System with a Lyapunov function over $\mathbb{R}^n$ but not globally asymptotically stable 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 $\mathbb{R}^n$.
 A: 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.
A: Interesting question!
The issue is that some trajectories may tend to infinity.
I don't think I've ever seen an explicit example, but (with some effort!) I managed to come up with one myself.
Take
$$
V(x,y) = \frac{x^2}{1+x^2} + y^2
,
$$
which is clearly positive definite, but doesn't tend to infinity as $\sqrt{x^2+y^2}\to \infty$ (indeed, $V(x,0)\to 1$ as $x \to \pm\infty$).
The level set $\{ V(x,y)=1 \}$ is given by $y=\pm 1/\sqrt{1+x^2}$, a pair of curves which extend out to infinity, approaching the $x$ axis. Between these two curves, you have level sets $\{ V(x,y)=c \}$ for $c \in (0,1)$ which are closed curves encircling the origin, and $\{ V(x,y)=0 \}$ which is just the origin.

Now consider the system
$$
\dot x = x \, \frac{3x^2y^2-1}{x^2 y^2+1}
,\qquad
\dot y = -y
.
$$
The idea here is that on the curve $xy=1$ (which for large positive $x$ lies just above the level curve $y=1/\sqrt{1+x^2}$) the system becomes $\dot x=x$, $\dot y=-y$, which has a solution that stays on that curve. That is,
$$
x(t)=e^t
,\qquad
y(t)=e^{-t}
$$
is a particular solution of our system, and it doesn't tend to the equilibrium $(x,y)=(0,0)$, so the system is not globally asymptotically stable.
And on the other hand, the system becomes $\dot x \approx -x$, $\dot y=-y$ when we are close to the origin, so it's locally asymptotically stable.
It remains to show that $\dot V$ is really negative definite:
$$
\begin{split}
\dot V
&
= \frac{\partial V}{\partial x} \, \dot x + \frac{\partial V}{\partial y} \, \dot y
\\ &
= \frac{2x}{(1+x^2)^2} \, x \, \frac{3x^2 y^2-1}{x^2 y^2+1} + 2y \, (-y)
\\ &
= \frac{-2}{(1+x^2)^2 (1+x^2 y^2)}
\biggl(
x^2 (1-3x^2 y^2) + y^2 (1+x^2)^2 (1+x^2 y^2)
\biggr)
\\ &
= \frac{-2}{(1+x^2)^2 (1+x^2 y^2)}
\biggl(
x^2
- 2 x^4 y^2
+ x^6 y^4
+ y^2
+ 2 x^2 y^2 
+ x^2 y^4
+ 2 x^4 y^4
\biggr)\\ &
= \frac{-2}{(1+x^2)^2 (1+x^2 y^2)}
\biggl(
x^2 (1 - x^2 y^2)^2
+ y^2 (1+2x^2)(1+x^2 y^2)
\biggr)
,
\end{split}
$$
which is clearly negative away from the origin. Done!
(Remark: The rewriting of $\dot V$ in the last step is due to a helpful comment by @SampleTime, which simplified the argument greatly. See the edit history for my own original version, which was much uglier!)
