Prove the mean value property of harmonic functions

[Ryker]

Homework Statement

Let $A \subset \mathbb{R}^{2}$ be an open connected set, and g: A → ℝ a C² function. Show that if g is harmonic, i.e. $\frac{\partial ^{2} g}{\partial {x_{1}}^{2}} + \frac{\partial ^{2} g}{\partial {x_{2}}^{2}} = 0$, then $g(x) = \frac{1}{2 \pi r} \int_{\partial B_{r}(x)}gds$.

Homework Equations

We're to use Green's theorem in the form $\int_{\partial A} G \cdot x = \int_{A} (\frac{\partial G_{2}}{\partial {x_{1}}} - \frac{\partial G_{1}}{\partial {x_{2}}})dx$.

The Attempt at a Solution

OK, I've been struggling to wrap my head around this one for at least three hours now, and after failing to make any sense of it, looking for help online, as well. I've stumbled upon many proofs, but they all use different versions of Green's theorem, and I just really don't get what's going on.

All I gathered is that I need to somehow show that the derivative of the RHS is 0, implying it's constant, and then use that by taking the limit r → 0 and argue that due to continuity it is equal to the LHS, i.e. g(x).

Any help would be greatly appreciated, and I'm sorry I don't have more of an attempt to show, but I have zero idea on how to do this. I tried taking the derivative of the RHS myself, but I can't seem to do that, either...

 

[mighty2000]

First of all, don't worry, Green's theorem can be quite confusing at first. But once you understand it, it becomes a powerful tool in solving problems like this one. Let me try to guide you through it.

To start off, let's define the function G(x) = \frac{1}{2 \pi r} \int_{\partial B_{r}(x)}gds. This is the function we want to show is constant. To do this, we will take the derivative of G with respect to r, and show that it is equal to 0. This will imply that G is constant, and thus equal to g(x).

So, let's begin. Using Green's theorem, we can rewrite G as:

G(x) = \frac{1}{2 \pi r} \int_{\partial B_{r}(x)}gds = \frac{1}{2 \pi} \int_{B_{r}(x)} \frac{\partial g}{\partial x_{1}} - \frac{\partial g}{\partial x_{2}} dx

Notice that we have switched the order of the integrals, because we can do that according to Fubini's theorem. Now, we can take the derivative of G with respect to r:

\frac{\partial G}{\partial r}(x) = \frac{1}{2 \pi} \int_{B_{r}(x)} \frac{\partial}{\partial r}(\frac{\partial g}{\partial x_{1}} - \frac{\partial g}{\partial x_{2}}) dx

Using the chain rule, we can simplify this to:

\frac{\partial G}{\partial r}(x) = \frac{1}{2 \pi} \int_{B_{r}(x)} \frac{\partial^{2} g}{\partial x_{1}^{2}} + \frac{\partial^{2} g}{\partial x_{2}^{2}} dx

But wait, we know that g is harmonic, so \frac{\partial^{2} g}{\partial x_{1}^{2}} + \frac{\partial^{2} g}{\partial x_{2}^{2}} = 0. Therefore, we have:

\frac{\partial G}{\partial r}(x) = \frac{1}{2 \pi} \int_{B