Statistical solutions and Onsager's conjecture

TL;DR

This paper proves a version of Onsager's conjecture for statistical solutions of the incompressible Euler equations, providing new insights into energy conservation and offering a novel proof for weak solutions.

Contribution

It introduces a statistical solutions framework for Onsager's conjecture and establishes energy conservation results within this context, extending previous work on weak solutions.

Findings

Proves Onsager's conjecture for statistical solutions.

Provides a new proof for the conservative direction of Onsager's conjecture.

Enhances understanding of energy conservation in fluid dynamics.

Abstract

We prove a version of Onsager's conjecture on the conservation of energy for the incompressible Euler equations in the context of statistical solutions, as introduced recently by Fjordholm et al. As a byproduct, we also obtain a new proof for the conservative direction of Onsager's conjecture for weak solutions. Dedicated to Edriss S. Titi on the occasion of his 60th birthday.

Equations

$$\begin{array}{rl}{\partial }_{t}v+\sum _k{\partial }_{x^k}(v{v}^k)+\nabla p=0& x\in D,t>0\\ \nabla \cdot v=0& x\in D,t>0\\ v(0,x)=v_0(x)& x\in D.\end{array}$$

$${\int }_{{\mathbb{R}}_{+}}{\int }_{D}v{\partial }_t\phi +\sum _{k}v{v}^k{\partial }_{x^k}\phi +p\nabla \phi dxdt+{\int }_D{v}_0(x)\phi (0,x)dx=0$$

$${\partial }_t\frac{|v{|}^2}{2}+\sum _k{\partial }_{x^k}\left(v_k\left(\frac{|v{|}^2}{2}+p\right)\right)=0.$$

$${\int }_D\frac{|v(T){|}^2}{2}dx={\int }_D\frac{|v_0{|}^2}{2}dx.$$

$${\partial }_t{v}^i+\sum _k{\partial }_{x^k}(v^i{v}^k)+{\partial }_{x^i}p=0.$$

$$\begin{array}{r}{\partial }_t\left(v^i(x)v^j(y)\right)+\sum _k{\partial }_{x^k}\left(v^i(x)v^k(x)v^j(y)\right)+\sum _k{\partial }_{y^k}\left(v^i(x)v^k(y)v^j(y)\right)\\ +{\partial }_{x^i}\left(p(x)v^j(y)\right)+{\partial }_{y^j}\left(v^i(x)p(y)\right)=0\end{array}$$

$$\begin{array}{r}{\int }_0^T{\int }_D{\int }_D{v}^i(x)v^j(y){\partial }_t\phi +\sum _k{v}^i(x)v^k(x)v^j(y){\partial }_{x^k}\phi +\sum _k{v}^i(x)v^k(y)v^j(y){\partial }_{y^k}\phi \\ +p(x)v^j(y){\partial }_{x^i}\phi +v^i(x)p(y){\partial }_{y^j}\phi dxdydt+{\int }_D{\int }_D{v}_0^i(x)v_0^j(y)\phi (0,x,y)dxdy=0\end{array}$$

$$-\Delta p=\sum _{k,l}{\partial }_{x^k,x^l}^2(v^l{v}^k),$$

$$\liminf_{\varepsilon\to 0}\frac{1}{\varepsilon}\int_{0}^{T}\int_{D}\mathchoice{{\vbox{\hbox{$\textstyle-$ }}\kern-7.31105pt}}{{\vbox{\hbox{$\scriptstyle-$ }}\kern-5.50664pt}}{{\vbox{\hbox{$\scriptscriptstyle-$ }}\kern-4.4333pt}}{{\vbox{\hbox{$\scriptscriptstyle-$ }}\kern-3.96664pt}}\!\int_{B_{\varepsilon}(x)}\big{|}v(x)-v(y)\big{|}^{3}\,dydxdt=0.$$

$$\begin{array}{r}{\int }_0^T{\int }_D{\int }_{D}v(x)\cdot v(y){\partial }_t\phi +v(x)\cdot v(y)\left(v(x)\cdot {\nabla }_x\phi +v(y)\cdot {\nabla }_y\phi \right)+p(x)v(y)\cdot {\nabla }_x\phi \\ +p(y)v(x)\cdot {\nabla }_y\phi dxdydt+{\int }_D{\int }_D{v}_0(x)\cdot v_0(y)\phi (0,x,y)dxdy=0.\end{array}$$

$${\partial }_t\phi ={\rho }_{\epsilon }{\partial }_t\psi ,{\nabla }_x\phi =\psi \nabla {\rho }_{\epsilon }+{\rho }_{\epsilon }\nabla \psi ,{\nabla }_y\phi =-\psi \nabla {\rho }_{\epsilon }.$$

$$\begin{split}\int_{0}^{T}\int_{D}\int_{D}v(x)\cdot v(y)\rho_{\varepsilon}\partial_{t}\psi+v(x)\cdot v(y)\bigl{(}v(x)-v(y)\bigr{)}\cdot\nabla\rho_{\varepsilon}\psi\\ +v(x)\cdot v(y)v(x)\cdot\nabla\psi\rho_{\varepsilon}+\big{(}p(x)v(y)-v(x)p(y)\big{)}\cdot\nabla\rho_{\varepsilon}\psi\\ +p(x)v(y)\cdot\nabla\psi\rho_{\varepsilon}\,dxdydt\end{split}$$

