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1-D Wave Eqn

  1. Oct 21, 2007 #1
    [SOLVED] 1-D Wave Eqn

    Alright, so this problem is giving me troubles, and I must just be missing the trick. The equation to solve is the one dimensional wave equation with isotropic, homogeneous, etc. (i.e. wave in a vacuum). Which means the PDE is

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

    with the boundary conditions that each end is fixed at zero, [itex]u(0,t) = 0[/itex] and [itex]u(L, t)[/itex]. I also know the initial conditions for the initial position and initial velocity are functions of x [itex] u(x,0) = f(x)[/itex] and [itex]\frac{\partial u}{\partial t}(x, 0) = g(x) [/itex]. My job is to show that the solution follows d'Alembert's solution (which I can prove), that the wave equation can be written as [itex]u(x,t) = R(x-ct) + S(x+ct)[/itex] where R and S are some functions.

    The solution is easily found

    [tex]u(x,t) = \sum_{n=1}^\infty \left(A_n sin\frac{n \pi x}{L} cos\frac{n \pi c t}{L} + B_n sin\frac{n \pi x}{L} sin\frac{n \pi c t}{L} \right)[/tex]

    Mostly what I am having trouble with is getting rid of the summations. The first initial condition is helpful because

    [tex] \sum_{n=1}^\infty A_n sin \frac{n \pi x}{L} = f(x)[/tex]

    which can be directly substituted in (though I don't know if I want or need to). The other IC is not so helpful

    [tex] \sum_{n=1}^\infty \frac{n \pi c}{L} B_n sin\frac{n \pi x}{L} = g(x)[/tex]

    and I cannot really do much other. The other thing I know I have to do is use trig identities, which will rewrite the solution as

    [tex]u(x,t) = \sum \left(\frac{A_n}{2} [sin \frac{n \pi}{L} (x + ct) + sin \frac{n \pi}{L} (x - ct)] + \frac{B_n}{2}[cos \frac{n \pi}{L} (x - ct) - cos\frac{n \pi}{L} (x + ct)] \right)[/tex]

    This starts to towards making some functions, and I converted this all into exponentials to get (after some grouping)

    [tex] u(x,t) = \sum (\frac{A'_n}{2i} - \frac{B'_n}{2}) Exp(i\frac{n\pi}{L}(x+ct)) + (-\frac{A'_n}{2i} - \frac{B'_n}{2}) Exp(-i\frac{n\pi}{L}(x+ct)) + (\frac{A'_n}{2i} + \frac{B'_n}{2}) Exp(i\frac{n\pi}{L}(x-ct)) + (-\frac{A'_n}{2i} + \frac{B'_n}{2}) Exp(-i\frac{n\pi}{L}(x-ct))[/tex]

    Where [itex]A'_n = A_n/2[/itex] and [itex]B'_n = B_n/2[/itex].

    Unfortunately, I can't rewrite these back into sines and cosines as two functions. Does someone know the trick, specifically to this problem, or in general, that gets from the superposed answer to the functional answer?
     
    Last edited: Oct 21, 2007
  2. jcsd
  3. Oct 21, 2007 #2
    Ok

    make a bijective coordinate change

    [tex]
    \begin{gathered}
    \xi \left( {x,t} \right) \hfill \\
    \psi (x,t) \hfill \\
    \end{gathered}
    [/tex]

    Using the chain rule for differentiation you get

    [tex]
    \begin{gathered}
    \frac{{\partial ^2 u}}
    {{\partial t^2 }} = \frac{{\partial ^2 u}}
    {{\partial \psi ^2 }}\left( {\frac{{\partial \psi }}
    {{\partial t}}} \right)^2 + 2\frac{{\partial ^2 u}}
    {{\partial \xi \partial \psi }}\frac{{\partial \xi }}
    {{\partial t}}\frac{{\partial \psi }}
    {{\partial t}} + \frac{{\partial u}}
    {{\partial \psi }}\frac{{\partial ^2 \psi }}
    {{\partial t^2 }} \hfill \\
    + \frac{{\partial ^2 u}}
    {{\partial \xi ^2 }}\left( {\frac{{\partial \xi }}
    {{\partial t}}} \right)^2 + \frac{{\partial u}}
    {{\partial \xi }}\frac{{\partial ^2 \xi }}
    {{\partial t^2 }} \hfill \\
    \end{gathered}
    [/tex]

    do the same thing for

    [tex]
    \frac{{\partial ^2 u}}
    {{\partial x^2 }}
    [/tex]

    plug all this crap into the wave equation, and bring everything to the same side, and you get.

    [tex]
    \begin{gathered}
    \left( {\left( {\frac{{\partial \psi }}
    {{\partial t}}} \right)^2 - c^2 \left( {\frac{{\partial \psi }}
    {{\partial x}}} \right)^2 } \right)\frac{{\partial ^2 u}}
    {{\partial \psi ^2 }} + 2\frac{{\partial ^2 u}}
    {{\partial \xi \partial \psi }}\left( {\frac{{\partial \xi }}
    {{\partial t}}\frac{{\partial \psi }}
    {{\partial t}} - c^2 \frac{{\partial \xi }}
    {{\partial x}}\frac{{\partial \psi }}
    {{\partial x}}} \right) \hfill \\
    \left( {\left( {\frac{{\partial \xi }}
    {{\partial t}}} \right)^2 - c^2 \left( {\frac{{\partial \xi }}
    {{\partial x}}} \right)^2 } \right)\frac{{\partial ^2 u}}
    {{\partial \xi ^2 }} + \left( {\frac{{\partial ^2 \psi }}
    {{\partial t^2 }} - c^2 \frac{{\partial ^2 \psi }}
    {{\partial x^2 }}} \right)\frac{{\partial u}}
    {{\partial \psi }} + \left( {\frac{{\partial ^2 \xi }}
    {{\partial t^2 }} - c^2 \frac{{\partial ^2 \xi }}
    {{\partial x^2 }}} \right)\frac{{\partial u}}
    {{\partial \xi }} = 0 \hfill \\
    \end{gathered}
    [/tex]

    Now, you can pick [tex]\xi ,\psi[/tex] to be whatever you want as long as it's a bijective map. Lets pick:

    [tex]
    \begin{gathered}
    \psi = x + ct \hfill \\
    \xi = x - ct \hfill \\
    \end{gathered}
    [/tex]

    This simplifies everything to [tex]
    - 4c^2 \frac{{\partial ^2 u}}
    {{\partial \psi \partial \xi }} = 0
    [/tex]

    so

    [tex]
    \frac{{\partial ^2 u}}
    {{\partial \psi \partial \xi }} = 0
    [/tex]

    This is easy enough to integrate

    namely:

    [tex]
    \begin{gathered}
    \frac{{\partial u}}
    {{\partial \psi }} = \frac{{\partial S(\psi )}}
    {{\partial \psi }} \hfill \\
    u\left( {\xi ,\psi } \right) = S\left( \psi \right) + R\left( \xi \right) \hfill \\
    \therefore u\left( {x,t} \right) = S\left( {x + ct} \right) + R\left( {x - ct} \right) \hfill \\
    \end{gathered}
    [/tex]
     
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