= Solution
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> contributes
$$
a\int_{x_1}^{x_2}\int_0^{2\pi}\sigma_{rx}(a,\theta,x)d\theta dx.
$$
At each remote section the <Hagen-Poiseuille flow> has $u_x$ independent of $x$, so $\sigma_{xx}=-p$. The downstream and upstream caps therefore contribute $-\pi a^2p(x_2)+\pi a^2p(x_1)$. 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>:
$$
\boxed{a\int_{x_1}^{x_2}\int_0^{2\pi}\sigma_{rx}(a,\theta,x)d\theta dx
=\pi a^2[p(x_2)-p(x_1)].}
$$
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.
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