Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 3 / 14E / iii / Solution

When $a=0$, define the Hamiltonian functionThen $\dot{x}=-H_y$ and $\dot{y}=H_x$, so $\dot{H}=0$. At $(\sqrt{b},0)$ the Hessian of $H$ is $2\sqrt{b}I$, a strict nondegenerate minimum; at $(-\sqrt{b},0)$ it is $-2\sqrt{b}I$, a strict maximum. Small regular level curves around either point are closed curves, and the vector field is nonzero on them. Motion around each curve therefore returns after a finite period, proving the existence of periodic orbits and classifying both equilibria as center equilibria.

The separatrix level is $H=0$, consisting of $x=0$ and $x^2/3+y^2=b$. The two arcs of this ellipse run from the lower saddle equilibrium to the upper one, while the vertical segment runs from the upper saddle to the lower one. They enclose the two families of periodic orbits, as shown in the last panel of the phase portrait. Outside the loops, regular levels are unbounded.

For $a>0$ the divergence of the original vector field is exactly $-a$. If a periodic orbit existed, uniqueness of trajectories would make it a simple closed curve, tangent everywhere to the vector field. The flux across its boundary would be zero, whereas the divergence theorem would give $-a$ times its positive enclosed area. This contradiction is the Bendixson-Dulac criterion. Hence $\boxed{\text{periodic orbits exist if and only if }a=0}$.