Solving the recurrence $a_{n+1}=(1-\alpha b_n)a_n$, $b_{n+1}=(1-\beta+\alpha a_n)b_n$

Question

Can anybody solve this recurrence relation?

$$\begin{align} a_1 &= p \\ b_1 &= 1 \\[4pt] a_{n+1} &=(1-\alpha b_n)a_n\\ b_{n+1} &=(1-\beta+\alpha a_n)b_n \end{align}$$

I think the fact that $a$ and $b$ are interrelated makes the problem extremely hard (or not). If you are interested please feel free to answer :-)

Answer 1

Initial remark: a particular case is treated here.

The first thing to do for such systems is to look for possible limits.

By continuity, if $\lim_{n\to +\infty}(a_n;b_n)=(a;b)$, we must have

$$\begin{cases}a&=&(1- \alpha b)a\\b&=&(1-\beta+\alpha a)b\end{cases} \iff \begin{cases}ab &=&0\\b(\alpha a-\beta)&=&0\end{cases}$$

(if we assume $\alpha \ne 0$).

As a consequence, if there is a limit $(a,b)$, necessarily, we must have $b=0$.

The fact that convergence occurs towards this type of solution is attested by simulation. See on the figure below the result of different simulations in the $(a;b)$ coordinate's plane.

Now that you have a neat idea of what happens, can you take it from here in a rigorous way ?

Fig. 1: Case $\alpha=0.1$ and $\beta=0.3$ for different values of $p$. For example the leftmost broken line begins at $(a_1;b_1)=(p;1)=(1/4;1)$ and connects the different $(a_n;b_n)$ issued from this initial seed with limit $(a,0)$. One notes a "backtracking" of the limit when $p>3$.

Please note that for different combinations, for example for $p=3, \ \alpha=0.7, \beta=0.3$ the convergence is towards $(a;b)=(0;0)$.

If you are interested in the future to write a simulation program, here is the one I have created for generating Fig. 1 (language: Matlab):

clear all;close all;hold on;
p=3;alp=0.3;bet=0.7; % conv to 0
p=3;alp=0.1;bet=0.3;
set(gcf,'color','w');
for q=1:2:19
    p=q/4;
    a=p,b=1;
    text(a,b+0.05,[num2str(q),'/4']);
    for k=1:24;
       X(k)=a;Y(k)=b;
       aprovi=(1-alp*b)*a;
       b=(1-bet+alp*a)*b;
       a=aprovi;
    end;
    plot(X,Y,'-ob');
end;