For a stationary vertical plane, set $U=0$, $\sin\theta=1$, and $\cos\theta=0$. One integration of the steady equation, using $h\to h_0$ and vanishing derivatives far above the pool, gives
$$
\gamma h^3h'''=\rho g(h_0^3-h^3).
$$
Define
$$
\boxed{H=\frac h{h_0}},
\qquad
\boxed{X=\frac{\epsilon x}{h_0}},
\qquad
\epsilon=\left(\frac{\rho g h_0^2}{\gamma}\right)^{1/3}.
$$
Then
$$
\boxed{H^3H'''=1-H^3}.
$$

Put $H=1+y$ and linearize to obtain $y'''=-3y$. The two characteristic roots with positive real part are $3^{1/3}e^{\pm i\pi/3}$. These are the modes that decay as $X\to-\infty$, so
$$
\boxed{H-1\sim C e^{3^{1/3}X/2}
\cos\left(\frac{\sqrt3,3^{1/3}}2X+\delta\right)}.
$$
The oscillations grow as one moves from the uniform film toward the pool.


Solution

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

 Solution

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