Change of variables in Belousov-Zhabotinsky reaction

Question

Here I am referring to http://demonstrations.wolfram.com/TheBelousovZhabotinskyReaction/, but not only this. If you press "Download demonstration as CDF" (implying you have the necessary tools to open it) and then look inside at the equations for that reaction. I have those myself.

A+Y -> X+P     with rate constant k1
X+Y -> 2P      with rate constant k2
A+X -> 2X+2Z   with rate constant k3
2X -> A+P      with rate constant k4
B+Z -> (1/2)fZ with rate constant k5

I have formulated these into differential equations, which is based on autocatalysis, for each of the intermediates X, Y and Z... Under here it is x, y and z. I use * as multiplication, because I think it simplifies it a bit. This is done by treating the A and B concentrations as constants.

dx/dt = k1*A*y-k2*x*y+k3*A*x-2*k4*x^2
dy/dt = -k1*A*y-k2*x*y+(1/2)*f*k5*B*z
dz/dt = 2*k3*A*x-k5*B*z

Now, my question is: how are the above 3 differential equations converted to the dimensionless "3x3 system" following the Law of Mass Action (as what I referred to in the start of this post says)... What is done... what are the steps... how do you do this..!!?!?

I have researched the Law of Mass Action and all myself but I can't seem to find out what is done to reach the following from the above 3 differential equations:

ε* dx/dτ = q*y-x*y+x(1-x)
δ* dy/dτ = -q*y-x*y+f*z
dz/dτ = x-z

I realize t is replaced by τ, but what about the rest? Can anyone give me steps of what is happening?! I will appreciate it a lot! I included a picture to this post. The picture shows what I'm referring :)!

Answer 1

You need these replacements:

$\varepsilon \to \frac{B k_5}{A k_3},$ $\delta \to \frac{2 B k_4 k_5}{A k_2 k_3},$ $\tau \to B k_5 t,$ $x \to \frac{2 k_4}{k_3 A} x,$ $y \to \frac{k_2}{A k_3} y,$ $z \to \frac{k_4 k_5 B}{k_3^2 A^2} z,$ $q \to \frac{2 k_1 k_4}{k_2 k_3}$

I found them here. But there is an error in $y$ replacement: $k_2$ instead of $k_4$.