ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedprecheck passverified 2026-10-23 (gpt-6.1-sol)

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Periodic channel Poiseuille flow uses a body force

Example

Let $L_{1}, L_{2}, h, ρ, η, G > 0$ and use the periodic flat channel $Ω = (R / L_{1} Z) \times (0, h) \times (R / L_{2} Z)$ . For the steady incompressible equations $ρ (u \cdot \nabla) u = - \nabla p + η Δ u + ρ b$ , $div u = 0$ , prescribe no-slip walls and constant body acceleration $b = (G / ρ, 0, 0)$ . Then $u (x, y, z) = (\frac{G}{2 η} y (h - y), 0, 0), p = p_{0}$ is an exact solution and its volume flux per unit transverse width, defined for $u = (U (y), 0, 0)$ by $Q = \int_{0}^{h} U (y) d y$ , is $G h^{3} / (12 η)$ . The formal pressure $p = p_{0} - G x$ is not a single-valued periodic pressure. On the infinite flat channel $R \times (0, h) \times R$ , the same velocity instead solves the zero-body-force equation with $p = p_{0} - G x$ . Among classical solutions of the form $\tilde{u} (x, y, z) = (\tilde{U} (y), 0, 0)$ with $\tilde{U} \in C^{2} ([0, h]; R)$ , a single-valued pressure $p \in C^{1} (Ω; R)$ , the same body force, and wall traces $\tilde{U} (0) = \tilde{U} (h) = 0$ , the velocity profile is unique. With SI units, $x, y, z, L_{1}, L_{2}, h$ are metres, $ρ$ is kg/m $^{3}$ , $η$ is Pa $\cdot$ s, $G$ is Pa/m, $b$ is m/s $^{2}$ , $u$ is m/s, pressure is Pa, and flux per unit width is m $^{2}$ /s.

Example references

Flat periodic and infinite channel domains

David Tong, Fluid Mechanics, Chapter 2, §§3.2.1–3.2.3

Verification

Given: The channels, coefficients, equations, and boundary data in the Example.

[F1] The periodic and infinite flat channels have the coordinate domains and wall faces specified in the flat-channel definition (Flat periodic and infinite channel domains).

[A1] The steady incompressible momentum equation and divergence constraint displayed in the Example are the system being checked; no empirical claim is made.

1.1givenF1A1algebra

If pressure were $p_{0} - G x$ on the periodic channel, its lift would satisfy $p (x + L_{1}, y, z) - p (x, y, z) = - G L_{1} \neq 0$ , contradicting periodicity. The constant periodic pressure together with $b = (G / ρ, 0, 0)$ preserves the same streamwise forcing in the stated equation.

2.1givenstep 1.1A1algebra

Write $U (y) = G y (h - y) / (2 η)$ . Then $U (0) = U (h) = 0$ , $U^{''} = - G / η$ , $div (U e_{x}) = 0$ , and $(U e_{x} \cdot \nabla) (U e_{x}) = U \partial_{x} (U e_{x}) = 0$ . The right side of [A1] is $0 - 0 + η U^{''} e_{x} + ρ (G / ρ) e_{x} = - G e_{x} + G e_{x} = 0$ , so the periodic profile solves the system.

3.1givenstep 2.1A1algebra

Direct integration gives $\int_{0}^{h} U (y) d y = G (2 η)^{- 1} \int_{0}^{h} (h y - y^{2}) d y = G h^{3} / (12 η)$ . Let $\tilde{u} = (\tilde{U} (y), 0, 0)$ and its single-valued periodic pressure $p$ be any competing classical shear solution in the class stated in the Example. The transverse components of the steady momentum equation give $\partial_{y} p = \partial_{z} p = 0$ , so $p = p (x)$ . The streamwise component gives $\partial_{x} p = η {\tilde{U}}^{''} (y) + G$ . The left side depends only on $x$ and the right side only on $y$ , so both equal a constant $c$ . Periodicity and single-valuedness give $0 = p (x + L_{1}) - p (x) = \int_{x}^{x + L_{1}} \partial_{x} p (s) d s = c L_{1}$ , hence $c = 0$ and ${\tilde{U}}^{''} = - G / η = U^{''}$ . The common endpoint values imply that $\tilde{U} - U$ is affine with zero values at $0$ and $h$ , so it vanishes identically.

4.1step 2.1F1A1algebra∎

On the infinite channel, $p = p_{0} - G x$ is single-valued and $- \nabla p = G e_{x}$ ; with $b = 0$ , this term cancels $η Δ (U e_{x}) = - G e_{x}$ . The same divergence, convection, and wall calculations as step 2.1 apply, proving the infinite-channel solution.

Periodic channel Poiseuille flow uses a body force

Example

Example references

Verification

Depends on

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