TITLE: Is the problem that Prof Otelbaev proved exactly the one stated by Clay Mathematics Institute?
QUESTION [50 upvotes]: 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)

REPLY [6 votes]: 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 $t=t_0$).