Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 4 / 32A (Dynamical Systems)

Table of contents
- a 32A
  - Solution a
- b 32A
  - Solution b
- c 32A
  - i c
    - Solution i
  - ii c
    - Solution ii

a

Solution

Order a period-three orbit as

a < b < c

. Its images produce two subintervals

I_{0}, I_{1}

with covering relations

I_{0} \to I_{0}, I_{1}

and

I_{1} \to I_{0}

. Any closed walk in this directed graph gives, by the interval-covering lemma and the intermediate value theorem, a fixed point of the corresponding iterate. Primitive closed walks of every length exist, giving an orbit of every positive period; this is the period-three case of Sharkovsky theorem.

b

Solution

The covering graph has at least one primitive cycle of lengths two and four and two inequivalent primitive cycles of length five. Thus the guaranteed minimum numbers of distinct orbits are respectively

1, 1, 2

.

c

i

Solution

For the stated map, the graph cycles give the period-two orbit

1/3 \mapsto 5/6 \mapsto 1/3,

and the period-four orbit

2/9 \mapsto 13/18 \mapsto 5/9 \mapsto 8/9 \mapsto 2/9.

ii

Solution

The two period-five orbits are

1/9 \to 11/18 \to 7/9 \to 4/9 \to 17/18 \to 1/9,

2/15 \to 19/30 \to 11/15 \to 8/15 \to 14/15 \to 2/15.

Direct use of the two affine branches verifies every arrow.

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