Past exam of the mathematics course of the University of Cambridge / 2010 / ii / Paper 4 / 14D / Solution

A fixed point is locally attracting when its multiplier satisfies $|F^{\prime }(x_{\ast })|<1$ and repelling when $|F^{\prime }(x_{\ast })|>1$. A multiplier crossing $+1$ can create or destroy fixed points in a fold bifurcation, or exchange their stability when branches cross, as in a transcritical bifurcation. A multiplier crossing $-1$ gives a flip, or period-doubling bifurcation, in which a two-cycle can emerge. These are the two multiplier mechanisms; additional symmetry can distinguish special cases at $+1$.

For the logistic map, the fixed points areThe second lies in $[0,1]$ only for $\mu \ge 1$. Their multipliers are $\mu$ and $2-\mu$. Thus zero attracts for $0<\mu <1$; the branches meet at $\mu =1$ and exchange stability, with the nonzero branch attracting for $1<\mu <3$. At $\mu =1$, positive orbits decrease to zero since $F(x)=x-x^2$. At $\mu =3$, put $x=2/3+z$. Then $z\mapsto -z-3z^2$ and its second iterate isso the fixed point is still weakly attracting at the bifurcation, although linear attraction has disappeared.

The supplied factorization gives the nonfixed two-cycle points for $\mu >3$:They are interchanged by $F$. Using their sum and product in $F^{\prime }(x_{+})F^{\prime }(x_{-})$ givesThe two-cycle is attracting exactly forIt is born from the fixed point at multiplier $+1$ for the second iterate, and at $\mu =1+\sqrt{6}$ its multiplier reaches $-1$, giving the next period doubling. The boundary cycles are nonhyperbolic; the strict multiplier test is not a proof of attraction there.

Beyond this threshold the logistic map undergoes further period doublings, with attracting cycles of periods $4,8,16,\dots$ and then chaotic parameter ranges, interspersed with periodic windows. It is not correct to say that every parameter above this threshold is chaotic. At $\mu =4$, the substitution $x={\sin }^2(\pi \theta )$ gives $F(x)={\sin }^2(2\pi \theta )$, illustrating its chaotic doubling-map behavior.