I'm trying to solve a second-order, non-linear, non-homogenous difference equation. I realize these can be very complicated (or impossible) to solve on paper, but I'm trying to characterize some properties of it. In particular, parametric conditions under which the solution is increasing and concave.

The system itself is as follows:

\begin{eqnarray} x_{n+1} &=& a+ x_{n} + F(y_n) \\ y_n &=&-\pi_n+(r+\delta_f)x_n+n\delta_c(x_n-x_{n-1}) \end{eqnarray}

where $F$ is the concave and increasing mapping $F:y_n\mapsto \big({y_n}^\rho + b^\rho\big)^{1/\rho}$ with $\frac{1}{2}<\rho<1$; $\pi_n>0$ is some increasing sequence of $n$; and $a,b,r,\delta_c,\delta_f>0$ are all parameters.

The two boundary conditions are the following. On the left, $x_0=0$ and $x_1 = a+T(\kappa)$ for some $\kappa>0$. On the right, $\lim_{n\rightarrow\infty} y_n = 0$ (or, equivalently, $\lim_{n\rightarrow\infty} (x_n-x_{n-1}) = a+b$).

I'm interested in cases where $x_{n+1}>x_{n}$ [i.e. $\{x_n\}$ increasing] and $(x_{n+1}-x_{n})<(x_{n}-x_{n-1})$ [i.e. $\{x_n\}$ concave]. Equivalently stated in terms of $y_n$, I need $\{y_n\}$ to be a decreasing sequence of positive numbers.

I've been painfully trying to check algebraically when these cases emerge conditional on parameters, but haven't got very far with pen and paper or in the computer.

I understand it's a complicated system, but I'd really appreciate if anyone could point me in the right direction to study these type of systems.

Thank you in advance!

  • $\begingroup$ If we let $\rho=1$ and $\pi_n=0$, the system becomes linear and has an explicit solution. It can probably also be solved for $\rho=1$ and arbitrary $\pi_n$ without too much difficulty. Are these limiting cases of any interest? $\endgroup$ – Kajelad Jul 18 '17 at 17:51
  • $\begingroup$ Hi, thanks so much for your answer. The case $\rho \rightarrow 1$ is something I had already thought about, and I should be able to say some things about the process in that case. But ideally I'd like to have something to say about the $\rho<1$ case as well. It'd be great to hear your thoughts. Thanks again $\endgroup$ – Pau Jul 18 '17 at 18:34

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