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I was using Fourier transform method to solve a non-linear differential equation and with $x(t)$ as the function I need to solve. Given that $F(x(t))=f(w)$, can we obtain a Fourier inverse of $F^{-1}(\frac{1}{f(w)})$ in terms of $x(t)$.

mrp
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Basham G
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  • I don't understand why $F^-1(f)$ is not why you need. – Fabrice NEYRET Nov 03 '15 at 16:13
  • Yes,but when i take the fourier transform of the entire differential equation,i end on one side second derivative of dirac delta function and on the other side F^-1(( f(w-a) + f(w+a))/f(w)). I used the convolution rule and found out the inverse of (f(w-a) + f(w+a)) but i had no clue about what could be F^-1(1/f(w)) which is needed to solve it. Help appreciated! – Basham G Nov 03 '15 at 16:59
  • f() is any possible function ? otherwise dividing by f in Fourier is deconvolving is regular space, generaly an ill-posed operation (see debluring). But if you know f, it might me that f(w-a) + f(w+a))/f(w) simplifies. – Fabrice NEYRET Nov 03 '15 at 18:13
  • We can assume that f(w) is well behaved and periodic but i cant take more thant i guess,isnt any generic comments possible ? – Basham G Nov 04 '15 at 14:26
  • For a gauss filter (the most lovely easy function of all), $F^{-1}(\frac 1 {f(w)})$ is the strange filter that would unblur the signal. Good luck to get a close form ;-) . It could work only for very special f well adapted to the operation. – Fabrice NEYRET Nov 04 '15 at 14:30
  • Okay!. I was trying to solve,x''=x cos(wt),using fourier transform method and ended up with above step,any hints? – Basham G Nov 04 '15 at 15:47
  • what was the fourier transform you obtained, and the next steps you did ? (may be you should update the initial question with that). – Fabrice NEYRET Nov 04 '15 at 16:11

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