Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 65 / 4 / c

The displayed orbit has the normalization $\alpha =1$. To verify it put $z=\tanh (\tau /\sqrt{2})$ and $f=1-z^2={\mathrm{sech}}^2(\tau /\sqrt{2})$. Then $z^{\prime }=f/\sqrt{2}$, $f^{\prime }=-\sqrt{2}fz$, soAlso $u=-2+3z^2$, and $(u^2-1)=3(1-z^2)(1-3z^2)$, exactly the same expression as $v^{\prime }$. Therefore the conservative equations hold. As $\tau \to \pm \infty$, $(u,v)\to (1,0)$; at $\tau =0$ the orbit reaches $(-2,0)$. Its conserved energy is $2/3$, so this is the required homoclinic orbit.

For a general positive $\alpha$ set $s=\sqrt{\alpha }$. Rescaling the verified solution givesDirect differentiation gives $u^{\prime }=v$ and $v^{\prime }=u^2-\alpha$, and the endpoints are $(s,0)$. This restores the parameter dependence implicit in the normalized printed orbit.