Put $b=\Delta\rho\,g>0$, so $p_{\rm ext}=bh$. Balancing the viscous and pressure terms in the <axisymmetric viscous-sheet stretching equations>, together with fixed <volume>, suggests a <similarity solution>
$$
h=t^{-1}H(\eta),\qquad u=t^{-1/2}U(\eta),\qquad \eta=r/\sqrt t.
$$
The <conservation of mass> equation becomes
$$
-H-\frac\eta2H'+\frac1\eta(\eta HU)'=0,
\qquad [\eta H(U-\eta/2)]'=0.
$$
There 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> gives
$$
\left(\frac{3\mu}{t}-bh\right)h_r=0,
\qquad (3\mu-bH)H'=0.
$$
A differentiable $H$ cannot have nonzero derivative on an interval while being fixed there at $3\mu/b$. Hence $H'=0$: the spreading sheet has uniform thickness.

At the material edge, $R'=u(R,t)=R/(2t)$. Since $\sigma_{rr}=-bh+3\mu/t$, the edge condition $h\sigma_{rr}=-bh^2/2$ fixes $h=6\mu/(bt)$. Fixed <volume> $V=\pi R^2h$ then fixes the radius. The <self-similar spreading of a viscous oil slick> is
$$
\boxed{h(r,t)=\frac{6\mu}{\Delta\rho\,g\,t},\quad
u(r,t)=\frac r{2t},\quad
R(t)=\left(\frac{\Delta\rho\,g\,Vt}{6\pi\mu}\right)^{1/2},\qquad 0\leq r<R(t).}
$$
The point-release idealization is singular at $t=0$; a finite initial uniform slick gives the same solution with a time shift.

Solution

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

For external pressure $\Delta\rho\,gh$ and edge condition $h\sigma_{rr}=-\Delta\rho\,gh^2/2$, the fixed-volume <similarity solution> is a uniformly thick disc:
$$
h=\frac{6\mu}{\Delta\rho\,gt},\qquad u=\frac r{2t},\qquad R^2=\frac{\Delta\rho\,gVt}{6\pi\mu}.
$$
<Conservation of mass> fixes the similarity velocity. The radial equation then forces a differentiable thickness profile to be constant, and the edge <stress> fixes that constant. The singular point-release solution describes the spreading scaling; a finite initial uniform disc corresponds to shifting time.

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