TITLE: Which result guarantees convergence of solution of an ODE to a set of non-compact, non-isolated equilibrium?
QUESTION [5 upvotes]: Consider a continuous ODE,
$$\dot x = f(x), f \in C^1$$
$\dot x = 0$ for all $x \in K \subset \mathbb{R}^n$, where we assume that $K$ is a closed but unbounded set of non-isolated equilibrium. For example, $K$ could be a line, or a half space, or a ray, etc.

Suppose that this ODE admits an energy function $E$ such that $E(x) > 0$
for all $x \neq K$, $E(x) = 0$ for all $x \in K$, and, $\dot E(x) < 0$ for
all $x \neq K$, $\dot E(x) = 0$ for all $x\in K$.

Does there exist any theorem or result that allow us to conclude the converge of $x$ towards $K$ or some point in $K$ starting from any point $x(0) \in \mathbb{R}^n \backslash K$?
The challenge here is, unlike what has been suggested here: Conditions for convergence to non-isolated fixed points $K$ is assumed to be unbounded. This eliminates the existence of some compact set which contains $\{x| \dot E = 0\}$, hence the Krasovsky-LaSalle theorem does not apply.

REPLY [3 votes]: The extension to the unbounded setting is due to Hale 1969: Dynamical systems and stability, where he states your statement provided that the positive orbit $\mathscr{O}^+(x_0)$ of the initial point $x_0$ is contained in a compact set. See Theorem 1.
This might sounds a bit unsatisfactory, but I don't think that some assumption of compactness, which might depend on the initial point, can be avoided in this generality without using any other special structure of the equations.
Note, that the missing compactness of the orbit is also the essential bit in the counter-example of AVK.