Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 78 (Slow Viscous Flow)

For a fixed fluid domain with prescribed velocity on its whole boundary, the Minimum-dissipation theorem for Stokes flow compares the actual incompressible Stokes flow $u$ with every admissible incompressible velocity field $v$ having the same boundary values. The trial fields need not themselves satisfy the momentum equation. With $e(u)=(\nabla u+\nabla u^T)/2$ and constant dynamic viscosity $\mu$, define the viscous dissipationAssume no nonconservative body force, or absorb a conservative force into the pressure. Set $w=v-u$, so $\nabla \cdot w=0$ and $w=0$ on the boundary. The Stokes stress tensor $\sigma =-pI+2\mu e(u)$ obeys $\nabla \cdot \sigma =0$. Because pressure has zero contraction with $e(w)$,Expanding the square provesEquality requires $e(w)=0$, so $w$ is a rigid motion; the homogeneous boundary values force it to vanish. The theorem concerns fixed boundary velocities, not fixed applied forces or arbitrary comparisons between different fluid domains.

For the concentric-sphere motion, seek $u=F(r)\mathbf{\Omega }\times x$. Incompressibility is automatic. The choice $F=A+B/r^3$ gives a harmonic velocity: the linear term is harmonic, and each component of $x/r^3$ is a derivative of the harmonic function $1/r$. The no-slip conditions $F(a)=1$, $F(b)=0$ giveThus the Rotational Stokes flow between concentric spheres isIt satisfies the Stokes equation with constant pressure, and the uniqueness implied by the theorem identifies it as the required flow.

In the Unscaled Papkovich–Neuber representation, take $\Phi =-u/2$ and $\chi =0$. Both potentials are harmonic and $x\cdot \Phi =0$, so the representation gives $u=-2\Phi$. Alsoand therefore $p=0$, after choosing the arbitrary pressure constant.

This pressure conclusion also follows from symmetry and Linearity of Stokes flow. Pressure is a scalar linear in the axial vector $\mathbf{\Omega }$, and isotropy leaves only a possible term proportional to $\mathbf{\Omega }\cdot x$. That is a pseudoscalar, forbidden by reflection symmetry of the concentric spherical domain. Hence the pressure perturbation vanishes. Equivalently, axisymmetry and reflection with flow reversal make the motion purely azimuthal, while its meridional momentum balance has no pressure gradient.

Let $v=\mathbf{\Omega }\times x$ in the following stress calculation. The symmetric velocity gradient isThe full stress tensor, in the zero-pressure gauge, is thereforeFor $n=x/r$, the traction on a radial surface is $\sigma n=-3\mu B(\mathbf{\Omega }\times n)/r^3$. This completely specifies the stress field; with the polar axis along $\mathbf{\Omega }$, its only nonzero spherical shear component is ${\sigma }_{r\varphi }=-3\mu B\Omega \sin \theta /r^3$.

The fluid couple on the inner sphere is the integral of $x\times \sigma n$ with the normal pointing from the solid into the fluid. Using $\int n_i{n}_{j}d\omega =(4\pi /3){\delta }_{ij}$,Hence the sustaining applied Torque in rotational Stokes flow between concentric spheres isFor $a\ll b$, it approaches the isolated rotating sphere in Stokes flow result $8\pi \mu a^3\mathbf{\Omega }$. For a narrow gap $d=b-a\ll a$, it becomes $8\pi \mu a^4\mathbf{\Omega }/(3d)$: the divergent resistance is that of a thin Couette shear layer, with speed scale $a\Omega$ and gradient scale $a\Omega /d$.

Now add the force-free, couple-free rigid particles and denote the resulting velocity by $u_p$. Extend $u_p$ through each particle as its rigid translational and rotational velocity. No slip makes this extension continuous across every particle surface; it is incompressible and has zero rate-of-strain tensor inside each particle. It therefore gives an admissible trial velocity in the original particle-free annulus, with exactly the same sphere boundary velocities. Its full-annulus dissipation is the actual fluid dissipation ${\mathcal{D}}_p$, since the added interior integrals are zero. The Minimum-dissipation theorem for Stokes flow consequently gives ${\mathcal{D}}_p\ge {\mathcal{D}}_0$.

