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
$$
\sigma_{rr}^{\rm out}=2\mu\frac{\partial u_r}{\partial r}\bigg|_{r=a}
=-\frac{2\mu a_t}{a}.
$$
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
$$
-P-\sigma_{rr}^{\rm out}=-\frac\gamma a.
$$
Thus
$$
\boxed{P(z,t)=\frac{2\mu}{a}\frac{\partial a}{\partial t}+\frac\gamma a}.
$$

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-329/2/solution