在数学生物学中, 如何准确地描述物种间关系是一个有意义的话题. 在现实的生活中, 物种的演化都可能被其他的物种或者物种自身的演化所影响, 在数学上一般用确定性(偏)微分方程来描述. 注意到噪声总是存在, 加上其它不确定性因素的影响, 随机(偏)微分方程可以更好的刻画物种的演化. 最早的双物种模型是著名的Lotka-Volterra捕食模型, 本文考虑三物种在随机环境下的演化行为

$ \begin{equation*} \left\{ \begin{array}{ll} dx_{1}(t)=x_{1}(t)[a_{1}-c_{11}x_{1}(t)+c_{12}x_{2}(t-\tau_{1})-c_{13}x_{3}(t-\tau_{2})]dt, \\ dx_{2}(t)=x_{2}(t)[a_{2}+c_{21}x_{1}(t-\tau_{3})-c_{22}x_{2}(t)-c_{23}x_{3}(t-\tau_{4})]dt, \\ dx_{3}(t)=x_{3}(t)[a_{3}+c_{31}x_{1}(t-\tau_{5})+c_{32}x_{2}(t-\tau_{6})-c_{33}x_{3}(t)]dt, \end{array} \right. \end{equation*} $

初始值为$ x_{i}(\theta) $, 并且

$ \begin{eqnarray*} x_{i}(\theta)=\phi_{i}(\theta), \ \theta\in[-\tau, 0], \ \tau=\max\{\tau_{1}, \tau_{2}, \tau_{3}, \tau_{4}, \tau_{5}, \tau_{6}\}, \ i=1, 2, 3, \end{eqnarray*} $

这里的$ \phi_{i}(\theta) $是定义在区间$ [-\tau, 0] $上的连续函数并且大于零, $ i=1, 2, 3 $. 换句话说, $ x_{0}=\phi=(\phi_{1}, \phi_{2}, \phi_{3})^{T}\in C([-\tau, 0], R_{+}^{3}) $, 其中$ R_{+}^{3}=\{(x_{1}, x_{2}, x_{3})\in R^{3}:x_{i}>0, 1\leq i\leq 3\} $. $ x_{3} $以$ x_{1} $和$ x_{2} $为食物, $ x_{1} $和$ x_{2} $是合作关系. $ a_{1}>0 $和$ a_{2}>0 $分别表示$ x_{1} $和$ x_{2} $的生长率. $ -a_{3}>0 $则表示$ x_{3} $的死亡率. $ c_{11} $, $ c_{22} $, $ c_{33} $分别表示物种$ x_{1} $, $ x_{2} $, $ x_{3} $内部的竞争系数(大于零). $ c_{12} $和$ c_{21} $为物种$ x_{1} $和$ x_{2} $的合作关系系数(大于零). $ c_{13} $和$ c_{23} $代表损失率(大于零). $ c_{31} $和$ c_{32} $代表俘获率(大于零). $ \tau_{1} $, $ \tau_{2} $, $ \tau_{3} $, $ \tau_{4} $, $ \tau_{5} $, $ \tau_{6} $代表时滞(大于零).

考虑到随机扰动

$ \begin{eqnarray*} a_{i}\rightarrow a_{i}+\sigma_{i}\dot{B}_{i}(t), \ i=1, 2, 3, \end{eqnarray*} $

其中$ \{B_{i}(t)\}_{t\geq0}, i=1, 2, 3, $是定义在具有滤子$ \mathcal {F}_{t\geq0} $的完备概率空间$ (\Omega, \mathcal {F}, \{\mathcal {F}\}_{t\geq0}, \mathbb{P}) $上的标准独立Wiener过程, $ \sigma_{i}^{2}, \ i=1, 2, 3, $代表噪声的强度. 因此, 我们有:

$ \begin{equation} \left\{ \begin{array}{ll} dx_{1}(t)=x_{1}(t)[a_{1}-c_{11}x_{1}(t)+c_{12}x_{2}(t-\tau_{1})-c_{13}x_{3}(t-\tau_{2})]dt+\sigma_{1}x_{1}(t)dB_{1}(t), \\ dx_{2}(t)=x_{2}(t)[a_{2}+c_{21}x_{1}(t-\tau_{3})-c_{22}x_{2}(t)-c_{23}x_{3}(t-\tau_{4})]dt+\sigma_{2}x_{2}(t)dB_{2}(t), \\ dx_{3}(t)=x_{3}(t)[a_{3}+c_{31}x_{1}(t-\tau_{5})+c_{32}x_{2}(t-\tau_{6})-c_{33}x_{3}(t)]dt+\sigma_{3}x_{3}(t)dB_{3}(t), \\ \end{array} \right. \end{equation} $

(1.1)

随机模型中, 一般是通过寻求正平衡状态来研究解的平稳分布, 由以往得到的结果知道可通过运用F-P方程的显示解法^{[1, 2]}、马尔科夫半群理论法^[3-5]、以及Lyapunov函数法^[6-9]来进行研究. 其中, F-P方程的显示解法可以解决不具有时滞的模型; 马尔科夫半群理论法需要一个标准测度, 而这样的标准测度有的很难找到; Lyapunov函数法则需要解的马尔科夫性质, 时滞的模型是不具有这种性质的. 另外, Hu-Wang所得到结果中的条件需要满足方程系数的线性增长^[10], 时滞的模型是不能满足的. 本文采用的方法则避免了以上提到的问题, 适用于具有时滞的模型, 对种群的动力学分析更具有现实意义.

最近, Liu等^[11]研究了接种疫苗和双疾病的随机时滞SIS传染病模型并且得到了这个模型的阈值. 在文献[12] 中, Liu和Jiang考虑了一类具有不完善疫苗接种的随机分阶进展HIV模型的平稳分布. 随机传染病模型的长期行为是一个重要的问题. 一类具有分布时滞和非线性扰动的随机逻辑斯蒂方程的长时间行为已经由Liu等^[13]所研究. 此外, 具有时滞的三物种模型也已得到研究^[14]. 近来, Liu-Fan^[15]建立了一类具有时滞的三物种随机级联捕食-被捕食系统的平稳分布. 关于平稳分布, 见参考文献[16-18].

在证明的过程中, 需要以下的引理

引理1.1^[19]    令$ g(t)\in C[\Omega\times[0, +\infty), R_{+}] $,

$ (1) $如果存在两个正常数$ T $和$ \lambda_{0} $使得对于所有的$ t\geq T $,

$ \begin{eqnarray*} \ln g(t)\leq \lambda t-\lambda_{0}\int_{0}^{t}g(s)ds+\sum\limits_{i=1}^{n}\alpha_{i}B_{i}(t), \end{eqnarray*} $

那么, 我们有

$ \begin{equation} \left\{ \begin{array}{ll} \overline{g}^{*}=\limsup\limits_{t\rightarrow +\infty}t^{-1}\int_{0}^{t}g(s)ds\leq\frac{\lambda}{\lambda_{0}} \ \ a.s., \ \ \ \ \ \ \lambda\geq 0, \\ \lim\limits_{t\rightarrow +\infty}g(t)=0 \ \ a.s., \ \ \ \ \ \ \lambda< 0.\\ \end{array} \right. \end{equation} $

(1.2)

$ (2) $如果存在三个正常数$ T $, $ \lambda $和$ \lambda_{0} $使得对于所有的$ t\geq T $,

$ \begin{eqnarray*} \ln g(t)\geq \lambda t-\lambda_{0}\int_{0}^{t}g(s)ds+\sum\limits_{i=1}^{n}\alpha_{i}B_{i}(t), \end{eqnarray*} $

那么, 有

$ \begin{eqnarray*} \overline{g}_{*}=\liminf\limits_{t\rightarrow +\infty}t^{-1}\int_{0}^{t}g(s)ds\geq\frac{\lambda}{\lambda_{0}}, \ \ a.s. \end{eqnarray*} $

引理1.2^[20]    令$ \mu $是关于$ P_{t} $的不变测度, 其中$ t\geq 0 $. 假定在弱意义下, 下面极限成立

$ \begin{eqnarray*} \lim\limits_{t\rightarrow +\infty}P_{t}(x, \cdot)=\mu \ \ \ \ x\in E, \end{eqnarray*} $

则$ \mu $是强混合的.

