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Logistic growth, population, limits

  1. Mar 23, 2008 #1
    The number of bacteria in a lab is model by the function M that satisfies the logistic differential equation dM/dt = 0.6M (1 - (M/200) ), where t is the time in days and M(0) = 50. What is the limit of M(t) as t approach infinity?

    Do i use the fundamental theorem of calculus?
  2. jcsd
  3. Mar 23, 2008 #2


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    You certainly could "use the fundamental theorem of calculus" and integrate that. But if you understand what the logistic equation means the answer is almost trivial. Can you see that if M is less than 200, the derivative is positive (so M is increasing)?
    Can you see that if M is larger than 200, the derivatiove is negative (so M is decreasing)?

    Now, what do you think the limit is?
  4. Mar 23, 2008 #3
    I understand the concept,
    but the answer key i have shows that a specific integral is the answer.
    Which mean the answer cannot be 'increasing toward infinity' or things like that.
    I wonder if anyone can list the steps how to get that specific number.

    Appreciate any help.
  5. Mar 23, 2008 #4


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    Well to do it the way your professor probably wants you to do it, you would divided both sides by the RHS, and multiply both sides by dt, so that you have f(M)dM = dt... carry out the integration and solve using the initial condition, then take the limit where t->infinity. However, there are much easier ways to show that the answer is 200.
  6. Mar 24, 2008 #5


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    How could your answer key show "that a specific integral is the answer" when the question is clearly answered by a number?

    In a logistic equation such as dx/dt= kx(1- x/A) the number "A" is referred to as the "carrying capacity". As I said before, if 0< x< A, dx/dt> 0 so x is increasing toward A. If x> A, dx/dt< 0 so x is decreasing toward A. More to the point, x= A is itself a solution to the equation and since the hypotheses of the "existance and uniqueness" theorem hold, no other solution can cross it. x approaches A as t goes to infinity. No "integration" is required.
  7. Mar 24, 2008 #6
    Obviously i don't know the formula dx/dt= kx(1- x/A) that well..
    My professor told me all i have to do is factor out the 1/200 from M/200,
    which makes (200-M). The number in front is the limit.
    I wonder can someone give me some extra explanation on that.
    Thank you .
  8. Mar 25, 2008 #7


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    Yes, obviously if you have 1- x/A, you can write that as (1/A)(A- x). In particular,
    [tex]0.6M\left(1- \frac{M}{200}\right)= \frac{0.6M}{200}\left(200- M\right)[/itex]

    If M(t)= 200 for all t (so M is a constant function) dM/dt= 0 and 200- M= 0 so the right hand side of your differential equation is also 0. That is, M(t)= 200 is a solution to the differential equation.

    It should also be obvious that M(t)= 0 is a constant solution to the equation. M(t)= 200 and M(t)= 0 are called the equilibrium solutions.

    Since solutions to this equation are unique, the graphs of two solutions cannot cross. If, for some t, M is between 0 and 200, M(t) must always be between 0 and 200.
    But if M is between 0 and 200, the right hand side of your differential equation is positive so the derivative is positive: M must be increasing. As t increases, M(t) increases toward 200 but never reaches it.
  9. Mar 26, 2008 #8


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    To find, as shown you, where dM/dt = 0 is usually the first thing to do in handling differential equations, especially nonlinear. I should add that you should consider you could be starting in the lab with M > 200, and the prediction is ?

    I suggest you see the logic more easily when you 'scale parameters' i.e. in this case express it in terms of a variable x = M/B. More eloquently that is M/Mmax. You will find you get rid of a nuisance constant, one less thing to complicate formulas.

    Whether you do that or not, you ought to notice how the growth rate is maximal at half maximal growth, i.e. at M = Mmax/2, or x = 1/2. And that that rate is equal, at M (or x) an equal distance either side of the halfway point. Or, equivalently, at M and (Mmax - M) or at x and (1 - x). Always look for symmetries in such problems. I don't know how to transmit my intuition that that symmetry implies the predicted growth curve itself against t has a symmetry, of rotation about the halfway point. Well maybe after all I can but you try. :smile:

    Especially of you put it in terms of x, it is not very difficult to solve the d.e. (simple partial fractions). I'd suggest when working out that last point of last para above using the solutions you again make it easier by choosing your time origin so that you make t = 0 the time of half maximal growth when x = 1/2, so times before that are negative. With this choice, using the solutions, but hopefully also otherwise, you can demonstrate that x(t) = 1 - x(-t) . (Biologically meaningless when you are starting at x > 1 though.)

    I think that's right.:uhh: Anyway I think it usually best to change parameters so as to make it easiest for yourself with less inessential distractions and error possibilities. :biggrin:
    Last edited: Mar 26, 2008
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