What is the process of nondimensionalizing an equation?

Question

Question: I need to scale time by $\frac{1}{I}$ and species by $P$

for the following equation

$\frac{dS}{dt}=I(1-\frac{S}{P})-\frac{ES}{P}$

where

P - Size of the source pool of species on the mainland

S - number of species on the island

E - Extinction Rate for species

I - Immigration Rate for species

t - time

My Attempt:

So I need to introduce new variables, called m and n, to make this equation dimensionless.

$m = \frac{S}{S^*}, n = \frac{t}{t^*}$

$mS^*=S, nt^*=t$

So substituting these values into the equation, we have

$\frac{d(mS^*)}{d(nt^*)}=I(1-\frac{(mS^*)}{P})-\frac{E(mS^*)}{P}$

Diving $\frac{S^*}{t^*} $ throughout the equation, we have

$\frac{dm}{dn}=\frac{t^*I}{S^*}-\frac{Imt^*}{P}-\frac{Emt^*}{P}$

Ok. This is where the easy part ends. So, my $t^*$ is by itself and that's a unit of time, so since the time is already given why not let $t^*=\frac{1}{I}$

$\frac{dm}{dn}=\frac{(\frac{1}{I})I}{S^*}-\frac{Im(\frac{1}{I})}{P}-\frac{Em(\frac{1}{I})}{P}$

Except I have distributed by accident, so my equation is really

$\frac{dm}{dn}=I(\frac{1-mS^*}{P})\frac{t^*}{S^*}-\frac{Emt^*}{P}$

Now let $t^*=\frac{1}{I}$

$\frac{dm}{dn}=I(\frac{1-mS^*}{P})\frac{(\frac{1}{I})}{S^*}-\frac{Em(\frac{1}{I})}{P}$

$\frac{dm}{dn}=\frac{1-mS^*}{S^*}-\frac{Em}{IP}$

And then let $S =P$ because S is species and P is species so the units match.

$\frac{dm}{dn}=(1-\frac{mP}{P})-\frac{Em}{IP}$

$\frac{dm}{dn}=(\frac{1-\frac{mP}{P}}P)-\frac{Em}{IP}$

$\frac{dm}{dn}=(\frac{1-m}{P})-\frac{Em}{IP}$

If we factor out $\frac{1}{P}$, we have

$\frac{dm}{dn}=\frac{1}{P}((1-m)-\frac{Em}{I})$

which is almost close, but apparently a better substitution would be to let $t^*=\frac{I}{P}$ so that more parameters would cancel out. So that would mean let my $t^*$ be $\frac{unit-of-time}{species}$

Would it be ok to just let $t^*=\frac{1}{I}$ because that was given in the first place. And all the units will need to match first before I begin. I noticed that when I distributed the I, I have

$\frac{dS}{dt}=(I-\frac{IS}{P})-\frac{ES}{P}$

So basically that's

$\frac{d[species]}{d[time]}=([time]-\frac{[time][species]}{[species]})-\frac{[species][species]}{[species]}$

Don't I need a $\frac{1}{I}$ or $\frac{1}{time}$ for the last part of the equation where there are species all over the place? ANd $\frac{species}{(time)^2}$ is needed for the only time in the first section? If that's the case then the middle part needs a $\frac{species}{(time)^2}$ before I continue.

Answer 1

The instructions for nondimensionalizing an equation are as follows:

1. Identify all the independent and dependent variables.
2. Replace each of them with a quantity scaled relative to a characteristic unit of measure to be determined.
3. Divide through by the coefficient of the highest order polynomial or derivative term.
4. Choose judiciously the definition of the characteristic unit for each variable so that the coefficients of as many terms as possible become 1.
5. Rewrite the system of equations in terms of their new dimensionless quantities.

So given $$ \frac{dS}{dT}=I\left(1-\frac SP\right)-\frac{ES}P\tag{1} $$ Choose $S=\alpha s$, $T=\beta t$, $P=\gamma p$, $E=\delta e$ and $I=\epsilon i$. With some work, you can get $$ \frac{ds}{dt}=\beta\epsilon i\left(\frac1\alpha-\frac{s}{p\gamma}\right)-\frac{\beta\delta}{\gamma}\frac{es}{p} $$ Since $S$ and $P$ are both populations of species, we can probably expect that $\alpha=\gamma$. Similarly, $\delta=\epsilon$ makes sense since $I$ and $E$ are rates of changes of the populations. Thus, we get $$ \frac{ds}{dt}=\frac{\beta\epsilon}\alpha i\left(1-\frac{s}{p}\right)-\frac{\beta\epsilon}{\alpha}\frac{es}{p}\tag{2} $$ If you choose that $\beta\epsilon=\alpha$ (so you only need to define two of these three scales, since the third is given by the two), then (2) becomes $$ \frac{ds}{dt}=i\left(1-\frac{s}{p}\right)-\frac{es}{p}\tag{1*} $$ which, unless I've made a mistake somewhere, appears to be a dimensionless form of (1).