x \in G $, 否则$ I_G(x) = 0 $;

● 规定空集$ \emptyset $的下确界为$ \infty $;

● 定义非负闭区域$ \bar{\mathbb{R}}_+^d \triangleq [0, \infty)^d $.

考虑如下定义在时间区间$ [0, T] $上的$ d $维非线性随机微分方程：

$ \begin{align} dx(t)=f(x(t))dt+g(x(t))dB(t), \end{align} $

(2.1)

其中初始值$ x(0) = x_0 \in \mathbb{R}_{+}^d $, $ T $为固定正数. 漂移项$ f : \mathbb{R}^d \rightarrow \mathbb{R}^d $与扩散项$ g : \mathbb{R}^d \rightarrow \mathbb{R}^{d \times m} $均为Borel可测函数. 本文设定以下三个基本假设条件:

$ ({\rm{A}} _1) $ 假设漂移系数满足局部Lipschitz: 存在实数$ K_1 > 0 $, 以及$ \alpha > 0, \beta > 0 $使得对任意$ x, y \in \mathbb{R}_{+}^d $, 有

$ \Vert f(x)-f(y)\Vert\leqslant K_1\left( 1 + \Vert x\Vert^\alpha + \Vert y\Vert^\alpha + \Vert x^{-1}\Vert^{\beta} + \Vert y^{-1}\Vert^{\beta}\right)\Vert x-y\Vert. $

$ ({\rm{A}} _2 ) $ 存在正实数$ x^{*} > 0, p > 1, q > 0 $以及$ K_2 > 0 $使得对任意$ x \in {\mathbb{R}}_+^{d} $, 有

$ x^{T}f(x) + \frac{p - 1}{2}\Vert g(x)\Vert^2 \leqslant K_2(1 + \Vert x \Vert^2), $

并且对$ i = 1, \ldots, d $和任意$ x_i \in (0, x^*), x_j\in \mathbb{R}_+ $, 其中$ j\neq i $, 有

$ x_{i}f_{i}(x) - \frac{q + 1}{2} \Vert g_{i}(x) \Vert^2 \geqslant 0. $

$ ({\rm{A}} _3) $ 存在一对正实数$ r^{*} > 2 $和$ H > 0 $使得对任意$ x, y \in \mathbb{R}^d_{+} $, 有

$ {(x - y)}^{T}(f(x) - f(y)) + \frac{r^{*} - 1}{2}\Vert g(x) - g(y)\Vert^2 \leqslant H \Vert x - y\Vert^2. $

注释2.1  由($ A_1 $)可得

$ \Vert f(x)\Vert \leqslant\Vert f(x)-f({\bf 1})\Vert + \Vert f({\bf 1})\Vert \leqslant 3(\sqrt{d})^{\alpha\vee\beta}K_1\left(1+\Vert x\Vert ^\alpha + \left \Vert x^{-1}\right \Vert^{\beta}\right)\Vert x-{\bf 1}\Vert +\Vert f({\bf 1})\Vert , $

其中$ {\bf 1} = {(1, \ldots , 1)}^{T} $, 同时由Young不等式有

$ \begin{align*} \Vert f(x)\Vert &\leqslant 3(\sqrt{d})^{\alpha\vee\beta}K_1\left(1+\Vert x \Vert^\alpha + \left \Vert x^{-1}\right \Vert^{\beta}\right)(\Vert x \Vert + \Vert {\bf 1} \Vert) + \Vert f({\bf 1})\Vert\\ &\leqslant 3(\sqrt{d})^{(\alpha\vee\beta)+1}K_1\left(1+\Vert x\Vert + \Vert x\Vert ^{\alpha} + \left \Vert x^{-1}\right \Vert^{\beta} + \Vert x \Vert^{\alpha+1} + \Vert x\Vert \cdot\left \Vert x^{-1}\right \Vert^{\beta}\right) + \Vert f({\bf 1})\Vert\\ &\leqslant 9(\sqrt{d})^{(\alpha\vee\beta)+1}K_1 \left(1 + \Vert x\Vert^{\alpha+1} + \Vert x \Vert^{\beta+1} + \left \Vert x^{-1}\right \Vert^{\beta+1}\right) + \Vert f({\bf 1})\Vert. \end{align*} $

进一步, 有

$ \begin{align*} \Vert f(x)\Vert &\leqslant \left\{ \begin{aligned} & 27(\sqrt{d})^{(\alpha\vee\beta)+1}K_1 \left(1 + \left \Vert x^{-1}\right \Vert^{\beta+1}\right) + \Vert f({\bf 1})\Vert , \quad \Vert x\Vert \leqslant 1\\ & 18(\sqrt{d})^{(\alpha\vee\beta)+1}K_1 \left(1 + \Vert x\Vert^{(\alpha\vee\beta)+1} + \left \Vert x^{-1}\right \Vert^{\beta+1}\right) + \Vert f({\bf 1})\Vert , \quad \Vert x \Vert> 1. \end{aligned} \right. \end{align*} $

因此, 对任意$ x \in {\mathbb{R}}_+^{d} $有

$ \Vert f(x)\Vert \leqslant (27(\sqrt{d})^{(\alpha\vee\beta)+1}K_1 \vee \Vert f({\bf 1})\Vert) \left(1 + \Vert x\Vert^{(\alpha\vee\beta)+1} + \left \Vert x^{-1}\right \Vert^{\beta+1}\right). $

同理, 由($ A_2 $)可得, 对任意$ x \in {\mathbb{R}}_+^{d} $有

$ \begin{align*} \Vert g(x)\Vert^2&\leqslant\frac{2}{p-1}\left(\Vert x\Vert \cdot\Vert f(x)\Vert +K_2(1+\Vert x\Vert ^2)\right)\\ &\leqslant\frac{2}{p-1}\left(M\left(\Vert x\Vert + \Vert x\Vert^{(\alpha\vee\beta)+2} + \Vert x \Vert\cdot\left \Vert x^{-1}\right \Vert^{\beta+1}\right)+K_2(1 + \Vert x\Vert^2)\right), \end{align*} $

其中$ M=27(\sqrt{d})^{(\alpha\vee\beta)+1}K_1 \vee \Vert f({\bf 1})\Vert $. 因此, 存在实数$ C $, 使得

$ \begin{align*} \Vert f(x)\Vert\leqslant C\left(1 + \Vert x\Vert^{(\alpha\vee\beta)+1} + \left \Vert x^{-1}\right \Vert ^{\beta+1}\right), \quad \Vert g(x)\Vert^2\leqslant C\left(1 + \Vert x\Vert^{(\alpha\vee\beta)+2} + \left \Vert x^{-1}\right \Vert^{\beta+2}\right). \end{align*} $

注意到, 上式对$ \alpha=0 $或$ \beta = 0 $也成立. 另外若$ x \in \mathbb{R}_+^d $, 则由Hölder不等式有

$ \begin{align*} \left \Vert x^{-1}\right \Vert^{q} = \left( \sum\limits_{i = 1}^{d}\left(x_i^{-2}\right) \right)^{\frac{q}{2}} \leqslant \left( \left(\sum\limits_{i = 1}^{d}1^{\frac{q}{q-2}}\right)^{1-\frac{2}{q}} \left(\sum\limits_{i = 1}^{d}x_i^{-q}\right)^{\frac{2}{q}} \right)^{\frac{q}{2}} \leqslant d^{\frac{q}{2}-1}\sum\limits_{i = 1}^{d}x_i^{-q}, \end{align*} $

其中, $ q \geqslant 2 $.

下述引理表明, 随机微分方程(2.1)在时间区间$ [0, T] $上存在唯一的强解, 且该解具有严格正性, 即满足

$ \mathbb{P}\left(x(t)\in \mathbb{R}_+^d, \forall t\in[0, T]\right)=1. $

引理2.1  在($ A_1 $)、($ A_2 $)、($ A_3 $) 及$ (\alpha + 2)\vee(\beta+2) \leqslant p \wedge q $成立的条件下, 随机微分方程(2.1)在区间$ [0, T] $上存在唯一的强解. 进一步地, 存在常数$ C > 0 $, 使得以下估计成立：

$ \sup\limits_{t\in[0, T]}\mathbb{E}\Vert x(t\wedge\theta)\Vert ^{p}<C\quad \text{和} \quad \sup\limits_{t\in[0, T]}\mathbb{E}\Vert x(t\wedge\theta)^{-1}\Vert ^{q}<C, $

其中$ \theta $是任意停时, 进一步我们有

$ \mathbb{P}\left(x(t)\in \mathbb{R}_+^d, \forall t\in[0, T]\right)=1. $

证  在本文证明过程中, 令$ C $表示一般正常数(其具体数值在不同位置可能有所不同). 设$ k $为正整数, 并令

$ \tilde{\pi}_{k}(x_i)=k^{-1}I_{\{x_i<k^{-1}\}}+x_{i}I_{\{k^{-1}\leqslant x_i\leqslant k\}}+kI_{\{k<x_i\}}, $

以及

$ \tilde{\pi}_k(x) = \left(\tilde{\pi}_k(x_1), \ldots, \tilde{\pi}_k(x_d)\right)^T, $

$ \tilde{f}_k(x)=f\left(\tilde{\pi}_k(x)\right)\quad \text{和} \quad \tilde{g}_k(x)=g\left(\tilde{\pi}_k(x)\right). $

则由注释2.1, 有

$ \Vert \tilde{f}_k(x) - \tilde{f}_k(y)\Vert \leqslant L\Vert x - y\Vert \quad \text{和} \quad \Vert \tilde{f}_k(x)\Vert ^2 \vee \Vert \tilde{g}_k(x)\Vert ^2 \leqslant C(1 + \Vert x\Vert ^2), $

因此$ \tilde{f}_k(x) $和$ \tilde{g}_k(x) $满足全局Lipschitz以及线性增长. 那么, 由[19]中2.3节可得方程

$ d\tilde{x}_k(t)=\tilde{f}_k\left(\tilde{x}_k(t)\right)dt + \tilde{g}_k\left(\tilde{x}_k(t)\right)dB(t) $

在$ [0, T ] $存在唯一解. 现在定义停时

$ \tau_k=\inf\{t\in[0, T]\mid \tilde{x}_k(t)\not\in(\frac{1}{k}, k)^d\}. $

由于$ \tilde{x}_{k}(t) $的唯一性, 那么在$ t \in [0, T\wedge\tau_{n}] $时$ \tilde{x}_{m}(t) = \tilde{x}_{n}(t) $, 其中$ m > n $且足够大. 那么, $ \tau_{n} $非减, 同时定义$ \tau_{\infty} = \lim_{n \rightarrow \infty}\tau_{n} $.

设$ \omega \in \Omega $. 对任意$ t < \tau_{n}(\omega) $, 存在$ k(\omega) > 0 $使得$ t< \tau_{k}(\omega) \leqslant \tau_{\infty}(\omega) $. 现在定义$ x(t, \omega) = \tilde{x}_{k}(t, \omega) $, 对于一个固定的正整数$ n $和任意的$ t \in [0, T] $, 有

$ \begin{align*} x(t\wedge\tau_{n})= & \tilde{x}_n(t\wedge\tau_n) \\ = &x_{0}+\int_{0}^{t\wedge\tau_{n}}\tilde{f}_{n}(\tilde{x}_{n}(s))ds+\int_{0}^{t\wedge\tau_{n}}\tilde{g}_{n}(\tilde{x}_{n}(s))dB(s) \\ = &x_{0}+\int_{0}^{t\wedge\tau_{n}}f(x(s))ds+\int_{0}^{t\wedge\tau_{n}}g(x(s))dB(s). \end{align*} $

应用Itô公式、($ A_2 $)以及注释2.1, 对任意$ t \in [0, T ] $有

$ \begin{align*} \mathbb{E}\Vert x (t\wedge\tau_{n})\Vert ^{p} \leqslant &\Vert x_{0}\Vert ^{p} +p\mathbb{E}\int_{0}^{t\wedge\tau_{n}}\Vert x(s)\Vert ^{p-2}\left(x(s)^{T}f(x(s))+\frac{p-1}{2}\Vert g(x(s))\Vert ^{2}\right)ds \\ \leqslant &\Vert x_{0}\Vert ^{p} +C\mathbb{E}\int_{0}^{t\wedge\tau_{n}}\Vert x(s)\Vert ^{p-2} \left(1 + \Vert x(s)\Vert ^2\right)ds \\ \leqslant &\Vert x_{0}\Vert ^{p}+C\mathbb{E}\int_{0}^{t}\left(1+\Vert x(s\wedge\tau_{n})\Vert ^{p}\right)ds, \end{align*} $

结合Gronwall不等式, 存在常数$ C $使得

$ \sup\limits_{t\in[0, T]}\mathbb{E}\left(\Vert x(t\wedge\tau_n)\Vert ^{p}\right)<C. $

