Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Wave on a string and the chain rule Argh

  1. Jul 24, 2008 #1
    Wave on a string and the chain rule....Argh

    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

    Ben Niehoff

    User Avatar
    Science Advisor
    Gold Member

    Re: Wave on a string and the chain rule....Argh

    Hmm...I think there is an error at the beginning. It seems to be that you ought to have

    [tex]u = f(\alpha) + g(\beta)[/tex]

    as this would correspond to the general solution of the one-dimensional wave equation,

    [tex]u = f(x - ct) + g(x + ct)[/tex]

    Then, taking derivatives, you get

    [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]

    and then

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

    I think this is where you made your mistake. You have to apply the chain rule twice in order to take the second derivative!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Threads - Wave string chain Date
A Wave equation in free space Jan 7, 2018
B Wave equation, psi with dots and things like that... Dec 29, 2017
A Spheroidal Wave Equation Dec 15, 2017
Wave Equation String BVP Feb 3, 2009
Wave equation for a string Feb 6, 2005