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

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