引理1.3^[21]   假定$ y=y(x)\in(a, b) $和$ y=Y(x)\in(a, b) $分别是以下两个方程的解

$ \begin{equation*} \left\{ \begin{array}{ll} \frac{dy}{dx}=f(x, y), \\ y(x, 0)=y_{0}, \\ \end{array} \right. \end{equation*} $

和

$ \begin{equation*} \left\{ \begin{array}{ll} \frac{dY}{dx}=F(x, Y), \\ y(x, 0)=y_{0}, \\ \end{array} \right. \end{equation*} $

其中, $ f(x, y) $和$ F(x, y) $是连续的且满足局部Lipschitz条件, 并且$ (x_{0}, y_{0})\in R\times R $. 若$ f(x, y)<F(x, y) $, 则有$ y(x)<Y(x), \ x_{0}<x<b;\, y(x)>Y(x), \ a<x<x_{0}. $

引理1.4^[22]    假定$ n\geq2 $, 那么以下等式成立：

$ \begin{eqnarray*} \sum\limits_{i, j=1}^{n}m_{i}a_{ij}G_{i}(x_{i})=\sum\limits_{i, j=1}^{n}m_{i}a_{ij}G_{j}(x_{j}), \end{eqnarray*} $

其中$ G_{i}(x_{i}) $是任意的函数, $ 1\leq i\leq n $.

引理1.5^[20]    若$ \mu\in \mathcal{M}_{1}(E) $是半群$ P_{t}, \ t>0 $的唯一不变测度, 那么$ \mu $是遍历的.

引理1.6^[20]    假定$ \varphi\in L^{2}(E, \mu) $, 则存在$ \eta^{*}\in L^{2}(\Omega_{1}, \mathcal{F}_{1}, \mathbb{P}_{1}) $使得在$ L^{2}(\Omega_{1}, \mathcal{F}_{1}, \mathbb{P}_{1}) $有

$ \begin{eqnarray*} \lim\limits_{T\rightarrow +\infty}\frac{1}{T}\int_{0}^{T}\varphi(Z(s))ds=\eta^{*}, \ \ \mathbb{P}_{1}-a.s. \end{eqnarray*} $

此外, 若$ \mu $是遍历的, 则

$ \begin{eqnarray*} \lim\limits_{T\rightarrow +\infty}\frac{1}{T}\int_{0}^{T}\varphi(Z(s))ds=\int_{E}\varphi(x)\mu(dx)=\langle \mu, \varphi\rangle, \ \ \mathbb{P}_{1}-a.s. \end{eqnarray*} $

为方便本文中的计算, 我们有以下定义:

$ \begin{eqnarray*} b_{i}=a_{i}-\frac{\sigma_{i}^{2}}{2}, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Gamma=c_{11}c_{22}-c_{12}c_{21}, \\ \Gamma_{1}=a_{1}c_{22}+a_{2}c_{12}, \ \ \ \ \ \tilde{\Gamma}_{1}=(c_{22}\sigma_{1}^{2}+c_{12}\sigma_{2}^{2})/2, \\ \Gamma_{2}=a_{1}c_{21}+a_{2}c_{11}, \ \ \ \ \ \ \tilde{\Gamma}_{2}=(c_{21}\sigma_{1}^{2}+c_{11}\sigma_{2}^{2})/2, \\ \Gamma_{3}=a_{2}c_{33}-a_{3}c_{23}, \ \ \ \ \ \ \tilde{\Gamma}_{3}=(c_{33}\sigma_{2}^{2}-c_{23}\sigma_{3}^{2})/2, \\ \Gamma_{4}=a_{3}c_{22}+a_{2}c_{32}, \ \ \ \ \ \ \tilde{\Gamma}_{4}=(c_{22}\sigma_{3}^{2}+c_{32}\sigma_{2}^{2})/2, \\ \Gamma_{5}=a_{1}c_{33}-a_{3}c_{13}, \ \ \ \ \ \ \tilde{\Gamma}_{5}=(c_{33}\sigma_{1}^{2}-c_{13}\sigma_{3}^{2})/2, \\ \Gamma_{6}=a_{3}c_{11}+a_{1}c_{31}, \ \ \ \ \ \ \tilde{\Gamma}_{6}=(c_{11}\sigma_{3}^{2}+c_{31}\sigma_{1}^{2})/2, \\ \overline{f(t)}=\frac{1}{t}\int_{0}^{t}f(s)ds, \ f^{*}=\limsup\limits_{t\rightarrow \infty}f(t), \ f_{*}=\liminf\limits_{t\rightarrow \infty}f(t); \end{eqnarray*} $

以及

$ \begin{eqnarray*} \Lambda= \left| \begin{array}{ccc} c_{11} & -c_{12} & c_{13} \\ -c_{21} & c_{22} & c_{23} \\ -c_{31} & -c_{32} & c_{33} \\ \end{array} \right|. \end{eqnarray*} $

$ \Lambda_{i} $表示$ \Lambda $中第$ i $列用$ (a_{1}, a_{2}, a_{3})^{T} $代替, $ \tilde{\Lambda}_{i} $表示$ \Lambda $中第$ i $列用$ (\frac{\sigma_{1}^{2}}{2}, \frac{\sigma_{2}^{2}}{2}, \frac{\sigma_{3}^{2}}{2})^{T} $代替, 其中$ i=1, 2, 3 $. 可以很容易得到

$ \begin{eqnarray*} \Gamma_{1}>0, \ \tilde{\Gamma}_{1}>0, \ \Gamma_{2}>0, \ \tilde{\Gamma}_{2}>0, \ \Gamma_{3}>0, \ \tilde{\Gamma}_{4}>0, \ \Gamma_{5}>0, \ \tilde{\Gamma}_{6}>0. \end{eqnarray*} $

假设

$ \begin{eqnarray} c_{11}>c_{12}+c_{13}, \ c_{22}>c_{21}+c_{23}, \ c_{33}>c_{31}+c_{32}. \end{eqnarray} $

(1.3)

在这一节中，我们将建立解的存在唯一性.

引理2.1   给定任意的初始值$ (\phi_{1}, \phi_{2}, \phi_{3})\in C([-\tau, 0], R_{+}^{3}) $, 模型(1.1) 几乎处处有唯一全局解$ x(t)=(x_{1}(t), x_{2}(t), x_{3}(t))^{T}\in R_{+}^{3} $.

证   令$ z_{i}(t)=\ln x_{i}(t) $. 由Itô's公式以及(1.1), 可以得到

$ \begin{equation} \left\{ \begin{array}{ll} dz_{1}(t)=[b_{1}-c_{11}e^{z_{1}(t)}+c_{12}e^{z_{2}(t-\tau_{1})}-c_{13}e^{z_{3}(t-\tau_{2})}]+\sigma_{1}dB_{1}(t), \\ dz_{2}(t)=[b_{2}+c_{21}e^{z_{1}(t-\tau_{3})}-c_{22}e^{z_{2}(t)}-c_{23}e^{z_{3}(t-\tau_{4})}]+\sigma_{2}dB_{2}(t), \\ dz_{3}(t)=[b_{3}+c_{31}e^{z_{1}(t-\tau_{5})}+c_{32}e^{z_{2}(t-\tau_{6})}-c_{33}e^{z_{3}(t)}]+\sigma_{3}dB_{3}(t), \\ \end{array} \right. \end{equation} $

(2.1)

则$ z_{i}(\theta)=\ln \phi_{i}(\theta), \theta\in [-\tau, 0], \ i=1, 2, 3 $. 很显然, 这是满足局部Lipschitz条件. 那么我们根据存在性和唯一性定理可以知道(2.1) 在区间$ [0, \tau_{e}) $有一个局部解$ z_{t} $, 这里$ \tau_{e} $是爆破时刻. 也就是说, $ x(t)=(e^{z_{1}(t)}, e^{z_{2}(t)}, e^{z_{3}(t)})^{T} $是(1.1) 的唯一局部正解.