$$\begin{split}\int_{0}^{T}\int_{D}\int_{D}\partial_{t}\psi(t,x)\rho_{\varepsilon}(z)v(x)\cdot v(x-z)\,dxdzdt\\ +\int_{0}^{T}\int_{D}\int_{D}\psi(t,x)\nabla\rho_{\varepsilon}(z)\cdot\bigl{(}v(x)-v(x-z)\bigr{)}\bigl{(}v(x)\cdot v(x-z)\bigr{)}\,dxdzdt\\ +\int_{0}^{T}\int_{D}\int_{D}\nabla\psi(t,x)\rho_{\varepsilon}(z)\cdot v(x)\bigl{(}v(x)\cdot v(x-z)\bigr{)}\,dxdzdt\\ +\int_{0}^{T}\int_{D}\int_{D}\psi(t,x)\nabla\rho_{\varepsilon}(z)\cdot(p(x)v(x-z)-v(x)p(x-z))\,dxdzdt\\ +\int_{0}^{T}\int_{D}\int_{D}\nabla\psi(t,x)\rho_{\varepsilon}(z)\cdot v(x-z)p(x)\,dxdzdt=0.\end{split}$$

$${\mathcal{A}}_1$$

$${\mathcal{A}}_3+{\mathcal{A}}_5$$

$${\mathcal{A}}_{2,1}$$

$${\mathcal{A}}_{2,2}$$

$$\displaystyle\frac{\|\psi\|_{\infty}}{2}\int_{0}^{T}\int_{D}\int_{D}|\nabla\rho_{\varepsilon}(z)|\big{|}v(x)-v(x-z)\big{|}^{3}\ dzdxdt$$

$$\displaystyle\leqslant C\frac{1}{\varepsilon}\int_{0}^{T}\int_{D}\mathchoice{{\vbox{\hbox{$\textstyle-$ }}\kern-7.31105pt}}{{\vbox{\hbox{$\scriptstyle-$ }}\kern-5.50664pt}}{{\vbox{\hbox{$\scriptscriptstyle-$ }}\kern-4.4333pt}}{{\vbox{\hbox{$\scriptscriptstyle-$ }}\kern-3.96664pt}}\!\int_{B_{\varepsilon}(0)}\big{|}v(x)-v(x-z)\big{|}^{3}\ dzdxdt$$

$$\mathcal{A}_{2,2}=\frac{1}{2}\int_{0}^{T}\int_{D}\int_{D}\nabla\rho_{\varepsilon}(z)\cdot|v(x)|^{2}\bigl{(}v(x)-v(x-z)\bigr{)}\big{(}\psi(x)+\psi(x-z)\big{)}\,dxdzdt.$$

$$\mathcal{A}_{2,2}=\frac{1}{2}\int_{0}^{T}\int_{D}\int_{D}\rho_{\varepsilon}(z)|v(x)|^{2}\nabla\psi(t,x-z)\cdot\bigl{(}v(x)-v(x-z)\bigr{)}\,dxdzdt.$$

$$\mathcal{A}_{4}=\int_{0}^{T}\int_{D}\int_{D}p(x)\nabla\rho_{\varepsilon}(z)\cdot\big{(}\psi(x)v(x-z)-\psi(x+z)v(x+z)\big{)}\,dxdzdt.$$

$$\mathcal{A}_{4}=\int_{0}^{T}\int_{D}\int_{D}p(x)\rho_{\varepsilon}(z)\nabla\psi(x+z)\cdot v(x+z)\big{)}\,dxdzdt$$

$$\int_{D}\bigl{\langle}\nu^{1}_{x},\,|\xi|^{q}\bigr{\rangle}\,dx<\infty.$$

$$d_{\varepsilon}^{q}(\nu^{2}):=\left(\int_{D}\mathchoice{{\vbox{\hbox{$\textstyle-$ }}\kern-7.31105pt}}{{\vbox{\hbox{$\scriptstyle-$ }}\kern-5.50664pt}}{{\vbox{\hbox{$\scriptscriptstyle-$ }}\kern-4.4333pt}}{{\vbox{\hbox{$\scriptscriptstyle-$ }}\kern-3.96664pt}}\!\int_{B_{\varepsilon}(x)}\bigl{\langle}\nu^{2}_{x,y},\,|\xi_{1}-\xi_{2}|^{q}\bigr{\rangle}\,dydx\right)^{1/q}.$$

