Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 66 (Slow Viscous Flow)

Work in the frame in which the distant fluid is stationary, and take the pressure at infinity as zero. In the Unscaled Papkovich–Neuber representation, choose the harmonic potentialsBoth are harmonic away from the sphere's centre: $1/r$ is harmonic there and $\mathbf{V}\cdot \mathbf{x}/r^3=-\mathbf{V}\cdot \nabla (1/r)$. Substitution into $\mathbf{u}=\nabla (\mathbf{x}\cdot \mathbf{\Phi }+\chi )-2\mathbf{\Phi }$, $p=2\mu \nabla \cdot \mathbf{\Phi }$, gives the translating sphere in Stokes flow:Here $\mathbf{n}=\mathbf{x}/r$. At $r=a$ the coefficients of $\mathbf{I}$ sum to one and those of $\mathbf{nn}$ sum to zero, so the no-slip boundary condition is satisfied. The field decays at infinity. The Papkovich–Neuber representation enforces the Stokes equation and incompressibility, and Uniqueness of Stokes flow identifies the physical solution.

Differentiate radially at fixed direction $\mathbf{n}$:On the surface, multiplying the specified Newtonian fluid stress tensor by the outward normal from the solid givesThus the surface traction in translating-sphere Stokes flow is uniform and the fluid's force on the sphere isThis is the Stokes drag law; an external force of the opposite sign maintains the motion.

Let $\mathbf{F}$ denote the applied force, and take the normal ${\mathbf{n}}_S$ outward from the solid. At leading order the Stokes drag law and the torque on a rotating sphere in Stokes flow giveThe exact no-slip boundary condition and the exact resultant-load conditions areThe stress condition specifies the total force and torque, not a pointwise prescribed traction.

For the boundary perturbation of a nearly spherical particle, evaluate the no-slip condition at $\mathbf{x}=(a+\epsilon f)\mathbf{n}$. Taylor expansion givesThere is no $f{\mathbf{\Omega }}_0\times \mathbf{n}$ term because ${\mathbf{\Omega }}_0=0$. Therefore

To expand the integral conditions correctly, use surface independence of Stokes force and torque integrals. The Stokes equation implies $\nabla \cdot \mathbf{\sigma }=0$, and symmetry of the stress tensor implies that the angular-momentum flux is divergence-free as well. Evaluate the exact force and torque on a fixed sphere of radius $R$ enclosing the perturbed particle. There are then no geometric terms: the $O(\epsilon )$ integrals of ${\mathbf{\sigma }}_1$ on that fixed sphere vanish because the external force and torque have no $O(\epsilon )$ corrections. Each perturbation field solves the homogeneous exterior Stokes equations, so the divergence theorem transfers those integrals back to $r=a$. HenceIf the integrals are expanded directly on the displaced surface, terms involving the displaced evaluation of ${\mathbf{\sigma }}_0$, the changed normal and the changed area do appear individually. Their sum vanishes by the same surface-independence identity. Thus their absence in the final equations is a cancellation, not permission to omit geometric terms independently.

Write the first-order surface velocity as ${\mathbf{u}}_1={\mathbf{U}}_1+{\mathbf{\Omega }}_1\times \mathbf{x}+\mathbf{g}$, where the previous derivative givesAs a test field, choose a sphere translating with arbitrary velocity $\hat{\mathbf{V}}$. Its surface velocity is $\hat{\mathbf{V}}$ and its traction is $\hat{\mathbf{\sigma }}\mathbf{n}=-3\mu \hat{\mathbf{V}}/(2a)$. The Lorentz reciprocal theorem for Stokes flow givesTerms at infinity vanish because both velocities decay. Since $\int \mathbf{n}dS=0$, the rotation term has zero surface average. The arbitrary test velocity therefore gives the first-order mobility of a nearly spherical particle:

For the ellipsoidal perturbation, the second spherical moment gives $\int fdS=(4\pi a^3/3)\mathrm{tr}\mathbf{D}=0$. The supplied fourth moment gives, componentwise,Substitution yieldswhere the last notation uses symmetry of the tensor.

One can also use a rotating-sphere test with surface traction $-3\mu \hat{\mathbf{\Omega }}\times \mathbf{n}$. Its reciprocal integral, together with zero first-order torque, givesHere $f(-\mathbf{n})=f(\mathbf{n})$, so the last integrand is odd and its integral vanishes. Thus ${\mathbf{\Omega }}_1=0$. Inversion symmetry forbids a translation-to-rotation coupling for this centred ellipsoidal perturbation.

