Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 77 (Slow Viscous Flow)

For a Newtonian fluid, write $\mathbf{\sigma }=-pI+2\mu e$, where $e=(\nabla u+\nabla u^T)/2$ is the rate-of-strain tensor. The Stokes equation without body force gives ${\partial }_j{\sigma }_{ij}=0$. Since the stress tensor is symmetric, its contraction with the antisymmetric part of the velocity gradient is zero. ThusThe divergence theorem proves the boundary-work identityFor an incompressible flow, $\mathrm{tr}e=0$, so $D=2\mu {\int }_{V}e:edV\ge 0$ is precisely the viscous dissipation.

Apply this identity to the particle-free volume and then to the fluid outside the added particle. On the rigid particle, the velocity has the form $V+\Omega \times x$. Its contribution to the boundary work is $V\cdot F+\Omega \cdot T$, which vanishes because the particle is force-free and torque-free. The outer velocity is unchanged. Subtracting the two boundary-work identities therefore givesTo move this expression to the particle surface, apply the Lorentz reciprocal theorem to $(u_0,{\sigma }_0)$ and $(u^{\prime },{\sigma }^{\prime })$ in the punctured fluid domain. For completeness, its local proof isOn the outer boundary $u^{\prime }=0$. Let $N=-n$ be the particle-outward normal, opposite to the fluid-outward normal there. Integrating the preceding divergence identity givesThis expression includes the removal of the original fluid inside the particle; that removal must not be counted separately again.

For a sphere in a uniform straining Stokes flow, $E$ is symmetric and trace free, and $u_0=Ex$. Symmetry gives zero rigid translation and rotation. Put $H=x\cdot Ex$ and $r=|x|$. Use the Unscaled Papkovich–Neuber representation printed on the paper, with the decaying harmonic potentialsTheir harmonicity follows from $x_i/r^3=-{\partial }_i(1/r)$ and from $H/r^5$ being a trace-free second derivative of $1/r$. Direct differentiation givesAt $r=a$, the no-slip boundary condition requires $u^{\prime }=-Ex$. Matching the independent $Ex$ and $Hx$ terms fixes $B=-1/2$ and $C=5/6$. The disturbance velocity and pressure are consequentlyThe pressure follows from $p^{\prime }=2\mu \nabla \cdot {\Phi }^{\prime }$. The disturbance decays, the total velocity vanishes at the sphere, and the harmonic representation verifies the Stokes equation. The boundary traction has zero total force by odd symmetry, and zero total torque because $E$ is symmetric, confirming the assumed force-free and torque-free motion.

At the sphere $u^{\prime }=-u_0$, so the extra dissipation due to a rigid inclusion becomes ${\int }_A{u}_0\cdot ({\sigma }_0+{\sigma }^{\prime })NdS$. Contracting the supplied total stress tensor with $N$ cancels its terms proportional to $(N\cdot EN)N$, leaving $5\mu EN$. ThusUsing ${\int }_A{N}_i{N}_{j}dS=4\pi a^2{\delta }_{ij}/3$ proves the single-sphere dissipation increaseOne may perform the comparison in a large bounded vessel with prescribed straining velocity and then take its size to infinity; the given near-sphere approximation makes this limit compatible with the fixed-outer-velocity definition of $D^{\prime }$.

The sphere number density is $n_s=\varphi /(4\pi a^3/3)=3\varphi /(4\pi a^3)$. Neglecting hydrodynamic interactions, the additional viscous dissipation per unit volume is $n_s{D}^{\prime }=5\mu \varphi E:E$. Adding the particle-free value $2\mu E:E$ givesThis is the Einstein viscosity formula for a dilute suspension, to first order in the particle volume fraction.

Use the Unscaled Papkovich–Neuber representation with $e_k=e^{ikx-ky}$ and harmonic potentialsTo see how these coefficients arise, start with $\Phi =(0,A{e}_k)$ and $\chi =B{e}_k$. The representation givesAt $y=0$, the prescribed tangential velocity requires $ikB=U$, while zero normal velocity requires $A=-kB=iU$. Hence the decaying half-space flow isThe pressure follows from $2\mu {\partial }_y{\Phi }_y$. These harmonic potentials verify the Stokes equation and incompressibility; the polynomial factors do not prevent decay as $y\to \infty$.

