Alphabeta Math
DefinitionDefinition: AI-adaptedProof: Not applicablePipeline-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.

Constant-density incompressible Newtonian model

Physical scope

A smooth, unforced, constant-density incompressible Newtonian branch in a Cartesian inertial chart, with a pressure multiplier and constant nonnegative shear viscosity. Boundary and initial data are separate, and no solution existence or empirical applicability is asserted.

Definition

Let I⊂R be an open time interval. Use the Cartesian inertial-chart convention of the adopted continuum balances on Newtonian spacetime I×E: E is an oriented Euclidean affine three-space, and the event-coordinate map is (t,X)↦(t,xt(X)) with xt(X)=b(X)+tw, where b:E→R3 is an orientation-preserving affine isometry and w∈R3 is constant. Time is absolute and measured in seconds, spatial coordinates in metres, and w in metres per second. All derivatives below use this fixed chart. Let Ω⊂I×R3 be open. A smooth constant-density incompressible Newtonian branch has a constant ρ0>0 in kg m−3, a constant dynamic shear viscosity μ≥0 in Pa s, a velocity field u∈C2(Ω,R3) measured in m s−1, and a pressure multiplier p∈C1(Ω,R) measured in Pa. Define Lij=∂xjui, D=(L+LT)/2, and Dtu=∂tu+(u⋅∇)u. The branch is specified by ρ=ρ0,div⁡u=0,σ=−pI3+2μD,ρ0Dtu=div⁡σ, where (div⁡σ)i=∑j=13∂xjσij. Equivalently, for constant μ, its momentum equation is ρ0Dtu+∇p=μΔu,Δui=∑j=13∂xjxjui. Indeed, div⁡(2μD)=μΔu+μ∇(div⁡u)=μΔu. This is the incompressible specialization of the adopted Newtonian closure: θ=div⁡u=0 implies D0=D, so the bulk-viscosity and Fourier-flux terms do not enter this mechanical equation. The pressure is a constraint multiplier and is not identified with a compressible equation-of-state pressure. Define the kinematic viscosity ν=μ/ρ0, in m2 s−1. Boundary and initial data are not part of this definition, and the item asserts no solution existence.

Depends on

Dependency tree · 2 levels

Sources