Solving Laplace’s equation on a disk

Why care about solving Laplace’s equation

\Delta u = 0

on a disk?

Laplace’s equation is important in its own right—for example, it’s important in electrostatics—and understanding Laplace’s equation is a stepping stone to understanding many other PDEs.

Why care specifically about a disk? An obvious reason is that you might need to solve Laplace’s equation on a disk! But there are two less obvious reasons.

First, a disk can be mapped conformally to any simply connected proper open subset of the complex plane. And because conformal equivalence is transitive, two regions conformally equivalent to the disk are conformally equivalent to each other. For example, as I wrote about here, you can map a Mickey Mouse silhouette

Mickey Mouse

to and from the Batman logo

Batman logo

using conformal maps. In practice, you’d probably map Mickey Mouse to a disk, and compose that map with a map from the disk to Batman. The disk is a standard region, and so there are catalogs of conformal maps between the disk and other regions. And there are algorithms for computing maps between a standard region, such as the disk or half plane, and more general regions. You might be able to lookup a mapping from the disk to Mickey, but probably not to Batman.

In short, the disk is sort of the hub in a hub-and-spoke network of cataloged maps and algorithms.

Secondly, Laplace’s equation has an analytical solution on the disk. You can just write down the solution, and we will shortly. If it were easy to write down the solution on a triangle, that might be the hub, but instead its a disk.

Suppose u is a real-valued continuous function on the the boundary of the unit disk. Then u can be extended to a harmonic function, i.e. a solution to Laplace’s equation on the interior of the disk, via the Poisson integral formula:

u(z) = \frac{1}{2\pi} \int_0^{2\pi} u(e^{\i\theta})\, \text{Re}\left( \frac{e^{i\theta} + z}{e^{i\theta} - z}\right)\, d\theta

Or in terms of polar coordinates:

u(re^{i\varphi}) = \frac{1}{2\pi} \int_0^{2\pi} \frac{u(e^{\i\theta}) (1 - r^2)}{1 - 2r\cos(\theta-\varphi) + r^2}\, d\theta

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