众所周知,接种疫苗是一种常见疾病控制方法,比如天花通过接种疫苗的措施在自然界已经基本消失.近年来,学者们广泛研究了接种疫苗在抑制传染病传播中的作用,参见文献[1, 8, 10-11].在包含接种疫苗的传染病模型中,大部分假设已经接种疫苗的个体不会被感染,但是有一些疫苗的作用只是为了减少接种疫苗的个体被感染的可能性,参见文献[15-16]. 2003年, Gumel和Moghadas提出了一类带接种和非线性发生率的传染病模型,参见文献[3].此外,许多传染病在个体被感染后和个体具有感染能力之间有一段时间的潜伏期,参见文献[4-7, 9, 12-14, 18].例如,狂犬病的潜伏期为数月至数年.

通常在传染病模型中考虑潜伏期时,引入变量感染持续时间$ \theta $. $ y(t, \theta) $表示$ t $时刻感染时间为$ \theta $的被感染者的人口密度. $ S(t), V(t), E(t), I(t) $和$ R(t) $分别表示易感者类,接种者类,潜伏者类,染病者类和恢复者类在$ t $时刻的人口数量.当$ t\in[0, \tau] $有

(1.1) $ \begin{eqnarray} \frac{\partial y(t, \theta)}{\partial t}+\frac{\partial y(t, \theta)}{\partial \theta} = -\mu y(t, \theta), \end{eqnarray} $

其中$ \mu $表示人口的自然死亡率.假设感染个体的平均潜伏期为$ \tau $,即易感者个体或者接种者个体被感染者感染后经过$ \tau $时间后才具有感染性,进入染病者类$ I(t) $.换而言之,潜伏者个体在被感染后的$ \tau $时间内无法感染其他易感者个体,潜伏者类人口总数可由等式$ E(t) = \int_{0}^{\tau}y(t, \theta){\rm d}\theta $表示.对方程(1.1)两边分别从0到$ \tau $积分,得

(1.2) $ \begin{equation} \frac{{\rm d}E(t)}{{\rm d}t} = -\mu E(t)+y(t, 0)-y(t, \tau). \end{equation} $

考虑到模型的现实意义,假设$ y(t, \infty) = 0 $.易感者个体或接种者个体和染病者个体接触后才被感染,我们用饱和发生率刻画由于接触增加的感染时间为$ 0 $的人口密度

(1.3) $ \begin{equation} y(t, 0) = \big[\beta_{1}S(t)+\beta_{2}V(t)\big]g(I(t)), \end{equation} $

这里$ g(I) = \frac{I}{1+\alpha I} $.进而由方程(1.1)可得

(1.4) $ \begin{equation} y(t, \tau) = y(t-\tau, 0)e^{-\mu\tau} = [\beta_{1}S(t-\tau)+\beta_{2}V(t-\tau)]g(I(t-\tau))e^{-\mu\tau}, \end{equation} $

其中$ e^{-\mu\tau} $代表潜伏期类的存活率, $ \beta_{1} $和$ \beta_{2} $分别是易感者类和接种者类的传播率,特别地, $ \beta_{1}<\beta_{2} $.将等式(1.3)和(1.4)带入等式(1.2),得

(1.5) $ \begin{equation} \frac{{\rm d}E(t)}{{\rm d}t} = (\beta_1S+\beta_2V)g(I)-e^{-\mu \tau}\big(\beta_1S(t-\tau)+\beta_2V(t-\tau)\big)g(I(t-\tau))-\mu E(t). \end{equation} $

根据上述的分析,本文提出一类具有接种和非线性感染率的时滞传染病模型

(1.6) $ \begin{equation} \left\{\begin{array}{ll} \frac{{\rm d}S(t)}{{\rm d}t} = \pi-(\mu+\xi)S-\beta_1Sg(I), \\ \frac{{\rm d}V(t)}{{\rm d}t} = \xi S-\mu V-\beta_2Vg(I), \\ \frac{{\rm d}E(t)}{{\rm d}t} = (\beta_1S+\beta_2V)g(I)-e^{-\mu \tau}\big(\beta_1S(t-\tau)+\beta_2V(t-\tau)\big)g(I(t-\tau))-\mu E, \\ \frac{{\rm d}I(t)}{{\rm d}t} = e^{-\mu \tau}\big(\beta_1S(t-\tau)+\beta_2V(t-\tau)\big)g(I(t-\tau))-(\mu+\gamma) I, \\ \frac{{\rm d}R(t)}{{\rm d}t} = \gamma I -\mu R, \end{array}\right. \end{equation} $

这里$ \pi $表示易感者类的增加率, $ \xi $是易感者类的接种率, $ \gamma $是染病者类的恢复率.其他参数的意义和前文提到的意义一致,并且所有参数都是正数.显而易见$ E(t) $和$ R(t) $在模型(1.6)中独立于其他方程,我们只需要研究如下子系统的性质即可

(1.7) $ \begin{equation} \left\{\begin{array}{ll} \frac{{\rm d}S(t)}{{\rm d}t} = \pi-(\mu+\xi)S-\beta_1Sg(I), \\ \frac{{\rm d}V(t)}{{\rm d}t} = \xi S-\mu V-\beta_2Vg(I), \\ \frac{{\rm d}I(t)}{{\rm d}t} = e^{-\mu \tau}\big(\beta_1S(t-\tau)+\beta_2V(t-\tau)\big)g(I(t-\tau))-(\mu+\gamma) I. \end{array}\right. \end{equation} $

考虑模型的现实意义,模型(1.7)的初值条件如下

(1.8) $ \begin{eqnarray} \left\{\begin{array}{ll} S(\theta) = \varphi_1(\theta)\geq0, \ \ \ S(0)>0, \\ V(\theta) = \varphi_2(\theta)\geq0, \\ I(\theta) = \varphi_3(\theta)\geq0, \end{array}\right. \end{eqnarray} $

其中$ \theta \in [-\tau, 0] $.定义$ \Phi = (\varphi_1, \varphi_2, \varphi_3) $表示在Banach空间中从$ [-\tau, 0] $映射到$ {\Bbb R}_+^3 $中的连续函数,这里$ {\Bbb R}_+^3 = \{(x_1, x_2, x_3):x_i\geq 0, \ i = 1, 2, 3\} $.

