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ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck pendingverified 2026-10-23 (gpt-6.1-sol)
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Pressure normalization carries density units

Physical scope

Symbolic conversion of a dimensionless smooth-MINI pressure observation to physical pressure using constant density and declared length, velocity, and time scales. It reports no simulation or measurement.

Example

Assume the full hypotheses of the preceding no-slip continuum CFD theorem, including its Countable Choice, strong-solution regularity, the explicitly adopted distributional weak extension of the physical momentum balance at that regularity (the continuous exact-field equation of the MINI interface), mean-zero pressure gauge, projected initialization, actual mixed backward-Euler solution, and time-step restriction. Take the dimensionless physical domain D∗=(0,1)3, constant density ρ0>0 kilograms per cubic metre, and positive scales ℓ0 metres and U0 metres per second. Use the mean-zero pressure weight ξ(y)=y1−12 and the normalized test η∈Cc∞(0,T∗) with ∥η′∥L1(0,T∗)=1. Let pphys=ρ0U02p∗ and let pˉphys,h be its corresponding right-constant physical pressure reconstruction. The physically weighted space-time pressure error

Ephys:=∫0(ℓ0/U0)T∗∫ℓ0D∗ξ(x/ℓ0)(pphys(x,t)−pˉphys,h(x,t))η(U0t/ℓ0) dx dt

has the symbolic bound ∣Ephys∣≤ρ0U0ℓ04(K+1)H/12, in pascal cubic-metre seconds. The factor ρ0 converts kinematic pressure pphys/ρ0, measured in square metres per square second, to physical pressure in pascals; the remaining factors convert the dimensionless pressure-time-volume integral. No measurement or computed numerical value is asserted.

Facts & Assumptions

Given: The full hypotheses and notation of the preceding no-slip continuum CFD theorem; D∗=(0,1)3; ρ0,ℓ0,U0>0 with the stated SI units; the mean-zero weight ξ(y)=y1−12; and η∈Cc∞(0,T∗) with ∥η′∥L1=1.

[F1]

The preceding theorem gives the dimensionless time-tested pressure estimate with bound (K+1)H∥η′∥L1∥ξ∥Q under the stated smooth-solution and MINI hypotheses. No-slip continuum CFD observable error

Verification

Given: The unit-cube domain, normalized pressure sensor and test, and the theorem supplier [F1].

1.1givenalgebra

The sensor has mean zero because ∫01(y1−12) dy1=0. Its squared L2(D∗) norm is ∫01(y1−12)2 dy1=1/12, since the other two unit-length coordinates contribute factors one. Thus ξ∈Q and ∥ξ∥Q=1/12.

1.2givenconstruct

Choose 0<a<b<T∗ and let η^(t)=exp⁡(−1/((t−a)(b−t))) for a<t<b, with η^(t)=0 otherwise. This is a nonconstant smooth compactly supported function, so ∥η^′∥L1>0. Setting η=η^/∥η^′∥L1 gives ∥η′∥L1=1.

2.1F1step 1.1step 1.2

Apply [F1] with these sensors. By step 1.1 and step 1.2, the dimensionless time-tested pressure error is at most (K+1)H/12.

3.1step 2.1algebra∎

Under x=ℓ0y and t=(ℓ0/U0)t∗, the physical pressure is ρ0U02 times the dimensionless pressure, the volume element is ℓ03dy, and the time element is (ℓ0/U0)dt∗. Their product is ρ0U0ℓ04. Multiplying the bound in step 2.1 by this positive factor proves the stated physical-unit estimate. This conversion is symbolic and supplies no empirical observation.

Depends on

Dependency tree · 2 levels