Steady states/systems of differential equations/Phase portraits

sid9221
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http://dl.dropbox.com/u/33103477/2007%2010b.png

I'm can't get my head around this question, there doesn't seem to be enough information to compute a steady state ?

Any ideas on how to begin ??
 
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Hi sid9221,

First of all, can't you solve explicitly for p(t) and r(t)? The ODE for r(t) should look very familiar.

This problem seems a bit strange. Typically to find the steady states you set ##\frac{dp}{dt} = 0## and ##\frac{dr}{dt} = 0##. These two conditions give you steady states ##(p_*, r_*)##. What do you get in this case?

It seems very fishy to me that ##\frac{dp}{dt}## and ##\frac{dr}{dt}## don't seem to depend on p at all...

Could we see part a) of the question, too?
 
Part a) is completely unrelated.

It's some nonsense about the population of fish...
 
sid9221 said:
Part a) is completely unrelated.

It's some nonsense about the population of fish...

The situations may be different, but are the models similar?

Did you solve for p(t) and r(t) or find the condition given by setting the derivatives to zero?
 
http://dl.dropbox.com/u/33103477/Untitled.png
 
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Thanks for posting the full problem.

Anyway, if you set ##\frac{dp}{dt}=0## and ##\frac{dr}{dt} = 0##, what must r equal?

Also, if ##r'(t) = - \beta \, r(t)##, what function must ##r(t)## be?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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