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Wave Equation, stuck on a partial calculation

  1. Aug 30, 2008 #1
    Hey everybody,

    My professor started our PDE I class in Chapter six, so I am having a hard time with the really basic stuff to get the theory down.

    One of my questions to answer is to verify a solution by using direct substitution.

    [tex]u(x,t) \ = \ \frac{1}{2}\left[\phi(x+t) \ + \ \phi(x-t) \right] \ + \ \frac{1}{2} \int^{x+t}_{x-t}\Psi(s)ds [/tex]

    With initial conditions

    [tex] u(x,t_{0}) = \phi(x) \ , \ \frac{\partial u}{\partial t} (x,t_{0}) = \Psi(x), \ and \ t_{0} = 0 [/tex]

    satisfies [tex]\frac{\partial^{2}u}{\partial x^{2}} - \frac{\partial^{2}u}{\partial t^{2}} = 0 [/tex]

    It was easy for me to plug and chug to show that [tex] u(x,t_{0}) = \phi(x) \ and \ \frac{\partial u}{\partial t} (x,t_{0}) = \Psi(x) [/tex]

    Clearly my next step is to find [tex] \frac{\partial^{2}u}{\partial x^{2}} [/tex]

    but that's the step on which I'm stuck. Can someone get me started? If someone can show me how to do the second partial w.r.t x, it would be a good exercise for me to figure out the second partial w.r.t t. I apply the chain rule and just get a bunch of garbage back, which means I'm messing it up somewhere.

    Any input is appreciated.
  2. jcsd
  3. Aug 30, 2008 #2


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    You need "Leibniz' formula":
    [tex]\frac{d}{dx}\int_{\alpha(x)}^{\beta(x)} \phi(x,t)dt= \frac{d\beta}{dx}\phi(x,\beta(x))- \frac{d\alpha}{dx}\phi(x,\alpha(x))+ \int_{\alpha(x)}^{\beta(x)} \frac{\partial \phi}{\partial x}dt[/tex]

    It's really just applying the chain rule correctly.
  4. Aug 31, 2008 #3
    \frac{d}{dx}\int_{\alpha(x)}^{\beta(x)} \phi(x,t)dt= \frac{d\beta}{dx}\phi(x,\beta(x))- \frac{d\alpha}{dx}\phi(x,\alpha(x))+ \int_{\alpha(x)}^{\beta(x)} \frac{\partial \phi}{\partial x}dt

    I'm having trouble understanding this. It looks like the integrand you gave is a function of 2 variables, but the integrand I have is a function of one variable. I'm not sure what to do with that.

    [tex] \frac{\partial}{\partial x} (\phi(x+t)) = ? [/tex]

    is it [tex] = \phi '(x + t), \ or \ \phi_{x}(x+t) \ ? [/tex]

    In the notation of a function, how do I write that? I guess I'm having a notational brain fart.
    Last edited: Aug 31, 2008
  5. Aug 31, 2008 #4
    I found this worked out in Walter Strauss's book, so I guess I don't need help anymore. Thanks for looking :)
  6. Sep 1, 2008 #5


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    In that case,
    [tex]\frac{\partial\phi}{\partial x}= 0[/tex]
    and the formula becomes
    [tex] \frac{d}{dx}\int_{\alpha(x)}^{\beta(x)} \phi(x,t)dt= \frac{d\beta}{dx}\phi(x,\beta(x))- \frac{d\alpha}{dx}\phi(x,\alpha(x)) [/tex]

    Use the chain rule. If [itex]\phi'(u)[/itex] is the derivative of [itex]\phi[/itex] as a function of the single variable u, then
    [tex]\frac{\partial\phi(x+t)}{\partial x}= \phi'(x+t)(1)= \phi'(x+t)[/tex]
  7. Sep 3, 2008 #6
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