Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 329 (Slow Viscous Flow)

The Papkovich–Neuber representation writes a homogeneous incompressible Stokes flow in terms of a harmonic vector field $\mathbf{\Phi }$ and harmonic scalar $\chi$:with ${\nabla }^2\mathbf{\Phi }=0$ and ${\nabla }^2\chi =0$. The representation satisfies incompressibility because ${\nabla }^2(\mathbf{x}\cdot \mathbf{\Phi })=2\nabla \cdot \mathbf{\Phi }$, and substitution then verifies the Stokes equation.

For a sphere translating with constant vector velocity $\mathbf{U}$, rotational covariance and decay at infinity suggest a harmonic vector monopole and scalar dipole:Substitution gives the translating sphere in Stokes flowAt $r=a$ the radial tensor terms cancel and $\mathbf{u}=\mathbf{U}$, while $\mathbf{u}\to 0$ as $r\to \infty$, so the no-slip boundary condition and far-field condition hold. The resulting traction integrates to the Stokes drag law $6\pi \mu a\mathbf{U}$ in magnitude.

To the requested order, sphere 1 has velocity $\mathbf{V}$ and creates the translating-sphere field from part a. At the centre of sphere 2, a vector distance $\mathbf{R}$ away, this incident field isSphere 2 is force free, so Faxén's first law givesThe source-dipole term is harmonic, whileThereforeThis is the two-sphere Rotne--Prager mobility through order $a^3/R^3$.

The incident strain at sphere 2 is $O(Va/R^2)$. A force-free sphere in this strain creates a stresslet of size $O(\mu a^{3}Va/R^2)=O(\mu V{a}^4/R^2)$, whose velocity back at sphere 1 is $O(V{a}^4/R^4)$. HenceThe nearly uniform returned flow merely advects sphere 1 and does not change its fixed Stokeslet strength, because its applied force remains fixed. The next scattered disturbance is therefore generated by the returned velocity gradient, of order $V{a}^4/R^5$. It induces a stresslet of size $O(\mu V{a}^7/R^5)$ at sphere 1 and hence velocity $O(V{a}^7/R^7)$ at sphere 2. This method of reflections for Stokes flow explains both the absence of an $O(V{a}^5/R^5)$ term and the next order $O(V{a}^7/R^7)$.

At leading order, $X_1=Vt$, $d{X}_1/dt=V$, $Y_1=0$, and the much smaller motion of sphere 2 may be neglected when evaluating the separation:The $y$ component of the leading $a/R$ term in the mobility from part b isThusIntegrating from the initial position $X_1=0$ to infinity gives the hydrodynamic displacement of a force-free sphere

The leading horizontal velocity isConsequently $X_2\sim (3a/2)\log (X_1/Y_0)$ andlogarithmically. The sphere is carried arbitrarily far downstream even though its transverse displacement approaches a finite limit.

The long-wave approximation makes $u$ independent of $z$ at leading order. Mass conservation and symmetry about $z=0$ then giveFor an incompressible Newtonian fluid,The leading normal-stress balance on either surface isSince ${\sigma }_{xx}-{\sigma }_{zz}=4\mu u_x$, it follows that

For a slice of length $\delta x$, the axial forces are the integrated normal stresses $2h{\sigma }_{xx}$ on its vertical ends, ambient pressure on the varying end height, and the horizontal components of surface tension on its two sloping faces. Expanding their difference to first order in $\delta x$ cancels the uniform $p_a$ terms and the lower-order capillary terms, leavingThe kinematic condition on $z=h(x,t)$ is $h_t+u{h}_x=w$. Substituting $w=-h{u}_x$ gives the second required relation,

The gas pressure exceeds the distant liquid pressure by the capillary pressure $\gamma /a$ associated with each cylindrical bubble. In the flat film the interface curvature is nearly zero, so its liquid pressure is close to the gas pressure and therefore exceeds the external liquid pressure by approximately $\gamma /a$. This pressure excess drives liquid out through the two transition regions.

Inside the flat region $h=h_0(t)$, so $h_{xxx}=0$. The extensional-force equation gives $(h_0{u}_x{)}_x=0$, hence $u_x$ is independent of $x$. Symmetry gives $u(0,t)=0$ and define $u(L,t)=U(t)$, soThe mass-conservation equation becomesIt contains no $x$ dependence, so an initially uniform film remains uniform within the flat region.

