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I Ode using Fourier Transform

  1. Jan 2, 2018 #1
    Hi, let's take this ode:
    y''(t) = f(t),y(0)=0, y'(0)=0.
    using the FT it becomes:
    -w^2 Y(w) = F(w)
    Y(w)=( -1/w^2 )F(w)

    so i can say that -1/w^2 is the fourier transorm of the green's function(let's call it G(w)).
    then
    y(t) = g(t) * f(t)
    where
    g(t) = F^-1 (G(w))
    (inverse fourier transorm)
    how can i solve the integral to find g(t)?
    if f(t)=0 for t<0 and f(t)=1 for t>=0 how can i say that y(t)= 1/2t^2?



     
  2. jcsd
  3. Jan 2, 2018 #2

    Delta²

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    Last edited: Jan 2, 2018
  4. Jan 2, 2018 #3
    First of all thank you for the answer :)
    My problem is that i don't know why the inverse fourier tranform of -1/w^2 is tsgn(t).
    using the definition of inverse fourier transform I have to calculate an integral from -infinity to +infinity of a function that has some issue in w=0.
    I can use Jordan's lemma and the residue theorem but w=0 lies on the integration's curve and it's a second order pole.. how can i overcome this problem?
     
  5. Jan 2, 2018 #4

    Delta²

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    Don't use the definition of inverse fourier transform. Use the properties of fourier transform. We know that the fourier transform of sgn(t) is ##\frac{1}{iw}## (you can prove this by the definition of fourier transform). Therefore by a known property of fourier transform (property 107 here https://en.wikipedia.org/wiki/Fourier_transform#Tables_of_important_Fourier_transforms)

    it follows that the fourier transform of ##t~sgn(t)## will be

    ##i\frac{d}{dw}(\frac{1}{iw})=-\frac{1}{w^2}##
     
    Last edited by a moderator: Jan 18, 2018
  6. Jan 2, 2018 #5
    Thank you very much, this is a very simple and smart method to calculate some fourier transforms :) i will keep that in mind.
     
  7. Jan 2, 2018 #6

    jasonRF

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    If you are using the Fourier transform to solve this kind of problem then you need to be a little more careful. The solution of
    [tex]
    -\omega^2 Y(\omega) = F(\omega)
    [/tex]
    is
    [tex]
    Y(\omega) = - \frac{1}{\omega^2} F(\omega) + a \, \delta(\omega) + b \, \delta^\prime(\omega),
    [/tex]
    where ##a## and ##b## are arbitrary constants that you determine by using your initial conditions.

    By the way, the Laplace transform is a more straightforward way of solving these kind of problems, in my opinion.

    Jason
     
  8. Jan 2, 2018 #7
    Thank you :) yes i know that i have to add the solution of the associated homogeneus equation to find the complete solution..I wanted to use ft to solve this ode because i haven't studied laplace transform yet, thanks for the advice!
     
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