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PutThe equilibrium points of a dynamical system form the three branch familiesFor a simple equilibrium, fixed point stability for an autonomous differential equation is determined by the sign of $f_x$. On the three families,Thus solid branches in the figure are stable and dashed branches are unstable.

If $a<0$, the pair $x=\pm \sqrt{\mu -a}$ emerges stably from $x=0$ at $\mu =a$ in a pitchfork bifurcation. At $\mu =0$, the pair $x=\pm \sqrt{2\mu }$ meets $x=0$ with unstable nonzero branches; after reversing the parameter orientation this is a subcritical pitchfork bifurcation. At $\mu =-a$, the two positive branches cross and exchange stability, as do the two negative branches, giving two simultaneous transcritical bifurcations at $x=\pm \sqrt{-2a}$.

If $a=0$, every branch meets at $(\mu ,x)=(0,0)$. For $\mu <0$ only the stable trivial branch exists. For $\mu >0$, $x=0$ is stable, the inner branches $x=\pm \sqrt{\mu }$ are unstable, and the outer branches $x=\pm \sqrt{2\mu }$ are stable. This is a degenerate fifth-order bifurcation rather than a generic pitchfork.

If $a>0$, a supercritical pitchfork bifurcation occurs at $\mu =0$: the trivial branch loses stability and the stable branches $x=\pm \sqrt{2\mu }$ emerge. At $\mu =a$, the unstable branches $x=\pm \sqrt{\mu -a}$ meet the trivial branch, which regains stability; this is the parameter-reversed form of a subcritical pitchfork bifurcation.

These cases comprise the bifurcations of the 2020 Cambridge quintic system.

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.