For the Newtonian fluid stress tensor, ${\sigma }_{xy}=\mu (u_y+v_x)$ and ${\sigma }_{yy}=-p+2\mu v_y$. On the boundary $u_y=-2kU{e}^{ikx}$, $v_x=0$ and $v_y=-ikU{e}^{ikx}$. ThereforeTake real parts for a physical sinusoidal flow. Together with the normal-velocity result supplied in the question, these formulas give the Fourier traction map for a viscous half-space; for signed wave numbers its resistance coefficient is $2\mu |k|$.

Under the plug-like extensional viscosity approximation, incompressibility gives $w_z=-u_x$. Symmetry about $z=0$ fixes $w=-z{u}_x$. Remove the reference hydrostatic stress of the middle fluid, and write $S$ for its leading vertical normal stress. Vertical traction balance at the upper surface givesThe middle fluid's horizontal normal stress exceeds $S$ by $4\mu u_x$: the horizontal strain is $u_x$ and the vertical strain is $-u_x$. HenceAt a slightly inclined upper interface, tangential traction continuity gives, to the order retained in the small-outer-viscosity extensional approximation,The slope term is needed even though the interface is nearly horizontal: dropping it would miss the derivative of $h$ in the extensional resultant. Terms involving the outer deviatoric normal-stress difference times the slope are smaller in this approximation. Integrating horizontal force balance over the symmetric layer, or over its upper half, now givesSubstitution yields the extensional-flow equationThe full layer thickness is $2h$, so its conservation of mass is $(2h{)}_t+(2hu{)}_x=0$, orThis explains the factor $4\mu$ in the extensional stress rather than the shear viscosity coefficient alone.

For a Fourier mode of wave number $k>0$, set $h=h_0+\eta e^{ikx-st}$ and $u=\hat{u}{e}^{ikx-st}$. Neglect ${\sigma }_{zz}^{+}$ as stipulated. The Fourier traction map for a viscous half-space, with outer dynamic viscosity $\lambda \mu$, gives ${\sigma }_{xz}^{+}=-2\lambda \mu k\hat{u}{e}^{ikx-st}$. The two linear equations areEliminating $\hat{u}$ proves the positive decay rate for relaxation of a symmetric viscous layer:When $\lambda \ll \kappa$, inner longitudinal extension dominates the viscous dissipation, and $s\sim \Delta \rho g{h}_0/(4\mu )$. When $\kappa \ll \lambda$ while the plug approximation remains valid, tangential motion of the semi-infinite outer fluids supplies the main resistance, and $s\sim \Delta \rho g{h}_0\kappa /(2\lambda \mu )$. The PDF's second inequality is $\kappa \ll \lambda$; the TeX converts it incorrectly to $\kappa \ll 1$. At much larger outer viscosity the plug assumption fails, as the next calculation shows.

Applying the stated long-wave expression formally in the intended high-viscosity regime, the numerator is dominated by $2\lambda {\kappa }^2/3$ and the displayed denominator by $\lambda$. It givesThe mechanism is inner transverse shear: the outer fluids almost immobilize the interfaces tangentially, so the middle layer has a pressure-driven parabolic velocity instead of the plug-like extensional velocity of part (b). Indeed, with nearly stationary faces at $z=\pm h$, its integrated flux is $-2h^3{p}_x/(3\mu )$. With $p_x=\Delta \rho g{h}_x$, conservation of mass gives precisely the displayed decay rate. The leading viscous dissipation is then in the middle layer, even though its neighbours have much greater dynamic viscosity.

