= Solution
Use <pressure> with the ambient <hydrostatic pressure> removed, and let $\Delta P$ be the <pressure> below the falling <sphere> minus that above it. Away from the <sphere> the tube flow is <Hagen-Poiseuille flow>. The upper and lower clear lengths add to $L$ to leading order in $a/L$, so their <pressure> drops add:
$$
\boxed{Q=\frac{\pi a^4\Delta P}{8\mu L}.}
$$
The large exterior reservoir returns this flux with negligible <pressure> loss compared with the long narrow tube. The <sphere> is assumed far enough from the ends that entrance effects and its occupied length give only lower-order corrections.
Now use a frame translating downwards with the <sphere>, and take $z$ positive upwards from its center. The <sphere> is stationary and the tube wall moves upwards with speed $U$. The narrow gap is
$$
h(z)=a-\sqrt{a^2(1-\epsilon/2)^2-z^2}\simeq\frac a2\left[\epsilon+(z/a)^2\right].
$$
It has axial length scale $a\sqrt\epsilon$, making <lubrication theory> appropriate. The <sphere-frame flux in a tube> is constant: the upward relative flux through the whole section is $\pi a^2U-Q$, so its flux per unit circumference is
$$
\boxed{q=\frac{\pi a^2U-Q}{2\pi a}.}
$$
With $y=0$ on the <sphere> and $y=h$ on the wall, <Couette-Poiseuille flow in a thin gap> gives
$$
u(y)=\frac{p_z}{2\mu}y(y-h)+\frac{Uy}{h},\qquad q=-\frac{h^3p_z}{12\mu}+\frac{Uh}{2},\qquad p_z=\frac{6\mu U}{h^2}-\frac{12\mu q}{h^3}.
$$
Since $\Delta P=-\int p_z\,dz$, the pressure-jump relation is
$$
\boxed{\frac{\Delta P}{6\mu}=2qI_3-UI_2,\qquad I_n=\int_{-\infty}^{\infty}\frac{dz}{h(z)^n}.}
$$
The parabolic inner gap can be extended to infinite $z$ because these integrals are dominated by $|z|=O(a\sqrt\epsilon)$. The <lubrication resistance integrals for a nearly occluding sphere> are
$$
I_n=2^n a^{1-n}\epsilon^{1/2-n}J_n,\qquad J_n=\int_{-\infty}^{\infty}(1+\xi^2)^{-n}\,d\xi.
$$
Setting $\xi=\tan\theta$ evaluates $J_1=\pi$, $J_2=\pi/2$, $J_3=3\pi/8$, hence
$$
\boxed{I_1=2\pi\epsilon^{-1/2},\quad I_2=\frac{2\pi}{a}\epsilon^{-3/2},\quad I_3=\frac{3\pi}{a^2}\epsilon^{-5/2}.}
$$
For the <force>, enclose the fluid between sections just above and below the <sphere>, including the annular layer. The net upward <pressure> <force> is $\pi a^2\Delta P$. The tube wall exerts upward shear traction $\mu u_y(h)$ on this fluid; the <sphere> exerts a downward <force> equal to the upward hydrodynamic <force> $F$ on itself. From the gap solution,
$$
u_y(h)=\frac{p_zh}{2\mu}+\frac Uh=\frac{4U}{h}-\frac{6q}{h^2}.
$$
The axial <control volume> <force> balance therefore gives the <control-volume drag formula for a sphere in a tube>:
$$
\boxed{F=\pi a^2\Delta P+2\pi a\mu(4UI_1-6qI_2).}
$$
Using a <control volume> avoids having to integrate the curved-sphere normal <pressure> and tangential shear separately.
