CounterexampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck pendingverified 2026-10-23 (gpt-6.1-sol)

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Restricted collision support need not force Maxwellians

Statement refuted

Let $v, w, v', w' \in R^{3}$ be velocities measured in metres per second, let $d v$ be three-dimensional Lebesgue measure, and set $Q_{e l} : = {(v, w, v', w') : v + w = v' + w', ∣ v ∣^{2} + ∣ w ∣^{2} = ∣ v' ∣^{2} + ∣ w' ∣^{2}}$ . Let $ν$ be a nonnegative countably additive measure on $Q_{e l}$ invariant under exchange of either velocity pair and interchange of incoming and outgoing pairs. For a positive finite-valued Lebesgue-measurable velocity density $g \in L^{1} (R^{3})$ whose four coordinate evaluations on $Q_{e l}$ are $ν$ -measurable, set $A_{g} : = g (v) g (w)$ and $B_{g} : = g (v') g (w')$ , and define the possibly infinite collision entropy production by $P_{ν} (g) : = \frac{1}{4} \int_{Q_{e l}} (A_{g} - B_{g}) \log (A_{g} / B_{g}) d ν$ . A Maxwellian means a function $C \exp (- ∣ v - U ∣^{2} / (2 θ))$ for a density scale $C > 0$ with the units of $g$ , a velocity $U \in R^{3}$ , and squared speed $θ > 0$ . Full support means every nonempty relatively open subset of $Q_{e l}$ has positive $ν$ -measure. The proposed implication that $P_{ν} (g) = 0$ forces $g$ to be Maxwellian for every such symmetric $ν$ and positive integrable $g$ is false without a full-support assumption on $ν$ .

Statement references

Tong, §2.3, full lecture notes

Golse, §3.2, full chapter

Facts & Assumptions

Given: The elastic collision set and weak collision identity from the collision-measure definition.

[F1]

Elastic collision measure and weak collision operator: A symmetric nonnegative measure may have restricted support; the weak identity is required only on its declared test subspace.

[A1]

Let $ν$ be the unit Dirac measure at $(0, 0, 0, 0) \in Q_{e l}$ , let $D_{ν}$ be the positive finite-valued Borel-measurable velocity densities, set $Q_{ν} (g) = 0$ for every $g \in D_{ν}$ , and for each $g \in D_{ν}$ let $S_{g}$ be the complex linear span of $C_{c}^{\infty} (R^{3}; C)$ together with $1, v_{1}, v_{2}, v_{3}, ∣ v ∣^{2}, \log (g / f_{*})$ , where $f_{*} > 0$ is the reference density in [A2].

[A2]

Let $v_{*} = 1 m s^{- 1}$ , fix a density scale $f_{*} > 0$ with the units of $f$ , and use $f_{r e f} = f_{*}$ in the logarithmic test. Set $f (v) : = f_{*} e^{- (∣ v ∣ / v_{*})^{2}} (2 + \frac{1}{2} \sin (v_{1} / v_{*})) .$

Counterexample

1.1A1F1algebra

The origin quadruple is elastic and is fixed by particle exchange and incoming/outgoing interchange, so its Dirac measure is symmetric and locally finite. Every positive finite-valued Borel-measurable density has Borel-measurable collision evaluations on $Q_{e l}$ . At the only point in the support, $v = w = v^{'} = w^{'} = 0$ , hence $A = B$ and $δ φ = 0$ for every compact smooth test and every polynomial/logarithmic test in $S_{f}$ . The collision integral is absolutely zero; the left side is zero because $Q_{ν} (f) = 0$ . Thus the declared collision model satisfies the weak identity on the test subspace required by the elastic-collision definition, and its entropy production is exactly zero.

2.1A2algebra∎

The density in [A2] is positive, smooth, and integrable with every polynomial velocity weight because its oscillatory factor is bounded above and below by positive constants and the Gaussian factor decays at infinity. Its logarithm has second derivative in $v_{1}$ equal to the constant $- 2 / v_{*}^{2}$ plus the second derivative of $\log (2 + \frac{1}{2} \sin s)$ divided by $v_{*}^{2}$ , where $s = v_{1} / v_{*}$ . The latter second derivative is $- 1 / 16$ at $s = 0$ and $- 1 / 5$ at $s = π / 2$ . Therefore the Hessian of $\log f$ is not constant. The logarithm of any Maxwellian has constant Hessian, so this $f$ is not Maxwellian despite zero production.

Depends on

Elastic collision measure and weak collision operator