There is a nonuniformity in taking this conclusion literally for every $\lambda \gg {\kappa }^{-1}$. A long-wave remainder written for fixed $\lambda$ cannot be assumed uniform as $\lambda$ diverges. Here this matters physically because sufficiently viscous outer fluids also resist normal interface motion. The distinction can be resolved directly, without an unspecified remainder, from the Stokes equation. For a symmetric thickness mode, the middle vertical velocity has the formIt is odd in $z$, and the middle horizontal velocity is even. At $z=h_0$, the exact outer traction map is ${\sigma }_{xz}^{+}=-2\lambda \mu ku$ and ${\sigma }_{zz}^{+}=-2\lambda \mu kw$. Tangential continuity and the normal hydrostatic jump therefore giveThe latter uses $p=\mu (w_{zzz}-k^2{w}_z)/k^2$, obtained from horizontal force balance. The kinematic condition is $w(h_0)=-s\eta$. Writing $S=\sinh \kappa$ and $C=\cosh \kappa$, tangential continuity fixesEliminating $A,B$ in the remaining two equations gives the exact symmetric-layer rateFor $\kappa \ll 1$ and $\lambda \kappa \gg 1$, this proves the uniform high-viscosity limit of viscous-layer relaxationThus the intended inner-shear answer is valid in the distinguished range ${\kappa }^{-1}\ll \lambda \ll {\kappa }^{-3}$. If $\lambda {\kappa }^3\gg 1$, the main resistance is instead outer normal motion andBoth ranges satisfy the printed $\lambda \gg {\kappa }^{-1}\gg 1$. For example $\lambda ={\kappa }^{-4}$ lies in the latter range and has $s\sim [\Delta \rho g{h}_0/(2\mu )]{\kappa }^3$, rather than a rate proportional to ${\kappa }^2$. The exact denominator contains a ${\lambda }^2{\kappa }^3$ contribution hidden by a fixed-parameter remainder. The inner-shear limit is the intended result, but an additional upper restriction on $\lambda$ is needed to make it a uniform conclusion.

Choose the axial origin at the sphere's centre. Reflection $x\mapsto -x$ leaves the straight tube, the spherical geometry and the transverse displacement of the centre unchanged. It reverses the prescribed axial Hagen-Poiseuille flow, while it leaves a hypothetical transverse particle velocity unchanged. On the other hand, kinematic reversibility of Stokes flow says that reversing the imposed flow reverses every particle velocity, including that transverse component. The force-free and torque-free constraints are preserved by both operations. Uniqueness of the coupled Stokes flow and rigid-body motion therefore makes the transverse velocity equal to its own negative. It must vanish.

Consequently $\dot{c}=0$ as the particle is advected. This is no lateral migration of a single sphere in a uniform tube in Stokes flow. The argument uses the fore-aft symmetry of a rigid sphere and a uniform tube; it does not extend automatically to inertial migration or deformable particles.

Integrate axial force balance $\nabla \cdot \sigma =0$ over the fluid between the two remote tube sections, excluding the particle. The wall's axial traction contributesAt each remote section the Hagen-Poiseuille flow has $u_x$ independent of $x$, so ${\sigma }_{xx}=-p$. The downstream and upstream caps therefore contribute $-\pi a^{2}p(x_2)+\pi a^{2}p(x_1)$. The particle contributes zero net axial force because it is force-free. Adding the three contributions proves the wall-shear and pressure-drop balance in a tube:Here $p$ is the physical pressure. The sign follows from the outward normals on the two caps; no convention for a positive pressure-drop magnitude has yet been introduced.

In this continuation $\lambda$ measures eccentricity, not the viscosity ratio used in Question 2. Take $ϵ\to 0$ with $0\le \lambda <1$ fixed. For the nearly occluding sphere in a cylindrical tube, the narrow gap has thickness $H\sim ϵa$ and axial length $\ell \sim a\sqrt{ϵ}$, obtained by balancing $h_0$ and $x^2/(2a)$. In the sphere frame the tangential velocity scale is $U$. The lubrication approximation then gives the gap scalesThe large local pressure excursion is distinct from the eventual net pressure jump. Away from the sphere the velocity varies across a radius $a$, so the far Hagen-Poiseuille flow hasThese scales also show why the gap pressure can be much larger than its shear stress while the leading net jump can be smaller than that pressure.

