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D'Alembert's Solution to wave equation

  1. Dec 30, 2013 #1
    Hello,

    How does the change of variables ## \alpha = x + at , \quad \beta = x - at ## change the differential equation

    $$ a^2 \frac{ \partial ^2 y}{ \partial x^2 } = \frac{ \partial ^2 y} {\partial t ^2} $$

    to

    $$ \frac{ \partial ^2 y}{\partial \alpha \partial \beta } = 0$$

    ? I'm having a hard time following the proofs on wolfram alpha, etc
     
  2. jcsd
  3. Dec 30, 2013 #2
    I think I've got it, although I don't think this is a legitimate proof:

    $$ \frac{ \partial \alpha}{\partial t} = a, \quad \frac{\partial \beta}{\partial t} = -a \implies \frac{1}{-a^2} \frac{ \partial \alpha \partial \beta}{\partial t^2 } = 1 $$

    By the same logic,

    $$ \frac{ \partial \alpha \partial \beta}{ \partial x^2} = 1$$

    From the wave equation:

    $$ a^2 \frac{ \partial ^2 y}{\partial x^2} \frac{\partial x^2}{\partial \alpha \partial \beta } = \frac{ \partial ^2 y}{\partial t^2} \frac{\partial t^2}{\partial \alpha \partial \beta} (-a^2) $$

    $$ \implies \frac{ \partial ^2 y}{\partial \alpha \partial \beta } = 0 $$

    Is that OK? It's cool to just treat everything like a fraction?
     
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