Dynamics and asymptotic profiles of endemic equilibrium for SIS epidemic patch models

Abstract

In this paper, we perform qualitative analysis to two SIS epidemic models in a patchy environment, without and with linear recruitment. The model without linear recruitment was proposed and studied by Allen et al. (SIAM J Appl Math 67(5):1283–1309, 2007). This model possesses a conserved total population number, whereas the model with linear recruitment has a varying total population. However, both models have the same basic reproduction number. For both models, we establish the global stability of endemic equilibrium in a special case, which partially solves an open problem. Then we investigate the asymptotic behavior of endemic equilibrium as the mobility of infected and/or susceptible population tends to zero. Though the basic reproduction number is a well-known critical index, our theoretical results strongly suggest that other factors such as the variation of total population number and individual movement may also play vital roles in disease prediction and control. In particular, our results imply that the variation of total population number can cause infectious disease to become more threatening and difficult to control.

Appendix

Here we shall discuss the global dynamics of the ODE problem (6.2) with initial data $S(0) = S_0 \ge 0$ and $I(0) = I_0 >0$. Our main result reads as follows.

Theorem A

The following statement holds.

• (i) If $\beta \le \theta \gamma$, then $\lim _{t\rightarrow \infty }(S(t),I(t)) = (S^*,0)$, where $S^*$ is some positive constant;

• (ii) If $\beta >\theta \gamma$, then $\lim _{t\rightarrow \infty }(S(t),I(t)) = (0,0).$

Proof

Clearly, (S, I) exists globally, and $S(t)>0$, $I(t)>0$ for all $t>0$. One also observes that

$$\begin{aligned} \frac{d}{dt}(S+I) = (\theta - 1)\gamma I<0,\quad \forall t>0. \end{aligned}$$

(B3)

As a result, $(S+I)(t)$ is strictly decreasing and the limit $\lim _{t\rightarrow \infty }(S+I)(t)$ exists. Obviously, $S'(t)$ and $I'(t)$ remain bounded for $t \in (0,\infty )$. In view of Peng and Zhao (2012, Lemma 4.1) and (B3), we must have

$$\begin{aligned} \lim _{t\rightarrow \infty }I(t) = 0. \end{aligned}$$

(B4)

Thus, $\lim _{t\rightarrow \infty }S(t)$ exists and we denote it by $S^*$.

• (i) By virtue of $\beta \le \theta \gamma$, from the S-equation it follows that

  $$\begin{aligned} \frac{dS}{dt}\ge (\theta \gamma - \beta )I\ge 0,\quad \forall t>0. \end{aligned}$$

  That is, S(t) is nondecreasing and hence $S^* = \lim _{t\rightarrow \infty } S(t) >0.$

• (ii) We first consider the case of $\beta >\gamma$. Then there exists some $\epsilon _0>0$ such that $\beta - \gamma - \epsilon _0>0$. Now suppose $S^*>0$. Since $\frac{\beta S}{S+I} \rightarrow \beta$ as $t\rightarrow \infty$ by (B4), there exists some $T>0$ fulfilling $\frac{\beta S}{S+I} > \beta - \epsilon _0$ for all $t>T$. Notice that

  $$\begin{aligned} \frac{d I}{dt} = I\left( \frac{\beta S}{S+I} - \gamma \right)> (\beta - \gamma - \epsilon _0)I>0,\quad \forall t>T, \end{aligned}$$

  from which it readily follows that $I(t) \rightarrow \infty$ as $t\rightarrow \infty$, contradicting (B4). Therefore, $S^* = 0$ in this case.

