I learned the following theorem in Folland's Introduction to Partial Differential Equations(p.69 Chapter 2):
Suppose $u$ is harmonic on an open set $\Omega\subset{\mathbb R}^n$. If $x\in\Omega$ and $r>0$ is small enough so that $\overline{B_r(x)}\subset\Omega$, then
$$u(x)=\frac{1}{r^{n-1}\omega_n}\int_{S_r(x)}u(y)d\sigma(y)=\frac{1}{\omega_n}\int_{S_1(0)}u(x+ry)d\sigma(y)，$$
where $$\omega_n=\frac{2\pi^{n/2}}{\Gamma(n/2)}.$$
I found that I could not immediately reconstruct a proof for the theorem. A key point is that one needs to use the Green's identity, which is a basic property of harmonic functions. But I don't see any "clue" that how people actually come up with this theorem and such proof. (Maybe this is the common problem, at least for me, for most of the textbooks.) A curious search in Google returns nothing satisfactory to me. Since this is a basic property of harmonic functions, I am wondering that if one needs to know this history of harmonic functions in order to know this theorem well.
Here is my question:
- Can any one here come up with a motivation of this theorem in PDE?
My second question may be more vague:
- How can I approach the proof of this theorem more "naturally" instead of just remembering bunch of facts? (In the language of Polya, any heuristics here?)
Best Answer
I can tell you my favourite proof of the mean value property, which I find more intuitive than the one via Green's identity. To make the proof rigorous, you need to know about integration over the manifold $O(n)$ of all orthogonal matrices, but you can just depict what I do as averaging over all orthogonal matrices. Let's take $x=0$ for simplicity. Heuristically, you'd expect that $$ \frac{1}{r^{n-1}\sigma_{n-1}}\int_{S_r(0)}u(y)\,d\sigma(y) = {\rlap{\;\bar{}}\int_{O(n)}} u(Az)\,d\sigma(A) =: f(z) $$ for $z\in\mathbb R^n$ with $|z|=r$. On the left hand side you take the average of $u(y)$ over all $y$ with $|y|=r$, on the right hand side it's the average of $u(Az)$ over all the orthogonal matrices $A$, which should be and is indeed the same.
Now you can differentiate the right hand side under the integral sign: take the Laplacian with respect to $z$. Then, due to $\Delta u=0$, you see that $f$ is harmonic. Moreover, $f$ is radially symmetric, i.e., $f(z)$ depends on $|z|$ only. Finally, use the fact that a radially symmetric harmonic function defined on all of $\mathbb R^n$ is constant. This yields $f(z) = f(0) = u(0)$, and we're done.