Alphabeta Math
Pipeline-generated
How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Weak fields, boundary power and dissipative uniqueness — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)Open item page →

Weak Maxwell does not authorize quadratic self-energy

Physical scope

The ideal point-charge Coulomb field in the distributional pure electrostatic model. It is locally integrable but its ordinary quadratic electric-energy density is not locally integrable for nonzero charge, so the field lies outside the finite-energy boundary model. No claim is made about regularized or renormalized fields, particle mass, self-force, measurements, or every distributional Maxwell solution.

Statement refuted

Every distributional vacuum Maxwell solution with a point-charge source has a locally integrable ordinary electric-energy density uE=ε02∣E∣2 near the charge.

Facts & Assumptions

Given: A nonempty open time interval I, a bounded Lipschitz polyhedron Ω⊆R3 containing x0, a point charge q≠0 in coulombs, a position x0 in metres, and ε0>0 in F m−1. The ideal field Eq has units V m−1.

Eq(x)=q(x−x0)4πε0∣x−x0∣3(x≠x0).

[F1]

For a stationary point source, Eq is locally integrable, smooth off x0, and satisfies div⁡Eq=(q/ε0)δx0 and curl⁡Eq=0 distributionally; the value assigned at x0 is immaterial. The localized Newtonian potential and the ideal point-charge Coulomb field

[F2]

For 0<δ<R, the truncated ordinary electric-field energy is

ε02∫{δ≤∣x−x0∣≤R}∣Eq(x)∣2 dλ3(x)=q28πε0(1δ−1R),

and this tends to +∞ as δ↓0 when q≠0. Divergence of the point-charge field energy and the limits of mass inference

[F3]

The finite-energy boundary model requires E(t),H(t)∈L2(Ω;R3) and defines its energy by the corresponding finite quadratic integral. Finite-energy PEC and impedance boundary formulation

[A1]

The adopted distributional vacuum Maxwell equations apply to the point-source model with div⁡E=ρ/ε0, div⁡B=0, curl⁡E=−∂tB, and curl⁡B=μ0J+c−2∂tE; μ0 has units H m−1 and c has units m s−1. Distributional vacuum Maxwell equations for the declared point and stationary-sheet sources Framework-local SI units, the vacuum constants and the exact/measured status of their values

[F4]

A stationary point charge has ρ=qδx0 and J=0; q is measured in coulombs under the framework's point-source and SI conventions. Distributional Maxwell equations with point and stationary-sheet sources Framework-local SI units, the vacuum constants and the exact/measured status of their values

[F5]

E and B are the electric and magnetic spatial vector fields of the framework, measured in V m−1 and tesla. The electric and magnetic fields as fixed-time spatial vector fields Framework-local SI units, the vacuum constants and the exact/measured status of their values

Counterexample

Given: The hypotheses above. Since x0∈Ω and Ω is open, choose R>0 metres with B‾R(x0)⊂Ω; in the energy estimate let δ range over 0<δ<R/2 metres.

1.1A1F1F4F5givenalgebra

Set E(x,t)=Eq(x) time-independently and put B=0, J=0. The field convention is [F5], and by [F4] the source is ρ=qδx0; [F1] gives local integrability, div⁡Eq=(q/ε0)δx0, and curl⁡Eq=0. Since ∂tE=∂tB=0, the other three equations listed in [A1] hold with both sides zero. This is a distributional vacuum Maxwell solution with the stated stationary point source.

1.2F2givenalgebra

The shell in [F2] with outer radius R/2 lies inside BR(x0). Nonnegativity of ∣Eq∣2 gives ε02∫BR(x0)∣Eq∣2 dλ3 ≥ q28πε0(1δ−2R) for every 0<δ<R/2. This lower bound is unbounded as δ↓0 because q2/ε0>0; thus each ball around x0 has infinite ordinary electric energy and uE∉L1 near the charge.

2.1F3step 1.1step 1.2contradiction∎

By [F3], a field with nonintegrable ordinary quadratic energy is outside the finite-energy boundary model, even though step 1.1 supplies its distributional Maxwell identities. This contradicts the refuted claim and proves the counterexample.

5 · Examples, experiments and counterexamples

ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passverified 2026-10-22 (gpt-6.1-sol)Open item page →

Positive admittance removes energy

Physical scope

A local sign check for the impedance boundary condition in the finite-energy model, using an orthonormal tangent basis on a flat face. It records outward power and the corresponding sign of the energy balance; it is not an outgoing-wave model or an experiment.

Example

Use the boundary geometry of the preceding finite-energy model, with Γ=∂Ω. On a flat boundary face with outward normal n=ez, take a right-handed tangent basis ex,ey, with ez×ex=ey, and prescribe tangential boundary data Eτ=E0ex for E0∈R. For surface admittance Y≥0, the boundary condition Hτ=Y(n×Eτ) gives Hτ=YE0ey and outward power

n⋅(Eτ×Hτ)=YE02≥0.

If the sign in the boundary law is reversed, Hτ=−Y(n×Eτ), the outward power is −YE02; it is strictly inward when YE02>0.

