On the Energy Dissipation Rate of Ensemble Eddy Viscosity Models of Turbulence: Shear Flows

Abstract

Classical eddy viscosity models add a viscosity term with a turbulent viscosity coefficient developed beginning with the Kolmogorov–Prandtl parameterization. Approximations of unknown accuracy of the unknown mixing lengths and turbulent kinetic energy are typically constructed by solving associated systems of nonlinear convection–diffusion-reaction equations with nonlinear boundary conditions. These often over-diffuse, so additional fixes are added such as wall laws, or different approximations are used in different regions (which must also be specified). Alternately, one can solve an ensemble of NSEs with perturbed data, compute the ensemble mean and fluctuation, and simply directly compute the turbulent viscosity parameterization. This idea is recent. From previous work it seems to be of a lower complexity and greater accuracy. It also produces parameterizations with the correct near-wall asymptotic behavior. The question then arises: Does this ensemble eddy viscosity approach over-diffuse solutions? This question is addressed herein.

1. Introduction

One failure mode of eddy viscosity models of turbulence is over-dissipation, leading to a lower Reynolds number (even laminar) solution. Over-dissipation can be caused by too large a turbulent viscosity ${\nu }_{turb}$ in a flow interior and by too large a ${\nu }_{turb}$ in the near-wall boundary layer region where $\nabla u$ is large. The first possibility was analyzed in a previous paper [1]. Here, we extend the analysis of energy dissipation of ensemble eddy viscosity models to shear flows, addressing the near-wall region.

Turbulent flow simulations with uncertain data require computing velocity and pressure ensembles [2], $u_j=u(x,t;{\omega }_j),p_j=p(x,t;{\omega }_j),j=1,\cdots ,J$, of (typically) an eddy viscosity model. (Notation: To reduce non-essential notation, we have replaced the mechanically correct deformation tensor in the eddy viscosity term with the full gradient. Also, with slight abuse of notation, x denotes $(x,y,z)$ and its first component.)

$$\begin{array}{c}{u}_t+u·\nabla u-\nu ▵u-\nabla ·\left({\nu }_{turb}(·)\nabla u\right)+\nabla p=0,\mathrm{in}\mathrm{\Omega },j=1,\dots ,J,\\ \nabla ·u=0,\mathrm{and}u(x,0;{\omega }_j)=u_j^0\left(x\right),\mathrm{in}\mathrm{\Omega }\mathrm{and}u=0,\mathrm{on}\partial \mathrm{\Omega }.\end{array}$$

(1)

Here $\nu$ is the kinematic viscosity and ${\omega }_j$ are the sampled values determining the ensemble data. The domain is $\mathrm{\Omega }={(0,L)}^3$ with $L$-periodic boundary conditions in $x,y$. The ensemble mean ${〈u〉}_e$, fluctuation $u^{\prime }$, its magnitude $|u^{\prime }{|}_e$, and induced turbulent kinetic energy (TKE) density $k^{\prime }$ are

$$\begin{array}{c}\mathrm{ensemble}\mathrm{average}:{〈u〉}_e:={\displaystyle \frac{1}{J}}\sum _{j=1}^{J}u(x,t;{\omega }_j),\mathrm{fluctuation}:u^{\prime }(x,t;{\omega }_j):=u-{〈u〉}_e,\\ |u^{\prime }{|}_e^2:={\displaystyle \frac{1}{J}}\sum _{j=1}^J|u^{\prime }{|}^2\mathrm{and}\mathrm{TKE}\mathrm{density}:k^{\prime }(x,t):={〈{\displaystyle \frac{1}{2}}{\left|u^{\prime }\right|}^2〉}_e={\displaystyle \frac{1}{2}}{\left|u^{\prime }\right|}_e^2.\end{array}$$

The eddy viscosity is given by the Kolmogorov–Prandtl relation in terms of the turbulence length scale $l(x,t)$ and the turbulent kinetic energy $k^{\prime }(x,t)$:

$${\nu }_{turb}(·)=\mu l\sqrt{k^{\prime }},\mathrm{for}\mu \mathrm{an}\mathcal{O}\left(1\right)\mathrm{parameter}.$$

where $\tau$ is a selected turbulence time scale chosen smaller than the large-scale turnover time $T^{*}$, $\tau \le T^{*}$. If $\tau \simeq T^{*}$, the model can by interpreted as time evolving to a RANS approximation, while if $\tau \simeq$ numerical time-step, it is related to an LES model in time.

The shear boundary conditions at $z=0,L$ are no slip for, respectively, a fixed and moving wall:

$$u(x,y,0,t;{\omega }_j)=(0,0,0)\mathrm{and}u(x,y,L,t;{\omega }_j)=(U,0,0).$$

(2)

The global length and velocity scales are L, U and the usual Reynolds number is $\mathcal{R}e=LU/\nu$. From the energy inequality ((3) below), the energy dissipation rate ${\epsilon }_{\mathrm{model}}$ is

$${\epsilon }_{\mathrm{model}}:={〈{\displaystyle \frac{1}{\left|\mathrm{\Omega }\right|}}{\int }_{\mathrm{\Omega }}\nu |\nabla u(x,t;{\omega }_j){|}^2+{\nu }_{turb}(·){|\nabla u(x,t;{\omega }_j)|}^{2}dx〉}_e.$$

With an ensemble of solutions, $k^{\prime }$ can simply be calculated as above without approximation or modeling [3,4]. The length scale $l=\sqrt{2k^{\prime }}\tau$ (and in [5,6,7]) is determined from the use of a turbulence model for numerical simulations with fixed time or space resolution. Thus, we assume that in an under-resolved simulation, we are given a small time scale (related to the time step) denoted $\tau$. We select the length scale $l(x,t)$ to be the distance a fluctuation travels in time $\tau$

$$l=\sqrt{2k^{\prime }(x,t)}\tau \mathrm{so}\mathrm{that}{\nu }_{turb}=\mu {\left|u^{\prime }\right|}_e^2\tau$$

This choice was mentioned by Prandtl in 1926 and independently used by Kolmogorov in his URANS model [7]. The shear boundary conditions mean that the near-wall region, where velocity gradients are large, will have a different characteristic than the interior of the flow. There are different ways this different characteristic can be reflected in the model and in the analysis. For example, the final result (below) will take a different value of the parameter $\mu$ in the near-wall region and in the flow interior.

The primary failure model of eddy viscosity is over-dissipation of solutions, e.g., [8]. There are significant phenomenological reasons, Section 1.1, to hope that ${\nu }_{turb}=\mu {\left|u^{\prime }\right|}_e^2\tau$ (1) does not over-dissipate model solutions, i.e., that its energy dissipation rate is comparable uniformly in Reynolds number to the $\mathcal{O}(U^3/L)$ energy input rate. To study dissipation due to near-wall effects, we follow the plan in Doering and Constantin [9] and study shear flows in the simplest geometry. Extension of the NSE results to general shear flows in general domains was accomplished in Wang [10]; we conjecture that similar extensions of the results herein are possible.

1.1. The Energy Inequality

The analysis herein treats weak solutions to the model satisfying a standard energy inequality (3) below. To explain, let $\varphi (x,y,z)$ denote a divergence-free function with $\varphi ,\nabla \varphi \in L^2\left(\mathrm{\Omega }\right)$ with the shear boundary conditions (2) satisfied. A standard energy equality is obtained (for sufficiently regular solutions) by integrating the dot product of the model with $u-\varphi$. Motivated by the resulting energy equality, we assume that for any divergence-free $u_0, f\in L^2$, a weak solution of the model (1) with shear boundary conditions (2) exists (Existence theory for (2) is not yet established. For periodic boundary conditions, it is developed in Section 4 of [1].) for each realization and satisfies the associated energy inequality. Specifically, for any divergence-free function $\varphi (x,y,z)$ with $\varphi ,\nabla \varphi \in L^2\left(\mathrm{\Omega }\right)$ and satisfying (2), we assume

$$\begin{array}{c}{\displaystyle \frac{1}{2}}{\displaystyle \frac{d}{dt}}{\left|\right|v\left|\right|}^2+{\int }_{\mathrm{\Omega }}[\nu +{\nu }_{turb}]{|\nabla v|}^{2}dx\le \\ (v_t,\varphi )+{\int }_{\mathrm{\Omega }}[\nu +{\nu }_{turb}]\nabla v:\nabla \varphi dx+(v·\nabla v,\varphi ).\end{array}$$

(3)

This implies, in particular, the ensemble averaged energy inequality

$$\begin{array}{c}{〈{\displaystyle \frac{1}{2}}{\displaystyle \frac{1}{\left|\mathrm{\Omega }\right|}}{\int }_{\mathrm{\Omega }}{\left|u(x,T;{\omega }_j)\right|}^{2}dx〉}_e\\ +{\int }_0^T{〈{\displaystyle \frac{1}{\left|\mathrm{\Omega }\right|}}{\int }_{\mathrm{\Omega }}\nu |\nabla u(x,t;{\omega }_j){|}^2+\mu \tau |u^{\prime }(x,t){|}_e^2{|\nabla u(x,t;{\omega }_j)|}^{2}dx〉}_{e}dt\le \\ {\displaystyle \frac{1}{2}}{〈{\displaystyle \frac{1}{\left|\mathrm{\Omega }\right|}}{\int }_{\mathrm{\Omega }}{\left|u_0(x;{\omega }_j)\right|}^{2}dx〉}_e+{\int }_0^T{〈{\displaystyle \frac{1}{\left|\mathrm{\Omega }\right|}}{\int }_{\mathrm{\Omega }}f(x,t;{\omega }_j)·u(x,t;{\omega }_j)dx〉}_{e}dt.\end{array}$$

(4)

This assumption of energy inequality for weak solutions based on energy equality for sufficiently smooth strong solutions is motivated by results for the Navier–Stokes equations, e.g., [11] for an early example, and by conjectures on anomalous dissipation. Since $|u^{\prime }{|}_e^2$ is independent of ${\omega }_j$, the turbulent viscosity ${\nu }_{turb}$ will be the same for all realizations (i.e., independent of ${\omega }_j$). This property was referred to as universality in Carati, Roberts and Wray [12]. It was fundamental to reducing the computational cost of solving for an ensemble of model solutions in the algorithms developed in [4].

