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ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedprecheck passverified 2026-10-23 (gpt-6.1-sol)
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Exact constant-density compressible viscous shear

Example

Assume Countable Choice. Let ℓ,ρ∗,η,κ>0, γ>1, ζ≥0, and A∈R. On Tℓ3=(R/2πℓZ)3 with coordinates (x,y,z), set ν=η/ρ∗, p∗=κρ∗γ, and V(t,y)=Ae−νt/ℓ2sin⁡(y/ℓ). Then ρ=ρ∗ and u=(V,0,0) give a global smooth solution with no external force of ρt+div⁡(ρu)=0,(ρu)t+div⁡(ρu⊗u)+∇(κργ)=div⁡ ⁣(2ηD0+ζ(div⁡u)I), where D0=(∇u+∇uT)/2−(div⁡u)I/3. This is the first sine heat mode of the compatible channel shear evolution at channel height h=πℓ, periodically extended in y. It is one exact solution of the compressible viscous system, not a general nonconstant-density existence result. With SI units, x,y,z,ℓ are metres, t is seconds, A,u are m/s, ρ∗ is kg/m3, p∗ is Pa, η,ζ are Pa⋅s, κ is Pa⋅(m3/kg)γ, and ν is m2/s.

Verification

Given: Countable Choice, the torus, constants, constitutive stress, and compressible equations in the Example.

[F1] For channel height h=πℓ, the sine-mode heat evolution has terms bne−ν(nπ/h)2tsin⁡(nπy/h) and is smooth for compatible odd periodic data (Global compatible no-slip shear evolution).

[A1] Countable Choice is carried from the compatible shear theorem's Fourier-basis input (The Axiom of Countable Choice (ACω)).

[A2] The conservative mass and momentum equations and Newtonian stress tensor displayed in the Example are the mathematical system being checked; no empirical constitutive assertion is made.

1.1givenF1A1algebra

Take F(y)=Asin⁡(y/ℓ) on 0≤y≤πℓ. Its odd extension has period 2πℓ and is smooth. Under Countable Choice [A1], the channel Fourier formula [F1] has all sine coefficients zero except b1=A, because 2/(πℓ)∫0πℓAsin⁡2(y/ℓ) dy=A. The resulting V=Ae−νt/ℓ2sin⁡(y/ℓ) is smooth and 2πℓ-periodic in y, so it defines the stated torus field and satisfies Vt=νVyy.

2.1givenstep 1.1A2algebra

Since u=(V(y),0,0) is independent of x and ρ is constant, ρt=0 and div⁡(ρu)=ρ∗∂xV=0. Thus the mass equation in [A2] holds.

3.1givenstep 1.1A2algebra∎

We have div⁡u=0, u⋅∇u=V∂xu=0, and ∇(κρ∗γ)=0. The symmetric gradient has only xy and yx entries, each Vy/2, so it is trace-free and D0=D. Therefore div⁡(2ηD0)=ηΔu=(ηVyy,0,0) and the bulk-viscosity term is zero. The momentum left side in [A2] is (ρ∗Vt,0,0)=(ηVyy,0,0) by step 1.1 and ν=η/ρ∗, so the momentum equation holds. This verifies the exact solution for all t≥0.

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