Non-isolated equilibrium points and the Lasalle invariance prinicple Hi!
Let's say that we have a dynamical system described by
$\dot{x} = f(x)$,
where f is some nonlinear function, which has several
equilibria. Assume that we have found a continously 
differentiable Liapunov function V such that
$\dot{V} = 0 \Rightarrow \dot{x} = 0$.
Then, assuming that V is radially unbounded, by the LaSalle
invariance principle we should be able to say that the system 
always converges to an equilibrium point. However, in some works 
I have seen the additional requirement that in order to show convergence,
all equilibrium points must be isolated, otherwise the system
could move indefinitely inside a connected set of equilibrium points.
Can that really be the case for the situation described above?
Doesn't $\dot{x} = 0$ mean that the system has "stopped" (assuming that 
$x$ completely describes the state of the system)? It seems to me that
in my case, the assumption about isolated equilibria is unnecessary.
Kind regards
Olav
 A: To illustrate Michael's answet, take the system
$$\dot x=(1-x^2-y^2)x-zy,\qquad \dot y=(1-x^2-y^2)y+zx,\qquad \dot z=-z^2.$$
In cylindrical coordinates, it writes
$$\dot r=(1-r^2)r,\qquad\dot\theta=z,\qquad \dot z=-z^2.$$
The first equation tells that $r(t)\rightarrow1$ as $t\rightarrow+\infty$. Besides, 
$$z(t)=\frac{z_0}{1+tz_0}.$$
If $z(0)=z_0$ is positive, the trajectory is defined for all $t>0$, and $\dot\theta$ is not integrable at $+\infty$, so that the solution spins infinitely many times towards the unit circle.
Upon Didier's request. here is a similar example, in the plane. Take two functions $r\mapsto h(r),k(r)$. Consider the system
$$\dot r=h(r),\qquad \dot\theta=k(r),$$
which rewrites
$$\dot x=\frac{h(r)}{r}x-k(r)y,\qquad\dot y=\frac{h(r)}{r}y+k(r)x.$$
Assume that $(r-1)h(r)<0$ for $r\ne 0,1$. Then you have a Lyapunov function $V(r)$, minimal at $r=1$. The rest points form the circle $\{r=1\}$. If $h$ is flat enough at $r=1$, the convergence $r\rightarrow1$ as $t\rightarrow+\infty$ is algebraic. Then if $k=r^2-1$, a trajectory spins infinitely many times.
A: There are also gradient flows with limit cycles. That is
$$
\dot x = - \nabla V(x)
$$
meaning that $V$ itself is a Lyapunov function by construction.
Such an example in polar form $V(r \cos(\theta),r\sin(\theta))=v(r,\theta)$ is given by
$$
v(r,\theta) = 
\begin{cases}
   \exp\Bigl(-\tfrac{1}{1-r^2}\Bigr) , & r < 1 \\
   \exp\Bigl(-\tfrac{1}{r^2-1}\Bigr) \sin\Bigl(\frac{1}{r-1} - \theta\Bigr) , & r>1  \\
0 & r=0
\end{cases}
$$
This is a non-analytic function (at $r=1$). The solution to the gradient flow initially at the origin stays there, for initial points with $r<1$ it converges to $r=1$ with $\theta$ staying constant. For initial $r=1$ solutions are stuck again. For initial $r>1$ there are two cases: For generic initial $\theta$ it converges to an outwards spiraling solution. However, for one specific initial choice $\theta^*=\theta^*(r)$, there is an inwards spiraling solution having the unit circle as limit cycle. See also the picture below.
I'm not sure if one could fix the example to have for generic initial values limit cycles, or if those for gradient systems are somehow special.

A: The system stops only in the limit as time goes to infinity. You could, for instance, have a circle consisting of equilibrium point, and as time goes to infinity, solutions spiral towards the circle with continually decreasing speed.
