Let $y$ measure distance normal to the plane. The normal momentum balance and capillary pressure condition give
$$
p=\rho g\cos\theta\,(h-y)-\gamma h_{xx},
\qquad
p_x=\rho g\cos\theta\,h_x-\gamma h_{xxx}.
$$
The downslope lubrication equation is
$$
\mu u_{yy}=p_x-\rho g\sin\theta.
$$
Apply no slip $u(0)=-U$ and zero tangential stress $u_y(h)=0$. Integration gives the flux
$$
q=-Uh+\frac{h^3}{3\mu}
\left(\rho g\sin\theta-
ho g\cos\theta\,h_x+\gamma h_{xxx}\right).
$$
The <thin-film equation> $h_t+q_x=0$ is therefore
$$
\boxed{
h_t-Uh_x+\frac{\rho g\sin\theta}{3\mu}(h^3)_x
=\frac1{3\mu}\frac\partial{\partial x}
\left[h^3\left(\rho g\cos\theta\,h_x-\gamma h_{xxx}\right)\right]}.
$$

Long-wave information near a uniform film propagates with the kinematic speed
$$
c=q'(h_0)=-U+\frac{\rho g h_0^2\sin\theta}{\mu}.
$$
When $c>0$, disturbances travel from $x=-\infty$ into the domain, so $h\to h_0$ is an admissible upstream boundary condition. When $c<0$, information travels toward $x=-\infty$ from the pool, so the same condition cannot independently be imposed there.

Solution

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

 Solution

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