因此, 对于$ i = 1, \ldots, d $, 有

$ \sup\limits_{t\in[0, T]}\mathbb{E}\left({x_i(t\wedge\tau_n)}^{p}\right)<C. $

同理, 对$ i = 1, \ldots, d $, 使用上述估计, 当$ t\in [0, T] $有

$ \begin{align*} &\mathbb{E}\left(x_i(t\wedge\tau_{n})^{-q}\right) \\ =& x_{i}(0)^{-q} -q\mathbb{E}\int_{0}^{t\wedge\tau_{n}}x_i(s)^{-(q+2)}\left(x_i(s)f_i(x(s))-\frac{q+1}{2}\Vert g_i(x(s))\Vert ^{2}\right)ds\\ \leqslant& x_{i}(0)^{-q} + C\mathbb{E}\int_{0}^{t\wedge\tau_{n}} x_i(s)^{-(q+2)}\left(1+\Vert x(s)\Vert ^{(\alpha +2) \vee (\beta + 2)} + \left \Vert x(s)^{-1}\right \Vert ^{\beta+2}\right)I_{\{x_i(s)\geqslant x^*\}}ds\\ \leqslant& C + C\mathbb{E}\int_{0}^{t\wedge\tau_{n}} \left(1+\Vert x(s)\Vert ^{(\alpha +2) \vee (\beta + 2)} + \left \Vert x(s)^{-1}\right \Vert ^{\beta+2}\right)ds\\ \leqslant& C + C\mathbb{E}\int_{0}^{t\wedge\tau_{n}} \Vert x(s)\Vert ^{(\alpha +2) \vee (\beta + 2)} ds + C\mathbb{E}\int_{0}^{t\wedge\tau_{n}} \left \Vert x(s)^{-1}\right \Vert ^{\beta+2}ds\\ \leqslant& C + C\mathbb{E}\int_{0}^{t\wedge\tau_{n}} \sum\limits_{j = 1}^{d}x_j(s)^{-(\beta+2)}ds. \end{align*} $

那么,

$ \mathbb{E}\sum\limits_{i = 1}^{d}\left(x_i(t\wedge\tau_{n})^{-q}\right) \leqslant C + C\mathbb{E}\int_{0}^{t\wedge\tau_{n}} \sum\limits_{j = 1}^{d}x_j(s)^{-(\beta+2)}ds \leqslant C + C\mathbb{E}\int_{0}^{t\wedge\tau_{n}} \sum\limits_{j = 1}^{d}x_j(s)^{-q}ds. $

应用Gronwall不等式, 存在常数$ M $使得

$ \sup\limits_{t\in[0, T]}\sum\limits_{i=1}^{d}\mathbb{E}\left(x_i(t\wedge\tau_{n})^{-q}\right)<M. $

进一步, 存在常数$ C $使得

$ \sup\limits_{t\in[0, T]}\mathbb{E}\left \Vert x(t\wedge \tau_n)^{-1}\right \Vert ^q < C. $

因此

$ \sup\limits_{t \in [0, T]} \mathbb{E} \left( x_i(t \wedge \tau_n)^p + x_i(t \wedge \tau_n)^{-q} \right) < C. $

同样地, 如果$ \mathbb{P}(\tau_{\infty} \leqslant T) > 0 $, 则使$ n $足够大时

$ \sum\limits_{i=1}^{d}\mathbb{E}\left(x_i(T\wedge\tau_n)^{p} + x_i(T\wedge\tau_n)^{-q}\right)\geqslant n^{p\wedge q}\mathbb{P}(\tau_\infty\leqslant T), $

无界. 从而矛盾. 因此$ \mathbb{P}(\tau_{\infty} > T ) = 1 $. 这意味这SDE (2.1) 在$ [0, T ] $上有唯一强解并且

$ \mathbb{P}\left(x(t)\in \mathbb{R}_+^d, \forall t\in[0, T]\right)=1. $

按照上述类似的讨论, 存在常数$ C $使得

$ \sup\limits_{t\in[0, T]}\mathbb{E}\Vert x(t\wedge\theta)\Vert ^{p}<C\quad \text{和} \quad \sup\limits_{t\in[0, T]}\mathbb{E}\Vert x(t\wedge\theta)^{-1}\Vert ^{q}<C, $

其中$ \theta $表示任意停时.

为了定义对数截断EM数值解, 我们首先对于每个分量进行对数变换, 定义$ y = \left(\ln x_1, \ldots, \ln x_d\right)^T \triangleq \ln x, x \in \mathbb{R}_+^d. $另外使用逆变换, 定义$ x = \left(e^{y_1}, \ldots, e^{y_d}\right)^T. $应用Itô公式, 得到新的SDE

$ dy(t) = F(y)dt + G(y)dB(t), $

其中

$ F(y) = \text{diag}\left(e^{-y_1}, \ldots, e^{-y_d}\right)f(e^{y}) - \frac{1}{2}\text{diag}\left(e^{-2y_1}, \ldots, e^{-2y_d}\right){\left(\Vert g_1(e^y)\Vert ^2, \ldots, \Vert g_d(e^y)\Vert ^2\right)}^T, $

且

$ G(y) = \text{diag}\left(e^{-y_1}, \ldots, e^{-y_d}\right)g(e^{y}), $

对任意$ y\in \mathbb{R}^d $, 由注释2.1有

$ \begin{align*} \Vert F(y)\Vert \leqslant& C_1\left(\left \Vert e^{-y}\right \Vert \cdot\left \Vert f(e^y)\right \Vert + \left \Vert e^{-2y}\right \Vert \cdot\left \Vert g(e^y)\right \Vert ^2\right)\\ \leqslant & C_2\left(\left \Vert e^{-y}\right \Vert \cdot\left \Vert f(e^y)\right \Vert + \left \Vert e^{-y}\right \Vert ^2\cdot\left \Vert g(e^y)\right \Vert ^2\right)\\ \leqslant & C_3\left \Vert e^{-y}\right \Vert \left(1 + \left \Vert e^y\right \Vert ^{(\alpha\vee\beta)+1} + \left \Vert e^{-y}\right \Vert ^{\beta+1}\right) + C_3\left \Vert e^{-y}\right \Vert ^2\left(1 + \left \Vert e^y\right \Vert ^{(\alpha\vee\beta)+2} + \left \Vert e^{-y}\right \Vert ^{\beta+2}\right)\\ \leqslant & C_4\left(1+\left \Vert e^y\right \Vert ^{(\alpha\vee\beta)+4} + \left \Vert e^{-y}\right \Vert ^{(\alpha\vee\beta)+4}\right)\\ \leqslant & C_5\left(1+e^{((\alpha\vee\beta)+4)\Vert y\Vert}\right). \end{align*} $

同样地,

$ \Vert G(y)\Vert ^2 \leqslant C_6\left(1+e^{((\alpha\vee\beta)+4)\Vert y\Vert}\right). $

因此, 存在某个常数$ C_0 > 1 $及与参数$ \alpha $、$ \beta $相关的常数$ C(\alpha, \beta) \leqslant (\alpha\vee\beta)+4 $, 使得

$ \begin{align} \Vert F(y)\Vert \vee \Vert G(y)\Vert ^2 \leqslant C_0(1 + e^{C(\alpha, \beta)\Vert y\Vert }), \end{align} $

(3.1)

现引入严格递增的连续函数$ \varphi(r) = C_0(2 + e^{C(\alpha, \beta)r}) $, 使得

$ \sup\limits_{\Vert y\Vert <r}\Vert F(y)\Vert \vee \Vert G(y)\Vert ^2 \leqslant \varphi(r), \quad \forall r > 0. $

记$ \varphi $的反函数为$ \varphi^{-1} $, 显然$ \varphi^{-1}: [3C_0, \infty) \rightarrow \bar{\mathbb{R}}_+ $也是一个严格递增的连续函数. 此外, 我们引入一个严格递减函数$ h:(0, 1] \rightarrow [1, \infty) $, 该函数对$ \forall \Delta\in(0, 1] $满足

$ \begin{align} \lim\limits_{\Delta\to0}h(\Delta)=+\infty, \quad{\Delta}^{\frac{1}{3}} h(\Delta)\leqslant4C_0\vee\varphi(\Vert \ln x_0\Vert ) \quad\text{和}\quad h(1)>3C_0\vee\varphi(\Vert \ln x_0\Vert ). \end{align} $

(3.2)

对于给定的步长$ \Delta \in (0, 1] $, 我们定义截断映射$ \pi_{\Delta}:\mathbb{R} \rightarrow \mathbb{R} $为

$ \pi_{\Delta}(y) = \left(\Vert y\Vert \wedge\varphi^{-1}(h(\Delta))\right)\frac{y}{\Vert y\Vert }, $

其中约定当$ y = 0 $时, $ \frac{y}{\Vert y\Vert } = 0 $. 我们进一步定义截断函数

$ F_\Delta(y)=F(\pi_\Delta(y))\quad\text{和}\quad G_\Delta(y)=G(\pi_\Delta(y)), $

其中$ y \in \mathbb{R}^d $. 由此可得以下一致界：

$ \begin{align} \Vert F_\Delta(y)\Vert \vee\Vert G_\Delta(y)\Vert ^2\leqslant\varphi\left(\varphi^{-1}(h(\Delta))\right)=h(\Delta), \end{align} $

(3.3)

该不等式对所有$ y \in \mathbb{R}^d $均成立.

基于上述构造, 我们可以建立变换后SDE的离散时间对数截断EM数值解. 设时间节点$ t_{k}=k\Delta $处的数值解为$ Y_{\Delta}(t_{k})\approx y(t_{k}) $, 其计算过程如下：首先取初始值$ Y_{\Delta}(0) = y_0 = \ln(x_0) $然后通过递推公式

$ Y_{\Delta}(t_{k+1})=Y_{\Delta}(t_{k})+F_{\Delta}\left(Y_{\Delta}(t_{k})\right)\Delta+G_{\Delta}\left(Y_{\Delta}(t_{k})\right)\Delta B_{k}, $

进行计算, 其中$ k = 0, 1, \ldots, $且$ \Delta B_{k} = B(t_{k+1}) - B(t_k) $表示布朗运动增量. 在此基础上, 我们将构造变换后SDE的两种连续时间对数截断EM数值解格式. 第一种格式定义为：

$ \bar{y}_{\Delta}(t)=\sum\limits_{k=0}^{\infty}Y_{\Delta}(t_{k})I_{[t_{k}, t_{k+1})}(t), \quad t \geqslant 0. $

这是一个简单过程, 其样本路径不连续. 我们将其称为连续时间过程对数截断EM解. 另一种格式定义为：

$ y_{\Delta}(t)=y_{0}+\displaystyle \int_{0}^{t}F_{\Delta}\left(\bar{y}_{\Delta}(s)\right)ds+\displaystyle \int_{0}^{t}G_{\Delta}\left(\bar{y}_{\Delta}(s)\right)dB(s), \quad t \geqslant 0. $

我们将其称为连续时间连续样本对数截断EM解. 对于所有$ k \geqslant 0 $, 都有$ y_{\Delta}(t_{k})=\bar{y}_{\Delta}(t_{k})=Y_{\Delta}(t_{k}) $成立. 此外, $ y_{\Delta}(t_{k}) $是一个Itô过程, 其微分形式为

$ dy_{\Delta}(t)=F_{\Delta}\left(\bar{y}_{\Delta}(t)\right)dt+G_{\Delta}\left(\bar{y}_{\Delta}(t)\right)dB(t). $

最后, 通过变换

$ \bar{x}_{\Delta}(t) = \left(e^{\bar{y}_{\Delta, 1}(t)}, \ldots, e^{\bar{y}_{\Delta, d}(t)}\right)^T \triangleq e^{\bar{y}_{\Delta}(t)}, \quad x_{\Delta}(t) = \left(e^{y_{\Delta, 1}(t)}, \ldots, e^{y_{\Delta, d}(t)}\right)^T \triangleq e^{y_{\Delta}(t)} $

我们得到原随机微分方程的数值解$ \bar{x}_{\Delta}(t) $和$ x_{\Delta}(t) $.

引理3.1  设$ Z_1, \ldots, Z_m\stackrel{\mathrm{i.i.d.}}{\sim}\mathcal N(0, 1) $, $ S_m \triangleq \sum_{i=1}^{m}Z_i^2\sim\chi_m^{2} $为自由度$ m $的卡方变量. 则对任意$ \beta>0 $有,

$ \mathbb{E}\exp(\beta \Vert \Delta B_k\Vert )\leqslant2^{\frac{m}{2}}\exp\left(\beta^2\Delta\right). $

证  利用Young不等式和卡方矩母函数有

$ \begin{align*} \mathbb{E}\exp(\beta \Vert \Delta B_k\Vert )=& \mathbb{E}\exp\left(\beta\sqrt{\Delta}\sqrt{S_m}\right)\\ \leqslant& \mathbb{E}\exp\left(\beta^2\Delta+\frac{1}{4} S_m\right)\\ =& \exp(\beta^2\Delta)(1-\frac{1}{2})^{-\frac{m}{2}}\\ =& 2^{\frac{m}{2}}\exp\left(\beta^2\Delta\right). \end{align*} $

为简洁起见, 我们作如下定义：

$ \text{diag}\left(\frac{x_{\Delta , 1}(t)}{\bar{x}_{\Delta , 1}(t)}, \ldots, \frac{x_{\Delta , d}(t)}{\bar{x}_{\Delta , d}(t)}\right) \triangleq \text{diag}\left(\frac{x_{\Delta}(t)}{\bar{x}_{\Delta}(t)}\right). $

引理3.2  对于任意实数$ \bar{p} $, 存在依赖于$ \bar{p} $的常数$ C_1(\bar{p}) $使得

$ \begin{align} \sup\limits_{\Delta\in(0, 1]}\sup\limits_{t\in[0, T]}\mathbb{E}{\left(\frac{\Vert x_\Delta(t)\Vert }{\Vert \bar{x}_\Delta(t)\Vert }\right)}^{\bar{p}} \leqslant C_1(\bar{p}). \end{align} $

(3.4)

当$ \bar{p}\geqslant 2 $且满足上述条件时, 存在依赖于$ \bar{p} $的常数$ C_2(\bar{p}) $满足对$ \Delta \in (0, 1] $有

$ \begin{align} \sup\limits_{t\in[0, T]}\mathbb{E}\left \Vert \text{diag}\left(\frac{x_{\Delta}(t)}{\bar{x}_{\Delta}(t)}\right)-E\right \Vert ^{\bar{p}} \leqslant C_2(\bar{p})\Delta^{\frac{\bar{p}}{2}}h(\Delta)^{\frac{\bar{p}}{2}}. \end{align} $

(3.5)

其中$ E $表示单位矩阵.

