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