Scaling Differential Equations

[rudders93]

Homework Statement

Not exactly a problem, more an example that has me confused :s

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

This describes a population model where N is the population, \kappa 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:

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

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

Thanks!

 

[HallsofIvy]

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 \kappa are 1.

However, what you have here,
\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)
does not give the result you cite. You must mean
\frac{dN}{dt}= \kappa N- \left(\kappa/a\right)N^2= \kappa N(1- N/a)
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 "\tau" independently.

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

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

Now, replacing N by au(\tau) makes it
a\frac{du}{d\tau}= au(\tau)\left(1- \frac{au(\tau)}{a}\right)
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
\frac{du}{d\tau}= u(\tau)(1- u(\tau))

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.)

 

[rudders93]

Hi,

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

I think I understand now

 

[Ray Vickson]

Homework Statement

Not exactly a problem, more an example that has me confused :s

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

This describes a population model where N is the population, \kappa 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:

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

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

Thanks!

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.

RGV