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iterated map (differential equations)

Consider the iterated map x_n+1=rx_n(1-x_n²).
Show that for 0<r<3√3/2 if 0 ≤x_n≤1 then 0≤x_n+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?

Reply 1

Dilzo999

Hi there from Warwick, my name is Jesse a handsome man, I'm sure you've seen me in your lectures :wink:

, I'm stuck on the exact same question would you believe it.

(edited 9 years ago)

Reply 2

ellemay96

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 :wink:

, 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 :-(

Reply 3

TeeEm

Original post by ellemay96

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 u_n = u_n+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)

Reply 4

ghostwalker

Original post by ellemay96

Consider the iterated map x_n+1=rx_n(1-x_n²).
Show that for 0<r<3√3/2 if 0 ≤x_n≤1 then 0≤x_n+1≤1.

For the first part, for r in the range specified, in order for

x_{n+1}

to be in [0,1], what is the restriction on the value of

x_n(1-x_n^2)

, i.e. what must it's range be restricted to?

Now show that

x_n\in[0,1]

implies

x_n(1-x_n^2)

lies in that range, and hence

x_{n+1}\in[0,1]

Reply 5

Hasufel

stability implies that

\left | f ' (x_{0}) \right |<1

(UNstability implies that

\left | f ' (x_{0}) \right |>1

) for fixed point

x_{0}

here, stability condition gives:

\left | r(1-3x^{2}) \right |<1

r<1

then

x

MUST

=0

for stability to hold (since we KNOW r is +ve and then we have: .

r \left | (1-3x^{2}) \right |<1

, which must mean that x=0)

if you solve the equation:

x=rx(1-x^{2})

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:

\left |3-2r \right | <1 => 1<r<2

(3root(3)/2 approx = 2.6)

2). UNstability equation (simplifies (verify this) to:

\left |3-2r \right | >1 => r>2

r<1

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

This plot is just 30 iterations for