# iterated map (differential equations)Watch

#1
Consider the iterated map xn+1=rxn(1-xn2).
Show that for 0<r<3√3/2 if 0 ≤xn≤1 then 0≤xn+1≤1. Show that if
r<1 then the only fixed point in (0,1) is zero, and that this is stable. When
r>1 there is another fixed point in (0,1). Find the value of this fixed point
(as a function of r). For which values of r is it stable, and for which values is it unstable? What would you expect to happen when r>2?
0
4 years ago
#2
Hi there from Warwick, my name is Jesse a handsome man, I'm sure you've seen me in your lectures , I'm stuck on the exact same question would you believe it.
0
#3
(Original post by Dilzo999)
Hi there from Warwick, my name is Jesse a handsome man, I'm sure you've seen me in your lectures , I'm stuck on the exact same question would you believe it.
Everyone that I've asked is stuck on it and I was hoping I would be that one person with the brilliant answer .. But nope. I hate this module so much :-(
0
4 years ago
#4
(Original post by ellemay96)
Consider the iterated map xn+1=rxn(1-xn2).
Show that for 0<r<3√3/2 if 0 ≤xn≤1 then 0≤xn+1≤1. Show that if
r<1 then the only fixed point in (0,1) is zero, and that this is stable. When
r>1 there is another fixed point in (0,1). Find the value of this fixed point
(as a function of r). For which values of r is it stable, and for which values is it unstable? What would you expect to happen when r>2?
I have not done difference equations since I was at university and sometimes the vocabulary is different, but my suggestion is

in a difference equation you can find limits, say L, by setting un = un+1 =L

say here

L=rL(1-L2)

either L = 0

or 1=r(1-L2)
L=+/- ROOT(1 - 1/r)

but if r is between 0 and 1 there will be no more solutions except L =0

etc

(PS: I stand corrected if I am suggesting non sensible things)
0
4 years ago
#5
(Original post by ellemay96)
Consider the iterated map xn+1=rxn(1-xn2).
Show that for 0<r<3√3/2 if 0 ≤xn≤1 then 0≤xn+1≤1.
For the first part, for r in the range specified, in order for to be in [0,1], what is the restriction on the value of , i.e. what must it's range be restricted to?

Now show that implies lies in that range, and hence
4 years ago
#6
stability implies that

(UNstability implies that ) for fixed point

here, stability condition gives:

if then MUST for stability to hold (since we KNOW r is +ve and then we have: . , which must mean that x=0)

if you solve the equation: you get the fixed points - one of which is zero, the other is the +ve root TeeEm has mentioned, (both in the range we wish) of x in terms of r.

The +ve root value of x (fixed point) you have found as a function of r goes into the:

1). stability equation (simplifies (verify this) to: (3root(3)/2 approx = 2.6)

2). UNstability equation (simplifies (verify this) to: OR for THIS point. We reject the latter value, as anything

satisfying it gives us, if we substitute back into x=sqrt(1-1/r).., a complex number.

I will admit, its a while since i have done this, and the last question kind of confuses me - its apparently leading to chaotic iterations! (although i would have said i expect it to diverge to infinity)

This plot is just 30 iterations for
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