ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck pendingverified 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.

Adopted Poiseuille and Couette channel predictions

Physical scope

Exact steady profiles in the stated periodic incompressible Newtonian channel model with no-slip walls; the Poiseuille pressure-drop notation is replaced by a constant body acceleration on the periodic cell. No measured flow or claim about a named material is included.

Example

Assume Countable Choice, meaning every countable family of nonempty sets has a choice function.

Let $L_{x}, L_{z}, h > 0$ be lengths in metres and let $Ω = (R / L_{x} Z) \times (0, h) \times (R / L_{z} Z)$ have periodic streamwise coordinate $x$ and spanwise coordinate $z$ , with stationary walls $y = 0, h$ ; all three coordinates are measured in metres. Let $ρ > 0$ be constant density in kilograms per cubic metre, $η > 0$ constant dynamic viscosity in pascal-seconds, and $p_{0}$ a constant pressure in pascals. Use the incompressible Newtonian model $ρ (\partial_{t} u + (u \cdot \nabla) u) = div σ + ρ b$ , $div u = 0$ , $σ = - p I + 2 η D (u)$ , where $D (u)_{i j} = (\partial_{j} u_{i} + \partial_{i} u_{j}) / 2$ and divergence is rowwise. For matrices $A, B \in R^{3 \times 3}$ define the Frobenius contraction $A : B : = \sum_{i, j = 1}^{3} A_{i j} B_{i j}$ and the squared norm $∣ D (u) ∣^{2} : = D (u) : D (u) = \sum_{i, j = 1}^{3} D (u)_{i j}^{2}$ . In the verification, $D$ abbreviates $D (u)$ . In the Poiseuille case let $G > 0$ be a pressure-gradient scale in pascals per metre, set the body acceleration $b = (G / ρ, 0, 0)$ in metres per second squared and $p = p_{0}$ , and define

$u_{P} (x, y, z) = (\frac{G y (h - y)}{2 η}, 0, 0) .$

In the Couette case let $V \in R$ be a tangential wall speed in metres per second, set $b = 0$ , $p = p_{0}$ , and define

$u_{C} (x, y, z) = (\frac{V y}{h}, 0, 0),$

with no-slip wall velocity $g = 0$ at $y = 0$ and $g = (V, 0, 0)$ at $y = h$ . Both fields are exact steady smooth solutions with impermeable no-slip walls. Their volumetric fluxes per unit span are $Q_{P} = G h^{3} / (12 η)$ and $Q_{C} = V h / 2$ , each in square metres per second, and their vorticities $ω : = \nabla \times u$ in inverse seconds are $ω_{P} = (0, 0, - G (h - 2 y) / (2 η))$ and $ω_{C} = (0, 0, - V / h)$ . In the Poiseuille case the body-force power per unit wall area equals viscous dissipation and stationary walls do no work. In the Couette case the moving top wall supplies power equal to viscous dissipation. A periodic pressure cannot contain the nonperiodic term $- G x$ ; the stated constant pressure and body acceleration are the periodic-cell formulation. These are conditional predictions of the adopted equations, not observations or claims about all fluids.

Example references

David Tong, Fluid Mechanics, Chapter 2, §§3.2.2–3.2.3

Joseph M. Powers, Lecture Notes on Intermediate Fluid Mechanics, §5.4.1

Facts & Assumptions

Given: Countable Choice, the periodic channel, constants, body accelerations, wall velocities, fields and SI units in the Example.

[A1]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): every countable family of nonempty sets has a choice function; the depended-on mechanical-power theorem states this hypothesis explicitly.

[F1]

Mechanical power balance under adopted wall laws: under its smooth constant-density incompressible equations, domain-preserving material motion, and adopted material momentum balance, the no-slip branches have the mechanical-power identities proved in steps 1.1–5.1 of that item. Its separate uniqueness clause is proved there but is not used in this example; no existence claim is imported.

Verification

Given: The fields and model assumptions in the Example.

