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Consider the Dirichlet boundary value problem for the Poisson-equation $$ -\Delta u=f\text{ in }B_R(0)\subset\mathbb{R}^3,~~~~~~~~~~u=0\text{ on }S_R(0) $$ with $f\in L^{\infty}(B_R(0))\cap C^{0,\lambda}(B_R(0))$ and $0<\lambda<1$.

(a) Give the solution $u\in C^{2,\lambda}(B_R(0))\cap C(\overline{B_R(0)})$ by use of the Green function $G(x,y)$ of $B_R(0)$.

(b) Give $u(0)$ for the case that $f(x)=f(\lVert x\rVert)$.

Concerning (a) I used a theorem of our lecture, saying that the solution is given by $$ u(x)=\int_{\Omega}G(x,y)f(y)\, dy. $$ So here it is $$ \frac{1}{4\pi}\left(\int_{B_R(0)}\frac{f(y)}{\lVert x-y\rVert}\, dy-\int_{B_R(0)}\frac{f(y)}{\left\lVert\frac{\lvert y\rVert}{R}x-\frac{R}{\lVert y\rVert}y\right\rVert}\, dy\right). $$

Concerning (b), I have (using polar coordinates) $$ u(0)=\frac{1}{4\pi}\left(\int_{B_R(0)}\frac{f(\lVert y\rVert)}{\lVert y\rVert}\, dy-\frac{1}{R}\int_{B_R(0)}f(\lVert y\rVert)\, dy\right)\\=\frac{1}{4\pi}\left(2\pi\int_0^R\int_0^{\pi}f(r)r\sin\theta\, d\theta\, dr-\frac{2\pi}{R}\int_0^R\int_0^{\pi}f(r)r^2\sin\theta\, d\theta\, dr\right)\\=\int_0^R f(r) r\, dr-\frac{1}{R}\int_0^R f(r)\, dr $$

Would be great the know, if I am right.

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I believe you are correct except for your use of the term "polar coordinates" when you mean "spherical coordinates". I checked that, for $\|x\|=R$, $$ \left\|\frac{\|y\|}{R}x-\frac{R}{\|y\|}y\right\|^{2}=\frac{\|y\|^{2}}{R^{2}}\|x\|^{2}-2x\cdot y + \frac{R^{2}}{\|y\|^{2}}\|y\|^{2}=\|y\|^{2}-2x\cdot y+R^{2}=\|y-x\|^{2}, $$ which means that $$ \frac{1}{\|x-y\|}-\frac{1}{\|\frac{\|y\|}{R}x-\frac{R}{\|y\|}y\|} $$ vanishes for $\|x\|=R$. And you've correctly used spherical coordinates. I cannot see any error.

Disintegrating By Parts
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  • Thank you very much! I only wonder if one could maybe simplify my results or maybe calculate a little but more - but because the function f is not known I think not. –  Dec 23 '13 at 06:13