## 具有年龄结构的麻疹传染病模型的稳定性分析

1 新疆农业大学数理学院 乌鲁木齐 830052

2 新疆大学数学与系统科学学院 乌鲁木齐 830046

3 长安大学理学院 西安 710064

## Analysis of the Stability for Measles Epidemic Model with Age-Structured

Sun Dandan,1, Li Yingke,1, Teng Zhidong2, Zhang Tailei3

1 School of Mathematics and Physics, Xinjiang Agriculture University, Urumqi 830052

2 School of Mathematics and System Science, Xinjiang University, Urumqi 830046

3 School of Sciences, Changan University, Xi an 710064

 基金资助: 中国博士后科学基金资助项目.  2020M683714XB新疆维吾尔自治区自然科学计划（自然科学基金）面上项目.  2021D01A98陕西省自然科学基础研究计划项目.  2021JM-445

 Fund supported: the China Postdoctoral Science Foundation.  2020M683714XBthe NSF of Xinjiang.  2021D01A98the Natural Science Basic Research Plan in Shaanxi Province.  2021JM-445

Abstract

In this paper, a kind of SVEIR measles epidemic model with age structure is established. Firstly, the model is transformed into Volterra integral equation and the well-possdness of solutions of the model is obtained, including non-negativity, boundedness, asymptotic smoothness, etc. Then the equilibria and the basic reproduction number ${{\cal R}}_{0}$ of the model is derived, and it is proved that the epidemic is uniformly persistent when ${{\cal R}}_{0}>1$. Further by analyzing the characteristic equations and selecting suitable Lyapunov functions, we get the model only has the disease-free equilibrium that is globally asymptotically stable if ${{\cal R}}_{0}<1$; if ${{\cal R}}_{0}>1$, the disease-free equilibrium is unstable, the endemic disease equilibrium exist and is globally asymptotically stable. These main theoretical results are applied in the analysis of the trend in data on measles infectious diseases across the country.

Keywords： Age-structured measles model ; Well-posedness of solutions ; Basic reproduction number ; Uniform persistence ; Stability of equilibrium

Sun Dandan, Li Yingke, Teng Zhidong, Zhang Tailei. Analysis of the Stability for Measles Epidemic Model with Age-Structured. Acta Mathematica Scientia[J], 2021, 41(6): 1950-1968 doi:

## 2 模型的建立

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

(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

## 3 解的适定性

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

$\lambda>0$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

${\rm (ii)}$模型(2.1) 的解最终有界, 即对任意$x_{0}\in{\mathbb Y} $$\limsup\limits_{t\rightarrow \infty}\|F(t, x_0)\|_{{\mathbb Y}} \leq\frac{\Lambda}{\mu}. 对任意初值 x_{0}\in{{\mathbb Y}} , 对模型(2.1) 的解范数 \|F(t, x_0)\|_{{\mathbb Y}} 求导数. 由于 那么从模型(2.1) 可得 由此进一步得到 $$\|F(t, x_0)\|_{{\mathbb Y}} \leq\frac{\Lambda}{\mu}-e^{-\mu t}\Big(\frac{\Lambda}{\mu}-\|x_0\|_{{\mathbb Y}}\Big).$$ x_{0}\in\Pi 时, 由式(3.4) 得知对任意 t>0$$ 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.

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

${\rm (i)}$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 $$u(t, h)\rightarrow0 , 且当 \|x_{0}\|_{\mathbb Y}\leq h 时, 对任意 t\geq0$$ \|\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.

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

显然对任何$h>0 $$\lim\limits_{t\rightarrow \infty}u(t, h) = 0 . 由式(3.2) 和(3.3), 对任意 x_0\in{\mathbb Y} , 当 \|x_0\|_{\mathbb Y}\leq h 时可以得到 注意到对任意 l\geq0 , 有 \varepsilon_{i}(l)\geq\mu\;(i = 1, 2) , 进一步得到 命题3.1得证. 为了验证引理3.2的条件 {\rm (iii)} , 首先引进如下判定空间 L^{1}_{+} 中的有界闭集紧性的结果. 引理3.3[11] 常数 p\geq1 , 设集合 K\subset L^{p}[0, \infty) 是有界闭集. 如果下述条件成立: {\rm (i)} 对任意 h\in K , \lim\limits_{s\rightarrow0}\int_{0}^{\infty}|h(z+s)-h(z)|^{p}{\rm d}z = 0 ; {\rm (ii)}$$ \lim\limits_{s\rightarrow \infty}\int_{s}^{\infty}|h(z)|^{p}{\rm d}z = 0 $$h\in K 一致成立. K$$ L^{p}[0, \infty)$中是紧的.

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$S^{0}-\epsilon_{0}>0$

$$$E(t)\geq \tilde{E}(t), \; I(t)\geq \tilde{I}(t), \; r(t, b)\geq \tilde{r}(t, b),$$$

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

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

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

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

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

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} $$I(t)\leq M . 因此, 从模型(2.1) 的第一个方程可得 考虑下述比较系统 它具有全局渐近稳定的平衡点 \nu^* = \frac{\Lambda}{\mu+\xi+\beta M} . 由比较原理得到 \liminf\limits_{t\to\infty}S(t)\geq\nu^* , 这表明模型(2.1) 中的 S(t) 是一致持续的. 对于任意初值 (S_{0}, v_{0}(\cdot), E_{0}, I_{0}, r_{0}(\cdot))\in{\mathbb Z} , 且 E_{0}>0, I_{0}>0, r_{0}(\cdot)\not\equiv0 , 由式(3.3) 得知, 对于所有 t>0$$ \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 $$E(t)>0, \; I(t)>0 . 因此, 集合 {\mathbb Z} 为模型(2.1) 的正向不变集. 定义集合 M_{1} = \{T_{0}\} , \omega(x_{0}) 为解 F(t, x_{0})$$ \omega$极限集. 由于对于所有$t\geq0 $$F(t, T_{0}) = T_{0}\in\partial{{\mathbb Z}} , 我们有 M_{1}\subset\bigcup\limits_{x_{0}\in M_{\partial}}\omega(x_{0}). 接下来证明 \bigcup\limits_{x_{0}\in M_{\partial}}\omega(x_{0})\subset M_{1} . 对任意 x_{0}\in M_{\partial}$$ F(t, x_{0})\in\partial{{\mathbb Z}}$对所有$t\geq 0$均成立. 进一步有$E(t)\equiv0 $$I(t)\equiv0$$ \int_{0}^{\infty}r(t, b){\rm d}b\equiv0$.$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

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

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

## 5 平衡点的稳定性

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

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

$x_{1}(t) = x_{1}^{0}e^{\lambda t}, x_{2}(t, a) = x_{2}^{0}(a)e^{\lambda t}, x_{3}(t) = x_{3}^{0}e^{\lambda t}, x_{4}(t) = x_{4}^{0}e^{\lambda t}, x_{5}(t, b) = x_{5}^{0}(b)e^{\lambda t}$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

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

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

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

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

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

选择Lyapunov函数如下

选择Lyapunov函数如下

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