DOI:10.12677/PM.2019.94068,PDF,HTML,XML,被引量下载: 861浏览: 1,172国家自然科学基金支持
作者:郭 奥,丁孝全^*：河南科技大学数学与统计学院，河南 洛阳
关键词:随机互惠系统；全局吸引性；持续性；灭绝性；周期解；Stochastic Mutualism System；Global Attractivity；Permanence；Extinction；Periodic Solution

文章引用：郭奥, 丁孝全. 一类非自治随机互惠系统的渐近性态[J]. 理论数学, 2019, 9(4): 514-526. https://doi.org/10.12677/PM.2019.94068

1. 引言

在自然界中，种群之间的互惠关系是非常普遍的，如传粉昆虫与开花植物、豆科植物与根瘤菌、海葵与小丑鱼等。许多学者对互惠种群模型进行了深入研究，并获得了丰富的研究成果 [1] [2] [3] [4] 。而现实世界中，种群的演化不可避免地受到环境噪声的影响，而且这些噪声是不容忽视的，更合理的种群模型应包括随机因素 [5] - [11] 。近年来，随机互惠模型受到学术界的广泛关注 [11] - [21] 。

2006年，Gravesa等人 [3] 提出以下互惠模型

$\{\begin{array}{l}\frac{\text{d}{x}_1\left(t\right)}{\text{d}t}=x_1\left(t\right)\left[r_1-b_1{\text{e}}^{-k_1{x}_2\left(t\right)}-c_1{x}_1\left(t\right)\right],\hfill \\ \frac{\text{d}{x}_2\left(t\right)}{\text{d}t}=x_2\left(t\right)\left[r_2-b_2{\text{e}}^{-k_2{x}_1\left(t\right)}-c_2{x}_2\left(t\right)\right],\hfill \end{array}$(1.1)

其中 $x_i\left(t\right)\left(i=1,2\right)$是第i个种群在时刻t的种群密度； $r_i,b_i,c_i,k_i$为正常数，相应的生物意义见文献 [3] 。2009年，向等人 [4] 在系统(1.1)的基础上，考虑以下非自治互惠模型

$\{\begin{array}{l}\frac{\text{d}{x}_1\left(t\right)}{\text{d}t}=x_1\left(t\right)\left[r_1\left(t\right)-b_1\left(t\right){\text{e}}^{-k_1\left(t\right)x_2\left(t\right)}-c_1\left(t\right)x_1\left(t\right)\right],\hfill \\ \frac{\text{d}{x}_2\left(t\right)}{\text{d}t}=x_2\left(t\right)\left[r_2\left(t\right)-b_2\left(t\right){\text{e}}^{-k_2\left(t\right)x_1\left(t\right)}-c_2\left(t\right)x_2\left(t\right)\right],\hfill \end{array}$(1.2)

建立了系统(1.2)的持续性、周期解的存在性和全局吸引性。2011年，为探讨随机扰动对系统(1.1)的影响，吕 [11] 考虑以下随机互惠模型

$\{\begin{array}{c}\text{d}{x}_1\left(t\right)=x_1\left(t\right)\left[r_1-b_1{\text{e}}^{-k_1{x}_2\left(t\right)}-c_1{x}_1\left(t\right)\right]\text{d}t+{\sigma }_1{x}_1\left(t\right)\text{d}{B}_1\left(t\right),\\ \text{d}{x}_2\left(t\right)=x_2\left(t\right)\left[r_2-b_2{\text{e}}^{-k_2{x}_1\left(t\right)}-c_2{x}_2\left(t\right)\right]\text{d}t+{\sigma }_2{x}_2\left(t\right)\text{d}{B}_2\left(t\right),\end{array}$(1.3)

给出了系统(1.3)的持续性、灭绝性和全局吸引性。

本文在文献 [3] [4] [11] 基础上，研究以下具有随机干扰的非自治互惠模型

$\{\begin{array}{l}\text{d}{x}_1\left(t\right)=x_1\left(t\right)\left[r_1\left(t\right)-b_1\left(t\right){\text{e}}^{-k_1\left(t\right)x_2\left(t\right)}-c_1\left(t\right)x_1\left(t\right)\right]\text{d}t+{\sigma }_1\left(t\right)x_1\left(t\right)\text{d}{B}_1\left(t\right),\\ \text{d}{x}_2\left(t\right)=x_2\left(t\right)\left[r_2\left(t\right)-b_2\left(t\right){\text{e}}^{-k_2\left(t\right)x_1\left(t\right)}-c_2\left(t\right)x_2\left(t\right)\right]\text{d}t+{\sigma }_2\left(t\right)x_2\left(t\right)\text{d}{B}_2\left(t\right),\end{array}$(1.4)

其中对 $r_i\left(t\right),b_i\left(t\right),k_i\left(t\right),c_i\left(t\right),{\sigma }_i\left(t\right)\left(i=1,2\right)$是连续函数，具有正的上、下界； ${\left(B_1\left(t\right),B_2\left(t\right)\right)}^{\text{T}}$是定义在完备概率空间 $\left(\Omega ,\mathcal{F},{\left\{\mathcal{F}\right\}}_{t\ge 0},\mathbb{P}\right)$上的二维标准Brown运动。本文旨在利用随机微分方程理论 [22] [23] [24] [25] ，探讨系统(1.4)的全局正解的存在唯一性、持续性、灭绝性、全局吸引性和周期解的存在性。

为方便讨论，给出以下记号：

1) $R_{+}^2=\left\{\left(x_1,x_2\right):x_1>0,x_2>0\right\}$；

2) 对 $\left[0,\infty \right)$上的有界函数f，定义 $\stackrel{⌣}{f}=\underset{t\in \left[0,\infty \right)}{\sup }f\left(t\right),\stackrel{⌢}{f}=\underset{t\in \left[0,\infty \right)}{\inf }f\left(t\right)$；

3) 给定 $\theta >0$，对 $\left[0,\theta \right]$上的可积函数g，定义 ${\langle g\rangle }_{\theta }=\frac{1}{\theta }{\displaystyle {\int }_0^{\theta }g\left(s\right)\text{d}s}$。

2. 正解的存在唯一性

本节建立系统(1.4)的全局正解的存在唯一性，这是本文后续工作的基础。

定理2.1：对任意初值 $x\left(0\right)={\left(x_1\left(0\right),x_2\left(0\right)\right)}^{\text{T}}\in R_{+}^2$，系统(1.4)存在唯一的全局解 $x\left(t\right)={\left(x_1\left(t\right),x_2\left(t\right)\right)}^{\text{T}}$，并且该解以概率1停留在 $R_{+}^2$中。

证明：易知系统(1.4)的系数满足局部Lipschitz条件。因此，对于任意初值 $x\left(0\right)\in R_{+}^2$，系统(1.4)在区间 $\left[0,{\tau }_e\right)$上存在唯一的局部解 $x\left(t\right)={\left(x_1\left(t\right),x_2\left(t\right)\right)}^{\text{T}}$，其中 ${\tau }_e$是爆破时刻。下面证明 $x\left(t\right)$是全局的，即证明 ${\tau }_e=\infty$几乎必然成立。取充分大的正整数 $m_0$，使 $x_1\left(0\right)\in \left(\frac{1}{m_0},m_0\right)$和 $x_2\left(0\right)\in \left(\frac{1}{m_0},m_0\right)$。对任意正整数 $m>m_0$，定义停时：

