# Is the problem that Prof Otelbaev proved exactly the one stated by Clay Mathematics Institute?

Recently, mathematician Mukhtarbay Otelbaev published a paper Existence of a strong solution of the Navier-Stokes equations, in which he claim that he solved one of the Millennium Problems: existence and smoothness of the Navier-Stokes Equation. Unfortunately, the paper is in Russian but I cannot read Russian. There is a summary in English at the end of the paper, in which I found that the problem he proved was

Let $Q \equiv (0,2\pi)^3\subseteq \mathbb{R}^3$ be a 3-dim domain, $\Omega=(0,a)\times Q$, a>0.

$\textbf{Navier-stokes problem}$ is to find unknowns: a speed vector $u(t) = (u_1(t,x), u_2(t,x), u_3(t,x))$ and a scalar pressure function $p(t,x)$ at the points $x\in Q$ and time $t\in (0,a)$ satisfying the system of the equations

...

initial $$u(t,x)\vert_{t=0} =0$$

...

But the problem that stated by the Clay Mathematics Institute was, see http://www.claymath.org/sites/default/files/navierstokes.pdf,

$\textbf{(B) Existence and smoothness of Navier–Stokes solutions in$\mathbb{R}^3/\mathbb{Z}^3$.}$ ... Let $u^0$ be any smooth, divergence-free vector field satisfying $u^0(x+e_j) = u^0(x)$; we take $f(x, t)$ to be identically zero. Then there exist smooth functions $p(x,t)$, $u_i(x,t)$ on $\mathbb{R}^3 \times[0,\infty)$. that ...

So my question is:

1) Can the $a>0$ in his proof be $\infty$? Or is it enough to prove the problem for arbitrary $a>0$?

2) Is there any theory that can turn the problem with arbitrary initial value $u^0$ (of course satisfying some condition) into a problem with initial value being $0$?

3) The problem that formulated by the Clay Institute assumes $f\equiv 0$. Prof Otelbaev proved his result for all $f\in L_2(\Omega)$. Is this result much stronger?

Update: There is an article stating that the $L_2$ estimate is not enough to solve the problem. (in Spanish)

1. Proving $\|\Delta u\|_{L^2(\Omega)} < \infty$ for any $a>0$ is sufficient to solve the Millennium problem. In fact $\|(-\Delta)^{3/4} u\|_{L^2(\Omega)} < \infty$ is enough. Also there is no problem if $C$ and $l$ depend upon $a$, as long as they are finite for every $a$.
2. Setting $\nu = 1$ is not a big deal - this is easily overcome by putting the equation into dimensionless units. (Since the units of length are fixed, you can rescale the velocity and time units, but that is enough degrees of freedom to get the Reynold's number equal to 1.)
3. The assumption $u(x,0) = 0$ - this one I am not sure about. I don't immediately see an argument that you can deduce the $u(x,0) \ne 0$, $f \equiv 0$ case from the $u(x,0) = 0$, $f \in L_2(\Omega)$ case. Even if such an argument isn't available, this paper, if correct, still seems to be an enormous advance.

Update: Actually 3 is also not an issue. See user121805's answer.

So if this paper is correct, it solves the Clay Millennium problem.

• Could you please add a ref paper for "Proving $\|\Delta u\|_{L^2(\Omega)} < \infty$ for any a>0 is sufficient to solve the Millennium problem. In fact $\|(-\Delta)^{3/4} u\|_{L^2(\Omega)} < \infty$ is enough"? Thanks a lot! – mike Sep 19 '18 at 0:09
• @mike I don't know an exact reference, but you could get it by following a proof of the Prodi-Serrin conditions. Sorry for the bad answer. – Stephen Montgomery-Smith Sep 19 '18 at 2:18

This is a very good point and I tend to agree that the problem solved in the Otelbaev's paper is different from the problem statement from Clay Institute.

Let me repeat the key claims of the paper referred below as NS2013.

First recall the statement of the boundary value problem for the Navier-Stokes equation as formulated in NS2013.

Let $Q\subset R^3$ be a region of a 3-dimensional space, Let $\Omega=(0,a)\times Q$. Below we assume $Q=(-\pi,\pi)^3$ a cube with edges of length $2\pi$. Inside $\Omega$ the following equations hold. $$\left\{ \begin{array}{l} \frac{\partial u}{\partial t} + u\cdot\nabla u = \Delta u - \nabla p + f,\\ \nabla \cdot u = 0 \end{array} \right.$$ Initial conditions are given as $u(t,x) |_{t=0} = 0$, for $x\in \bar Q$ and the boundary conditions are periodic, i.e. for $k=1,2,3$ $$\begin{array}{l} u(t,x) |_{x_k=-\pi} = u(t,x) |_{x_k=\pi},\\ p(t,x) |_{x_k=-\pi} = p(t,x) |_{x_k=\pi},\\ \left.\frac{\partial u(t,x)}{\partial x_k}\right|_{x_k=-\pi} = \left.\frac{\partial u(t,x)}{\partial x_k}\right|_{x_k=\pi} \end{array}$$ Additionally $$\int_Q p(t,x)dx = p_0 > 0$$

For $f\in L_2(\Omega)$ we call a solution $(u;p)$ of the problem defined above a strong solution iff the following functions are in $L_2(\Omega)$.

$$\frac{\partial u}{\partial t}, \Delta u, u\cdot \nabla u, \mathop{\textrm{grad}} p$$

And the main result in NS2013 is Theorem 1 on page 10.

