# Finding the stress tensor?

Consider the flow $\vec u=(uy,0,0)$ between two plates $y=0$ and $y=1$ (chosen out of simplicity). I want to find the stress tensor of such a flow given by: $$\sigma_{ij}=-\delta_{ij}p+\eta \left( \frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_k}\right)$$ Although this seems quite easy (e.g. use Navier-Stokes to find $p$ and then simply sub $\vec u$ into the above) I can find no examples (at all) of a calculation of the stress tensor for any flow $\vec u$. Is there a reason behind this? Am I missing something? If not please can you provide such an example.

There are many exact solutions of the Navier-Stokes equation, which you can find in standard text books. In practice we don't usually specify the velocity field. Instead, we have an equation of state $P(\rho,T)$, and a set of boundary and/or initial conditions.
The specific example you provide is very simple (this is known as Poiseulle flow). In order to satisfy no-slip boundary conditions, the top surface has to move at speed $(u,0,0)$. Then your velocity field is a solution to the static continuity and Navier-Stokes equation for a constant pressure. This is the case because the fluid velicity is linear, so the stress tensor, which is build from first derivatives of $\vec{u}$, is constant. $$\sigma_{xx}=\sigma_{yy}=\sigma_{zz}=-P, \; \; \; \; \sigma_{xy}=\sigma_{yx}=\eta u.$$ Then the derivative of the stress tensor (the RHS of the Navier-Stokes equation) is zero. The LHS is zero because the flow is static, $\partial_t \vec{u}=0$, and the comoving derivative $(\vec{u}\cdot\vec\nabla)\vec{u}$ also vanishes. As a result the Navier-Stokes equation is indeed satisfied.