Laplace's Equation and Harmonic Functions

Harmonic functions and Laplace's equation are foundational across mathematics and its applications. In complex analysis, the real and imaginary parts of any holomorphic function are harmonic (a consequence of the Cauchy–Riemann equations), linking Laplace's equation to conformal mapping and the theory of analytic functions. In electromagnetism, electrostatic potentials in charge-free regions and magnetostatic scalar potentials both solve Laplace's equation, with boundary conditions on conductors set by the physical geometry. In fluid dynamics, the velocity potential of an incompressible, irrotational flow is harmonic, underlying classical aerodynamic theory and the study of ideal fluids. In probability, harmonic functions correspond exactly to the values left invariant, in expectation, by Brownian motion — the mean value property is the continuous analogue of a martingale property, connecting PDE theory to stochastic processes. In numerical analysis and engineering, finite-difference and finite-element methods for Laplace's and Poisson's equations underlie simulations of heat conduction, electrostatic fields, and structural stress. In computer graphics and geometry processing, discrete harmonic functions and the graph Laplacian are used for mesh smoothing, image inpainting, and manifold learning. This chain of ideas continues directly into the Maximum Principle for Parabolic and Elliptic Equations, which generalizes the argument given here to time-dependent and more general elliptic operators, and into Green's Functions for Boundary Value Problems, which builds explicit solution formulas for the Dirichlet and Neumann problems atop the uniqueness result proved above.

Ask the tutor: How does the mean value property generalize to prove existence of solutions via Perron's method, and how does that connect to Green's functions?

Imagine a thin, uniform metal plate whose edge is held at fixed but spatially varying temperatures, and wait until the interior temperature stops changing. The resulting steady-state temperature $u(x,y)$ satisfies the two-dimensional heat equation $u_t = k\Delta u$ with $u_t = 0$, i.e. exactly $\Delta u = 0$. The condition $\Delta u = 0$ says, pointwise, that $u_{xx} = -u_{yy}$: any concavity in the $x$-direction is exactly compensated by convexity in the $y$-direction. There is no room left for the function to have an interior "bump" — a local max would require $u_{xx} \le 0$ and $u_{yy}\le 0$ simultaneously (with at least one strict), forcing $\Delta u < 0$, contradicting harmonicity. This is the seed of the maximum principle, made rigorous below. A second, more geometric way to see it: harmonicity means the value at the center of any small ball equals the average of the values on its boundary sphere. A harmonic function has no preferred direction to "lean" — every direction of averaging gives the same answer as being at the center. Electrostatic potentials in charge-free regions, and steady fluid-flow potentials in source-free, irrotational, incompressible flow, both obey this same averaging law, which is why Laplace's equation appears across so many branches of physics.

Example 1: Dirichlet problem on a disk with asymmetric boundary data. Solve $\Delta u = 0$ on the open unit disk $\Omega = \{(x,y): x^2+y^2<1\}$, with boundary condition, in polar coordinates $(r,\theta)$,

In polar coordinates the Laplacian is

Separation of variables $u = R(r)\Theta(\theta)$ with $\Theta$ periodic of period $2\pi$ gives the standard family of bounded-at-the-origin solutions

each individually harmonic (direct substitution: $\Delta(r^n \cos n\theta) = n(n-1)r^{n-2}\cos n\theta + n r^{n-2}\cos n\theta - n^2 r^{n-2}\cos n\theta = 0$, and similarly for $\sin n\theta$). By linearity of $\Delta$, any finite sum of such terms is harmonic. Match the boundary data directly, since it is already a finite trigonometric sum:

This matches $a_0 = 3$ ($n=0$ term), $a_1 = 5$ ($n=1$ cosine term), $b_2 = -2$ ($n=2$ sine term), all other coefficients zero. Hence

Check. In Cartesian coordinates, $r\cos\theta = x$ and $r^2 \sin 2\theta = 2r^2\sin\theta\cos\theta = 2xy$, so

Direct differentiation: $u_x = 5-4y$, $u_{xx}=0$; $u_y = -4x$, $u_{yy}=0$. So $\Delta u = 0+0=0$, confirming harmonicity independently of the polar-coordinate computation. At the center $u(0,0) = 3$. The average of $u$ over the boundary circle is

since $\int_0^{2\pi}\cos\theta\, d\theta = \int_0^{2\pi}\sin2\theta\, d\theta = 0$. This matches $u(0,0)=3$ exactly, confirming the mean value property (A) in this concrete case. By the maximum principle (C), the extreme values of $u$ on the closed disk occur on the boundary; since $u(1,\theta) = 3+5\cos\theta - 2\sin2\theta$ is a smooth function of $\theta$ on a compact interval, its extrema can be found by calculus on the circle rather than by searching the whole disk — a genuine computational saving the maximum principle guarantees is valid.

Example 2: A genuinely 3-D harmonic function from electrostatics — the Newtonian/Coulomb potential. Let $\Omega = \mathbb{R}^3 \setminus \{0\}$ and consider

This is the potential (up to a physical constant) of a point charge at the origin. We verify $\Delta u = 0$ for $\rho \ne 0$ directly, without symmetry shortcuts, differentiating $u = (x^2+y^2+z^2)^{-1/2}$ term by term.

By the symmetry of the formula under permuting $x,y,z$ (the algebra is identical for each variable, so we only need write it once and cyclically permute):

Summing,

So $u = 1/\rho$ is harmonic on $\mathbb{R}^3\setminus\{0\}$, matching the physical fact that the Coulomb potential satisfies $\Delta u=0$ away from the charge itself (at the origin $u$ blows up and is not even defined, let alone $C^2$ — this is precisely why the domain excludes the origin, and why a point charge appears as a singular source, or delta function, on the right side of Poisson's equation $\Delta u = -4\pi q\,\delta_0$ in the distributional sense). Note this example shows the mean value property must be applied only to balls entirely contained in the domain of harmonicity: the spherical average of $u=1/\rho$ over a sphere of radius $r$ centered at the origin equals $1/r$ (by symmetry, since $u=1/\rho=1/r$ everywhere on that sphere), but $u(0)$ is undefined — there is no contradiction with Theorem (A) because $x_0=0 \notin \Omega$.