Help Finding Saddle Connections of a System I am having a hard time finding saddle connections for a problem I'm working on.
Starting with a conservative system $u''-u+2u^3=0$, I showed that the orbits in the phase plane $x=u,y=u'$ ($x'=y, y'=x-2x^3)$ are given by $y^2=x^2-x^4+2E$ for some constant E. 
There is a saddle point at $(0,0)$ (when $E=0$) and I'm trying to show that there is a saddle connection on $y^2=x^2-x^4, x= \pm \text{sech}(t)$ that connects the saddle point to itself, but I don't know how to go about doing this as I cannot really find any examples dealing with this.
 A: Let's see if this helps, although it is a qualitative approach.
Your system has 3 equilibrium points:


*

*$p_1=(0,0)$

*$p_2=(0,\sqrt{1/2})$

*$p_3=(0,-\sqrt{1/2})$


$p_1$ is a saddle, as you said, but $p_2$ and $p_3$ are centres. Near such points you know the dynamics from the linearisation, it looks like in the following picture (qualitatively)

Next note that for $y=0$ $y'\neq 0$ for all $x$ apart from the 3 equilibrium points. This means that all trajectories must cross the $x$-axis orthogonally. Actually all trajectories must cross the $x$-axis twice due to the symmetry of the problem. Next, how to argue that the invariant manifolds of the saddle are connected? First we observe that there are no sinks not sources. Which means that the invariant manifold of the saddle cannot approach, in forward or in backward time, any other equilibrium point. Therefore, the phase portrait must be as shown below.

So, let's suppose you are convinced that qualitatively speaking you have a saddle connection. It also explains why $x(t)=\pm \text{sech}(t)$ as $\lim_{|t|\to\infty}\text{sech}(t)=0$ just as in your saddle connection.
I hope this helps.
