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Transport Equation

  1. Jan 10, 2012 #1
    how do you use the change of variable:
    [itex]\xi[/itex] = x - ct

    on the PDE
    ut + c ux = 0

    where u(t,x) = v(t,x-ct) = v(t,[itex]\xi[/itex])

    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. jcsd
  3. Jan 12, 2012 #2
    first use [itex]\xi=x-ct[/itex] to eliminate x:
    [itex]u(t,x)=v(t,\xi(t))[/itex]
    Olver mentions the multivariate calculus rule, meaning you should apply the following:
    [itex]\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}[/itex]

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

    do something similar for [itex]u_x[/itex] and substitute.
     
  4. Jan 12, 2012 #3

    LCKurtz

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  5. Jan 13, 2012 #4
    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.
     
    Last edited: Jan 13, 2012
  6. Jan 13, 2012 #5
    Thanks for the reply, I got the first step of what you did, and got the equation

    [itex]\frac{\partial u}{\partial t} = \frac{\partial v}{\partial t} - c \frac{\partial v}{\partial \xi}[/itex]

    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

    [itex]\frac{\partial u}{\partial x} = \frac{ \partial v}{\partial\xi}[/itex]

    which is the step i cannot figure out
     
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