## fredag 16 maj 2014

### Towards Computational Solution of Clay Navier-Stokes Problem 2

This is a continuation of a previous post: The basic energy estimate which is easily proved analytically by multiplying the momentum equation by the velocity $u_\epsilon$ and integrating, reads for $T>0$ with $Q =\Omega\times (0,T)$:
• $\int_\Omega\vert u_\epsilon (x,T)\vert^2\, dx +\int_{Q}\epsilon\vert\nabla u_\epsilon (x,t)\vert^2\, dxdt =\int_\Omega\vert u^0(x)\vert^2\, dx$
or in short notation with obvious meaning:
• $U(T) + D_\epsilon (U) = U(0)$,
which expresses a balance of kinetic energy $U(T)$ at time $T$ and dissipation $D_\epsilon (U)$ over the time interval $(0,T)$ summing up to initial kinetic energy $U(0)$.

Computations with small $\epsilon$ (compared to data as $\Omega$ and $U(0)$) produce turbulent solutions characterized by
•  $D_\epsilon (U) =\alpha U(0)$ where $\alpha$ is not small,
that is solutions with substantial (turbulent) dissipation. For turbulent solutions $\vert \nabla u\vert$ is large, typically scaling with $\epsilon^{-\frac{1}{2}}$, even if initial data is smooth, which can be viewed as an expression of non-smoothness.

The basic energy estimate can thus be used to signify non-smoothness by substantial turbulent dissipation. The Clay problem can thus be reduced to the question of proving that the dissipation term is  substantial in the basic energy estimate.

Evidence to this effect is given by computation. Analytical evidence can be given by the following argument: Smooth laminar solutions have small dissipation but smooth laminar solutions are all unstable. If the dissipation remained small it would mean that an unstable solution would remain smooth and unstable, which is not possible under perturbation.

The dissipation therefore must be substantial in the basic energy estimate and only a non-smooth solution can exist (and does exist by computation). An answer to the Clay problem may thus be possible along the following lines, assuming the viscosity is small and data are smooth:
1. Solutions exist for all time and do not cease to exist by blow-up.
2. Solutions become non-smooth (turbulent) in finite time.
3. Solutions cannot stay smooth for all time, because any smooth solution is unstable.
4. Solutions are weakly well-posed in the sense that solution mean-values are stable to perturbations, because of a cancellation effect in turbulent solutions which is not present for smooth solutions.
The group of mathematicians in charge of the problem (Fefferman, Constantin and Tao) do not answer my repeated requests to open a discussion about the formulation of the problem and possible approaches to solution. This is not helpful to progress. Mathematicians apparently want to have a heaven of their own, where they can explain phenomena which have no scientific relevance, but this is a dangerous strategy in the long run, because without connection to science funding may cease.