 One-charged-wall electro-osmotic channel

With $\psi (0)=\zeta$, $\psi \prime (d)=0$ and no slip at either wall, linear screening gives $\psi /\zeta =\cosh [\kappa (d-x)]/\cosh (\kappa d)$. The velocity is $v/U=1-(1-\mathrm{sech}\kappa d)x/d-\cosh [\kappa (d-x)]/\cosh (\kappa d)$. This electrically neutral-wall boundary assumes no external normal field; it differs from a grounded wall.