= Solution
Let $z$ measure distance across the narrow gap and let $\theta$ be polar angle about the tube axis. In <lubrication theory>, radial velocity and radial pressure variation are negligible. Axisymmetric <incompressible flow> is obtained from
$$
u_\theta(\theta,z)=\frac{\phi(z)}{\sin\theta},
$$
because $\partial_\theta(u_\theta\sin\theta)=0$. The tangential <Stokes equation> then separates:
$$
\sin\theta\frac{\partial p}{\partial\theta}
=\mu R\frac{d^2\phi}{dz^2}=a,
$$
and hence, after choosing an irrelevant pressure constant,
$$
\boxed{p(\theta)=\frac a2
\log\frac{1-\cos\theta}{1+\cos\theta}}.
$$
In the sphere frame, $u_z=-u_\theta\sin\theta=-\phi$. The <no-slip boundary condition> gives $\phi(0)=0$ on the sphere and $\phi(\delta)=U$ on the membrane translating backward relative to it. The sphere-frame volume flux inherited from the narrow remote tube is $-\pi r_0^2U$, so
$$
2\pi R\int_0^\delta\phi(z)\,dz=\pi r_0^2U.
$$
Under the asymptotic condition $r_0^2\ll R\delta$, the right-hand side is negligible at leading order. Solving the quadratic profile subject to the two wall values and zero leading-order integral gives
$$
\boxed{\phi(z)=U\left[3\left(\frac z\delta\right)^2
-2\frac z\delta\right]},
\qquad
\boxed{a=\frac{6\mu RU}{\delta^2}}.
$$
Changing the chosen positive tube direction reverses both signs but leaves the drag magnitude unchanged.
The pressure scale is $p\sim\mu UR/\delta^2$, whereas the viscous shear scale is $\tau\sim\mu U/\delta$. After multiplication by comparable areas, pressure drag exceeds shear drag by $R/\delta\gg1$, an instance of <lubrication pressure dominates shear stress>. Put $c=\cos\Delta\theta$. The axial pressure force is
$$
F_p=2\pi R^2\int_{\Delta\theta}^{\pi-\Delta\theta}
p(\theta)\sin\theta\cos\theta\,d\theta
$$
$$
=\pi aR^2\left[
-2c+(1-c^2)\log\frac{1+c}{1-c}
\right].
$$
As $\Delta\theta\to0$, the bracket tends to $-2$, and therefore
$$
F_p\sim-\frac{12\pi\mu R^3}{\delta^2}U.
$$
The resulting <confined-sphere drag coefficient> is
$$
\boxed{\zeta_{\rm tube}=\frac{12\pi\mu R^3}{\delta^2}
=\left(6\pi\mu R\right)\frac{2R^2}{\delta^2}}.
$$
Thus it exceeds the free <Stokes drag law> coefficient by $2R^2/\delta^2$. The <Stokes–Einstein relation> then gives
$$
\boxed{D_{\rm tube}=\frac{k_BT}{\zeta_{\rm tube}}
=D_0\frac{\delta^2}{2R^2}},
\qquad
D_0=\frac{k_BT}{6\pi\mu R}.
$$
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