The boundary-work identity follows by integrating $\nabla \cdot (\sigma u)$: dissipation equals the mechanical power supplied at all rigid boundaries. The outer sphere is stationary; the inner sphere has no translation, so any force needed to hold its centre does no work; and each added particle contributes zero power because its net force and couple are zero. Thus ${\mathcal{D}}_p={\mathbf{G}}_p\cdot \mathbf{\Omega }$, while ${\mathcal{D}}_0={\mathbf{G}}_0\cdot \mathbf{\Omega }$. ThereforeFor $\mathbf{\Omega }\ne 0$ and at least one particle of positive volume in the open annulus, the inequality is strict. Equality would make the extended velocity identical to the particle-free solution, yet that solution has nonzero strain on every open region away from the rotation axis and cannot be rigid throughout any particle interior. Hence the applied couple component along the imposed angular velocity increases. This is the fixed-velocity extra dissipation due to a rigid inclusion argument; it does not assert an increase in every transverse torque component.

For a slender viscous bubble, approximate the interface normal by the radial direction and its curvature by $1/a$. Axial curvature and axial derivatives of the leading external motion are smaller by the slenderness factors. The exterior leading Stokes flow is locally radial. Incompressibility gives ${\partial }_r(r{u}_r)=0$, while the kinematic boundary condition gives $u_r(a)=a_t$ to leading order, henceThe radial viscous Laplacian of this $1/r$ velocity vanishes, so exterior pressure is constant to leading order; set its reference value to zero. Its radial normal stress at the interface is $2\mu {\partial }_r{u}_r{|}_a=-2\mu a_t/a$. Inner viscous normal stress is smaller by the viscosity ratio $\lambda$, and the bubble's leading normal stress is $-P$. The interfacial stress balance with variable surface tension, here with constant $\gamma$, consequently givesThese are local leading-order relations: the outer axial flow and tangential traction enter at higher slenderness order. The sign also has a useful check: a uniform cylinder with no imposed excess pressure contracts, $a_t=-\gamma /(2\mu )$, under surface tension.

When pressure is uniform along the bubble, its normal-stress relation is $2\mu a_t=Pa-\gamma$. Use the PDF's amplitude convention $a=f+\sqrt{2}g\sin kz$; the square root acts on $2$ only, not on $2g$. Comparing constant and sinusoidal terms givesThe mean cross-sectional area of an incompressible periodic segment is proportional to $⟨a^2⟩=f^2+g^2$ and is conserved. ThusIts time derivative vanishes, and substitution of the preceding two equations gives $P{\alpha }_0=\gamma f$. Therefore the uniform-pressure evolution of a slender viscous bubble isFor a small disturbance to radius $a_0$, $f=a_0+O(g^2)$ and ${\alpha }_0=a_0^2+O(g^2)$, so its amplitude grows as $g\propto e^{st}$ withThe leading long-wave growth rate is independent of $k$ because axial curvature has been neglected; this is not a finite-wavelength dispersion relation.

While $f>\sqrt{2}g>0$, the minimum radius is $a_{\min }=f-\sqrt{2}g$. Its derivative isThus the minimum keeps decreasing in the nonlinear uniform-pressure model, with no need to solve $f,g$ explicitly. Finite internal viscosity eventually invalidates this model close enough to the neck.

Let $s=a_{\min }$ be a small neck radius in the preceding sinusoidal shape. Expanding around a minimum gives $a\simeq s+(g{k}^2/\sqrt{2})(z-z_{\ast }{)}^2$, so its local axial length is $\ell \sim [s/(g{k}^2){]}^{1/2}$. The uniform-pressure model has $|a_t|\sim \gamma /\mu$ near the neck. Local conservation of mass therefore demands axial inner flux $Q\sim s(\gamma /\mu )\ell$. Internal Hagen-Poiseuille flow of viscosity $\lambda \mu$ would require a pressure differenceIf $sk\ll \lambda$, while $gk\ll 1$ by slenderness, this ratio is very large. Hence internal pressure variation certainly cannot be neglected when $a_{\min }k\ll \lambda$. This is a sufficient breakdown condition; appreciable variation can arise earlier, especially because the radial viscous and capillary terms nearly cancel in the uniform-pressure regime.

For the inner flow, the leading axial pressure is uniform across a section. Neglect the much smaller interfacial axial velocity relative to the internal pressure-driven motion. The lubrication approximation solves $\lambda \mu r^{-1}(r{u}_{z,r}{)}_r=P_z$, with regularity at zero and $u_z(a)=0$ at leading order. ThusThe area balance $(\pi a^2{)}_t+Q_z=0$, combined with the exterior normal-stress relation, gives the lubrication equation for a bubble with viscous exterior:The neglected inner viscous normal stress is smaller by $\lambda$ in the similarity scales obtained next; the external axial interfacial speed is likewise smaller than the large pressure-driven inner speed.

