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Mean and fluctuation energy exchange

Statement

Let $I \subseteq R$ be an open interval, let $d \geq 2$ and $N \geq 1$ be integers, and define the flat torus $T^{d} : = (R / 2 π Z)^{d}$ , with integration using Lebesgue measure on a periodic cell $[0, 2 π]^{d}$ . Fix constants $a_{j} \geq 0$ with $\sum_{j = 1}^{N} a_{j} = 1$ and a common constant $ν \geq 0$ . For $j = 1, \dots, N$ , let $u_{j} : I \times R^{d} \to R^{d}$ be continuously differentiable in $(t, x)$ with continuous second spatial derivatives, and let $p_{j} : I \times R^{d} \to R$ be continuously differentiable. Require these fields to be $2 π$ -periodic in each spatial coordinate, so they descend to $T^{d}$ . Suppose they are unforced and satisfy pointwise on $I \times R^{d}$

$div u_{j} = 0, \partial_{t} u_{j} + div (u_{j} \otimes u_{j}) + \nabla p_{j} = ν Δ u_{j} .$

Use $(u \otimes v)_{r s} = u_{r} v_{s}$ , $(div A)_{r} = \sum_{s} \partial_{x_{s}} A_{r s}$ for a matrix field $A$ , and componentwise Laplacian. For $A \in R^{d \times d}$ set $tr A : = \sum_{r} A_{r r}$ . Define $\overset{ˉ}{u} = \sum_{j} a_{j} u_{j}$ , $v_{j} = u_{j} - \overset{ˉ}{u}$ , $R = \sum_{j} a_{j} v_{j} \otimes v_{j}$ , and

$E_{m} (t) : = \frac{1}{2} \int_{T^{d}} ∣ \overset{ˉ}{u} (t, x) ∣^{2} d x, E_{f} (t) : = \frac{1}{2} \int_{T^{d}} tr R (t, x) d x = \frac{1}{2} \int_{T^{d}} \sum_{j} a_{j} ∣ v_{j} (t, x) ∣^{2} d x .$

For a matrix $A$ and vector field $q$ , set $A : \nabla q : = \sum_{r, s = 1}^{d} A_{r s} \partial_{x_{s}} q_{r}$ and $∣ \nabla q ∣^{2} : = \sum_{r, s = 1}^{d} ∣ \partial_{x_{s}} q_{r} ∣^{2}$ . Then for every $t \in I$ ,

$\frac{d E_{m}}{d t} = - ν \int_{T^{d}} ∣ \nabla \overset{ˉ}{u} ∣^{2} d x + \int_{T^{d}} R : \nabla \overset{ˉ}{u} d x,$

$\frac{d E_{f}}{d t} = - ν \int_{T^{d}} \sum_{j} a_{j} ∣ \nabla v_{j} ∣^{2} d x - \int_{T^{d}} R : \nabla \overset{ˉ}{u} d x .$

For every $t_{0} < t_{1}$ in $I$ , the endpoint identities are

$E_{m} (t_{1}) - E_{m} (t_{0}) = - ν \int_{t_{0}}^{t_{1}} \int_{T^{d}} ∣ \nabla \overset{ˉ}{u} ∣^{2} d x d t + \int_{t_{0}}^{t_{1}} \int_{T^{d}} R : \nabla \overset{ˉ}{u} d x d t,$

$E_{f} (t_{1}) - E_{f} (t_{0}) = - ν \int_{t_{0}}^{t_{1}} \int_{T^{d}} \sum_{j} a_{j} ∣ \nabla v_{j} ∣^{2} d x d t - \int_{t_{0}}^{t_{1}} \int_{T^{d}} R : \nabla \overset{ˉ}{u} d x d t .$

The transfer term has opposite signs in the two balances and cancels in their sum. No sign is asserted for $\int_{T^{d}} R : \nabla \overset{ˉ}{u} d x$ .

Statement references

Eyink, Johns Hopkins University Turbulence Notes, Chapter VIb, §(B) Reynolds Averaging and kinetic-energy balances, pp. 17–19

Flierl and Ferrari, MIT OCW 12.820, Chapter 3, §3.7, full lecture notes

Facts & Assumptions

[F1]

Exact finite Reynolds averaged equation: For fixed nonnegative normalized weights and a smooth finite family of divergence-free solutions with common viscosity, the weighted mean satisfies the exact averaged equation with covariance term $- div R$ .

[F2]

Exact finite Reynolds averaged equation: The fixed-weight fluctuations have zero weighted mean, $\sum_{j} a_{j} v_{j} = 0$ .

[F3]

Exact finite Reynolds averaged equation: The pointwise kinetic-energy split is $\sum_{j} a_{j} ∣ u_{j} ∣^{2} = ∣ \overset{ˉ}{u} ∣^{2} + tr R$ .

[A1]

The periodic lifts are maps on $R^{d}$ satisfying the Statement’s regularity, fixed-weight, common-viscosity, divergence, and unforced equation hypotheses. Thus [F1] applies with its open spatial domain $Ω = R^{d}$ ; the periodic cell is used only for the energy integrals.

Given: The data in the Statement, including [A1] and the finite Reynolds averaged equation and energy split [F1].

Proof

1.1givenA1F1algebra

By [F1], the mean field satisfies $\partial_{t} \overset{ˉ}{u} + div (\overset{ˉ}{u} \otimes \overset{ˉ}{u}) + \nabla \overset{ˉ}{p} = ν Δ \overset{ˉ}{u} - div R$ with $div \overset{ˉ}{u} = 0$ . Differentiate $E_{m}$ under the integral, which is valid by the stated regularity and compactness of the periodic cell. For every continuously differentiable periodic scalar $g$ , $\int_{T^{d}} \partial_{x_{s}} g d x = 0$ , since integration over $x_{s} \in [0, 2 π]$ gives equal endpoint values. Thus the convection and pressure terms integrate to zero because they are respectively $div (∣ \overset{ˉ}{u} ∣^{2} \overset{ˉ}{u} / 2)$ and $div (\overset{ˉ}{p} \overset{ˉ}{u})$ ; periodic integration by parts gives $\int \overset{ˉ}{u} \cdot Δ \overset{ˉ}{u} = - \int ∣ \nabla \overset{ˉ}{u} ∣^{2}$ and $- \int \overset{ˉ}{u} \cdot div R = \int R : \nabla \overset{ˉ}{u}$ . This proves the mean-energy identity.

2.1givenA1step 1.1algebra

Dot the $j$ th equation in [A1] with $u_{j}$ , integrate, and multiply by $a_{j}$ . The time term is $\frac{1}{2} a_{j} \frac{d}{d t} \int ∣ u_{j} ∣^{2}$ . The convection and pressure terms are periodic divergences, respectively $div (∣ u_{j} ∣^{2} u_{j} / 2)$ and $div (p_{j} u_{j})$ , because $div u_{j} = 0$ ; their integrals vanish by the periodic-cell argument in step 1.1. Periodic integration by parts gives the viscous contribution $- ν a_{j} \int ∣ \nabla u_{j} ∣^{2}$ . Summing over $j$ yields $\frac{d}{d t} (\frac{1}{2} \int \sum_{j} a_{j} ∣ u_{j} ∣^{2}) = - ν \int \sum_{j} a_{j} ∣ \nabla u_{j} ∣^{2}$ .

3.1F2F3step 1.1step 2.1algebra

Fixed weights and [F2] give $\sum_{j} a_{j} \nabla v_{j} = \nabla (\sum_{j} a_{j} v_{j}) = 0$ , while [F3] gives $\frac{1}{2} \int \sum_{j} a_{j} ∣ u_{j} ∣^{2} = E_{m} + E_{f}$ . Expanding $\nabla u_{j} = \nabla \overset{ˉ}{u} + \nabla v_{j}$ and summing squared norms therefore gives $\sum_{j} a_{j} ∣ \nabla u_{j} ∣^{2} = ∣ \nabla \overset{ˉ}{u} ∣^{2} + \sum_{j} a_{j} ∣ \nabla v_{j} ∣^{2}$ . Subtracting step 1.1's mean-energy balance from step 2.1's total balance yields the stated fluctuation-energy balance with the transfer term's opposite sign.

4.1givenA1step 1.1step 3.1algebra∎

Both energies and the transfer integrand are continuous in time under [A1]. Integrating the two derivative identities from step 3.1 and step 1.1 between any $t_{0} < t_{1}$ in $I$ by the fundamental theorem of calculus gives the stated finite-time identities. Adding them cancels the transfer integrals exactly; no sign restriction on either transfer integral was used.

Depends on

Exact finite Reynolds averaged equation