In 3 spatial dimensions, the incompressible Navier-Stokes equations are: $$ \begin{split} \frac{\partial u_i}{\partial t} + \sum_{j=1}^3 u_j \frac{\partial u_i}{\partial x_j} &= - \frac{\partial p}{\partial x_i} + \nu \sum_{j=1}^3 \frac{\partial^2 u_i}{\partial x_j^2}\qquad\text{ for } i=1,2,3\\ \sum_{j=1}^3 \frac{\partial u_j}{\partial x_j} & = 0 \end{split} $$ Let me consider only general planar Couette boundary conditions, force periodic boundary conditions (in two of the three dimensions), and normalize without loss of generality (so $1/\nu$ is the Reynolds number): $$ \begin{split} u_2(x_1,1,x_3,t)&=u_2(x_1,-1,x_3,t)=u_3(x_1,1,x_3,t)=u_3(x_1,-1,x_3,t)=0\\ u_1(x_1,1,x_3,t)&=-u_1(x_1,-1,x_3,t)=1\\ \frac{\partial p}{\partial x_2}(x_1,1,x_3,t)&=\frac{\partial p}{\partial x_2}(x_1,-1,x_3,t)=0\\ \\ u_i(x_1,x_2,x_3,t)&=u_i(x_1+L,x_2,x_3,t)=u_i(x_1,x_2,x_3)\qquad\text{ for }i=1,2,3\\ p(x_1,x_2,x_3,t)&=p(x_1+L,x_2,x_3,t)=p(x_1,x_2,x_3+W,t) \end{split} $$ A lot of numerical research goes into finding the largest $\nu$ which allows a forever-turbulent solution (i.e., allows a solution with no steady limit as $t\to\infty$). This maximum value seems to be about 1/350. If I force $W=L=5$, I too can simulate this turbulence for $\nu$=1/625.
In 2D, however, most experts seem to believe that any initial condition (no matter how "swirly" or how singular, as long as the derivatives written above exist) and any non-zero positive $\nu$ (and any $L$), will decay to the laminar Couette solution as $t\to\infty$: $$ u_1=x_2~~~~~~~u_2=0~~~~~~~u_3=0\\ p=p_\infty~~\text{(some constant, independent of position and time)} $$ In other words, the 2D critical Reynolds number is infinite. What is the best argument for this? I don't believe a proof has ever been given.
