Transport Equation

[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 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

[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.

[LCKurtz]

It is frowned upon in these forums to multi-post your questions. This thread is apparently a continuation of:

http://www.physicsforums.com/showthread.php?t=566713

[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.

Sorry again...Didn't mean to waste your time reading two posts.

[Stalker_VT]

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.

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