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Homework Help: Wave on a string, and the chain rule argh

  1. Jul 24, 2008 #1
    So, I am working through the wave equation for a review before my friend and I go off to grad school. It has been a couple of years since we both graduated with our BS in Physics.

    So, here is the question:

    Suppose I want to solve the wave equation using a change of variables. Lets use [tex]\alpha = x+ct[/tex], and [tex]\beta = x-ct, and\: u = \alpha + \beta[/tex]

    The wave equation is
    [tex]\frac{\partial^{2} u}{\partial t^{2}}=c^{2}\frac{\partial^{2} u}{\partial x^{2}}[/tex]

    Now, if we take the partial derivative of u with respect to x and applying the chain rule one gets:

    [tex]\frac{\partial u}{\partial x}=\frac{\partial u}{\partial \alpha}\frac{\partial \alpha}{\partial x}+\frac{\partial u}{\partial \beta}\frac{\partial \beta}{\partial x}[/tex]

    Now if we evaluate [tex]\frac{\partial \alpha}{\partial x}, and \frac{\partial \beta}{\partial x} [/tex] we get
    [tex]\frac{\partial u}{\partial x} = \frac{\partial u}{\partial \alpha} + \frac {\partial u}{\partial \beta} [/tex]

    So, what and how do I evaluate the second partial differential with respect to x? I get

    [tex] \frac{\partial^{2}u}{\partial x^{2}} = \frac{\partial^{2}u}{\partial \alpha^{2}}\ \frac{\partial \alpha}{\partial x}+\frac{\partial^{2}u}{\partial \beta^{2}}\ \frac{\partial \beta}{\partial x} [/tex]

    Now, I know this isn't quite right. I am supposed to get:
    [tex] \frac{\partial^{2}u}{\partial x^{2}} = \frac{\partial^{2}u}{\partial \alpha^{2}}\ \frac{\partial \alpha}{\partial x}+ \frac{\partial^{2}u}{\partial \alpha \partial \beta}}\ \frac{\partial \beta}{\partial x} + \frac{\partial^{2}u}{\partial \beta \partial \alpha}}\ \frac{\partial \alpha}{\partial x} +\frac{\partial^{2}u}{\partial \beta^{2}}\ \frac{\partial \beta}{\partial x} [/tex]

    Can anyone help me? Thanks.
  2. jcsd
  3. Jul 24, 2008 #2
    just think about what d/dx is from your first chain rule equation (factor out the u on which the differential acts on). then use this to find the second derivative and u'll get the mixed terms.
  4. Jul 24, 2008 #3


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    When you are going for the second derivative, apply the operator:

    \frac{\partial}{\partial x} = \frac{\partial \alpha}{\partial x} \frac{\partial}{\partial \alpha} + \frac{\partial \beta}{\partial x} \frac {\partial }{\partial \beta}

    to [tex]\frac{\partial u}{\partial x}[/tex].
    Last edited: Jul 24, 2008
  5. Jul 24, 2008 #4
    Thanks for the input guys. That works. I just didn't think to look at it that way and treat the chain rule as an operator.
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