Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics IV / 6 / b

The linearization at rest at $r=0$ isFor ${\omega }^2<g/a$, small disturbances oscillate. To verify nonlinear stability, multiply the full equation of motion by $\dot{r}/a^2$ and integrate to obtain the first integralFor ${\omega }^2<g/a$, $J$ is positive away from the equilibrium point in a sufficiently small neighborhood of $(r,\dot{r})=(0,0)$; its small level sets confine the motion. ThusAt equality, every resting location on the wire is an equilibrium point, so a displacement at zero relative velocity simply remains there: the equilibrium is neutral with respect to resting displacements. It is unstable with respect to general perturbations in phase space, however. An arbitrarily small outward initial velocity $v_0>0$ gives $\dot{r}=v_0/\sqrt{1+r^2/a^2}>0$ from the first integral, eventually taking the bead outside any prescribed small neighborhood. Thus strict phase space stability holds only for ${\omega }^2<g/a$.