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Change of Variables and Chain Rule

  1. Jan 11, 2012 #1
    1. The problem statement, all variables and given/known data
    I am trying to solve the transport PDE using a change of variables and the chain rule, and my problem seems to be with the chain rule. The PDE is:

    [itex]\frac{\partial u}{\partial t}[/itex]+c[itex]\frac{\partial u}{\partial x}[/itex] = 0 ...................(1)

    The change of variables (change of reference frame) is:

    [itex]\xi[/itex] = x - ct......................(2)

    From this we know that

    u(t,x) = v(t, x - ct) = v(t, [itex]\xi[/itex])........................(3)

    3. The attempt at a solution

    Taking the total derivative of both sides of (3)

    [itex]\frac{d}{dt}[/itex] [u(t,x) = v(t,[itex]\xi[/itex])]

    using chain rule yields

    [itex]\frac{\partial u}{\partial t}[/itex] = [itex]\frac{\partial v}{\partial t}[/itex] - c[itex]\frac{\partial u}{\partial x}[/itex].......................(4)

    I think the next step is to PROVE that

    [itex]\frac{\partial u}{\partial x}[/itex] = [itex]\frac{\partial v}{\partial \xi}[/itex].....................(5)

    and then substitute (4) into (5) to get

    [itex]\frac{\partial v}{\partial t}[/itex] = 0

    but i am not sure how to do this...Any help Greatly Appreciated

    Thanks!
     
    Last edited: Jan 11, 2012
  2. jcsd
  3. Jan 11, 2012 #2

    LCKurtz

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    If I understand correctly, I think you are basically done. You have $$
    \frac{\partial v(t,\xi)}{\partial t}=0$$Take the anti-partial derivative of this with respect to t. Here your constant of integration is an arbitrary function of the other variable:$$
    v(t,\xi) = f(\xi) = f(x-ct)$$so you have$$
    u(x,t)=v(t,\xi)=f(\xi)=f(x-ct)$$Unless, of course, you already understood that and I misunderstand what your question is.
     
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