= Solution
For the stated <axisymmetric flow>, the diagonal components of the <rate-of-strain tensor> are
$$
e_{rr}=u_r,\qquad e_{\theta\theta}=u/r,\qquad e_{zz}=w_z.
$$
Their 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 $\sigma_{zz}=-p_{\rm ext}$, so the <Newtonian fluid stress tensor> gives
$$
p=p_{\rm ext}+2\mu w_z=p_{\rm ext}-2\mu(u_r+u/r).
$$
Therefore
$$
\boxed{\sigma_{rr}=-p_{\rm ext}+4\mu u_r+2\mu u/r,\qquad
\sigma_{\theta\theta}=-p_{\rm ext}+2\mu u_r+4\mu u/r.}
$$
For 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 $2\delta\theta\,r p_{\rm ext}h_r\delta r$.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-329-sheet-forces.png]
{title=Forces on an annular viscous-sheet sector and capillary traction at a hole edge}
{description=The left panel shows the radial and hoop tractions on the four vertical faces. The right panel shows the two surface-tension forces pulling the rounded hole edge into the sheet. The separate radial pressure force on the sloping broad surfaces is proportional to $p_{\rm ext}h_r$.}
{height=450}
Neglecting inertia, <force balance> is thus
$$
\partial_r(rh\sigma_{rr})-h\sigma_{\theta\theta}+rp_{\rm ext}h_r=0.
$$
Substituting the two <stresses> cancels the terms involving $p_{\rm ext}h_r$ and gives the <axisymmetric viscous-sheet stretching equations>:
$$
\boxed{2\mu\left[\partial_r(2rhu_r+hu)-h(2u/r+u_r)\right]=rh\,\partial_rp_{\rm ext}.}
$$
Finally, <conservation of mass> in the sector gives
$$
\boxed{h_t+\frac1r\partial_r(rhu)=0.}
$$
Back to article page
