Past exam of the mathematics course of the University of Cambridge / 2011 / ii / Paper 4 / 7C (Dynamical Systems)

For a nearly Hamiltonian system, the unperturbed Hamiltonian is conserved and its level curves typically contain a family of periodic orbits. A weak perturbation changes the energy slowly. The energy balance method integrates $\dot{H}$ over one unperturbed period to obtain the leading energy change $\Delta H(H)$. A zero of this change selects a possible persistent periodic orbit. A simple zero persists for sufficiently small perturbation by a Poincaré return map and the implicit function theorem. A negative slope of the averaged energy drift gives attraction; a positive slope gives repulsion. A multiple zero requires higher-order analysis, and a zero-free leading drift rules out periodic motion in the corresponding perturbative energy range.

Set $H=({\dot{x}}^2+x^2)/2$. ThenOn an unperturbed circle write $x=r\cos t$, $\dot{x}=-r\sin t$. The supplied integrals giveEquivalently, the leading averaged amplitude equation isWith $s=r^2$, its nonzero stationary amplitudes solve $\beta s^2-2\alpha s+8=0$. If ${\alpha }^2>8\beta$, both roots are positive and simple:The averaged drift is positive below $r_{-}$, negative between $r_{-}$ and $r_{+}$, and positive above $r_{+}$. Thus the inner orbit is attracting and the outer orbit repelling, with leading displacement amplitudes $r_{-}$ and $r_{+}$ and period $2\pi +O(ϵ)$. The simple zeros of the Poincaré return map persist for sufficiently small $ϵ$.

If ${\alpha }^2<8\beta$, the leading quadratic is strictly positive for all $s\ge 0$; the energy increases on every nonzero unperturbed orbit, so the energy balance method yields no periodic orbits in any fixed compact range of nonzero amplitudes for sufficiently small $ϵ$. This perturbative calculation does not itself establish a uniform global exclusion at amplitudes growing as $ϵ$ tends to zero. At equality, the leading drift has a double zero at $r^2=\alpha /\beta$. It is positive on both sides, with a semistable candidate at that amplitude. First-order averaging alone cannot decide whether the exact system has zero, one, or two nearby periodic orbits at equality: higher-order drift determines the unfolding of this degenerate zero. The original PDF has ${\alpha }^2=8\beta$; the extra exponent on $\beta$ in the converted TeX is an OCR error.