= Solution
The fixed-point conditions are
$$
0=N(1-S)-\beta IS,
\qquad
0=I\{(\beta-1)S-I\}.
$$
On $I=0$ 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>
$$
\boxed{(S_*,I_*)=\left(\frac1\beta,1-\frac1\beta\right).}
$$
The <Jacobian matrix> is
$$
J(S,I)=
\begin{pmatrix}
1-2S-(1+\beta)I&1-(1+\beta)S\\
(\beta-1)I&(\beta-1)S-2I
\end{pmatrix}.
$$
At $(1,0)$ 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 $N(t)\to1$.
For the <phase portrait>, the line $I=0$ is invariant and points toward $(1,0)$. Also
$$
\dot I=I\{(\beta-1)S-I\},
\qquad
\dot N=N(1-N),
$$
so 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>.
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