Integrating $\psi'''=-1$ and using $\psi(0)=\psi'(0)=0$ gives
$$
\psi(\eta)=\frac C2\eta^2-\frac16\eta^3.
$$
The conditions at $\eta_0>0$ imply $C=\eta_0/3$ and $-\eta_0^2/6=-1$. Hence
$$
\boxed{\eta_0=\sqrt6},
\qquad
\boxed{\psi(\eta)=\frac{\eta^2(\sqrt6-\eta)}6}.
$$

Let the final crest have $H=B\psi((X-X_c)/L)$. The crest balance $H'''=-1$ gives $B=L^3$. Matching its terminal slope to the final-trough slope from part (e) gives
$$
L^2=A\delta^{-1/10},
\qquad
L=A^{1/2}\delta^{-1/20},
\qquad
B=A^{3/2}\delta^{-3/20}.
$$
Therefore
$$
\boxed{H(X)\sim A^{3/2}\delta^{-3/20}
\psi\left(\frac{X-X_c}{A^{1/2}\delta^{-1/20}}\right)}.
$$
At the end matching to the penultimate trough,
$$
H''\sim L\psi''(0)
=A^{1/2}\delta^{-1/20}\frac{\sqrt6}{3}
=\boxed{\left(\frac{2A}{3}\right)^{1/2}(\epsilon/2)^{-1/20}}.
$$

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 has
$$
\boxed{H_{\rm penultimate}\asymp\epsilon^{3/100}}
$$
up to numerical constants. The exponent $3/100$ is extremely small, so attaining a clean asymptotic separation would require unrealistically tiny $\epsilon$. Finite geometry, nonzero outer effects, molecular forces, and eventual rupture can intervene before experiments display many members of the predicted wave hierarchy.


Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-329/3/f/solution