9
$\begingroup$

I have an integral that looks like a convolution across the radial direction in a cylindrical coordinate system: $$h(r) = \int_{0}^{\infty}f(r')g(r-r')r'\, \mathrm dr'.$$ Both $f$ and $g$ are radially symmetric functions (i.e. not functions of $\theta$). After some reading on Hankel Transforms, I assumed this convolution would have a Fourier transform like property for the Hankel transform, that is: $$H(\rho)=F(\rho)G(\rho)$$ where $H$, $F$, and $G$ are the zeroth order Hankel transforms of $h$, $f$, and $g$ respectively where the Hankel transform is defined as: $$F(\rho)=\int_{0}^{\infty}f(r)J_0(r\rho)r\, \mathrm dr.$$ However, it appears this is not the case. One very helpful paper, "Operational and convolution properties of two-dimensional Fourier transforms in polar coordinates", addresses the convolution problems of the Hankel transform directly. The author's conclusion is that because the Hankel transform represents a 2D Fourier transform for a radially symmetric function, the convolution property holds only for a 2D convolution. The author even points out my integral exactly to say it should not be confused with a 2D convolution, which involves both the $\theta$ and $r$ coordinates.

In my reading on Fourier transforms in cylindrical coordinates, I can only find applications where a 2D or 3D Fourier transform is required. I'm wondering if there is there a reason why a 1D Fourier transform across the radial coordinate is not valid.

Is anyone familiar with a convolution property for my integral? I'd be very interested in a transform for which my first and second equations are valid, but it appears the Hankel transform is not the one.

$\endgroup$
  • 1
    $\begingroup$ Ping me here or in hbar in two days if you don't have an answer for this. $\endgroup$ – Emilio Pisanty May 26 '18 at 0:49
  • $\begingroup$ Howdy Emilio, I have not found an answer yet. After a bit more reading I feel I understand a bit more from the convolution theorem why my integral might not have a convenient transform, but I'd still be very interested in a way to simplify integrals of this form. $\endgroup$ – Kthaxt May 29 '18 at 21:41
  • 1
    $\begingroup$ There you go =) $\endgroup$ – Emilio Pisanty May 29 '18 at 21:59
  • $\begingroup$ Hello! There's a mistake in the integral for $h$: the integral measure should be $dr'$ not $dr$. Are the other $r$s ok? In particular is the free $r$ next to the measure in the integral primed or not? $\endgroup$ – John Donne May 30 '18 at 12:24
  • 1
    $\begingroup$ Oh you're right, it should be r'. The free r should also be r'. $\endgroup$ – Kthaxt May 30 '18 at 17:24
2
+100
$\begingroup$

There is no integral transform which satisfies the required properties. The convolution as defined by the OP is not symmetric in $f$ and $g$, while the product of the two transforms is symmetric. While uniqueness of transforms is generally non trivial, the property in this case fails very badly.

Moreover, the argument of the function $g$ becomes negative despite its interpretation as a radius, which may or may not need addressing based on the meaning of the functions involved. This is what the author of the paper meant: if $g(r)$ is a function of $r>0$ then it's meaningless to take a (1D) radial convolution precisely because of this.

As to how to deal with such integrals, we can try to do some calculations. If we define $$S(x)=\begin{cases}x && x\geq 0\\0 &&x<0\end{cases}$$ then $$h(r) = \int_{0}^{\infty}f(r')g(r-r')r'\, \mathrm dr'=\int_{-\infty}^{\infty}S(r')f(r')g(r-r')\, \mathrm dr'=(Sf)*g$$ where $*$ denotes the usual (1D) convolution.

If $f$ decays sufficiently quickly, we can apply the standard convolution theorem for Fourier transforms. However $S$ does not have a Fourier transform, so we need more info to make progress.

Otherwise we can take a different route. We assume that $f(r)=g(r)=0$ for $r<0$, which fixes the issue mentioned above, and we take a Laplace transform. Using the convolution theorem: $$\tilde h(\rho)=\tilde g \,\mathcal{L}(Sf)=\tilde g (\tilde S * \tilde f)= \int_{0}^{\rho} \frac{\tilde g(\rho)\tilde f(\rho-\rho')}{(\rho')^2}d\rho'=\tilde g(\rho) \int_{1/\rho}^{\infty} \tilde f(\rho-1/t)dt$$ where in the last step we substituted $t=1/\rho'$. I think this is the closest you can get in general. Whether it is useful or not depends on the specific problem.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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