接下来我们要证明这个解是全局的. 换句话说, 我们需要证明$ \tau_{e}=+\infty $. 我们可以考虑如下模型:

$ \begin{equation} \left\{ \begin{array}{ll} dy_{1}(t)=y_{1}(t)[a_{1}-c_{11}y_{1}(t)]dt+\sigma_{1}y_{1}(t)dB_{1}(t), \\ dy_{2}(t)=y_{2}(t)[a_{2}+c_{21}y_{1}(t-\tau_{3})-c_{22}y_{2}(t)]dt+\sigma_{2}y_{2}(t)dB_{2}(t), \\ dy_{3}(t)=y_{3}(t)[a_{3}+c_{31}y_{1}(t-\tau_{5})+c_{32}y_{2}(t-\tau_{6})-c_{33}y_{3}(t)]dt+\sigma_{3}y_{3}(t)dB_{3}(t), \\ \end{array} \right. \end{equation} $

(2.2)

其中$ y_{i}(\theta)=\phi_{i}(\theta)>0, \ \theta\in [-\tau, 0], \ i=1, 2, 3 $. 根据引理1.3我们可以得到$ t\in [0, \tau_{e}) $时,

$ \begin{eqnarray} x_{i}(t)\leq y_{i}(t), \ \ a.s., \ \ i=1, 2, 3. \end{eqnarray} $

(2.3)

下面, 考虑(2.2) 中的$ y_{i}(t), \ i=1, 2, 3 $.

$ \begin{equation*} \left\{ \begin{array}{ll} dy_{1}(t)=y_{1}(t)[a_{1}-c_{11}y_{1}(t)]dt+\sigma_{1}y_{1}(t)dB_{1}(t), \\ y_{1}(0)=\phi_{1}(0), \\ \end{array} \right. \end{equation*} $

令$ u_{1}(t)=\ln y_{1}(t) $, 通过运用Itô's公式可以得到

$ \begin{eqnarray*} du_{1}(t)=[b_{1}-c_{11}e^{u_{1}(t)}]dt+\sigma_{1}dB_{1}(t). \end{eqnarray*} $

令$ v_{1}(t)=u_{1}(t)-\sigma_{1}B_{1}(t) $, 再次运用Itô's公式可以得到

$ \begin{eqnarray*} dv_{1}(t)=[b_{1}-c_{11}e^{v_{1}(t)+\sigma_{1}B_{1}(t)}]dt. \end{eqnarray*} $

再令$ w_{1}(t)=v_{1}(t)-b_{1}t $, 我们可以得到$ w_{1}^{'}(t)=v_{1}^{'}(t)-b_{1} $. 简单计算可得

$ \begin{eqnarray*} de^{-w_{1}(t)}=c_{11}e^{\sigma_{1}B_{1}(t)+b_{1}t}, \end{eqnarray*} $

对上面式子的左右两端同时从0到t进行积分可得:

$ \begin{eqnarray*} e^{-w_{1}(t)}-e^{-w_{1}(0)}=c_{11}\int_{0}^{t}e^{\sigma_{1}B_{1}(s)+b_{1}s}ds. \end{eqnarray*} $

又因为$ e^{-w_{1}(t)}=e^{-\ln y_{1}(t)+\sigma_{1}B_{1}(t)+b_{1}t}, \ e^{-w_{1}(0)}=\phi_{1}(0)^{-1}, $因此, 可以得到

$ \begin{eqnarray*} y_{1}(t)=\frac{e^{\sigma_{1}B_{1}(t)+b_{1}t}}{\phi_{1}(0)^{-1}+c_{11}\int_{0}^{t}e^{\sigma_{1}B_{1}(s)+b_{1}s}ds}. \end{eqnarray*} $

那么, 运用同样的方法我们可以得到

$ \begin{eqnarray*} y_{2}(t)=\frac{e^{\sigma_{2}B_{2}(t)+b_{2}t+c_{21}\int_{0}^{t}y_{1}(s-\tau_{3})ds}}{\phi_{2}(0)^{-1}+c_{22}\int_{0}^{t}e^{\sigma_{2}B_{2}(s)+b_{2}s+c_{21}\int_{0}^{s}y_{1}(u-\tau_{3})du}ds}, \end{eqnarray*} $

$ \begin{eqnarray*} y_{3}(t)=\frac{e^{\sigma_{3}B_{3}(t)+b_{3}t+c_{31}\int_{0}^{t}y_{1}(s-\tau_{5})ds+c_{32}\int_{0}^{t}y_{2}(s-\tau_{6})ds}}{\phi_{3}(0)^{-1}+c_{33}\int_{0}^{t}e^{\sigma_{3}B_{3}(s)+b_{3}s+c_{31}\int_{0}^{s}y_{1}(u-\tau_{5})du+c_{32}\int_{0}^{s}y_{2}(u-\tau_{6})du}ds}. \end{eqnarray*} $

因此, 当$ t\geq 0 $时, $ y_{i}(t), \ i=1, 2, 3 $是存在的. 也就是说$ \tau_{e}=+\infty $.

注   利用文献[23], 我们可以得到

$ \begin{eqnarray} \limsup\limits_{t\rightarrow \infty}\frac{1}{t}\ln x_{i}(t)<0 \ \ a.s., \ \ i=1, 2, 3, \end{eqnarray} $

(2.4)

并且存在一个正常数$ K $使得下式成立

$ \begin{eqnarray} \limsup\limits_{t\rightarrow \infty}E(x_{i}(t))\leq K, \ \ i=1, 2, 3. \end{eqnarray} $

(2.5)

引理3.1    对于模型(2.2) 而言,

$ (1) $若$ b_{i}<0, \ i=1, 2, 3 $, 则

$ \begin{eqnarray*} \lim\limits_{t\rightarrow +\infty}y_{i}(t)=0, \ \ a.s., \ \ i=1, 2, 3; \end{eqnarray*} $

$ (2) $若$ b_{1}\geq0, \; b_{2}+\frac{b_{1}c_{21}}{c_{11}}<0, \; b_{3}+\frac{b_{1}c_{31}}{c_{11}}<0 $, 则

$ \begin{eqnarray*} \lim\limits_{t\rightarrow +\infty}\overline{y_{1}(t)}=\frac{b_{1}}{c_{11}}, \ \ \lim\limits_{t\rightarrow +\infty}y_{i}(t)=0, \ \ a.s., \ \ i=2, 3; \end{eqnarray*} $

$ (3) $若$ b_{1}\geq0, \; b_{2}+\frac{b_{1}c_{21}}{c_{11}}\geq0, \; b_{3}+\frac{b_{1}c_{31}}{c_{11}}+\frac{b_{2}+\frac{b_{1}c_{21}}{c_{11}}}{c_{22}}c_{32}<0 $, 则

$ \begin{eqnarray*} \lim\limits_{t\rightarrow +\infty}\overline{y_{1}(t)}=\frac{b_{1}}{c_{11}}, \ \ \lim\limits_{t\rightarrow +\infty}\overline{y_{2}(t)}=\frac{b_{2}+\frac{b_{1}c_{21}}{c_{11}}}{c_{22}}, \ \ \lim\limits_{t\rightarrow +\infty}y_{3}(t)=0, \ \ a.s.; \end{eqnarray*} $

$ (4) $若$ b_{1}\geq0, \; b_{2}+\frac{b_{1}c_{21}}{c_{11}}\geq0, \; b_{3}+\frac{b_{1}c_{31}}{c_{11}}+\frac{b_{2}+\frac{b_{1}c_{21}}{c_{11}}}{c_{22}}c_{32}\geq0 $, 则

