Analytical solution for Navier-Stokes where non-linear term is important In general, analytical solution of the Navier-Stokes is difficult to obtain. Simple cases where analytical solutions are known include Poiseuille flow and barometric pressure variation for compressible fluids. In all the cases I know the non-linear term in fluid velocity turns out to be zero because of the geometry of the problem. Is there any case where the analytical solution is wrong when the non-linear term is neglected?
 A: There are several well known exact solutions of the Navier-Stokes equations widely used to test numerical algorithms, for example,
C. ROSS ETHIER and D. A. STEINMAN. Exact fully 3d Navier-Stokes solutions for benchmarking. INTERNATIONAL JOURNAL FOR NUMERICAL METHODS IN FLUJDS, 19:369–375, 1994.
\begin{equation}\label{benchmarc_solution}
    \begin{array}{cc}
u_e=-a \exp(-d^2 t) (\exp(a x) \sin(a y + d z) + \exp(a z) \cos(a x + d y)),\\
v_e =-a \exp(-d^2 t) (\exp(a y) \sin(a z + d x) + \exp(a x) \cos(a y + d z)),\\
w_e = -a \exp(-d^2 t) (\exp(a z) \sin(a x + d y) + \exp(a y) \cos(a z + d x)),\\
p_e = -\frac{a^2}{2}\exp(-2 d^2 t) (\exp(2 a x) + \exp(2 a y) + \exp(2 a z) +\\
   2 \sin(a x + d y) \exp(a (y + z)) \cos(a z + d x) +\\ 
   2 \sin(a y + d z) \exp(a (x + z)) \cos(a x + d y) +\\ 
   2 \sin(a z + d x) \exp(a (y + x)) \cos(a y + d z))
    \end{array}
\end{equation}
here $a, d$ are free parameters. In this example $(\vec{u}.\nabla)\vec{u} \ne 0$, where $\vec {u}=(u_e,v_e,w_e)$. The usage of this solution can be found   here.
A: In cylindrical Couette flow the non-linear term is non-zero: in fact it is the centripetal acceleration $$\mathbf{ u}\cdot\nabla\mathbf{u}=
-\frac{u^2}{r}\mathbf{\hat r}.$$