We present several results (summarized next and proven in Section 3) and a collection of open problems (in Section 4). The effective viscosity, ${\nu }_{eff}$, and effective Reynolds number are defined in a standard manner, Definition 2 in Section 3. The (upper) near-wall region ${\mathcal{S}}_{\beta }$ is denoted as

$${\mathcal{S}}_{\beta }=\left\{(x,y,z):0\le x\le L,0\le y\le L,(1-\beta )L<z<L\right\},\beta ={\displaystyle \frac{1}{8}}\mathcal{R}{e}_{eff}^{-1}.$$

The first theorem in Section 3 is that energy dissipation is governed by the ratio of viscosities in the near-wall region.

Theorem 1.

Any weak solution of the eddy viscosity model (1) satisfying the energy inequality (3) has its model energy dissipation bounded as

$$\begin{array}{c}\lim \underset{T\to \infty }{\sup }{\displaystyle \frac{1}{T}}{\int }_0^T{\epsilon }_{model}dt\\ \le \left({\displaystyle \frac{5}{2}}+16{\displaystyle \frac{\nu }{{\nu }_{eff}}}+32\lim \underset{T\to \infty }{\sup }{\displaystyle \frac{1}{T}}{\int }_0^T{\displaystyle \frac{1}{|{\mathcal{S}}_{\beta }|}}{\int }_{{\mathcal{S}}_{\beta }}{\displaystyle \frac{{\nu }_{turb}}{{\nu }_{eff}}}dxdt\right){\displaystyle \frac{U^3}{L}}.\end{array}$$

(5)

We expect $u^{\prime }$ and thus ${\nu }_{turb}$ to be small in ${\mathcal{S}}_{\beta }$ (being constrained by the wall), so this estimate is promising. The second theorem in Section 3 reflects the anisotropic nature of the near-wall region. It also suggests an anisotropic turbulent viscosity may be appropriate in the near-wall region with a smaller coefficient in the wall normal direction.

Theorem 2.

Let $T^{*}$ denote the large-scale turnover time. Any weak solution of the eddy viscosity model (1) satisfying the energy inequality (11) has its model energy dissipation bounded as

$$\begin{array}{c}\lim \underset{T\to \infty }{\sup }{\displaystyle \frac{1}{T}}{\int }_0^T{\epsilon }_{model}dt\le \left({\displaystyle \frac{5}{2}}+16{\displaystyle \frac{\nu }{{\nu }_{eff}}}\right){\displaystyle \frac{U^3}{L}}\\ +\left({\displaystyle \frac{\sqrt[3]{3}}{6}}{C}_{26}^2\right)\mu {\displaystyle \frac{\tau }{T^{*}}}\lim \underset{T\to \infty }{\sup }{\displaystyle \frac{1}{T}}{\int }_0^T\left({\displaystyle \frac{1}{|{\mathcal{S}}_{\beta }|}}{\int }_{{\mathcal{S}}_{\beta }}{\nu }_{eff}{〈{\left|{\displaystyle \frac{\partial u^{\prime }}{\partial z}}\right|}^2〉}_{e}dx\right)dt\end{array}$$

(6)

To obtain an estimate independent of the model solution, the integral over ${\mathcal{S}}_{\beta }$ on the RHS must be replaced by one over $\mathrm{\Omega }$, ${\displaystyle \frac{\partial u}{\partial z}}$ replaced by $\nabla u$, and the resulting term subsumed in the LHS. The (pessimistic) result (concluding Section 3) is the following wherein ${\displaystyle \frac{\nu }{{\nu }_{eff}}}\le 1.$

Theorem 3.

Suppose μ is one constant value in the flow interior and $\mu ={\mu }_{\beta }$ in ${\mathcal{S}}_{\beta }$. If

$${\mu }_{\beta }\le 0.27064\mathcal{R}{e}^{-1}$$

then

$${〈{\epsilon }_{model}〉}_{\infty }\le \left(5+32{\displaystyle \frac{\nu }{{\nu }_{eff}}}\right){\displaystyle \frac{U^3}{L}}.$$

1.2. Related Work

We cannot over-stress the importance of the work of Constantin, Doering and Foias [9,13] on the NSE to the analysis herein. Their work (building on [14,15,16]) has been developed in many important directions for the NSE in subsequent years. For turbulence models, upper bounds for energy dissipation rates, inspired by [9,13], address the most important failure mode of model over-dissipation, e.g., in [17,18,19,20].

Many studies have computed flow ensembles of various turbulence models for various applications (e.g., [5,12,21,22]). Herein, eddy viscosity models using ensemble data for parameterization, as proposed in Carati, Roberts and Wray [12] and developed in [3,4,23], are analyzed. In these, the turbulent viscosity term replaces the Reynolds stresses that have the near-wall behavior $\mathcal{O}\left(d^2\right)$ as the wall-normal distance $d\to 0.$ Since gradients are large in the boundary layer region, the near-wall $\mathcal{O}\left(d^2\right)$ should be replicated in ${\nu }_{turb}$ to reduce the chance of over-dissipation of solutions (Pakzad [17]). For the EEV model, near-wall asymptotics of ${\nu }_{turb}$ depend on $|u^{\prime }{(x,t)|}_e^2$. Since $u=0$ at the wall, a formal Taylor series expansion suggests the model studied herein may satisfy this requirement.

2. Notation and Preliminaries

In the analysis herein, the flow domain is the open box $\mathrm{\Omega }={(0,0,L)}^3$ in ${\mathbb{R}}^3$. The $L^2\left(\mathrm{\Omega }\right)$ norm and the inner product are $\parallel ·\parallel$ and $(·,·)$. Likewise, the $L^p\left(\mathrm{\Omega }\right)$ norms and the Sobolev $W_p^k\left(\mathrm{\Omega }\right)$ norms are ${\parallel ·\parallel }_{L^p}$ and ${\parallel ·\parallel }_{W_p^k}$ respectively. $H^k\left(\mathrm{\Omega }\right)$ is the Sobolev space $W_2^k\left(\mathrm{\Omega }\right)$, with norm ${\parallel ·\parallel }_k$. C represents a generic positive constant independent of $\nu ,U,L,$ and other model parameters. Its value may vary from situation to situation. For $v=v(x,t;{\omega }_j),$ recall that ${\left|v\right|}_e^2(x,t):={\displaystyle \frac{1}{J}}{\sum }_{j=1}^J{\left|v\right|}^2.$

Three kinds of averaging, ensemble, finite time and infinite time, are used. Ensemble averaging, already introduced, is ${〈\varphi 〉}_e:={\displaystyle \frac{1}{J}}{\sum }_{j=1}^J\varphi \left({\omega }_j\right)$. The short and long time averages of a function $\varphi \left(t\right)$ are respectively

$${〈\varphi 〉}_T={\displaystyle \frac{1}{T}}{\int }_0^T\varphi \left(t\right)dt.\mathrm{and}{〈\varphi 〉}_{\infty }=\lim \underset{T\to \infty }{\sup }{\displaystyle \frac{1}{T}}{\int }_0^T\varphi \left(t\right)dt.$$

The next lemma follows by inserting and rearranging the averages.

Lemma 1.

All three averages, ${〈\varphi 〉}_e,{〈\varphi 〉}_T$ and ${〈\varphi 〉}_{\infty }$, satisfy

$$〈\varphi \psi 〉\le {〈{\left|\varphi \right|}^2〉}^{1/2}{〈{\left|\psi \right|}^2〉}^{1/2},{〈{〈\varphi 〉}_e〉}_T={〈{〈\varphi 〉}_T〉}_e,{〈{〈\varphi 〉}_e〉}_{\infty }={〈{〈\varphi 〉}_{\infty }〉}_e.$$

To develop the results, some scaling constants are needed.

Definition 1.

The large velocity scale for shear flow is clearly the lid’s shear velocity (denoted U). The fluctuation scale $U^{\prime }$, large scale turnover time $T^{*}$, and Reynolds number $\mathcal{R}e$ are

$$U^{\prime }={〈{〈{\displaystyle \frac{1}{\left|\mathrm{\Omega }\right|}}\left|\right|u^{\prime }{\left|\right|}^2〉}_e〉}_{\infty }^{{\displaystyle \frac{1}{2}}},T^{*}={\displaystyle \frac{L}{U}}\mathit{and}\mathcal{R}e={\displaystyle \frac{LU}{\nu }}$$

(7)

Under the assumption of the energy inequality, the above quantities are well defined and finite.

