$$u_1=x_1^2x_2$$ $$u_2=A+Bx_1x_2^2$$ $$p=cosx_1$$
The fluid is an incompressible Newtonian fluid $\implies$ $u_i,i=0$ and $tor_ij=-p\delta_ij+\mu u_ij$
Fluid bounded by a stationary rigid plate at $x_2=0$
Find A and B
So far, I have bounded by rigid plate at $x_2=0$ $\implies$ $u(x_2=0)=0$ so $A=0$ I'm trying to formulate $tor_ij$ to find B but getting nowhere. Any ideas?
I then have to define the traction the fluid exerts onto the plate...
I presume this is viscous fluid with viscosity $\mu$. You are right about constant $A$ by using the "no-slip" condition on the boundary $x_2=0$. To get $B$, you may just use the incompressibility condition, i.e.
$$ \nabla \cdot \vec u=\frac{\partial u_1}{\partial x_1}+\frac{\partial u_2}{\partial x_2}=0 $$
The equation should require that $B=-1$.
To get the traction the fluid exerts on the plate, you should be introduced the concept of stress tensor:
$$ \sigma_{ij}=-p\delta_{ij}+2\mu e_{ij} $$ where $$ e_{ij}=\frac12 \left( \frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i} \right) $$
which is the symmetric strain rate tensor.
Having calculated the $2 \times 2$ matrix $\sigma_{ij}$, and remembering that the force per unit area (stress) exerted on the fluid by a surface with unit normal $\vec n$ pointing out of the fluid is $\sigma \times \vec n$, you should then be able to calculate the force, depending on which side it is.
Hope this helps.
