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}.


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.

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