Only closed homoclinic orbits?

Question

Given a fixed vector field $\dot{\mathbb{x}}(t)$ and a fixed point $\mathbb{x}^{*}$. Is it necessary that the orbit that satisfies $\lim_{t \to \pm \infty} \mathbb{\phi} (t) = \mathbb{x}^{*} $ (where $\phi (t)$ is the trajectory) be closed? I mean, Should an homoclinic be a closed orbit in the phase space?

Answer 1

Well, technically a homoclinc orbit doesn't include the limit $x^*$ and therefore is not closed (nor open for phase spaces of dimension $>1$).

However, the set $H=\phi(\mathbb{R}) \cup \{x^*\}$ is closed (sketch of proof):

Assume that there exists $y \neq x^*, \not \in H$ that is a limit point of $H$ and pick $\epsilon < \frac{1}{2} d(y, x^*)$. Since $\lim_{t \to \pm \infty} \phi(t) = x^*$, there exists $R> 0$ such that for all $|t|\geq R$ we have that $d(\phi(t), x^*)\leq \epsilon$. Therefore $y$ is not a limit point of $\phi \left( (-\infty, -R] \cup [R, \infty) \right)$. Since $\phi(t)$ is continuous (as the solution of a differential equation) and $[-R, R]$ is compact, we have that $\phi([-R,R])$ is compact and therefore contains all of its limit points. Thus we have that $y$ is not a limit point of $\phi([-R, R])$.

This leaves nowhere for a sequence $\{t_n\}_{n=1}^\infty$ to exist such that $\lim_{n \to \infty} \phi(t_n) = y$. In other words, $H \subseteq \phi([-R, R]) \cup \overline{B_\epsilon (x^*)}$.