\sqrt{\eps} \; t,$$ and denoting the derivatives with respect to $\tau$ by a prime, in the leading order (\[vp4\]) can be written as \^ &=& f\_[2,0]{}, \[vp5\]\
\^ &=& f\_[1,0]{} - \^. In (\[vp5\]), we used a notation $f_{1,0} = f_1\left( 0, \sg, \th \right)$ and $f_{2,0} =
f_2\left( 0, \sg, \th \right)$. We can expand $f_{1,0}$ as $$f_{1,0} = a(\sg) + b(\sg, \th),$$ where $a(\sg)$ is the average value of $f_{1,0}$ over a period: $$a = \frac1{2\pi} \int_0^{2\pi} f_{1,0} \: d\th,$$ and $b(\sg, \th)$ is the remaining, zero-average part. Introduce a potential energy for $f_{1,0}$: $$f_{1,0} = -\frac{\pt V\left( \sg, \th\right)}{\pt \th}.$$ The function $V$ contains a periodic and a linear in $\th$ parts: $$V = a(\sg) \th + \int b(\sg, \th) \; d\th = a(\sg) \th + \ovl{b}(\sg, \th).$$ Define the resonance energy: H\_R = 12 (\^)\^2 + V \[vp5a\] and denote by $h$ the value of $H_R$. Note, that at the crossing $\th^{\pr} = 0$.

The relation between $a$ and $\max\left|b(\th)\right|$ defines the properties of the phase portrait on $(\th, \th^{\pr})$ phase plane. If $$\max\left|b(\th)\right| > \left| a \right|,$$ the phase portrait looks like the one shown in Fig. 1a. For the sake of simplicity, we assumed 