Convergence for a non-linear second order difference equation

Question

In my work, I need to study the convergence of sequence defined by the non-linear recurrence relation $$ u_0,u_1>0, \qquad \forall n\in \mathbb N, \; u_{n+2}=a\ln(1+u_n)+b\ln(1+u_{n+1}) $$ with fixed $a,b> 0$.

For $a=b=1$ Wolfram gives the limit $-1-2 W_{-1}\left(-\frac{1}{2\sqrt{e}}\right)$. (W is the Lambert function) https://en.m.wikipedia.org/wiki/Lambert_W_function. I'm looking for ideas to study convergence of this sequence. I would suspect that non-classical arguments are needed to do so.

Note: I have already asked the question on Mathematics Stack Exchange, but it was closed quickly for a reason that I do not understand.

Answer 1

I would suspect that non-classical arguments are needed to do so.

All you need to know is that $t\mapsto \frac 1{1+t}$ is a decreasing function, so for $0<x\le x'$ we have $\frac{\log(1+x')}{\log(1+x)}\le \frac {x'}x$. This immediately implies that the mapping $T:(x,y)\mapsto (y,a\log(1+x)+b\log(1+y))$ is non-expanding in the metric $d((x,y),(x',y'))=max(|\log x-\log x'|,|\log y-\log y'|)$ on $(0,+\infty)^2$ and $T^2$ is a weak contraction ($d(T^2p,T^2q)<d(p,q)$ if $p\ne q$).

The next step is to consider the equation $x=(a+b)\log(1+x)$ and notice that either $a+b\le 1$ (in which case the iterations trivially converge to $(0,0)$, i.e., "escape to infinity" in our metric) or $a+b>1$ in which case there is a positive solution $x_0$ of that equation and $T$ maps every compact ball $B(p_0,r)$ into itself where $p_0=(x_0,x_0)$ is a fixed point of $T$. If $(u_0,u_1)$ lies in that ball, we can apply the usual result about weak contractions on compact sets to conclude that we have convergence to $p_0$.

but it was closed quickly for a reason that I do not understand.

The closure reason cited on MSE is totally ridiculous IMHO. The very fact that you ask a well-posed mathematical question is a sufficient proof of "relevance to you" and nobody is obliged to verify the "relevance to the community" (whatever it might mean) when asking. The only possible reason for closure I see is that the question is rather trivial, but, given the usual amount of total junk floating on MSE, I doubt that it was what determined its fate. So, please, accept my apologies for the MSE users behavior, ignore this incident and keep asking :-)

Answer 2

If $f(x,y) = \ln(1+x) + \ln(1+y)$, and $p = -1 - 2 W_{-1}(-1/(2 \sqrt{e}))$, it is easy to verify that $f(p,p) = p$. Moreover, it appears numerically that $\|(x_3,x_4) - (p,p)\| < \|(x_1, x_2) - (p,p)\|$ where $x_3 = f(x_1, x_2)$ and $x_4 = f(x_2, x_3)$ and $(x_1, x_2)$ is sufficiently close to $(p,p)$. Here is a plot of $\|(x_3,x_4) - (p,p)\|^2/\|(x_1,x_2) - (p,p)\|^2$ as a function of $(x_1,x_2)$ for $0.1 \le x_1 \le 5$, $0.1 \le x_2 \le 5$.

If $(x_1,x_2)$ is in some circle centred at $(p,p)$ which is contained in the region where $\|(x_3,x_4) - (p,p)\| < \| (x_1, x_2) - (p,p)\|$, we will have $x_n \to p$ as $n \to \infty$.