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Let $X:= \mathbb{R}^2 \backslash(0,0)$ and $f: X \rightarrow X,\\ \;\;\;\;\; p \mapsto p + |p|e^{i|p|}$ be a continuous map on $X$ with the euclidean metric. According to [Devaney's definition of chaos][1], a chaotic dynamical system must fulfill the following:

  1. $f$ is transitive
  2. the periodic points points of $f$ are dense in $X$

I did a numerical simulation of $f$ for various starting values and it seems to be chaotic, but this chaos might be due to the finite precision computers offer in such simulations. Still I think that $f$ is chaotic. Think of what $f$ is doing to a point $p$: it moves p relative to p in some direction which is proportional to the distance between $d(p,\mathcal{O})$, a length of $d(p,\mathcal{O})$. The direction part is sometimes stable and sometimes chaotic (as is shown in this Youtube video of the trajectory for some initial starting value). I think it is stable when $p$ is close to [some sort of spiral][3] and unstable when farthest away from any point on the spiral. I also made a density map: density map of f but I'm not entirely sure if the map is exactly the same. I may be taking the square root of the distance, something like $p \mapsto p + \sqrt{|p|} e^{i \sqrt{|p|}}$.

Anyways, I would like to prove or disprove the two [conditions of Devaney][1]. As I understand it, you get the third for free if 1 and 2 are fulfilled. Transitivity of f is given if for all non empty open subsets $U,V$ of $X$ there exists some $k \in \mathbb{N}$ such that $f^k(U) \cap V \neq \emptyset$ in words: for any starting point you will eventually get as close as you want to any other point after a finitely many applications of $f$. Well maybe not, because it's about open sets and not points. The problem is, I haven't the slightest clue how to prove this. Worry about 2 later.

Sorry had to take out Devaney's paper, since I can only post two links if my reputation is below 10... but a google search will give you it. It was published in JSTOR.

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Mark I.O.
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  • Just checking: Is it possible the mapping is defined by $f(p) = |p| e^{i|p|}$? As written, $f(\pi) = \pi + |\pi|e^{i|\pi|} = 0$, so that $f$ does not map $X$ to $X$. – Andrew D. Hwang Jul 07 '17 at 12:52
  • Yes, I think you're right. I think there is a sequence in $$X$$, arranged on a spiral, of which each point maps to $0$. In other words, there is at least another point $v_0 \in X$ such that $f(v_0) = \pi$ but also another point $v_1$ which maps $f(v_1) = v_0$. On a side-note, I'm no mathematician, I do this for fun so this could be total bull manure. – Mark I.O. Aug 04 '17 at 13:12

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