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Fourier Transform plus Shift

  1. Jun 4, 2006 #1
    Find the solution (in integral form) of the equation:

    [tex]
    u(x+1,t) - 2u(x,t) + u(x-1,t) = u_t
    [/tex]
    [tex]u(x,0) = f(x)[/tex]

    Hint: Use the shift formula

    [tex]
    F[f(ax-b)] = \frac{\exp{i\omega b/a}}{|a|} \overline{f}(\omega/a)
    [/tex]

    So I took the Fourier transform of each term using the shift formula:

    [tex]
    \exp{(-i\omega)} \overline{u} - 2\overline{u} + \exp{(i\omega)}\overline{u} = \overline{u}_t
    [/tex]

    But I don't think this is correct thus far...
     
    Last edited: Jun 4, 2006
  2. jcsd
  3. Jun 4, 2006 #2
    Why don't you think it is right? Also, that expression is not fully simplified. Maybe you were expecting something with sinusoids?
     
  4. Jun 4, 2006 #3
    Well I guess I get stuck on how to solve the ODE:

    [tex]
    (\exp{(-i\omega)} - 2 + \exp{(i\omega)})\overline{u} = \overline{u}_t
    [/tex]

    So [tex](\exp{(-i\omega)} - 2 + \exp{(i\omega)})[/tex] can be treated as a constant with respect to t. But when I solve I get exponents raised to exponents...


    The answer is suppose to be (limits of integration -inf to +inf):

    [tex]
    \frac{1}{2\pi}\int{e^{-i\omega x-2(1-\cos{\omega})t} \int{e^{i\omega\xi}f(\xi)d\xi}d\omega}
    [/tex]
     
    Last edited: Jun 4, 2006
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