Poisson’s equation and the promise of uniqueness
Use an energy identity to decide what an electrostatic solution must satisfy.
On this page
The question for today
What you’ll understand
- Derive Poisson’s equation with the correct sign.
- Prove uniqueness for complete Dirichlet data.
- Separate uniqueness from existence and identify missing boundary information.
Finding a potential that has the right curvature is only part of an electrostatic problem. The same charge distribution can coexist with many externally imposed fields. Boundary data select which field is realized. A uniqueness theorem turns a plausible construction into a decisive solution: once the equation, sources, and complete boundary conditions agree, there is no hidden alternative. We will prove that claim for a bounded vacuum region and see exactly which assumptions the proof uses.
Sources determine curvature
In electrostatics the electric field is conservative and can be written as the negative gradient of a scalar potential. Combining this statement with Gauss’s law produces Poisson’s equation. Positive charge gives negative potential curvature when summed over all coordinate directions. A positive second derivative in one direction does not alone imply negative charge, because the other directions also contribute. Reducing the Laplacian to one dimension therefore requires translational symmetry, not merely choosing an axis.
A particular solution accounts for the specified volume charge. Adding any harmonic function leaves that volume charge unchanged, while changing the field and usually the boundary potential. A uniform applied field is a simple example of this freedom. Thus the Coulomb integral over charges inside a finite region is generally incomplete: external electrodes contribute a harmonic correction inside that region. The correction encodes the apparatus rather than an additional interior source.
The difference has no source
Suppose two sufficiently regular potentials solve the same Poisson problem in one bounded, connected region, with equal potential everywhere on its boundary. Their difference has zero Laplacian and zero boundary value. Multiply its Laplace equation by the difference itself and integrate by parts. The remaining volume integral is the squared magnitude of its gradient. Every contribution is nonnegative, so cancellation cannot conceal a nonzero difference field.
The boundary term vanishes because the difference potential is zero there. Consequently its gradient vanishes throughout the connected region, making the difference constant. The boundary condition fixes that constant to zero. Connectedness matters: a constant could otherwise be chosen separately in each disconnected component. For point sources one compares solutions with the same singularities and applies the argument to their regular difference, rather than integrating divergent point self-energies.
What the theorem licenses
Uniqueness is conditional: it says at most one solution satisfies these requirements. It does not construct that solution or establish that arbitrary inconsistent data admit one. Likewise, checking boundary values only at a few points is insufficient. Exterior problems also require an admissible condition at infinity; otherwise a growing harmonic field can slip into a proposed answer. These distinctions make uniqueness useful when validating images, separated series, or numerical approximations.
Inspect the field lab as a diagnostic of the conservative-field assumption. Reset and set source strength to 1 and rotation to 0; predict radial arrows. For a fresh comparison, Reset and set source strength to 1 and rotation to 1; predict an added tangential component. Rotation introduces circulation and disqualifies the field from this electrostatic gradient argument. The softened planar model neither solves three-dimensional Poisson boundary data nor proves uniqueness.
Work it through
A uniformly charged slab between electrodes
A one-dimensional vacuum gap extends from x=0 to x=2 m. The constant ratio ρ/ε₀ is 4 V/m², and both end potentials are zero. Determine V(x), its maximum, and the field at x=1.5 m.
Integrate V″=−4 twice. The general potential is −2x²+Ax+B with volt and metre units understood. The left boundary sets B=0; it is a physical boundary condition, not a guessed symmetry constant.
The right boundary requires −8+2A=0, giving A=4 V/m. Thus V=2x(2−x). Its derivative vanishes at x=1 m, and the negative second derivative confirms a maximum of 2 V there.
The field is −V′=4x−4 V/m. At 1.5 m it is +2 V/m. The derivative of this field is +4 V/m², consistent with the positive specified charge. Complete end data make this the unique one-dimensional solution.
The maximum potential is 2 V at the midpoint; the field at x=1.5 m is +2 V/m toward increasing x.
The living notebook
Explore the model
Interactive lab · fields
Test the conservative-field assumption
Reset and set q=1, Ω=0. Predict radial arrows. Then Reset and set q=1, Ω=1; predict a tangential addition. Controls retain time when changed, so begin each comparison with Reset.
- 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 forThis softened planar source-plus-rotation model detects an assumption to check; it has no electrodes or Poisson solver.
Pause & reason
Check your understanding
Your turn
Try it yourself
A charge-free one-dimensional gap has length 3 m, with V(0)=2 V and V(3 m)=8 V. Find its potential and field, and explain why a quadratic correction is excluded.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
A charge-free potential has zero second derivative, and both end values must hold.
Compare with the solution
The linear solution is V=2 V+(2 V/m)x, giving E=−2 V/m. Any genuine quadratic correction has nonzero second derivative and introduces volume charge. A linear correction that preserves both end values must also vanish. This illustrates uniqueness within the stated one-dimensional geometry.
Keep the idea
Take it with you
- Poisson’s equation fixes source-dependent curvature.
- Complete Dirichlet data eliminate every harmonic ambiguity.
- Uniqueness is conditional and does not itself establish existence.
Read further
Original sources and deeper explanations to follow at your own pace.
Optional progress, saved in this browser when available. Revisit any lesson at any time.
Lesson not marked complete.