我们提出一个具有校正隔离率的易感者-感染者-隔离者模型, 该模型用常微分方程描述如下

$ \begin{equation}\label{eq1} \left\{ \begin{array}{llll} \dot{S}(t) = A-d_{1}S(t)-\frac{\beta S(t)I(t)}{1+\alpha_{1}S(t)+\alpha_{2}I(t)}+\gamma I(t)+\rho Q(t), \\ \dot{I}(t) = \frac{\beta S(t)I(t)}{1+\alpha_{1}S(t)+\alpha_{2}I(t)}-(d_{2}+\gamma+\delta)I(t), \\[2mm] \dot{Q}(t) = \delta I(t)-(d_{3}+\rho)Q(t), \end{array} \right. \end{equation} $

(1.1)

其中, $S(t)$ 为易感者在 $t$ 时刻的密度, $I(t)$ 为没有接受隔离的感染者的密度, $Q(t)$ 为 $t$ 时刻已接受隔离的隔离者的密度.参数 $A, d_1, d_2, d_3$ 为正数, 且 $\delta, \rho, \gamma$ , $\alpha_i$ 是非负常数.常数 $A$ 是易感者的出生及移民的输入率; $\beta$ 是易感者与感染者(即, 与非隔离者)之间的平均接触率; $d_{1}$ 为易感者的自然死亡率; $d_{2}$ 为感染者由疾病引起的死亡率; $d_{3}$ 为隔离者由疾病诱发的死亡率; $\delta$ 是被隔离的感染者的比率; $\gamma$ 和 $\rho$ 分别是经过治疗、隔离后, 感染者与隔离者重新回到易感者的比率; $\frac{\beta SI}{1+\alpha_{1}S+\alpha_{2}I}$ 刻画了不含隔离者, 且依赖易感者和感染者的校正隔离率(见图 1).

为了研究流行病, Kermack与McKendrick从简单的SIR模型开始, 于1927年首次提出了双线性传染率^[1].借鉴Kermack与McKendrick的双线性传染率 $\beta SI$ , 改进后的转移率更接近种群自身的实际情况, 且其有利于描述该种群的动力学性质^[1-15].当 $\alpha_{1}=\alpha_{2}=0$ 时, 模型(1.1)的校正隔离率 $\frac{\beta SI}{1+\alpha_{1}S+\alpha_{2}I}$ 退化为双线性传染率 $\beta SI$ , 相关的流行病最新研究结果见文献[7-9]及其引文.当 $\alpha_{1}=0$ 时, 模型(1.1)的校正隔离率转化为饱和传染率 $\frac{\beta SI}{1+\alpha I}$ (见文献[10-12]).具有一般传染率(即, 非线性传染率)的模型 $\frac{\beta SI}{\varphi(I)}$ , 其相关结果请参考[13-15]及其引文.

对于一般的随机流行病模型, 例如:易感者-感染者-隔离者模型, 其疾病的持久性与绝灭性等更多细节见文献[2].最近, 蒋达清等^[7]提出了DI SIR模型, 通过对双线性传染率 $\beta $ 进行随机干扰, 利用构造的合适的Lyapunov函数, 他们分别考查了无病平衡点及地方并平衡点附近解的渐近行为.进而, 刘红等^[12]在随机DI SIR流行病模型中引入饱和传染率 $\frac{\beta_j S I_j}{1+\alpha_j I_j}$ , 他们的结果表明:若干扰较大, 感染者指数衰减到零, 而易感者弱收敛到平稳分布, 且不依赖于 $R_0$ .若干扰较小且 $R_0\leq1$ , 则有相同的指数稳定性及弱收敛发生.若干扰较小且 $R_0>1$ , 通过构造一系列Lyapunov函数, 则得到解的遍历性及正常返性.同时, 对于一类具有接种的随机SIS模型, 赵亚男等^[8]发现, 该模型的感染者指数减少到零, 且存在一个均值意义下的持久解.类似地, 林玉国等^[9]考查了SIR模型, 并得到种群的密度收敛到一个不变密度, 或者收敛到一个类似的测度, 更多的细节请见引文.

学者May^[4]曾指出:当考虑到环境的噪音时, 模型的参数将会展现出随机扰动.现在, 我们对确定的易感者-感染者-隔离者模型中的参数引入随机干扰

$ d_{i}\rightarrow d_i+\sigma_i\dot{B}_i(t) \ (i=1, 2, 3), \quad \beta\rightarrow \beta+\sigma_4\dot{B}_4(t), $

其中 $B_{i}(t)\ (i=1, 2, 3, 4)$ 为相互独立的一维布朗运动, $\sigma_{i}\ (i=1, 2, 3, 4)$ 表示白噪声的强度.于是, 我们得到一个具有校正隔离率的随机易感者-感染者-隔离者模型

$ \begin{equation}\label{eq2} \left\{ \begin{array}{llll} \mbox{d}S(t) &=& \Big[A-d_{1}S(t)-\frac{\beta S(t)I(t)}{1+\alpha_{1}S(t)+\alpha_{2}I(t)}+\gamma I(t)+\rho Q(t)\Big]\mbox{d}t\\ &&-\sigma_{1}S(t)\mbox{d}B_{1}(t) -\frac{\sigma_{4}S(t)I(t)}{1+\alpha_{1}S(t)+\alpha_{2}I(t)}\mbox{d}B_{4}(t), \\ \mbox{d}I(t)& =&\Big[\frac{\beta S(t)I(t)}{1+\alpha_{1}S(t)+\alpha_{2}I(t)}-(d_{2}+\gamma+\delta)I(t)\Big]\mbox{d}t\\ &&-\sigma_{2}I(t)\mbox{d}B_{2}(t) +\frac{\sigma_{4}S(t)I(t)}{1+\alpha_{1}S(t)+\alpha_{2}I(t)}\mbox{d}B_{4}(t), \\[2mm] \mbox{d}Q(t) &=&[\delta I(t)-(d_{3}+\rho)Q(t)]\mbox{d}t-\sigma_{3}Q(t)\mbox{d}B_{3}(t). \end{array} \right. \end{equation} $

(1.2)

我们将在第二节证明随机模型(1.3)解的存在唯一性.在第三节讨论解沿无病平衡点绝灭的充分性条件, 并证明当时间 $t$ 充分大时, 感染者的密度指数衰减这一结论.结合Lyapunov函数、伊藤公式及数值模拟, 正解的平稳分布及正态分布将在第四节探讨.

由于模型(1.2)的系数满足局部Lipschitz条件, 不满足线性增长条件, 本节我们将利用文献[5]中的研究方法, 证明模型(1.2)存在唯一的全局正解.

定理2.1 对任意初值 $(S(0), I(0), Q(0))\in {\cal R}_{+}^{3}$ 及 $t\geq0$ , 模型 $(1.2)$ 存在唯一一个解 $(S(t), I(t), Q(t))$ , 且该解以概率1位于区域 ${\cal R}_{+}^{3}$ 内.

