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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

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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.

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Is there an obvious Lyapunov function for the system? – Andrey Rekalo Jan 21 '11 at 14:13
Denis: What would be a simple $2$-dimensional version of your example? – Did Jan 21 '11 at 14:30
You can make a $C^\infty$ vector field in the plane that is 0 on the unit disk, and where all orbits outside the disk spiral counterclockwise inward toward the disk. There's a $C^{infty} Lyapunov function that is identically 0 on the disk, and a monotone function of distance from the origin outside the disk. – Bill Thurston Jan 21 '11 at 14:44
@Andrey. Take any smooth function $V(X)=g(|X|^2)$ with the property that $g'<0$ over $(0,1)$ and $g'>0$ over $(1,+\infty)$. It may be flat at $0$. – Denis Serre Jan 21 '11 at 15:06
@Denis Serre: Yes, but $V$ should also satisfy the conditions $V(0)=0$ and $V(X)>0$ for $X\neq 0$. These are not compatible with the property that $g'(x)<0$ for $x\in(0,1)$. – Andrey Rekalo Jan 21 '11 at 15:18

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.

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