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TheoremStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck pendingverified 2026-10-23 (gpt-6.1-sol)
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Exact homogeneous BGK fast relaxation

Statement

Adopt Countable Choice, denoted ACω, as in the kinetic-density definition. Let v∗=1 m s−1, ω3(v):=∑j=03(∣v∣/v∗)j, let τ>0 be a relaxation time in seconds, and let ϵ>0 be dimensionless. Let f0:R3→[0,∞) be a finite-valued measurable velocity density with units fixed by the kinetic-density definition, with ∫ω3f0 dv<∞, number density n>0, and strictly positive centered second moment. Let M=M[f0] be its positive-temperature Gaussian from the single-species BGK definition. Fix a constant reference density fref>0 with the same units as f0 and assume ∫f0∣log⁡(f0/fref)∣ dv<∞, with 0log⁡0:=0 and the integrand f0∣log⁡(f0/fref)∣ assigned value zero where f0=0. Define fϵ(t,v):=M(v)+e−t/(ϵτ)(f0(v)−M(v)),t≥0 in seconds. Then fϵ≥0, has the same mass, momentum, and kinetic energy as f0 for every t, belongs to C1([0,∞);L1(ω3 dv)), and solves, as an equality in L1(ω3 dv), ∂tfϵ=M[fϵ]−fϵϵτ,fϵ(0)=f0. A solution means a nonnegative C1([0,∞);L1(ω3 dv)) function satisfying this equality at each time. The solution is unique among nonnegative solutions in this function class whose density and temperature remain strictly positive. Moreover, ∫R3ω3∣fϵ(t)−M∣ dv=e−t/(ϵτ)∫R3ω3∣f0−M∣ dv, and for every T∗>0, ∫0T∗∫R3ω3∣fϵ−M∣ dv dt≤ϵτ∫R3ω3∣f0−M∣ dv. Thus the equilibrium limit is exact for each fixed t>0 and in spacetime weighted L1, but need not be uniform at t=0. This branch has stationary homogeneous Euler moments; it gives no evolving inhomogeneous Euler limit.

Facts & Assumptions

Given: Countable Choice and nonnegative f0 with finite weighted third moment and positive variance, its moment-matched Maxwellian M, and ϵ,τ>0.

[F1]

Single-species BGK relaxation operator: QBGK(f)=(M[f]−f)/τ.

[F2]

Gaussian Maxwellian normalization, stress, and heat flux: under the adopted Countable Choice, M[f0] has exactly the mass, momentum, and energy moments used to define it, and belongs to L1(ω3 dv).

[A1]

The initial data and parameters satisfy the hypotheses in the Statement.

Proof

1.1F1A1algebra

Put a(t)=e−t/(ϵτ). Then 0<a(t)≤1 and fϵ=(1−a(t))M+a(t)f0, so fϵ≥0. Since M,f0∈L1(ω3 dv), this formula is continuously differentiable in that Banach space and ∂tfϵ=−a(t)ϵτ(f0−M)=M−fϵϵτ.

1.2A1F1F2algebra

Let f~ be any other solution in the stated class. For each invariant polynomial ψ of degree at most two, the functional Λψ(g)=∫ψg dv is bounded on L1(ω3 dv) because ∣ψ(v)∣≤Cψω3(v). Continuity of this linear functional and the defining Banach-space derivative give (Λψ∘f~)′(t)=Λψ(∂tf~). Integrating the equation against these invariants shows each derivative is zero, because M[f~] and f~ have the same defining moments by [F2]. Its moments are therefore those of f0, so M[f~(t)]=M. The equation reduces to ∂t(f~−M)=−(f~−M)/(ϵτ) in L1(ω3 dv). Multiplication by et/(ϵτ) has zero derivative, hence f~(t)=M+e−t/(ϵτ)(f0−M)=fϵ(t).

2.1A1F1F2step 1.1algebra

By [F2], the invariant moments of M and f0 agree. The affine formula for fϵ therefore preserves them. In particular its positive number density and positive centered second moment stay fixed, so M[fϵ(t)]=M for every t. Step 1.1 is consequently the stated nonlinear BGK equation with the correct moment-matched target.

3.1A1step 1.1algebra∎

Homogeneity of the weight and the exact formula give ∫ω3∣fϵ(t)−M∣ dv=a(t)∫ω3∣f0−M∣ dv. Integrating in t yields the stated bound because ∫0T∗a(t) dt=ϵτ(1−e−T∗/(ϵτ))≤ϵτ. For fixed t>0, a(t)→0 as ϵ→0, whereas at t=0 the distance is ∫ω3∣f0−M∣.

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