On the upper half-filament, write $\mathbf r(s)=(h(s),s)$ with $0\leq s\leq L$. To first order in slope, <resistive-force theory> supplies the uniform transverse load $f_x=\zeta_\perp U$, and Euler--Bernoulli balance gives
$$
A h''''=\zeta_\perp U.
$$
Midpoint symmetry and clamping impose $h(0)=h'(0)=0$; the free-end force and moment conditions are $h'''(L)=h''(L)=0$. With $H=h/L$ and $\xi=s/L$,
$$
H''''(\xi)=\operatorname{Sp},
\qquad
\operatorname{Sp}=\frac{\zeta_\perp L^3U}{A}.
$$
Four integrations give
$$
\boxed{H(\xi)=\frac{\operatorname{Sp}}{24}
(\xi^4-4\xi^3+6\xi^2)}.
$$
In particular, $H(1)=\operatorname{Sp}/8$ and $H'(1)=\operatorname{Sp}/6$. The small-slope approximation first fails when the end slope is order one, giving the estimate $\operatorname{Sp}\sim6$; parametrically, breakdown occurs at <Sperm number> of order one.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 355 / 1 / a / Solution

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Resistive-force theory approximates the local viscous force on a slender filament by separate drag coefficients multiplying its velocity components parallel and perpendicular to the tangent.


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