一类n-种群捕食–竞争随机系统持久性研究

Research on Persistence of a Class of n-Species Predator-Competition Stochastic System

摘要: 本文研究一类n种群捕食–竞争随机系统的几乎必然持久性。通过构造适当的Liapunov函数，运用Itô公式得到该系统全局唯一正解的存在性和最终随机有界性，利用非负半鞅收敛理论得到该随机系统的持久性充分条件，并将该持久性充分条件拓展到该随机非线性系统，并得到该随机非线性系统满足特定的条件依概率1灭绝。最后，通过数值模拟验证我们所得结论的正确性。

Abstract: This paper studies the almost inevitable persistence of a class of N-population predator-competitive stochastic systems. By constructing an appropriate Liapunov function and applying the Itô formula, the existence of the globally unique positive solution and the ultimate random boundedness of the system are obtained. The sufficient condition for the persistence of the random system is obtained by using the non-negative half-martingales convergence theory, and this sufficient condition for persistence is extended to the random nonlinear system. And it is obtained that the stochastic nonlinear system satisfies specific conditions and becomes extinct with a probability of 1. Finally, the correctness of the conclusion we reached was verified through numerical simulation.

1. 引言

在实际的生态系统中，不同种群之间存在竞争关系是很普遍的，也往往存在着一些种群以另一些相互竞争的种群为食的现象。在生态学理论研究中，多物种群落相互作用的动力系统是揭示生物多样性维持机制、种群动态规律及生态系统持久性的核心课题。持久性作为一个关键的全局性概念，其研究具有首要的理论与现实意义。持久性，即所有种群在初始存在的情况下，其密度长期远离零灭绝状态，是生态系统得以延续和维持其功能的前提，它比局部渐近稳定性更能反映生物群落抵抗干扰和物种长期共存的能力。因此，探究n种群捕食–竞争系统在何种条件下能够实现持久性，不仅是数学生态学中的一个深刻的理论问题，对于理解生物多样性的起源、评估生态系统的脆弱性以及制定科学的保护策略也具有重要的指导价值。

许多学者对n-种群的研究以及其持久性方面已经做了许多可喜的成果，例如，Li等研究了一类非同步扩散的n种群捕食竞争系统的持久性，通过构造Lyapunov函数以及微分不等式原理得到该系统中n种群永久持续生存的充分条件[1]。Shi等通过研究一类非线性捕食–竞争系统，当系数是连续的概周期函数时，增长率可能为负的情形下，得到了该系统的周期性解和持久性充分性条件[2]。Wang等研究了一类具有时滞的两种群非线性竞争系统，得到了第二个种群灭绝的充分条件，由微分方程比较定理和一些分析技巧，通过一系列引理最终获得第二个种群灭绝性的证明[3]。Chen等针对有界域内的n物种捕食–猎物模型展开研究，通过构造正解或拟解建立了时间依赖解与稳态解的关联条件，并通过数值模拟验证了理论结果[4]。Tuerxun Nafeisha等[5]构建了一类n物种随机食物链模型，引入白噪声与Lévy跳跃，通过发展辅助函数法与随机微分方程理论，建立系统的全局随机动力学，数值模拟验证了理论结果[5]。Jiang Zhao等研究了一类n物种Lotka-Volterra合作模型，引入反馈控制项与连续时滞，利用积分不等式技术、比较原理及Lyapunov-Razumikhin方法，推导出系统全局吸引性、持久性及周期解存在性的条件，并通过数值案例验证结果的实用性[6]。Li通过研究一类n种群捕食–竞争系统，利用构造Lyapunov函数以及微分不等式得到了该系统的概周期解的全局稳定性充分条件[7]。Seno H研究了一类n猎物-1捕食者的Lotka-Volterra系统，研究猎物间通过共享捕食者产生的表观竞争效应，推导可行平衡点渐近稳定的充要条件，明确其决定猎物灭绝或存续[8]。于勇和于红两位学者提出的一类多种群非自治捕食–竞争系统：

$\{\begin{array}{l}\text{d}{X}_n\left(t\right)=X_n\left(t\right)\left[-b_n\left(t\right)+{\displaystyle \sum _{j=1}^{n-1}{a}_{nj}}\left(t\right)X_j\left(t\right)-a_{nn}\left(t\right)X_n\left(t\right)\right]\text{d}t,\hfill \\ \text{d}{X}_i\left(t\right)=X_i\left(t\right)\left[b_i\left(t\right)-{\displaystyle \sum _{j=1}^n{a}_{ij}}\left(t\right)X_j\left(t\right)\right]\text{d}t\left(1\le i\le n-1\right),\hfill \end{array}$ (1)

其中$X_n$ 为捕食者种群在时刻$t$ 的密度；$X_i\left(1\le i\le n-1\right)$ 为食饵种群在时刻$t$ 的密度；通过构造Liapunov函数，得到了该系统的一致持续生存和全局渐近稳定的充分条件[9]。尽管随机n种群模型研究已取得丰硕成果，但对于随机噪声特异性作用于捕食过程的捕食–竞争模型，其动力学行为的理论研究尚不充分。

在实际情况中，除了生物种间相互作用外，环境噪声在种群动态中也起着关键作用。由于生态系统中存在长期而复杂的环境波动，模型参数常常呈现出随机扰动特性。在本文中我们假设环境主要影响增长率和死亡率。在系统(1)的基础上，假设：

$b_i\left(t\right)\to b_i\left(t\right)+{\sigma }_i\left(t\right)\dot{B}\left(t\right).$

因而可得到随机模型(SDE)表示为：

$\{\begin{array}{l}\text{d}{X}_n\left(t\right)=X_n\left(t\right)\left[-b_n\left(t\right)+{\displaystyle \sum _{j=1}^{n-1}{a}_{nj}}\left(t\right)X_j\left(t\right)-a_{nn}\left(t\right)X_n\left(t\right)\right]\text{d}t-{\sigma }_n\left(t\right)X_n\left(t\right)\text{d}{B}_n\left(t\right),\hfill \\ \text{d}{X}_i\left(t\right)=X_i\left(t\right)\left[b_i\left(t\right)-{\displaystyle \sum _{j=1}^n{a}_{ij}}\left(t\right)X_j\left(t\right)\right]\text{d}t+{\sigma }_i\left(t\right)X_i\left(t\right)\text{d}{B}_i\left(t\right)\left(1\le i\le n-1\right),\hfill \end{array}$ (2)

其中$B_i\left(t\right)\left(i=1,\cdots ,n\right)$ 表示定义在一个完备的概率空间$\left(\Omega ,F,{\left\{F_t\right\}}_{t\ge 0},P\right)$ 上的独立标准布朗运动；$b_i\left(t\right),a_{ij}\left(t\right),{\sigma }_i\left(t\right)\left(i,j=1,2,\cdots ,n\right)$ 是正值连续有界函数；$b_i\left(t\right)$ 表示第$i$ 个种群增长率$\left(i=1,2,\cdots ,n-1\right)$ ；$b_n\left(t\right)$ 表示第$n$ 个种群死亡率；$a_{ij}\left(t\right)$ 表示第$i$ 和$j$ 个种群的种间作用系数；${\sigma }_i\left(t\right)$ 第$i$ 个种群噪声强度。

本文将对系统(2)做出的研究在章节2给出。

符号说明：$\stackrel{\vee }{a_{ij}}=\underset{1\le i,j\le n}{\max }{a}_{ij},0<\stackrel{\wedge }{{\displaystyle a_{ij}}}=\underset{1\le i,j\le n}{\min }{a}_{ij}$ ，$\stackrel{\vee }{a}=\underset{1\le i,j\le n}{\max }\stackrel{\vee }{{\displaystyle a_{ij}}},0<\stackrel{\wedge }{a}=\underset{1\le i,j\le n}{\min }\stackrel{\wedge }{{\displaystyle a_{ij}}}$ ，$R_{+}^n=\left\{X\in R^n:X_i>0,1\le i\le n\right\}$ 。

2. 主要结论和结果

2.1. 全局唯一正解存在性

在研究复杂动态系统的过程中，唯一全局正解的存在性是理解随机系统行为的关键基石。因此，这里先给出该系统全局唯一正解的存在性证明，为后续分析提供了稳固的起点。

定理1：对任意给定的初值$X\left(0\right)\in {ℝ}_{+}^n$ ，系统(2)依概率1存在唯一的全局正解。

证明：由文献[9]易知，系统(2)的解为正，且该系统的系数满足局部Lipschitz条件，因此对任意给定的初始值$X\left(0\right)\in {ℝ}_{+}^n$ ，存在唯一的局部正解$X\left(t\right)$ 在$t\in \left[0,{\tau }_e\right)$ 上，其中${\tau }_e$ 为爆炸时间。要证明该解是全局的，需证${\tau }_e=\infty$ 几乎必然成立。

设$k_0>0$ 足够大，使得$X\left(0\right)$ 的每个分量都落在区间$\left[\frac{1}{k_0},k_0\right]$ 内。对每个整数$k\ge k_0$ ，定义停时：

${\tau }_k=\inf \left\{t\in \left[0,{\tau }_e\right):x_i\left(t\right)\notin \left(\frac{1}{k},k\right),i=1,\cdots ,n\right\},$

定义$\inf \varnothing =\infty$ (其中$\varnothing$ 表示空集)。显然，当$k\to \infty$ 时，${\tau }_k$ 单调递增。令${\tau }_{\infty }=\underset{k\to \infty }{\lim }{\tau }_k$ ，则${\tau }_{\infty }\le {\tau }_e$ 几乎必然成立[10]。若能证${\tau }_{\infty }=\infty$ 几乎必然成立，则${\tau }_e=\infty$ 几乎必然成立，且对所有$t\ge 0$ ，$X\left(t\right)\in {ℝ}_{+}^n$ 几乎必然成立。下证${\tau }_{\infty }=\infty$ 几乎必然成立。

