Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2008 / 9 / c / Solution

Suppose a nonconstant periodic orbit exists for the continuously differentiable planar vector field. Uniqueness implies that its image over a minimal period is a simple closed curve, enclosing a region $\Omega$ of positive area. Along the orbit, the differential form $fdy-gdx$ vanishes because $dx=fdt$ and $dy=gdt$. Green's theorem, with positive orientation, therefore givesa contradiction. This is the Bendixson-Dulac criterion in the whole plane; the fact that the enclosed region also lies in the domain is essential.

For the second-order equation set $y=\dot{x}$. The equivalent system is $\dot{x}=y$, $\dot{y}=-(1+x^2)y-\cos x$. Its divergence is $\boxed{-(1+x^2)<0}$ throughout the plane. Hence there are no nonconstant periodic solutions.

The nonconstant qualification is necessary: the equilibrium solutions $x(t)\equiv \pi /2+k\pi$, $k\in \mathbb{Z}$, do exist and are periodic in the literal sense of satisfying every positive period. The source's usual dynamical interpretation of “periodic solution” here is a nonconstant periodic orbit.