Let $\varphi=\pi/2-\phi\to0$ near a shoulder of the barrier. Since $\cos\phi\sim\varphi$, the solution from part c predicts
$$
H\sim(R\varphi)^{1/4},
\qquad
\Delta\sim R^{1/4}\varphi^{-3/4}.
$$
Thus the supposedly thin radial region broadens while its large-thickness assumption eventually weakens, and azimuthal derivatives become singular; the approximation cannot remain uniform at the shoulder.

The radial and azimuthal derivative scales of $h^3$ are
$$
\frac{\partial h^3}{\partial r}sim\frac{H^3}{\Delta},
\qquad
\frac1R\frac{\partial h^3}{\partial\phi}
\sim\frac{H^3}{R\varphi}.
$$
They become comparable when $\Delta\sim R\varphi$. Using the expression above,
$$
R^{1/4}\varphi^{-3/4}\sim R\varphi,
$$
so
$$
\boxed{\varphi=O(R^{-3/7})},
\qquad a=\frac37.
$$
At this transition,
$$
\boxed{H=O(R^{1/7})},
\qquad
\boxed{\Delta=O(R^{4/7})}.
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 329 / 3 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-329/3/d/solution