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Math Help - Population

  1. #1
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    Population

    A population p(t) satisfies dp/dt = f(p), where f(p) = kp (1-p/a)(1-p/b)

    The population lies in the interval 0<=p<=b

    k,a,b are positive constants with b>a


    Determine all equilibrium points.

    For this part f(p)=0 so p=0, p=a, p=b

    Then I have to sketch f as a function of p. I know by expanding f(p) that is a cubic equation but I don't have any data about k,a,b except that they are all >0.
    I found f'(p) and solved it to be equal to zero to find the stationary points, i found an answer but was a complete mess. I don't know what we are expected to sketch exactly without any further information.
    I will appreciate any help. Thanks in advance!
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  2. #2
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    Without specific values for k, a, and b, all you can say is that the graph passes through (0, 0), rises to a maximum somewhere between x= 0 and x= a, goes down to (a, 0), then down to a minimum somewhere between a and b, the back up to (b, 0).
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  3. #3
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    Thank you very much, I deduced that but how do I justify my sketch concerning the stationary points? What can I say to justify as much as possible my sketch?
    I know it's a cubic equation and everyone knows that it goes like that, but I want a more appropriate mathematical justification if it is possible. Thanks again for the help!
    And another two questions if it is possible. I have to
    1) find for which values of p(0) does p(t)->a as t-> infinity
    2) sketch p as a function of t for the following initial conditions: p(0)=0, p(0)=1/2*a, p(0) = a, p(0)=1/2*(a+b), p(0)=b

    My question especially for 2) is how to sketch p without solving explicitly the differential equation, which I don't know how to solve since it is a third degree non separable differential equation. Thank you very much again! Appreciate it!
    Last edited by Darkprince; March 10th 2011 at 04:46 PM.
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  4. #4
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    1) when p(0)=0 then dp/dt=0 and the solution is p=0.
    y&#39; - y &#40;1-y&#41;&#40;1-y&#47;2&#41; &#61; 0, y&#40;0&#41;&#61;0 - Wolfram|Alpha
    2)when 0<p(0)<a then p goes to a because dp/dt>0.
    y&#39; - y &#40;1-y&#41;&#40;1-y&#47;2&#41; &#61; 0, y&#40;0&#41;&#61;0.5 - Wolfram|Alpha
    3)when a<p(0)<b then p goes to a because dp/dt<0.
    3)when p(0)>b then p goes to infinity because dp/dt>0.
    You can try various initial conditions.
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  5. #5
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    Thank you for your reply, but how can I obtain the solution of a nonlinear differential equation? Also is there a way to sketch them without solving explicitly the diff. equation? Thanks in advance!
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