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

For a slender viscous bubble, approximate the interface normal by the radial direction and its curvature by $1/a$. Axial curvature and axial derivatives of the leading external motion are smaller by the slenderness factors. The exterior leading Stokes flow is locally radial. Incompressibility gives ${\partial }_r(r{u}_r)=0$, while the kinematic boundary condition gives $u_r(a)=a_t$ to leading order, henceThe radial viscous Laplacian of this $1/r$ velocity vanishes, so exterior pressure is constant to leading order; set its reference value to zero. Its radial normal stress at the interface is $2\mu {\partial }_r{u}_r{|}_a=-2\mu a_t/a$. Inner viscous normal stress is smaller by the viscosity ratio $\lambda$, and the bubble's leading normal stress is $-P$. The interfacial stress balance with variable surface tension, here with constant $\gamma$, consequently givesThese are local leading-order relations: the outer axial flow and tangential traction enter at higher slenderness order. The sign also has a useful check: a uniform cylinder with no imposed excess pressure contracts, $a_t=-\gamma /(2\mu )$, under surface tension.

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.

Let $s=a_{\min }$ be a small neck radius in the preceding sinusoidal shape. Expanding around a minimum gives $a\simeq s+(g{k}^2/\sqrt{2})(z-z_{\ast }{)}^2$, so its local axial length is $\ell \sim [s/(g{k}^2){]}^{1/2}$. The uniform-pressure model has $|a_t|\sim \gamma /\mu$ near the neck. Local conservation of mass therefore demands axial inner flux $Q\sim s(\gamma /\mu )\ell$. Internal Hagen-Poiseuille flow of viscosity $\lambda \mu$ would require a pressure differenceIf $sk\ll \lambda$, while $gk\ll 1$ by slenderness, this ratio is very large. Hence internal pressure variation certainly cannot be neglected when $a_{\min }k\ll \lambda$. This is a sufficient breakdown condition; appreciable variation can arise earlier, especially because the radial viscous and capillary terms nearly cancel in the uniform-pressure regime.

For the inner flow, the leading axial pressure is uniform across a section. Neglect the much smaller interfacial axial velocity relative to the internal pressure-driven motion. The lubrication approximation solves $\lambda \mu r^{-1}(r{u}_{z,r}{)}_r=P_z$, with regularity at zero and $u_z(a)=0$ at leading order. ThusThe area balance $(\pi a^2{)}_t+Q_z=0$, combined with the exterior normal-stress relation, gives the lubrication equation for a bubble with viscous exterior:The neglected inner viscous normal stress is smaller by $\lambda$ in the similarity scales obtained next; the external axial interfacial speed is likewise smaller than the large pressure-driven inner speed.

Put $\tau =t_{\ast }-t$, $a\sim {\tau }^p$ and axial length $\ell \sim {\tau }^q$. The radial viscous normal stress scales as $\mu a_t/a\sim {\tau }^{-1}$ and capillary pressure as $\gamma /a\sim {\tau }^{-p}$. Retaining both requires $p=1$. The left side $a{a}_t$ scales as ${\tau }^{2p-1}$, while the axial-flux divergence scales as $a^{4}P/{\ell }^2\sim {\tau }^{4p-1-2q}$. Equating them gives $q=p=1$. Thus both local radius and axial length shrink linearly in the remaining time.

To obtain the specified dimensionless coefficients, choose the linear pinch-off scaling of a slender viscous bubbleThen $a_t=(\gamma /\mu )(\zeta A^{\prime }-A)$ andThe constant $-2$ has no axial derivative. Substituting into the flux equation and cancelling its dimensional factor givesBoth axial derivatives on the right are essential; the outer derivative is present in the PDF but lost in the converted TeX expression. The physical slope is $a_z=\sqrt{\lambda }{A}^{\prime }$, so an order-one similarity profile remains slender for $\lambda \ll 1$. No solution of this similarity ordinary differential equation is required.