本文的其余部分安排如下.在第二节中,得到模型的基本再生数,并研究了地方病平衡点的存在性和系统解的非负性以及有界性.第三节利用特征方程研究了两个平衡点的局部稳定性.在第四节中,我们讨论了系统(1.7)的一致持续性.在第五节中,通过构建合适的Lyapunov泛函和应用Lyapunov-LaSalle不变性原理,我们分别获得了无病平衡点和地方病平衡点的全局渐近稳定性.最后,我们给出了一些数值拟合对理论结果进行了检验.

易知模型(1.7)总存在一个无病平衡点$ E_0(S_0, V_0, 0) $,这里$ S_0 = \frac{\pi}{\mu+\xi} $, $ V_0 = \frac{\pi \xi}{(\mu+\xi)\mu} $.下面证明模型(1.7)唯一地方病平衡点的存在性.运用文献[2]中的方法,可得模型(1.7)的基本再生数为

(2.1) $ \begin{equation} R_0 = e^{-\mu \tau}\pi \frac{\beta_1\mu+\beta_2\xi}{\mu(\mu+\xi)(\mu+\gamma)}. \end{equation} $

定理2.1  如果$ R_0>1 $,模型(1.7)存在唯一的地方病平衡点$ E^*(S^*, V^*, I^*) $,这里

$S^* = \frac{\pi}{\mu+\xi+\beta_1g(I^*)}, \quad V^* = \frac{\xi \pi}{\big(\mu+\xi+\beta_{1}g(I^*)\big)(\mu+\beta_2g(I^*))}. $

证  由模型(1.7)的第一个和第二个方程知,可将$ V^* $和$ S^* $用$ I^* $表示

(2.2) $ \begin{eqnarray} &&S^* = \frac{\pi}{\mu+\xi+\beta_1g(I^*)}, \end{eqnarray} $

(2.3) $ \begin{eqnarray} &&V^* = \frac{\xi \pi}{(\mu+\xi+\beta_{1}g(I^*))(\mu+\beta_2g(I^*))} . \end{eqnarray} $

将(2.2)和(2.3)式代入模型(1.7)的第三个方程,得

(2.4) $ \begin{equation} F(I^*) = A{I^*}^2+BI^*+C, \end{equation} $

这里

$ \begin{eqnarray*} &&A = (\mu \alpha+\beta_2)\big((\mu+\xi)\alpha+\beta_1\big)>0, \\ &&B = (\mu+\xi)(2\mu\alpha+\beta_2)+\mu\beta_1-\frac{\alpha(\beta_1\mu+\beta_2\xi)+\beta_1\beta_2}{\mu+\gamma}e^{-\mu \tau}\pi, \\ && C = \mu(\mu+\xi)(1-R_0). \end{eqnarray*} $

当$ R_0>1 $,注意$ C = \mu(\mu+\xi)(1-R_0)<0 $和$ A>0 $,易知$ F(I^*) = 0 $有一个正根和一个负根.

当$ R_0\leq 1 $,注意

$ \begin{eqnarray*} B& = &(\mu+\xi)(2\mu\alpha+\beta_2)+\mu\beta_1-\frac{\alpha(\beta_1\mu+\beta_2\xi)+\beta_1\beta_2}{\beta_1 \mu+\beta_2 \xi}\mu(\mu+\xi)R_0 \\ &\geq&(\mu+\xi)(2\mu\alpha+\beta_2)+\mu\beta_1-\frac{\alpha(\beta_1\mu+\beta_2\xi)+\beta_1\beta_2}{\beta_1 \mu+\beta_2 \xi}\mu(\mu+\xi) \\ & = &\frac{(\mu+\xi)\big(\mu\alpha(\mu\beta_1+\xi\beta_2)+\xi\beta_2^2 \big)}{\mu\beta_1+\xi\beta_2}+\mu\beta_1>0 \end{eqnarray*} $

和

$C = \mu^2(\mu+\xi)(1-R_0)\geq0. $

易知$ F(I^*) = 0 $不可能有正根的.证毕.

定理2.2  模型(1.7)在初始条件(1.8)下的解是非负和有界的.

证  首先,我们证明$ S(t) $和$ V(t) $是有界的,并且$ I(t) $有一个上界.显然

(2.5) $ \begin{equation} \pi-(\mu+\xi)S-\frac{\beta_1}{\alpha}S<S'<\pi-(\mu+\xi)S. \end{equation} $

根据比较原理有

(2.6) $ \begin{equation} m_{1} = \frac{\pi}{\mu+\xi+\frac{\beta_1}{\alpha}} \leq S \leq \frac{\pi}{\mu+\xi} = M_{1}. \end{equation} $

对$ V $用和$ S $相同的方法有

(2.7) $ \begin{equation} \xi m_{1}-\mu V-\frac{\beta_2}{\alpha}V<V'<\xi m_{1}-\mu V \end{equation} $

和

(2.8) $ \begin{equation} m_{2} = \frac{\xi m_{1}}{\mu+\frac{\beta_2}{\alpha}} \leq V \leq \frac{\xi M_{1}}{\mu} = M_{2}. \end{equation} $

对$ I $有

(2.9) $ \begin{equation} I'<(\beta_1 M_1+\beta_2 M_2)e^{-\mu \tau}\frac{1}{\alpha}-(\mu+\gamma) I, \end{equation} $

再次利用比较原理

(2.10) $ \begin{equation} I\leq \frac{(\beta_1 M_1+\beta_2 M_2)e^{-\mu \tau}}{\alpha (\mu+\gamma)} = M_{3}. \end{equation} $

接下来证明$ I\geq 0 $.利用反证法,假设$ I\geq 0 $不成立.那么一定存在$ T \in (-\tau, +\infty) $有$ I(T) = 0 $.令

$ T^{+} = \inf\{t:t\in(-\tau, T), I(t) = 0\}, $

易知$ I(T^+) = 0 $和$ I'(T^+) < 0 $.然而

(2.11) $ \begin{equation} I'(T^+)>(\beta_1 m_1+\beta_2 m_2)e^{-\mu \tau}\frac{I(T^{+}-\tau)}{1+\alpha I(T^{+}-\tau) }-(\mu+\gamma) I(T^+) \geq 0, \end{equation} $

与假设矛盾.因此$ I(t)\geq 0 $.证毕.

