Pitchfork bifurcation normal form

The supercritical pitchfork normal form is $\dot{x}=\mu x-x^3$: the trivial branch is stable for $\mu <0$ and two stable nonzero branches emerge for $\mu >0$.

For $k>1$, the rational equilibrium point curve in the display has expansion $r=1+q+[1-q(k-1)]u+qk(k-1)u^2+\cdots$. Its cubic pitchfork bifurcation coefficient vanishes at $q=1/(k-1)$, leaving $r-(1+q)=k{u}^2+\cdots$. For larger $q$, the nonzero branch folds at $u=[\sqrt{q(k-1)}-1]/k$ and $r=[k-1+q+2\sqrt{q(k-1)}]/k$. If the other eigenvalues remain stable, the two folds bound a bistable wedge between stable zero and stable nonzero states, separated by an unstable branch. If another eigenvalue is critical, a higher-dimensional centre manifold is required instead.

The polar form of the cubic confinement model has an origin pitchfork on ${\mu }^2+{\sigma }^2=1$ when $\mu \ne 0$. Its small squared amplitude satisfies $S\sim 2\mu \delta \mu /(\mu +\sigma )$ along fixed $\sigma$. Thus the leading cubic coefficient vanishes when $\mu +\sigma =0$, at $(\sigma ,\mu )=(\pm 1/\sqrt{2},\mp 1/\sqrt{2})$. At either point the exact radius equation gives $\delta \mu =-S^2/(4{\mu }_0)+o(S^2)$, so amplitude scales as $|\delta \mu {|}^{1/4}$. The lower point has a stabilizing quintic coefficient $1/(4{\mu }_0)<0$; folds of nonzero equilibrium pairs meet the pitchfork curve there. The upper point has the opposite quintic sign and an already unstable transverse eigenvalue.

In the displayed pitchfork bifurcation normal form, zero is stable for $\mu <0$ and unstable for $\mu >0$. At $\mu =0$, a symmetric stable pair $x=\pm \sqrt{\mu }$ appears on the side where zero is unstable. Reversing the parameter orientation changes the side, while retaining the supercritical relationship between the new stable branches and the destabilized central branch.

The normal form $\dot{x}=\mu x+x^3$ has a stable trivial equilibrium for $\mu <0$ and two unstable nonzero equilibria for $\mu <0$. They collide with the trivial branch at $\mu =0$, after which that branch is unstable.

Reflection symmetry $x\mapsto -x$ forces the reduced vector field to be odd in $x$, naturally producing paired nonzero equilibrium branches.

Adding a small symmetry-breaking term to a pitchfork bifurcation disconnects its two symmetric daughter branches and generically leaves one or more saddle-node bifurcations. This unfolding is called an imperfect pitchfork.

For $C>0$, nonzero equilibrium points satisfy $\eta =g(a)=a-C{a}^2+a^4$ and are stable exactly when $a{g}^{\prime }(a)>0$. The origin is stable for $\eta <0$. The quadratic term removes the odd symmetry and changes the local branch crossing into a transcritical bifurcation, with $a=\eta +O({\eta }^2)$. The negative quintic term bounds trajectories. There is one negative-amplitude fold, supplying a stable finite-amplitude branch and hysteresis with the stable origin. Positive-amplitude folds occur precisely when $C>3/2$: $g^{\prime }(a)=1-2Ca+4a^3$ has two positive roots then, a double positive root at $C=3/2$, and none below it. All branches and stability assignments therefore depend on $C$; a single symmetric pitchfork sketch cannot describe this family.