0
$\begingroup$

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?

$\endgroup$
4
  • $\begingroup$ 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

1
$\begingroup$

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).

$\endgroup$
5
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.