麻疹、肺结核、艾滋病等许多传染病的感染者在疾病发展的某个阶段会呈现与年龄相关联的特征, 这是影响疾病传播的重要因素之一. 麻疹是由麻疹病毒引起的呼吸道传染病, 病毒存在于患者口、鼻、咽、眼结合膜分泌物内, 通过喷嚏、咳嗽和说话等由飞沫传播, 传染性极强, 易感者接触后其发病率可高达95% 以上, 通常发病年龄为1–5周岁, 约占总发病数的80% 左右. 目前避免发病传染最有效的办法是接种疫苗, 接种疫苗后7–12天就可产生抗体, 比直接感染后产生抗体的时间短, 因此对易感者进行应急接种可控制疫情蔓延或终止流行. 国内规定初种年龄为8个月左右, 但仍有10%–30% 的人群接种后其免疫力随时间的推移而逐年下降最后成为易感者^{[1, 2]}. 麻疹潜伏期多为7–14天, 最长可达21天.麻疹痊愈与患者年龄有关, 患者痊愈后, 由于防护不力会感染麻疹病毒而再次发病. 麻疹主要是在儿童间发生的传染病, 故年龄特征在麻疹传染病的传播动力学中扮演着非常重要的角色. 近年来研究麻疹疾病的文章已有许多, 如姜翠翠等^[3]对具有部分免疫和潜伏期的麻疹传染病模型的稳定性进行了分析; 佘文兵^[4]建立了一类具有接种的麻疹模型; 靖晓洁等^[5]考虑了部分免疫和环境传播的麻疹传染病模型的全局稳定性; Huang等^[6]考虑了一个有周期性传播率的麻疹模型, 并分析其动力学行为, 提高了人们对麻疹的认识; Zhou等^[7]验证了具有年龄结构的离散型SIR流行病模型的稳定性, 并将这些理论结果应用在麻疹疫苗的接种策略中. 然而考虑年龄结构的麻疹传染病模型的研究工作相对较少, 故建立疾病接种、复发的年龄结构麻疹模型, 并对其动力学性质进行研究有一定的理论意义和实际应用价值.

本文内容安排如下: 在第2节, 建立了一类年龄结构的麻疹传染病动力学模型; 在第3节, 讨论模型解的适定性; 在第4节, 证明疾病的一致持续性; 在第5节, 建立平衡态的局部和全局稳定性的判别准则; 在第6节, 给出数值例子和模拟; 最后, 第7节是一个简短的总结.

结合上述麻疹的主要传播特征和机理, 我们建立一类具有接种、复发的年龄结构SVEIR麻疹传染病模型. 将人群分为互不重合的五类, 设$ S(t) $、$ v(t, a) $、$ E(t) $、$ I(t) $和$ r(t, b) $分别表示$ t $时刻易感者的数量、接种年龄为$ a $的人群密度、潜伏者的数量、感染疾病的数量和处在疾病复发年龄为$ b $的治愈人群密度. 对于时刻$ t $接种年龄为$ a $的人群密度$ v(t, a) $和时刻$ t $复发年龄为$ b $的移出者人群密度$ r(t, b) $, 从接种仓室移出到易感仓室与从移出仓室复发到染病仓室的年龄依赖比率分别为$ \omega_{1}(a) $和$ \omega_{2}(b) $, 故时刻$ t $从接种仓室和从移出仓室转出的总量分别为$ \int_{0}^{\infty}\omega_{1}(a)v(t, a){\rm d}a $和$ \int_{0}^{\infty}\omega_{2}(b)r(t, b){\rm d}b $, 建立如下微分方程模型:

(2.1) $ \begin{equation} \left\{ \begin{array}{ll} \frac{{\rm d} S}{{\rm d} t} = \Lambda-(\mu+\xi)S-\beta SI+\int_{0}^{\infty}\omega_1(a)v(t, a){\rm d}a, \\ \frac{\partial v(t, a)}{\partial t}+\frac{\partial v (t, a)}{\partial a} = -(\omega_1(a)+\mu)v(t, a), \\ \frac{{\rm d} E}{{\rm d} t} = \beta SI-(\varepsilon+\mu)E, \\ \frac{{\rm d} I}{{\rm d} t} = \int_{0}^{\infty}\omega_2(b)r(t, b){\rm d}b+\varepsilon E-(\mu+\delta+k)I, \\ \frac{\partial r(t, b)}{\partial t}+\frac{\partial r(t, b)}{\partial b} = -(\omega_2(b)+\mu)r(t, b), \\ v(t, 0) = \xi S, r(t, 0) = kI, t\geq0, \\ S(0) = S_{0}, v(0, a) = v_{0}(a), E(0) = E_{0}, I(0) = I_{0}, r(0, b) = r_{0}(b), \end{array} \right. \end{equation} $

其中$ S_{0}, E_{0}, I_{0}\in{\mathbb R_{+}} = [0, \infty) $和$ v_{0}(\cdot), r_{0}(\cdot)\in L^{1}_{+} $, $ L^{1} = L^{1}((0, \infty), {\mathbb R}) $为映$ \phi:(0, \infty)\rightarrow {\mathbb R} $上的所有Lebesgue绝对可积函数所构成的空间, 而$ L^{1}_{+} = L^{1}_{+}((0, \infty), {\mathbb R_{+}}) $为$ L^{1} $的正锥. 正常数$ \Lambda $、$ \beta $、$ \xi $、$ \varepsilon $、$ \mu $、$ \delta $和$ k $分别表示易感者的输入率、疾病感染率、易感者的接种率、潜伏者的发病率、自然死亡率、因病死亡率和移出者的复发率. 对于函数$ \omega_i(l)\;(i = 1, 2) $, 全文假设如下条件成立.

(A) 函数$ \omega_i(l)\in L^{1}_{+} $, 具有上界$ \bar{\omega_i} $且Lipschitz连续, Lipschitz常数为$ M_{\omega_{i}}(i = 1, 2) $, 其中

$ L^{1}_{+} = \{\phi\in L^{1}[0, \infty): \phi(l)\geq 0, \;\forall\;l\geq 0\}. $

模型(2.1) 的相空间定义为$ {\mathbb Y} = {\mathbb R_{+}}\times L^{1}_{+}\times{\mathbb R_{+}}\times{\mathbb R_{+}}\times L^{1}_{+} $. 对任意$ x = (x_1, x_2, x_3, x_4, x_5) $$ \in{\mathbb Y} $, 其范数定义为

$ \|x\|_{{\mathbb Y}} = |x_1|+\int_{0}^{\infty}|x_2(a)|{\rm d}a+ |x_3|+|x_4|+\int_{0}^{\infty}|x_5(b)|{\rm d}b, $

此范数表示各仓室人口总和. 模型(2.1) 解的初始值为$ x_{0} = (S_{0}, v_{0}(\cdot), E_{0}, I_{0}, r_{0}(\cdot))\in {\mathbb Y} $, 且由模型(2.1) 的兼容性条件得到$ v(0, 0) = \xi S_{0} = v_{0}(0) $和$ r(0, 0) = kI_{0} = r_{0}(0) $.

由文献[8]可知模型(2.1) 在时刻$ t = 0 $通过初始值$ x_{0} $的解是存在唯一且非负, 并且可延拓为一个饱和解. 为了叙述方便, 记这个解为$ F(t, x_0) = (S(t), v(t, \cdot), E(t), I(t), r(t, \cdot)) $, 显然有$ F(0, x_0) = x_0\in{\mathbb Y} $和

$ \|F(t, x_0)\|_{{\mathbb Y}} = S(t)+\int_{0}^{\infty}v(t, a){\rm d}a+E(t)+I(t)+\int_{0}^{\infty}r(t, b){\rm d}b. $

进一步, 给出如下记号:

$ \varepsilon_{i}(l) = \omega_{i}(l)+\mu, \; \rho_{i}(l) = e^{-\int_{0}^{l}\varepsilon_{i}(\tau){\rm d}\tau}, \theta_{i} = \int_{0}^{\infty}\omega_{i}(l)\rho_{i}(l){\rm d}l, \; i = 1, 2. $

由假设(A) 可以推出$ 0\leq\rho_{i}(s)\leq1 $, $ 0\leq\theta_{i}\leq1 $和

(3.1) $ \begin{equation} \frac{{\rm d}\rho_{i}(s)}{{\rm d}s} = -\varepsilon_{i}(s)\rho_{i}(s), \; \forall \; s\geq0, \; i = 1, 2. \end{equation} $

另外, 对任意复数$ \lambda $, 定义如下函数

$ \theta_{i}(\lambda) = \int_{0}^{\infty}\omega_{i}(l) e^{[-(\lambda+\mu)l- \int_{0}^{l}\omega_{i}(\tau){\rm d}\tau]}{\rm d}l, \; i = 1, 2. $

当$ \lambda>0 $为实数时, 我们得到$ \theta_{i}(\lambda)\leq\theta_{i}(0)\triangleq\theta_{i} $. 进一步定义函数

$ \pi_{i}(l) = \int_{l}^{\infty}\omega_{i}(\tau)e^ {-\int_{l}^{\tau}\varepsilon_{i}(s){\rm d}s}{\rm d}\tau, \;i = 1, 2. $

可知$ \pi_{i}(l)>0, \pi_{i}(0) = \theta_{i} $和$ \frac{{\rm d}\pi_{i}(l)}{{\rm d}l} = \pi_{i}(l)\varepsilon_{i}(l)-\omega_{i}(l), \; \forall \; l\geq0, \; i = 1, 2. $应用Volterra公式(参见文献[9]), 沿着$ t-a = const $求解模型(2.1) 的第2、5两个方程可得

(3.2) $ \begin{equation} v(t, a) = \left\{ \begin{array}{ll} v(t-a, 0)e^{-\int_{0}^{a}\varepsilon_{1}(s){\rm d}s} = \xi S(t-a)\rho_{1}(a), \;&t>a\geq0, \\ v_{0}(a-t)e^{-\int_{a-t}^{a}\varepsilon_{1}(s){\rm d}s} = v_{0}(a-t)\frac{\rho_{1}(a)}{\rho_{1}(a-t)}, \;&a\geq t, \end{array}\right. \end{equation} $

(3.3) $ \begin{equation} r(t, b) = \left\{ \begin{array}{ll} r(t-b, 0)e^{-\int_{0}^{b}\varepsilon_{2}(s){\rm d}s} = kI(t-b)\rho_{2}(b), \;&t>b\geq0, \\ r_{0}(b-t)e^{-\int_{b-t}^{b}\varepsilon_{2}(s){\rm d}s} = r_{0}(b-t)\frac{\rho_{2}(b)}{\rho_{2}(b-t)}, \;&b\geq t. \end{array}\right. \end{equation} $

将模型(2.1) 的第1、3、4方程和式(3.2)、(3.3) 放在一起，则模型(2.1) 被表示为所谓的Volterra型积分方程.

