Find the equilibrium for a logistic recurrence relation.

**alyosha2** · September 24th 2011, 01:18 AM

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

say

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

Find the equilibrium for a logistic recurrence relation.-ql_bf5e677d64f5a7bc68e51ac64ca44ce9_l3.png

?

Thanks.

**chisigma** · September 24th 2011, 02:59 AM

Originally Posted by alyosha2

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

say

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

Click image for larger version.

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?

Thanks.

The images show two different equations...

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

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

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

Kind regards

$\chi$ $\sigma$

**alyosha2** · September 24th 2011, 03:57 AM

Oh, sorry. It's (1) I just messed the subscript up on (2).

**chisigma** · September 24th 2011, 04:20 AM

Very well!...we write the difference equation as...

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

... and neceesary it is 'started' by an 'initial condition' $p_{0}=p$ . 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 $x_{0}=8333.33...$ but, because is $f^{'}(x_{0})>0$ , $x_{0}$ is a repulsive fixed point and that means that $x_{0}$ is a unstable equilibrium point. In other words, for $p=x_{0}$ the solution of (1) is the constant sequence $p_{n}=x_{0}$ , but for all other p the solution will diverge...

Kind regards

$\chi$ $\sigma$

**alyosha2** · September 24th 2011, 05:01 AM

Originally Posted by chisigma

Very well!...we write the difference equation as...

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

... and neceesary it is 'started' by an 'initial condition' $p_{0}=p$ . 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 $x_{0}=8333.33...$ but, because is $f^{'}(x_{0})>0$ , $x_{0}$ is a repulsive fixed point and that means that $x_{0}$ is a unstable equilibrium point. In other words, for $p=x_{0}$ the solution of (1) is the constant sequence $p_{n}=x_{0}$ , but for all other p the solution will diverge...

Kind regards

$\chi$ $\sigma$

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

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

thanks.

**chisigma** · September 24th 2011, 06:24 AM

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

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

thanks.

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

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

... and Your task is to 'convert' it in the form...

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

Isn't it?...

Kind regards

$\chi$ $\sigma$

**alyosha2** · September 24th 2011, 06:39 AM

Originally Posted by chisigma

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

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

... and Your task is to 'convert' it in the form...

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

Isn't it?...

Kind regards

$\chi$ $\sigma$

yes, that's correct.

**chisigma** · September 24th 2011, 06:48 AM

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

$\chi$ $\sigma$

**alyosha2** · September 24th 2011, 06:56 AM

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

$\chi$ $\sigma$

So it's impossible to find the equilibrium?

**chisigma** · September 24th 2011, 07:19 AM

Originally Posted by alyosha2

So it's impossible to find the equilibrium?

... that seems to be

...

Kind regards

$\chi$ $\sigma$

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