## The Question :

*104 people think this question is useful*

**Moderator Notice**: I am unilaterally closing this question for three reasons.

- The discussion here has turned too chatty and not suitable for the MSE framework.
- Given the recent pre-print of T. Tao (see also the blog-post here), the continued usefulness of this question is diminished.
- The final update on this answer is probably as close to an “answer” an we can expect.

Eminent Kazakh mathematician

Mukhtarbay Otelbaev, Prof. Dr. has published a full proof of the Clay Navier-Stokes Millennium Problem.

Is it correct?

See http://bnews.kz/en/news/post/180213/

A link to the paper (in Russian):

http://www.math.kz/images/journal/2013-4/Otelbaev_N-S_21_12_2013.pdf

Mukhtarbay Otelbaev has published over 200 papers, had over 70 PhD students, and he is a member of the Kazak Academy of Sciences. He has published papers on Navier-Stokes and Functional Analysis.

please confine answers to any actual mathematical error found!

thanks

*The Question Comments :*

## The Answer 1

*65 people think this answer is useful*

This web page has Theorem 6.1. It is written in Spanish, but actually is rather easy to follow even if (like me) you don’t know any Spanish. However it is not made clear on this web site that the statement of Theorem 6.1 is “**If** $\|A^\theta \overset 0u\| \le C_\theta\|$, **then** $\| \overset0u \| \le C_1(1+\|\overset0f\|+\|\overset0f\|^l)$.”

http://francis.naukas.com/2014/01/18/la-demostracion-de-otelbaev-del-problema-del-milenio-de-navier-stokes/

This is the proposed counterexample to Theorem 6.1 given at http://dxdy.ru/topic80156-60.html. I used google translate, and then cleaned it up. I also added details here and there.

Let $\hat H = \ell_2$.

Let the operator $A$ be defined by $ Ae_i = e_i $ for $ i < 50$, $ Ae_i = ie_i $ for $ i \ge $ 50

Define the bilinear operator $ L $ to be nonzero only on two-dimensional subspaces $ L (e_ {2n}, e_ {2n +1}) = \frac1n (e_ {2n} + e_ {2n +1}) $, with $ n \ge 25 $.

Check conditions:

U3. Even with a margin of 50 .

U2. $ (e_i, L (e_i, e_i)) = 0 $ for $ i \ge $ 50. This is also true for eigenvectors $ u $ with $ \lambda = 1 $, because for them $ L (u, u) = 0 $.

U4. $ L (e, u) = 0 $ for the eigenvectors $ e $ with $ \lambda = 1 $ also trivial. (Stephen’s note: he also needs to check $L_e^*u = L_u^*e = 0$, but that looks correct to me.)

U1. $ (Ax, x) \ge (x, x) $. Also $$ \| L (u, v) \| ^ 2 = \sum_{n \ge 25} u ^ 2_ {2n} v ^ 2_ {2n +1} / n ^ 2 \le C\|(u_n/\sqrt n)\|_4^2 \|(v_n/\sqrt n)\|_4^2 \le C\|(u_n/\sqrt n)\|_2^2 \|(v_n/\sqrt n)\|_2^2 = C \left (\sum u ^ 2_ {n} / n \right) \left (\sum v ^ 2_ {n} / n \right) $$ so we can take $\beta = -1/2$.

And now consider the elements $ u_n =-n (e_ {2n} + e_ {2n +1}) $. Their norms are obviously rising.

Let $ \theta = -1 $. Then the $A^\theta $-norms of all these elements are constant. But, $ f_n = u_n + L (u_n, u_n) = 0 $.

**Update:** Later on in http://dxdy.ru/topic80156-90.html there is a response relayed from Otelbaev in which he asserts he can fix the counterexample by adding another hypothesis to Theorem 6.1, namely the existence of operators $P_N$ converging strongly to the identity such that one has good solvability properties for $u + P_N L(P_N u,P_N u) = f$, in that if

$\| f \|$ is small enough then $\| u \|$ is also small.

Terry Tao communicated to me that he thinks a small modification of the counterexample also defeats this additional hypothesis.

**Update 2:** Terry Tao modified his example to correct for that fact that the statement of Theorem 6.1 is that a bound on $u \equiv \overset0u$ implies a lower bound on $f \equiv \overset0f$ rather than the other way around (i.e. we had a translation error for Theorem 6.1 that I point out above).

