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

On the upper half-filament, write $\mathbf{r}(s)=(h(s),s)$ with $0\le s\le L$. To first order in slope, resistive-force theory supplies the uniform transverse load $f_x={\zeta }_{\perp }U$, and Euler--Bernoulli balance givesMidpoint symmetry and clamping impose $h(0)=h^{\prime }(0)=0$; the free-end force and moment conditions are $h^{\prime \prime \prime }(L)=h^{\prime \prime }(L)=0$. With $H=h/L$ and $\xi =s/L$,Four integrations giveIn particular, $H(1)=\mathrm{Sp}/8$ and $H^{\prime }(1)=\mathrm{Sp}/6$. The small-slope approximation first fails when the end slope is order one, giving the estimate $\mathrm{Sp}\sim 6$; parametrically, breakdown occurs at Sperm number of order one.

Choose tangent and normal vectorsso ${\hat{\mathbf{t}}}_s=-\kappa \hat{\mathbf{n}}$, ${\hat{\mathbf{n}}}_s=\kappa \hat{\mathbf{t}}$, and $\kappa ={\theta }_s$. Thenwhile the drag components are $f_n=-{\zeta }_{\perp }U\cos \theta$ and $f_t=({\zeta }_{\perp }U/2)\sin \theta$.

LetSince $N=A{\kappa }_s$, one has $\mathcal{N}=K^{\prime }$. Force balance becomesAt the free end, $\mathcal{T}(1)=\mathcal{N}(1)=K(1)=0$; at the held midpoint, position and orientation impose $\theta (0)=0$ in this convention. Reversing the normal reverses the corresponding signs without changing the shape.

Far from the midpoint at $\mathrm{Sp}\gg 1$, the filament aligns with the flow: $\theta =\pi /2$, $K=\mathcal{N}=0$. The tangential equation and free-end condition give the outer tension

Near $\xi =0$, the outer tension is $\mathrm{Sp}/2$ to leading order. The normal equation is dominated by bending and tension,because the direct normal drag is smaller by ${\mathrm{Sp}}^{-1/2}$ in the turning layer. Hence the decaying inner curvature iswhere matching and the midpoint angle determine $C$. The elastohydrodynamic boundary layer of a filament therefore has scaled arclength