I am struggling to solve this recurrence equation,
$$a_{n}(1-sa_{n-1}^{2})+sa_{n-1}^2-a_{n-1}=0$$
where the parameter $s\in[0,1]$ and the initial condition $a_{0} > 0$ is close to 1. I have graphed it in matlab and it decreases really fast even if $a_{0} = 0.9$. Any idea on how to solve this will be much appreciated. Thanks!
Note: if $a_{0}=1$, then $a_{n}=1, \forall n$.
You have $a_n = \dfrac{a_{n-1} - sa_{n-1}^2}{1- sa_{n-1}^{2}} = a_{n-1} \left(\dfrac{1-sa_{n-1}}{1- sa_{n-1}^{2}}\right)$ providing that $sa_{n-1}^2 \not = 1$
If $0 \le a_{n} \le 1$ and $0 \le s \le 1$ and $sa_{n-1}^2 \not = 1$, then the sequence is decreasing (possibly weakly) and bounded below by $0$, so there must be a limit
Solving $a(1-sa^{2})+sa^2-a=0$ leads to $a=0$ or $1$, so these are the only possible fixed points. So, given the sequence is decreasing,
if $0 \le a_{n} \lt 1$ and $0 \le s \le 1$ then the limit must be $0$
if $a_{n} = 1$ and $0 \le s \lt 1$ then the limit must be $1$
