# Difference equations and stability

1. Oct 9, 2005

### Benny

Hi I am unsure about stability of fixed points here is an example.

$$x_{n + 1} = x_n$$

There are fixed points at x = 0 and x = 1. In general when talking about difference equations and whether a fixed point is stable or unstable, does this refer to points in a neighbourhood of those points? For example if for some difference equation there is a fixed point at x = 0.12345678, and it is unstable. Does this mean that if I repeatedly 'apply' the recurrence relation(an example is the one I provided although it probably isn't the best example for my question) to a point near that fixed point for example x = 0.12, then successive values that I obtain will 'diverge' from the initial value of 0.12?

I just wanted to check because I need to know this in order to complete my assignment. The assignment questions I have are completely different to my example. Basically I just need to verify that I have the correct definition for 'stable' and 'unstable' fixed points for a recurrence relation. Any help appreciated.

2. Oct 9, 2005

### HallsofIvy

Are you sure of your example?? xn+1= xn has every point as fixed point, not just 1 and 0!
In order to determine the fixed points of of a difference equation, replace each xn, xn+1, etc by x and solve for x. Your example just gives x= x.

Yes, the concept of "stable" and "unstable" fixed points depends on what happens to points close to the fixed points.

A difference equation that does have 0 and 1 as fixed points is
xn+1= xn2. If x is a fixed point then setting xn= x will give xn+1= x so x= x2 which has solutions x= 0 and x= 1. If we look at points close to 0, we see that repeatedly squaring a number close to 0 gives a sequence that converges to 0: 0 is a stable fixed point. If, however, we repeatedly square a number close to 0 then either: (a) x< 1 so we get a sequence that converges to 0 or:(b) x> 1 so we get a sequence that diverges. Either way, the sequence does not converge to or stay near 1. 1 is an unstable fixed point.

3. Oct 9, 2005

### Benny

Yeah my example should have more than just 1 and zero as fixed points.

I also have the result that if $$x_{n + 1} = f\left( {x_n } \right),\left| {f'\left( p \right)} \right| < 1$$ where p is a fixed point then x = p is a stable point. So if x_(n+1) = 1.5x_(n) then any fixed points(if there are any) will be unstable because (1.5x)' = 1.5 > 1 for all x? Another thing I don't really understand is the concept of unstable. From the material that I've got, there's a mention of intervals being streched if a point is unstable. But what interval is this?

I just have one more question. What is a periodic point? I've tried looking at a few websites but they don't really give a definition of periodic and non-periodic points.

4. Oct 9, 2005

### Physics Monkey

Consider the map $$x_{n+1} = |x_n -1|$$. If you start out at x=0 then you jump to x=1 then back to x=0, etc. In this case x=0 is a periodic point of period 2 because if $$x_n = 0$$ then $$x_{n+2} = 0$$. Are there any other periodic points?

The alternating behavior you see here is called a limit cycle. A limit cycle is stable if, as you perturb away from it, the system moves back towards the limit cycle. Is the limit cycle $$0,1,0$$ stable?

Edit: The example I gave is kind of trivial, but you can try out the new definitions on a slightly less trivial map like $$x_{n+1} = (x_n - 1)^2$$.

Last edited: Oct 9, 2005
5. Oct 9, 2005

### HallsofIvy

Yes. And it is easy to see that the only fixed point is x= 0. If x is exactly 0, no matter how many times you multiply it by 1.5, you still get 0! However, if x is not 0, no matter how close it is, multiplying by 1.5 steadily moves it away from 0.

Some small interval around your fixed point. In the example above, if you take $(-\delta, \delta)$, after one iteration of xn+1= 1.5xn applied to every point in that interval, you have $(-1.5\delta, 1.5\delta)$- the interval has been stretched by exactly a factor of 1.5- points are moving away from the fixed point.

Consider this iteration on [0,1]: xn+1= 2xn mod 1. That "mod 1" means "if the result of multiplying by 2 is larger than or equal to 1, drop the interger part". For example, if x0= 2/5, then x1= 2(2/5)= 4/5. To find x2, multiply by 2 again: 8/5= 1.6 so x2= 0.6= 3/5. To find x3, multiply by 2 again: 6/5= 1.2 so x3= 0.2= 1/5. x4= 2(1/5)= 2/5, the number we started with. Since we will now do exactly the same thing again, all numbers past here will be the same:
2/5, 4/5, 3/5, 1/5, 2/5, 4/5, 3/5, 1/5, 2/5, 4/5, 3/5, 1/5, 2/5, ....
That sequence is "periodic with period 4" since every 4th number repeats. We say that the starting point, 2/5, is a "periodic point". Of course, if we had started with 4/5, 3/5, or 1/5, we would have gotten the same sequence. If a sequence is periodic, every point in it is a periodic point.

Last edited by a moderator: Oct 9, 2005
6. Oct 9, 2005

### Benny

HallsofIvy - Thanks again for your explanations.

Physics monkey - I'm thinking that one would be another periodic point. The point x = 0.5 seems to just stay there so I'm not sure about that one. As for the second one, what happens if the starting point is x_0 = 12. Apply the recurrence relation over and over again it results in values which get closer to zero and eventually reach zero if I am reading it correctly. After that, further applications of the recurrence relation results in values which alternate between zero and one like before. Starting at x_0 = 12, the values appear to converge to zero rather than 'move further away' from zero.

7. Oct 9, 2005

### Physics Monkey

You are right, 1 is another periodic point of the first map, in fact it is part of the limit cycle $$0,1,0$$. Of course x=.5 is a fixed point, so you can call that a periodic point of period 1 if you like. There are still more periodic points for the first map.

Also, on the second map, how did you decide that x_0 = 12 heads towards zero? Doesn't the sequence go like 12, (12-1)^2 = 121, (121-1)^2 = 14400, etc. It seems to me that it diverges.

Last edited: Oct 9, 2005
8. Oct 9, 2005

### saltydog

$$x_{n+1}=ax_n(1-x_n)$$

from the perspective of it's Feigenbaum plot? You know, a is a parameter you vary from 1 to 4 and then run the iterator for some random starting values between 0 and 1. Sometimes it settles to a single point, sometimes to a set of "Periodic Points", sometimes it's chaotic.

9. Oct 9, 2005

### Benny

Physics Monkey - I was a little unclear when I said "the second one", I replied at around the same time as your edit was made. I was looking at x_(n+1) = |x_n - 1| and the the part where you mentioned limit cycles being stable. I figured that 12 is reasonably far away from zero and one but as I repeatedly apply the recurrence relation to it, the values I obtain decrease from zero. From there, oscillation between zero and one occurs. I wasn't really sure about that one. It was just in response to the limit cycle question you asked me.

Saltydog - I've mainly been trying to just understand the basics (such as definitions). It is for an assignment where the associated material is covered in only three classes. But even when I look up the most basic things, such as what "chaotic" means I find things on "transitivity, density" and other terms which I've rarely come across or never even heard of before. So while I'm grateful for your help, I don't really understand the last bit.

10. Oct 9, 2005

### Physics Monkey

Benny, glad to hear the mistake was mine. x0 = 12 does indeed end up in the 0,1 limit cycle, but what about x0 = 12.1? The first example I gave is a little pathological in that it has an infinite number of limit cycles (try to find them all) but none of them are stable. In fact when you perturb one limit cycle, you just end up on another limit cycle! I would suggest playing with the second map I gave, which has only one limit cycle and trying to figure out if the cycle is stable.

11. Oct 9, 2005

### Benny

Ok I'll try the second one at another time and see if I can find some interesting behaviour. But right now I need to get some sleep. :zzz: