Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 3 / 7A / Solution

The map normal forms at multiplier $+1$ are given in the numbered subparts below. For the multiplier $-1$ problem, the second iterate instead exposes the emerging period-two cycle.

For the multiplier $-1$ problem, write $x=X+\alpha /2$. To the required weighted order, with $X=O({\mu }^{1/2})$, the shifted map is $X_{n+1}=-X_n+b{X}_n+\gamma X_n^2+\delta X_n^3$, where $b=\beta +\alpha \gamma$. Composing it with itself cancels the quadratic term. The cubic contribution is $-2\delta X^3-2{\gamma }^2{X}^3$, while the linear correction is $-2bX$. Consequently the quadratic-cubic flip criticality calculation givesThus the second iterate has normal form $X_{n+2}=X_n+\hat{\mu }{X}_n-A{X}_n^3$, up to terms of weighted order at least four. This is the asymptotic interpretation of the stated remainder under $X=O({\mu }^{1/2})$; a uniformly valid local expansion about the exact fixed point also retains mixed parameter remainders.

Assuming $\hat{\mu }$ crosses zero transversely and $A\ne 0$, the small nonzero fixed points of the second iterate satisfy $X^2\sim \hat{\mu }/A$. They form a period-two orbit. For $A>0$ they exist on the side $\hat{\mu }>0$, where the original fixed point has become unstable, and the second-iterate multiplier is $1-2\hat{\mu }+o(\hat{\mu })$, so the cycle is stable. The period-doubling bifurcation is therefore supercritical precisely when $\delta +{\gamma }^2>0$. Equality is a degenerate case requiring higher-order terms.