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

At leading order in the small axial slope, the outer flow is locally the two-dimensional radial incompressible flow

u_{r} (r, z, t) = \frac{a a _{t}}{r},

because

(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 is

σ_{rr}^{out} = 2 μ \frac{\partial u _{r}}{\partial r}_{r = a} = - \frac{2 μ a _{t}}{a} .

Neglecting the

O (a_{z})

tangential corrections, axial curvature, and the small internal viscous normal stress, the Young–Laplace equation with cylindrical curvature

1/ a

gives

- P - σ_{rr}^{out} = - \frac{γ}{a} .

Thus

P (z, t) = \frac{2 μ}{a} \frac{\partial a}{\partial t} + \frac{γ}{a} .

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