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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 :)!

enter image description here

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    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`2012-12-15
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    Take a look at [Tikhonov's theorem](http://en.wikipedia.org/wiki/Tikhonov's_theorem_(dynamical_systems))2012-12-15
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    Hmm, I don't know exactly how to use that on my stuff!2012-12-15
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    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?2012-12-15
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    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.2012-12-15
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    Well just from looking on the 3rd equation I can say, that $z$ was changed to $\frac{z}{k_5 B}$, $t$ to $\frac{\tau}{B k_5}$ and $x$ to $\frac{x}{2 A k_3}$. The rest is easy to do.2012-12-15
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    Or $z -> \frac{z}{B}$, $t -> \frac{\tau}{B k_5}$, $x -> \frac{k_5 x}{2 k_3 A }$2012-12-15
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    Excuse me for asking so much this is very foreign to me! Which 3rd equation do you refer to? And by what means were they changed? From where do you get the "idea": "Oh this has to be changed to that", what's behind it.. I'm really sorry when you say "the rest is easy to do" because I can't see it. You say z was changed to z/(k5B).. hmm. is that the rescaling... oh how do I say this... -part of the nondimensionalization? But 'how do you know you had to rescale by exactly k5B'?2012-12-15
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    Well, that's what you do when you nondimensionalize your equations. It's a chemistry, so $x,y,z$ are some concentrations so as $A,B$. Then to nondimensionalize a variable $z$ you replace it with something like $\frac{z}{B}$.2012-12-15
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    So... you substitute z with z/(k5*B)? and x with x/(2A*k5) and hmm... t with Tau/(B*k5)? What about Y? Oh and one last thing that troubles me... how do you replace t when the only place t shows itself is in the "d/dt"2012-12-15
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    Oh, and from where do the epsilon and delta parameters originate? I appreciate your replies very much, but I just need a bit more help. Thank you!2012-12-15
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    Could you please change you title to something more specific? Thanks2012-12-15
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    Dear God, I am having such a flashback. I worked on models of the BZ reaction back in the 1980s.2012-12-15

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