The energy dissipation rate. The model’s ensemble averaged energy dissipation rate, from the energy inequality (3) above, is

$$\epsilon \left(t\right):={〈{\left|\mathrm{\Omega }\right|}^{-1}{\int }_{\mathrm{\Omega }}\nu |\nabla u(x,t;{\omega }_j){|}^2+\mu \tau |u^{\prime }(x,t){|}_e^2{|\nabla u(x,t;{\omega }_j)|}^{2}dx〉}_e$$

It is convenient to decompose the energy dissipation rate by

$$\begin{array}{c}\epsilon \left(t\right)={\epsilon }_{viscous}\left(t\right)+{\epsilon }_{turb}\left(t\right)\mathrm{where}\\ {\epsilon }_{viscous}\left(t\right)={〈{\left|\mathrm{\Omega }\right|}^{-1}{\int }_{\mathrm{\Omega }}\nu {|\nabla u(x,t;{\omega }_j)|}^{2}dx〉}_e,\\ {\epsilon }_{turb}\left(t\right)={〈{\left|\mathrm{\Omega }\right|}^{-1}{\int }_{\mathrm{\Omega }}\mu \tau |u^{\prime }(x,t;{\omega }_j){|}_e^2{|\nabla u(x,t;{\omega }_j)|}^{2}dx〉}_e.\end{array}$$

The Hardy inequality. With shear boundary conditions imposed at the top ($z=L$) and bottom ($z=0$), the key idea in the analysis of shear flows is to use $L^p-L^q$ generalizations of the Hardy inequality (responding to the dominant nonlinearity in the problem) to connect the near-wall solution behavior to the eddy viscosity coefficient. The basic, 1925, Hardy inequality is that for any $F\left(x\right)$ with all integrals finite and $F\left(0\right)=0$, and any $1<p<\infty$,

$${\int }_0^{\infty }{\left|{\displaystyle \frac{F\left(x\right)}{x}}\right|}^{p}dx\le {\left({\displaystyle \frac{p}{p-1}}\right)}^p{\int }_0^{\infty }{\left|F_x\left(x\right)\right|}^{p}dx.$$

(8)

There have been many extensions and generalizations of the Hardy inequality. In the analysis of Section 3, we use the following $L^p-L^q$ extension (example 1.1 equation (1.6) in Person and Samko [24], also [25]). For the parameters used, the optimal constant $C_{pq}$, (17) below, was derived in Bliss [26]. Person and Samko [24] establish that, for all (non-negative) measurable functions, $1<p\le q<\infty$, and for parameters satisfying ${\displaystyle \frac{\alpha +1}{q}}={\displaystyle \frac{1}{p}}-1$,

$${\left({\int }_0^{\infty }{x}^{\alpha }{\left({\int }_0^{x}f\left(t\right)dt\right)}^{q}dx\right)}^{1/q}\le C_{pq}{\left({\int }_0^{\infty }{f}^p\left(t\right)dx\right)}^{1/p}.$$

(9)

The choice of $\varphi$ in the energy inequality. One critical step in the proof is a precise choice of $\varphi$ in the energy inequality. The choice made is related to the Hopf extension [16] (from 1955) and follows the pioneering analysis of Doering and Constantin [9]. The $\varphi =\varphi \left(z\right)$ chosen interpolates the boundary conditions: ${\varphi \left(z\right)|}_{z=0}={[0,0,0]}^T$ and ${\varphi \left(z\right)|}_{z=L}={[U,0,0]}^T$. It is zero in the flow interior, piecewise linear near the moving wall, in ${\mathcal{S}}_{\beta }$, and divergence-free. Specifically, $\varphi \left(z\right)={[\tilde{\varphi }\left(z\right),0,0]}^T$ where

$$\tilde{\varphi }\left(z\right)=\left\{\begin{array}{cc}0,& z\in [0,L-\beta L]\\ {\displaystyle \frac{U}{\beta L}}(z-(L-\beta L)),& z\in [L-\beta L,L]\end{array}\right.\beta ={\displaystyle \frac{1}{8}}\mathcal{R}{e}_{eff}^{-1}.$$

The following values are easily calculated from the explicit formula for $\varphi \left(z\right):$

$$\begin{array}{cc}{\left|\right|\varphi \left|\right|}_{L^{\infty }\left(\mathrm{\Omega }\right)}=U,& {\left|\right|\nabla \varphi \left|\right|}_{L^{\infty }\left(\mathrm{\Omega }\right)}={\displaystyle \frac{U}{\beta L}},\\ {\left|\right|\varphi \left|\right|}^2={\displaystyle \frac{1}{3}}{U}^2\beta L^3,& {\left|\right|\nabla \varphi \left|\right|}^2={\displaystyle \frac{U^{2}L}{\beta }}.\end{array}$$

3. Shear Flows

Over-dissipation is caused by incorrect values of ${\nu }_{turb}$ in regions of small scales, i.e., where $\nabla v$ is large. These small scales are generated in the boundary layer and in the interior by breakdown of large scales through the nonlinearity. This section considers those generated predominantly in the turbulent boundary layer, studied via shear boundary conditions. Shear flows can develop in several ways. Inflow boundary conditions can emulate a jet of water entering a vessel. A body force $f(·)$ can be specified to be non-zero, large and tangential at a fixed wall. The simplest (chosen herein and inspired by [9,10]) is a moving wall modeled by a boundary condition $v=g$ on the boundary where $g·\widehat{n}=0$. This setting includes flows between rotating cylinders. We impose $L$-periodic boundary conditions in $x,y$, a fixed-wall no-slip condition at $z=0$, and a wall at $z=L$ moving with velocity $(U,0,0)$:

$$\begin{array}{cc}Boundary& Conditions:\\ \mathrm{moving}\mathrm{top}\mathrm{lid}:& u(x,y,L,t;{\omega }_j)=(U,0,0)\\ \mathrm{fixed}\mathrm{bottom}\mathrm{wall}:& u(x,y,0,t;{\omega }_j)=(0,0,0)\\ \mathrm{periodic}\mathrm{side}\mathrm{walls}:& \begin{array}{c}u(x+L,y,z,t;{\omega }_j)=u(x,y,z,t;{\omega }_j),\\ u(x,y+L,z,t;{\omega }_j)=u(x,y,z,t;{\omega }_j).\end{array}\end{array}$$

(10)

Herein, we assume that a weak solution of the model (1) with shear boundary conditions (10) exists for each realization and satisfies the usual energy inequality. Specifically, for any divergence-free function $\varphi (x,y,z)$ with $\varphi ,\nabla \varphi \in L^2\left(\mathrm{\Omega }\right)$ and satisfying the shear boundary conditions (10),

$$\begin{array}{c}{\displaystyle \frac{1}{2}}{\displaystyle \frac{d}{dt}}{\left|\right|u\left|\right|}^2+{\int }_{\mathrm{\Omega }}[\nu +{\nu }_{turb}]{|\nabla u|}^{2}dx\le \\ (u_t,\varphi )+{\int }_{\mathrm{\Omega }}[\nu +{\nu }_{turb}]\nabla u:\nabla \varphi dx+(u·\nabla u,\varphi ).\end{array}$$

(11)

To prepare the proof of the main result, we recall from, e.g., equations (20) and (22) in Doering and Constantin [9], that uniform bounds follow from (11) for both the NSE and (1).

Proposition 1 (Uniform Bounds).

Consider the model (1) with shear boundary conditions (10). For a weak solution satisfying (11), the following are finite and bounded uniformly in T:

$$\begin{array}{c}{\left|\right|u\left(T\right)\left|\right|}^2,{\int }_{\mathrm{\Omega }}{\nu }_{turb}(·,T)dx,{\displaystyle \frac{1}{T}}{\int }_0^T\left({\int }_{\mathrm{\Omega }}{|\nabla u|}^{2}dx\right)dt,\\ \mathit{and}{\displaystyle \frac{1}{T}}{\int }_0^T\left({\int }_{\mathrm{\Omega }}[\nu +{\nu }_{turb}]{|\nabla u|}^{2}dx\right)dt.\end{array}$$

To formulate our first main result we recall the definition of the effective viscosity ${\nu }_{eff}$ ($\ge \nu$) and a few related quantities. The limit superiors in the infinite time averages ${〈·〉}_{\infty }$ are finite due to the uniform bounds above.

Definition 2.

The effective viscosity ${\nu }_{eff}$ is

$${\nu }_{eff}:={\displaystyle \frac{{〈{〈\frac{1}{\left|\mathrm{\Omega }\right|}{\int }_{\mathrm{\Omega }}[\nu +{\nu }_{turb}]{|\nabla u|}^{2}dx〉}_e〉}_{\infty }}{{〈{〈\frac{1}{\left|\mathrm{\Omega }\right|}{\int }_{\mathrm{\Omega }}{|\nabla u|}^{2}dx〉}_e〉}_{\infty }}}.$$

The large-scale turnover time is $T^{*}=L/U$. The Reynolds number and effective Reynolds number are $\mathcal{R}e=UL/\nu$ and $\mathcal{R}{e}_{eff}=UL/{\nu }_{eff}.$ Let $\beta ={\displaystyle \frac{1}{8}}\mathcal{R}{e}_{eff}^{-1}$ and denote the near-wall region ${\mathcal{S}}_{\beta }$ by

$${\mathcal{S}}_{\beta }=\left\{(x,y,z):0\le x\le L,0\le y\le L,(1-\beta )L<z<L\right\}.$$

We can now present and prove the first result.

