具有年龄结构的SVIQR传染病模型的稳定性分析

Stability Analysis of SVIQR Infectious Disease Model with Age Structure

摘要: 讨论一类具有疫苗接种和隔离的年龄结构SVIQR传染病模型，得出基本再生数R0的表达式。证明了当R0 < 1时，无病平衡点全局渐近稳定；当R0 > 1时，无病平衡点不稳定；此时系统存在地方病平衡点，并且证明了地方病平衡点是唯一的并且局部渐进稳定。最终得出，隔离是消除传染病的有效方式，如果不能进一步降低传播率，那么就需要最大限度地提高疫苗接种率。

Abstract: A class of age-structured SVIQR infectious disease models with vaccination and isolatio-n was discussed. The expressions of basic regeneration number R0 were derived. It was proved that when R0 < 1, the disease-free equilibrium point was globally asymptotically stable; when R0 > 1, the dis-ease-free equilibrium point was unstable. And there exists endemic equilibrium state, and it was proved that endemic equilibrium point was unique and locally asymptotically stable. Ultimately, isolation was an effective way to eliminate infectious diseases; if transmission rates cannot be re-duced in future, then vaccination rates need to be maximized.

1. 引言

传染病是人类历史上极易造成重大伤亡的疾病，近年来，艾滋病、肺结核、麻疹等疾病从未淡出过人们的视野，这就警示我们，传染病一直存在，我们必须不断提出新的方法来面对新出现或者反复出现的传染病。年龄是研究传染病模型中的重要因素，最早建立和研究年龄结构的流行病模型是Hoppenstendt [1] 。此后，研究年龄结构的流行病模型的工作不断出现，大多数作者讨论的是SIS，SIR，SEIR模型 [2] [3] [4] [5] [6] 。在突发传染病暴发的早期阶段，非药物预防和控制措施往往在遏制疾病传播和暴发方面发挥着关键作用。隔离是发现传染病病人或疑似者，将其安置在一定场所，使之不与易感人群接触的一种措施。隔离是控制传染源的重要措施之一。隔离的方式，应根据疾病的传播途径而定。某些传染病可在家隔离，如麻疹等，有些则应在临时隔离室或住院隔离，许多学者在模型中考虑了隔离措施，从而让模型更加符合实际情况 [7] [8] [9] [10] 。文献 [7] 中提出了一类具有接种和隔离治疗的肺结核模型，研究了模型的全局稳定性。文献 [8] 中研究了易感人群软隔离行为对COVID-19在武汉的传播影响。文献 [9] 中研究了由确诊病例驱动跟踪隔离的时滞传染病数学模型。文献 [10] 中研究了具有隔离和不完全治疗的传染病模型的全局稳定性。受以上文章启发，加入了年龄结构，研究了具有疫苗接种和隔离的年龄结构SVIQR传染病模型，相比传统的模型，更加具有实际意义。

2. 模型建立

本文将所研究的人群分为五类，分别为易感者类、疫苗接种类、染病者类、隔离者类和康复者类，并用 $S\left(a,t\right),V\left(a,t\right),I\left(a,t\right),Q\left(a,t\right),R\left(a,t\right)$ 表示在t时年龄a易感者类、疫苗接种者类、染病者类、隔离者类和康复者类的年龄分布函数。种群的全体成员的年龄分布函数为 $N\left(a,t\right)$ ，即 $N\left(a,t\right)=S\left(a,t\right)+V\left(a,t\right)+I\left(a,t\right)+Q\left(a,t\right)+R\left(a,t\right)$ ，根据传染病建模方法，建立如下具有年龄结构的传染病模型，

$\{\begin{array}{l}\frac{\partial S}{\partial a}+\frac{\partial S}{\partial t}=-S\left(a,t\right){\displaystyle {\int }_0^{a^{+}}\frac{\beta \left(a,a^{\prime }\right)I\left(a^{\prime },t\right)}{N\left(a^{\prime },t\right)}\text{d}{a}^{\prime }-\mu \left(a\right)S\left(a,t\right)-\delta \left(a\right)S\left(a,t\right),}\\ \frac{\partial V}{\partial a}+\frac{\partial V}{\partial t}=\delta \left(a\right)S\left(a,t\right)-\left[\mu \left(a\right)+m\right]V\left(a,t\right),\\ \frac{\partial I}{\partial a}+\frac{\partial I}{\partial t}=S\left(a,t\right){\displaystyle {\int }_0^{a^{+}}\frac{\beta \left(a,a^{\prime }\right)I\left(a^{\prime },t\right)}{N\left(a^{\prime },t\right)}\text{d}{a}^{\prime }-\left[\mu \left(a\right)+\rho \left(a\right)+\sigma \left(a\right)\right]I\left(a,t\right)},\\ \frac{\partial Q}{\partial a}+\frac{\partial Q}{\partial t}=\rho \left(a\right)I\left(a,t\right)-\left[\mu \left(a\right)+\gamma \left(a\right)\right]Q\left(a,t\right),\\ \frac{\partial R}{\partial a}+\frac{\partial R}{\partial t}=\gamma \left(a\right)Q\left(a,t\right)+mV\left(a,t\right)+\sigma \left(a\right)I\left(a,t\right)-\mu \left(a\right)R\left(a,t\right),\end{array}$ (1)

初值条件

$S\left(a,0\right)=S_0\left(a\right),V\left(a,0\right)=V_0\left(a\right),I\left(a,0\right)=I_0\left(a\right),Q\left(a,0\right)=Q_0\left(a\right),R\left(a,0\right)=R_0\left(a\right)$ ，

边界条件

$S\left(0,t\right)=A,V\left(0,t\right)=0,I\left(0,t\right)=0,Q\left(0,t\right)=0,R\left(0,t\right)=0.$

各参数表示的意义如下表1：

Table 1. Each parameter and its significance

表1. 各参数及意义

基本假设

1) 具有一定的初始人口；

2) 不考虑因病死亡；

3) 接种疫苗之后不在染病。

为了简化系统(1)，对系统(1)进行归一变换，令

$s\left(a,t\right)=\frac{S\left(a,t\right)}{N\left(a,t\right)},v\left(a,t\right)=\frac{V\left(a,t\right)}{N\left(a,t\right)},i\left(a,t\right)=\frac{I\left(a,t\right)}{N\left(a,t\right)},q\left(a,t\right)=\frac{Q\left(a,t\right)}{N\left(a,t\right)},r\left(a,t\right)=\frac{R\left(a,t\right)}{N\left(a,t\right)}.$

