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Exact kinetic moment balances and unclosed fluxes

Statement

Let $Ω = \prod_{j = 1}^{3} (R / L_{j} Z)$ with $L_{j} > 0$ , let $I \subset R$ be open, and fix $m > 0$ . Let $F : Ω \times R^{3} \times I \to [0, \infty)$ be $C^{1}$ , with $n (r, t) = \int F (r, v, t) d v > 0$ . For every compact $J \subset I$ , assume some $C_{J} < \infty$ and $N > 6$ satisfy

$∣ F ∣ + ∣ \partial_{t} F ∣ + ∣ \nabla_{r} F ∣ + ∣ \nabla_{v} F ∣ \leq C_{J} (1 + ∣ v ∣)^{- N} (t \in J) .$

Let $a : Ω \times I \to R^{3}$ be $C^{1}$ and independent of $v$ . Let $Q (r, \cdot, t)$ be measurable with $\int (1 + ∣ v ∣^{2}) ∣ Q ∣ d v < \infty$ and

$\int Q d v = 0, \int v Q d v = 0, \int ∣ v ∣^{2} Q d v = 0$

at every $(r, t)$ . Suppose pointwise that

$\partial_{t} F + v \cdot \nabla_{r} F + a \cdot \nabla_{v} F = Q .$

Define $ρ, u, Π, e, q$ as in Kinetic density, velocity, covariance stress and heat flux. Then the exact balances are

$\partial_{t} ρ + \nabla_{r} \cdot (ρ u) = 0,$

$\partial_{t} (ρ u) + \nabla_{r} \cdot (ρ u \otimes u + Π) = ρ a,$

$\partial_{t} ⁣ (\frac{1}{2} ρ ∣ u ∣^{2} + e) + \nabla_{r} \cdot ⁣ [(\frac{1}{2} ρ ∣ u ∣^{2} + e) u + Π u + q] = ρ a \cdot u,$

and

$\partial_{t} e + \nabla_{r} \cdot (e u + q) + Π : \nabla_{r} u = 0.$

These identities are exact moment equations under the stated generic conservation assumptions on $Q$ . They do not supply a constitutive relation for $Π$ or $q$ , a closure, or a hydrodynamic limit.

In the SI physical interpretation, $t$ is measured in seconds, $a$ is acceleration in $m s^{- 2}$ , and $Q$ has the units of $F$ per second.

Facts & Assumptions

Given: $Ω$ , $I$ , $m$ , $F$ , $a$ , and $Q$ satisfying the Statement.

[A1]

$F$ is periodic and $C^{1}$ in $(r, v, t)$ with the stated uniform velocity-decay bounds on compact time slabs; $a$ is $C^{1}$ and independent of velocity.

[A2]

The source $Q$ has finite absolute moments through order two and its number, momentum and squared-speed moments vanish.

[F1]

Kinetic density, velocity, covariance stress and heat flux: $ρ = m n$ , $u = n^{- 1} \int v F d v$ , $Π = m \int (v - u) \otimes (v - u) F d v$ , $e = \frac{1}{2} tr Π$ , and $q = \frac{m}{2} \int ∣ v - u ∣^{2} (v - u) F d v$ .

[F2]

Dominated convergence: an integrable majorant on a measure space permits differentiation or limits under an integral when the corresponding difference quotients converge pointwise and share that majorant.

[F3]

Fubini's theorem for L^1 functions on a sigma-finite product: absolutely integrable functions on sigma-finite products may be integrated in either order.

Proof

Given: [A1], [A2], the displayed kinetic equation, and [F1]–[F3].

Proof technique: direct.

1.1A1A2F2F3algebra

Fix $(r, t)$ and multiply the equation by one of $ψ (v) = 1$ , $ψ (v) = m v_{i}$ , or $ψ (v) = m ∣ v ∣^{2} / 2$ . Since $N > 6$ , polar integration gives $\int_{∣ v ∣ > 1} ∣ v ∣^{3} (1 + ∣ v ∣)^{- N} d v < \infty$ , so the moments and fluxes through order three are absolutely integrable; the energy integration-by-parts boundary term on $∣ v ∣ = R$ is $O (R^{4 - N}) \to 0$ . The decay in [A1] and [F2] then justify differentiating each velocity moment and commuting spatial derivatives with its integral; [F3] justifies the absolutely convergent rearrangements. Integrating the transport term gives $\nabla_{r} \cdot \int v ψ F d v$ . Integrating the acceleration term by parts in velocity gives $- \int F a \cdot \nabla_{v} ψ d v$ ; its boundary term vanishes by the preceding estimate. The $Q$ integral is zero for each selected $ψ$ by [A2]. The resulting raw identities are $\partial_{t} n + \nabla_{r} \cdot \int v F d v = 0$ , $\partial_{t} (m \int v F d v) + \nabla_{r} \cdot (m \int v \otimes v F d v) = ρ a$ , and $\partial_{t} (\frac{m}{2} \int ∣ v ∣^{2} F d v) + \nabla_{r} \cdot (\frac{m}{2} \int ∣ v ∣^{2} v F d v) = ρ a \cdot u$ .

2.1F1step 1.1algebra

Write $c = v - u$ , so $\int c F d v = 0$ . Expanding $v = u + c$ gives $m \int v \otimes v F d v = ρ u \otimes u + Π$ . Expanding $∣ v ∣^{2} v = (∣ u ∣^{2} + 2 u \cdot c + ∣ c ∣^{2}) (u + c)$ and discarding only terms whose centered first moment is zero gives $\frac{m}{2} \int ∣ v ∣^{2} v F d v = (\frac{1}{2} ρ ∣ u ∣^{2} + e) u + Π u + q .$ Substitution into the raw moment equations of step 1.1 yields the first three displayed balances.

3.1F1step 2.1algebra∎

Dot the momentum equation from step 2.1 with $u$ . The continuity equation gives $u \cdot [\partial_{t} (ρ u) + \nabla_{r} \cdot (ρ u \otimes u)] = \partial_{t} (\frac{1}{2} ρ ∣ u ∣^{2}) + \nabla_{r} \cdot (\frac{1}{2} ρ ∣ u ∣^{2} u) .$ Since $Π$ is symmetric by [F1], the product rule gives $u \cdot (\nabla_{r} \cdot Π) = \nabla_{r} \cdot (Π u) - Π : \nabla_{r} u$ . Subtracting this kinetic-energy balance from the total-energy balance in step 2.1 cancels $ρ a \cdot u$ and leaves $\partial_{t} e + \nabla_{r} \cdot (e u + q) + Π : \nabla_{r} u = 0$ .

Exact kinetic moment balances and unclosed fluxes

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources