TITLE: System with a Lyapunov function over $\mathbb{R}^n$ but not globally asymptotically stable
QUESTION [19 upvotes]: I'd like to find an example of a system $\dot{\mathbf{x}} = F(\mathbf{x})$, where $\mathbf{x} = \mathbf{0}$ is an equilibrium point, with a corresponding Lyapunov function $V(\mathbf{x})$ that satisfies:
(1) $V(\mathbf{x}) > 0 \; \forall \mathbf{x} \in \mathbb{R}^n - \{\mathbf{0}\}$, $V(\mathbf{0}) = 0$
(2) $\dot V (\mathbf{x}) < 0 \; \forall \mathbf{x} \in \mathbb{R}^n - \{\mathbf{0}\}$, $\dot V(\mathbf{0}) = 0$
But where the origin is only locally asymptotically stable.
Context:
According to every text book that I've read on the subject, in order to guarantee global asymptotic stability, (1) and (2) are not enough: I also need 
$\lim_{||\mathbf{x}|| \to \infty} V(\mathbf{x}) = \infty$ (or, equivalently, that every subset of V is bounded). However, when I read the proof of local asymptotic stability, it seems to me that if (1) and (2) apply to all of $\mathbb{R}^n$, then asymptotic stability should also apply to all of $\mathbb{R}^n$.

REPLY [4 votes]: Just for the record, here's another example that I just found on p. 109 of the book Stability of Motion by Wolfgang Hahn (1967), where it is credited to a 1952 paper in Russian by Barbashin & Krasovskii:
$$
\dot x = -\frac{6x}{(1+x^2)^2} + 2y
,\qquad
\dot y = -\frac{2(x+y)}{(1+x^2)^2}.
$$
With the same(!) $V$ as in the example I came up with in my other answer,
$$
V(x,y) = \frac{x^2}{1+x^2} + y^2
,
$$
one computes
$$
\dot V = -\frac{4}{(1+x^2)^2} \left( \frac{3x^2}{(1+x^2)^2} + y^2 \right)
,
$$
so $V$ is positive definite and $\dot V$ is negative definite, and hence $V$ is a strong Lyapunov function on all of $\mathbf{R}^2$.
But Hahn shows that trajectories of this system cannot cross the curve $$
y=\frac{2}{x-\sqrt2}
\quad
(x>\sqrt2)
$$
in the direction towards the origin.
(He compares $\dot y/\dot x$ to the slope of the curve.)
Hence not all solutions tend to the origin.