则系统(1)变为如下形式

$\{\begin{array}{l}\frac{\partial s}{\partial a}+\frac{\partial s}{\partial t}=-\left[\lambda \left(a,t\right)+\delta \left(a\right)\right]s\left(a,t\right),\\ \frac{\partial v}{\partial a}+\frac{\partial v}{\partial t}=\delta \left(a\right)s\left(a,t\right)-mv\left(a,t\right),\\ \frac{\partial i}{\partial a}+\frac{\partial i}{\partial t}=\lambda \left(a,t\right)s\left(a,t\right)-\left[\rho \left(a\right)+\sigma \left(a\right)\right]i\left(a,t\right),\\ \frac{\partial q}{\partial a}+\frac{\partial q}{\partial t}=\rho \left(a\right)i\left(a,t\right)-\gamma \left(a\right)q\left(a,t\right),\\ \frac{\partial r}{\partial a}+\frac{\partial r}{\partial t}=\gamma \left(a\right)q\left(a,t\right)+mv\left(a,t\right)+\sigma \left(a\right)i\left(a,t\right),\end{array}$ (2)

初始条件

$\begin{array}{l}s\left(a,0\right)=\frac{S_0\left(a\right)}{N_0\left(a\right)}=s_0\left(a\right),v\left(a,0\right)=\frac{V_0\left(a\right)}{N_0\left(a\right)}=v_0\left(a\right),i\left(a,0\right)=\frac{I_0\left(a\right)}{N_0\left(a\right)}=i_0\left(a\right),\\ q\left(a,0\right)=\frac{Q_0\left(a\right)}{N_0\left(a\right)}=q_0\left(a\right),r\left(a,0\right)=\frac{R_0\left(a\right)}{N_0\left(a\right)}=r_0\left(a\right),\end{array}$

边界条件

$\text{s}\left(0,t\right)=1,v\left(0,t\right)=0,i\left(0,t\right)=0,q\left(0,t\right)=0,r\left(0,t\right)=0,$

其中

$\lambda \left(a,t\right)=k\left(a\right){\displaystyle {\int }_0^{a^{+}}\tilde{\beta }\left(a^{\prime }\right)i\left(a^{\prime },t\right)\text{d}{a}^{\prime }}.$

可以获得

$s\left(a,t\right)+v\left(a,t\right)+i\left(a,t\right)+q\left(a,t\right)+r\left(a,t\right)=1.$

3. 无病平衡点的存在性和稳定性

3.1. 无病平衡点的存在性

当系统(2)达到稳定的年龄分布时， $s,v,i,q,r$ 只与年龄a有关，所以系统变为如下形式

$\{\begin{array}{l}\frac{\text{d}s}{\text{d}a}=-\left[\lambda \left(a\right)+\delta \left(a\right)\right]s\left(a\right),\\ \frac{\text{d}v}{\text{d}a}=\delta \left(a\right)s\left(a\right)-mv\left(a\right),\\ \frac{\text{d}i}{\text{d}a}=\lambda \left(a\right)s\left(a\right)-\left[\rho \left(a\right)+\sigma \left(a\right)\right]i\left(a\right),\\ \frac{\text{d}q}{\text{d}a}=\rho \left(a\right)i\left(a\right)-\gamma \left(a\right)q\left(a\right),\\ \frac{\text{d}r}{\text{d}a}=\gamma \left(a\right)q\left(a\right)+mv\left(a\right)+\sigma \left(a\right)i\left(a\right),\end{array}$ (3)

初始条件

$\text{s}\left(0\right)=1,v\left(0\right)=0,i\left(0\right)=0,q\left(0\right)=0,r\left(0\right)=0,$

其中

$\lambda \left(a\right)=k\left(a\right){\displaystyle {\int }_0^{a^{+}}\tilde{\beta }\left(a^{\prime }\right)i\left(a^{\prime }\right)\text{d}{a}^{\prime }}.$

r在上面4个方程均为出现，所以只考虑前4个方程，得到其无病平衡点

$E^0=\left(s^0\left(a\right),v^0\left(a\right),i^0\left(a\right),q^0\left(a\right)\right).$

显然， $i^0\left(a\right)=q^0\left(a\right)=0,$

$s^0\left(a\right)={\text{e}}^{-{\displaystyle {\int }_0^a\delta \left(s\right)\text{d}s}},v^0\left(a\right)={\displaystyle {\int }_0^a\delta \left(\tau \right){\text{e}}^{-{\displaystyle {\int }_0^a\delta \left(\xi \right)\text{d}\xi }}{\text{e}}^{-{\displaystyle {\int }_{\tau }^{a}m}\text{d}s}\text{d}\tau }.$

综上，我们可知无病平衡点存在且唯一，即 $E^0=\left(s^0\left(a\right),v^0\left(a\right),0,0\right)$ 。

3.2. 无病平衡点的局部稳定性

为了研究无病平衡点的局部稳定性，我们对系统(3)进行平移变换，令

$\begin{array}{l}s\left(a,t\right)=s^0\left(a\right)+\bar{s}\left(a,t\right),v\left(a,t\right)=v^0\left(a\right)+\bar{v}\left(a,t\right),\\ i\left(a,t\right)=i^0\left(a\right)+\bar{i}\left(a,t\right),q\left(a,t\right)=q^0\left(a\right)+\bar{q}\left(a,t\right).\end{array}$

则系统变为如下形式

$\{\begin{array}{l}\frac{\partial \bar{s}}{\partial a}+\frac{\partial \bar{s}}{\partial t}=-\bar{\lambda }\left(a,t\right)\bar{s}\left(a,t\right)-\delta \left(a\right)\bar{s}\left(a,t\right)-\bar{\lambda }\left(a,t\right)s^0\left(a\right),\\ \frac{\partial \bar{v}}{\partial a}+\frac{\partial \bar{v}}{\partial t}=\delta \left(a\right)\bar{s}\left(a,t\right)+\delta \left(a\right)s^0\left(a\right)-m\bar{v}\left(a,t\right)-m{v}^0\left(a\right),\\ \frac{\partial \bar{i}}{\partial a}+\frac{\partial \bar{i}}{\partial t}=\bar{\lambda }\left(a,t\right)\bar{s}\left(a,t\right)+\bar{\lambda }\left(a,t\right)s^0\left(a\right)-\left[\rho \left(a\right)+\sigma \left(a\right)\right]\bar{i}\left(a,t\right),\\ \frac{\partial \bar{q}}{\partial a}+\frac{\partial \bar{q}}{\partial t}=\rho \left(a\right)\bar{i}\left(a,t\right)-\gamma \left(a\right)\bar{q}\left(a,t\right),\end{array}$ (4)

边界条件

$\bar{s}\left(0,t\right)=0,\bar{v}\left(0,t\right)=0,\bar{i}\left(0,t\right)=0,\bar{q}\left(0,t\right)=0,$

其中

$\bar{\lambda }\left(a,t\right)=k\left(a\right){\displaystyle {\int }_0^{a^{+}}\tilde{\beta }\left(a^{\prime }\right)\bar{i}\left(a^{\prime },t\right)\text{d}{a}^{\prime }}.$

