how to compute the convolution of two measures explicitly

Question

Here is my example:u and v are the surface measures on the spheres {${x;|x|=a}$} and {${x;|x|=b}$} in $\mathbb{R}^{3}$.Then what's $u\ast v$ ? And what if in $\mathbb{R}^{n}$?

Answer 1

Let $x=rs$ with $r\geqslant0$ and $s$ in $S^{n-1}$ denote the radial representation of $x$ in $\mathbb R^n$. Then, $(u\ast v)(\mathrm dx)=r\,\kappa_n(r)\,\mathrm dr\sigma_{n-1}(\mathrm ds)$ with $$\kappa_n(r)=c_n\cdot (4a^2b^2-(r^2-a^2-b^2)^2)^{(n-3)/2}\cdot[|a-b|\lt r\lt a+b].$$ Equivalently, $$(u\ast v)(\mathrm dx)=\kappa_n(|x|)\,|x|^{2-n}\mathrm dx. $$

To show this, first note that $u\ast v$ is radially symmetric and that its radial part $\kappa_n$ solves, for every function $\varphi$, $$ \int\varphi(r^2)\kappa_n(r)\,\mathrm dr=\int\varphi(|x|^2)\,(\mu\ast\nu)(\mathrm dx)=\iint\varphi(|as+bt|^2)\,a^{n-1}\sigma_{n-1}(\mathrm ds)b^{n-1}\sigma_{n-1}(\mathrm dt). $$ By rotational invariance, this is $$ \int\varphi(|as+b|^2)\,(ab)^{n-1}\sigma_{n-1}(\mathrm ds)=\int\varphi(a^2+b^2+2ab\langle s,1\rangle)\,(ab)^{n-1}\sigma_{n-1}(\mathrm ds). $$ One knows that $$ \sigma_{n-1}(\mathrm ds)=\sin^{n-2}\phi_1\cdot\sin^{n-3}\phi_2\cdots\sin\phi_{n-2}\cdot\mathrm d\phi_1\cdots\mathrm d\phi_{n-1}, $$ for some angles $\phi_k$ independent and uniform on $[0,\pi]$ except $\phi_{n-1}$ which is uniform on $[0,2\pi]$. Hence, $\langle s,1\rangle=\cos\theta_1$ and $$ \int\varphi(r^2)\kappa(r)\,\mathrm dr=c_n\int_0^\pi\varphi(a^2+b^2+2ab\cos\theta)\,(ab)^{n-1}\sin^{n-2}\theta\cdot\mathrm d\theta. $$ Finally, the change of variables $r^2=a^2+b^2+2ab\cos\theta$ in the last integral yields the result.