Put $\tau =t_{\ast }-t$, $a\sim {\tau }^p$ and axial length $\ell \sim {\tau }^q$. The radial viscous normal stress scales as $\mu a_t/a\sim {\tau }^{-1}$ and capillary pressure as $\gamma /a\sim {\tau }^{-p}$. Retaining both requires $p=1$. The left side $a{a}_t$ scales as ${\tau }^{2p-1}$, while the axial-flux divergence scales as $a^{4}P/{\ell }^2\sim {\tau }^{4p-1-2q}$. Equating them gives $q=p=1$. Thus both local radius and axial length shrink linearly in the remaining time.

To obtain the specified dimensionless coefficients, choose the linear pinch-off scaling of a slender viscous bubbleThen $a_t=(\gamma /\mu )(\zeta A^{\prime }-A)$ andThe constant $-2$ has no axial derivative. Substituting into the flux equation and cancelling its dimensional factor givesBoth axial derivatives on the right are essential; the outer derivative is present in the PDF but lost in the converted TeX expression. The physical slope is $a_z=\sqrt{\lambda }{A}^{\prime }$, so an order-one similarity profile remains slender for $\lambda \ll 1$. No solution of this similarity ordinary differential equation is required.

An insoluble surfactant is advected with the surface fluid and has no exchange with the bulk. With negligible surface diffusion, its amount on a material surface strip is conserved. Under the small-slope lubrication approximation, surface arclength equals projected horizontal length to leading order, givingThis is the leading planar form of conservation of insoluble surfactant on a moving interface.

Let $y$ measure height above the substrate. The normal-stress and vertical momentum conditions give $p=p_{\mathrm{atm}}-\gamma h_{xx}+\rho g(h-y)$, so $p_x=\rho g{h}_x-(\gamma h_{xx}{)}_x$. Horizontal Stokes flow obeys $\mu u_{yy}=p_x$, with no slip at $y=0$ and Marangoni stress $\mu u_y(h)={\gamma }_x$. ThusUse $h_t+q_x=0$ and ${\gamma }_x=-A{C}_x$ to obtain the thin-film equations with insoluble surfactant:The differing flux coefficients are essential: the surfactant travels at the surface velocity, not at the depth-averaged liquid velocity.

Neglect the two pressure-gradient terms for the spreading outer region. If its half-length is $\ell$, fixed surfactant mass gives $C\sim M/\ell$, $C_x\sim M/{\ell }^2$. The surface velocity is then $u_s\sim AM{h}_0/(\mu {\ell }^2)$. Equating this to $\ell /t$ gives $\ell \propto t^{1/3}$. For the precise finite-mass Marangoni spreading on a liquid film solution choosewith symmetry about zero and support $0\le \eta \le {\eta }_N$ on the positive half-line. The reduced differential equations areTheir boundary conditions are zero flux at the centre and zero concentration at the moving front. The two integral constraints express conserved surfactant mass and zero net change of liquid volume relative to the original uniform layer:The second follows by integrating $h_t+q_x=0$ across the whole layer: the localized release adds surfactant but no liquid, and the flux vanishes at infinity. Outside the spreading region the leading depth remains $h_0$.

Integrating the concentration equation from the centre gives $H\Gamma {\Gamma }^{\prime }=-\eta \Gamma /3$. Where $\Gamma >0$, this is $H{\Gamma }^{\prime }=-\eta /3$. Substituting into the height equation givesTherefore $H=c\eta$, and ${\Gamma }^{\prime }=-1/(3c)$. With $\Gamma ({\eta }_N)=0$, the linear similarity profiles for surfactant spreading are $\Gamma =({\eta }_N-\eta )/(3c)$. The liquid-volume constraint gives $c{\eta }_N=2$, while the surfactant constraint gives ${\eta }_N^2/(3c)=1$. Thus ${\eta }_N^3=6$, andIn physical variables, the leading profiles on the whole line areOutside this interval $h=h_0$ and $C=0$. The formal outer solution reaches zero depth at the central point, and height $2h_0$ immediately inside the front. It is not a microscopic description of the central dry region or of the front transition. As a conservation check, the front velocity equals both the incoming surface velocity and the liquid Rankine–Hugoniot speed $q/(2h_0-h_0)=2AM{h}_0/(\mu x_N^2)={\dot{x}}_N$.

