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Tensor test algebra and positive tempered moment data — Examples

1 · Prerequisites

None. This page is self-contained.

2 · Summary

The counterexample checks the radial power printed for the Yukawa expression in the cited source. Its $r^{- 2} e^{- μ r}$ term has nonzero screened Laplacian at every $r > 0$ , while $r^{- 1} e^{- μ r}$ solves the homogeneous equation off the origin. No delta normalization or full Green-function claim is made.

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

None yet.

5 · Examples, experiments and counterexamples

CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedverified 2026-10-15 (gpt-6-luna, gpt-6.1-sol, and delegated Codex review/repair agents; see completed run evidence)Open item page →

Radial Yukawa source power fails the homogeneous equation

Statement refuted

For every $μ \geq 0$ , the function $u_{μ} (x) : = \frac{e^{- μ ∣ x ∣}}{∣ x ∣^{2}} (x \in R^{3} ∖ {0})$ solves the homogeneous screened equation $(Δ - μ^{2}) u_{μ} = 0$ at every point of $R^{3} ∖ {0}$ , where $Δ : = \partial_{1}^{2} + \partial_{2}^{2} + \partial_{3}^{2}$ is the Euclidean Laplacian. The cited source uses the same off-origin Yukawa radial expression up to a constant factor; no distributional statement at the origin is included here.

Facts & Assumptions

Given: $μ \geq 0$ , $x \in R^{3} ∖ {0}$ , $r : = ∣ x ∣ > 0$ , and the Euclidean Laplacian specified in the refuted statement.

[A1]

The refuted claim is the pointwise identity $(Δ - μ^{2}) u_{μ} (x) = 0$ for every $x \neq 0$ .

Counterexample

1.1givenalgebra

For any twice differentiable radial function $w (x) = q (r)$ on $r > 0$ , $\partial_{j} w = q^{'} (r) x_{j} / r$ and $\sum_{j = 1}^{3} \partial_{j}^{2} w = q^{''} (r) \sum_{j = 1}^{3} \frac{x_{j}^{2}}{r^{2}} + q^{'} (r) \sum_{j = 1}^{3} (\frac{1}{r} - \frac{x_{j}^{2}}{r^{3}}) = q^{''} (r) + \frac{2}{r} q^{'} (r) .$ Thus the radial Laplacian formula follows directly from the chain rule on $R^{3} ∖ {0}$ .

2.1A1step 1.1algebra

For $q (r) = e^{- μ r} r^{- 2}$ , differentiation gives $q^{'} (r) = e^{- μ r} (- \frac{μ}{r^{2}} - \frac{2}{r^{3}}), q^{''} (r) = e^{- μ r} (\frac{μ^{2}}{r^{2}} + \frac{4 μ}{r^{3}} + \frac{6}{r^{4}}) .$ Substituting into step 1.1 yields $(Δ - μ^{2}) u_{μ} (x) = 2 e^{- μ r} (\frac{μ}{r^{3}} + \frac{1}{r^{4}}) > 0 (r > 0),$ including $μ = 0$ . This contradicts [A1] at every point off the origin.

3.1step 1.1algebra∎

For comparison, set $v_{μ} (x) = e^{- μ r} / r$ for $r > 0$ . Direct differentiation gives $v_{μ}^{'} (r) = e^{- μ r} (- \frac{μ}{r} - \frac{1}{r^{2}}), v_{μ}^{''} (r) = e^{- μ r} (\frac{μ^{2}}{r} + \frac{2 μ}{r^{2}} + \frac{2}{r^{3}}) .$ Using step 1.1, the terms $2 μ / r^{2} + 2 / r^{3}$ cancel against $2 v_{μ}^{'} / r$ , leaving $Δ v_{μ} = μ^{2} v_{μ}$ on $r > 0$ . Thus the $r^{- 1}$ power passes the off-origin homogeneous check, but this does not prove its full distributional Green-function normalization.