Consider
\begin{align} \frac{dS}{dt} &= \mu N -\frac{\beta S I}{N} - \nu S\\[2ex] \frac{dI}{dt} &= \frac{\beta S I}{N} -\nu I \end{align} Where $N=S+I$ is the total population.
By substituting $S=N-I$ into the above we have:
\begin{align} \frac{dI}{dt} &= (\beta - \nu)I - \beta I^2 \end{align}
Solving the above with initial condition $I(0)= I_0$ analytically, we have the solution to the system:
\begin{align} I(t) &= \frac{I_0 (\beta - \nu)}{\beta I_0 - e^{-\left(\beta - \nu\right)t} \left[ \beta I_0 - \left(\beta - \nu\right)\right]}.\\[2ex] S(t) &= 1-I(t). \end{align}
We can make inferences of the long term behaviour of this model by examining the possible values of $\left(\beta -\nu\right)$, that is, of course when $I(t)$ is feasible. We have two cases \begin{align*} \beta - \nu & < 0 \\[2ex] \beta - \nu & > 0 \end{align*}
If $\beta - \nu < 0$, then $e^{-\left(\beta - \nu\right)t} \rightarrow \infty \text{ as } t \rightarrow \infty$ hence
\begin{align} \lim_{t \to \infty} I(t) = 0. \end{align}
If $\beta - \nu > 0$, then $e^{-\left(\beta - \nu\right)t} \rightarrow 0 \text{ as } t \rightarrow \infty$ hence we have the limit
\begin{align} \lim_{t \to \infty} I(t) = \frac{I_0\left(\beta-\nu\right)}{\beta I_0} = 1-\frac{\nu}{\beta}. \end{align}
The equilibrium points:
\begin{align*} e_1 : \left( S_1^*, I_1^*\right)&= \left(1, 0\right), \\[2ex] e_2 : \left( S_2^*, I_2^*\right)&= \left(\frac{\nu }{\beta}, \frac{\nu}{\beta}\left(\frac{\beta}{\nu}-1 \right)\right) \end{align*}
where $\mathcal{R}_0 = \beta/\nu$.
Using the monotonicity theorem, we can prove global stability of the equilibrium of interest on the line without ends by showing that the rhs is non-positive in the intervals to both sides of the equilibrium. How can I do this explicitly?