= Solution
Adding the three equations of the <SIRS model> gives
$$
\frac d{dt}(S+I+R)=0,
$$
so
$$
\boxed{S(t)+I(t)+R(t)=N.}
$$
At a fixed point, $I(bS-c)=0$. The disease-free equilibrium is
$$
\boxed{(S,I,R)=(N,0,0).}
$$
For an endemic equilibrium $I^*>0$, one has $S^*=c/b$ and $aR^*=cI^*$. Conservation then gives
$$
\boxed{S^*=\frac cb,
\qquad I^*=\frac{a(bN-c)}{b(a+c)},
\qquad R^*=\frac{c(bN-c)}{b(a+c)}.}
$$
It exists with positive infected population exactly when
$$
\boxed{bN>c,}
$$
equivalently when the <basic reproduction number> $bN/c$ exceeds one.
Eliminating $R=N-S-I$ yields
$$
\dot S=a(N-S-I)-bSI,
\qquad
\dot I=(bS-c)I.
$$
At the endemic equilibrium its <Jacobian matrix> is
$$
J=\begin{pmatrix}
-a-bI^*&-(a+c)\\
bI^*&0
\end{pmatrix}.
$$
Its trace and determinant are
$$
T=-\frac{a(a+bN)}{a+c}<0,
\qquad D=a(bN-c)>0.
$$
Therefore the endemic equilibrium is locally asymptotically stable. It is a stable node when
$$
\boxed{\frac{a^2(a+bN)^2}{(a+c)^2}\geq4a(bN-c),}
$$
and a stable focus when the inequality is reversed. Near a node, perturbations are sums of two nonoscillatory decaying exponentials. Near a focus, they execute damped oscillations with decay rate $-T/2$ and angular frequency $\sqrt{4D-T^2}/2$.
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