定义$C^2$ 函数$V:{ℝ}_{+}^n\to {ℝ}_{+}$ ：$V\left(t,X\right)={\displaystyle \sum _{i=1}^n\left[X_i\left(t\right)-1-\log \left(X_i\left(t\right)\right)\right]}$ 。由$u-1-\log \left(u\right)\ge 0$ (当$u>0$ 时)，可知$V\left(X\right)$ 非负。对$V\left(t,X\right)$ 应用Itô公式得：

$\begin{array}{l}\text{d}V=\left[{\displaystyle \sum _{i=1}^{n-1}\left(X_i\left(t\right)-1\right)}\left(b_i\left(t\right)-{\displaystyle \sum _{j=1}^n{a}_{ij}}\left(t\right)X_j\left(t\right)\right)+\left(X_n\left(t\right)-1\right)\left(-b_n\left(t\right)+{\displaystyle \sum _{j=1}^{n-1}{a}_{nj}}\left(t\right)X_j\left(t\right)-a_{nn}\left(t\right)X_n\left(t\right)\right)\right]\text{d}t\\ +\left[\frac{1}{2}{\displaystyle \sum _{i=1}^n{\sigma }_i^2}\left(t\right)\right]\text{d}t+{\displaystyle \sum _{i=1}^{n-1}\left(X_i\left(t\right)-1\right){\sigma }_i\left(t\right)\text{d}{B}_i\left(t\right)}-\left(X_n\left(t\right)-1\right){\sigma }_n\left(t\right)\text{d}{B}_n\left(t\right).\end{array}$

记

$\begin{array}{c}LV\left(X\right)={\displaystyle \sum _{i=1}^{n-1}\left(X_i\left(t\right)-1\right)}\left(b_i\left(t\right)-{\displaystyle \sum _{j=1}^n{a}_{ij}}\left(t\right)X_j\left(t\right)\right)\\ +\left(X_n\left(t\right)-1\right)\left(-b_n\left(t\right)+{\displaystyle \sum _{j=1}^{n-1}{a}_{nj}}\left(t\right)X_j\left(t\right)-a_{nn}\left(t\right)X_n\left(t\right)\right)+\frac{1}{2}{\displaystyle \sum _{i=1}^n{\sigma }_i^2\left(t\right)}\\ ={\displaystyle \sum _{i=1}^{n-1}{X}_i}\left(t\right)b_i\left(t\right)-{\displaystyle \sum _{i=1}^{n-1}{X}_i}\left(t\right){\displaystyle \sum _{j=1}^n{a}_{ij}}\left(t\right)X_j\left(t\right)-{\displaystyle \sum _{i=1}^{n-1}{b}_i}\left(t\right)+{\displaystyle \sum _{i=1}^{n-1}{\displaystyle \sum _{j=1}^n{a}_{ij}}}\left(t\right)X_j\left(t\right)-b_n\left(t\right)X_n\left(t\right)\\ +{\displaystyle \sum _{j=1}^{n-1}{a}_{nj}}\left(t\right)X_j\left(t\right)X_n\left(t\right)-a_{nn}\left(t\right)X_n{\left(t\right)}^2+b_n\left(t\right)-{\displaystyle \sum _{j=1}^{n-1}{a}_{nj}}\left(t\right)X_j\left(t\right)+a_{nn}\left(t\right)X_n\left(t\right)+\frac{1}{2}{\displaystyle \sum _{i=1}^n{\sigma }_i^2}\left(t\right).\end{array}$ (3)

从而有

$\text{d}V=LV\left(X\left(t\right)\right)\text{d}t+{\displaystyle \sum _{i=1}^{n-1}{\sigma }_i}\left(t\right)X_i\text{d}{B}_i\left(t\right)-{\sigma }_n\left(t\right)X_n\text{d}{B}_n\left(t\right).$

取期望并积分得

$EV\left(X\left({\tau }_k\wedge T\right)\right)=V\left(X_0\right)+E{\displaystyle {\int }_0^{T\wedge {\tau }_k}\left(LV\left(X\left(t\right)\right)\right)\text{d}s},$

在(3)式中，显然有

$\begin{array}{l}-{\displaystyle \sum _{j=1}^{n-1}{a}_{nj}}\left(t\right)X_j\left(t\right)-{\displaystyle \sum _{i=1}^{n-1}{b}_i}\left(t\right)+a_{nn}\left(t\right)X_n\left(t\right)-a_{nn}{X}_n{\left(t\right)}^2+b_n\left(t\right)<0,\\ {\displaystyle \sum _{j=1}^{n-1}{a}_{nj}}\left(t\right)X_j\left(t\right)X_n\left(t\right)-{\displaystyle \sum _{i=1}^{n-1}{X}_i}\left(t\right){\displaystyle \sum _{j=1}^n{a}_{ij}}\left(t\right)X_j\left(t\right)<0.\end{array}$

因此对$LV\left(X\left(t\right)\right)$ 利用放缩法可得

$LV\left(X\left(t\right)\right)\le \stackrel{\vee }{b}{\displaystyle \sum _{i=1}^n{X}_i}\left(t\right)+n\stackrel{\vee }{a}{\displaystyle \sum _{i=1}^n{X}_i}\left(t\right)+\frac{1}{2}{\displaystyle \sum _{i=1}^n{\sigma }_i^2}=\left(\stackrel{\vee }{b}+n\stackrel{\vee }{a}\right){\displaystyle \sum _{i=1}^n{X}_i\left(t\right)}+\frac{1}{2}{\displaystyle \sum _{i=1}^n{\sigma }_i^2}\left(t\right).$

注意到当$u>0$ 时$u\le 2\left[u-1-\log \left(u\right)\right]+2$ ，从而

$\begin{array}{c}LV\left(X\left(t\right)\right)\le \left(\stackrel{\vee }{b}+n\stackrel{\vee }{a}\right){\displaystyle \sum _{i=1}^n{X}_i\left(t\right)}+\frac{1}{2}{\displaystyle \sum _{i=1}^n{\sigma }_i^2\left(t\right)}\le 2\left[\stackrel{\vee }{b}+n\stackrel{\vee }{a}\right]\left[V\left(X\right)+n\right]+\frac{1}{2}n\stackrel{\vee }{{\sigma }_i^2}\\ =2\left[\stackrel{\vee }{b}+n\stackrel{\vee }{a}\right]V\left(X\right)+n\left[2\stackrel{\vee }{b}+2n+n\stackrel{\vee }{a}+\frac{1}{2}n\stackrel{\vee }{{\sigma }_i^2}\right]=K_{1}V\left(X\right)+K_2.\end{array}$

其中$K_1=2\left[\stackrel{\vee }{b}+n\stackrel{\vee }{a}\right]>0$，$K_2=n\left[2\stackrel{\vee }{b}+2n+n\stackrel{\vee }{a}+\frac{1}{2}n\stackrel{\vee }{{\sigma }_i^2}\right]>0$为正常数，因为$X\left(t\right)\in {ℝ}_{+}^n$ ，因此可得到

$\mathbb{E}V\left(X\left({\tau }_k\wedge T\right)\right)\le V\left(0\right)+K_2\mathbb{E}\left({\tau }_k\wedge T\right)+K_1\mathbb{E}{\displaystyle {\int }_0^{{\tau }_k\wedge T}V}\left(X\left(t\right)\right)\text{d}t\le V\left(0\right)+K_{2}T+K_1{\displaystyle {\int }_0^T\mathbb{E}}V\left(X\left({\tau }_k\wedge t\right)\right)\text{d}t.$

由Gronwall不等式得

$\mathbb{E}V\left(X\left({\tau }_k\wedge T\right)\right)\le \left[V\left(0\right)+K_{2}T\right]{\text{e}}^{K_{1}T}.$ (4)

注意到当$\omega =\left\{{\tau }_k\le T\right\}$ 时，存在某个$i$ 使得$x_i\left({\tau }_k,\omega \right)$ 等于$k$ 或$\frac{1}{k}$ ，因此，由(4)得

$\left[V\left(0\right)+K_{2}T\right]{\text{e}}^{K_T}\ge E\left[1_{\left\{{\tau }_k\le T\right\}}\left(\omega \right)V\left(X\left({\tau }_k,\omega \right)\right)\right]\ge P\left\{{\tau }_k\le T\right\}\left[\left(k-1-\log k\right)\wedge \left(\frac{1}{k}-1+\log k\right)\right].$

其中$1_{\left\{{\tau }_k\le T\right\}}$ 是$\left\{{\tau }_k\le T\right\}$ 上的指标函数．令$k\to \infty$ 则

$\underset{k\to \infty }{\lim }P\left\{{\tau }_k\le T\right\}=0.$

因为$T>0$ 是任意的，则$P\left\{{\tau }_{\infty }<\infty \right\}=0$ ，所以$P\left\{{\tau }_{\infty }=\infty \right\}=1$ 。证毕。

2.2. 随机最终有界

当我们深入探究系统在随机环境下的长期表现时，一个自然的问题随之而来：在随机扰动的情况下，系统的解是否依然能够保持某种有界性？这就引出了我们接下来要探讨的随机最终有界性。

定义1：若对任意小的$\epsilon >0$ ，存在正常数$T=T\left(\epsilon \right)$ 和$C=C\left(\epsilon \right)$ ，使得当$t\ge T$ 时，系统(2)的解$X\left(t\right)$ 满足有$P\left(\left|X\left(t\right)\right|\ge C\right)<\epsilon$ 成立，则称系统(2)随机最终有界。

定理2：若满足$a_{in}>a_{ni}>0,\left(1\le i\le n-1\right)$ 时，则系统(2)的解是随机最终有界的。

证明：定义$U\left(X\left(t\right)\right)={\text{e}}^t{\displaystyle \sum _{i=1}^n{X}_i\left(t\right)}:={\text{e}}^{t}V\left(X\left(t\right)\right)$ 时，对$U\left(X\left(t\right)\right)$ 应用Itô公式得

