Laplace's Equation on a Disk

A disk is simply a filled-in circle, like a round metal plate, and solving Laplace's equation on it means finding the settled temperature everywhere inside once you've fixed the temperature around the rim. It's good for anything round in a steady state: pipes, cylinders, a wire's electric field, or medical scanners that read voltages around a patient's chest. All we're really saying is that the inside is a blurred copy of the rim, blurrier the closer you get to the middle. The formula $A_0 + \sum \left(\frac rR\right)^n(A_n\cos n\theta + B_n\sin n\theta)$ looks long, but it's just three pieces: $A_0$ (the rim's average temperature), $A_n\cos n\theta + B_n\sin n\theta$ (the rim's ups and downs, with $n$ of them around the circle), and $\left(\frac rR\right)^n$ (a dimmer switch that's fully on at the rim and turns down as you move inward, fastest for the bumpiest ripples). That's the whole picture; bear with the polar coordinates that get us there.

Why polar coordinates? A rectangle's edges are lines $x =$ const and $y =$ const, which is why $X(x)Y(y)$ worked. A disk's edge is $r = R$, so we want the boundary to be a coordinate line again. Writing the Laplacian in polar coordinates gives the equation above, and then we separate as usual.

Separation in polar form. With $u = R(r)\Theta(\theta)$: $\Theta'' + \lambda\Theta = 0$ must be $2\pi$-periodic, because going once around the circle brings you back to the same point. That's exactly the circular ring from earlier, so $\lambda = n^2$ and $\Theta$ is $\cos n\theta$ or $\sin n\theta$. The radial equation $r^2R'' + rR' - n^2R = 0$ is an Euler equation with solutions $r^n$ and $r^{-n}$ (and $1$, $\ln r$ for $n = 0$).

Boundedness. $r^{-n}$ and $\ln r$ blow up at the center, and a real plate doesn't have infinite temperature in the middle. So inside a disk only $r^n$ survives. (Outside a disk, it's the reverse: $r^n$ blows up far away, so $r^{-n}$ is the one you keep.)

What $r^n$ does. Think about what the factor $\left(\frac rR\right)^n$ means. At the rim it's 1, so you get the boundary data back. Moving inward it shrinks, and the bigger $n$ is, the faster it shrinks. Fast wiggles around the rim get smoothed out almost immediately, while slow ones reach deeper in. It's the same smoothing you saw on the rectangle, where high modes stayed near the hot edge.

The mean value property falls right out: at $r = 0$ every term except $A_0$ vanishes, and $A_0$ is the boundary average. There's nothing special about the unit disk or about this point. Any point inside any region is the center of some small disk, so the value there is the average around that circle. That's the property the last lesson promised.

In Cartesian form, $r^n\cos n\theta$ and $r^n\sin n\theta$ are the real and imaginary parts of $(x + iy)^n$: $x$, $y$, $x^2 - y^2$, $2xy$, $x^3 - 3xy^2$, and so on. So the harmonic function $x^2 - y^2$ you may have checked by hand is really just $r^2\cos 2\theta$.

A common mistake. It's tempting to think the center temperature depends on how the heat is laid out along the rim, say, that a hot spot near the top should warm the center more if it's concentrated. It doesn't. The center only sees $A_0$, the plain average around the rim. Two rims with the same average give the same center temperature, however differently they're arranged.

Temperatures in pipes and cylinders, the electric field around a charged wire, and the flow of fluid around a cylinder are Laplace problems in polar coordinates.