Well these days, everybody is talking about the epidemic model SIR
which is given by the following differential equation:
$$
\begin{cases}
S'(t)=-aS(t)I(t)\\
I'(t)=aS(t)I(t)-bI(t)\\
R'(t)=bI(t)\\
S(0)>0,I(0)>0
\end{cases}
$$
We can see that $S'(t)+I'(t)+R'(t)=0$, so $S(t)+I(t)+R(t)=N$ is
a constant. Thus this model reduces to two differential equations:
$$
\begin{cases}
S'(t)=-aS(t)I(t)\\
I'(t)=aS(t)I(t)-bI(t)\\
S(0)>0,I(0)>0
\end{cases}
$$
All the studies I saw start with the fact that a solution of this
model is global (defined on $[0,\infty)$) and positive. So I wanted
to prove that this differential equation has a global solution. However
I see that this cannot be done by the global Lipschitz Picard-Lindelöf/contraction
theorem, since the function
$$
f(x,y)=\left(-axy,axy-by\right)
$$
is not globally Lipschitz. How can we prove the existence of a global
solution and its positivity?
Best Answer
As you noted, $S+I+R=N$ is constant.
We assume we begin with $S, I, R \ge 0$, since negative populations are not of interest. Moreover the constants $a, b \ge 0$. To show global existence, we just need to show that the solution stays in the compact set $D = \{(s,i,r): s+i+r=N, s \ge 0, i \ge 0, r \ge 0\}$. But $R' \ge 0$ while $S' \ge - aN S$ and $I' \le -b I$ when $(S,I,R) \in D$, which implies $S$ and $I$ never reach $0$.