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Modeling of a Single Species Population (F.O. D.F.Q.)

  1. Mar 29, 2014 #1
    First off, this is a typed paper project, and I am not asking for anyone to solve my work or to give me the answers. Just please point me in the right direction or help me through any parts I am not understanding. This is also a 7 part project, and I would like help on this first problem to get the ball rolling. I have also done my research on this too.

    1. The problem statement, all variables and given/known data
    There is a fish population in a lake. By setting up a differential equation, we will investigate the dynamics of this population and show that the population will eventually approach the so called carrying capacity of the environment if the initial population is larger than a "threshold" and become extinct if it is smaller than the threshold. For the purpose of protection of this population, we will set up a scheme for fishing.

    1) Denote by P(t) the fish population at time t. Assume the birth rate β(P) and the death rate δ(P) of P(t) per individual per year are given by...

    β(P) = a
    δ(P) = bP+(c/P)

    respectively, where a, b, c are postitive constants such that a2-4bc > 0. Apply the principle

    dP/dt = [β(P)-δ(P)]P

    to show that the P(t) satisfies the differential equation

    dP/dt = k(M-P)(P-m) (1)

    where k = b, m = (a-√(a2-4bc))/(2b) and M = (a+√(a2-4bc))/(2b). We observe that the logistic equation is the special form of (1) when m = 0. Let the initial population at t0 be given by

    P(t0)=P0




    2. Relevant equations
    They were stated above, but here they are again.

    β(P) = a
    δ(P) = bP+(c/P)
    dP/dt = [β(P)-δ(P)]P

    dP/dt = k(M-P)(P-m) (1)
    k = b
    m = (a-√(a2-4bc))/(2b)
    M = (a+√(a2-4bc))/(2b)

    P(t0)=P0




    3. The attempt at a solution
    Well my attempt in getting is by separation...

    dP/([β(P)-δ(P)]P) = dt

    and then to integrate it...

    ∫1/([β(P)-δ(P)]P) dP = ∫ dt

    and then partial fractions I thought, but it doesn't come out correct. I have also tried wolfram alpha and to check my work. I throw in the equation to be differentiated and it give me something that is sort of similar, but still far off.

    Could someone please help me get on the right track here?
     
    Last edited by a moderator: Mar 29, 2014
  2. jcsd
  3. Mar 29, 2014 #2

    Ray Vickson

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    What solution P(t) do you (or Wolfram Alpha) get, and what do you expect to get? Whys do you say your soltion is "far off"?
     
  4. Mar 29, 2014 #3

    Ray Vickson

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    What solution P(t) do you (or Wolfram Alpha) get, and what do you expect to get? Whys do you say your solution is "far off"?
     
  5. Mar 29, 2014 #4
    Well the question is asking to differentiate the equation...
    dP/dt = [β(P)-δ(P)]P

    And shouldn't it come out to be the equation...
    dP/dt = k(M-P)(P-m)


    OR am completely over thinking this problem and that I should be looking at the second equation to differentiate. And then explain how the principle...
    dP/dt = [β(P)-δ(P)]P relate to the differentiable equation... ?

    It would make a lot more sense if thats the case lol.
     
  6. Mar 29, 2014 #5

    pasmith

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    No. The question asks you to substitute the given [itex]\beta(P) = a[/itex] and [itex]\delta(P) = bP + c/P[/itex] into the equation
    [tex]
    \frac{dP}{dt} = (\beta(P) - \delta(P))P
    [/tex]
    and then to explain why the right hand side is equal to [itex]k(M - P)(P - m)[/itex] where [itex]k[/itex], [itex]M[/itex] and [itex]m[/itex] take the values given in the question.

    My reading of the question is that actually solving the ODE, or at least discovering how the long-term behaviour of its solution depends on the initial population, will be done over the course of the next six parts of your project. However the present question is really more concerned with quadratic functions than calculus.
     
  7. Mar 29, 2014 #6
    So I am supposed to set the two equations equal to each other and solve for P? I'm currently doing this right now.

    [itex] (a - (bP + \frac{c}{P}))P = k(M-P)(P-m)[/itex]
     
  8. Mar 30, 2014 #7

    Ray Vickson

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    Yes, that is what you are supposed to do. The original rhs ##[a - (bP + c/P)]P = -c + aP - bP^2## is a quadratic in ##P##, so can be re-written as ##k(M-P)(P-m)## for some M, m and k.
     
  9. Mar 30, 2014 #8
    Solving the quadratic equation looks like so below...

    [itex]\frac{dP}{dt} = -bP^{2}+aP-c[/itex]

    [itex]P = \frac{-a \pm\sqrt{a^{2} - 4bc}}{-2b}[/itex]

    But now I am confused since M and m are slightly different from the equation above. Am I supposed to now solve for P? Or plug in the new found equation into P? Because it seems like I am supposed to derived the differential equation that is given, correct?

    In all honesty I am not sure what I am supposed to do next.

    To everyone that has replied thank you for the help and thanks for anyone who replies afterwords. Just wanted to say that in case I forget.
     
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