Past exam of the mathematics course of the University of Cambridge / 2011 / ii / Paper 3 / 14C / iv / Solution

The bifurcation diagram has boundary branches $y=0$ at $(1,0)$ and $y=\mu$ at $(0,\mu )$. The former is stable for $0<\mu <1$ and a saddle for $\mu >1$; the latter is a saddle for $0<\mu <1$ and stable for $\mu >1$. The origin is an unstable equilibrium for every $\mu >0$; its $y=0$ branch coincides in this projection with the $(1,0)$ branch, so it is shown separately in the caption.

The stable interior branch $y_{-}$ rises from $0$ at $\mu =1$ to $1/2$ at $\mu =5/4$. The unstable interior branch $y_{+}$ falls from $1$ to $1/2$ over the same interval. These meet at the saddle-node bifurcation; at $\mu =1$ they meet the transcritical and pitchfork points respectively. There is bistability between the lower interior node and $(0,\mu )$ for $1<\mu <5/4$.