${\tau }_m=\inf \left\{t\in \left[0,{\tau }_e\right):x_1\left(t\right)\notin \left(\frac{1}{m},m\right)或者x_2\left(t\right)\notin \left(\frac{1}{m},m\right)\right\}.$

对于空集 $\varnothing$，规定 $\inf \varnothing =\infty$。易知 $\left\{{\tau }_m\right\}$是一个单调递增序列。令 ${\tau }_{\infty }=\underset{m\to \infty }{\lim }{\tau }_m$，此时 ${\tau }_{\infty }\le {\tau }_e$，若证明 ${\tau }_{\infty }=\infty$，则 ${\tau }_e=\infty$。

下面用反证法证明 ${\tau }_{\infty }=\infty$几乎必然成立。若该结论不成立，则存在常数 $T>0$和 $\epsilon \in \left(0,1\right)$使得

$\mathbb{P}\left\{{\tau }_{\infty }\le T\right\}>\epsilon$.

从而，存在正整数 $m_1>m_0$，使得对任意正整数 $m\ge m_1$，有

$\mathbb{P}\left\{{\tau }_m\le T\right\}\ge \epsilon$.(2.1)

定义Lyapunov函数：

$V\left(x_1,x_2\right)=\left(x_1-1-\ln x_1\right)+\left(x_2-1-\ln x_2\right).$

显然，对任意 ${\left(x_1,x_2\right)}^{\text{T}}\in R_{+}^2$， $V\left(x_1,x_2\right)>0$。由Itô公式 [22] 得

$\text{d}V\left(x_1,x_2\right)=LV\left(x_1,x_2\right)+\left(x_1-1\right){\sigma }_1\left(t\right)\text{d}{B}_1\left(t\right)+\left(x_2-1\right){\sigma }_2\left(t\right)\text{d}{B}_2\left(t\right),$(2.2)

其中

$\begin{array}{c}LV\left(x_1,x_2\right)=\left(x_1-1\right)\left[r_1\left(t\right)-b_1\left(t\right){\text{e}}^{-k_1\left(t\right)x_2\left(t\right)}-c_1\left(t\right)x_1\left(t\right)\right]\\ +\left(x_2-1\right)\left[r_2\left(t\right)-b_2\left(t\right){\text{e}}^{-k_2\left(t\right)x_1\left(t\right)}-c_2\left(t\right)x_2\left(t\right)\right]+\text{0}\text{.5}\left({\sigma }_1^2\left(t\right)+{\sigma }_2^2\left(t\right)\right)\\ \le x_1\left({\stackrel{⌣}{r}}_1+{\stackrel{⌣}{c}}_1-{\stackrel{⌢}{c}}_1{x}_1\right)+x_2\left({\stackrel{⌣}{r}}_2+{\stackrel{⌣}{c}}_2-{\stackrel{⌢}{c}}_2{x}_2\right)+{\stackrel{⌣}{b}}_1+{\stackrel{⌣}{b}}_2+0.5\left({\stackrel{⌣}{\sigma }}_1^2+{\stackrel{⌣}{\sigma }}_2^2\right)\\ =:G\left(x_1,x_2\right).\end{array}$(2.3)

显然， $G\left(x_1,x_2\right)$在 $R_{+}^2$上有正的上界，记为M。

对(2.2)两边从0到 ${\tau }_m\wedge T$积分，然后取数学期望，再结合(2.3)得

$\mathbb{E}V\left(x_1\left({\tau }_m\wedge T\right),x_2\left({\tau }_m\wedge T\right)\right)=V\left(x_1\left(0\right),x_2\left(0\right)\right)+\mathbb{E}{\displaystyle {\int }_0^{{\tau }_m\wedge T}LV\text{d}t}\le V\left(x_1\left(0\right),x_2\left(0\right)\right)+MT.$(2.4)

对 $m\ge m_1$，记 ${\Omega }_m=\left\{{\tau }_m\le T\right\}$。由 ${\tau }_m$的定义可知，对每个 $\omega \in {\Omega }_m$，存在 $x\left({\tau }_m,\omega \right)$的某个分量 $x_i\left({\tau }_m,\omega \right)\left(i=1或2\right)$等于m或者1/m。结合(2.1)，有

$\mathbb{E}V\left(x_1\left({\tau }_m\wedge T\right),x_2\left({\tau }_m\wedge T\right)\right)\ge \mathbb{E}\left[I_{{\Omega }_m}V\left(x_1\left({\tau }_m\wedge T\right),x_2\left({\tau }_m\wedge T\right)\right)\right]\ge \epsilon \left[m-1-\ln m\right]\wedge \left[\ln m-1+\frac{1}{m}\right],$(2.5)

其中 $I_{{\Omega }_m}$是 ${\Omega }_m$的示性函数。由(2.4)和(2.5)得

$V\left(x_1\left(0\right),x_2\left(0\right)\right)+MT\ge \epsilon \left[m-1-\ln m\right]\wedge \left[\ln m-1+\frac{1}{m}\right].$

令 $m\to \infty$，有

$V\left(x_1\left(0\right),x_2\left(0\right)\right)+MT\ge \infty ,$

矛盾。所以必有 ${\tau }_{\infty }=\infty$。证毕。

3. 全局吸引性

本节讨论系统(1.4)的解得全局吸引性。为此，需要下面三个引理。

引理3.1：( [22] )如果 $\left[0,\infty \right)$上的实值随机过程 $x\left(t\right)$满足

$\mathbb{E}{\left|x\left(t\right)-x\left(s\right)\right|}^{\alpha }\le c{\left|t-s\right|}^{1+\beta },0\le s,t<+\infty ,$

其中 $\alpha$、 $\beta$和c均为正常数，那么存在 $x\left(t\right)$的一个连续修正 $\tilde{x}\left(t\right)$，使得对于任意的 $\gamma \in \left(0,\frac{\beta }{\alpha }\right)$，存在正随机变量 $h\left(\omega \right)$满足

$P\left\{\omega :\underset{0<\left|t-s\right|<h\left(\omega \right),0\le s,t<+\infty }{\sup }\frac{\left|\tilde{x}\left(t,\omega \right)-\tilde{x}\left(s,\omega \right)\right|}{{\left|t-s\right|}^{\gamma }}\le \frac{2}{1-2^{-\gamma }}\right\}=1.$

换言之， $\tilde{x}\left(t\right)$几乎所有的样本轨道都以指数 $\gamma$局部但一致Hölder连续。

引理3.2：( [26] )若非负函数 $f\left(t\right)$在 $\left[0,\infty \right)$一致连续且可积，则 $\underset{t\to \infty }{\lim }f\left(t\right)=0$。

引理3.3：对于任意初值 $x\left(0\right)\in R_{+}^2$，系统(1.4)的解 $x\left(t\right)$几乎所有的样本轨道都是一致Hölder连续的。

证明：对任意 $p>1$，定义Lyapunov函数： $V_1\left(x_1\right)={\text{e}}^t{x}_1^p$。由Itô公式得

$\begin{array}{l}\text{d}\left({\text{e}}^t{x}_1^p\right)={\text{e}}^t{x}_1^p\left[1+p\left(r_1\left(t\right)-b_1\left(t\right){\text{e}}^{-k_1\left(t\right)x_2}-c_1\left(t\right)x_1\right)+0.5p\left(p-1\right){\sigma }_1^2\left(t\right)\right]\text{d}t+p{\text{e}}^t{x}_1^p{\sigma }_1\text{d}{B}_1\left(t\right)\\ \le {\text{e}}^t{x}_1^p\left[1+p{\stackrel{⌣}{r}}_1+0.5p\left(p-1\right){\stackrel{⌣}{\sigma }}_1^2-p{\stackrel{⌢}{c}}_1{x}_1\right]\text{d}t+p{\text{e}}^t{x}_1^p{\sigma }_1\left(t\right)\text{d}{B}_1\left(t\right)\end{array}$(3.1)

令 $H\left(x\right)=x^p\left[1+p{\stackrel{⌣}{r}}_1+0.5p\left(p-1\right){\stackrel{⌣}{\sigma }}_1^2-p{\stackrel{⌢}{c}}_{1}x\right]$。易知 $H\left(x\right)$在上 $\left(0,\infty \right)$有正上界，记为 $K\left(p\right)$。对(3.1)两边从0到t积分，并取数学期望得

${\text{e}}^t\mathbb{E}{x}_1^p\left(t\right)-\mathbb{E}{x}_1^p\left(0\right)\le {\displaystyle {\int }_0^t{\text{e}}^{s}H\left(x_1\left(s\right)\right)\text{d}s}\le K\left(p\right){\text{e}}^t.$

从而，

$\underset{t\to \infty }{\lim \sup }\mathbb{E}{x}_1^p\left(t\right)\le K\left(p\right).$

于是，存在与p和 $x_1\left(0\right)$有关的正常数 $\tilde{K}\left(p,x_1\left(0\right)\right)$，使得对任意 $t\ge 0$，有

$\mathbb{E}{x}_1^p\left(t\right)\le \tilde{K}\left(p,x_1\left(0\right)\right).$(3.2)

考虑系统(1.4)的第一个方程的随机积分形式：

$x_1\left(s\right)=x_1\left(0\right)+{\displaystyle {\int }_0^t{f}_1\left(s\right)\text{d}s}+{\displaystyle {\int }_0^t{g}_1\left(s\right)\text{d}{B}_1\left(s\right)},$(3.3)

其中

$f_1\left(s\right)=x_1\left(s\right)\left[r_1\left(s\right)-b_1\left(s\right){\text{e}}^{-k_1\left(s\right)x_2\left(s\right)}-c_1\left(s\right)x_1\left(s\right)\right],g_1\left(s\right)={\sigma }_1\left(s\right)x_1\left(s\right).$

利用均值不等式，由(3.2)得

$\begin{array}{c}\mathbb{E}{\left|f_1\left(s\right)\right|}^p=\mathbb{E}\left(x_1^p\left(s\right){\left|r_1\left(s\right)-b_1\left(s\right){\text{e}}^{-k_1\left(s\right)x_2\left(s\right)}-c_1\left(s\right)x_1\left(s\right)\right|}^p\right)\\ \le 0.5\mathbb{E}{x}_1^{2p}\left(s\right)+0.5\mathbb{E}{\left|r_1\left(s\right)-b_1\left(s\right){\text{e}}^{-k_1\left(s\right)x_2\left(s\right)}-c_1\left(s\right)x_1\left(s\right)\right|}^{2p}\\ \le 0.5\mathbb{E}{x}_1^{2p}\left(s\right)+2^{2p-2}{\left({\stackrel{⌣}{r}}_1+{\stackrel{⌣}{b}}_1\right)}^{2p}+2^{2p-2}{\stackrel{⌣}{c}}_1\mathbb{E}{x}_1^{2p}\left(s\right)\\ \le \left(0.5+2^{2p-2}{\stackrel{⌣}{c}}_1\right)\tilde{K}\left(2p,x_1\left(0\right)\right)+2^{2p-2}{\left({\stackrel{⌣}{r}}_1+{\stackrel{⌣}{b}}_1\right)}^{2p},\end{array}$(3.4)

以及

$\mathbb{E}{\left|g_1\left(s\right)\right|}^p=\mathbb{E}\left({\sigma }_1^p\left(s\right)x_1^p\left(s\right)\right)\le {\stackrel{⌣}{\sigma }}_1{}^p\mathbb{E}{x}_1^p\left(s\right)\le {\stackrel{⌣}{\sigma }}_1^p\tilde{K}\left(p,x_1\left(0\right)\right).$(3.5)

对任意 $0\le t_1<t_2\le t_1+1$和 $p>2$，利用随机积分的矩不等式 [23] ，由(3.5)得

$\begin{array}{c}\mathbb{E}{\left|{\displaystyle {\int }_{t_1}^{t_2}{g}_1\left(s\right)\text{d}{B}_1\left(s\right)}\right|}^p\le {\left[\frac{p\left(p-1\right)}{2}\right]}^{\frac{p}{2}}{\left(t_2-t_1\right)}^{\frac{p-2}{2}}{\displaystyle {\int }_{t_1}^{t_2}\mathbb{E}{\left|g_1\left(s\right)\right|}^p\text{d}s}\\ \le {\stackrel{⌣}{\sigma }}_1^p\tilde{K}\left(p,x_1\left(0\right)\right){\left[\frac{p\left(p-1\right)}{2}\right]}^{\frac{p}{2}}{\left(t_2-t_1\right)}^{\frac{p}{2}}.\end{array}$(3.6)

令

$\frac{1}{p}+\frac{1}{q}=1,$

利用Hölder不等式，由(3.4)得

$\begin{array}{c}\mathbb{E}{\left({\displaystyle {\int }_{t_1}^{t_2}\left|f_1\left(s\right)\right|\text{d}s}\right)}^p\le \mathbb{E}{\left\{{\left({\displaystyle {\int }_{t_1}^{t_2}{1}^q\text{d}s}\right)}^{\frac{1}{q}}\cdot {\left({\displaystyle {\int }_{t_1}^{t_2}{\left|f_1\left(s\right)\right|}^p\text{d}s}\right)}^{\frac{1}{p}}\right\}}^p={\left(t_2-t_1\right)}^{\frac{p}{q}}\cdot {\displaystyle {\int }_{t_1}^{t_2}\mathbb{E}{\left|f_1\left(s\right)\right|}^p\text{d}s}\\ \le \left[\left(0.5+2^{2p-2}{\stackrel{⌣}{c}}_1\right)\tilde{K}\left(2p,x_1\left(0\right)\right)+2^{2p-2}{\left({\stackrel{⌣}{r}}_1+{\stackrel{⌣}{b}}_1\right)}^{2p}\right]{\left(t_2-t_1\right)}^p.\end{array}$(3.7)

