Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 329 / 3 / f

Integrating ${\psi }^{\prime \prime \prime }=-1$ and using $\psi (0)={\psi }^{\prime }(0)=0$ givesThe conditions at ${\eta }_0>0$ imply $C={\eta }_0/3$ and $-{\eta }_0^2/6=-1$. Hence

Let the final crest have $H=B\psi ((X-X_c)/L)$. The crest balance $H^{\prime \prime \prime }=-1$ gives $B=L^3$. Matching its terminal slope to the final-trough slope from part (e) givesThereforeAt the end matching to the penultimate trough,

The trough scaling from part (e) says that a trough matched to curvature $K$ has thickness scale $K^{-3/5}$. With $K\asymp {\delta }^{-1/20}$, the penultimate trough therefore hasup to numerical constants. The exponent $3/100$ is extremely small, so attaining a clean asymptotic separation would require unrealistically tiny $ϵ$. Finite geometry, nonzero outer effects, molecular forces, and eventual rupture can intervene before experiments display many members of the predicted wave hierarchy.