$$d_{\epsilon }^q({\nu }^2)$$

$$\displaystyle\leqslant\left(\int_{D}\mathchoice{{\vbox{\hbox{$\textstyle-$ }}\kern-7.31105pt}}{{\vbox{\hbox{$\scriptstyle-$ }}\kern-5.50664pt}}{{\vbox{\hbox{$\scriptscriptstyle-$ }}\kern-4.4333pt}}{{\vbox{\hbox{$\scriptscriptstyle-$ }}\kern-3.96664pt}}\!\int_{B_{\varepsilon}(x)}\bigl{\langle}\nu^{2}_{x,y},\,|\xi_{1}|^{q}\bigr{\rangle}\,dydx\right)^{1/q}+\left(\int_{D}\mathchoice{{\vbox{\hbox{$\textstyle-$ }}\kern-7.31105pt}}{{\vbox{\hbox{$\scriptstyle-$ }}\kern-5.50664pt}}{{\vbox{\hbox{$\scriptscriptstyle-$ }}\kern-4.4333pt}}{{\vbox{\hbox{$\scriptscriptstyle-$ }}\kern-3.96664pt}}\!\int_{B_{\varepsilon}(x)}\bigl{\langle}\nu^{2}_{x,y},\,|\xi_{2}|^{q}\bigr{\rangle}\,dydx\right)^{1/q}$$

$$\displaystyle=2\left(\int_{D}\bigl{\langle}\nu^{1}_{x},\,|\xi|^{q}\bigr{\rangle}\,dx\right)^{1/q}<\infty,$$

$$d_{\epsilon }^q({\nu }^2)={\left({\int }_D\text{vbox}-{\int }_{B_{\epsilon }(x)}|u(x)-u(y){|}^{q}dydx\right)}^{1/q},$$

$$\int_{D}\bigl{\langle}\nu^{1}_{t,x},\,|v|^{q}\bigr{\rangle}\,dx\leqslant c\qquad\text{for a.e.\ }t\in{\mathcal{T}}.$$

$$d_{\varepsilon}^{q}(\nu^{2}):=\left(\int_{0}^{T}\int_{D}\mathchoice{{\vbox{\hbox{$\textstyle-$ }}\kern-7.31105pt}}{{\vbox{\hbox{$\scriptstyle-$ }}\kern-5.50664pt}}{{\vbox{\hbox{$\scriptscriptstyle-$ }}\kern-4.4333pt}}{{\vbox{\hbox{$\scriptscriptstyle-$ }}\kern-3.96664pt}}\!\int_{B_{\varepsilon}(x)}\bigl{\langle}\nu^{2}_{t,x,y},\,|\xi_{1}-\xi_{2}|^{q}\bigr{\rangle}\,dydxdt\right)^{1/q}.$$

$$\begin{cases}\displaystyle\int_{D}\bigl{\langle}\nu^{1}_{t,x},\,|v|^{2}\bigr{\rangle}\,dx\leqslant c&\text{for a.e.\ }t\in[0,T]\\ \displaystyle\int_{0}^{T}\int_{D}\bigl{\langle}\nu^{1}_{t,x},\,|v|^{3}+|p|^{3/2}\bigr{\rangle}\,dxdt<\infty\end{cases}$$

$$\lim_{\varepsilon\to 0}\int_{0}^{T}\int_{D}\mathchoice{{\vbox{\hbox{$\textstyle-$ }}\kern-7.31105pt}}{{\vbox{\hbox{$\scriptstyle-$ }}\kern-5.50664pt}}{{\vbox{\hbox{$\scriptscriptstyle-$ }}\kern-4.4333pt}}{{\vbox{\hbox{$\scriptscriptstyle-$ }}\kern-3.96664pt}}\!\int_{B_{\varepsilon}(x)}\bigl{\langle}\nu^{2}_{t,x,y},\,|v_{1}-v_{2}|^{2}+|v_{1}-v_{2}|^{3}+|p_{1}-p_{2}|^{3/2}\bigr{\rangle}\,dydxdt=0,$$

Full text

Dedicated to Edriss S. Titi on the occasion of his 60th birthday.

1 Introduction

We consider the $d$-dimensional incompressible Euler equations: Find a function $v=(v^{1},\dots,v^{d}):\mathbb{R}_{+}\times D\to\mathbb{R}^{d}$ and a function $p:\mathbb{R}_{+}\times D\to\mathbb{R}$ such that

[TABLE]

Here and below, the summation limits, when not specified, are always from $k=1$ to $k=d$. The initial data $v_{0}$ is assumed to lie in $L^{2}(D)$. The spatial parameter $x$ takes values in a set $D$, which we will take as either $\mathbb{R}^{d}$ or the ($d$-dimensional) torus $\mathbb{T}^{d}$111On domains with boundaries, the situation is more subtle: While one can show the local version of the energy equality with almost no further effort, it is no longer clear whether this implies also the global conservation of energy under the usual Besov regularity assumption.. The temporal domain is $[0,T]$ for some $T>0$. By a solution of the Euler equations we will mean a weak solution of (1.1), i.e. a function v\in L^{2}_{\textit{loc}}\big{(}[0,T]\times D;\mathbb{R}^{d}\big{)} such that

[TABLE]

for all $\varphi\in C_{c}^{\infty}([0,T)\times\mathbb{R}^{d})$, as well as satisfying the divergence free condition in the sense of distributions.