证 反证法.若结论不成立, 假设模型(1.2)在区间 $[0, \tau_{e})$ 上, 存在一个局部解 $(S(t), I(t), $ $ Q(t))$ , 其中 $\tau_{e}$ 为爆破时间, $(S(0), I(0), Q(0))\in {\cal R}_{+}^{3}$ 为任意的初值.为了证明模型(1.2)的解是全局正解, 我们需要证明 $\tau_{e}=\infty$ 几乎处处成立.定义停时

$ \tau_{+}=\inf\{t\in[0, \tau_{e}): S(t)\leq0~\mbox{或}~I(t)\leq 0 ~\mbox{或}~Q(t)\leq 0\}. $

贯穿全文, 令 $\inf\emptyset=\infty$ .下面, 我们将证明 $\tau_{+}=\infty$ 几乎处处成立.若不成立, 即 $\tau_{+}<\infty$ , 于是, 存在时间 $T$ 及任意小的正数 $\varepsilon$ 使得 $\mathit{\boldsymbol{P}}\{\tau_{+}<T\}>\varepsilon$ .定义一个 $C^{2}$ -函数

$ V(S(t), I(t), Q(t))=\ln S(t)I(t)Q(t), $

根据伊藤公式, 得到

$ \begin{eqnarray}\label{eq3} &&\mbox{d}V(S(t), I(t), Q(t)) \\ &=& \Big[\frac{A}{S(t)}-\frac{\beta I(t)}{1+\alpha_{1}S(t)+\alpha_{2}I(t)}-d_{1}+\frac{\gamma I(t)}{S(t)}+\frac{\rho Q(t)}{S(t)}-\frac{1}{2}\sigma_{1}^{2}\\ &&-\frac{1}{2}\sigma_{4}^{2}\Big(\frac{I(t)}{1+\alpha_{1}S(t)+\alpha_{2}I(t)}\Big)^{2} +\frac{\beta S(t)}{1+\alpha_{1}S(t)+\alpha_{2}I(t)}-(d_{2}+\gamma+\delta)\\ &&-\frac{1}{2}\sigma_{2}^{2} -\frac{1}{2}\sigma_{4}^{2}\Big(\frac{S(t)}{1+\alpha_{1}S(t)+\alpha_{2}I(t)}\Big)^{2} +\frac{\delta I(t)}{Q(t)}-(d_{3}+\rho)-\frac{1}{2}\sigma_{3}^{2}\Big]\mbox{d}t\\ &&-\sigma_{1}\mbox{d}B_{1}(t)-\sigma_{2}\mbox{d}B_{2}(t) -\sigma_{3}\mbox{d}B_{3}(t) -\sigma_{4}\frac{I(t)}{1+\alpha_{1}S(t)+\alpha_{2}I(t)}\mbox{d}B_{4}(t)\\ &&+\sigma_{4}\frac{S(t)}{1+\alpha_{1}S(t)+\alpha_{2}I(t)}\mbox{d}B_{4}(t) \\ &\geq&K\mbox{d}t-\sigma_{1}\mbox{d}B_{1}(t)-\sigma_{2}\mbox{d}B_{2}(t)-\sigma_{3}\mbox{d}B_{3}(t) +\sigma_{4}\frac{S(t)-I(t)}{1+\alpha_{1}S(t)+\alpha_{2}I(t)}\mbox{d}B_{4}(t), \end{eqnarray} $

(2.1)

其中

$ K=-d_{1}-d_{2}-d_{3}-\gamma-\delta-\rho-\frac{1}{2}\sigma_{1}^{2} -\frac{1}{2}\sigma_{2}^{2}-\frac{1}{2}\sigma_{3}^{2}-\frac{\beta}{\alpha_{2}}-\frac{1}{2\alpha_{1}^2}\sigma_{4}^{2} -\frac{1}{2\alpha_{2}^2}\sigma_{4}^{2}. $

(2.1)式两侧由 $0$ 到 $t$ 积分, 于是

$ \begin{eqnarray}\label{eq4} V(S(t), I(t), Q(t))&\geq&V(S(0), I(0), Q(0)) +Kt-\sigma_{1}B_{1}(t)-\sigma_{2}B_{2}(t)-\sigma_{3}B_{3}(t)\nonumber\\ && +\sigma_{4}\int_{0}^{t}\frac{S(u)-I(u)}{1+\alpha_{1}S(u)+\alpha_{2}I(u)}\mbox{d}B_{4}(u). \end{eqnarray} $

(2.2)

根据 $V(S(t), I(t), Q(t))$ 的定义, 我们注意到

$ \begin{array}{llll} \lim\limits_{t \to\tau_{+}}V(S(t), I(t), Q(t))=-\infty, \end{array} $

于是, (2.2)式两端令 $t \to\tau_{+}$ , 得到

$ \begin{eqnarray}\label{eq5} -\infty&\geq&V(S(0), I(0), Q(0))+K\tau_{+} -\sigma_{1}B_{1}(\tau_{+})-\sigma_{2}B_{2}(\tau_{+})-\sigma_{3}B_{3}(\tau_{+})\\ && +\sigma_{4}\int_{0}^{\tau_{+}}\frac{S(u)-I(u)}{1+\alpha_{1}S(u)+\alpha_{2}I(u)}\mbox{d}B_{4}(u) \\ &>&-\infty, \end{eqnarray} $

(2.3)

这与假设矛盾.于是, $\tau_{+}=\infty$ .证毕.

显然, $E_{0}=(\frac{A}{d_{1}}, 0, 0)$ 为模型(1.2)的解, 称为无病平衡点.本节我们将研究 $E_0$ 附近解的渐近性质, 即, 定理3.1-3.2将给出模型(1.2)的解在无病平衡点附近, 疾病消失、感染者密度趋于零的充分性条件.