定义集合

$ \Pi = \Big\{(S, v(\cdot), E, I, r(\cdot))\in {\mathbb Y}: S+\int_{0}^{\infty}v(a){\rm d}a+E+I+\int_{0}^{\infty}r(b){\rm d}b\leq\frac{\Lambda}{\mu}\Big\}, $

有如下结果.

定理3.1  $ {\rm (i)} $$ \Pi $是模型(2.1) 的正向不变集, 即若$ x_{0}\in\Pi $, 当$ t>0 $时, 则$ F(t, x_{0})\in\Pi $.

$ {\rm (ii)} $模型(2.1) 的解最终有界, 即对任意$ x_{0}\in{\mathbb Y} $有$ \limsup\limits_{t\rightarrow \infty}\|F(t, x_0)\|_{{\mathbb Y}} \leq\frac{\Lambda}{\mu}. $

证  对任意初值$ x_{0}\in{{\mathbb Y}} $, 对模型(2.1) 的解范数$ \|F(t, x_0)\|_{{\mathbb Y}} $求导数. 由于

$ \int_{0}^{\infty}\frac{\partial v(t, a)}{\partial t}{\rm d}a = \int_{0}^{\infty}\Big[-(\omega_{1}(a)+\mu)v(t, a)-\frac{\partial v(t, a)}{\partial a}\Big]{\rm d}a\leq -\int_{0}^{\infty}\varepsilon_{1}(a)v(t, a){\rm d}a+\xi S, $

$ \int_{0}^{\infty}\frac{\partial r(t, b)}{\partial t}{\rm d}b = \int_{0}^{\infty}\Big[-(\omega_{2}(b)+\mu)r(t, b)-\frac{\partial r(t, b)}{\partial b}\Big]{\rm d}b \leq -\int_{0}^{\infty}\varepsilon_{2}(b)r(t, b){\rm d}b+kI. $

那么从模型(2.1) 可得

$ \begin{eqnarray*} \frac{{\rm d}\|F(t, x_0)\|_{{\mathbb Y}}}{{\rm d}t} & = & \frac{{\rm d}S}{{\rm d}t}+\int_{0}^{\infty}\frac{\partial v(t, a)}{\partial t}{\rm d}a+\frac{{\rm d}E}{{\rm d}t}+\frac{{\rm d}I}{{\rm d}t}+\int_{0}^{\infty}\frac{\partial r(t, b)}{\partial t}{\rm d}b\\ &\leq& \Lambda-(\mu+\xi)S- \beta SI+\int_{0}^{\infty}\omega_1(a)v(t, a){\rm d}a\\ &&-\int_{0}^{\infty}\varepsilon_1(a)v(t, a){\rm d}a+\xi S+\beta SI -(\varepsilon+\mu)E\\ &&+\int_{0}^{\infty}\omega_{2}(b)r(t, b){\rm d}b+\varepsilon E-(\mu+\delta+k)I -\int_{0}^{\infty}\varepsilon_{2}(b)r(t, b){\rm d}b+kI\\ &\leq& \Lambda-\mu\|F(t, x_0)\|_{{\mathbb Y}}. \end{eqnarray*} $

由此进一步得到

(3.4) $ \begin{equation} \|F(t, x_0)\|_{{\mathbb Y}} \leq\frac{\Lambda}{\mu}-e^{-\mu t}\Big(\frac{\Lambda}{\mu}-\|x_0\|_{{\mathbb Y}}\Big). \end{equation} $

当$ x_{0}\in\Pi $时, 由式(3.4) 得知对任意$ t>0 $有$ F(t, x_{0})\in\Pi $, 这表明$ \Pi $集合是模型(2.1) 的正向不变集. 因此结论$ ({\rm{i}})$成立.

进一步由式(3.4) 得知, 对任意$ x_{0}\in{\mathbb Y} $, $ \|F(t, x_0)\|\leq\frac{\Lambda}{\mu}+\|x_0\|_{{\mathbb Y}} $成立. 这表明解$ F(t, x_0) $在其定义区间上有界. 因此$ F(t, x_0) $能被延拓到整个$ [0, \infty) $上. 进一步有$ \limsup\limits_{t\rightarrow \infty}\|F(t, x_0)\|_{{\mathbb Y}} \leq\frac{\Lambda}{\mu} $, 这表明$ F(t, x_{0}) $最终有界, 因此结论(ii) 成立. 定理3.1得证.

由定理3.1, 根据半流定义, 模型(2.1) 的解$ F(t, x_{0}) $在$ t\geq0 $时构成一个解半流.

引理3.1  设常数$ M \geq \frac{\Lambda}{\mu} $, 初值$ x_{0}\in{\mathbb Y} $. 如果$ \|x_0\|_{{\mathbb Y}}\leq M $, 则对任意$ t\geq0 $, 解$ F(t, x_0) = (S(t), v(t, a), E(t), I(t), r(t, b)) $满足如下结论:

$ {\rm (i)} $$ 0\leq S(t), \; \int_{0}^{\infty}v(t, a){\rm d}a, \; E(t), \; I(t), \; \int_{0}^{\infty}r(t, b){\rm d}b\leq M; $

$ {\rm (ii)} $$ v(t, 0)\leq\xi M, \; r(t, 0)\leq k M. $

证  若$ \|x_{0}\|_{{\mathbb Y}}\leq\frac{\Lambda}{\mu} $, 则对任意$ t\geq0 $有$ \|F(t, x_{0})\|_{{\mathbb Y}}\leq\frac{\Lambda}{\mu}-e^{-\mu t}\Big(\frac{\Lambda}{\mu}-\|x_{0}\|_{{\mathbb Y}}\Big)\leq\frac{\Lambda}{\mu}\leq M. $若$ \frac{\Lambda}{\mu}\leq\|x_{0}\|_{{\mathbb Y}}\leq M $, 则有$ \|F(t, x_{0})\|_{{\mathbb Y}}\leq\frac{\Lambda}{\mu}-e^{-\mu t}\Big(\frac{\Lambda}{\mu}-\|x_{0}\|_{{\mathbb Y}}\Big)\leq\frac{\Lambda}{\mu}-\Big(\frac{\Lambda}{\mu}-\|x_{0}\|_{{\mathbb Y}}\Big) = \|x_{0}\|_{{\mathbb Y}}\leq M. $进一步也有$ v(t, 0) = \xi S\leq\xi\|F(t, x_{0})\|_{{\mathbb Y}}\leq\xi M $和$ r(t, 0) = kI\leq k\|F(t, x_{0})\|_{{\mathbb Y}}\leq kM $成立. 引理3.1得证.

定义3.1^[10]  对任意非空有界闭集$ B\subset{\mathbb Y} $且满足$ F(t, B)\subset B $, 如果存在一个紧集$ B_{0}\subset{\mathbb Y} $, 当$ t\rightarrow \infty $时有$ F(t, B)\rightarrow B_{0} $, 那么称解半流$ F(t, x_{0}):{\mathbb R}_{+}\times{\mathbb Y}\rightarrow {\mathbb Y} $渐近光滑.

引理3.2^[10]  对于解半流$ F(t, x_{0}):{\mathbb R}_{+}\times {\mathbb Y}\rightarrow {\mathbb Y} $, 如果下述条件成立:

$ {\rm (i)} $解半流$ F(t, x_{0}) $可分解为$ F(t, x_0) = \phi(t, x_0)+\varphi(t, x_0) $;

$ {\rm (ii)} $存在函数$ u:{\mathbb R}^{2}_{+}\rightarrow {\mathbb R}_{+} $满足, 对任意$ h>0 $, 当$ t\rightarrow \infty $时$ u(t, h)\rightarrow0 $, 且当$ \|x_{0}\|_{\mathbb Y}\leq h $时, 对任意$ t\geq0 $有$ \|\varphi(t, x_0)\|_{\mathbb Y}\leq u(t, h) $成立;

$ {\rm (iii)} $对任意$ t\geq0 $, $ x_{0}\in {\mathbb Y} $, $ \phi(t, x_0) $完全连续.

则解半流$ F(t, x_{0}) $在$ {\mathbb Y} $上渐近光滑.

下面我们将使用引理3.2来证明模型(2.1) 的解半流$ F(t, x_{0}) $是渐近光滑的. 首先将模型(2.1) 的解半流$ F(t, x_{0}) $分解成两个算子$ \varphi(t, x_{0}), \; \phi(t, x_{0}):{\mathbb R}_{+}\times {\mathbb Y}\rightarrow {\mathbb Y} $之和. 为此设$ \varphi(t, x_0) = (0, x_2(t, \cdot), 0, 0, x_5(t, \cdot)) $和$ \phi(t, x_{0}) = (x_{1}(t), \tilde{x}_2(t, \cdot), x_{3}(t), x_{4}(t), \tilde{x}_5(t, \cdot)), $其中

(3.5) $ \begin{equation} \begin{array}{ll} x_2(t, a) = \left\{ \begin{array}{ll} 0\; \; \; , \;&t>a\geq0, \\ v(t, a), \;&a\geq t\geq0. \end{array}\right. \qquad \tilde x_2(t, a) = \left\{ \begin{array}{ll} v(t, a), \;&t>a\geq0, \\ 0\; \; \; \; , \;&a\geq t\geq0. \end{array}\right.\\ x_5(t, b) = \left\{ \begin{array}{ll} 0\; \; , \;&t>b\geq0, \\ r(t, b), \;&b\geq t\geq0. \end{array}\right.\qquad \tilde x_5(t, b) = \left\{ \begin{array}{ll} r(t, b), \;&t>b\geq0, \\ 0\; \; \; , \;&b\geq t\geq0. \end{array}\right. \end{array} \end{equation} $

于是对任意$ t\geq0 $, 可得$ F(t, x_{0}) = \phi(t, x_{0})+\varphi(t, x_{0}) $.