Assume that $v$ is a smooth solution of (1.1). Multiplying the first equation of (1.1) by $v$ gives

[TABLE]

Integrating this local energy identity over $x\in D$ and $t\in[0,T]$, we obtain the global energy identity

[TABLE]

In 1949 Lars Onsager conjectured [17] that if $v$ is Hölder continuous with exponent greater than $1/3$ then the above calculations can be made rigorous:

In fact it is possible to show that the velocity field in such “ideal” turbulence cannot obey any Lipschitz condition of the form (…) for any order $n$ greater than $1/3$; otherwise the energy is conserved.222As a historical sidenote, Onsager wrote down a formal proof of his own conjecture—close in spirit to the later proof by Duchon and Robert—which he never published; see [9]. Thus, his rather cryptic “it is possible to show” should in fact be interpreted literally rather than hypothetically.

As was shown in 1994 by Constantin, E and Titi [3], the conjecture is indeed true when $\frac{1}{3}$-Hölder continuity is replaced by $B_{3}^{\alpha,\infty}$ (Besov) regularity for any $\alpha>\frac{1}{3}$. The proof uses a regularization of (1.1) together with some basic estimates in Besov spaces. Independently, Eyink [8] proved the conjecture in Fourier space under a stronger assumption. Duchon and Robert [7] employed the regularization technique from [3] to quantify the anomalous energy dissipation $\mathcal{E}(u)$ of an arbitrary solution $u$—the amount by which equality in (1.3) fails to hold. The sharp exponent $\alpha=\frac{1}{3}$ was shown to suffice for energy conservation at the cost of a slightly stronger summability assumption for the Besov space in [2], see also [18]. More recently, related results were given for density-dependent Euler models in [16, 10, 6]. On the other hand, there is the question whether energy can be dissipated for any Hölder or Besov regularity below the exponent $\frac{1}{3}$. This difficult problem was solved only very recently [14, 1].

In [11] the authors developed the concept of statistical solutions of hyperbolic conservation laws, which are solutions of an evolution equation which incorporate uncertainty in the solution. The uncertainty can be due to errors in the initial or boundary data or—as is more relevant in the present setting—modeling in the context of turbulent flows. Statistical solutions were formulated as maps from time $t$ to probability measures $\mu_{t}$ on $L^{2}(D)$. The authors showed that any probability measure $\mu\in\mathcal{P}(L^{2}(D))$ can be described equivalently as a correlation measure, a hierarchy $\boldsymbol{\nu}=(\nu^{1},\nu^{2},\dots)$ in which each element $\nu^{k}$ gives the joint probability distribution $\nu^{k}_{x_{1},\dots,x_{k}}$ of the values $v(x_{1}),\dots,v(x_{k})$ for any choice of spatial points $x_{1},\dots,x_{k}\in D$. The evolution equation for $\mu_{t}$ is most naturally described in terms of its corresponding correlation measure, yielding an infinite hierarchy of evolution equations. In particular, the equation for the one-point distribution $\nu^{1}_{x}$ coincides with DiPerna’s definition of measure-valued solutions, see [4, 5, 19].

The purpose of the present paper is twofold. First, in Section 2 we provide an alternative, rather elementary proof of (local) energy conservation of solutions of the Euler equation which is close in spirit to the regularization technique in [7], but is also reminiscent of Kruzhkov’s doubling of variables technique [15]. Second, in Section 3 we formulate the concept of statistical solutions of the incompressible Euler equations (1.1), and we show that the proof of energy conservation is readily generalized to statistical solutions. We end by comparing the concepts of Besov regularity of functions and of correlation measures in Section 4.

Our main result (Theorem 3.8) says that an uncertain fluid flow—realized as a statistical solution of (1.1)—conserves energy provided it has more than $\frac{1}{3}$ of a derivative. The ease by which statements about regularity can be formulated using correlation measures indicates to us that statistical solutions—and not measure-valued solutions—are the right notion of solutions for uncertain (or unsteady) fluid flows. We refer the interested reader to the upcoming paper [12] where we discuss statistical solutions of the Navier–Stokes and Euler equations and the connection to Kolmogorov’s theory of turbulence (cf. also Remark 3.10).

2 Onsager’s conjecture

We first write the Euler equation in component form:

[TABLE]

It is straightforward to see that if $(v,p)$ is any smooth solution of the above equation, then the function $(t,x,y)\mapsto v^{i}(t,x)v^{j}(t,y)$ satisfies

[TABLE]

(where we suppress the dependence on $t$). The proof of this claim consists of evaluating (2.1) at $x$ and at $y$, multiplying the former by $v^{j}(y)$ and the latter by $v^{i}(x)$, and then summing the two. With only a bit more work, one shows that if $v$ is any weak solution of the Euler equation then (2.2) is satisfied in the distributional sense.

Lemma 2.1**.**

If $v:(0,T)\times D\to\mathbb{R}^{d}$ is a weak solution of the Euler equation then

[TABLE]

for every test function $\varphi=\varphi(t,x,y)\in C_{c}^{\infty}([0,T)\times D^{2})$ and every $i,j=1,\dots,d$.

If the local energy inequality (1.3) is to hold for a weak solution, then we need in addition v\in L^{3}_{loc}\big{(}[0,T)\times D;\mathbb{R}^{d}\big{)} and p\in L_{loc}^{3/2}\big{(}[0,T)\times D;\mathbb{R}^{d}\big{)} for the equality to make sense distributionally. The Besov regularity of $v$ required to show the equality entails in particular $v\in L^{3}$, and this in turn implies $p\in L^{3/2}$. Indeed, taking the divergence of the momentum equation in (1.1) we obtain

[TABLE]

and thus $v\in L^{3}$ implies $p\in L^{3/2}$ by standard elliptic theory.

