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FT to solve 2nd order ODE; only one solution

  1. Sep 16, 2011 #1
    If I solve a simple 2nd order ODE using a Fourier transform, I only get one solution. E.g.:
    [tex] \frac{d^2f}{dx^2}=\delta [/tex]
    [tex] (2\pi ik)^2\tilde{f}=1 [/tex]
    [tex] \tilde{f}=\frac{1}{(2\pi ik)^2} [/tex]
    [tex] f = \frac{1}{2}xsgn(x) [/tex]

    However, the general solution is

    [tex] f = \frac{1}{2}xsgn(x) + Cx + D [/tex]

    Why do I only get one of the solutions? Are the solutions with C and D non-zero not also valid distributions whose second derivatives are the delta distribution?
  2. jcsd
  3. Sep 17, 2011 #2
    My current thinking on this is that we could start with
    [tex] f=\frac{1}{2}xsgn(x) + x + 1 [/tex]
    but as soon as we take the derivatives, we lose the last two terms, and the Fourier transform is then effectively operating on the C=D=0 solution. So it's not really surprising that we only get one solution from the Fourier transform method.
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