ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck pass

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.

Poiseuille and Couette channel flows of the viscous branch: explicit profiles, flux and wall data

Physical scope

A worked steady plane channel problem of the adopted incompressible Newtonian branch: the Poiseuille profile with pressure gradient 6, height 2 and viscosity 3 is verified against the model's momentum equation with zero body force, its flux per unit width 4/3 and centre speed 1 are computed, the Couette profile is verified with its nonzero curl, and the no-slip wall conditions are recorded as adopted ideal boundary data. The example reports no measurement.

Example

Adopt the incompressible Newtonian branch of Adopted fluid constitutive branches: incompressible Newtonian, barotropic inviscid, and the periodic-cell idealization with constant density $ρ > 0$ , dynamic viscosity $η = 3 P a s$ and channel height $h = 2 m$ , whose SI coordinate domain is $O = R \times (0, 2) \times R$ with coordinates $(x, y, z)$ , in the steady description and with no body force, and explicitly adopt the steady incompressible model equation $ρ (u \cdot \nabla) u = - \nabla p + η Δ u$ together with $div u = 0$ (Incompressible Newtonian and barotropic inviscid fluid constitutive models). Here displayed coordinate profiles and numbers are SI numerical values: $v$ has units $m s^{- 1}$ , the flux per width has units $m^{2} s^{- 1}$ , $x, y, z$ are in metres and $p$ is in pascals. Deriving the adopted differential equation from integral balances would additionally require the mass and force premises of The momentum equation of the continuum model: local form rho D_t u = rho b + div sigma; those are not inferred from a constitutive law alone. For a $C^{1}$ spatial field $w = (w_{1}, w_{2}, w_{3})$ , define $curl w : = (\partial_{y} w_{3} - \partial_{z} w_{2}, \partial_{z} w_{1} - \partial_{x} w_{3}, \partial_{x} w_{2} - \partial_{y} w_{1})$ ; irrotational means this vector vanishes, equivalently the spatial derivative matrix is symmetric. Then:

(Poiseuille.) With pressure gradient $G = 6 P a m^{- 1}$ , i.e. $p (x) = p_{0} - 6 x$ , the profile $u = (v (y), 0, 0)$ , $v (y) = y (2 - y)$ (the value $G y (h - y) / (2 η)$ of the general formula with $h = 2, η = 3$ ) satisfies $div u = 0$ and $η v^{''} (y) = - 6 = - G$ , hence the steady momentum equation with no body force; it satisfies the no-slip data $v (0) = v (2) = 0$ ; its flux per unit width is $\int_{0}^{2} v (y) d y = 4 / 3$ and its centre speed is $v (1) = 1$ .
(Couette.) With constant pressure and wall speed $V = 2 m s^{- 1}$ , the profile $v (y) = V y / 2 = y$ satisfies the same equations with zero body force and the boundary data $v (0) = 0$ , $v (2) = 2$ ; its velocity gradient $D u = (\begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array})$ is not symmetric, so the flow is not irrotational: $curl u = (0, 0, - V / h) = (0, 0, - 1)$ .
(Uniqueness in the ansatz.) Within the ansatz $u = (v (y), 0, 0)$ , $p = p_{0} - G x$ , the difference of two profiles with the same $G$ and the same no-slip data satisfies $w^{''} = 0$ with $w (0) = w (2) = 0$ , hence $w \equiv 0$ : the Poiseuille profile is the unique steady solution in this class with those data.
(Reading.) The wall conditions are adopted ideal boundary data: the model is used with the no-slip assumption $v = 0$ at the fixed walls, which is the idealization recorded in Adopted fluid constitutive branches: incompressible Newtonian, barotropic inviscid, and the periodic-cell idealization; the profiles are exact solutions of the model and are not approximations within it. Dimensions: $[G] = μ ℓ^{- 2} τ^{- 2}$ , so $[G h^{3} / η] = ℓ^{2} τ^{- 1}$ (flux per unit width) and $[G h^{2} / η] = ℓ τ^{- 1}$ (speed), consistently with Continuum body data, the fixed inertial frame and the dimensions of the fields, Dimensions and SI unit conventions of the non-relativistic classical mechanics framework.

Facts & Assumptions

Given: the branch data $ρ > 0$ , $η = 3$ ; the channel height $h = 2$ ; the gradients $G = 6$ and $V = 2$ ; the profiles.

[F1]

Branch and model equation (Adopted fluid constitutive branches: incompressible Newtonian, barotropic inviscid, and the periodic-cell idealization, Incompressible Newtonian and barotropic inviscid fluid constitutive models, The momentum equation of the continuum model: local form rho D_t u = rho b + div sigma, The continuity equation of the continuum model: local form of mass balance, the material form, and the incompressibility reading): for the incompressible Newtonian branch with no body force, steady fields satisfy $div u = 0$ and $ρ (u \cdot \nabla) u = - \nabla p + η Δ u$ , where the stress is $σ = - p I_{3} + η (D u + D u^{T})$ .

