= 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 $\mu u''=G$: the pressure force opposes positive $x$ motion. The <no-slip boundary condition> imposes $u(0)=0$, $u(h)=U$. Integrating twice gives
$$
\boxed{u(y)=-\frac G{2\mu}y(h-y)+\frac{Uy}{h}.}
$$
The <velocity gradient> is $u'=G(y-h/2)/\mu+U/h$, an increasing function of $y$. It is negative somewhere in the channel exactly when its lower-wall limit is negative:
$$
\boxed{G>\frac{2\mu U}{h^2}.}
$$
Under this condition the negative-gradient interval is $0<y<h/2-\mu U/(Gh)$.
The <volume flux> per unit span is
$$
Q=\int_0^h u(y)\,dy=-\frac{Gh^3}{12\mu}+\frac{Uh}{2}.
$$
Thus the <zero-flux Couette-Poiseuille flow> requires
$$
\boxed{G=\frac{6\mu U}{h^2},\qquad \frac uU=3s^2-2s,\quad s=y/h.}
$$
The 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.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2014/ib/paper-1-zero-flux-channel.png]
{title=Zero-net-flux channel velocity, showing reverse flow below two-thirds depth, its minimum at one-third depth, and the moving upper wall}
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