Heteroclinic orbit

Remarks

Example: for $x' = 1 - x^2$ on $\R$, the equilibria are $-1$ and $1$. The solution with $x(0) = 0$ is $x(t) = \tanh t$, since $(\tanh t)' = 1 - \tanh^2 t$. It exists for all $t$, tends to $-1$ as $t \to -\infty$ and tends to $1$ as $t \to +\infty$. So $(-1, 1)$ is a heteroclinic orbit from $-1$ to $1$.

Non-example: for the same equation, the solution with $x(0) = 2$ is $x(t) = \coth(t + c)$, where $c > 0$ and $\coth c = 2$. It tends to $1$ as $t \to +\infty$, but it blows up at $t = -c$, so clause 1 fails.

A heteroclinic orbit from $p$ to $q$ lies in the unstable set of $p$ and in the stable set of $q$. If $p = q$ in the same construction, the orbit is called a homoclinic orbit. Some authors also allow $p$ and $q$ to be periodic orbits instead of equilibria.