Theorem 4.

Any weak solution of the eddy viscosity model (1) satisfying the energy inequality (11) has its model energy dissipation bounded as

$${〈{\epsilon }_{model}〉}_{\infty }\le \left({\displaystyle \frac{5}{2}}+16{\displaystyle \frac{\nu }{{\nu }_{eff}}}+32{〈{\displaystyle \frac{1}{|{\mathcal{S}}_{\beta }|}}{\int }_{{\mathcal{S}}_{\beta }}{\displaystyle \frac{{\nu }_{turb}}{{\nu }_{eff}}}dx〉}_{\infty }\right){\displaystyle \frac{U^3}{L}}.$$

(12)

Remark 1.

Before beginning the proof, we record a few observations. First, the result shows the critical importance of the behavior of the turbulent viscosity in the near-wall region ${\mathcal{S}}_{\beta }$. Note that the average value over Ω is

$${〈{\displaystyle \frac{1}{\left|\mathrm{\Omega }\right|}}{\int }_{\mathrm{\Omega }}{\displaystyle \frac{{\nu }_{turb}}{{\nu }_{eff}}}dx〉}_{\infty }\le 1.$$

If the average value of ${\nu }_{turb}/{\nu }_{eff}$ in ${\mathcal{S}}_{\beta }$ (not Ω) is bounded uniformly in the Reynolds number, then non-over-dissipation of the model follows.

Let $\tilde{z}$ denote the distance to the top wall $z=L$. A formal Taylor expansion shows that in ${\mathcal{S}}_{\beta }:$$u^{\prime }\left(z\right)=\mathcal{O}\left(\tilde{z}\right)$. Thus, ${\nu }_{turb}=\mu \tau {\left|u^{\prime }(x,t)\right|}_e^2=\mu \tau \mathcal{O}({\tilde{z}}^2)$ there. This leads to the (incorrect) heuristic prediction that

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{|{\mathcal{S}}_{\beta }|}}{\int }_{{\mathcal{S}}_{\beta }}{\displaystyle \frac{{\nu }_{turb}}{{\nu }_{eff}}}dx& =\mu \tau {\nu }_{eff}^{-1}{\displaystyle \frac{1}{|{\mathcal{S}}_{\beta }|}}{\int }_{{\mathcal{S}}_{\beta }}\mathcal{O}({\tilde{z}}^2)dxdydx=\mu \tau {\nu }_{eff}^{-1}{\displaystyle \frac{1}{|{\mathcal{S}}_{\beta }|}}{L}^3\mathcal{O}\left({\beta }^3\right)\hfill \\ \hfill & =\mu \tau {\nu }_{eff}^{-1}\mathcal{O}\left({\beta }^2\right)=\mathcal{O}\left(\mathcal{R}{e}_{eff}^{-1}\right).\hfill \end{array}$$

The reason this argument is incorrect is that the constant in “$\mathcal{O}$” involves $u_z$, which (plausibly) grows like $\mathcal{O}\left(\mathcal{R}{e}_{eff}\right)$ in the near-wall region. Accounting for this, we obtain the heuristic prediction ${\nu }^{-1}{\displaystyle \frac{1}{|{\mathcal{S}}_{\beta }|}}{\int }_{{\mathcal{S}}_{\beta }}{\displaystyle \frac{{\nu }_{turb}}{{\nu }_{eff}}}dx=\mathcal{O}\left(\mathcal{R}{e}_{eff}^{+1}\right)$. This $\mathcal{R}{e}_{eff}^{+1}$ dependence is the physical justification for choosing a small value of μ in the near-wall region.

The idea of the analysis. The model uses a standard eddy viscosity formulation without modeling the TKE. There are 2 natural choices of the length scale: the wall distance and the one herein. We address the question: With the common and natural length scale herein, how does the model dissipation scale with respect to the Reynolds number? The analysis addressing that question uses a simplified approach in the usual NSE terms. The critical term (the one that may cause over-dissipation) is the nonlinear, non-monotone eddy viscosity term. Standard estimates seem to fail, so the analysis herein aims to estimate that term with the largest possible power of the small parameter β. That term is multiplied and divided by the highest power of $(L-z)$ where the “bad” negative powers can be incorporated into the turbulent viscosity. The resulting power ${(L-z)}^{\pm 4}$ and $L^6$ norm are calculated in the next lemma. The remainder of the proof is unwinding the consequences on concrete estimates of these choices.

We now give the proof of the theorem.

Proof. 

We begin by recalling the choice of $\varphi$ from Section 2. Following Doering and Constantin [9], choose $\varphi \left(z\right)={[\tilde{\varphi }\left(z\right),0,0]}^T$ in the energy inequality (3) where

$$\tilde{\varphi }\left(z\right)=\left\{\begin{array}{cc}0,& z\in [0,L-\beta L]\\ {\displaystyle \frac{U}{\beta L}}(z-(L-\beta L)),& z\in [L-\beta L,L]\end{array}\right.\beta ={\displaystyle \frac{1}{8}}\mathcal{R}{e}_{eff}^{-1}.$$

This function $\varphi \left(z\right)$ is piecewise linear, continuous, divergence-free and satisfies the boundary conditions. The following are easily calculated values

$$\begin{array}{cc}{\left|\right|\varphi \left|\right|}_{L^{\infty }\left(\mathrm{\Omega }\right)}=U,& {\left|\right|\nabla \varphi \left|\right|}_{L^{\infty }\left(\mathrm{\Omega }\right)}={\displaystyle \frac{U}{\beta L}},\\ {\left|\right|\varphi \left|\right|}^2={\displaystyle \frac{1}{3}}{U}^2\beta L^3,& {\left|\right|\nabla \varphi \left|\right|}^2={\displaystyle \frac{U^{2}L}{\beta }}.\end{array}$$

This $\varphi \left(z\right)$ interpolates the boundary conditions ${\varphi \left(z\right)|}_{z=0}={[0,0,0]}^T$ and ${\varphi \left(z\right)|}_{z=L}={[U,0,0]}^T$. It is zero in the flow interior, piecewise linear near the moving wall, in ${\mathcal{S}}_{\beta }$, and divergence-free.

With this choice of $\varphi$, time averaging the energy inequality (3) over $[0,T]$ and normalizing by $\left|\mathrm{\Omega }\right|=L^3$ gives

$$\begin{array}{c}{\displaystyle \frac{1}{2T{L}^3}}{\left|\right|v\left(T\right)\left|\right|}^2+{\displaystyle \frac{1}{T}}{\int }_0^T\left({\displaystyle \frac{1}{L^3}}{\int }_{\mathrm{\Omega }}[\nu +{\nu }_{turb}]{|\nabla v|}^{2}dx\right)dt\le {\displaystyle \frac{1}{2T{L}^3}}{\left|\right|v\left(0\right)\left|\right|}^2\\ +{\displaystyle \frac{1}{T{L}^3}}(v\left(T\right)-v\left(0\right),\varphi )+{〈{\displaystyle \frac{1}{L^3}}(v·\nabla v,\varphi )〉}_T\\ +{\displaystyle \frac{1}{T}}{\int }_0^T\left({\displaystyle \frac{1}{L^3}}{\int }_{\mathrm{\Omega }}[\nu +{\nu }_{turb}]\nabla v:\nabla \varphi dx\right)dt.\end{array}$$

(13)

Due to the above uniform-in-T bounds, the time-averaged energy inequality (3) can be expressed as

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{T}}{\int }_0^T{\epsilon }_{\mathrm{model}}dt& \le \mathcal{O}\left({\displaystyle \frac{1}{T}}\right)+{\displaystyle \frac{1}{T}}{\int }_0^T\left({\displaystyle \frac{1}{L^3}}(v·\nabla v,\varphi )\right)dt+\hfill \\ \hfill & +{\displaystyle \frac{1}{T}}{\int }_0^T\left({\displaystyle \frac{1}{L^3}}{\int }_{\mathrm{\Omega }}[\nu +{\nu }_{turb}]\nabla v:\nabla \varphi dx\right)dt.\hfill \end{array}$$

(14)