$\begin{array}{c}\text{d}U={\text{e}}^t{\displaystyle \sum _{i=1}^n{X}_i\left(t\right)}+{\text{e}}^t{\displaystyle \sum _{i=1}^{n-1}{X}_i\left(t\right)}\left[b_i\left(t\right)-{\displaystyle \sum _{j=1}^n{a}_{ij}}\left(t\right)X_j\left(t\right)\right]\text{d}t\\ +{\text{e}}^t{X}_n\left(t\right)\left[-b_n\left(t\right)+{\displaystyle \sum _{j=1}^{n-1}{a}_{nj}}\left(t\right)X_j\left(t\right)-a_{nn}\left(t\right)X_n\left(t\right)\right]\text{d}t\\ +{\text{e}}^t{\displaystyle \sum _{i=1}^{n-1}{\sigma }_i}\left(t\right)X_i\left(t\right)\text{d}{B}_i\left(t\right)-{\text{e}}^t{\sigma }_n\left(t\right)X_n\left(t\right)\text{d}{B}_n\left(t\right).\end{array}$

记$F\left(X\left(t\right)\right)={\text{e}}^t{\displaystyle \sum _{i=1}^n{X}_i}+{\text{e}}^t{\displaystyle \sum _{i=1}^{n-1}{X}_i\left(t\right)}\left[b_i\left(t\right)-{\displaystyle \sum _{j=1}^n{a}_{ij}}\left(t\right)X_j\left(t\right)\right]\text{d}t+{\text{e}}^t{X}_n\left(t\right)\left[-b_n\left(t\right)+{\displaystyle \sum _{j=1}^{n-1}{a}_{nj}}\left(t\right)X_j\left(t\right)-a_{nn}\left(t\right)X_n\left(t\right)\right]$ 。从而

$\text{d}U=F\left(X\left(t\right)\right)\text{d}t+{\displaystyle \sum _{i=1}^{n-1}{\sigma }_i}\left(t\right)X_i\left(t\right)\text{d}{B}_i\left(t\right)-{\sigma }_n\left(t\right)X_n\left(t\right)\text{d}{B}_n\left(t\right).$ (5)

对每个整数$k\ge \left|X\left(0\right)\right|$ ，定义停时：${\tau }_k=\inf \left\{t\in R_{+}:\left|X\left(t\right)\right|\ge k\right\}$ ，由

$\begin{array}{c}F\left(X\left(t\right)\right)\le {\text{e}}^t{\displaystyle \sum _{i=1}^n{X}_i}\left(t\right)+{\text{e}}^t{\displaystyle \sum _{i=1}^{n-1}{X}_i\left(t\right)}\left[b_i\left(t\right)-{\displaystyle \sum _{j=1}^{n-1}{a}_{ij}}\left(t\right)X_j\left(t\right)\right]\\ -{\text{e}}^t{\displaystyle \sum _{i=1}^{n-1}{X}_i}\left(t\right)X_n\left(t\right)\left(a_{in}\left(t\right)-a_{ni}\left(t\right)\right)-{\text{e}}^t{a}_{nn}\left(t\right)X_n^2\left(t\right)\\ \le {\text{e}}^t\left(b_i\left(t\right)+1\right){\displaystyle \sum _{i=1}^n{X}_i\left(t\right)}-{\text{e}}^t\stackrel{\wedge }{a}{\left({\displaystyle \sum _{i=1}^n{X}_i\left(t\right)}\right)}^2\le K.\end{array}$

对(5)积分取期望可得到

${\text{e}}^{{\tau }_k\wedge t}EV\left(X\left({\tau }_k\wedge t\right)\right)\le V\left(X_0\right)+K{\displaystyle {\int }_0^{{\tau }_k\wedge t}{\text{e}}^s}\text{d}t.$

令$k$ 趋于无穷大，可得

${\text{e}}^{t}EV\left(X\left(t\right)\right)\le V\left(X_0\right)+K{\text{e}}^t.$ (6)

由范数定义，即若$X\in {ℝ}^n$ ，其范数记为$\left|X\right|={\left({\displaystyle \sum _{i=1}^n{X}_i^2}\right)}^{\frac{1}{2}}$ 。从而式(6)可化简得出

$E\left|X\left(t\right)\right|\le EV\left(X\left(t\right)\right)\le V\left(x_0\right){\text{e}}^{-t}+K.$

显然

$\underset{t\to \infty }{\lim \sup }E\left|X\left(t\right)\right|\le K.$

对任意$\eta >0$ ，存在$T_1>0$ ，使得当$t\ge T_1$ 时，$\mathbb{E}\left|X\left(t\right)\right|\le K+\eta$ 。取$C=\frac{2\left(K+\eta \right)}{\epsilon }$ ，根据Markov不等式得：

$P\left(\left|X\left(t\right)\right|\ge C\right)\le \frac{\mathbb{E}\left|X\left(t\right)\right|}{C}\le \frac{K+\eta }{C}=\frac{K+\eta }{\frac{2\left(K+\eta \right)}{\epsilon }}\le \frac{\epsilon }{2}.$

综上，对任意 $\epsilon >0$ ，取 $T=T_1$ 、$C=\frac{2\left(K+\eta \right)}{\epsilon }$ ，则当$t\ge T$ 时，

$P\left(\left|X\left(t\right)\right|\ge C\right)<\epsilon ,$

证毕。

2.3. 随机持久性

随机最终有界性为我们理解系统在随机干扰下的稳定性提供了重要依据。然而，仅仅知道系统状态是有界的还远远不够，我们更关心系统能否在随机环境中持续生存并保持一定的活力。这就促使我们进一步研究系统(2)的随机持久性。

定义2 [11]：对随机系统(2)，若任意初始值$X_0\in {ℝ}_{+}^n$ ，其解$X\left(t\right)=\left(X_1\left(t\right),\cdots ,X_n\left(t\right)\right)$ 满足

$0<\underset{t\to \infty }{\lim \inf }{X}_i\left(t\right)\le \underset{t\to \infty }{\lim \sup }{X}_i\left(t\right)<\infty \text{a}\text{.s}\text{.},i=1,2,\cdots ,n.$

则称该随机系统是几乎必然随机持久的。

引理2 [11] 设$A\left(t\right)$ 和$U\left(t\right)$ 是两个连续的适应递增过程($A\left(0\right)=U\left(0\right)=0$ 几乎必然成立)，$M\left(t\right)$ 是一个实值连续局部鞅($M\left(0\right)=0$ ，$\mathbb{E}\left[M\left(t\right)\right]=0$ )。定义非负过程$X\left(t\right)=X_0+A\left(t\right)-U\left(t\right)+M\left(t\right)$ ($X_0$ 为非负随机变量)。若$X\left(t\right)$ 非负，则

$\left\{\underset{t\to \infty }{\lim }A\left(t\right)<\infty \right\}\subset \left\{\underset{t\to \infty }{\lim }U\left(t\right)<\infty \right\}\cap \left\{\underset{t\to \infty }{\lim }X\left(t\right)<\infty \right\}\text{a}\text{.s}\text{.}$

定理3 设初始值$X\left(0\right)=X_0>0$ 。若参数满足$\underset{t\to \infty }{\lim \inf }\left(a_{ij}\left(t\right)\right)>0,\left(1\le i,j\le n\right)$ ，$\underset{t\to \infty }{\lim \inf }\left(A-{\sigma }_n^2\left(t\right)-b_n\left(t\right)\right)>0$ ，$\underset{t\to \infty }{\lim \inf }\left(b_i\left(t\right)-{\sigma }_i^2\left(t\right)\right)>0,\left(1\le i\le n-1\right)$ ，其中${\displaystyle \sum _{j=1}^{n-1}\stackrel{\wedge }{a_{nj}}\left(t+T\right)\stackrel{\wedge }{D_i}}=A>0$ ，$0<\stackrel{\wedge }{D_i}=\underset{1\le i\le n}{\min }{X}_i\left(t+T\right)$ ，则系统(2)的解$X\left(t\right)$ 几乎必然具有随机持久性，即

$0<\underset{t\to \infty }{\lim \inf }{X}_i\left(t\right)\le \underset{t\to \infty }{\lim \sup }{X}_i\left(t\right)<\infty \text{a}\text{.s}\text{.}$

证明：由$\underset{t\to \infty }{\lim \inf }\left(b_i\left(t\right)-{\sigma }_i^2\left(t\right)\right)>0$ ，$\underset{t\to \infty }{\lim \inf }\left(A-{\sigma }_n^2\left(t\right)-b_n\left(t\right)\right)>0$ ，和$\underset{t\to \infty }{\lim \inf }\left(a_{ij}\left(t\right)\right)>0$ ，由极限的定义可知：对任意的$\epsilon >0$ ，存在$T>0$ ，使得当$t\ge T$ 时，有$b\left(t+T\right)-{\sigma }^2\left(t+T\right)>\epsilon >0$ ，$A-{\sigma }_n^2\left(t+T\right)-b_n\left(t+T\right)>\epsilon >0$ 和$a_{ij}\left(t+T\right)>\epsilon >0$ 。通过时间变换，可得

$\{\begin{array}{l}\text{d}{\tilde{X}}_n\left(t\right)={\tilde{X}}_n\left(t\right)\left[-b_n\left(t+T\right)+{\displaystyle \sum _{j=1}^{n-1}{a}_{nj}}\left(t+T\right){\tilde{X}}_j\left(t\right)-a_{nn}\left(t+T\right){\tilde{X}}_n\left(t\right)\right]\text{d}t-{\sigma }_n\left(t+T\right){\tilde{X}}_n\left(t\right)\text{d}{\tilde{B}}_n\left(t\right),\hfill \\ \text{d}{\tilde{X}}_i\left(t\right)={\tilde{X}}_i\left(t\right)\left[b_i\left(t+T\right)-{\displaystyle \sum _{j=1}^n{a}_{ij}}\left(t+T\right){\tilde{X}}_j\left(t\right)\right]\text{d}t+{\sigma }_i\left(t+T\right){\tilde{X}}_i\left(t\right)\text{d}{\tilde{B}}_i\left(t\right)\left(1\le i\le n-1\right),\hfill \end{array}$ (7)

