# Homework Help: 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!