定理3.1 对任意初值 $(S(0), I(0), Q(0))\in {\cal R}_{+}^{3}$ , 令 $(S(t), I(t), Q(t))$ 为模型 $(1.2)$ 的解, 若白噪声的强度满足

$ \sigma_{1}^{2}<\frac{d_{1}}{2}, ~ \sigma_{2}^{2}<2d_{2}+\delta-\rho, ~ \sigma_{3}^{2}<2d_{3}+\rho-\frac{\rho^2}{d_{1}}-\delta, $

则该解有如下性质

$ \limsup\limits_{t\to \infty}\frac{1}{t}\, E\int_{0}^{t}\Big[\Big(S(u)-\frac{A}{d_{1}}\Big)^{2}+I^{2}(u)+Q^{2}(u)\Big]\mbox{d}u \leq\sigma_{1}\frac{A^{2}}{\xi d_{1}^{2}}+ \frac{A(d_{1}+d_{2}+\delta)}{\xi d_{1}\alpha_{2}}, $

其中

$ \xi=\min\{\xi_{1}, \xi_{2}, \xi_{3}\}, ~ \xi_{1}=\frac{d_{1}}{2}-\sigma_{1}^{2}, ~ \xi_{2}=d_{2}+\frac{\delta}{2}-\frac{\rho}{2}-\frac{\sigma_{2}^{2}}{2}, ~ \xi_{3}=d_{3}+\frac{\rho}{2}-\frac{\delta}{2}-\frac{\rho^{2}}{2d_{1}}-\frac{\sigma_{3}^{2}}{2}. $

证 令 $u=S-\frac{A}{d_{1}}$ , $v=I$ , $w=Q$ , 则模型(1.2)改写为

$ \begin{equation}\label{eq6} \left\{ \begin{array}{llll} \mbox{d}u(t)&=&\Big[-d_{1}u-\frac{\beta (u+\frac{A}{d_{1}})v}{1+\alpha_{1}(u+\frac{A}{d_{1}})+\alpha_{2}v}+\gamma v+\rho w\Big]\mbox{d}t \\ &&-\sigma_{1}(u+\frac{A}{d_{1}})\mbox{d}B_{1}(t) -\sigma_{4}\frac{(u+\frac{A}{d_{1}})v}{1+\alpha_{1}(u+\frac{A}{d_{1}})+\alpha_{2}v}\mbox{d}B_{4}(t), \\ \mbox{d}v(t)&=&\Big[\frac{\beta (u+\frac{A}{d_{1}})v}{1+\alpha_{1}(u+\frac{A}{d_{1}})+\alpha_{2}v}-(d_{2}+\gamma+\delta)v\Big]\mbox{d}t \\ &&-\sigma_{2}v\mbox{d}B_{2}(t) +\sigma_{4}\frac{ (u+\frac{A}{d_{1}})v}{1+\alpha_{1}(u+\frac{A}{d_{1}})+\alpha_{2}v}\mbox{d}B_{4}(t), \\ \mbox{d}w(t)&=&[\delta v-(d_{3}+\rho)w]\mbox{d}t-\sigma_{3}w\mbox{d}B_{3}(t). \end{array} \right. \end{equation} $

(3.1)

定义由 ${\cal R}_+^3$ 映射到 ${\cal R}_+^3$ 的三个 $C^{2}$ -函数

$ V_{1}(u, v, w)=\frac{1}{2}(u+v)^{2}, ~ V_{2}(u, v, w)=v, ~ V_{3}(u, v, w)=\frac{1}{2}w^{2}. $

根据高维伊藤公式, 有

$ \begin{eqnarray*} {\cal L}V_{1}(u, v, w)&=&(u+v)[-d_{1}u-(d_{2}+\delta)v+\rho w] +\frac{1}{2}\sigma_{1}^{2}\Big(u+\frac{A}{d_{1}}\Big)^{2}+\frac{1}{2}\sigma_{2}^{2}v^{2}\\ &\leq&-(d_{1}-\sigma_{1}^{2})u^{2}-\Big(d_{2}+\delta-\frac{1}{2}\sigma_{2}^{2}\Big)v^{2}+\rho uw+\rho vw\\ &&-(d_{1}+d_{2}+\delta)uv+\sigma_{1}^{2}\frac{A^{2}}{d_{1}^{2}}\\ &\leq&-\Big(\frac{d_{1}}{2}-\sigma_{1}^{2}\Big)u^{2}-\Big(d_{2}+\delta-\frac{\rho}{2}-\frac{1}{2}\sigma_{2}^{2}\Big)v^{2} +\Big(\frac{\rho^{2}}{2d_{1}}+\frac{\rho}{2}\Big)w^{2}\\ &&-(d_{1}+d_{2}+\delta)uv+\sigma_{1}^{2}\frac{A^{2}}{d_{1}^{2}}, \end{eqnarray*} $

$ \begin{eqnarray*} {\cal L}V_{2}(u, v, w)&=&\Big[\frac{\beta (u+\frac{A}{d_{1}})v}{1+\alpha_{1}(u+\frac{A}{d_{1}})+\alpha_{2}v}-(d_{2}+\gamma+\delta)v\Big] \\ & \leq&\beta uv+ \frac{\beta A}{d_{1}\alpha_{2}}, \\ {\cal L}V_{3}(u, v, w)&=&w[\delta v-(d_{3}+\rho)w]+\frac{1}{2}\sigma_{3}^{2}w^{2} \\ & =&-\Big(d_{3}+\rho-\frac{1}{2}\sigma_{3}^{2}\Big)w^{2}+\delta vw\\ &\leq&-\Big(d_{3}+\rho-\frac{1}{2}\sigma_{3}^{2}-\frac{\delta}{2}\Big)w^{2}+\frac{\delta}{2}v^{2}. \end{eqnarray*} $

构造

$ V(u, v, w)=V_{1}(u, v, w)+\frac{d_{1}+d_{2}+\delta}{\beta}V_{2}(u, v, w)+V_{3}(u, v, w), $

则有

$ \begin{eqnarray}\label{eq7} \mbox{d}V(u, v, w)&=&{\cal L}V(u, v, w)\mbox{d}t-(u+v)\Big[\sigma_{1}\Big(u+\frac{A}{d_{1}}\Big)\mbox{d}B_{1}(t) +\sigma_{2}v\mbox{d}B_{2}(t)\Big]\\ &&-\frac{d_{1}+d_{2}+\delta}{\beta}\Big[\sigma_{2}v\mbox{d}B_{2}(t) +\frac{\sigma_{4} (u+\frac{A}{d_{1}})v}{1+\alpha_{1}(u+\frac{A}{d_{1}})+\alpha_{2}v}\mbox{d}B_{4}(t)\Big]\\ &&-\sigma_{3}w^{2}\mbox{d}B_{3}(t), \end{eqnarray} $

(3.2)

其中

$ \begin{eqnarray*} {\cal L}V(u, v, w) &=& -\Big(\frac{d_{1}}{2}-\sigma_{1}^{2}\Big)u^{2} -\Big(d_{2}+\frac{\delta}{2}-\frac{\rho}{2}-\frac{1}{2}\sigma_{2}^{2}\Big)v^{2} \nonumber\\ && -\Big(d_{3}+\frac{\rho}{2}-\frac{\delta}{2} -\frac{\rho^{2}}{2d_{1}} -\frac{1}{2}\sigma_{3}^{2}\Big)w^{2}+\sigma_{1}^{2}\frac{A^{2}}{d_{1}^{2}}+ \frac{A(d_{1}+d_{2}+\delta)}{d_{1}\alpha_{2}}.\nonumber \end{eqnarray*} $

