Threshold Behavior in a Stochastic SIS Epidemic Model with Standard Incidence

Abstract

In this paper, we consider a stochastic SIS epidemic model with perturbed disease transmission coefficient. We discuss the threshold behavior in the sense that if $\mathcal {R}_0^s<1$, the disease dies out in probability; whereas if $\mathcal {R}_0^s>1$, the densities of the distributions of the solution can converge in $L^1$ to an invariant density.

Appendix

In this appendix, we prove that

$$\begin{aligned} E\left[ \int _0^T\frac{S_tI_t^aX_t^{-b}Y_t^{-b}}{S_t+I_t} dB_t\right] =0,\quad \text {for}\; \text {any }\ T>0. \end{aligned}$$

(7.1)

In the following, whenever the formula of random variables is used it is understood in the almost sure sense, and writing “almost surely” or “a.s.” is omitted.

By Itô’s formula, we get

$$\begin{aligned} d\log S_t&= \left( \frac{\Lambda }{S_t}-\frac{\beta I_t}{S_t+I_t}-\mu +\phi \frac{I_t}{S_t}-\frac{\sigma I_t^2}{2(S_t+I_t)^2}\right) dt-\frac{\sigma I_t}{S_t+I_t}dB_t\\&\ge \left( -\beta -\mu +\phi \frac{I_t}{S_t}-\frac{\sigma }{2}\right) dt-\frac{\sigma I_t}{S_t+I_t}dB_t,\\ \frac{d\log X_t}{dt}&= -\mu +\frac{\alpha I_t}{X_t}\quad \text {and}\quad \frac{d\log Y_t}{dt} =-(\alpha +\mu )+\frac{\alpha S_t}{Y_t}. \end{aligned}$$

Integrating the above three formulas from $0$ to $T$ and taking $T\rightarrow \infty$, we get

$$\begin{aligned} \lim _{T\rightarrow \infty }\frac{1}{T}\int _0^T\frac{I_t}{S_t}dt=\frac{1}{\phi }\lim _{T\rightarrow \infty }\left( \beta +\mu +\frac{\sigma }{2}+\frac{\log S_T}{T}\right)&\le \frac{\beta +\mu +\sigma /2}{\phi },\end{aligned}$$

(7.2)

$$\begin{aligned} \lim _{T\rightarrow \infty }\frac{1}{T}\int _0^T\frac{I_t}{X_t}dt=\frac{1}{\alpha }\lim _{T\rightarrow \infty }\left( \mu +\frac{\log X_T}{T}\right)&\le \frac{\mu }{\alpha }, \end{aligned}$$

(7.3)

and

$$\begin{aligned} \lim _{T\rightarrow \infty }\frac{1}{T}\int _0^T\frac{S_t}{Y_t}dt=\frac{1}{\alpha }\lim _{T\rightarrow \infty }\left( \alpha +\mu +\frac{\log Y_T}{T}\right) \le 1+\frac{\mu }{\alpha }, \end{aligned}$$

(7.4)

where in the first inequality we use the fact that $\lim _{T\rightarrow \infty }\frac{1}{T}\int _0^T\frac{I_t}{S_t+I_t}dB_t=0$. By employing Hölder’s Inequality, (7.2)–(7.4) guarantee that

$$\begin{aligned} \int _0^T\left( \frac{S_tI_t^aX_t^{-b}Y_t^{-b}}{S_t+I_t}\right) ^2 dt=&\int _0^T\frac{S_t^{2}I_t^{2a-4b}}{(S_t+I_t)^2}\left( \frac{I_t}{S_t}\right) ^{2b}\left( \frac{S_t}{Y_t}\right) ^{2b}\left( \frac{I_t}{X_t}\right) ^{2b} dt\\ \le&\left( \frac{\Lambda }{\mu }\right) ^{2a-4b}\int _0^T\left( \frac{I_t}{S_t}\right) ^{2b}\left( \frac{S_t}{Y_t}\right) ^{2b}\left( \frac{I_t}{X_t}\right) ^{2b} dt \le M_T, \end{aligned}$$

where $M_T$ is a positive constant depending on $T$. Consequently,

$$\begin{aligned} \int _0^TE\left( \frac{S_tI_t^aX_t^{-b}Y_t^{-b}}{S_t+I_t}\right) ^2 dt\le M_T. \end{aligned}$$

Thus, (7.1) holds.