[proofplan] We pass to frozen action-angle variables depending on the slow time $s=\varepsilon t$. The time-dependence of this canonical change of variables contributes an order-$\varepsilon$ angle-dependent perturbation to the Hamiltonian. Since the unperturbed angle velocity is bounded away from zero, we solve a homological equation that removes the oscillatory part of the perturbation from the action equation. The transformed action changes only at order $\varepsilon^2$, so over times of length $O(\varepsilon^{-1})$ it changes by $O(\varepsilon)$; the original action differs from it by another $O(\varepsilon)$. [/proofplan]

[step:Write the Hamiltonian in slow action-angle variables] Set $s=\varepsilon t$. For each $s\in[0,C]$, define $\mu(s):=\lambda(s)$. By hypothesis, the map \begin{align*} \Phi_{\mu(s)}:J\times\mathbb{T}\to\mathbb{R}^2 \end{align*} is a symplectic action-angle chart for the frozen Hamiltonian. Thus, along the solution, we write \begin{align*} (q_\varepsilon(t),p_\varepsilon(t))=\Phi_{\mu(\varepsilon t)}(I(t),\theta(t)). \end{align*} By the exact moving-chart hypothesis, the transformed Hamiltonian is the smooth map \begin{align*} \mathcal{H}_\varepsilon:J\times\mathbb{T}\times[0,C]\to\mathbb R \end{align*} defined by \begin{align*} \mathcal{H}_\varepsilon(I,\theta,s)=K(I,\mu(s))+\varepsilon R(I,\theta,s), \end{align*} where $R:J\times\mathbb{T}\times[0,C]\to\mathbb{R}$ is smooth and $2\pi$-periodic in $\theta$. This exactness assumption rules out a non-Hamiltonian flux term from the moving annulus of charts. Hamilton's equations in these variables are therefore \begin{align*} \dot I(t)=-\varepsilon\,\partial_\theta R(I(t),\theta(t),\varepsilon t). \end{align*} Also, \begin{align*} \dot\theta(t)=\partial_I K(I(t),\mu(\varepsilon t))+\varepsilon\,\partial_I R(I(t),\theta(t),\varepsilon t). \end{align*} By assumption, $I(t)\in J_0$ for all $0\le t\le C/\varepsilon$. Choose once and for all a compact interval $J_1$ determined by $J_0$ and $J$, with $J_0\subset J_1^\circ\subset J$. All estimates below are made on $J_1\times\mathbb{T}\times[0,C]$. [/step]

[step:Solve the homological equation for the oscillatory perturbation]Define the frozen frequency \begin{align*} \omega(I,s):=\partial_I K(I,\mu(s)). \end{align*} By hypothesis, $|\omega(I,s)|\ge\omega_0>0$ on $J_1\times[0,C]$. Define the angle average of $R$ by \begin{align*} \overline R(I,s):=\frac{1}{2\pi}\int_0^{2\pi}R(I,\theta,s)\,d\mathcal{L}^1(\theta). \end{align*} Since $R-\overline R$ has zero average in $\theta$, define \begin{align*} S:J_1\times\mathbb{T}\times[0,C]\to\mathbb{R} \end{align*} to be the unique smooth $2\pi$-periodic function with zero angle average satisfying \begin{align*} \omega(I,s)\,\partial_\theta S(I,\theta,s)=R(I,\theta,s)-\overline R(I,s). \end{align*} This definition is valid because division by $\omega(I,s)$ is uniformly bounded by $\omega_0^{-1}$, and the right-hand side has zero average in $\theta$.[/step]

[guided]The action equation contains the oscillatory term $-\varepsilon\partial_\theta R$. If we integrate this directly over time $0\le t\le C/\varepsilon$, the factor $\varepsilon$ is not enough by itself, since the time interval has length of order $\varepsilon^{-1}$. We need to use the rapid rotation in $\theta$. The fast rotation is controlled by \begin{align*} \omega(I,s):=\partial_I K(I,\mu(s)). \end{align*} The non-separatrix hypothesis is encoded in the bound $|\omega(I,s)|\ge\omega_0>0$ on $J_1\times[0,C]$. This is the place where the proof would fail near a separatrix: the period can become unbounded, equivalently the frequency can approach zero, and the division below would no longer be uniformly controlled. We separate $R$ into its average and oscillatory parts. Define \begin{align*} \overline R(I,s):=\frac{1}{2\pi}\int_0^{2\pi}R(I,\theta,s)\,d\mathcal{L}^1(\theta). \end{align*} The function $R(I,\theta,s)-\overline R(I,s)$ has zero average over $\theta$. Therefore it has a periodic primitive in $\theta$. Because $\omega(I,s)$ is bounded away from zero, we can define a smooth periodic function \begin{align*} S:J_1\times\mathbb{T}\times[0,C]\to\mathbb{R} \end{align*} with zero angle average by requiring \begin{align*} \omega(I,s)\,\partial_\theta S(I,\theta,s)=R(I,\theta,s)-\overline R(I,s). \end{align*} The zero-average normalization removes the additive constant ambiguity. Smoothness follows from the smoothness of $R$, $K$, and $\lambda$, together with the uniform lower bound on $|\omega|$.[/guided]