系统(4)线性部分如下

$\{\begin{array}{l}\frac{\partial \bar{s}}{\partial a}+\frac{\partial \bar{s}}{\partial t}=-\delta \left(a\right)\bar{s}\left(a,t\right)-\bar{\lambda }\left(a,t\right)s^0\left(a\right),\\ \frac{\partial \bar{v}}{\partial a}+\frac{\partial \bar{v}}{\partial t}=\delta \left(a\right)\bar{s}\left(a,t\right)-m\bar{v}\left(a,t\right),\\ \frac{\partial \bar{i}}{\partial a}+\frac{\partial \bar{i}}{\partial t}=\bar{\lambda }\left(a,t\right)s^0\left(a\right)-\left[\rho \left(a\right)+\sigma \left(a\right)\right]\bar{i}\left(a,t\right),\\ \frac{\partial \bar{q}}{\partial a}+\frac{\partial \bar{q}}{\partial t}=\rho \left(a\right)\bar{i}\left(a,t\right)-\gamma \left(a\right)\bar{q}\left(a,t\right),\end{array}$ (5)

边界条件 $\bar{s}\left(0,t\right)=0,\bar{v}\left(0,t\right)=0,\bar{i}\left(0,t\right)=0,\bar{q}\left(0,t\right)=0.$

考虑系统(5)的指数解形式，令

$\bar{s}\left(a,t\right)=\bar{s}\left(a\right){\text{e}}^{\lambda t},\bar{v}\left(a,t\right)=\bar{v}\left(a\right){\text{e}}^{\lambda t},\bar{i}\left(a,t\right)=\bar{i}\left(a\right){\text{e}}^{\lambda t},\bar{q}\left(a,t\right)=\bar{q}\left(a\right){\text{e}}^{\lambda t}.$

则系统(5)变为如下形式

$\{\begin{array}{l}\frac{\text{d}\bar{s}}{\text{d}a}=-\hat{\lambda }\left(a\right)s^0\left(a\right)-\left[\delta \left(a\right)+\lambda \right]\bar{s}\left(a\right),\\ \frac{\text{d}\bar{v}}{\text{d}a}=\delta \left(a\right)\bar{s}\left(a\right)-\left[\lambda +m\right]\bar{v}\left(a\right),\\ \frac{\text{d}\bar{i}}{\text{d}a}=\hat{\lambda }\left(a\right)s^0\left(a\right)-\left[\lambda +\rho \left(a\right)+\sigma \left(a\right)\right]\bar{i}\left(a\right),\\ \frac{\text{d}\bar{q}}{\text{d}a}=\rho \left(a\right)\bar{i}\left(a\right)-\gamma \left(a\right)\bar{q}\left(a\right),\end{array}$ (6)

初始条件 $\bar{s}\left(0\right)=0,\bar{e}\left(0\right)=0,\bar{i}\left(0\right)=0,\bar{q}\left(0\right)=0,$

其中

$\hat{\lambda }\left(a\right)=k\left(a\right){\displaystyle {\int }_0^{a^{+}}\tilde{\beta }\left(a^{\prime }\right)\bar{i}\left(a^{\prime }\right)\text{d}{a}^{\prime }}.$

解系统(6)可以得

$\{\begin{array}{l}\bar{s}\left(a\right)=-{\displaystyle {\int }_0^a\hat{\lambda }\left(\tau \right)s^0\left(\tau \right){\text{e}}^{-{\displaystyle {\int }_{\tau }^a\left[\lambda +\delta \left(s\right)\right]\text{d}s}}\text{d}\tau },\\ \bar{v}\left(a\right)={\displaystyle {\int }_0^a\left[\delta \left(\tau \right)\bar{s}\left(\tau \right)\right]{\text{e}}^{-{\displaystyle {\int }_{\tau }^a\left(\lambda +m\right)\text{d}s}}\text{d}\tau ,}\\ \bar{i}\left(a\right)={\displaystyle {\int }_0^a\hat{\lambda }\left(\tau \right)s^0\left(\tau \right){\text{e}}^{-{\displaystyle {\int }_{\tau }^a\left[\lambda +\rho \left(s\right)+\sigma \left(s\right)\right]\text{d}s}}\text{d}\tau },\\ \bar{v}\left(a\right)={\displaystyle {\int }_0^a\left[\rho \left(\tau \right)\bar{i}\left(\tau \right)\right]{\text{e}}^{-{\displaystyle {\int }_{\tau }^a\gamma \left(s\right)\text{d}s}}\text{d}\tau .}\end{array}$ (7)

令 $\Lambda ={\displaystyle {\int }_0^{a^{+}}\tilde{\beta }\left(a\right)\bar{i}\left(a\right)\text{d}a}$ ，有 $\hat{\lambda }\left(a\right)=\Lambda k\left(a\right)$ 。

将 $\bar{i}\left(a\right)$ 的表达式代入，可以获得

$\Lambda ={\displaystyle {\int }_0^{a^{+}}\tilde{\beta }\left(a\right)}{\displaystyle {\int }_0^a\Lambda k\left(\tau \right)s^0\left(\tau \right){\text{e}}^{-{\displaystyle {\int }_{\tau }^a\left[\lambda +\rho \left(s\right)+\sigma \left(s\right)\right]\text{d}s}}}\text{d}\tau \text{d}a.$

产生以下关于 $\lambda$ 的特征方程

$1={\displaystyle {\int }_0^{a^{+}}\tilde{\beta }\left(a\right)}{\displaystyle {\int }_0^{a}k\left(\tau \right)s^0\left(\tau \right){\text{e}}^{-{\displaystyle {\int }_{\tau }^a\left[\lambda +\rho \left(s\right)+\sigma \left(s\right)\right]\text{d}s}}}\text{d}\tau \text{d}a.$

利用 $F\left(\lambda \right)$ 来表示右侧

$F\left(\lambda \right)={\displaystyle {\int }_0^{a^{+}}\tilde{\beta }\left(a\right)}{\displaystyle {\int }_0^{a}k\left(\tau \right)s^0\left(\tau \right){\text{e}}^{-{\displaystyle {\int }_{\tau }^a\left[\lambda +\rho \left(s\right)+\sigma \left(s\right)\right]\text{d}s}}}\text{d}\tau \text{d}a.$

基本再生数指的是一个病人在平均患病期内所传染的人数，是考量病毒能否流行的指标性因素，可以将基本再生数定义为 $R_0=F\left(0\right)$ ，即

$R_0={\displaystyle {\int }_0^{a^{+}}\tilde{\beta }\left(a\right)}{\displaystyle {\int }_0^{a}k\left(\tau \right)s^0\left(\tau \right){\text{e}}^{-{\displaystyle {\int }_{\tau }^a\left[\rho \left(s\right)+\sigma \left(s\right)\right]\text{d}s}}}\text{d}\tau \text{d}a.$ (8)