(3.2)式的两端由0到 $t$ 积分并取期望, 得到

$ \begin{eqnarray*} 0 &\leq & EV(u(t), v(t), w(t))\\ &\leq &V(u(0), v(0), w(0))\\ && + E\int_{0}^{t}\Big(-\xi_{1}u^{2}(s) -\xi_{2}v^{2}(s) -\xi_{3}w^{2}(s)+\sigma_{1}^{2}\frac{A^{2}}{d_{1}^{2}} + \frac{A(d_{1}+d_{2}+\delta)}{d_{1}\alpha_{2}}\Big)\mbox{d}s.\nonumber \end{eqnarray*} $

选择正常数 $\xi=\min\{\xi_{1}, \xi_{2}, \xi_{3}\}$ , 因此, 不等式

$ \begin{equation}\label{eq8} \begin{array}{llll} \limsup\limits_{t\to \infty}\frac{1}{t}\, E\int_{0}^{t}(u^{2}(s)+v^{2}(s)+w^{2}(s))\mbox{d}s \leq\sigma_{1}^{2}\frac{A^{2}}{\xi d_{1}^{2}}+ \frac{A(d_{1}+d_{2}+\delta)}{\xi d_{1}\alpha_{2}} \end{array} \end{equation} $

(3.3)

成立.证毕.

定理3.2 若充分接触率 $\beta$ 满足如下性质

$ \beta^{2} <2\sigma_{4}^{2}\Big(d_{2}+\gamma+\delta+\frac{1}{2}\sigma_{2}^{2}\Big), $

则感染者的密度将几乎处处指数趋于零, 即

$ \begin{eqnarray} \limsup\limits_{t\rightarrow\infty}\frac{\ln I(t)}{t} \leq\frac{\beta^{2}}{2\sigma_{4}^{2}}-\Big(d_{2}+\gamma+\delta+\frac{1}{2}\sigma_{2}^{2}\Big)<0~~\mbox{a.s.}.\nonumber \end{eqnarray} $

证 利用伊藤公式, 我们得到

$ \begin{eqnarray}\label{eq9} \mbox{d}\ln I(t) &=&\frac{\mbox{d}I(t)}{I(t)}-\frac{(\mbox{d}I(t))^{2}}{2I^{2}(t)} \\ &=&\Big[\frac{\beta S(t)}{1+\alpha_{1}S(t)+\alpha_{2}I(t)}-(d_{2}+\gamma+\delta)\Big]\mbox{d}t\nonumber\\ &&-\sigma_{2}\mbox{d}B_{2}(t)+\sigma_{4}\frac{S(t)}{1+\alpha_{1}S(t)+\alpha_{2}I(t)}\mbox{d}B_{4}(t)\nonumber\\ &&-\frac{1}{2I^{2}(t)}\Big[\sigma_{2}^{2}I^{2}(t)+\sigma_{4}^{2}\frac{S^{2}(t)I^{2}(t)}{(1+\alpha_{1}S(t)+\alpha_{2}I(t))^{2}}\Big]\mbox{d}t\nonumber\\ &=&\Big[\frac{\beta S(t)}{1+\alpha_{1}S(t)+\alpha_{2}I(t)} -\frac{\sigma_{4}^{2}S^{2}(t)}{2(1+\alpha_{1}S(t)+\alpha_{2}I(t))^{2}}\Big]\mbox{d}t\\ &&-\Big(d_{2}+\gamma+\delta+\frac{1}{2}\sigma_{2}^{2}\Big)\mbox{d}t-\sigma_{2}\mbox{d}B_{2}(t)\\ &&+\sigma_{4}\frac{S(t)}{1+\alpha_{1}S(t)+\alpha_{2}I(t)}\mbox{d}B_{4}(t), \end{eqnarray} $

(3.4)

(3.4)式两侧由 $0$ 到 $t$ 积分, 得到

$ \begin{eqnarray*} \ln I(t) &=& \int_{0}^{t}\frac{\beta S(u)}{1+\alpha_{1}S(u)+\alpha_{2}I(u)}\mbox{d}u -\frac{1}{2}\int_{0}^{t}\frac{\sigma_{4}^{2}S^{2}(u)}{(1+\alpha_{1}S(u)+\alpha_{2}I(u))^{2}}\mbox{d}u\nonumber\\ &&+\ln I(0) -\Big(d_{2}+\gamma+\delta+\frac{1}{2}\sigma_{2}^{2}\Big)t-\sigma_{2}B_{2}(t)+M(t), \nonumber \end{eqnarray*} $

其中局部鞅为

$ M(t)=\int_{0}^{t}\frac{\sigma_{4}S(u)}{1+\alpha_{1}S(u)+\alpha_{2}I(u)}\mbox{d}B_{4}(u), \nonumber $

其二次变分为

$ \langle M(t), M(t)\rangle =\int_{0}^{t} \Big[\frac{\sigma_{4}S(u)}{1+\alpha_{1}S(u)+\alpha_{2}I(u)}\Big]^{2}\mbox{d}u.\nonumber $

我们选择 $\delta=2, \nu_{k}=\nu>0$ 和 $\tau_{k}=k$ (见文献[5, 定理7.4]), 对几乎所有的 $\omega\in\Omega$ , 存在一个随机整数 $k_{0}(\omega)$ 使得对 $k>k_{0}(w)$ 及 $t\in[0, k]$ , 我们有

$ \begin{eqnarray}\label{eq10} M(t)-\frac{1}{2}\langle M(t), M(t)\rangle \leq-\frac{1-\nu}{2}\int_{0}^{t} \Big[\frac{\sigma_{4}S(u)}{1+\alpha_{1}S(u)+\alpha_{2}I(u)}\Big]^{2}\mbox{d}u+\frac{2}{\nu}\ln{k}. \end{eqnarray} $

(3.5)

另外, 由基本不等式 $(a+b)^2\leq2a^2+2b^2$ , 有

$ \begin{eqnarray}\label{eq11} \frac{\beta S(u)}{1+\alpha_{1}S(u)+\alpha_{2}I(u)}-\frac{1-\nu}{2}\Big[\frac{\sigma_{4}S(u)}{1+\alpha_{1}S(u)+\alpha_{2}I(u)}\Big]^{2} \leq\frac{\beta^{2}}{2(1-\nu)\sigma_{4}^{2}}. \end{eqnarray} $

(3.6)

对 $k-1\leq t\leq k$ , 由表达式( $3.5$ )和( $3.6$ ), 得到

$ \begin{eqnarray} \frac{\ln I(t)}{t} \leq\frac{\ln I(0)}{t} +\Big[\frac{\beta^{2}}{2(1-\nu)\sigma_{4}^{2}}-\Big(d_{2}+\gamma+\delta+\frac{1}{2}\sigma_{2}^{2}\Big)\Big] -\sigma_{2}\frac{B_2(t)}{t}+\frac{2\ln{k}}{\nu(k-1)}.\nonumber \end{eqnarray} $

令 $t\rightarrow\infty$ , 根据文献[5, 引理2.6]知,

$ \frac{B_{2}(t)}{t}\to 0, ~~\frac{\ln{k}}{k-1}\to 0. $

当 $\nu$ 趋于0时, 定理结论成立.证毕.

例3.3 考虑模型

$ \begin{equation}\label{eq12} \left\{ \begin{array}{llll} \mbox{d}S(t) &=& \Big[0.1-0.1S(t)-\frac{0.2S(t)I(t)}{1+S(t)+I(t)}+0.1 I(t)+0.004 Q(t)\Big]\mbox{d}t\\ &&-0.2 S(t)\mbox{d}B_{1}(t) -\frac{0.4 S(t)I(t)}{1+S(t)+I(t)}\mbox{d}B_{4}(t), \\ \mbox{d}I(t) &=& \Big[\frac{0.2 S(t)I(t)}{1+S(t)+I(t)}-(0.3+0.1+0.1)I(t)\Big]\mbox{d}t\\ &&-0.7 I(t)\mbox{d}B_{2}(t) +\frac{0.4 S(t)I(t)}{1+S(t)+I(t)}\mbox{d}B_{4}(t), \\[2mm] \mbox{d}Q(t) &=& [0.1 I(t)-(0.2+0.004)Q(t)]\mbox{d}t-0.8Q(t)\mbox{d}B_{3}(t). \end{array} \right. \end{equation} $

(3.7)

根据Milstein高阶方法^[6], 我们选择初值 $(S(0), I(0), Q(0))=(2.4, 2.4, 1.2)$ , 定理3.1-3.2的条件均满足, 且无病平衡点为 $(\frac{A}{d_1}, 0, 0)=(1, 0, 0)$ , 模型(1.2)的解的绝灭性如图 2-3所示.

对于易感者-感染者-隔离者模型(1.2), 其地方病平衡点为 $E^{\ast}=(S^{\ast}, I^{\ast}, Q^{\ast})$ , 对应的分量分别为

$ \begin{eqnarray} & I^{\ast}= \frac{A\beta-(\alpha_{1}A+d_{1})(d_{2}+\gamma+\delta)}{(d_{2}+\gamma+\delta) [d_{1}\alpha_{2}-d_{2}\alpha_{1}-\alpha_{1}\delta+\alpha_{1}\rho\delta(d_3+\rho)^{-1}]+\beta[d_{2}+\delta-\rho\delta(d_3+\rho)^{-1}]}, \nonumber\\[10pt] & S^{\ast} =\frac{(d_{2}+\gamma+\delta)(\alpha_{2}I^{\ast}+1)}{\beta-\alpha_{1} (d_{2}+\gamma+\delta)}, ~~ Q^{\ast}= \frac{\delta}{d_{3}+\rho}I^{\ast}.\nonumber \end{eqnarray} $

在适当的充分条件下, 本节将讨论模型(1.2)解的平稳分布的存在性, 及该解的遍历性(见定理4.1).若地方病平衡点 $E^{\ast}$ 是稳定的, 则进一步研究解的正态分布(见定理4.2).

定理4.1 若白噪声的强度满足

$ \sigma_{1}^{2}<d_{1}, ~ \sigma_{2}^{2}<d_{2}, ~ \sigma_{3}^{2}<d_{3} $

且

$ \eta < \min\{\eta_{1}S^{\ast 2}, \eta_{2}I^{\ast 2}, \eta_{3}Q^{\ast 2}\}, $

则模型 $(1.2)$ 存在一个平稳分布, 且该解是遍历的, 其中

$ \begin{eqnarray*} &&\eta_{1}=(c_{1}+c_{4})d_{1}-(c_{1}+c_{4})\sigma_{1}^{2}, \nonumber\\ &&\eta_{2}=(c_{1}+c_{4})d_{2}-(c_{1}+c_{4})\sigma_{2}^{2}+c_{4}\delta, \nonumber\\ &&\eta_{3}=(c_{1}+c_{2})d_{3}-(c_{1}+c_{2})\sigma_{3}^{2}+c_{2}\rho, \nonumber\\ &&\eta=(c_{1}+c_{4})S^{\ast2}\sigma_{1}^{2}+\Big(c_{1}+c_{4}+\frac{c_{3}}{2I^{\ast}}\Big)I^{\ast 2}\sigma_{2}^{2}+ (c_{1}+c_{2})Q^{\ast2}\sigma_{3}^{2}+\frac{c_{3}}{2\alpha_{1}^{2}}I^{\ast}\sigma_{4}^{2}, \nonumber \\ &&c_{1}=\rho, ~ c_{2}=\frac{\rho}{\delta}(d_{2}-d_{1}), ~\nonumber\\ &&c_{3}=\frac{1+\alpha_{1}S^{\ast}+\alpha_{2}I^{\ast}}{\beta}[\rho(d_{1}+d_{2})+(d_{1}+d_{3})(d_{1}+d_{2}+\delta)], ~ c_{4}=d_{1}+d_{3}.\nonumber \end{eqnarray*} $

证 地方病平衡点 $(S^{\ast}, I^{\ast}, Q^{\ast})$ 满足等式

$ A=d_{1}S^{\ast}+\frac{\beta S^{\ast}I^{\ast}}{1+\alpha_{1}S^{\ast}+\alpha_{2}I^{\ast}} -\gamma I^{\ast}-\rho Q^{\ast}, $

$ \frac{\beta S^{\ast}I^{\ast}}{1+\alpha_{1}S^{\ast}+\alpha_{2}I^{\ast}}=(d_{2}+\gamma+\delta)I^{\ast}, ~~~\delta I^{\ast}=(d_{3}+\rho)Q^{\ast}. $

定义由 ${\cal R}_{+}^{3}$ 映射到 ${\cal R}_{+}$ 的 $C^{2}$ -函数

$ \begin{equation}\label{eq13} \begin{array}{lllll} & V_1(S, I, Q)=\frac{1}{2}(S-S^{\ast} +I-I^{\ast}+Q-Q^{\ast})^{2}, ~~~ V_2(S, I, Q)=\frac{1}{2}(Q-Q^{\ast})^{2}, \\ & V_3(S, I, Q)= I-I^{\ast}-I^{\ast}\ln\frac{I}{I^{\ast}}, ~~~ V_4(S, I, Q)=\frac{1}{2}(S-S^{\ast}+I-I^{\ast})^{2}, \end{array} \end{equation} $

(4.1)

根据高维伊藤公式, 得到

$ \begin{eqnarray} {\cal L}V_{1}(S, I, Q) &\leq&-(d_{1}-\sigma_{1}^{2})(S-S^{\ast})^{2}-(d_{2}-\sigma_{2}^{2})(I-I^{\ast})^{2}-(d_{3}-\sigma_{3}^{2})(Q-Q^{\ast})^{2}\\ &&-(d_{1}+d_{2})(S-S^{\ast})(I-I^{\ast})-(d_{1}+d_{3})(S-S^{\ast})(Q-Q^{\ast})\\ &&-(d_{2}+d_{3})(I-I^{\ast})(Q-Q^{\ast})+\sigma_{1}^{2}S^{\ast2}+\sigma_{2}^{2}I^{\ast 2}+\sigma_{3}^{2}Q^{\ast2}, \end{eqnarray} $

