PDEs: Diffusion Equation Change of Variables

[Master1022]

TL;DR Summary

Question about change of variables for the Diffusion PDE

Hi,

I understand the underlying concept of changing variables in PDEs (so that we can reduce it to a simpler form), however, I am just not completely clear on the mathematics of it so I have a quick question about it.

For example, if we have the transmission line equation $$\frac{\partial V}{\partial t} = \frac{L^2}{RC} \frac{\partial ^2 V}{\partial x^2}$$
with initial condition: $V(0,x) = 0$ and boundary conditions: $V(t,0) = V_0$ and $V(t,L) = 0$. Now we want to change variables to reduce the problem into a simpler form without any free parameters.

The Method:
1) Rescale the length variable so that it ranges over one
I let $\xi = \frac{x}{L}$

2) Rescale the other independent variable to remove free parameters from the general expression
$$\frac{\partial V}{\partial t} = \frac{L^2}{RC} \frac{\partial ^2 V}{\partial x^2}$$ $$\frac{\partial V}{\partial t} = \frac{L^2}{RC} \frac{\partial ^2 V}{\partial \left( L \xi \right)^2}$$ $$RC \frac{\partial V}{\partial t} = \frac{\partial ^2 V}{\partial \xi^2}$$
then I let $\tau = \frac{t}{RC}$, therefore $$\frac{\partial V}{\partial \tau} = \frac{\partial ^2 V}{\partial \xi^2}$$

So far, I am happy with this. Now this is where I start to have some questions

3) Redefine the independent variable to non-dimensionalize and simplify B/ICs
We want to simplify the boundary condition $V(t, 0) = V_0$ and we therefore let $U = \frac{V(t, 0)}{V_0}$. So now that boundary condition is = 1.

However, does making this change not introduce a new free parameter back into the geral expression?
$$\frac{\partial \left(U V_0 \right)}{\partial \tau} = \frac{\partial ^2 \left(U V_0 \right)}{\partial \xi^2}$$
The source I am using seems to suggest that step 3) does not change the GE in the way I suggested, but instead changes V to U directly, but I cannot see why.

I see that we are non-dimensionalising it, but we did the same thing with both of the other variables and we subsituted their new values into the GE.

Thank you, any help is appreciated.

 

[Master1022]

I have solved my misunderstanding, when we apply the partial derivative twice on the RHS, we are not squaring the numerator and hence there will only be one $V_0$. Therefore, the $V_0$ (to the power 1) in the numerator of both the LHS and RHS of the equation will cancel out.