Past exam of the mathematics course of the University of Cambridge 2022 iii Paper 329 2 Solution 2026-09-28
Write the perturbation as and use the same factor for all velocity and pressure amplitudes. At the upper interface, the linearized kinematic boundary condition, zero tangential traction, and normal-stress balance areThe first term in the normal stress is the linearization of the attractive disjoining pressure , while the second is the stabilizing capillary pressure. Symmetry about makes even and odd, and supplies the lower-interface conditions.
For a two-dimensional Fourier mode, the Papkovich–Neuber representation is equivalently expressed by the odd Biharmonic stream function for planar Stokes flowThe tangential-stress condition at , with , giveswhereas the kinematic condition gives . The associated normal traction isEquating this with the linearized interfacial traction yields the dispersion relationThis is the Van der Waals rupture instability of a viscous sheet.
For , the growth rate is positive, starts from as , and decreases to zero like as . For , it has the same long-wave limit, vanishes at , and is negative for : surface tension damps wavelengths shorter than the cutoff. Long waves feel the attractive interaction but require coherent flow over a large distance; at short wavelengths viscous resistance suppresses the clean-film instability, while capillarity adds direct decay. A finite film size, fluid inertia, surrounding-fluid stresses, gravity, surface viscosity, and failure of the continuum disjoining pressure law can shift the observable most unstable wavelength.
During a growing thin spot, interfacial flow stretches the surface and dilutes its surfactant, thereby increasing the local surface tension above . Adjacent less-stretched regions retain more surfactant and lower tension. The resulting surface-tension gradient pulls toward the thin spot and opposes the outward flow that drives thinning. This is surfactant stabilization of film rupture; in the strong limit the surfaces behave almost as immobile boundaries.
For strong surfactant and , instability requires . The Taylor expansionreduces the supplied relation toIt is maximal atThus the characteristic rupture time is . Surfactant greatly extends the life of a soap bubble while its film is moderately thick, but the growth-rate dependence predicts rapid final rupture after drainage has made the film sufficiently thin.
Past exam of the mathematics course of the University of Cambridge 2022 iii Paper 344 1 a Solution 2026-09-28
For a binary fluid mixture, the scalar compositional order parameter may be taken as the local concentration difference between its two molecular species. Its spatial integral is fixed by their separately fixed total amounts:Locally, can change only through transport and therefore obeys a continuity equation. By contrast, is the local polar order parameter measuring the mean tail-to-head orientation of the surfactant molecules. Individual molecules can rotate in place, so the integral of need not be conserved.