 Solution

Take a steady parallel velocity $(u(y),0)$ and interpret the specified gradient as $p_x=G>0$. The horizontal Navier-Stokes equation reduces to $\mu u^{\prime \prime }=G$: the pressure force opposes positive $x$ motion. The no-slip boundary condition imposes $u(0)=0$, $u(h)=U$. Integrating twice givesThe velocity gradient is $u^{\prime }=G(y-h/2)/\mu +U/h$, an increasing function of $y$. It is negative somewhere in the channel exactly when its lower-wall limit is negative:Under this condition the negative-gradient interval is $0<y<h/2-\mu U/(Gh)$.

The volume flux per unit span isThus the zero-flux Couette-Poiseuille flow requiresThe profile starts at zero, reaches $u=-U/3$ at $y=h/3$, crosses zero at $y=2h/3$, and reaches $U$ at the moving wall. Negative and positive flux contributions cancel, even though the local velocity is nonzero.