对于无病平衡点 $E^0$ 的局部稳定性，我们有以下结果。

定理1 如果 $R_0<1$ ， $E^0$ 是局部稳定的。如果 $R_0>1$ ， $E^0$ 是不稳定的。

证明 从 $F\left(\lambda \right)$ 的表达式中，可以得到以下的性质

$F^{\prime }\left(\lambda \right)<0,\underset{\lambda \to -\infty }{\lim }F\left(\lambda \right)=+\infty ,\underset{\lambda \to +\infty }{\lim }F\left(\lambda \right)=0.$

则存在唯一的实根 ${\lambda }^{*}$ ，使得 $F\left({\lambda }^{*}\right)=1$ 。

Figure 1. The approximate graph of $F(\; \lambda \; )$

图1. $F\left(\lambda \right)$ 的近似图

情况一 $F\left(\lambda \right)=1$ 有一个实根。如果 $R_0<1$ ，即 $F\left(0\right)<1$ ，则 ${\lambda }^{*}<0$ 。因此，如果 $R_0<1$ ， $F\left({\lambda }^{*}\right)=1$ 有唯一的负实根。

情况二 $F\left(\lambda \right)=1$ 有一个复根。令 $\lambda =\alpha +\beta i$ 是满足 $F\left(\lambda \right)=1$ 的根。则 $\mathrm{Re}F\left(\lambda \right)=1$ ， $\mathrm{Im}F\left(\lambda \right)=0$ ，并且

$\mathrm{Re}\left({\text{e}}^{\lambda }\right)=\mathrm{Re}\left({\text{e}}^{\alpha +\beta i}\right)=\mathrm{Re}\left[{\text{e}}^{\alpha }\left(\cos \beta +i\sin \beta \right)\right]={\text{e}}^{\alpha }\cos \beta \le {\text{e}}^{\alpha }={\text{e}}^{\mathrm{Re}\lambda },$

$\begin{array}{c}F\left(\lambda \right)={\displaystyle {\int }_0^{a^{+}}\tilde{\beta }\left(a\right)}{\displaystyle {\int }_0^{a}k\left(\tau \right)s^0\left(\tau \right){\text{e}}^{-{\displaystyle {\int }_{\tau }^a\left[\lambda +\rho \left(s\right)+\sigma \left(s\right)\right]\text{d}s}}}\text{d}\tau \text{d}a\\ \le {\displaystyle {\int }_0^{a^{+}}\tilde{\beta }\left(a\right)}{\displaystyle {\int }_0^{a}k\left(\tau \right)s^0\left(\tau \right){\text{e}}^{-\left[\mathrm{Re}\lambda \left(a-\tau \right)\right]}{\text{e}}^{-{\displaystyle {\int }_{\tau }^a\left[\rho \left(s\right)+\sigma \left(s\right)\right]\text{d}s}}}\text{d}\tau \text{d}a=F\left(\mathrm{Re}\lambda \right).\end{array}$

所以，可以获得

$1\le F\left(\mathrm{Re}\lambda \right)\iff F\left({\lambda }^{*}\right)\le F\left(\mathrm{Re}\lambda \right)\iff \mathrm{Re}\lambda \le {\lambda }^{*}<0.$

$F\left(\lambda \right)$ 的近似图如图1所示，因此，若 $F\left({\lambda }^{*}\right)=1$ 有一个复根，则复根必须要有负实部。从以上可知，当 $R_0<1$ 时， $F\left({\lambda }^{*}\right)=1$ 总有负实部，即 $E_0$ 是局部稳定的。反之，当 $R_0>1$ 时， $E_0$ 是不稳定的。

3.3. 无病平衡点的全局稳定性

在本节中讨论了当 $R_0<1$ 时无病稳态 $E_0$ 的全局稳定性，并得到如下结果。

定理2 如果 $R_0<1$ ，则无病平衡点 $E_0$ 是全局渐近稳定的。

证明 只需要证明当 $t\to \infty$ 时，下列极限成立

$\{\begin{array}{l}\underset{t\to \infty }{\lim }s\left(a,t\right)=s^0\left(a\right)={\text{e}}^{-{\displaystyle {\int }_0^a\delta \left(\tau \right)\text{d}\tau }},\\ \underset{t\to \infty }{\lim }v\left(a,t\right)=v^0\left(a\right)={\displaystyle {\int }_0^a\delta \left(\tau \right)s\left(\tau \right){\text{e}}^{-{\displaystyle {\int }_{\tau }^{a}m}\text{d}s}\text{d}\tau },\\ \underset{t\to \infty }{\lim }i\left(a,t\right)=i^0\left(a\right)=0,\\ \underset{t\to \infty }{\lim }q\left(a,t\right)=q^0\left(a\right)=0.\end{array}$ (9)

利用特征线法求解 $s\left(a,t\right),v\left(a,t\right),i\left(a,t\right),q\left(a,t\right)$ 。因为我们想研究 $t\to \infty$ 时的解，所以只考虑当 $t>a$ 时的解。可以得到

$\{\begin{array}{l}s\left(a,t\right)={\text{e}}^{-{\displaystyle {\int }_0^a\left[\lambda \left(\tau +t-a,\tau \right)+\delta \left(\tau \right)\right]\text{d}\tau }},\\ v\left(a,t\right)={\displaystyle {\int }_0^a\delta \left(\tau \right)s\left(\tau +t-a,\tau \right){\text{e}}^{-{\displaystyle {\int }_{\tau }^{a}m\text{d}s}}\text{d}\tau ,}\\ i\left(a,t\right)={\displaystyle {\int }_0^a\lambda \left(\tau +t-a\right)s\left(\tau +t-a,\tau \right){\text{e}}^{-{\displaystyle {\int }_{\tau }^a\left[\rho \left(s\right)+\sigma \left(s\right)\right]\text{d}s}}\text{d}\tau ,}\\ q\left(a,t\right)={\displaystyle {\int }_0^a\rho \left(\tau \right)i\left(\tau +t-a,\tau \right){\text{e}}^{-{\displaystyle {\int }_{\tau }^a\gamma \left(s\right)\text{d}s}}\text{d}\tau .}\end{array}$ (10)

当 $t>a$ 时，将 $s\left(a,t\right)$ 的表达式代入 $i\left(a,t\right)$ ，可以获得

$i\left(a,t\right)={\displaystyle {\int }_0^a\lambda \left(\tau +t-a\right)s\left(\tau +t-a,\tau \right){\text{e}}^{-{\displaystyle {\int }_{\tau }^a\left[\rho \left(s\right)+\sigma \left(s\right)\right]\text{d}s}}\text{d}\tau }$ 。

令 $\chi \left(t\right)={\displaystyle {\int }_0^{a^{+}}\tilde{\beta }\left(a\right)i\left(t,a\right)\text{d}a}$ ，则 $\lambda \left(a,t\right)$ 可以表示为

