Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 1 / 7A / ii / Solution

For general $k$, the same energy gives $\dot{E}=-k{y}^2$. For $k=0.1$ the eigenvalues at the origin are $-0.05\pm i\sqrt{3.99}/2$, so it is a stable focus. At $(1,0)$ they are $(-0.1\pm \sqrt{4.01})/2$, so it remains a saddle equilibrium. The vector field has constant divergence $-0.1$: there are no periodic orbits, by the planar divergence criterion. Nor can the former homoclinic orbit survive, since $E$ strictly decreases along every nonconstant orbit except at isolated crossings of $y=0$.

The stable manifold of the saddle separates escape to positive $x$ from convergence to the origin. Its two branches approach the saddle in forward time with tangent slope $(-0.1-\sqrt{4.01})/2$. The left unstable branch spirals into the origin, whereas the right unstable branch escapes. The shaded component on the origin's side of the stable manifold in the right panel is the basin of attraction. The boundary itself is excluded: its trajectories converge to the saddle. This boundary is the damped stable manifold, rather than the conservative energy contour. energy decrease prevents bounded recurrent motion, and every bounded orbit off the saddle's stable manifold consequently converges to the origin. The plotted manifold was obtained by integrating backwards from the saddle's stable eigendirection.