证  证明过程中, 约定用$ C_1(\bar{p}) $和$ C_2(\bar{p}) $表示仅依赖于$ \bar{p} $的正实数, 这些常数与$ \Delta $和$ k $无关, 且在不同位置可能出现不同的取值. 根据$ x_{\Delta}(t) $和$ y_{\Delta}(t) $, 的定义, 可得：

$ \Vert x_{\Delta}(t)\Vert = \left \Vert\begin{pmatrix}\bar{x}_{\Delta , 1}(t)\exp\left(F_{\Delta, 1}(\bar{y}_{\Delta}(t))(t-t_k) + G_{\Delta, 1}(\bar{y}_{\Delta}(t))(B(t)-B(t_k))\right)\\ \vdots\\\bar{x}_{\Delta , d}(t)\exp\left(F_{\Delta, d}(\bar{y}_{\Delta}(t))(t-t_k) + G_{\Delta, d}(\bar{y}_{\Delta}(t))(B(t)-B(t_k))\right)\end{pmatrix}\right \Vert , $

由此可得以下估计式：

$ \begin{align*} \Vert x_{\Delta}(t)\Vert =&\left(\sum\limits_{1=1}^{d}\bar{x}_{\Delta, i}^2(t)\exp\left(2F_{\Delta, i}(\bar{y}_{\Delta}(t))(t-t_k) + 2G_{\Delta, i}(\bar{y}_{\Delta}(t))(B(t)-B(t_k))\right)\right)^{\frac{1}{2}}\\ \leqslant& \left(\sum\limits_{1=1}^{d}\bar{x}_{\Delta, i}^2(t)\right)^{\frac{1}{2}}\exp\left(\max\left(F_{\Delta, i}(\bar{y}_{\Delta}(t))(t-t_k) + G_{\Delta, i}(\bar{y}_{\Delta}(t))(B(t)-B(t_k))\right)\right)\\ \leqslant& \Vert \bar{x}_{\Delta}(t)\Vert \exp\left(\Vert F_{\Delta}(\bar{y}_{\Delta}(t))(t-t_k) + G_{\Delta}(\bar{y}_{\Delta}(t))(B(t)-B(t_k))\Vert \right), \end{align*} $

以及

$ \begin{align*} \Vert x_{\Delta}(t)\Vert &\geqslant\left(\sum\limits_{1=1}^{d}\bar{x}_{\Delta, i}^2(t)\right)^{\frac{1}{2}}\exp\left(\min\left(F_{\Delta, i}(\bar{y}_{\Delta}(t))(t-t_k) + G_{\Delta, i}(\bar{y}_{\Delta}(t))(B(t)-B(t_k))\right)\right)\\ &\geqslant\Vert \bar{x}_{\Delta}(t)\Vert \exp\left(-\Vert F_{\Delta}(\bar{y}_{\Delta}(t))(t-t_k) + G_{\Delta}(\bar{y}_{\Delta}(t))(B(t)-B(t_k))\Vert \right), \end{align*} $

其中$ t \in [t_k, t_{k+1}) $. 结合引理3.1可得对$ t \in [t_k, t_{k+1}) $有

$ \begin{align*} \mathbb{E}\left(\frac{\Vert x_{\Delta}(t)\Vert }{\Vert \bar{x}_{\Delta}(t)\Vert }\right)^{\bar{p}} &\leqslant \mathbb{E}\left(\exp\Vert F_{\Delta}(\bar{y}_{\Delta}(t))(t-t_k) + G_{\Delta}(\bar{y}_{\Delta}(t))(B(t)-B(t_k))\Vert \right)^{| \bar{p}| }\\ & \leqslant \mathbb{E}\exp\left(|\bar{p}|\cdot\Vert F_{\Delta}(\bar{y}_{\Delta}(t))(t-t_k)\Vert + |\bar{p}|\cdot\Vert G_{\Delta}(\bar{y}_{\Delta}(t))(B(t)-B(t_k))\Vert \right)\\ & \leqslant \mathbb{E}\exp\left(|\bar{p}|h(\Delta)\Delta + |\bar{p}|h(\Delta)^{\frac{1}{2}}\Vert B(t) - B(t_k)\Vert \right)\\ & \leqslant 2^{\frac{m}{2}}\mathbb{E}\exp\left(|\bar{p}|h(\Delta)\Delta + {\bar{p}}^2h(\Delta)\Delta\right), \end{align*} $

由于$ {\Delta}^{\frac{1}{3}}h(\Delta) \leqslant 4C_0\vee\varphi(\Vert \ln x_0\Vert ) $, 因此存在仅依赖于$ \bar{p} $的常数$ C_{1}(\bar{p}) $, 使得对所有$ \Delta \in (0, 1] $都有$ \mathbb{E}\left(\frac{\Vert x_\Delta(t)\Vert }{\Vert \bar{x}_\Delta(t)\Vert }\right)^{\bar{p}}\leqslant C_1(\bar{p}), $类似地, 对于每个分量$ i = 1, \ldots, d, $我们可推得

$ \begin{align*} \mathbb{E}\left(\frac{x_{\Delta, i}(t)}{\bar{x}_{\Delta, i}(t)}\right)^{\bar{p}} \leqslant& \mathbb{E}\exp\left(|\bar{p}|\cdot|F_{\Delta, i}(\bar{y}_{\Delta}(t))(t-t_k)| + |\bar{p}|\cdot\Vert G_{\Delta, i}(\bar{y}_{\Delta}(t))(B(t)-B(t_k))\Vert \right)\\ \leqslant& \mathbb{E}\exp\left(|\bar{p}|h(\Delta)\Delta + |\bar{p}|h(\Delta)^{\frac{1}{2}}\Vert B(t) - B(t_k)\Vert \right)\\ \leqslant& C_1(\bar{p}). \end{align*} $

对$ e^{y_{\Delta}(t)} $的各个分量应用Itô公式, 我们得到如下分量表达式：对于$ i = 1, \ldots, d $且$ t \in [t_k, t_{k+1}) $,

$ \begin{align*} x_{\Delta, i}(t) =& \bar{x}_{\Delta, i}(t) + \displaystyle \int _{t_k}^{t}x_{\Delta, i}(s)\left(F_{\Delta, i}(\bar{y}_{\Delta}(s)) + \frac{1}{2}\left\Vert G_{\Delta, i}(\bar{y}_{\Delta}(s))\right\Vert^2\right)ds +\displaystyle \int _{t_k}^{t}x_{\Delta, i}(s)G_{\Delta, i}(\bar{y}_{\Delta}(s))dB(s), \end{align*} $

现考虑$ \bar{p} \geqslant 2 $的情形. 由于$ x_{\Delta}(t), \bar{x}_{\Delta}(t) \in \mathbb{R}_+^d $, 注意到当$ t\in[t_k, t_{k+1}) $时$ \bar x_{\Delta, i}(t)=\bar x_{\Delta, i}(t_k) $, 结合式(3.2)、(3.3)、Hölder不等式以及文献[19] 中的定理7.1, 可推导得

$ \begin{align*} &\mathbb{E}\left| {\frac{x_{\Delta, i}(t)}{\bar{x}_{\Delta, i}(t)}}-1\right| ^{\bar{p}}\\ =&\mathbb{E}\left| \displaystyle \int _{t_{k}}^{t}\frac{x_{\Delta, i}(s)}{\bar{x}_{\Delta, i}(s)}\left(F_{\Delta, i}(\bar{y}_{\Delta}(s))+\frac{1}{2}\Vert G_{\Delta, i}(\bar{y}_{\Delta}(s))\Vert ^{2}\right)ds +\displaystyle \int _{t_{k}}^{t}\frac{x_{\Delta, i}(s)}{\bar{x}_{\Delta, i}(s)}G_{\Delta, i}(\bar{y}_{\Delta}(s))dB(s)\right|^{\bar{p}}\\ \leqslant& C_{2}(\bar{p})(t-t_{k})^{\bar{p}-1}\mathbb{E}\displaystyle \int _{t_{k}}^{t}\left|\frac{x_{\Delta, i}(s)}{\bar{x}_{\Delta, i}(s)}\right|^{\bar{p}}\left|F_{\Delta, i}(\bar{y}_{\Delta}(s)) +\frac{1}{2}\Vert G_{\Delta, i}(\bar{y}_{\Delta}(s))\Vert ^{2}\right|^{\bar{p}}ds\\ &+C_{2}(\bar{p})(t-t_{k})^{\frac{\bar{p}}{2}-1}\mathbb{E}\displaystyle \int _{t_{k}}^{t}\left|\frac{x_{\Delta, i}(s)}{\bar{x}_{\Delta, i}(s)}\right|^{\bar{p}}\left \Vert G_{\Delta, i}(\bar{y}_{\Delta}(s))\right \Vert ^{\bar{p}}ds\\ \leqslant& C_{2}(\bar{p})\Delta^{\frac{\bar{p}}{2}-1}(\Delta^{\frac{\bar{p}}{2}}h(\Delta)^{\bar{p}}+h(\Delta)^{\frac{\bar{p}}{2}})\mathbb{E}\displaystyle \int _{t_{k}}^{t}\left|\frac{x_{\Delta, i}(s)}{\bar{x}_{\Delta, i}(s)}\right|^{\bar{p}}ds\\ \leqslant& C_{2}(\bar{p})\Delta^{\frac{\bar{p}}{2}}h(\Delta)^{\frac{\bar{p}}{2}}, \end{align*} $

其中$ t \in [t_k, t_{k+1}) $. 由注释2.1即可得所需结论(3.5).

下文将采用以下记法

$ \frac{1}{x_{\Delta}(t)} = \left(\frac{1}{x_{\Delta, 1}(t)}, \ldots, \frac{1}{x_{\Delta, d}(t)}\right)^{T} , \quad \frac{1}{\bar{x}_{\Delta}(t)} = \left(\frac{1}{\bar{x}_{\Delta, 1}(t)}, \ldots, \frac{1}{\bar{x}_{\Delta, d}(t)}\right)^{T} $

以及分量平方差向量

$ x_{\Delta}(t)^2 - \bar{x}_{\Delta}(t)^2 = \left(x_{\Delta, 1}(t)^2 - \bar{x}_{\Delta, 1}(t)^2, \ldots, x_{\Delta, d}(t)^2 - \bar{x}_{\Delta, d}(t)^2\right)^T. $

引理3.3  在假设条件($ A_1 $)、($ A_2 $)、($ A_3 $) 成立且满足$ (\alpha + 2) \vee (\beta+2) < p\wedge q $以及$ p\geqslant4 $的条件下, 对于任意停时$ \theta $, 存在常数$ C $使得下列矩估计一致成立

$ \sup\limits_{\Delta \in (0, 1]}\sup\limits_{t\in[0, T]}\mathbb{E}\Vert x_{\Delta}(t\wedge\theta)\Vert ^{p}<C \quad \text{和} \quad \sup\limits_{\Delta \in (0, 1]}\sup\limits_{t\in[0, T]}\mathbb{E}\left \Vert \frac{1}{x_{\Delta}(t\wedge\theta)}\right \Vert ^q <C. $

证  在下述证明过程中, 我们约定用$ C $表示与步长$ \Delta $无关的通用正常数, 其具体值在不同位置可能不同. 取步长$ \Delta \in (0, 1] $并定义停时$ \tau_n = \inf\{t \in [0, T]|x_{\Delta}(t) \notin (\frac{1}{n}, n)^{d}\} $应用Itô公式并取数学期望, 可得