命题3.1  设$ u(t, h) = he^{-\mu t} $, 任意$ h>0 $且$ \|x_0\|_{\mathbb Y}\leq h $, 则对任意$ t\geq0 $有$ \|\varphi(t, x_0)\|_{\mathbb Y}\leq u(t, h) $.

证  显然对任何$ h>0 $有$ \lim\limits_{t\rightarrow \infty}u(t, h) = 0 $. 由式(3.2) 和(3.3), 对任意$ x_0\in{\mathbb Y} $, 当$ \|x_0\|_{\mathbb Y}\leq h $时可以得到

$ \begin{eqnarray*} \|\varphi(t, x_0)\|_{\mathbb Y}& = &|0|+\int_{0}^{\infty}|x_{2}(t, a)|{\rm d}a+|0|+|0|+\int_{0}^{\infty}|x_{5}(t, b)|{\rm d}b\\ & = &\int_{t}^{\infty}\Big|v_{0}(a-t)\frac{\rho_{1}(a)}{\rho_{1}(a-t)}\Big|{\rm d}a+\int_{t}^{\infty}\Big|r_{0}(b-t)\frac{\rho_{2}(b)} {\rho_{2}(b-t)}\Big|{\rm d}b\\ & = &\int_{0}^{\infty}\Big|v_{0}(s)\frac{\rho_{1}(t+s)}{\rho_{1}(s)}\Big|{\rm d}s+\int_{0}^{\infty}\Big|r_{0}(s)\frac{\rho_{2}(t+s)} {\rho_{2}(s)}\Big|{\rm d}s\\ & = &\int_{0}^{\infty}\Big|v_{0}(s)e^{-\int_{s}^{t+s}\varepsilon_{1}(\tau){\rm d}\tau}\Big|{\rm d}s+\int_{0}^{\infty}\Big|r_{0}(s) e^{-\int_{s}^{t+s}\varepsilon_{2}(\tau){\rm d}\tau}\Big|{\rm d}s. \end{eqnarray*} $

注意到对任意$ l\geq0 $, 有$ \varepsilon_{i}(l)\geq\mu\;(i = 1, 2) $, 进一步得到

$ \begin{eqnarray*} \|\varphi(t, x_0)\|_{{\mathbb Y}}&\leq&e^{-\mu t}\Big(|0|+\int_{0}^{\infty}|v_{0}(s)|{\rm d}s+|0|+|0|+\int_{0}^{\infty}|r_{0}(s)|{\rm d}s\Big)\\ & = &e^{-\mu t}\|x_0\|_{\mathbb Y}\leq he^{-\mu t} = u(t, h). \end{eqnarray*} $

命题3.1得证.

为了验证引理3.2的条件$ {\rm (iii)} $, 首先引进如下判定空间$ L^{1}_{+} $中的有界闭集紧性的结果.

引理3.3^[11]  常数$ p\geq1 $, 设集合$ K\subset L^{p}[0, \infty) $是有界闭集. 如果下述条件成立:

$ {\rm (i)} $对任意$ h\in K $, $ \lim\limits_{s\rightarrow0}\int_{0}^{\infty}|h(z+s)-h(z)|^{p}{\rm d}z = 0 $;

$ {\rm (ii)} $$ \lim\limits_{s\rightarrow \infty}\int_{s}^{\infty}|h(z)|^{p}{\rm d}z = 0 $对$ h\in K $一致成立.

则$ K $在$ L^{p}[0, \infty) $中是紧的.

命题3.2  对任意$ t\geq0 $, $ \phi(t, x_{0}) $在$ x_{0}\in {\mathbb Y} $上完全连续.

证  对任意常数$ M\geq\frac{\Lambda}{\mu} $和任意初值$ x_{0}\in {\mathbb Y} $且$ \|x_0\|\leq M $. 考虑解$ F(t, x_0) = (S(t), v(t, a), E(t), I(t), r(t, b)) $, 由引理3.1知$ S(t) $, $ E(t) $, $ I(t) $均存在于紧集$ [0, M] $中. 为此只需要验证$ \tilde x_2(t, a) $和$ \tilde x_{5}(t, b) $位于空间$ L_{+}^{1} $的某个准紧子集中, 且这个准紧子集仅与$ M $的选取有关. 由式(3.2) 和式(3.5) 得到

$ 0\leq\tilde{x}_2(t, a) = \left\{ \begin{array}{ll} \xi S(t-a)\rho_{1}(a), \;&t>a\geq0, \\ 0, \; \; \; \; \; \; \; \; \; \; \; &a\geq t\geq0. \end{array}\right. $

注意到式(3.1) 以及引理3.1的结论(ⅰ) 可以得到: $ \tilde{x}_{2}(t, a)\leq\xi Me^{-\mu a} $, 这表明引理3.3中条件(ⅱ) 成立.

下面证明条件(ⅰ) 成立, 对充分小的$ s\in(0, t) $, 我们得到

$ \begin{eqnarray*} & &\int_{0}^{\infty}|\tilde x_{2}(t, a+s)-\tilde x_{2}(t, a)|{\rm d}a = \int_{0}^{t}|v(t, a+s)-v(t, a)|{\rm d}a\\ & = &\int_{0}^{t}\xi|S(t-a-s)\rho_{1}(a+s)-S(t-a)\rho_{1}(a)|{\rm d}a\\ & = &\int_{0}^{t-s}\xi|S(t-a-s)\rho_{1}(a+s)-S(t-a)\rho_{1}(a)|{\rm d}a+\int_{t-s}^{t}\xi|S(t-a)\rho_{1}(a)|{\rm d}a\\ &\leq & B_{1}+B_{2}+B_{3}, \end{eqnarray*} $

其中

$ \begin{eqnarray*} B_{1}& = &\int_{0}^{t-s}\xi S(t-a-s)|\rho_{1}(a)-\rho_{1}(a+s)|{\rm d}a\\ & \leq&\xi M\Big(\int_{0}^{t-s}\rho_{1}(a){\rm d}a-\int_{0}^{t-s}\rho_{1}(a+s){\rm d}a\Big)\\ & = &\xi M\Big(\int_{0}^{t-s}\rho_{1}(a){\rm d}a-\int_{s}^{t}\rho_{1}(a){\rm d}a\Big)\\ & = &\xi M\Big(\int_{0}^{t-s}\rho_{1}(a){\rm d}a-\int_{s}^{t-s}\rho_{1}(a){\rm d}a-\int_{t-s}^{t}\rho_{1}(a){\rm d}a\Big)\\ & = &\xi M\Big(\int_{0}^{s}\rho_{1}(a){\rm d}a-\int_{t-s}^{t}\rho_{1}(a){\rm d}a\Big)\leq\xi Ms, \\ B_{2}& = &\int_{0}^{t-s}\xi\rho_{1}(a)|S(t-a-s)-S(t-a)|{\rm d}a\\ &\leq&\xi s(\Lambda+(\mu+\xi)M+\beta M^{2}+\bar{\omega}_{1}M)\int_{0}^{t-s}e^{-\mu a}{\rm d}a\\ &\leq&\frac{\xi s}{\mu}(\Lambda+(\mu+\xi)M+\beta M^{2}+\bar{\omega}_{1}M), \\ B_{3}& = &\int_{t-s}^{t}\xi |S(t-a)\rho_{1}(a)|{\rm d}a\leq\xi Ms. \end{eqnarray*} $

于是最终得到

$ \int_{0}^{\infty}|\tilde x_{2}(t, a+s)-\tilde x_{2}(t, a)|{\rm d}a\leq\Big(2M+\frac{1}{\mu}(\Lambda+(\mu+\xi)M+\beta M^{2}+\bar{\omega}_{1}M)\Big)\xi s. $

由此进一步得到: $ s\rightarrow0 $时, $ \int_{0}^{\infty}|\tilde x_{2}(t, a+s)-\tilde x_{2}(t, a)|{\rm d}a $一致收敛于0. 这表明引理3.3中条件(ⅰ) 成立. 因此由引理3.3推出$ \tilde{x}_{2}(t, a) $位于$ L^{1}_{+} $的某个准紧子集$ B_{\tilde{x}_{2}} $中. 类似可知$ \tilde{x}_{5}(t, b) $也位于$ L^{1}_{+} $的某个准紧子集$ B_{\tilde{x}_{5}} $中. 因此$ \phi(t, x_{0})\subseteq [0, M]\times B_{\tilde{x}_{2}}\times[0, M]\times[0, M]\times B_{\tilde{x}_{5}} $在$ {\mathbb Y} $中是紧的. 由上述讨论和引理3.3可得$ \phi(t, x_{0}) $在$ x_{0}\in {\mathbb Y} $上是完全连续的. 命题3.2得证.

由命题3.1、命题3.2和引理3.2, 最终得到如下结论:

定理3.2  由模型(2.1) 生成的解半流$ \{F(t, x_{0})\}_{t\geq0} $是渐近光滑的.

作为定理3.2的结论, 我们得到关于模型(2.1) 全局吸引子存在性的结论.

推论3.1  模型(2.1) 生成的解半流$ F(t, x_{0}) $有一个紧的全局吸引子.

证  由定理3.2, 模型的解半流$ F(t, x_{0}) $是渐近光滑的, 且根据引理3.1得模型(2.1) 解$ F(t, x_{0}) $的最终有界性, 也就是解半流$ F(t, x_{0}) $是点耗散的. 因此, 由文献[12]中的定理2.33, 可得模型(2.1) 的连续半流$ F(t, x_{0}) $具有一个紧的全局吸引子. 推论3.1得证.

