Interperate Jacobian Determinant - Stability of Equilibriums

Question

In my SIR model, I have the following Jacobian Matrix

\begin{align*} J =\begin{bmatrix} -\alpha I & -\alpha S & \zeta & 0 \\ \alpha I & \alpha S - \beta - \rho & 0 & 0 \\ 0 & \beta & -\zeta & 0 \\ 0 & \rho & 0 & 0\end{bmatrix}\end{align*}

Now I have two equilibriums given by the following:

Disease-Free Equilibrium

Consider the case when $I=0$: $$(\bar{S},\bar{I},\bar{R},\bar{D}) = (N,0,0,0)$$ The Jacobian at the Disease-Free Equilibrium is thus \begin{align*}J_1(N,0,0,0) = \begin{bmatrix}0 & -\alpha N & \zeta & 0 \\ 0 & \alpha N - \beta - \rho & 0 &0 \\ 0 & \beta & -\zeta & 0 \\ 0 & \rho & 0 & 0 \end{bmatrix} \end{align*}

We have $\det(J_1 - \lambda \textbf{I} ) = \lambda ^2 (\zeta +\lambda ) (-\text{$\alpha $N}+\beta +\lambda +\rho )$

EDIT:

Here are the ODE's

$$\frac{dS}{dt} =-\alpha SI + \zeta R$$ $$\frac{dI}{dt} = \alpha SI - \beta I - \rho I$$ $$\frac{dR}{dt} = \beta I - \zeta R$$ $$\frac{dD}{dt} = \rho I$$

where

1. $\alpha = $ rate at which susceptibles become infected
2. $\beta = $ rate at which infected becomes recovered
3. $\zeta = $ rate at which members of recovery become susceptible again
4. $\rho = $ rate at which infected pass away

Answer 1

You've factored the characteristic polynomial of $J_1$, so you know the eigenvalues: $0, 0, -\zeta, N \alpha - \beta - \rho$. $-\zeta < 0$, but $N \alpha - \beta - \rho$ could go either way. The criterion for the disease to "take off", i.e. for a small number of infectives to start an epidemic, is $N \alpha - \beta - \rho > 0$ (the zero eigenvalues are just associated with eigenvectors $(1,0,0,0)$ and $(0,0,0,1)$ which just increase the population of susceptibles or deceased).

Your Doomsday scenario is not an equilibrium. It is easy to see that there are no equilibria where $I > 0$, because $D$ will be increasing as long as $I > 0$.