Homoclinic orbit

Remarks

A homoclinic orbit lies in the intersection of the stable and the unstable set of $x_0$, and it does not contain $x_0$. Example: for $x' = y$, $y' = x - x^2$, the function $\gamma(t) = (x(t), x'(t))$ with $x(t) = \frac{3}{2} \operatorname{sech}^2(t/2)$ is a solution. Indeed $x'' = \frac{3}{2} s - \frac{9}{4} s^2 = x - x^2$ with $s = \operatorname{sech}^2(t/2)$, and $x(t) \to 0$, $x'(t) \to 0$ as $t \to \pm\infty$, so $\gamma$ is homoclinic to the origin. Its trajectory is the part of the curve $y^2 = x^2 - \frac{2}{3} x^3$ with $x > 0$. Non-example: for the pendulum $x' = y$, $y' = -\sin x$, the solution with $x(t) \to -\pi$ as $t \to -\infty$ and $x(t) \to \pi$ as $t \to +\infty$ joins two different equilibria in $\R^2$. So it is a heteroclinic orbit, not a homoclinic one. A periodic orbit is also not homoclinic, since it does not tend to an equilibrium.