0
$\begingroup$

I'm studying some basic behaviour of the logistic map $x_{n+1}=r x_n(1-x_n)$, with $x_0\in (0,1]$ and $r\in (0,4]$, for a project. I can't seem to figure out why the map does not converge (i.e. has no fixed points) for $r>3$ or how I could prove that it doesn't.

Edit: striked a part of the text that, apparently, wasn't really what I meant to ask but to make sure it can still be read, to potentially make sense of some answers which might have been given to the original question.

  • 1
    One way to see this: compute the fixpoints, i.e. solutions to $x = xr(1-x)$ and then compute $f'(x)$ for the fixpoints. Note that if $|f'(x)| > 1$ then the fixpoints are repellers. Show that for $r>3$ both fixpoint are repellers. – 2017-01-26

1 Answers 1

1

It still has a fixed point in $(0,1]$ for $r > 3$, namely $(r-1)/r$. But if $f(x) = r x (1-x)$, $f'((r-1)/r) = 2 - r$. If $r > 3$ this is less than $-1$, which implies that the fixed point is unstable. The other fixed point $0$ is unstable if $r > 1$.

Since both fixed points are unstable, the sequence $x_n$ can only converge if it hits a fixed point exactly. This can happen, e.g. if $x_0 = 1/r$ you get $x_1 = 1 - 1/r$, and then $x_n = 1 - 1/r$ for all $n \ge 1$.

  • 0
    Thank you for your answer! However, I do not understand why the bound on the derivative imply instability. – 2017-01-26
  • 0
    This is the basic result on stability for fixed points. For a recursion $x_{n+1} = f(x_n)$, a fixed point $p$ is stable if $|f'(p)| < 1$ and unstable if $|f'(p)| > 1$. This is because $f(x) - p = f'(p) (x - p) + o(x - p)$. Thus if $|f'(p)| > 1 + \epsilon > 1$, $|f(x) - p| \ge (1+\epsilon) |x-p|$ when $|x-p|$ is sufficiently small, while if $|f'(p)| < 1 - \epsilon < 1$, $|f(x) - p| \le (1-\epsilon) |x-p|$ when $|x-p|$ is sufficiently small. – 2017-01-27