定义集合$ \hat{{\mathbb Y}} = {\mathbb R}_{+}\times{\mathbb R}_{+}\times L_{+}^{1} $, $ {\mathbb Z} = {\mathbb R}_{+}\times L_{+}^{1}\times\hat{{\mathbb Z}} $, 并且

$ \hat{{\mathbb Z}} = \Big\{(E(t), I(t), r(\cdot))^{T}\in\hat{{\mathbb Y}}: E>0, \; I>0, \; \int_{0}^{\infty}r(b){\rm d}b>0\Big\}. $

我们有$ \partial\hat{{\mathbb Z}} = \hat{{\mathbb Y}}\setminus\hat{{\mathbb Z}} $和$ \partial {\mathbb Z} = {\mathbb Y}\backslash {\mathbb Z}. $

参考文献[13], 模型(2.1) 的基本再生数$ {\cal R}_{0} $定义为

$ {{\cal R}}_{0} = \frac{\beta S^{0}\varepsilon+(\mu+\varepsilon)k\theta_{2}}{(\mu+\delta+k)(\mu+\varepsilon)}. $

可将上式写为

$ {\mathcal R}_{0} = \frac{1}{\mu+\delta+k}\big(\frac{\beta S^{0}\varepsilon}{\mu+\varepsilon}+k\theta_{2}\big), $

其中$ \frac{1}{\mu+\varepsilon} $表示被感染后在潜伏仓室停留的平均时间, $ \frac{\varepsilon}{\mu+\varepsilon} $表示成为染病者的概率, $ S^{0} $表示传染病初期接种疫苗后处于易感人群的总数, $ \theta_{2} $表示由于复发而从移出仓室到染病仓室的转化概率, 那么$ {\cal R}_{0} $表示一个病人在一个染病期内感染成新的染病者的平均数量, 这种感染一方面由于接种、潜伏后感染, 另一方面由于复发再感染而引起.

显然, 模型(2.1) 总有无病平衡点$ T_{0} = (S^{0}, v^{0}(a), 0, 0, 0) $, 其满足如下平衡点方程

$ \left\{ \begin{array}{ll} \Lambda-(\mu+\xi)S^{0}+\int_{0}^{\infty}\omega_{1}(a)v^{0}(a){\rm d}a = 0, \\ \frac{{\rm d} v^{0}(a)}{{\rm d} a} = -(\omega_{1}(a)+\mu)v^{0}(a), \\ v^{0}(0) = \xi S^{0}. \end{array} \right. $

引理4.1  若$ {\cal R}_{0}>1 $, 则存在一个常数$ \epsilon>0 $, 使得对任意初值$ x_{0}\in{\mathbb Y} $且$ E_{0}>0, \; I_{0}>0, $$ r_{0}(\cdot)\not\equiv0 $, 模型(2.1) 的解半流$ F(t, x_{0}) $满足$ \limsup\limits_{t\rightarrow \infty}\|F(t, x_{0})-T_{0}\|_{{\mathbb Y}}\geq\epsilon $.

证  首先由$ {\cal R}_{0}>1 $, 选择足够小常数$ \epsilon_{0}>0 $和足够大常数$ \bar{b}>0 $使得$ S^{0}-\epsilon_{0}>0 $和

$ \frac{1}{\mu+\delta+k}\Big(\frac{\beta (S^{0}-\epsilon_{0})\varepsilon}{\mu+\varepsilon}+k\int_0^{\bar{b}}\omega_2(l)\rho_2(l){\rm d} l\Big)>1. $

若结论不成立, 则存在模型(2.1) 的解$ F(t, x_{0}) = (S(t), v(t, \cdot), E(t), I(t), r(t, \cdot)) $使得

$ \limsup\limits_{t\rightarrow \infty}\|F(t, x_{0})-T_{0}\|_{{\mathbb Y}}<\epsilon_{0}, $

其中$ x_{0}\in{\mathbb Y} $且$ E_{0}>0, \; I_{0}>0, \; r_{0}(\cdot)\not\equiv0 $. 于是存在充分大的$ T $使得对任意$ t\geq T $, 有

$ 0<S^{0}-\epsilon_{0}<S(t)<S^{0}+\epsilon_{0}. $

由此, 对任意$ t\geq T $, 我们有

$ \left\{ \begin{array}{ll} \frac{{\rm d} E(t)}{{\rm d}t}\geq\beta (S^0-\epsilon_{0})I(t)-(\varepsilon+\mu)E(t), \\ \frac{{\rm d} I(t)}{{\rm d}t}\geq\int_{0}^{\bar{b}}\omega_2(b)r(t, b){\rm d}b+\varepsilon E(t)-(\mu+\delta+k)I(t), \\ \frac{\partial r(t, b)}{\partial t}+\frac{\partial r(t, b)}{\partial b} = -(\omega_2(b)+\mu)r(t, b), \\ r(t, 0) = kI(t), \; t\geq T. \end{array} \right. $

参考文献[14]中年龄依赖偏微分方程的比较原理, 对于所有$ t\geq T $和$ b\in[0, \bar{b}] $有

(4.1) $ \begin{equation} E(t)\geq \tilde{E}(t), \; I(t)\geq \tilde{I}(t), \; r(t, b)\geq \tilde{r}(t, b), \end{equation} $

其中$ (\tilde{E}(t), \tilde{I}(t), \tilde{r}(t, b)) $是如下线性比较系统的解

(4.2) $ \begin{equation} \left\{ \begin{array}{ll} \frac{{\rm d} \tilde{E}(t)}{{\rm d}t} = \beta (S^0-\epsilon_{0})\tilde{I}(t)-(\varepsilon+\mu)\tilde{E}(t), \\ \frac{{\rm d} \tilde{I}(t)}{{\rm d}t} = \int_{0}^{\bar{b}}\omega_2(b)\tilde{r}(t, b){\rm d}b+\varepsilon \tilde{E}(t)-(\mu+\delta+k)\tilde{I}(t), \\ \frac{\partial \tilde{r}(t, b)}{\partial t}+\frac{\partial \tilde{r}(t, b)}{\partial b} = -(\omega_2(b)+\mu)\tilde{r}(t, b), \\ \tilde{r}(t, 0) = k\tilde{I}(t), \; t\geq T, \;b\in[0, \bar{b}], \end{array} \right. \end{equation} $

且满足初始条件$ \tilde{E}(T) = E(T), \; \tilde{I}(T) = I(T), \; \tilde{r}(T, b) = r(T, b), \;\forall\;b\in[0, \bar{b}] $. 求解系统(4.2) 如下形式的解$ E^*(t) = E^*_{1}e^{\lambda(t-T)}, \; I^*(t) = I^*_{1}e^{\lambda(t-T)}, \; r^*(t, b) = r^*_{1}(b)e^{\lambda(t-T)} $, 其中$ E^*_{1}, \; I^*_{1}, \; r^*_{1}(b) $是特征函数且不全为$ 0 $, $ \lambda $为特征值. 将这种形式的解带入系统(4.2), 得到如下线性特征值问题

(4.3) $ \begin{equation} \left\{ \begin{array}{ll} { } (\lambda+\varepsilon+\mu)E^*_{1} = \beta (S^{0}-\epsilon_{0})I^*_{1}, \\ { } (\lambda+\mu+\delta+k)I^*_{1} = \int_{0}^{\bar{b}}\omega_2(b)r^*_{1}(b){\rm d}b+\varepsilon E^*_{1}, \\ \frac{{\rm d}r^*_{1}(b)}{{\rm d} t} = -(\lambda+\omega_2(b)+\mu)r^*_{1}(b), \\ r^*_{1}(0) = kI^*_{1}. \end{array} \right. \end{equation} $

由系统(4.3) 的第3个方程我们得

(4.4) $ \begin{equation} r^*_{1}(b) = kI^*_{1}e^{-\int_{0}^{b}(\lambda+\varepsilon_{2}(s)){\rm d}s}. \end{equation} $

将系统(4.3) 的第一个方程和(4.4) 代入(4.3) 的第二个方程可得

$ \int_{0}^{\bar{b}}\omega_{2}(b)kI^*_{1}e^{-\int_{0}^{b}(\lambda+\varepsilon_{2}(s)){\rm d}s}{\rm d}b +\frac{\beta(S-\epsilon_{0})\varepsilon}{\lambda+\mu+\varepsilon}I^*_{1}-(\lambda+\mu+\delta+k)I^*_{1} = 0. $

由此得到系统(4.3) 的特征方程为

(4.5) $ \begin{equation} H(\lambda)\triangleq \frac{1}{\lambda+\mu+\delta+k}\Big(k\int_{0}^{\bar{b}}\omega_{2}(b)e^{-\int_{0}^{b}(\lambda+\varepsilon_{2}(s)){\rm d}s}{\rm d}b +\frac{\beta(S-\epsilon_{0})\varepsilon}{\lambda+\mu+\varepsilon}\Big) = 1. \end{equation} $

显然$ H(0) = \frac{1}{\mu+\delta+k}\big(k\int_0^{\bar{b}}\omega_2(l)\rho_2(l){\rm d} l+\frac{\beta(S-\epsilon_{0})\varepsilon}{\mu+\varepsilon}\big)>1 $和$ \lim\limits_{\lambda\rightarrow +\infty}H(\lambda) = 0 $. 因此, 方程$ H(\lambda)-1 = 0 $至少有一个实根$ \lambda_{0}>0 $. 这表明方程(4.2) 有如下形式解

(4.6) $ \begin{equation} E^*(t) = E^*_{1}e^{\lambda_{0}(t-T)}, \; I^*(t) = I^*_{1}e^{\lambda_{0}(t-T)}, \; r^*(t, b) = r^*_{1}(b)e^{\lambda_{0}(t-T)}, \end{equation} $

其中$ E^*_1 = \frac{\beta(S^0-\epsilon_0)I^*_1}{\lambda_0+\mu+\varepsilon}>0 $, $ r^*_{1}(b) = kI^*_1e^{-\int_0^b(\lambda_0+\varepsilon_2(s)){\rm d}s}>0 $, 且$ I^*_1>0 $. 进一步选取$ 0<I^*_1<\tilde{I}(T) $使得$ E^*_1<\tilde{E}(T) $和$ r^*_{1}(b)<\tilde{r}(T, b), \;\forall\;b\in[0, \bar{b}] $, 由比较原理得到

(4.7) $ \begin{equation} \tilde{E}(t)>E^*(t), \;\tilde{I}(t)>I^*(t), \;\tilde{r}(t, b)>r^*(t, b), \end{equation} $

对任何$ t\geq T $成立. 最后, 由(4.1), (4.6) 和(4.7) 式可以推出$ E(t)+I(t)+\int_{0}^{\infty}r(t, b){\rm d}b $在$ t\geq T $上无界. 这与$ F(t, x_{0}) $的有界性矛盾. 引理4.1得证.

