Why Normalize Boundary Conditions in PDEs?

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garcijon
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Hi All,

This is my first post on these forums. I am not looking for a solution to this problem but more interested in seeing if someone can point me to a resource that can explain the following. Thanks in advance for any help.

I'm trying to solve a pde which gives a temperature profile.

Heat equation written as rho C_p times partial T over partial t equals k times the Laplacian of T, with the Laplacian written as nabla squared T.

We end up changing over to spherical coordinates where our boundary conditions are not completely clear although our initial condition is.

Handwritten PDE set-up: dG/dt = (k / (rho C_p)) d^2G/dr^2 = alpha d^2G/dr^2, with G(r = R) = R T_s and G(t = 0) = r T_0 at infinity, equals r dot T_0 = ?


The professor writes something like this. basically it normalizes the boundary conditions to (0,1). I can't wrap my head around this and googling for a while hasn't given me any results that clarify this.

Handwritten definition of a normalized temperature T' = (T - T0)/(Ts - T0), with the note that T' equals 0 at t=0 or x→∞ and 1 at x=0.
 
garcijon said:
The professor writes something like this. basically it normalizes the boundary conditions to (0,1). I can't wrap my head around this and googling for a while hasn't given me any results that clarify this.

View attachment 75728

This is a straightforward linear rescaling of the form [tex] f : [a,b] \to [0,1] : x \mapsto \frac{x - a}{b - a}.[/tex] Observe that [itex]f[/itex] is strictly increasing with [itex]f(a) = 0[/itex] and [itex]f(b) = 1[/itex].

If what you are struggling with is "why does he bother to do this?" then the concept you are looking for is nondimensionalization. Also it's easier to work with [itex]G(t = 0) = G(r \to \infty) = 0[/itex] rather than [itex]G(t = 0) = rT_0[/itex] and [itex]G(r \to \infty) = \infty[/itex].