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

For $k=0$, the first integralhas $\dot{E}=0$. The equilibrium points are $(0,0)$ and $(1,0)$. The linearization has eigenvalues $\pm i$ at the origin and $\pm 1$ at $(1,0)$. The origin is a center equilibrium: the sufficiently small positive level sets of $E$ are closed curves, since its Hessian matrix there is positive-definite. The other equilibrium is a saddle equilibrium.

At the saddle, $E=1/6$, and the separatrix satisfiesIts two branches with $-1/2\le x\le 1$ form a homoclinic orbit through the saddle; its remaining branches extend to $x>1$. Inside the loop, the closed trajectories run clockwise: above the horizontal axis $\dot{x}=y>0$. The saddle directions have slopes $+1$ and $-1$. Outside the loop there are unbounded trajectories. The left panel shows the closed orbits and all saddle branches.