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Here I am referring to, 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 :)!

enter image description here

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migrated from Dec 15 '12 at 14:29

This question came from our site for users of Mathematica.

By the way, I also realize just now that instead of: ε* dx/dτ = q*y-x*y+x(1-x) δ* dy/dτ = -q*y-x*y+f*z You can use: dx/dτ = (q*y-x*y+x(1-x))/ε dy/dτ = (-q*y-x*y+f*z)/δ – Jensjakob Kristiansen Dec 15 '12 at 11:01
Take a look at Tikhonov's theorem – swish Dec 15 '12 at 11:02
Hmm, I don't know exactly how to use that on my stuff! – Jensjakob Kristiansen Dec 15 '12 at 11:18
It can be used for further simplification of the system , to get rid if some equations. But I just realized that it's not the question. Do you just need to know what replacements were used? – swish Dec 15 '12 at 11:23
Yes, I can't puncture the process... Like, where do I start if I want to go from the 3 differential equations dx/dt, dy/dt and dz/dt to those of tau? >and probably also what steps are next, because this is so way out of the league compared to what I should be learning... My situation is, that I've gotten a project on Oscillating Reactions that my teachers didn't know much about. They think it's "just" to plot those 3 differential equations I have, and nothing about making them dimensionless... but I have to do it now that I am in this. – Jensjakob Kristiansen Dec 15 '12 at 11:24

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 $k_4$.

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