How to find the basic reproductive number of a discrete SIS epidemic model

Question

I have been following a textbook called Mathematical Models in Population Biology and Epidemiology.

The SIS model is given by the system \begin{aligned} S_{n+1} &= \Lambda + S_n e^{-\mu} e^{-\alpha I_n} + I_n e^{-\mu}[1 - e^{-\sigma}], \\ I_{n+1} &= S_n e^{-\mu}[1 - e^{-\alpha I_n}] + I_n e^{-\mu} e^{-\sigma}. \end{aligned}

The probability of death due to natural causes, the probability of recovering, and the probability of not becoming infected are given, respectively, by the following equations \begin{align} & 1 - e^{-\mu}, \\ & 1 - e^{-\sigma}, \\ & e^{-\alpha I_n}, \end{align} where $\mu$, $\sigma$, and $\alpha$ are positive constants.

The total population is $T_{n} = S_{n} + I_{n}$. By saying the initial population size starts at its asymptotic limit $T_\infty$, we can say $T_0 = T_\infty$. Then, we can substitute $S_n$ by $S_n = T_\infty - I_n$ to get the single equation $$ I_{n+1} = \left(T_{\infty} - I_n \right) e^{-\mu} \left[ 1 - e^{-\alpha I_n} \right] + I_n e^{-(\mu + \sigma)} $$

In a course I am currently in, the basic reproductive number was found by finding the threshold quantity of $I_{n+1}$. The book states that $$R_0 = \frac{\alpha T_\infty e^{-\mu}}{1 - e^{-(\mu + \sigma)} }.$$

So I started by trying to factor out $I_n$ but the equation is too complicated for me to do so. Can anyone help me figure out how the authors derived $R_0$?

Answer 1

Assume indeed that $S_n+I_n=T$ for every $n$, which is equivalent to $\Lambda=T(1-\mathrm e^{-\mu})$, then the evolution of $(S_n)$ is given by $S_{n+1}=g(S_n)$, where, for every $S$, $$g(S)=T(1-\mathrm e^{-\mu-\sigma})+\mathrm e^{-\mu-\alpha T}S\mathrm e^{\alpha S}-\mathrm e^{-\mu}(1-\mathrm e^{-\sigma})S.$$ The function $g$ is strictly convex (being the sum of an affine function and a positive multiple of the function $S\mapsto S\mathrm e^{\alpha S}$ which is strictly convex on $S\geqslant0$) and such that $g(T)=T$, thus, there are two regimes:

• Either $g'(T)\gt1$: Then $g(S)\gt S$ for every $S$ in $(0,S_*)$ and $g(S)\lt S$ for every $S$ in $(S_*,T)$ for some $S_*$ in $(0,T)$ such that $g(S_*)=S_*$, hence $S_n\to S_*$ for every initial population in $[0,T)$ and the population $I_n$ of infected individuals survives (and converges to $I_\infty=T-S_*$).
• Or, $g'(T)\leqslant1$: Then $g(S)\gt S$ for every $S$ in $(0,T)$, hence $S_n\to T$ for every initial population in $[0,T]$. The population $I_n$ of infected individuals dies out.

A simple computation shows that $g'(T)\leqslant1$ (extinction of the infected) if and only if $$\alpha T\leqslant\mathrm e^\mu-\mathrm e^{-\sigma}.$$ As is customary in epidemiology models, one can call basic reproductive number every quantity $R_0$ depending on the parameters of the model such that $R_0\lt1$ implies the extinction of the infected population and $R_0\gt1$ implies its survival. Here, a suitable choice for $R_0$ is indeed $$R_0=\frac{\alpha T}{\mathrm e^\mu-\mathrm e^{-\sigma}}.$$