定理4.1  如果$ {\cal R}_{0}>1 $, 则存在常数$ \epsilon_{1}>0 $使得对任意初始值$ x_{0}\in {\mathbb Y} $且$ E_{0}>0, I_{0}>0, $$ r_{0}(\cdot)\not\equiv0 $, 模型(2.1) 的解$ F(t, x_{0}) $满足$ \liminf\limits_{t\rightarrow \infty}S(t)\geq\epsilon_{1}, $$ \liminf\limits_{t\rightarrow \infty}\|v(t, \cdot)\|_{L^1}\geq\epsilon_{1}, $$ \liminf\limits_{t\rightarrow \infty}E(t)\geq\epsilon_{1}, \liminf\limits_{t\rightarrow \infty}I(t)\geq\epsilon_{1}, $$ \liminf\limits_{t\rightarrow \infty}\|r(t, \cdot)\|_{L^1}\geq\epsilon_{1}. $

证  由引理3.1, 存在常数$ M>0 $使得对模型(2.1) 的任意解$ (S(t), v(t, \cdot), E(t), I(t), r(t, \cdot)) $, $ \exists\; t_{0}>0 $, 对任意$ t\geq t_{0} $有$ I(t)\leq M $. 因此, 从模型(2.1) 的第一个方程可得

$ \frac{{\rm d}S}{{\rm d}t}\geq\Lambda-(\mu+\xi)S-\beta SM, \; t\geq t_{0}. $

考虑下述比较系统

$ \frac{{\rm d}\nu}{{\rm d}t} = \Lambda-(\mu+\xi)\nu-\beta \nu M, \; t\geq t_{0}. $

它具有全局渐近稳定的平衡点$ \nu^* = \frac{\Lambda}{\mu+\xi+\beta M} $. 由比较原理得到$ \liminf\limits_{t\to\infty}S(t)\geq\nu^* $, 这表明模型(2.1) 中的$ S(t) $是一致持续的.

对于任意初值$ (S_{0}, v_{0}(\cdot), E_{0}, I_{0}, r_{0}(\cdot))\in{\mathbb Z} $, 且$ E_{0}>0, I_{0}>0, r_{0}(\cdot)\not\equiv0 $, 由式(3.3) 得知, 对于所有$ t>0 $有$ \int_{0}^{\infty}r(t, b){\rm d}b>0 $. 由模型(2.1) 的第三和第四个两个方程分别得到, 对于任意$ t>0 $有$ E(t)>0, \; I(t)>0 $. 因此, 集合$ {\mathbb Z} $为模型(2.1) 的正向不变集. 定义集合

$ M_{\partial} = \Big\{x_{0} = (S_{0}, v_{0}(\cdot), E_{0}, I_{0}, r_{0}(\cdot))\in{\mathbb Y}:F(t, x_{0}) \in\partial{{\mathbb Z}}, \;\forall\;t\geq0\Big\}. $

设$ M_{1} = \{T_{0}\} $, $ \omega(x_{0}) $为解$ F(t, x_{0}) $的$ \omega $极限集. 由于对于所有$ t\geq0 $有$ F(t, T_{0}) = T_{0}\in\partial{{\mathbb Z}} $, 我们有$ M_{1}\subset\bigcup\limits_{x_{0}\in M_{\partial}}\omega(x_{0}). $

接下来证明$ \bigcup\limits_{x_{0}\in M_{\partial}}\omega(x_{0})\subset M_{1} $. 对任意$ x_{0}\in M_{\partial} $有$ F(t, x_{0})\in\partial{{\mathbb Z}} $对所有$ t\geq 0 $均成立. 进一步有$ E(t)\equiv0 $或$ I(t)\equiv0 $或$ \int_{0}^{\infty}r(t, b){\rm d}b\equiv0 $. 若$ E (t)\equiv0 $, 则由$ S(t) $的一致持续性, 从模型(2.1) 的第三个方程得知$ I(t)\equiv0 $. 进一步得到$ r(t, 0) = kI(t)\equiv0 $, 于是$ \int_{0}^{\infty}r(t, b){\rm d}b\equiv0 $. 这样, 模型(2.1) 退化为如下子系统

(4.8) $ \begin{equation} \left\{ \begin{array}{ll} \frac{{\rm d} S(t)}{{\rm d} t} = \Lambda-(\mu+\xi)S(t)+\int_{0}^{\infty}\omega_{1}(a)v(t, a){\rm d}a, \\ \frac{\partial v(t, a)}{\partial t}+\frac{\partial v (t, a)}{\partial a} = -(\omega_1(a)+\mu)v(t, a). \end{array} \right. \end{equation} $

显然, 由系统(4.8) 得到$ \lim\limits_{t\rightarrow \infty}S(t) = S^{0} $和$ \lim\limits_{t\rightarrow \infty}v(t, a) = v^{0}(a) $. 这表明$ \omega(x_{0}) = \{T_0\} $. 同样的, 若$ I(t)\equiv0 $或$ \int_{0}^{\infty}r(t, b){\rm d}b\equiv0 $, 对所有$ t\geq 0 $和$ b\geq 0 $也可以得到, $ E(t)\equiv 0 $, $ I(t)\equiv 0 $和$ \int_{0}^{\infty}r(t, b){\rm d}b\equiv 0 $, 则模型(2.1) 也退化成子系统(4.8). 因此, $ \lim\limits_{t\rightarrow \infty}S(t) = S^{0} $和$ \lim\limits_{t\rightarrow \infty}v(t, a) = v^{0}(a) $. 这表明$ \omega(x_{0}) = T_{0} $, 因此$ \bigcup\limits_{x_{0}\in M_{\partial}}\omega(x_{0})\subset M_{1} $. 从而得到$ \bigcup\limits_{x_{0}\in M_{\partial}}\omega(x_{0}) = M_{1} $.

由$ \bigcup\limits_{x_{0}\in M_{\partial}}\omega(x_{0}) = M_{1} $, 当$ t\rightarrow \infty $时, 在$ M_{\partial} $上的任意解$ F(t, x_{0}) $都趋向于$ T_{0} $. 由引理4.1, 得到$ P_{0} $是$ {\mathbb Y} $上一个孤立的不变集, 且$ W^{s}(T_{0})\cap{\mathbb Z} = \emptyset $, 其中$ W^{s}(T_{0}) $是$ T_{0} $的稳定集.

此外, 从上述的讨论中, 注意到$ M_{1} $在$ \partial{{\mathbb Z}} $中没有形成环. 参见文献[15, 16]和推论3.1中的动力系统的持续理论, 我们最终得到模型(2.1) 的解半流$ F(t, x_0) $是一致持续的. 定理4.1得证.

由模型(2.1) 解半流$ F(t, x_{0}) $的一致持续性得知, 当$ {\cal R}_{0}>1 $时, 模型(2.1) 存在一个地方病平衡点$ T^{\ast} = (S^{\ast}, v^{\ast}(a), E^{\ast}, I^{\ast}, r^{\ast}(b)) $, 其满足如下平衡点方程

(4.9) $ \begin{equation} \left\{ \begin{array}{ll} \Lambda-(\mu+\xi)S^{\ast}-\beta S^{\ast}I^{\ast}+\int_{0}^{\infty}\omega_{1}(a)v^{\ast}(a){\rm d}a = 0, \\ \frac{{\rm d} v^{\ast}(a)}{{\rm d} a} = -(\omega_{1}(a)+\mu)v^{\ast}(a), \\ \beta S^{\ast}I^{\ast}-(\varepsilon+\mu)E^{\ast} = 0, \\ \int_{0}^{\infty}\omega_2(b)r^{\ast}(b){\rm d}b+\varepsilon E^{\ast}-(\mu+\delta+k)I^{\ast} = 0, \\ \frac{{\rm d} r^{\ast}(b)}{{\rm d} b} = -(\omega_2(b)+\mu)r^{\ast}(b), \\ v^{\ast}(0) = \xi S^{\ast}, r^{\ast}(0) = kI^{\ast}. \end{array} \right. \end{equation} $

定理5.1  如果$ {\cal R}_{0}<1 $, 则无病平衡点$ T_{0} $是局部渐近稳定的; 如果$ {\cal R}_{0}>1 $, 则$ T_{0} $不稳定.

证  令$ x_{1}(t) = S(t)-S^{0}, x_{2}(t, a) = v(t, a)-v^{0}(a), x_{3}(t) = E(t), x_{4}(t) = I(t), x_{5}(t, b) = r(t, b) $, 将模型(2.1) 在$ T_{0} $线性化可得

(5.1) $ \begin{equation} \left\{ \begin{array}{l} \frac{{\rm d}x_{1}(t)}{{\rm d}t} = -(\mu+\xi)x_{1}(t)-\beta S^{0}x_{4}(t)+\int_{0}^{\infty}\omega_{1}(a)x_{2}(t, a){\rm d}a, \\ \Big(\frac{\partial }{\partial t}+\frac{\partial }{\partial a}\Big)x_{2}(t, a) = -\varepsilon_{1}(a)x_{2}(t, a), \\ \frac{{\rm d}x_{3}(t)}{{\rm d}t} = \beta S^{0}x_{4}(t)-(\mu+\varepsilon)x_{3}(t), \\ \frac{{\rm d}x_{4}(t)}{{\rm d}t} = \int_{0}^{\infty}\omega_{2}(b)x_{5}(t, b){\rm d}b+\varepsilon x_{3}(t)-(\mu+\delta+k)x_{4}(t), \\ \Big(\frac{\partial }{\partial t}+\frac{\partial }{\partial b}\Big)x_{5}(t, b) = -\varepsilon_{2}(b)x_{5}(t, b), \\ x_{2}(t, 0) = \xi x_{1}(t), \; \; x_{5}(t, 0) = k x_{4}(t). \end{array} \right. \end{equation} $

设$ x_{1}(t) = x_{1}^{0}e^{\lambda t}, x_{2}(t, a) = x_{2}^{0}(a)e^{\lambda t}, x_{3}(t) = x_{3}^{0}e^{\lambda t}, x_{4}(t) = x_{4}^{0}e^{\lambda t}, x_{5}(t, b) = x_{5}^{0}(b)e^{\lambda t} $是方程(5.1) 的解, 其中$ (x_{1}^{0}, x_{2}^{0}(a), x_{3}^{0}, x_{4}^{0}, x_{5}^{0}(b)) $为初始值, $ \lambda $是一个复数. 从方程组(5.1) 各方程分别有

(5.2) $ \begin{equation} \lambda x_{1}^{0} = -(\mu+\xi)x_{1}^{0}-\beta S^{0}x_{4}^{0}+\int_{0}^{\infty}\omega_{1}(a)x_{2}^{0}(a){\rm d}a, \end{equation} $

