Invariant set for SIRS model A SIRS model is a set of ODE used to describe the evolution of epidemic/pandemic. It is defined as follows:
$$\begin{cases}
\dot{s} = -\beta s i + \omega r\\
\dot{i} = \beta s i - \gamma i \\
\dot{r} = \gamma i - \omega r
\end{cases},$$
where $s$, $i$ and $r$ represent the share of susceptible, infected and recovered individuals, respectively, and $\beta$, $\omega$ and $\gamma$ are positive parameters.
First of all, one notice that:
$$\dot{s} + \dot{i} + \dot{r} = 0.$$
As a direct consequence, if $s(0) + i(0) + r(0) = 1,$ then $s(t) + i(t) + r(t) = 1 ~\forall t > 0.$
Anyway, this does not ensure that no state variable can assume negative value or ca be bigger than $1$ for some time instant $t' > 0$. This is what I want to prove.
In other words, how can I prove that the set:
$$S = \left\{(s(t), i(t), r(t)) : s(t) \geq 0 \wedge i(t) \geq 0 \wedge r(t) \geq 0 \wedge s(t) + i(t) + r(t) = 1 ~\forall t > 0\right\}$$
is invariant for the ODE system reported at the beginning of this post?
Thanks in advance.
 A: Let $z(t,z(0))=(s(t),i(t),r(t))$ be the solution to the SIRS with initial condition $z(0)\in S$ with $z(0)\overset{\Delta}=(s(0), i(0), r(0))\neq 0$ (since $s(0)+i(0)+r(0)=1$).
Let $T\overset{\Delta}=\inf\limits_{t\geq 0}\left\{z(t,z(0))\in S^{c}\right\}$ be the hitting time to the complement of $S$. Assume $T<\infty$. Since $S$ is closed, at the instant $T$, the solution lies at the boundary of $S$: either i) $s(T)=0$ or ii) $i(T)=0$ or iii) $r(T)=0$. Note that $z(T,z(0))\neq 0$ ($z(t,0)=0$ is the unique $0$ solution), therefore some state-variable should be positive. Say $i(T)>0$ is the only positive one. Then, $r(T)=0$, but $\dot{r}(T)=\gamma i(T)>0$. Further, $s(T)=0$ and $\dot{s}(T)=0$, but $\ddot{s}(T)=-\beta\dot{s}(T)i(T)-\beta s(T) \dot{i}(T)+\omega \dot{r}(T)=\omega \dot{r}(T)>0$. Since $z(t,z(0))$ is smooth, there exists $\epsilon>0$ so that $i(t)>0$ and $s(t)>0$ and $r(t)>0$ for all $t\in (T,T+\epsilon)$. This conclusion can be obtained in the other cases as well when $i(T)=0$. This contradicts the assumption $T<\infty$, therefore $T=\infty$.
