﻿ 一类传染病模型的稳定性研究

# 一类传染病模型的稳定性研究Local Stability of a Class of Epidemic Model

Abstract: In this paper, the local stability condition of a class of bilinear incidence epidemic model with ver-tical transmission and contact transmission is presented.

1. 引言

$\left\{\begin{array}{l}\frac{\text{d}S\left(t\right)}{\text{d}t}=r\left(1-\frac{S\left(t\right)+I\left(t\right)}{k}\right)S\left(t\right)-dS\left(t\right)-\beta I\left(t\right)S\left(t\right)+qr\left(1-\frac{S\left(t\right)+I\left(t\right)}{k}\right)I\left(t\right)\\ \frac{\text{d}I\left(t\right)}{\text{d}t}=\beta I\left(t\right)S\left(t\right)-\left(d+\mu \right)I\left(t\right)+\left(1-q\right)r\left(1-\frac{S\left(t\right)+I\left(t\right)}{k}\right)I\left(t\right)\end{array}$ (1)

$\left\{\begin{array}{l}\frac{\text{d}N\left(t\right)}{\text{d}t}=\left(r-d\right)N\left(t\right)-\frac{r}{k}{N}^{2}\left(t\right)-\mu I\left(t\right)\\ \frac{\text{d}I\left(t\right)}{\text{d}t}=\left[\left(1-q\right)r-\left(d+\mu \right)\right]I\left(t\right)+\frac{\beta k-\left(1-q\right)r}{k}N\left(t\right)I\left(t\right)-\beta {I}^{2}\left(t\right)\end{array}$ (2)

${R}_{0}=\frac{r}{d}$ , ${N}_{0}=\frac{k\left(r-d\right)}{r}$ , $\stackrel{¯}{N}=\frac{k\left[\left(1-q\right)r-\left(d+\mu \right)\right]}{\left(1-q\right)r-\beta k}$ , ${N}_{\Delta }=\frac{\beta k\left(r-d\right)+\mu \left[\left(1-q\right)r-\beta k\right]}{2\beta r}$,

$R=\frac{\stackrel{¯}{N}}{{N}_{0}}=\frac{r\left[\left(1-q\right)r-\left(d+\mu \right)\right]}{\left(r-d\right)\left[\left(1-q\right)r-\beta k\right]}$ , ${\sigma }_{1}=\frac{\left(1-q\right)r}{\beta k}$ , ${\sigma }_{2}=\frac{\left(1-q\right)r}{d+\mu }$,

2. 结论

1) ${R}_{0}>1>{\sigma }_{i}$$R<1$$2{N}_{*}>{N}_{0}$$i=1,2$

2) ${R}_{0}>{\sigma }_{i}>1$$R<1$$\beta r\left(2{N}_{*}-{N}_{0}\right)>\mu \left[\left(1-q\right)r-\beta k\right]$$i=1,2$ ，系统(2)的地方病平衡点 $E\left({N}_{*},{I}_{*}\right)$ 是局部稳定的。

$\begin{array}{l}\stackrel{˙}{x}=Ax+f\left(x,y\right)\hfill \\ \stackrel{˙}{y}=By+g\left(x,y\right)\hfill \end{array}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(x,y\right)\in {R}^{c}×{R}^{s}$ (3)

$f\left(0,0\right)=0$ , $Df\left(0,0\right)=0$ ; $g\left(0,0\right)=0$ , $Dg\left(0,0\right)=0$ .

