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Normalizing the PDE to make BC's homogeous 
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#1
Dec412, 04:23 PM

P: 660

Folks,
Given the pde ## \displaystyle k\frac{\partial^2 T}{\partial x^2}=\rho c_0 \frac{\partial T}{\partial t}## and the BC ##T(0,t)=T_\infty## and ##T(L,t)=T_\infty## for ##t>0## and the initial condition ##T(x,0)=T_0## The author proceeds to 'normalize' the PDE in order to make the BC homogeneous. He has the following ## \displaystyle \alpha=\frac {k}{\rho c_0}##, ## \bar x = x/L##, ##\displaystyle \bar t = \frac {\alpha t}{L^2}##, ##\displaystyle u=\frac{TT_\infty}{T_0T_\infty}## This leads to ##\displaystyle \frac{\partial^2 u}{\partial x^2}+ \frac{\partial u}{\partial t}=0## ##u(0,t)=0##, ##u(1,t)=0## and ##u(x,0)=1## 1)How did he arrive at the first and third line from the bottom? 2) Why does he make the BC's homogeneous? 


#2
Dec412, 10:50 PM

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From the physics it's clear that T_{∞} is the asymptotic temperature everywhere, so it's natural to work in terms difference from there. (You then expect a factor e^{λt} in the answer.) Replacing quantities that have dimension by a fraction of some natural unit of that dimension (like, fraction of the way along a rod) also helps to unclutter the algebra. 


#3
Dec512, 01:21 PM

P: 660

For instance, if I was given a different set of IC's and BC's I wouldn't know how to proceed.. Thanks 


#4
Dec512, 02:21 PM

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P: 1,024

Normalizing the PDE to make BC's homogeous
Suppose
[tex] k\frac{\partial^2 T}{\partial x^2} = \rho c_0 \frac{\partial T}{\partial t} [/tex] for [itex]a < x < b[/itex] and [itex]t > 0[/itex] subject to the boundary conditions [itex]T(a,t)=T_\infty[/itex] and [itex]T(b,t)=T_\infty[/itex] and the initial condition [itex]T(x,0) = T_0[/itex]. The idea is to set [itex]x = a + L\tilde x[/itex] so that [itex]\tilde x = 0[/itex] when [itex]x = a[/itex] and [itex]\tilde x = 1[/itex] when [itex]x = b[/itex]. Clearly this requires [itex]L = b  a[/itex]. We also set [itex]t = S\tilde t[/itex], where S is to be chosen later. We now set [itex]T(x,t) = T_\infty + (T_0  T_\infty)u(\tilde x, \tilde t)[/itex]. With these substitutions we have the boundary conditions that [itex]u(0,\tilde t) = u(1,\tilde t) = 1[/itex] and the initial condition [itex]u(\tilde x,0) = 0[/itex]. We have, by the chain rule, [tex] \frac{\partial T}{\partial t} = \frac{(T_0  T_\infty)}{S}\frac{\partial u}{\partial \tilde t}, \\ \frac{\partial^2 T}{\partial x^2} = \frac{(T_0  T_\infty)}{L^2}\frac{\partial^2 u}{\partial \tilde x^2} [/tex] so that, substituting these into our intital PDE, [tex] k \frac{(T_0  T_\infty)}{L^2}\frac{\partial^2 u}{\partial \tilde x^2} = \rho c_0 \frac{(T_0  T_\infty)}{S}\frac{\partial u}{\partial \tilde t}. [/tex] Now it is convenient to choose S so that [tex] k\frac{(T_0  T_\infty)}{L^2} = \rho c_0 \frac{(T_0  T_\infty)}{S} [/tex] which requires [tex]S = \frac{\rho c_0 L^2}{k}[/tex] and our PDE is now [tex] \frac{\partial^2 u}{\partial \tilde x^2} = \frac{\partial u}{\partial \tilde t} [/tex] At this point it is conventional to drop the tildes. This is only possible because the boundary conditions at each end are equal; in general, if we have boundary conditions [itex]T(a,t) = T_1[/itex] and [itex]T(b,t) = T_2[/itex], then we can set [itex]T = T_{1} + (T_{0}  T_{1})u[/itex] as before, but our boundary conditions become [itex]u(0,t) = 1[/itex] and [itex]u(1,t) = \theta[/itex] where [itex]\theta = (T_2  T_1)/(T_0  T_1)[/itex]. This is an example of Nondimensionalization. 


#5
Dec512, 04:58 PM

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#6
Dec612, 10:50 AM

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#7
Dec712, 01:01 PM

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