Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 86 (Slow Viscous Flow)

Work in the laboratory frame, with fluid at rest at infinity and the instantaneous sphere centre at the origin. In the Unscaled Papkovich–Neuber representation, take the harmonic potentialsDifferentiating the two terms gives the translating sphere in Stokes flow:Here $\mathbf{n}=\mathbf{x}/r$. The potentials are harmonic outside the sphere, so they satisfy the Stokes equations. At $r=a$, the two coefficients of $\mathbf{nn}$ cancel and the coefficient of $I$ is one, proving the no-slip boundary condition $\mathbf{u}=\mathbf{V}$; the field decays at infinity. Its leading part is a Stokeslet of force $6\pi \mu a\mathbf{V}$ exerted on the fluid.

Let $\hat{\mathbf{z}}$ point upwards, $\delta =V_1-V_2$, and $D=a_2{V}_2-a_1{V}_1$. Each isolated sphere moves with velocity $-V_i\hat{\mathbf{z}}$. To leading order its correction is the other's incident Stokeslet velocity:The forces generating these fields are the fixed sedimenting loads, so their leading strengths are set by the isolated speeds. Subtraction yields the leading interaction of two sedimenting spheres:Since $\hat{\mathbf{z}}=\cos \theta \hat{\mathbf{r}}-\sin \theta \hat{\mathbf{\theta }}$, this reduces toThe stipulated higher-order errors can be appended to both equations; they do not enter this leading calculation.

For the same-sign case, interchange labels if necessary to arrange $\delta >0,D>0$, and put $c=3D/4$. The vertical separation $z=r\cos \theta$ satisfiesIf sphere one is initially above sphere two, its vertical order reverses; in every case $z\to -\infty$, implying $r\to \infty$. For a noncollinear trajectory there is no collision in the point-force equations. Indeed, divide the two polar equations and integrate:At side-by-side passage, $\theta =\pi /2$, the closest separation is the positive root of $\delta r^2+2cr=C$. Before passage $\dot{r}<0$ and after passage $\dot{r}>0$, proving overtaking and unbounded subsequent separation. This passing invariant for unequal point-force spheres describes a physical far-field passage only when the closest separation also greatly exceeds the sphere radii. Exactly collinear approach instead reaches the near-contact regime, where this leading approximation cannot describe two finite spheres passing through one another.

For opposite signs, choose $\delta >0,D<0$ and consider the invariant configuration with sphere one vertically above sphere two, $\theta =0$. The radial equation isIf $r>r_{\ast }$, separation decreases; if $r<r_{\ast }$, it increases. Thus it approaches this equilibrium along the vertical line. Linearization gives $\dot{\delta r}=-(\delta /r_{\ast })\delta r$. But a small tilt obeys $\dot{\theta }=(\delta /(2r_{\ast }))\theta$, so the vertical bound pair in the point-force sedimentation model is stable only against vertical perturbations and is a saddle in the full separation dynamics. The opposite vertical orientation has the reversed stability. Reversing gravity reverses both $\delta$ and $D$, leaves $r_{\ast }$ unchanged, and reverses every trajectory; radial attraction becomes repulsion. This is fully consistent with kinematic reversibility of Stokes flow.

There is an important restriction in the printed data. Its hierarchy $|D|\ll (a_1+a_2)|\delta |$ implies $r_{\ast }\ll a_1+a_2$, so the proposed equilibrium would overlap the physical spheres and lie outside the well-separated approximation. With that hierarchy there is no justified physical far-field bound pair. The intended far-field equilibrium instead requires the first inequality reversed, $|D|\gg (a_1+a_2)|\delta |$. The formal equilibrium and its vertical stability above show the intended mechanism, while also identifying the missing validity and transverse-stability qualifications.