${W}^{c}\left(0\right)=\left\{\left(x,y\right)\in {R}^{c}×{R}^{s}|y=h\left(x\right),|x|<\delta ,h\left(0\right)=0,Dh\left(0\right)=0,0<\delta \ll 1\right\}$,

$\stackrel{˙}{\nu }=A\nu +f\left(\nu ,h\left(\nu \right)\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}\nu \in {R}^{c}$ . (4)

3. 定理证明

1) ${R}_{0}>{\sigma }_{2}>1>{\sigma }_{1}$$\beta k\left(r-d\right)=2\sqrt{\beta k\mu r\left[\left(1-q\right)r-\left(d+\mu \right)\right]}+\mu \left[\beta k-\left(1-q\right)r\right]$

$\beta k\left(r-d\right)+\mu \left[\left(1-q\right)r-\beta k\right]>0$

2) ${R}_{0}>1>{\sigma }_{i}$$R<1$$i=1,2$

3) ${R}_{0}>{\sigma }_{i}>1$$R<1$$i=1,2$

4) ${R}_{0}>{\sigma }_{i}>1$$R>1$$\beta k\left(r-d\right)-\mu \left[\left(1-q\right)r-\beta k\right]>0$$i=1,2$

${\left[\beta k\left(r-d\right)\right]}^{2}=4\beta k\mu r\left[\left(1-q\right)r-\left(d+\mu \right)\right]-\mu \left[\left(1-q\right)r-\beta k\right]\left\{\left[\left(1-q\right)r-\beta k\right]\mu +2\beta k\left(r-d\right)\right\}$

${J}_{E}=\left(\begin{array}{cc}r-d-\frac{2r}{k}{N}_{*}& -\mu \\ \frac{\beta k-\left(1-q\right)r}{k}{I}_{*}& -\beta {I}_{*}\end{array}\right)$ .

${\lambda }^{2}+{b}_{1}\lambda -{b}_{0}=0$ .

${\lambda }_{1}=\frac{-{b}_{1}-\sqrt{{b}_{1}^{2}+4{b}_{0}}}{2}$ , ${\lambda }_{2}=\frac{-{b}_{1}+\sqrt{{b}_{1}^{2}+4{b}_{0}}}{2}$ .

${N}_{*}=\frac{\beta k\left(r-d\right)+\mu \left[\left(1-q\right)r-\beta k\right]}{2\beta r}$,

$\begin{array}{c}{I}_{*}=\frac{\left(1-q\right)r-\left(d+\mu \right)}{\beta }-\frac{\left(r-d\right)\left[\left(1-q\right)r-\beta k\right]}{2\beta r}-\frac{\mu {\left[\left(1-q\right)r-\beta k\right]}^{2}}{2{\beta }^{2}kr}\\ =\frac{k{\left(r-d\right)}^{2}}{4\mu r}-\frac{\mu {\left[\left(1-q\right)r-\beta k\right]}^{2}}{4{\beta }^{2}kr}\end{array}$ .

${b}_{0}=\frac{\beta \left[k\left(r-d\right)-2r{N}_{*}\right]+\mu \left[\left(1-q\right)r-\beta k\right]}{k}{I}_{*}=0$ .

$\left\{\begin{array}{l}\frac{\text{d}x\left(t\right)}{\text{d}t}=\left(r-d\right)x\left(t\right)-\frac{2r}{k}{N}_{*}x\left(t\right)-\mu y\left(t\right)-\frac{r}{k}{x}^{2}\left(t\right)\\ \frac{\text{d}y\left(t\right)}{\text{d}t}=\frac{\beta k-\left(1-q\right)r}{k}{I}_{*}x\left(t\right)-\beta {I}_{*}y\left(t\right)+\frac{\beta k-\left(1-q\right)r}{k}x\left(t\right)y\left(t\right)-\beta {y}^{2}\left(t\right)\end{array}$

$\left(\begin{array}{c}{x}^{\prime }\\ {y}^{\prime }\end{array}\right)=\left(\begin{array}{cc}\left(r-d\right)-\frac{2r}{k}{N}_{\ast }& -\mu \\ \frac{\beta k-\left(1-q\right)r}{k}{I}_{*}& -\beta {I}_{\ast }\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)+\left(\begin{array}{c}-\frac{r}{k}{x}^{2}\\ \frac{\beta k-\left(1-q\right)r}{k}xy-\beta {y}^{2}\end{array}\right)$ .

