Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 329 / 2

For the stated axisymmetric flow, the diagonal components of the rate-of-strain tensor areTheir sum is $u_r+u/r+w_z=0$, the incompressibility condition. In the leading thin-sheet approximation, vanishing tangential traction makes $u$ independent of $z$. The normal stress boundary condition is ${\sigma }_{zz}=-p_{\mathrm{ext}}$, so the Newtonian fluid stress tensor givesThereforeFor the small annular sector, the inner and outer radial faces contribute $2\delta \theta {\partial }_r(rh{\sigma }_{rr})\delta r$ in the radial direction. The two azimuthal faces contribute $-2\delta \theta h{\sigma }_{\theta \theta }\delta r$: their hoop tractions have inward radial components. The combined radial force of the external pressure on the sloping upper and lower surfaces is $2\delta \theta r{p}_{\mathrm{ext}}{h}_r\delta r$.

Neglecting inertia, force balance is thusSubstituting the two stresses cancels the terms involving $p_{\mathrm{ext}}{h}_r$ and gives the axisymmetric viscous-sheet stretching equations:Finally, conservation of mass in the sector gives

Put $b=\Delta \rho g>0$, so $p_{\mathrm{ext}}=bh$. Balancing the viscous and pressure terms in the axisymmetric viscous-sheet stretching equations, together with fixed volume, suggests a similarity solutionThe conservation of mass equation becomesThere is no ongoing source at the origin, so the integration constant vanishes and $U=\eta /2$ wherever $H>0$. Thus $u=r/(2t)$. Substitution into radial force balance givesA differentiable $H$ cannot have nonzero derivative on an interval while being fixed there at $3\mu /b$. Hence $H^{\prime }=0$: the spreading sheet has uniform thickness.

At the material edge, $R^{\prime }=u(R,t)=R/(2t)$. Since ${\sigma }_{rr}=-bh+3\mu /t$, the edge condition $h{\sigma }_{rr}=-b{h}^2/2$ fixes $h=6\mu /(bt)$. Fixed volume $V=\pi R^{2}h$ then fixes the radius. The self-similar spreading of a viscous oil slick isThe point-release idealization is singular at $t=0$; a finite initial uniform slick gives the same solution with a time shift.

Each broad face supplies surface tension $\gamma$ pulling the rounded hole edge into the remaining sheet. Their resultant is $2\gamma$ per unit circumference. At the inner boundary the fluid's outward normal is $-{\mathbf{e}}_r$, so the boundary traction is outward in the radial direction when $h{\sigma }_{rr}=-2\gamma$. The right panel of the preceding diagram shows these two capillary pulls.

With uniform thickness and $p_{\mathrm{ext}}=0$, the axisymmetric viscous-sheet stretching equations reduce toThis Euler-Cauchy equation gives $u=Ar+B/r$. The fixed outer rim imposes $u(R_0)=0$, hence $B=-A{R}_0^2$. Because $(ru{)}_r/r=2A$, conservation of mass gives $\dot{h}=-2Ah$, independent of $r$. Thus a uniform sheet remains uniform.

The radial stress is ${\sigma }_{rr}=6\mu A-2\mu B/r^2$. Its value at the hole edge determinesThe edge is material, so $\dot{R}=u(R)$ andNeglecting the initially tiny hole's volume, conservation of mass gives $\pi h(R_0^2-R^2)=\pi h_0{R}_0^2$, or $h=h_0/(1-x^2)$ with $x=R/R_0$. Therefore the capillary growth of a hole in a viscous sheet obeysFor a finite initial hole $x_i$, replace $h_0$ in the denominator by $h_0(1-x_i^2)$. A nonzero seed is needed: the exact initial condition $x(0)=0$ gives the stationary solution of this differential equation. For a small positive seed, $x$ initially grows exponentially at rate $\gamma /(\mu h_0)$.