First estimate transverse relaxation without assuming a thin cross-section. With both depth and width of order $h_0$, a transverse surface-height variation of order $h_0$ gives a pressure gradient of order $\rho g$. The Stokes equation then gives transverse horizontal and vertical velocities of order $g{h}_0^2/\nu$. Moving the surface through distance $h_0$ takesFor longitudinal variations over $L\gg h_0$, the hydrostatic pressure gradient is of order $\rho g{h}_0/L$, so the longitudinal velocity is $u\sim g{h}_0^3/(\nu L)$. Incompressibility gives a vertical velocity $w\sim h_{0}u/L$, henceThus cross-channel variations relax rapidly compared with the longitudinal evolution. These are viscous estimates, requiring inertial relaxation to be negligible; the slow-flow regime in particular excludes a large transverse inertial response. After that initial relaxation the free surface is nearly horizontal across the channel, so its height is $h_0(x,t)$ and its local vertical depth is $h_0-|y|\cot \alpha$.

Retain the full cross-sectional geometry for a fixed $0<\alpha <\pi /2$; no small-angle approximation is used. With $|h_{0x}|\ll 1$, the leading hydrostatic pressure is $p=p_a+\rho g(h_0-z)$. Axial viscous derivatives are small relative to cross-sectional ones. The axial Stokes equation is thereforeIntroduce $Y=y/h_0$, $Z=z/h_0$, and the triangular domainLet the dimensionless cross-sectional Poisson equation beThe two lower sloping sides impose the no-slip boundary condition; the upper horizontal side imposes the stress-free boundary condition. Symmetry would additionally give ${\varphi }_Y=0$ at $Y=0$ if only half the domain were used. The full domain needs no extra midline boundary condition. Define $Q(\alpha )={\int }_{{\mathcal{D}}_{\alpha }}\varphi dYdZ$. ThenThe cross-sectional area is $A=h_0^2\tan \alpha$. Conservation of mass, $A_t+q_x=0$, gives the V-shaped channel gravity current equationIn particular, neither replacing this triangular Poisson problem by a vertically parabolic film nor taking an extreme angle is necessary.

Put $C=gQ(\alpha )/(\nu \tan \alpha )$. For a symmetric fixed-volume similarity solution, write $h_0=t^{-b}f(x{t}^{-d})$. Volume conservation gives $d=2b$, and balancing $(h_0^2{)}_t=C(h_0^4{h}_{0x}{)}_x$ gives $2b+1=5b+2d$. Hence $b=1/7$, $d=2/7$. Integrating the similarity equation once, with zero flux at the centre, yieldsIntegrating again and imposing $h_0=0$ at $x=\pm X(t)$ givesThe constant-volume similarity in a V-shaped channel has a compact support. If $H(t)=h_0(0,t)$, its volume isEliminating $H$ therefore givesThis describes the large-time spreading from a localized release; the point-release profile is singular at $t=0$, and the idealized front has a narrow region where its divergent slope violates the long-wave approximation. Neither feature changes the leading bulk similarity law.

At a fixed nonzero $x^{\ast }$, the depth is initially zero until $X(t)=|x^{\ast }|$. After arrival it rises and then falls. Because $X^2\propto t^{4/7}$, differentiation of $h_0(x^{\ast },t{)}^3=3(X^2-x^{\ast 2})/(7Ct)$ givesThus its unique maximum occurs at $X^2=7x^{\ast 2}/3$; thereafter the depth tends to zero as $t^{-1/7}$. If $t_a$ is the arrival time, $t_{\max }/t_a=(7/3{)}^{7/4}$. At the singular release point $x^{\ast }=0$, the point-source similarity profile instead decreases from the start; a finite initial release regularizes its early behaviour.

The right-hand sketch shows the triangular bulk at half its previous maximum depth, with thin draining films on the previously wetted portions of both walls. These residual wall films in a receding gravity current are not captured by a perfectly dry-wall triangular section. They drain rapidly compared with bulk longitudinal spreading. For an order-one angle, a wall film of thickness $b$ and wetted length $\ell$ has speed $O(g{b}^2/\nu )$ by falling film flow; its drainage time gives $b\sim (\nu \ell /(gt){)}^{1/2}$. At a fixed nonzero station the previous maximum wetted height is finite, so film area decays as $t^{-1/2}$, faster than the bulk area $h_0^2\sim t^{-2/7}$. In stations whose arrival and peak times are of order the observation time, the typical residual film fraction is $O(\sqrt{{\tau }_y/t})\to 0$. Therefore the films modify the detailed receding cross-section but do not hold a leading-order fraction of the volume or invalidate the large-time nose law. Microscopic contact-line physics is beyond this ideal viscous model.

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.

