 Solution

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 $\gamma /(\mu h_0)$.