How does this difference equation (discrete function map) behave?

Question

Consider the iterated map

$x_{n+1} = rx_n(1 - x_n^2)$

As this is non-linear, I'm a bit confused on how to approach it's analysis. I'd like to find out/justify the following things I found out with plotting and numeric testing:

1. Is it true that $[x_n \in (0,1) \implies x_{n+1} \in (0,1)], \ \forall r \in (0,\frac{3\sqrt{3}}{2})$. Why?

2. This iterated map has two fixed points, which are $x_0=0$ and $x_0=\sqrt{1-\frac{1}{r}}$. Why does this second fixed point only exist when $r \gt 1$? How do I determine it's stability, which depends on r?

3. How do I show that $x_0=0$ is an attractive fixed point?

4. What happens when $r \gt 2$?

How must I procede? Either hints or solutions are welcome.

Answer 1

Clearly, $x_{n+1}>0$. On the other hand, by the inequality between geometric and quadratic means, $$x_{n+1}<\frac {3\sqrt 3}{2}x_n(1-x_n^2)=\frac {3\sqrt 3}{2\sqrt{2}}\cdot \sqrt{2}x_n\cdot\sqrt{1-x_n^2}\cdot \sqrt{1-x_n^2}\le $$ $$\frac {3\sqrt 3}{2\sqrt{2}}\left(\frac{2x_n^2+1-x_n^2+1-x_n^2} 3 \right)^{3/2}= \frac {3\sqrt 3}{2\sqrt{2}}\left(\frac23 \right)^{3/2}=1.$$

Fixed point of the iterated map are exactly real roots of the equation $x=rx(1-x^2)$. For any $r$ the equation has a root $0$. Since $f’(0)=r$, this point is attractive when $|r|<1$ and repelling when $|r|>1$. Moreover, when $|r|\le 1$ and $|x_n|\le 1$ then $|x_{n+1}|\le |x_n|$, so the sequence $|\{x_n\}|$ is non-increasing and thus it converges to some number $a$. Since $a=ra(1-a^2)$, we have $a=0$.

It is easy to check that two other roots exist iff $1/r<1$ and the are equal $\pm \sqrt{1-\frac{1}{r}}$. The value of $f’(x_0)$ at any of these roots $x_0$ equals $3-2r$, so $x_0$ is attractive when $1<r<2$ and repelling when $r<0$ and $r>2$. When $r=2$ then put $f(x)=rx(1-x^2)$.

Since $f(x_0)=x_0$ and $f’(x_0)=0$, by Taylor's formula in Lagrange’s form we have $$x_{n+1}-x_0=f(x_n)-x_0=f(x_0)+(x_n-x_0)f’(x_0)+\frac{f’’(c)}{2}(x_n-x_0)^2-x_0=$$ $$\frac{f’’(c)}{2}(x_n-x_0)^2=-3c(x_n-x_0)^2,$$ for some $c$ between $x_0$ and $x_n$. Since $x_0=\pm \frac{1}{\sqrt{2}}$, we have $|x_{n+1}-x_0|\le |x_n-x_0|/2$ provided $|x_n-x_0|\le\frac 16$, so the point $x_0$ is attractive.

Answer 2

Let's draw the functions $$ \left\{ \matrix{ f(x) = x\left( {1 - x^2 } \right) \hfill \cr g(x) = x/r \hfill \cr} \right. $$ both of which are odd, and the crossing point $C$ on the first quadrant $$ x/r = x\left( {1 - x^2 } \right)\quad \Rightarrow \quad C = \left( {\sqrt {{{r - 1} \over r}} ,\,{1 \over r}\sqrt {{{r - 1} \over r}} } \right) $$ the other being the origin and the symmetric of $C$, then the maximum $M$ of $f(x)$ $$ M = \left( {{{\sqrt 3 } \over 3},\;2{{\sqrt 3 } \over 9}} \right) $$ and the point $L$ at which $x/r = M_y$ $$ L = \left( {2r{{\sqrt 3 } \over 9},\;2{{\sqrt 3 } \over 9}} \right) $$ and finally the point $S$ at which $f(x) = - M_y$ for positive $x$ $$ S = \left( {2{{\sqrt 3 } \over 3},\; - 2{{\sqrt 3 } \over 9}} \right) $$

The recurrence is represented by the path shown in blue.

Then it is easy to understand what happens when, at varying $r$ the point $L$:

• lies before the point $I$ , i.e. $r \le 1$, the stable point is the origin;
• $L \in (I,M]$, i.e. $1 < r \le 3/2$, the stable point is $M$;
• $L \in (M,S]$, i.e. $3/2< r \le 2$, the stable point is $C$;
• $L \in (S, \infty )$, i.e. $2< r $, the sequence diverges, the last terms oscillating in sign;