0

I have the following map defined on the unit square $0\leq x,y \leq 1$: $$B(x,y)=\begin{cases}(2x,y/3) & 0\leq x< 1/2\\ \\(2x-1,(2+y)/3)& 1/2\leq x < 1 \end{cases}$$

How do I find two points in this unit square which form a period-2 orbit of $B$?

John Hughes
  • 93,729
FlyGuy
  • 11
  • Why do you think that there is a period-2 orbit for $B$? – John Hughes Jul 04 '19 at 17:50
  • I never said I did. That's what I want to check and I'm unsure how. Do I just check for points $(x,y)$ where operating the map twice gives them back? – FlyGuy Jul 04 '19 at 17:54
  • @FlyGuy Yes, you do. Just also note that fixed points (solutions of $f(z) = z$) are also solutions of $f(f(z)) = z$ and you should exclude them. – Evgeny Jul 04 '19 at 17:56
  • Hmmm...asking how to find something generally suggests to me that you think the thing is there to be found. Perhaps you meant to ask "How do I determine whether there's a period-2 orbit, and if there is one, how do I find it?" – John Hughes Jul 04 '19 at 17:57
  • @JohnHughes To be precise, it is stated in the following way in the question "Find points $y_1=(y_{11},y_{12})$ and $y_2=(y_{21},y_{22})$ which form a period-2 orbit of $B$, $y_j=y_{2n+j}$ with $n$ integer. Is this periodic orbit stable?". So suppose there do exist two points which are a period-2 orbit. – FlyGuy Jul 04 '19 at 18:00
  • Ah, now I understand. This is just an exercise of some sort. Typically, if you want help with such things, it's a good idea to include the ideas you've had and the things you've tried in the question itself, so that we don't just think you're trying to have us do your homework for you. – John Hughes Jul 04 '19 at 18:12
  • @JohnHughes I understand. My apologies. I'm trying right now by simply checking for $B^2(x^\prime,y^\prime)=(x^\prime,y^\prime)$. I'll update on that matter. Thanks! – FlyGuy Jul 04 '19 at 18:14
  • OK, I think I found them. The points are: $y_1=(1/3,3/4)$ and $y_2=(2/3,1/4)$. Clearly, I also got the two fixed points in the process. – FlyGuy Jul 04 '19 at 18:27
  • Two questions if I may: (1) The way the question is worded is kind of bothering me. Why is it stated in the question that $y_{2n+j}=y_j$? Doesn't that make it looking for a period-$2n$ orbit? Or now that I found the "fundamental" period-2 orbit, then all of its multiples are periodic as well? If so, why? $(2)$ Assuming my period-2 points are right, is it weird that the sum of each of the $x$ values and $y$ values equal to 1? What does that mean? – FlyGuy Jul 04 '19 at 18:43
  • I think that it means "$y_{i} = y_{i + 2n}$ for every integer $n$." Written less formally, it means that $y_0 = y_2 = y_4 = \ldots$ and $y_1 = y_3 = y_5 = \ldots$, i.e., it's just a restatement of the idea that this is an order with a period that's either $1$ or $2$. (Presumably the fact that it's not $1$ is meant to be covered by asking you to find two points that constitute a period-2 orbit. As for the second question: mere coincidence. You could easily restructure the problem so that this is not true. – John Hughes Jul 04 '19 at 20:02
  • Thanks John. Lastly, I'd like to ask how do I check the stability of a periodic orbit in a discrete system? – FlyGuy Jul 04 '19 at 20:39

0 Answers0