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

  1. The discussion here has turned too chatty and not suitable for the MSE framework.
  2. Given the recent pre-print of T. Tao (see also the blog-post here), the continued usefulness of this question is diminished.
  3. 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

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Wel, he could be a crackpot about this particular problem, or perhaps he's only wrong. On the other hand, perhaps he really proved the conjecture. Looks like we shall wait a little, uh? –  DonAntonio Jan 11 at 17:48
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In mathematics, the correctness of proof does not depend on the credentials of the mathematician but by the quality of the argument. So can you say something about the proof? –  user44197 Jan 11 at 17:54
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I found a mistake: The formula on top of page 76 reads $$\int_0^t<\omega'+A\omega+B(\omega,\omega),\omega>_H=\ldots$$ while it probably should read $$\int_0^t\langle \omega'+A\omega+B(\omega,\omega),\omega\rangle_H=\ldots$$ :) –  Hagen von Eitzen Jan 11 at 18:25
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This is a Reddit-style announcement / call for a discussion, not really on-topic on Math.SE (cf. old meta thread and another one). –  Grigory M Jan 11 at 20:34
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To quote the proposed policy there: (Q) Is the new claimed proof of (conjecture) correct? (A) If it's at all credible, be assured that people are reading it and trying to determine that. In fact, why not read it yourself! But don't post this question on this site. –  Grigory M Jan 11 at 20:39
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closed as too broad by Willie Wong Feb 18 at 10:14

There are either too many possible answers, or good answers would be too long for this format. Please add details to narrow the answer set or to isolate an issue that can be answered in a few paragraphs.If this question can be reworded to fit the rules in the help center, please edit the question.

8 Answers

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.

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What I have read of the paper (which is up to Lemma 6.8) is correct and readable. So my feeling is that there is a specific place where a mistake was made. If I were in Otelbaev's position, I would go through the argument again very carefully, and try to find where the mistake is. And then spend some time trying to fix it, and checking and rechecking the fix. My gut feeling is that the mistake is quite subtle. Because what I have read so far does make me think highly of Otelbaev's competency in mathematics. –  Stephen Montgomery-Smith Jan 25 at 0:46
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I have also had some scary situations with papers on NS. I had a paper accepted for publication. It was only after I got the galley proofs, that I realized I had made a big mistake. And my mistake wasn't subtle at all! –  Stephen Montgomery-Smith Jan 25 at 0:50
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Also, maybe the analogy with chess isn't quite right. It is not really a competition, but more of a group effort to search for the truth. I know that 99.99% of mathematicians treat it as competition (and I have to include myself in that group), but maybe we shouldn't be doing that. –  Stephen Montgomery-Smith Jan 25 at 0:55
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Maybe it should be more like being one of the team of astronauts selected by NASA to go to the moon. They all hope that they are the one that will be chosen to make the first footstep on the moon, knowing that history will remember them as vividly as Columbus is remembered. But when someone else is selected for that honor, you stand on the sidelines and hope for their success. –  Stephen Montgomery-Smith Jan 25 at 16:11
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I have to admit that I have stopped reading the paper. Partly because of the counterexample, but also because our school semester has just started. I think Otelbaev is very precise in his writing, In my opinion, his paper has many less errors than other papers I have read. But without comprehending the big idea, it is very hard going. The Navier-Stokes equation has been worked on so hard by so many people, and I think there has to be some breakthrough insight before it will be solved. And a paper that claims to solve the problem should probably say up front what the new insight is. –  Stephen Montgomery-Smith Jan 26 at 19:20
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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

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I think I mostly understand the paper up until page 24. I am not quite sure where he is going with $\widehat{B_P}(u)$, but everything else up until then is rather standard. But on page 24, he introduces something I have never seen before. In particular, do you know what $\alpha(\xi)$ and $\beta(\xi)$ are in equation (5.3)? –  Stephen Montgomery-Smith Jan 15 at 13:24
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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.

