We know that if we have a function $f(x)$, and we call $g(\omega)$ its Fourier transform, then the Fourier transform of $x f(x)$ is

$$\imath \frac{\mathrm{d} g(\omega))}{\mathrm{d}\omega} $$

and viceversa, ${\mathrm{d}f(x)}/{\mathrm{d}x}$ becomes $\imath \omega g(\omega)$

How can I transform an operator, for example

$$L(u):= u+x^2 u+\frac{\mathrm{d}u}{\mathrm{d}x}$$

so that it automatically transforms the variable $x$ in a derivative, and $x^n$ in a derivative of order $n$, and viceversa?

The specific problem I'm solving is the following:

ds[u_]:= A.{D[u,s],D[u,a]}.{1,0} 
da[u_]:= A.{D[u,s],D[u,a]}.{0,1}   (*where A is a constant matrix*)
G1[t_,x_,y_]:= G0[t,x,y] Integrate[K[tt,x,y],{tt,t,T}]

where G0[t,s,a] is a function already defined and K[tt,x,y] is a polinomial in the variable $x$ and $y$ (so its integral is still a polinomial in $x$ and $y$.

I know the inverse Fourier transform of G0, that is F0[t,s,a]. What I need is to do is to substitute all the the $x^n$ terms in the polinomial with the operator ds[u] applied $n$ times and similarly, $y^m$ with the the operator da[u] applied $m$ times. thanks

  • 3
    $\begingroup$ It took a lot of effort to edit your post. Please see the editing help and learn to format your post correctly. Your future posts and updates should not be unformatted like this one and should use proper punctuations and spelling. Take the time to format and typeset it properly. You're not in any race to post the question as quick as you can. Often times, it greatly delays getting an answer to the question because people are put off by the lack of formatting. Taking that extra 2 minutes to format it will be fully worth it. $\endgroup$
    – rm -rf
    Commented Dec 14, 2012 at 15:32
  • $\begingroup$ @rm-rf is right. The sloppy formulation of the question makes it hard to answer and unhelpful for future readers. I think your title is wrong because you say you already have the FT (and I answered how to do it in general). All you want is two pattern substitutions. $\endgroup$
    – Jens
    Commented Dec 14, 2012 at 16:06

2 Answers 2


I am not sure I understand your operator. Can't read it well.

But I thought to show one way to convert a time domain second order differential equation to Fourier domain and also to Laplace domain. may be that will help.

eq = y''[t] - y'[t] + y[t];
r = (FourierTransform[#, t, ω] & ) /@ List @@ eq;
Total[r /. FourierTransform[y[t],t, ω] -> ℱ[ω]]

and for Laplace

eq = y''[t] - y'[t] + y[t];
r = LaplaceTransform[eq, t, s];
r /. { LaplaceTransform[y[t], t, s] -> Υ[s],y[0] -> 0, y'[0] -> 0}

which gives

enter image description here

If you could clarify your operator more, may be someone can give better help

  • $\begingroup$ And the viceversa? for example eq=t y[t] and its transform is a derivative operator? is it can be done like that? thanks $\endgroup$
    – andrea
    Commented Dec 14, 2012 at 13:15

The starting point is the statement that if you want the differential operator to satisfy

$$\mathscr{L}u(x) = \psi(x)$$

Then the Fourier components of $u(x)$ have to be related to the Fourier components of $\psi(x)$ in the way we're trying to calculate here.

Here is a way to get the relation you're after by using Mathematica's Fourier transform capabilities:

Clear[u, x];
Assuming[{κ ∈ Reals},
     (* Here follows the differential operator: *)
      u + x^2 u + D[u, x]
     (* it acts on a Fourier component k: *)
    1/Sqrt[2 Pi] E^(I k x) φ[k]],
   (* the inverse transform integrates over x: *)
   x, κ],
  (* the Fourier components are now integrated up again: *)
  {k, -∞, ∞}]

$-\phi ''(\kappa )+i \kappa \phi (\kappa )+\phi (\kappa )$

Here $\phi(\kappa)$ is the Fourier component of the function you called $u$. I used coordinate $x$ and Fourier variables $k$ and $\kappa$ in the above calculation.

The trick is an interchange of integration order (which is tantamount to dropping the boundary term in an integration by parts when moving the differentiations onto the Fourier component).


The last result is the Fourier transform of the operator, with $\phi$ representing the dummy argument that must be replaced by the function you'd like to operate on. I.e., with the above result as an operator and a target function s[u], you could then do

%/.φ -> u

$-u''(\kappa )+i \kappa u(\kappa )+u(\kappa )$


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