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Solve D'Alembert problem.

  1. Mar 24, 2010 #1
    1. The problem statement, all variables and given/known data

    [tex] \frac{\partial u^2}{\partial t^2} = c^2 \frac{\partial u^2}{\partial x^2} \;\;,\;\; u(0,t)=u(L,t)=0 \;\;,\;\; u(x,0)=f(x) \;\;,\;\; \frac{\partial u}{\partial t}(x,0)=g(x)[/tex]

    [tex]f(x) \;and\; g(x) \;are\; symmetric\; about\;\; x=\frac{L}{2} \;\Rightarrow f(L-x)=f(x) \;\;and\;\; g(L-x)=g(x)[/tex]

    Show [tex]u(x,t+\frac{L}{c})=-u(x,t)[/tex]

    2. Relevant equations

    [tex]u(x,t)=\frac{1}{2}[f(x+ct)+f(x-ct)]+\frac{1}{2}[G(x+ct)-G(x-ct)] \;\;\;where\;\;\; G(x)=\frac{1}{c}[G(x+ct)-G(x-ct)][/tex]

    [tex]u(-x,t)=-u(x,t) \;\;,\;\; u(x+2L,t)=u(x,t) \;\;,\;\; u(x-L,t)=u(x+L,t)[/tex]

    3. The attempt at a solution

    u(x,t) is periodic with T=2L.

    [tex]u(x, t+\frac{L}{c} ) =\frac{1}{2}[f(x+c (t+\frac{L}{c}) )+f(x-c(t+\frac{L}{c}) )]+\frac{1}{2}[G(x+c(t+\frac{L}{c}) )-G(x-c(t+\frac{L}{c}) )][/tex]

    [tex]\Rightarrow u(x, t+\frac{L}{c} ) =\frac{1}{2}[ f((x+L)+ct )+f((x-L)-ct)]+\frac{1}{2}[G((x+L)+ct)-G((x-L)-ct)][/tex]

    [tex]u(x-L,t)=u(x+L,t) \Rightarrow \; u(x, t+\frac{L}{c} ) =\frac{1}{2}[ f((x+L)+ct )+f((x+L)-ct)]+\frac{1}{2}[G((x+L)+ct)-G((x+L)-ct)][/tex]

    I can see odd and even function with symmetric at the middle of the period like sin(x) and cos(x) resp. That [tex]sin(x+\pi)=-sin(x) \;and\; cos(x+\pi)=-cos(x)[/tex]

    I just don't know how to express in mathametical terms. Can someone at least get me hints or answer?

  2. jcsd
  3. Mar 25, 2010 #2
  4. Mar 28, 2010 #3
    Can anyone at least give me some opinion even you might not have the answer?
  5. Mar 28, 2010 #4


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    Try expressing u(x,t) in terms of the normal modes.
  6. Mar 28, 2010 #5
    You mean in fouries series expansion? I'll look into this and post back. Thanks
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