Let $\hat H$ be $N$-dimensional Euclidean space, with $N \ge 50$. Let $\theta = -1$ and $\beta = -1/100$. Take

$$ A e_n = \begin{cases} e_n & \text{for $n<50$} \\

50\ 2^{n-50} e_n & \text{for $50 \le n \le N$.}\end{cases}$$

and

$$L(e_n, e_n) = – 2^{-(n-1)/2} e_{n+1} \quad\text{for $50 \le n < N$,}$$

and all other $L(e_i,e_j)$ zero.

Axioms (Y.2) and (Y.4) are easily verified. For (Y.1), observe that

if $u = \sum_n c_n e_n$ and $v = \sum_n d_n e_n$, then for a universal constant $C$, we have

$$ \| L(u,v) \|^2 \le C \sum_n 2^{-n} c_n^2 d_n^2, \\

|c_n| \le C 2^{n/100} \| A^\beta u \| ,\\

|d_n| \le C 2^{n/100} \| A^\beta v \| ,$$

and the claim (Y.1) follows from summing geometric series.

Finally, set

$$

u = \sum_{n=50}^N 2^{n/2} e_n

$$

then one calculates that

$$

\| A^\theta u \| < C

$$

for an absolute constant C, and

$$

u + L(u,u) = 2^{50/2} e_{50}

$$

so

$$

\| u + L(u,u) \| \le C

$$

but that

$$

\| u \| \ge 2^{N/2}.

$$

Since $N$ is arbitrary, this gives a counterexample to Theorem 6.1.

By writing the equation $u+L(u,u)=f$ in coordinates we obtain $f_n = u_n$ for $n \le 50$, and $f_n = u_n + 2^{-n/2} u_{n-1}^2$ if $50<n\le N$. Hence we

see that $u$ is

uniquely determined by $f$. From the inverse function theorem we see

that if $\| f \|$ is sufficiently small then $\| u \| < 1/2$, so the

additional axiom Otelbaev gives to try to fix Theorem 6.1 is also

obeyed (setting $P_N$ to be the identity).

**Update 3:** on Feb 14, 2014, Professor Otelbaev sent me this message, which I am posting with his permission:

Dear Prof. Montgomery-Smith,

To my shame, on the page 56 the inequality (6.34) is incorrect therefore the proposition 6.3 (p. 54) isn’t proved. I am so sorry.

Thanks for goodwill.

Defects I hope to correct in English version of the article.

## The Answer 2

*44 people think this answer is useful*

I have started to translate the paper so that English speakers can explore it. I’ve only had time for the abstract, introduction, and main result statement, but that already gives an important part of the picture. Any further contributions are welcome. https://github.com/myw/navier_stokes_translate

## The Answer 3

*27 people think this answer is useful*

OK, I spent an afternoon getting help with Russian, and I think I understand a lot more.

So first he actually proves a rather abstract theorem (Theorem 2), and strong solutions of the Navier-Stokes is merely a corollary. He shows the existence of solutions satisfying certain bounds to

$$ \dot u + Au + B(u,u) = f , \quad u(0) = 0,$$

where $A$ and $B$ satisfy rather mild hypotheses that, for example, the replacements $A = E-\Delta$, and $B(u,v) = e^t u \cdot \nabla v + \nabla p$ where $p$ is a scalar chosen so that $B(u,v)$ is divergence free.

(I always thought the proof or counterexample would use the special structure of $B(u,v)$ that comes with the Navier-Stokes equation.)

In Chapter 5, he outlines how he will turn it into a different abstract problem, explaining that it is sufficient to find a bound on $\overset{0}{v} = \dot u + Au$. He constructs an equation for a quantity $v(\xi) \equiv v(\xi,t,x)$, so that in effect it is a time dependent velocity field described by a parameter $\xi$. He creates a differential equation in $\xi$, which morphs $v(0) = \overset 0v$ into $v(\xi_1)$, where $\|v(\xi)\| = \|\overset0v\|$, but $v(\xi_1)$ is easier to work with. This equation is given by equations (5.2) and (5.3).

So far, the only part of equation (5.3) that I am beginning to understand is the $-\alpha(\xi) R(v(\xi))$ part. $R(v)$ measures how far $v$ is from being an eigenvector of $A^\theta$. And so the differential equation

$$ \frac{dv}{d\xi} = -\alpha(\xi) R(v(\xi)) $$

pushes $v$ into becoming closer to become an eigenvector.