$ \begin{eqnarray*} \lim\limits_{t\rightarrow +\infty}\overline{y_{1}(t)}=\frac{b_{1}}{c_{11}}, \ \ \lim\limits_{t\rightarrow +\infty}\overline{y_{1}(t)}=\frac{b_{2}+\frac{b_{1}c_{21}}{c_{11}}}{c_{22}}, \\ \lim\limits_{t\rightarrow +\infty}\overline{y_{3}(t)}=\frac{b_{3}+\frac{b_{1}c_{31}}{c_{11}}+\frac{b_{2}+\frac{b_{1}c_{21}}{c_{11}}}{c_{22}}c_{32}}{c_{33}}, \ \ a.s.; \end{eqnarray*} $

$ (5) $若$ b_{1}<0, \; b_{2}\geq0, \; b_{3}+\frac{b_{2}c_{32}}{c_{22}}\geq0 $, 则

$ \begin{eqnarray*} \lim\limits_{t\rightarrow +\infty}y_{1}(t)=0, \ \lim\limits_{t\rightarrow +\infty}\overline{y_{2}(t)}=\frac{b_{2}}{c_{22}}, \ \lim\limits_{t\rightarrow +\infty}\overline{y_{3}(t)}=\frac{b_{3}+\frac{b_{2}c_{32}}{c_{22}}}{c_{33}}, \ \ a.s.; \end{eqnarray*} $

$ (6) $若$ b_{1}\geq0, \; b_{2}+\frac{b_{1}c_{21}}{c_{11}}<0, \; b_{3}+\frac{b_{1}c_{31}}{c_{11}}\geq0 $, 则

$ \begin{eqnarray*} \lim\limits_{t\rightarrow +\infty}\overline{y_{1}(t)}=\frac{b_{1}}{c_{11}}, \ \lim\limits_{t\rightarrow +\infty}y_{2}(t)=0, \ \lim\limits_{t\rightarrow +\infty}\overline{y_{3}(t)}=\frac{b_{3}+\frac{b_{1}c_{31}}{c_{11}}}{c_{33}}, \ \ a.s.; \end{eqnarray*} $

$ (7) $若$ b_{1}<0, \; b_{2}\geq0, \; b_{3}+\frac{b_{2}c_{32}}{c_{22}}<0 $, 则

$ \begin{eqnarray*} \lim\limits_{t\rightarrow +\infty}y_{1}(t)=0, \ \lim\limits_{t\rightarrow +\infty}\overline{y_{2}(t)}=\frac{b_{2}}{c_{22}}, \ \lim\limits_{t\rightarrow +\infty}y_{3}(t)=0, \ \ a.s. \end{eqnarray*} $

证   对模型(2.2) 运用Itô's公式并从$ 0 $到$ t $进行积分, 再除以$ t $可以得到

$ \begin{eqnarray} t^{-1}\ln\frac{y_{1}(t)}{y_{1}(0)}&=&b_{1}-c_{11}\overline{y_{1}(t)}+t^{-1}\sigma_{1}B_{1}(t), \end{eqnarray} $

(3.1)

$ \begin{eqnarray} t^{-1}\ln\frac{y_{2}(t)}{y_{2}(0)}&=&b_{2}+c_{21}\overline{y_{1}(t)}-t^{-1}c_{21}[\int_{t-\tau_{3}}^{t}y_{1}(s)ds-\int_{-\tau_{3}}^{0}y_{1}(s)ds]\\ &&-c_{22}\overline{y_{2}(t)}+t^{-1}\sigma_{2}B_{2}(t), \end{eqnarray} $

(3.2)

$ \begin{eqnarray} t^{-1}\ln\frac{y_{3}(t)}{y_{3}(0)}&=&b_{3}+c_{31}\overline{y_{1}(t)}-t^{-1}c_{31}[\int_{t-\tau_{5}}^{t}y_{1}(s)ds-\int_{-\tau_{5}}^{0}y_{1}(s)ds]\\ &&+c_{32}\overline{y_{2}(t)}-t^{-1}c_{32}[\int_{t-\tau_{6}}^{t}y_{2}(s)ds-\int_{-\tau_{6}}^{0}y_{2}(s)ds]\\ &&-c_{33}\overline{y_{3}(t)}+t^{-1}\sigma_{3}B_{3}(t). \end{eqnarray} $

(3.3)

(1) 由(3.1) 以及引理1.1可得$ \lim_{t\rightarrow +\infty}y_{1}(t)=0 $几乎处处成立. 因此我们有, 对于任意的$ \epsilon>0 $, 存在$ T>0 $, 当$ t>T $时使得

$ \begin{eqnarray*} \overline{y_{1}(t)}\leq\frac{\epsilon}{c_{21}}, \end{eqnarray*} $

代入(3.2), 则有

$ \begin{eqnarray*} t^{-1}\ln\frac{y_{2}(t)}{y_{2}(0)}&\leq&b_{2}+c_{21}\overline{y_{1}(t)}-c_{22}\overline{y_{2}(t)}+t^{-1}\sigma_{2}B_{2}(t)\\ &\leq&b_{2}+\epsilon-c_{22}\overline{y_{2}(t)}+t^{-1}\sigma_{2}B_{2}(t), \end{eqnarray*} $

其中, t足够大并且$ \epsilon $足够小使得$ 0<\epsilon<-b_{2} $, 由引理1.1可得$ \lim_{t\rightarrow +\infty}y_{2}(t)=0 $几乎处处成立. 同理得$ \lim_{t\rightarrow +\infty}y_{3}(t)=0 $几乎处处成立.

(2) 由(3.1) 以及引理1.1可得

$ \begin{eqnarray*} \overline{y_{1}}^{*}=\limsup\limits_{t\rightarrow +\infty}t^{-1}\int_{0}^{t}y_{1}(s)ds\leq\frac{b_{1}}{c_{11}}, \ \ \overline{y_{1}}_{*}=\liminf\limits_{t\rightarrow +\infty}t^{-1}\int_{0}^{t}y_{1}(s)ds\geq\frac{b_{1}}{c_{11}}, \ \ a.s., \end{eqnarray*} $

也就是说

$ \begin{eqnarray*} \lim\limits_{t\rightarrow +\infty}\overline{y_{1}(t)}=\frac{b_{1}}{c_{11}}, \ \ a.s. \end{eqnarray*} $

因此，有

$ \begin{eqnarray*} \lim\limits_{t\rightarrow +\infty}t^{-1}\int_{t-\tau_{3}}^{t}y_{1}(s)ds=\lim\limits_{t\rightarrow +\infty}t^{-1}\left[\int_{0}^{t}y_{1}(s)ds-\int_{0}^{t-\tau_{3}}y_{1}(s)ds\right]=0, \ \ a.s. \end{eqnarray*} $

对于任意的$ \epsilon>0 $, 存在$ T>0 $, 当$ t>T $时使得

$ \begin{eqnarray*} \overline{y_{1}(t)}\leq\frac{b_{1}}{c_{11}}+\frac{\epsilon}{2c_{21}}, \ \ t^{-1}\left[\int_{t-\tau_{3}}^{t}y_{1}(s)ds-\int_{-\tau_{3}}^{0}y_{1}(s)ds\right]\geq-\frac{\epsilon}{2c_{21}}, \end{eqnarray*} $

把上面的式子代入(3.2) 有

$ \begin{eqnarray} t^{-1}\ln\frac{y_{2}(t)}{y_{2}(0)}&\leq&b_{2}+\frac{b_{1}c_{21}}{c_{11}}+\epsilon-c_{22}\overline{y_{2}(t)}+t^{-1}\sigma_{2}B_{2}(t), \end{eqnarray} $

(3.4)