(5.3) $ \begin{equation} \lambda x_{2}^{0}(a)+\frac{{\rm d}x_{2}^{0}(a)}{{\rm d}a} = -\varepsilon_{1}(a)x_{2}^{0}(a), \quad x_{2}^{0}(0) = \xi x_{1}^{0}, \end{equation} $

(5.4) $ \begin{equation} \lambda x_{3}^{0} = \beta S^{0}x_{4}^{0}-(\mu+\varepsilon)x^{0}_{3}, \end{equation} $

(5.5) $ \begin{equation} \lambda x_{4}^{0} = \int_{0}^{\infty}\omega_{2}(b)x_{5}^{0}(b){\rm d}b+\varepsilon x^{0}_{3}-(\mu+\delta+k)x_{4}^{0}, \end{equation} $

(5.6) $ \begin{equation} \lambda x_{5}^{0}(b)+\frac{{\rm d}x_{5}^{0}(b)}{{\rm d}b} = -\varepsilon_{2}(b)x_{5}^{0}(b), \quad x_{5}^{0}(0) = k x_{4}^{0}. \end{equation} $

求解(5.3), (5.6) 式后同(5.4) 式一起代入(5.2) 和(5.5)式中联立可得

$ \Big(\lambda+\mu+\xi-\xi\int_{0}^{\infty}\omega_{1}(a) e^{[-(\lambda+\mu)a-\int_{0}^{a}\omega_{1}(\tau){\rm d}\tau]}{\rm d}a\Big)x_{1}^{0}+\beta S^{0}x_{4}^{0} = 0, $

$ \Big(\lambda+\mu+\delta+k-\frac{\beta S^{0}\varepsilon}{\lambda+\mu+\varepsilon}-k\int_{0}^{\infty}\omega_{2}(b) e^{[-(\lambda+\mu)b-\int_{0}^{b}\omega_{2}(\tau){\rm d}\tau]}{\rm d}b\Big)x_{4}^{0} = 0. $

由于$ x^{0}_{1}, \; x^{0}_{4} $不全为零, 因此从上式知$ \lambda $满足方程$ H_{1}(\lambda)H_{2}(\lambda) = 0 $, 其中

$ H_{1}(\lambda)\triangleq\lambda+\mu+\delta+k-\frac{\beta S^{0}\varepsilon}{\lambda+\mu+\varepsilon}-k\int_{0}^{\infty}\omega_{2}(b) e^{[-(\lambda+\mu)b-\int_{0}^{b}\omega_{2}(\tau){\rm d}\tau]}{\rm d}b, $

$ H_{2}(\lambda)\triangleq\lambda+\mu+\xi-\xi\int_{0}^{\infty}\omega_{1}(a) e^{[-(\lambda+\mu)a-\int_{0}^{a}\omega_{1}(\tau){\rm d}\tau]}{\rm d}a. $

由于

$ H'_{1}(\lambda) = 1+\frac{\beta S^{0}\varepsilon}{(\lambda+\mu+\varepsilon)^{2}}+bk\int_{0}^{\infty}\omega_{2}(b) e^{[-(\lambda+\mu)b-\int_{0}^{b}\omega_{2}(\tau){\rm d}\tau]}{\rm d}b>0, $

并且$ \lim\limits_{\lambda\rightarrow -\infty} H_{1}(\lambda) = -\infty $和$ \lim\limits_{\lambda\rightarrow +\infty} H_{1}(\lambda) = +\infty $, 因此$ H_{1}(\lambda) $有唯一的实根$ \bar{\lambda} $. 注意到

$ H_{1}(0) = \mu+\delta+k-\frac{\beta S^{0}\varepsilon}{\mu+\varepsilon}-k\theta_{2} = (\mu+\delta+k)(1-{\mathcal R}_{0}), $

于是如果$ {\cal R}_{0}<1 $, 则$ \bar{\lambda}<0 $; 如果$ {\cal R}_{0}>1 $, 则$ \bar{\lambda}>0 $. 现令$ \lambda = \alpha+{\rm i}\beta $是方程$ H_{1}(\lambda) = 0 $的复根, 则$ H_{1}(\alpha)\leq0 $, 这表明$ \bar{\lambda}>\alpha $. 因此, 当且仅当$ {\cal R}_{0}>1 $时$ H_{1}(\lambda) = 0 $至少有一个正实部的根, 当且仅当$ {\cal R}_{0}<1 $时$ H_{1}(\lambda) = 0 $的所有根具有负实部.

假设$ \lambda = \alpha+{\rm i}\beta $是$ H_{2}(\lambda) = 0 $的满足$ \alpha\geq0 $的任意根. 由于

$ \alpha+\mu+\xi-\xi\int_0^{\infty}\omega_{1}(a) e^{[-(\alpha+\mu)a-\int_0^a\omega_{1}(\tau){\rm d}\tau]}\cos(\beta a){\rm d}a = 0, $

并且

$ \Big|\int_{0}^{\infty}\omega_{1}(a) e^{-(\alpha+\mu)a-\int_{0}^{a}\omega_{1}(\tau){\rm d}\tau}\cos(\beta a){\rm d}a\Big|\leq\int_{0}^{\infty}\omega_{1}(a)e^{-\int_{0}^{a}\omega_{1}(\tau){\rm d}\tau}{\rm d}a\leq 1. $

则有$ \alpha+\mu+\xi-\xi\int_0^{\infty}\omega_{1}(a) e^{[-(\alpha+\mu)a-\int_0^a\omega_{1}(\tau){\rm d}\tau]}\cos(\beta a){\rm d}a>0 $, 得到矛盾, 因此$ H_{2}(\lambda) = 0 $的所有根具有负实部. 定理5.1得证.

定理5.2  如果$ {\cal R}_{0}<1 $, 则无病平衡点$ T_{0} $全局渐近稳定.

证  选择Lyapunov函数如下

$ L_{0}(t) = L^{0}_{s}(t)+L^{0}_{v}(t)+\frac{\varepsilon}{\mu+\varepsilon}E(t)+I(t)+\int_{0}^{\infty}\pi_{2}(b)r(t, b){\rm d}b, $

这里$ L^{0}_{s}(t) = \frac{\varepsilon}{\mu+\varepsilon}S^{0}\Phi\Big(\frac{S}{S^{0}}\Big) $和$ L^{0}_{v}(t) = \frac{\varepsilon}{\mu+\varepsilon}\int_{0}^{\infty}v^{0}(a)\Phi\Big(\frac{v(t, a)}{v^{0}(a)}\Big){\rm d}a, $其中$ \Phi(u) = u-1-\ln u $. 由于$ \pi_{2}(0) = \theta_{2} $, 得到

$ \int_{0}^{\infty}\pi_{2}(b)\frac{\partial r(t, b)}{\partial t}{\rm d}b = -\pi_{2}(b)r(t, b)\Big|^{\infty}+\theta_{2}kI -\int_{0}^{\infty}\omega_{2}(b)r(t, b){\rm d}b. $

由于$ \mu+\xi = \frac{1}{S^{0}}\Big(\Lambda+\int_{0}^{\infty}\omega_{1}(a)v^{0}(a){\rm d}a\Big) $, 并且

$ \begin{eqnarray*} \frac{{\rm d}L^{0}_{s}(t)}{{\rm d}t}& = & \frac{\varepsilon}{\mu+\varepsilon}\Big(1-\frac{S^{0}}{S}\Big)\frac{{\rm d}S}{{\rm d}t}\\ & = & -\frac{\varepsilon}{\mu+\varepsilon}(\beta SI-\beta S^{0}I)-\frac{\varepsilon}{\mu+\varepsilon}\Lambda\Big(\frac{S}{S^{0}}+\frac{S^{0}}{S}-2\Big)\\ && +\frac{\varepsilon}{\mu+\varepsilon}\int_{0}^{\infty}\omega_{1}(a)v^{0}(a) \Big[\frac{v(t, a)}{v^{0}(a)}-\frac{S}{S^{0}}-\frac{S^{0}v(t, a)}{Sv^{0}(a)}+1\Big]{\rm d}a, \end{eqnarray*} $

$ \frac{{\rm d}L^{0}_{v}(t)}{{\rm d}t} = \frac{\varepsilon}{\mu+\varepsilon}\Big[-v^{0}(a)\Phi\Big(\frac{v(t, a)}{v^{0}(a)}\Big)\Big|^{\infty}-\xi S^{0}\Phi\Big(\frac{S}{S^{0}}\Big)- \int_{0}^{\infty}\varepsilon_{1}(a)v^{0}(a)\Phi\Big(\frac{v(t, a)}{v^{0}(a)}\Big){\rm d}a\Big]. $

因此

$ \begin{eqnarray*} \frac{{\rm d} L_{0}(t)}{{\rm d}t}& = & \frac{{\rm d}L^{0}_{s}(t)}{{\rm d}t}+\frac{{\rm d} L^{0}_{v}(t)}{{\rm d}t}+\frac{\varepsilon}{\mu+\varepsilon}\frac{{\rm d} E(t)}{{\rm d}t}+\frac{{\rm d} I(t)}{{\rm d}t}+\int_{0}^{\infty}\pi_{2}(b)\frac{\partial r(t, b)}{\partial t}{\rm d}b\\ & = & -\frac{\varepsilon\Lambda}{\mu+\varepsilon} \Big(\frac{S}{S^{0}}+\frac{S^{0}}{S}-2\Big)-\frac{\varepsilon}{\mu+\varepsilon}v^{0}(a) \Phi\Big(\frac{v(t, a)}{v^{0}(a)}\Big)\Big|^{\infty}-\pi_{2}(b)r(t, b)\Big|^{\infty}\\ && -\frac{\xi\varepsilon S^{0}}{\mu+\varepsilon}\Phi\Big(\frac{S}{S^{0}}\Big) -\frac{\mu\varepsilon}{\mu+\varepsilon}\int_{0}^{\infty}v^{0}(a) \Phi\Big(\frac{v(t, a)}{v^{0}(a)}\Big){\rm d}a+\frac{\varepsilon}{\mu+\varepsilon}\Sigma\\ && +\frac{\beta\varepsilon S^{0}I}{\mu+\varepsilon}+k\theta_{2}I-(k+\delta+\mu)I, \end{eqnarray*} $