Anyway, it looks like Chapter 6 is the meat of the paper. Theorem 6.1 seems to be the main result. However it has a rather odd condition, namely that the dimension of the eigenspace corresponding to the smallest eigenvalue of $A$ should be at least 20. So I will be interested to see how he converts the Navier-Stokes into an equation with this property.

## The Answer 4

*13 people think this answer is useful*

A full translation of the main theorem (Theorem 6.1) and conditions (Y.1)-(Y.4), due to Sergei Chernyshenko, can be found at

http://go.warwick.ac.uk/jcrobinson/lf/otelbaev

There is also a brief discussion of the method of proof used by Otelbaev.

[Responses below are to an earlier version of this post in which I thought I had found an error in the Galerkin argument used in the final part of the proof of Theorem 6.1.]

## The Answer 5

*6 people think this answer is useful*

what do you make of Definition 2? He defines a strong solution so that all terms of NSE are required to live in $L^2$. Global regularity of NSE calls for pressure and velocity fields in $C^\infty$. Is it just me or are we speaking of a very, very, very weak notion of strong solution, which has nothing to do with the millenium problem?

## The Answer 6

*6 people think this answer is useful*

I read on the Russian side dxdy a comment from a mathematician in Almaty, KZ, that sheds some light on the process. It’s the comment on page 13 by MAnvarbek.

Otelbaev presented his proof 1 year ago, and immediately they found large errors, and no new ideas. The problem was with all the parameters.

Otelbaev worked on correcting the errors, and then published—without again showing his result. Editors in Almaty did not approve publishing the paper.

The author of this statement wrote also that there is a committee of the Institute of Mathematics analysing the paper by Otelbaev. They thank the user **sup** at dxdy for proposing the example and Tao for improving it. The example saves a lot of work for the Committee, but they are sure that they would find the mistake anyway.

## The Answer 7

*5 people think this answer is useful*

On the Spanish site

http://francis.naukas.com/2014/01/18/la-demostracion-de-otelbaev-del-problema-del-milenio-de-navier-stokes/#comment-21031

the following info appeared

A young guy in Russia seems to have found a concrete gap in the proof.

This concerns Statement 6.3. In the â€˜proofâ€™, on p.56, the passage from

(6.33) to (6.34) is made by saying â€˜using this and that and also thatâ€™

. However no reasons are visible where does the extra ||z|| on the

right hand side come from. At least some very detailed explanation

for this is needed.

## The Answer 8

*3 people think this answer is useful*

I favor the Terry Tao version he uses Von Neuman criteria, this is a bit better and more elegant in my opinion.

http://terrytao.wordpress.com/2014/02/04/finite-time-blowup-for-an-averaged-three-dimensional-navier-stokes-equation/

I see from Otelbaev’s http://enu.kz/repository/repository2013/articlemmf1.pdf, that he’s made improvements. I see where he comes from Hilbert (Banach) and Cauchy problem and uses Sobolev and Galerkins, still he makes the case for weak solutions, and then reiterates to claim strong solution. However, if he’s relying on boundedness and trilinears a,b to find optimal boundedness using strong-weak uniqueness using Serrin criterion this was done by Lemarie already. Paraproduct issues aside Serrin criteria assumes Navier Stokes does not blow up, however, that is based on log inequalities from Wong which obtained them for earlier scholars.

I used Navier Stokes (NS) during my MSc thesis at Rice. At NASA-JSC though we applied different corrections to Navier Stokes though.

So I’d care to see merely a strong solution not just a Strong-Weak Uniqueness as other arguments have been claiming for years.

Which by the way Magnetohydrodynamic work with embedded theory, very much published already proves. Littlewood-Paley conjecture and Soboloev Embedding, Young’s Inequalities, all these do break down you know! So then the key is has he solved the remaining open problem???? If so where are those answers.

What was provided is a repeat of all the MHD and electrolyte theory already in publication since 2004.

I mean in that case other groups have obtained prior answers that run around the same lines.

Is there an English translation as there are Chinese and Americans and myself that have similar findings from a Physics perspective.

Also, what about the Terry Tao version with the Von Neuman criteria, this is a bit more elegant in my opinion.

Is Perelman contributing to this answer, I wonder where his comments would be? I’m assuming he would also start discussing the need for saddle criteria, where is that in this paper here.

Betty Rostro, PhD