Past exam of the topics in applied mathematics course of the University of Oxford / 2010 / B / 5

For $f(x)=rx(1-x)$, the fixed points are $x=0$ and $x_{\ast }=1-1/r$ when the latter is defined. On the physical interval the nonzero branch exists for $r>1$, meeting zero at $r=1$. Their multipliers are $r$ and $2-r$, respectively. Thus zero is linearly attracting for $0\le r<1$ and repelling for $r>1$; the positive point is linearly attracting for $1<r<3$ and repelling for $r>3$. At $r=1$ and $r=3$ the relevant multiplier has magnitude one, so linearization is inconclusive. At $r=1$, $x_{n+1}=x_n-x_n^2$ attracts small positive points to zero nonlinearly.

Factoring $f^2(x)-x$ and removing its fixed-point factors gives $r^2{x}^2-r(r+1)x+(r+1)=0$. Its distinct roots areThey map to each other, giving a genuine two-cycle only beyond $r_0=3$. The cycle multiplier isSolving $|M|<1$ in $r>3$ givesAt $r=3$, the two roots coalesce with $x_{\ast }=2/3$, where the fixed-point multiplier is $-1$: a supercritical period-doubling bifurcation creates the attracting two-cycle. At the upper endpoint its multiplier reaches $-1$, marking the next loss of linear stability. The two-cycle is superstable at $r=1+\sqrt{5}$, where $M=0$. At exactly $r=3$ the fixed point remains weakly attracting through higher-order terms: for $\eta =x-2/3$, the second iterate is $\eta \mapsto \eta -18{\eta }^3-27{\eta }^4$, showing alternating perturbations decay slowly.

For a nearby orbit, linearized separation satisfies ${\delta }_{i+1}=f^{\prime }(x_i){\delta }_i$. Repeated use of the chain rule gives ${\delta }_n={\delta }_0{\prod }_{i=0}^{n-1}{f}^{\prime }(x_i)$ to first order. Taking the logarithm of its magnitude and dividing by $n$ gives the exponential growth rateThe infinitesimal-perturbation limit is taken before interpreting long-time growth; finite separations eventually leave the linear regime. The limit is assumed to exist, with a limsup used when needed.

On a $p$-cycle, blocks of $p$ successive derivatives repeat. If $M=(f^p{)}^{\prime }(x_0)={\prod }_{i=0}^{p-1}{f}^{\prime }(x_i)$, the remainder of fewer than $p$ terms has vanishing contribution to the long-time average andLinear attraction means $|M|<1$, giving $\lambda <0$. A superstable cycle has $M=0$, meaning at least one derivative in the cycle vanishes. Its infinitesimal return multiplier is zero and the corresponding logarithmic exponent is $-\infty$. This concerns the derivative cocycle; actual finite perturbations can converge by higher-order rather than literally vanishing in one step.

In the two-symbol shift representation, $\boxed{\dots 000.000\dots }$ is fixed under a shift and codes the fixed point labelled $0$. The constant all-ones sequence supplies the analogous example. The dot only marks the present position; shifting it advances one iterate. The assertion uses the faithful orbit coding assumed in the question.

Use the primitive repeating word $\boxed{\dots 001001.001001\dots }$. One shift changes the sequence and three shifts return it, so the least period is three. A word such as $000$ repeated would instead have least period one; primitivity is what distinguishes a true three-cycle from a periodic notation for a fixed point.

The sequence $\boxed{\dots 000010000.00000\dots }$ has exactly one $1$ and otherwise zero entries. Under forward shifts that exceptional symbol moves out of every fixed observation window; the same happens under backward shifts in the other direction. Thus both temporal limits are the all-zero fixed sequence, while the orbit itself is nonconstant: it is homoclinic to that fixed point. The bi-infinite coding specifies a chosen full orbit even when the original interval map is not invertible.

Choose $\boxed{\dots 00000.11111\dots }$, with zeros on the entire left tail and ones on the entire right tail. Forward shifts converge in the symbolic metric to the all-one fixed point, while backward shifts converge to the all-zero fixed point. Its two limiting invariant states differ, so it is a heteroclinic orbit. Reversing the two tails reverses these asymptotic roles.

One explicit future itinerary is obtained by concatenating all binary words, ordered first by length and then lexicographically:Extend it arbitrarily to negative indices, for example by zeros. Every finite binary block occurs in its forward itinerary, and indeed occurs arbitrarily far out as a subword of longer words. Its shift orbit is therefore dense in the full two-symbol shift space. It is not eventually periodic: arbitrarily long runs of zeros and ones occur, together with all other finite words.

The full shift has sensitive dependence because sequences agreeing on an arbitrarily large initial window can differ just beyond it, and shifting brings that disagreement into view. Thus this itinerary supplies a dense aperiodic orbit in the chaotic symbolic system assumed by the question. A merely nonrepeating sequence would not be enough: restricted aperiodic sequences can have much simpler, nonchaotic orbit closures. Nor does symbolic description alone assign a numerical Lyapunov exponent without information about the underlying map and its coding.