其中$\tilde{X}\left(t\right)=X\left(t+T\right)$ ，${\tilde{B}}_i\left(t+T\right)=B_i\left(t+T\right)-B_i\left(T\right)$ 。显然${\tilde{B}}_i\left(t\right)$ 是一个具有关于新滤波的性质布朗运动。

(1) 首先证明$\underset{t\to \infty }{\lim \sup }{\tilde{X}}_i\left(t\right)<\infty \text{a}\text{.s}\text{.},\left(1\le i\le n\right)$。由(7)可知：

$\text{d}{\tilde{X}}_i\left(t\right)={\tilde{X}}_i\left(t\right)\left[b_i\left(t+T\right)-{\displaystyle \sum _{j=1}^n{a}_{ij}}\left(t+T\right){\tilde{X}}_j\left(t\right)\right]\text{d}t+{\sigma }_i\left(t+T\right){\tilde{X}}_i\left(t\right)\text{d}{\tilde{B}}_i\left(t\right)\left(1\le i\le n-1\right),$

设函数$f\left(t,\tilde{X}\right)={\text{e}}^t{\tilde{X}}_i\left(t\right)$，根据Itô公式得：

$\text{d}\left({\text{e}}^t{\tilde{X}}_i\left(t\right)\right)={\text{e}}^t{\tilde{X}}_i\left(t\right)\text{d}t+{\text{e}}^t\left[{\tilde{X}}_i\left(t\right)\left(b_i\left(t+T\right)-{\displaystyle \sum _{j=1}^n{a}_{ij}}\left(t+T\right){\tilde{X}}_j\left(t\right)\right)\text{d}t+{\sigma }_i\left(t+T\right){\tilde{X}}_i\left(t\right)\text{d}{\tilde{B}}_i\left(t\right)\right].$

积分化简得

${\text{e}}^t{\tilde{X}}_i\left(t\right)\le {\tilde{X}}_i\left(0\right)+{\displaystyle {\int }_0^t{\text{e}}^s}\left[\left(b_i\left(s+T\right)+1\right){\tilde{X}}_i\left(s\right)-a_{ii}\left(s+T\right){\tilde{X}}_i^2\left(s\right)\right]\text{d}s+M_i\left(t\right),$

其中$M_i\left(t\right)={\displaystyle {\int }_0^t{\sigma }_i}\left(s+T\right){\text{e}}^s{\tilde{X}}_i\left(s\right)\text{d}{\tilde{B}}_i\left(s\right)$。因而有

${\text{e}}^t{\tilde{X}}_i\left(t\right)\le {\tilde{X}}_i\left(0\right)+C\left({\text{e}}^t-1\right)+M_i\left(t\right).$

两边除以${\text{e}}^t$ 可得

${\tilde{X}}_i\left(t\right)\le {\tilde{X}}_i\left(0\right){\text{e}}^{-t}+C\left(1-{\text{e}}^{-t}\right)+M_i\left(t\right){\text{e}}^{-t}.$

可得

$X_i\left(t\right)\le X_i\left(0\right)+C+M_i\left(t\right).$

记$Z_i\left(t\right)={\tilde{X}}_i\left(0\right)+C+M_i\left(t\right)$ ，由引理2得

$\underset{t\to \infty }{\lim \sup }{\tilde{X}}_i\left(t\right)\le \underset{t\to \infty }{\lim \sup }{Z}_i\left(t\right)<\infty \text{a}\text{.s}.\left(1\le i\le n-1\right).$ (8)

由(7)可知：

$\text{d}{\tilde{X}}_n\left(t\right)={\tilde{X}}_n\left(t\right)\left[-b_n\left(t+T\right)+{\displaystyle \sum _{j=1}^{n-1}{a}_{nj}}\left(t+T\right){\tilde{X}}_j\left(t\right)-a_{nn}\left(t+T\right){\tilde{X}}_n\left(t\right)\right]\text{d}t-{\sigma }_n\left(t+T\right){\tilde{X}}_n\left(t\right)\text{d}{\tilde{B}}_n\left(t\right),$

应用Itô公式到${\text{e}}^t{X}_n\left(t\right)$ 可得

$\text{d}\left({\text{e}}^t{\tilde{X}}_n\left(t\right)\right)={\text{e}}^t{\tilde{X}}_n\left(t\right)\left[-b_n\left(t+T\right)+1+{\displaystyle \sum _{j=1}^{n-1}{a}_{nj}}\left(t+T\right){\tilde{X}}_j\left(t\right)-a_{nn}\left(t\right){\tilde{X}}_n\left(t\right)\right]\text{d}t+{\sigma }_n\left(t+T\right){\text{e}}^t{\tilde{X}}_n\left(t\right)\text{d}{\tilde{B}}_n.$

积分得

${\text{e}}^t{\tilde{X}}_n\left(t\right)={\tilde{X}}_n\left(0\right)+{\displaystyle {\int }_0^t{\text{e}}^s}{\tilde{X}}_n\left(s\right)\left[-b_n\left(s+T\right)+1+{\displaystyle \sum _{j=1}^{n-1}{a}_{nj}}\left(s+T\right){\tilde{X}}_j\left(s\right)-a_{nn}\left(s+T\right){\tilde{X}}_n\left(s\right)\right]\text{d}s+M_n\left(t\right).$

由(8)可知，${\tilde{X}}_i\left(t\right)\left(1\le i\le n-1\right)$ 有上界，从而令${\displaystyle \sum _{j=1}^{n-1}\stackrel{\vee }{a_{nj}}}{\tilde{X}}_j=B$ ，因此有

${\tilde{X}}_n\left(t\right)\le {\tilde{X}}_n\left(0\right){\text{e}}^{-t}+{\displaystyle {\int }_0^t{\text{e}}^{s-t}}\left[{\tilde{X}}_n\left(s\right)\cdot B-\stackrel{\wedge }{{\displaystyle a_{nn}}}{\tilde{X}}_n^2\left(s\right)\right]\text{d}s+{\text{e}}^{-t}{M}_n\left(t\right).$

从而有

${\tilde{X}}_n\left(t\right)\le {\tilde{X}}_n\left(0\right)+B^{\prime }+M_n\left(t\right),$ (9)

其中$B,B^{\prime }$ 都是正常数。对(8)应用引理2可得

$\underset{t\to \infty }{\lim \sup }{\tilde{X}}_n\left(t\right)<\infty \text{a}\text{.s}\text{.}$ (10)

结合(8)，(10)对$\left(1\le i\le n\right)$ 有

$\underset{t\to \infty }{\lim \sup }{\tilde{X}}_i\left(t\right)<\infty \text{a}\text{.s}\text{.}$ (11)

(2) 再证$\lim \inf {\tilde{X}}_i\left(t\right)>0,\text{a}\text{.s}.$

对系统(3)的第二个方程进行变形：令$y_i\left(t\right)={\tilde{X}}_i{\left(t\right)}^{-1}$ ，${\tilde{X}}_i\left(t\right)>0,\left(1\le i\le n-1\right)$ ，$f\left(t,y\right)={\text{e}}^{ct}{\displaystyle \sum _{i=1}^{n-1}{y}_i}\left(t\right)$ ，对$f\left(t,y\right)$ 应用Itô公式得

$\text{d}f={\text{e}}^{ct}{\displaystyle \sum _{i=1}^{n-1}\left[\left(c-b_i\left(t+T\right)+{\displaystyle \sum _{j=1}^n{a}_{ij}}\left(t+T\right)y_j^{-1}\left(t\right)+{\sigma }_i^2\left(t+T\right)\right)y_i\left(t\right)\text{d}t-{\sigma }_i\left(t+T\right)y_i\left(t\right)\text{d}{\tilde{B}}_i\left(t\right)\right]}.$

对上式从0到$t$ 积分

$\begin{array}{c}{\displaystyle {\int }_0^t\text{d}}f={\displaystyle {\int }_0^t{\text{e}}^{cs}}{\displaystyle \sum _{i=1}^{n-1}{y}_i}\left(s\right)\left[-\left(b_i\left(s+T\right)-{\sigma }_i^2\left(s+T\right)-c\right)+{\displaystyle \sum _{j=1}^n{a}_{ij}}\left(s+T\right)y_j^{-1}\left(s\right)\right]\text{d}s\\ -{\displaystyle \sum _{i=1}^{n-1}{\displaystyle {\int }_0^t{\text{e}}^{cs}}}{\sigma }_i\left(t+T\right)y_i\left(s\right)\text{d}{\tilde{B}}_i\left(s\right).\end{array}$

其中$c$ 取充分小的正常数，$c{M}_i\left(t\right)={\displaystyle \sum _{i=1}^{n-1}{\displaystyle {\int }_0^t{\sigma }_i}}\left(s+T\right){\text{e}}^{cs}{y}_i\left(s\right)\text{d}{\tilde{B}}_i\left(s\right)$，由$\underset{t\to \infty }{\lim \inf }\left(b_i\left(t\right)-{\sigma }_i^2\left(t\right)\right)>0,\left(1\le i\le n-1\right)$ ，因而有

