## Green representation and a special representation

$\mathbf{Problem:}$ Suppose $\Delta u=f(x)$, where $f$ is a radial function and has compact support in $\mathbb{R}^n$, $n\geq 3$. Prove that

$\displaystyle u=\frac{1}{n(n-2)w_n}\int_{\mathbb{R}^n}f(y)(\max\{|x|,|y|\})^{-n+2}dy$

Compare with $u=f\ast \Gamma$, where $\displaystyle \Gamma(x)=\frac{1}{n(n-2)w_n}|x|^{-n+2}$.

$\mathbf{Proof:}$

Since $f$ has compact support, by Green’s representation

$\displaystyle u=f\ast \Gamma=\frac{1}{n(n-2)w_n}\int_{\mathbb{R}^n}f(y)|x-y|^{2-n}dy$

So $u$ is a radial function. We know that

$\displaystyle \frac{1}{n(n-2)w_n}\int_{\mathbb{R}^n}f(y)(\max\{|x|,|y|\})^{-n+2}dy=\int_{|y|\leq |x|}\frac{f(y)}{|x|^{n-2}}dy+\int_{|y|>|x|}\frac{f(y)}{|y|^{n-2}}dy$

By the divergence theorem

$\displaystyle \int_{|y|\leq |x|}\frac{f(y)}{|x|^{n-2}}dy=\frac{1}{|x|^{n-2}}\int_{|y|\leq |x|}\Delta u(y)dy=\frac{1}{|x|^{n-2}}\int_{\partial B}\frac{\partial u}{\partial \nu}ds$

here $B$ is a centered ball with radius $|x|$ and positive orientation.

By the Green’s second identity,

$\displaystyle \int_{|y|>|x|}\frac{f(y)}{|y|^{n-2}}dy=-\int_{\partial B}\frac{1}{|y|^{n-2}}\frac{\partial u}{\partial \nu}-u\frac{\partial}{\partial \nu}\left(\frac{1}{|y|^{n-2}}\right)ds=-\frac{1}{|x|^{n-2}}\int_{\partial B}\frac{\partial u}{\partial \nu}ds+u\int_{\partial B}\frac{\partial}{\partial \nu}\frac{1}{|y|^{n-2}}ds$

So $\displaystyle \frac{1}{n(n-2)w_n}\int_{\mathbb{R}^n}f(y)(\max\{|x|,|y|\})^{-n+2}dy=u\int_{\partial B}\frac{\partial}{\partial \nu}\frac{1}{|y|^{n-2}}ds=n(n-2)w_nu$

$\text{Q.E.D}\hfill \square$

$\mathbf{Remark:}$ Almut Burchard. A short course on Rearrangement Inequalities.