为了研究模型(1.7)的局部稳定性,首先对模型(1.7)在平衡点$ E(\widetilde{S}, \widetilde{V}, \widetilde{I}) $处线性化,得如下特征方程

(3.1) $ \begin{eqnarray} \left|\begin{array}{ccc} -\big(\mu+\xi+\beta_1g(\tilde{I})\big)-\lambda & 0 & -\frac{\beta_1 \tilde{S}}{(1+\alpha\tilde{I})^2} \\ \xi & -(\mu+\beta_2 g(\tilde{I}))-\lambda & -\frac{\beta_2 \tilde{V}}{(1+\alpha\tilde{I})^2} \\ e^{-(\mu+\lambda)\tau}\beta_1g(\tilde{I}) & e^{-(\mu+\lambda)\tau}\beta_2 g(\tilde{I}) & e^{-(\mu+\lambda)\tau}\frac{\beta_1 \tilde{S}+\beta_2 \tilde{V}}{(1+\alpha\tilde{I})^2}-\mu-\gamma-\lambda \end{array}\right| = 0. \end{eqnarray} $

定理3.1  如果$ R_0 < 1 $,无病平衡点$ E_0 $局部渐近稳定;如果$ R_0>1 $,无病平衡点$ E_0 $是不稳定的.

证  特征方程(3.1)在$ E_0 $处有

(3.2) $ \begin{eqnarray} \big(\mu+\xi+\lambda \big)(\mu+\lambda)\big( e^{-(\mu+\lambda)\tau}(\beta_1 S_0 +\beta_2 V_0)-\mu-\gamma-\lambda \big) = 0. \end{eqnarray} $

方程(3.2)有三个根$ \lambda_1 = -\big(\mu+\xi\big)<0 $, $ \lambda_2 = -\mu<0 $和$ \lambda_3 $,其中$ \lambda_3 $是方程$ K(\lambda_3) = 0 $的解,这里

(3.3) $ \begin{eqnarray} K(\lambda_3)& = e^{-(\mu+\lambda_3)\tau}(\beta_1 S_0 +\beta_2 V_0 )-\mu-\gamma-\lambda_3 = -\lambda_3-(\mu+\gamma)(1-e^{-\lambda_3\tau}R_0). \end{eqnarray} $

假设$ \lambda_3 = x_1+y_1{\rm i} $,有

$ \begin{eqnarray*} K(\lambda_3) = -x_1-{\rm i} y_1-(\mu+\gamma)+(\mu+\gamma)R_0e^{-\tau x}\big(\cos{(\tau y)}-{\rm i}\sin{(\tau y)}\big). \end{eqnarray*} $

分离上面方程右端的实部和虚部,然后平方相加得

(3.4) $ \begin{eqnarray} (x_1+\mu+\gamma)^2+y_1^2 = (\mu+\gamma)^2(R_0e^{-\tau x_1})^2. \end{eqnarray} $

显然当$ R_0<1 $时$ x_1<0 $一定成立,否则(3.4)式不可能成立.于是我们断言$ \Re(\lambda_3)<0 $,故$ E_0 $局部渐近稳定.

当$ R_0>1 $,易知$ K(0) = -(\mu+\gamma)(1-R_0)>0 $和$ \lim\limits_{\lambda_3 \rightarrow +\infty} K( \lambda_3 ) = -\infty $.因此$ K(\lambda_3) = 0 $在区间$ (0, +\infty) $一定存在正根.此时$ E_0 $是不稳定的.证毕.

定理3.2  如果$ R_0 > 1 $,地方病平衡点$ E^* $局部渐近稳定.

证  在$ E^* $处,特征方程(3.1)为

(3.5) $ \begin{eqnarray} &&(\lambda+\mu+\xi+\beta_1 g(I^*))(\lambda +\mu +\beta_2 g(I^*))(\lambda+\mu+\gamma) \\ & = &\frac{e^{-(\mu+\lambda)\tau}(\beta_1 S^*+\beta_2 V^*)}{(1+\alpha I^*)^2}(\lambda+\mu)\bigg( \lambda+\mu+\xi+\frac{\beta_1\beta_2g(I^*)(S^*+V^*)}{\beta_1 S^*+\beta_2 V^*} \bigg). \end{eqnarray} $

利用(2.2)和(2.3)式得

(3.6) $ \begin{equation} \beta_1 S^*+\beta_2 V^* = \frac{\big(\mu \beta_1+\xi \beta_2+\beta_1 \beta_2 g(I^*)\big)\pi}{(\mu+\xi+\beta_1 g(I^*))(\mu+\beta_2 g(I^*))}, \end{equation} $

(3.7) $ \begin{equation} S^*+V^* = \frac{\big( \mu+\xi+\beta_2 g(I^*) \big)\pi}{(\mu+\xi+\beta_1 g(I^*))(\mu+\beta_2 g(I^*))}. \end{equation} $

注意

(3.8) $ \begin{equation} \mu+\gamma = e^{-\mu \tau}(\beta_1S^*+\beta_2V^*)\frac{1}{1+\alpha I^*}, \end{equation} $

利用(3.7)式, (3.8)式和$ \beta_2<\beta_1 $,有

$\frac{e^{-(\mu+\lambda)\tau}(\beta_1 S^*+\beta_2 V^*)}{(1+\alpha I^*)^2} = e^{-\lambda\tau}\frac{\mu+\gamma}{1+\alpha I^*}, $

$ \frac{\beta_1\beta_2g(I^*)(S^*+V^*)}{\beta_1 S^*+\beta_2 V^*}-\beta_1g(I^*) = \frac{\beta_1g(I^*)S^*(\beta_2-\beta_1)}{\beta_1S^*+\beta_2V^*}<0. $

不妨假设$ \lambda = x_2+{\rm i} y_2 $ ($ x_2\geq 0 $).由于$ \beta_2<\beta_1 $,易知

$ \mid \lambda+\mu+\beta_2g(I^*) \mid>\mid \lambda+\mu \mid, $

$ \mid \lambda+\mu+\gamma \mid >\mid \mu+\gamma \mid>\bigg| e^{-\lambda\tau}\frac{\mu+\gamma}{1+\alpha I^*} \bigg|, $

$\mid \lambda+\mu+\xi+\beta_1g(I^*) \mid>\bigg| \lambda+\mu+\xi+\frac{\beta_1\beta_2g(I^*)(S^*+V^*)}{\beta_1 S^*+\beta_2 V^*}\bigg|, $

则有

$ \begin{eqnarray*} &&\big| (\lambda+\mu+\xi+\beta_1 g(I^*))(\lambda +\mu +\beta_2 g(I^*))(\lambda+\mu+\gamma) \big|\\ &>&\bigg| \frac{e^{-(\mu+\lambda)\tau}}{(1+\alpha I^*)^2(\beta_1 S^*+\beta_2 V^*)}(\lambda+\mu)\big( \lambda+\mu+\xi+\frac{\beta_1\beta_2g(I^*)(S^*+V^*)}{\beta_1 S^*+\beta_2 V^*} \big) \bigg|, \end{eqnarray*} $

这和等式(3.5)矛盾,意味着特征方程(3.5)的根不可能有非负实部,于是$ E^* $局部渐近稳定.证毕.

定义4.1^[17]   系统(1.7)是一致持续的是指：如果正存在$ m_i(i = 1, 2, 3) $和$ M_i(i = 1, 2, 3) $使得系统(1.7)的任意正解$ \big(S(t), V(t), I(t)\big) $都存在$ T>0 $,当$ t\geq T $时有$ m_1\leq S(t) \leq M_1 $, $ m_2\leq V(t) \leq M_2 $和$ m_3\leq I(t) \leq M_3 $,即

$ \begin{eqnarray*} &&m_1\leq \liminf\limits_{t\rightarrow \infty}S(t) \leq \limsup\limits_{t\rightarrow \infty}S(t)\leq M_1, \\& &m_2\leq \liminf\limits_{t\rightarrow \infty}V(t) \leq \limsup\limits_{t\rightarrow \infty}V(t)\leq M_2, \\& &m_3\leq \liminf\limits_{t\rightarrow \infty}I(t) \leq \limsup\limits_{t\rightarrow \infty}I(t)\leq M_3. \end{eqnarray*} $

定理4.1  如果$ R_0 > 1 $,模型(1.7)在初始条件(1.8)下是一致持续的,即存在正的常数$ m_i (i = 1, 2, 3) $使得对任意解$ (S(t), V(t), I(t)) $,有

$ \begin{eqnarray*} &&m_1\leq \liminf\limits_{t\rightarrow \infty}S(t) \leq \limsup\limits_{t\rightarrow \infty}S(t)\leq M_1, \\ &&m_2\leq \liminf\limits_{t\rightarrow \infty}V(t) \leq \limsup\limits_{t\rightarrow \infty}V(t)\leq M_2, \\ &&m_3\leq \liminf\limits_{t\rightarrow \infty}I(t) \leq \limsup\limits_{t\rightarrow \infty}I(t)\leq M_3. \end{eqnarray*} $

证  根据定理2.2易知

$ \begin{eqnarray*} & &m_1\leq \liminf\limits_{t\rightarrow \infty}S(t) \leq \limsup\limits_{t\rightarrow \infty}S(t)\leq M_1, \\ &&m_2\leq \liminf\limits_{t\rightarrow \infty}V(t) \leq \limsup\limits_{t\rightarrow \infty}V(t)\leq M_2 \end{eqnarray*} $

和$ \limsup\limits_{t\rightarrow \infty}I(t)\leq M_3 $.因此主要证明$ \liminf\limits_{t\rightarrow \infty}I(t)\geq m_3 $即可.

当$ t\geq 0 $,定义一个可微函数

(4.1) $ \begin{equation} W(t) = e^{\mu \tau}I(t)+\beta_1 \int_{t-\tau}^{t}S(\theta)g(I(\theta)){\rm d}\theta+\beta_2 \int_{t-\tau}^{t}V(\theta)g(I(\theta)){\rm d}\theta. \end{equation} $

$ W(t) $沿着模型(1.7)解的方向求导,有

(4.2) $ \begin{equation} W'(t) = (\beta_1S+\beta_2V)g(I)-(\mu+\gamma)e^{\mu\tau}. \end{equation} $

对于任意$ 0<q<1 $,易知

$S^* = \frac{\pi}{\mu+\xi+\beta_1 g(I^*)}<\frac{\pi}{\mu+\xi+\beta_1 g(qI^*)} = k_1, $

于是存在一个足够大的常数$ \rho_1 \geq 1 $使得

$ S^*<k_1(1-e^{-\frac{\pi}{k_1}\rho_1 \tau}) = S^{\triangle}(\rho_1). $

同理可得

$V^* = \frac{\xi S^*}{\mu+\beta_2 g(I^*)}<\frac{\xi S^{\triangle}(\rho_1)}{\mu+\beta_2 g(q I^*)} = k_2. $

存在一个足够大的常数$ \rho_2 \geq 1 $使得

$ V^*<k_2(1-e^{-\frac{\xi S^{\triangle}(\rho_1)}{k_2}\rho_2 \tau}) = V^{\triangle}(\rho_2). $

令$ \rho = \max\{\rho_1, \rho_2\} $,则有

(4.3) $ \begin{equation} S^*<S^{\triangle}(\rho_1)<k_1(1-e^{-\frac{\pi}{k_1}\rho \tau}) = S^{\triangle} \end{equation} $

和

(4.4) $ \begin{equation} V^*<V^{\triangle}(\rho_2)<k_2(1-e^{-\frac{\xi S^{\triangle}}{k_2}\rho \tau}) = V^{\triangle}. \end{equation} $

我们断言$ I(t)\leq qI^* $对所有的$ t>\rho \tau $是不可能的.利用反证法,假设$ I(t)\leq qI^* $对所有$ t>\rho \tau $成立.由模型(1.7)第一个方程易知,当$ t>\rho \tau $,有

(4.5) $ \begin{equation} S'(t) \geq \pi-(\mu+\xi+\beta_1 g(qI^*))S = \pi-\frac{\pi}{k_1} g(q I^*))S, \end{equation} $

于是

(4.6) $ \begin{equation} S(t)\geq k_1(1-e^{-\frac{\pi}{k_1}\rho \tau}) = S^{\triangle}. \end{equation} $

类似地,当$ t>\rho \tau $,有

(4.7) $ \begin{equation} V(t)\geq k_2(1-e^{-\frac{\xi S^{\triangle}}{k_2}\rho \tau}) = V^{\triangle}. \end{equation} $

当$ t>\rho \tau +\tau $,注意

$\frac{(\beta_1S^*+\beta_2V^*)g(qI^*)}{qI^*}>\frac{(\beta_1S^*+\beta_2V^*)g(I^*)}{I^*} = (\mu+\gamma) e^{\mu \tau}, $

由(4.2)式得

(4.8) $ \begin{eqnarray} W'(t)&\geq&\bigg[(\beta_1S^{\triangle}+\beta_2V^{\triangle})\frac{g(qI^*)}{qI^*}-(\mu+\gamma) e^{\mu\tau}\bigg]I \\ & = &\big[\beta_1(S^{\triangle}-S^*)+\beta_2(V^{\triangle}-V^*)\big]\frac{g(qI^*)}{qI^*}I + \bigg[\frac{(\beta_1S^*+\beta_2V^*)g(qI^*)}{qI^*}-(\mu+\gamma) e^{\mu\tau}\bigg]I \\ & \geq & \big[\beta_1(S^{\triangle}-S^*)+\beta_2(V^{\triangle}-V^*)\big]\frac{g(qI^*)}{qI^*}I. \end{eqnarray} $

令

$ \underline{i} = \min\limits_{\theta\in [-\tau, 0]} I(\rho \tau +2 \tau +\theta), $

我们断言$ I(t)>\underline{i} $对所有$ t\geq\rho \tau +\tau $成立.利用反证法,假设存在$ T_1\geq0 $使得当$ \rho \tau +\tau \leq t < \rho \tau +2\tau+T_1 $有$ I(t)\geq \underline{i} $,且$ I(\rho \tau +2\tau+T_1) = \underline{i} $和$ I'(\rho \tau +2\tau+T_1)\leq 0 $,利用方程(4.6), (4.7)和模型(1.7)的第三个方程,对于$ t_1 = \rho \tau +2\tau+T_1 $,有

$ \begin{eqnarray*} I'(t_1)&\geq & e^{-\mu\tau}(\beta_1 S^{\triangle}+\beta_2 V^{\triangle} )g(I(t_1-\tau))-(\mu+\gamma) \underline{i} \\ & = &\bigg(\frac{e^{-\mu\tau}(\beta_1 S^{\triangle}+\beta_2 V^{\triangle} )g(I(t_1-\tau))}{\underline{i}}-(\mu+\gamma)\bigg)\underline{i} \\ &>&\bigg(e^{-\mu\tau}(\beta_1 S^{\triangle}+\beta_2 V^{\triangle} ) \frac{g(\underline{i})}{\underline{i}}- (\mu+\gamma) \bigg) \underline{i} \\ &>&\bigg(e^{-\mu\tau}(\beta_1 S^{\triangle}+\beta_2 V^{\triangle} ) \frac{g(I^*)}{I^*}- (\mu+\gamma) \bigg) \underline{i}>0, \end{eqnarray*} $

和假设矛盾.因此,对所有的$ t\geq\rho \tau +\tau $, $ I(t)>\underline{i} $成立.当$ t\geq \rho \tau +2\tau $,根据不等式(4.8),有

(4.9) $ \begin{equation} W'(t)\geq \big[\beta_1(S^{\triangle}-S^*)+\beta_2(V^{\triangle}-V^*)\big]\frac{g(qI^*)}{qI^*}\underline{i}>0, \end{equation} $

这意味着当$ t\rightarrow +\infty $, $ W(t) \rightarrow +\infty $.由(4.1)式易知

$W(t)\leq e^{\mu \tau }M_3+(\beta_1M_1+\beta_2 M_2)\tau g(M_3) $

和(4.9)式矛盾.因此,对所有$ t>\rho \tau $, $ I(t)\leq qI^* $不可能都成立.

对于足够大的$ t $, $ I(t) $有两种可能性. (ⅰ)对所有足够大的$ t>\rho \tau $, $ I(t)\geq qI^* $都成立; (ⅱ)对足够大的$ t $, $ I(t) $在$ qI^* $附近震荡.接下来证明对所有足够大的$ t $, $ I(t)\geq m_3 = qI^*e^{-\tau (\mu+\gamma)} $都成立.如果是第(ⅰ)种情况,和我们要证明的内容是一致的.对于第(ⅱ)种情况,存在足够大的$ t_1<t_2 $使得

$ I(t_1) = I(t_2) = qI^*, \ \ \ \ I(t)<qI^*, \ \ \ t_1<t<t_2. $

如果$ t_2-t_1\leq \tau $,由模型(1.7)第三个方程易知

$ I'(t)>- (\mu+\gamma) I(t), $

则当$ t_1<t<t_2 $,有

$ I(t)>I(t_1)e^{-\mu(t-t_1)} = qI^*e^{-(\mu+\gamma) \tau} = m_3. $

如果$ t_2-t_1> \tau $,由模型(1.7)第三个方程可得,当$ t\in[t_1, t_1+\tau] $, $ I(t)>m_3 $成立.接下来我们证明当$ t\in[t_1+\tau, t_2] $, $ I(t)>m_3 $也成立.利用反证法,假设存在$ T_2>0 $,当$ t\in[t_1+\tau, t_1+\tau+T_2) $,有$ I(t)>m_3 $且$ I(t_1+\tau+T_2) = m_3 $和$ I'(t_1+\tau+T_2)\leq 0 $成立.另一方面,令$ t_2 = t_1+\tau+T_2 $,利用模型(1.7)的第三个方程和不等式$ m_3<qI^*<I^* $以及

$\frac{e^{-\mu \tau}(\beta_1 S^{\triangle}+\beta_2 V^{\triangle})}{1+\alpha m_3}>\frac{e^{-\mu \tau}(\beta_1 S^{*}+\beta_2 V^{*})}{1+\alpha I^*} = \mu+\gamma, $

有

$ \begin{eqnarray*} I'(t_2)& = &e^{-\mu \tau}(\beta_1 S(t_2-\tau)+\beta_2V(t_2-\tau))g(I(t_2-\tau))-(\mu+\gamma) I(t_2)\\ &\geq &e^{-\mu \tau}(\beta_1 S^{\triangle}+\beta_2 V^{\triangle})g(m_3)-(\mu+\gamma) m_3 \\ &\geq& \bigg[\frac{e^{-\mu \tau}(\beta_1 S^{\triangle}+\beta_2 V^{\triangle})}{1+\alpha m_3}-(\mu+\gamma) \bigg]m_3>0, \end{eqnarray*} $

这和以上假设矛盾,意味着对所有的$ t\in[t_1, t_2] $且$ t>\rho \tau $, $ I(t)\geq m_3 $成立.由于区间$ [t_1, t_2] $选择是任意的,因此已证明在第(ⅱ)种情况下,所有足够大的$ t $满足$ I(t)\geq m_3 $.同时又由于$ 0<q<1 $是任意选择的,于是$ \liminf\limits_{t\rightarrow +\infty}I(t) \geq m_3 $成立.证毕.

定理5.1  如果$ R_{0}\leq 1 $,模型(1.7)的无病平衡点$ E_{0} = (S_{0}, V_{0}, 0) $全局渐近稳定.

证  定义如下的Lyapunov泛函

$ V_{1}(t) = S_{0}\bigg(\frac{S}{S_{0}}-1-\ln\frac{S}{S_{0}} \bigg)+V_{0}\bigg(\frac{V}{V_{0}}-1-\ln\frac{V}{V_{0}}\bigg)+e^{\mu\tau}I+U^{-}(t), $

这里

$ U^{-}(t) = \int_{0}^{\tau}\beta_{1}S(t-\theta)g(I(t-\theta)){\rm d}\theta+\int_{0}^{\tau}\beta_{2}V(t-\theta)g(I(t-\theta)){\rm d}\theta. $

易得

$ \begin{eqnarray*} \frac{{\rm d}U^{-}}{{\rm d}t}& = &\frac{\rm d}{{\rm d}t}\int_{0}^{\tau}\beta_{1}S(t-\theta)g(I(t-\theta)){\rm d}\theta+\frac{\rm d}{{\rm d}t}\int_{0}^{\tau}\beta_{2}V(t-\theta)g(I(t-\theta)){\rm d}\theta\\ & = &\beta_{1}S(t)g(I(t))-\beta_{1}S(t-\tau)g(I(t-\tau))+\beta_{2}V(t)g(I(t))-\beta_{2}V(t-\tau)g(I(t-\tau)). \end{eqnarray*} $

$ V_{1}(t) $沿模型(1.7)解的方向求导,得

$ \begin{eqnarray*} \frac{{\rm d}V_{1}}{{\rm d}t}& = &(1-\frac{S_{0}}{S})\big(\pi-(\mu+\xi)S-\beta_{1}S g(I(t))\big)+(1-\frac{V_{0}}{V})\big(-\mu V+\xi S-\beta_{2}Vg(I(t))\big)\\ &&+e^{\mu\tau}\big(-(\mu+\gamma) I(t)+e^{-\mu\tau}\beta_{1}S(t-\tau)g(I(t-\tau))\\ &&+e^{-\mu\tau}\beta_{2}V(t-\tau)g(I(t-\tau))\big)+\frac{{\rm d}U^{-}}{{\rm d}t}\\ & = &(1-\frac{S_{0}}{S})\big(\pi-(\mu+\xi)S-\beta_{1}S g(I(t))\big)+(1-\frac{V_{0}}{V})\big(-\mu V+\xi S-\beta_{2}Vg(I(t))\big)\\ &&-e^{\mu\tau}(\mu+\gamma) I(t)+\beta_{1}S g(I(t))+\beta_{2}Vg(I(t)). \end{eqnarray*} $

利用等式$ (\mu+\xi)S_{0} = \pi $, $ \mu V_{0} = \xi S_{0} $,有

$ \begin{eqnarray*} \frac{{\rm d}V_{1}}{{\rm d}t}& = &(1-\frac{S_{0}}{S})\big((\mu+\xi)S_{0}(1-\frac{S}{S_{0}})-\beta_{1}S g(I)\big)\\ &&+(1-\frac{V_{0}}{V})\big(\xi S_{0}(\frac{S}{S_{0}}-\frac{V}{V_{0}}\big)-\beta_{2}V g(I))-e^{\mu\tau}(\mu+\gamma) I(t)+\beta_{1}S g(I(t))+\beta_{2}Vg(I(t))\\ & = &\mu S_{0}(2-\frac{S}{S_{0}}-\frac{S_{0}}{S})+\xi S_{0}(3-\frac{S_{0}}{S}-\frac{V}{V_{0}}-\frac{SV_{0}}{S_{0}V})+(R_{0}-e^{\mu\tau}(1+\alpha I))(\mu+\gamma) g(I)\\ &\leq &\mu S_{0}(2-\frac{S}{S_{0}}-\frac{S_{0}}{S})+\xi S_{0}(3-\frac{S_{0}}{S}-\frac{V}{V_{0}}-\frac{SV_{0}}{S_{0}V})+(R_{0}-(1+\alpha I))(\mu+\gamma) g(I)\\ & = &\mu S_{0}(2-\frac{S}{S_{0}}-\frac{S_{0}}{S})+\xi S_{0}(3-\frac{S_{0}}{S}-\frac{V}{V_{0}}-\frac{SV_{0}}{S_{0}V})-\frac{\alpha(\mu+\gamma) I^{2}}{1+\alpha I}+(R_{0}-1)(\mu+\gamma) g(I). \end{eqnarray*} $

