The Probabilistic Approach

Here we will describe a probabilistic approach to solving equation (1.6). For simplicity let us consider the case when the forcing term ${\mathbf{f}}= 0$. We will not be rigorous.

We will suppose that we have found ${\mathbf{u}}$ using equation (1.5). Now let ${\mathbf{b}}_t$ be a Brownian motion in 3 dimensions, starting at the origin. Define $\tilde{\mathbf{u}}({\mathbf{x}},t) = {\mathbf{u}}({\mathbf{x}}+ 2\nu {\mathbf{b}}_t,t)$. Let $\widetilde {\mathbf{m}}$ be a random vector field that satisfies the equations

$$\begin{array}{c} \displaystyle \frac {\partial \widetilde{\m... ...\mathbf{x}}) \quad {\rm in \ } \mbox{\Bbbb R}^3\, . \end{array}$$ (6.1)

Now let $\overline{\mathbf{m}}({\mathbf{x}},t) = {\mathbf{m}}(x-2\nu{\mathbf{b}}_t,t)$. Then ${\mathbf{m}}({\mathbf{x}},t) = E(\overline{\mathbf{m}}({\mathbf{x}},t))$ satisfies equation (1.6). (Here $E(\cdot)$ represents the expected value.)

The reason why this works is because of the Itô formula. We have that

$$\frac {\partial \overline{\mathbf{m}}}{\partial t} + {\mathbf... ...l {\mathbf{b}}}{\partial t} \cdot \nabla \overline{\mathbf{m}},$$

and taking expectations the result follows.

The solution to equation (6.1) can be computed as follows. Suppose that the initial value of ${\mathbf{m}}$ satisfies equation (1.3). Then if

$$\begin{array}{c} \displaystyle \frac {\partial \alpha_i}{\pa... ...tial t} + \tilde{\mathbf{u}}\cdot \nabla \beta_i =0 \end{array}$$

then

$$\widetilde {\mathbf{m}}({\mathbf{x}},t) = \sum_{i=1}^R \beta_i({\mathbf{x}},t) \nabla \alpha_i({\mathbf{x}},t)$$

is the solution to system (6.1). But the transport equations are easily solved by $\alpha_i({\mathbf{x}},t) = \alpha_i(\tilde\mathbf{A}({\mathbf{x}},t),0)$ and $\beta_i({\mathbf{x}},t) = \beta_i(\tilde\mathbf{A}({\mathbf{x}},t),0)$, where $\tilde\mathbf{A}$ is the back to coordinates map induced by the flow $\tilde {\mathbf{u}}$.

This can be used to obtain the following plausibility argument for the regularity of the Navier-Stokes equations. Let $W^{-1,BMO}$ denote the space of functions from $\mbox{\Bbbb R}^3$ for which minus one derivative is in the space of functions of bounded mean oscillation. It is known that the space $L^{\infty}(I; W^{-1,BMO})$ is a critical space for proving regularity for the Navier-Stokes equations (see below). That is, if one can show that the solution to the Navier-Stokes equations is uniformly in time in any space better than $W^{-1,BMO}$ (such as $W^{-1+\epsilon,BMO}$ for any $\epsilon>0$), then the solution is regular.

Now if the initial data are very nice, then by using some partition of unity argument, we may suppose that indeed the initial value of ${\mathbf{m}}$ does satisfy equation (1.3) for some finite value of $R$, where the initial values of $\alpha_i$ and $\beta_i$ are compactly supported smooth functions. Then it is easy to see that the solutions for $\alpha_i$ and $\beta_i$ provided by the transport equations stay uniformly in $L^\infty$. Thus it follows that $\nabla \alpha_i$ is uniformly in the space $W^{-1,BMO}$.

Thus $\widetilde {\mathbf{m}}$ is a finite sum of a product of functions uniformly in $L^\infty$ and functions uniformly in $W^{-1,BMO}$. Thus it might seem that we are close to showing that ${\mathbf{u}}$ (which is the Leray projection of an average of translations of $\tilde {\mathbf{m}}$) is in a space that is critical for proving regularity.

There are some large, probably insurmountable problems with this approach. The lesser problem is that we need a space that is better than critical. The bigger problem is that the space created by taking the convex closure of products of bounded functions and functions in $W^{-1,BMO}$ is not really a well defined space, in that it encompasses every function.

Criticality of $L^{\infty}(I; W^{-1,BMO})$: Let us present a formal proof of this fact, in the case of the Cauchy problem with zero right-hand side. Let ${\mathbf{u}}$ be the solution to the Navier-Stokes equations which belongs to the space $L^{\infty}(I; W^{-1,BMO})$. Multiply equation (1.5)$_1$ by $\Delta {\mathbf{u}}$ and integrate over $\mbox{\Bbbb R}^3$. Notice also that

$$\left\vert \int \Delta {\mathbf{u}}\cdot ({\mathbf{u}}\cdot ... ...athbf{u}}\right\vert \le \Vert \nabla {\mathbf{u}}\Vert _3^3 .$$

Then

$$\frac 12 \frac{d}{dt} \Vert\nabla {\mathbf{u}}\Vert _2^2 + \... ...athbf{u}}\Vert _2^2 \leq \Vert \nabla {\mathbf{u}}\Vert _3^3 .$$

Using the inequality

$$\Vert\nabla {\mathbf{u}}\Vert _3 \leq C \Vert{\mathbf{u}}\Ve... ...,BMO}^{\frac 13} \Vert\nabla^2 {\mathbf{u}}\Vert _2^{\frac 23}$$

(see [9]) we get

$$\frac{d}{dt} \Vert\nabla {\mathbf{u}}\Vert _2^2 + \nu \Vert... ...athbf{u}}\Vert _{-1,BMO} \Vert\nabla^2 {\mathbf{u}}\Vert _2^2$$ (6.2)

and if $\Vert{\mathbf{u}}\Vert _{-1,BMO}$ is sufficiently small, the solution is smooth. The proof can be done rigorously using the fact that for smooth initial condition there exists a local smooth solution; on this interval we obtain estimate (6.2) and therefore the solution can not blow up.