Laplace equation and boundaries
Use mean values, flux compatibility, and boundary data to understand harmonic equilibria and their uniqueness.
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The question for today
What you’ll understand
- Interpret a harmonic function as a source-free equilibrium.
- Apply the maximum principle to a bounded Dirichlet problem.
- Check Neumann compatibility and identify its constant ambiguity.
At a source-free equilibrium, a field can vary from point to point without accumulating locally. The Laplace equation describes that balance for potentials, steady temperatures, and other linear models with uniform coefficients. Its solutions are constrained across a region by boundary information rather than marched forward from one instant. Asking whether boundary values or boundary fluxes are known changes both the existence conditions and the uniqueness statement, so the boundary is part of the equation's meaning.
Harmonic means locally balanced
A twice continuously differentiable function is harmonic where Δu=0. In a two-dimensional Cartesian picture, its second derivatives in perpendicular directions sum to zero: upward curvature in one direction must be balanced by downward curvature in another. This is stronger than merely having zero slope at selected points. A harmonic function equals its average over every sufficiently small sphere centered in its domain, and also over the enclosed ball. That mean-value property explains why a strict interior peak is incompatible with source-free balance.
For a connected bounded domain and a harmonic function continuous up to its boundary, the maximum and minimum are controlled by boundary values. If an interior maximum is attained, the strong maximum principle makes the function constant. Apply this statement to the difference of two functions sharing the same Dirichlet values: its boundary values are zero, so both its maximum and minimum are zero. This proves uniqueness. It does not by itself prove that every prescribed boundary function produces a solution on every possible rough domain; existence and boundary regularity are separate questions.
Values and normal fluxes impose different constraints
Dirichlet data prescribe the value of u. Neumann data prescribe the outward normal derivative, which is related to physical flux after inserting the appropriate sign and material coefficient. For −Δu=f with ∂_n u=g, integration over the domain gives the compatibility condition ∫f=−∫g. The minus sign follows from the chosen Poisson convention. A purported solution that violates this balance cannot be rescued by a more accurate numerical grid: the source and boundary data demand contradictory total flux.
Pure Neumann data also leave an additive constant undetermined, because derivatives cannot detect shifting u by a constant. On a connected suitable bounded domain, normalization such as zero spatial mean removes that ambiguity. Dirichlet values instead fix the constant directly. A mixed problem needs a specified partition of the boundary and a meaningful trace on its Dirichlet part. These distinctions reappear in linear systems: a discretized pure Neumann Laplacian has a constant null vector, whereas imposing an adequate Dirichlet condition removes that particular nullspace.
Equilibrium is global even when the equation is local
A local differential equation can encode a global dependency. Changing a Dirichlet value along one boundary portion generally changes the harmonic solution throughout the domain. There is no evolution time in the equilibrium problem and therefore no finite-speed propagation claim to apply to it. For example, on a disk, separating angular Fourier modes shows that the nth boundary oscillation extends inward with a radial factor r^n after scaling the radius to one. Short angular scales are consequently concentrated near the boundary, while the mean reaches the center unchanged.
The fields lab lets you change a softened source q and an independent rotational component Ω. With Ω set to zero, inspect radial directions, then restore Ω to see why a field with circulation need not be the gradient of a globally defined scalar potential. The softened radial field is not asserted to be harmonic: its divergence is generally nonzero near the source, and the lab has no domain boundary or Dirichlet/Neumann solver. It provides a flux-and-direction comparison; the polynomial calculation below is the actual harmonic boundary-value example.
Work it through
Verify a harmonic potential and its boundary flux
On the unit disk, verify that solves the Dirichlet problem with boundary value cos(2θ). Check its total outward normal derivative.
Compute u_xx=2 and u_yy=−2. Their sum is zero, so u is harmonic throughout the disk, including its center.
In polar coordinates u=r²cos(2θ). At r=1 its boundary value is cos(2θ), exactly as prescribed. The maximum principle makes this continuous harmonic solution unique for those Dirichlet data.
The outward normal derivative at the unit circle is u_r=2cos(2θ). Integrating from 0 to 2π gives zero net normal derivative, as required for a source-free harmonic field.
At r=1/2 and θ=0 the value is 1/4, compared with boundary value one along the same ray. The radial factor r² suppresses this nonconstant angular mode toward the center.
The polynomial is harmonic, realizes the given boundary values, and has zero net outward derivative despite nonzero local flux.
The living notebook
Explore the model
Interactive lab · fields
Separate radial response from circulation
Set rotation Ω to zero and vary source q in the softened field q r/(r²+ε). Then increase Ω to add Ω(−y,x). Inspect the changed direction field. This is not a harmonic-potential or boundary-value solver; softening introduces distributed divergence and the rotational part has circulation.
- Fx at (1, 0)
- 0.833
- Fy at (1, 0)
- 0.5
- Field magnitude there
- 0.972
The softened illustration is q·r/(r² + 0.2) + Ω(−y, x). Moving markers indicate local direction, not integrated particle paths or a Maxwell solution. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forSource balance and potential structure require mathematical checks beyond the arrows in a field picture.
Pause & reason
Check your understanding
Your turn
Try it yourself
On 0<x<1, solve −u''=2 with u'(0)=1 and u'(1)=−1. Explain the role of outward normals and normalize to zero mean.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
At the left endpoint ∂_n u=−u′(0), while at the right endpoint ∂_n u=u′(1).
Compare with the solution
Integration gives u'=1−2x and u=x−x²+C. The outward normal derivatives are both −1, whose sum is −2, matching the negative source integral. Since ∫₀¹(x−x²)dx=1/6, zero mean fixes C=−1/6. Thus . Without normalization every additive constant is allowed.
Keep the idea
Take it with you
- The maximum principle proves uniqueness for an appropriate Dirichlet problem.
- Pure Neumann data require total-flux compatibility.
- A Neumann solution is determined only up to constants unless normalized.
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