0
$\begingroup$

Assume $f(r)=\delta(r-R)$ where $\delta(\cdot)$ is a ring delta function. In other word, $f$ is a circular delta function on a circle with radius $R$. I want to do the convolution of $f$ with itself ($f*f$).

How can I do convolution in polar(spherical) coordinates. I read that I can use Hankel and inverse Fourier transforms to obtain the convolution, however I am not into the subject and I will appreciate if some could help me with this.

$\endgroup$
  • 2
    $\begingroup$ As far as I am aware, there is not much theory on the product of two distributions as things get very delicate. I'm not sure that there is a well-defined answer to your question. $\endgroup$ – Cameron Williams May 30 '14 at 23:29
1
$\begingroup$

Although convolutions of distributions are not defined in general, you can do it in this case. Let $$ f_\epsilon(x) = \epsilon^{-1} I_{|x| \in [R, R+\epsilon]} .$$ Then compute $f_\epsilon*f_\epsilon$, and then let $\epsilon \to 0$.

What you will see is that $f*f(x)$ is the "area" of the intersection of the boundaries of the circles of radius $R$, one centered at $0$, and the other centered at $x$. Obviously this area isn't properly defined, but you could think of thickening the boundaries of the circles so that their thicknesses are $\epsilon$, computing the area of this intersection, dividing by $\epsilon^2$, and then letting $\epsilon \to 0$. You will realize that the answer depends only on the angle at which the two circles intersect each other.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Can we say the convolution is the area of a circle with radius $R$ and amplitude $\frac{1}{(2\pi R)^2}$? because as we move one circle along a line from $0$ to $2R$, the two circles intersect each other in two points. Therefore we can expect the answer is a cylinder centred in $0$, radius $2R$ and height $\frac{1}{(2\pi R)^2}$. In fact if $f(r)$ is the pdf of a random variable uniformly distributed on the circumference, the sum of two independent such random variables is a uniform distribution on a circle with radius $2r$ $\endgroup$ – M.X May 31 '14 at 10:09
  • $\begingroup$ No. If you think about it, $f*f(r)$ should blow up to infinity as $r \to 0^+$ and $r \to 2R^-$, and it should be zero if $r > R$. The infinities should be sufficiently regulated so that $\int_0^R f(r) r \, dr = 2\pi R^2$. $\endgroup$ – Stephen Montgomery-Smith May 31 '14 at 14:11

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.