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Homework Help: Scaling Differential Equations

  1. Oct 7, 2011 #1
    1. The problem statement, all variables and given/known data
    Not exactly a problem, more an example that has me confused :s

    [itex]\frac{dN}{dt}=\kappa N-(\kappa/a^{2})N[/itex]

    This describes a population model where N is the population, [itex]\kappa[/itex] is the net births (ie: births less deaths) and it doesn't tell you what a is (this in all the logistic population model). According to http://en.wikipedia.org/wiki/Logistic_function it says its the carrying capacity which is the populations 'maximum' given some constraints.

    Then it says it's possible to scale the problem with the following:

    [itex]t=\frac{\tau}{\kappa}, N(t)=au(\tau)[/itex] and therefore [itex]\frac{du}{d\tau}=u(1-u)[/itex]

    My question is how did they decide upon this way of scaling and how did they get from setting [itex]t=\frac{\tau}{\kappa}[/itex] to [itex]\frac{du}{d\tau}=u(1-u)[/itex]?

  2. jcsd
  3. Oct 7, 2011 #2


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    First, deciding how to "scale" is often the result of "hindsight"- that is, someone solved the equation and then, based on what the solution looked like, realized the problem would be easier in a different form. Here, the point is to choose "units of measure" so that a and [itex]\kappa[/itex] are 1.

    However, what you have here,
    [tex]\frac{dN}{dt}= \kappa N- \left(\kappa/a^2\right)N= \kappa N(1- 1/a^2)= \kappa N\left(\frac{a^2- 1}{a^2}\right)[/tex]
    does not give the result you cite. You must mean
    [tex]\frac{dN}{dt}= \kappa N- \left(\kappa/a\right)N^2= \kappa N(1- N/a)[/tex]
    That is, you want the square on the "N", not the "a" in the denominator.

    You can then change "N" to "u" and "t" to "[itex]\tau[/itex]" independently.

    You change "t" to "[itex]\tau[/itex]" using the "chain rule".With [itex]t= \tau/\kappa[/itex], or [itex]\tau= \kappa t[/itex] we have [itex]d\tau/dt= \kappa[/itex]. So
    [tex]\frac{dN}{dt}= \frac{dN}{d\tau}\frac{d\tau}{dt}= \kappa \frac{dN}{d\tau}[/tex].

    That makes the equation
    [tex]\kappa\frac{dN}{d\tau}= \kappa N\left(1- \frac{N}{a}\right)[/tex]
    and the "[itex]\kappa[/itex]"s cancel.

    Now, replacing N by [itex]au(\tau)[/itex] makes it
    [tex]a\frac{du}{d\tau}= au(\tau)\left(1- \frac{au(\tau)}{a}\right)[/tex]
    and the two "a"s in the fraction on the right cancel, as well as the "a" on either side of the equation. That leaves
    [tex]\frac{du}{d\tau}= u(\tau)(1- u(\tau))[/tex]

    By the way, "a" is the "carrying capacity". The population can't be negative, of course, and if N>a, 1- N/a will be negative causing a decrease in population, while for any N< a, there is an increase in population- the population will eventually stabilize at the "equilibrium value", a. (Strictly speaking, there are two values of N that make dN/dt 0- N= 0 and N= a- those are the two "equilibrium values". Obviously, if the population is 0, it won't change. "a" is the only positive population that has that property.)
    Last edited by a moderator: Oct 7, 2011
  4. Oct 7, 2011 #3

    Thanks for the super detailed response! Yep, I incorrectly copied the question :(

    I think I understand now
  5. Oct 7, 2011 #4

    Ray Vickson

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    If this is supposed to be (related to) the logistic equation, you have it all wrong: you should have k*N - (k/a^2)*N^2 [N^2, not N!]. If you have N in the second term your right-hand side is just r*N, where r = k - k/a^2.

    Last edited: Oct 7, 2011
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