# Normalizing the PDE to make BC's homogeous

by bugatti79
Tags: homogeous, normalizing
 P: 652 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{T-T_\infty}{T_0-T_\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?
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P: 8,880
 Quote by bugatti79 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{T-T_\infty}{T_0-T_\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?
You mean, how did he choose those substitutions?
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.
 2) Why does he make the BC's homogeneous?
Just to simplify writing and reading the equations. The analysis thereafter could have been done without any of these substitutions, but it would have been a bit harder to follow.
P: 652
 Quote by haruspex You mean, how did he choose those substitutions? 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. Just to simplify writing and reading the equations. The analysis thereafter could have been done without any of these substitutions, but it would have been a bit harder to follow.
ok, I get the idea of what your are saying. Is there a mathematical procedure of where he arrived?

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

P: 679

## Normalizing the PDE to make BC's homogeous

Suppose
$$k\frac{\partial^2 T}{\partial x^2} = \rho c_0 \frac{\partial T}{\partial t}$$
for $a < x < b$ and $t > 0$ subject to the boundary conditions $T(a,t)=T_\infty$ and $T(b,t)=T_\infty$ and the initial condition $T(x,0) = T_0$.

The idea is to set $x = a + L\tilde x$ so that $\tilde x = 0$ when $x = a$ and $\tilde x = 1$ when $x = b$. Clearly this requires $L = b - a$.

We also set $t = S\tilde t$, where S is to be chosen later.

We now set $T(x,t) = T_\infty + (T_0 - T_\infty)u(\tilde x, \tilde t)$. With these substitutions we have the boundary conditions that $u(0,\tilde t) = u(1,\tilde t) = 1$ and the initial condition $u(\tilde x,0) = 0$.

We have, by the chain rule,
$$\frac{\partial T}{\partial t} = \frac{(T_0 - T_\infty)}{S}\frac{\partial u}{\partial \tilde t}, \\ \frac{\partial^2 T}{\partial x^2} = \frac{(T_0 - T_\infty)}{L^2}\frac{\partial^2 u}{\partial \tilde x^2}$$
so that, substituting these into our intital PDE,
$$k \frac{(T_0 - T_\infty)}{L^2}\frac{\partial^2 u}{\partial \tilde x^2} = \rho c_0 \frac{(T_0 - T_\infty)}{S}\frac{\partial u}{\partial \tilde t}.$$
Now it is convenient to choose S so that
$$k\frac{(T_0 - T_\infty)}{L^2} = \rho c_0 \frac{(T_0 - T_\infty)}{S}$$
which requires
$$S = \frac{\rho c_0 L^2}{k}$$
and our PDE is now
$$\frac{\partial^2 u}{\partial \tilde x^2} = \frac{\partial u}{\partial \tilde t}$$
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 $T(a,t) = T_1$ and $T(b,t) = T_2$, then we can set $T = T_{1} + (T_{0} - T_{1})u$ as before, but our boundary conditions become $u(0,t) = 1$ and $u(1,t) = \theta$ where $\theta = (T_2 - T_1)/(T_0 - T_1)$.

This is an example of Nondimensionalization.
P: 652
 Quote by pasmith Suppose $$k\frac{\partial^2 T}{\partial x^2} = \rho c_0 \frac{\partial T}{\partial t}$$ for $a < x < b$ and $t > 0$ subject to the boundary conditions $T(a,t)=T_\infty$ and $T(b,t)=T_\infty$ and the initial condition $T(x,0) = T_0$. The idea is to set $x = a + L\tilde x$ so that $\tilde x = 0$ when $x = a$ and $\tilde x = 1$ when $x = b$. Clearly this requires $L = b - a$. We also set $t = S\tilde t$, where S is to be chosen later. We now set $T(x,t) = T_\infty + (T_0 - T_\infty)u(\tilde x, \tilde t)$. With these substitutions we have the boundary conditions that $u(0,\tilde t) = u(1,\tilde t) = 1$ and the initial condition $u(\tilde x,0) = 0$. We have, by the chain rule, $$\frac{\partial T}{\partial t} = \frac{(T_0 - T_\infty)}{S}\frac{\partial u}{\partial \tilde t}, \\ \frac{\partial^2 T}{\partial x^2} = \frac{(T_0 - T_\infty)}{L^2}\frac{\partial^2 u}{\partial \tilde x^2}$$ so that, substituting these into our intital PDE, $$k \frac{(T_0 - T_\infty)}{L^2}\frac{\partial^2 u}{\partial \tilde x^2} = \rho c_0 \frac{(T_0 - T_\infty)}{S}\frac{\partial u}{\partial \tilde t}.$$ Now it is convenient to choose S so that $$k\frac{(T_0 - T_\infty)}{L^2} = \rho c_0 \frac{(T_0 - T_\infty)}{S}$$ which requires $$S = \frac{\rho c_0 L^2}{k}$$ and our PDE is now $$\frac{\partial^2 u}{\partial \tilde x^2} = \frac{\partial u}{\partial \tilde t}$$ 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 $T(a,t) = T_1$ and $T(b,t) = T_2$, then we can set $T = T_{1} + (T_{0} - T_{1})u$ as before, but our boundary conditions become $u(0,t) = 1$ and $u(1,t) = \theta$ where $\theta = (T_2 - T_1)/(T_0 - T_1)$. This is an example of Nondimensionalization.
Thanks for this excellent post. This nondimensionalisation I never heard before. Regards
P: 679
 Quote by pasmith We now set $T(x,t) = T_\infty + (T_0 - T_\infty)u(\tilde x, \tilde t)$. With these substitutions we have the boundary conditions that $u(0,\tilde t) = u(1,\tilde t) = 1$ and the initial condition $u(\tilde x,0) = 0$.
This should read "we have the boundary conditions that $u(0,\tilde t) = u(1,\tilde t) = 0$ and the initial condition $u(\tilde x,0) = 1$".

 This is only possible because the boundary conditions at each end are equal; in general, if we have boundary conditions $T(a,t) = T_1$ and $T(b,t) = T_2$, then we can set $T = T_{1} + (T_{0} - T_{1})u$ as before, but our boundary conditions become $u(0,t) = 1$ and $u(1,t) = \theta$ where $\theta = (T_2 - T_1)/(T_0 - T_1)$.
This should read "our boundary conditions become $u(0,t) = 0$ and $u(1,t) = \theta$".
P: 652
 Quote by pasmith This should read "we have the boundary conditions that $u(0,\tilde t) = u(1,\tilde t) = 0$ and the initial condition $u(\tilde x,0) = 1$". This should read "our boundary conditions become $u(0,t) = 0$ and $u(1,t) = \theta$".
Noted. Thanks for the update.

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