I have a function of three variables $f(x,y,z)$, and it obeys a linear partial differential equation. I'm checking my by-hand calculations with Mathematica. I want to convert the PDE into the Fourier-spectral domain, but Mathematica isn't playing nice. I find the following behavior baffling.

FourierTransform[D[f[x, y, z], x], {x, y, z}, {kx, ky, kz}]

-I kx FourierTransform[FourierTransform[FourierTransform[f[x,y,z],x,kx],y,ky],z,kz]

FourierTransform[D[f[x, y, z], y], {x, y, z}, {kx, ky, kz}]

FourierTransform[Derivative[0, 1, 0][f][x, y, z],{x,y,z},{kx,ky,kz}]

FourierTransform[D[f[x, y, z], z], {x, y, z}, {kx, ky, kz}]

FourierTransform[Derivative[0, 0, 1][f][x, y, z],{x,y,z},{kx,ky,kz}]

Essentially, it's like Mathematica uses $\mathcal{F}\left[\frac{\partial f}{\partial x}\right]=-i k_x \mathcal{F}\left[f\right]$ for the first argument $x$, but does not use the same rules/identities for partials w.r.t. second and third arguments, i.e. $\mathcal{F}\left[\frac{\partial f}{\partial y}\right]$ and $\mathcal{F}\left[\frac{\partial f}{\partial z}\right]$ are not simplified.

What's going on here? How can I force Mathematica to do this simplification for the other variables?


1 Answer 1


I also would expect Mathematica to simplify all Fourier transformed derivatives equally, but it may be understandable that the simplifications are harder to see when the derivative is not taken with respect to the innermost Fourier transform variable.

To work around this problem, you could change the order of integrations for the Fourier transform to always have the differentiation variable as the innermost integration variable, too. For example,

FourierTransform[D[f[x, y, z], y], {y, x, z}, {ky, kx, kz}]

==> -I ky FourierTransform[
  FourierTransform[FourierTransform[f[x, y, z], y, ky], x, kx], z, kz]

Here I've made use of the assumption that the above order of integrations is equal to

FourierTransform[D[f[x, y, z], y], {x, y, z}, {kx, ky, kz}]

To explain why the different treatment happens, one would have to know more about the internals of the FourierTransform algorithm. That's not so trivial because this includes dealing with functions that may not be square-integrable (which Mathematica is able to do). For square-integrable functions, the relation you're expecting comes from the fact that the derivative can be removed using an integration by parts, where the boundary term at infinity is zero:

$$\begin{array}{ccl}&&\int_{-\infty}^{\infty}e^{ik_{y}y}\int_{-\infty}^{\infty}e^{ik_{x}x}\frac{d}{dx}f(x,y)\, dx\, dy\\&=&\int_{-\infty}^{\infty}e^{ik_{y}y}\left\{ \underbrace{\left[e^{ik_{x}x}f(x,y)\right]_{-\infty}^{\infty}}_{=0}-\int_{-\infty}^{\infty}\left(\frac{d}{dx}e^{ik_{x}x}\right)f(x,y)\, dx\right\} \, dy\\&=&-ik_{x}\int_{-\infty}^{\infty}e^{ik_{y}y}\int_{-\infty}^{\infty}e^{ik_{x}x}f(x,y)\, dx\, dy.\end{array}$$

However, when the differentiation is not with respect to the innermost integration variable, then the corresponding boundary term is itself a Fourier integral (say over $x$ in the above example), which must be evaluated in the limits for $y \to \pm \infty$:

$$\begin{array}{ccl}& & \int_{-\infty}^{\infty}e^{ik_{y}y}\int_{-\infty}^{\infty}e^{ik_{x}x}\frac{d}{dy}f(x,y)\, dx\, dy\\ & = & \left[e^{ik_{y}y}\int_{-\infty}^{\infty}e^{ik_{x}x}f(x,y)\, dx\right]_{-\infty}^{\infty}-ik_{y}\int_{-\infty}^{\infty}e^{ik_{y}y}\int_{-\infty}^{\infty}e^{ik_{x}x}f(x,y)\, dx\, dy\end{array}$$

This involves an assumption not about the square-integrability of $f$ alone, but of its Fourier integral with respect to a different variable. So maybe this is why Mathematica doesn't like to make that simplification. As I said, things are more complicated if I want to include non-normalizable functions whose transforms involve Dirac delta functions etc. Therefore, this isn't a complete explanation.

  • $\begingroup$ enlightening (+1) $\endgroup$
    – Stefan
    Commented Jul 15, 2013 at 19:36
  • $\begingroup$ Yes, thank you for the exposition. Any ideas about "Assumptions->" on the integrals in question to force Mathematica to recognize that the boundary terms vanish? $\endgroup$
    – rajb245
    Commented Jul 16, 2013 at 13:46
  • $\begingroup$ I think any kind of assumptions or rules that one could come up with would be much more complex than the simple manual approach I suggested in the answer. One may as well just take the replacement of D[__,y] by - I ky as a definition in itself to get around this limitation of FourierTransform. $\endgroup$
    – Jens
    Commented Jul 16, 2013 at 21:35

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