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3D Navier-Stokes equations: Otelbaev's proof


Here is a brief outline of Otelbaev's proof. This page is the result of ongoing discussion with Mikolaj Sierzega (Warwick), Sergei Chernyshenko (Imperial), and Andrew Wynn (Imperial).

Counterexample

A counterexample to the main theorem (Theorem 6.1, see below) was posted by "sup" on the noticeboard dxdy, which can be found here. An English translation of the original post (due to Sergei) can be downloaded here.

This was cleaned up by Stephen Montgomery-Smith and posted on the Mathematics Stack Exchange forum here. This same thread also includes an improved version of the counterexample, due to Terry Tao, that is somewhat simpler and addresses the proposed fix by Otelbaev.

In order to understand the counterexample, an English translation of Otelbaev's Theorem 6.1 and its assumptions can be found below, along with a brief outline of his argument.

There is an error in the paper. This was spotted by a "young Russian guy" and has been confirmed by Otelbaev. You can find more details on the Mathematics Stack Exchange post above. The problem comes on page 56 the inequality (6.34) does not follow from (6.33), since an additional factor of $\|z\|$ has been introduced.

 

Outline of Otelbaev's approach

Otelbaev recasts the Navier-Stokes equations in the familiar abstract form

$du/dt+Au+B(u,u)=f(t),$

where $A:H\to H$ is a linear operator and $B:H\times H\to H$ is bilinear.


He treats the case $f\in L^2(0,T;H)$ and $u(0)=0$. This case is sufficient to solve the equation

$du/dt+Au+B(u,u)=0$

with $u(0)=u_0$ provided that $u_0$ is sufficiently smooth that $Au_0+B(u_0,u_0)\in L^2$, by considering $u(t)=tu_0$ and setting

$f(t)=u_0+Au(t)+B(u(t),u(t))$ for $t\in[0,1]$ and $f(t)=0$ for $t>1$.


Then he recasts the equation by setting $v=u'+Au$ (where $u'=du/dt$) and considering

$v+L(v,v)=f$,

an equation for $v\in\hat H=L^2(0,T;H)$ (note that the Hilbert space $\hat H$ corresponds to a space-time solution of the original abstract setup).


Here $L(v,v)=B(Tv,Tv)$ and $T$ is an operator such that $T(u'+Au)=u$.

($Tv$ is defined by the integral solution of the heat equation $u'+Au=g$ with $g=u'+Au$.)


The properties of $A$ and $B$ that can be derived for the abstract form of the Navier-Stokes equations can be used to derive various properties of $A$ and $L$ as operators on $\hat H$.


Here is a full translation from Sergei (in italics) of the assumptions and conclusion of the main theorem, Theorem 6.1 (Section 6, page 27):


Let $\hat H$ be a Hilbert space with scalar product $\langle\cdot,\cdot\rangle$, $A$ a self-adjoint linear operator on $\hat H$, and $L(\cdot,\cdot)$ a bilinear continuous transform. That is, if $g,v\in\hat H$, then $L(g,v)$ is a linear transformation with respect to each of its arguments with the other argument held fixed. For $u\in\hat H$ we denote by $L_u$ an operator acting on $g\in\hat H$ according to the formula

$L_ug=L(u,g)+L(g,u)$

and by $L_u^*$ we denote the adjoint of $L_u$.


We will use the conditions (Y.1)-(Y.4) as follows:


(Y.1) for some $-\infty<\beta<0$ and for any $u,v\in\hat H$,

$\langle Au,u\rangle\ge\|u\|^2,$

$\|L(u,v)\|\le C_\beta\|A^\beta u\|\,\|A^\beta v\|$

for some $C_\beta$ that does not depend on $u$ and $v$.


(Y.2) If $u$ is an eigenvector of the operator $A$ (that is $Au=\lambda u$) then $\langle u,L(u,u)\rangle=0$.


