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Two cylindrical bubbles separated by a uniform flat liquid film of half-length $L$ drain through short curvature-transition regions at the ends. The elevated gas pressure implies a film liquid pressure approximately $\gamma /a$ above the outer liquid pressure. The transition length is $\delta \sim \sqrt{a{h}_0}$, and its volume flux is nearly constant because $\delta /L\ll 1$. Write $uh=U{h}_0$ and $\xi =(x-L)/\delta$. The planar viscous-sheet stretching equations reduce to $h{h}^{\prime \prime \prime }=C(\log h{)}^{\prime \prime }$, with $C=8\mu U{h}_0\delta /\gamma$. Two integrations, matched to a flat film at the inner end, give $h^{\prime }/\sqrt{h}=(16\mu U\sqrt{a}/(3\gamma ))[1-(h_0/h{)}^{3/2}]$. Matching the curvature to $1/a$ gives the displayed speed. Uniform-film conservation of mass then gives ${\dot{h}}_0=-U{h}_0/L$, hence $h_0=[h_i^{-1/2}+3\gamma t/(16\mu L\sqrt{a}){]}^{-2}$. There is no finite-time contact in this ideal continuum model.

In a slender symmetric viscous sheet, the two broad faces have zero tangential traction. A cross-thickness variation of $w$ on the thickness scale would produce leading shear $\mu w_x$, whereas $\mu u_z$ is smaller by the square of the aspect ratio. The leading transverse momentum balance and the shear-free conditions therefore give $w_x=0$: write $w=w(z,t)$.

Incompressibility gives $u_x=-w_z$. Symmetry about $x=0$ sets the integration constant to zero, soThe normal traction on either nearly vertical face matches the ambient hydrostatic pressure, giving ${\sigma }_{xx}=-p_a$ to leading order. The Newtonian fluid stress tensor then yieldsThe factor four is the planar extensional resistance, as in the planar viscous-sheet stretching equations, rather than the factor three for uniaxial extension with two contracting transverse directions.

The following slice diagram shows all axial-force contributions per unit $y$-width; the end arrows indicate positive tensile stress and reverse if that signed stress is negative.

The two end tractions have net $z$-component ${\partial }_z(h{\sigma }_{zz})\delta z$. The normal pressure on the inclined side faces contributes $p_a{h}_z\delta z$, and gravitational acceleration gives the weight $\rho gh\delta z$. Hence force balance isSubstitute the stress expression and $p_a^{\prime }={\rho }_{a}g$:soKeeping the side-pressure component is essential: dropping it would retain an incorrect dependence on the arbitrary ambient pressure level. The remaining equation from conservation of mass isIt follows either by integrating incompressibility across the moving faces or directly from the face kinematic boundary condition.

When effective gravitational acceleration vanishes, $4\mu h{w}_z$ is independent of $z$. It equals the applied excess tensile force $F(t)$, with ambient pressure removed from the end loading. Along a material element labeled by its initial position $z_0$,Integration gives uniform material thinning under planar-sheet tension:A material interval conserves its area: $h(z,t)dz=h_0(z_0)d{z}_0$. ThusWith the specified fixed material origin, integrate the latter relation to obtainThese formulas hold while the material thickness remains positive.

For the quadratic initial profile, put $q=H-\Delta >0$. If $k\ne 0$, evaluating the integral at $z_0=L_0$ givesFor $k=0$ it instead givesFor a constant positive pulling force, $\Delta =Ft/(4\mu )$ first reaches the minimum initial thickness $H$ atBoth expressions for $L$ diverge there. Write $q=F(t^{\ast }-t)/(4\mu )$. If $k\ne 0$, the inverse tangent tends to $\pi /2$ and $\Delta \to H$, givingFor $k=0$,This is finite-time extension of a quadratically thickened sheet. With nonzero $k$, the dominant extension comes from the small material region $|z_0|=O(\sqrt{q/H}/|k|)$ near the thickness minimum, where the integrand is proportional to $1/(q+H{k}^2{z}_0^2)$. Its width shrinks like $\sqrt{q}$, giving an integrated divergence $q^{-1/2}$. The fixed ends are asymptotically far away in those local coordinates, so $A$ does not depend on $L_0$. For a uniform sheet the entire material interval has denominator $q$, giving the stronger $q^{-1}$ divergence and retaining its length in $A$. The singularity is a prediction of the ideal slender-sheet model as its minimum thickness tends to zero.

