Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 1 / 32B (Dynamical Systems)

For the unperturbed Hamiltonian system, choose a conserved energy $H(x,y)$. Along the perturbed system,The energy balance method integrates this expression around each unperturbed periodic orbit. A zero of the resulting $O(\epsilon )$ energy change is a necessary leading-order condition for a nearby periodic orbit. A simple zero at which the drift changes from positive inside to negative outside predicts a stable orbit; the opposite change predicts an unstable one.

At $\epsilon =0$,is conserved, and its orbit of amplitude $a>0$ may be writtenThe energy change over one period, to first order in $\epsilon$, isHere the stated trigonometric integrals, together with
${\int }_0^{2\pi }{\cos }^2\theta {\sin }^4\theta d\theta =\pi /8$, were used. The two candidate leading-order amplitudes are therefore

For $0<k<6$, the roots satisfy $\sqrt{k/3}<\sqrt{2}$. The averaged drift is positive below the first root, negative between the roots, and positive above the second. Hence the orbit of amplitude $\sqrt{k/3}$ is stable and the orbit of amplitude $\sqrt{2}$ is unstable.

For $k>6$, their order reverses. The same sign calculation shows that the orbit of amplitude $\sqrt{2}$ is stable and that of amplitude $\sqrt{k/3}$ is unstable. Thus in either case the smaller-amplitude candidate is stable and the larger one unstable.

At $k=6$, the two zeros coalesce andThe first-order drift has a double zero and does not change sign there. The energy balance calculation is therefore degenerate: it neither proves persistence of a periodic orbit nor determines its stability near $k=6$. Higher-order terms are needed.