$ \begin{align*} \mathbb{E}\left \Vert x_{\Delta}(t \wedge \tau_n \wedge \theta)\right \Vert ^p \leqslant & \left \Vert x_0\right \Vert ^p + \mathbb{E}\displaystyle \int _{0}^{t \wedge \tau_n \wedge \theta} p\left \Vert x_{\Delta}(s)\right \Vert ^{p-2} \left(x_{\Delta, 1}(s)^2, \ldots, x_{\Delta, d}(s)^2\right)\\ &\times\left( F_{\Delta}(\bar{y}_{\Delta}(s)) + \frac{1}{2}\left(\left \Vert G_{\Delta, 1}(\bar{y}_{\Delta}(s))\right \Vert ^{2}, \ldots, \left \Vert G_{\Delta, d}(\bar{y}_{\Delta}(s))\right \Vert ^{2}\right)^T \right)ds\\ &+\mathbb{E}\displaystyle \int _{0}^{t \wedge \tau_n \wedge \theta}\frac{1}{2}p(p-1)\left \Vert x_{\Delta}(s)\right \Vert ^{p-2} \left \Vert \text{diag}\left(x_{\Delta, 1}(s), \ldots, x_{\Delta, d}(s)\right)G_{\Delta}(\bar{y}_{\Delta}(s)) \right \Vert ^2 ds\\ \leqslant & \left \Vert x_0\right \Vert ^p + \mathbb{E}\displaystyle \int _{0}^{t \wedge \tau_n \wedge \theta}p\left \Vert x_{\Delta}(s)\right \Vert ^{p-2} \left(x_{\Delta, 1}(s)^2, \ldots, x_{\Delta, d}(s)^2\right)\\ &\times\left( F_{\Delta}(\bar{y}_{\Delta}(s)) + \frac{p}{2}\left(\left \Vert G_{\Delta, 1}(\bar{y}_{\Delta}(s))\right \Vert ^{2}, \ldots, \left \Vert G_{\Delta, d}(\bar{y}_{\Delta}(s))\right \Vert ^{2}\right)^T \right)ds\\ = & \left \Vert x_0\right \Vert ^p + \mathbb{E}\displaystyle \int _{0}^{t \wedge \tau_n \wedge \theta}p\left \Vert x_{\Delta}(s)\right \Vert ^{p-2} \left(x_{\Delta}(s)^2 - \bar{x}_{\Delta}(s)^2\right)^T\\ &\times\left( F_{\Delta}(\bar{y}_{\Delta}(s)) + \frac{p}{2}\left(\left \Vert G_{\Delta, 1}(\bar{y}_{\Delta}(s))\right \Vert ^{2}, \ldots, \left \Vert G_{\Delta, d}(\bar{y}_{\Delta}(s))\right \Vert ^{2}\right)^T \right)ds\\ & + \mathbb{E}\displaystyle \int _{0}^{t \wedge \tau_n \wedge \theta}p\left \Vert x_{\Delta}(s)\right \Vert ^{p-2} \left(\bar{x}_{\Delta, 1}(s)^2, \ldots, \bar{x}_{\Delta, d}(s)^2\right)\\ &\times\left( F_{\Delta}(\bar{y}_{\Delta}(s)) + \frac{p}{2}\left(\left \Vert G_{\Delta, 1}(\bar{y}_{\Delta}(s))\right \Vert ^{2}, \ldots, \left \Vert G_{\Delta, d}(\bar{y}_{\Delta}(s))\right \Vert ^{2}\right)^T \right)ds\\ \triangleq & \left \Vert x_0\right \Vert ^p + A + B. \end{align*} $

由Itô公式可得：

$ \begin{align*} x_{\Delta, i}(t)^2 - \bar{x}_{\Delta, i}(t)^2 =& \displaystyle \int _{t_k}^{t}2x_{\Delta, i}(s)^2\left(F_{\Delta, i}(\bar{y}_{\Delta}(s)) + \Vert G_{\Delta, i}(\bar{y}_{\Delta}(s))\Vert ^2\right)ds\\ &+ \displaystyle \int _{t_k}^{t}2x_{\Delta, i}(s)^2G_{\Delta, i}(\bar{y}_{\Delta}(s))dB(s). \end{align*} $

于是,

$ \begin{align*} &x_{\Delta}(t)^2 - \bar{x}_{\Delta}(t)^2 \\ =& 2\displaystyle \int _{t_k}^{t}\text{diag}(x_{\Delta, 1}(s)^2, \ldots, x_{\Delta, d}(s)^2)\left(F_{\Delta}(\bar{y}_{\Delta}(s)) + \left(\Vert G_{\Delta, 1}(\bar{y}_{\Delta}(s))\Vert ^2, \ldots, \Vert G_{\Delta, d}(\bar{y}_{\Delta}(s))\Vert ^2\right)^T\right)ds\\ & + 2\displaystyle \int _{t_k}^{t}\text{diag}(x_{\Delta, 1}(s)^2, \ldots, x_{\Delta, d}(s)^2)G_{\Delta}(\bar{y}_{\Delta}(s))dB(s). \end{align*} $

综合应用Itô公式、式(3.2)、(3.3)、Hölder不等式以及文献[19] 中定理7.1, 可推得以下估计：对于$ t \in [t_k, t_{k+1}) $,

$ \begin{align*} &\mathbb{E}\left \Vert x_{\Delta}(t)^2 - \bar{x}_{\Delta}(t)^2\right \Vert ^{\frac{p}{2}} \\ \leqslant & C \mathbb{E}\Big \Vert \displaystyle \int _{t_k}^{t}\text{diag}(x_{\Delta, 1}(s)^2, \ldots, x_{\Delta, d}(s)^2) \\ &\times\left(F_{\Delta}(\bar{y}_{\Delta}(s)) + \left(\Vert G_{\Delta, 1}(\bar{y}_{\Delta}(s))\Vert ^2, \ldots, \Vert G_{\Delta, d}(\bar{y}_{\Delta}(s))\Vert ^2\right)^T\right)ds\Big \Vert ^{\frac{p}{2}}\\ &+ C \mathbb{E}\left \Vert \displaystyle \int _{t_k}^{t}\text{diag}(x_{\Delta, 1}(s)^2, \ldots, x_{\Delta, d}(s)^2)G_{\Delta}(\bar{y}_{\Delta}(s))dB(s)\right \Vert ^{\frac{p}{2}}\\ \leqslant & C(t - t_k)^{\frac{p}{2}-1}\mathbb{E}\displaystyle \int _{t_k}^{t}\Vert x_{\Delta}(s)\Vert ^p\left \Vert F_{\Delta}(\bar{y}_{\Delta}(s)) + \left(\Vert G_{\Delta, 1}(\bar{y}_{\Delta}(s))\Vert ^2, \ldots, \Vert G_{\Delta, d}(\bar{y}_{\Delta}(s))\Vert ^2\right)^T\right \Vert ^{\frac{p}{2}}ds\\ &+ C(t - t_k)^{\frac{p}{4}-1}\mathbb{E}\displaystyle \int _{t_k}^{t}\Vert x_{\Delta}(s)\Vert ^p\left \Vert G_{\Delta}(\bar{y}_{\Delta}(s))\right \Vert ^{\frac{p}{2}}ds\\ \leqslant& C \Delta^{\frac{p}{4}-1}(\Delta^{\frac{p}{4}}h(\Delta)^{\frac{p}{2}} + h(\Delta)^{\frac{p}{4}})\mathbb{E}\displaystyle \int _{t_k}^{t}\Vert x_{\Delta}(s)\Vert ^p ds\\ \leqslant& C \Delta^{\frac{p}{4}-1}h(\Delta)^{\frac{p}{4}}\mathbb{E}\displaystyle \int _{t_k}^{t}\Vert x_{\Delta}(s)\Vert ^p ds, \end{align*} $

基于Young不等式, 我们进一步得到

$ \begin{align*} A \leqslant& \mathbb{E}\displaystyle \int _{0}^{t \wedge \tau_n \wedge \theta}p\Vert x_{\Delta}(s)\Vert ^{p - 2}\left\Vert x_{\Delta}(s)^2 - \bar{x}_{\Delta}(s)^2\right\Vert \\ &\times\left\Vert F_{\Delta}(\bar{y}_{\Delta}(s)) + \frac{p}{2}\left(\left\Vert G_{\Delta, 1}(\bar{y}_{\Delta}(s))\right\Vert ^{2}, \ldots, \left\Vert G_{\Delta, d}(\bar{y}_{\Delta}(s))\right\Vert ^{2}\right)^T\right\Vert ds \\ \leqslant& \mathbb{E}\displaystyle \int _{0}^{t \wedge \tau_n \wedge \theta}(p - 2)\Vert x_{\Delta}(s)\Vert ^{p}ds + \mathbb{E}\displaystyle \int _{0}^{t \wedge \tau_n \wedge \theta}\frac{2}{p}\left\Vert x_{\Delta}(s)^2 - \bar{x}_{\Delta}(s)^2\right\Vert ^{\frac{p}{2}} \\ &\times\left\Vert F_{\Delta}(\bar{y}_{\Delta}(s)) + \frac{p}{2}\left(\left\Vert G_{\Delta, 1}(\bar{y}_{\Delta}(s))\right\Vert ^{2}, \ldots, \left\Vert G_{\Delta, d}(\bar{y}_{\Delta}(s))\right\Vert ^{2}\right)^T\right\Vert ^{\frac{p}{2}}ds\\ \leqslant& C\mathbb{E}\displaystyle \int _{0}^{t \wedge \tau_n \wedge \theta}\Vert x_{\Delta}(s)\Vert^{p}ds + C\mathbb{E}\displaystyle \int _{0}^{t \wedge \tau_n \wedge \theta}\left\Vert x_{\Delta}(s)^2 - \bar{x}_{\Delta}(s)^2\right\Vert^{\frac{p}{2}}h(\Delta)^{\frac{p}{2}}ds\\ \leqslant& C\mathbb{E}\displaystyle \int _{0}^{t \wedge \tau_n \wedge \theta}\Vert x_{\Delta}(s)\Vert^{p}ds + C\displaystyle \int _{0}^{t \wedge \tau_n \wedge \theta}\Delta^{\frac{p}{4}-1}h(\Delta)^{\frac{p}{2}}h(\Delta)^{\frac{p}{4}}\displaystyle \int _{t_k}^{s}\mathbb{E}\Vert x_{\Delta}(u)\Vert^{p}du ds \end{align*} $

$ \begin{align*} \leqslant& C\mathbb{E}\displaystyle \int _{0}^{t \wedge \tau_n \wedge \theta}\Vert x_{\Delta}(s)\Vert^{p}ds + C\displaystyle \int _{0}^{t \wedge \tau_n \wedge \theta}\Delta^{\frac{p}{4}}h(\Delta)^{\frac{3p}{4}}\sup\limits_{r\in[0, s]}\mathbb{E}\Vert x_{\Delta}(r)\Vert^{p} ds\\ \leqslant& C\displaystyle \int _{0}^{t \wedge \tau_n \wedge \theta}\sup\limits_{r \in [0, s]}\mathbb{E}\Vert x_{\Delta}(r)\Vert^{p}ds, \end{align*} $

其中$ s \in [t_k, t_{k+1}) $. 根据假设条件($ A_2 $)和引理3.2, 我们可得到以下估计式

$ \begin{align*} B =& \mathbb{E}\displaystyle \int _{0}^{t \wedge \tau_n \wedge \theta}p\left\Vert x_{\Delta}(s)\right\Vert^{p-2} \left(\bar{x}_{\Delta, 1}(s)^2, \ldots, \bar{x}_{\Delta, d}(s)^2\right)\\ &\times\left( F_{\Delta}(\bar{y}_{\Delta}(s)) + \frac{p}{2}\left(\left\Vert G_{\Delta, 1}(\bar{y}_{\Delta}(s))\right\Vert^{2}, \ldots, \left\Vert G_{\Delta, d}(\bar{y}_{\Delta}(s))\right\Vert^{2}\right)^T \right)ds\\ =&\mathbb{E}\displaystyle \int _{0}^{t \wedge \tau_n \wedge \theta}p\left\Vert x_{\Delta}(s)\right\Vert^{p-2}\left(\bar{x}_{\Delta}(s)^{T}f_{\Delta}(\bar{x}_{\Delta}(s)) + \frac{p - 1}{2}\Vert g_{\Delta}(\bar{x}_{\Delta}(s))\Vert^2 \right)ds\\ \leqslant& C\mathbb{E}\displaystyle \int _{0}^{t \wedge \tau_n \wedge \theta}\Vert x_{\Delta}(s)\Vert^{p-2}(1+\Vert\bar{x}_{\Delta}(s)\Vert^2)ds\\ \leqslant& C + C\mathbb{E}\displaystyle \int _{0}^{t \wedge \tau_n \wedge \theta}\Vert x_{\Delta}(s)\Vert^{p}ds + C\mathbb{E}\displaystyle \int _{0}^{t \wedge \tau_n \wedge \theta}\Vert\bar{x}_{\Delta}(s)\Vert^{p}ds\\ \leqslant& C + C\displaystyle \int _{0}^{t \wedge \tau_n \wedge \theta}\left(\sup\limits_{r\in[0, s]}\mathbb{E}\Vert x_{\Delta}(r)\Vert^{p}\right)ds. \end{align*} $

由此可得存在常数$ C $使得对所有$ r \in [0, T] $成立：

$ \sup\limits_{t \in [0, r]}\mathbb{E}\Vert x_{\Delta}(t \wedge \tau_n \wedge \theta)\Vert^p \leqslant C + C\displaystyle \int _{0}^{r}\sup\limits_{t \in [0, s]}\mathbb{E}\Vert x_{\Delta}(t \wedge \tau_n \wedge \theta)\Vert^pds, $

应用Gronwall不等式可知存在常数$ C $满足：

$ \sup\limits_{t\in[0, T]}\mathbb{E}\Vert x_{\Delta}(t\wedge\tau_n\wedge\theta)\Vert^{p}<C. $

类似地, 结合注释2.1、引理3.2和条件$ (\alpha + 2) \vee (\beta+2) < p\wedge q $, 可选取充分小的$ \varepsilon > 0 $使得$ ((\alpha \vee \beta) + 2)(1+\varepsilon) < p\wedge q $, 从而对每个分量$ i = 1, \ldots, d $有

