= Solution
For a steady unidirectional field $\mathbf u=w\mathbf k$, <incompressibility> gives $\partial_z w=0$. Thus $w=w(x,y)$ and the convective acceleration vanishes. The transverse momentum equations give $p_x=p_y=0$, so $p=p(z)$. The axial equation is
$$
\mu\Delta_\perp w=p'(z).
$$
Its left side is independent of $z$ and its right side independent of $x,y$, so both equal a constant $G$. No slip against the translating wall gives $w=U$ on the boundary. Thus the asserted Poisson problem follows.
For the circular pipe, the axisymmetric solution is
$$
\boxed{w(r)=U+\frac G{4\mu}(r^2-R^2).}
$$
It satisfies the boundary value and $\Delta_\perp r^2=4$. Integrating over the cross-section gives
$$
Q=2\pi\int_0^Rw(r)r\,dr=\pi R^2U-\frac{\pi R^4G}{8\mu}.
$$
Therefore the laboratory-frame <volume flux> vanishes precisely when
$$
\boxed{U=\frac{R^2}{8\mu}\frac{dp}{dz}.}
$$
For a <pressure> decrease in the positive axial direction, this requires upstream wall translation.
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