Past exam of the nonlinear systems course of the University of Oxford / 2023 / 3

For the left shift, ${\sigma }^p(s)=s$ precisely when the binary sequence is obtained by repeating its first $p$ symbols. There are exactly $2^p$ such sequences, although some have smaller least period. Hence the set of all periodic sequences is a countable union of finite sets and is at most countable. Passing from periodic points to periodic orbits cannot increase this cardinality.

To prove infinitude, for every integer $p\ge 1$ take the repeated word $(0^{p-1}1{)}^{\infty }$. Its least period is $p$, because its ones are separated by exactly $p$ positions; a smaller positive period would carry a one to a zero. Distinct least periods give distinct periodic orbits. Thus there are countably infinitely many periodic orbits.

The space $\Sigma$ of all infinite binary sequences is uncountable by the Cantor diagonal argument: given any list $s^{(1)},s^{(2)},\dots$, the sequence $t_i=1-s_i^{(i)}$ differs from its $i$th entry at position $i$, so it is absent from the list. Part (i) shows that only countably many sequences are periodic. Their complement is therefore uncountable.

Each forward orbit $\{{\sigma }^n(s):n\ge 0\}$ is countable. If there were only countably many nonperiodic forward orbits, their union together with the periodic points would be countable, contradicting the preceding argument. Hence there are uncountably many nonperiodic orbits, rather than merely uncountably many nonperiodic starting points.

One can also exclude eventual periodicity. An eventually periodic sequence is specified by a finite initial word and a finite repeating word, so only countably many exist. Removing all of them still leaves an uncountable set. This proves the assertion even if “nonperiodic” is understood to exclude orbits that eventually enter a periodic orbit.

Construct one infinite binary sequence $u$ by concatenating every binary word of length one, then every word of length two, then every word of length three, and so on, in lexicographic order within each length. Every finite binary word occurs somewhere in $u$.

Given any target sequence $s$ and any $\eta >0$, choose $N$ such that $2^{-N}<\eta$. The word $s_1\cdots s_N$ occurs in $u$, beginning after some $k$ symbols. Thus ${\sigma }^{k}u$ agrees with $s$ in its first $N$ positions, and the prescribed metric givesEvery open open ball about every point therefore meets the forward orbit of $u$. This forward orbit is dense in $\Sigma$. The argument identifies finite-prefix cylinder sets as arbitrarily small neighborhoods in this metric; no binary-expansion convention for real numbers is needed.

For the real logistic map, $f(x)=x$ factors asThere is always the fixed point $x_0=0$. If $\mu \ne 0$ there is also $x_{\ast }=1-1/\mu$, distinct from zero unless $\mu =1$. At $\mu =0$ the map is constant and zero is its only fixed point. Negative $\mu$ must be retained because the paper specifies $x\in \mathbb{R}$.

Since $f^{\prime }(x)=\mu (1-2x)$, the fixed-point multipliers areThe fixed point stability for an iteration criterion gives $x_0$ attracting for $-1<\mu <1$, and $x_{\ast }$ attracting for $1<\mu <3$. The respective fixed points are repelling when $|\mu |>1$ and $|2-\mu |>1$. At the unit-modulus boundaries the linear criterion needs a nonlinear supplement.

At $\mu =1$, the two fixed point branches meet and exchange stability: this is a transcritical bifurcation. Explicitly, with $\delta =\mu -1$,which has the required nonzero parameter and quadratic terms. At the critical value itself, $f(x)=x-x^2$ attracts small positive points and pushes negative points away, so the merged fixed point is not stable from both sides.

