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Differential Equations Model Help

  1. Sep 6, 2008 #1
    1. The problem statement, all variables and given/known data
    A certain bacterium, given plenty of nutrient and room, is known to grow according to the Malthusian model with reproductive rate r. Suppose that the biologist working with the culture harvests the bacteria at a constant rate of h bacteria per hour. Use qualitative analysis to discuss the fate of the culture.


    2. Relevant equations
    [tex] P(t) = P_{0}e^{rt}[/tex]


    3. The attempt at a solution
    I have added [tex] P(t) = P_{0}e^{rt} - h*t[/tex] to my calculator and picked some values, however they don't match with the answer in the back of the book. I am wanting help understanding the answer in the back...The answer makes sense...but I can't figure out why...

    The answer in the back of the book is:

    The long-time activity of the population depends entirely on the sign of P(0) - (h/r). If it is negative, then the population will die out. If it is positive, the population will grow exponentially. If it is zero, then the population will remain at a constant (h/r).

    Like I said, the answer makes some sense, however, I can't figure out why. Any help with the why is greatly appreciated. Thanks.
     
  2. jcsd
  3. Sep 6, 2008 #2

    Dick

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    You can't assume P(t)=P0*exp(rt) if the bacteria are being harvested. You need to solve the rate equation P'(t)=r*P(t)-h with the initial condition P(0)=P0. Can your calculator do that? Or can you reason out why the conclusions are true?
     
  4. Sep 6, 2008 #3
    Thanks for the clarification. My calculator can't do that, however I believe I can reason from the P'(t) = r*P(t) - h why the conclusions are true. Shouldn't it be h*t ?
     
  5. Sep 6, 2008 #4

    Defennder

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    Note that when you have a constant harvesting rate of h/hour, you have to re-solve the DE and not simply make use of the answer to the Malthusian growth model.

    So, the original DE was [tex]\frac{dP}{dt} = rP[/tex]. The constant harvesting rate is r, which means the DE becomes: [tex]\frac{dP}{dt} = rP - h[/tex]. Even without solving the DE you should be able to analyze it qualitatively. Just ask yourself what happens to the baterial culture at t=0, just by looking at the DE. Supposing you are evaluating dP/dt at t=0. Then it is evident that we need to know if [tex]rP_0 - h[/tex] is negative or positive. If it is negative, then this implies that [tex]P_0 = \frac{h}{r}[/tex] is negative and it is evident that once the rate is negative, the baterium population would be sloping down all the way till extinction.

    On the other hand, if it is positive, then the original sample would increase and in the long run the exponential growth rate r would make neglible the constant harvesting rate h, and the culture will grow at an expotential rate. On the other hand, if P(0)-h/r is zero, it means right from the start there is no change in the population since all the birth and death factors cancel out exactly to give the same population. So the population remains constant throughtout.
     
  6. Sep 6, 2008 #5
    Oh right...I hate doing anything homework related and being sick. Thanks again for your help. Do you know of a good numerical solver. In my next problem I am asked to plot various solution trajectories for the following:

    P' = .38p(1 - P/1000) - h(t). I assume the lowercase p here is a typo and it should be P to match the form. In addition to that h(t) = 200 if t<3 and h(t) = 0 otherwise. I don't know where I could find a free solver to graph various solution trajectories... Is there a program you can suggest? Also, do you know if anything on this website would work?
     
  7. Sep 6, 2008 #6

    Defennder

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    You don't need a numerical solver. The DE is solvable by hand. Just plug in the values and graph them.
     
  8. Sep 6, 2008 #7
    I know that while h(t) = 200 the graph plummets for many solutions. As soon as h(t) = 0 it rises again. I suppose I should just skip to part b and solve for the exact value of P that is between it crashing and rising... I am to find the critical value where once P is slightly below it P crashes completely...and the if P is slightly above this value it falls, but then hits a point where it recovers.

    Thanks again for your help.
     
  9. Sep 7, 2008 #8
    I am feeling like a moron lately...I am having issues with the above problem. To find that critical number don't I want the derivative to be zero? Like I said, being sick doesn't help when it comes to trying to focus.

    I have tried plugging in 200 for h and solving when the derivative is equal to zero...

    0 = .38p(1 - P/1000) - 200 but I either wind up with something that is wrong or something that doesn't cross the x axis when I graph that...

    I figured I'd work from a graph first, then go through the hand solving.
     
    Last edited: Sep 7, 2008
  10. Sep 7, 2008 #9
    Would it be best to start a new thread for my second question, or should I just continue on with this thread? Thanks.
     
  11. Sep 7, 2008 #10

    Dick

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    If you've been out sick, are you sure that the rest of the class wasn't told how to access software to integrate this? As Defennder said, you can write down a solution (sort of, you have to match the t<3 and t>3 solutions) by hand and plot it, but it's pretty messy.
     
  12. Sep 7, 2008 #11
    I haven't been away from class. I got sick this weekend and luckily it's going away. It's just hard to focus. The exact directions are:

    "Use an analytic method to determine the exact value of the 'critical' initial population found in part (a). Justify your answer."

    Part a requires a numerical solver. I know the initial population needs to be between 414 and 415 based off the back of the book ...but I am not certain how to get to the exact value. I thought that for that to be true the derivative at the critical value would be zero because it would be switching from a declining population to a rising population.

    0 = .38p(1 - P/1000) - 200 <---when I solve that for P I get an answer outside of the 414-415 range

    Also... if I solve:

    0 = .38p(1 - P/1000)

    I get an answer outside of the range as well. I figure it's something wrong with how I am looking at the problem.
     
  13. Sep 7, 2008 #12

    Dick

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    Ok. It looks like you want to get a steady state solution in the t>3 region. That means P(3)=0 or P(3)=1000 (since those are the values which will give you zero derivative). So now you want figure out what value of P(0) will evolve to give one of those values. So integrate the ODE (as I said, it's kind of messy) and set P(3)=0. Once you have that set t=0 to figure out what P(0) is. I did it and I got 414-415.
     
  14. Sep 7, 2008 #13
    Ok...it does look messy. Thanks for the help. I'll give that a go.
     
    Last edited: Sep 7, 2008
  15. Sep 7, 2008 #14

    Dick

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    Work carefully. I'll give you a hint. You have one over a quadratic in P. Complete the square and use a tangent substitution. It's not hard in principal, it's just a mess because you've got all those stupid numbers 0.38, 1000, 200 flying around.
     
  16. Sep 7, 2008 #15
    Ok so I will be having:

    [tex] 1/(.38p + .38p^{2}/1000)dp = 200dt[/tex]
     
  17. Sep 7, 2008 #16

    Dick

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    No. dP/(0.38P-0.38P^2/1000-200)=dt. You made a sign error and you can't leave that 200 dangling on the right. Now can you? If you are going to do a lot of algebra it pays to be really careful. Otherwise you'll wind up doing a lot of work for nothing.
     
  18. Sep 7, 2008 #17
    I have no clue where the + came from. Wow...I can't believe I made those two mistakes. I am clearly not thinking straight...thanks for being so patient with me.
     
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