(4.2)

$ \begin{eqnarray} {\cal L}V_{2}(S, I, Q) &=&(Q-Q^{\ast})[\delta (I-I^{\ast})-(d_{3}+\rho)(Q-Q^{\ast})]+\frac{1}{2}\sigma_{3}^{2}(Q-Q^{\ast}+Q^{\ast})^{2} \\ &\leq&-(d_{3}+\rho-\sigma_{3}^{2})(Q-Q^{\ast})^{2}+\delta(I-I^{\ast})(Q-Q^{\ast})+\sigma_{3}^{2}Q^{\ast2}, \end{eqnarray} $

(4.3)

$ \begin{eqnarray} {\cal L}V_{3}(S, I, Q) &=&(I-I^{\ast})\Big[\frac{\beta S}{1+\alpha_{1}S+\alpha_{2}I}-\frac{\beta S}{1+\alpha_{1}S^{\ast}+\alpha_{2}I^{\ast}} \\ &&+\frac{\beta}{1+\alpha_{1}S^{\ast}+\alpha_{2}I^{\ast}}(S-S^{\ast})\Big] +\frac{1}{2}\sigma_{2}^{2}I^{\ast} +\frac{1}{2}\sigma_{4}^{2}\frac{S^{2}I^{\ast}}{(1+\alpha_{1}S+\alpha_{2}I)^{2}}\\ &\leq&\frac{\beta}{1+\alpha_{1}S^{\ast}+\alpha_{2}I^{\ast}}(S-S^{\ast})(I-I^{\ast}) +\frac{1}{2}\Big(\sigma_{2}^{2}+\frac{\sigma_{4}^{2}}{\alpha_{1}^{2}}\Big)I^{\ast}, \end{eqnarray} $

(4.4)

$ \begin{eqnarray} {\cal L}V_{4}(S, I, Q) &\leq&(S-S^{\ast}+I-I^{\ast})[A-d_{1}S-(d_{2}+\delta)I+\rho Q] +\frac{1}{2}\sigma_{1}^{2}S^{2}+\frac{1}{2}\sigma_{2}^{2}I^{2}\\ &\leq&-(d_{1}-\sigma_{1}^{2})(S-S^{\ast})^{2}-(d_{2}+\delta-\sigma_{2}^{2})(I-I^{\ast})^{2}\\ &&-(d_{1}+d_{2}+\delta)(S-S^{\ast})(I-I^{\ast}) +\rho(S-S^{\ast})(Q-Q^{\ast})\\ &&+\rho(I-I^{\ast})(Q-Q^{\ast})+\sigma_{1}^{2}S^{\ast2}+\sigma_{2}^{2}I^{\ast 2}. \end{eqnarray} $

(4.5)

于是, 由表达式(4.2)-(4.5), 得到

$ \begin{eqnarray}\label{eq18} {\cal L}V(S, I, Q)&=&c_1{\cal L}V_1(S, I, Q)+c_2{\cal L}V_2(S, I, Q)+c_3{\cal L}V_3(S, I, Q)+c_4{\cal L}V_4(S, I, Q)\\ &\leq&-(c_{1}+c_{4})(d_{1}-\sigma_{1}^{2})(S-S^{\ast})^{2} -[c_{1}(d_{2}-\sigma_{2}^{2})+c_{4}(d_{2}+\delta-\sigma_{2}^{2})](I-I^{\ast})^{2}\\ &&-[c_{1}(d_{3}-\sigma_{3}^{2})+c_{2}(d_{3}+\rho-\sigma_{3}^{2})](Q-Q^{\ast})^{2}\\ &&+\Big[-c_{1}(d_{1}+d_{2})-c_{4}(d_{1}+d_{2}+\delta) +\frac{c_{3}\beta}{1+\alpha_{1}S^{\ast}+\alpha_{2}I^{\ast}}\Big](S-S^{\ast})(I-I^{\ast})\\ &&+[-c_{1}(d_{2}+d_{3})+c_{2}\delta+c_{4}\rho](I-I^{\ast})(Q-Q^{\ast})\\ && +[-c_{1}(d_{1}+d_{3})+c_{4}\rho](S-S^{\ast})(Q-Q^{\ast})\\ && +c_{1}(\sigma_{1}^{2}S^{\ast2}+\sigma_{2}^{2}I^{\ast 2}+\sigma_{3}^{2}Q^{\ast2})+c_{2}\sigma_{3}^{2}Q^{\ast2}\\ && +c_{4}(\sigma_{1}^{2}S^{\ast2}+\sigma_{2}^{2}I^{\ast 2})+ \frac{c_{3}}{2} \Big(\sigma_{2}^{2}+\frac{\sigma_{4}^{2}}{\alpha_{1}^{2}}\Big)I^{\ast}. \end{eqnarray} $

(4.6)

选择

$ c_{1}=\rho, c_{2}=\frac{\rho}{\delta}(d_{2}-d_{1}), $

$ c_{3}=\frac{1+\alpha_{1}S^{\ast}+\alpha_{2}I^{\ast}}{\beta}[\rho(d_{1}+d_{2})+(d_{1}+d_{3})(d_{1}+d_{2}+\delta)], ~~c_{4}=d_{1}+d_{3}, $

使得

$ {\cal L}V(S, I, Q) \leq -\eta_{1}(S-S^{\ast})^{2}-\eta_{2}(I-I^{\ast})^{2}-\eta_{3}(Q-Q^{\ast})^{2}+\eta. $

由文献[15, 引理3.1.2], 结合定理4.1的条件, 椭球

$ \eta_{1}(S-S^{\ast})^{2}+\eta_{2}(I-I^{\ast})^{2} +\eta_{3}(Q-Q^{\ast})^{2}=\eta $

位于区域 ${\cal R}_{+}^{3}$ 内, 且

$ {\cal L}V(S, I, Q)\leq -c<0 $

成立, 其中 $U$ 为椭球的某邻域, $\bar{U}\subseteq {\cal R}_{+}^{3}$ , $c$ 是一个正常数且 $(S, I, Q)\in {\cal R}_{+}^{3}\setminus U$ .