For the conditional energy consequence, suppose an actual finite-energy model solution exists on a nonempty open interval I and bounded Lipschitz polyhedron Ω: E,H∈C1(I;L2(Ω;R3))∩C(I;H(curl⁡;Ω)), J∈C(I;L2(Ω;R3)), the two evolution equations of the preceding boundary model hold as L2 identities, and tangential boundary fields Eτ,Hτ∈L2(Γ×I;R3) satisfy its Green compatibility identity. Take Y∈L∞(Γ), Y≥0, as in that model. If the reversed law Hτ=−Y(n×Eτ) holds almost everywhere on all of Γ×I, then the model energy W(t)=12∫Ω(E⋅εE+H⋅μH) dλ3 obeys

W(t)−W(s)=−∫st∫ΩE⋅J dλ3 dr+∫st∫ΓY∣Eτ∣2 dS dr(s<t).

In particular, if J=0 and the last integral is positive, the stored energy increases. This conclusion is conditional on existence of a solution satisfying the reversed law.

Source convention

Chen's boundary relation is n×H−λEt=g. Here Et denotes the tangential electric component, identified with Eτ, and g is prescribed tangential boundary data. For comparison with the homogeneous law of this example, set g=0 and use Hτ=Y(n×Eτ). Since ∣n∣=1 and n⋅Eτ=0, the right-handed cross product gives n×Hτ=Yn×(n×Eτ)=−YEτ. Thus our law corresponds to Chen's homogeneous relation with λ=−Y. This is a conversion of conventions; the cited passage does not stipulate a sign for λ or supply a passivity claim.

Diagram

The drawing is a projection of the tangent plane: ez points out of the page, and the right-handed convention is ex×ey=ez. The arrows depict the representative case E0>0 and Y>0. If E0<0 while Y>0, both tangential arrows reverse; if E0=0, both vectors vanish; and if Y=0, Hτ vanishes. The verification below supplies the sign calculation.

\begin{tikzpicture}[>=stealth,scale=1.1]
  \draw[->] (-0.2,0) -- (2.8,0) node[below] {$e_x$};
  \draw[->] (0,-0.2) -- (0,2.5) node[left] {$e_y$};
  \draw[->,thick,blue] (0,0) -- (1.7,0) node[below=4pt] {$E_\tau=E_0e_x$};
  \draw[->,thick,red] (0,0) -- (0,1.7) node[left=4pt] {$H_\tau=YE_0e_y$};
  \node at (2.2,1.4) {$n=e_z\ \odot$};
\end{tikzpicture}

Facts & Assumptions

Given: The impedance boundary law, n=ez, Eτ=E0ex, an amplitude E0 in V m−1, and an admittance Y≥0 in siemens. The outward power density is measured in W m−2.

[A1]

For the conditional energy conclusion only, the finite-energy model solution class has E,H∈C1(I;L2(Ω;R3))∩C(I;H(curl⁡;Ω)), J∈C(I;L2(Ω;R3)), the displayed evolution equations as L2 identities, and a tangential L2(Γ×I) pair satisfying the Green compatibility identity in the model definition. The reversed boundary law is an additional assumption on all of Γ×I; it is not part of the model's original nonnegative-admittance branch. Finite-energy PEC and impedance boundary formulation

[F1]

In the impedance branch, Hτ=Y(n×Eτ), and the outward boundary power is the scalar n⋅(Eτ×Hτ). Finite-energy PEC and impedance boundary formulation

[F2]

For an admissible impedance solution, the energy identity includes the boundary term −∫∂ΩY∣Eτ∣2 dS in the change of field energy. Energy balance and zero-data forward uniqueness for admissible weak Maxwell fields

[F3]

The cross product is the bilinear right-handed coordinate product of The cross product in R3. For a unit normal n and tangential vector e with e⋅n=0, substituting that coordinate formula gives n⋅(e×(n×e))=∣n∣2∣e∣2−(e⋅n)2=∣e∣2; in the displayed tangent basis, ez×ex=ey and ex×ey=ez. The dot product and norm are those of The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn.

Verification

Given: The values and conventions of the Example.

1.1givenF1F3algebra

Apply the impedance law [F1]: Hτ=Y(ez×E0ex)=YE0ey by ez×ex=ey from [F3].

2.1step 1.1F3givenalgebra

Using bilinearity and ex×ey=ez from [F3], n⋅(Eτ×Hτ)=ez⋅(E0ex×YE0ey)=YE02, which is nonnegative because Y≥0 and E02≥0. Replacing the boundary law by its negative changes Hτ to −YE0ey, so the same calculation gives −YE02, strictly negative when YE02>0.

3.1A1F2F3step 2.1algebra∎

For the conditional solution in [A1], differentiate the bounded symmetric quadratic form W in the real L2 norm and substitute the evolution equations. For almost every t, this gives W′(t)=(curl⁡H,E)L2−(J,E)L2−(H,curl⁡E)L2. The Green compatibility identity then yields W′(t)=−(J,E)L2−∫Γn⋅(Eτ×Hτ) dS. Under the reversed law, [F3] gives n⋅(Eτ×Hτ)=−Y∣Eτ∣2, so integration in time gives the stated positive boundary term. For the original law the same calculation agrees with [F2]. The boundary sign conclusion uses the Green identity and assumes an admissible solution; the example itself asserts no solution existence.

Sources