其中$ \epsilon $足够小使得$ 0<\epsilon<-b_{2}-\frac{b_{1}c_{21}}{c_{11}} $, 引理1.1暗含了$ \lim_{t\rightarrow +\infty}y_{2}(t)=0 $几乎处处成立. 则对于任意的$ \epsilon>0 $, 存在$ T>0 $, 当$ t>T $时使得

$ \begin{eqnarray*} &&\overline{y_{1}(t)}\leq\frac{b_{1}}{c_{11}}+\frac{\epsilon}{4c_{31}}, \ \ t^{-1}\left[\int_{t-\tau_{6}}^{t}y_{2}(s)ds-\int_{-\tau_{6}}^{0}y_{2}(s)ds\right]\geq-\frac{\epsilon}{4c_{32}}, \\ &&t^{-1}\left[\int_{t-\tau_{5}}^{t}y_{1}(s)ds-\int_{-\tau_{5}}^{0}y_{1}(s)ds\right]\geq-\frac{\epsilon}{4c_{31}}, \ \overline{y_{2}(t)}\leq\frac{\epsilon}{4c_{32}}, \end{eqnarray*} $

把上面的式子代入(3.3) 有

$ \begin{eqnarray*} t^{-1}\ln\frac{y_{3}(t)}{y_{3}(0)}&\leq&b_{3}+\frac{b_{1}c_{31}}{c_{11}}+\epsilon-c_{33}\overline{y_{3}(t)}+t^{-1}\sigma_{3}B_{3}(t), \end{eqnarray*} $

其中, $ \epsilon $足够小使得$ 0<\epsilon<-b_{3}-\frac{b_{1}c_{31}}{c_{11}} $, 由于引理1.1得到

$ \begin{eqnarray*} \lim\limits_{t\rightarrow +\infty}y_{3}(t)=0, \ \ a.s. \end{eqnarray*} $

(3) 由上面的结果可知

$ \begin{eqnarray*} \lim\limits_{t\rightarrow +\infty}\overline{y_{1}(t)}=\frac{b_{1}}{c_{11}}, \ \ a.s. \end{eqnarray*} $

因为$ b_{2}+\frac{b_{1}c_{21}}{c_{11}}\geq0 $, (3.4) 以及引理1.1, 则有

$ \begin{eqnarray*} \overline{y_{2}}^{*}=\limsup\limits_{t\rightarrow +\infty}t^{-1}\int_{0}^{t}y_{2}(s)ds\leq\frac{b_{2}+\frac{b_{1}c_{21}}{c_{11}}+\epsilon}{c_{22}}, \ \ a.s., \end{eqnarray*} $

又由于$ \epsilon $的任意性得到

$ \begin{eqnarray*} \overline{y_{2}}^{*}=\limsup\limits_{t\rightarrow +\infty}t^{-1}\int_{0}^{t}y_{2}(s)ds\leq\frac{b_{2}+\frac{b_{1}c_{21}}{c_{11}}}{c_{22}}, \ \ a.s. \end{eqnarray*} $

若$ b_{2}+\frac{b_{1}c_{21}}{c_{11}}=0 $, 则

$ \begin{eqnarray*} \lim\limits_{t\rightarrow +\infty}\overline{y_{2}(t)}=\frac{b_{2}+\frac{b_{1}c_{21}}{c_{11}}}{c_{22}}=0, \ \ a.s. \end{eqnarray*} $

若$ b_{2}+\frac{b_{1}c_{21}}{c_{11}}>0 $, 由极限的定义，类似于(2) 的证明，借助于引理1.1我们有

$ \begin{eqnarray*} \overline{y_{2}}_{*}=\liminf\limits_{t\rightarrow +\infty}t^{-1}\int_{0}^{t}y_{2}(s)ds\geq\frac{b_{2}+\frac{b_{1}c_{21}}{c_{11}}}{c_{22}}, \ \ a.s. \end{eqnarray*} $

从而可得

$ \begin{eqnarray*} \lim\limits_{t\rightarrow +\infty}\overline{y_{2}(t)}=\frac{b_{2}+\frac{b_{1}c_{21}}{c_{11}}}{c_{22}}, \ \ a.s., \end{eqnarray*} $

则对于任意的$ \epsilon>0 $, 存在$ T>0 $, 当$ t>T $时使得

$ \begin{eqnarray*} &&\overline{y_{1}(t)}\leq\frac{b_{1}}{c_{11}}+\frac{\epsilon}{4c_{31}}, \ \ t^{-1}\left[\int_{t-\tau_{5}}^{t}y_{1}(s)ds-\int_{-\tau_{5}}^{0}y_{1}(s)ds\right]\geq-\frac{\epsilon}{4c_{31}}, \\ &&\overline{y_{2}(t)}\leq\frac{b_{2}+\frac{b_{1}c_{21}}{c_{11}}}{c_{22}}+\frac{\epsilon}{4c_{32}}, \ \ t^{-1}\left[\int_{t-\tau_{6}}^{t}y_{2}(s)ds-\int_{-\tau_{6}}^{0}y_{2}(s)ds\right]\geq-\frac{\epsilon}{4c_{32}}, \end{eqnarray*} $

代入到(3.3) 得到

$ \begin{eqnarray} t^{-1}\ln\frac{y_{3}(t)}{y_{3}(0)}&\leq&b_{3}+\frac{b_{1}c_{31}}{c_{11}}+\epsilon+\frac{b_{2}+\frac{b_{1}c_{21}}{c_{11}}}{c_{22}}c_{32}-c_{33}\overline{y_{3}(t)}+t^{-1}\sigma_{3}B_{3}(t), \end{eqnarray} $

(3.5)

此处$ \epsilon $足够小使得$ 0<\epsilon<-b_{3}-\frac{b_{1}c_{31}}{c_{11}}-\frac{b_{2}+\frac{b_{1}c_{21}}{c_{11}}}{c_{22}}c_{32} $, 由引理1得到$ \lim_{t\rightarrow +\infty}y_{3}(t)=0 $几乎必然成立.

(4)–(7)的证明类似于(3), 我们把细节留给感兴趣的读者. 证毕.

类似地, 我们可得如下的结论.

注    由引理1.1我们可以得到, $ 1\leq i\leq 3 $, $ \lim\limits_{t\rightarrow +\infty}\overline{y_{i}(t)}=J, \ \ a.s, $这里$ J $为大于零的常数. 所以, 对于任意的$ \tilde{\tau}\geq 0 $, 有

$ \begin{eqnarray*} \lim\limits_{t\rightarrow +\infty}t^{-1}\int_{t-\tilde{\tau}}^{t}y_{i}(s)ds=\lim\limits_{t\rightarrow +\infty}t^{-1}[\int_{0}^{t}y_{i}(s)ds-\int_{0}^{t-\tilde{\tau}}y_{i}(s)ds]=0, \ \ a.s., \end{eqnarray*} $

由于(2.3), 可得

$ \begin{eqnarray} \lim\limits_{t\rightarrow +\infty}t^{-1}\int_{t-\tilde{\tau}}^{t}x_{i}(s)ds=0, \ \ a.s., \ \ i=1, 2, 3, \ \tilde{\tau}\geq 0. \end{eqnarray} $

(3.6)

引理3.2   假设$ (1.3) $成立, 则对于模型(1.1) 而言

$ (1) $若$ b_{i}<0, \ i=1, 2, 3 $, 则

$ \begin{eqnarray*} \lim\limits_{t\rightarrow +\infty}x_{i}(t)=0, \ \ a.s., \ \ i=1, 2, 3; \end{eqnarray*} $

$ (2) $若$ b_{1}\geq0, \; b_{2}+\frac{b_{1}c_{21}}{c_{11}}<0, \; b_{3}+\frac{b_{1}c_{31}}{c_{11}}<0 $, 则

$ \begin{eqnarray*} \lim\limits_{t\rightarrow +\infty}\overline{x_{1}(t)}=\frac{b_{1}}{c_{11}}, \ \ \lim\limits_{t\rightarrow +\infty}x_{i}(t)=0, \ \ a.s., \ \ i=2, 3; \end{eqnarray*} $

