= Solution
For planar <incompressible flow>, $u_x+w_z=0$. Locally this permits a <streamfunction> with $u=\psi_z$, $w=-\psi_x$, because equality of the mixed derivatives enforces <incompressibility>. The remaining component of the <curl> is $u_z-w_x=\psi_{zz}+\psi_{xx}$, so <irrotational flow> requires <Laplace's equation> for $\psi$.
Put $K=\pi/b$. The sinusoidal contribution to the supplied <streamfunction> has $\psi_{xx}=U_0aK^2\cos(Kx)e^{-Kz}$ and $\psi_{zz}=-U_0aK^2\cos(Kx)e^{-Kz}$; the linear term is also harmonic. Hence
$$
u=U_0[1+aK\cos(Kx)e^{-Kz}],\qquad
w=-U_0aK\sin(Kx)e^{-Kz}.
$$
As $z\to\infty$, these approach the specified <uniform flow>, $(u,w)=(U_0,0)$.
The impermeability <kinematic boundary condition> is $w=u\,z_s'(x)$ on $z=z_s(x)$, equivalently $\psi$ is constant on the wall. For $aK\ll1$, evaluating the perturbed velocities at $z=0$ gives $w=-U_0aK\sin(Kx)=U_0z_s'$ to first order. The residual is of order $U_0(aK)^2$. Equivalently $\psi(x,z_s)=U_0a\cos(Kx)[1-e^{-Ka\cos(Kx)}]=O(U_0Ka^2)$. The surface velocity is
$$
\boxed{u_s=U_0[1+aK\cos(Kx)e^{-Ka\cos(Kx)}]
=U_0[1+aK\cos(Kx)]+O(U_0(aK)^2).}
$$
The first expression is the supplied approximate field evaluated exactly at the wall; its accuracy as a wall-flow solution is still only first order in $a/b$.
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