Past exam of the mathematics course of the University of Cambridge / 2021 / ia / Paper 2 / 7A / Solution

Put $\omega =\dot{\theta }$. The first-order system isWriteOn $[0,\pi ]$, $h$ is strictly increasing. Moreover,when $3\lambda <4\mu$. Thus there is one ${\theta }_{\ast }\in (0,\pi )$ with $h({\theta }_{\ast })=0$, in addition to the fixed points at $\theta =0,\pi$. Explicitly,

Let $g(\theta )=\sin \theta h(\theta )$. Linearization at $({\theta }_0,0)$ has eigenvalues satisfyingAt zero, $g^{\prime }(0)=h(0)<0$, and at $\pi$, $g^{\prime }(\pi )=-h(\pi )<0$, so both endpoints are saddles. At the interior point,so $({\theta }_{\ast },0)$ is a center. On the full angular interval there is a second center at $(-{\theta }_{\ast },0)$.

The system has the conserved energy of a conservative planar phase portrait,For $\lambda =1$, $\mu =3/2$,The phase portrait on the cylinder consists of centers at $\theta =\pm \arccos (-11/16)$ surrounded by closed periodic-energy curves, with saddles at $\theta =0$ and the identified point $\theta =\pi =-\pi$. The saddle-energy contours form the separatrices between librations in the potential wells and trajectories crossing the lower barrier.

If $3\lambda >4\mu$, then $h(0)>0$ and monotonicity gives no interior zero on $(0,\pi )$. The point $(0,0)$ becomes a center, while $(\pi ,0)$ remains a saddle. At $3\lambda =4\mu$, the two off-axis centers coalesce with the origin and the linearization there is degenerate.

Solved by gpt-5.6-sol high.