Suppose I have a discrete dynamical system with a finite set X of states, and suppose I want to prove that every state of X ends up, sooner or later, in a subset Z under the dynamics of the system. Then a natural proof strategy is to break this "convergence" statement into two parts, by first showing that every state of X eventually ends up in a subset Y, and then showing that every state in Y ends up in Z (with Z $\subset$ Y $\subset$ X). This might simplify the problem quite a bit.

My question is,

Under what conditions will the above proof strategy work for a continuous dynamical system $\dot{x}=f(x)$ ($x \in R^n$)?

The main issue I have is that it might take infinite time to reach Y. To see how things may go wrong if $f$ is not continuous, consider the following example:

  • $\dot{x} = -1-x$, $\quad$ for $x<-1$,
  • $\dot{x}=-x$, $\quad\quad$ for $-1 \le x \le 1$, $\quad$ and
  • $\dot{x}=1-x$, $\quad$ for $x>1$.

In this case every $x$ in $[-1,1]$ converges to zero, and every point outside $[-1,1]$ converges to either $-1$ or $1$, but it takes infinite time to do so and it is not true that all points in $R$ converge to zero.

Smoothness of $f$ might be enough to rule out this behavior; I am unable to construct a counterexample with $f$ continuous. Is it actually enough, and if so, why?

  • $\begingroup$ I don't understand the question. The "strategy" you use, depends only in the group property, so, for flows it holds too. Do you have a concrete example showing what you want? $\endgroup$
    – rpotrie
    Commented Nov 23, 2010 at 22:08
  • $\begingroup$ It is not entirely clear what you are asking for. Maybe you can provide an outrageous example for when the intuition you describe fails? $\endgroup$ Commented Nov 23, 2010 at 22:11
  • $\begingroup$ I tried to clarify the question. Thanks for bearing with me while I learn more basic dynamical systems theory! $\endgroup$
    – Vincenzo
    Commented Nov 24, 2010 at 10:15
  • $\begingroup$ In your example it is possible to converge to a point in $Y = [-1,1]$ without entering $Y$. If you require $Y$ to be an open set, this can not happen. Perhaps this is what you are getting at. $\endgroup$
    – Mike Hall
    Commented Nov 24, 2010 at 15:02

1 Answer 1


If the equation is defined in $\mathbb{R}$, then $f$ continuous is enough. It is clear that if a point of $X$ does not reach the set $Y$ in finite time, then $f$ vanishes in this limit point. Since every point of $Y$ converges to $Z$, you get that this limit point must thus belong to $Z$.

In higher dimensions, I would say that the same holds only that there is the subtelty that you don't have a priori a dynamical system defined (due to lack of uniqueness of solutions, the group condition fails) however, if $\dot x = f(x)$ integrates into a flow (which must then be a continuous flow, this will hold for example if $f$ is locally Lipchitz) an analogous argument works (now $f$ may not vanish at the omega-limit set of a trayectory, but if $x$ does not enter $Y$, then it will converge to a point in $Z$ anyway).

  • $\begingroup$ Thanks rpotrie. For the (high-dimensional) locally Lipschitz case, do you know if the argument you just sketched is formalized in some standard text, or maybe some standard theorem from which it descends immediately? $\endgroup$
    – Vincenzo
    Commented Nov 25, 2010 at 10:04
  • 1
    $\begingroup$ Let me try to be more specific: If the equation integrates into a flow (that is, $\varphi_{t+s}(x)=\varphi_t(\varphi_s(x))$) we get that we can define the omega-limit of a point (as the set of points to which the orbit converges) which is an invariant set and the omega limit set of a omega limit set is contained in itself. Consider $x\in X$, then its omega limit intersects $Y$, but since it is invariant and every point of $Y$ has its omega limit intersecting $Z$ we conclude that the omega limit set of $x$ must intersect $Z$ also. $\endgroup$
    – rpotrie
    Commented Nov 25, 2010 at 10:11
  • 1
    $\begingroup$ If you prefer to put the hipotesis that the omega limit of every point of $X$ is contained (instead of intersects) in $Y$ the same argument works. You should look at the kew word omega limit, see en.wikipedia.org/wiki/Limit_set $\endgroup$
    – rpotrie
    Commented Nov 25, 2010 at 10:13
  • $\begingroup$ Still thinking about this, sorry... $\omega(\omega(X))$ is certainly contained in $\omega(X)$, but is it always the case that $\omega(\omega(X))=\omega(X)$? If yes, why? And if no, then your argument only proves that $\omega(X)$ and $Z$ intersect, while I need to argue that $\omega(X)$ is contained in $Z$... $\endgroup$
    – Vincenzo
    Commented Dec 10, 2010 at 17:54
  • $\begingroup$ It is easy to construct an example where $\omega(\omega(X))$ is strictly contained in $\omega(X)$ (consider a rotation of the circle wich attracts a neighborhood in the plane, and multiply the vector field by a bump function which vanishes in a point of the circle). So, if the question is stated as: Does $\omega(X) \subset Y$ and $\omega(Y) \subset Z$ imply $\omega(X) \subset Z$? The answer is no. I guess I had misunderstood the question when I answered it. $\endgroup$
    – rpotrie
    Commented Dec 18, 2010 at 17:24

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