[step:Introduce an averaged action that differs from the frozen action by order $\varepsilon$] Define the corrected action function \begin{align*} \mathcal J_\varepsilon:[0,C/\varepsilon]\to\mathbb R \end{align*} by \begin{align*} \mathcal J_\varepsilon(t):=I(t)+\varepsilon\,\partial_\theta S(I(t),\theta(t),\varepsilon t). \end{align*} Since $S$ is smooth on the compact set $J_1\times\mathbb{T}\times[0,C]$, the quantity $\partial_\theta S$ is bounded there. Let \begin{align*} A_0:=\sup_{(I,\theta,s)\in J_1\times\mathbb{T}\times[0,C]}|\partial_\theta S(I,\theta,s)|. \end{align*} Then, whenever $I(t)\in J_1$, \begin{align*} |\mathcal J_\varepsilon(t)-I(t)|\le \varepsilon A_0. \end{align*} [/step]

[step:Differentiate the corrected action and cancel the order $\varepsilon$ oscillation] Differentiate $\mathcal J_\varepsilon(t)$ using the chain rule: \begin{align*} \dot{\mathcal J}_\varepsilon(t)=\dot I(t)+\varepsilon\,\partial_{I\theta}S(I(t),\theta(t),s)\dot I(t)+\varepsilon\,\partial_{\theta\theta}S(I(t),\theta(t),s)\dot\theta(t)+\varepsilon^2\,\partial_{s\theta}S(I(t),\theta(t),s), \end{align*} where $s=\varepsilon t$. Substitute the equations of motion. The order-$\varepsilon$ part is \begin{align*} -\varepsilon\,\partial_\theta R(I,\theta,s)+\varepsilon\,\omega(I,s)\,\partial_{\theta\theta}S(I,\theta,s). \end{align*} Differentiate the homological equation in $\theta$. Since $\omega$ is independent of $\theta$, \begin{align*} \omega(I,s)\,\partial_{\theta\theta}S(I,\theta,s)=\partial_\theta R(I,\theta,s). \end{align*} Thus the order-$\varepsilon$ terms cancel. All remaining terms contain the factor $\varepsilon^2$ and are smooth functions of $(I,\theta,s)$ on the compact set $J_1\times\mathbb{T}\times[0,C]$. Hence there is a constant $A_1>0$ such that \begin{align*} |\dot{\mathcal J}_\varepsilon(t)|\le A_1\varepsilon^2 \end{align*} for every $t$ for which $I(t)\in J_1$. [/step]

[step:Integrate the averaged equation over the long time interval] For $0\le t\le C/\varepsilon$, integrate the previous estimate with respect to one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on the time interval: \begin{align*} |\mathcal J_\varepsilon(t)-\mathcal J_\varepsilon(0)|\le \int_0^t A_1\varepsilon^2\,d\mathcal{L}^1(\tau). \end{align*} Since $t\le C/\varepsilon$, this gives \begin{align*} |\mathcal J_\varepsilon(t)-\mathcal J_\varepsilon(0)|\le A_1C\varepsilon. \end{align*} [/step]

[step:Transfer the estimate back to the frozen action] Using $|I(t)-\mathcal J_\varepsilon(t)|\le A_0\varepsilon$ and $|I(0)-\mathcal J_\varepsilon(0)|\le A_0\varepsilon$, we obtain \begin{align*} |I(t)-I(0)|\le |I(t)-\mathcal J_\varepsilon(t)|+|\mathcal J_\varepsilon(t)-\mathcal J_\varepsilon(0)|+|\mathcal J_\varepsilon(0)-I(0)|. \end{align*} Therefore \begin{align*} |I(t)-I(0)|\le (2A_0+A_1C)\varepsilon. \end{align*} Set $M:=2A_0+A_1C$. Since $J_1$ was fixed as a choice determined by $J_0$ and $J$, this constant depends only on the compact action interval $J_0$, the frozen Hamiltonian family, the slow path $\lambda$, the chosen exact action-angle chart, and $C$. Hence \begin{align*} I(t)=I(0)+O(\varepsilon) \end{align*} uniformly for $0\le t\le C/\varepsilon$. This is the claimed adiabatic invariance of the action. [/step]