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I have a sequence $\{x_n\}_{n\ge 0}$ with $x_0>0$, controlled by the difference inequality: $$x_{n+1}\le ax_n^2+b$$ where, $a,b>0$. Had $b$ been $0$ and $a<1$, I would find $x_n\to 0$ as $n\to \infty$.

However, the presence of $b$ makes finding closed form next to impossible, except maybe for some specialized values of $b$. But I am not interested in closed forms, I am interested only in the necessary conditions on $a, b$ for the convergence or divergence of the sequence, or at least finding an upper bound for $\lim_{n\to \infty}x_n$, if the limit exists. It seems that if $a,b<1$, the sequence becomes bounded, and an upper bound is possible (although not sure if a closed form of the upper bound exists), and that if $a>1$, the sequence might diverge to infinity, at least the right hand side of the inequality seems to do so; but what about the following cases:

1) $a<1,b>1$,

2) $a=1,b>0$

Please direct me to references. I think probably the literature of nonlinear dynamics would be helpful in answering questions like this, but it would be really helpful to get pointers for specific topics in that field that might help answering this question. Thanks in advance.

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  • $\begingroup$ If there is no restriction that ensures that $x_n>0$ for each $n$, there is no upper bound: you can just choose $x_1<-2/a$ and then $x_{n+1}=2^{n+1}/a$ for example. If you impose $x_n>0$, then iterating $f(x)=ax^2+b$, you obtain monotonic convergence to a fixed point (possibly the fixed point at $\infty$). This fixed point is the upper bound. $\endgroup$
    – Anthony Quas
    Commented Jan 11, 2017 at 7:33
  • $\begingroup$ Yes, I forgot to add, each of $x_n, n\ge 1$ is positive as well. Yes, I know that the upper bound is going to follow the forward orbit of the map $f$, but I want to ask, is there any way to predict how the orbit is going to behave if we have knowledge about $a,b$? In other words, I should have really been asking about how does the forward orbit of the map $f$ behaves, given conditions on $a,b$. $\endgroup$
    – Samrat Mukhopadhyay
    Commented Jan 11, 2017 at 7:47
  • $\begingroup$ I think this really isn't the right forum for this question, but you should look up "staircase diagram". $\endgroup$
    – Anthony Quas
    Commented Jan 11, 2017 at 7:57
  • $\begingroup$ You forgot to say what $x_0$ is. Even when $b=0$, the sequence sometimes converges, sometimes does not depending on $x_0$. $\endgroup$
    – Alexandre Eremenko
    Commented Jan 11, 2017 at 16:00
  • $\begingroup$ @AlexandreEremenko, but doesn't $a\in(0,1),b=0$, imply that $\lim_{n\to \infty}x_n\to 0$ (assuming each of $x_n>0$) for all positive $x_0$? $\endgroup$
    – Samrat Mukhopadhyay
    Commented Jan 11, 2017 at 18:33

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