Physics theoremStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck pendingverified 2026-10-23 (gpt-6.1-sol)

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Plane and pipe Couette–Poiseuille verification — conditional physical prediction

Physical scope

Infinite/fully developed channel or pipe ansatz with the stated regularity; exact stationary Navier–Stokes and Stokes solutions for every parameter choice in the Statement, without a small-inertia approximation; uniqueness only within the stated ansätze. Worked example is the actual explicit field/parameter selection in that section. B leaf cannot supply any later item.

Statement

Let $h, a, ρ_{0}, μ > 0$ , $p_{0}, G, V_{0}, V_{1} \in R$ , and $ν = μ / ρ_{0}$ . Use SI units: lengths in metres, $p_{0}$ in pascals, velocities in $m s^{- 1}$ , density in $k g m^{- 3}$ , viscosity in $P a s$ , pressure in pascals, and $G$ in pascals per metre. Let $e_{x}, e_{y}, e_{z}$ be the standard Cartesian unit vectors. In Cartesian coordinates, $div u = \partial_{x} u_{x} + \partial_{y} u_{y} + \partial_{z} u_{z}$ and $(Δ u)_{i} = \partial_{x x} u_{i} + \partial_{y y} u_{i} + \partial_{z z} u_{i}$ . The stress is $σ = - p I + 2 μ D$ , with $D = (\nabla u + \nabla u^{T}) / 2$ and $(\nabla u)_{i j} = \partial_{x_{j}} u_{i}$ . The selected constant-density incompressible Newtonian model has $\partial_{t} u + (u \cdot \nabla) u + ρ_{0}^{- 1} \nabla p - ν Δ u = 0, div u = 0.$

For the infinite plane channel $Ω_{c h} = {(x, y, z) : x, z \in R, 0 < y < h}$ , set $u = (v (y), 0, 0)$ , $p = p_{0} - G x$ , with $v \in C^{2} ([0, h])$ and wall values $v (0) = V_{0}$ , $v (h) = V_{1}$ . The unique profile within this one-dimensional ansatz is $v (y) = V_{0} + (V_{1} - V_{0}) \frac{y}{h} + \frac{G}{2 μ} y (h - y) .$ Its volume flux per unit width in $z$ , measured in $m^{2} s^{- 1}$ , and channel shear values are $Q_{c h} = \int_{0}^{h} v (y) d y = \frac{h}{2} (V_{0} + V_{1}) + \frac{G h^{3}}{12 μ}, σ_{x y} (0) = μ \frac{V_{1} - V_{0}}{h} + \frac{G h}{2}, σ_{x y} (h) = μ \frac{V_{1} - V_{0}}{h} - \frac{G h}{2} .$

For the infinite circular pipe $Ω_{a} = {(x, y, z) : x \in R, y^{2} + z^{2} < a^{2}}$ , take $x$ as axial coordinate, $r = (y^{2} + z^{2})^{1 / 2}$ , $u = (w (r), 0, 0)$ , and $p = p_{0} - G x$ . Require $w (a) = 0$ and require the radial field $(y, z) \mapsto w ((y^{2} + z^{2})^{1 / 2})$ to be $C^{2}$ on the closed disk. The unique profile within this ansatz is $w (r) = \frac{G}{4 μ} (a^{2} - r^{2}), Q_{p i p e} = \int_{0}^{a} 2 π r w (r) d r = \frac{π G a^{4}}{8 μ} .$ The pipe volume flux is in $m^{3} s^{- 1}$ . For $r > 0$ let $e_{r} = (0, y / r, z / r)$ and define the radial-axial stress component by $σ_{r x} = e_{r} \cdot σ e_{x}$ ; it is $σ_{r x} (r) = μ w^{'} (r) = - G r / 2$ .

Both fields satisfy the exact stationary Navier–Stokes equations for every allowed parameter choice and also the corresponding steady Stokes equations because their time derivatives and convective accelerations vanish. Uniqueness is only within the stated ansätze; pipe uniqueness uses full-disk regularity at the axis. These infinite or fully developed geometries give no finite entrance or exit error estimate, stability or transition claim, observation, or empirical-validity claim.

Statement references

Stewart, MIT 8.09, Fluid Mechanics, §6.7, printed pp. 130–133

Fitzpatrick, Fluid Mechanics, UT Austin, flow between parallel plates

Fitzpatrick, Fluid Mechanics, UT Austin, pipe Poiseuille flow

Facts & Assumptions

Given: Positive $h, a, ρ_{0}, μ$ ; real $p_{0}, G, V_{0}, V_{1}$ ; the stated infinite geometries, one-dimensional ansätze, pressure fields and boundary data.

[A1]

Single-component nonpolar continuum and material balances: the adopted continuum framework includes the mechanical momentum balance; it is a physical model input, not empirical proof.

[A2]

Specified wall, thermal, periodic and initial data: no-slip and pressure data are selected separately and imply no solution-existence result.

[A3]

Newtonian stress and Fourier conductive closure: the selected mechanical closure is $σ = - p I + 2 μ D$ in this incompressible constant-viscosity subcase; heat flux is unused.

[F1]

Constant-density incompressible Newtonian model: the local branch has $div u = 0$ and constant-density Newtonian momentum equation, equivalent to $\partial_{t} u + (u \cdot \nabla) u + ρ_{0}^{- 1} \nabla p - ν Δ u = 0$ .

[F2]

Plane and pipe Couette–Poiseuille verification: the channel and fixed-wall pipe profiles are unique within their stated one-dimensional ansätze; pipe uniqueness assumes full-disk regularity, and both fields solve NS and Stokes because acceleration vanishes.

Proof

1.1givenA1A2A3F1algebra

In both fields velocity has only an axial component independent of the axial coordinate; hence $div u = 0$ and $(u \cdot \nabla) u = 0$ . The plane momentum equation reduces to $μ v^{''} = - G$ , and the pipe equation to $μ (w^{''} + r^{- 1} w^{'}) = - G$ for $r > 0$ .

2.1givenA2F2step 1.1algebra

Integrating the plane equation twice gives $v (y) = - G y^{2} / (2 μ) + C_{1} y + C_{0}$ . The wall values determine $C_{0} = V_{0}$ and $C_{1} = (V_{1} - V_{0}) / h + G h / (2 μ)$ , yielding the stated profile uniquely. Its integral gives $Q_{c h}$ ; differentiation and $σ_{x y} = μ v^{'}$ give both wall shear values.

2.2givenA2A3F2step 1.1algebra

Multiplication of the pipe equation by $r$ gives $(r w^{'})^{'} = - G r / μ$ . Integration yields $w = - G r^{2} / (4 μ) + C \log r + D$ on $r > 0$ . Full-disk $C^{2}$ regularity forces $C = 0$ , and $w (a) = 0$ fixes $D = G a^{2} / (4 μ)$ , proving uniqueness within this ansatz. Integration with area element $2 π r d r$ gives $Q_{p i p e}$ ; differentiation gives $σ_{r x} = - G r / 2$ .

3.1A1F1F2step 2.1step 2.2∎

Time derivatives and convective accelerations vanish, so each field solves the stationary Navier–Stokes and corresponding steady Stokes equations for every allowed parameter choice, without a limiting or small-parameter hypothesis. This says nothing about finite-device behavior or empirical validity.

Plane and pipe Couette–Poiseuille verification — conditional physical prediction

Statement

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources