Past exam of the nonlinear systems course of the University of Oxford / 2023 / 2 / d / Solution

Use $y$ vertically and $\mu$ horizontally in the bifurcation diagram; the other coordinate is recovered from $x=\mu y$. The origin is the horizontal branch $y=0$. Draw it solid on $-1<\mu <0$, where it is asymptotically stable, and dashed on the remaining hyperbolic intervals, where it is unstable. Its three nonhyperbolic fixed points in parameter space are marked at $\mu =-1,0,1$; open markers distinguish these boundary cases from the strict linear stability classifications on the intervening intervals.

The nonzero fixed point branches are $y=\pm \sqrt{\mu (\mu -1)}$ for $\mu <0$ and $\mu >1$. Each is dashed because it is a saddle fixed point of a map, as proved in part (c). Near zero their shape is $y\sim \pm \sqrt{-\mu }$: the unstable branches lie on the same parameter side as the attracting origin, displaying the subcritical pitchfork bifurcation. The branches issuing from $\mu =1$ have $y\sim \pm \sqrt{\mu -1}$. At $\mu =-1$ no additional fixed point branch appears; the Floquet multipliers cross the unit circle at $\pm i$.