模型(1.2)在地方病平衡点 $(S^{\ast}, I^{\ast}, Q^{\ast})$ 的扩散矩阵为

$ \begin{eqnarray*} A&=&(a_{ij})_{3\times 3}\\ &=&\left( \begin{array}{ccc} \sigma_{1}^{2}S^{\ast2}+\sigma_{4}^{2}\frac{S^{\ast2}I^{\ast2}}{(1+\alpha_{1}S^{\ast}+\alpha_{2}I^{\ast})^{2}} & -\sigma_{4}^{2}\frac{S^{\ast2}I^{\ast2}}{(1+\alpha_{1}S^{\ast}+\alpha_{2}I^{\ast})^{2}}& 0\\ -\sigma_{4}^{2}\frac{S^{\ast2}I^{\ast2}}{(1+\alpha_{1}S^{\ast}+\alpha_{2}I^{\ast})^{2}} & ~~\sigma_{2}^{2}I^{\ast2}+\sigma_{4}^{2}\frac{S^{\ast2}I^{\ast2}}{(1+\alpha_{1}S^{\ast}+\alpha_{2}I^{\ast})^{2}} ~~& 0 \\[2mm] 0&0&\sigma_{3}^{2}Q^{\ast2} \end{array}\right), \end{eqnarray*} $

于是

$ \begin{eqnarray*} \sum\limits_{i, j=1}^{3}a_{ij}\lambda_{i}\lambda_{j} &=&\sigma_{1}^{2}S^{\ast2}\lambda_{1}^{2}+\sigma_{2}^{2}I^{\ast2}\lambda_{2}^{2} +\sigma_{3}^{2}Q^{\ast2}\lambda_{3}^{2} +\sigma_{4}^{2}\frac{S^{\ast2}I^{\ast2}}{(1+\alpha_{1}S^{\ast}+\alpha_{2}I^{\ast})^{2}} (\lambda_1-\lambda_2)^2 \\ %&&+2\sigma_{1}^{2}S^{\ast}Q^{\ast}\lambda_{1}\lambda_{3} +2\sigma_{1}^{2}I^{\ast}Q^{\ast}\lambda_{2}\lambda_{3}+\sigma_{2}^{2}I^{\ast2}(\lambda_{2}-\lambda_{3})^{2}\\ &\geq&\sigma_{1}^{2}S^{\ast2}\lambda_{1}^{2} +\sigma_{2}^{2}I^{\ast2}\lambda_{2}^{2} +\sigma_{3}^{2}Q^{\ast2}\lambda_{3}^{2}\geq M|\lambda|^{2}, \end{eqnarray*} $

对固定的常数 $M=\min\{\sigma_{1}^{2}S^{\ast2}, \sigma_{2}^{2}I^{\ast2}, \sigma_{3}^{2}Q^{\ast2}\}>0$ 及任意的 $U\subset{\cal R}_{+}^{3}$ , $\lambda\in{\cal R}_{+}^{3}$ .证毕.

定理4.2 对任意初值 $(S(0)$ , $I(0)$ , $Q(0))\in {\cal R}_{+}^{3}$ , 若地方病平衡点 $(S^{\ast}, I^{\ast}, Q^{\ast})$ 是稳定的, 则模型 $(1.2)$ 的解渐近服从一个三维正态分布, 其均值为 $(S^{\ast}, I^{\ast}, Q^{\ast})$ , 方差为

$ C_{(S, I, Q)}(0)= \frac{1}{2\pi}\int_{-\infty}^{+\infty} [\mu'(X^{\ast})-{\rm i}wI]^{-1}\sigma(X^{\ast})\sigma^{\mbox{ T}}(X^{\ast}) [(\mu'(X^{\ast})+{\rm i}wI)^{\mbox{ T}}]^{-1}\mbox{d}w, $

其中 $\mu'(X^{\ast})$ 和 $\sigma'(X^{\ast})$ 见(4.7)和(4.8)式在 $X^{\ast}$ 点的值.

证 令 $X(t)=(S(t), I(t), Q(t))^{\mbox{ T}}$ 和 $\mbox{d}B(t)=(\mbox{d}B_1(t), \mbox{d}B_2(t), \mbox{d}B_3(t), \mbox{d}B_4(t))^{\mbox{ T}}$ , 模型(1.2)可改写为

$ \begin{equation}\label{eq19} \mbox{d}X(t)=\mu(X(t))\mbox{d}t+\sigma(X(t))\mbox{d}B(t), \end{equation} $

(4.7)

其中

$ \mu(X(t)) = \left( \begin{array}{c} A-d_{1}S(t)-\frac{\beta S(t)I(t)}{1+\alpha_{1}S(t)+\alpha_{2}I(t)}+\gamma I(t)+\rho Q(t)\\ \frac{\beta S(t)I(t)}{1+\alpha_{1}S(t)+\alpha_{2}I(t)}-(d_{2}+\gamma+\delta)I(t)\\ [2mm] \delta I(t)-(d_{3}+\rho)Q(t) \end{array} \right), $

$ \sigma(X(t))=\left( \begin{array}{ c c c c } -\sigma_{1}S(t)&0&0 & -\sigma_{4}\frac{S(t)I(t)}{1+\alpha_{1}S(t)+\alpha_{2}I(t)}\\ 0&-\sigma_{2}I(t)&0 & \sigma_{4}\frac{S(t)I(t)}{1+\alpha_{1}S(t)+\alpha_{2}I(t)}\\[2mm] 0&0&-\sigma_{3}Q(t) &0 \end{array}\right). $

令 $u(t)=X(t)-X^{\ast}$ , 沿地方病平衡点 $X^{\ast}$ 作泰勒展开, (4.7)式成为

$ \begin{eqnarray}\label{eq20} \mbox{d}u(t) &=& [\mu(X^{\ast}) +\mu'(X^{\ast})u(t)+\mbox{o}(|u(t)|^{2})]\mbox{d}t\\ &&+[\sigma(X^{\ast})+\sigma'(X^{\ast})\mbox{diag}(u(t)) +\mbox{o}(\mbox{diag}(u(t)))]\mbox{d}B(t), \end{eqnarray} $

(4.8)

其中

$ \mu'(X^{\ast})= \left( \begin{array}{cccc} -\frac{\beta I^{\ast}+\beta\alpha_{2}I^{\ast2}}{(1+\alpha_{1}S^{\ast}+\alpha_{2}I^{\ast})^{2}}-d_{1} &~~ -\frac{\beta S^{\ast}+\beta\alpha_{1}S^{\ast2}}{(1+\alpha_{1}S^{\ast}+\alpha_{2}I^{\ast})^{2}}+\gamma ~~& \rho\\ \frac{\beta I^{\ast}+\beta\alpha_{2}I^{\ast2}}{(1+\alpha_{1}S^{\ast}+\alpha_{2}I^{\ast})^{2}} & \frac{\beta S^{\ast}+\beta\alpha_{1}S^{\ast2}}{(1+\alpha_{1}S^{\ast}+\alpha_{2}I^{\ast})^{2}}-(d_{2}+\gamma+\delta) & 0\\[2mm] 0&\delta&-(d_{3}+\rho) \end{array}\right). $

