Clay Institute Navier Stokes I'm trying to understand the Navier-Stokes Problem from the Clay Institute text here: http://www.claymath.org/sites/default/files/navierstokes.pdf . I've done an example problem to see what the problem is, but I don't run into any.
We are trying to satisfy:
\begin{equation}
\frac{\partial \textbf{u}}{\partial t} + (\textbf{u}\cdot\nabla)\textbf{u}=-\frac{\nabla P}{\rho} + \nu\nabla^{2}\textbf{u}+\textbf{f},
\end{equation}
\begin{equation}\label{incompr}
\nabla\cdot\textbf{u}=0.
\end{equation}
We get to pick a velocity (u) and pressure (P) function to satisfy these equations. We take $n=3$ by putting our velocity and pressure vectors in the 3D Cartesian plane with $x,y,z$. We let $\textbf{f}$ be 0 according to the problem description, and assume kinematic viscosity (nu) to be greater than 0. Let
\begin{equation}
     \textbf{u}(x,t)=\begin{bmatrix}
                x+t \\
                -2y+t \\
                z+t 
                \end{bmatrix}           
\end{equation}
Then $\textbf{u}(x,t)$ satisfies the divergence free condition because 
\begin{equation}
\nabla \cdot \textbf{u} = \frac{\partial u_1}{\partial x}+\frac{\partial u_2}{\partial y}+\frac{\partial u_3}{\partial z}
\end{equation}
which is
\begin{equation}
\nabla \cdot \textbf{u} = 1-2+1=0
\end{equation}
Then 
\begin{equation}\frac{\partial \textbf{u}}{\partial t}=\begin{bmatrix}
                1 \\
                1 \\
                1 
                \end{bmatrix}
\end{equation}
And
\begin{equation}
\textbf{u} \cdot \nabla = (x+t)\frac{\partial}{\partial x} + (-2y+t)\frac{\partial}{\partial y}+(z+t)\frac{\partial}{\partial z}
\end{equation}
And
\begin{equation}
\begin{split}
(\textbf{u} \cdot \nabla)\textbf{u} &= (x+t)\frac{\partial \textbf{u}}{\partial x} + (-2y+t)\frac{\partial \textbf{u}}{\partial y}+(z+t)\frac{\partial \textbf{u}}{\partial z} \\
           &=(x+t)\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}+(-2y+t)\begin{bmatrix} 0 \\ -2 \\ 0 \end{bmatrix}+(z+t)\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \\
           &= \begin{bmatrix} x+t \\ 4y-2t \\ z+t \end{bmatrix}
\end{split}
\end{equation}
Since laplacian of $\textbf{u}$ is 0 the whole kinematic velocity term goes to 0. And the final Navier-Stokes expression is:
\begin{equation}
\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}+\begin{bmatrix} x+t \\ 4y-2t \\ z+t \end{bmatrix}=-\frac{\nabla P}{\rho}
\end{equation}
So we set $\rho$ equal to 1 for easy calculation. Bring the negative to the left hand side and combine the left hand side.
\begin{equation}
\begin{bmatrix} -x-t-1 \\ -4y+2t-1 \\ -z-t-1 \end{bmatrix}=\nabla P
\end{equation}
and we solve for a solution for P
\begin{equation}
P=\frac{-x^2}{2}-tx-x-2y^2+2ty-y-\frac{z^2}{2}-tz-z
\end{equation}
So we now have infinitely differentiable functions for u and P and they are smooth since they are polynomial. Where did I make a mistake/misunderstand the problem?
 A: In the context of the Clay Institute's presentation of the Navier-Stokes equations (as subject for the Millennium Problem Prizes), your attempt most closely resembles the challenge to give a proof of statement (A):

(A) Existence and smoothness of Navier–Stokes solutions on $\mathbb{R}^3$.
  Take $\nu \gt 0$ and $n = 3$. Let $u^\circ(x)$ be any smooth, divergence-free vector field satisfying (4).  Take $f(x, t)$ to be identically zero. Then there exist smooth functions $p(x, t), u_i(x, t)$ on $\mathbb{R}^3 × [0,\infty)$ that satisfy (1), (2), (3), (6), (7).

The notation of the linked PDF conforms closely to the notation of the Question above.  The component version of (1) there has been recast here as a vector equation above, $\textbf{u}=(u_1,u_2,u_3)$, and pressure $p$ there has been labelled $P$ here.  Also the mentions of $x$ in the PDF were intended to mean points in $\mathbb{R}^n$ with $n=2,3$. For consistency we will adopt the notation of the Question, asking Readers to supply any necessary rewriting for comparison with the PDF.  In particular, except where quoting the PDF, our $x$ is only the first coordinate of $\mathbb{R}^3$, supplemented by variable $y,z$ as used in the Question.
As already noted by Commenters RRL and Winther, the N-S solution provided in the Question satisfies (1) and (2) but only for a specific initial condition (3):
$$ \textbf{u}^\circ(x,y,z) = (x,-2y,z) $$
and this linear initial condition is not "physically reasonable" as defined in the PDF's "bounded energy" restriction (7):
$$ \int_{\mathbb{R}^3} |u(x,t)|^2 \textrm{d}x \lt C \;\;\text{ for all }\; t \ge 0 $$
NB: This is one of the cases where, in quoting the PDF, $x$ has a multidimensional meaning.
Indeed a fair reading of the linked PDF would require the proof of statement (A) to apply to all initial conditions satisfying (4) for some choice of $\alpha,K \gt 0$:
$$ | \partial_x^\alpha u^\circ(x) | \le C_{\alpha K} (1+|x|)^{-K} \;\;\text{ on }\mathbb{R}^n, \text{ for some positive } \alpha \text{ and } K $$
The PDF's bound (4) involves both smoothness and growth of the initial condition and leaves something to the Reader's imagination to interpret how $\alpha$ and $K$ are quantified.  However the "bounded energy" inequality (7) must also be satisfied by the initial condition by taking $t=0$, so we may confidently excluded the solution offered above from what would satisfy the Millennium Problem requirements.

For some background and perspective on the incompressible Navier-Stokes in the context of more general fluid flow problems, the Wikipedia article Navier-Stokes equations is a reasonable jumping off point.
In general it should be noted that the two-dimensional problems corresponding to (A) and (B) (periodic solution) were settled by Ladyzhenskaya.
Much literature on the incompressible N-S equations is concerned with numerical approximation schemes, esp. finite elements.  In this context the analysis of discrete versions of the N-S equations can be understood as parallels to the continuous problem.  In that vein I would recommend my adviser's book, Finite Element Methods for Viscous Incompressible Flows: A Guide to Theory, Practice, and Algorithms by Max Gunzburger (2012).