$\begin{array}{c}{\displaystyle {\int }_0^t\text{d}}f\le {\displaystyle {\int }_0^t{\text{e}}^{cs}}{\displaystyle \sum _{i=1}^{n-1}{y}_i}\left(s\right)\left[{\displaystyle \sum _{j=1}^n{a}_{ij}}\left(s+T\right)y_j^{-1}\left(s\right)\right]\text{d}s-c{M}_i\left(t\right)\\ \le {\displaystyle {\int }_0^t{\text{e}}^{cs}}{\displaystyle \sum _{i=1}^n{y}_i}\left(s\right)\left[{\displaystyle \sum _{j=1}^n{a}_{ij}}\left(s+T\right)y_j^{-1}\left(s\right)\right]\text{d}s-c{M}_i\left(t\right)\\ \le {\displaystyle {\int }_0^t{\text{e}}^{cs}}{\displaystyle \sum _{i=1}^n{y}_i}\left(s\right)\left[{\displaystyle \sum _{j=1}^n\stackrel{\vee }{a_{ij}}}{y}_j^{-1}\left(s\right)\right]\text{d}s-c{M}_i\left(t\right).\end{array}$

由${\text{e}}^{ct}{y}_i\left(t\right)<{\text{e}}^{ct}{\displaystyle \sum _{i=1}^{n-1}{y}_i\left(t\right)}$ 化简得

可得

$y_i\left(t\right)\le y_i\left(0\right)+C-c{\text{e}}^{-t}{M}_i\left(t\right).$

其中C为正常数，记$Y_i\left(t\right)=y_i\left(0\right)+C-c{\text{e}}^{-t}{M}_i\left(t\right)$ ，由引理2可得

$\underset{t\to \infty }{\lim \sup }{y}_i\left(t\right)\le \underset{t\to \infty }{\lim \sup }{Y}_i\left(t\right)<\infty .\text{a}\text{.s}\text{.}$

从而

$\underset{t\to \infty }{\lim \inf }{\tilde{X}}_i\left(t\right)=\underset{t\to \infty }{\lim \inf }\frac{1}{y_i\left(t\right)}\ge \frac{1}{\underset{t\to \infty }{\lim \sup }{y}_i\left(t\right)}>0.\text{a}\text{.s}\text{.}$ (12)

对(7)第一个方程：构造倒数辅助函数：$y_n\left(t\right)={\tilde{X}}_n{\left(t\right)}^{-1}$ ，对$g\left(t,y_n\right)={\text{e}}^{c_{n}t}{y}_n\left(t\right)$ 应用Itô公式，推导得微分形式：

$\begin{array}{c}\text{d}\left({\text{e}}^{c_{n}t}{y}_n\left(t\right)\right)={\text{e}}^{c_{n}t}{y}_n\left(t\right)\cdot \left[c_n+b_n\left(t+T\right)-{\displaystyle \sum _{j=1}^{n-1}{a}_{nj}}\left(t+T\right)y_j{\left(t\right)}^{-1}+a_{nn}\left(t+T\right)y_n{\left(t\right)}^{-1}+{\sigma }_n{\left(t+T\right)}^2\right]\text{d}t\\ -{\text{e}}^{c_{n}t}{\sigma }_n\left(t+T\right)y_n\left(t\right)\text{d}{\tilde{B}}_n\left(t\right).\end{array}$

由(8)，(12)可知，${\tilde{X}}_i\left(t\right)\left(1\le i\le n-1\right)$ 有界，因此存在常数$\stackrel{\wedge }{D_i}>0$ ，有$y_i{\left(t\right)}^{-1}={\tilde{X}}_i\left(t\right)\ge \stackrel{\wedge }{D_i}$ 。令${\displaystyle \sum _{l=1}^{n-1}\stackrel{\wedge }{a_{nl}}\left(t+T\right)\stackrel{\wedge }{D_i}}=A$ ，对$\text{d}\left({\text{e}}^{c_{n}t}{y}_n\left(t\right)\right)$ 积分，并化简得

${\text{e}}^{c_{n}t}{y}_n\left(t\right)\le y_n\left(0\right)+{\displaystyle {\int }_0^t{\text{e}}^{c_{n}s}}{y}_n\left(s\right)\cdot \left[c_n+b_n\left(s+T\right)-A+a_{nn}\left(s+T\right)y_n{\left(s\right)}^{-1}+{\sigma }_n{\left(s+T\right)}^2\right]\text{d}s-c_n{M}_n\left(t\right).$

取$c_n$ 为足够小的正常数，由$\underset{t\to \infty }{\lim \inf }\left(A-{\sigma }_n^2\left(t\right)-b_n\left(t\right)\right)>0$ ，从而有 ${\text{e}}^{c_{n}t}{y}_n\left(t\right)\le y_n\left(0\right)+{\displaystyle {\int }_0^t{\text{e}}^{c_{n}s}}\stackrel{\vee }{a_{nn}}\text{d}s-c_n{M}_n\left(t\right)$。

因此可得

$y_n\left(t\right)\le y_n\left(0\right)+C^{\prime }-c_n{M}_n\left(t\right),$

其中$C_2,C^{\prime }$ 为正常数，由引理2可得

$\underset{t\to \infty }{\lim \sup }{y}_n\left(t\right)<\infty \text{a}\text{.s}\text{.}\Rightarrow \underset{t\to \infty }{\lim \inf }{\tilde{X}}_n\left(t\right)=\underset{t\to \infty }{\lim \inf }\frac{1}{y_n\left(t\right)}>0.\text{a}\text{.s}\text{.}$ (13)

结合(12)，(13)对$\left(1\le i\le n\right)$ 有

$\lim \inf {\tilde{X}}_i\left(t\right)>0.$ (14)

由(11)、(14)可得$0<\underset{t\to \infty }{\lim \inf }{\tilde{X}}_i\left(t\right)\le \underset{t\to \infty }{\lim \sup }{\tilde{X}}_i\left(t\right)<\infty \text{a}\text{.s}\text{.}$，而${\tilde{X}}_i\left(t\right)=X_i\left(t+T\right)$ ，因而有

$0<\underset{t\to \infty }{\lim \inf }{X}_i\left(t\right)\le \underset{t\to \infty }{\lim \sup }{X}_i\left(t\right)<\infty \text{a}\text{.s}\text{.}$

证毕。

2.4. 结论延拓

为更真实反映自然种群在复杂环境中的动态平衡与波动和刻画种内或种间非线性竞争，下面将随机模型(2)拓展到相应的Gilpin-Ayala竞争随机模型：

$\{\begin{array}{l}\text{d}{X}_n\left(t\right)=X_n\left(t\right)\left[-b_n\left(t\right)+{\displaystyle \sum _{j=1}^{n-1}{a}_{nj}}\left(t\right)X_j^{\theta }\left(t\right)-a_{nn}\left(t\right)X_n^{\theta }\left(t\right)\right]\text{d}t-{\sigma }_n\left(t\right)X_n\left(t\right)\text{d}{B}_n\left(t\right),\\ \text{d}{X}_i\left(t\right)=X_i\left(t\right)\left[b_i\left(t\right)-{\displaystyle \sum _{j=1}^n{a}_{ij}}\left(t\right)X_j^{\theta }\left(t\right)\right]\text{d}t+{\sigma }_i\left(t\right)X_i\left(t\right)\text{d}{B}_i\left(t\right)\left(1\le i\le n-1\right).\end{array}$ (15)

其中$\theta$ 为正常数，参数条件同随机模型(2)一致，若该模型的参数满足相关条件，可得以下定理。

定理4 设初始值$X\left(0\right)=X_0>0$ 。若参数满足$\underset{t\to \infty }{\lim \inf }\left(a_{ij}\left(t\right)\right)>0$ ，$\underset{t\to \infty }{\lim \inf }\left(A\left(\theta \right)-\frac{\theta +1}{2}{\sigma }_n^2\left(t\right)-b_n\left(t\right)\right)>0$ ，$\underset{t\to \infty }{\lim \inf }\left(b_i\left(t\right)-\frac{\theta +1}{2}{\sigma }_i^2\left(t\right)\right)>0,\left(1\le i\le n-1\right)$ ，其中${\displaystyle \sum _{j=1}^{n-1}\stackrel{\wedge }{{\displaystyle a_{nj}}}\left(t+T\right)\stackrel{\wedge }{D_i^{\theta }}}=A\left(\theta \right)>0$ ，$0<\stackrel{\wedge }{D_i}=\underset{1\le i\le n}{\min }{X}_i\left(t+T\right)$ 。则系统(15)的解$X\left(t\right)$ 几乎必然具有随机持久性，即

$0<\underset{t\to \infty }{\lim \inf }{X}_i\left(t\right)\le \underset{t\to \infty }{\lim \sup }{X}_i\left(t\right)<\infty \text{a}\text{.s}\text{.}$

证明：和定理3类似，通过时间变换可得到：

$\{\begin{array}{l}\text{d}{\tilde{X}}_n\left(t\right)={\tilde{X}}_n\left(t\right)\left[-b_n\left(t+T\right)+{\displaystyle \sum _{j=1}^{n-1}{a}_{nj}}\left(t+T\right){\tilde{X}}_j^{\theta }\left(t\right)-a_{nn}\left(t+T\right){\tilde{X}}_n^{\theta }\left(t\right)\right]\text{d}t-{\sigma }_n\left(t\right){\tilde{X}}_n\left(t\right)\text{d}{\tilde{B}}_n\left(t\right),\\ \text{d}{\tilde{X}}_i\left(t\right)={\tilde{X}}_i\left(t\right)\left[b_i\left(t+T\right)-{\displaystyle \sum _{j=1}^n{a}_{ij}}\left(t+T\right){\tilde{X}}_j^{\theta }\left(t\right)\right]\text{d}t+{\sigma }_i\left(t+T\right){\tilde{X}}_i\left(t\right)\text{d}{\tilde{B}}_i\left(t\right)\left(1\le i\le n-1\right).\end{array}$ (16)

(1) 首先证明$\underset{t\to \infty }{\lim \sup }{\tilde{X}}_i\left(t\right)<\infty \text{a}\text{.s}\text{.},\left(1\le i\le n\right)$。由(16)可知：