The jump from $2h_0$ to $h_0$ cannot persist as a smooth solution with both hydrostatic and capillary gradients absent. In an edge region of width $\Delta$, a depth change of order $h_0$ produces $h_x\sim h_0/\Delta$ and $h_{xxx}\sim h_0/{\Delta }^3$. Keep the stipulated concentration-gradient scale $M/x_N^2$. The three liquid-flux scales areFor $g=0$, balancing $q_M$ and $q_{\gamma }$ gives the capillary smoothing of a Marangoni front width

Gravity dominates capillarity when $q_g/q_{\gamma }\sim \rho g{\Delta }^2/{\gamma }_0\gg 1$. In physical units the condition is ${\Delta }^2\gg {\gamma }_0/(\rho g)$: the printed ${\gamma }_0/g$ omits density and is dimensionally incomplete unless surface tension has already been divided by density. Balancing the gravity flux with $q_M$ yields the gravity smoothing of a Marangoni front:This layer grows faster than the spreading half-length. It reaches the whole pool whenOnly scaling constants can be fixed by this balance, so an exact numerical coefficient in $t_{\ast }$ is not implied. The gravity-dominant assumption must hold at this crossover as well.

For $t\gg t_{\ast }$, the sharp-front linear-depth outer profile is replaced by a gravity-levelled surfactant film: the height approaches $h_0$, with a small depression under the surfactant and redistributed liquid outside. The rapid hydrostatic return flow almost cancels the net Marangoni liquid flux. Setting $q\simeq 0$ and $h\simeq h_0$ gives $h_x\simeq -3A{C}_x/(2\rho g{h}_0)$ and fractional depth variation $O(AM/(\rho g{h}_0^2{x}_N))\ll 1$. The surface still moves outward, with $u_s\simeq -A{h}_0{C}_x/(4\mu )$, so leveling the bulk liquid does not stop surfactant spreading.

Choose transverse $x$ along the cusp from the line of cylinder contact, and $y$ across its narrow gap. Expanding the two circular boundaries for $x\ll a$ gives $y=\pm x^2/(2a)$, so the local full gap is $d(x)=x^2/a$. At the meniscus, $x\simeq w$, a semicircle fits across this gap with radius $d(w)/2=w^2/(2a)$. Thus the leading area and transverse curvature in the cusp flow between touching cylinders areThe meniscus area correction is $O(w^4/a^2)$ and is small relative to $w^3/a$. The curvature is stated as a positive suction magnitude; the liquid pressure relative to air is $p=-2\gamma a/w^2$. Axial curvature is negligible compared with this transverse curvature under the slow-variation approximation.

At fixed $x$, axial lubrication theory gives a local slit Poiseuille flow between rigid walls $y=\pm d/2$:Integrating across the cusp givesThe local meniscus edge correction occupies a transverse width small compared with $w$ and does not change this leading flux. Since $p_z=4\gamma a{w}^{-3}{w}_z$, the axial volume flux is $Q=-\gamma w^4{w}_z/(21\mu a^2)$. Area conservation ${\mathcal{A}}_t+Q_z=0$ now givesWriting $u=w^3$ also puts this in porous medium equation form $u_t=\gamma (u^{5/3}{)}_{zz}/(35\mu a)$, explaining the finite-support spreading profile.

Set $K=\gamma /(7\mu a)$. Fixed volume means $\int w^{3}dz=3aV$. If $w\propto t^{-\beta }$ and axial extent is proportional to $t^{\alpha }$, this conservation gives $\alpha =3\beta$. The evolution equation gives $3\beta +1=5\beta +2\alpha$, hence $\beta =1/8$, $\alpha =3/8$.

For the symmetric source-type similarity solution, write $w=t^{-1/8}f(\eta )$, $\eta =z/t^{3/8}$. The differential equation becomesIntegrate from the centre using zero flux and symmetry. This gives $K{f}^4{f}^{\prime }=-3\eta f^3/8$, so on the positive support $f{f}^{\prime }=-3\eta /(8K)$. With $f({\eta }_N)=0$,Therefore the fixed-volume capillary spreading in a cylindrical cusp profile isLet $B=21\mu a/(8\gamma t)$. Its volume isWith $s=\sin \theta$, the last integral is ${\int }_{-\pi /2}^{\pi /2}{\cos }^4\theta d\theta =3\pi /8$, equal to the supplied sine integral by symmetry. Consequently $V=\pi B^{3/2}{z}_N^4/(8a)$, yieldingThe other tip is at $-z_N$. The maximum width decays as $t^{-1/8}$. This is the leading slender outer solution: the ideal point injection and the steep microscopic nose require local descriptions outside its uniform lubrication range.

Finally, for vertical cylinders let $\rho$ denote the liquid density, and take the bath pressure at $z=0$ as the air-pressure reference. At rest the liquid pressure is $p(z)=-\rho gz$. Equating hydrostatic suction with the wetting-meniscus capillary pressure gives the hydrostatic rise in a cylindrical cusp:The cusp can therefore support an indefinitely high, increasingly narrow ideal wetting tail rather than a fixed terminal rise height. Large $z$ makes $w\ll a$ and strengthens the transverse cusp approximation, until microscopic physics limits the continuum description.