= Solution
The potential factors as $V=x^2(3-2x)$, so its zeros are $0$ and $3/2$, with a double zero at the origin. Its <derivatives> are $V'=6x(1-x)$ and $V''=6-12x$. Thus there is a local minimum $(0,0)$, a local maximum $(1,1)$ and an inflection at $(1/2,1/2)$. The potential tends to $+\infty$ on the far left and $-\infty$ on the far right. These features determine the left-hand sketch.
Put $v=\dot x$. <Newton's second law> gives the <phase plane> system $\dot x=v$, $\dot v=-V'=6x(x-1)$. Its conserved <mechanical energy> and trajectories are
$$
\boxed{E=\frac12v^2+3x^2-2x^3,\qquad
v=\pm\sqrt{2[E-V(x)]}.}
$$
The <linearization> at $(0,0)$ has <eigenvalues> $\pm i\sqrt6$, and the surrounding energy curves are closed, so it is a <center equilibrium>. At $(1,0)$ the <eigenvalues> are $\pm\sqrt6$, giving a <saddle equilibrium>. The figure shows the <cubic potential barrier phase portrait>; arrows point right when $v>0$ and left when $v<0$.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2007/ia/paper-4-potential-phase.png]
{title=Cubic potential, trapped oscillations, escaping trajectories and the barrier separatrix}
For \b[$0<E<1$], there are three turning-point roots of $V(x)=E$. The two leftmost enclose a closed orbit around the origin: the particle oscillates between them. There is also a separate open component to the right of the third root, where an incoming particle turns and escapes to $+\infty$. The closed orbits run clockwise in the $(x,v)$ plane.
For \b[$E=1$], the factorization $1-V(x)=(x-1)^2(2x+1)$ gives the <separatrix>
$$
v^2=2(x-1)^2(2x+1).
$$
The branch with $-1/2\le x<1$ is a <homoclinic orbit>: it leaves the saddle asymptotically, turns at $x=-1/2$ and returns asymptotically. On $x>1$, one branch approaches the saddle and the other departs towards infinity. Approaching the saddle takes infinite time. The saddle itself is also an <equilibrium> trajectory.
For \b[$E>1$], the particle can cross the barrier. There is one left <turning point> below $-1/2$; an incoming particle passes through the well, turns on the left and escapes to the right. For \b[$E<0$], only a right-hand open component exists, with its <turning point> beyond $3/2$. At \b[$E=0$], the stationary centre coexists with a right-hand open trajectory turning at $3/2$. Thus negative energy does not imply trapping here: the cubic potential is unbounded below on the right.
Back to article page