The main issue is the RHS model term, $\int {\nu }_{turb}\nabla v:\nabla \varphi dx$. Before treating that, we recall the analysis of Doering and Constantin [9] and Wang [10] for the two terms shared by the NSE, $(v·\nabla v,\varphi )$ and $\int \nu \nabla v:\nabla \varphi dx$. For the nonlinear term $NLT$,

$$NLT:={\displaystyle \frac{1}{T}}{\int }_0^T{\displaystyle \frac{1}{L^3}}(v·\nabla v,\varphi )dt,$$

we write $v·\nabla v$ as $v·\nabla v=[v-\varphi ]·\nabla v+\varphi ·\nabla v$ and split the integral into two terms. Since $\varphi$ is zero outside ${\mathcal{S}}_{\beta }$, we have

$$\begin{array}{c}NLT={\displaystyle \frac{1}{T}}{\int }_0^T{\displaystyle \frac{1}{L^3}}([v-\varphi ]·\nabla v,\varphi )+(\varphi ·\nabla v,\varphi )dt\\ \le {\displaystyle \frac{1}{T}}{\int }_0^T\left({\displaystyle \frac{1}{L^3}}{\int }_{{\mathcal{S}}_{\beta }}{|v-\varphi ||\nabla v|\left|\varphi \right|+\left|\varphi \right|}^2|\nabla v|dx\right)dt\\ \le {\displaystyle \frac{1}{T}}{\int }_0^T\left({\displaystyle \frac{1}{L^3}}{∥{\displaystyle \frac{v-\varphi }{L-z}}∥}_{L^2\left({\mathcal{S}}_{\beta }\right)}{\left|\right|\nabla v\left|\right|}_{L^2\left({\mathcal{S}}_{\beta }\right)}{\left|\right|(L-z)\varphi \left|\right|}_{L^{\infty }\left({\mathcal{S}}_{\beta }\right)}\right)dt\\ +{\displaystyle \frac{1}{L^3}}{\displaystyle \frac{1}{T}}{\int }_0^T{\left|\right|\varphi \left|\right|}_{L^{\infty }\left({\mathcal{S}}_{\beta }\right)}^2{\left|\right|\nabla v\left|\right|}_{L^1\left({\mathcal{S}}_{\beta }\right)}dt.\end{array}$$

In the third step, we have multiplied and divided by $\left(L-z\right)$ before applying Hölder’s inequality. On the RHS, ${\left|\right|\varphi \left|\right|}_{L^{\infty }\left({\mathcal{S}}_{\beta }\right)}^2=\varphi {\left(L\right)}^2=U^2$ and ${\left|\right|(L-z)\varphi \left|\right|}_{L^{\infty }\left({\mathcal{S}}_{\beta }\right)}={\displaystyle \frac{1}{4}}\beta LU.$ Since $v-\varphi$ vanishes on the $z=L$ boundary of $\partial {\mathcal{S}}_{\beta }$, Hardy’s inequality, (8) in Section 2, the triangle inequality and a calculation imply

$$\begin{array}{cc}\hfill {∥{\displaystyle \frac{v-\varphi }{L-z}}∥}_{L^2\left({\mathcal{S}}_{\beta }\right)}& \le 2{∥\nabla (v-\varphi )∥}_{L^2\left({\mathcal{S}}_{\beta }\right)}\le 2{∥\nabla v∥}_{L^2\left({\mathcal{S}}_{\beta }\right)}+2{∥\nabla \varphi ∥}_{L^2\left({\mathcal{S}}_{\beta }\right)}\hfill \\ \hfill & \le 2{∥\nabla v∥}_{L^2\left({\mathcal{S}}_{\beta }\right)}+2U\sqrt{{\displaystyle \frac{L}{\beta }}}.\hfill \end{array}$$

Thus, we have the estimate

$$\begin{array}{cc}\hfill NLT& \le {\displaystyle \frac{\beta LU}{4}}{\displaystyle \frac{1}{L^3}}{\displaystyle \frac{1}{T}}{\int }_0^T\left({2\left|\right|\nabla v\left|\right|}_{L^2\left({\mathcal{S}}_{\beta }\right)}^2+2U\sqrt{{\displaystyle \frac{L}{\beta }}}{\left|\right|v\left|\right|}_{L^2\left({\mathcal{S}}_{\beta }\right)}\right)dt\hfill \\ \hfill & +{\displaystyle \frac{1}{L^3}}\left({\displaystyle \frac{1}{T}}{\int }_0^T{\displaystyle \frac{U^2}{L^3}}{\left|\right|\nabla v\left|\right|}_{L^1\left({\mathcal{S}}_{\beta }\right)}dt\right).\hfill \end{array}$$

(15)

For the last term on the RHS, Hölders inequality in space then in time implies

$$\begin{array}{cc}\hfill {\displaystyle \frac{U^2}{L^3}}{\displaystyle \frac{1}{T}}{\int }_0^T{\int }_{{\mathcal{S}}_{\beta }}|\nabla v|·1dxdt& \le {\displaystyle \frac{U^2}{L^3}}{\displaystyle \frac{1}{T}}{\int }_0^T\sqrt{{\int }_{{\mathcal{S}}_{\beta }}{|\nabla v|}^{2}dx}\sqrt{\beta L^3}dt\hfill \\ \hfill & \le {\displaystyle \frac{U^2\sqrt{\beta }}{L^{3/2}}}{\displaystyle \frac{1}{T}}{\int }_0^{T}1·\sqrt{{\int }_{{\mathcal{S}}_{\beta }}{|\nabla v|}^{2}dx}dt\hfill \\ \hfill & \le {\displaystyle \frac{U^2\sqrt{\beta }}{L^{3/2}}}{\left({\displaystyle \frac{1}{T}}{\int }_0^T{\int }_{{\mathcal{S}}_{\beta }}{|\nabla v|}^{2}dxdt\right)}^{1/2}.\hfill \end{array}$$

Increase the integral from ${\mathcal{S}}_{\beta }$ to $\mathrm{\Omega }$, and use $\beta ={\displaystyle \frac{1}{8}}\mathcal{R}{e}_{eff}^{-1}.$ Rearranging and using the arithmetic-geometric mean inequality gives an estimate useful for the last term in (15):

$$\begin{array}{c}{\displaystyle \frac{U^2}{L^3}}{\displaystyle \frac{1}{T}}{\int }_0^T{\left|\right|\nabla v\left|\right|}_{L^1\left({\mathcal{S}}_{\beta }\right)}dt\le U^2\sqrt{\beta }{\left({\displaystyle \frac{1}{T}}{\int }_0^T{\displaystyle \frac{1}{L^3}}{\int }_{\mathrm{\Omega }}{|\nabla v|}^{2}dxdt\right)}^{1/2}\\ \le U^2\sqrt{{\displaystyle \frac{1}{8}}{\displaystyle \frac{1}{LU}}}{\left({\displaystyle \frac{1}{T}}{\int }_0^T{\displaystyle \frac{1}{L^3}}{\int }_{\mathrm{\Omega }}{\nu }_{eff}{|\nabla v|}^{2}dxdt\right)}^{1/2}\\ \le {\left({\displaystyle \frac{U^3}{L}}\right)}^{1/2}{\left({\displaystyle \frac{1}{8}}{\displaystyle \frac{1}{T}}{\int }_0^T{\displaystyle \frac{1}{L^3}}{\int }_{\mathrm{\Omega }}{\nu }_{eff}{|\nabla v|}^{2}dxdt\right)}^{1/2}\\ \le {\displaystyle \frac{1}{2}}{\displaystyle \frac{U^3}{L}}+{\displaystyle \frac{1}{16}}{\displaystyle \frac{1}{T}}{\int }_0^T\left({\displaystyle \frac{1}{L^3}}{\int }_{\mathrm{\Omega }}{\nu }_{eff}{|\nabla v|}^{2}dx\right)dt.\end{array}$$

Similar manipulations (using $\beta ={\displaystyle \frac{1}{8}}{\displaystyle \frac{{\nu }_{eff}}{LU}}$) yield an estimate for the second term on the RHS in (15):

$$\begin{array}{c}{\displaystyle \frac{\beta LU}{4}}{\displaystyle \frac{1}{L^3}}{\displaystyle \frac{1}{T}}{\int }_0^T\left(2U\sqrt{{\displaystyle \frac{L}{\beta }}}{\left|\right|v\left|\right|}_{L^2\left({\mathcal{S}}_{\beta }\right)}\right)dt\le {\displaystyle \frac{\beta LU}{2}}{\displaystyle \frac{1}{T}}{\int }_0^T{\displaystyle \frac{1}{L^3}}{\left|\right|\nabla v\left|\right|}_{L^2\left({\mathcal{S}}_{\beta }\right)}^{2}dt+{\displaystyle \frac{1}{8}}{\displaystyle \frac{U^3}{L}}\\ \le {\displaystyle \frac{1}{8}}{\displaystyle \frac{1}{T}}{\int }_0^T{\displaystyle \frac{1}{L^3}}{\nu }_{eff}{\left|\right|\nabla v\left|\right|}_{L^2\left(\mathrm{\Omega }\right)}^{2}dt+{\displaystyle \frac{1}{8}}{\displaystyle \frac{U^3}{L}}.\end{array}$$

The first term on the RHS is simplest:

$$\begin{array}{cc}\hfill {\displaystyle \frac{\beta LU}{4}}{\displaystyle \frac{1}{L^3}}{\displaystyle \frac{1}{T}}{\int }_0^T{2\left|\right|\nabla v\left|\right|}_{L^2\left({\mathcal{S}}_{\beta }\right)}^{2}dt& ={\displaystyle \frac{1}{16}}{\displaystyle \frac{1}{L^3}}{\displaystyle \frac{1}{T}}{\int }_0^T{\nu }_{eff}{\left|\right|\nabla v\left|\right|}_{L^2\left({\mathcal{S}}_{\beta }\right)}^{2}dt\hfill \\ \hfill & \le {\displaystyle \frac{1}{16}}{\displaystyle \frac{1}{T}}{\int }_0^T{\displaystyle \frac{{\nu }_{eff}}{L^3}}{\left|\right|\nabla v\left|\right|}_{L^2\left(\mathrm{\Omega }\right)}^{2}dt.\hfill \end{array}$$

