$\left| t_{1,2} - t_*\right|
\sim 1 /\eps$ and $t_*$ is the only moment of crossing of resonance on the time interval $(t_1,
t_2)$. Then the jump of $\Ffi$ on the resonance is $$\Dt \Ffi = \int_{t_1}^{t_2} \dot{\Ffi} \; dt.$$ In the first approximation, we can replace in the integral the limits $t_{2,1}$ with $\pm\infty$ (far from the resonance $\Ffi$ does not change much). We get = \_[-]{}\^ dt = 2 \_[-s ]{}\^[\_\*]{} 1[\^]{} d = -2 s \_[-s ]{}\^[\_\*]{} d . \[vp5b\] where $s = \sign (a)$ and $\th_*$ and $h_*$ are the values of $\th$ and $h$, respectively, when the exact trajectory hits ${\mit R}$. $\ovl{\th}_*$ is the resonance phase, that is equal to the shifted value of $\th$: $\ovl{\th}_* = \ovl{\th}_* - 2\pi K$, where the value of $K$ is chosen in such a way, that $0 < -\ovl{\th}_* - (\ovl{b}/a) \sin \ovl{\th}_* < 2 \pi$.

If the quantity $\Ffi$ can be calculated explicitly as a function of $I_1$ and $I_2$, we can directly substitute $\dot{\Ffi} = \dot{\Ffi}(I_1, I_2, \th)$ into (\[vp5b\]) and obtain the expression for $\Dt \Ffi$. This is the path we take in the subsequent part of the paper, when we consider a particular example 