3
$\begingroup$

I want to do a Fourier transform to the below function by Mathematica. How can I do it? Here $c$, $d$, $a$, $L$ are constants.

$$ w(r)= \left\{ \begin{array}{ll} -\frac{c}{\epsilon r} & r> L \\ -\frac{c}{\epsilon d}\left[\frac{a}{r}-\ln\left(\frac{r}{L}\right)\right] & a < r\leq L \\ -\frac{c}{\epsilon d}\left[\ln\left(\frac{a}{L}\right)\right] & r \leq a \\ \end{array} \right. $$

$\endgroup$

1 Answer 1

7
$\begingroup$

It looks like Mathematica can do this. Define something like this (I've simplified your version slightly)

w[r_] := Piecewise[{{-1/r, r > L}, {-1/d (a/r - Log[r/L]), 
    a < r <= L}, {-1/d Log[a/L], r <= a}}]

Mathematica will return a continuous Fourier transform

Assuming[0 < a < L, FourierTransform[w[r], r, k]]

I suggest you check the help for FourierTransform to ensure that the definition used is the one you want.

$\endgroup$
2
  • $\begingroup$ Thank you @mikado! Is this 'Assuming[0 < a < L, FourierTransform[w[r], r, k]]' code only for the travel from 0 to L, exclude the r>L and r<= L? $\endgroup$
    – aluuzz
    Sep 8, 2019 at 22:50
  • $\begingroup$ No, I'm just saying that a is positive and less than L $\endgroup$
    – mikado
    Sep 9, 2019 at 5:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.