$ \begin{align*} &\mathbb{E}(x_{\Delta, i}(t\wedge\tau_n\wedge\theta))^{-q}\\ =& x_{\Delta, i}(0)^{-q} - q\mathbb{E}\displaystyle \int _{0}^{t \wedge \tau_n \wedge \theta}x_{\Delta, i}(s)^{-q}\left(F_{\Delta, i}(\bar{y}_{\Delta}(s))-\frac{q}{2}\Vert G_{\Delta, i}(\bar{y}_{\Delta}(s))\Vert^2\right)ds\\ =& x_{\Delta, i}(0)^{-q} - q\mathbb{E}\displaystyle \int _{0}^{t \wedge \tau_n \wedge \theta}x_{\Delta, i}(s)^{-q} \left(\frac{\bar{x}_{\Delta, i}(s)f_{\Delta, i}(\bar{x}_{\Delta}(s)) - \frac{q+1}{2}\Vert g_{\Delta, i}(\bar{x}_{\Delta}(s))\Vert^2}{\bar{x}_{\Delta, i}(s)^2}\right)ds\\ \leqslant& C + C\mathbb{E}\displaystyle \int _{0}^{t \wedge \tau_n \wedge \theta}x_{\Delta, i}(s)^{-q} \left( \frac{|\bar{x}_{\Delta, i}(s)|| f_{\Delta, i}(\bar{x}_{\Delta}(s))| + \Vert g_{\Delta, i}(\bar{x}_{\Delta}(s))\Vert^2}{\bar{x}_{\Delta, i}(s)^2} \right) I_{\left\{\bar{x}_{\Delta, i}(s) \geqslant x^*\right\}}ds\\ \leqslant& C + C\mathbb{E}\displaystyle \int _{0}^{t \wedge \tau_n \wedge \theta}\left(\frac{{x_{\Delta, i}(s)}}{{\bar{x}_{\Delta, i}(s)}}\right)^{-q}{\bar{x}_{\Delta, i}(s)}^{-q-2} \\ &\times(1 + \Vert\bar{x}_{\Delta}(s)\Vert^{(\alpha + 2)\vee(\beta+2)} + \left\Vert\bar{x}_{\Delta}(s)^{-1}\right\Vert^{\beta + 2}) I_{\left\{\bar{x}_{\Delta, i}(s) \geqslant x^*\right\}}ds \end{align*} $

$ \begin{align*} \leqslant& C + C\mathbb{E}\displaystyle \int _{0}^{t \wedge \tau_n \wedge \theta}\left(\frac{{x_{\Delta, i}(s)}}{{\bar{x}_{\Delta, i}(s)}}\right)^{-q}(1 + \Vert\bar{x}_{\Delta}(s)\Vert^{(\alpha + 2)\vee(\beta+2)} + \left\Vert\bar{x}_{\Delta}(s)^{-1}\right\Vert^{\beta + 2})ds\\ \leqslant& C + C\mathbb{E}\displaystyle \int _{0}^{t \wedge \tau_n \wedge \theta}\left(\frac{{x_{\Delta, i}(s)}}{{\bar{x}_{\Delta, i}(s)}}\right)^{-q}ds + C\mathbb{E}\displaystyle \int _{0}^{t \wedge \tau_n \wedge \theta}\left(\frac{{x_{\Delta, i}(s)}}{{\bar{x}_{\Delta, i}(s)}}\right)^{-q(1+\varepsilon^{-1})}\\ &+ \Vert\bar{x}_{\Delta}(s)\Vert^{((\alpha\vee\beta)+2)(1+\varepsilon)}+\left\Vert\bar{x}_{\Delta}(s)^{-1}\right\Vert^{(\beta + 2)(1+\varepsilon)}ds \\ \leqslant& C + C\mathbb{E}\displaystyle \int _{0}^{t \wedge \tau_n \wedge \theta}\left\Vert\bar{x}_{\Delta}(s)^{-1}\right\Vert^{(\beta + 2)(1+\varepsilon)}ds \\ \leqslant& C + C\mathbb{E}\displaystyle \int _{0}^{t \wedge \tau_n \wedge \theta}\sum\limits_{i=1}^{d} \bar{x}_{\Delta, i}(s)^{-(\beta + 2)(1+\varepsilon)}ds, \end{align*} $

因此有

$ \sup\limits_{t \in [0, r]}\mathbb{E}\sum\limits_{i=1}^{d}(x_{\Delta, i}(t\wedge\tau_n\wedge\theta))^{-q} \leqslant C+C\displaystyle \int _{0}^{r}\sup\limits_{t \in [0, s]}\mathbb{E}\sum\limits_{i=1}^{d}(x_{\Delta, i}(t\wedge\tau_n\wedge\theta))^{-q}ds. $

再次应用Gronwall不等式即得

$ \sup\limits_{t\in[0, T]}\mathbb{E}\sum\limits_{i=1}^{d}(x_{\Delta, i}(t\wedge\tau_n\wedge\theta))^{-q} < C. $

进一步有

$ \sup\limits_{t\in[0, T]}\mathbb{E}\left\Vert{x_{\Delta}(t\wedge\tau_n\wedge\theta)}^{-1}\right\Vert^{q} < C. $

最终令$ n \to \infty $即得所需结论.

在下文中, 我们记$ e_{\Delta}(t) = x(t) - x_{\Delta}(t) $并取实数$ R > \Vert\ln x_0\Vert $. 定义两个停时

$ \begin{align} \tau_{R}=\inf\left\{t\in[0, T]\mid\Vert y(t)\Vert\geqslant R\right\}\quad\text{和}\quad\tau_{R}^{\Delta}=\inf\left\{t\in[0, T]\mid \Vert y_{\Delta}(t)\Vert\geqslant R\right\}, \end{align} $

(3.6)

其中$ y(t) = \ln(x(t)) $. 进一步设$ \tau = \tau_{R} \wedge \tau_{R}^{\Delta} $. 在后续推导中, 我们用$ C $表示依赖于$ x_0, T $等参数, 但与步长$ \Delta $和$ R $无关的通用正常数, 其具体值在不同位置可能不同.

引理3.4  在假设条件($ A_1 $)、($ A_2 $)和($ A_3 $)成立且满足参数约束$ \frac{pr^*}{p-r^*}<\frac{p\wedge q}{(\alpha + 2)\vee(\beta+2)} $的前提下, 给定阈值$ R > \Vert \ln x_0\Vert $, 设$ \tau $为式(3.6)定义的停时. 当步长$ \Delta $充分小使得$ \varphi^{-1}(h(\Delta)) \geqslant R $时, 对任意$ 2 \leqslant r < r^{*} $, 存在与步长$ \Delta $无关的常数$ C $使得

$ \sup\limits_{t\in[0, T]}\mathbb{E}\Vert e_{\Delta}(t\wedge\tau)\Vert^{r}<C\Delta^{\frac{r}{2}}h(\Delta)^{\frac{r}{2}}. $

证  首先观察到, 对于所有$ s\in[0, T\wedge\tau] $, 有$ \Vert y_{\Delta}(s)\Vert< R $. 根据假设条件$ \varphi^{-1}(h(\Delta)) \geqslant R $, 可以推得在$ s \in [0, T\wedge\tau] $范围内, $ F_{\Delta}(\bar{y}_{\Delta}(s)) = F(\bar{y}_{\Delta}(s)) $和$ G_{\Delta}(\bar{y}_{\Delta}(s)) = G(\bar{y}_{\Delta}(s)) $成立.

对$ e^{y_{\Delta}(t)} $应用Itô公式, 得到

$ \begin{align*} &x_{\Delta}(t\wedge\tau) \\ =& x_0 + \displaystyle \int _{0}^{t\wedge\tau}\text{diag}\left(x_{\Delta, 1}(s), \ldots, x_{\Delta, d}(s)\right)\left(F_{\Delta}(\bar{y}_{\Delta}(s)) + \frac{1}{2}\left(\left\Vert G_{\Delta, 1}(\bar{y}_{\Delta}(s))\right\Vert^2, \ldots, \left\Vert G_{\Delta, d}(\bar{y}_{\Delta}(s))\right\Vert^2\right)^T\right)ds \\ &+ \displaystyle \int _{0}^{t\wedge\tau}\text{diag}\left(x_{\Delta, 1}(s), \ldots, x_{\Delta, d}(s)\right)G_{\Delta}(\bar{y}_{\Delta}(s))dB(s). \end{align*} $

进一步地

$ \begin{align*} &e_{\Delta}(t\wedge\tau)\\ =& \displaystyle \int _{0}^{t\wedge\tau}f(x(s))\\ &- \text{diag}\left(x_{\Delta, 1}(s), \ldots, x_{\Delta, d}(s)\right)\Big(F_{\Delta}(\bar{y}_{\Delta}(s)) + \frac{1}{2}\left(\Vert G_{\Delta, 1}(\bar{y}_{\Delta}(s))\Vert^2, \ldots, \Vert G_{\Delta, d}(\bar{y}_{\Delta}(s))\Vert^2\right)^T\Big)ds\\ &+ \displaystyle \int _{0}^{t\wedge\tau}g(x(s)) - \text{diag}\left(x_{\Delta, 1}(s), \ldots, x_{\Delta, d}(s)\right)G_{\Delta}(\bar{y}_{\Delta}(s))dB(s)\\ =& \displaystyle \int _{0}^{t\wedge\tau}f(x(s)) - \text{diag}\left(\frac{x_{\Delta, 1}(s)}{\bar{x}_{\Delta, 1}(s)}, \ldots, \frac{x_{\Delta, d}(s)}{\bar{x}_{\Delta, d}(s)}\right)f(\bar{x}_{\Delta}(s))ds\\ &+ \displaystyle \int _{0}^{t\wedge\tau}g(x(s)) - \text{diag}\left(\frac{x_{\Delta, 1}(s)}{\bar{x}_{\Delta, 1}(s)}, \ldots, \frac{x_{\Delta, d}(s)}{\bar{x}_{\Delta, d}(s)}\right)g(\bar{x}_{\Delta}(s))dB(s). \end{align*} $

由Young不等式

$ \begin{align*} 2ab \leqslant \frac{2\varepsilon}{2} a^2 + \frac{2}{2\varepsilon} b^2 = \varepsilon a^2 + \frac{1}{\varepsilon} b^2, \quad a, b>0, \end{align*} $

取$ \varepsilon = \frac{r^*-r}{r-1} $, 对$ \Vert x(t) - x_{\Delta}(t)\Vert^r $应用Itô公式, 则可得到以下估计式：

$ \begin{align*} \mathbb{E}\Vert e_{\Delta}(t\wedge\tau)\Vert^r \leqslant& \mathbb{E}\displaystyle \int _{0}^{t\wedge\tau}r\Vert e_{\Delta}(s)\Vert^{r-2}e_{\Delta}(s)^T\left(f(x(s)) - \text{diag}\left(\frac{x_{\Delta}(s)}{\bar{x}_{\Delta}(s)}\right)f(\bar{x}_{\Delta}(s))\right)ds\\ & + \mathbb{E}\displaystyle \int _{0}^{t\wedge\tau}\frac{r(r-1)}{2}\Vert e_{\Delta}(s)\Vert^{r-2}\left\Vert g(x(s)) - \text{diag}\left(\frac{x_{\Delta}(s)}{\bar{x}_{\Delta}(s)}\right)g(\bar{x}(s))\right\Vert^2ds\\ \leqslant& J_1 + J_2, \end{align*} $

其中

$ \begin{align*} J_1 =& r\mathbb{E}\displaystyle \int _{0}^{t\wedge\tau}\Vert e_{\Delta}(s)\Vert^{r-2}\left(e_{\Delta}(s)^T\left(f(x(s)) - f(x_{\Delta}(s))\right) + \frac{r^*-1}{2}\left\Vert g(x(s)) - g(x_{\Delta}(s))\right\Vert^2\right)ds \end{align*} $

以及

$ \begin{align*} J_2 =& r\mathbb{E}\displaystyle \int _{0}^{t\wedge\tau}\Vert e_{\Delta}(s)\Vert^{r-2}e_{\Delta}(s)^T\left(f(x_{\Delta}(s)) - \text{diag}\left(\frac{x_{\Delta}(s)}{\bar{x}_{\Delta}(s)}\right)f(\bar{x}_{\Delta}(s))\right)ds\\ &+\frac{r(r-1)(r^*-1)}{2(r^*-r)}\mathbb{E}\displaystyle \int _{0}^{t\wedge\tau}\Vert e_{\Delta}(s)\Vert^{r-2}\left\Vert g(x_{\Delta}(s)) - \text{diag}\left(\frac{x_{\Delta}(s)}{\bar{x}_{\Delta}(s)}\right)g(\bar{x}_{\Delta}(s))\right\Vert^2ds. \end{align*} $

根据假设条件($ A_3 $), 我们首先得到：$ J_1 \leqslant rH\mathbb{E} \int _{0}^{t\wedge\tau}\Vert e_{\Delta}(s)\Vert^rds $. 进一步应用Young不等式, 可推导出以下估计：

$ \begin{align*} J_2 \leqslant& C\mathbb{E}\displaystyle \int _{0}^{t\wedge\tau}\Vert e_{\Delta}(s)\Vert ^{r-1}\left\Vert f(x_{\Delta}(s)) - f(\bar{x}_{\Delta}(s)) + f(\bar{x}_{\Delta}(s)) - \text{diag}\left(\frac{x_{\Delta}(s)}{\bar{x}_{\Delta}(s)}\right)f(\bar{x}_{\Delta}(s))\right\Vert ds\\ &+ C\mathbb{E}\displaystyle \int _{0}^{t\wedge\tau}\Vert e_{\Delta}(s)\Vert^{r-2}\left\Vert g(x_{\Delta}(s)) - g(\bar{x}_{\Delta}(s)) + g(\bar{x}_{\Delta}(s)) - \text{diag}\left(\frac{x_{\Delta}(s)}{\bar{x}_{\Delta}(s)}\right)g(\bar{x}_{\Delta}(s))\right\Vert^2ds\\ \end{align*} $

$ \begin{align*} \leqslant& C\mathbb{E}\displaystyle \int _{0}^{t\wedge\tau}\Vert e_{\Delta}(s)\Vert^{r} + \left\Vert f(x_{\Delta}(s)) - f(\bar{x}_{\Delta}(s)) + f(\bar{x}_{\Delta}(s)) - \text{diag}\left(\frac{x_{\Delta}(s)}{\bar{x}_{\Delta}(s)}\right)f(\bar{x}_{\Delta}(s))\right\Vert^rds\\ &+ C\mathbb{E}\displaystyle \int _{0}^{t\wedge\tau}\Vert e_{\Delta}(s)\Vert^{r} + \left\Vert g(x_{\Delta}(s)) - g(\bar{x}_{\Delta}(s)) + g(\bar{x}_{\Delta}(s)) - \text{diag}\left(\frac{x_{\Delta}(s)}{\bar{x}_{\Delta}(s)}\right)g(\bar{x}_{\Delta}(s))\right\Vert^rds\\ \leqslant& C\mathbb{E}\displaystyle \int _{0}^{t\wedge\tau}\Vert e_{\Delta}(s)\Vert^{r}ds \\ &+ C\mathbb{E}\displaystyle \int _{0}^{t\wedge\tau}\left\Vert f(x_{\Delta}(s)) - f(\bar{x}_{\Delta}(s))\right\Vert^rds + C\mathbb{E}\displaystyle \int _{0}^{t\wedge\tau}\left\Vert E - \text{diag}\left(\frac{x_{\Delta}(s)}{\bar{x}_{\Delta}(s)}\right)\right\Vert^r\Vert f(\bar{x}_{\Delta}(s))\Vert^rds\\ &+C\mathbb{E}\displaystyle \int _{0}^{t\wedge\tau}\left\Vert g(x_{\Delta}(s)) - g(\bar{x}_{\Delta}(s))\right\Vert^rds + C\mathbb{E}\displaystyle \int _{0}^{t\wedge\tau}\left\Vert E - \text{diag}\left(\frac{x_{\Delta}(s)}{\bar{x}_{\Delta}(s)}\right)\right\Vert^r\Vert g(\bar{x}_{\Delta}(s))\Vert^rds. \end{align*} $

基于假设条件($ A_1 $)、注释2.1以及Hölder不等式, 可得以下估计式：

$ \begin{align*} J_2 \leqslant& C\mathbb{E}\displaystyle \int _{0}^{t\wedge\tau}\Vert e_{\Delta}(s)\Vert^{r}ds+ C\mathbb{E}\displaystyle \int _{0}^{t\wedge\tau}\Big(\mathbb{E}\Big(1 + \Vert x_{\Delta}(s)\Vert^{(1+\varepsilon)\alpha r} + \left\Vert x_{\Delta}(s)^{-1}\right\Vert^{(1+\varepsilon)\beta r} + \Vert\bar{x}_{\Delta}(s)\Vert^{(1+\varepsilon)\alpha r} \\ &+ \left\Vert\bar{x}_{\Delta}(s)^{-1}\right\Vert^{(1+\varepsilon)\beta r}\Big)\Big)^{\frac{1}{1+\varepsilon}}\left(\mathbb{E}\Vert x_{\Delta}(s) - \bar{x}_{\Delta}(s)\Vert^{\frac{(1+\varepsilon)r}{\varepsilon}}\right)^{\frac{\varepsilon}{1+\varepsilon}}ds \\ &+C\mathbb{E}\displaystyle \int _{0}^{t\wedge\tau}\Big(\mathbb{E}\Big(1 + \Vert x_{\Delta}(s)\Vert^{(1+\varepsilon)\alpha r/2} + \left\Vert x_{\Delta}(s)^{-1}\right\Vert^{(1+\varepsilon)\beta r/2} + \Vert\bar{x}_{\Delta}(s)\Vert^{(1+\varepsilon)\alpha r/2} \\ & + \left\Vert\bar{x}_{\Delta}(s)^{-1}\right\Vert^{(1+\varepsilon)\beta r/2}\Big)\Big)^{\frac{1}{1+\varepsilon}}\left(\mathbb{E}\Vert x_{\Delta}(s) - \bar{x}_{\Delta}(s)\Vert^{\frac{(1+\varepsilon)r}{\varepsilon}}\right)^{\frac{\varepsilon}{1+\varepsilon}}ds \\ &+C\mathbb{E}\displaystyle \int _{0}^{t\wedge\tau}\left(\mathbb{E}\left\Vert E - \text{diag}\left(\frac{x_{\Delta}(s)}{\bar{x}_{\Delta}(s)}\right)\right\Vert^{\frac{(1+\varepsilon)r}{\varepsilon}}\right)^{\frac{\varepsilon}{1+\varepsilon}} \\ &\times\left(\mathbb{E}\left(1 + \Vert\bar{x}_{\Delta}(s)\Vert^{(1+\varepsilon)((\alpha\vee\beta)+1)r} + \left\Vert\bar{x}_{\Delta}(s)^{-1}\right\Vert^{(1+\varepsilon)(\beta+1)r}\right)\right)^{\frac{1}{1+\varepsilon}}ds. \\ &+C\mathbb{E}\displaystyle \int _{0}^{t\wedge\tau}\left(\mathbb{E}\left\Vert E - \text{diag}\left(\frac{x_{\Delta}(s)}{\bar{x}_{\Delta}(s)}\right)\right\Vert^{\frac{(1+\varepsilon)r}{\varepsilon}}\right)^{\frac{\varepsilon}{1+\varepsilon}} \\ &\times\left(\mathbb{E}\left(1 + \Vert\bar{x}_{\Delta}(s)\Vert^{(1+\varepsilon)((\alpha\vee\beta)+2)r/2} + \left\Vert\bar{x}_{\Delta}(s)^{-1}\right\Vert^{(1+\varepsilon)(\beta+2)r/2}\right)\right)^{\frac{1}{1+\varepsilon}}ds. \end{align*} $

在$ \frac{pr^*}{p-r^*}<\frac{p}{(\alpha + 2)\vee(\beta+2)}\wedge\frac{q}{(\alpha + 2)\vee(\beta+2)} $成立的条件下, 存在$ \varepsilon > 0 $使得

$ \frac{p}{(\alpha + 2)\vee(\beta+2)}\wedge\frac{q}{(\alpha + 2)\vee(\beta+2)} > (1+\varepsilon)r^*>\frac{pr^*}{p-r^*}. $

由此可得以下参数约束关系：

$ \frac{(1+\varepsilon)((\alpha\vee\beta)+2)r^*}{2}<(1+\varepsilon)((\alpha\vee\beta)+2)r^* < p\wedge q , $

由于$ r^* > r \geqslant 2 $, 可得$ p>4 $, 结合$ (1+\varepsilon)(p-r^*)>p $, 有$ 2<\frac{(1+\varepsilon)r}{\varepsilon}<\frac{(1+\varepsilon)r^*}{\varepsilon}<p $. 综合应用式(3.2)、(3.3)、引理3.2、Hölder不等式以及文献[19]中的定理7.1, 我们最终得到

$ \begin{align*} &\mathbb{E}\left\Vert x_{\Delta}(t)-\bar{x}_{\Delta}(t)\right\Vert^{\frac{(1+\varepsilon)r}{\varepsilon}}\\ \leqslant& C\mathbb{E}\bigg\Vert\displaystyle \int _{t_k}^{t}\text{diag}\left(x_{\Delta, 1}(s), \ldots, x_{\Delta, d}(s)\right)\\ &\times\left(F_{\Delta}(\bar{y}_{\Delta}(s)) + \frac{1}{2}\left(\left\Vert G_{\Delta, 1}(\bar{y}_{\Delta}(s))\right\Vert^{2}, \ldots, \left\Vert G_{\Delta, d}(\bar{y}_{\Delta}(s))\right\Vert^{2}\right)^T\right)ds\bigg\Vert^{\frac{(1+\varepsilon)r}{\varepsilon}}\\ &+ C\mathbb{E}\left\Vert\displaystyle \int _{t_k}^{t}\text{diag}\left(x_{\Delta, 1}(s), \ldots, x_{\Delta, d}(s)\right)G_{\Delta}(\bar{y}_{\Delta}(s))dB(s)\right\Vert^{\frac{(1+\varepsilon)r}{\varepsilon}}\\ \leqslant& C(t - t_k)^{\frac{(1+\varepsilon)r}{\varepsilon}-1} \\ &\times\mathbb{E}\displaystyle \int _{t_k}^{t}\Vert x_{\Delta}(s)\Vert^{\frac{(1+\varepsilon)r}{\varepsilon}} \left\Vert F_{\Delta}(\bar{y}_{\Delta}(s)) + \frac{1}{2}\left(\left\Vert G_{\Delta, 1}(\bar{y}_{\Delta}(s))\right\Vert^{2}, \ldots, \left\Vert G_{\Delta, d}(\bar{y}_{\Delta}(s))\right\Vert^{2}\right)^T\right\Vert^{\frac{(1+\varepsilon)r}{\varepsilon}}ds\\ &+ C(t - t_k)^{\frac{(1+\varepsilon)r}{2\varepsilon}-1}\mathbb{E}\displaystyle \int _{t_k}^{t}\Vert x_{\Delta}(s)\Vert^{\frac{(1+\varepsilon)r}{\varepsilon}}\left\Vert G_{\Delta}(\bar{y}_{\Delta}(s))\right\Vert^{\frac{(1+\varepsilon)r}{\varepsilon}}ds\\ \leqslant& C{\Delta}^{\frac{(1+\varepsilon)r}{2\varepsilon}-1}\left({\Delta}^{\frac{(1+\varepsilon)r}{2\varepsilon}}h(\Delta)^{\frac{(1+\varepsilon)r}{\varepsilon}} + h(\Delta)^{\frac{(1+\varepsilon)r}{2\varepsilon}}\right)\mathbb{E}\displaystyle \int _{t_k}^{t}\Vert x_{\Delta}(s)\Vert^{\frac{(1+\varepsilon)r}{\varepsilon}}ds\\ \leqslant& C{\Delta}^{\frac{(1+\varepsilon)r}{2\varepsilon}}h(\Delta)^{\frac{(1+\varepsilon)r}{2\varepsilon}}. \end{align*} $

综合引理3.2和3.3的结论, 我们得到以下估计：$ J_2 \leqslant C \int _{0}^{t\wedge\tau}\mathbb{E}\Vert e_{\Delta}(s)\Vert^r + C{\Delta}^{\frac{r}{2}}h(\Delta)^{\frac{r}{2}} $. 应用Gronwall不等式可得$ \sup_{t\in[0, T]}\mathbb{E}\Vert e_{\Delta}(t\wedge\tau)\Vert^{r}<C\Delta^{\frac{r}{2}}h(\Delta)^{\frac{r}{2}} $.

现在, 我们给出关于收敛速率的主要结果.

定理4.1  在假设条件($ A_1 $)、($ A_2 $)和($ A_3 $)成立, 且参数满足$ \frac{pr^*}{p-r^*}<\frac{p\wedge q}{(\alpha + 2)\vee(\beta+2)} $的前提下, 对于任意$ 2 \leqslant r < r^* $和步长$ \Delta \in (0, 1] $, 存在常数$ C $使得以下强收敛估计成立：

$ \sup\limits_{t\in[0, T]}\mathbb{E}\Vert e_{\Delta}(t)\Vert^{r}<C\Delta^{\frac{(p-r)(p\wedge q)r}{3\sqrt{d}(2(p-r)(p\wedge q)+C(\alpha, \beta)pr)}}, $

其中$ h(\Delta)=\left(3.5C_0\vee\varphi(\Vert\ln x_0\Vert)\right)\Delta^{-\frac{C(\alpha, \beta)pr}{3(2(p-r)(p\wedge q)+C(\alpha, \beta)pr)}} $.