$\lambda \left(a,t\right)=k\left(a\right)\chi \left(t\right)$ 。

将 $i\left(a,t\right)$ 代入 $\chi \left(t\right)$ 中，可以获得

$\begin{array}{c}\chi \left(t\right)={\displaystyle {\int }_0^{a^{+}}\tilde{\beta }\left(a\right){\displaystyle {\int }_0^a\lambda \left(\tau +t-a\right)s\left(\tau +t-a,\tau \right){\text{e}}^{-{\displaystyle {\int }_{\tau }^a\left[\rho \left(s\right)+\sigma \left(s\right)\right]\text{d}s}}\text{d}\tau }}\text{d}a\\ ={\displaystyle {\int }_0^t\tilde{\beta }\left(a\right){\displaystyle {\int }_0^a\lambda \left(\tau +t-a\right)s\left(\tau +t-a,\tau \right){\text{e}}^{-{\displaystyle {\int }_{\tau }^a\left[\rho \left(s\right)+\sigma \left(s\right)\right]\text{d}s}}\text{d}\tau }}\text{d}a+{\displaystyle {\int }_t^{a^{+}}\tilde{\beta }\left(a\right)i\left(t,a\right)\text{d}a}.\end{array}$

显然，当t很大时，上面最后一项为零。

此外，由于 $\lambda \left(a,t\right)>0$ ，所以

$s\left(\tau +t-a,\tau \right)={\text{e}}^{-{\displaystyle {\int }_0^a\left[\lambda \left(\tau +t-a,\tau \right)+\delta \left(\tau \right)\right]\text{d}\tau }}\le {\text{e}}^{-{\displaystyle {\int }_0^a\delta \left(\tau \right)\text{d}\tau }}=s^0\left(\tau \right)$ 。

因此

$\begin{array}{c}\chi \left(t\right)={\displaystyle {\int }_0^{a^{+}}\tilde{\beta }\left(a\right){\displaystyle {\int }_0^{a}k\left(\tau \right)\chi \left(\tau +t-a\right)s\left(\tau +t-a,\tau \right){\text{e}}^{-{\displaystyle {\int }_{\tau }^a\left[\rho \left(s\right)+\sigma \left(s\right)\right]\text{d}s}}\text{d}\tau }}\text{d}a\\ \le {\displaystyle {\int }_0^{a^{+}}\tilde{\beta }\left(a\right){\displaystyle {\int }_0^a{\text{e}}^{-{\displaystyle {\int }_{\tau }^a\left[\rho \left(s\right)+\sigma \left(s\right)\right]\text{d}s}}{s}^0\left(\tau \right)k\left(\tau \right)\chi \left(\tau +t-a\right)\text{d}\tau }}\text{d}a.\end{array}$

又因为

$R_0={\displaystyle {\int }_0^{a^{+}}\tilde{\beta }\left(a\right)}{\displaystyle {\int }_0^a{\text{e}}^{-{\displaystyle {\int }_{\tau }^a\left[\rho \left(s\right)+\sigma \left(s\right)\right]\text{d}s}}k\left(\tau \right)s^0\left(\tau \right)\text{d}\tau \text{d}a},$

对 $\chi \left(t\right)$ 两边取极限可得

$\underset{t\to \infty }{\lim }\sup \chi \left(t\right)\le R_0\underset{t\to \infty }{\lim }\sup \chi \left(t\right)$ ，

当 $R_0<1$ 时

$\underset{t\to \infty }{\lim }\sup \chi \left(t\right)=0$ 。

因为 $\chi \left(t\right)\ge 0,\chi \left(t\right)={\displaystyle {\int }_0^{a^{+}}\tilde{\beta }\left(a\right)i\left(t,a\right)\text{d}a}$ ，可以获得

$\begin{array}{l}\underset{t\to \infty }{\lim }\sup i\left(a,t\right)=0=i^0\left(a\right),\\ \underset{t\to \infty }{\lim }\sup q\left(a,t\right)=0=q^0\left(a\right).\end{array}$

把 $s\left(t,a\right)$ 的表达式代入得

$\begin{array}{l}\underset{t\to \infty }{\lim }\sup s\left(t,a\right)\\ =\underset{t\to \infty }{\lim }\sup {\text{e}}^{-{\displaystyle {\int }_0^a\left[\lambda \left(\tau +t-a,\tau \right)+\delta \left(\tau \right)\right]\text{d}\tau }}\\ =\underset{t\to \infty }{\lim }\sup {\text{e}}^{-{\displaystyle {\int }_0^a\left[k\left(\xi \right)\chi \left(\xi +t-a\right)+\delta \left(\xi \right)\right]\text{d}\xi }}\\ ={\text{e}}^{-{\displaystyle {\int }_0^a\delta \left(\xi \right)\text{d}\xi }}=s^o\left(a\right),\end{array}$

同理 $\underset{t\to \infty }{\lim }\sup v\left(t,a\right)=v^0\left(a\right)$ 。

即 $R_0<1$ ，则无病平衡点 $E^0$ 是全局渐近稳定的。

4. 地方病平衡点的存在性和稳定性

4.1. 地方病平衡点的存在性

定理3 如果 $R_0>1$ ，则系统存在唯一的地方病平衡点 $E^{*}=\left(s^{*}\left(a\right),v^{*}\left(a\right),i^{*}\left(a\right),q^{*}\left(a\right)\right)$ 。

证明 当系统达到稳定的年龄分布时，系统只与年龄有关，则系统的稳态解满足如下形式

$\{\begin{array}{l}\frac{\text{d}{s}^{*}}{\text{d}a}=-\left[{\lambda }^{*}\left(a\right)+\delta \left(a\right)\right]s^{*}\left(a\right),\\ \frac{\text{d}{v}^{*}}{\text{d}a}=\delta \left(a\right)s^{*}\left(a\right)-m{v}^{*}\left(a\right),\\ \frac{\text{d}{i}^{*}}{\text{d}a}={\lambda }^{*}\left(a\right)s^{*}\left(a\right)-\left[\rho \left(a\right)+\sigma \left(a\right)\right]i^{*}\left(a\right),\\ \frac{\text{d}{q}^{*}}{\text{d}a}=\rho \left(a\right)i^{*}\left(a\right)-\gamma \left(a\right)q^{*}\left(a\right),\end{array}$ (11)

初始条件 ${\text{s}}^{*}\left(0\right)=1,v^{*}\left(0\right)=0,i^{*}\left(0\right)=0,q^{*}\left(0\right)=0,$ 其中 ${\lambda }^{*}\left(a\right)=k\left(a\right){\displaystyle {\int }_0^{a^{+}}\tilde{\beta }\left(a\right)i^{*}\left(a\right)\text{d}a}=k\left(a\right){\Lambda }^{*},{\Lambda }^{*}={\displaystyle {\int }_0^{a^{+}}\tilde{\beta }\left(a\right)i^{*}\left(a\right)\text{d}a}$ 。

