 Quintic saturation of a symmetry-broken 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.