Transport Equation

1. Jan 10, 2012

Stalker_VT

how do you use the change of variable:
$\xi$ = x - ct

on the PDE
ut + c ux = 0

where u(t,x) = v(t,x-ct) = v(t,$\xi$)

to get

vt = 0

I am trying to follow the steps of Peter J. Olver in this document http://www.math.umn.edu/~olver/pd_/lnw.pdf [Broken] starting on pg 4 but get stuck here...i think i am missing an important calculus concept but cannot figure out what. He says to use the chain rule but i cannot figure out how that helps.

Any help GREATLY appreciated

Last edited by a moderator: May 5, 2017
2. Jan 12, 2012

bigfooted

first use $\xi=x-ct$ to eliminate x:
$u(t,x)=v(t,\xi(t))$
Olver mentions the multivariate calculus rule, meaning you should apply the following:
$\frac{\partial u(a(t),b(t))}{\partial t}=\frac{\partial u}{\partial a}\frac{\partial a(t)}{\partial t} + \frac{\partial u}{\partial b}\frac{\partial b(t)}{\partial t}$

In this case
$a(t)=t$
and
$b(t)=\xi(t)$
$\frac{\partial u(t,\xi(t))}{\partial t}=\frac{\partial u}{\partial t}1 + \frac{\partial u}{\partial \xi}\frac{\partial \xi}{\partial t}$
and note that
$\frac{\partial \xi}{\partial t}=-c$

do something similar for $u_x$ and substitute.

3. Jan 12, 2012

LCKurtz

4. Jan 13, 2012

Stalker_VT

O I am sorry, I am new to these forums and was not sure which section to post in because it was "like" a homework question, but I really was just looking for the concept behind that one step.

Last edited: Jan 13, 2012
5. Jan 13, 2012

Stalker_VT

Thanks for the reply, I got the first step of what you did, and got the equation

$\frac{\partial u}{\partial t} = \frac{\partial v}{\partial t} - c \frac{\partial v}{\partial \xi}$

And I realized (by trying do to do it) that when you do a similar thing for ux you get an equation very similar to the first. It does not seem to give you the relation

$\frac{\partial u}{\partial x} = \frac{ \partial v}{\partial\xi}$

which is the step i cannot figure out