= Solution
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 to
$$
\boxed{w=-zu_x.}
$$
Vertical 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> gives
$$
\boxed{\sigma_{zz}=-p_a+\frac\gamma2h_{xx}.}
$$
The <Newtonian fluid stress tensor> also gives $\sigma_{zz}=-p+2\mu w_z=-p-2\mu u_x$. Hence $p=p_a-\gamma h_{xx}/2-2\mu u_x$ and
$$
\boxed{\sigma_{xx}=-p_a+\frac\gamma2h_{xx}+4\mu u_x.}
$$
The factor four is the planar extension factor: both the longitudinal viscous <stress> and the <pressure> adjustment from transverse contraction contribute.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2010/iii/paper-66-sheet-forces.png]
{title=End tractions, ambient pressure and the four surface-tension pulls on a planar liquid-sheet slice}
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 tension
$$
\mathcal T=h(\sigma_{xx}+p_a)+2\gamma\cos\theta
=4\mu h u_x+\frac\gamma2hh_{xx}+2\gamma-\frac\gamma4h_x^2
$$
to the retained small-slope order. The ambient-pressure contributions cancel in this excess tension. Differentiating cancels the two $h_xh_{xx}$ terms, leaving the <planar viscous-sheet stretching equations>
$$
\boxed{(4\mu h u_x)_x+\frac\gamma2h h_{xxx}=0.}
$$
The upper-face kinematic condition is $w(x,h/2,t)=(h_t+uh_x)/2$. Combining it with $w=-zu_x$ gives the second equation,
$$
\boxed{h_t+(hu)_x=0,}
$$
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> $p_{\rm film}=p_g-2\mu u_x$. The extensional correction is small compared with $\gamma/a$ because the transition scale will satisfy $\sqrt{ah_0}/L\ll1$. Thus $p_{\rm film}\simeq p_g>p_e$, 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 $x=0$ selects
$$
\boxed{u(x,t)=\frac{U(t)}Lx.}
$$
Here $U$ 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 $h_0(t)$,
$$
\boxed{\dot h_0=-\frac{U(t)}Lh_0.}
$$
The end transitions determine $U$, as derived in the following parts.
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