Past exam of the mathematics course of the University of Cambridge / 2007 / ii / Paper 4 / 14E / i / Solution

The fixed-point equation is $x[a{x}^2+bx+\mu -1]=0$. Thuswhen the discriminant is nonnegative. At zero the multiplier is $F^{\prime }(0)=\mu$, so it is linearly stable for $-1<\mu <1$ and unstable for $|\mu |>1$. At a nonzero fixed point, eliminate $\mu$ to obtain the branch and its multiplier:The branch is stable precisely when $-2<x(b+2ax)<0$. These formulas specify stability throughout the diagram, including any extra flip points outside the region requested.

If $b=0$ and $a>0$, the two nonzero branches $x=\pm \sqrt{(1-\mu )/a}$ exist for $\mu <1$ and have multiplier $3-2\mu >1$. Thus $\mu =1$ is a subcritical pitchfork bifurcation. If $b\ne 0$, the branch through zero has $x\sim (1-\mu )/b$ and $M\sim 2-\mu$; its stability exchanges with zero at $\mu =1$, a transcritical bifurcation. The two nonzero roots meet atwhere $M=1$. The graph of $\mu (x)$ has nonzero second derivative $-2a$, proving a saddle-node bifurcation. At zero and $\mu =-1$, the multiplier is $-1$, indicating a period-doubling bifurcation; generically its nondegeneracy coefficient is $a+b^2$. If $a+b^2=0$, necessarily $b\ne 0$ here, composition at $\mu =-1$ instead gives $F^2(x)-x=4b^4{x}^5+O(x^6)$. For $\delta =\mu +1>0$ small, the leading return equation is $-2\delta x+4b^4{x}^5=0$, producing a two-cycle with leading amplitudes $x\sim \pm [\delta /(2b^4){]}^{1/4}$. Its return multiplier is $1+8\delta +o(\delta )>1$. Thus this special case still has a flip bifurcation, but it is degenerate and subcritical rather than the generic cubic normal form.

For $a,b>0$, the fold lies above $\mu =1$. Between it and the transcritical point, the branch $-b/(2a)<x<0$ is stable provided its multiplier remains above $-1$, while the other branch is unstable. For $a,b<0$, the fold lies below $\mu =1$: immediately above it the branch $x<-b/(2a)$ is stable and the branch between $-b/(2a)$ and zero is unstable; the branch crossing zero becomes stable for $\mu >1$. All these statements are subject to the displayed exact multiplier criterion if the diagram is extended. The sketch uses representative parameter values and stops before additional flips, as stipulated.