Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 78 / 2 / a

When pressure is uniform along the bubble, its normal-stress relation is $2\mu a_t=Pa-\gamma$. Use the PDF's amplitude convention $a=f+\sqrt{2}g\sin kz$; the square root acts on $2$ only, not on $2g$. Comparing constant and sinusoidal terms givesThe mean cross-sectional area of an incompressible periodic segment is proportional to $⟨a^2⟩=f^2+g^2$ and is conserved. ThusIts time derivative vanishes, and substitution of the preceding two equations gives $P{\alpha }_0=\gamma f$. Therefore the uniform-pressure evolution of a slender viscous bubble isFor a small disturbance to radius $a_0$, $f=a_0+O(g^2)$ and ${\alpha }_0=a_0^2+O(g^2)$, so its amplitude grows as $g\propto e^{st}$ withThe leading long-wave growth rate is independent of $k$ because axial curvature has been neglected; this is not a finite-wavelength dispersion relation.

While $f>\sqrt{2}g>0$, the minimum radius is $a_{\min }=f-\sqrt{2}g$. Its derivative isThus the minimum keeps decreasing in the nonlinear uniform-pressure model, with no need to solve $f,g$ explicitly. Finite internal viscosity eventually invalidates this model close enough to the neck.