由(3.3)、(3.6)和(3.7)得

$\begin{array}{l}\mathbb{E}{\left|x_1\left(t_2\right)-x_1\left(t_1\right)\right|}^p\\ \le 2^{p-1}\mathbb{E}{\left({\displaystyle {\int }_{t_1}^{t_2}\left|f_1\left(s\right)\right|\text{d}s}\right)}^p+2^{p-1}\mathbb{E}{\left|{\displaystyle {\int }_{t_1}^{t_2}{g}_1\left(s\right)\text{d}{B}_1\left(s\right)}\right|}^p\\ \le 2^{p-1}\left\{\left(0.5+2^{2p-2}{\stackrel{⌣}{c}}_1\right)\tilde{K}\left(2p,x_1\left(0\right)\right)+2^{2p-2}{\left({\stackrel{⌣}{r}}_1+{\stackrel{⌣}{b}}_1\right)}^{2p}+{\stackrel{⌣}{\sigma }}_1^p\tilde{K}\left(p,x_1\left(0\right)\right){\left[\frac{p\left(p-1\right)}{2}\right]}^{\frac{p}{2}}\right\}{\left(t_2-t_1\right)}^{\frac{p}{2}}.\end{array}$

由引理3.1，对任意 $\gamma \in \left(0,\frac{p-2}{2p}\right)$时， $x_1\left(t\right)$的几乎所有的样本轨道关于指数 $\gamma$是一致Hölder连续的。对 $x_2\left(t\right)$，同理可证。证毕。

下面给出系统(1.4)的解的全局吸引性结果。

定理3.1：若 $\min \left\{\underset{t\ge 0}{\inf }\left[c_1\left(t\right)-b_2\left(t\right)k_2\left(t\right)\right],\underset{t\ge 0}{\inf }\left[c_2\left(t\right)-b_1\left(t\right)k_1\left(t\right)\right]\right\}>0$，则对任意初值 $x\left(0\right)\in R_{+}^2$，系统(1.4)的解 $x\left(t\right)$是全局吸引的。

证明：记

$\mu =\min \left\{\underset{t\ge 0}{\inf }\left[c_1\left(t\right)-b_2\left(t\right)k_2\left(t\right)\right],\underset{t\ge 0}{\inf }\left[c_2\left(t\right)-b_1\left(t\right)k_1\left(t\right)\right]\right\},$

则 $\mu >0$。设 $x\left(t\right)=\left(x_1\left(t\right),x_2\left(t\right)\right)$和 $\bar{x}\left(t\right)=\left({\bar{x}}_1\left(t\right),{\bar{x}}_2\left(t\right)\right)$分别系统(1.4)的初值为 $x\left(0\right)$和 $\bar{x}\left(0\right)$的两个解，由Itô公式和Lagrange中值定理，可得

$\begin{array}{c}d\left(\ln x_1\left(t\right)-\ln {\bar{x}}_1\left(t\right)\right)=\left\{-b_1\left(t\right)\left[{\text{e}}^{-k_1\left(t\right)x_2\left(t\right)}-{\text{e}}^{-k_1\left(t\right){\bar{x}}_2\left(t\right)}\right]-c_1\left(t\right)\left[x_1\left(t\right)-{\bar{x}}_1\left(t\right)\right]\right\}\text{d}t\\ =\left\{b_1\left(t\right)k_1\left(t\right){\text{e}}^{-k_1\left(t\right){\xi }_1\left(t\right)}\left[x_2\left(t\right)-{\bar{x}}_2\left(t\right)\right]-c_1\left(t\right)\left[x_1\left(t\right)-{\bar{x}}_1\left(t\right)\right]\right\}\text{d}t,\end{array}$(3.8)

$\begin{array}{c}d\left(\ln x_2\left(t\right)-\ln {\bar{x}}_2\left(t\right)\right)=\left\{-b_2\left(t\right)\left[{\text{e}}^{-k_2\left(t\right)x_1\left(t\right)}-{\text{e}}^{-k_2\left(t\right){\bar{x}}_1\left(t\right)}\right]-c_2\left(t\right)\left[x_2\left(t\right)-{\bar{x}}_2\left(t\right)\right]\right\}\text{d}t\\ =\left\{b_2\left(t\right)k_2\left(t\right){\text{e}}^{-k_2\left(t\right){\xi }_2\left(t\right)}\left[x_1\left(t\right)-{\bar{x}}_1\left(t\right)\right]-c_2\left(t\right)\left[x_2\left(t\right)-{\bar{x}}_2\left(t\right)\right]\right\}\text{d}t,\end{array}$(3.9)

其中 ${\xi }_1\left(t\right)$位于 $x_2\left(t\right)$和 ${\bar{x}}_2\left(t\right)$之间，而 ${\xi }_2\left(t\right)$位于 $x_1\left(t\right)$和 ${\bar{x}}_1\left(t\right)$之间。令

$V_2\left(t\right)=\left|\ln x_1\left(t\right)-\ln {\bar{x}}_1\left(t\right)\right|+\left|\ln x_2\left(t\right)-\ln {\bar{x}}_2\left(t\right)\right|,$

利用Tanaka-Meyer公式 [22] ，由(3.9)得

$\begin{array}{l}{V}_2\left(t\right)-V_2\left(0\right)\\ ={\displaystyle {\int }_0^t\mathrm{sgn}\left(\ln x_1\left(s\right)-\ln {\bar{x}}_1\left(s\right)\right)\text{d}\left(\ln x_1\left(s\right)-\ln {\bar{x}}_1\left(s\right)\right)}\\ +{\displaystyle {\int }_0^t\mathrm{sgn}\left(\ln x_2\left(s\right)-\ln {\bar{x}}_2\left(s\right)\right)\text{d}\left(\ln x_2\left(s\right)-\ln {\bar{x}}_2\left(s\right)\right)}\\ ={\displaystyle {\int }_0^t\mathrm{sgn}\left(x_1\left(s\right)-{\bar{x}}_1\left(s\right)\right)\left\{b_1\left(s\right)k_1\left(s\right){\text{e}}^{-k_1\left(s\right){\xi }_1\left(s\right)}\left[x_2\left(s\right)-{\bar{x}}_2\left(s\right)\right]-c_1\left(s\right)\left[x_1\left(s\right)-{\bar{x}}_1\left(s\right)\right]\right\}\text{d}s}\\ +{\displaystyle {\int }_0^t\mathrm{sgn}\left(x_2\left(s\right)-{\bar{x}}_2\left(s\right)\right)\left\{b_2\left(s\right)k_2\left(s\right){\text{e}}^{-k_2\left(s\right){\xi }_2\left(s\right)}\left[x_1\left(s\right)-{\bar{x}}_1\left(s\right)\right]-c_2\left(s\right)\left[x_2\left(s\right)-{\bar{x}}_2\left(s\right)\right]\right\}\text{d}s}\end{array}$

$\begin{array}{l}\le {\displaystyle {\int }_0^t\left[b_1\left(s\right)k_1\left(s\right)\left|x_2\left(s\right)-{\bar{x}}_2\left(s\right)\right|-c_1\left(s\right)\left|x_1\left(s\right)-{\bar{x}}_1\left(s\right)\right|\right]\text{d}s}\\ +{\displaystyle {\int }_0^t\left[b_2\left(s\right)k_2\left(s\right)\left|x_1\left(s\right)-{\bar{x}}_1\left(s\right)\right|-c_2\left(s\right)\left|x_2\left(s\right)-{\bar{x}}_2\left(s\right)\right|\right]\text{d}s}\\ \le -{\displaystyle {\int }_0^t\left\{\left[c_1\left(s\right)-b_2\left(s\right)k_2\left(s\right)\right]\left|x_1\left(s\right)-{\bar{x}}_1\left(s\right)\right|+\left[c_2\left(s\right)-b_1\left(s\right)k_1\left(s\right)\right]\left|x_2\left(t\right)-{\bar{x}}_2\left(t\right)\right|\right\}\text{d}s}\\ \le -\mu {\displaystyle {\int }_0^t\left[\left|x_1\left(s\right)-{\bar{x}}_1\left(s\right)\right|+\left|x_2\left(s\right)-{\bar{x}}_2\left(s\right)\right|\right]\text{d}s}.\end{array}$

即

$V_2\left(t\right)+\mu {\displaystyle {\int }_0^t\left[\left|x_1\left(s\right)-{\bar{x}}_1\left(s\right)\right|+\left|x_2\left(s\right)-{\bar{x}}_2\left(s\right)\right|\right]\text{d}t}\le V_2\left(0\right).$

于是， $\left|x_1\left(t\right)-{\bar{x}}_1\left(t\right)\right|+\left|x_2\left(t\right)-{\bar{x}}_2\left(t\right)\right|$在上 $\left[0,\infty \right)$可积。另一方面，由引理3.3， $\left|x_1\left(t\right)-{\bar{x}}_1\left(t\right)\right|+\left|x_2\left(t\right)-{\bar{x}}_2\left(t\right)\right|$在 $\left[0,\infty \right)$上一致连续。由引理3.2可知

$\underset{t\to \infty }{\lim }\left[\left|x_1\left(t\right)-{\bar{x}}_1\left(t\right)\right|+\left|x_2\left(t\right)-{\bar{x}}_2\left(t\right)\right|\right]=0.$

从而，对任意初值 $x\left(0\right)\in R_{+}^2$，系统(1.4)的解 $x\left(t\right)$是全局吸引的。证毕。

4. 持续性与灭绝性

本节讨论系统(1.4)的持续性与灭绝性。为此，介绍如下非自治随机Logistic方程的持续性结果。

考虑非自治随机Logistic方程

$\text{d}u\left(t\right)=u\left(t\right)\left[r\left(t\right)-a\left(t\right)u\left(t\right)\right]\text{d}t+\sigma \left(t\right)u\left(t\right)\text{d}B\left(t\right),$(4.1)

其中 $a\left(t\right)$、 $b\left(t\right)$和 $\sigma \left(t\right)$连续，具有正的上、下界。对方程(4.1)，由文献 [8] 可知：

引理4.1：若 $\underset{t\to +\infty }{\lim \inf }\left[r\left(t\right)-0.5{\sigma }^2\left(t\right)\right]>0$，则方程(4.1)是随机持续的，即对任意 $\epsilon \in \left(0,1\right)$，存在正数 $\alpha$和 $\beta$，使得对任意初值 $u\left(0\right)>0$，方程(4.1)的解 $u\left(t\right)$满足

$\underset{t\to \infty }{\lim \inf }\mathbb{P}\left\{u\left(t\right)\le \alpha \right\}\ge 1-\epsilon ,\underset{t\to \infty }{\lim \inf }\mathbb{P}\left\{u\left(t\right)\ge \beta \right\}\ge 1-\epsilon .$

引理4.2：若 $\underset{t\to +\infty }{\lim \inf }\frac{1}{t}{\displaystyle {\int }_0^t\left[r\left(s\right)-0.5{\sigma }^2\left(s\right)\right]\text{d}s}>0$，则方程(4.1)是平均持续的，即对任意初值 $u\left(0\right)>0$，方程(4.1)的解 $u\left(t\right)$以概率1满足

$\underset{t\to \infty }{\lim \inf }\frac{1}{t}{\displaystyle {\int }_0^{t}u\left(s\right)\text{d}s}>0.$

下面依次给出系统(1.4)的随机持续性、平均持续性和灭绝性结果。

定理4.1：若 $\min \left\{\underset{t\to \infty }{\lim \inf }\left[r_1\left(t\right)-b_1\left(t\right)-0.5{\sigma }_1^2\left(t\right)\right],\underset{t\to \infty }{\lim \inf }\left[r_2\left(t\right)-b_2\left(t\right)-0.5{\sigma }_2^2\left(t\right)\right]\right\}>0$，则系统(1.4)是随机持续的，即对任意 $\epsilon \in \left(0,1\right)$，存在正数 $\alpha$和 $\beta$，使得对任意初值 $x\left(0\right)\in R_{+}^2$，系统(1.4)的解 ${\left(x_1\left(t\right),x_2\left(t\right)\right)}^{\text{T}}$满足

$\underset{t\to \infty }{\lim \inf }P\left\{x_i\left(t\right)\le \alpha \right\}\ge 1-\epsilon ,\underset{t\to \infty }{\lim \inf }P\left\{x_i\left(t\right)\ge \beta \right\}\ge 1-\epsilon ,i=1,2.$

证明：由定理3.1知，对于任意初值 $x\left(0\right)\in R_{+}^2$，系统(1.4)存在全局唯一解，且以概率1停留在 $R_{+}^2$中。对系统(1.4)的第一个方程，有

$\begin{array}{c}\text{d}{x}_1\left(t\right)=x_1\left(t\right)\left[r_1\left(t\right)-b_1\left(t\right){\text{e}}^{-k_1\left(t\right)x_2\left(t\right)}-c_1\left(t\right)x_1\left(t\right)\right]\text{d}t+{\sigma }_1\left(t\right)x_1\left(t\right)\text{d}{B}_1\left(t\right)\\ \le x_1\left(t\right)\left[r_1\left(t\right)-c_1\left(t\right)x_1\left(t\right)\right]\text{d}t+{\sigma }_1\left(t\right)x_1\left(t\right)\text{d}{B}_1\left(t\right),\end{array}$(4.2)

以及

$\begin{array}{c}\text{d}{x}_1\left(t\right)=x_1\left(t\right)\left[r_1\left(t\right)-b_1\left(t\right){\text{e}}^{-k_1\left(t\right)x_2\left(t\right)}-c_1\left(t\right)x_1\left(t\right)\right]\text{d}t+{\sigma }_1\left(t\right)x_1\left(t\right)\text{d}{B}_1\left(t\right)\\ \ge x_1\left(t\right)\left[r_1\left(t\right)-b_1\left(t\right)-c_1\left(t\right)x_1\left(t\right)\right]\text{d}t+{\sigma }_1\left(t\right)x_1\left(t\right)\text{d}{B}_1\left(t\right).\end{array}$(4.3)

考虑随机微分方程

$\text{d}\Lambda \left(t\right)=\Lambda \left(t\right)\left[r_1\left(t\right)-c_1\left(t\right)\Lambda \left(t\right)\right]\text{d}t+{\sigma }_1\left(t\right)\Lambda \left(t\right)\text{d}{B}_1\left(t\right),$(4.4)

和

$\text{d}\lambda \left(t\right)=\lambda \left(t\right)\left[r_1\left(t\right)-b_1\left(t\right)-c_1\left(t\right)\lambda \left(t\right)\right]\text{d}t+{\sigma }_1\left(t\right)\lambda \left(t\right)\text{d}{B}_1\left(t\right).$(4.5)

设 $\Lambda \left(t\right)$和 $\lambda \left(t\right)$分别是方程(4.4)和(4.5)的初值为 $\Lambda \left(0\right)=x_1\left(0\right)$和 $\lambda \left(0\right)=x_1\left(0\right)$的解。根据随机微分方程的比较定理 [22] ，由(4.2)和(4.3)得

$\lambda \left(t\right)\le x_1\left(t\right)\le \Lambda \left(t\right),\text{a}\text{.s}\text{.}$(4.6)

由引理4.1，当 $\underset{t\to \infty }{\lim \inf }\left[r_1\left(t\right)-b_1\left(t\right)-0.5{\sigma }_1^2\left(t\right)\right]>0$时，系统(4.4)和(4.5)都是随机持续的。因此，对任意 $\epsilon \in \left(0,1\right)$，存在正数 ${\alpha }_1$和 ${\beta }_1$，对任意 $x_1\left(0\right)>0$，有

$\underset{t\to \infty }{\lim \inf }P\left\{\Lambda \left(t\right)\le {\alpha }_1\right\}\ge 1-\epsilon$和 $\underset{t\to \infty }{\lim \inf }P\left\{\lambda \left(t\right)\ge {\beta }_1\right\}\ge 1-\epsilon$(4.7)

由(4.6)和(4.7)得

$\underset{t\to \infty }{\lim \inf }P\left\{x_1\left(t\right)\le {\alpha }_1\right\}\ge 1-\epsilon$和 $\underset{t\to \infty }{\lim \inf }P\left\{x_1\left(t\right)\ge {\beta }_1\right\}\ge 1-\epsilon$(4.8)

同理可证，对任意 $\epsilon \in \left(0,1\right)$，存在正数 ${\alpha }_2$和 ${\beta }_2$，对任意 $x_2\left(0\right)>0$，有

$\underset{t\to \infty }{\lim \inf }P\left\{x_2\left(t\right)\le {\alpha }_2\right\}\ge 1-\epsilon$和 $\underset{t\to \infty }{\lim \inf }P\left\{x_2\left(t\right)\ge {\beta }_2\right\}\ge 1-\epsilon$(4.9)

取 $\alpha =\max \left\{{\alpha }_1,{\alpha }_2\right\}$和 $\beta =\min \left\{{\beta }_1,{\beta }_2\right\}$，由(4.8)和(4.9)可知，对任意 $i=1,2$，有

$\underset{t\to \infty }{\lim \inf }P\left\{x_i\left(t\right)\le \alpha \right\}\ge 1-\epsilon ,\underset{t\to \infty }{\lim \inf }P\left\{x_i\left(t\right)\ge \beta \right\}\ge 1-\epsilon .$

证毕。

定理4.2：若 $\min \left\{\underset{t\to \infty }{\lim \inf }\frac{1}{t}{\displaystyle {\int }_0^t\left[r_1\left(s\right)-b_1\left(s\right)-0.5{\sigma }_1^2\left(s\right)\right]\text{d}s},\underset{t\to \infty }{\lim \inf }\frac{1}{t}{\displaystyle {\int }_0^t\left[r_2\left(s\right)-b_2\left(s\right)-0.5{\sigma }_2^2\left(s\right)\right]\text{d}s}\right\}>0$，则系统(1.4)是平均持续的，即对任意初值 $x\left(0\right)\in R_{+}^2$，系统(1.4)的解 ${\left(x_1\left(t\right),x_2\left(t\right)\right)}^{\text{T}}$以概率1满足

$\underset{t\to \infty }{\lim \inf }\frac{1}{t}{\displaystyle {\int }_0^t{x}_i\left(t\right)\text{d}t}>0,\text{a}\text{.s}\text{.,}i=1,2.$

证明：根据引理4.2，由

$\underset{t\to \infty }{\lim \inf }\frac{1}{t}{\displaystyle {\int }_0^t\left[r_1\left(s\right)-b_1\left(s\right)-0.5{\sigma }_1^2\left(s\right)\right]\text{d}s}>0$

可知，对方程(4.5)的初值为 $\lambda \left(0\right)=x_1\left(0\right)$的解 $\lambda \left(t\right)$，以概率1成立

$\underset{t\to \infty }{\lim \inf }\frac{1}{t}{\displaystyle {\int }_0^t\lambda \left(t\right)\text{d}t}>0.$(4.10)

由(4.6)和(4.10)可知，以概率1成立

$\underset{t\to \infty }{\lim \inf }\frac{1}{t}{\displaystyle {\int }_0^t{x}_1\left(s\right)\text{d}s}\ge \underset{t\to \infty }{\lim \inf }\frac{1}{t}{\displaystyle {\int }_0^t\lambda \left(t\right)\text{d}t}>0.$

同理可证，以概率1成立

$\underset{t\to \infty }{\lim \inf }\frac{1}{t}{\displaystyle {\int }_0^t{x}_2\left(s\right)\text{d}s}>0.$

证毕。

定理4.3：若 $\underset{t\to +\infty }{\lim \sup }\frac{1}{t}{\displaystyle {\int }_0^t\left[r_i\left(s\right)-0.5{\sigma }_i^2\left(s\right)\right]\text{d}s}<0$，则对任意初值 $x\left(0\right)\in R_{+}^2$，系统(1.4)的第i个种群以概率1指数灭绝，即以概率1成立

$\underset{t\to \infty }{\lim \sup }\frac{\ln x_i\left(t\right)}{t}<0,i=1,2.$

证明：只对 $i=1$的情况证明， $i=2$的情况类似。由Itô公式得

$\text{d}\ln x_1\left(t\right)=\left[r_1\left(t\right)-b_1\left(t\right){\text{e}}^{-k_1\left(t\right)x_2}-c_1\left(t\right)x_1-0.5{\sigma }_1^2\left(t\right)\right]\text{d}t+{\sigma }_1\left(t\right)\text{d}{B}_1\left(t\right).$(4.11)

对(4.11)两边从0到t积分得

$\ln x_1\left(t\right)\le \ln x_1\left(0\right)+{\displaystyle {\int }_0^t\left[r_1\left(s\right)-0.5{\sigma }_1^2\left(s\right)\right]\text{d}s}+M\left(t\right),$(4.12)

其中 $M\left(t\right)={\displaystyle {\int }_0^t{\sigma }_1\left(s\right)\text{d}B\left(s\right)}$是局部鞅，其二次变差为

${\langle M,M\rangle }_t={\displaystyle {\int }_0^t{\sigma }_1^2\left(s\right)\text{d}s}\le {\stackrel{⌣}{\sigma }}_1^{2}t.$

由局部鞅的强大数定律 [23]

