D.3 Some properties of harmonic functions

The mean value theorem says that if you take a sphere around some point, the average of $u$ on the surface of that sphere is the value of $u$ at the center of the sphere. That is true as long as $u$ satisfies the Laplace equation inside the sphere.

To prove the mean value theorem, take the origin of your coordinate system at the center of the sphere. Then integrate the Laplace equation over the volume of a sphere. Use the divergence theorem to get

$$0=\int_{\Omega} \nabla\cdot\nabla u { \rm d}V = \int_{\... ... \int_{\delta\Omega} \frac{\partial u}{\partial n} { \rm d}S$$

But the normal direction is the radial direction, so

$$0 = \int_{\delta\Omega} \frac{\partial u}{\partial r} { \rm d}S$$

This holds for a spherical surface of any radius $r$ around the origin as long as the Laplace equation is applicable in the sphere. Divide by the total spherical surface:

$$0 = \int_{\delta\Omega} \frac{\partial u}{\partial r} \frac{{\rm d}S}{S}$$

The ratio ${\rm d}{S}/S$ does not depend on the radius. So you can take the derivative out of the integral to get

$$0 = \frac{{\rm d}}{{\rm d}r} \int_{\delta\Omega} u \frac{{\rm d}S}{S}$$

The integral is by definition the average of $u$ on the spherical surface. So it does not depend on the radius of the surface. That means it remains the same when you let the radius of the spherica; surface go to zero. But when $r$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, you get the average of $u$ at the origin, which is simply $u$ at the origin, since the origin is a single point.

The minimum and maximum properties follow immediately from the mean value theorem. Note that the minimum property implies the maximum property: the maximum of a harmonic function $u$ is the minimum of the harmonic function $-u$.

To show the minimum property, consider an arbitrary point in the interior of the domain. Put a sphere of a sufficiently small radius around the point; the sphere must stay in the domain. Now $u$ is at the considered point equals the average value of $u$ on the spherical surface. And an average is always in between the minimum and maximum values. So $u$ at the considered point cannot be less than the smallest $u$ value on the spherical surface. So $u$ at the considered point cannot be a unique minimum, lower than all other $u$ values.

You may wonder whether $u$ might be a minimum that is not unique. But for the average of $u$ on the spherical surface to equal the lowest value of $u$ requires that $u$ is everywhere the lowest value on the surface. If $u$ would be above the minimum anywhere, the average would be above the minimum. So any spherical surface around the considered point has the same value of $u$ as the considered point. At least as long as the sphere stays inside the domain. In other word, $u$ is constant within some sphere around the considered point that goes up to the boundary. And for every point inside that sphere there is again a surrounding sphere in which $u$ is constant. You can then readily see from a sketch that this means that $u$ will have to be the minimum everywhere. In other words, $u$ must be a constant for there to be a nonunique minimum in the interior of the region.