$ (3) $若$ \Lambda_{i}-\tilde{\Lambda}_{i}>0, \ i=1, 2, 3 $, 则

$ \begin{eqnarray*} \lim\limits_{t\rightarrow +\infty}\overline{x_{i}(t)}=\frac{\Lambda_{i}-\tilde{\Lambda}_{i}}{\Lambda}, \ \ a.s., \ \ i=1, 2, 3; \end{eqnarray*} $

$ (4) $若$ \Gamma_{1}-\tilde{\Gamma}_{1}>0, \; \Gamma_{2}-\tilde{\Gamma}_{2}>0, \; \Lambda_{3}-\tilde{\Lambda}_{3}<0 $, 则

$ \begin{eqnarray*} \lim\limits_{t\rightarrow +\infty}\overline{x_{i}(t)}=\frac{\Gamma_{i}-\tilde{\Gamma}_{i}}{\Gamma}, \ \ i=1, 2, \ \ \lim\limits_{t\rightarrow +\infty}x_{3}(t)=0, \ \ a.s.; \end{eqnarray*} $

$ (5) $若$ b_{1}<0, \; b_{3}c_{22}+b_{2}c_{32}>0, \; \Gamma_{3}-\tilde{\Gamma}_{3}>0 $, 则

$ \begin{eqnarray*} \lim\limits_{t\rightarrow +\infty}x_{1}(t)=0, \ \lim\limits_{t\rightarrow +\infty}\overline{x_{2}(t)}=\frac{\Gamma_{3}-\tilde{\Gamma}_{3}}{c_{22}c_{33}+c_{23}c_{32}}, \ \lim\limits_{t\rightarrow +\infty}\overline{x_{3}(t)}=\frac{\Gamma_{4}-\tilde{\Gamma}_{4}}{c_{22}c_{33}+c_{23}c_{32}}, \ \ a.s.; \end{eqnarray*} $

$ (6) $若$ b_{2}<0, \; b_{3}c_{11}+b_{1}c_{31}>0, \; b_{1}c_{33}-b_{3}c_{13}>0 $, 则

$ \begin{eqnarray*} \lim\limits_{t\rightarrow +\infty}\overline{x_{1}(t)}=\frac{\Gamma_{5}-\tilde{\Gamma}_{5}}{c_{11}c_{33}+c_{13}c_{31}}, \ \lim\limits_{t\rightarrow +\infty}x_{2}(t)=0, \ \lim\limits_{t\rightarrow +\infty}\overline{x_{3}(t)}=\frac{\Gamma_{6}-\tilde{\Gamma}_{6}}{c_{11}c_{33}+c_{13}c_{31}}, \ \ a.s.; \end{eqnarray*} $

$ (7) $若$ b_{1}<0, \; b_{3}c_{22}+b_{2}c_{32}<0, \; b_{2}\geq0 $, 则

$ \begin{eqnarray*} \lim\limits_{t\rightarrow +\infty}x_{1}(t)=0, \ \ \lim\limits_{t\rightarrow +\infty}\overline{x_{2}(t)}=\frac{ b_{2}}{c_{22}}, \ \ \lim\limits_{t\rightarrow +\infty}x_{3}(t)=0, \ \ a.s. \end{eqnarray*} $

引理3.2的证明类似于引理3.1, 在此省略.

引理4.1    假设$ (1.3) $成立, 则模型(1.1) 是依分布稳定的.

证    假定模型(1.1) 有两个解, $ x(t)=x(t;x(\theta)) $与$ \tilde{x}(t)=x(t;\tilde{x}(\theta)) $, 并且$ x(\theta)\in C([-\tau, 0], R_{+}^{3}), \ \tilde{x}(\theta)\in C([-\tau, 0], R_{+}^{3}) $. 令

$ \begin{eqnarray*} K=(k_{ij})_{3\times3}= \left( \begin{array}{ccc} c_{11} & c_{12} & c_{13} \\ c_{21} & c_{22} & c_{23} \\ c_{31} & c_{32} & c_{33} \\ \end{array} \right), \ \ L_{K}= \left( \begin{array}{ccc} c_{12}+c_{13} & -c_{12} & -c_{13} \\ -c_{21} & c_{21}+c_{23} & -c_{23} \\ -c_{31} & -c_{32} & c_{31}+c_{32} \\ \end{array} \right), \end{eqnarray*} $

且$ m_{1}, \ m_{2}, \ m_{3} $分别是$ L_{K} $对角元素的代数余子式. 因为K是不可约的, 由文献[2] 可知$ m_{i}>0, \ i=1, 2, 3 $. 其次, 我们有如下定义

$ \begin{eqnarray*} V(t)&=&\sum\limits_{i=1}^{3}m_{i}|\ln x_{i}(t)-\ln\tilde{x}_{i}(t)|+m_{1}c_{12}\int_{t-\tau_{1}}^{t}|x_{2}(s)-\tilde{x}_{2}(s)|ds\\ &&+m_{1}c_{13}\int_{t-\tau_{2}}^{t}|x_{3}(s)-\tilde{x}_{3}(s)|ds+m_{2}c_{21}\int_{t-\tau_{3}}^{t}|x_{1}(s)-\tilde{x}_{1}(s)|ds\\ &&+m_{2}c_{23}\int_{t-\tau_{4}}^{t}|x_{3}(s)-\tilde{x}_{3}(s)|ds+m_{3}c_{31}\int_{t-\tau_{5}}^{t}|x_{1}(s)-\tilde{x}_{1}(s)|ds\\ &&+m_{3}c_{32}\int_{t-\tau_{6}}^{t}|x_{2}(s)-\tilde{x}_{2}(s)|ds. \end{eqnarray*} $

通过运用Itô's公式得到

$ \begin{eqnarray*} &&d^{+}V(t)\\ &\leq&-m_{1}c_{11}|x_{1}(t)-\tilde{x}_{1}(t)|dt+m_{1}c_{12}|x_{2}(t-\tau_{1})-\tilde{x}_{2}(t-\tau_{1})|dt\\ &&+m_{1}c_{13}|x_{3}(t-\tau_{2})-\tilde{x}_{3}(t-\tau_{2})|dt+m_{2}c_{21}|x_{1}(t-\tau_{3})-\tilde{x}_{1}(t-\tau_{3})|dt\\ &&-m_{2}c_{22}|x_{2}(t)-\tilde{x}_{2}(t)|dt+m_{2}c_{23}|x_{3}(t-\tau_{4})-\tilde{x}_{3}(t-\tau_{4})|dt\\ &&+m_{3}c_{31}|x_{1}(t-\tau_{5})-\tilde{x}_{1}(t-\tau_{5})|dt+m_{3}c_{32}|x_{2}(t-\tau_{6})\\ &&-\tilde{x}_{2}(t-\tau_{6})|dt-m_{3}c_{33}|x_{3}(t)-\tilde{x}_{3}(t)|dt\\ &&+m_{1}c_{12}|x_{2}(t)-\tilde{x}_{2}(t)|dt-m_{1}c_{12}|x_{2}(t-\tau_{1})-\tilde{x}_{2}(t-\tau_{1})|dt\\ &&+m_{1}c_{13}|x_{3}(t)-\tilde{x}_{3}(t)|dt-m_{1}c_{13}|x_{3}(t-\tau_{2})-\tilde{x}_{3}(t-\tau_{2})|dt\\ &&+m_{2}c_{21}|x_{1}(t)-\tilde{x}_{1}(t)|dt-m_{2}c_{21}|x_{1}(t-\tau_{3})-\tilde{x}_{1}(t-\tau_{3})|dt\\ &&+m_{2}c_{23}|x_{3}(t)-\tilde{x}_{3}(t)|dt-m_{2}c_{23}|x_{3}(t-\tau_{4})-\tilde{x}_{3}(t-\tau_{4})|dt\\ &&+m_{3}c_{31}|x_{1}(t)-\tilde{x}_{1}(t)|dt-m_{3}c_{31}|x_{1}(t-\tau_{5})-\tilde{x}_{1}(t-\tau_{5})|dt\\ &&+m_{3}c_{32}|x_{2}(t)-\tilde{x}_{2}(t)|dt-m_{3}c_{32}|x_{2}(t-\tau_{6})-\tilde{x}_{2}(t-\tau_{6})|dt\\ &=&-\sum\limits_{i=1}^{3}m_{i}c_{ii}|x_{i}(t)-\tilde{x}_{i}(t)|dt+\sum\limits_{i=1}^{3}\sum\limits_{j\neq i}m_{i}k_{ij}|x_{j}(t)-\tilde{x}_{j}(t)|dt. \end{eqnarray*} $