$\underset{t\to \infty }{\lim }\frac{M\left(t\right)}{t}=0,\text{a}\text{.s}.$(4.13)

由(4.12)和(4.13)得

$\underset{t\to \infty }{\lim \sup }\frac{\ln x_1\left(t\right)}{t}\le \underset{t\to \infty }{\lim \sup }\frac{1}{t}{\displaystyle {\int }_0^t\left[r_1\left(s\right)-\frac{1}{2}{\sigma }_1^2\left(s\right)\right]\text{d}s}<0,\text{a}\text{.s}.$

证毕。

5. 周期解

本节假定系统(1.4)的系数都是 $\theta$-周期的。下面介绍Khasminskii [24] 的随机周期解理论，并据此建立系统(1.4)的 $\theta$-周期解的存在性。

定义5.1：随机过程 $\xi \left(t\right)$称为是 $\theta$-周期的，如果对任意有限个非负实数 $t_1,t_2,\cdots ,t_n$，随机向量 $\left(\xi \left(t_1+k\theta \right),\xi \left(t_2+k\theta \right),\cdots ,\xi \left(t_n+k\theta \right)\right)$的联合分布与 $k\left(k=0,1,2,\cdots \right)$无关。

考虑随机微分方程

$\text{d}x\left(t\right)=f\left(t,x\left(t\right)\right)\text{d}t+g\left(t,x\left(t\right)\right)\text{d}{B}_t,$(5.1)

其中 $f\in L^1\left(R_{+};R^d\right)$， $g\in L^2\left(R_{+};R^{d\times m}\right)$。给定函数 $V\in C^{1,2}\left(R_{+}\times R^d,R\right)$，定义算子

$LV\left(t,x\right)=V_t\left(t,x\right)+V_x\left(t,x\right)f\left(t,x\right)+0.5\text{trace}\left[g^{\text{T}}\left(t,x\right)V_{xx}\left(t,x\right)g\left(t,x\right)\right].$

引理5.1：( [24] )设 $f\left(t,x\right)$和 $g\left(t,x\right)$关于t是 $\theta$-周期的。若方程(5.1)有全局唯一解，并且存在关于t是 $\theta$-周期的函数 $V\in C^{1,2}\left(R_{+}\times R^d,R_{+}\right)$，使得：

1) $\underset{k\to \infty }{\lim }\underset{\left|x\right|>k}{\inf }V\left(t,x\right)\to \infty$；

2) $\underset{k\to \infty }{\lim }\underset{\left|x\right|>k}{\sup }LV\left(t,x\right)=-\infty$；

则方程(5.1)存在一个 $\theta$-周期解。

定理5.1：若 $\min \left\{{\langle r_1-b_1-0.5{\sigma }_1^2\rangle }_{\theta },{\langle r_2-b_2-0.5{\sigma }_2^2\rangle }_{\theta }\right\}>0$，则系统(1.4)存在一个正的 $\theta$-周期解。

证明：由定理2.1，对于任意初值 $x\left(0\right)\in R_{+}^2$，系统(1.4)存在唯一的全局正解 ${\left(x_1\left(t\right),x_2\left(t\right)\right)}^{\text{T}}$。令

$u_1\left(t\right)=\ln x_1\left(t\right),u_2\left(t\right)=\ln x_2\left(t\right).$

利用Itô公式，由(1.4)得

$\{\begin{array}{l}\text{d}{u}_1\left(t\right)=\left[r_1\left(t\right)-0.5{\sigma }_1^2\left(t\right)-b_1\left(t\right){\text{e}}^{-k_1\left(t\right){\text{e}}^{u_2\left(t\right)}}-c_1\left(t\right){\text{e}}^{u_1\left(t\right)}\right]\text{d}t+{\sigma }_1\left(t\right)\text{d}{B}_1\left(t\right),\\ \text{d}{u}_2\left(t\right)=\left[r_2\left(t\right)-0.5{\sigma }_2^2\left(t\right)-b_2\left(t\right){\text{e}}^{-k_2\left(t\right){\text{e}}^{u_1\left(t\right)}}-c_2\left(t\right){\text{e}}^{u_2\left(t\right)}\right]\text{d}t+{\sigma }_2\left(t\right)\text{d}{B}_2\left(t\right).\end{array}$(5.2)

定义Lyapunov函数：

$V_3\left(t,u_1,u_2\right):\left[0,+\infty \right)\times R^2\to R_{+}$, $V_3\left(t,u_1,u_2\right)={\displaystyle \underset{i=1}{\overset{2}{\sum }}\left\{{\text{e}}^{u_i}+{\text{e}}^{q\left[w_i\left(t\right)-u_i\right]}\right\}}$,

其中 $w_i\left(t\right)$是微分方程

${w^{\prime }}_i\left(t\right)=r_i\left(t\right)-b_i\left(t\right)-0.5{\sigma }_i^2\left(t\right)-{\langle r_i-b_i-0.5{\sigma }_i^2\rangle }_{\theta }$(5.3)

的初值为 $w_i\left(0\right)=0$的解，而q是满足

${\langle r_i-b_i-0.5{\sigma }_i^2\rangle }_{\theta }-0.5q{\stackrel{⌣}{\sigma }}_i^2>0$(5.4)

的充分小的正常数。对(5.3)两边从t到 $t+\theta$积分得

$w_i\left(t+\theta \right)-w_i\left(t\right)={\displaystyle {\int }_t^{t+\theta }\left[r_i\left(s\right)-b_i\left(s\right)-0.5{\sigma }_i^2\left(s\right)-{\langle r_i-b_i-0.5{\sigma }_i^2\rangle }_{\theta }\right]\text{d}s}=0.$

于是， $w_i\left(t\right)$是 $\theta$-周期函数。显然，

$\underset{k\to \infty }{\lim }\underset{\left|u\right|>k}{\inf }{V}_3\left(t,u_1,u_2\right)\to \infty ,$