Matching the flat film to a cylindrical interface gives the local parabolic lubrication gapThe thickness changes by $O(h_0)$ across the transition, soIn the extensional-force equation, viscous and capillary terms have scalesTheir balance gives

The transition adjusts on time $\delta /U$, whereas the flat film drains on time $L/U$. Since $\delta \ll L$, the transition is quasi-steady. Its thin-film mass flux therefore satisfies

LetFlux conservation gives $u=U/H$. Substitution into the extensional-force equation yields the third-order equationSinceone integration, using $H\to 1$ and its derivatives tending to zero on the flat-film side, givesPut $q(H)=H_{\xi }$. After division by $q$, this becomesThe integrating factor $H^{-1/2}$ givesA second integration and $q(1)=0$ therefore give

On the bubble side, matching to its cylindrical shape gives $H\sim {\xi }^2/2$, so $H^{-1/2}{H}_{\xi }\to \sqrt{2}$. Taking $H\to \infty$ in the integrated equation yields $8V/3=\sqrt{2}$. Hence

Insert the velocity from part d into the uniform-film equation:If $h_i=h_0(0)$, integration givesThusThe film thins algebraically and does not reach zero thickness in finite time within this continuum model; rupture would require physics omitted here, such as intermolecular forces.

Take $x$ downslope, $y$ across the slope, and $z$ normal to the plane. The leading normal momentum balance givesThe tangential lubrication theory equations, with no slip at $z=0$ and zero tangential stress at $z=h$, then give the depth-integrated fluxFor $\alpha \ll 1$, defineDropping tildes, steady mass conservation $\nabla \cdot \mathbf{q}=0$ becomeswith $h\to 1$ as $x\to -\infty$.

With no $y$ dependence, integrate the dimensionless equation once. The upstream condition fixes the constant:Write $h=1+\eta$ upstream. Linearization gives ${\eta }_x=3\eta$, soFor $A>0$, the thickness increases monotonically. At large $h$,Integration gives $x-x_0=h-frac12h^{-2}+O(h^{-5})$, and inversion yieldsIn dimensional variables, $d{h}_{\dim }/d{x}_{\dim }\to \alpha$. The increasing thickness therefore cancels the plane's downward slope, so the free surface becomes asymptotically horizontal. The profile represents the upslope edge of a deep viscous pool or pond held back by an obstruction.

The outward normal from the semicircular barrier is ${\mathbf{e}}_r=(-\cos \varphi ,\sin \varphi )$. The leading radial flux in the thick region isForone has $h_s=-H/\Delta$. Negligible radial flux therefore requiresThis expresses local hydrostatic blocking: the free-surface gradient opposes the downslope gravitational flux. Streamlines arriving from upslope divide at $\varphi =0$ and run around the two sides of the barrier toward $\varphi =\pm \pi /2$.

At leading order the azimuthal flux density is $q_{\varphi }\simeq h^3\sin \varphi$. Its integral across the thick region isThis must equal the upstream unit flux intercepted between the symmetry axis and that polar angle, namely $R\sin \varphi$. HenceCombining this with $H=\Delta \cos \varphi$ gives

Let $\phi =\pi /2-\varphi \to 0$ near a shoulder of the barrier. Since $\cos \varphi \sim \phi$, the solution from part c predictsThus the supposedly thin radial region broadens while its large-thickness assumption eventually weakens, and azimuthal derivatives become singular; the approximation cannot remain uniform at the shoulder.

The radial and azimuthal derivative scales of $h^3$ areThey become comparable when $\Delta \sim R\phi$. Using the expression above,soAt this transition,

Downslope of the barrier, the two side streams turn inward under the transverse hydrostatic pressure gradient. They enclose a thin wake or thickness deficit immediately behind the barrier, while excess fluid initially remains concentrated near its shoulders. Farther downstream, lateral spreading fills the wake and restores the uniform layer.

To estimate the recovery distance, set $h=1+\eta$ with $|\eta |\ll 1$ in the governing equation. In a far wake that varies slowly in $x$,which is a diffusion equation with downslope coordinate acting as time. A transverse disturbance of width $R$ therefore spreads over downslope distanceBefore this scale the two-dimensional wake retains the barrier's cross-slope structure; after it, transverse leveling has mixed that structure across its full width and the approach to the uniform film changes character.