Conjugate Maps in Chaos Theory

Overview

In chaos theory, the conjugacy of maps is akin to isometry and isomorphism, and indeed, in a more general context of dynamics, it is synonymous with homeomorphism itself.

¹ Although not exactly the same according to some textbooks, their purpose is precisely the same. As is the case with mathematics, the idea is to verify a property in a simpler computation context first and then preserve that property where the actual proof is necessary.

Definition²

For two maps defined in $X$, if there exists a continuous bijection $C$ such that $C \circ f = g \circ C$ is satisfied, then $f$ and $g$ are said to be conjugate.

Theorems³

Let’s assume for all $x$, it holds that $g \left( C(x) \right) = C \left( f(x) \right)$.

• [1]: If $x$ is a period-$k$ point of $f$, then $C(x)$ is a period-$k$ point of $g$.
• [2]: If in the period-$k$ orbit of $f$, $C’ \ne 0$ holds $$\left( g^{k} \right) ' \left( C (x) \right) = \left( f^{k} \right) ' (x)$$

Explanation

These two theorems imply that conjugacy maintains irrespective of iterating the maps or taking derivatives. This means if it is easier to calculate the Lyapunov exponent in one system, it should be equally feasible in the system conjugate to it.

Examples⁴

A good example is showing that the logistic map has a chaotic orbit.

For example, the tent map $T : [0,1] \to [0,1]$ is defined as $T(x) = 1 - | 1 - 2x|$, and the logistic map $G$ is defined as part of the logistic family as $G (x) := g_{4} (x) = 4x(1-x)$. In this case

$$C(x) : = {{ 1- \cos \pi x} \over { 2 }}$$

is a continuous bijection that exists such that $T$ and $G$ are conjugate. It can be easily verified by actually calculating that

$$G(C(x)) = \sin^2 \pi x = C ( T(x) )$$

The pictures above show the graphs obtained by repeatedly applying the tent map and the process of finding the period-$k$ points of $T$ by finding the intersection points with $y=x$. This demonstrates that for all $k \in \mathbb{N}$, there exists a period-$k$ orbit for $T$, and due to the existence of $C$ and theorem [1], it can also be shown that for all $k \in \mathbb{N}$, there exists a period-$k$ orbit for $G$.

Meanwhile, as $[0,1]$ almost everywhere $\ln \left| \left( T^{k} (x) \right)' \right| = \ln 2 > 0$, the period point $x$ of $T$ is a source, and the period point $C(X)$ of $G$ is also a source. Therefore, the period orbit corresponding to the period orbit of $T$ $\left\{ x , \cdots , \right\}$ cannot be asymptotically periodic for $G$ $\left\{ C(x) , \cdots \right\}$, and according to theorem [2], it guarantees that the Lyapunov exponent is positive.

This ultimately means that the logistic map has a chaotic orbit. While this indirect proof might feel difficult, it’s significantly easier than directly proving that the logistic map has a chaotic orbit.

Proof

Proof of theorem [1]

Assuming that $C \left( f^{k-1} (x) \right) = g^{k-1} \left( C (x) \right)$ holds,

$$ \begin{align*} C \left( f^{k} (x) \right) =& g \left[ C \left( f^{k-1} (x) \right) \right] \\ =& g \left[ g^{k-1} \left( C (x) \right) \right] \\ =& g^{k} \left( C (x) \right) \end{align*} $$

Meanwhile, when $k=1$, since $f^{1} (x) = x$,

$$ g \left( C(x) \right) = C \left( f(x) \right) = C (x) $$

By mathematical induction for all $k \in \mathbb{N}$,

$$ C \left( f^{k} (x) \right) = g^{k} \left( C (x) \right) $$

If $x$ is a period-$k$ point of $f$, since $f^{k} (x) = x$,

$$ g^{k} \left( C (x) \right) = C(x) $$

Thus, $C(x)$ is a period-$k$ point of $g$.

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Proof of theorem [2]

Let’s consider $x$ is a period-$k$ point of $f$.

From the proof of theorem [1],

$$ g^{k} \left( C (x) \right) = C \left( f^{k} (x) \right) $$

By the chain rule,

$$ \left( g^{k} \right)' \left( C(x) \right) C’ (x) = C ' (x) \left( f^{k} \right)' (x) $$

Since $C ' (x) \ne 0$, by cancelling the terms on both sides,

$$ \left( g^{k} \right)' \left( C(x) \right) = \left( f^{k} \right)' (x) $$

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Code

Below is code written in R to draw graphs of the tent map $T$, the logistic map $G$, and $T^{k}$.

tent<-function(x) {1 - abs(1-2*x)}
logistic<-function(x) {4*x*(1-x)}
win.graph(8,4); par(mfrow=c(1,2))
plot(tent,main='Tent Map T')
plot(logistic,main='Logistic Map G\')
 
win.graph(9,3.5); par(mfrow=c(1,3))
plot(tent,main='T',xlab='x',ylab='y'); abline(0,1)
plot(seq(0,1,len=1000),tent(tent(seq(0,1,len=1000))),main='T^2',type='l',xlab='x',ylab='y');abline(0,1)
plot(seq(0,1,len=1000),tent(tent(tent(tent(tent(seq(0,1,len=1000))))))
     ,main='T^k',type='l',xlab='x',ylab='y');abline(0,1)