## 275 -Average values of harmonic functions

I want to mention here an important property of harmonic functions, the mean value property, and some of its consequences. I restrict myself to functions of two variables for clarity.

Many important properties of harmonic functions (and, by extension, of analytic functions) can be established solely in the basis of the mean value property. I don’t know how to prove this property (or that it characterizes harmonicity) without appealing to Stokes’s theorem, or one of its immediate consequences (the topic of Chapter 14 of the book); in fact, I doubt such an approach is possible. It is a good exercise to see, at least formally, how this result gives the mean value property, but a rigorous treatment tends to be somewhat involved. Unfortunately, the arguments that show that the statements below hold tend to require techniques that are beyond the scope of Calculus III, so I will skip them.

1. Let $f(x,y)$ be a harmonic function defined on an open region ${\mathcal R}$ of the plane, so $f$ is twice continuously differentiable, and the equation $\displaystyle \frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2}=0$

holds at all points $(x,y)$ in ${\mathcal R}$.

Then $f$ satisfies the average (or mean) value property, namely, for any closed disk $D$ contained in ${\mathcal R}$, $\displaystyle f(x_0,y_0)=\frac1{\iint_D\,dA}\iint_D f\,dA,$

where $(x_0,y_0)$ is the center of the disk $D$ and $\iint_D\,dA$ is of course the area of $D$, $\pi R^2$, where $R$ is the radius of $D$.

If one wants to explicitly compute double integrals over disks, like $\iint_D f\,dA$, it tends to be the case that introducing polar coordinates is rather helpful, if one parametrizes $D$ appropriately. This usually means one represents an arbitrary point in $D$ as $(x_0,y_0)+(r\cos\theta,r\sin\theta)$ for some $\theta\in[0,2\pi]$ and $r\in[0,R]$, so $\iint_D f\,dA=\int_0^{2\pi}\int_0^R f(x_0+r\cos\theta,y_0+r\sin\theta)\,r\,dr\,d\theta.$

For example, let $f(x,y)=x^2-y^2$ on ${\mathcal R}={\mathbb R}^2$, so $f$ is clearly harmonic since $f_{xx}+f_{yy}=2-2=0$. Consider the disk $D$ of radius $R$ centered at $(x_0,y_0)$. We have that $f(x_0+r\cos\theta,y_0+r\sin\theta)=(x_0+r\cos\theta)^2-(y_0+r\sin\theta)^2$ $=(x_0^2-y_0^2)+2r(x_0\cos\theta-y_0\sin\theta)+r^2(\cos^2\theta-\sin^2\theta)$ $=(x_0^2-y_0^2)+2r(x_0\cos\theta-y_0\sin\theta)+r^2\cos2\theta,$ so $\displaystyle\iint_D f\,dA=\int_0^{2\pi}\int_0^R [(x_0^2-y_0^2)+2r(x_0\cos\theta-y_0\sin\theta)$ $\displaystyle +r^2\cos2\theta]r\,dr\,d\theta=(x_0^2-y_0^2)\pi R^2$ $\displaystyle +2\int_0^R \left[r^2x_0\int_0^{2\pi}\cos\theta\,d\theta -r^2y_0\int_0^{2\pi}\sin\theta\,d\theta\right]\,dr$ $displaystyle+\int_0^R r^3\int_0^{2\pi}\cos2\theta\,d\theta\,dr=(x_0^2-y_0^2)\pi R^2,$

as claimed by the mean value property.

2. Conversely, if $f$ is a continuous function defined on an open region ${\mathcal R}$ and $f$ satisfies the mean value property, i.e., for all closed disks $D$ contained in ${\mathcal R}$ one has that $\displaystyle f(x_0,y_0)=\frac 1{\iint_D\,dA} \iint_D f\,dA$

where $(x_0,y_0)$ is the center of $D$, then $f$ is harmonic.

3. One of the first consequences one may notice of this fact is that it gives us that the function $f$, originally only assumed to be continuous, is in fact twice continuously differentiable. Intuitively, one should expect this since $f$ is expressed as a double integral of a continuous function. In fact, we can iterate this idea: Since $f$ is twice continuously differentiable, then $f$ can be expressed as a double integral of a twice continuously differentiable function, and so it is four times continuously differentiable, etc: Any harmonic function has continuous derivatives of all orders.

4. Another important consequence is that harmonic functions on (connected) open regions do not have any local maxima or minima, unless they are actually constant (in which case, of course, any point is a local maximum and a local minimum. This is because if $(x_0,y_0)$ is a local maximum or minimum of the harmonic function $f$, then the average value of $f$ on any small disk around $(x_0,y_0)$ is just $f(x_0,y_0)$. If $f$ is not constant on the disk, then for its average to be $f(x_0,y_0)$, one must have points on the disk where the value of $f$ is larger, and points where it is smaller. But then, since this holds no matter how small the disk one consider is, it follows that there are points arbitrarily close to $(x_0,y_0)$ where $f$ takes values larger than $f(x_0,y_0)$, and also arbitrarily close points where $f$ takes values smaller than $f(x_0,y_0)$. But this means that after all, $(x_0,y_0)$ is neither a local maximum nor a local minimum.

[Notice that if $f$ is harmonic, then $f_{xx}=-f_{yy}$; why can’t we simply appeal to the second derivative test to conclude that all critical points of $f$ must be saddle points?]

5. One can also show that if $f$ is constant on a small disk $D$ then in fact it is constant through the whole region ${\mathcal R}$ (or at least, through the connected component of ${\mathcal R}$ containing $D$). Although the proof of this is not too difficult, I omit it since it requires a slightly subtler analysis of regions in the plane than we have covered in lecture.