即 $V_3$满足引理5.1的条件(1)。由(5.3)和(5.4)得

$\begin{array}{c}L\left[{\text{e}}^{u_1}+{\text{e}}^{q\left[w_1\left(t\right)-u_1\right]}\right]={\text{e}}^{u_1}\left[r_1\left(t\right)-b_1\left(t\right){\text{e}}^{-k_1\left(t\right)e^{u_2}}-c_1\left(t\right){\text{e}}^{u_1}\right]\\ +q{e}^{q\left[w_1\left(t\right)-u_1\right]}\left[{w^{\prime }}_1{}_1\left(t\right)-r_1\left(t\right)+0.5{\sigma }_1^2\left(t\right)+b_1\left(t\right){\text{e}}^{-k_1\left(t\right){\text{e}}^{u_2}}+c_1\left(t\right){\text{e}}^{u_1}+0.5q{\sigma }_1^2\left(t\right)\right]\\ \le {\text{e}}^{u_1}\left[{\stackrel{⌣}{r}}_1-{\stackrel{⌢}{c}}_1{\text{e}}^{u_1}\right]+q{\text{e}}^{q\left[w_1\left(t\right)-u_1\right]}\left[{w^{\prime }}_1\left(t\right)-r_1\left(t\right)+0.5{\sigma }_1^2\left(t\right)+b_1\left(t\right)+{\stackrel{⌣}{c}}_1{\text{e}}^{u_1}+0.5q{\stackrel{⌣}{\sigma }}_1^2\right]\\ ={\text{e}}^{u_1}\left[{\stackrel{⌣}{r}}_1-{\stackrel{⌢}{c}}_1{\text{e}}^{u_1}\right]+q{\text{e}}^{q\left[w_1\left(t\right)-u_1\right]}\left[{\stackrel{⌣}{c}}_1{\text{e}}^{u_1}-{\langle r_1-b_1-0.5{\sigma }_1^2\rangle }_{\theta }+0.5q{\stackrel{⌣}{\sigma }}_1^2\right]\\ \le {\text{e}}^{u_1}\left[{\stackrel{⌣}{r}}_1+q{\stackrel{⌣}{c}}_1{\text{e}}^{q\left[{\stackrel{⌣}{w}}_1-u_1\right]}-{\stackrel{⌢}{c}}_1{\text{e}}^{u_1}\right]-q\left[{\langle r_1-b_1-0.5{\sigma }_1^2\rangle }_{\theta }-0.5q{\stackrel{⌣}{\sigma }}_1^2\right]{\text{e}}^{q\left[{\stackrel{⌢}{w}}_1-u\right]}.\end{array}$

同理，

$L\left[{\text{e}}^{u_2}+{\text{e}}^{q\left[w_2\left(t\right)-u_2\right]}\right]\le {\text{e}}^{u_2}\left[{\stackrel{⌣}{r}}_2+q{\stackrel{⌣}{c}}_2{\text{e}}^{q\left[{\stackrel{⌣}{w}}_2-u_1\right]}-{\stackrel{⌢}{c}}_2{\text{e}}^{u_2}\right]-q\left[{\langle r_2-b_2-0.5{\sigma }_2^2\rangle }_{\theta }-0.5q{\stackrel{⌣}{\sigma }}_2^2\right]{\text{e}}^{q\left[{\stackrel{⌢}{w}}_2-u_2\right]}.$

于是，

$L{V}_3\le {\displaystyle \underset{i=1}{\overset{2}{\sum }}\left\{{\text{e}}^{u_i}\left[{\stackrel{⌣}{r}}_i+q{\stackrel{⌣}{c}}_i{\text{e}}^{q\left[{\stackrel{⌣}{w}}_i-u_i\right]}-{\stackrel{⌢}{c}}_i{\text{e}}^{u_i}\right]-q\left[{\langle r_i-b_i-0.5{\sigma }_i^2\rangle }_{\theta }-0.5q{\stackrel{⌣}{\sigma }}_i^2\right]{\text{e}}^{q\left[{\stackrel{⌢}{w}}_i-u_i\right]}\right\}}.$(5.5)

由(5.4)和(5.5)，易知

$\underset{k\to \infty }{\lim }\underset{\left|u\right|>k}{\inf }L{V}_3\left(t,u_1,u_2\right)\to -\infty$,

即 $V_3$满足引理5.1的条件(2)。于是，系统(5.2)存在一个 $\theta$-周期解。从而，系统(1.4)存在一个正的 $\theta$-周期解。证毕。

6. 数值模拟

为验证理论分析结果，本节采用Milstein方法 [25] 对随机系统(1.4)进行数值模拟。

例6.1 在系统(1.4)中，取 $r_1\left(t\right)=8+0.3\sin t$， $r_2\left(t\right)=9+0.5\sin t$， $b_1\left(t\right)=3+0.2\sin t$， $b_2\left(t\right)=2+0.1\sin t$， $c_1\left(t\right)=2+0.1\sin \left(\sqrt{2}t\right)$， $c_2\left(t\right)=1+0.3\sin \left(\sqrt{3}t\right)$， ${\sigma }_1\left(t\right)=1+0.1\sin t$， ${\sigma }_2\left(t\right)=1+0.2\sin t$， $k_1\left(t\right)=0.1$， $k_2\left(t\right)=0.2$，则

$\min \left\{\underset{t\to \infty }{\lim \inf }\left[r_1\left(t\right)-b_1\left(t\right)-0.5{\sigma }_1^2\left(t\right)\right],\underset{t\to \infty }{\lim \inf }\left[r_2\left(t\right)-b_2\left(t\right)-0.5{\sigma }_2^2\left(t\right)\right]\right\}>0,$

满足定理4.1的条件。从图1可知系统(1.4)是随机持续的。

例6.2在系统(1.4)中，取 $r_1\left(t\right)=8+0.3\sin t$， $r_2\left(t\right)=9+0.5\sin t$， $b_1\left(t\right)=3+0.2\sin t$， $b_2\left(t\right)=2+0.1\sin t$， $c_1\left(t\right)=2+0.1\sin \left(\sqrt{2}t\right)$， $c_2\left(t\right)=1+0.3\sin \left(\sqrt{3}t\right)$， ${\sigma }_1\left(t\right)=1+0.1\sin t$， ${\sigma }_2\left(t\right)=5+0.2\sin t$， $k_1\left(t\right)=0.1$， $k_2\left(t\right)=0.2$，则

$\underset{t\to \infty }{\lim \inf }\left[r_1\left(t\right)-b_1\left(t\right)-0.5{\sigma }_1^2\left(t\right)\right]>0,\underset{t\to \infty }{\lim \inf }\left[r_2\left(t\right)-b_2\left(t\right)-0.5{\sigma }_2^2\left(t\right)\right]<0,$

满足定理4.3的条件。从图2可知系统(1.4)的种群1是随机持续的，但种群2是灭绝的。

例6.3在系统(1.4)中，取 $r_1\left(t\right)=8+0.3\sin t$， $r_2\left(t\right)=9+0.5\sin t$， $b_1\left(t\right)=3+0.2\sin t$， $b_2\left(t\right)=2+0.1\sin t$， $c_1\left(t\right)=2+0.1\sin \left(t\right)$， $c_2\left(t\right)=1+0.3\sin \left(t\right)$， ${\sigma }_1\left(t\right)=1+0.1\sin t$， ${\sigma }_2\left(t\right)=5+0.2\sin t$， $k_1\left(t\right)=0.1$， $k_2\left(t\right)=0.2$，则

$\begin{array}{l}\min \left\{{\langle r_1-b_1-0.5{\sigma }_1^2\rangle }_{\text{2π}},{\langle r_2-b_2-0.5{\sigma }_2^2\rangle }_{\text{2π}}\right\}>0,\\ \min \left\{\underset{t\in \left[0,2\text{π}\right]}{\min }\left[c_1\left(t\right)-b_2\left(t\right)k_2\left(t\right)\right],\underset{t\in \left[0,2\text{π}\right]}{\min }\left[c_2\left(t\right)-b_1\left(t\right)k_1\left(t\right)\right]\right\}>0,\end{array}$

满足定理3.1和定理5.1的条件。从图3可知系统(1.4)存在唯一的全局吸引的 $2\text{π}$-周期正解。

基金项目

本文得到国家自然科学基金项目(11271110)和河南省教育厅科技攻关项目(15A120009)的支持。

NOTES

^*通讯作者。