 Solution

Use the Unscaled Papkovich–Neuber representation with $e_k=e^{ikx-ky}$ and harmonic potentialsTo see how these coefficients arise, start with $\Phi =(0,A{e}_k)$ and $\chi =B{e}_k$. The representation givesAt $y=0$, the prescribed tangential velocity requires $ikB=U$, while zero normal velocity requires $A=-kB=iU$. Hence the decaying half-space flow isThe pressure follows from $2\mu {\partial }_y{\Phi }_y$. These harmonic potentials verify the Stokes equation and incompressibility; the polynomial factors do not prevent decay as $y\to \infty$.

For the Newtonian fluid stress tensor, ${\sigma }_{xy}=\mu (u_y+v_x)$ and ${\sigma }_{yy}=-p+2\mu v_y$. On the boundary $u_y=-2kU{e}^{ikx}$, $v_x=0$ and $v_y=-ikU{e}^{ikx}$. ThereforeTake real parts for a physical sinusoidal flow. Together with the normal-velocity result supplied in the question, these formulas give the Fourier traction map for a viscous half-space; for signed wave numbers its resistance coefficient is $2\mu |k|$.