Chaos Theory: Topological conjugation

Sorry about the late posts all of a sudden, just started work and have little to no time most days!

In this post (on chaos theory) we’re going to be looking at the relationship between logistic map and tent map; both maps we have explored previously in this category. We’ll be proving that the maps are identical under iteration (topologically conjugate).

This occurs for the case \mu = 2 and r = 4 for the tent map and logistic map respectively, denoting that x_n = \dfrac{2}{\pi}\sin^{-1} \sqrt{y_n} then given the tent map (see below) we can procede to prove this relationship the two maps share.The tent map is given by

T(x)=    \begin{cases}    2 x_{n}, \qquad \qquad 0 \leq x \leq \frac{1}{2}\\    2 - 2 x_{n}, \qquad \frac{1}{2} \leq x \leq 1    \end{cases}

Case 1: x_{n+1} = 2 x_n

\dfrac{2}{\pi}\sin^{-1} \sqrt{y_{n+1}} = \dfrac{4}{\pi}\sin^{-1} \sqrt{y_n} \Longrightarrow \sqrt{y_{n+1}} = \sin(2sin^{-1} \sqrt{y_n}).

Using \theta = \sin^{-1} \sqrt{y_n} then it follows that \sin \theta = \sqrt{y_n} and cos \theta = \sqrt{1-y_n}.

Substituting gives \sqrt{y_{n+1}} = \sin 2 \theta = 2 \sin \theta \cos \theta = 2 y_n (\sqrt{1-y_n}).

Squaring the result gives the logistic map y_{n+1} = 4 y_n (1 - y_n).

Case 2: x_{n+1} = 2 - 2 x_n

\dfrac{2}{\pi}\sin^{-1} \sqrt{y_{n+1}} = 2 - \dfrac{4}{\pi}\sin^{-1} \sqrt{y_n} \Longrightarrow \sqrt{y_{n+1}} = \sin \pi - \sin(sin^{-1} \sqrt{y_n}).

We know that \sin \pi = 0. Similarly, we’ll use the same substitution \theta.

Substituting gives \sqrt{y_{n+1}} = - \sin 2 \theta = - 2 \sin \theta \cos \theta = - 2 y_n (\sqrt{1-y_n}).

Squaring the result gives the logistic map y_{n+1} = 4 y_n (1 - y_n).

With this result we can conclude that if the tent map has chaotic orbits then the logistic map must also have chaotic orbits.

P.S. I’m aware that sometimes the mathematics being explained here doesn’t view properly, just refresh the page and it should fix the problem. Thanks!