Build a Green function with a declared convention
Track normalization, boundary terms, and the Neumann zero mode.
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The question for today
What you’ll understand
- Declare a consistent Green-function sign and normalization.
- Derive the Dirichlet representation including nonzero boundary values.
- Recognize why a closed-domain Neumann kernel needs special treatment.
A Green function is the response to a normalized point source with specified boundary behavior. Once that response is known, superposition handles an arbitrary source distribution. The phrase “the Green function” is incomplete until the operator, delta normalization, and boundary conditions are stated. Different textbooks choose opposite signs or factors of four pi. Both conventions work; mixing pieces from them produces an answer with the wrong charge or the wrong boundary voltage.
Normalize the elementary response
Here choose the positive operator minus the Laplacian, and define its Dirichlet Green function by a unit delta source. In free three-dimensional space the corresponding kernel is positive one over four pi times separation. Applying the operator produces a unit delta, not four pi times a delta. The electrostatic potential then includes an additional factor of inverse permittivity. The kernel has inverse-length units, while its convolution with charge density has potential units.
A finite-domain kernel adds a harmonic correction that sets the kernel to zero on the boundary. For a grounded plane, subtracting the reflected free-space kernel does exactly this. That correction is not an optional refinement: it encodes the electrodes. Some references instead define the kernel for the positive Laplacian and use a negative free-space kernel. Their representation formula changes signs consistently. Always compare the defining differential equation before borrowing a formula.
The boundary term carries imposed voltage
Apply Green’s second identity to the potential and the kernel in the source coordinate. The delta selects the potential at the observation point, while Poisson’s equation supplies the volume charge integral. Because the Dirichlet kernel vanishes on the boundary, the term involving the unknown normal derivative of the potential disappears. The remaining boundary integral involves the known boundary potential and the normal derivative of the kernel, with a minus sign in our convention.
If every boundary is grounded, that remaining term vanishes too. If there is no volume charge but the boundary is biased, the boundary term alone creates the solution. Omitting it would falsely predict zero potential everywhere. Symmetry of the Dirichlet kernel under exchanging its arguments follows from the same identity and homogeneous boundary conditions. This reciprocity statement relies on the self-adjoint operator and appropriate domain, rather than on any generic pair of response functions.
An inverse fails on constants
A closed-domain Neumann Laplacian annihilates constants. Consequently it cannot possess an ordinary inverse on all sources. One consistent generalized kernel satisfies minus its Laplacian equal to delta minus reciprocal domain volume, has zero normal derivative, and has zero volume mean. Subtracting the uniform term removes the incompatible constant source. Another convention keeps a pure delta and uses a nonzero constant boundary derivative. These conventions must not be combined with each other’s formulas.
Use the field lab only to test linear scaling of a source response. Reset and set q=0.5 and rotation to 0, then Reset and set q=1 and rotation to 0. Predict that the numerical Fx readout at (1,0) increases from 0.5/1.2≈0.417 to 1/1.2≈0.833, while Fy stays zero. The mathematical field doubles; visible arrow lengths use a compressed scale. This supports the superposition step behind a kernel integral. The softened planar field has neither a delta normalization nor conducting boundaries, so it cannot validate a Green function’s normalization, boundary term, or Neumann mean constraint.
Work it through
Check a one-dimensional kernel
For −d²/dx² on 0<x<L with zero endpoint values, use G(x,s)=min(x,s)[L−max(x,s)]/L. Let L=2 m and solve −V″=2 V/m² at x=1 m.
At the midpoint observation x=1 m, the kernel is s/2 for 0≤s≤1 m and (2 m−s)/2 for 1≤s≤2 m, with consistent metre units. Both branches vanish at the appropriate source endpoint.
The derivative with respect to s jumps from +1/2 to −1/2 at s=1 m. Its jump is −1, so minus the second derivative gives a unit delta. This explicitly verifies the sign and normalization rather than inferring them from a sketch.
Integrate the two triangles: their total kernel area is 0.5 m². Multiplication by the source 2 V/m² gives 1 V. Direct integration yields V=x(2 m−x)(1 V/m²), confirming the midpoint and both grounded endpoints.
The midpoint potential is 1 V. The one-dimensional Green kernel has length units, unlike the inverse-length three-dimensional kernel.
The living notebook
Explore the model
Interactive lab · fields
Check linear source response
Reset and set q=0.5, Ω=0; predict Fx at (1,0)≈0.417 and Fy=0. Reset and set q=1, Ω=0; predict Fx≈0.833 and Fy=0. The field doubles, while visible arrow lengths use a compressed scale.
- 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 is a softened planar response and illustrates linearity only. It cannot check delta normalization or impose a Dirichlet or Neumann boundary.
Pause & reason
Check your understanding
Your turn
Try it yourself
On 0<x<2 m with grounded ends, a source −V″=4 V/m² occupies only 0<x<1 m. Use the midpoint kernel above to find V(1 m).
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Only the left kernel branch contributes; do not integrate the source over the empty half.
Compare with the solution
At observation x=1 m, integrate (s/2)(4 V/m²) from s=0 to 1 m. The result is 1 V. A uniform source of 4 V/m² over the entire interval would give 2 V, so the half-filled source gives half that midpoint value by the left-right symmetry of the kernel, despite its nonuniform spatial distribution.
Keep the idea
Take it with you
- A Green function includes an operator, normalization, and boundary specification.
- Nonzero Dirichlet voltages contribute through a boundary term.
- The Neumann constant mode prevents an unrestricted ordinary inverse.
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