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How would find the equilibriums of the following system:

$$ \frac{d I}{dt}=-\frac{\beta S I}{N} \qquad (1)$$

$$\frac{d S}{dt}=\frac{\beta S I}{N} -\gamma I \qquad (2)$$

$$\frac{d R}{dt}=\gamma I \qquad (3) $$

where $S+I+R=N$ and $\beta,\gamma >0$.

I know one such equilibrium is \begin{align} e_1 : \left( S_1^*, I_1^*, R_1^*\right)&= \left(N, 0, 0\right), \\[2ex] \end{align}

How do I find the other(if it exists)? I believe it should be similar to the problem below unless I'm mistaken.

I have solved the more complicated case:

$$ \frac{d S}{dt}=\mu N -\frac{\beta S I}{N} - \nu S \qquad (1)$$

$$\frac{d I}{dt}=\frac{\beta S I}{N} -\gamma I - \nu I \qquad (2)$$

$$\frac{d R}{dt}=\gamma I - \nu R \qquad (3) $$

where $S+I+R=N$, $\quad$ $\mu =\nu$, $\quad$ $\mu, \beta, \nu,\gamma >0$.

Solutions are:

\begin{align} e_1 : \left( S_1^*, I_1^*, R_1^*\right)&= \left(N, 0, 0\right), \\[2ex] e_2 : \left( S_2^*, I_2^*, R_2^*\right)&= \left(\frac{N\left(\gamma+\nu\right)}{\beta}, N\nu\left(\frac{1}{\gamma +\nu}-\frac{1}{\beta}\right), N\gamma\left(\frac{1}{\gamma +\nu}-\frac{1}{\beta}\right)\right) \end{align}

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Notice that a necessary condition for an equilibrium is given by: $$\frac{dR}{dt} = 0 \implies I = 0.$$ Since the condition $I=0$ is enough to have $\frac{dI}{dt} = \frac{dS}{dt} = 0$, you have a sufficient condition. Thus the equilibriums are given by: $$(S^*,I^*,R^*) = (K, 0, N-K),$$ for any $0\leq K \leq N$.

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  • $\begingroup$ Mhm, I was being stupid again. Anyway thanks :) $\endgroup$ Commented Mar 10, 2021 at 15:49
  • $\begingroup$ Hmm technically your solution is correct however since this is an epidemic model, we cant have zero infected individuals with positive recovered individuals simultaneously because this would be nonsensical. So I believe the only equilibrium is the disease-free one. $\endgroup$ Commented Mar 10, 2021 at 16:01
  • $\begingroup$ @Math Why not? If your epidemic vanishes, i.e. $I(t)\rightarrow 0$ as $t\rightarrow\infty$, but $I(t_0)>0$ you would then have had a nonzero recovered population. $\endgroup$ Commented Mar 10, 2021 at 16:54
  • $\begingroup$ because $I=0$ implies $R=0$ $\endgroup$ Commented Mar 15, 2021 at 12:08

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