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For a nonrotating force-free sphere moving with speed $U$ in a stationary tube, use its moving frame: the tube wall has velocity $-U$. In the parabolic lubrication gap $h=h_0+x^2/(2a)$, $h_0=ϵa(1+\lambda \cos \theta )$, moving-boundary lubrication flux gives $q=-h^3{p}_x/(12\mu )-Uh/2$. The wall-shear and pressure-drop balance in a tube rules out an $O({ϵ}^{-3/2})$ net pressure jump. Applying this common pressure recovery to each angular strip gives $q=-2U{h}_0/3$. The next balance yields the displayed positive localized pressure drop. Meanwhile the integrated leading shear stress on the sphere vanishes in each strip; this is consistent with negligible leading torque, and does not mean the local shear or pressure force vanishes.

Integrate axial force balance $\nabla \cdot \sigma =0$ over the fluid between the two remote tube sections, excluding the particle. The wall's axial traction contributesAt each remote section the Hagen-Poiseuille flow has $u_x$ independent of $x$, so ${\sigma }_{xx}=-p$. The downstream and upstream caps therefore contribute $-\pi a^{2}p(x_2)+\pi a^{2}p(x_1)$. The particle contributes zero net axial force because it is force-free. Adding the three contributions proves the wall-shear and pressure-drop balance in a tube:Here $p$ is the physical pressure. The sign follows from the outward normals on the two caps; no convention for a positive pressure-drop magnitude has yet been introduced.

Apply axial momentum balance to the fluid between the remote sections $x_1$ and $x_2$. The pressure forces on the end discs, the integrated tube-wall shear, and the force exerted by the sphere are the only resultant axial forces. The sphere is force-free, so its contribution vanishes andThis is the wall-shear and pressure-drop balance in a tube.