系统的解为

$\{\begin{array}{l}{s}^{*}\left(a\right)={\text{e}}^{-{\displaystyle {\int }_0^a\lambda \left(\tau \right)+\delta \left(\tau \right)\text{d}\tau }},\\ v^{*}\left(a\right)={\displaystyle {\int }_0^a\delta \left(\tau \right)s^{*}\left(\tau \right){\text{e}}^{-{\displaystyle {\int }_{\tau }^{a}m\text{d}s}}\text{d}\tau },\\ i^{*}\left(a\right)={\Lambda }^{*}{\displaystyle {\int }_0^{a}k\left(\tau \right){\text{e}}^{-{\displaystyle {\int }_0^a\left[k\left(\tau \right){\Lambda }^{*}+\delta \left(\tau \right)\right]\text{d}\tau }}{\text{e}}^{-{\displaystyle {\int }_{\tau }^a\left[\rho \left(s\right)+\sigma \left(s\right)\right]\text{d}s}}\text{d}\tau },\\ q^{*}\left(a\right)={\displaystyle {\int }_0^a\rho \left(\tau \right)i^{*}\left(\tau \right){\text{e}}^{-{\displaystyle {\int }_{\tau }^a\gamma \left(s\right)}ds}\text{d}\tau }.\end{array}$ (12)

由此可见，当 ${\Lambda }^{*}=0$ 时，地方病平衡点 $E^{*}=\left(s^{*}\left(a\right),v^{*}\left(a\right),i^{*}\left(a\right),q^{*}\left(a\right)\right)$ 转变为无病平衡点 $E^0=\left(s^0\left(a\right),v^0\left(a\right),i^0\left(a\right),q^0\left(a\right)\right)$ 。

将 $i^{*}\left(a\right)$ 的表达式代入 ${\Lambda }^{*}={\displaystyle {\int }_0^{a^{+}}\tilde{\beta }\left(a\right)i^{*}\left(a\right)\text{d}a}$ ，又因为 ${\Lambda }^{*}\ne 0$ ，可得

$1={\displaystyle {\int }_0^{a}k\left(\tau \right){\text{e}}^{-{\displaystyle {\int }_0^a\left[k\left(\tau \right){\Lambda }^{*}+\delta \left(\tau \right)\right]\text{d}\tau }}{\text{e}}^{-{\displaystyle {\int }_{\tau }^a\left[\rho \left(s\right)+\sigma \left(s\right)\right]\text{d}s}}\text{d}\tau .}$

用 $H\left({\Lambda }^{*}\right)$ 来表示右侧

$H\left({\Lambda }^{*}\right)={\displaystyle {\int }_0^{a}k\left(\tau \right){\text{e}}^{-{\displaystyle {\int }_0^a\left[k\left(\tau \right){\Lambda }^{*}+\delta \left(\tau \right)\right]\text{d}\tau }}{\text{e}}^{-{\displaystyle {\int }_{\tau }^a\left[\rho \left(s\right)+\sigma \left(s\right)\right]\text{d}s}}\text{d}\tau }.$

Figure 2. The approximate graph of $H(\; \Lambda \; *\; )$

图2. $H\left({\Lambda }^{*}\right)$ 的近似图

从 $H\left({\Lambda }^{*}\right)$ 的表达式中，可以得到以下的性质

$H^{\prime }\left({\Lambda }^{*}\right)<0,\underset{{\Lambda }^{*}\to -\infty }{\lim }H\left({\Lambda }^{*}\right)=+\infty ,H\left(0\right)=0,$

$\underset{{\Lambda }^{*}\to +\infty }{\lim }H\left({\Lambda }^{*}\right)=\underset{{\Lambda }^{*}\to +\infty }{\lim }\frac{1}{{\Lambda }^{*}}{\displaystyle {\int }_0^{a^{+}}\tilde{\beta }\left(a\right)i^{*}\left(a\right)\text{d}a}\le \underset{{\Lambda }^{*}\to +\infty }{\lim }\frac{1}{{\Lambda }^{*}}{\displaystyle {\int }_0^{a^{+}}\tilde{\beta }\left(a\right)\text{d}a}=0$ 。

关于 $H\left({\Lambda }^{*}\right)$ 的近似图如图2所示，由此表明，当 $R_0>1$ 时， $H\left({\Lambda }^{*}\right)=1$ 有唯一的正实根 ${\Lambda }_{~}^{*}$ 。则系统具有唯一的正地方性平衡点 $E^{*}=\left(s^{*}\left(a\right),v^{*}\left(a\right),i^{*}\left(a\right),q^{*}\left(a\right)\right)$ 。

4.2. 地方病平衡点的局部稳定性

为了分析 $R_0>1$ 时，地方病平衡点的局部稳定性，还需要讨论在 $E^{*}$ 处的线性化系统，令

$\begin{array}{l}s\left(a,t\right)=s^{*}\left(a\right)+\tilde{s}\left(a,t\right),v\left(a,t\right)=v^{*}\left(a\right)+\tilde{v}\left(a,t\right),\\ i\left(a,t\right)=i^{*}\left(a\right)+\tilde{i}\left(a,t\right),q\left(a,t\right)=q^{*}\left(a\right)+\tilde{q}\left(a,t\right).\end{array}$

取其线性部分，系统变为如下形式

$\{\begin{array}{l}\frac{\partial \tilde{s}}{\partial a}+\frac{\partial \tilde{s}}{\partial t}=-{\lambda }^{*}\left(a\right)\tilde{s}\left(a,t\right)-\delta \left(a\right)\tilde{s}\left(a,t\right)-\tilde{\lambda }\left(a,t\right)s^{*}\left(a\right),\\ \frac{\partial \tilde{v}}{\partial a}+\frac{\partial \tilde{v}}{\partial t}=\delta \left(a\right)\tilde{s}\left(a,t\right)-m\tilde{v}\left(a,t\right),\\ \frac{\partial \tilde{i}}{\partial a}+\frac{\partial \tilde{i}}{\partial t}={\lambda }^{*}\left(a\right)\tilde{s}\left(a,t\right)+\tilde{\lambda }\left(a,t\right)s^{*}\left(a\right)-\left[\rho \left(a\right)+\sigma \left(a\right)\right]\tilde{i}\left(a,t\right),\\ \frac{\partial \tilde{q}}{\partial a}+\frac{\partial \tilde{q}}{\partial t}=\rho \left(a\right)\tilde{i}\left(a,t\right)-\gamma \left(a\right)\tilde{q}\left(a,t\right),\end{array}$ (13)