Use $y=a-r$, so $y=0$ is the moving tube wall and $y=h$ is the locally stationary sphere. The no-slip boundary conditions are $u(0)=-U$, $u(h)=0$. Integrating $\mu u_{yy}=p_x$, with pressure independent of $y$ to leading order, gives the Couette-Poiseuille flow in a thin gapDifferentiate at both surfaces and eliminate $p_x$ using this moving-boundary lubrication flux:The coordinate reversal means ${\sigma }_{rx}=-{\sigma }_{xy}$ at the tube wall.

At leading order $q$ is independent of $x$ in each angular strip, though it may depend on $\theta$. Both boundaries are stationary as shapes in the sphere frame and impermeable, so the integrated continuity equation is $q_x+a^{-1}(q_{\theta }^{\mathrm{circ}}{)}_{\theta }=0$. The pressure changes along $x$ over $a\sqrt{ϵ}$ but along the circumference over $a$. Thus circumferential velocity is only $O(U\sqrt{ϵ})$, and its flux is $O(Ua{ϵ}^{3/2})$. Circumferential flux divergence is smaller than the possible leading axial divergence by a factor $ϵ$. Consequently $q_x=0$ to the order needed. This is not an assumption of equal flux at different angular positions.

For fixed $\theta$ define the lubrication resistance integrals for a nearly occluding sphereIn particular, $J_1=\pi \sqrt{2a/h_0}$, $J_2=J_1/(2h_0)$ and $J_3=3J_1/(8h_0^2)$. The local pressure relation givesA putative generic value $q=O(U{h}_0)$ makes this jump $O(\mu U/(a{ϵ}^{3/2}))$. However, the integrated wall shear in part (ii) is only $O(\mu Ua/\sqrt{ϵ})$, so the force-free global balance permits a common net pressure jump only of order $\mu U/(a\sqrt{ϵ})$. The outer flow has circumferentially uniform pressures on both sides. Therefore the $O({ϵ}^{-3/2})$ pressure jump must cancel in every angular strip, not merely after angular averaging. The pressure recovery condition in lubrication flow determinesThis explains how the leading-order part of the global force balance selects the local flux.

The total sphere-frame volume flux is nowTransforming back to the laboratory adds $\pi a^{2}U$, so $Q_{\mathrm{lab}}=\pi a^{2}U(1-4ϵ/3)$ to this order. For physical pressure, Hagen-Poiseuille flow gives $Q_{\mathrm{lab}}=-\pi a^4{p}_x/(8\mu )$. ThusThe printed positive expression is the pressure-drop gradient in the direction of the laboratory flow; the Poiseuille hint omits the minus sign when $p_x$ is read as the derivative of physical pressure. With the specified wall velocity $-U$, the laboratory flow is positive and its physical pressure gradient is negative.

Next insert the leading flux into the wall shear stress:Let $\Delta p>0$ be the localized extra pressure drop, after subtracting the regular far Poiseuille gradient. Contributions outside the neck are only $O(\mu U/a)$ and are smaller than the $O({ϵ}^{-1/2})$ jump. Since ${\sigma }_{rx}=-{\sigma }_{xy}$, part (ii) gives $\pi a^2\Delta p=a\int \int {\sigma }_{xy}(0)dxd\theta$. Hence the leading pressure drop isFor example, the concentric limit is $\Delta p\sim 4\pi \mu U\sqrt{2/ϵ}/a$. The integral grows as the minimum gap shrinks; the fixed-$\lambda <1$ qualification prevents treating a further near-contact limit as uniformly covered.

On the sphere the leading shear stress is ${\sigma }_{xy}(h)=-2\mu U/h+4\mu U{h}_0/h^2$, soPositive and negative tangential stresses cancel within each angular strip. Thus the leading tangential resultant, and its leading torque contribution, vanish: negligible rotation is consistent with the torque-free condition. The shear is not zero pointwise, and the pressure forces are not zero. Pressure on a sphere cannot produce a torque about its centre because its normal is radial; its force contributions and the matched far pressure jump remain essential to the force-free transport of a nearly occluding sphere.