$\left(\begin{array}{c}\text{1}\\ \text{1}-{\sigma }_{\text{1}}\end{array}\right)$$\left(\begin{array}{c}1\\ A\end{array}\right)$ ，这里 $A=\frac{\beta {I}_{\ast }}{\mu }$ 。记 $T=\left(\begin{array}{cc}1& 1\\ 1-{\sigma }_{1}& A\end{array}\right)$ ，通过变换 $\left(\begin{array}{c}x\\ y\end{array}\right)=T\left(\begin{array}{c}m\\ n\end{array}\right)$ ，可以得到：

$\left(\begin{array}{c}{m}^{\prime }\\ {n}^{\prime }\end{array}\right)=\left(\begin{array}{cc}0& 0\\ 0& -{b}_{1}\end{array}\right)\left(\begin{array}{c}m\\ n\end{array}\right)+{T}^{-1}\left(\begin{array}{c}-\frac{r}{k}{m}^{2}-\frac{r}{k}{n}^{2}-\frac{2r}{k}mn\\ \beta \left(1-{\sigma }_{1}\right)\left(1-{\sigma }_{1}-A\right)mn+\beta A\left(1-{\sigma }_{1}-A\right){n}^{2}\end{array}\right)$ .

$\left\{\begin{array}{l}{m}^{\prime }=-\frac{Ar}{\left(A+{\sigma }_{1}-1\right)k}{m}^{2}-\frac{Ar}{\left(A+{\sigma }_{1}-1\right)k}{n}^{2}-\frac{2Ar}{\left(A+{\sigma }_{1}-1\right)k}mn+\beta \left(1-{\sigma }_{1}\right)mn+\beta A{n}^{2}\\ {n}^{\prime }=-{b}_{1}n-\frac{r\left({\sigma }_{1}-1\right)}{\left(A+{\sigma }_{1}-1\right)k}{m}^{2}-\frac{r\left({\sigma }_{1}-1\right)}{\left(A+{\sigma }_{1}-1\right)k}{n}^{2}-\frac{2r\left({\sigma }_{1}-1\right)}{\left(A+{\sigma }_{1}-1\right)k}mn-\beta \left(1-{\sigma }_{1}\right)mn-\beta A{n}^{2}\end{array}$

$a=\frac{r\left(1-{\sigma }_{1}\right)}{\left(A+{\sigma }_{1}-1\right){b}_{1}k}$ , $b=-\frac{\beta r{\left(1-{\sigma }_{1}\right)}^{2}}{\left(A+{\sigma }_{1}-1\right){b}_{1}^{2}k}+\frac{2{r}^{2}\left({\sigma }_{1}-1\right)\left({\sigma }_{1}-1-A\right)}{{\left(A+{\sigma }_{1}-1\right)}^{2}{b}_{1}^{2}{k}^{2}}$ .

${N}_{*}=\frac{\beta k\left(r-d\right)+\mu \left[\left(1-q\right)r-\beta k\right]+\sqrt{{\left\{\beta k\left(r-d\right)+\mu \left[\left(1-q\right)r-\beta k\right]\right\}}^{2}-4\beta k\mu r\left[\left(1-q\right)r-\left(d+\mu \right)\right]}}{2\beta r}$${I}_{*}=\frac{\beta k-\left(1-q\right)r}{\beta k}{N}_{*}+\frac{\left(1-q\right)r-\left(d+\mu \right)}{\beta }$ .

${R}_{0}>{\sigma }_{i}>1$$R>1$$\beta k\left(r-d\right)-\mu \left[\left(1-q\right)r-\beta k\right]>0$ , $i=1,2$,

${\left[\beta k\left(r-d\right)\right]}^{2}=4\beta k\mu r\left[\left(1-q\right)r-\left(d+\mu \right)\right]-\mu \left[\left(1-q\right)r-\beta k\right]\left\{\left[\left(1-q\right)r-\beta k\right]\mu +2\beta k\left(r-d\right)\right\}$,

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