Mean value formula for the laplace equation on a disk 

Use the mean value formulas to prove that (for $n\geq 3$) $$u(0) = \frac{1}{n \alpha(n)r^{n-1}}\int_{\partial B\left(0,r\right)} g \, dS + \frac{1}{n\left(n-2\right) \alpha \left(n\right)} \int_{B\left(0,r\right)} \left(\frac{1}{\left|x\right|^{n-2}}-\frac{1}{r^{n-2}} \right) f \, dx,$$ if $u$ satisfies the PDE $$\begin{cases} -\Delta u=f & \text{in } B(0,r) \\ u=g & \text{on } \partial B(0,r), \end{cases}$$ ? where $B(0,r)$ denotes the ball of radius rr centerned at the origion, and $\alpha(n)$ is the volume of the unit ball in $\mathbb{R}^n$.

Answer

For $r>s>0$, define $$\varphi(s):=\frac{1}{n\alpha(n)s^{n-1}}\int_{\partial B(0,s)} u(y) \, ds.$$ Let the average of the function ff over the ball of radius ss be denoted by $\bar{f}(s)$ , i.e. $$\bar{f}(s)=\frac{1}{\alpha(n)s^n}\int_{B(0,s)} f \, dx.$$ Taking the derivative of $\varphi(s)$ with respect to $s$ yields (see proof of the mean value formulas in Section 2.2 in Evans' PDE book) $$\varphi '(s) = \frac{s}{n} \left(\frac{1}{\alpha(n)s^n}\int_{B(0,s)} \Delta u\, dx\right) = - \frac{s}{n} \left(\frac{1}{\alpha(n)s^n}\int_{B(0,s)} f \, dx \right)=- \frac{s}{n} \tilde{f}(s).$$ Now integrate both sides from $0$ to $r$ and use the mean value property: $$\varphi (r) - \varphi (0) = - \int_{0}^{r} \frac{s}{n} \bar{f}(s) ds \Rightarrow \varphi (0) - \varphi(r) = \int_{0}^{r} \frac{s}{n} \bar{f}(s) ds.$$ Hence $$u (0) - \varphi(r) = \int_{0}^{r} \frac{s}{n} \bar{f}(s) ds.$$ Note that $$\varphi(r)=\frac{1}{n\alpha(n)r^{n-1}}\int_{\partial B(0,r)} g(y) \, ds. $$ So it remains to show $$\int_{0}^{r} \frac{s}{n} \bar{f}(s) ds= \frac{1}{n\left(n-2\right) \alpha \left(n\right)} \int_{B\left(0,r\right)} \left(\frac{1}{\left|x\right|^{n-2}}-\frac{1}{r^{n-2}} \right) f \, dx.$$ Rewrite the right hand side as follows $$\int_{0}^{r} \frac{s}{n} \bar{f}(s) ds=\int_{0}^{r} \frac{s}{n} \frac{\int_{B(0,s)} f \, dx}{\alpha (n) s^n} \, ds = \int_{0}^{r} \frac{1}{n \alpha (n) s^{n-1}} \int_{B(0,s)} f \, dx \, ds.$$ Apply integration by parts with $u=\int_{B(0,s)} f \, dx$ and $dv=\frac{1}{ s^{n-1}} \, ds$. This gives $$du=\int_{\partial B(0,s)} f \, dS$$ by the Coarea Formula, and $v=\frac{s^{2-n}}{2-n}$. Thus $$\int_{0}^{r} \frac{s}{n} \bar{f}(s) ds=\frac{1}{n \alpha (n)} \left[ \frac{s^{2-n}}{2-n} \int_{B(0,s)} f \, dx \bigg|_{0}^{r} - \int_{0}^{r} \frac{s^{2-n}}{2-n} \int_{\partial B(0,s)} f \, dS \, ds \right]. $$ Note that for $n \geq 3$, $$\lim_{s \rightarrow 0^{+}} \frac{s^{2-n}}{2-n} \int_{B(0,s)} f \, dx = 0.$$ Hence $$\int_{0}^{r} \frac{s}{n} \bar{f}(s) ds=\frac{1}{n \alpha (n)} \left[ \frac{r^{2-n}}{2-n} \int_{B(0,r)} f \, dx - 0 - \frac{1}{2-n} \int_{0}^{r} s^{2-n} \int_{\partial B(0,s)} f \, dS \, ds \right] \\ =\frac{1}{n (n-2) \alpha (n)} \left[ \int_{0}^{r} s^{2-n} \left( \int_{\partial B(0,s)} f \, dS \right) \, ds - \int_{B(0,r)} r^{2-n} f(x) \, dx \right] \\ =\frac{1}{n (n-2) \alpha (n)} \left[ \int_{0}^{r} \left( \int_{\partial B(0,s)} \frac{1}{s^{n-2}} f \, dS(y) \right) \, ds - \int_{B(0,r)} \frac{f(x)}{r^{2-n}} \, dx \right] \\ =\frac{1}{n (n-2) \alpha (n)} \left[ \int_{B(0,x)} \frac{f(x)}{\left|x\right|^{n-2}} \, dx - \int_{B(0,r)} \frac{f(x)}{r^{2-n}} \, dx \right] \\ =\frac{1}{n (n-2) \alpha (n)} \int_{B(0,x)} \left( \frac{1}{\left|x\right|^{n-2}} - \frac{1}{r^{2-n}} \right) f(x) \, dx.$$ To obtain the fourth equality above, we have used integration in Polar coordinates in $\mathbb{R}^n$ which is a special case of the coarea formula.

The answer is accepted.