[Math] Proving a function is chaotic on an interval

chaos theorydynamical systems

I'm self studying from the book 'First course in chaotic dynamical systems' and am having a hard time grasping how to prove that a function is chaotic.

For the function we have $T(x) = 2x$ for $x \leq \frac{1}{2}$ and $T(x) = 2 – 2x$ for $x > \frac{1}{2}$. I need to prove that it is chaotic on $[0, 1]$.

The book says the three conditions we need to prove are as follows:
1. Periodic points for $F$ are dense
2. $F$ is transitive
3. $F$ depends sensitively on initial conditions*

*The book notes that #3 follows from the first two.

Now I couldn't figure out a way to even tackle this problem via proof so I thought a visual aid would help. So I created the following graph. Each $x$ value is a seed and each $y$ value is a point on an orbit. 200 points on the orbit of each seed were computed.

enter image description here

I think the plot might show proof that $T$ is transitive since it's looking like we can always find an orbit that gets near two arbitrary points.

However, I don't see how we could have a set of periodic points dense in $[0, 1]$ since all we have are the (period 1) fixed points $0$ and $\frac{2}{3}$. Hence, all this allows me to say is that $T$ is chaotic for period 1, right?

My questions:
How do I go about proving that $T$ is chaotic?
What is going on in my plot? There seem to be some pretty interesting patterns/trends in there and I can't figure out what they mean.

Plot code: 

import matplotlib.pyplot as plt

plt.axis([0, 1, 0, 1])
plt.xlabel('x0')
plt.ylabel('y')

x = 0
n = 1

while x <= 1:
    for i in range(0, 500):
        if x >= 0 and x <= 1:
           plt.plot([.005625*(n - 1)], [x], 'o', markersize = .0625, 
                     markerfacecolor = 'green', markeredgecolor = 'green')
        print 'x: ' + str(.005625*(n - 1)) + ' y: ' + str(x)
        if x <= .5:
            x = 2*x
        if x > .5:
            x = 2 - 2*x

    x = n*.005625
    n += 1

Note that .005625 is essentially arbitrary.

Best Answer

An encoding of $T$ uses the fact that every number $x$ in $[0,1]$ can be represented as $$x=\frac12-\frac12\sum_{n\geqslant1}\frac{s_n(x)}{2^n},\qquad s_n(x)=\pm1,$$ where the sequence $(s_n(x))$ is unique except for the dyadic numbers $x$ in $(0,1)$. Then, $$s_n(Tx)=s_1(x)s_{n+1}(x)\qquad (n\geqslant1),$$ hence, for every $k\geqslant1$, $$s_n(T^kx)=s_k(x)s_{n+k}(x)\qquad (n\geqslant1).$$ Several properties of $T$ are easily seen on this representation:

  • The periodic points of period $1$ are such that $s_n(x)=(s_1(x))^n$ for every $n$, that is, $x=0$ or $x=\frac23$.
  • The periodic points of period $2$ but not $1$ are such that $s_{n+2}(x)=s_2(x)s_{n}(x)$ for every $n$ but $s_2(x)\ne1$, that is, $x=\frac25$ or $x=\frac45$.
  • There are exactly $2^k$ periodic points of period $k$.
  • If the expansion $(s_n(x))$ is ultimately periodic in the sense that $s_k(x)=1$ and $s_{n+k}(x)=s_n(x)$ for some fixed $k$ and every $n$ large enough, then $x$ is ultimately periodic for $T$ in the sense that $T^{n+k}x=T^nx$ for every $n$ large enough.

To get a periodic point $z$ close to some $x$, fix some $k\geqslant1$ and define $(s_n(z))$ as follows:

  • $s_n(z)=s_n(x)$ for every $1\leqslant n\leqslant k-1$
  • $s_{n+k}(z)=s_k(z)s_n(z)$ for every $n\geqslant1$

The first condition ensures that $|x-z|\leqslant1/2^{k-1}$, the second condition that $T^{k}z=z$.

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