3
$\begingroup$

I was recently reading about chaos theory. Chaos is commonly defined to exhibit 3 properties:

  1. Sensitivity to initial conditions/Lyapunov coefficient is positive
  2. Exhibits topological mixing
  3. Dense periodic orbits

It can also be proven that 3+2 implies 1.

We also knew there are systems that satisfy only 1 but are non-chaotic, such as scaling $x_{n+1} \mapsto kx_n$ where $k \neq 0$ because every point diverges to infinty.

We also have systems which satisfy only 3 and thus also non-chaotic

An easy example of a non-chaotic system satisfying 1 and 3 is considering a phase space where there are periodic orbits of different sizes centered round rational points, such that no two periodic orbits intersect each other. Then the property of rationals ensures for arbitrarily small deviations in initial conditions, one ends up in completely different orbits and hence diverging trajectories without a given point ending up in all open sets.

However, is it possible to have topological mixing, such that every open set in the domain can end up intersecting every other open set, but it either does not have dense periodic orbits nor sensitivity to initial conditions. I don't see when one is allowed to have open sets smeared against each other, how can arbitrarily close trajectories not diverge exponentially.

  • Are there non-chaotic systems satisfy only 2?

  • Are there non-chaotic systems satisfy 1 and 2?

$\endgroup$
  • 1
    $\begingroup$ Some dynamical systems exhibit shearing ; that is, the differentials are transvections. Then they can stretch open sets, and wrap them around so as to get topological mixing, while still having zero Lyapunov exponents. An example would be the horocycle flow on a compact surface of negative curvature, which exhibits 2, but neither 1 nor 3. $\endgroup$ – D. Thomine Aug 6 '18 at 10:38
0
$\begingroup$

Let $S$ be the unit circle, leaving out the points $e^{2\pi i\alpha}$ with $\alpha$ rational, let $f(e^{2\pi ix})=e^{4\pi ix}$. Then $f$ satisfies 1 and 2 but has no periodic orbits.

Vellekoop and Berglund, On intervals: transitivity$\rightarrow$chaos, Amer Math Monthly 101 (1994) 353-355, proved that if $f$ is a continuous map on an interval, then 2 implies chaos.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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