At $\mu =-1$ the origin’s fixed-point multiplier crosses $-1$, while at $\mu =3$ the nonzero fixed-point multiplier crosses $-1$. Both are period-doubling bifurcations. To check existence and stability of the new periodic orbit, factor the second iteration of a map:The two non-fixed roots areThey form a real period-two orbit for $\mu <-1$ or $\mu >3$. Its periodic-orbit multiplier isIt is less than one in absolute value just below $-1$ and just above $3$, respectively. Thus these are supercritical bifurcations in the direction where the parent fixed point loses stability. At the critical values the fixed points themselves are weakly attracting: at $\mu =-1$, $f^2(x)=x-2x^3+x^4$; at $\mu =3$, setting $u=x-2/3$ gives $u\mapsto u-18u^3-27u^4$ after two steps. Both shrink sufficiently small nonzero absolute values. The bifurcation values for finite fixed points are thereforeThe disappearance of the formula for $x_{\ast }$ at $\mu =0$ corresponds to that branch going to infinity, not a local bifurcation at a finite fixed point.

Solving $\mu x(1-x)=1$ by the quadratic formula givesFor $\mu >4$, $0<x_{-}<1/2<x_{+}<1$. On $L=[0,x_{-}]$ the derivative $f^{\prime }(x)=\mu (1-2x)$ is strictly positive, while on $R=[x_{+},1]$ it is strictly negative. Thus the restrictions are strictly monotonic functions. Their endpoint values areBy continuity and strict monotonicity, each restriction maps its interval bijectively onto $[0,1]$, with opposite orientations. The corresponding inverse functions are

Write $G=g_{s_1}\circ \cdots \circ g_{s_{n-1}}$. Because $g_{s_n}([0,1])\subseteq [0,1]$, taking the image under the same function composition givesThis uses preservation of set inclusion by every function, so decreasing inverse branches cause no difficulty. Since $f\circ g_L=f\circ g_R={\mathrm{id}}_{[0,1]}$, cancellation of the first inverse branch gives the equality of imagesThese compact intervals are itinerary cylinders: their symbols record successive positions under iteration of a map. In particular,Conversely these three memberships allow successive substitution of the appropriate inverse branches, and therefore characterize $I_{LLR}$ exactly.

The set $I^n$ consists precisely of points whose first $n$ positions $x,f(x),\dots ,f^{n-1}(x)$ all lie in $L\cup R$. HenceDeleting the first position preserves this condition, proving $f(\Lambda )\subseteq \Lambda$. For the reverse inclusion, if $y\in \Lambda$, take $x=g_L(y)$. Then $x\in L$ and $f^k(x)=f^{k-1}(y)\in L\cup R$ for every $k\ge 1$. Thus $x\in \Lambda$ and $f(x)=y$. In fact both inverse branches give preimages in $\Lambda$. Thereforewhich proves full invariance, not merely forward invariance.

For any infinite symbolic sequence $s$, the itinerary cylinders $I_{s_1\cdots s_n}$ are nonempty compact nested intervals. The nested interval theorem gives at least one point in their intersection. Every $x\in \Lambda$ has a unique itinerary because $L\cap R=\varnothing$. To justify that the itinerary also determines a unique point, it suffices to show the interval lengths shrink. The fact that logistic inverse branches contract in an angular coordinate works for every $\mu >4$; a direct claim that $|g_L^{\prime }|,|g_R^{\prime }|<1$ in the original coordinate would fail for some $\mu$ just above four.

Use the homeomorphism $h(u)={\sin }^2(\pi u/2)$ of $[0,1]$. The transformed inverse branches areTheir derivatives satisfyThe inequality follows by squaring and using $4/\mu <1$. Thus every length-$n$ transformed itinerary cylinder has length at most ${\mu }^{-n/2}$. Since $|h^{\prime }|\le \pi /2$, its original interval has length at most $(\pi /2){\mu }^{-n/2}$ and shrinks to a point. This gives the claimed one-to-one identification with the one-sided full shift.

Finally part (ii) givesThus $f$ acts by the left shift. The identification is a topological conjugacy: matching long prefixes forces closeness by the shrinking estimate, while distinct finite itinerary cylinders are disjoint compact sets and so their inverse coding is continuous. In particular $f$ need not be one-to-one on $\Lambda$, just as a one-sided left shift has two possible first symbols before any given tail.