Using the last three estimates in the $NLT$ upper bound (15), we obtain (term by term)

$$\begin{array}{c}NLT\le {\displaystyle \frac{1}{16}}{\displaystyle \frac{1}{T}}{\int }_0^T{\displaystyle \frac{{\nu }_{eff}}{L^3}}{\left|\right|\nabla v\left|\right|}_{L^2\left(\mathrm{\Omega }\right)}^{2}dt+{\displaystyle \frac{1}{8}}{\displaystyle \frac{1}{T}}{\int }_0^T{\displaystyle \frac{{\nu }_{eff}}{L^3}}{\left|\right|\nabla v\left|\right|}_{L^2\left(\mathrm{\Omega }\right)}^{2}dt\\ +{\displaystyle \frac{1}{8}}{\displaystyle \frac{U^3}{L}}+{\displaystyle \frac{1}{2}}{\displaystyle \frac{U^3}{L}}+{\displaystyle \frac{1}{16}}{\displaystyle \frac{1}{T}}{\int }_0^T{\displaystyle \frac{{\nu }_{eff}}{L^3}}{\left|\right|\nabla v\left|\right|}_{L^2\left(\mathrm{\Omega }\right)}^{2}dt\\ or:NLT\le {\displaystyle \frac{1}{4}}{\displaystyle \frac{1}{T}}{\int }_0^T{\displaystyle \frac{{\nu }_{eff}}{L^3}}{\left|\right|\nabla v\left|\right|}_{L^2\left(\mathrm{\Omega }\right)}^{2}dt+{\displaystyle \frac{5}{8}}{\displaystyle \frac{U^3}{L}}.\end{array}$$

Thus, from (14),

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{T}}{\int }_0^T{\epsilon }_{\mathrm{model}}dt& \le \mathcal{O}({\displaystyle \frac{1}{T}})+{\displaystyle \frac{1}{4}}{\displaystyle \frac{1}{T}}{\int }_0^T{\displaystyle \frac{{\nu }_{eff}}{L^3}}{\left|\right|\nabla v\left|\right|}_{L^2\left(\mathrm{\Omega }\right)}^{2}dt+{\displaystyle \frac{5}{8}}{\displaystyle \frac{U^3}{L}}\hfill \\ \hfill & +{\displaystyle \frac{1}{T}}{\int }_0^T\left({\displaystyle \frac{1}{L^3}}{\int }_{\mathrm{\Omega }}[\nu +{\nu }_{turb}]\nabla v:\nabla \varphi dx\right)dt.\hfill \end{array}$$

Consider now the last term on the RHS. Since $\varphi$ is zero off ${\mathcal{S}}_{\beta }$ and $\nabla \varphi ={\displaystyle \frac{U}{\beta L}}$ on ${\mathcal{S}}_{\beta }$, we have

$$\begin{array}{c}{\displaystyle \frac{1}{T}}{\int }_0^T{\displaystyle \frac{1}{L^3}}{\int }_{\mathrm{\Omega }}[\nu +{\nu }_{turb}]\nabla v:\nabla \varphi dxdt={\displaystyle \frac{1}{T}}{\int }_0^T{\displaystyle \frac{1}{L^3}}{\int }_{{\mathcal{S}}_{\beta }}[\nu +{\nu }_{turb}]\nabla v:\nabla \varphi dxdt\\ \le {\displaystyle \frac{1}{2}}{\displaystyle \frac{1}{T}}{\int }_0^T{\epsilon }_{\mathrm{model}}dt+{\displaystyle \frac{1}{2}}{\displaystyle \frac{1}{T}}{\int }_0^T\left({\displaystyle \frac{1}{L^3}}{\int }_{{\mathcal{S}}_{\beta }}[\nu +{\nu }_{turb}]{\left({\displaystyle \frac{U}{\beta L}}\right)}^{2}dx\right)dt\\ \le {\displaystyle \frac{1}{2}}{\displaystyle \frac{1}{T}}{\int }_0^T{\epsilon }_{\mathrm{model}}dt+{\displaystyle \frac{1}{2}}{\left({\displaystyle \frac{U}{\beta L}}\right)}^2\beta {\displaystyle \frac{1}{T}}{\int }_0^T\left({\displaystyle \frac{1}{\beta L^3}}{\int }_{{\mathcal{S}}_{\beta }}\nu +{\nu }_{turb}dx\right)dt.\end{array}$$

Thus, as $\beta ={\displaystyle \frac{1}{8}}\mathcal{R}{e}_{eff}^{-1}$ implies $2\beta \mathcal{R}{e}_{eff}=1/4$,

$$\begin{array}{c}{\displaystyle \frac{1}{2}}{\displaystyle \frac{1}{T}}{\int }_0^T{\epsilon }_{\mathrm{model}}dt\le \mathcal{O}({\displaystyle \frac{1}{T}})+{\displaystyle \frac{1}{4}}{\displaystyle \frac{1}{T}}{\int }_0^T{\displaystyle \frac{{\nu }_{eff}}{L^3}}{\left|\right|\nabla v\left|\right|}^{2}dt\\ +{\displaystyle \frac{5}{8}}{\displaystyle \frac{U^3}{L}}+{\displaystyle \frac{\beta }{2}}{\left({\displaystyle \frac{U}{\beta L}}\right)}^2{\displaystyle \frac{1}{T}}{\int }_0^T{\displaystyle \frac{1}{\beta L^3}}{\int }_{{\mathcal{S}}_{\beta }}\nu +{\nu }_{turb}dxdt.\end{array}$$

As $T_j\to \infty$,

$${\displaystyle \frac{1}{T}}{\int }_0^T{\displaystyle \frac{{\nu }_{eff}}{L^3}}{\left|\right|\nabla v\left|\right|}^{2}dt\to {〈{\epsilon }_{\mathrm{model}}〉}_{\infty },$$

and we calculate

$${\displaystyle \frac{\beta }{2}}{\left({\displaystyle \frac{U}{\beta L}}\right)}^2{\displaystyle \frac{1}{T}}{\int }_0^T{\displaystyle \frac{1}{\beta L^3}}{\int }_{{\mathcal{S}}_{\beta }}\nu dxdt=\nu {\displaystyle \frac{\beta }{2}}{\left({\displaystyle \frac{U}{\beta L}}\right)}^2=4{\displaystyle \frac{\nu }{{\nu }_{eff}}}{\displaystyle \frac{U^3}{L}}.$$

Thus,

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{2}}{\displaystyle \frac{1}{T}}{\int }_0^T{\epsilon }_{\mathrm{model}}dt& \le \mathcal{O}({\displaystyle \frac{1}{T}})+{\displaystyle \frac{1}{4}}{\displaystyle \frac{1}{T}}{\int }_0^T{\displaystyle \frac{{\nu }_{eff}}{L^3}}{\left|\right|\nabla v\left|\right|}_{L^2\left(\mathrm{\Omega }\right)}^{2}dt\hfill \\ \hfill & +(4{\displaystyle \frac{\nu }{{\nu }_{eff}}}+{\displaystyle \frac{5}{8}}){\displaystyle \frac{U^3}{L}}+\beta {\left({\displaystyle \frac{U}{\beta L}}\right)}^2{\displaystyle \frac{1}{T}}{\int }_0^T{\displaystyle \frac{1}{\beta L^3}}{\int }_{{\mathcal{S}}_{\beta }}{\nu }_{turb}dxdt.\hfill \end{array}$$

The last term on the RHS is rearranged to be

$$\beta {\left({\displaystyle \frac{U}{\beta L}}\right)}^2{\displaystyle \frac{1}{T}}{\int }_0^T{\displaystyle \frac{1}{\beta L^3}}{\int }_{{\mathcal{S}}_{\beta }}{\nu }_{turb}dxdt=8\left[{\displaystyle \frac{1}{T}}{\int }_0^T{\displaystyle \frac{1}{|{\mathcal{S}}_{\beta }|}}{\int }_{{\mathcal{S}}_{\beta }}{\displaystyle \frac{{\nu }_{turb}}{{\nu }_{eff}}}dxdt\right]{\displaystyle \frac{U^3}{L}},$$

The proof is completed by taking the limit superior as $T\to \infty$. This gives

$${〈{\epsilon }_{\mathrm{model}}〉}_{\infty }\le \left({\displaystyle \frac{5}{2}}+16{\displaystyle \frac{\nu }{{\nu }_{eff}}}+32{〈{\displaystyle \frac{1}{|{\mathcal{S}}_{\beta }|}}{\int }_{{\mathcal{S}}_{\beta }}{\displaystyle \frac{{\nu }_{turb}}{{\nu }_{eff}}}dx〉}_{\infty }\right){\displaystyle \frac{U^3}{L}}.$$

□

The above theorem shows that the behavior of the turbulent viscosity in the near-wall region is a critical factor in model dissipation. The near-wall term is estimated by a precise use of the following lemma which is an application of the $L^p-L^q$ Hardy inequality (9).

Lemma 2.

For $F(x,y,z)\in H^1\left({\mathcal{S}}_{\beta }\right)$ with $F(x,y,L)=0$, there holds

$${\left({\int }_{(1-\beta )L}^L{|L-z|}^{-4}{\left|F(x,y,z)\right|}^{6}dz\right)}^{1/6}\le C_{26}{\left({\int }_{(1-\beta )L}^L{\left|F_z(x,y,z)\right|}^{2}dz\right)}^{1/2}$$

(16)

Proof. 

