ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck passverified 2026-10-23 (gpt-6.1-sol)

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Vanishing scalar density leaves velocity undetermined by momentum

Example

At a fixed point, let $ρ \in [0, \infty)$ be mass density and let $u \in R^{3}$ be velocity. In SI units, $ρ$ is measured in $k g m^{- 3}$ , $u$ in $m s^{- 1}$ , and the momentum-density vector $M = ρ u$ in $k g m^{- 2} s^{- 1}$ . For the field interpretation, use the restricted Lebesgue spaces, measurable-map and almost-everywhere conventions of the conservative weak-form definition under Countable Choice; the pointwise map $u \mapsto ρ u$ below is purely algebraic. To use momentum density as a field in a conservative weak form, let $Ω \subseteq R^{3}$ be open and $T > 0$ , take measurable maps $ρ : Ω \times [0, T) \to [0, \infty)$ and $u : Ω \times [0, T) \to R^{3}$ , and require their product $U (x, t) : = ρ (x, t) u (x, t)$ to be locally integrable, including near $t = 0$ . The vector $M$ above is a pointwise value of such a product, not itself a field. At $ρ = 0$ , distinct vectors $u_{1} \neq u_{2}$ both give $M = 0$ , so the momentum density does not determine velocity. At $ρ > 0$ , the relation has the unique inverse $u = M / ρ$ . This is a pointwise algebraic example and makes no claim that any chosen fields solve a differential equation.

Example references

Locally integrable conservative weak and initial forms

Facts & Assumptions

Given: A fixed point, a scalar $ρ \geq 0$ , and a vector $u \in R^{3}$ . The example tests whether the algebraic map $u \mapsto M = ρ u$ is injective.

[F1]

Locally integrable conservative weak and initial forms: a conserved field $U$ is a locally integrable vector-valued field whose conservative equation is formulated against compactly supported test functions; here $M$ is a pointwise vector, while $U (x, t) = ρ (x, t) u (x, t)$ is an admissible field only under the stated measurability and local-integrability requirements (with three components).

Verification

1.1givenF1algebra

Choose $u_{1} = 0$ and $u_{2} = e_{1}$ , where $e_{1} = (1, 0, 0)$ . They are distinct, but at $ρ = 0$ their momentum densities satisfy $ρ u_{1} = 0 = ρ u_{2}$ . Thus the map $u \mapsto ρ u$ is not injective when $ρ = 0$ .

2.1step 1.1givenalgebra∎

If $ρ > 0$ and $M = ρ u$ , division by the positive scalar gives the unique recovery $u = M / ρ$ . Hence the failure in step 1.1 is exactly the zero-density degeneracy, not a claim about positive density.

Depends on

Locally integrable conservative weak and initial forms