 Solution

The Hamilton equations are $\dot{q}=p/m$ and $\dot{p}=-V_q$. Applying the chain rule to $H(q,p,\lambda (t))$, their two phase-space contributions cancel, giving $\dot{H}=V_{\lambda }\dot{\lambda }=H_{\lambda }\dot{\lambda }$.

For fixed $\lambda$, a closed periodic orbit of energy $E$ has $p=\pm \sqrt{2m(E-V)}$. Thus its action variable isThe sign matches the direction on each half of the orbit, making the enclosed phase-plane area positive. Since $dq=\dot{q}dt=pdt/m$, differentiating under the integral givesFor slowly varying $\lambda$, the exact energy equation averaged over a nearly frozen cycle gives $d⟨H⟩/dt=⟨V_{\lambda }⟩\dot{\lambda }$ to the retained adiabatic order. Using $E=⟨H⟩$, the two chain-rule terms therefore cancel:Here the angle brackets differentiate the explicit parameter dependence along the frozen cycle, not an independently chosen family of energies. The action is an adiabatic invariant, with errors beyond the slow-cycle approximation.

For $V=\lambda q^{2n}$, scale $q=(E/\lambda {)}^{1/(2n)}s$. ThenThe integral is a positive constant depending only on $n$. Constancy of $I$ gives $E^{n+1}/\lambda$ constant, hence