由引理1.4可以得到

$ \begin{eqnarray*} \sum\limits_{i=1}^{3}\sum\limits_{j\neq i}m_{i}c_{ij}|x_{j}(t)-\tilde{x}_{j}(t)|=\sum\limits_{i=1}^{3}\sum\limits_{j\neq i}m_{i}c_{ij}|x_{i}(t)-\tilde{x}_{i}(t)|. \end{eqnarray*} $

因此

$ \begin{eqnarray*} &&d^{+}V(t)\\ &\leq&-\sum\limits_{i=1}^{3}m_{i}c_{ii}|x_{i}(t)-\tilde{x}_{i}(t)|dt+\sum\limits_{i=1}^{3}\sum\limits_{j\neq i}m_{i}k_{ij}|x_{i}(t)-\tilde{x}_{i}(t)|dt\\ &=&-m_{1}(c_{11}-c_{12}-c_{13})|x_{1}(t)-\tilde{x}_{1}(t)|dt-m_{2}(c_{22}-c_{21}-c_{23})|x_{2}(t)-\tilde{x}_{2}(t)|dt\\ &&-m_{3}(c_{33}-c_{31}-c_{32})|x_{3}(t)-\tilde{x}_{3}(t)|dt. \end{eqnarray*} $

将上式的左右两端同时从0到t进行积分并取期望有

$ \begin{eqnarray*} E(V(t))&\leq&V(0)-m_{1}(c_{11}-c_{12}-c_{13})\int_{0}^{t}E|x_{1}(s)-\tilde{x}_{1}(s)|ds\\ &&-m_{2}(c_{22}-c_{21}-c_{23})\int_{0}^{t}E|x_{12}(s)-\tilde{x}_{2}(s)|ds\\ &&-m_{3}(c_{33}-c_{31}-c_{32})\int_{0}^{t}E|x_{3}(s)-\tilde{x}_{3}(s)|ds, \end{eqnarray*} $

由此可知

$ \begin{eqnarray*} &&m_{1}(c_{11}-c_{12}-c_{13})\int_{0}^{t}E|x_{1}(s)-\tilde{x}_{1}(s)|ds+m_{2}(c_{22}-c_{21}-c_{23})\int_{0}^{t}E|x_{12}(s)-\tilde{x}_{2}(s)|ds\\ &&+m_{3}(c_{33}-c_{31}-c_{32})\int_{0}^{t}E|x_{3}(s)-\tilde{x}_{3}(s)|ds\leq V(0)<+\infty, \end{eqnarray*} $

从而得到

$ \begin{eqnarray} E|x_{i}(t)-\tilde{x}_{i}(t)|\in L^{1}[0, +\infty). \end{eqnarray} $

(4.1)

由模型(1.1) 可以得到

$ \begin{eqnarray*} E(x_{1}(t))&=&x_{1}(0)+\int_{0}^{t}[a_{1}E(x_{1}(s))-c_{11}E((x_{1}(s)))^{2}\\ &&+c_{12}E(x_{1}(s)x_{2}(s-\tau_{1}))-c_{13}E(x_{1}(s)x_{3}(s-\tau_{2}))]ds, \end{eqnarray*} $

也就是说$ E(x_{1}(t)) $是可微的. 由(2.5) 可知

$ \begin{eqnarray*} \frac{dE(x_{1}(t))}{dt}&=&a_{1}E(x_{1}(t))-c_{11}E((x_{1}(t)))^{2}\\ &&+c_{12}E(x_{1}(t)x_{2}(t-\tau_{1}))-c_{13}E(x_{1}(t)x_{3}(t-\tau_{2}))\\ &\leq&a_{1}E(x_{1}(t))+c_{12}E(x_{1}(t)x_{2}(t-\tau_{1}))\\ &\leq&a_{1}K+c_{12}E(x_{1}(t)x_{2}(t-\tau_{1}))\\ &\leq&D, \end{eqnarray*} $

其中D是一个正常数. 那么, $ E(x_{1}(t)) $是一致连续的. 同理得$ E(x_{2}(t)) $与$ E(x_{3}(t)) $也是一致连续的. 由(4.1) 知

$ \begin{eqnarray} \lim\limits_{t\rightarrow +\infty}E|x_{i}(t)-\tilde{x}_{i}(t)|=0, \ i=1, 2, 3. \end{eqnarray} $

(4.2)

设$ \mathbb{P}(R_{+}^{3}) $是定义在$ R_{+}^{3} $上的所有概率测度并且$ P(t, x(\theta), Q) $是$ x(t;x(\theta)) $的概率, 其中$ x(t;x(\theta))\in Q $. 对于$ \forall P_{1}\in\mathbb{P}, \ \forall P_{2}\in\mathbb{P} $, 我们有如下定义

$ \begin{eqnarray*} M=\{h:R^{3}\rightarrow R||h(x)-h(y)|\leq\|x-y\|, |h(\cdot)|\leq1\}, \end{eqnarray*} $

$ \begin{eqnarray*} d_{M}(P_{1}, P_{2})=\sup\limits_{h\in M}|\int_{R_{+}^{3}}h(x)P_{1}(dx)-\int_{R_{+}^{3}}h(x)P_{2}(dx)|. \end{eqnarray*} $

于是, 有$ \forall h\in M, t>0, s>0 $,

$ \begin{eqnarray*} &&|Eh(x(t+s;x(\theta)))-Eh(x(t;x(\theta)))|\\ &=&|E[E(h(x(t+s;x(\theta)))|\mathcal {F}_{s})]-Eh(x(t;x(\theta)))|\\ &=&|\int_{R_{+}^{3}}Eh(\tilde{x}(t;\tilde{x}(\theta)))P(s, x(\theta), d\tilde{x}(\theta))-Eh(x(t;x(\theta)))|\\ &\leq&\int_{R_{+}^{3}}|Eh(\tilde{x}(t;\tilde{x}(\theta)))-Eh(x(t;x(\theta)))|P(s, x(\theta), d\tilde{x}(\theta)). \end{eqnarray*} $

又因(4.2) 有对于任意的$ \epsilon>0 $, 存在$ T>0 $, 当$ t>T $时使得$ \sup_{h\in M}|Eh(\tilde{x}(t;\tilde{x}(\theta)))-Eh(x(t;x(\theta)))|\leq\epsilon, $因此

$ \begin{eqnarray*} |Eh(x(t+s;x(\theta)))-Eh(x(t;x(\theta)))|\leq\epsilon. \end{eqnarray*} $

由$ h $的任意性可知

$ \begin{eqnarray*} \sup\limits_{h\in M}|Eh(x(t+s;x(\theta)))-Eh(x(t;x(\theta)))|\leq3\epsilon. \end{eqnarray*} $

换言之, 对于$ \forall t\geq T $, $ s>0 $有

$ \begin{eqnarray*} d_{M}(P(t+s, x(\theta), \cdot), P(t, x(\theta), \cdot))\leq\epsilon, \end{eqnarray*} $