易知当$ R_{0}\leq 1 $,对所有$ S(t), V(t), I(t) \geq 0 $, $ \frac{{\rm d} V_{1}}{{\rm d}t} \leq 0 $.并且$ \frac{{\rm d}V_{1}}{{\rm d}t} = 0 $当且仅当$ S = S_{0} $, $ V = V_{0} $, $ I = 0 $.于是$ \{E_0\} $是$ \{(S(t), V(t), I(t))\mid \frac{{\rm d } V_{1}}{{\rm d}t} = 0 \} $中最大不变集.根据Lyapunov-LaSalle不变性原理, $ E_0 $是全局渐近稳定的.证毕.

定理5.2  如果$ R_{0}>1 $,模型(1.7)的地方病平衡点$ E^{*} = (S^{*}, V^{*}, I^{*}) $全局渐近稳定.

证  定义如下的Lyapunov泛函

$ \begin{eqnarray*} V_{2}(t)& = &S^{*} \bigg(\frac{S}{S^{*}}-1-\ln\frac{S}{S^{*}}\bigg)+V^{*} \bigg(\frac{V}{V^{*}}-1-\ln\frac{V}{V^{*}}\bigg)\\ &&+e^{\mu\tau} \bigg(I-I^{*}-\int_{I^{*}}^{I(t)}\frac{g(I*)}{g(\eta)}{\rm d}\eta\bigg)+U^{+}(t), \end{eqnarray*} $

这里

$ \begin{eqnarray*} U^{+}(t)& = &\beta_{1}S^{*}g(I^{*})\int_{0}^{\tau} \bigg(\frac{S(t-\theta)g(I(t-\theta))}{S^{*}g(I)^{*}}-1-\ln\frac{S(t-\theta)g(I(t-\theta))}{S^{*}g(I)^{*}}\bigg){\rm d}\theta\\ &&+\beta_{2}V^{*}g(I^{*} )\int_{0}^{\tau} \bigg(\frac{V(t-\theta)g(I(t-\theta))}{V^{*}g(I)^{*}}-1-\ln\frac{V(t-\theta)g(I(t-\theta))}{V^{*}g(I)^{*}}\bigg){\rm d}\theta. \end{eqnarray*} $

易得

$ \begin{eqnarray*} \frac{{\rm d} U^{+}}{{\rm d}t} & = &\beta_{1}Sg(I)-\beta_{1}S(t-\tau)g(I(t-\tau))+\beta_{1}S^{*}g(I^{*})\ln\frac{S(t-\tau)g(I(t-\tau))}{Sg(I)}\\ &&+\beta_{2}Vg(I)-\beta_{2}V(t-\tau)g(I(t-\tau))+\beta_{2}V^{*}g(I^{*})\ln\frac{V(t-\tau)g(I(t-\tau))}{Vg(I)}. \end{eqnarray*} $

$ V_{2}(t) $沿着模型(1.7)解的方向求导,得

$ \begin{eqnarray*} \frac{{\rm d}V_{2}}{{\rm d}t}& = &(1-\frac{S^{*}}{S})(\pi-(\xi+\mu)S-\beta_{1}Sg(I))+(1-\frac{V^{*}}{V})(\xi S-\mu V-\beta_{2}Vg(I))\\ &&+(1-\frac{g(I^{*})}{g(I)})\big(\beta_{1}S(t-\tau)g(I(t-\tau))+\beta_{2}V(t-\tau)g(I(t-\tau))-e^{\mu\tau}(\mu+\gamma) I\big)+\frac{{\rm d} U^{+}}{{\rm d}t}\\ & = &(1-\frac{S^{*}}{S})\big(\pi-(\xi+\mu)S-\beta_{1}Sg(I)\big)+(1-\frac{V^{*}}{V})\big(\xi S-\mu V-\beta_{2}Vg(I)\big)\\ &&+(1-\frac{g(I^{*})}{g(I)})\big(\beta_{1}S(t-\tau)g(I(t-\tau))+\beta_{2}V(t-\tau)g(I(t-\tau))-e^{\mu\tau}(\mu+\gamma) I\big)\\ &&+\beta_{1}Sg(I)-\beta_{1}S(t-\tau)g(I(t-\tau))+\beta_{1}S^{*}g(I^{*})\ln\frac{S(t-\tau)g(I(t-\tau))}{Sg(I)}\\ &&+\beta_{2}Vg(I)-\beta_{2}V(t-\tau)g(I(t-\tau))+\beta_{2}V^{*}g(I^{*})\ln\frac{V(t-\tau)g(I(t-\tau))}{Vg(I)}. \end{eqnarray*} $

利用等式

$(\mu+\xi)S^{*}+\beta_{1}S^{*}g(I^{*}) = \pi, \qquad \mu V^{*}+\beta_{2}V^{*}g(I^{*}) = \xi S^{*} $

和

$\beta_{1}S^{*}g(I^{*})+\beta_{2}V^{*}g(I^{*}) = e^{\mu\tau}(\mu+\gamma) I^{*}, $

