 Solution

The fixed-point conditions areOn $I=0$ they give $(S,I)=(0,0)$ and the disease-free equilibrium $(1,0)$. An interior fixed point requires $I=(\beta -1)S$, so it exists only for $\beta >1$; substitution in the first equation gives the endemic equilibrium

The Jacobian matrix isAt $(1,0)$ its eigenvalues are $-1$ and $\beta -1$. Hence $(1,0)$ is asymptotically stable for $\beta <1$ and a saddle point for $\beta >1$. At the endemic equilibrium, the equivalent $(N,\theta )$ coordinates give eigenvalues $-1$ and $1-\beta$, so it is a stable node whenever it exists. The origin is unstable because every solution with $N(0)>0$ has $N(t)\to 1$.

For the phase portrait, the line $I=0$ is invariant and points toward $(1,0)$. Alsoso trajectories move toward the line $S+I=1$. If $\beta <1$, then $\dot{I}<0$ throughout the interior and every nonzero solution approaches $(1,0)$. If $\beta >1$, the additional infective nullcline is the ray $I=(\beta -1)S$; $I$ increases below this ray and decreases above it, and every trajectory with $I(0)>0$ approaches $(1/\beta ,1-1/\beta )$. At $\beta =1$ the disease-free and endemic branches exchange stability in a transcritical bifurcation.