  We now consider the case of $\beta = \gamma$. Let $U(t)= S(t)/I(t)$. Then direct calculations show that $\frac{d U}{dt} = \beta \theta$. Consequently, it holds

  $$\begin{aligned} U(t) = \beta \theta t + \frac{S_0}{I_0} \quad \text{ and } \text{ so }\quad S(t) = \left( \beta \theta t+ \frac{S_0}{I_0} \right) I(t),\quad \forall t \ge 0. \end{aligned}$$

  Then using the I-equation, we have

  $$\begin{aligned} \frac{dI}{dt} = \beta I \left( \frac{S}{S+I} - 1\right) = \beta I \left( \frac{\beta \theta t + S_0/{I_0}}{\beta \theta t+S_0/{I_0} +1 } -1 \right) = -\frac{\beta I}{\beta \theta t + S_0/{I_0} + 1},\quad \forall t>0. \end{aligned}$$

  Solving this ODE, we obtain

  $$\begin{aligned} I(t) = I_0 \left( \frac{S_0/{I_0} + 1}{\beta \theta t + S_0/{I_0} + 1} \right) ^{1/\theta },\quad \forall t\ge 0, \end{aligned}$$

  and in turn,

  $$\begin{aligned} S(t) =I_0 \left( \beta \theta t+ \frac{S_0}{I_0} \right) \left( \frac{S_0/{I_0} + 1}{\beta \theta t + S_0/{I_0} + 1} \right) ^{1/\theta },\quad \forall t\ge 0. \end{aligned}$$

  Thus, $S^* = \lim _{t\rightarrow \infty }S(t) = 0.$

  It remains to handle the case of $\theta \gamma<\beta < \gamma$. We pick $\delta _0>0$ such that $\theta \gamma - \beta +\delta _0<0$. We again suppose that $S^*>0$. Then there exists some $T_1>0$ satisfying

  $$\begin{aligned} \beta - \delta _0< \beta \frac{S(t)}{S(t)+I(t)} < \beta +\delta _0,\quad \forall t>T_1. \end{aligned}$$

  It follows from the I-equation that

  $$\begin{aligned} \frac{dI}{dt}= I \left( \beta \frac{S}{S+I} -\gamma \right)> I(\beta -\gamma - \delta _0), \quad \forall t>T_1. \end{aligned}$$

  Then

  $$\begin{aligned} I(t) > I(T_1) e^{-(\gamma - \beta +\delta _0)(t-T_1)},\quad \forall t\ge T_1. \end{aligned}$$

  Making use of the S-equation, we deduce that

  $$\begin{aligned} \frac{dS}{dt} = I \left( \theta \gamma - \beta \frac{S}{S+I} \right)< I(\theta \gamma - \beta +\delta _0) < I(T_1) (\theta \gamma - \beta +\delta _0) e^{-(\gamma - \beta +\delta _0)(t-T_1)},\quad \forall t>T_1. \end{aligned}$$

  Since $S(t) \rightarrow S^*>0$ as $t\rightarrow \infty$, then for any $\epsilon$ satisfying

  $$\begin{aligned} 0< \epsilon < -I(T_1) \frac{\theta \gamma - \beta +\delta _0}{\gamma - \beta +\delta _0}, \end{aligned}$$

  (B5)

  there exists some $T_2>0$ such that $S(t)<S^*+\epsilon$ for all $t\ge T_2.$ We now take $T=\max \{T_1,T_2\}$. Consider the following auxiliary problem

  $$\begin{aligned} \left\{ \begin{array}{ll} \displaystyle \frac{d u}{dt} = I(T_1) (\theta \gamma - \beta +\delta _0) e^{-(\gamma - \beta +\delta _0)(t-T_1)},\quad t>T,\\ \displaystyle u(T)=S^*+\epsilon , \end{array}\right. \end{aligned}$$

  (B6)

  It can be easily checked that the unique solution u(t) of (B6) satisfies

  $$\begin{aligned} \lim _{t\rightarrow \infty } u(t) = S^*+\epsilon + I(T_1) \frac{\theta \gamma -\beta +\delta _0}{\gamma - \beta +\delta _0}<S^*, \end{aligned}$$

  (B7)

  thanks to (B5). On the other hand, a simple comparison of S and u indicates that $S(t) < u(t)$ for all $t>T$, contradicting (B7) if we send $t\rightarrow \infty$. Thus, we must have $S^*=0.$

$\square$