Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 4 / 32A (Dynamical Systems)

For $x\in [0,1]$, the nonnegative function $x(1-x^2)$ has derivative $1-3x^2$, so its maximum occurs at $x=1/\sqrt{3}$ and equalsThe map preserves the interval exactly when this maximum is at most one:Thus the upper limit is

A fixed point satisfiesso either $x=0$ or $x^2=1-1/\lambda$. When $0<\lambda <1$, the latter equation has no real solution, and the origin is the only fixed point. Sincethe criterion for fixed point stability for an iteration shows that the origin is stable for $\lambda <1$.

For $\lambda >1$, the new fixed point in $[0,1]$ isThe origin has multiplier $F_x(0,\lambda )=\lambda >1$ and is therefore unstable. At the new fixed point,HenceThe point $x^{\ast }$ is stable in that interval and unstable for $\lambda >2$. At $\lambda =2$ its multiplier is $-1$, the threshold for a period-doubling bifurcation.

Let $x^{\ast }(\lambda )=\sqrt{1-1/\lambda }$ and center the map at this parameter-dependent fixed point:A Taylor polynomial in $X$, with the coefficients expanded at $(x,\lambda )=(1/\sqrt{2},2)$, givesunder the period-doubling scaling $X=O(\sqrt{\mu })$. The coefficients areHere we used $F_x(x^{\ast }(\lambda ),\lambda )=3-2\lambda$, $F_{xx}=-6\lambda x$, and $F_{xxx}=-6\lambda$.

Composing the local map with itself and retaining the terms of order $\mu X$ and $X^3$ givesBesides the fixed solution $X=0$, the two leading solutions therefore satisfyThus the period-two points born at $\lambda =2$ areto leading order. With the coefficients above, the radicand is positive for $\mu >0$.

The exact second-iterate equation factors asThe positive nonfixed period-two points come from the final factor. If $y=x^2$, their squared values areThey are real and distinct for $\lambda >2$.

The multiplier of a periodic orbit of an iteration isFor $\lambda =2+\mu$ with $0<\mu \ll 1$,whose modulus is less than one. The new two-cycle is therefore locally asymptotically stable.