$\text{d}{\tilde{X}}_i\left(t\right)={\tilde{X}}_i\left(t\right)\left[b_i\left(t+T\right)-{\displaystyle \sum _{j=1}^n{a}_{ij}}\left(t+T\right){\tilde{X}}_j^{\theta }\left(t\right)\right]\text{d}t+{\sigma }_i\left(t+T\right){\tilde{X}}_i\left(t\right)\text{d}{\tilde{B}}_i\left(t\right)\left(1\le i\le n-1\right).$

设函数$f\left(t,\tilde{X}\right)={\text{e}}^t{\tilde{X}}_i^{\theta }\left(t\right)$，根据Itô公式得：

$\begin{array}{c}\text{d}\left({\text{e}}^t{\tilde{X}}_i^{\theta }\left(t\right)\right)={\text{e}}^t{\tilde{X}}_i^{\theta }\left(t\right)\text{d}t+{\text{e}}^t[{\tilde{X}}_i^{\theta }\left(t\right)\left(\theta b_i\left(t+T\right)+\frac{\theta \left(\theta -1\right)}{2}{\sigma }_i^2\left(t+T\right)-\theta {\displaystyle \sum _{j=1}^n{a}_{ij}}\left(t+T\right){\tilde{X}}_j^{\theta }\left(t\right)\right)\text{d}t\\ \begin{array}{c}\\ \end{array}+\theta {\sigma }_i\left(t+T\right){\tilde{X}}_i^{\theta }\left(t\right)\text{d}{\tilde{B}}_i\left(t\right)]\\ \le {\text{e}}^t[{\tilde{X}}_i^{\theta }\left(t\right)\left(1+\theta b_i\left(t+T\right)+\frac{\theta \left(\theta -1\right)}{2}{\sigma }_i^2\left(t+T\right)-\theta a_{ii}\left(t+T\right){\tilde{X}}_i^{\theta }\left(t\right)\right)\text{d}t\\ \begin{array}{c}\\ \end{array}+\theta {\sigma }_i\left(t+T\right){\tilde{X}}_i^{\theta }\left(t\right)\text{d}{\tilde{B}}_i\left(t\right)].\end{array}$

则有

$\text{d}\left({\text{e}}^t{\tilde{X}}_i^{\theta }\left(t\right)\right)\le {\text{e}}^{t}K+{\text{e}}^t\theta {\sigma }_i\left(t+T\right)X_i^{\theta }\left(t\right)\text{d}{\tilde{B}}_i\left(t\right).$

积分化简得

$\begin{array}{c}{\text{e}}^t{\tilde{X}}_i^{\theta }\left(t\right)\le {\tilde{X}}_i^{\theta }\left(0\right)+{\displaystyle {\int }_0^t{\text{e}}^{s}K\text{d}s}+{\displaystyle {\int }_0^t{\text{e}}^s\theta {\sigma }_i\left(s+T\right){\tilde{X}}_i^{\theta }\left(s\right)\text{d}{\tilde{B}}_i\left(s\right)}\\ ={\tilde{X}}_i^{\theta }\left(0\right)+\left({\text{e}}^t-1\right)K+{\displaystyle {\int }_0^t{\text{e}}^s\theta {\sigma }_i\left(s+T\right){\tilde{X}}_i^{\theta }\left(s\right)\text{d}{\tilde{B}}_i}\left(s\right).\end{array}$

两边除以${\text{e}}^t$ 可得

${\tilde{X}}_i^{\theta }\left(t\right)\le {\tilde{X}}_i^{\theta }\left(0\right){\text{e}}^{-t}+K\left(1-{\text{e}}^{-t}\right)+{\displaystyle {\int }_0^t{\text{e}}^{s-t}\theta {\sigma }_i\left(s+T\right){\tilde{X}}_i^{\theta }\left(s\right)\text{d}{\tilde{B}}_i}\left(s\right),$

可得

${\tilde{X}}_i^{\theta }\left(t\right)\le {\tilde{X}}_i^{\theta }\left(0\right)+K+{\displaystyle {\int }_0^t{\text{e}}^{s-t}\theta {\sigma }_i\left(s+T\right){\tilde{X}}_i^{\theta }\left(s\right)\text{d}{\tilde{B}}_i}\left(s\right).$

记$Z_i^{\theta }\left(t\right)={\tilde{X}}_i^{\theta }\left(0\right)+K+{\displaystyle {\int }_0^t{\text{e}}^{s-t}\theta {\sigma }_i\left(s+T\right){\tilde{X}}_i^{\theta }\left(s\right)\text{d}{\tilde{B}}_i\left(s\right)}$，由引理2得$\underset{t\to \infty }{\lim \sup }{\tilde{X}}_i^{\theta }\left(t\right)\le \underset{t\to \infty }{\lim \sup }{Z}_i^{\theta }\left(t\right)<\infty \text{a}\text{.s}.\left(1\le i\le n-1\right)$。

从而有

$\underset{t\to \infty }{\lim \sup }{\tilde{X}}_i\left(t\right)<\infty \text{a}\text{.s}.\left(1\le i\le n-1\right).$ (17)

由(16)可知

$\text{d}{\tilde{X}}_n\left(t\right)={\tilde{X}}_n\left(t\right)\left[-b_n\left(t+T\right)+{\displaystyle \sum _{j=1}^{n-1}{a}_{nj}}\left(t+T\right){\tilde{X}}_j^{\theta }\left(t\right)-a_{nn}\left(t+T\right){\tilde{X}}_n^{\theta }\left(t\right)\right]\text{d}t-{\sigma }_n\left(t\right){\tilde{X}}_n\left(t\right)\text{d}{\tilde{B}}_n\left(t\right),$

应用Itô公式到${\text{e}}^t{X}_n^{\theta }\left(t\right)$ ：

$\begin{array}{c}\text{d}\left({\text{e}}^t{\tilde{X}}_n^{\theta }\left(t\right)\right)={\text{e}}^t{\tilde{X}}_n^{\theta }\left(t\right)\left[-\theta b_n\left(t+T\right)+1+\theta {\displaystyle \sum _{j=1}^{n-1}{a}_{nj}}\left(t+T\right){\tilde{X}}_j^{\theta }\left(t\right)-\theta a_{nn}\left(t+T\right){\tilde{X}}_n^{\theta }\left(t\right)+\frac{\theta \left(\theta -1\right)}{2}{\sigma }_n^2\left(t+T\right)\right]\text{d}t\\ +\theta {\sigma }_n(t+T)e^t{\tilde{X}}_n{}^{\theta }(t)d{\tilde{B}}_n\\ \le {\text{e}}^t{\tilde{X}}_n^{\theta }\left(t\right)\left[-\theta b_n\left(t+T\right)+1+\theta {\displaystyle \sum _{j=1}^{n-1}\stackrel{\vee }{a_{nj}}}\left(t+T\right)\stackrel{\vee }{{\tilde{X}}_j^{\theta }}\left(t\right)-\theta a_{nn}\left(t+T\right){\tilde{X}}_n^{\theta }\left(t\right)+\frac{\theta \left(\theta -1\right)}{2}{\sigma }_n^2\left(t+T\right)\right]\text{d}t\\ +\theta {\sigma }_n(t+T)e^t{\tilde{X}}_n{}^{\theta }(t)d{\tilde{B}}_n.\end{array}$

由(17)可知，${\tilde{X}}_i^{\theta }\left(t\right)\left(1\le i\le n-1\right)$ 有上界，从而令${\displaystyle \sum _{j=1}^{n-1}\stackrel{\vee }{a_{nj}}\stackrel{\vee }{{\tilde{X}}_j^{\theta }}}=B_1$ ，因此积分化简得

因此有

${\tilde{X}}_n{}^{\theta }(t)\le {\tilde{X}}_n{}^{\theta }(0)+{B^{\prime }}_1+{\displaystyle {\int }_0^t\theta {\sigma }_n(s+T)e^{s-t}{\tilde{X}}_n{}^{\theta }(s)d{\tilde{B}}_n(s)},$ (18)

其中$B_1,{B^{\prime }}_1$ 都是正常数。对(18)应用引理2可得$\underset{t\to \infty }{\lim \sup }{\tilde{X}}_n^{\theta }\left(t\right)<\infty \text{a}\text{.s}\text{.}$因此有

$\underset{t\to \infty }{\lim \sup }{\tilde{X}}_n\left(t\right)<\infty \text{a}\text{.s}\text{.}$ (19)

结合(17)，(19)对$\left(1\le i\le n\right)$ 有

$\underset{t\to \infty }{\lim \sup }{\tilde{X}}_i\left(t\right)<\infty \text{a}\text{.s}\text{.}$ (20)

(2) 再证$\lim \inf X_i\left(t\right)>0\text{a}\text{.s}.$

对(16)的第二个方程进行变形：令$y_i^{\theta }\left(t\right)={\tilde{X}}_i{\left(t\right)}^{-\theta }$ ，${\tilde{X}}_i\left(t\right)>0,\left(1\le i\le n-1\right)$ ，$f\left(t,y\right)={\text{e}}^{ct}{\displaystyle \sum _{i=1}^{n-1}{y}_i^{\theta }\left(t\right)}$ ，对$f\left(t,y\right)$ 应用Itô公式得

$\text{d}f={\text{e}}^{ct}{\displaystyle \sum _{i=1}^{n-1}\left[\left(c-\theta b_i\left(t+T\right)+\theta {\displaystyle \sum _{j=1}^n{a}_{ij}}\left(t+T\right)y_j^{-\theta }\left(t\right)+\frac{\theta \left(\theta +1\right)}{2}{\sigma }_i^2\left(t+T\right)\right)y_i^{\theta }\left(t\right)\text{d}t-\theta {\sigma }_i\left(t+T\right)y_i^{\theta }\left(t\right)\text{d}{\tilde{B}}_i\left(t\right)\right]}.$