证  设$ R = \varphi^{-1}(h(\Delta)) $. 当参数满足$ \frac{pr^*}{p-r^*}<\frac{p\wedge q}{(\alpha + 2)\vee(\beta+2)} $时, 可推得$ (\alpha\vee\beta)+2 < p\wedge q $. 综合应用Young不等式、Markov不等式及引理3.3, 我们得到：

$ \begin{align*} \sup\limits_{t\in[0, T]}\mathbb{E}(\Vert e_\Delta(t)\Vert^rI_{\{\tau\leqslant T\}}) =&\sup\limits_{t\in[0, T]}\mathbb{E}(\Vert e_\Delta(t)\Vert^r{\delta}^{\frac{r}{p}}I_{\{\tau\leqslant T\}}{\delta}^{-\frac{r}{p}})\\ \leqslant& \frac{r}{p}\sup\limits_{t\in[0, T]}\mathbb{E}\Vert e_\Delta(t)\Vert^p\delta + \frac{p-r}{p}\mathbb{E}(I_{\left\{\tau\leqslant T\right\}}^{\frac{p}{p-r}}){\delta}^{-\frac{r}{p-r}}\\ =& \frac{r}{p}\sup\limits_{t\in[0, T]}\mathbb{E}\Vert e_\Delta(t)\Vert^p\delta + \frac{p-r}{p}\mathbb{P}(\tau\leqslant T){\delta}^{-\frac{r}{p-r}}\\ \end{align*} $

$ \begin{align*} \leqslant& C\delta + C\left(\frac{\sum\nolimits_{i=1}^{d}\mathbb{E}[x_i^p(T\wedge\tau)+x_{\Delta, i}^p(T\wedge\tau)]}{e^{pR/\sqrt{d}}}+\frac{\sum\nolimits_{i=1}^{d}\mathbb{E}[x_i^{-q}(T\wedge\tau)+x_{\Delta, i}^{-q}(T\wedge\tau)]}{e^{qR/\sqrt{d}}}\right){\delta}^{-\frac{r}{p-r}}\\ \leqslant& C\delta + Ce^{-(p\wedge q)\varphi^{-1}(h(\Delta))/\sqrt{d}}{\delta}^{-\frac{r}{p-r}}. \end{align*} $

取$ \delta = e^{-\left((p-r)(p\wedge q)\varphi^{-1}(h(\Delta))\right)/\sqrt{d}p} $, 则可推得以下估计式：

$ \begin{align*} \sup\limits_{t\in[0, T]}\mathbb{E}(\Vert e_\Delta(t)\Vert^rI_{\{\tau\leqslant T\}}) \leqslant& Ce^{-\left((p-r)(p\wedge q)\varphi^{-1}(h(\Delta))\right)/\sqrt{d}p} + Ce^{-(p\wedge q)\varphi^{-1}(h(\Delta))/\sqrt{d}}e^{\left(r(p\wedge q)\varphi^{-1}(h(\Delta))\right)/\sqrt{d}p}\\ \leqslant& Ce^{-\left((p-r)(p\wedge q)\varphi^{-1}(h(\Delta))\right)/\sqrt{d}p}. \end{align*} $

综合上述结果与引理3.4, 我们最终得到以下收敛性估计：

$ \begin{align*} \sup\limits_{t\in[0, T]}\mathbb{E}(\Vert e_\Delta(t)\Vert^r) =& \sup\limits_{t\in[0, T]}\mathbb{E}(\Vert e_\Delta(t)\Vert^rI_{\{\tau > T\}}) + \sup\limits_{t\in[0, T]}\mathbb{E}(\Vert e_\Delta(t)\Vert^rI_{\{\tau\leqslant T\}})\\ =&\sup\limits_{t\in[0, T]}\mathbb{E}\Vert e_\Delta(t\wedge\tau)\Vert^r + \sup\limits_{t\in[0, T]}\mathbb{E}(\Vert e_\Delta(t)\Vert^rI_{\{\tau\leqslant T\}})\\ \leqslant&C\Delta^{\frac{r}{2}}h(\Delta)^{\frac{r}{2}} + Ce^{-\left((p - r)(p\wedge q)\varphi^{-1}(h(\Delta))\right)/\sqrt{d}p}. \end{align*} $

由于$ h(\Delta) \geqslant h(1) > 3C_0 $且$ \varphi(r) = C_0(2 + e^{C(\alpha, \beta)r}) $有

$ e^{-\left((p-r)(p\wedge q)\varphi^{-1}(h(\Delta))\right)/\sqrt{d}p} = \left(\frac{h(\Delta)}{C_0} - 2\right)^{-\frac{(p-r)(p\wedge q)}{\sqrt{d}pC(\alpha, \beta)}} \leqslant \left(\frac{h(\Delta)}{3C_0}\right)^{-\frac{(p-r)(p\wedge q)}{\sqrt{d}pC(\alpha, \beta)}}. $

现取$ h(\Delta)=\left(3.5C_0\vee\varphi(\Vert\ln x_0\Vert)\right)\Delta^{-\frac{C(\alpha, \beta)pr}{3(2(p-r)(p\wedge q)+C(\alpha, \beta)pr)}}, $则存在常数$ C $使得最终强收敛估计成立：

$ \sup\limits_{t\in[0, T]}\mathbb{E}\Vert e_{\Delta}(t)\Vert^{r}<C\Delta^{\frac{(p-r)(p\wedge q)r}{3\sqrt{d}(2(p-r)(p\wedge q)+C(\alpha, \beta)pr)}}. $

注释4.1  在引理3.4中取定$ \varepsilon = 1/2 $. 若进一步假设参数满足$ 1.5r^* < \frac{p\wedge q}{(\alpha + 6)\vee(\beta+6)} $, 则可推得$ \frac{p\wedge q}{(\alpha\vee\beta)+6} > (1+\varepsilon)r^* = \frac{(1+\varepsilon)r^*}{2\varepsilon}. $因此,

$ (1+\varepsilon)((\alpha\vee\beta)+6)r^*<p\wedge q, \quad \frac{((\alpha\vee\beta)+6)(1+\varepsilon)r^*}{2\varepsilon}<p\wedge q. $

基于式(3.2)、(3.3)、注释2.1、引理3.2和3.3, 结合Hölder不等式及文献[19] 中定理7.1, 我们可建立如下估计：

$ \begin{align*} &\mathbb{E}\left|{\frac{x_{\Delta, i}(t)}{\bar{x}_{\Delta, i}(t)}}-1\right|^{\frac{(1+\varepsilon)r}{\varepsilon}} \\ \leqslant& C(t-t_{k})^{\frac{(1+\varepsilon)r}{\varepsilon}-1}\mathbb{E}\displaystyle \int _{t_{k}}^{t}\left|\frac{x_{\Delta, i}(s)}{\bar{x}_{\Delta, i}(s)}\right|^{\frac{(1+\varepsilon)r}{\varepsilon}}\left| F_{\Delta, i}(\bar{y}_{\Delta}(s)) +\frac{1}{2}\Vert G_{\Delta, i}(\bar{y}_{\Delta}(s))\Vert^{2}\right|^{\frac{(1+\varepsilon)r}{\varepsilon}}ds \\ &+C(t-t_{k})^{\frac{(1+\varepsilon)r}{2\varepsilon}-1}\mathbb{E}\displaystyle \int _{t_{k}}^{t}\left|\frac{x_{\Delta, i}(s)}{\bar{x}_{\Delta, i}(s)}\right|^{\frac{(1+\varepsilon)r}{\varepsilon}}\left\Vert G_{\Delta}(\bar{y}_{\Delta}(s))\right\Vert^{\frac{(1+\varepsilon)r}{\varepsilon}}ds\\ \leqslant& C\Delta^{\frac{(1+\varepsilon)r}{\varepsilon}-1}h(\Delta)^{\frac{(1+\varepsilon)r}{2\varepsilon}}\mathbb{E}\displaystyle \int _{t_{k}}^{t}\left|\frac{x_{\Delta, i}(s)}{\bar{x}_{\Delta, i}(s)}\right|^{\frac{(1+\varepsilon)r}{\varepsilon}}\left(\left\Vert F_{\Delta}(\bar{y}_{\Delta}(s))\right\Vert^{\frac{(1+\varepsilon)r}{2\varepsilon}} + \left\Vert G_{\Delta}(\bar{y}_{\Delta}(s))\right\Vert^{\frac{(1+\varepsilon)r}{\varepsilon}}\right)ds \end{align*} $

$ \begin{align*} &+C\Delta^{\frac{(1+\varepsilon)r}{2\varepsilon}-1}\mathbb{E}\displaystyle \int _{t_{k}}^{t}\left|\frac{x_{\Delta, i}(s)}{\bar{x}_{\Delta, i}(s)}\right|^{\frac{(1+\varepsilon)r}{\varepsilon}}\left\Vert G_{\Delta}(\bar{y}_{\Delta}(s))\right\Vert^{\frac{(1+\varepsilon)r}{\varepsilon}}ds \\ \leqslant& C\Delta^{\frac{5(1+\varepsilon)r}{6\varepsilon}-1}\displaystyle \int _{t_{k}}^{t}\left(\mathbb{E}\left|\frac{x_{\Delta, i}(s)}{\bar{x}_{\Delta, i}(s)}\right|^{\frac{(1+\varepsilon)(1+\eta)r}{\varepsilon\eta}}\right)^{\frac{\eta}{1+\eta}} \\ &\times\left(\mathbb{E}\left(1+\Vert\bar{x}_{\Delta}(s)\Vert^{\frac{((\alpha\vee\beta)+4)(1+\varepsilon)(1+\eta)r}{2\varepsilon}} + \left\Vert\bar{x}_{\Delta}(s)^{-1}\right\Vert^{\frac{((\alpha\vee\beta)+4)(1+\varepsilon)(1+\eta)r}{2\varepsilon}}\right)\right)^{\frac{1}{1+\eta}}ds \\ &+C\Delta^{\frac{(1+\varepsilon)r}{2\varepsilon}-1}\displaystyle \int _{t_{k}}^{t}\left(\mathbb{E}\left|\frac{x_{\Delta, i}(s)}{\bar{x}_{\Delta, i}(s)}\right|^{\frac{(1+\varepsilon)(1+\eta)r}{\varepsilon\eta}}\right)^{\frac{\eta}{1+\eta}} \\ &\times\left(\mathbb{E}\left(1+\Vert\bar{x}_{\Delta}(s)\Vert^{\frac{((\alpha\vee\beta)+4)(1+\varepsilon)(1+\eta)r}{2\varepsilon}} + \left\Vert\bar{x}_{\Delta}(s)^{-1}\right\Vert^{\frac{((\alpha\vee\beta)+4)(1+\varepsilon)(1+\eta)r}{2\varepsilon}}\right)\right)^{\frac{1}{1+\eta}}ds \\ \leqslant& C\Delta^{\frac{(1+\varepsilon)r}{2\varepsilon}}. \end{align*} $

同样地,

$ \begin{align*} &\mathbb{E}\left\Vert x_{\Delta}(t)-\bar{x}_{\Delta}(t)\right\Vert^{\frac{(1+\varepsilon)r}{\varepsilon}}\\ \leqslant& C(t - t_k)^{\frac{(1+\varepsilon)r}{\varepsilon}-1} \\ &\times\mathbb{E}\displaystyle \int _{t_k}^{t}\Vert x_{\Delta}(s)\Vert^{\frac{(1+\varepsilon)r}{\varepsilon}}\left\Vert F_{\Delta}(\bar{y}_{\Delta}(s)) + \frac{1}{2}\left(\left\Vert G_{\Delta, 1}(\bar{y}_{\Delta}(s))\right\Vert^{2}, \ldots, \left\Vert G_{\Delta, d}(\bar{y}_{\Delta}(s))\right\Vert^{2}\right)^T\right\Vert^{\frac{(1+\varepsilon)r}{\varepsilon}}ds \\ &+ C(t - t_k)^{\frac{(1+\varepsilon)r}{2\varepsilon}-1}\mathbb{E}\displaystyle \int _{t_k}^{t}\Vert x_{\Delta}(s)\Vert^{\frac{(1+\varepsilon)r}{\varepsilon}}\left\Vert G_{\Delta}(\bar{y}_{\Delta}(s))\right\Vert^{\frac{(1+\varepsilon)r}{\varepsilon}}ds \\ \leqslant& C{\Delta}^{\frac{(1+\varepsilon)r}{\varepsilon}-1}h(\Delta)^{\frac{(1+\varepsilon)r}{2\varepsilon}} \\ &\times\mathbb{E}\displaystyle \int _{t_k}^{t}\Vert x_{\Delta}(s)\Vert^{\frac{(1+\varepsilon)r}{\varepsilon}} \left(1+\Vert\bar{x}_{\Delta}(s)\Vert^{\frac{((\alpha\vee\beta)+4)(1+\varepsilon)r}{2\varepsilon}} + \left\Vert\bar{x}_{\Delta}(s)^{-1}\right\Vert^{\frac{((\alpha\vee\beta)+4)(1+\varepsilon)r}{2\varepsilon}}\right)ds \\ &+ C{\Delta}^{\frac{(1+\varepsilon)r}{2\varepsilon}-1}\mathbb{E}\displaystyle \int _{t_k}^{t}\Vert x_{\Delta}(s)\Vert^{\frac{(1+\varepsilon)r}{\varepsilon}}\left(1+\Vert\bar{x}_{\Delta}(s)\Vert^{\frac{((\alpha\vee\beta)+4)(1+\varepsilon)r}{2\varepsilon}} + \left\Vert\bar{x}_{\Delta}(s)^{-1}\right\Vert^{\frac{((\alpha\vee\beta)+4)(1+\varepsilon)r}{2\varepsilon}}\right)ds \\ \leqslant& C{\Delta}^{\frac{(1+\varepsilon)r}{2\varepsilon}-1}\mathbb{E}\displaystyle \int _{t_k}^{t}\Vert x_{\Delta}(s)\Vert^{\frac{(1+\varepsilon)r}{\varepsilon}}ds \\ &+ C{\Delta}^{\frac{(1+\varepsilon)r}{2\varepsilon}-1}\displaystyle \int _{t_k}^{t}\left(\mathbb{E}\Vert x_{\Delta}(s)\Vert^{\frac{(1+\varepsilon)((\alpha\vee\beta)+6)r}{2\varepsilon}}\right)^{\frac{2}{(\alpha\vee\beta)+6}}\left(\mathbb{E}\left\Vert\bar{x}_{\Delta}(s)\right\Vert^{\frac{((\alpha\vee\beta)+6)(1+\varepsilon)r}{2\varepsilon}}\right)^{\frac{(\alpha\vee\beta)+4}{(\alpha\vee\beta)+6}}ds \\ &+ C{\Delta}^{\frac{(1+\varepsilon)r}{2\varepsilon}-1}\displaystyle \int _{t_k}^{t}\left(\mathbb{E}\Vert x_{\Delta}(s)\Vert^{\frac{(1+\varepsilon)((\alpha\vee\beta)+6)r}{2\varepsilon}}\right)^{\frac{2}{(\alpha\vee\beta)+6}}\left(\mathbb{E}\left\Vert\bar{x}_{\Delta}(s)^{-1}\right\Vert^{\frac{((\alpha\vee\beta)+6)(1+\varepsilon)r}{2\varepsilon}}\right)^{\frac{(\alpha\vee\beta)+4}{(\alpha\vee\beta)+6}}ds \\ \leqslant& C{\Delta}^{\frac{(1+\varepsilon)r}{2\varepsilon}}. \end{align*} $

现设$ h(\Delta) = \left(4C_0\vee\varphi(\Vert\ln x_0\Vert)\right)\Delta^{-\frac{1}{3}} $. 通过引理3.4和定理4.1的类似论证方法, 可得到以下强收敛估计：

$ \sup\limits_{t\in[0, T]}\mathbb{E}\Vert e_{\Delta}(t)\Vert^r < C{\Delta}^{\frac{r}{2}} + C{\Delta}^{\frac{(p-r)(p\wedge q)}{3\sqrt{d}C(\alpha, \beta)p}} < C{\Delta}^{\frac{r}{2}\wedge \frac{(p-r)(p\wedge q)}{3\sqrt{d}C(\alpha, \beta)p}}. $

在第四节中, 我们建立了对数截断EM方法的一般收敛性理论, 所得收敛速率表达式较为复杂. 本节将通过一个典型算例具体展示该方法的收敛性能. 值得注意的是, 对于下列具体的随机微分方程模型, 该方法实际可获得略优于$ 1/2 $阶的收敛速率.

