I have the following system with kinetic energy: $T = \frac{m\dot{x}^2}{2}$ and potential energy: $V=ax-x^3$. I want to find if there are any separatrices and if there are, is there a condition on $a$.
THIS IS WHAT I HAVE DONE:
Hamilton's Equations give: $\begin{cases}\dot{q} = \frac{\partial H}{\partial p}=\frac{p}{m} \\ \dot{p} = \frac{\partial H}{\partial q} = a-3q^2 \end{cases}$
The fixed points are: $\left(-\sqrt{\frac{a}{3}}, 0\right)$ and $\left(\sqrt{\frac{a}{3}}, 0\right)$.
Taking the double derivative wrt $q$ of the potential we have:$V'' = -6q$, therefore only $\left(\sqrt{\frac{a}{3}}, 0\right)$ corresponds to a saddle point. From my understanding, the separatrices are meeting at the sadde point but I don't know how to find them.
Here is the contour plot for the hamiltonian:
Any ideas?