对上式从0到$t$ 积分：

$\begin{array}{c}{\displaystyle {\int }_0^t\text{d}}f={\displaystyle {\int }_0^t{\text{e}}^{cs}}{\displaystyle \sum _{i=1}^{n-1}{y}_i^{\theta }}\left(s\right)\left[-\left(\theta b_i\left(s+T\right)-\frac{\theta \left(\theta +1\right)}{2}{\sigma }_i^2\left(s+T\right)-c\right)+\theta {\displaystyle \sum _{j=1}^n{a}_{ij}}\left(s+T\right)y_j^{-\theta }\left(s\right)\right]\text{d}s\\ -\theta {\displaystyle \sum _{i=1}^{n-1}{\displaystyle {\int }_0^t{\text{e}}^{cs}}}{\sigma }_i\left(s+T\right)y_i^{\theta }\left(s\right)\text{d}{\tilde{B}}_i\left(s\right).\end{array}$

其中$c$ 取充分小的正常数，${M^{\prime }}_i\left(t\right)=\theta {\displaystyle \sum _{i=1}^{n-1}{\displaystyle {\int }_0^t{\text{e}}^{cs}}}{\sigma }_i\left(s+T\right)y_i^{\theta }\left(s\right)\text{d}{\tilde{B}}_i\left(s\right)$，由$\underset{t\to \infty }{\lim \inf }\left(b_i\left(t\right)-\frac{\theta +1}{2}{\sigma }_i^2\left(t\right)\right)>0,\left(1\le i\le n-1\right)$ 。因而有

$\begin{array}{c}{\displaystyle {\int }_0^t\text{d}f}\le {\displaystyle {\int }_0^t{\text{e}}^{cs}}{\displaystyle \sum _{i=1}^{n-1}{y}_i^{\theta }}\left(s\right)\left[\theta {\displaystyle \sum _{j=1}^n{a}_{ij}}\left(s+T\right)y_j^{-\theta }\left(s\right)\right]\text{d}s-{M^{\prime }}_i\left(t\right)\\ \le {\displaystyle {\int }_0^t{\text{e}}^{cs}}{\displaystyle \sum _{i=1}^n{y}_i^{\theta }}\left(s\right)\left[\theta {\displaystyle \sum _{j=1}^n{a}_{ij}}\left(s+T\right)y_j^{-\theta }\left(s\right)\right]\text{d}s-{M^{\prime }}_i\left(t\right)\\ \le {\displaystyle {\int }_0^t{\text{e}}^{cs}}{\displaystyle \sum _{i=1}^n{y}_i^{\theta }}\left(s\right)\left[\theta {\displaystyle \sum _{j=1}^n\stackrel{\vee }{a_{ij}}}{y}_j^{-\theta }\left(s\right)\right]\text{d}s-{M^{\prime }}_i\left(t\right).\end{array}$

由${\text{e}}^{ct}{y}_i^{\theta }\left(t\right)<{\text{e}}^{ct}{\displaystyle \sum _{i=1}^{n-1}{y}_i^{\theta }\left(t\right)}$ 化简得

(21)

可得

$y_i^{\theta }\left(t\right)\le y_i^{\theta }\left(0\right)+C-{\text{e}}^{-ct}{M^{\prime }}_i\left(t\right).$

记$Y_i^{\theta }\left(t\right)=y_i^{\theta }\left(0\right)+C-{\text{e}}^{-ct}{M^{\prime }}_i\left(t\right)$ ，由引理2可得$\underset{t\to \infty }{\lim \sup }{y}_i^{\theta }\left(t\right)\le \underset{t\to \infty }{\lim \sup }{Y}_i^{\theta }\left(t\right)<\infty \text{a}\text{.s}\text{.}$ 从而可得

$\underset{t\to \infty }{\lim \inf }{\tilde{X}}_i^{\theta }\left(t\right)=\underset{t\to \infty }{\lim \inf }\frac{1}{y_i^{\theta }\left(t\right)}\ge \frac{1}{\underset{t\to \infty }{\lim \sup }{y}_i^{\theta }\left(t\right)}>0\text{a}\text{.s}\text{.}$

因而有

$\underset{t\to \infty }{\lim \inf }{\tilde{X}}_i(t)>0(1\le i\le n-1)\text{a}\text{.s}\text{.}$ (22)

对(16)第一个方程：构造倒数辅助函数：$y_n{}^{\theta }(t)={\tilde{X}}_n{(t)}^{-\theta }$ ，对$g\left(t,y_n\right)={\text{e}}^{c_{n}t}{y}_n^{\theta }\left(t\right)$ 应用Itô公式，推导得微分形式：

$\begin{array}{l}d\left(e^{c_{n}t}{y}_n{}^{\theta }(t)\right)=e^{c_{n}t}{y}_n{}^{\theta }(t)\cdot \left[c_n+\theta b_n(t+T)-\theta {\displaystyle \sum _{j=1}^{n-1}{a}_{nj}}(t+T)y_j{(t)}^{-\theta }+\theta a_{nn}(t+T)y_n{(t)}^{-\theta }+\frac{\theta \left(\theta +1\right)}{2}{\sigma }_n{(t+T)}^2\right]dt\\ -e^{c_{n}t}\theta {\sigma }_n(t+T)y_n{}^{\theta }(t)d{\tilde{B}}_n(t).\end{array}$

由(17)，(22)可知，${\tilde{X}}_i\left(t\right)\left(1\le i\le n-1\right)$ 有界，则存在常数$\stackrel{\wedge }{D_i}>0$ ，有$y_i{(t)}^{-1}={\tilde{X}}_i(t)\ge \stackrel{\wedge }{D_i}$ 。令${\displaystyle \sum _{l=1}^{n-1}\stackrel{\wedge }{a_{nl}}\left(t+T\right)\stackrel{\wedge }{D_i^{\theta }}}=A\left(\theta \right)$ ，对$\text{d}\left({\text{e}}^{c_{n}t}{y}_n\left(t\right)\right)$ 从0到$t$ 积分，并化简得

$e^{c_{n}t}{y}_n(t)\le y_n(0)+{\displaystyle {\int }_0^t{e}^{c_{n}s}}{y}_n{}^{\theta }(s)\cdot \left[c_n+\theta b_n(s+T)-\theta A(\theta )+\theta a_{nn}(s+T)y_n{(s)}^{-\theta }+\frac{\theta \left(\theta +1\right)}{2}{\sigma }_n{(s+T)}^2\right]ds-M_n{}^{\prime }(t).$

取$c_n$ 为充分小的正常数，记$M_n{}^{\prime }(t)={\displaystyle {\int }_0^t{e}^{c_{n}s}\theta {\sigma }_n(s+T)y_n{}^{\theta }(s)d{\tilde{B}}_n(s)}$ 。由$\underset{t\to \infty }{\lim \inf }\left(A\left(\theta \right)-\frac{\theta +1}{2}{\sigma }_n^2\left(t\right)-b_n\left(t\right)\right)>0$ ，从而有

$e^{c_{n}t}{y}_n{}^{\theta }(t)\le y_n{}^{\theta }(0)+{\displaystyle {\int }_0^t{e}^{c_{n}s}}\theta \stackrel{\vee }{a_{nn}}(t+T)ds-M_n{}^{\prime }(t).$

从而可得

$y_n^{\theta }\left(t\right)\le y_n^{\theta }\left(0\right)+C^{\prime }-{M^{\prime }}_n\left(t\right),$

其中$C^{\prime }$ 为正常数，由引理2可得$\underset{t\to \infty }{\lim \sup }{y}_n^{\theta }\left(t\right)<\infty \text{a}\text{.s}\text{.}$ 因此可得

$\underset{t\to \infty }{\lim \inf }{\tilde{X}}_n^{\theta }\left(t\right)=\underset{t\to \infty }{\lim \inf }\frac{1}{y_n^{\theta }\left(t\right)}>0\text{a}\text{.s}$,

从而有

$\lim \inf {\tilde{X}}_n\left(t\right)>0.$ (23)

结合(22)，(23)对$\left(1\le i\le n\right)$ 有

$\lim \inf {\tilde{X}}_i\left(t\right)>0.$ (24)

由(20)，(24)可得$0<\underset{t\to \infty }{\lim \inf }{\tilde{X}}_i\left(t\right)\le \underset{t\to \infty }{\lim \sup }{\tilde{X}}_i\left(t\right)<\infty \text{a}\text{.s}\text{.}$，由${\tilde{X}}_i\left(t\right)=X_i\left(t+T\right)$ 因此有$0<\underset{t\to \infty }{\lim \inf }{X}_i\left(t\right)\le \underset{t\to \infty }{\lim \sup }{X}_i\left(t\right)<\infty \text{a}\text{.s}\text{.}$ 证毕。

至于种群的灭绝性，本文给出简要探讨证明：

定理5 设初始值$X\left(0\right)=X_0>0$ 。若参数满足$\underset{t\to \infty }{\lim \inf }\left(a_{ij}\left(t\right)\right)>0$ ，$\underset{t\to \infty }{\lim \inf }\left(A\left(\theta \right)-\frac{\theta +1}{2}{\sigma }_n^2\left(t\right)-b_n\left(t\right)\right)<0$ ，$\underset{t\to \infty }{\lim \inf }\left(b_i\left(t\right)-\frac{\theta +1}{2}{\sigma }_i^2\left(t\right)\right)<0,\left(1\le i\le n-1\right)$ ，其中${\displaystyle \sum _{j=1}^{n-1}\stackrel{\wedge }{a_{nj}}\left(t+T\right)\stackrel{\wedge }{D_i^{\theta }}}=A\left(\theta \right)>0$ ，$0<\stackrel{\wedge }{D_i}=\underset{1\le i\le n}{\min }{X}_i\left(t+T\right)$ 。则系统(15)的解$X\left(t\right)$ 依概率1灭绝，即

