Past exam of the topics in applied mathematics course of the University of Oxford / 2009 / B / 4 / a / iii / Solution

For $\nu =1,\kappa =0$, the first integral isIts minimum $-1/4$ occurs at the two centres. Levels between $-1/4$ and $0$ give a periodic orbit around each centre equilibrium; positive levels give larger periodic orbits surrounding both wells. The zero level is the figure-eight separatrix through the saddle, with two homoclinic loops.

On $E=0$, $y^2=x^2(1-x^2/2)$. The printed parametrization $x=\pm \sqrt{2}\mathrm{sech}z$, $y=\pm \sqrt{2}\mathrm{sech}z\tanh z$, taking matching signs, satisfies this identity because $1-{\mathrm{sech}}^{2}z={\tanh }^{2}z$. It also has $dx/dz=-y$ and $dy/dz=-(x-x^3)$, so its parameter is physical reversed time, $z=-t+\text{constant}$, rather than $z=t$. It is nonetheless exactly a parametrization of the requested trajectory. In physical forward time one can use $x(t)=\sqrt{2}\mathrm{sech}t$, $y(t)=-\sqrt{2}\mathrm{sech}t\tanh t$ on the right loop.

Approach to the origin requires $z\to \pm \infty$, hence infinite physical time. Equivalently $|y|\sim x$ near the right-loop endpoint, so the elapsed time contains the divergent integral ${\int }_0^{\delta }dx/x$. The point at $x=\sqrt{2}$ is reached in finite time from any interior point; it is the asymptotic saddle endpoints, not that turning point, that cause the infinite total traversal time.