MHB Dirichlet Problem on Sphere: Proving Mean Value of f

[Dustinsfl]

For the Dirichlet problem on a sphere of radius a with the boundary condition
$$
u(a,\theta,\phi) = f(\theta,\phi),
$$
show that the value of $u$ at the origin $(r = 0)$ is equal to the average value of $f$ over the surface of the sphere.

I know that the max and min occur on the boundaries and that is because the origin is the mean value but I don't know how to show it is the mean value.

 

[Jhoneine]

Let $V$ be the volume of the sphere of radius $a$. Then, by the divergence theorem,\begin{align*}\int_V \nabla \cdot \mathbf{u}\, dV &= \int_{\partial V} \mathbf{u}\cdot \hat{\mathbf{n}}\, dA \\&= \int_S u(a,\theta,\phi)\, dA \\&= \int_S f(\theta,\phi)\, dA \\&= a^2 \int_S f(\theta,\phi)\, d\Omega \\&= a^2 \langle f \rangle_S\end{align*}where $\langle f \rangle_S = \frac{1}{4\pi}\int_S f(\theta,\phi)\, d\Omega$ is the average value of $f$ over the surface of the sphere. On the other hand, since $\nabla \cdot \mathbf{u} = 0$ in the interior of the sphere, we have\begin{align*}\int_V \nabla \cdot \mathbf{u}\, dV &= 0 \\\implies a^2 \langle f \rangle_S &= 0 \\\implies \langle f \rangle_S &= 0\end{align*}Therefore, the value of $u$ at the origin must be equal to the average value of $f$ over the surface of the sphere, i.e. $u(0,0,0) = \langle f \rangle_S$.