Laplace on a Disk
On a circular plate the natural coordinates are polar, (r, \theta), and
the Laplacian takes the form
\nabla^2 u = u_{rr} + \frac{1}{r}u_r + \frac{1}{r^2}u_{\theta\theta} = 0.
Separating u = R(r)\,\Theta(\theta), the angular part must be
2\pi-periodic, which forces \Theta = \cos(n\theta), \sin(n\theta)
with integer n. The radial equation is then a Cauchy–Euler equation
whose solution bounded at the centre is R = r^n.
The solution and the Poisson kernel
Superposing the circular harmonics r^n\cos(n\theta) and
r^n\sin(n\theta) gives, on the unit disk,
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),
where a_n, b_n are the Fourier coefficients of the boundary data
f(\theta) = u(1, \theta). Because the centre value is
u(0) = a_0/2 — exactly the average of f — this
recovers the mean-value property directly.
Summing the series in closed form gives the elegant Poisson integral formula,
which writes the interior value at any point 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.
A four-lobed harmonic
With boundary data f(\theta) = \cos 2\theta, the solution is the single
circular harmonic u = r^2\cos 2\theta (which equals
x^2 - y^2). Warm and cool lobes alternate around the rim and fade toward
the centre, where the value is the boundary average — here zero.