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

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.

In the short end transition, the full thickness changes by $O(h_0)$ while the curvature of each interface changes to $O(1/a)$. Since each interface has height $h/2$, the curvature scale is $O(h_0/{\delta }^2)$, with order-one factors irrelevant to this estimate. ThereforeThis is the geometric scale of the capillary drainage of a film between two bubbles. The assumed regime additionally requires $h_0\ll a$ and $\sqrt{a{h}_0}\ll L$, so the transition is long compared with its thickness and short compared with the flat sheet length.

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.

The conservation of mass equation is $h_t+(uh{)}_x=0$. The flat region changes on time scale $L/U$, whereas a transition has advection time $\delta /U$. Hence $h_t=O(U{h}_0/L)$, and its flux change across the transition is $O(U{h}_0\delta /L)$, small compared with the entering flux $U{h}_0$. To leading order the transition is quasisteady and

Eliminate $u=q/h$ in the longitudinal equation. Then $h{u}_x=-q(\log h{)}_x$, soUse $\xi =(x-L)/\delta$, with $\delta =\sqrt{a{h}_0}$, and use a prime for differentiation with respect to $\xi$. This gives the third-order equationHere $h$ remains a dimensional thickness; only the longitudinal coordinate is scaled.

Match at the inner end to $h\to h_0$, $h^{\prime }\to 0$, $h^{\prime \prime }\to 0$. Since $(h{h}^{\prime \prime }-h^{\prime 2}/2{)}^{\prime }=h{h}^{\prime \prime \prime }$, the first integral isDivide by $h^{3/2}$ to obtain $(h^{\prime }/\sqrt{h}{)}^{\prime }=B{h}^{\prime }/h^{5/2}$. Its second integral, using the flat-film limit to fix the constant, isLet $A=16\mu U\sqrt{a}/(3\gamma )>0$. In the overlap $h\gg h_0$, the equation gives $h^{\prime }\sim A\sqrt{h}$, hence $h^{\prime \prime }\sim A^2/2$. The curvature of each physical interface is $h_{xx}/2\sim A^2/(4a{h}_0)$. Matching to the outer bubble curvature $1/a$ therefore requires $A=2\sqrt{h_0}$ and fixes the numerical coefficient:

Finally the uniform-film evolution becomes ${\dot{h}}_0=-3\gamma h_0^{3/2}/(8\mu L\sqrt{a})$. If $h_0(0)=h_i$ at the start of the drainage regime, its solution isThus capillary drainage of a film between two bubbles gives an algebraic $t^{-2}$ decrease, and the two bubbles do not touch in finite time in this ideal model. As $h_0$ decreases, $h_0/\delta$ and $\delta /L$ become still smaller, so the geometric scale separation improves. At molecular thicknesses, disjoining pressure and changes in the assumed stress-free boundary condition can invalidate the clean continuum-sheet description; the algebraic law alone makes no claim about that physical endpoint.