其中

$ \begin{eqnarray*} \Sigma& = & \int_{0}^{\infty}\omega_{1}(a)v^{0}(a)\Big[\frac{v(t, a)}{v^{0}(a)}-\frac{S}{S^{0}}-\frac{S^{0}v(t, a)}{Sv^{0}(a)}+1\Big] {\rm d}a-\int_{0}^{\infty}\omega_{1}(a)v^{0}(a)\Phi\Big(\frac{v(t, a)}{v^{0}(a)}\Big){\rm d}a\\ & = & \int_{0}^{\infty}\omega_{1}(a)v^{0}(a)\Big[\ln\frac{v(t, a)}{v^{0}(a)}-\frac{S}{S^{0}}-\frac{S^{0}v(t, a)}{Sv^{0}(a)}+2\Big]{\rm d}a\\ & = & -\int_{0}^{\infty}\omega_{1}(a)v^{0}(a)\Big[\Phi\Big(\frac{S^{0}v(t, a)}{Sv^{0}(a)}\Big)+\Phi\Big(\frac{S}{S^{0}}\Big)\Big]{\rm d}a\leq0. \end{eqnarray*} $

由于$ \frac{\beta\varepsilon S^{0}I}{\mu+\varepsilon}+k\theta_{2}I-(k+\delta+\mu)I = (k+\delta+\mu)( {\mathcal R}_{0}-1)I $, 则当$ {\cal R}_{0}<1 $时有$ \frac{{\rm d}L^{0}(t)}{{\rm d}t}\leq0 $，并且$ \frac{{\rm d}L^{0}(t)}{{\rm d}t} = 0 $表明$ S(t)\equiv S^{0} $和$ v(t, a)\equiv v^{0}(a) $. 进一步由模型(2.1) 的第一、三和四个方程得到$ I(t)\equiv 0 $, $ E(t)\equiv 0 $和$ r(t, b)\equiv 0 $. 因此由LaSalle不变集原理^[17], 得知平衡点$ T_{0} $全局渐近稳定性.定理5.2得证.

定理5.3  如果$ {\cal R}_{0}>1 $, 则地方病平衡点$ T^{\ast} $局部渐近稳定.

证  令$ x_{1}(t) = S(t)-S^{*}, x_{2}(t, a) = v(t, a)-v^{*}(a), x_{3}(t) = E(t)-E^{*}, $$ x_{4}(t) = I(t)-I^{*}, $$ x_{5}(t, b) = r(t, b)-r^{*}(b) $, 将模型(2.1) 在$ T^{\ast} $线性化可得

(5.7) $ \begin{equation} \left\{ \begin{array}{l} \frac{{\rm d}x_{1}(t)}{{\rm d}t} = -(\mu+\xi)x_{1}(t)-\beta I^{*}x_{1}(t)-\beta S^{*}x_{4}(t)+\int_{0}^{\infty}\omega_{1}(a)x_{2}(t, a){\rm d}a, \\ \Big(\frac{\partial }{\partial t}+\frac{\partial }{\partial a}\Big)x_{2}(t, a) = -\varepsilon_{1}(a)x_{2}(t, a), \\ \frac{{\rm d}x_{3}(t)}{{\rm d}t} = -(\mu+\varepsilon)x_{3}(t)+\beta I^{\ast}x_{1}(t)+\beta S^{\ast}x_{4}(t), \\ \frac{{\rm d}x_{4}(t)}{{\rm d}t} = \int_{0}^{\infty}\omega_{2}(b)x_{5}(t, b){\rm d}b+\varepsilon x_{3}(t) -(\mu+\delta+k)x_{4}(t), \\ \Big(\frac{\partial }{\partial t}+\frac{\partial }{\partial b}\Big)x_{5}(t, b) = -\varepsilon_{2}(b)x_{5}(t, b), \\ x_{2}(t, 0) = \xi x_{1}(t), \; x_{5}(t, 0) = k x_{4}(t). \end{array} \right. \end{equation} $

设$ x_{1}(t) = x_{1}^{0}e^{\lambda t}, x_{2}(t, a) = x_{2}^{0}(a)e^{\lambda t}, x_{3}(t) = x_{3}^{0}e^{\lambda t}, x_{4}(t) = x_{4}^{0}e^{\lambda t}, x_{5}(t, b) = x_{5}^{0}(b)e^{\lambda t} $是方程(5.7) 的解, 其中$ (x_{1}^{0}, x_{2}^{0}(a), x_{3}^{0}, x_{4}^{0}, x_{5}^{0}(b)) $为初始值, $ \lambda $是一个复数. 将此解代入方程(5.7) 得

(5.8) $ \begin{equation} \lambda x_{1}^{0} = -(\mu+\xi)x^{0}_{1}-\beta I^{*}x_{1}^{0}-\beta S^{*}x_{4}^{0}+\int_{0}^{\infty}\omega_{1}(a)x_{2}^{0}(a){\rm d}a, \end{equation} $

(5.9) $ \begin{equation} \lambda x_{2}^{0}(a)+\frac{{\rm d}x_{2}^{0}(a)}{{\rm d}a} = -\varepsilon_{1}(a)x_{2}^{0}(a), \quad x_{2}^{0}(0) = \xi x_{1}^{0}, \end{equation} $

(5.10) $ \begin{equation} \lambda x_{3}^{0} = \beta I^{*}x_{1}^{0}+\beta S^{*}x_{4}^{0}-(\mu+\varepsilon)x^{0}_{3}, \end{equation} $

(5.11) $ \begin{equation} \lambda x_{4}^{0} = \int_{0}^{\infty}\omega_{2}(b)x_{5}^{0}(b){\rm d}b+\varepsilon x^{0}_{3}-(\mu+\delta+k)x_{4}^{0}, \end{equation} $

(5.12) $ \begin{equation} \lambda x_{5}^{0}(b)+\frac{{\rm d}x_{5}^{0}(b)}{{\rm d}b} = -\varepsilon_{2}(b)x_{5}^{0}(b), \quad x_{5}^{0}(0) = k x_{4}^{0}. \end{equation} $

求解(5.9) 和(5.12) 式后, 将其代入其他方程联立可得

$ \begin{eqnarray*} && \Big(\lambda+k+\delta+\mu-k\theta_{2}(\lambda)-\frac{\beta\varepsilon S^{\ast}}{\lambda+\varepsilon+\mu}\Big)x^{0}_{4} = \frac{\beta\varepsilon I^{\ast}}{\lambda+\varepsilon+\mu}x^{0}_{1}\\ & = &\frac{-\beta S^{*}\varepsilon\beta I^{\ast}}{(\lambda+\mu+\varepsilon)(\lambda+\mu+\xi-\xi\theta_{1}(\lambda)+\beta I^{\ast})}x^{0}_{4}. \end{eqnarray*} $

由于$ x^{0}_{4} $不为零, 整理上式可得

(5.13) $ \begin{equation} H_{3}(\lambda) = \frac{\beta S^{*}\varepsilon(\lambda+\mu+\xi-\xi\theta_{1}(\lambda))} {(\lambda+\mu+\xi-\xi\theta_{1}(\lambda)+\beta I^{*})(\lambda+\mu+\varepsilon)(\lambda+\mu+\delta+k)}+\frac{k\theta_{2}(\lambda)}{\lambda+\mu+\delta+k} = 1. \end{equation} $

假设方程(5.13) 有一个根$ \lambda_{1} = a_{1}+{\rm i}b_{1} $, 且$ a_{1}\geq0 $. 由方程(4.9) 的第三至六方程联立可得$ \frac{\beta S^{*}\varepsilon}{\mu+\varepsilon}+k\theta_{2}-(\mu+\delta+k) = 0 $, 并且由式(3.1) 得到$ |\theta_{i}(\lambda_1)|\leq\theta_{i}, $其中$ \theta_{i} = \theta_{i}(0), i = 1, 2, $则(5.13) 式变形为

$ \begin{eqnarray*} |H_{3}(\lambda_{1})|&\leq&\Big|\frac{\beta S^{*}\varepsilon(\lambda_{1}+\mu+\xi-\xi\theta_{1}(\lambda_{1}))} {(\lambda_{1}+\mu+\xi-\xi\theta_{1}(\lambda_{1})+\beta I^{*})(\lambda_{1}+\mu+\varepsilon)(\lambda_{1}+\mu+\delta+k)}\Big|+\Big| \frac{k\theta_{2}(\lambda_{1})}{(\lambda_{1}+\mu+\delta+k)}\Big|\\ &\leq&\frac{1}{|a_{1}+{\rm i}b_{1}+\mu+\delta+k|}\Big(\Big|\frac{\beta S^{*}\varepsilon}{a_{1}+{\rm i}b_{1}+\mu+\varepsilon}\Big|+|k\theta_{2}|\Big)\\ &\leq&\frac{1}{\mu+\delta+k}\Big(\frac{\beta S^{*}\varepsilon}{\mu+\varepsilon}+k\theta_{2}\Big) = 1. \end{eqnarray*} $

则产生矛盾. 则(5.13) 式的所有特征根具有严格负实部, 即$ T^{\ast} $在$ {\cal R}_{0}>1 $时局部渐近稳定. 定理5.3得证.

定理5.4  如果$ {\cal R}_{0}>1 $, 则地方病平衡点$ T^{\ast} $是全局渐近稳定的.