Take $p$ as excess pressure relative to the fluid far from the gap, and let ${\dot{h}}_0<0$ for approach. The lower surface of the sphere is $h_0+a-\sqrt{a^2-r^2}$ above the plane. In the dominant region $r=O(\sqrt{a{h}_0})\ll a$, this is the parabolic lubrication gapIts slope is $O(\sqrt{h_0/a})\ll 1$, justifying lubrication theory. Both rigid surfaces have zero radial velocity, so the no-slip boundary condition and radial Stokes equation give the radial volume flux per unit circumferenceAxisymmetric conservation of mass gives $h_t+r^{-1}(rq{)}_r=0$. Symmetry at $r=0$ therefore yields $q=-r{\dot{h}}_0/2$, hence $p_r=6\mu {\dot{h}}_{0}r/h^3$. Matching to $p\to 0$ outside the pressure-bearing region givesExtending the parabolic profile to infinity in this integral is a leading-order matching device; the main pressure and force come from $r=O(\sqrt{a{h}_0})$.

The radial scale $\sqrt{a\hat{h}}$ balances the two contributions to the gap. The time scale $\hat{h}/\hat{v}$ is the time for a typical gap change, and the lubrication pressure scale $\mu a\hat{v}/{\hat{h}}^2$ follows either from the formula above or from $p\sim \mu \hat{v}{\ell }^2/{\hat{h}}^3$ with ${\ell }^2=a\hat{h}$. The elastic surface's upward displacement is $-\eta p$, so the actual gap is increased by $+\eta p$. Thus the dimensionless gap in weakly compliant sphere-plane squeeze flow isA dot now means differentiation with respect to $T$. The dimensionless Reynolds lubrication equation isEven without fluid inertia, the elastic surface has a pressure-dependent normal velocity, so ${\mathcal{H}}_T$ includes $\epsilon P_T$; omitting it would wrongly discard the acceleration term below.

Write $d=H_0+R^2/2$ and abbreviate $H=H_0$, $v={\dot{H}}_0$, $b={\ddot{H}}_0$ during the calculation. The rigid term isAt first order the Reynolds equation givesBecause $RdR=dd$, one integration givesSet $\xi =R^2/(2H)$, so $d=H(1+\xi )$. Dividing and changing variables yieldsThe first three terms are derivatives of $I_3,I_4,I_5$, respectively. Integrating from infinity to enforce $P_1\to 0$ givesThe rational expansion of $I_n$ follows by differentiating $\log [\xi /(1+\xi )]+{\sum }_{j=1}^{n-1}[j(1+\xi {)}^j{]}^{-1}$, obtaining $1/[\xi (1+\xi {)}^n]$, and fixing its value to zero at infinity.

Near $\xi =0$ every $I_n$ has the same singular term $\log \xi$. Its coefficient must vanish to keep the pressure regular, soEquivalently, the integrated radial flux above must vanish at $R=0$; otherwise $P_{1R}$ would have a $1/R$ singularity. Using $I_4=I_3+[3(1+\xi {)}^3{]}^{-1}$ and $I_5=I_4+[4(1+\xi {)}^4{]}^{-1}$ cancels all logarithms and gives the simplified regular expression

The upward hydrodynamic force has scale $\mu a^2\hat{v}/\hat{h}$. Its leading contribution is pressure integrated over horizontal projected area; vertical shear forces are smaller in the lubrication limit. Thus $F=2\pi {\int }_0^{\infty }(P_0+\epsilon P_1+\cdots )RdR$. Use $RdR=H_{0}d\xi$ and ${\int }_0^{\infty }(1+\xi {)}^{-n}d\xi =1/(n-1)$. The two contributions areTherefore

For weight-controlled sedimentation, let the fixed effective load be $F=6\pi A$, $A>0$, and neglect particle inertia as well as fluid inertia. At leading order ${\dot{H}}_0=-A{H}_0$ and ${\ddot{H}}_0=A^2{H}_0$. The first-order bracket evaluated on this trajectory is $-(7/5)A^2/H_0^2$. Solving the force balance perturbatively therefore givesHence weak compliance initially increases the sedimentation speed in the regime where the perturbation is valid. The squeeze pressure indents the coating away from the sphere, widens the flow passage and lowers resistance. This is a load-controlled conclusion on the slow sedimenting branch, not a claim about an arbitrary externally imposed acceleration at startup.

The load-controlled breakdown of weak squeeze-flow compliance occurs when the relative correction $O(\epsilon A/H_0^2)$ becomes order one:The central rigid pressure has size $P_0(0)\sim A/H_0$, so the relative elastic gap change is also $\epsilon P_0(0)/H_0\sim \epsilon A/H_0^2$. Thus breakdown means that elastic deflection is comparable with the nominal undeformed gap. The regular expansion about a rigid plane then ceases to be uniform and the coupled elastic-lubrication problem must be retained. It does not by itself imply contact, fluid inertia or failure of the thin-gap geometry.