Laplace on a Disk

Picture a circular drumhead whose rim is clamped at different, unchanging temperatures around its edge — or a circular metal plate whose boundary is held at a fixed electric potential. Once things settle down, the temperature (or potential) inside satisfies Laplace's equation. We already solved this on a rectangle. But a disk isn't a rectangle — it has no corners, no straight edges, and its natural "coordinate lines" are circles and spokes, not a grid of horizontal and vertical lines.

Trying to force x, y coordinates onto a circular problem is like describing a pizza slice using north-south, east-west directions — technically possible, but awkward. The fix is to switch to polar coordinates (r, \theta), and when we do, separation of variables produces a genuinely different flavour of solution: sines and cosines for the angular direction, but simple powers of the radius for the radial direction. No exponentials, no hyperbolic functions — just r^n.

Setting up the disk problem

In polar coordinates, the Laplacian takes the form

\nabla^2 u = u_{rr} + \frac{1}{r}u_r + \frac{1}{r^2}u_{\theta\theta} = 0,

and we look for u(r,\theta) on the unit disk 0 \le r \le 1, given boundary data u(1,\theta) = f(\theta) around the rim. As before, we try a separated solution

u(r,\theta) = R(r)\,\Theta(\theta).

Substituting and dividing through by R\Theta splits the equation into a radial piece and an angular piece, tied together by a separation constant. Call it n^2:

\Theta'' + n^2 \Theta = 0, \qquad r^2 R'' + r R' - n^2 R = 0.

Worked example: solving the two halves

Step 1 — the angular equation. The equation \Theta'' + n^2\Theta = 0 has general solution \Theta = A\cos(n\theta) + B\sin(n\theta) for any real n. But \theta and \theta + 2\pi describe the exact same point on the disk — walk all the way around and you're back where you started. So we need \Theta(\theta) = \Theta(\theta + 2\pi), which forces n to be an integer. Non-integer n would make \Theta disagree with itself after one lap — not a valid function on the disk at all.

Step 2 — the radial equation. The equation r^2 R'' + rR' - n^2 R = 0 is a Cauchy–Euler equation: try R = r^m and the algebra collapses to m^2 - n^2 = 0, so m = \pm n. The two radial solutions are therefore R = r^n and R = r^{-n} (with \log r replacing them both when n = 0).

Step 3 — keep only what's physical. Every point of a solid disk includes the centre, r = 0. Only R = r^n stays finite there — the other branch is discarded (more on why below). Pairing the surviving radial part with each angular piece gives the building blocks, the circular harmonics:

r^n\cos(n\theta), \qquad r^n\sin(n\theta), \qquad n = 0, 1, 2, \dots

Three concrete cases

Superposing the circular harmonics with coefficients matched to the boundary data gives the general solution:

u(r,\theta) = \frac{a_0}{2} + \sum_{n=1}^{\infty} r^n\big(a_n\cos n\theta + b_n\sin n\theta\big).

A few small cases make this concrete. Suppose the rim of the disk is held at temperature f(\theta).

Constant rim, f(\theta) = 20. Only the n=0 term survives: u(r,\theta) = 20 everywhere. A uniformly heated rim gives a uniformly heated plate — no surprises.
Single wiggle, f(\theta) = \cos\theta. Only n=1 contributes: u(r,\theta) = r\cos\theta = x. That's just a plane tilted linearly across the disk — you can check directly that u_{xx} + u_{yy} = 0 since u=x is linear.
A mixed rim, f(\theta) = 5 + 3\cos 2\theta - \sin 3\theta. Reading off the coefficients term by term gives, immediately, u(r,\theta) = 5 + 3r^2\cos 2\theta - r^3\sin 3\theta — no integrals needed, because the boundary data was already a short trig sum.

A four-lobed harmonic

Take the pure n=2 harmonic on its own: f(\theta) = \cos 2\theta, so u = r^2\cos 2\theta (which is exactly x^2 - y^2 in Cartesian coordinates). Warm and cool lobes alternate four times around the rim, and the colour fades to nothing at the centre — because the centre value here is 0, precisely the average of \cos 2\theta around the full circle.

It's tempting to keep both radial solutions, r^n and r^{-n}, the way you'd keep both exponentials e^{kx} and e^{-kx} on a rectangle. Don't. As r \to 0, the term r^{-n} shoots off to infinity for every n \ge 1 — drag the slider below towards higher n and watch how violently the orange curve diverges near r = 0, while the blue curve calmly settles to zero.

A physical temperature or potential can't actually be infinite at an interior point, so on a solid disk — one that includes its own centre — every r^{-n} term must be thrown out. This is a boundedness condition, and it is specific to full disks. If the domain were instead an annulus (a ring-shaped washer with a hole cut out of the middle, so r = 0 is never actually reached), the r^{-n} solutions become perfectly legal again, because there's no centre point left to blow up at.

The payoff: the mean-value property

Look again at the general solution and set r = 0. Every term with n \ge 1 carries a factor of r^n, which vanishes at the centre. All that survives is

u(0,\theta) = \frac{a_0}{2},

and a_0/2 is nothing but the average of the boundary data f(\theta) around the whole rim. That's a strikingly clean and genuinely useful fact:

If u is harmonic on a disk and continuous up to its boundary, then the value of u at the centre equals the average of its values around the entire boundary circle:

u(0) = \frac{1}{2\pi}\int_0^{2\pi} u(1,\theta)\,d\theta.

Check it against the mixed-rim example above: the average of 5 + 3\cos 2\theta - \sin 3\theta over a full period is just 5 (the cosine and sine terms average to zero), and indeed u(0,\theta) = 5 — exactly what the formula predicts, with no integration required.

Summing the whole series in closed form gives the elegant Poisson integral formula, which writes the value at any interior point — not just the centre — as a weighted average of the boundary values:

u(r, \theta) = \frac{1}{2\pi}\int_0^{2\pi} \frac{1 - r^2}{1 - 2r\cos(\theta - \phi) + r^2}\,f(\phi)\,d\phi.

The fraction is the Poisson kernel — the precise weight each boundary point contributes to each interior point. Set r = 0 and the kernel becomes the flat constant 1, recovering the plain average from above.

The mean-value property looks like a cute computational shortcut, but it is one of the deepest facts in the whole subject. If the value at every interior point equals the average of a circle of neighbours around it, that point can never be strictly bigger than all of its neighbours at once — averages don't work that way. So a harmonic function can have no strict local maximum or minimum tucked away inside its domain; every hottest and coldest point must live on the boundary.

This single observation is the seed of the maximum principle, and the mean-value property itself resurfaces throughout complex analysis and potential theory — it's the reason the real and imaginary parts of any complex-differentiable function are automatically harmonic, and why "harmonic" and "no interior extremes" turn out to be almost the same idea.