$\underset{t\to \infty }{\lim }{X}_i\left(t\right)=0.$

证明：对食饵的灭绝性，和定理4的(2)再证$\lim \inf X_i\left(t\right)>0\text{a}\text{.s}.$ 证明过程类似，且函数选取都不变，令$-\lambda =c=\underset{t\to \infty }{\lim \inf }\left(b_i\left(t\right)-\frac{\theta +1}{2}{\sigma }_i^2\left(t\right)\right)<0,\left(1\le i\le n-1\right)$ 可同理得(21)，因而有

$y_i^{\theta }\left(t\right)\le y_i^{\theta }\left(0\right){\text{e}}^{\lambda t}+\frac{\theta K\left({\text{e}}^{\lambda t}-1\right)}{\lambda }-{\text{e}}^{\lambda t}{M^{\prime }}_i\left(t\right)\le y_i^{\theta }\left(0\right){\text{e}}^{\lambda t}+\frac{\theta K\left({\text{e}}^{\lambda t}\right)}{\lambda }-{\text{e}}^{\lambda t}{M^{\prime }}_i\left(t\right).$

其中$K={\displaystyle \sum _{i=1}^n{\displaystyle \sum _{j=1}^n\stackrel{\vee }{a_{ij}}}}$ ，由$y_i^{\theta }\left(t\right)={\tilde{X}}_i{\left(t\right)}^{-\theta }>0$ ，所以$\underset{t\to \infty }{\lim }{\tilde{X}}_i^{\theta }\left(t\right)=0$ ，$\underset{t\to \infty }{\lim }{\tilde{X}}_i\left(t\right)=0$ 。$\left(1\le i\le n-1\right)$ ，从而食饵种群依概率1灭绝，从生态系统的角度出发，若所有食饵灭绝，则捕食者会因失去食物来源而导致灭绝。对捕食者自身的灭绝性，同理和定理4的(2)再证$\lim \inf X_i\left(t\right)>0\text{a}\text{.s}.$ 证明过程类似，且函数选取都不变，令$\underset{t\to \infty }{c_n=\lim \inf }\left(A\left(\theta \right)-\frac{\theta +1}{2}{\sigma }_n^2\left(t\right)-b_n\left(t\right)\right)<0$ ，最终可得$\underset{t\to \infty }{\lim }{\tilde{X}}_n\left(t\right)=0$ 。证毕

2.5. 数值模拟

本节对系统(2)和(15)通过数值模拟来说明白噪声的影响。

对系统(15)，取适当的参数满足定理4的条件(见表1、表2)，通过MATLAB数值仿真可得该系统的随机持久性(如图1和图2)：$0<\underset{t\to \infty }{\lim \inf }{X}_i\left(t\right)\le \underset{t\to \infty }{\lim \sup }{X}_i\left(t\right)<\infty$ 几乎必然成立。

对线性系统(2)，参数条件满足定理3的条件(见表3)，通过MATLAB数值仿真可得该系统的随机持久性(如图3和图4)：$0<\underset{t\to \infty }{\lim \inf }{X}_i\left(t\right)\le \underset{t\to \infty }{\lim \sup }{X}_i\left(t\right)<\infty$ 几乎必然成立。

对系统(15)，参数条件满足定理4的条件(见表4)，通过MATLAB数值仿真可得该系统的随机持久性(如图5和图6)：$0<\underset{t\to \infty }{\lim \inf }{X}_i\left(t\right)\le \underset{t\to \infty }{\lim \sup }{X}_i\left(t\right)<\infty$ 几乎必然成立。

Table 1. Model parameter settings

表1. 模型参数设置

参数类型	表达式	说明
$b\left(t\right)$ ：内禀增长率/死亡率	$b_1\left(t\right)=0.9+0.05\sin \left(0.06t\right)+0.05\cos \left(\sqrt{3}\cdot 0.04t\right)$	$X_1$ 的增长率
	$b_2\left(t\right)=0.7+0.08\sin \left(0.09t\right)+0.06\cos \left(\sqrt{3}\cdot 0.05t\right)$	$X_2$ 的增长率
	$b_3\left(t\right)=0.15+0.07\sin \left(0.08t\right)+0.03\cos \left(\sqrt{5}\cdot 0.04t\right)$	$X_3$ 死亡率
$A_1\left(t\right)$ ：交互系数	$a_{11}\left(t\right)=0.3+0.03\sin \left(0.05t\right)$	$X_1$ 的交互系数
	$a_{12}\left(t\right)=0.06+0.02\sin \left(0.03t\right)$
	$a_{13}\left(t\right)=0.18+0.04\sin \left(0.05t\right)$
$A_2\left(t\right)$ ：交互系数	$a_{21}\left(t\right)=0.08+0.02\sin \left(0.03t\right)$	$X_2$ 的交互系数
	$a_{22}\left(t\right)=0.15+0.03\sin \left(0.06t\right)$
	$a_{23}\left(t\right)=0.21+0.02\sin \left(0.04t\right)$
$A_3\left(t\right)$ ：交互系数	$a_{31}\left(t\right)=0.16+0.04\sin \left(0.05t\right)$	$X_3$ 的交互系数
	$a_{32}\left(t\right)=0.2+0.02\sin \left(0.04t\right)$
	$a_{33}\left(t\right)=0.25+0.03\sin \left(0.05t\right)$
${\sigma }_i$ ：噪声强度	${\sigma }_1\left(t\right)=0.08+0.01\sin \left(0.08t\right)+0.01\cos \left(\sqrt{3}\cdot 0.06t\right)$
	${\sigma }_2\left(t\right)=0.08+0.01\sin \left(0.08t\right)+0.01\cos \left(\sqrt{3}\cdot 0.06t\right)$
	${\sigma }_3\left(t\right)=0.08+0.01\sin \left(0.08t\right)+0.01\cos \left(\sqrt{3}\cdot 0.06t\right)$

Table 2. $\theta =0.5$ test system result data

表2. $\theta =0.5$ 试验系统结果数据

$X_1\left(0\right)=2.5,X_2\left(0\right)=2.5,X_3\left(0\right)=1$	$\Delta t=0.01$	$\theta =0.5$
食饵1条件	$\underset{t\to \infty }{\lim \inf }\left(b_1\left(t\right)-\frac{\theta +1}{2}{\sigma }_1^2\left(t\right)\right)$	0.7928
食饵2条件	$\underset{t\to \infty }{\lim \inf }\left(b_2\left(t\right)-\frac{\theta +1}{2}{\sigma }_2^2\left(t\right)\right)$	0.5534
食饵种群下确界最小值之和	$A\left(\theta =0.5\right)$	1.4373
捕食者条件	$\underset{t\to \infty }{\lim \inf }\left(A\left(\theta \right)-\frac{\theta +1}{2}{\sigma }_n^2\left(t\right)-b_n\left(t\right)\right)$	0.1743

Figure 1. $\theta =0.5$ numerical simulation of stochastic persistence of a three-population random solution

图1. $\theta =0.5$ 时三种群随机解的随机持久性数值模拟图

Figure 2. $\theta =0.5$ stationary distribution histogram of the three populations

图2. $\theta =0.5$ 时三种群平稳分布直方图

Table 3. $\theta =1$ test system result data

表3. $\theta =1$ 试验系统结果数据

$X_1\left(0\right)=2.5,X_2\left(0\right)=2.5,X_3\left(0\right)=1$	$\Delta t=0.01$	$\theta =1$
食饵1条件	$\underset{t\to \infty }{\lim \inf }\left(b_1\left(t\right)-\frac{\theta +1}{2}{\sigma }_1^2\left(t\right)\right)$	0.7903
食饵2条件	$\underset{t\to \infty }{\lim \inf }\left(b_2\left(t\right)-\frac{\theta +1}{2}{\sigma }_2^2\left(t\right)\right)$	0.5509
食饵种群下确界最小值之和	$A\left(\theta =1\right)$	1.1759
捕食者条件	$\underset{t\to \infty }{\lim \inf }\left(A\left(\theta \right)-\frac{\theta +1}{2}{\sigma }_n^2\left(t\right)-b_n\left(t\right)\right)$	0.0933

Figure 3. $\theta =1$ numerical simulation of stochastic persistence of a three-population random solution

图3. $\theta =1$ 时三种群随机解的随机持久性数值模拟图

Figure 4. $\theta =1$ stationary distribution histogram of the three populations

图4. $\theta =1$ 时三种群平稳分布直方图

Table 4. $\theta =1.5$ test system result data

表4. $\theta =1.5$ 试验系统结果数据

$X_1\left(0\right)=2.5,X_2\left(0\right)=2.5,X_3\left(0\right)=1$	$\Delta t=0.01$	$\theta =1.5$
食饵1条件	$\underset{t\to \infty }{\lim \inf }\left(b_1\left(t\right)-\frac{\theta +1}{2}{\sigma }_1^2\left(t\right)\right)$	0.7878
食饵2条件	$\underset{t\to \infty }{\lim \inf }\left(b_2\left(t\right)-\frac{\theta +1}{2}{\sigma }_2^2\left(t\right)\right)$	0.5484
食饵种群下确界最小值之和	$A\left(\theta =1.5\right)$	0.9311
捕食者条件	$\underset{t\to \infty }{\lim \inf }\left(A\left(\theta \right)-\frac{\theta +1}{2}{\sigma }_n^2\left(t\right)-b_n\left(t\right)\right)$	0.0174

Figure 5. $\theta =1.5$ numerical simulation of stochastic persistence of a three-population random solution

图5. $\theta =1.5$ 时三种群随机解的数值模拟图

Figure 6. $\theta =1.5$ stationary distribution histogram of the three populations

图6. $\theta =1.5$ 时三种群平稳分布直方图

NOTES

^*第一作者。

^#通讯作者。