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Deriving d'Alemberts solution - Boundary conditions

  1. Feb 5, 2013 #1
    Hi,

    I shall show (using Fourier transform) that the solution to

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

    is

    [tex]
    u(x,t) = (f(x+t) + f(x-t))/2
    [/tex]

    I got it almost: Taking the Fourier transform in the variable x (leaving t constant) and the derivative theorem gives:

    [tex]
    \hat{u}_{tt}(s,t) + 4\pi^2 s^2 \hat{u}(s,t) = 0
    [/tex]

    And the solution to this differential equation is:

    [tex]
    \hat{u}(s,t) = A e^{j2\pi s t} + B e^{-j2\pi s t}
    [/tex]

    However, now I am stuck in incorporating the boundary conditions and I do not see why A and B should become [itex]f(x)=u(x,0)[/itex].

    My approach was: Since u(x,t) was transferred to Fourier domain via x rather than t, [itex]u(s,0) = \hat{u}(x,0)[/itex]. Derivating the solution for t and filling t=0 gives A=B. But I neither see the factor of 1/2 there, nore why A and B should be f(x). Can anyone give me a pointer?

    Thanks!
     
  2. jcsd
  3. Feb 6, 2013 #2
    Got it; finally!
    It is so trivial from where I was: A=B was correct, it is a constant regarding t but not s, so A(s)=B(s).

    The other "boundary" condition is u(x,0)=f(x), so when transforming back my solution: u(x,t) = A(x+t)+B(x-t) with t=0: u(x,0)=A(x)+B(x) which means that A(x)=B(x)=f(x)/2
     
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