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A disease-free equilibrium is an equilibrium of an epidemic model at which every infected compartment is zero.

An endemic equilibrium is an equilibrium at which infection persists, so at least one infected compartment is positive.

If a population $N$ has per-capita birth rate $b{e}^{-aN}$ and per-capita death rate $d$, thenWhen $b>d$, its positive equilibrium isand the derivative of the scalar vector field there is $-ad{N}^{\ast }<0$, so the equilibrium is locally asymptotically stable.

At a positive equilibrium, division by $N$ givesSince $b>d$, this has the positive solution

Write the vector field as $F(N)=N(b{e}^{-aN}-d)$. Its derivative isAt $N=N^{\ast }$, the first two terms cancel and $b{e}^{-a{N}^{\ast }}=d$, soThe one-dimensional linear stability analysis therefore shows that $N^{\ast }$ is locally asymptotically stable.

With $s=0$, the equilibria satisfyBesides $g=0$, the two positive equilibria arewhere $g_{-}{g}_{+}=1$. Sincethe nonzero equilibrium relation $k=(1+g^2)/g$ givesThe fixed point stability for an autonomous differential equation therefore shows that $g=0$ and $g_{+}$ are stable, while $g_{-}$ is unstable. The graph of $f(g,0)$ starts at zero with negative slope, crosses upward at $g_{-}$, crosses downward at $g_{+}$, and tends to $-\infty$ as $g\to \infty$.

For a constant input $s$, write the equilibrium equation asThe low stable equilibrium and the intervening unstable equilibrium coalesce in a saddle-node bifurcation. More precisely, let $g_c$ be the first positive solution ofand defineEquivalently, $f(g_c,s_c)=0$ and ${\partial }_{g}f(g_c,s_c)=0$, with $g_c$ on the low-concentration branch. If $s_1>s_c(k)$, that branch no longer exists. Holding the input long enough carries the trajectory into the basin of attraction of the high state. When the input returns to zero, the concentration converges toThis is the saturating autocatalytic switch.

For $k\gg 1$, the threshold occurs at $g=O(k^{-1})$, soThe approximate double-root conditions areThus $g_c\simeq 1/(2k)$ andHence the constant in $s_c(k)\simeq C{k}^{-1}$ is $\boxed{C=1/4}$, as recorded by the strong-autocatalysis switching threshold.