# Find the equilibrium for a logistic recurrence relation.

• Sep 24th 2011, 01:18 AM
alyosha2
Find the equilibrium for a logistic recurrence relation.
How would I go about showing the reccurence relation fits the logisitcal model and hecne finding the equilibrium?

say

http://quicklatex.com/cache3/ql_2fa0...84ee8f7_l3.png

I can see the first part seems like it fits but how do i got about making a constant figure like the one above into Attachment 22388?

Thanks.
• Sep 24th 2011, 02:59 AM
chisigma
Re: Find the equilibrium for a logistic recurrence relation.
Quote:

Originally Posted by alyosha2
How would I go about showing the reccurence relation fits the logisitcal model and hecne finding the equilibrium?

say

http://quicklatex.com/cache3/ql_2fa0...84ee8f7_l3.png

I can see the first part seems like it fits but how do i got about making a constant figure like the one above into Attachment 22388?

Thanks.

The images show two different equations...

$\displaystyle p_{n+1}-p_{n}=.24\ p_{n}-2000$ (1)

$\displaystyle p_{n} +1-p_{n}=.24\ p_{n}-2000$ (2)

Which is the 'right' equation?... (1) or (2)?...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Sep 24th 2011, 03:57 AM
alyosha2
Re: Find the equilibrium for a logistic recurrence relation.
Oh, sorry. It's (1) I just messed the subscript up on (2).
• Sep 24th 2011, 04:20 AM
chisigma
Re: Find the equilibrium for a logistic recurrence relation.
Very well!...we write the difference equation as...

$\displaystyle \Delta_{n}= p_{n+1}-p_{n}= .24\ p_{n}-2000= f(p_{n})$ (1)

... and neceesary it is 'started' by an 'initial condition' $\displaystyle p_{0}=p$. The procedure for solving a difference equation of the type (1) is described in...

http://www.mathhelpforum.com/math-he...-i-188482.html

In Your case the f(x) has only one zero in $\displaystyle x_{0}=8333.33...$ but, because is $\displaystyle f^{'}(x_{0})>0$ , $\displaystyle x_{0}$ is a repulsive fixed point and that means that $\displaystyle x_{0}$ is a unstable equilibrium point. In other words, for $\displaystyle p=x_{0}$ the solution of (1) is the constant sequence $\displaystyle p_{n}=x_{0}$, but for all other p the solution will diverge...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Sep 24th 2011, 05:01 AM
alyosha2
Re: Find the equilibrium for a logistic recurrence relation.
Quote:

Originally Posted by chisigma
Very well!...we write the difference equation as...

$\displaystyle \Delta_{n}= p_{n-1}-p_{n}= .24\ p_{n}-2000= f(p_{n})$ (1)

... and neceesary it is 'started' by an 'initial condition' $\displaystyle p_{0}=p$. The procedure for solving a difference equation of the type (1) is described in...

http://www.mathhelpforum.com/math-he...-i-188482.html

In Your case the f(x) has only one zero in $\displaystyle x_{0}=8333.33...$ but, because is $\displaystyle f^{'}(x_{0})>0$ , $\displaystyle x_{0}$ is a repulsive fixed point and that means that $\displaystyle x_{0}$ is a unstable equilibrium point. In other words, for $\displaystyle p=x_{0}$ the solution of (1) is the constant sequence $\displaystyle p_{n}=x_{0}$, but for all other p the solution will diverge...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

I'm sorry, I'm not really sure I understand. What I'm trying to do is convert the given recurrence relation into the form for the logistic model

http://quicklatex.com/cache3/ql_eb38...dc35743_l3.png

I didin't think you would need the first term but in this case it is $\displaystyle P_0$ = 25000.

thanks.
• Sep 24th 2011, 06:24 AM
chisigma
Re: Find the equilibrium for a logistic recurrence relation.
Quote:

Originally Posted by alyosha2
I'm sorry, I'm not really sure I understand. What I'm trying to do is convert the given recurrence relation into the form for the logistic model

http://quicklatex.com/cache3/ql_eb38...dc35743_l3.png

I didin't think you would need the first term but in this case it is $\displaystyle P_0$ = 25000.

thanks.

If I understand correctly You have the difference equation...

$\displaystyle p_{n+1}-p_{n}= .24\ p_{n}-2000$ (1)

$\displaystyle p_{n+1}-p_{n}= r\ p_{n}\ (1-\frac{p_{n}}{E})$ (2)

Isn't it?...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Sep 24th 2011, 06:39 AM
alyosha2
Re: Find the equilibrium for a logistic recurrence relation.
Quote:

Originally Posted by chisigma
If I understand correctly You have the difference equation...

$\displaystyle p_{n+1}-p_{n}= .24\ p_{n}-2000$ (1)

$\displaystyle p_{n+1}-p_{n}= r\ p_{n}\ (1-\frac{p_{n}}{E})$ (2)

Isn't it?...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

yes, that's correct.
• Sep 24th 2011, 06:48 AM
chisigma
Re: Find the equilibrium for a logistic recurrence relation.
Quote:

Originally Posted by alyosha2
yes, that's correct.

All right!... the only problem is that doesn't exist any pair r and E which 'convert' the equation (1) into equation (2) or vice versa...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Sep 24th 2011, 06:56 AM
alyosha2
Re: Find the equilibrium for a logistic recurrence relation.
Quote:

Originally Posted by chisigma
All right!... the only problem is that doesn't exist any pair r and E which 'convert' the equation (1) into equation (2) or vice versa...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

So it's impossible to find the equilibrium?
• Sep 24th 2011, 07:19 AM
chisigma
Re: Find the equilibrium for a logistic recurrence relation.
Quote:

Originally Posted by alyosha2
So it's impossible to find the equilibrium?

... that seems to be (Thinking)...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$