Write $\delta=\epsilon/2$. In the final trough, use
$$
H=C\phi(\xi),
\qquad
\xi=\frac{X-X_0}{L}.
$$
The trough equation $H^3H'''=1$ requires $C^4/L^3=1$, so $L=C^{4/3}$. Matching the limiting curvature $H''\to\delta^{-1/2}$ to $\phi''\to1$ requires
$$
\frac C{L^2}=C^{-5/3}=\delta^{-1/2}.
$$
Thus
$$
\boxed{C=\delta^{3/10}},
\qquad
\boxed{L=\delta^{2/5}},
$$
and the final trough is
$$
\boxed{H(X)\sim\delta^{3/10}
\phi\left(\frac{X-X_0}{\delta^{2/5}}\right)}.
$$
If $\phi(\xi)\sim-A\xi$ as $\xi\to-\infty$, its slope on the crest side is
$$
\boxed{H'\sim-A\frac CL=-A\delta^{-1/10}
=-A(\epsilon/2)^{-1/10}}.
$$


Solution

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

 Solution

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