Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 4 / 31A / b / Solution

For $F(x)=x(\mu -x^2)$, the fixed points are zero for all $\mu$ and $\pm \sqrt{\mu -1}$ for $\mu >1$. Their multipliers are respectively $\mu$ and $3-2\mu$, giving attraction for $-1<\mu <1$ and $1<\mu <2$ respectively.

Remove these fixed points from the roots of $F^2(x)=x$. The factor $\mu +1-x^2$ gives the genuine two-cycleIts multiplier is $(2\mu +3{)}^2>1$, so it is always unstable. It collapses to zero at $\mu =-1$, the subcritical period-doubling bifurcation of zero.

The remaining equation is $x^4-\mu x^2+1=0$, whose positive roots in $x^2$ exist only for $\mu \ge 2$. At $\mu =2$ they give the fixed points $\pm 1$, not genuine two-cycles. For $\mu >2$, put $r=\sqrt{(\mu +\sqrt{{\mu }^2-4})/2}>1$. Since $\mu =r^2+r^{-2}$, $F(r)=r^{-1}$ and $F(r^{-1})=r$. There are exactly two further cycles,Their common multiplier is $(\mu -3r^2)(\mu -3r^{-2})=9-2{\mu }^2$. They are attracting for $2<\mu <\sqrt{5}$ and repelling for $\mu >\sqrt{5}$, with marginal multiplier $-1$ at $\sqrt{5}$. Thus there are no cycles for $\mu \le -1$, one for $-1<\mu \le 2$, and three for $\mu >2$.

The nonzero fixed points emerge in a supercritical pitchfork bifurcation at one and each flips to an attracting two-cycle at two. Beyond $\sqrt{5}$ further period doubling to four-cycles is expected, followed by additional bifurcations and potentially chaotic dynamics. This local expectation does not assert chaos for every larger parameter; periodic windows and escape are possible.