Fixing the boundary condition like [1] does on the wave equation constrains the type of function which can be a solution to the wave equation. In this case making it a sine to meet the boundary condition. My question is for the Navier Stokes equation in general how does the no slip boundary condition constrain the type of function that can be a solution? Is this question even answerable without a general solution like in the wave equation case?
[1] https://www.math.ubc.ca/~feldman/m267/separation.pdf