For any $f\in L_2(\Omega)$ there exists a strong solution, such that $$\left\|\frac{\partial u}{\partial t}\right\| + \|\Delta u\| + \|u\cdot \nabla u\|+ \mathop{\textrm{grad}} p \leq C (1+ \|f\|+\|f\|^l).$$ Here $\|\cdot\|$ is the $L_2$ norm and the constants $C>0$ and $l\geq 1$ do not depend on $f\in L_2$.

This looks like a difficult result yet it does not claim the existence of a smooth solution, i.e. $u\in C^\infty$. The existence of a solution in $L_2(\Omega)$ solution with a bounded norm and bounded derivatives does not imply that the solution is smooth and even that the solution does not contain discontinuities.

I reckon this is an important result but is not a solution of the Millenium problem. Possibly due to a misinterpretation of the Millenium problem statement. However my confidence level in this conclusion is far below 100%.

Update: there is a community project to translate the paper initiated by Misha Wolfson, available on GitHub

Update 2: The search for solutions of Navier-Stokes equations in $L_2$ follows the manifest stated by Olga Ladyzhenskaya in her 2003 introduction to the Sixth problem of the millennium . She writes that the key problem regarding Navier-Stokes is whether the equations together with initial and boundary conditions provide a deterministic description of the dynamics of incompressible fluid. And in answering this question the researcher should not be constrained by infinite or any other differentiability restrictions on the solution.

: O.A. Ladyzhenskaya, "Sixth problem of the millennium: Navier-Stokes equations, existence and smoothness", Russ. Math. Surv. 58 251

• As far as I known, we can improve the smoothness if we have some estimate. But I do know little about this problem. I find a remark in his paper on page 10: Этот результат позволяет использовать теорию возмущений, и повышать гладкость решения при повышении гладкости данных задачи. (I use OCR to get this sentence and use Google to translate it) – Fang Hung-chien Jan 12 '14 at 19:18
• @Unknown The comment says that Theorem 1 makes it possible to use perturbation theory to increase the differentiability class when the problem data has higher differentiability class (here it is the function $f$ only). – Dmitri Chubarov Jan 13 '14 at 5:31
• I wonder if his constants C and l depend on a. It doesn't say in the English abstract. I suspect they do, in which case the extension to infinity may not hold. – user121255 Jan 14 '14 at 13:49

1) if the estimates do not saturate at t=a, then the solution cam be extended past that time. am open,closed argument then gives infinite time existence.

2) subtract a smooth extension v of the smooth boundary data. u-v has zero initial data, and satisfies a similar system with added lower order terms of linear type, with smooth coefficients.

3) i think he means: apply the difference quotient method (see ex ladyzhenskya, or lieberman books) to get the same estimates for all order difference quotients, hence for all weak derivatives.

penny smith

This is my thought on $u(0,x)=0$ (I am no way specialist in this field). Take $u(t,x)=u(x)\times s(t)$, such that $s(0)=0$ and $s(t_0>0)=1$ and $u(x)$ is divergence free. Since $u(t,x)$ is divergence free we can substitute it in the momentum eqution. Putting $p(x,t)=0$ we get $f(x,t)$ with all necessary properties. So, we can bring the liquid to any desirable initial data at $t_0$ in this way. Putting $f(t,x)=0$ for $t>t_0$ we come to the Millenium problem (that starts from $t=t_0$).

• Please don't remove the $s: they make MathJax work. – Shaun Jan 16 '14 at 16:16 • Yes, of course. I should have thought of that. Note that a lot of smoothness might be needed on$u(x)$to make it work. But the Clay Millennium problem allows you to put as much smoothness on the initial data as you like. – Stephen Montgomery-Smith Jan 16 '14 at 23:13 • But one minor correction. You cannot set$p = 0$. But you can compute$p$from$u(x)$be applying the Leray projection (also called the Hodge decomposition) to$u \cdot \nabla u$. In other words, pick$p$to be such that$u\cdot\nabla u + \nabla p$is divergence free. – Stephen Montgomery-Smith Jan 16 '14 at 23:17 • And actually my minor correction isn't needed, because you simply absorb the pressure term into$f$. – Stephen Montgomery-Smith Jan 17 '14 at 3:22 Just read it for and hour. What really bugs me is that$u(0,\mathbf x)=0$(can this be corrected on$\mathbb{T}^3=\mathbb{R}^3 / \mathbb{Z}^3$via$\mathbf{f}(\mathbf{x},t) \in L_2$?), and$\nu=1$. There is no parameter on the$\mathbb{A}$self-adjoint operator, so one may derive lighter estimates for quadratic form$\mathbb{B}$. It can be shown that the attractor dimension$Y$of 3D NS grows as$Y \propto \nu^{-9/4}$and that is bad for such prove concept. • the viscosity can be normalized to one by a rescaling of time and space. But, is this compatible with his given periodic boundary conditions--that is, does his proof still work? – electronp Jan 14 '14 at 18:39 • Normalizing$\nu=1$will result in very high velocity physically. You either introduce Reynolds$R=\frac{L u}{\nu}$number or you change the whole equation. One way or another you cannot escape the parameter on$\mathbb{A}\$ in the real problem. As a matter of fact NS equations are derived from integral conservation laws and those are correlations between inertia and viscosity in conservation of momentum. – User1611161 Jan 14 '14 at 19:06
• This seems more like a comment than an answer, but I can't figure out to which post. – robjohn Jan 14 '14 at 19:40