边界条件

$\text{<mover accent="true"> s <mo> ˜ </mo> </mover>}\left(0,t\right)=0,\tilde{v}\left(0,t\right)=0,\tilde{i}\left(0,t\right)=0,\tilde{q}\left(0,t\right)=0,$

其中

$\tilde{\lambda }\left(a,t\right)=k\left(a\right){\displaystyle {\int }_0^{a^{+}}\tilde{\beta }\left(a\right)\tilde{i}\left(a,t\right)\text{d}a},{\lambda }^{*}\left(a\right)=k\left(a\right){\Lambda }_{~}^{*},{\Lambda }_{~}^{*}={\displaystyle {\int }_0^{a^{+}}\tilde{\beta }\left(a\right)i^{*}\left(a\right)\text{d}a}$ 。

现在考虑系统的非零指数解，令

$\tilde{s}\left(a,t\right)=\tilde{s}\left(a\right){\text{e}}^{\omega t},\tilde{v}\left(a,t\right)=\tilde{v}\left(a\right){\text{e}}^{\omega t},\tilde{i}\left(a,t\right)=\tilde{i}\left(a\right){\text{e}}^{\omega t},\tilde{q}\left(a,t\right)=\tilde{q}\left(a\right){\text{e}}^{\omega t}.$

则系统变为如下形式

$\{\begin{array}{l}\frac{\text{d}\tilde{s}}{\text{d}a}=-\omega \tilde{s}\left(a\right)-{\Lambda }_{~}^{*}k\left(a\right)\tilde{s}\left(a\right)-\delta \left(a\right)\tilde{s}\left(a\right)-\tilde{\Lambda }k\left(a\right)s^{*}\left(a\right),\\ \frac{\text{d}\tilde{v}}{\text{d}a}=-\omega \tilde{v}\left(a\right)+\delta \left(a\right)\tilde{s}\left(a\right)-m\tilde{v}\left(a\right),\\ \frac{\text{d}\tilde{i}}{\text{d}a}=-\omega \tilde{i}\left(a\right)-\left[\rho \left(a\right)+\sigma \left(a\right)\right]\tilde{i}\left(a\right)+{\Lambda }_{~}^{*}k\left(a\right)\tilde{s}\left(a\right)+\tilde{\Lambda }k\left(a\right)s^{*}\left(a\right),\\ \frac{\text{d}\tilde{q}}{\text{d}a}=-\omega \tilde{q}\left(a\right)+\rho \left(a\right)\tilde{i}\left(a\right)-\gamma \left(a\right)\tilde{q}\left(a\right),\end{array}$ (14)

初始条件

$\text{<mover accent="true"> s <mo> ˜ </mo> </mover>}\left(0\right)=0,\tilde{v}\left(0\right)=0,\tilde{i}\left(0\right)=0,\tilde{q}\left(0\right)=0,$

其中

$\tilde{\Lambda }={\displaystyle {\int }_0^{a^{+}}\tilde{\beta }\left(a\right)\tilde{i}\left(a\right)\text{d}a}.$

因为 $\tilde{\Lambda }\ne 0$ ，令 $\hat{s}=\frac{\tilde{s}}{\tilde{\Lambda }},\hat{v}=\frac{\tilde{v}}{\tilde{\Lambda }},\hat{i}=\frac{\tilde{i}}{\tilde{\Lambda }},\hat{q}=\frac{\tilde{q}}{\tilde{\Lambda }}$ 。这个系统变为

$\{\begin{array}{l}\frac{\text{d}\hat{s}}{\text{d}a}=-\omega \hat{s}\left(a\right)-{\Lambda }_{~}^{*}k\left(a\right)\hat{s}\left(a\right)-\delta \left(a\right)\hat{s}\left(a\right)-k\left(a\right)s^{*}\left(a\right),\\ \frac{\text{d}\hat{v}}{\text{d}a}=-\omega \hat{v}\left(a\right)+\delta \left(a\right)\hat{s}\left(a\right)-m\hat{v}\left(a\right),\\ \frac{\text{d}\hat{i}}{\text{d}a}=-\omega \hat{i}\left(a\right)-\left[\rho \left(a\right)+\sigma \left(a\right)\right]\hat{i}\left(a\right)+{\Lambda }_{~}^{*}k\left(a\right)\hat{s}\left(a\right)+k\left(a\right)s^{*}\left(a\right),\\ \frac{\text{d}\hat{q}}{\text{d}a}=-\omega \hat{q}\left(a\right)+\rho \left(a\right)\hat{i}\left(a\right)-\gamma \left(a\right)\hat{q}\left(a\right),\end{array}$ (15)

其中

$1={\displaystyle {\int }_0^{a^{+}}\tilde{\beta }\left(a\right)\hat{i}\left(a\right)\text{d}a}.$

系统(15)的解为

$\{\begin{array}{l}\hat{s}\left(a\right)=-{{\displaystyle {\int }_0^a\text{e}}}^{-{\displaystyle {\int }_{\eta }^a\left[\omega +k\left(s\right){\Lambda }_{~}^{*}+\delta \left(s\right)\right]\text{d}s}}{s}^{*}\left(\eta \right)k\left(\eta \right)\text{d}\eta ,\\ \hat{v}\left(a\right)={{\displaystyle {\int }_0^a\text{e}}}^{-{\displaystyle {\int }_{\eta }^a\left(\omega +m\right)\text{d}s}}\left[\delta \left(\eta \right)\hat{s}\left(\eta \right)\right]\text{d}\eta ,\\ \hat{i}\left(a\right)={{\displaystyle {\int }_0^a\text{e}}}^{-{\displaystyle {\int }_{\eta }^a\left[\omega +\rho \left(s\right)+\sigma \left(s\right)\right]\text{d}s}}\left[{\Lambda }_{~}^{*}k\left(\eta \right)\hat{s}\left(\eta \right)+k\left(\eta \right)s^{*}\left(\eta \right)\right]\text{d}\eta ,\\ \hat{q}\left(a\right)={{\displaystyle {\int }_0^a\text{e}}}^{-{\displaystyle {\int }_{\eta }^a\left[\omega +\gamma \left(s\right)\right]\text{d}s}}\rho \left(\eta \right)\hat{i}\left(\eta \right)\text{d}\eta .\end{array}$ (16)

令 $\phi \left(\omega \right)={\displaystyle {\int }_0^{a^{+}}\tilde{\beta }\left(a\right)\hat{i}\left(a\right)\text{d}a}$ ，当 $R_0>1$ 时，我们想要证明 $\phi \left(\omega \right)=1$ 具有负实部。

