The SIR epidemic model presents three differential equations
for three time-dependent variables $s(t), i(t), r(t)$:
$$\begin{align}
\frac{ds}{dt} & = – \beta i s \\
\frac{di}{dt} & = \beta i s – \gamma i \\
\frac{dr}{dt} & = \gamma i
\end{align}$$
It is assumed that the variables are non-negative, $s(t) + i(t) + r(t) = 1$,
and the coefficients $\beta, \gamma$ are positive.
In the literature it is claimed or taken as self-evident that
$$
\lim_{t \to \infty} i(t) = 0
$$
How can this behavior be proven rigorously?
Best Answer
First notice that we only need to look at the phase diagram $s\times i$, since $r=1-s-i$. Looking at the phase diagram we will ignore time (different parts of the trajectory will be realized within different velocities), it is possible because the system is autonomous.
First, we need to prove that the system converges to some point, then we'll find the candidates. Let $V(s,i) = s$, we have $\frac{d}{dt}V((s(t), i(t))) = -\beta s(t)i(t) \leq 0$, then, by LaSalle’s theorem: $$ \lim_{t\to\infty} d( (s(t), i(t)), E) = 0$$ where $E = \{ (s,i) : -\beta si = 0 \} = \{ (s,i) : s = 0\text{ or }i=0 \}$.
Now we can study the trajectories and find the exact limit.
Notice that if $\exists t^*$ such that $i(t^*)=0$ or $s(t^*)=0$ we are done. The case $i(t^*) = 0$ has null derivative and the case $s(t^*) = 0$ implies that $i(t)$ has exponential decay.
Since the SEIR is an autonomous system we have, by the implicit function theorem, on the points such that $\frac{ds}{dt} \neq 0$: $$ \frac{di}{ds} = \frac{ \frac{di}{dt} }{ \frac{ds}{dt} } = \frac{\gamma}{\beta s}-1 \Rightarrow i(s) = \frac{\gamma}{\beta}\ln{\frac{s}{s_0}} - s + i_0 + s_0 $$
We have $\frac{ds}{dt} \neq 0 \Leftrightarrow s> 0\text{ and }i>0$. And it is also true that $s$ is decreasing with $t$ on $s>0$ and $i>0$. So: $$ \lim_{t\to\infty} i(t) = \inf_{\substack{i>0\\s_0\geq s>0}} \left\{\frac{\gamma}{\beta}\ln{\frac{s}{s_0}} - s + i_0 + s_0 \right\} = 0$$
$$ \lim_{t\to\infty} s(t) = \inf \{ s\geq 0 : i(s) \geq 0 \}$$
The inf above is well-defined since $\lim_{s\to 0} i(s) = \lim_{s\to \infty} i(s) = -\infty$ (just for fun: it also has a maximum on $s=\frac{\gamma}{\beta}$, it means that from any starting point with $s> \frac{\gamma}{\beta}$ de infection will grows until $s = \frac{\gamma}{\beta}$ and then starts to decay, when we reach $s\leq \frac{\gamma}{\beta}$ we have the so called "cattle immunity").
Plotting all we've got, we have: