Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 4 / 32E (Dynamical Systems)

An interval map $F:I\to I$ has a horseshoe when there are an open interval $J\subseteq I$ and two disjoint open intervals $K_0,K_1\subseteq J$ such that

The map is chaotic in Glendinning's sense when some positive iterate $F^n$ has a horseshoe.

The graph of $F$ is the symmetric tent with verticesIf $0<\mu <1$, then $F\ge 0$ andso $F^2$ is an upward-pointing V with its minimum at $(0,1-\mu )$ and equal endpoint values $1-\mu +{\mu }^2$.

If $1<\mu <2$, the zeros of $F$ are $\pm 1/\mu$ andThus its five vertices are

For $0<\mu <1$, one iterate sends the whole interval into $[1-\mu ,1]$, whereThis is a contraction mapping with unique fixed pointEvery orbit converges to $x_{\ast }$, alternating around it because the multiplier is $-\mu$.

For $\mu =1$, the first iterate lies in $[0,1]$ and $F(x)=1-x$ there. Hence $F^2$ is the identity on $[0,1]$: $1/2$ is fixed, while every other point in that interval belongs to the period-two orbit $\{x,1-x\}$.

The map $F$ itself has a horseshoe only for $\mu =2$. Indeed, its Lipschitz constant is $\mu$. If two intervals $J_0,J_1$ form a horseshoe and their convex hull has length $L$, thenwhich forces $\mu \ge 2$. At $\mu =2$, take $J=(-1,1)$, $K_0=(-1,0)$ and $K_1=(0,1)$; both $K_i$ map onto $J$.

The Lipschitz constant of $F^2$ is ${\mu }^2$, so the same length argument makes $\mu \ge \sqrt{2}$ necessary for an $F^2$ horseshoe. It is sufficient. PutFor $\mu \ge \sqrt{2}$, one has $a<1/\mu \le b\le 1$. Withthe two relevant monotone branches satisfyThus $K_0,K_1\subset J$ are disjoint and $F^2(K_0)=F^2(K_1)=J$. Therefore

For $0<\mu <1$, all orbits converge to one fixed point, and for $\mu =1$ every orbit is eventually fixed or period two, so no iterate has a horseshoe. For the symmetric tent map with slope magnitude $\mu >1$, the topological entropy is $\log \mu >0$. The interval-map positive-entropy horseshoe theorem says that a continuous interval map has positive topological entropy exactly when some iterate has a horseshoe. HenceFor $\mu \ge \sqrt{2}$ this already follows from part (iii); for $1<\mu <\sqrt{2}$, a higher iterate supplies the horseshoe.