Theorem 2.2**.**

Let v\in L^{\infty}\big{(}(0,T);L^{2}\big{(}D;\mathbb{R}^{d}\big{)}\big{)}\cap L^{3}((0,T)\times D;\mathbb{R}^{d}) be a weak solution of the Euler equations, and accordingly $p\in L^{3/2}((0,T)\times D;\mathbb{R}^{d})$. Assume that

[TABLE]

Then $(v,p)$ satisfies the local energy identity (1.3).

Proof.

Set $j=i$ in (2.3) and sum over $i$. We can then write the resulting identity as

[TABLE]

Fix a number $\varepsilon>0$. We choose now the test function $\varphi(t,x,y)=\rho_{\varepsilon}(x-y)\psi\left(t,x\right)$, where $\rho_{\varepsilon}(z)=\varepsilon^{-d}\rho(\varepsilon^{-1}z)$ for a nonnegative, rotationally symmetric mollifier $\rho\in C_{c}^{\infty}(D)$ with unit mass and support in $B_{0}(1)$ (the unit ball in $\mathbb{R}^{d}$) and $\psi\in C_{c}^{\infty}((0,T)\times D)$. Then

[TABLE]

Continuing from (2.6), we now have

[TABLE]

Making the change of variables $z=x-y$ gives

[TABLE]

Decompose the above into a sum of five terms $\mathcal{A}_{1}+\dots+\mathcal{A}_{5}$. By applying the Lebesgue differentiation theorem, it is easy to see that

[TABLE]

as $\varepsilon\to 0$. For $\mathcal{A}_{2}$ we can write $\mathcal{A}_{2}=\mathcal{A}_{2,1}+\mathcal{A}_{2,2}$, where

[TABLE]

The first term can be bounded by

[TABLE]

the inequality following from the fact that $\|\nabla\rho_{\varepsilon}\|_{L^{1}(D)}\leqslant C\varepsilon^{-1}$. By the Besov regularity assumption (2.5), the above vanishes along a subsequence $\varepsilon^{\prime}\to 0$. In the second term $\mathcal{A}_{2,2}$ we make the change of variables $x\mapsto x+z$ in the term $v(x-z)$ and then the change of variables $z\mapsto-z$ to obtain

[TABLE]

Writing $\psi\nabla\rho_{\varepsilon}=\nabla(\psi\rho_{\varepsilon})-\rho_{\varepsilon}\nabla\psi$ and using the divergence constraint, we obtain

[TABLE]

The Lebesgue differentiation theorem now implies that $\mathcal{A}_{2,2}\to 0$ as $\varepsilon\to 0$.

For $\mathcal{A}_{4}$, the change of variables $x\mapsto x+z$ in the second term gives

[TABLE]

The divergence constraint implies that the first term is zero, while the second term gives

[TABLE]

which converges to $\int_{0}^{T}\int_{D}p(x)\nabla\psi\cdot v\,dxdt$. Summing up all the terms, we conclude that in the limit $\varepsilon\to 0$ we obtain the local energy identity (1.3) in distributional form. ∎

3 Statistical solutions of the Euler equations

In this section we prove that statistical solutions of Euler’s equation are energy conservative under a Besov-type regularity assumption. In Section 3.1 we introduce the necessary technical machinery, in Section 3.2 we define statistical solutions of Euler’s equation and in Section 3.3 we carry out the proof of energy conservation. The proof closely follows the proof of energy conservation for weak solutions in Section 2. In particular, there are direct analogues of the weak formulation(s), the divergence constraint, the Besov regularity assumption, and the Lebesgue differentiation theorem.

3.1 Correlation measures

Definition 3.1**.**

Let $d,N\in\mathbb{N}$, $q\in[1,\infty)$ and let $D\subset\mathbb{R}^{d}$ be an open set (the “space domain”) and denote $\mathbf{U}=\mathbb{R}^{N}$ (“phase space”). A correlation measure from $D$ to $\mathbf{U}$ is a collection $\boldsymbol{\nu}=(\nu^{1},\nu^{2},\dots)$ of maps satisfying:

1. (i)

$\nu^{k}$ is a Young measure from $D^{k}$ to $\mathbf{U}^{k}$. 2. (ii)

Symmetry: if $\sigma$ is a permutation of $\{1,\dots,k\}$ and $f\in C_{0}(\mathbf{U}^{k})$ then \bigl{\langle}\nu^{k}_{\sigma(x)},\,f(\sigma(\xi))\bigr{\rangle}=\bigl{\langle}\nu^{k}_{x},\,f(\xi)\bigr{\rangle} for a.e. $x\in D^{k}$. 3. (iii)

Consistency: If $f\in C_{b}(\mathbf{U}^{k})$ is of the form $f(\xi_{1},\dots,\xi_{k})=g(\xi_{1},\dots,\xi_{k-1})$ for some $g\in C_{0}(\mathbf{U}^{k-1})$, then \bigl{\langle}\nu^{k}_{x_{1},\dots,x_{k}},\,f\bigr{\rangle}=\bigl{\langle}\nu^{k-1}_{x_{1},\dots,x_{k-1}},\,g\bigr{\rangle} for almost every $(x_{1},\dots,x_{k})\in D^{k}$. 4. (iv)

$L^{q}$* integrability:*

[TABLE] 5. (v)

Diagonal continuity (DC): $\lim_{\varepsilon\to 0}d_{\varepsilon}^{q}(\nu^{2})=0$, where

[TABLE]

We denote the set of all correlation measures by $\mathcal{L}^{q}(D,\mathbf{U})$.

Remark 3.2**.**

The “modulus of continuity” $d_{\varepsilon}^{q}(\nu^{2})$ is bounded irrespective of $\varepsilon>0$, due to the $L^{q}$ bound. Indeed,

[TABLE]

where we have used Minkowski’s inequality and then the consistency requirement.

Remark 3.3**.**

An example of a correlation measure is $\nu^{k}_{x_{1},\dots,x_{k}}=\delta_{u(x_{1})}\otimes\cdots\otimes\delta_{u(x_{k})}$, where $x_{1},\dots,x_{k}\in D$, $k\in\mathbb{N}$, $u\in L^{q}(D,\mathbb{R}^{N})$ is a measurable function and $\delta_{v}$ is the Dirac measure centered at $v\in\mathbb{R}^{N}$. The symmetry and consistency conditions (ii), (iii) follow immediately, and the $L^{q}$ integrability condition (iv) asserts that $u\in L^{q}(D,\mathbb{R}^{N})$. Moreover, the modulus of continuity $d_{\varepsilon}^{q}$ in condition (v) is

[TABLE]

which vanishes as $\varepsilon\to 0$ due to the Lebesgue differentiation theorem. Thus, diagonal continuity is the assertion that the Lebesgue differentiation theorem—which automatically holds for $L^{q}$ functions—also holds for $\boldsymbol{\nu}$. Correlation measures which are concentrated on a single function $u$ are called atomic.

Definition 3.4**.**

Let $d,N\in\mathbb{N}$ and let $D\subset\mathbb{R}^{d}$ be an open set (the “space domain”), let ${\mathcal{T}}\subset\mathbb{R}$ be an interval (the “time domain”) and denote $\mathbf{U}=\mathbb{R}^{N}$ (“phase space”). A time-dependent correlation measure from ${\mathcal{T}}\times D$ to $\mathbf{U}$ is a collection $\boldsymbol{\nu}=(\nu^{1},\nu^{2},\dots)$ of maps satisfying:

1. (i)

$\nu^{k}$ is a Young measure from ${\mathcal{T}}\times D^{k}$ to $\mathbf{U}^{k}$. 2. (ii)

Symmetry: if $\sigma$ is a permutation of $\{1,\dots,k\}$ and $f\in C_{0}(\mathbf{U}^{k})$ then \bigl{\langle}\nu^{k}_{t,\sigma(x)},\,f(\sigma(\xi))\bigr{\rangle}=\bigl{\langle}\nu^{k}_{t,x},\,f(\xi)\bigr{\rangle} for a.e. $(t,x)\in{\mathcal{T}}\times D^{k}$. 3. (iii)

Consistency: If $f\in C_{b}(\mathbf{U}^{k})$ is of the form $f(\xi_{1},\dots,\xi_{k})=g(\xi_{1},\dots,\xi_{k-1})$ for some $g\in C_{0}(\mathbf{U}^{k-1})$, then \bigl{\langle}\nu^{k}_{t,x_{1},\dots,x_{k}},\,f\bigr{\rangle}=\bigl{\langle}\nu^{k-1}_{t,x_{1},\dots,x_{k-1}},\,g\bigr{\rangle} for almost every $(t,x_{1},\dots,x_{k})\in{\mathcal{T}}\times D^{k}$. 4. (iv)

$L^{q}$* integrability:* There is a $c>0$ such that

[TABLE] 5. (v)

Diagonal continuity (DC): $\lim_{\varepsilon\to 0}d_{\varepsilon}^{q}(\nu^{2})=0$, where

[TABLE]

3.2 Statistical solutions

For the following definition, recall that the natural framework to study the local energy (in)equality for the incompressible Euler equations is $v\in L^{3}_{t,x}$, $p\in L^{3/2}_{t,x}$. Therefore, we need to distinguish the integrability conditions corresponding to the velocity and the pressure, respectively, which leads to the “mixed" integrability and diagonal continuity conditions (3.5) and (3.5) below. The integration variable of $\nu^{k}_{t,x}$ will be denoted $\xi=(v^{1},\dots,v^{d},q)\in\mathbf{U}^{k}$, with the interpretation of $v^{i}=(v^{i}_{1},\dots,v^{i}_{k})$ as the $i$-th component of velocity and of $p=(p_{1},\dots,p_{k})$ as the scalar pressure at $k$ different spatial points $x=(x_{1},\dots,x_{k})$. Here, our phase space is $\mathbf{U}=\mathbb{R}^{d+1}$.

Definition 3.5**.**

Let $D\subset\mathbb{R}^{d}$ be a spatial domain and let $T>0$. Let $\bar{\boldsymbol{\nu}}\in\mathcal{L}^{2}(D,\mathbb{R}^{d})$ be given initial data. By a statistical solution of the incompressible Euler equations, we will mean a time-dependent correlation measure $\boldsymbol{\nu}$ from $[0,T]\times D$ to $\mathbb{R}^{d+1}$, where the integrability condition (3.3) is to be understood as

[TABLE]

for some $c>0$, and diagonal continuity is to be understood as

[TABLE]

such that:

1. (i)

For all $k\in\mathbb{N}$, $\nu^{k}=\nu^{k}_{t,x_{1},\dots,x_{k}}$ satisfies

[TABLE]

for all $i_{1},\dots,i_{k}=1,\dots,d$ and for all \varphi\in C_{c}^{\infty}\big{(}(0,T)\times D^{k}\big{)}. (Here we abbreviate $v_{1}^{i_{1}}\cdots p_{l}\cdots v_{k}^{i_{k}}=v_{1}^{i_{1}}\cdots v_{l-1}^{i_{l-1}}p_{l}v_{l+1}^{i_{l+1}}\cdots v_{k}^{i_{k}}$, the $l$th component of $v$ being omitted.) 2. (ii)

$\boldsymbol{\nu}$ is divergence-free: For every $\varphi\in C_{c}^{\infty}(\mathbb{R}_{+}\times D^{k})$ and every $\kappa\in C(\mathbf{U}^{k-1})$ for which \bigl{\langle}\nu^{k-1},\,|\kappa|\bigr{\rangle}<\infty, we have

[TABLE]

Remark 3.6**.**

The first two instances of (3.7) are:

[TABLE]

for all \varphi\in C_{c}^{\infty}\big{(}(0,T)\times D\big{)} and all $i=1,\dots,k$, and

[TABLE]

for all \varphi\in C_{c}^{\infty}\big{(}(0,T)\times D^{2}\big{)} and all $i,j=1,\dots,d$. These are direct analogues of (1.2) and (2.3), respectively, and are the only instances of (3.7) which will be used in the remainder.

Remark 3.7**.**

By (3.5), all integrals in (3.9) and (3.10) are well-defined.

3.3 Energy conservation for statistical solutions

We are now ready to prove the “energy conservation” part of Onsager’s conjecture for statistical solutions. In the same vein as Duchon and Robert [7], we will prove a somewhat stronger result by quantifying the precise energy dissipation $\mathcal{E}(u)$, and prescribing a sufficient condition that ensures that $\mathcal{E}(u)\equiv 0$.

Theorem 3.8**.**

Let $\boldsymbol{\nu}$ be a statistical solution of the incompressible Euler equations on $[0,T]\times D$, where either $D=\mathbb{T}^{d}$ or $D=\mathbb{R}^{d}$. Let $\rho_{\varepsilon}(z)=\varepsilon^{-d}\rho(z/\varepsilon)$ be a rotationally symmetric mollifier. Then the distribution $\mathcal{E}(\boldsymbol{\nu})\in\mathcal{D}^{\prime}(D)$ given by

[TABLE]

is well-defined and independent of the choice of $\rho$, and $\boldsymbol{\nu}$ satisfies

[TABLE]

in the sense of distributions. If $\boldsymbol{\nu}$ satisfies the regularity condition

[TABLE]

then $\mathcal{E}(\boldsymbol{\nu})\equiv 0$.

Remark 3.9**.**

The left-hand side of (3.12) equals $\frac{1}{\varepsilon}d_{\varepsilon}^{3}(\nu^{2})^{3}$, cf. (3.4). Whereas the requirement of diagonal continuity merely requires that $d_{\varepsilon}^{3}(\nu^{2})$ vanishes as $\varepsilon\to 0$, the regularity assumption (3.12) imposes a rate at which it vanishes.

Remark 3.10**.**

In turbulence theory, the Kolmogorov four-fifths law states that in a homogeneous (but not necessarily isotropic) turbulent flow, the left-hand side of (3.11) equals

[TABLE]

where the angle brackets denote the expected value over an ensemble of turbulent flows; cf. [13, Section 6.2.5]. It is readily seen that $\mathcal{E}(\boldsymbol{\nu})$ is a distributional version of the above quantity. (See also [7, Section 5].)

Proof.

Set $j=i$ in (3.10) and sum over $i$:

[TABLE]

(where we suppress the dependence on $t$). Fix a number $\varepsilon>0$. We choose again the test function $\varphi(t,x,y)=\rho_{\varepsilon}(x-y)\psi(t,x)$, where $\rho_{\varepsilon}(z)=\varepsilon^{-d}\rho(\varepsilon^{-1}z)$ for a nonnegative, rotationally symmetric mollifier $\rho\in C_{c}^{\infty}(D)$ with unit mass and support in $B_{0}(1)$ and $\psi\in C_{c}^{\infty}((0,T)\times D)$. Then as before,

[TABLE]

Continuing from (3.13), we now have

[TABLE]

Making the change of variables $z=x-y$ yields

[TABLE]

Decompose the above into a sum of five terms $\mathcal{A}_{1}+\dots+\mathcal{A}_{5}$. We consider each in order.

