= Solution
Write $D=R_e-R_i$, $S=R_e+R_i$, $A=R_e^2-R_i^2=DS$, and $\Delta p=p_i-p_e$. <Incompressible flow> with <axial strain rate> $E$ gives
$$
u_r=-\frac E2r+\frac{C(t)}r.
$$
Its radial <Stokes equation> gives $p_r=0$, because $u_{r,rr}+u_{r,r}/r-u_r/r^2=0$. The radial <Newtonian fluid stress tensor> component is $\sigma_{rr}=-p-\mu E-2\mu C/r^2$. The outer and inner <Young–Laplace equation> conditions have opposite curvature signs:
$$
\sigma_{rr}(R_e)=-p_e-\frac\gamma{R_e},\qquad
\sigma_{rr}(R_i)=-p_i+\frac\gamma{R_i}.
$$
Subtracting and eliminating $C$ gives
$$
\boxed{C=\frac{R_i^2R_e^2}{2\mu A}\left[\Delta p-\gamma\left(\frac1{R_i}+\frac1{R_e}\right)\right],\qquad
p=p_e-\frac{\Delta p R_i^2}{A}+\frac{\gamma S}{A}-\mu E.}
$$
In particular, the boxed <pressure> is equivalent to the printed identity for $(-p-\mu E)A$. For equal gas <pressures>, $C=-\gamma R_iR_e/(2\mu D)$, so <surface tension> draws both interfaces inward in addition to the imposed extension.
The <kinematic boundary condition> is $\dot R_j=-ER_j/2+C/R_j$ at either interface. Taking the difference, and then the difference of squared radii, gives
$$
\boxed{\dot D=-\frac E2D+\frac\gamma{2\mu}-\frac{\Delta p R_iR_e}{2\mu S},\qquad
\dot A=-EA,\quad A(t)=A(0)e^{-Et}.}
$$
The cancellation of $C$ in the area equation is <mass conservation>: axial stretching reduces the liquid area while radial redistribution changes the hole size. For $E=0$, equal <pressures>, and $R_i(0)=b$, $R_e(0)=2b$, one has $A=3b^2$ and $D=b+\gamma t/(2\mu)$. Since $S=A/D$,
$$
R_i=\frac12\left(\frac A D-D\right),\qquad
R_e=\frac12\left(\frac A D+D\right).
$$
Closure occurs at $D=\sqrt A=\sqrt3b$, so
$$
\boxed{t_*=\frac{2\mu b}{\gamma}(\sqrt3-1).}
$$
This <capillary collapse of an annular viscous cylinder> has finite closure time for $\gamma>0$ and $b>0$. For $\gamma=0$ it does not close in this case. The annular formula is used only up to closure; the singular cylindrical inner curvature is not a <boundary condition> on the subsequent solid cylinder.
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