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Continue in the bistable regime $k>2$. At $s=0$ the reaction function factors as $f(g,0)=g(kg-1-g^2)/(1+g^2)$. Besides $g=0$, its positive equilibria areAt zero, $f_g=-1$, so it is stable. At a positive equilibrium, use $k=(1+g^2)/g$ to obtain $f_g=(1-g^2)/(1+g^2)$. Therefore $g_1$ is unstable and $g_2$ stable. The sign of $f$ is negative on $(0,g_1)$, positive on $(g_1,g_2)$ and negative beyond $g_2$; these signs determine the requested reaction-rate sketch.

The stationary points of $f(g,0)$ obey $(1+g^2{)}^2=2kg$. For $k>2$ its smaller solution $g_c<1$ is the local minimum, and its larger solution lies above one and is the local maximum. Since increasing $s$ translates the whole graph upward, the low and middle equilibria collide whenFor $s>s_c$, both stationary values of $f$ are positive, while $f(g,s)\to -\infty$ as $g\to \infty$; there is exactly one nonnegative equilibrium, on the descending high branch. At equality a double low equilibrium remains, so the uniqueness assertion requires strict exceedance.

For completeness, if the later positive-input request is considered for arbitrary $k>0$, the derivative is $f_g=2kg/(1+g^2{)}^2-1$. Its maximum is $9k/(8\sqrt{3})-1$, attained at $g=1/\sqrt{3}$. For $k\le 8\sqrt{3}/9$, the reaction graph is decreasing and every $s>0$ already has a unique steady state, so take $s_c=0$. For $k>8\sqrt{3}/9$, the smaller positive stationary point gives the same $s_c=g_c(1-g_c^2)/2$; above it the graph crosses zero only on its final descending branch. Persistent robust switching after input is removed uses the previously assumed bistable regime $k>2$.

Starting near zero, raise $s$ past $s_c$ and keep it there long enough for $g$ to pass the unstable threshold and approach the high branch. Decrease $s$ back to zero slowly, or after ensuring $g>g_1$. The trajectory then stays in the basin of the high stable equilibrium and ends at the displayed $g_2$. The persistent switched state is $g_2=(k+\sqrt{k^2-4})/2$. This is the saturating autocatalytic switch; the conclusion presupposes $k>2$ and enough forcing duration, not merely an arbitrarily short pulse of large amplitude.

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