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

Let $y$ measure distance normal to the plane. The normal momentum balance and capillary pressure condition giveThe downslope lubrication equation isApply no slip $u(0)=-U$ and zero tangential stress $u_y(h)=0$. Integration gives the fluxThe thin-film equation $h_t+q_x=0$ is therefore

Long-wave information near a uniform film propagates with the kinematic speedWhen $c>0$, 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.

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, givesDefineThen

Put $H=1+y$ and linearize to obtain $y^{\prime \prime \prime }=-3y$. The two characteristic roots with positive real part are $3^{1/3}{e}^{\pm i\pi /3}$. These are the modes that decay as $X\to -\infty$, soThe oscillations grow as one moves from the uniform film toward the pool.

In a static meniscus, hydrostatic pressure variation $\rho g\ell$ balances capillary pressure $\gamma /\ell$, giving the capillary lengthThe small parameter satisfiesIt is reasonable to match to a nearly static meniscus because the bulk pool has negligible thin-film viscous resistance, while its local capillary-hydrostatic shape approaches the vertical wall almost tangentially.

Since $h=h_{0}H$ and $x=h_{0}X/ϵ$, dimensional curvature isMatching it to $(2\rho g/\gamma {)}^{1/2}=\sqrt{2}/\ell$ gives

The dimensionless equation is $H^3{H}^{\prime \prime \prime }=1-H^3$. At a crest, $H\gg 1$, soThe imposed far-field flux is negligible there; the dominant physical balance is between local gravity-driven drainage and the capillary-pressure gradient.

At a trough, $H\ll 1$, soLocal gravity-driven flux is negligible; the capillary-pressure gradient drives the fixed flux through the trough against viscous resistance. These alternating balances generate the strongly nonlinear capillary waves.

Write $\delta =ϵ/2$. In the final trough, useThe trough equation $H^3{H}^{\prime \prime \prime }=1$ requires $C^4/L^3=1$, so $L=C^{4/3}$. Matching the limiting curvature $H^{\prime \prime }\to {\delta }^{-1/2}$ to ${\varphi }^{\prime \prime }\to 1$ requiresThusand the final trough isIf $\varphi (\xi )\sim -A\xi$ as $\xi \to -\infty$, its slope on the crest side is

Integrating ${\psi }^{\prime \prime \prime }=-1$ and using $\psi (0)={\psi }^{\prime }(0)=0$ givesThe conditions at ${\eta }_0>0$ imply $C={\eta }_0/3$ and $-{\eta }_0^2/6=-1$. Hence

Let the final crest have $H=B\psi ((X-X_c)/L)$. The crest balance $H^{\prime \prime \prime }=-1$ gives $B=L^3$. Matching its terminal slope to the final-trough slope from part (e) givesThereforeAt the end matching to the penultimate trough,

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 hasup to numerical constants. The exponent $3/100$ is extremely small, so attaining a clean asymptotic separation would require unrealistically tiny $ϵ$. Finite geometry, nonzero outer effects, molecular forces, and eventual rupture can intervene before experiments display many members of the predicted wave hierarchy.