Far from the midpoint at $\operatorname{Sp}\gg1$, the filament aligns with the flow: $\theta=\pi/2$, $K=\mathcal N=0$. The tangential equation and free-end condition give the outer tension
$$
\boxed{\mathcal T_{\rm out}(\xi)
=\frac{\operatorname{Sp}}2(1-\xi)}.
$$

Near $\xi=0$, the outer tension is $\operatorname{Sp}/2$ to leading order. The normal equation is dominated by bending and tension,
$$
K''-\frac{\operatorname{Sp}}2K=0,
$$
because the direct normal drag is smaller by $\operatorname{Sp}^{-1/2}$ in the turning layer. Hence the decaying inner curvature is
$$
\boxed{K(\xi)=C
\exp\!\left[-\sqrt{\frac{\operatorname{Sp}}2}\,\xi\right]},
$$
where matching and the midpoint angle determine $C$. The <elastohydrodynamic boundary layer of a filament> therefore has scaled arclength
$$
\boxed{\delta\xi=O(\operatorname{Sp}^{-1/2})}.
$$

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-355/1/c/solution