Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2011/ii/paper-3/14c/iv/solution

The bifurcation diagram has boundary branches at and at . The former is stable for and a saddle for ; the latter is a saddle for and stable for . The origin is an unstable equilibrium for every ; its branch coincides in this projection with the branch, so it is shown separately in the caption.
The stable interior branch rises from at to at . The unstable interior branch falls from to over the same interval. These meet at the saddle-node bifurcation; at they meet the transcritical and pitchfork points respectively. There is bistability between the lower interior node and for .
Figure 1.
Equilibrium branches and stability for Question 14; the always unstable origin also projects to y equals zero
. Details.

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