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.

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    $\begingroup$ 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

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.

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  • $\begingroup$ 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

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