Zone 1:
R > 1
both | r1 |  and | r2 | greater than 1.








When r1 and r2 are both positive, this case is quite simple : as soon as X(n) leaves the segment [B1;B2], we have | X(n+1) | > | X(n) |.
Moreover, it is sure (meaning with a probability equal to one) that if we wait long enough, one iterate X(n) will leave this interval.

The graph below shows this phenomenon.
 
 

Log(|X(n)|) vs n




Consequently, the system diverges absolutely (meaning independently of the probabilities p1 and p2) to infinity.

There is no attractor.
 

As it was said previously, the behaviour of the system is quite similar in the other "symetrical" domains.
However, few differences can be noticed.
This is why I thought it was useful to discuss these cases, just to convince ourselves that these differences were very small.
 
 

We can prove that in this case, there is also "pseudo-stable"  interval [B1-b1;B2+b2] in which the successive iterates may stay for a while (because of the alternance of the centers of repulsion), but out of which the modulus of X(n) increases in a monotonic way.
The boundaries of this pseudo-stable interval are defined through b1 and b2 :

b1=(B2-B1)*( | r2 | +1)/( | r1 | * | r2 | -1)
and
b2=(B2-B1)*( | r1 | +1)/( | r1 | * | r2 | -1).

Moreover, as in the first case, we are sure that if we wait long enough, X(n) will leave this segment and so, we can conclude that for any initial condition X(0), the system diverges and that there is no attractor.
 

The pictures below show an example for r1=r2=-1.1. For these values, with B1=-1 and B2=+1, we have b1=b2=20.

The first one clearly shows that after a finite number of iteration (~150), the growth of | X(n) | becomes monotonic.
 
 

Log(|X(n)|) vs n




We can see on the second graph what happends during the first iterations. The interaction between the two functions makes X(n) oscillating in the neighbourhood of B1 and B2.
However, as soon as | X(n) | becomes greater than 21, the system still oscillates (because r1 and r2 are negative) but we clearly have a monotonic increase of | X(n) |.
 
 

zoom on the first 130 iterations


This case is quite similar to the previous one, but this time the "pseudo-stability" interval is [B1-b1;B2] with:

b1=(B2-B1) * | r1 |.

We can remark that this case is less "stable" than when both r1 and r2 are negative and hence, there is of course no attractor.

The graph below shows what happends when r1=-1.2 and r2=1.05 (we have then b1=2.4):
A soon as the red curves leaves the interval [-3.4;1] the modulus of X(n) increases in a continuous way.
 
 

X(n) vs n

Everything happends like in the previous case, inverting r1 and r2.
There is still no attractor.
 
 

Conclusion:

We have clearly seen that in all these case ( | r1 | > 1 and | r2 | > 1 ), there can never be an attractor as the system diverges to inifinity in a monotonic way.
 
 



 

see you ...