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Homework Help: 1-D wave-equation and change of variables

  1. Aug 23, 2008 #1
    1. The problem statement, all variables and given/known data
    Hi all.

    I have the 1-D wave-equation, and I wish to make a change of variables, where a = x+ct and b = x-ct. I get:

    [tex]
    \begin{array}{l}
    c^2 \frac{{\partial ^2 u}}{{\partial x^2 }} = c^2 \left[ {\frac{{d^2 u}}{{da^2 }}\left( {\frac{{da}}{{dx}}} \right)^2 + \frac{{du}}{{da}}\frac{{d^2 a}}{{dx^2 }}} \right] + c^2 \left[ {\frac{{d^2 u}}{{db^2 }}\left( {\frac{{db}}{{dx}}} \right)^2 + \frac{{du}}{{db}}\frac{{d^2 b}}{{dx^2 }}} \right] = c^2 \frac{{d^2 u}}{{da^2 }} + c^2 \frac{{d^2 u}}{{db^2 }} \\
    \frac{{\partial ^2 u}}{{\partial t^2 }} = \left[ {\frac{{d^2 u}}{{da^2 }}\left( {\frac{{da}}{{dt}}} \right)^2 + \frac{{du}}{{da}}\frac{{d^2 a}}{{dt^2 }}} \right] + \left[ {\frac{{d^2 u}}{{db^2 }}\left( {\frac{{db}}{{dt}}} \right)^2 + \frac{{du}}{{db}}\frac{{d^2 b}}{{dt^2 }}} \right] = c^2 \frac{{d^2 u}}{{da^2 }} + c^2 \frac{{d^2 u}}{{db^2 }} \\
    \end{array}
    [/tex]

    For this I have used the chain rule for higher derivates (for second derivates, from Wikipedia: http://en.wikipedia.org/wiki/Chain_rule). The result I wish to get is:

    [tex]
    \frac{{\partial ^2 u}}{{\partial a\partial b}} = 0
    [/tex]

    I can't quite see how I would get this. Am I on the right track here?

    Cheers,
    Niles.


    EDIT: Ok, now this is the second thread in a row I am doing this: I preview the thread, and the title resets itself. The original title was: "1-D wave-equation and change of variables". If a moderator can insert the proper title, I would be grateful.
     
  2. jcsd
  3. Aug 23, 2008 #2

    cristo

    User Avatar
    Staff Emeritus
    Science Advisor

    I've changed the title. Thanks for letting us know that it was the preview that causes the problem.
     
  4. Aug 24, 2008 #3
    Ok, my first attempt is quite bad. Instead of using the expression from Wikipedia, I can just derive du/dx and du/dt with respect to dx and dt, respectively. But still, this gives me:

    [tex]
    c^2 \frac{{\partial ^2 u}}{{\partial x\partial a}} + c^2 \frac{{\partial ^2 u}}{{\partial x\partial b}} = c\frac{{\partial ^2 u}}{{\partial t\partial a}} - c\frac{{\partial ^2 u}}{{\partial t\partial b}}
    [/tex]

    Is it possible to go from this to

    [tex]

    \frac{{\partial ^2 u}}{{\partial a\partial b}} = 0

    [/tex]
    ?
     
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