From an abstract solution to an explicit one
By now you know a great deal about harmonic functions without ever having written one down. You know the Dirichlet problem — prescribe the temperature all around the boundary of a region, find the steady temperature inside — has exactly one solution, and that the mean-value property and the maximum principle guarantee it is unique and well-behaved. But existence proofs and uniqueness proofs are promissory notes: they swear an answer is out there without ever showing you its face. This guide cashes the note. On one special shape — the disk — we will write the solution down completely, as a single integral you could evaluate by hand.
Why the disk, and not a square or a blob? Because the disk has a symmetry — rotation — that perfectly matches the symmetry of Laplace's equation itself, which treats all directions alike. That match is exactly what separation of variables needs to bite. On a disk the natural coordinates are polar, (r, theta): a radius r out from the centre and an angle theta around. The boundary is then dead simple — it is just the circle r = 1, parameterized by the single angle theta. Prescribing boundary data means handing over one function f(theta) of the rim angle, and the prize at the end is a formula that turns that f into the interior temperature u(r, theta) directly.
Building the answer from a Fourier series
Here is the engine. In polar coordinates Laplace's equation u_xx + u_yy = 0 becomes u_rr + (1/r) u_r + (1/r^2) u_(theta theta) = 0. Try a product u = R(r) T(theta). Separation splits this into two ordinary problems: T must be periodic with period 2*pi (you return to where you started after one full turn), which forces T to be cos(n*theta) or sin(n*theta) for whole numbers n; and R then solves a simple equation whose solution that stays finite at the centre is R = r^n. So the harmonic building blocks on a disk are r^n cos(n*theta) and r^n sin(n*theta) — and any sum of them is again harmonic.
Now superpose. Write the interior solution as a sum over all these blocks, u(r, theta) = a_0/2 + sum over n of r^n (a_n cos(n*theta) + b_n sin(n*theta)). Set r = 1 and this must equal the boundary data f(theta). But that is precisely a Fourier series for f: the coefficients a_n and b_n are nothing but the Fourier coefficients of the rim function. So the recipe is clear in principle — Fourier-analyze f around the rim, then attach a factor r^n to the n-th mode and you are done. The factor r^n is the whole story of the interior: since 0 <= r < 1, each higher mode is multiplied by a smaller and smaller power of r, so the wiggly high-frequency detail of the boundary data is damped more and more strongly the deeper you go toward the centre.
Summing the series into one kernel
A Fourier series is a beautiful recipe but an awkward object — you must compute infinitely many coefficients and then sum infinitely many terms. The magic of the disk is that this infinite sum can be done once, in closed form, for all data at once. Substitute the integral formulas for a_n and b_n back into the series, swap the sum and the integral, and the geometric-looking series in r^n collapses. What drops out is the Poisson integral formula: a single integral that takes the boundary function f and returns the interior value u(r, theta) with no series left to sum.
u(r, theta) = (1 / (2*pi)) * integral_0^(2*pi) P(r, theta - phi) f(phi) d phi
Poisson kernel (unit disk, radius 1):
1 - r^2
P(r, psi) = -----------------------------
1 - 2 r cos(psi) + r^2
interior value = weighted average of the rim values f(phi),
the weight P leaning toward the nearest boundary point.Read the formula as a weighted average and it stops looking like algebra. The value u(r, theta) at an interior point is an average of all the rim values f(phi), but a biased one: each rim point votes with weight P(r, theta - phi), the Poisson kernel. That weight is always positive, and it is largest for the rim point nearest your location and smallest for the one diametrically across. So the temperature where you stand is mostly an echo of the boundary just over your shoulder, with fainter contributions from the far side — exactly the intuition that nearby boundary matters most. The 1 - r^2 in the numerator is the dial: at the centre r = 0 it makes the weight uniform, and as r climbs toward 1 the weight piles up ever more sharply on the single nearest rim point.
What the formula proves for free
The Poisson formula is not just a calculator; it pays off every promissory note from the earlier guides in one stroke. Set r = 0, the centre. There the kernel collapses to P(0, psi) = 1, a flat weight, and the integral becomes the plain average of f around the whole rim. That is the mean-value property — the centre value equals the boundary average — now derived, not assumed. The kernel being strictly positive means u(r, theta) is a genuine weighted average of the boundary values, so it can never exceed the largest of them nor fall below the smallest: there is the maximum principle, read straight off the formula.
It also settles existence for the disk's Dirichlet problem, which no abstract argument had yet done. Hand over any reasonable boundary function f — even a discontinuous one, say warm on the top semicircle and cold on the bottom — and the integral produces an honest harmonic function inside that approaches f as you walk out to the rim (at every point where f is continuous). Existence and uniqueness together make this a genuinely solved, well-posed problem on the disk: a solution exists, it is the only one, and it depends continuously on the data, since nudging f a little nudges its weighted average only a little.
- Given the rim data f(theta), feel free to skip the coefficients entirely — go straight to the integral.
- Pick the interior point (r, theta) where you want the temperature.
- Form the Poisson kernel P(r, theta - phi) = (1 - r^2) / (1 - 2 r cos(theta - phi) + r^2) as a function of the rim angle phi.
- Average f against this weight: u = (1/(2*pi)) integral over phi of P(r, theta - phi) f(phi). That number is the answer.
Limits, honesty, and what generalizes
Be clear about what this gift does and does not cover. First, the formula solves the Dirichlet problem — boundary values prescribed — and not the Neumann problem, where you prescribe the boundary flux instead; that one needs its own (closely related) kernel and the compatibility condition you met earlier. Second, the closed form above is the disk's privilege, bought with its rotational symmetry. On a square or an irregular region there is no such tidy P, and you fall back on a Fourier-type series, a numerical scheme, or the Green's function — of which the Poisson kernel is in fact the boundary trace, the disk being the one shape where that Green's function is known explicitly.
A subtle honesty about the boundary. The interior is gloriously smooth, but the formula only recovers f at the rim where f is continuous. At a jump — the seam between the warm and cold semicircles — the interior solution smoothly splits the difference, sliding to the average of the two one-sided values as you approach the seam, never actually equalling the jump itself. That is not a flaw; it is the smoothing again, telling you no harmonic function can carry a true discontinuity into its interior. And the whole construction rests on the data living on the boundary of the region: prescribing data on a curve inside the disk, Cauchy-style, would be the ill-posed setup that the well-posedness story of this rung warns against.