I am considering the $2$D system \begin{align} \dot{x} &= \frac{\sqrt{2}}{4}x(x-1)\sin \phi\\ \dot{\phi} &= \frac{1}{2\sqrt{2}}[1-\cos \phi - \frac{1}{8} x\cos \phi], \end{align} where $0 \leq x \leq 1$, $-\pi \leq \phi \leq \pi$. I wish to show there exist no closed orbits. For visualisation, the phase portrait on a cylinder appears below, with a saddle point on $x = 0$ and stable and unstable fixed nodes at $x = 1$. Though it appears to be visually true, I am unsure as to how to produce a more formal argument demonstrating that it is true.
