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Fourier's method, division by zero

  1. Feb 18, 2015 #1
    1. The problem statement, all variables and given/known data
    Solve the BVP for a vibrating string with Separation of Variables/Fourier's method.

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

    The string is of length L with each end fixed, ie u(0,t) = u(L,t) = 0

    3. The attempt at a solution

    I know how to do Fourier's method but have a query regarding separation of variables. I make the assumption u(x,t) = X(x)T(t) for some functions X and T and then divide the LHS and RHS by X(x)T(t) to find that both are constant etcetc, but this step has me confused.

    Right so X(x)T(t) can't be 0 along the whole string because that would just be a trivial solution so the division is warranted, or that's how I remember being taught it in any case. But won't the string pass u(x,t) = 0? [just thinking physically it kind of has to if the coordinate system is set at u(x,t) = 0 when the string is simply at rest and stretched out, and IIRC the solutions do allow for it]. In that case either X(x) or T(t) will be zero at certain x / certain t so aren't we disregarding any solution that has the string going past u(x,t) = 0 by doing this division?

    In short, we divide by X(x)T(t) which takes on the value zero at certain (x, t). Why can we do this?

    I'm guessing it's rather simple but can't wrap my head around it, would appreciate some explaining thank you.
     
    Last edited: Feb 18, 2015
  2. jcsd
  3. Feb 18, 2015 #2

    Orodruin

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  4. Feb 19, 2015 #3

    ehild

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    If you search the solution in separable form, u(x,t)=X(x)T(t) and do the differentiations, you get the equation T" X = c2 T X". You might want solutions for which T"=AT and X"=BX holds, with A and B constants. So ATX-c2BTX = TX(A-c2B)=0 for all x and t. TX might be zero, but not everywhere and all time. As TX is not zero for all t and x, A-c2B = 0 must hold.

    Now you have the equations T"(t)=AT(t) and X"(x)=BX(x) to solve.
     
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