证  选择Lyapunov函数如下

$ L^{\ast}(t) = L^{\ast}_{s}(t)+L^{\ast}_{v}(t)+E^{\ast}\Phi\Big(\frac{E(t)}{E^{\ast}}\Big)+\frac{\mu+\varepsilon}{\varepsilon}I^{\ast} \Phi\Big(\frac{I(t)}{I^{\ast}}\Big)+L^{\ast}_{r}(t), $

其中$ L^{\ast}_{s}(t) = S^{*}\Phi(\frac{S}{S^{*}}) $, $ L^{\ast}_{v}(t) = \int_{0}^{\infty}\pi_{1}(a)v^{*}(a)\Phi(\frac{v(t, a)}{v^{*}(a)}){\rm d}a $和

$ L^{\ast}_{r}(t) = \frac{\mu+\varepsilon}{\varepsilon}\int_{0}^{\infty}\pi_{2}(b)r^{*}(b)\Phi(\frac{r(t, b)}{r^{*}(b)}){\rm d}b. $

由于

$ \mu+\xi = \frac{1}{S^{*}}\Big(\Lambda-\beta S^{\ast}I^{\ast}+\int_{0}^{\infty}\omega_{1}(a)v^{\ast}(a){\rm d}a\Big) $

和

$ \begin{eqnarray*} \frac{{\rm d}L^{\ast}_{s}}{{\rm d}t} & = & -\Lambda\Big[\Phi \Big(\frac{S^{*}}{S}\Big)+\Phi\Big(\frac{S}{S^{*}}\Big)\Big]+\beta SI^{*}-\beta SI-\beta S^{\ast}I^{\ast}+\beta S^{\ast}I\\ && +\int_{0}^{\infty}\omega_{1}(a)v^{*}(a)\Big[\frac{v(t, a)}{v^{*}(a)}-\frac{S}{S^{*}}- \frac{S^{*}v(t, a)}{S v^{*}(a)}+1\Big]{\rm d}a, \end{eqnarray*} $

$ \frac{{\rm d}L^{\ast}_{v}}{{\rm d}t} = -\pi_{1}(a)v^{\ast}(a)\Phi\Big(\frac{v(t, a)}{v^{\ast}(a)}\Big)\Big|^{\infty}+ \theta_{1}\xi S^{\ast}\Phi\Big(\frac{S}{S^{\ast}}\Big)-\int_{0}^{\infty}\omega_{1}(a)v^{\ast}(a) \Phi\Big(\frac{v(t, a)}{v^{*}(a)}\Big){\rm d}a, $

$ \begin{eqnarray*} \frac{{\rm d}L^{\ast}_{r}}{{\rm d}t}& = & -\frac{\mu+\varepsilon}{\varepsilon}\pi_{2}(b)r^{*}(b)\Phi\Big(\frac{r(t, b)}{r^{*}(b)}\Big) \Big|^{\infty}+\frac{\mu+\varepsilon}{\varepsilon}\theta_{2}k I^{*}\Phi\Big(\frac{I}{I^{*}}\Big)\\ && -\frac{\mu+\varepsilon}{\varepsilon}\int_{0}^{\infty}\omega_{2}(b)r^{*}(b) \Phi\Big(\frac{r(t, b)}{r^{*}(b)}\Big){\rm d}b, \end{eqnarray*} $

由此可得

$ \begin{eqnarray*} \frac{{\rm d} L^{*}(t)}{{\rm d}t} & = & -\Lambda\Big[\Phi \Big(\frac{S^{*}}{S}\Big)+\Phi\Big(\frac{S}{S^{\ast}}\Big)\Big]-\pi_{1}(a)v^{*}(a)\Phi\Big(\frac{v(t, a)}{v^{*}(a)}\Big)\Big|^{\infty}\\ && -\frac{\mu+\varepsilon}{\varepsilon}\pi_{2}(b)r^{*}(b)\Phi\Big(\frac{r(t, b)}{r^{*}(b)}\Big) \Big|^{\infty}+\sum\limits_{l = 1}^{3}B_{l}, \end{eqnarray*} $

其中

$ \begin{eqnarray*} B_{1}& = & \int_{0}^{\infty}\omega_{1}(a)v^{*}(a)\Big[\frac{v(t, a)}{v^{*}(a)}-\frac{S}{S^{*}}- \frac{S^{*}v(t, a)}{S v^{*}(a)}+1-\Phi\Big(\frac{v(t, a)}{v^{*}(a)}\Big)\Big]{\rm d}a+\theta_{1}\xi S^{\ast}\Phi\Big(\frac{S}{S^{\ast}}\Big)\\ & = & -\int_{0}^{\infty}\omega_{1}(a)v^{*}(a)\Phi\Big(\frac{S^{\ast}v(t, a)}{Sv^{*}(a)}\Big){\rm d}a.\\ B_{2}& = & \beta SI^{\ast}-\beta SI-\beta S^{\ast}I^{\ast}+\beta S^{\ast}I+\beta SI -(\mu+\varepsilon)E-\frac{E^{\ast}}{E}\beta SI+(\mu+\varepsilon)E^{\ast}\\ && +(\mu+\varepsilon)E-(\mu+\varepsilon)\frac{IE^{\ast}}{I^{\ast}}+(\mu+\varepsilon)E^{\ast} -(\mu+\varepsilon)\frac{I^{\ast}E}{I}.\\ & = & \beta S^{\ast}I^{\ast}\Big[1+\frac{S}{S^{\ast}}-\frac{EI^{\ast}}{E^*I}-\frac{E^{\ast}SI}{ES^{\ast}I^{\ast}}\Big].\\ B_{3}& = & \frac{\mu+\varepsilon}{\varepsilon}\int_{0}^{\infty}\omega_{2}(b)r^{*}(b) \Big[\frac{r(t, b)}{r^{*}(b)}-\frac{I}{I^{*}}-\frac{I^{*}r(t, b)}{Ir^{*}(b)} +1-\Phi\Big(\frac{r(t, b)}{r^{*}(b)}\Big) \Big]{\rm d}b +\frac{\mu+\varepsilon}{\varepsilon}\theta_{2}kI^{*}\Phi\Big(\frac{I}{I^{*}}\Big)\\ & = & -\frac{\mu+\varepsilon}{\varepsilon}\int_{0}^{\infty}\omega_{2}(b)r^{*}(b)\Phi \Big(\frac{I^{\ast}r(t, b)}{Ir^{*}(b)}\Big){\rm d}b. \end{eqnarray*} $

因此有

$ \frac{{\rm d}L^{*}(t)}{{\rm d}t}\leq0, $

其中等号成立当且仅当$ S = S^{*} $, $ v(t, a) = v^{\ast}(a) $, $ E = E^{\ast} $, $ I = I^{\ast} $和$ r(t, b) = r^{\ast}(b) $, 那么由LaSalle不变集原理, 平衡点$ T^{\ast} $全局渐近稳定性. 定理5.4得证.

本节结合具体数据用Matlab进行数值模拟. 参见文献[18]中统计的麻疹数据, 得到2005–2013年全国麻疹病例按年龄段的发病率, 如图 1所示. 模型中所用到的参数值如下, 这些参数都参见文献[19-21], 对参数的具体值做如下进一步解释:

(1) 假设中国人的平均寿命为840月(70年), 故死亡率$ = \frac{1}{840} = 0.0012 $.

(2) 使用最小二乘法拟合数据得到参数$ \beta = 0.437 $.

(3) 选择模型参数$ \frac{1}{\varepsilon} = 0.5, \; \frac{1}{k} = 0.663, \; \delta = 0.003, \; \Lambda = 2 $. 那么$ \theta_{1} = 0.0042, \; \theta_{2} = 0.3739 $.

基本再生数是决定疾病是否流行的重要指标. 本文所建立模型的基本再生数

$ {\cal R}_{0} = \frac{\beta S^{0}\varepsilon+(\mu+\varepsilon)k\theta_{2}}{(\mu+\delta+k)(\mu+\varepsilon)}. $

偏置相关系数(PRCC) 已被广泛用于分析参数的敏感性, 由$ {\mathcal R}_{0} $表达式, $ \beta $对$ {\cal R}_{0} $影响是正向的, $ \delta $对$ {\cal R}_{0} $影响是反向的. 进一步取样本空间$ n = 1500 $, 把它作为输入变量, 而$ {\cal R}_{0} $的值作为输出参数, 可计算出影响$ {\cal R}_{0} $的四个参数的PRCC值, 这些值的绝对值排序决定了它们对$ {\cal R}_{0} $影响的差异, 而加号或减号表示影响为正或负. 由图 2可以看出, $ \beta, \varepsilon $对$ {\cal R}_{0} $有正的影响, 而$ k, \delta $对$ {\cal R}_{0} $有负的影响, $ \beta, \delta $是对$ {\cal R}_{0} $影响较大的参数, 其次是麻疹染病率$ \varepsilon $, 最后是麻疹的感染者进入恢复者的比率$ k $. 故有效降低麻疹感染的基本方法为: (1) 有效控制麻疹的感染率, 如疾病的易感者在生活中应多注意避免接触患者, 同时对染病者进行管控, 最终达到降低感染率的效果. (2) 控制染病率, 处在潜伏期时应及时就医. (3) 疾病治疗阶段切不可放松, 实施积极的治疗办法, 提高治愈率, 严格控制因病死亡.

取$ \xi = 0.23 $得$ {\cal R}_{0} = 0.8303<1 $, 由定理5.2知无病平衡点$ T_{0} = (S^{0}, v^{0}(a), 0, 0, 0) $是全局渐近稳定的, 其中$ S^{0}\approx8, v^{0}(a)\approx1.84\rho_{1}(a) $, 即解曲线收敛于无病平衡点, 由图 3知随着时间的推移, 各个仓室的人数震荡收敛于无病平衡点. 取$ \xi = 0.103 $得$ {\cal R}_{0} = 1.7934>1 $, 由定理5.4知地方病平衡点$ T^{\ast} = (S^{\ast}, v^{\ast}(a), E^{\ast}, I^{\ast}, r^{\ast}(b)) $是全局渐近稳定的, 其中$ S^{\ast}\approx1, \; v^{\ast}(a)\approx0.103\rho_{1}(a), \; E^{\ast}\approx0.66, \; I^{\ast}\approx3, \; r^{\ast}(b)\approx4.52\rho_{2}(b) $, 即解曲线收敛于地方病平衡点, 由图 4得到经过一段时间后, 各个仓室的人数震荡收敛于地方病平衡点.

本文建立了一类具有接种和复发的年龄结构SVEIR麻疹传染病模型. 计算得到模型的基本再生数为$ {\mathcal R}_{0} = \frac{\beta S^{0}\varepsilon+(\mu+\varepsilon)k\theta_{2}}{(\mu+\delta+k)(\mu+\varepsilon)} $. 根据Webb等^[9]在文中建立和使用的方法, 将模型(2.1) 中的偏微分方程沿着特征线积分, 从而模型转化成了所谓的Volterra型积分方程, 进一步得到了该模型解的适定性. 通过使用动力系统持续性理论, 得到了当$ {\cal R}_{0}>1 $时疾病一致持续. 通过构造适当的Volterra型的Lyapunov函数证明了当$ {\cal R}_{0}<1 $时, 无病平衡点$ T_{0} $全局渐近稳定, 当$ {\cal R}_{0}>1 $时无病平衡点不稳定, 并且地方病平衡点全局渐近稳定. 最后作为理论结果的一个应用, 参考全国2005-2013年全国麻疹病例年龄分布, 用Matlab进行数值模拟, 结合理论分析给出了控制麻疹的一些建议.