Nondimensionalization of diffusion equation

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Homework Statement


We let a dye diffuses into an environment of dimension L. We inject that dye into a box by one face, at t = 0 on x = 0. The linear density c follows that equation :
upload_2017-4-23_11-22-49.png


with the conditions :
upload_2017-4-23_11-22-40.png

Homework Equations

/ questions[/B]
a. nondimensionalize the equations and the conditions
b. reveal a term homogeneous to time, and its signification
c. compare the characteristic lenghts of these equation systems

The Attempt at a Solution


By nondimensionalize this equation, I found this :
upload_2017-4-23_11-24-30.png

But I think it's wrong... I use the "formal way" to nondimensionalize the equation as shown in the Khan academy video on youtube.
May I ask for help ?
Thanks a lot
 

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I think you should start by finding out the numbers ##\alpha_1 , \alpha_2 , \alpha_3##, ##\beta_1 , \beta_2 , \beta_3##, ##\gamma_1 , \gamma_2 , \gamma_3## so that the variables

##\tilde{x}=L^{\alpha_1}m_0^{\alpha_2}D^{\alpha_3}x##
##\tilde{t}=L^{\beta_1}m_0^{\beta_2}D^{\beta_3}t##
##\tilde{c}=L^{\gamma_1}m_0^{\gamma_2}D^{\gamma_3}c##

become dimensionless. ##L## is any characteristic length of the system you want to choose.
 
Hi, I've tried what you've advised me, here are my results :
upload_2017-4-23_16-42-1.png

We therefore have:
upload_2017-4-23_16-43-59.png


For the conditions I found:

upload_2017-4-23_16-42-42.png


I'm not quite sure about the integral term though..
 

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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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