Consider
\begin{align} \frac{dS}{dt} &= \mu N -\frac{\beta S I}{N} +\gamma I - \nu S\\[2ex] \frac{dI}{dt} &= \frac{\beta S I}{N} -\gamma I -\nu I \end{align} where $N=S+I$ is the total population.
If $\mu=\nu$, the above reduces to
\begin{align} \frac{dS}{dt} &= \nu -\beta S I +\gamma I- \nu S\\[2ex] \frac{dI}{dt} &= \beta S I -\nu I -\gamma I \end{align}
with 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{\gamma+\nu}{\beta}, \frac{\gamma+\nu}{\beta}\left(\frac{\beta}{\gamma+\nu} -1\right)\right) \end{align*}
I have analysed the global stability of $e_1$ however I haven't managed to find a global Lyapunov function to check stability for $e_2$. I know the Lyapunov function should composites of the Volterra function, any ideas?
EDIT
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.
Which reduces to
\begin{align} \frac{dS}{dt} &= \nu -\beta S I - \nu S\\[2ex] \frac{dI}{dt} &= \beta S I -\nu I \end{align}
The equilibrium points now read
\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)= \left(\frac{1}{\mathcal{R}_0}, \frac{1}{\mathcal{R}_0}\left(\mathcal{R}_0-1\right)\right). \end{align*}
The biological feasible region for this system is the simplex in the first quadrant, the set being $\Omega = \left\lbrace \left(S,I\right)\in \mathbb{R}_+^2 : S\geq 0, I \geq 0, S+I \leq 1 \right\rbrace$. This set is a positively invariant set for our system.
Theorem: If $\mathcal{R}_0 > 1$, then the endemic equilibrium $e_2$ is globally asymptotically stable in the interior of $\Omega$.
Proof: Consider the Lyapunov function
\begin{equation} V(S,I) = \left(S-S_2^*\right)+ \left( I-I_2^*\right) -S_2^* \ln \frac{S}{S_2^*} - I_2^* \ln \frac{I}{I_2^*}. \end{equation}
This function is positive definite since $V\left(S,I\right) \geq 0$ and $V\left(S,I\right)=0$ at the equilibrium point $e_2$. The derivative of $V$ along solutions of (2.3) and (2.4) we have
\begin{align*} \dot V &= \dot S +\dot I - \frac{S_2^*}{S}\dot S-\frac{I_2^*}{I}\dot I\\ &= \nu(1-S) -\beta S I + \beta S I -\nu I -\frac{\nu}{\beta S}\left( \nu -\nu S -\beta S I \right) - I_2^*\left(\beta S - \nu\right)\\ &= \nu(1-S) -\frac{\nu}{\beta S}\left( \nu -\nu S \right) - I_2^*\left(\beta S - \nu\right)\\ &= \nu(1-S) -\frac{\nu}{\beta S}\left( \nu -\nu S \right) - \left(1-S_2^*\right)\left(\beta S - \beta S_2^*\right)\\ &= \nu(1-S)\left[1-\frac{\nu}{\beta S}\right] - \left(1-S_2^*\right)\left(\beta S - \beta S_2^*\right)\\[1ex] &= \nu(1-S)\left[\frac{\beta S -\beta S_2^*}{\beta S}\right] - \left(1-S_2^*\right)\left(\beta S - \beta S_2^*\right)\\[1ex] &= \nu\beta\left(S-S_2^*\right)\left[\frac{1-S}{\beta S} -\frac{1-S_2^*}{\beta S_2^*} \right]\\[1ex] &= -\nu\beta\left(S_2^*-S\right)\left[\frac{\beta S_2^*\left(1-S\right) - \beta S\left(1-S_2^*\right)}{\beta S\left(\beta S_2^*\right)}\right]\\[1ex] &= -\beta\left(S_2^*-S\right)\left[\frac{\beta S_2^* -\beta S}{\beta S}\right]\\[1ex] &= -\beta\left(S_2^*-S\right)^2\left[\frac{1}{S}\right]\\[1ex] & \leq 0 \end{align*}
We notice $\dot V$ is always negative with the exception of the special case where $S$ and $I$ take on the the endemic equilibrium values, thus we can say $\dot V$ is semi positive definite. We see, when $S\rightarrow 0$ or $S \rightarrow \infty$, $\dot V \rightarrow \infty$. Similarly, when $I\rightarrow 0$ or $I \rightarrow \infty$, $\dot V \rightarrow \infty$. We can now conclude that $V$ is a Lyapunov function for our system and according to Lyapunov stability theorems, the endemic equilibrium $e_2$ is globally asymptotically stable in $\Omega$.
This is just in reference to the $\mu=\nu$ case. In this case one can check that $\dot{N}=0$ so we may consider it as a fixed variable. We now compute that $$\dot{I}=\frac{\beta}{N}(N-I)I-(\gamma+\nu)I=-\frac{\beta}{N}I^2+bI,$$
where we notate $b=\beta-\gamma-\nu$. We assume that $b\neq 0$, though this case can be dealt with similarly. This admits two stable points, $I=0$ and $I=bN/\beta$. This equation is solved by $I=0$ and $$I(t)=\frac{b\exp(bt)}{c+N^{-1}\beta\exp(bt)},$$
The behavior around $c\sim 0$ corrosponds to $I=bN/\beta$, and the behavior around $c^{-1}\sim 0$ corrosponds to $I=0$.
For $b<0$, the point $I=0$ is stable, and $I=bN/\beta$ is unstable. When $b>0$ the point $I=0$ is unstable, and $I=bN/\beta$ is stable. In both cases, the Lyponauv exponent away from the unstable point is $|b|$.