For $\mathcal{A}_{1}$ we can write $\mathcal{A}_{1}=\mathcal{A}_{1,1}+\mathcal{A}_{1,2}$, where

[TABLE]

The first term $\mathcal{A}_{1,1}$ converges as $\varepsilon\to 0$ to \int_{0}^{T}\int_{D}\partial_{t}\psi(t,x)\bigl{\langle}\nu^{1}_{t,x},\,|v|^{2}\bigr{\rangle}\,dxdt. Indeed,

[TABLE]

where we have changed variables $x\mapsto x+z$ in the second equality and used the consistency and symmetry properties of $\nu^{2}$ in the third equality. Letting $\varepsilon\to 0$ we obtain \mathcal{A}_{1,1}\to\int_{0}^{T}\int_{D}\partial_{t}\psi(t,x)\bigl{\langle}\nu^{1}_{t,x},\,|v|^{2}\bigr{\rangle}\,dxdt.

The second term $\mathcal{A}_{1,2}$ converges to zero as $\varepsilon\to 0$ by diagonal continuity. Thus,

[TABLE]

For $\mathcal{A}_{2}$ we can then write $\mathcal{A}_{2}=\mathcal{A}_{2,1}+\mathcal{A}_{2,2}$, where

[TABLE]

In the limit $\varepsilon\to 0$, the first term is precisely $\mathcal{E}(\boldsymbol{\nu})(\psi)$. (This limit is well-defined because all the remaining terms converge as $\varepsilon\to 0$.) Under the regularity assumption (3.12) this term can be bounded as

[TABLE]

as $\varepsilon\to 0$ (after choosing a suitable subsequence if necessary), where we also used the fact that $\|\nabla\rho_{\varepsilon}\|_{L^{\infty}(\mathbb{R}^{d})}\leqslant C\varepsilon^{-1-d}$.

For the second term $\mathcal{A}_{2,2}$ we use the transformation $x\mapsto x+z$ in the term with $|v_{2}|^{2}$ and then the transformation $z\mapsto-z$ and thus get

[TABLE]

But now the divergence constraint (3.8) implies

[TABLE]

and then the diagonal continuity condition implies $\lim_{\varepsilon\to 0}\mathcal{A}_{2,2}=0$.

Using an argument similar to the treatment of $\mathcal{A}_{1}$, it is not hard to see that diagonal continuity implies

[TABLE]

as $\varepsilon\to 0$, and similarly

[TABLE]

Finally, for $\mathcal{A}_{4}$ we translate $x\mapsto x+z$ in the second term and obtain

[TABLE]

Owing to the divergence condition, the first term is zero, whereas the second term (again invoking the divergence constraint) equals

[TABLE]

Once more invoking diagonal continuity yields

[TABLE]

Collecting all terms now gives the desired result.

∎

4 Probabilistic versus deterministic regularity

In this section we will compare the regularity of functions and of correlation measures, and we will show that if a correlation measure is concentrated on a family of $L^{2}$ functions (soon to be made precise), then this family is at least as regular as the correlation measure (also to be made precise). We will use the notation

[TABLE]

for a function $v:D\to\mathbb{R}^{d}$, and for a correlation measure $\boldsymbol{\nu}=(\nu^{1},\nu^{2},\dots)$ we write

[TABLE]

(For notational convenience we only look at space-dependent functions in this section.)

In [11] the authors proved that the set of correlation measures, as defined in Definition 3.4, are equivalent to the set $\mathcal{P}(L^{2}(D))$ of probability measures on $L^{2}(D)$. We make this duality more precise in the following definition:

Definition 4.1**.**

A probability measure $\mu\in\mathcal{P}(L^{2}(D;\mathbf{U}))$ is said to be dual to a correlation measure $\boldsymbol{\nu}$ from $D$ to $\mathbf{U}$ provided

[TABLE]

for every $k\in\mathbb{N}$ and for every Caratheodory function $g:D^{k}\to C(\mathbf{U}^{k})$.

Proposition 4.2**.**

Let $\mu\in\mathcal{P}(L^{2}(D;\mathbf{U}))$ be dual to a space-time correlation measure $\boldsymbol{\nu}$, and assume that $\boldsymbol{\nu}$ is $\alpha$-Besov regular in the sense that

[TABLE]

Then

[TABLE]

for $\mu$-almost every $v\in L^{2}(D;\mathbf{U})$. Conversely, if there is a common subsequence $\varepsilon_{n}\to 0$ such that

[TABLE]

for $\mu$-almost every $v\in L^{2}(D;\mathbf{U})$, then (4.2) holds.

Proof.

Using the duality between $\mu$ and $\nu$ and Fatou’s lemma yields

[TABLE]

The conclusion follows immediately. Conversely, if (4.4) holds then the dominated convergence theorem implies that (4.2) holds along the prescribed subsequence $\varepsilon_{n}$. ∎

Acknowledgments

U.S.F. was supported in part by the grant Waves and Nonlinear Phenomena (WaNP) from the Research Council of Norway.