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I wish to find the values for which the logistic map behaves as a contraction map

$$x_{n+1}=rx_n(1+x_n)\equiv F(x;r)$$

i.e, I wish to find for which $r$, the mapping above admits a unique fixed point $F(x^*;r)=x^*$ using the contraction map theorem.

It is easy to find the fixed points, and it exists in any text book, as well as Banach's fixed point theorem, however I have no idea on how to apply this theorem over the example of the logistic map.

The most helpful reference I've found is in here, but I still can't figure out what is required.

https://www.math.ucdavis.edu/~hunter/book/ch3.pdf

Can any one please suggest how to approach this problem?

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  • $\begingroup$ How about solving $x=rx(1+x)$ by hand? $\endgroup$ – John B May 9 at 21:13
  • $\begingroup$ @JohnB, I would like to show it via contraction mapping $\endgroup$ – jarhead May 10 at 6:48
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Banach's fixed-point theorem isn't sufficient to characterise all values of r. It claims only that if a map is a contraction map, then it has a unique fixed point. It does not speak of non-contraction maps.

So, provided that the logistic function $f \colon [0,1] \to [0,1]$ given by $f(x) = rx(1-x)$ is a contraction map, you can show it has a unique fixed point.

If $r < 1$, then $f$ as above is a contraction map (show that $f'(x) < 1$ on $[0,1]$). Hence for $r < 1$, it has a unique fixed point (it'll be $x = 0$, when the population dies out).

If $r \geq 1$, it is no longer a contraction map on $[0,1]$. For example, if $r = 2$, then $f(0) = 0$ and $f(1/4) = 3/8$. So $|f(0) - f(1/4)| = 3/8 > 1/4 = |0-1/4|$. Hence it is not a contraction map, and Banach tells us nothing. This shouldn't be too surprising, since for $r = 2$ it has two fixed points ($x = 0$ and $x = 1/2$).

I leave it as an exercise to show in general that it isn't a contraction map for $r \geq 1$. So for $r \geq 1$, the fixed point theorem doesn't tell you anything; alternate means are required.

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  • $\begingroup$ thank you for your answer. Can you please specify the alternate means you are referring to? $\endgroup$ – jarhead May 10 at 8:38
  • $\begingroup$ Consider that $rx(1-x) = x \iff rx(1-x)-x = 0$ - now you’re tasked with finding the roots of a quadratic polynomial. How many roots can a quadratic polynomial have and how can you decide whether they are in the interval [0,1]? You can do this with some basic high school algebra. $\endgroup$ – Chris May 11 at 12:01

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