First, note that for $F\left(x\right)$ vanishing at $x=0$, there holds

$${\left({\int }_0^{\infty }{\left|{\displaystyle \frac{F\left(x\right)}{x^{2/3}}}\right|}^{6}dx\right)}^{1/6}\le C_{26}{\left({\int }_0^{\infty }{\left|F_x\left(x\right)\right|}^{2}dx\right)}^{1/2}.$$

This follows from (9) by letting $F\left(x\right)={\int }_0^{x}f\left(t\right)dt$ and choosing $q=6, p=2$, and $\alpha =$$-4$. These choices satisfy $\left(\alpha +1\right)/6=(1/2)-1$ and $1<p\le q<\infty$ as required for (9). We apply this result where F is a function of $x,y,z$ defined on ${\mathcal{S}}_{\beta }$, the variable x is replaced by z and a linear change of variables is made so the zero boundary condition is imposed at $z=L$ rather than $z=0$. The result is

$${\left({\int }_{(1-\beta )L}^L{|L-z|}^{-4}{\left|F(x,y,z)\right|}^{6}dz\right)}^{1/6}\le C_{26}{\left({\int }_{(1-\beta )L}^L{\left|F_z(x,y,z)\right|}^{2}dz\right)}^{1/2}$$

□

This lemma is used to estimate the integral

$${\int }_{{\mathcal{S}}_{\beta }}{\nu }_{turb}dx=\mu \tau {\int }_{{\mathcal{S}}_{\beta }}{\left|u^{\prime }(x,t)\right|}_e^{2}dx.$$

Lemma 3.

We have

$$\begin{array}{cc}\hfill {\int }_{{\mathcal{S}}_{\beta }}{\left|u^{\prime }(x,y,z,t;{\omega }_j)\right|}^{2}dx& \le 3^{-2/3}{C}_{26}^2{\beta }^2{L}^2{\int }_{{\mathcal{S}}_{\beta }}{\left|{\displaystyle \frac{\partial u^{\prime }}{\partial z}}(x,y,z,t;{\omega }_j)\right|}^{2}dx,\hfill \\ \hfill {\int }_{{\mathcal{S}}_{\beta }}{\left|u^{\prime }(x,y,z,t)\right|}_e^{2}dx& \le 3^{-2/3}{C}_{26}^2{\beta }^2{L}^2{\int }_{{\mathcal{S}}_{\beta }}{〈{\left|{\displaystyle \frac{\partial u^{\prime }}{\partial z}}(x,y,z,t;{\omega }_j)\right|}^2〉}_{e}dx\hfill \end{array}$$

and thus

$${\int }_{{\mathcal{S}}_{\beta }}{\nu }_{turb}dx\le \mu \tau 3^{-2/3}{C}_{26}^2{\beta }^2{L}^2{\int }_{{\mathcal{S}}_{\beta }}{〈{\left|{\displaystyle \frac{\partial u^{\prime }}{\partial z}}(x,y,z,t;{\omega }_j)\right|}^2〉}_{e}dx$$

Proof. 

We have for each ${\omega }_j$, by Hölders inequality in the z integral,

$$\begin{array}{c}{\int }_{{\mathcal{S}}_{\beta }}{\left|u^{\prime }\right|}^{2}dx={\int }_0^L{\int }_0^L\left[{\int }_{(1-\beta )L}^L{\left(L-z\right)}^{4/3}{\left|{\displaystyle \frac{u^{\prime }(x,y,z,t;{\omega }_j)}{{\left(L-z\right)}^{2/3}}}\right|}^{2}dz\right]dxdy\le \\ {\int }_0^L{\int }_0^L{\left[{\int }_{(1-\beta )L}^L{\left({\left(L-z\right)}^{4/3}\right)}^{3/2}dz\right]}^{{\displaystyle \frac{2}{3}}}{\left[{\int }_{(1-\beta )L}^L{\left|{\displaystyle \frac{u^{\prime }(x,y,z,t;{\omega }_j)}{{\left(L-z\right)}^{2/3}}}\right|}^{6}dz\right]}^{{\displaystyle \frac{1}{3}}}dxdy\\ :=\int \int A·Bdxdy.\end{array}$$

The first integral (A) on the RHS is

$$A={\left[{\int }_{(1-\beta )L}^L{\left(L-z\right)}^{2}dz\right]}^{2/3}={\left[{\displaystyle \frac{{\left(\beta L\right)}^3}{3}}\right]}^{2/3}=3^{-2/3}{\beta }^2{L}^2.$$

For the second integral (B), we apply (16) from the last Lemma:

$$\begin{array}{cc}\hfill B& ={\left[{\int }_{(1-\beta )L}^L{\left(L-z\right)}^{-4}{\left|u^{\prime }(x,y,z,t;{\omega }_j)\right|}^{6}dz\right]}^{1/3}\hfill \\ \hfill & \le C_{26}^2{\int }_{(1-\beta )L}^L{\left|{\displaystyle \frac{\partial u^{\prime }}{\partial z}}(x,y,z,t;{\omega }_j)\right|}^{2}dz,\hfill \end{array}$$

In combination, we have

$${\int }_{{\mathcal{S}}_{\beta }}{\left|u^{\prime }(x,t;{\omega }_j)\right|}^{2}dx\le \left(3^{-2/3}{\beta }^2{L}^2\right)C_{26}^2{\int }_0^L{\int }_0^L{\int }_{(1-\beta )L}^L{\left|{\displaystyle \frac{\partial u^{\prime }}{\partial z}}(x,t;{\omega }_j)\right|}^{2}dzdxdy$$

which is the first estimate. The second follows by taking the ensemble average of the first. □

We can now estimate the last term on the RHS of (12).

Proposition 2.

We have

$$\begin{array}{c}32{〈{\displaystyle \frac{1}{|{\mathcal{S}}_{\beta }|}}{\int }_{{\mathcal{S}}_{\beta }}{\displaystyle \frac{{\nu }_{turb}}{{\nu }_{eff}}}dx〉}_{\infty }{\displaystyle \frac{U^3}{L}}\\ \le {\displaystyle \frac{\sqrt[3]{3}}{6}}{C}_{26}^2\mu {\displaystyle \frac{\tau }{T^{*}}}{〈{\displaystyle \frac{1}{|{\mathcal{S}}_{\beta }|}}{\int }_{{\mathcal{S}}_{\beta }}{\nu }_{eff}{〈{\left|{\displaystyle \frac{\partial u^{\prime }}{\partial z}}\right|}^2〉}_{e}dx〉}_{\infty }\\ \le {\displaystyle \frac{\sqrt[3]{3}}{6}}{C}_{26}^2\mu {\displaystyle \frac{\tau }{T^{*}}}{〈{\displaystyle \frac{1}{|{\mathcal{S}}_{\beta }|}}{\int }_{{\mathcal{S}}_{\beta }}{\nu }_{eff}{〈{\left|\nabla u\right|}^2〉}_{e}dx〉}_{\infty }.\end{array}$$

Proof. 

From the above estimates, we have

$$\begin{array}{c}32{〈{\displaystyle \frac{1}{|{\mathcal{S}}_{\beta }|}}{\int }_{{\mathcal{S}}_{\beta }}{\displaystyle \frac{{\nu }_{turb}}{{\nu }_{eff}}}dx〉}_{\infty }{\displaystyle \frac{U^3}{L}}\\ \le \left(32·3^{-2/3}{C}_{26}^2\right){\beta }^2{L}^2{\displaystyle \frac{\mu \tau }{{\nu }_{eff}}}{\displaystyle \frac{1}{|{\mathcal{S}}_{\beta }|}}{\int }_{{\mathcal{S}}_{\beta }}{〈{\left|{\displaystyle \frac{\partial u^{\prime }}{\partial z}}\right|}^2〉}_{e}dx{\displaystyle \frac{U^3}{L}}.\end{array}$$

Algebraic rearrangement of various multipliers in the above (using turnover time $T^{*}=L/U$) gives

$${\beta }^2{L}^2{\displaystyle \frac{\mu \tau }{{\nu }_{eff}}}{\displaystyle \frac{U^3}{L}}={\left({\displaystyle \frac{{\nu }_{eff}}{8LU}}\right)}^2{L}^2{\displaystyle \frac{1}{{\nu }_{eff}}}{\displaystyle \frac{\mu \tau }{T^{*}}}{\displaystyle \frac{L}{U}}{\displaystyle \frac{U^3}{L}}={\displaystyle \frac{1}{64}}{\nu }_{eff}{\displaystyle \frac{\mu \tau }{T^{*}}}.$$

Thus,

$$\begin{array}{c}32{〈{\displaystyle \frac{1}{|{\mathcal{S}}_{\beta }|}}{\int }_{{\mathcal{S}}_{\beta }}{\displaystyle \frac{{\nu }_{turb}}{{\nu }_{eff}}}dx〉}_{\infty }{\displaystyle \frac{U^3}{L}}\\ \le \left({\displaystyle \frac{32}{64}}·3^{-2/3}{C}_{26}^2\right)\mu {\displaystyle \frac{\tau }{T^{*}}}{\displaystyle \frac{1}{|{\mathcal{S}}_{\beta }|}}{\int }_{{\mathcal{S}}_{\beta }}{\nu }_{eff}{〈{\left|{\displaystyle \frac{\partial u^{\prime }}{\partial z}}\right|}^2〉}_{e}dx.\end{array}$$

The result now follows from

$${〈{\left|{\displaystyle \frac{\partial u^{\prime }}{\partial z}}\right|}^2〉}_e\le {〈{\left|{\displaystyle \frac{\partial u}{\partial z}}\right|}^2〉}_e\le {〈{\left|\nabla u\right|}^2〉}_e.$$

□

To develop the numerical value for ${\displaystyle \frac{\sqrt[3]{3}}{6}}{C}_{26}^2\simeq$ $0.23092$, we begin with ${\displaystyle \frac{\sqrt[3]{3}}{6}}\simeq 0.24037$ and, from Bliss [26], with $p=p^{\prime }=2,q=6$:

$$C_{pq}={\left({\displaystyle \frac{p^{\prime }}{q}}\right)}^{1/p}\left({\displaystyle \frac{\frac{q-p}{p}\mathrm{\Gamma }\left(\frac{pq}{q-p}\right)}{\mathrm{\Gamma }\left(\frac{p}{q-p}\right)\mathrm{\Gamma }\left(\frac{p(q-1)}{q-p}\right)}}\right)$$

(17)

so that, to 5 significant digits,

$$C_{26}={\displaystyle \frac{1}{\sqrt{3}}}\left({\displaystyle \frac{2\mathrm{\Gamma }\left(3\right)}{\mathrm{\Gamma }\left(\frac{1}{2}\right)\mathrm{\Gamma }\left(\frac{5}{2}\right)}}\right)\simeq 0.98014.$$

We can now give another result for the time-averaged energy dissipation rate.

Theorem 5.

Any weak solution of the eddy viscosity model (1) satisfying the energy inequality (3) has its model energy dissipation bounded as

$$\begin{array}{c}{〈{\epsilon }_{model}〉}_{\infty }\le \left({\displaystyle \frac{5}{2}}+16{\displaystyle \frac{\nu }{{\nu }_{eff}}}\right){\displaystyle \frac{U^3}{L}}\end{array}$$

$$\begin{array}{c}+\left({\displaystyle \frac{\sqrt[3]{3}}{6}}{C}_{26}^2\right)\mu {\displaystyle \frac{\tau }{T^{*}}}{〈{\displaystyle \frac{1}{|{\mathcal{S}}_{\beta }|}}{\int }_{{\mathcal{S}}_{\beta }}{\nu }_{eff}{〈{\left|\nabla u\right|}^2〉}_{e}dx〉}_{\infty }\\ and\\ {〈{\epsilon }_{model}〉}_{\infty }\le \left({\displaystyle \frac{5}{2}}+16{\displaystyle \frac{\nu }{{\nu }_{eff}}}\right){\displaystyle \frac{U^3}{L}}\end{array}$$

(18)

$$\begin{array}{c}+\left({\displaystyle \frac{\sqrt[3]{3}}{6}}{C}_{26}^2\right)\mu {\displaystyle \frac{\tau }{T^{*}}}{〈{\displaystyle \frac{1}{|{\mathcal{S}}_{\beta }|}}{\int }_{{\mathcal{S}}_{\beta }}{\nu }_{eff}{〈{\left|{\displaystyle \frac{\partial u^{\prime }}{\partial z}}\right|}^2〉}_{e}dx〉}_{\infty }\end{array}$$

(19)

Proof. 

This follows by combining the previous estimates. □

To obtain a closed estimate we now consider choosing a different $\mu$-value, $\mu ={\mu }_{\beta }$, in the near-wall region than away from walls. This becomes necessary because the integral over ${\mathcal{S}}_{\beta }$ must be increased to one over $\mathrm{\Omega }$.

Theorem 6.

Suppose μ is one constant value in the flow interior and $\mu ={\mu }_{\beta }$ in ${\mathcal{S}}_{\beta }$. If ${\mu }_{\beta }\le 0.27064\mathcal{R}{e}^{-1}$, then

$${〈{\epsilon }_{model}〉}_{\infty }\le \left(5+32{\displaystyle \frac{\nu }{{\nu }_{eff}}}\right){\displaystyle \frac{U^3}{L}}.$$

Proof. 

To begin, note that if $\mu$ is piecewise constant, the previous estimates hold with $\mu \left(x\right)$ inside the volume integral or the volume integral split into sub-regions ${\mathcal{S}}_{\beta }$ and $\mathrm{\Omega }-{\mathcal{S}}_{\beta }$. We estimate the last term on the RHS of (18) (increasing the integral to over $\mathrm{\Omega }$) as follows:

$$\begin{array}{c}\left({\displaystyle \frac{\sqrt[3]{3}}{6}}{C}_{26}^2\right){\mu }_{\beta }{\displaystyle \frac{\tau }{T^{*}}}{〈{\displaystyle \frac{1}{|{\mathcal{S}}_{\beta }|}}{\int }_{{\mathcal{S}}_{\beta }}{\nu }_{eff}{〈{\left|\nabla u\right|}^2〉}_{e}dx〉}_{\infty }\le \\ \left({\displaystyle \frac{8\sqrt[3]{3}}{6}}{C}_{26}^2\right){\mu }_{\beta }{\displaystyle \frac{\tau }{T^{*}}}\mathcal{R}{e}_{eff}{〈{\displaystyle \frac{1}{L^3}}{\int }_{\mathrm{\Omega }}{\nu }_{eff}{〈{\left|\nabla u\right|}^2〉}_{e}dx〉}_{\infty }\\ \le \left({\displaystyle \frac{8\sqrt[3]{3}}{6}}{C}_{26}^2\right){\mu }_{\beta }{\displaystyle \frac{\tau }{T^{*}}}\mathcal{R}{e}_{eff}{〈{\epsilon }_{\mathrm{model}}〉}_{\infty }=\left({\displaystyle \frac{8\sqrt[3]{3}}{6}}{C}_{26}^2\right){\mu }_{\beta }{\displaystyle \frac{\tau }{T^{*}}}\mathcal{R}e\left({\displaystyle \frac{\nu }{{\nu }_{eff}}}\right){〈{\epsilon }_{\mathrm{model}}〉}_{\infty }\end{array}$$

In the above calculation, the multipliers 8 and $\mathcal{R}{e}_{eff}$ arise because

$${\displaystyle \frac{1}{|{\mathcal{S}}_{\beta }|}}={\displaystyle \frac{1}{\beta L^3}}={\beta }^{-1}{\displaystyle \frac{1}{L^3}}=8\mathcal{R}{e}_{eff}{\displaystyle \frac{1}{L^3}}.$$

The multiplier (to 5 digits) ${\displaystyle \frac{8C_{26}^2\sqrt[3]{3}}{6}}\simeq 1.8474$. Note that

$${\displaystyle \frac{\nu }{{\nu }_{eff}}}\le 1\mathrm{and}{\displaystyle \frac{\tau }{T^{*}}}\le 1.$$

Thus, if ${\mu }_{\beta }$ is chosen so that

$${\displaystyle \frac{8C_{26}^2\sqrt[3]{3}}{6}}{\mu }_{\beta }\mathcal{R}e\le {\displaystyle \frac{1}{2}}\mathrm{implied}\mathrm{by}{\mu }_{\beta }\le 0.27064\mathcal{R}{e}^{-1}$$

it follows that

$${〈{\epsilon }_{\mathrm{model}}〉}_{\infty }\le \left(5+32{\displaystyle \frac{\nu }{{\nu }_{eff}}}\right){\displaystyle \frac{U^3}{L}}\left(\le 37{\displaystyle \frac{U^3}{L}}\right).$$

□

4. Conclusions

We have analyzed wall effects on energy dissipation rates for the ensemble eddy viscosity model. The results herein are only a first step. There are a number of important mathematical challenges (open problems) that remain:

• The constant multipliers in the results are large as a result of the number of estimates employed. Significant reduction would bring the analytical estimates closer to experimental data.
• The model is based on a turbulence length scale $l(x,t)=|u^{\prime }{(x,t)|}_e\tau$. Since this is an $L^2$ function, there is no mathematical reason preventing this length scale predicting (locally) eddies farther apart than the domain size! In [1], to prove the existence of a model solution, it was found necessary to cap l at the domain size by $l(x,t)=\min \left\{\right|u^{\prime }(x,t){|}_e\tau ,L\}.$ This can even be reasonably modified to $l(x,t)=\min \left\{\right|u^{\prime }(x,t){|}_e\tau ,d\left(x\right)\}$. Analytical evaluation of the effects of such caps on energy dissipation rates is another important open problem.
• We conjecture that the assumption ${\mu }_{\beta }\lesssim \mathcal{R}{e}^{-1}$ is not necessary. However, the work herein suggests that dropping ${\mu }_{\beta }\lesssim \mathcal{R}{e}^{-1}$ will require a new idea at some point.
• Based on the results in [12] (that $J=16$ was large enough to capture low-order flow statistics), we have considered J fixed. The analysis of the limit $J\to \infty$ is an important open problem as is testing and experiments with the model for turbulent channel flows.
• For shear flow, existence of weak solutions and their energy inequality are open problems in analysis.
• The model herein, like all EV models, is dissipative and thus cannot account for the intermittent transfer of energy from unresolved fluctuations back to the mean velocity. Attempts have been made based on negative viscosities, but these cannot be correct in principle. A more correct approach is through an exact equation for variance evolution, developed in [3]. Analysis of these model extensions to account for intermittence is an open problem.