= 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 $p_{\rm ext}=0$, the <axisymmetric viscous-sheet stretching equations> reduce to
$$
r u_{rr}+u_r-u/r=0.
$$
This <Euler-Cauchy equation> gives $u=Ar+B/r$. The fixed outer rim imposes $u(R_0)=0$, hence $B=-AR_0^2$. 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 determines
$$
\boxed{A=-\frac{\gamma R^2}{\mu h(R_0^2+3R^2)},\qquad
B=\frac{\gamma R^2R_0^2}{\mu h(R_0^2+3R^2)}.}
$$
The edge is material, so $\dot R=u(R)$ and
$$
\dot R=\frac{\gamma R(R_0^2-R^2)}{\mu h(R_0^2+3R^2)}.
$$
Neglecting the initially tiny hole's volume, <conservation of mass> gives $\pi h(R_0^2-R^2)=\pi h_0R_0^2$, or $h=h_0/(1-x^2)$ with $x=R/R_0$. Therefore the <capillary growth of a hole in a viscous sheet> obeys
$$
\boxed{\frac{dx}{dt}=\frac{\gamma}{\mu h_0}\frac{x(1-x^2)^2}{1+3x^2}.}
$$
For 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 $\gamma/(\mu h_0)$.
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