May 6th Reading + Thoughts: 1/r harmonic function, relation between harmonic and holomorphic functions, space of harmonic polynomials

May 7—2025

Goal 1: Geometrically show that $u(x, y, z) = \frac{1}{\sqrt{x^2 + y^2 + z^2}}$ is harmonic.

To show $u(x, y) = \frac{1}{\sqrt{x^2 + y^2 + z^2}}$ is harmonic, it suffices to show that div.grad( $\frac{1}{\sqrt{x^2 + y^2 + z^2}}$ ) = 0. The key is to study the function $f(x, y, z) = grad( \frac{1}{\sqrt{x^2 + y^2 + z^2}})$ . Notice that in polar coordinates, $f(x, y, z) = f(r)$ (where $r = \sqrt{x^2 + y^2 + z^2}$ ) depends only on the radial distance. Hence, the magnitude of this function on any point of a circle of fixed radius is a constant.

Given arbitrary $R > r > 0$ . Let the region between the circles of radius $R$ and $r$ be $D$ . We now compute the flux of ${grad}(\frac{1}{r})$ across the $\partial D$ . Since the gradient gives the direction of the greatest ascent, it follows that the gradient of $\frac{1}{r}$ at each point points toward the origin. From the previous page, we know that $U \propto \frac{1}{r}$ , where $U$ is the potential. Since $F = \nabla U$ , it follows that $F \propto \nabla \frac{1}{r} = grad(\frac{1}{r})$ , where $F$ is the corresponding force. Since $F \propto \frac{1}{r^2}$ , it follows that $grad (\frac{1}{r}) \propto \frac{1}{r^2}$ . Hence, we have:

$\frac{\text{flux across sphere of radius R}}{\text{flux across sphere of radius r}} = \frac{R^2}{r^2}$ .

We also know that:

$\frac{\text{surface area of circle of radius R}}{\text{surface area of circle of radius r}} = \frac{r^2}{R^2}$ .

Thus, the flux of $f$ across the surface of the two spheres are equal, and so the total flux across the region is 0.

It follows by the divergence theorem that:

$\iiint_{D} \nabla \cdot grad(\frac{1}{r}) dV = \iint_{\partial D} grad(\frac{1}{r}) \cdot \overrightarrow{n} dS = 0$ .

Since $r, R$ had been arbitrary, it follows that $\nabla \cdot grad(\frac{1}{r}) = 0$ . Hence, $\frac{1}{r}$ is harmonic.

Important: The function $f(x, y) = \frac{1}{sqrt(x^2 + y^2)}$ is not harmonic. What is difference?

The difference is that

$\frac{\text{flux across circle of radius R}}{\text{flux across circle of radius r}} = \frac{r^2}{R^2}$

but

$\frac{\text{circumference of circle of radius R}}{\text{circumference of circle of radius r}} = \frac{R}{r}$

and so the flux across the two boundaries are no longer the same. Hence, the total flux across the region is non-zero.

Goal 2: Suppose $u(x, y)$ is a harmonic function on $\text{D} \subset \mathbb{R}^2$ . Why do we require the condition that it be defined on a simply connected domain in order for it to be the real part of a holomorphic function $f(z) = u(z) + iv(z)$ ?

By the Cauchy-Riemann equations, it is not difficult to see that the real part of any holomorphic function is harmonic. (Note: proof of the converse will follow in a later post.)

We use the example of $f(x, y) = log(\sqrt{x^2+y^2})$ defined on the punctured plane $mathbb{R}^2 \setminus {(0,0)}$ . Note that $log(\sqrt{x^2 + y^2}) = Re(log(z))$ , but $log(z)$ is multi-valued, hence it is not holomorphic.

Claim: $S$ is a basis for the space of harmonic polynomials.

It suffices to show that the dimension of the space of homogeneous harmonic polynomial of fixed degree $r \geq 1$ is 2.

Given $n \in \mathbb{N}$ . Given a homogeneous harmonic polynomial of degree n, suppose that it contains $x^ry^{n-r}, where$ latex 0 \leq r \leq n$. Notice that it would then have to contain $x^{r-2}y^{n-r+2}, x^{r-4}y^{n-r+4}, ...$ and $x^{r+2}y^{n-r-2}, x^{r+4}y^{n-r-4}, ...$ , i.e. it will contain all terms of the form $x^{r+2k}$ where $k \in \mathbb{Z}$ and $0 \leq r+2k \leq n$ . And if the polynomial contains $x^{r+1}y^{n-r-1}$ it follows that it will contain all terms of the form $x^{r+2k-1}$ where $k \in \mathbb{Z}$ and $0 \leq r+2k-1 \leq n$ . Hence, the space of homogeneous harmonic polynomial of fixed degree $r \geq 1$ is at least 2. This concludes the proof.