= Solution
With primes denoting differentiation in $\tau$, the weighted scaling gives
$$
\boxed{u'=v,\qquad v'=-\alpha u-u^3+\varepsilon(\beta-u^2)v.}
$$
Indeed $\dot x=\varepsilon^2u'$ and $\dot y=\varepsilon^3v'$, so the damping terms carry one extra power of $\varepsilon$. Consequently
$$
H=\frac12v^2+\frac{\alpha}{2}u^2+\frac14u^4,\qquad
H'=\varepsilon(\beta-u^2)v^2.
$$
At $\varepsilon=0$ this proves conservation of $H$ directly.
For $\alpha=-a_0^2<0$, the potential is a double well. Its minima are at $u=\pm a_0$, where $H=-a_0^4/4$; these are centre <equilibria>. The origin is a <saddle equilibrium> and has $H=0$. Energies $-a_0^4/4<H<0$ give two separate closed contours, one in each well. Energies $H>0$ give a single closed contour surrounding both wells. The intervening figure-eight consists of two homoclinic loops:
$$
\boxed{H_{\rm hom}=0,\qquad v^2=a_0^2u^2-\frac12u^4.}
$$
Their outer turning points are $u=\pm\sqrt2a_0$. The positive loop, for example, is parametrized by $u=\sqrt2a_0\operatorname{sech}(a_0\tau)$, $v=u'$.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2002/iii/paper-56-double-well-contours.png]
{title=Conserved-energy contours and the two homoclinic loops}
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