(Y.3) The spectrum of the operator $A$ consists of the sequence of values

$1\equiv\lambda_1<\lambda_2<\lambda_3<\cdots<\lambda_n<\lambda_{n+1}<\cdots,$

where $16\le\lambda_2\le 100$, such that the space $\hat H$ may be represented as an orthogonal sum $\hat H=\oplus_{j=1}^M G_{\lambda_j}$ of the eigenspaces $G_{\lambda_j}$ ($G_{\lambda_j}=\{x:\ Ax=\lambda_jx\}$), which may be finite- or infinite-dimensional, and $M={\rm dim}\hat H$ may be finite or infinite. Here we require that

${\rm dim}\,G_{\lambda_1}\ge 20$


(Y.4) If $u\in\hat H$, $e\in G_{\lambda_1}=\{x:\ Ax=x\}$ then

$L(e,u)=L(u,e)=L_ue=L_u^*e=L_eu=L_e^*u=0.$


When the conditions (Y.2) and (Y.4) will be used the reader will notice that these conditions can be relaxed. To decrease the volume of work we take these conditions in the form written above.


THEOREM

Let conditions (Y.1)-(Y.4) be satisfied. Let us assumed that for some $u\in\hat H$ the weak estimate

$\|A^\theta u\|\le C_\theta$

holds for some $\theta<\min(-3/4,\beta)$, where $\beta$ is from (Y.1). Then for $u$ the following strong estimate holds:

$\|u\|\le C_1(1+\|f\|+\|f\|^l),$

where $f=u+L(u,u)$. Here $C_1$ and $l$ depend only on $C_\beta$ and $\beta$ from (Y.1), and $C_\theta$ and $\theta$.


We should immediately say that since we are interested in the a priori estimates for values of that are not small, for convenience sake we have added one in the expression in the brackets.


We will prove this theorem first in the case when $\hat H$ is a finite-dimensional space. In this case we will obtain estimates that are independent of the dimension of the space $\hat H$. After that we will obtain the proof of Theorem 6.1 in full. At the end of the section we will obtain Theorem 6.2, in which the condition (Y.4) will be eliminated.



The main part of the proof assumes that the solution $v\in\hat H$, and derives bounds on $v$ in $\hat H$ by defining an artificial flow on a parametrised function $v(\xi)$ with $v(0)=v$, where

$\|v(\xi)\|_{\hat H}=\|v(0)\|_{\hat H}=\|v\|_{\hat H}$ for every $\xi>0$;

$\|A^{-\theta}v(\xi)\|\le\|A^{-\theta}v(0)\|=\|A^{-\theta}v\|$ for every $\xi>0$;

and it is shown that there is a $\xi^*>0$ such that $\|v(\xi^*)\|_{\hat H}$ can be estimated in terms of $\|v(\xi^*)+L(v(\xi^*),v(\xi^*))\|_{\hat H}$.

Coupled with the above estimates, a bound on $d/d\xi \|\|v(\xi)+L(v(\xi),v(\xi))\|_{\hat H}$ allows one to bound $\|v(\xi^*)+L(v(\xi^*),v(\xi^*))\|_{\hat H}$ in terms of $\|f\|_{\hat H}$, $\|v\|_{\hat H}$, and $\|A^{-\theta}v\|_{\hat H}$; an application of Young's inequality then yields the bound on $\|v\|_{\hat H}$ in terms of $\|f\|_{\hat H}$.


Making this argument rigorous requires the use of a Galerkin approximation argument.

Since the Navier-Stokes equations have a weak solution, there is a solution $v$ of $v+L(v,v)=f$ that satisfies $\|A^{-\theta}v\|<\infty$.

Now let $P_n$ be a family of orthogonal projections in $\hat H$ such that $P_nx\to x$ in $\hat H$ for every $x\in\hat H$.

(Note that these are orthogonal projectors in the "space-time solution space" $\hat H$, not in the more familiar space $H$.)


Now consider $P_nv\in\hat H$. This satisfies the equation

$P_nv+P_nL(P_nv,P_nv)=f_n$,

where we define $f_n$ to be the left-hand side of this equation.


It is simple to show that $A_n=P_nA$ and $L_n(\cdot,\cdot)=P_nL(P_n\cdot,P_n\cdot)$ satisfy (Y.1)-(Y.4), and so certainly

$\|P_nv\|_{\hat H}\le C(1+\|f_n\|_{\hat H}+\|f_n\|_{\hat H}^l)$

for each $n$.


Of course, the idea now is to take $n\to\infty$ and deduce a bound on $v$ in terms of $f$. This requires that $f_n$ converges to $f$ in $\hat H$, and noting that $f_n$ is defined by

$f_n=P_nv+P_nL(P_nv,P_nv)$

this reduces to showing that $P_nv\to v$ and $P_nL(P_nv,P_nv)\to L(v,v)$; this follows from the assumption (Y.1) on $L$.