= Solution
Put
$$
f(x;\mu,a)=-x(x^2-2\mu)(x^2-\mu+a).
$$
The <equilibrium points of a dynamical system> form the three branch families
$$
x=0,qquad x=\pm\sqrt{2\mu}\quad(\mu\geq0),qquad x=\pm\sqrt{\mu-a}\quad(\mu\geq a).
$$
For a simple equilibrium, <fixed point stability for an autonomous differential equation> is determined by the sign of $f_x$. On the three families,
$$
\begin{array}{c|c}
\text{branch}&f_x\\ \hline
x=0&-2\mu(\mu-a)\\
x^2=2\mu&-4\mu(\mu+a)\\
x^2=\mu-a&2(\mu-a)(\mu+a).
\end{array}
$$
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 bifurcation>[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>.
Back to article page
