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Topological Conjugation between two dynamical systems

  1. Dec 10, 2013 #1
    1. The problem statement, all variables and given/known data

    Find a topological conjugation between g(x) and T(x) where g and T are mappings (both tent maps [graphically speaking])

    2. Relevant equations


    g:[-1, 1] → [-1,1]
    g(x) = 1-2|x|

    T:[0,1] → [0, 1]
    T(x) = 2x when x ≤ 1/2 and 2(1-x) when x ≥ 1/2

    h ° T = g ° h (homeomorphism)


    3. The attempt at a solution

    h:[0, 1] → [-1, 1]
    h(x) = cos(∏x)

    when T(x) = 2x and x ≤ 1/2:

    cos(∏*(2x)) = sin^2(∏x) - cos^(∏x) = 1 - 2cos^2(∏x) = 1 - 2|cos^2(∏x) = -cos(2∏x)

    when T(x) = 2(1-x) and x ≥ 1/2

    cos(2∏(1-x) = cos(2∏-2x∏) = -cos^2(∏x)+sin^2(∏x) = 1-2|cos^2(∏x) + sin^2(∏x) = 1-2|cos(2∏x)|

    The above attempt I know is incorrect because after I introduce the absolute value brackets I do not get the desired result.

    This is when I get stuck. I have tried many different variations of trig functions to act as the conjugator between g and T(x), however I have had no luck (after many hours.) I know for that it would be easy to find a homeomorphism if it wasnt for the |x| part of the 1-2|x| dynamical system (tent map.) I do not think that I am supposed to find a conjugation between two tent maps (explicit i,.e. conjugating two piecewise functions) because it seems that would be highly redundant. If anyone could provide some assistance that would be great.
     
    Last edited: Dec 10, 2013
  2. jcsd
  3. Dec 10, 2013 #2

    pasmith

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    Homework Helper

    You want [itex]h(0) = -1[/itex] (fixed points must map to fixed points) and [itex]h(1) = 1[/itex].

    Cosine is even: [itex]\cos(2\pi(1 - x)) = \cos (-2\pi x) = \cos (2\pi x)[/itex].

    EDIT: did you consider the straightforward [itex]h : [0,1] \to [-1,1]: x \mapsto 2x - 1[/itex]?
     
    Last edited: Dec 10, 2013
  4. Dec 10, 2013 #3
    Thanks for the reply. Good point on the cosine being even - Forgot about that. I tried your "straightforward" approach, however no luok so far. If I'm not mistaken the homeomoprhism has to be algebraic at this point. I don't see a way of simplifying trig functions with those | | being part of the DDS.
     
    Last edited: Dec 10, 2013
  5. Dec 11, 2013 #4
    I used the affine transform you provdd ie. 2x-1 and it did not yield a conjugation. my work:

    g(h(x)) = 1-2|2x-1|
    h(T(x)) =

    x <= 1/2: 2(2x) - 1 = 4x - 1 [itex]\neq[/itex] -4x+3 = 1-2(2x-1) = g(h(x))

    x >= 1/2: 2(2-2x) - 1 = 4 - 4x - 1 = -4x + 3 = 1-2(2x-1) = 1-4x + 3 = -4x + 3 = g(h(x))

    this is of course using h(x) = 2x-1

    where h(0) = -1 and h(1) = 1
     
    Last edited: Dec 12, 2013
  6. Dec 12, 2013 #5

    pasmith

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    Homework Helper

    You should have [itex]g(h(x)) = 1 + 2h(x)[/itex] when [itex]x \leq \frac 12[/itex], because [itex]h(x) \leq 0[/itex] and [itex]|h(x)| = -h(x)[/itex] when [itex]x \leq \frac12[/itex].
     
  7. Dec 12, 2013 #6
    THanks for the reply I will try it out.

    EDIT: It works now. However, out of self-interest I would like to understand a bit more abour your statement. Isnt |h(x)| = -h(x) when x < 0? In this case x <= 1/2, so how would it be true that |h(x)| = -h(x) (when the condition is when its x < 0?

    Sincerely,

    Selig
     
    Last edited: Dec 12, 2013
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