= Solution
Substitution of the weighted scaling and $\tau=\varepsilon t$ gives
$$
\boxed{u'=v,\qquad v'=u^2-\alpha+\varepsilon(\beta+u)v.}
$$
The derivative of the candidate <Hamiltonian> is
$$
H'=v[u^2-\alpha+\varepsilon(\beta+u)v]+(\alpha-u^2)v=\varepsilon(\beta+u)v^2.
$$
Thus $H$ is a <conserved quantity> at $\varepsilon=0$. For $\alpha>0$ put $a=\sqrt\alpha$. The conservative <equilibrium point> $(-a,0)$ is a center with $H=-2a^3/3$, and $(a,0)$ is a <saddle equilibrium> with \b[saddle-loop energy]
$$
\boxed{H_s={2\over3}\alpha^{3/2}.}
$$
The closed component of a level between these energies surrounds the center. At $H=H_s$,
$$
v^2={2\over3}(u-a)^2(u+2a),\qquad -2a\le u\le a,
$$
is the <homoclinic orbit>, with left turning point $u=-2a$. The upper branch is traversed to the right and the lower branch to the left. Levels above the barrier have escaping trajectories; the saddle level also has unbounded branches to the right. For $\alpha=0$ the two critical points coalesce; for $\alpha<0$ the potential is monotone and there is no center or saddle loop.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2001/iii/paper-51-hamiltonian-contours.png]
{title=Hamiltonian level curves and the conservative saddle loop for alpha equal to one}
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