由于 $X^{\ast}$ 是稳定的, $\mu(X^{\ast})=0$ , $\mu'(X^{\ast})<0$ .当 $X$ 沿着 $X^{\ast}$ 扰动, 且白噪声的强度 $\sigma_{i}$ $(i=1, 2, 3, 4)$ 充分小时, $\sigma'(X^{\ast})\mbox{diag}(u)$ 可以忽略不计.因此, (4.8)式的估计式

$ \begin{equation}\label{eq21} \begin{array}{ll} \mbox{d}u(t)-\mu'(X^{\ast})u(t)\mbox{d}t =\sigma(X^{\ast})\mbox{d}B(t) \end{array} \end{equation} $

(4.9)

为一个三维的O-U过程(细节见文献[3]的第6章和第11章), 方程(4.9)的解 $u(t)$ 服从正态分布 ${\cal N}(0, C_{u}(0))$ , 其中 $C_{u}(0)$ 为方差矩阵.根据自协方差 $C_{u, u}(\tau)$ 的如下性质

$ C_{\frac{du}{dt}, \frac{du}{dt}}(\tau)=-C''_{u, u}(\tau), ~~C_{u, \frac{du}{dt}}(\tau)=-C'_{u, u}(\tau), ~~C_{\frac{du}{dt}, u}(\tau)=C'_{u, u}(\tau), $

方程(4.9)的两端同时取自协方差运算, 得到

$ \begin{eqnarray}\label{eq22} &&\sigma(X^{\ast})\sigma^{\mbox{ T}}(X^{\ast})\delta I\\ &=& -C''_{u, u}(\tau)+\mu'(X^{\ast})C_{u, u}(\tau)\mu'^{\mbox{ T}}(X^{\ast})+\mu'(X^{\ast})C'_{u, u}(\tau)-C'_{u, u}(\tau)\mu'^{\mbox{ T}}(X^{\ast}), \end{eqnarray} $

(4.10)

其中 $C'_{u, u}(\tau)$ 和 $C''_{u, u}(\tau)$ 分别表示自协方差关于 $\tau$ 的一阶、二阶导数; $\delta$ 是一个Dirac-Delta函数, $I$ 是三阶单位阵, $\sigma(X^{\ast})\sigma^{\mbox{ T}}(X^{\ast})$ 是扩散矩阵.

方程(4.10)的两端取傅里叶变换, 向量 $u$ 的谱密度矩阵 $S_{u}(w)$ 满足方程

$ \begin{eqnarray*} &&\sigma(X^{\ast})\sigma^{\mbox{ T}}(X^{\ast})\delta I\\ &=& -({\rm i}w)^{2}S_{u}(w)+\mu'(X^{\ast})S_{u}(w)\mu'^{\mbox{ T}}(X^{\ast})+{\rm i}w\mu'(X^{\ast})S_{u}(w)-{\rm i}wS_{u}(w)\mu'^{\mbox{ T}}(X^{\ast}). \end{eqnarray*} $

由于 $\mu'(X^{\ast})\pm {\rm i}wI$ 是可逆矩阵, 则有

$ S_{u}(w)=(\mu'(X^{\ast})-{\rm i}wI)^{-1}\sigma(X^{\ast})\sigma^{\mbox{ T}}(X^{\ast}) [(\mu'(X^{\ast})+{\rm i}wI)^{\mbox{ T}}]^{-1}. $

经过傅里叶逆变换, 令 $\tau=0$ , 我们有

$ C_{u}(0) = \frac{1}{2\pi}\int_{-\infty}^{+\infty} [\mu'(X^{\ast})-{\rm i}wI]^{-1}\sigma(X^{\ast})\sigma^{\mbox{ T}}(X^{\ast}) [(\mu'(X^{\ast})+{\rm i}wI)^{\mbox{ T}}]^{-1}\mbox{d}w. $

事实上, $C_{X^{\ast}}(0)=0$ 推出 $C_{(S, I, Q)}(0)=C_{u}(0)$ .定理的结论成立.证毕.

例4.3 考虑模型

$ \begin{equation}\label{eq23} \left\{ \begin{array}{llll} \mbox{d}S(t) &=&\Big[6-0.016S(t)-\frac{0.4S(t)I(t)}{1+S(t)+I(t)}+0.001 I(t)+0.001 Q(t)\Big]\mbox{d}t\\ && -0.003 S(t)\mbox{d}B_{1}(t)-\frac{0.002 S(t)I(t)}{1+S(t)+I(t)}\mbox{d}B_{4}(t), \\ \mbox{d}I(t) &=&\Big[\frac{0.4 S(t)I(t)}{1+S(t)+I(t)}-(0.018+0.001+0.001)I(t)\Big]\mbox{d}t\\ && -0.01 I(t)\mbox{d}B_{2}(t)+\frac{0.002 S(t)I(t)}{1+S(t)+I(t)}\mbox{d}B_{4}(t), \\[2mm] \mbox{d}Q(t) &=&[0.001 I(t)-(0.022+0.001)Q(t)]\mbox{d}t-0.004Q(t)\mbox{d}B_{3}(t). \end{array} \right. \end{equation} $

(4.11)

在模型(4.11)中取初值 $(S(0), I(0), Q(0))=(1, 0.02, 0.03)$ , 易于验证定理4.1的条件成立, 于是, 地方病平衡点为 $(S^{\ast}, I^{\ast}, Q^{\ast})=(16.7, 318, 13.8)$ . 图 4说明正解在地方病平衡点 $E^{\ast}$ 附近存在一个平稳分布; 图 5-7分别显示了易感者、感染者、隔离者的频率直方图.而且, 在一些充分的条件下, 我们得到了模型解的正态分布.

本文中, 我们研究了一类具有饱和传染率的随机流行病模型, 通过构建 $C^2$ -函数及应用伊藤公式, 得到了模型(1.2)存在唯一全局解这一结论.同时, 研究表明:当白噪声的强度 $\sigma_i$ $ (i=1, 2, 3, 4)$ 充分大的时候, 疾病在模型(1.2)的无病平衡点 $E_0$ 附近趋于绝灭, 且感染者的密度将指数趋于零(见图 2-3).当白噪声的强度 $\sigma_i$ 足够小的时候, 模型(1.2)的正解沿地方病平衡点 $E^{\ast}$ 服从唯一的平稳分布, 相关的模拟图如图 4-7所示.若模型(1.2)的地方病平衡点 $E^{\ast}$ 是稳定的, 在一定的充分条件下, 模型(1.2)的解将渐近服从一个三维正态分布, 且得到了均值与方差的表达式.数值模拟的结果支持了定理的主要结论, 并对模型给出了较好的解释, 即, 在长期动力学行为中, 图 2-3意味着疾病将沿无病平衡点 $E_0$ 附近绝灭, 图 4-7展示了疾病将沿地方病平衡点附近传播.