Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2010 / 8

Here a periodic solution must mean a nonconstant periodic orbit: constant equilibria are not excluded by the stated divergence hypothesis. Suppose such an orbit exists. Since the vector field is continuously differentiable, uniqueness for the autonomous differential equation makes its trajectory a simple closed curve $C$, bounding a region $D$ by the Jordan curve theorem. The Green theorem givesAlong the orbit, $dx=fdt$ and $dy=gdt$, so the line integral is identically zero. Reversing orientation if necessary does not change this conclusion. The area integral is strictly positive because its continuous integrand is positive everywhere and $D$ has positive area. This contradiction proves the Bendixson-Dulac criterion in this case.

For the second-order equation set $y=x^{\prime }$, giving $x^{\prime }=y$, $y^{\prime }=-y-\epsilon y^3-\sin x$. Choose the Dulac function $\rho =-1$. ThenThus there is no nonconstant periodic solution for $\epsilon >0$. The equilibria $x(t)=k\pi$ do remain solutions. As a direct check, the mechanical energy $H=y^2/2+1-\cos x$ obeys $H^{\prime }=-y^2-\epsilon y^4$, whose integral over a period could vanish only if $y\equiv 0$.