# Mean-value formula for inhomogeneous harmonic functions

I am working on Evans' PDE textbook problems, but I am stuck with the following problem about modification of the proof of the mean-value formula for harmonic functions. I cannot really see how to derive the second term of RHS of the formula below. I would appreciate it if someone could help me derive this formula.

Modify the proof of the mean-value formulas to show for $n\ge 3$ that $$u(0)=\frac{1}{V(\partial B(0,r))}\int_{\partial B(0,r)}g \, dS +\frac{1}{n(n-2)\alpha(n)}\int_{B(0,r)}\left[\frac{1}{|x|^{n-2}}-\frac{1}{r^{n-2}}\right]f \,dx,$$ provided $$\begin{cases}-\Delta u=f & \text{in } B^0(0,r) \\ \quad \, \, \, u=g & \text{on } \partial B(0,r).\end{cases}$$

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Perhaps you can give an outline of the proof that is to be modified? – AD. Oct 5 '12 at 10:20
You are right. I should have done that. – Pooya Oct 5 '12 at 17:21

## 2 Answers

$\def\Vol{\operatorname{Vol}}$As Evans does in his proof of the mean value formula, define $\phi\colon [0,r] \to \mathbb R$ by $\phi(s) = \frac 1{\Vol(\partial B_s)} \int_{\partial B_s} u(x)\, dS(x) = \frac 1{\Vol(\partial B_1)} \int_{\partial B_1} u(sx)\, dS(x)$ We have \begin{align*} \phi'(s) &= \frac 1{\Vol(\partial B_1)} \int_{\partial B_1} Du(sx)x\, dS(x)\\ &= \frac 1{n\alpha(n)s^{n-1}} \int_{\partial B_s} Du(x)\, \frac xs\, dS(x)\\ &= \frac 1{n\alpha(n)s^{n-1}} \int_{\partial B_s} Du(x)\, \nu_{B_s}(x)\, dS(x)\\ &= \frac 1{n\alpha(n)s^{n-1}} \int_{\partial B_s} \frac{\partial u}{\partial \nu}(x)\,dS(x)\\ &= \frac{1}{n\alpha(n)s^{n-1}} \int_{B_s} \Delta u(x)\, dx\\ &= -\frac 1{n\alpha(n)s^{n-1}}\int_{B_s}f(x)\, dx \end{align*} As $\phi$ is differentiable on $(0,r)$ and continuous on $[0,r]$, we have \begin{align*} \phi(r) -\phi(0) &= \int_0^r \phi'(s)\, ds\\ &= -\int_0^r \frac 1{n\alpha(n)s^{n-1}} \int_{B_s} f(x)\, dx\; ds\\ &= -\frac 1{n\alpha(n)} \int_0^r s^{1-n}\int_0^s \int_{\partial B_\sigma} f(x)\, dS(x)\; d\sigma\; ds\\ &= \frac 1{n\alpha(n)} \int_0^r \int_\sigma^r s^{1-n}\int_{\partial B_\sigma} f(x)\, dS(x)\; ds\; d\sigma\\ &= -\frac 1{n(n-2)\alpha(n)} \int_0^r \int_{\partial B_\sigma} f(x)\, dS(x)\; d\sigma\\ &= -\frac 1{n(n-2)\alpha(n)}\int_0^r \int_{\partial B_\sigma}\left(\frac 1{\sigma^{n-2}} - \frac 1{r^{n-2}}\right) f(x)\, dS(x)\;d\sigma\\ &= -\frac 1{n(n-2)\alpha(n)}\int_0^r \int_{\partial B_\sigma}\left(\frac 1{|x|^{n-2}} - \frac 1{r^{n-2}}\right) f(x)\, dS(x)\;d\sigma\\ &= -\frac 1{n(n-2)\alpha(n)}\int_{B_r} \left(\frac 1{|x|^{n-2}} - \frac 1{r^{n-2}}\right) f(x)\, dx \end{align*} Now, as $\phi(0) = u(0)$ and $\phi(r) = \frac 1{\Vol(\partial B_r)}\int_{\partial B_r} g(x)\, dS(x)$ the desired result follows.

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Thank you for the detailed answer. This helps me out. I made mistake in the latter half of your answer. – Pooya Oct 5 '12 at 17:27
Why did the sign disappear in step 4 of evaluating $\phi(r)-\phi(0)$? – Cookie Dec 28 '14 at 0:10

Alternatively, look at the second term on the right side. It's the fundamental solution, shifted down to be $0$ on the boundary of $B_r$, integrated against $\Delta u$. Heuristically, integrate by parts and use that the Laplacian of the fundamental solution is $\delta_0$ to get the desired formula.

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