1.1givenF1constructalgebra

Let $\tilde{Ω} = R \times (0, h) \times R$ and $Λ = {(m L_{x}, 0, n L_{z}) : m, n \in Z}$ . For $U = U_{P}$ in the Poiseuille case and $U = U_{C}$ in the Couette case, define the lifted material map for each $t$ by ${\tilde{χ}}_{t}^{U} (x, y, z) = (x + t U (y), y, z)$ . Both profiles are smooth, so this map is $C^{\infty}$ ; its inverse is $(x - t U (y), y, z)$ . It maps $\tilde{Ω}$ onto itself, fixes the $y$ coordinate, and obeys ${\tilde{χ}}_{t}^{U} (a + λ) = {\tilde{χ}}_{t}^{U} (a) + λ$ for every $λ \in Λ$ . Its Jacobian matrix is $(\begin{array}{ccc} 1 & t U^{'} (y) & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array})$ , with determinant $1 > 0$ . Since $y$ is unchanged, $\partial_{t} {\tilde{χ}}_{t}^{U} = (U (y), 0, 0) = \tilde{u} (t, {\tilde{χ}}_{t}^{U} (x, y, z))$ , so each map descends to the periodic channel and has the example's velocity as its Eulerian velocity. Thus both profiles meet the material-motion hypothesis of [F1].

1.2givenF1algebra

For either profile $u = (U (y), 0, 0)$ , $div u = 0$ , $\partial_{t} u = 0$ and $(u \cdot \nabla) u = U (y) \partial_{x} u = 0$ . The only nonzero entries of $D$ are $D_{x y} = D_{y x} = U^{'} / 2$ , so the only nonzero viscous shear entries are $σ_{x y} = σ_{y x} = η U^{'}$ , and $(div σ)_{x} = - \partial_{x} p + η U^{''}$ .

2.1givenF1step 1.2algebra

In the Poiseuille case, $U_{P} (0) = U_{P} (h) = 0$ and $U_{P}^{''} = - G / η$ . Thus $(div σ)_{x} + ρ b_{x} = η U_{P}^{''} + G = 0$ , and the other momentum components vanish, proving the steady equations and no-slip compatibility. Direct integration gives $Q_{P} = \int_{0}^{h} U_{P} (y) d y = G h^{3} / (12 η)$ and $ω_{P, z} = - U_{P}^{'} = - G (h - 2 y) / (2 η)$ .

2.2givenF1step 1.2algebra

In the Couette case, $U_{C} (0) = 0$ , $U_{C} (h) = V$ , and $U_{C}^{''} = 0$ , so the momentum equation holds with constant pressure and no body force; the velocity traces equal the stated wall velocities. Direct integration gives $Q_{C} = \int_{0}^{h} U_{C} (y) d y = V h / 2$ and $ω_{C, z} = - U_{C}^{'} = - V / h$ .

3.1givenA1F1step 2.1algebra

By the defined squared norm $∣ D ∣^{2} = D : D$ , for either profile $∣ D ∣^{2} = (U^{'})^{2} / 2$ , hence viscous dissipation per unit streamwise/spanwise wall area is $η \int_{0}^{h} (U^{'})^{2} d y$ . In the Poiseuille case this is $\frac{G^{2}}{4 η} \int_{0}^{h} (h - 2 y)^{2} d y = G^{2} h^{3} / (12 η)$ , while body-force power is $\int_{0}^{h} ρ (G / ρ) U_{P} d y = G Q_{P} = G^{2} h^{3} / (12 η)$ ; stationary-wall power is zero. This agrees with the stationary no-slip case of [F1].

4.1givenA1F1step 2.2algebra∎

In the Couette case viscous dissipation per unit wall area is $η \int_{0}^{h} (V / h)^{2} d y = η V^{2} / h$ . At the top wall $n = (0, 1, 0)$ and $σ n$ has tangential component $η (V / h) (1, 0, 0)$ , so wall power per unit area is $g \cdot σ n = η V^{2} / h$ ; the lower wall has $g = 0$ . This agrees with the tangential no-slip case of [F1]. Finally, a purported periodic pressure $p_{0} - G x$ changes by $- G L_{x}$ under $x \mapsto x + L_{x}$ , so it is not single-valued on $Ω$ when $G > 0$ ; the body-force formulation in the Example avoids that incompatibility.

Adopted Poiseuille and Couette channel predictions

Example

Example references

Facts & Assumptions

Verification

Depends on

Dependency tree · 2 levels

Sources