也就是$ \{P(t, x(\theta), \cdot)\} $在$ \mathbb{P}(R_{+}^{3}) $中是Cauchy列. 从而$ \xi(\theta)=(\xi_{1}(\theta), \xi_{2}(\theta), \xi_{3}(\theta))^{T}, \ \xi_{i}(\theta)\equiv0.1, \ \theta\in[-\tau, 0), \; \{P(t, \xi(\theta), \cdot)\} $也是Cauchy列. 因此$ \exists \ ! \ \mu(\cdot)\in\mathbb{P}(R_{+}^{3}) $使得

$ \begin{eqnarray} \lim\limits_{t\rightarrow +\infty}d_{M}(P(t, \xi(\theta), \cdot), \mu(\cdot))=0. \end{eqnarray} $

(4.3)

因(4.2) 可知

$ \begin{eqnarray*} \lim\limits_{t\rightarrow +\infty}d_{M}(P(t, x(\theta), \cdot), P(t, \xi(\theta), \cdot))=0. \end{eqnarray*} $

因此, 可以得到

$ \begin{eqnarray*} &&\lim\limits_{t\rightarrow +\infty}d_{M}(P(t, x(\theta), \cdot), \mu(\cdot))\\ &\leq&\lim\limits_{t\rightarrow +\infty}d_{M}(P(t, x(\theta), \cdot), P(t, \xi(\theta), \cdot))+\lim\limits_{t\rightarrow +\infty}d_{M}(P(t, \xi(\theta), \cdot), \mu(\cdot))\\ &=&0. \end{eqnarray*} $

即得证.

定理4.1    对于模型(1.1) 而言, 假设(1.3) 成立有

$ (1) $若$ b_{i}<0, \ i=1, 2, 3 $, 则$ x_{1}(t), \ x_{2}(t), \ x_{3}(t) $是灭绝的.

$ (2) $若$ b_{1}\geq0, \; b_{2}+\frac{b_{1}c_{21}}{c_{11}}<0, \; b_{3}+\frac{b_{1}c_{31}}{c_{11}}<0 $, 则$ x_{2}(t) $和$ x_{3}(t) $是灭绝的, $ x_{1}(t) $的分布弱收敛到唯一的遍历不变分布$ \nu_{1} $, 且有

$ \begin{eqnarray*} \lim\limits_{t\rightarrow +\infty}\overline{x_{1}(t)}=\int_{R_{+}}z_{1}\nu_{1}(dz_{1})=\frac{b_{1}}{c_{11}}, \ \ a.s.; \end{eqnarray*} $

$ (3) $若$ \Lambda_{i}-\tilde{\Lambda}_{i}>0, \ i=1, 2, 3 $, 则$ (x_{1}(t), x_{2}(t), x_{3}(t))^{T} $的分布弱收敛到唯一的遍历不变分布$ \nu_{2} $, 且有

$ \begin{eqnarray*} \lim\limits_{t\rightarrow +\infty}\overline{x_{i}(t)}=\int_{R_{+}^{3}}z_{i}\nu_{2}(dz_{1}, dz_{2}, dz_{3})=\frac{\Lambda_{i}-\tilde{\Lambda}_{i}}{\Lambda}, \ \ a.s., \ \ i=1, 2, 3; \end{eqnarray*} $

$ (4) $若$ \Gamma_{1}-\tilde{\Gamma}_{1}>0, \; \Gamma_{2}-\tilde{\Gamma}_{2}>0, \; \Lambda_{3}-\tilde{\Lambda}_{3}<0 $, 则$ x_{3}(t) $灭绝, $ (x_{1}(t), x_{2}(t))^{T} $的分布弱收敛到唯一的遍历不变分布$ \nu_{3} $, 且有

$ \begin{eqnarray*} \lim\limits_{t\rightarrow +\infty}\overline{x_{i}(t)}=\int_{R_{+}^{2}}z_{i}\nu_{3}(dz_{1}, dz_{2})=\frac{\Gamma_{i}-\tilde{\Gamma}_{i}}{\Gamma}, \ \ a.s., \ \ i=1, 2; \end{eqnarray*} $

$ (5) $若$ b_{1}<0, \; b_{3}c_{22}+b_{2}c_{32}>0, \; \Gamma_{3}-\tilde{\Gamma}_{3}>0 $, 则$ x_{1}(t) $灭绝, $ (x_{2}(t), x_{3}(t))^{T} $的分布弱收敛到唯一的遍历不变分布$ \nu_{4} $, 且有

$ \begin{eqnarray*} \lim\limits_{t\rightarrow +\infty}\overline{x_{2}(t)}=\int_{R_{+}^{2}}z_{2}\nu_{4}(dz_{2}, dz_{3})=\frac{\Gamma_{3}-\tilde{\Gamma}_{3}}{c_{22}c_{33}+c_{23}c_{32}}, \ \ a.s., \\ \lim\limits_{t\rightarrow +\infty}\overline{x_{3}(t)}=\int_{R_{+}^{2}}z_{3}\nu_{4}(dz_{2}, dz_{3})=\frac{\Gamma_{4}-\tilde{\Gamma}_{4}}{c_{22}c_{33}+c_{23}c_{32}}, \ \ a.s.; \end{eqnarray*} $

$ (6) $若$ b_{2}<0, \; b_{3}c_{11}+b_{1}c_{31}>0, \; b_{1}c_{33}-b_{3}c_{13}>0 $, 则$ x_{2}(t) $灭绝, $ (x_{1}(t), x_{3}(t))^{T} $的分布弱收敛到唯一的遍历不变分布$ \nu_{5} $, 且有

$ \begin{eqnarray*} \lim\limits_{t\rightarrow +\infty}\overline{x_{1}(t)}=\int_{R_{+}^{2}}z_{1}\nu_{5}(dz_{1}, dz_{3})=\frac{\Gamma_{5}-\tilde{\Gamma}_{5}}{c_{11}c_{33}+c_{13}c_{31}}, \ \ a.s., \\ \lim\limits_{t\rightarrow +\infty}\overline{x_{3}(t)}=\int_{R_{+}^{2}}z_{3}\nu_{5}(dz_{1}, dz_{3})=\frac{\Gamma_{6}-\tilde{\Gamma}_{6}}{c_{11}c_{33}+c_{13}c_{31}}, \ \ a.s.; \end{eqnarray*} $

$ (7) $若$ b_{1}<0, \; b_{3}c_{22}+b_{2}c_{32}<0, \; b_{2}\geq0 $, 则$ x_{1}(t) $和$ x_{3}(t) $是灭绝的, $ x_{2}(t) $的分布弱收敛到唯一的遍历不变分布$ \nu_{6} $, 且有

$ \begin{eqnarray*} \lim\limits_{t\rightarrow +\infty}\overline{x_{2}(t)}=\int_{R_{+}}z_{2}\nu_{6}(dz_{2})=\frac{ b_{2}}{c_{22}}, \ \ a.s.; \end{eqnarray*} $

证   (1) 可以由引理3.2中(1) 得出.

(2) 可以由引理3.2中(2) 得出. 此外, 由模型(1.1) 是平稳分布知存在一个概率测度$ \nu_{1} $使得$ x_{1}(t) $的分布弱收敛到$ \nu_{1} $. 又由$ | \int_{R_{+}}\varphi(x)p(t, x(\theta), dx_{1})-\int_{R_{+}}\varphi(x)\nu_{1}(dx_{1})|\rightarrow 0, $可知$ \langle\varphi, P\circ x(\theta)\rangle\rightarrow \langle\varphi, \nu_{1}\rangle, $则$ \nu_{1} $是不变的. 根据引理1.2知$ \nu_{1} $是强混合的. 又由引理1.5知$ \nu_{1} $是遍历的. 最后根据引理1.6得到

$ \begin{eqnarray*} \lim\limits_{t\rightarrow +\infty}\overline{x_{1}(t)}=\int_{R_{+}}z_{1}\nu_{1}(dz_{1}), \ \ a.s.; \end{eqnarray*} $

同(2) 的证明过程我们可以得到(3), (4), (5), (6), (7) 的证明.