TheoremStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck passverified 2026-10-23 (gpt-6.1-sol)

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.

Exact finite Reynolds averaged equation

Statement

Let $I \subseteq R$ be an open interval, let $d, N \geq 1$ be integers, let $Ω \subseteq R^{d}$ be open, and fix constants $a_{j} \geq 0$ with $\sum_{j = 1}^{N} a_{j} = 1$ and a common $ν \geq 0$ . For $j = 1, \dots, N$ , let $u_{j} : I \times Ω \to R^{d}$ be continuously differentiable in $(t, x)$ with continuous second spatial derivatives, let $p_{j} : I \times Ω \to R$ be continuously differentiable, and let $f_{j} : I \times Ω \to R^{d}$ be continuous. Suppose pointwise that

$div u_{j} = 0, \partial_{t} u_{j} + div (u_{j} \otimes u_{j}) + \nabla p_{j} = ν Δ u_{j} + f_{j} .$

Here $(u \otimes v)_{r s} = u_{r} v_{s}$ , $div$ of a matrix field is rowwise, $(div A)_{r} = \sum_{s = 1}^{d} \partial_{x_{s}} A_{r s}$ , and $Δ$ acts componentwise. Define the weighted means $\overset{ˉ}{u} = \sum_{j} a_{j} u_{j}$ , $\overset{ˉ}{p} = \sum_{j} a_{j} p_{j}$ , and $\overset{ˉ}{f} = \sum_{j} a_{j} f_{j}$ , the fluctuations $v_{j} = u_{j} - \overset{ˉ}{u}$ , and the covariance tensor $R = \sum_{j} a_{j} v_{j} \otimes v_{j}$ . Then, pointwise on $I \times Ω$ ,

$div \overset{ˉ}{u} = 0, \sum_{j = 1}^{N} a_{j} v_{j} = 0, \sum_{j = 1}^{N} a_{j} u_{j} \otimes u_{j} = \overset{ˉ}{u} \otimes \overset{ˉ}{u} + R,$

and the exact averaged equation is

$\partial_{t} \overset{ˉ}{u} + div (\overset{ˉ}{u} \otimes \overset{ˉ}{u}) + \nabla \overset{ˉ}{p} = ν Δ \overset{ˉ}{u} + \overset{ˉ}{f} - div R .$

The tensor $R$ is symmetric positive semidefinite, with $z^{T} R z = \sum_{j} a_{j} (v_{j} \cdot z)^{2} \geq 0$ for every $z \in R^{d}$ . With $tr A : = \sum_{r = 1}^{d} A_{r r}$ for $A \in R^{d \times d}$ , the pointwise energy split is $\sum_{j} a_{j} ∣ u_{j} ∣^{2} = ∣ \overset{ˉ}{u} ∣^{2} + tr R$ . These are identities for the specified smooth functions; they assert neither existence of such solutions nor a statistical or physical interpretation.

Statement references

Flierl and Ferrari, MIT OCW 12.820, Chapter 3, §§3.2 and 3.6–3.7 (covariance, closure, and Reynolds equations), full lecture notes

Eyink, Johns Hopkins University Turbulence Notes, Chapter VIb, §(B) Reynolds Averaging and kinetic-energy balances, pp. 17–19

Facts & Assumptions

[F1]

Finite weighted ensemble and covariance: For fixed weights $a_{j} \geq 0$ summing to one, the weighted mean is $\overset{ˉ}{X} = \sum_{j} a_{j} X_{j}$ and the fluctuation is $X_{j}^{'} = X_{j} - \overset{ˉ}{X}$ .

[F2]

Finite weighted ensemble and covariance: The cross-covariance is $C (X, Y) = \sum_{j} a_{j} X_{j}^{'} \otimes Y_{j}^{'}$ ; in particular the covariance tensor is $C (X) = C (X, X)$ .

[A1]

The fields, fixed weights, common viscosity, divergence constraints, and pointwise equations satisfy every hypothesis in the Statement.

Given: The data in the Statement, including [A1] and the finite weighted mean and covariance convention [F1].

Proof

1.1givenF1A1algebra

Because the weights are fixed constants and all sums are finite, each displayed derivative of a weighted mean is the same weighted sum of the corresponding derivatives. Thus $div \overset{ˉ}{u} = \sum_{j} a_{j} div u_{j} = 0$ , while $\sum_{j} a_{j} v_{j} = \sum_{j} a_{j} u_{j} - \overset{ˉ}{u} \sum_{j} a_{j} = \overset{ˉ}{u} - \overset{ˉ}{u} = 0$ .

2.1F1F2step 1.1algebra

For each $j$ , write $u_{j} = \overset{ˉ}{u} + v_{j}$ and expand the outer product. Summing with weights gives $\sum_{j} a_{j} u_{j} \otimes u_{j} = \overset{ˉ}{u} \otimes \overset{ˉ}{u} + \overset{ˉ}{u} \otimes \sum_{j} a_{j} v_{j} + (\sum_{j} a_{j} v_{j}) \otimes \overset{ˉ}{u} + \sum_{j} a_{j} v_{j} \otimes v_{j} = \overset{ˉ}{u} \otimes \overset{ˉ}{u} + R$ , using step 1.1 and the covariance definition [F2].

3.1givenA1F1F2step 1.1step 2.1algebra

Multiply the $j$ th equation in [A1] by $a_{j}$ and sum over $j$ . Fixed weights commute with all indicated linear differential operators, so the time, pressure, viscosity, and forcing terms become their barred means; substituting step 2.1 into the summed nonlinear term yields $\partial_{t} \overset{ˉ}{u} + div (\overset{ˉ}{u} \otimes \overset{ˉ}{u}) + \nabla \overset{ˉ}{p} = ν Δ \overset{ˉ}{u} + \overset{ˉ}{f} - div R$ . The divergence constraint is step 1.1.

4.1F2step 1.1algebra∎

By [F2], $R = \sum_{j} a_{j} v_{j} \otimes v_{j}$ , so it is symmetric and $z^{T} R z = \sum_{j} a_{j} (v_{j} \cdot z)^{2} \geq 0$ . Also $tr R = \sum_{j} a_{j} ∣ v_{j} ∣^{2}$ , so expanding $\sum_{j} a_{j} ∣ u_{j} ∣^{2} = \sum_{j} a_{j} ∣ \overset{ˉ}{u} + v_{j} ∣^{2}$ and using step 1.1 gives $\sum_{j} a_{j} ∣ u_{j} ∣^{2} = ∣ \overset{ˉ}{u} ∣^{2} + tr R$ .

Depends on

Finite weighted ensemble and covariance