Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 85 / 4 / b / Solution

Assume $\hat{s}>0$, as required for the subcritical case. The zero state exists for every $\mu$. For a nonzero uniform real state, put $B=A_0^2>0$. Its equilibrium point condition is $\mu +3\hat{s}B-10B^2=0$, givingBoth are positive for $-9{\hat{s}}^2/40<\mu <0$. At $\mu =0$, the lower branch reaches zero and the upper has $B=3\hat{s}/10$. For $\mu >0$ only the upper branch is positive. Each positive $B$ represents real states $A_0=\pm \sqrt{B}$, as well as their constant complex wave phase rotations.

The two nonzero branches merge atThe radial linearization is ${\lambda }_a=\mu +9\hat{s}B-50B^2=2B(3\hat{s}-20B)$. It is negative on the upper branch and positive on the lower. The zero state is stable for $\mu <0$ and unstable for $\mu >0$. Thus a subcritical pitchfork bifurcation branch leaves zero towards negative $\mu$, and its fold connects to the stable finite-amplitude branch. Constant wave phase shifts are neutral for nonzero complex states; the stated attraction is in amplitude, with spatial wave phase perturbations damped except for that uniform symmetry mode.