With $Q^*=Q/(\pi a^2U)$, $\Delta P^*=a\Delta P/(6\mu U)$, $q^*=2q/(aU)$, $F^*=F/(6\pi\mu aU)$, $I_n^*=a^{n-1}I_n$ and $L^*=4L/(3a)$, the three flux/<pressure> equations become
$$
Q^*=\frac{\Delta P^*}{L^*},\qquad\Delta P^*=q^*I_3^*-I_2^*,\qquad q^*=1-Q^*.
$$
Solving these linear relations gives
$$
\boxed{Q^*=\frac{I_3^*-I_2^*}{L^*+I_3^*},\qquad q^*=\frac{L^*+I_2^*}{L^*+I_3^*},\qquad\Delta P^*=\frac{L^*(I_3^*-I_2^*)}{L^*+I_3^*}.}
$$
The <force> relation is $F^*=\Delta P^*+\tfrac43I_1^*-q^*I_2^*$. Substitution first yields
$$
F^*=\frac{L^*I_3^*-2L^*I_2^*+\tfrac43L^*I_1^*+\tfrac43I_1^*I_3^*-(I_2^*)^2}{L^*+I_3^*}.
$$
Because $I_2^*/I_3^*=2\epsilon/3$ and $I_1^*/I_3^*=2\epsilon^2/3$, the terms $-2L^*I_2^*$ and $\tfrac43L^*I_1^*$ are lower order than $L^*I_3^*$. Keeping the uniformly relevant leading terms gives
$$
\boxed{F^*\simeq\frac{L^*I_3^*+\tfrac43I_1^*I_3^*-(I_2^*)^2}{L^*+I_3^*}=\frac{3\pi L^*\epsilon^{-5/2}+4\pi^2\epsilon^{-3}}{L^*+3\pi\epsilon^{-5/2}}.}
$$
The squared $I_2^*$ is essential; replacing it by $I_2^*I_3^*$ would give the wrong sign and scaling. This establishes the <three drag regimes for a nearly occluding sphere>.
For $L^*\ll\epsilon^{-1/2}$,
$$
\boxed{F^*\sim\frac{4\pi}{3}\epsilon^{-1/2},\quad Q^*\sim1,\quad q^*\sim\frac23\epsilon,\quad\Delta P^*\sim L^*,\quad\frac{F^*}{\Delta P^*}\gg1.}
$$
The <sphere> nearly carries the tube's fluid down with it: the far-field flux is close to the piston displacement rate. The relative annular leakage is small in total volume, and the dominant <force> comes from the local annular shear term. The <pressure> <force> associated with the overall tube resistance is smaller.
For $\epsilon^{-1/2}\ll L^*\ll\epsilon^{-5/2}$,
$$
\boxed{F^*\sim L^*,\quad\Delta P^*\sim L^*,\quad Q^*\sim1,\quad q^*\simeq\frac23\epsilon+\frac{L^*\epsilon^{5/2}}{3\pi}\ll1,\quad\frac{F^*}{\Delta P^*}\sim1.}
$$
The flow remains approximately piston-like, but the dominant <force> is now the <pressure> needed to drive <Hagen-Poiseuille flow> through the long clear tube. Both displayed terms in $q^*$ are retained because their relative size changes inside this regime without producing an additional leading drag regime.
For $L^*\gg\epsilon^{-5/2}$,
$$
\boxed{F^*\sim\Delta P^*\sim3\pi\epsilon^{-5/2},\quad Q^*\sim\frac{3\pi\epsilon^{-5/2}}{L^*}\ll1,\quad q^*\sim1,\quad\frac{F^*}{\Delta P^*}\sim1.}
$$
The large tube resistance makes its net through-flow negligible. The <sphere>'s displaced fluid instead passes upwards relative to it through the thin annular gap, with gap <velocity> of order $U/\epsilon$. Its <pressure> drop and resulting <pressure> <force> dominate.
Finally consider two identical co-moving <spheres>, with nonoverlapping lubrication regions and the same total clear-tube length to leading order. The common through-flux is not doubled. In the middle regime it is still approximately $\pi a^2U$, so the same total tube <pressure> drop is shared equally between the two identical <spheres>. The <load sharing between co-moving spheres in a tube> gives \b[half the single-sphere <force> on each <sphere> in regime (ii)]. In regime (i), drag is set locally by the nearly pressure-balanced annular shear and is unchanged on each <sphere>. In regime (iii), each <sphere> must pass essentially its own displacement flux through its annular gap, giving the same local <pressure> drag as before; adding another <pressure> jump changes the already small tube flux but not either leading drag. \b[The <force> is unchanged in regimes (i) and (iii).]
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