有

$ \begin{eqnarray*} \frac{{\rm d} V_{2}}{{\rm d} t}& = &\mu S^{*}(2-\frac{S}{S^{*}}-\frac{S^{*}}{S})+\mu V^{*}(3-\frac{S^{*}}{S}-\frac{V}{V^{*}}-\frac{SV^{*}}{S^{*}V})\\ &&+\beta_{2}V^{*}g(I^{*})(3-\frac{S^{*}}{S}-\frac{V}{V^{*}}-\frac{SV^{*}}{S^{*}V})-\beta_{1}Sg(I)+\beta_{1}S^{*}g(I)\\ &&-\beta_{2}Vg(I)+\beta_{2}V^{*}g(I) +\beta_{1}S^{*}g(I^{*})-\beta_{1}S^{*}g(I^{*})\frac{S^{*}}{S}+\beta_{2}Vg(I^{*})-\beta_{2}V^{*}g(I^{*})\\ &&+\beta_{1}S(t-\tau)g(I(t-\tau))+\beta_{2}V(t-\tau)g(I(t-\tau))-e^{\mu\tau}(\mu+\gamma) I\\ &&-\frac{g(I^{*})}{g(I)}\beta_{1}S(t-\tau)g(I(t-\tau))-\frac{g(I^{*})}{g(I)}\beta_{2}V(t-\tau)g(I(t-\tau))+e^{\mu\tau}\frac{g(I^{*})}{g(I)}\mu I\\ &&+\beta_{1}Sg(I)-\beta_{1}S(t-\tau)g(I(t-\tau))+\beta_{1}S^{*}g(I^{*})\ln\frac{S(t-\tau)g(I(t-\tau))}{Sg(I)}\\ &&+\beta_{2}Vg(I)-\beta_{2}V(t-\tau)g(I(t-\tau))+\beta_{2}V^{*}g(I^{*})\ln\frac{V(t-\tau)g(I(t-\tau))}{Vg(I)}. \end{eqnarray*} $

$ \begin{eqnarray*} \frac{{\rm d} V_{2}}{{\rm d} t}& = &\mu S^{*}(2-\frac{S}{S^{*}}-\frac{S^{*}}{S})+\mu V^{*}(3-\frac{S^{*}}{S}-\frac{V}{V^{*}}-\frac{SV^{*}}{S^{*}V})\\ &&+\beta_{1}S^{*}g(I^{*})(-\frac{S(t-\tau)g(I(t-\tau))}{S^{*}g(I)}+1+\ln\frac{S(t-\tau)g(I(t-\tau))}{S^{*}g(I)})\\ &&+\beta_{1}S^{*}g(I^{*})(1-\frac{S^{*}}{S}+\ln\frac{S^{*}}{S})+\beta_{1}S^{*}g(I^{*})[(1-\frac{g(I)}{g(I^{*})})(\frac{I}{I^{*}}\frac{g(I^{*})}{g(I)}-1)]\\ &&+\beta_{2}V^{*}g(I^{*})(-\frac{V(t-\tau)g(I(t-\tau))}{V^{*}g(I)}+1+\ln\frac{V(t-\tau)g(I(t-\tau))}{V^{*}g(I)})\\ &&+\beta_{2}V^{*}g(I^{*})(1-\frac{SV^{*}}{S^{*}V}+\ln\frac{SV^{*}}{S^{*}V})+\beta_{2}V^{*}g(I^{*})(1-\frac{S^{*}}{S}+\ln\frac{S^{*}}{S})\\ &&+\beta_{2}V^{*}g(I^{*})[(1-\frac{g(I)}{g(I^{*})})(\frac{I}{I^{*}}\frac{g(I^{*})}{g(I)}-1)] \leq0. \end{eqnarray*} $

显然, $ \frac{{\rm d} V_{2}}{{\rm d}t} \leq 0 $,并且$ \frac{{\rm d}V_{2}}{{\rm d}t} = 0 $当且仅当$ S(t) = S^{*}, V(t) = V^{*}, I(t) = I^* $.易知$ \{E^*\} $是$ \{(S(t), V(t), I(t))\mid \frac{{\rm d}V_{2}}{{\rm d}t} = 0 \} $中最大不变集.根据Lyapunov-LaSalle不变性原理, $ E^* $是全局渐近稳定的.证毕.

对于系统(1.7),我们选取参数值: $ \mu = 0.04, \alpha = 0.05, \beta_1 = 0.05, \beta_2 = 0.015, \gamma = 0.3, \tau = 5 $.首先,选择$ \pi = 0.5 $和$ \xi = 0.1 $.通过简单的计算可得基本再生数$ R_0 = 0.7523<1 $,数值模拟结果如图 1所示.可以看出,系统(1.7)的平衡点$ E_{0} = (S_{0}, V_{0}, 0) = (3.5714, 8.9286, 0) $全局渐近稳定.这意味着如果易感者类$ S $,接种者类$ V $和染病者类$ I $的初始人口都满足条件(1.8),它们最终趋向于无病平衡点$ E_0 $,疾病将会消亡.数值模拟结果符合定理5.1.

其次,选择$ \pi = 1 $,并通过改变接种率参数$ \xi $来讨论模型的动力学性质.当$ \xi = 0.1 $,通过简单的计算可知,基本再生数$ R_0 = 1.5050>1 $,数值模拟结果如图 2(a)所示.系统(1.7)的平衡点$ E^{*} = (S^{*}, V^{*}, I^{*}) = (5.7508, 9.5225, 0.7010) $全局渐近稳定,意味着如果易感者群体$ S $,接种者群体$ V $和感染者者群体$ I $的初始人口都满足条件(1.8),那么它们最终趋向于地方病平衡点$ E^* $,疾病持续流行.当增加接种率$ \xi = 0.9 $,此时,基本再生数$ R_0 = 0.9927<1 $,数值模拟结果如图 2(b)所示.系统(1.7)的平衡点$ E_{0} = (S_{0}, V_{0}, I_{0}) = (1.0635, 23.9360, 0) $全局渐近稳定,意味着如果易感者群体$ S $,接种者群体$ V $和感染者者群体$ I $的初始人口都满足条件(1.8),那么它们最终趋向于地方病平衡点$ E_0 $,疾病会消亡.

比较图 2(a)和图 2(b),易知当易感者类的接种率$ \xi $增加,疾病可能从持续流行变为消亡.因为随着易感者人群接种率的增加,基本再生数减少,疫苗接种对疾病的控制产生积极的作用.因此,增加疫苗接种的覆盖率可以有效地控制疾病的传播.同样,我们也可以通过使用一些抑制剂延长潜伏期使得基本再生数减小,从而控制疾病.