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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.

For $k\gg 1$, the low-concentration saddle node of the saturating autocatalytic switch has