Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 2 / 33B (Dynamical Systems)

For $H=x^2+y^2$, $\dot{H}=2\mu x^2(1-a{x}^2-y^2)$. It points outward on a sufficiently small circle $r<1/\sqrt{a}$ and inward on a circle $r>1$. The intervening annulus is compact, forward invariant, and contains no equilibrium. Poincare-Bendixson theorem therefore gives a periodic orbit.

The nontrivial Floquet multiplier is the factor by which a transverse perturbation changes after one period. Liouville's formula makes its logarithm the integral of the divergence. Here $\nabla \cdot f=\mu (1-3a{x}^2-y^2)$, yielding the stated formula.

For small $\mu$, average $\dot{H}$ around $x=r\cos 2t$, $y=-r\sin 2t$. This gives $⟨\dot{H}⟩=\mu [r^2-(3a+1)r^4/4]$. The unique positive balance is $r=2/\sqrt{3a+1}$, which lies between $1/\sqrt{a}$ and $1$ for $a>1$.

On that leading orbit, averaging the divergence gives $\sigma =\mu [1-(3a+1)r^2/2]+O({\mu }^2)=-\mu +O({\mu }^2)<0$, so the cycle is stable.

The averaged energy drift is $F(r)=\mu [r^2-(3a+1)r^4/4]$. Linearizing radial energy balance at its nonzero zero gives decay rate $-\mu$ after converting $H=r^2$, exactly the Floquet exponent found from divergence.