Let $$\dot{x}=F(x)$$ be an autonomous (i.e. it does not depend on $t$) system with $F: \mathbb{R}^n \to \mathbb{R}^m$ a regular as you want vector field.
Suppose also that $0$ is an hyperbolic equilibrium point, meaning that the Jacobian $JF_{|0}$ only have eigenvalues with non-zero real part. In this case $W^+$ and $W^-$ as defined below are manifolds.
Denote $$W^+=\{x: \|\phi^tx\| \to 0, \ t \to +\infty\}$$ $$W^- =\{x: \|\phi^tx\| \to 0, \ t \to -\infty\}$$ where $\phi$ is the associated flux which exists by the existence and uniqueness of solutions for $\dot{x}=F(x).$ Consider $$W^+ \cap W^- = \{x: \|\phi^tx\| \to 0, \ t \to \pm \infty\}.$$
Question 1: I read from a trustworthy source that in this case it must be that $W^+ \cap W^- = \{0\},$ but I am failing to thoroughly convince myself of this fact. Can you show me a proof ?
Best Answer
There are many examples as you ask (in the comments). For example, for the equation $$ (x,y)'=(y, x(x-2)(x-3)) $$ you get the phase portrait
Note the homoclinic orbit of the equilibrium $(3,0)$.