Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 329 / 2

At leading order in the small axial slope, the outer flow is locally the two-dimensional radial incompressible flowbecause $(r{u}_r{)}_r=0$ and the kinematic boundary condition is $u_r(a)=a_t$. This field is harmonic as a vector field away from the axis, so the exterior pressure is spatially constant and may be set to zero. Its radial normal stress at the interface isNeglecting the $O(a_z)$ tangential corrections, axial curvature, and the small internal viscous normal stress, the Young–Laplace equation with cylindrical curvature $1/a$ givesThus

Interpret the stated ansatz asMultiplying the pressure relation by $a$ gives $2\mu a_t=Pa-\gamma$. Equating its constant and sinusoidal parts yieldsThe volume per wavelength is proportional to the mean of $a^2$, namely $f^2+g^2$. HenceChoosingthen gives exactly

For $g\ll f\simeq a_0$, the disturbance amplitude obeysso its long-wave linear growth rate isThe minimum radius is $a_{\min }=f-\sqrt{2}g$, andSince the trajectory follows the circle $f^2+g^2={\alpha }_0$ toward increasing $g$, it reaches $f=\sqrt{2}g$ and hence $a_{\min }=0$. Thus the nonlinear disturbance narrows monotonically to pinch-off within this approximation.

When internal pressure gradients matter, axial lubrication flow in the low-viscosity interior has volume fluxConservation of cross-sectional area, $(\pi a^2{)}_t+Q_z=0$, and the normal-stress expression for $P$ give

Let $\tau =t^{\ast }-t$. Balancing $2\mu a_t/a$ with $\gamma /a$ gives $a\sim \tau$, so $p=1$. Balancing either pressure contribution after two axial derivatives with $a{a}_t$ then gives $q=1$. DefineAt fixed $z$, $a_t=C(\zeta A^{\prime }-A)$, andSubstitution leaves the parameter-free third-order equationThus the similarity exponents are $\boxed{p=q=1}$; boundary and matching conditions would select a particular profile $A$.