Electrostatic Poisson solvers and boundaries
Solve a tiny Dirichlet system and identify the compatibility and gauge constraints of periodic or Neumann electrostatics.
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The question for today
What you’ll understand
- Form a signed finite-difference Poisson matrix with units.
- Compare algebraic residual with discretization error.
- Distinguish a missing compatibility condition from an arbitrary potential constant.
An electrostatic field solve is incomplete until its domain and boundary conditions are specified. Poisson's equation with fixed potentials can have a unique solution, whereas periodic or pure Neumann problems retain a constant-potential ambiguity and impose a source compatibility condition. These are physical and mathematical features, not solver inconveniences. A small finite-difference system lets us compute the potential, field and residual by hand, then distinguish solving those algebraic equations accurately from approximating the continuum electrostatic problem accurately.
Build a signed discrete boundary-value problem
For one-dimensional SI electrostatics, minus the second derivative of potential equals charge density divided by permittivity. A centered difference gives a positive diagonal and negative neighbor coefficients when written as a positive definite Dirichlet matrix. Boundary potential terms move to the right-hand side for interior unknowns. Multiplying the equation by the squared spacing makes every matrix-equation component have potential units; retaining the divided form instead gives a residual with potential per length squared units.
The electric field is minus the potential gradient. It may be stored at cell faces or computed by a specified centered node difference; these choices have different boundary and Gauss-law interpretations. Our hand calculation uses centered node fields only as a declared diagnostic. A physical electrode potential, an imposed normal flux and a periodic identification constrain different quantities. Choosing a zero potential on a boundary is not generally equivalent to choosing the mean potential as a gauge in a periodic domain.
Reduce the algebraic error without mistaking it for continuum error
Write the discrete equations as Aφ = b and an approximate solution as φtilde. Its residual is b minus Aφtilde, while its algebraic solution error is the inverse matrix applied to that residual. A small residual is useful only relative to consistent norms, units and the conditioning of A. Refining the mesh can worsen the matrix condition number, so the same unscaled residual threshold need not imply the same potential error on every grid.
Jacobi iteration updates each unknown from the previous iteration's neighbors. It is simple to inspect but often slow for long-wavelength error, motivating multigrid or other preconditioned methods on larger problems. Iteration count is not physical elapsed time. Once algebraic error is small compared with truncation error, further iterations do little to improve the continuum approximation. A separate grid study, with consistent boundary data, measures the error remaining from replacing derivatives and geometry by discrete representations.
Apply compatibility before fixing a gauge
Summing a periodic discrete Laplacian cancels all neighbor differences. Its source must therefore have zero mean. A pure Neumann continuum problem has a corresponding relation between integrated source and outward normal derivative; if it holds, an additive constant remains arbitrary. Fixing one value or the mean removes that nullspace but cannot make an incompatible source solvable. Silently subtracting the source mean changes the physical problem unless a neutralizing background was intentionally part of the model.
Residuals and electric fields can be insensitive to a constant potential shift, which is why the gauge should be stated in reproducibility records. Boundary discretization also affects force accuracy near electrodes even when an iterative solve is exact. The optimization lab illustrates a convergent or divergent scalar iteration with a selectable step parameter. It has no Poisson matrix, spatial mesh, electrodes, nullspace or charge source and therefore cannot verify this electrostatic boundary-value solver.
Work it through
Two interior potentials and a residual
Place nodes at x = 0,1,2,3 m with φ0 = φ3 = 0 V and ρ/ε0 = 1 V/m² at each interior node. Solve the two interior equations, compute centered fields there and compare with two Jacobi iterations started from zero.
The matrix is A = [[2,−1],[−1,2]] and b = (1,1) V. Symmetry and substitution give the exact discrete potentials φ1 = φ2 = 1 V.
Centered differentiation gives E1 = −(1 − 0)/2 = −0.5 V/m and E2 = −(0 − 1)/2 = +0.5 V/m. The signs reflect a positive potential hump.
Jacobi gives (0.5,0.5) V after one iteration and (0.75,0.75) V after two. At that point Aφtilde = (0.75,0.75) V, so the scaled residual is (0.25,0.25) V. The exact discrete solve removes this algebraic residual, not every continuum discretization error in a general problem.
The discrete interior potentials are (1,1) V, fields (−0.5,+0.5) V/m and the two-iteration scaled residual (0.25,0.25) V.
The living notebook
Explore the model
Interactive lab · optimization
Iteration stability without a field solve
Reset and set BOTH Step size η = 0.50 and Initial position x₀ = 2; predict geometric convergence to zero. Reset and set η = 2.20 and x₀ = 2; predict alternating growth and eventual Outside plot reporting. This is scalar gradient descent on x²/2, not Jacobi on a Poisson grid, a boundary-condition test or an electrostatic field solver.
- Position x0
- 2
- Objective x²/2
- 2
- Iteration behavior
- Stable step
Gradient descent on f(x) = x²/2: xₙ = (1 − η)ⁿx₀. Playback advances two discrete iterations per second. Values outside |x| ≤ 4 are explicitly reported. 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 forA stable algebraic iteration needs its own analysis, distinct from the physical boundary problem.
Pause & reason
Check your understanding
Your turn
Try it yourself
A uniform four-node periodic discrete system has scaled source b = (1,0,0,0) V. Test compatibility. If a declared uniform neutralizing background is introduced, what corrected source results? Explain why fixing one potential alone is insufficient.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Sum the four equations first. The periodic neighbor terms telescope to zero regardless of the potential values.
Compare with the solution
The source sum is 1 V, so the original periodic system is incompatible. Subtracting its mean, 0.25 V, produces bnew = (0.75,−0.25,−0.25,−0.25) V with zero sum. This new system is solvable up to an additive potential constant, which can then be fixed by a gauge condition. The subtraction represents a changed physical source and must be declared. Fixing one potential in the original incompatible problem cannot supply the missing charge balance; deleting an equation to force a result would instead hide an unsatisfied physical constraint.
Keep the idea
Take it with you
- Poisson boundary conditions determine uniqueness and compatibility.
- An algebraic residual is not a continuum discretization error.
- A gauge choice cannot neutralize an incompatible periodic source.
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