[F2]

Channel-flow theorem (Exact plane Poiseuille and Couette flows of the incompressible Newtonian model, with flux and ansatz uniqueness): with $h > 0$ , $G \in R$ , $v (y) = G y (h - y) / (2 η)$ , $p = p_{0} - G x$ , the field $u = (v, 0, 0)$ satisfies the steady equations, has $v (0) = v (h) = 0$ , flux $G h^{3} / (12 η)$ , and is unique in the ansatz $u = (v (y), 0, 0)$ , $p = p_{0} - G x$ with these boundary data; the Couette profile $v = V y / h$ with constant pressure satisfies the equations with $v (0) = 0, v (h) = V$ and nonsymmetric gradient for $V \neq 0$ .

[F3]

Calculus (Euclidean spatial calculus: gradient, divergence and material derivative): componentwise $(u \cdot \nabla) u$ and $Δ u$ , together with $curl$ as defined in the Example for fields of the form $u = (v (y), 0, 0)$ , so that $(u \cdot \nabla) u = 0$ and $Δ u = (v^{''}, 0, 0)$ .

[F4]

Mean value theorem (The mean value theorem, as the case $g (x) = x$ of Cauchy's: for $f$ continuous on $[a, b]$ with $a < b$ and differentiable on $(a, b)$ there is $c \in (a, b)$ with $f (b) - f (a) = f^{'} (c) (b - a)$ ): if a real function is continuous on a compact interval and differentiable in its interior, its increment equals its derivative at some interior point times the interval length. In particular a function with zero derivative is constant on each compact subinterval of its open interval of definition.

Verification

technique · direct; substitute the explicit profiles into the model's equations, integrate for the flux, and evaluate the uniqueness difference equation

1.1givenF1F2F3algebra

Poiseuille equations: $v (y) = 6 y (2 - y) / 6 = y (2 - y)$ gives $v (0) = v (2) = 0$ , $v^{'} (y) = 2 - 2 y$ , $v^{''} = - 2$ and, by [F3], $Δ u = (- 2, 0, 0) = (- η^{- 1} G, 0, 0)$ with $η = 3$ , $G = 6$ ; since $(u \cdot \nabla) u = 0$ and $\nabla p = (- 6, 0, 0)$ , the steady momentum equation $ρ (u \cdot \nabla) u = - \nabla p + η Δ u$ holds identically, and $div u = \partial_{x} v = 0$ .

1.2givenF1F2F3algebra

Couette: with $v (y) = y$ , $v^{''} = 0$ so $η Δ u = 0$ ; with constant pressure $\nabla p = 0$ ; and $v (0) = 0, v (2) = 2$ ; the gradient matrix has the single nonzero entry $\partial_{y} v = 1$ , which is not symmetric and gives $curl u = (0, 0, - \partial_{y} v) = (0, 0, - 1)$ .

2.1step 1.1F2algebra

Flux and centre speed: $\int_{0}^{2} y (2 - y) d y = [y^{2} - y^{3} / 3]_{0}^{2} = 4 - \frac{8}{3} = \frac{4}{3}$ , matching $G h^{3} / (12 η) = 6 \cdot 8 / 36 = \frac{4}{3}$ , and $v (1) = 1$ , matching $G h^{2} / (8 η) = 6 \cdot 4 / 24 = 1$ .

2.2step 1.1F2F4algebra

Uniqueness in the ansatz: let $v_{1}, v_{2} \in C ([0, 2]) \cap C^{2} ((0, 2))$ be two profiles with the same $G$ and no-slip data, as in [F2]. Their difference $w$ satisfies $w^{''} = 0$ on $(0, 2)$ and $w (0) = w (2) = 0$ . Apply [F4] to $w^{'}$ on every compact subinterval of $(0, 2)$ to obtain one constant $c = w^{'} (y)$ throughout the interior. Applying [F4] to $w (y) - c y$ gives $w (y) = c y + d$ there. Continuity extends this affine formula to both endpoints; $w (0) = 0$ gives $d = 0$ and $w (2) = 0$ gives $2 c = 0$ , hence $w \equiv 0$ . No derivative at a wall is assumed.

3.1step 2.1step 1.2F1algebra∎

Reading: the no-slip conditions are exactly the adopted wall data of the branch's idealization, and the two profiles are exact solutions of the model on the channel; the dimensional checks $[G] = μ ℓ^{- 2} τ^{- 2}$ , $[G h^{3} / η] = ℓ^{2} τ^{- 1}$ , $[G h^{2} / η] = ℓ τ^{- 1}$ hold by the rules of Dimensions and SI unit conventions of the non-relativistic classical mechanics framework. No measured viscosity, pressure or flux is asserted.

Remarks

The example exhibits the branch's structure: a parabolic profile driven by a constant pressure gradient, a linear profile driven by a moving wall, the nonzero vorticity of the latter, and the uniqueness of the former inside the stated ansatz. Nothing is asserted about stability of these flows, about entrance and exit regions, or about flows at higher rates where the laminar ansatz or the no-slip idealization may fail.
The channel data are chosen so that the arithmetic is exact ( $h = 2, η = 3, G = 6$ ); the same formulas appear in the cited treatments for general data, and no numerical coincidence carries physical meaning.

Poiseuille and Couette channel flows of the viscous branch: explicit profiles, flux and wall data

Example

Facts & Assumptions

Verification

Remarks

Depends on

Dependency tree · 2 levels

Sources