Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 66 / 3 / 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 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.