Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 1 / 32A (Dynamical Systems)

In Glendinning's definition, a continuous interval map $F$ is chaotic when $F^m$ has a horseshoe for some $m\ge 1$. Thus there are two disjoint subintervals whose images under $F^m$ each cover their union, producing full two-symbol itinerary dynamics.

A map is chaotic in Devaney's sense when it is topologically transitive, its periodic points are dense, and it has sensitive dependence on initial conditions. Explicitly, transitivity means that for every nonempty open $U,V$ some $n$ has $F^n(U)\cap V\ne \varnothing$, while sensitivity means that one $\delta >0$ works so that arbitrarily close initial points eventually separate by more than $\delta$.

Write the period-three points in increasing order and take the two intervals between consecutive points. For either possible cyclic ordering, the interval-covering transition graph is, after interchanging its vertices,The loop at the first vertex gives a fixed point. For every $n\ge 2$, the closed itineraryhas least period $n$. The interval-covering periodic-orbit theorem supplies a point with that itinerary and therefore a cycle of least period $n$. Thus $F$ has periodic orbits of every positive period. This is the period-three case of the Sharkovsky theorem.

For the Fibonacci transition matrix above,where $L_n$ is the $n$th Lucas number. It counts closed symbolic itineraries of length $n$. Since $7$ is prime, the itineraries fixed by the seventh shift have least period either $1$ or $7$. Hence the number of cyclic classes of primitive length-seven itineraries isDistinct primitive cyclic itineraries give distinct cycles, so $F$ has at least four distinct $7$-cycles. This is the fibonacci transition-graph cycle count.

An itinerary fixed by the eighth shift has least period dividing $8$. Those with period dividing $4$ are exactly the itineraries fixed by the fourth shift. Thus the primitive length-eight itineraries formpointed words. Each cycle has eight cyclic shifts, so they givedistinct $8$-cycles. Therefore at least five such cycles are forced.