Past exam of the mathematics course of the University of Oxford / 2015 / Prelims / Mathematics IV / 3 / b

Here equilibrium means fixed position relative to the rotating wire: $r=a$ and $z=f(a)$ are constant, while $\theta =\omega t$ continues to change. Substituting $\dot{r}=\ddot{r}=0$ into the equation givesFor the linear stability calculation, set $r=a+ϵ$ and retain terms of first order in $ϵ$ and $\dot{ϵ}$. The term ${\dot{r}}^2$ contributes only at second order, and variation of the factor multiplying $\ddot{ϵ}$ also contributes only at second order. ThusThe two strict sign cases are treated in the following parts. If $g{f}^{\prime \prime }(a)={\omega }^2$, this linearization has no restoring term and higher orders are needed to classify the equilibrium.

For the specified wire,Hence the complete set of positive equilibrium radii isAt $a_n$,Therefore odd $n$ give a stable equilibrium, and even $n$ give an unstable equilibrium. There is no degenerate sign case in this example. The corresponding heights are $f(a_n)={\omega }^2{n}^2{\pi }^2{b}^2/(2g)+b(-1{)}^n$.

If $g{f}^{\prime \prime }(a)>{\omega }^2$, defineThe linearized equation is $\ddot{ϵ}+{\nu }^2ϵ=0$, with bounded simple harmonic motionThus the strict stable equilibrium condition isThis is also a sufficient condition for local nonlinear Lyapunov stability. Indeed the equation for the bead on a uniformly rotating wire has the conserved quantityIts derivative is $\dot{r}$ times the governing equation moved to one side, so it vanishes. At $a$, the last two terms have zero first derivative and positive second derivative $g{f}^{\prime \prime }(a)-{\omega }^2$. Their strict local minimum traps sufficiently small perturbations. The motion is bounded rather than asymptotically returning to rest.

If $g{f}^{\prime \prime }(a)<{\omega }^2$, putThe linearized equation becomes $\ddot{ϵ}-{\lambda }^2ϵ=0$, soArbitrarily small initial perturbations can have $C_1\ne 0$ and grow exponentially. Hence the strict unstable equilibrium condition isFor the particular wire, this occurs precisely at $a=2j\pi b$, $j=1,2,\dots$; the alternating radii $a=(2j-1)\pi b$ satisfy the stable equilibrium condition.