Balance the two terms in the planar viscous-sheet stretching equations over the transition. The velocity changes by $O(U)$ on length $\delta$, and $h_{xxx}=O(h_0/{\delta }^3)=O(1/(a\delta ))$. ThusUsing the geometric transition length givesThis balances capillary forcing with planar extensional viscous stress, not the no-slip Poiseuille resistance appropriate to a film bounded by rigid walls.

Let the longitudinal variation scale be $\ell \gg h$. The stress-free boundary condition gives $\mu (u_z+w_x)=0$ at each face. Incompressibility implies $w=O(hu/\ell )$, so $w_x$ is smaller than a putative $u_z=O(u/h)$ by $O(h^2/{\ell }^2)$. The leading Stokes equation and the two shear-free conditions therefore make $u$ independent of $z$; its thickness-dependent correction is smaller-order. Symmetry gives $w=0$ at $z=0$, and incompressibility integrates toVertical momentum makes the leading normal stress independent of $z$. With the normal directed from liquid into gas, the upper-face curvature is $-h_{xx}/2$ to leading order. The Young–Laplace equation givesThe Newtonian fluid stress tensor also gives ${\sigma }_{zz}=-p+2\mu w_z=-p-2\mu u_x$. Hence $p=p_a-\gamma h_{xx}/2-2\mu u_x$ andThe factor four is the planar extension factor: both the longitudinal viscous stress and the pressure adjustment from transverse contraction contribute.

For force balance on the illustrated slice, the vertical cuts give $(h{\sigma }_{xx}{)}_x\delta x$. Pressure on the sloping broad faces gives $p_a{h}_x\delta x$ horizontally. Each of the two interfaces exerts a tangential surface tension pull at each cut, with horizontal component $\gamma \cos \theta$, where $\tan \theta =h_x/2$. Thus the horizontal force balance is the derivative of the effective tensionto the retained small-slope order. The ambient-pressure contributions cancel in this excess tension. Differentiating cancels the two $h_x{h}_{xx}$ terms, leaving the planar viscous-sheet stretching equationsThe upper-face kinematic condition is $w(x,h/2,t)=(h_t+u{h}_x)/2$. Combining it with $w=-z{u}_x$ gives the second equation,which is conservation of mass per unit transverse width.

For the two bubbles, let $p_g$ be the common gas pressure and $p_e$ the surrounding liquid pressure away from the sheet. A cylindrical bubble has one nonzero principal curvature, so the Young–Laplace equation gives $p_g-p_e\simeq \gamma /a$. In the nearly flat sheet the normal stress is $-p_g$, giving liquid pressure $p_{\mathrm{film}}=p_g-2\mu u_x$. The extensional correction is small compared with $\gamma /a$ because the transition scale will satisfy $\sqrt{a{h}_0}/L\ll 1$. Thus $p_{\mathrm{film}}\simeq p_g>p_e$, and this capillary pressure difference drives liquid out through both ends.

For a spatially uniform flat sheet, $h_x=h_{xxx}=0$, so the longitudinal equation gives $u_{xx}=0$. Reflection symmetry at $x=0$ selectsHere $U$ is the outward speed at the positive end. The mass equation then gives $h_t=-Uh/L$, independent of $x$, so initially uniform thickness remains uniform. Writing it as $h_0(t)$,The end transitions determine $U$, as derived in the following parts.