The Hamiltonian for an oscillating particle with one degree of freedom is

$$H=\frac{p^{2}}{2 m}+V(q, \lambda)$$

The mass $m$ is a constant, and $\lambda$ is a function of time $t$ alone. Write down Hamilton's equations and use them to show that

$$\frac{d H}{d t}=\frac{\partial H}{\partial \lambda} \frac{d \lambda}{d t}$$

Now consider a case in which $\lambda$ is constant and the oscillation is exactly periodic. Denote the constant value of $H$ in that case by $E$. Consider the quantity $I=$ $(2 \pi)^{-1} \oint p d q$, where the integral is taken over a single oscillation cycle. For any given function $V(q, \lambda)$ show that $I$ can be expressed as a function of $E$ and $\lambda$ alone, namely

$$I=I(E, \lambda)=\frac{(2 m)^{1 / 2}}{2 \pi} \oint(E-V(q, \lambda))^{1 / 2} d q$$

where the sign of the integrand alternates between the two halves of the oscillation cycle. Let $\tau$ be the period of oscillation. Show that the function $I(E, \lambda)$ has partial derivatives

$$\frac{\partial I}{\partial E}=\frac{\tau}{2 \pi} \quad \text { and } \quad \frac{\partial I}{\partial \lambda}=-\frac{1}{2 \pi} \oint \frac{\partial V}{\partial \lambda} d t$$

You may assume without proof that $\partial / \partial E$ and $\partial / \partial \lambda$ may be taken inside the integral.

Now let $\lambda$ change very slowly with time $t$, by a negligible amount during an oscillation cycle. Assuming that, to sufficient approximation,

$$\frac{d\langle H\rangle}{d t}=\frac{\partial\langle H\rangle}{\partial \lambda} \frac{d \lambda}{d t}$$

where $\langle H\rangle$ is the average value of $H$ over an oscillation cycle, and that

$$\frac{d I}{d t}=\frac{\partial I}{\partial E} \frac{d\langle H\rangle}{d t}+\frac{\partial I}{\partial \lambda} \frac{d \lambda}{d t}$$

deduce that $d I / d t=0$, carefully explaining your reasoning.

When

$$V(q, \lambda)=\lambda q^{2 n}$$

with $n$ a positive integer and $\lambda$ positive, deduce that

$$\langle H\rangle=C \lambda^{1 /(n+1)}$$

for slowly-varying $\lambda$, where $C$ is a constant.

[Do not try to solve Hamilton's equations. Rather, consider the form taken by $I$. ]