Past exam of the mathematics course of the University of Cambridge / 2015 / ii / Paper 1 / 28B / b / Solution

This vector field is the negative of the one in part (a). For the same positive-definite quadratic form,The ellipse $V=1$ is invariant and has no stationary point. The angular variable increases around it, with a finite return time ${\int }_0^{2\pi }d\theta /(2{\cos }^2\theta +{\sin }^2\theta )=\sqrt{2}\pi$. It therefore explicitly supplies a periodic solution. Alternatively, the Poincare-Bendixson theorem applies to any compact invariant annulus $c\le V\le C$ with $0<c<1<C$, because its boundaries point inward and there is no equilibrium in the annulus.

If a periodic orbit had a point with $V<1$, then $V$ would strictly increase along it and could not return; a point with $V>1$ would make it strictly decrease. A periodic orbit meeting $V=1$ remains on that ellipse by uniqueness of solutions. Consequently the ellipse is the unique periodic orbit, up to time translation. It is attracting, and all nonzero trajectories approach it.