Topological Conjugation between two dynamical systems

[selig5560]

Homework Statement

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

Homework 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)

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.

 

[pasmith]

Homework Statement

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

Homework 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)

The Attempt at a Solution

h:[0, 1] → [-1, 1]
h(x) = cos(∏x)

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

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.

Cosine is even: \cos(2\pi(1 - x)) = \cos (-2\pi x) = \cos (2\pi x).

EDIT: did you consider the straightforward h : [0,1] \to [-1,1]: x \mapsto 2x - 1?

 

[selig5560]

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.

 

[selig5560]

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 \neq -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

 

[pasmith]

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 \neq -4x+3 = 1-2(2x-1) = g(h(x))

You should have g(h(x)) = 1 + 2h(x) when x \leq \frac 12, because h(x) \leq 0 and |h(x)| = -h(x) when x \leq \frac12.

 

[selig5560]

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?
Selig