例5.1  考虑如下二维随机微分方程：

$ \begin{align} \left\{ \begin{aligned} &dx_1 = (a_{11}x_1^{-1}(t) + a_{12}x_2(t) - a_{13}x_1^{2}(t)) dt + \sigma_1 x_1(t)dB_1(t), \\ &dx_2 = (a_{21}x_2^{-1}(t) + a_{22}x_1(t) - a_{23}x_2^{2}(t)) dt + \sigma_2 x_2(t)dB_2(t). \end{aligned} \right. \end{align} $

(5.1)

因此, 其系数为

$ \begin{align*} f\left( \begin{bmatrix} x_1\\x_2 \end{bmatrix} \right) =\begin{bmatrix} a_{11}x_1^{-1} + a_{12}x_2- a_{13}x_1^{2}\\ a_{21}x_2^{-1} + a_{22}x_1 - a_{23}x_2^{2} \end{bmatrix} \quad \text{和}\quad g\left( \begin{bmatrix} x_1\\x_2 \end{bmatrix} \right) =\begin{bmatrix} \sigma_1 x_1 & 0\\ 0 & \sigma_2 x_2 \end{bmatrix}. \end{align*} $

另外,

$ \begin{align*} x^Tf(x) &= \begin{bmatrix} x_1 & x_2 \end{bmatrix} \begin{bmatrix} a_{11}x_1^{-1} + a_{12}x_2- a_{13}x_1^{2}\\ a_{21}x_2^{-1} + a_{22}x_1 - a_{23}x_2^{2} \end{bmatrix} = a_{11} + a_{12}x_2x_1 - a_{13}x_1^{3} + a_{21} + a_{22}x_1x_2 - a_{23}x_2^{3}, \\ \Vert g(x)\Vert^2 &= \text{trace} \left( \begin{bmatrix} \sigma_1 x_1 & 0\\ 0 & \sigma_2 x_2 \end{bmatrix} \begin{bmatrix} \sigma_1 x_1 & 0\\ 0 & \sigma_2 x_2 \end{bmatrix} \right)= \sigma_1^2x_1^2 + \sigma_2^2x_2^2. \end{align*} $

设$ r \geqslant 2 $为给定实数. 对于任意$ x , y \in\mathbb{R}_+^2 $, 存在常数$ M $, 使得

$ \begin{align*} \Vert f(x) - f(y)\Vert =& \left\Vert \begin{pmatrix} a_{11}(x_1^{-1} - y_1^{-1}) + a_{12}(x_2 - y_2) - a_{13}(x_1^2 - y_1^2) \\ a_{21}(x_2^{-1} - y_2^{-1}) + a_{22}(x_1 - y_1) - a_{23}(x_2^2 - y_2^2) \end{pmatrix} \right\Vert \\ =& \Big((a_{11}(x_1^{-1} - y_1^{-1}) + a_{12}(x_2 - y_2) - a_{13}(x_1^2 - y_1^2))^2\\ & + (a_{21}(x_2^{-1} - y_2^{-1}) + a_{22}(x_1 - y_1) - a_{23}(x_2^2 - y_2^2))^2\Big)^{\frac{1}{2}} \\ \leqslant& a_{11}|y_1 - x_1|(x_1^{-1}\cdot y_1^{-1}) + a_{12}|x_2 - y_2| + a_{13}|x_1 - y_1|(x_1 + y_1) \\ & + a_{21}|y_2 - x_2|(x_2^{-1}\cdot y_2^{-1}) + a_{22}|x_1 - y_1| + a_{23}|x_2 - y_2|(x_2 + y_2) \\ \leqslant& \frac{1}{2}a_{11}|y_1 - x_1|(x_1^{-2} + y_1^{-2}) + a_{12}|x_2 - y_2| + a_{13}|x_1 - y_1|(x_1 + y_1) \\ & + \frac{1}{2}a_{21}|y_2 - x_2|(x_2^{-2} + y_2^{-2}) + a_{22}|x_1 - y_1| + a_{23}|x_2 - y_2|(x_2 + y_2) \\ \leqslant& M\left(1 + \Vert x\Vert + \Vert y\Vert + \left\Vert x^{-1}\right\Vert^2 + \left\Vert y^{-1}\right\Vert^2\right) \Vert x - y\Vert, \end{align*} $

由此假设($ A_1 $)成立, 参数取值为$ \alpha = 1 $和$ \beta = 2 $.

对于$ p = 13r^*, r^* = 2r $和$ x \in \mathbb{R}_+^2 $, 利用Young不等式可证存在常数$ K $使得

$ \begin{align*} &x^Tf(x) + \frac{p - 1}{2}\Vert g(x)\Vert^2\\ =& a_{11} + a_{12}x_1x_2 - a_{13}x_1^3 + a_{12} + a_{22}x_1x_2 - a_{23}x_1^3 + \frac{p-1}{2}(\sigma_1^2x_1^2+\sigma_2^2x_2^2)\\ \leqslant& a_{11} + a_{12} + \frac{1}{2}a_{12}x_1^2 + \frac{1}{2}a_{12}x_2^2 + \frac{1}{2}a_{22}x_1^2 + \frac{1}{2}a_{22}x_2^2 + \frac{p-1}{2}(\sigma_1^2+\sigma_2^2)(x_1^2+x_2^2) \\ \leqslant& K(1 + \Vert x\Vert^2). \end{align*} $

取$ q = p $, 我们总能找到一个足够小的$ x^*> 0 $使得

$ \begin{align*} x_1f_1(x) - \frac{q+1}{2}\Vert g_1(x)\Vert^2 = a_{11} + a_{12}x_1x_2 - a_{13}x_1^3 - \frac{q+1}{2}\sigma_1^2x_1^2 > 0 , \quad x_1 \in (0, x^*), x_2\in \mathbb{R}_+\\ x_2f_2(x) - \frac{q+1}{2}\Vert g_2(x)\Vert^2 = a_{12} + a_{22}x_1x_2 - a_{23}x_2^3 - \frac{q+1}{2}\sigma_2^2x_2^2 > 0 , \quad x_2 \in (0, x^*), x_1\in \mathbb{R}_+. \end{align*} $

由此假设($ A_2 $)成立.

设$ x, y \in \mathbb{R}_+^2 $, 则

$ \begin{align*} &(x-y)^T\left(f(x)-f(y)\right)\\ =&(x-y)^T \begin{bmatrix} a_{11}(x_1^{-1}-y_1^{-1}) + a_{12}(x_2 - y_2) - a_{13}(x_1^2 - y_1^2)\\ a_{21}(x_2^{-1}-y_2^{-1}) + a_{22}(x_1 - y_1) - a_{23}(x_2^2 - y_2^2) \end{bmatrix}\\ =&a_{11}(x_1-y_1)(x_1^{-1}-y_1^{-1}) + a_{12}(x_1 - y_1)(x_2 - y_2) - a_{13}(x_1 - y_1)(x_1^2 - y_1^2) \\ &+a_{21}(x_2-y_2)(x_2^{-1}-y_2^{-1}) + a_{22}(x_1 - y_1)(x_2 - y_2) - a_{23}(x_2 - y_2)(x_2^2 - y_2^2) \\ \leqslant& (a_{12}+a_{22})(x_1 - y_1)(x_2 - y_2) \\ \leqslant& \frac{1}{2}(a_{12}+a_{22})\left((x_1 - y_1)^2 + (x_2 - y_2)^2\right). \end{align*} $

以及

$ \begin{align*} \Vert g(x)-g(y)\Vert^2 =& \text{trace} \left( \begin{bmatrix} \sigma_1( x_1 - y_1) & 0\\ 0 & \sigma_2( x_2 - y_2) \end{bmatrix} \begin{bmatrix} \sigma_1 (x_1 - y_1) & 0\\ 0 & \sigma_2 (x_2 - y_2) \end{bmatrix} \right)\\ \leqslant& (\sigma_1^2 + \sigma_2^2)((x_1 - y_1)^2 + (x_2 - y_2)^2). \end{align*} $

因此, 存在常数$ H $使得

$ (x-y)^T\left(f(x)-f(y)\right) + \frac{r^*-1}{2}\Vert g(x)-g(y)\Vert^2 \leqslant H\Vert x -y\Vert^2. $

因此, 假设($ A_3 $)满足.

由$ 1.5r^* < \frac{p\wedge q}{(\alpha + 6)\vee(\beta+6)} $. 故注释4.1的条件也满足. 进一步

$ \frac{(p - r)(p\wedge q)}{3\sqrt{2}C(\alpha\vee\beta)p} > \frac{(p - r)(p\wedge q)}{3\sqrt{2}((\alpha\vee\beta)+6)p} = \frac{(26r - r)}{24\sqrt{2}} > \frac{r}{2}. $

因此, 对于任意$ \Delta \in (0, 1] $有

$ \sup\limits_{t \in [0, T]}\mathbb{E}\Vert x(t) - x_{\Delta}(t)\Vert^r < C{\Delta}^{\frac{r}{2}}. $

在本节中, 我们将对第5节中的模型进行数值模拟, 以支持我们的理论结果. 我们现在用100个步长为$ \Delta = 2^{-19}, 2^{-18}, \ldots , 2^{-12} $的样本路径进行数值模拟. 我们使用步长为$ \Delta = 2^{-20} $的数值解作为精确解的良好近似值.

考虑例5.1中参数取值为$ a_{11} = 1, a_{12} = 2, a_{13} = 3, a_{21} = 1, a_{22} = 1, a_{23} = 4, \sigma_1 = 2, \sigma_2 = 1 $的情形, 有

$ \begin{align*} F(y) =& \begin{bmatrix} e^{-y_1} & 0\\ 0 & e^{-y_2} \end{bmatrix} \begin{bmatrix} e^{-y_1} + 2e^{y_2} - 3e^{2y_1}\\ e^{-y_2} + e^{y_1} - 4e^{2y_2} \end{bmatrix} -\frac{1}{2} \begin{bmatrix} e^{-2y_1} & 0\\ 0 & e^{-2y_2} \end{bmatrix} \begin{bmatrix} 4e^{2y_1}\\ e^{2y_2} \end{bmatrix}\\ =&\begin{bmatrix} e^{-2y_1} + 2e^{y_2-y_1}- 3e^{y_1}-2\\ e^{-2y_2} + e^{y_1-y_2} - 4e^{y_2}-\frac{1}{2} \end{bmatrix} \end{align*} $

以及

$ \begin{align*} G(y) = \begin{bmatrix} e^{-y_1} & 0\\ 0 & e^{-y_2} \end{bmatrix} \begin{bmatrix} 2e^{y_1} & 0\\ 0 & e^{y_2} \end{bmatrix} =\begin{bmatrix} 2 & 0\\ 0 & 1 \end{bmatrix}. \end{align*} $

进一步我们也有

$ \begin{align*} \Vert F(y)\Vert =& \left(F_1(y)^2+F_2(y)^2\right)^{\frac{1}{2}} \\ =&\left((e^{-2y_1} + 2e^{y_2-y_1}- 3e^{y_1}-2)^2 + (e^{-2y_2} + e^{y_1-y_2} - 4e^{y_2}-\frac{1}{2})^2\right)^{\frac{1}{2}} \\ \leqslant& 2.5 + 5e^{2\Vert y\Vert} + 7e^{\Vert y\Vert} \\ \leqslant&8.5(1+e^{2\Vert y\Vert}) \end{align*} $

以及

$ \begin{align*} \Vert G(y)\Vert^2 = \sum\limits_{i=1}^{2}\sum\limits_{j=1}^{2}G_{i, j}^2(y)= 4 + 1 = 5. \end{align*} $

设定终端时间$ T= 1 $且初始值$ x_0 = (2, 1)^T $, 然后设置$ C(\alpha, \beta) = 2 $, 和

$ \varphi(r) = 8.5(2+e^{2r}), \quad h(\Delta) = 51{\Delta}^{-\frac{1}{3}}. $

数值实验结果(见图(6.1))显示, 二阶矩的强收敛误差阶数略高于1阶, 这与注释4.1的理论证明一致. 图(6.2)和(6.3)通过随机选取的数值样本路径, 直观展示了该方法生成的数值解始终保持严格正性.

本文对对数截断EM方法进行了深入研究, 通过改进约束条件使该方法可适用于多维随机微分方程. 我们不仅证明了该方法在$ L^p $空间的具体收敛速率, 同时保证数值解的正性. 特别地, 对于某些特定随机微分方程模型, 其收敛阶数可达$ 1/2 $阶甚至略高于$ 1/2 $阶.