Lyapunov Exponent

1. Apr 29, 2013

mliuzzolino

1. The problem statement, all variables and given/known data

Calculate the Lyapunov exponent for the linear map xn+1= rxn.

2. Relevant equations

λ = Lyapunov Exponent

λ = $\lim_{n \rightarrow \infty} \begin{bmatrix}\dfrac{1}{n} \sum_{i = 0}^{n - 1} ln|f'(x_i)| \end{bmatrix}$

3. The attempt at a solution

f'(x) = r.

λ = $\lim_{n \rightarrow \infty} \begin{bmatrix}\dfrac{1}{n} \sum_{i = 0}^{n - 1} ln|r| \end{bmatrix}$

= $\lim_{n \rightarrow \infty} \begin{bmatrix}\dfrac{ln(r)}{n} \end{bmatrix}$

This is where I'm a bit lost. Is λ = ∞, or is λ = ln(r)?

In another example, with the tent function $f(x) =\begin{cases}rx, \hspace{4mm} 0 \leq x \leq \dfrac{1}{2} \\ r - rx, \hspace{4mm} \dfrac{1}{2} \leq x \leq 1 \end{cases}$

λ = ln(r).

Is the Lyapunov exponent for both of these systems the same?

2. Apr 29, 2013

mliuzzolino

Anyone?

3. Apr 29, 2013

fzero

You haven't properly accounted for the sum here.

4. Apr 29, 2013

mliuzzolino

I have no idea how to really go about doing that. Any advice for starting it?

5. Apr 29, 2013

fzero

Since $\ln |r|$ is independent of $n$, it can be factored out of the sum. What is left over? How many terms appear in the sum?

6. Apr 29, 2013

mliuzzolino

If it's factored out isn't there nothing remaining?

As of today in class the professor mentioned something about n being the period, and a particular function he gave had period 2, so it was 1/2 times the sum. I'm just completely lost on this and the professor blows through everything at lightning speed as if we've all had the same 30+ years of experience in the topic as he has.

My brain is completely fried at this point.

7. May 1, 2013

fzero

Factoring out the $\ln|r|$, we can write this as

$$\lambda = \lim_{n \rightarrow \infty} \frac{\ln |r|}{n} \sum_{i = 0}^{n - 1} 1.$$

You're left with computing the sum, which depends on $n$.

The Lyapunov exponent is defined for any orbit $\{x_0,\cdots\}$. A periodic orbit is an orbit $\{z_0,\cdots, z_{k-1}\}$ such that $z_{i+k-1}=z_i$ for all $i\geq 0$ ($k-1$ looks a bit weird here, but that's because we started counting at 0). This means that the orbit goes from

$$z_0\rightarrow z_1 \rightarrow \cdots \rightarrow z_{k-1} \rightarrow z_0 \rightarrow \cdots,$$

i.e. it is a loop of sorts. This is why we call the orbit periodic and the integer $k$ is the period. If the prof used $n$ for the period, I can see how that can be confusing.

For a periodic orbit, the Lyapunov exponent reduces because of the periodicity. We can write

$$\lambda = \lim_{n\rightarrow \infty} \frac{1}{n} \sum_{i=0}^{n-1} \ln | f'(z_i)|,$$

as you have in the OP. Now let $n = Nk$ and we take $n\rightarrow \infty$ by taking the limit $N\rightarrow \infty$. The convenience of this is that, because of the periodicity,

$$\sum_{i=0}^{n-1} \ln | f'(z_i)| = N \sum_{i=0}^{k-1} \ln | f'(z_i)|.$$

What's happened here is that, for $n$ much, much larger than $k$, we essentially have an integer number $N$ of repeats of the periodic orbit. The "round-off" error should vanish in the limit that $n$ and $N$ go to infinity.

Going back to the formula for the exponent, we then find that

$$\lambda = \lim_{N\rightarrow \infty} \frac{1}{Nk} N \sum_{i=0}^{k-1} \ln | f'(z_i)| =\frac{1}{k} \sum_{i=0}^{k-1} \ln | f'(z_i)| .$$

Which is equivalent to the formula you describe, but perhaps written in a less confusing way by using $k$ instead of $n$.

8. May 2, 2013

mliuzzolino

That drastically cleared things up for me. Thank you!