= Solution
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.
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