Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 1 / 32A / c

When $k=0$, the system is conservative andis constant. The fixed points areThe potential $U$ has a strict minimum $U(0)=0$, strict maximaand tends to zero as $|x|\to \infty$. The conservative planar phase portrait therefore has:

The portrait is symmetric under $(x,y)\mapsto (-x,-y)$ and under time reversal $(x,y,t)\mapsto (x,-y,-t)$.

For $0<k\ll 1$, the same three fixed points remain. At the origin the Jacobian isIts eigenvalues have negative real part; for small $k$ they are complex, so the origin is a stable spiral. At $x=\pm 1$, $U^{\prime \prime }(\pm 1)<0$, making the Jacobian determinant negative, so both points remain saddles.

Every nonstationary trajectory moves across the former energy contours in the direction of decreasingThe unstable branches of the two saddles directed into the central region spiral into the origin, while their outward branches escape toward $x=\pm \infty$. The stable manifolds of $(\pm 1,0)$ are the separatrices that form the boundary of the basin of attraction of the origin. Initial points between these stable separatrices lose enough energy to remain between the barriers and spiral to the origin; points outside escape to one of the two infinities.

This gives the weakly damped damped rational double-barrier phase portrait and identifies the requested domain of stability as the open region bounded by those stable manifolds.