# Brusselator equation

• Feb 14th 2010, 09:37 PM
dedust
Brusselator equation
The Brusselator equation is

$\displaystyle \frac{d[X]}{dt} = k_1[A] - k_2[b][X] - k_4[X] + k_3[X]^2[Y]$

$\displaystyle \frac{d[Y]}{dt} = k_2[b][X] - k_3[x]^2[Y]$

by linear scaling the variables and time, we have

$\displaystyle \frac{dx}{dt} = a - (b + 1)x + x^2y$

$\displaystyle \frac{dy}{dt} = bx - x^2y$

but how do we do it? what substitusion should I use?

thx for any help
• Feb 18th 2010, 07:10 AM
Jester
Quote:

Originally Posted by dedust
The Brusselator equation is

$\displaystyle \frac{d[X]}{dt} = k_1[A] - k_2[b][X] - k_4[X] + k_3[X]^2[Y]$

$\displaystyle \frac{d[Y]}{dt} = k_2[b][X] - k_3[x]^2[Y]$

by linear scaling the variables and time, we have

$\displaystyle \frac{dx}{dt} = a - (b + 1)x + x^2y$

$\displaystyle \frac{dy}{dt} = bx - x^2y$

but how do we do it? what substitusion should I use?

thx for any help

Let $\displaystyle [X] = \alpha x,\;\; [Y] = \beta y,\;\; t = \gamma \tau$ substitute and isolate

$\displaystyle \frac{d x}{d \tau}$ and $\displaystyle \frac{d y}{d \tau}$

Then compare with

$\displaystyle \frac{dx}{dt} = a - (b + 1)x + x^2y$

$\displaystyle \frac{dy}{dt} = bx - x^2y$

literally term by term. This will give you three equations for your unknowns $\displaystyle \alpha, \beta \; \text{and}\; \gamma$.