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General solution for a standing wave to the wave equation.

  1. Apr 26, 2015 #1
    1. The problem statement, all variables and given/known data

    If ##u(x,t)\in \mathbb R, x\in[-\pi,\pi] ## represents a standing wave with ##u(\pm\pi,t)=0## Then what is the most general solution u(x,t)?

    2. Relevant equations


    3. The attempt at a solution

    Using the separation of variable technique: u(x,t)=P(x)Q(t)
    I get ##P(x) =A\sin{(\lambda x)}+B\cos{(\lambda x)}##
    ##u(\pm\pi,t)=0## implies either (1) or (2);
    (1) A=0 and ##\lambda=\frac{2n-1}{2}## so ##P(x)=B\cos{(\frac{2n-1}{2}x)}##
    (2) B=0 and ##\lambda=n## so ##P(x)=A\sin{(nx)}##
    If (1) then ##Q(t)=C_n\sin{(\frac{2n-1}{2}ct+\phi)}## and ##u(x,t)=\sum_{n=1}^\infty b_n\sin{(\frac{2n-1}{2}ct+\phi)}\cos{(\frac{2n-1}{2}x)}##
    If (2) then ##Q(t)=C_n\sin{(nct+\phi)}## and ##u(x,t)=\sum_{n=1}^\infty a_n\sin{(nct+\phi)}\sin{(nx)}##
    So i would be inclined to say the most general solution is linear combinations of (1) and (2);
    ##u(x,t)=\sum_{n=1}^\infty a_n\sin{(nct+\phi)}\sin{(nx)} + \sum_{n=1}^\infty b_n\sin{(\frac{2n-1}{2}ct+\phi)}\cos{(\frac{2n-1}{2}x)}##

    However this is what the solution says;
    ##u(x,t)=\sum_{n=1}^\infty a_n\sin{(nx)}\sin{(nct+\phi)}+\sum_{n=1}^\infty b_n\cos{(\frac{2n-1}{2}x)}\sin{(nct+\phi)}##
    I don't quite understand this solution because in one part of u(x,t) it has quantized lamda as integers (##\lambda=n##) and in another part it has quantized lamda as odd multiples divided by two (##\lambda=\frac{2n-1}{2}##). Its as if the solution has taken linear combinations of my (1) and (2) and let lambda equal different values. Are you allowed to do this?
     
    Last edited: Apr 26, 2015
  2. jcsd
  3. Apr 26, 2015 #2

    BvU

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    I see nothing wrong with your reasoning, so I suspect you found an error in the book solutions.
     
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