Write $G=\Delta \rho g>0$ and take $\tau >0$ in the positive $x$ direction. Removing the ambient hydrostatic pressure leaves excess pressure $G(h-z)$, so the horizontal pressure gradient in the dense gravity current is $G\nabla h$. The lubrication approximation givesThe no-slip boundary condition at $z=0$ and the imposed upper shear stress require $u(0)=v(0)=0$, $\mu u_z(h)=\tau$ and $\mu v_z(h)=0$. Integration yieldsConsequently the depth-integrated volume flux isUsing conservation of mass, $h_t+\nabla \cdot \mathbf{q}=0$, proves the evolution equation for a shear-driven viscous gravity current:The gravity term smooths thickness gradients, while the imposed shear transports the current downstream.

The approximations need more than a small ordinary Reynolds number. With thickness $H$, horizontal scale $L$, dense-fluid density ${\rho }_c=\rho +\Delta \rho$ and typical horizontal speed $V$, sufficient small parameters areThe first neglects horizontal viscous derivatives relative to vertical ones. The second compares convective inertia with vertical viscous stress: it is the reduced Reynolds number, ${\mathrm{Re}}_H(H/L)$, rather than a requirement that ${\mathrm{Re}}_H$ itself be small. The third keeps viscous normal stress small relative to the excess hydrostatic pressure $GH$. Here $V$ is of order $\tau H/\mu +G{H}^3/(\mu L)$, so this last condition requires $\tau /(GL)\ll 1$ together with $(H/L{)}^2\ll 1$. Explicit sufficient inertia conditions are ${\rho }_c\tau H^3/({\mu }^{2}L)\ll 1$ and ${\rho }_{c}G{H}^5/({\mu }^2{L}^2)\ll 1$. If an independently imposed unsteady time scale is $T$, also require ${\rho }_c{H}^2/(\mu T)\ll 1$; for $T\sim L/V$ this is the same reduced-inertia condition.

Set $A=\tau /(2\mu )$ and $D=G/(3\mu )$. Far downstream, the transverse scale $Y=y_N$ is much smaller than $x$, so the downstream gravity-flux divergence is smaller than the transverse one by order $(Y/x{)}^2$. For the steady flow, away from the source,The latter expresses conservation of the prescribed point-source volume flux. In scaling form, $Q\sim A{H}^{2}Y$ and $A{H}^2/x\sim D{H}^4/Y^2$. Eliminating $H$ gives $Y^3\sim DQx/A^2$, proving $y_N\propto x^{1/3}$ and $H\propto x^{-1/6}$.

For the exact downstream similarity solution, put $w=h^2$. Since $h^3{h}_y=(w^2{)}_y/4$, the reduced equation is the porous medium equationSeek $w=x^{-1/3}F(\eta )$ with $\eta =y/x^{1/3}$. Substitution givesIntegrate once, using symmetry and zero transverse flux at $\eta =0$, to obtain $-A\eta F/3=DF{F}^{\prime }/2$. Where $F>0$, this gives $F^{\prime }=-2A\eta /(3D)$, hence $F=A({\eta }_N^2-{\eta }^2)/(3D)$. Vanishing thickness fixes its support. The volume-flux condition then readsThus the downstream profile and half-width arewith $h=0$ outside that interval. This is the downstream similarity of a shear-driven gravity current. The square-root edge is the formal outer thin-layer solution; arbitrarily close to that edge its slope is not small, so the local front requires a separate description. This does not change the downstream scaling and flux calculation.

Near the source the two horizontal scales are comparable, say $\ell$, so the far-downstream reduction cannot locate the upstream nose. In the upstream direction the downstream shear flux $A{H}^2$ must balance the upstream gravity flux of order $D{H}^4/\ell$. Hence $\ell \sim (D/A)H^2$. The total discharge through a region of width $\ell$ is $Q\sim A{H}^2\ell$, giving $H^4\sim Q/D$. Combining these estimates gives the upstream reach scaleThe shear advects liquid toward positive $x$, while the hydrostatic thickness gradient drives liquid back toward negative $x$; their balance stops further upstream spreading. A numerical prefactor requires the full two-dimensional near-source free-boundary solution and cannot be fixed by these scaling arguments alone.