$\begin{array}{c}\phi \left(\omega \right)={\displaystyle {\int }_0^{a^{+}}\tilde{\beta }\left(a\right)}{{\displaystyle {\int }_0^a\text{e}}}^{-{\displaystyle {\int }_{\eta }^a\left(\omega +\rho \left(s\right)+\sigma \left(s\right)\right)\text{d}s}}\left({\Lambda }_{~}^{*}k\left(\eta \right)\hat{s}\left(\eta \right)+k\left(\eta \right)s^{*}\left(\eta \right)\right)\text{d}\eta \text{d}a\\ ={\displaystyle {\int }_0^{a^{+}}\tilde{\beta }\left(a\right)}{{\displaystyle {\int }_0^a\text{e}}}^{-{\displaystyle {\int }_{\eta }^a\left(\omega +\rho \left(s\right)+\sigma \left(s\right)\right)\text{d}s}}\left[\tilde{\Lambda }k\left(\eta \right)s^{*}\left(\eta \right)-{\Lambda }_{~}^{*}k\left(\eta \right){{\displaystyle {\int }_0^{\eta }\text{e}}}^{-{\displaystyle {\int }_{\xi }^{\eta }\left(\omega +k\left(s\right){\Lambda }_{~}^{*}+\delta \left(s\right)\right)\text{d}s}}{s}^{*}\left(\xi \right)k\left(\xi \right)\text{d}\xi \right]\text{d}\eta \text{d}a,\end{array}$

并且

$\begin{array}{c}\phi \left(0\right)={\displaystyle {\int }_0^{a^{+}}\tilde{\beta }\left(a\right)}{{\displaystyle {\int }_0^a\text{e}}}^{-{\displaystyle {\int }_{\eta }^a\left[\rho \left(s\right)+\sigma \left(s\right)\right]\text{d}s}}\left({\Lambda }_{~}^{*}k\left(\eta \right)\hat{s}\left(\eta \right)+k\left(\eta \right)s^{*}\left(\eta \right)\right)\text{d}\eta \text{d}a\\ ={\displaystyle {\int }_0^{a^{+}}\tilde{\beta }\left(a\right)}{{\displaystyle {\int }_0^a\text{e}}}^{-{\displaystyle {\int }_{\eta }^a\left[\rho \left(s\right)+\sigma \left(s\right)\right]\text{d}s}}\left[k\left(\eta \right)s^{*}\left(\eta \right)-{\Lambda }_{~}^{*}k\left(\eta \right){{\displaystyle {\int }_0^{\eta }\text{e}}}^{-{\displaystyle {\int }_{\xi }^{\eta }\left(k\left(s\right){\Lambda }_{~}^{*}+\delta \left(s\right)\right)\text{d}s}}{s}^{*}\left(\xi \right)k\left(\xi \right)\text{d}\xi \right]\text{d}\eta \text{d}a.\end{array}$

根据 $H\left({\Lambda }_{~}^{*}\right)$ 的表达式

$H\left({\Lambda }_{~}^{*}\right)={\displaystyle {\int }_0^{a^{+}}\tilde{\beta }\left(a\right)}{\displaystyle {\int }_0^a{s}^{*}\left(\eta \right)k\left(\eta \right){\text{e}}^{-{\displaystyle {\int }_{\eta }^a\left[\rho \left(s\right)+\sigma \left(s\right)\right]\text{d}s}}\text{d}\eta \text{d}a}=1.$

这个 $\phi \left(0\right)$ 可以转变为

$\phi \left(0\right)=1-{\displaystyle {\int }_0^{a^{+}}\tilde{\beta }\left(a\right)}{{\displaystyle {\int }_0^a\text{e}}}^{-{\displaystyle {\int }_{\eta }^a\left[\rho \left(s\right)+\sigma \left(s\right)\right]\text{d}s}}{\Lambda }_{~}^{*}k\left(\eta \right){{\displaystyle {\int }_0^{\eta }\text{e}}}^{-{\displaystyle {\int }_{\xi }^{\eta }\left(k\left(s\right){\Lambda }_{~}^{*}+\delta \left(s\right)\right)\text{d}s}}{s}^{*}\left(\xi \right)k\left(\xi \right)\text{d}\xi \text{d}\eta \text{d}a<1.$

Figure 3. The approximate graph of $\phi (\; \omega \; )$

图3. $\phi \left(\omega \right)$ 的近似图

由于 ${\Lambda }_{~}^{*}>0$ ，所以

$\phi \left(\omega \right)<{\displaystyle {\int }_0^{a^{+}}\tilde{\beta }\left(a\right)}{\displaystyle {\int }_0^a{s}^{*}\left(\eta \right)k\left(\eta \right){\text{e}}^{-{\displaystyle {\int }_{\eta }^a\left[\omega +\rho \left(s\right)+\sigma \left(s\right)\right]\text{d}s}}}\text{d}\eta \text{d}a\triangleq \psi \left(\omega \right).$

关于 $\phi \left(\omega \right)$ 的近似图如图3所示。由于 $\psi \left(\omega \right)$ 是单调递减的，因此对于所有 $\mathrm{Re}\omega >0$ ，有 $\phi \left(\omega \right)<\psi \left(0\right)=H\left({\Lambda }_{~}^{*}\right)=1$ ，所以 $\phi \left(\omega \right)=1$ 只发生在 $\mathrm{Re}\omega <0$ 的区域。因此，如果 $R_0>1$ ，则 $\phi \left(\omega \right)=1$ 的所有根都有负实部。

定理4 如果 $R_0>1$ ，则地方性平衡点 $E^{*}=\left(s^{*}\left(a\right),v^{*}\left(a\right),i^{*}\left(a\right),q^{*}\left(a\right)\right)$ 局部渐近稳定。

证明 证明和上述定理1证明类似。

5. 总结

本文在文献 [5] 的基础上对原有模型进行改进，考虑了隔离类，并将疫苗接种者列为单独一类进行讨论，建立了一类具有接种和隔离的年龄结构SVIQR传染病模型。分析计算得到模型的基本再生数

$R_0={\displaystyle {\int }_0^{a^{+}}\tilde{\beta }\left(a\right)}{\displaystyle {\int }_0^{a}k\left(\tau \right)s^0\left(\tau \right){\text{e}}^{-{\displaystyle {\int }_{\tau }^a\left[\rho \left(s\right)+\sigma \left(s\right)\right]\text{d}s}}}\text{d}\tau \text{d}a$ 。证明了当 $R_0<1$ 时，无病平衡点 $E^0$ 全局渐近稳定；当 $R_0>1$ 时，

无病平衡点 $E^0$ 不稳定，此时系统存在地方病平衡点 $E^{*}$ ，证明了当 $R_0>1$ 时，地方病平衡点 $E^{*}$ 是局部渐进稳定。从基本再生数的表达式可以看出，隔离率 $\rho \left(a\right)$ 或疫苗接种率 $\delta \left(a\right)$ 增大，则基本再生数 $R_0$ 减小，即隔离和提高疫苗接种率是消除传染病的有效方式。

基金项目

国家自然科学基金(11961039)；甘肃省高等学校青年博士项目(2022QB-056)。

NOTES

^*通讯作者。