Homoclinic orbits of cubic potential

Question

I found in Carles Simo's 'Hamiltonian Systems with Three or More Degrees of Freedom', among other references, that the homoclinic orbit for the cubic potential $\frac{y^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{2}}{2}$ is $\Gamma=\{(x_{0}(t),y_{0}(t))\}$ where $x_{0}(t)=(\sqrt{3}/2)(cosh(t/2))^{-2}$, $y_{0}(u)=\dot{x}_{0}(u)$. Does anybody know how to obtain this or has a good reference? I don't even know where to start! Thanks a lot!

Answer 1

You've switched $x$ and $y$ in your description, since the function $x_0(t)$ does not obey $\frac{\dot{x}_0^2}{2} + \frac{\dot{x}_0^3}{3}- \frac{x_0^2}{2} = \text{constant}$. Moreover, if you want to interpret the conserved quantity (Hamiltonian? Lagrangian?) $E = \frac{y^2}{2}+\frac{y^3}{3}-\frac{x^2}{2}$ as one having a 'cubic potential', then this implies that $\dot{y} = x$, and not the other way around. Also, you've made a mistake in the prefactor of $x_0$, as this function does not obey $F = \frac{x_0^2}{2} + \frac{x_0^3}{3}- \frac{\dot{x}_0^2}{2} = \text{constant}$. To remedy this, substitute $x = \frac{a}{cosh^2 \frac{t}{2}}$ and try to see for which value of $a$ you do get a constant value (hint: the constant value is zero).

To get an idea of how to obtain such an explicit expression, you can solve the quadratic equation $\frac{x^2}{2} + \frac{x^3}{3}- \frac{\dot{x}^2}{2} = 0$ for $\dot{x}$ to obtain a first order ODE for $x(t)$, which you then can solve using standard methods.