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Given a set of differential equations for a predator-prey model (Lotka-Volterra): \begin{align*} \frac{dH}{dt} &= bH - sHP \\ \frac{dP}{dt} &= -dP + esHP \\ \end{align*} Where $P$ is the number of predators, $H$ the number of preys, $t$ time, $b$ the birth rate of the prey, $d$ the death rate of the predator, $s$ the searching efficiency of the predator and $e$ the efficiency in which food is turned into extra predators.

I can write this in nondimensional form using the following parameterization: \begin{align*} h &= \frac{Hes}{d} \\ p &= \frac{Ps}{b} \\ \tau &= \sqrt{bd}t \\ \rho &= \sqrt{\frac{b}{d}} \end{align*}

To become: \begin{align*} \frac{dh}{d\tau} &= \rho h (1-p) \\ \frac{dp}{d\tau} &= - \frac{1}{\rho} p (1 - h) \\ \end{align*}

My question is, how do you get this result? I understand the basics of nondimensionalization. First I find the dimensions of each variable and next I define new variables from products of variables with the same dimension. However, I think the dimensions of all variables/constants are as follows:

$H,P$ in units, $t$ in time and $b,s,d,e$ in inverse of time.

But this gets me nowhere.

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Is the question (a) how to get from the old to the new system given the change of variables, or (b) how to find the change of variables in the first place? – Hans Lundmark Oct 25 '11 at 14:24
Mainly how to find the new variables. Looking at these variables I see that the dimensions should be $s = [P^{-1} t^{-1}], d = [t^{-1}], e = [H^{-1} P], d = [t^{-1}]$ where $[x]$ means the dimension of $x$. I just have no idea how to find that. – Zagga Oct 25 '11 at 14:44
up vote 4 down vote accepted

Sometimes it's obvious what scales to use; for example, if there's a logistic growth term $rN(1-N/K)$, then it's natural to measure the population size $N$ in units of the carrying capacity $K$, and time in units of the reciprocal growth rate $1/r$.

In other cases, like here, it's not as obvious, but what you can always do is plug in undetermined scales (to be determined later) in the equations. Taking $$ H = c_1 h, \qquad P = c_2 p, \qquad t = c_3 \tau, $$ the system becomes $$ \begin{align*} \frac{c_1}{c_3} \frac{dh}{d\tau} &= b (c_1 h) - s (c_1 h) (c_2 p), \\ \frac{c_2}{c_3} \frac{dp}{d\tau} &= -d (c_2 p) + es (c_1 h) (c_2 p), \\ \end{align*} $$ which can be rearranged to $$ \begin{align*} \frac{dh}{d\tau} &= (c_3 b) h - (c_2 c_3 s h) p, \\ \frac{dp}{d\tau} &= -(c_3 d) p + (c_1 c_3 es) h p. \\ \end{align*} $$ Now you try to choose $c_1$, $c_2$, $c_3$ so that the coefficients become as simple as possible. Usually there is no canonical choice which gives the absolutely simplest result, but rather there are many choices which all lead to "equally simple" equations. (Sometimes the choice matters, depending on what you want to do with the equations later, but I don't think you need to worry about this here.)

There are four coefficients, but only three choices to be made, so you can't expect to make all coefficients equal to one, but you could for example choose $c_3=1/b$ and then choose $c_1$ and $c_2$ to make $c_1 c_3 es=1$ and $c_2 c_3 s h=1$. This would give the system $$ \begin{align*} \frac{dh}{d\tau} &= h - h p, \\ \frac{dp}{d\tau} &= -(d/b) p + h p, \\ \end{align*} $$ where you can call the remaining coefficient $d/b$ something (say $\alpha$) to get the final form of the equations.

Now, this choice wasn't the one made by whoever gave you your equations. Instead, they opted for making the coefficients of $h$ and $-hp$ equal in the first equation, and the coefficients of $-p$ and $hp$ equal in the second equation. This requires $b=c_2 s$ and $d=c_1 es$. (Compare this to your given formulas relating $P$ and $H$ to $p$ and $h$, remembering that we have $H=c_1 h$ and $P=c_2 p$.) Finally, $c_3$ was for some reason chosen to make the overall factor $\rho$ in the first equation equal to the reciprocal of the overall factor $1/\rho$ in the second equation. Again, this is not a canonical choice but rather arbitrary.

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Awesome, thanks! – Zagga Oct 25 '11 at 18:36
I think the parameterization was chosen so that the equations are zero at points $(h=0,p=0)$ and $(h=1,p=1)$. Now this is $(h=0,p=0)$ and $(h=\alpha,p=1)$. If I chance $c_1 = -(b/es)$ I obtain the point $(h=1,p=1)$ again, but doesn this fundamentally change the equations? – Zagga Oct 25 '11 at 20:32
No such alternative choices will cause any fundamental changes to equations, only cosmetic ones. The exact location of the equilibrium points will of course depend on the scales used, but the phase portrait will still look more or less the same. (Changing the scales for $P$ and $H$ will just stretch the picture horizontally and vertically, and a change of time scale isn't visible at all in the phase portrait since one draws only the solution curves without indicating the speed with which they are traversed.) – Hans Lundmark Oct 25 '11 at 20:41

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