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There is a typo in the statement of Lemma 6.2 - it is stated correctly at the end of the proof. The second $A^\beta$ should be $A^\theta$. (Lemma 6.2 is a standard type of result.) –  Stephen Montgomery-Smith Jan 16 at 23:39
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Do you plan on updating this post further as you continue to read the article? If so, I'd just like to say I appreciate it. –  Mike Jan 17 at 8:33
    
For anyone else trying to read the paper and who doesn't understand Russian: in Lemma 6.4, one hypothesis is that $\hat H$ is finite dimensional. (I was scratching my head over that one.) Today I'll look over Lemma 6.6. –  Stephen Montgomery-Smith Jan 17 at 11:35
    
@Mike Yes, I will. I did part (a) of Lemma 6.6 this morning, and so far things are still looking good. But I think I still have yet to get to the difficult part of the proof. –  Stephen Montgomery-Smith Jan 17 at 21:02
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I think the typo is insignificant. It should read $ {} = \lambda_2^{4\theta} \frac{\hat C}{250}$. And since $\lambda_2 \ge 16$, and $\theta<-\frac34$, you can then finish off with ${} \le \frac{\hat C}{250}$. –  Stephen Montgomery-Smith Jan 21 at 13:00
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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.]

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I was wrong with this. At some point I had lost track of the fact that $\beta<0$, which allows the convergence argument to work. I have edited the above webpage accordingly; it still contains a brief summary of the argument. –  James Robinson Jan 25 at 20:52
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Also the counterexamples to Theorem 6.1 are finite dimensional in nature, so it looks like the error is somewhere other than in Galerkin approximations. –  Stephen Montgomery-Smith Jan 26 at 1:09
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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?

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This is the answer I got from experts of the subject (without your "very, very, very"). Now the question is who is at the source of this whole PR operation. –  Did Jan 14 at 10:09
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I think definition 2 is enough to get the millennium problem. math.stackexchange.com/questions/635530/… –  Stephen Montgomery-Smith Jan 15 at 1:45
    
@did: Could you elaborate? Do you mean, "A PR operation to discredit the solution as weak?", or, "A PR operation to pass off a weak solution as a solution?", or something else entirely? –  KDN Jan 22 at 21:40
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My initial answer was based on ignorance; thanks to the kind help from the experts I've been convinced that $\triangle u \in L^2$ suffices for the solution to be strong. The PR side of the question remains interesting. I stumbled upon a very interesting thread in a Russian forum where an elegant counterexample to Theorem 2 has been presented, however, I really want to leave the authors the rights to decide when and whether to publish it. –  uvs Jan 23 at 10:20
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@KDN: Here we have a highly professional mathematician who has been working on this extremely tough problem for many years. Multiple times he has announced a proof in local conferences and presented it to the local experts, multiple times they have pointed out subtle problems with the proof, and every single time he has returned back to work without making any fuss. Suddenly he completes another attempt with fair chances of a mistake, but this time he pushes the paper to be published without any reviews whatsoever and follows up with messages to the general public. What has happened? –  uvs Feb 10 at 8:25
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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.

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"...they would find the mistake anyway." Do you mean a specific mistake, or are implying that there is likely one? –  NiftyKitty95 Jan 27 at 23:51
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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 ‘proof’, on p.56, the passage from (6.33) to (6.34) is made by saying ‘using this and that and also that’ . 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.

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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

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Please do not remove the Magnetohydrodynamics, Physicists correct Navier Stokes to make use of MHD ( embedded theory ), published since 2004. Littlewood-Paley conjecture and Soboloev Embedding, Young's Inequalities, 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. Is there an English translation as there are Chinese and Americans-and myself, that have similar findings from a Physics perspective. Betty Rostro, PHD –  Dr. Betty Rostro Feb 17 at 5:11
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Welcome to Math.SE, Dr. Rostro. You've picked a rather atypical Question to start with, so please have a look at the Help Center for a little context. We try to address Questions with substantive Answers, rather than inviting discussion-style posts as in many forums. –  hardmath Feb 18 at 9:34
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No I don't think Ostaelbev's answer is correct. I've been analyzing his math all week long, and have not found any new modifications to his prior paper. As such, I need to view the full manuscript in English to give a final answer, but from what I've seen, he has failed to include criteria that would prove strong smoothness, norms aside, that is not enough. –  Dr. Betty Rostro Feb 22 at 16:36
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