My question concerns part (c) of the following:
I have no problem answering parts (a) and (b), but I am a little confused as two things: 1.) to how to treat the case $u=0$ 2.) find the range of values of the momentum for which we can relative equilibrium.
My attempt was as follows:
In part (b), I found Hamilton's equations to be $$\dot{u}=\frac{\partial H}{\partial p_u}=\frac{1}{m}p_u$$ $$\dot{\theta}=\frac{\partial H}{\partial p_\theta}=\frac{1}{m(a^2+u^2)}p_\theta$$ $$\dot{p_u}=-\frac{\partial H}{\partial u}=\frac{u}{m(a^2+u^2)}p_{\theta}^2-ku$$ $$\dot{p_\theta}=-\frac{\partial H}{\partial \theta}=0$$ So in part (c), if $u$ is constant, then its time derivative is $0$. Then by our first Hamilton equation, the momentum $p_u$ must be $0$. Then from our third equation we must have $$0=\dot{p_u}=\frac{u}{m(a^2+u^2)}p_{\theta}^2-ku$$ If $u$ is non-zero, then we have the relation $$0=\frac{1}{m(a^2+u^2)}p_{\theta}^2-k$$
I am unsure if the above is correct, as I cannot see how to obtain a range of values of $p_\theta$ from it, or what happens when $u=0$. Any help clarifying would be much appreciated.
Continuing from your last formula, you have that
$$p_{\theta}=\pm\sqrt{km(a^2+u^2)} \; .$$
On the other hand, in the case $u=0$ there is no restriction on $p_{\theta}$ except it has to be constant in time by the fourth equation. But it can be any constant.