For a force $\mathbf{F}\delta (\mathbf{x})$ acting on an unbounded incompressible fluid, the Stokeslet and its pressure areThey solve $-\nabla p+\mu {\nabla }^2\mathbf{u}+\mathbf{F}\delta =0$ and $\nabla \cdot \mathbf{u}=0$. Approximate a slender body by a line distribution of these point forces on its centreline. At a point labelled by arclength $s$,To extract the large logarithm, write $s^{\prime }=s+\ell$ and approximate $\mathbf{X}(s+\ell )-\mathbf{X}(s)\simeq \ell \mathbf{t}$, $\mathbf{f}(s+\ell )\simeq \mathbf{f}(s)$, where $\mathbf{t}={\mathbf{X}}^{\prime }$ is the unit tangent vector. The near-line contribution has kernel $(I+\mathbf{tt})/(8\pi \mu |\ell |)$. Cut it off at the wire radius $R$ and a macroscopic length $L$. The two sides of the line giveThe far contribution and changes in the cutoff add terms without the leading logarithm. Since $(I+\mathbf{tt}{)}^{-1}=I-\mathbf{tt}/2$, inversion proves the leading slender-body force densityThis is force on the fluid; the fluid's force on the wire has the opposite sign. It gives twice the resistance to transverse motion as to tangential motion at this order.

The stated arclength describes the circular helix $\mathbf{X}=(a\cos \theta ,a\sin \theta ,b\theta )$. The PDF omits the second $a$ in the coordinates, so the literal printed curve and its arclength are inconsistent unless $a=1$. First use the circular interpretation consistent with the given arclength. Put $q=\sqrt{a^2+b^2}$ and $c=2\pi \mu /\ln (L/R)$. ThenRotation about $z$ has velocity $\mathbf{v}=\Omega (-a\sin \theta ,a\cos \theta ,0)$, so $\mathbf{t}\cdot \mathbf{v}=\Omega a^2/q$. The axial force density is $f_z=-c\Omega a^{2}b/q^2$. Integrating over one turn givesFor translation, $\mathbf{v}=W\hat{\mathbf{z}}$ and $\mathbf{f}=cW(2\hat{\mathbf{z}}-(b/q)\mathbf{t})$. Only the tangential term contributes to axial couple. Since $(\mathbf{X}\times \mathbf{t}{)}_z=a^2/q$, its density is $-cW{a}^{2}b/q^2$, givingThese are the off-diagonal coefficients of the axial resistance matrix of a slender helix. Their equality does not depend on this particular geometry: with symmetric local resistance $D=c(2I-\mathbf{tt})$, the rotation-induced axial force is $\int \hat{\mathbf{z}}\cdot D(\hat{\mathbf{z}}\times \mathbf{X})ds$, whereas the translation-induced axial torque is $\int (\hat{\mathbf{z}}\times \mathbf{X})\cdot D\hat{\mathbf{z}}ds$. They agree pointwise by symmetry of $D$. The full Stokes-flow statement is the Lorentz reciprocal theorem.

For completeness, retaining the literal elliptic helix instead gives $s^{\prime }=\sqrt{a^2{\sin }^2\theta +{\cos }^2\theta +b^2}$ and the axial coupling of an elliptic helixThus reciprocity holds for the literal curve as well; only the geometric coefficient and arclength change. In all these formulas the equal and opposite force or couple on the wire reverses the displayed sign.

The leading disturbance in sphere Stokes flow has size $u\sim Ua/r$. Consequently viscous diffusion has size $\nu u/r^2$, background advection has size $Uu/r$, and their ratio is $Ur/\nu$. Even if the sphere Reynolds number $Ua/\nu$ is small, this ratio becomes order one at $r\sim \nu /U$. Thus the Stokes approximation is nonuniform at distances of order $\nu /U$. At that outer distance the disturbance itself is small relative to $U$, so self-convection is smaller than background advection and the Oseen approximation is appropriate:Write $\mathbf{u}=\nabla \varphi +\nu \nabla \chi -\mathbf{U}\chi$. It is sufficient outside the origin to impose $\Delta \varphi =0$ and $(\nu \Delta -\mathbf{U}\cdot \nabla )\chi =0$: these give zero divergence, and substitution into momentum gives $p=-\rho \mathbf{U}\cdot \nabla \varphi$. With $\chi =h{e}^{\mathbf{U}\cdot \mathbf{x}/(2\nu )}$, differentiation yieldsA decaying radial choice is $h=C{e}^{-Ur/(2\nu )}/r$. ThereforeThe exponent is nonpositive, so this scalar decays at infinity, including as $1/r$ directly downstream. In the overlap $a\ll r\ll \nu /U$,Choosing $\varphi =-\nu C/r$ cancels the first term's gradient in $\nabla \varphi +\nu \nabla \chi$. The remaining leading velocity isMatching fixes $C=3a/2$. Thus the potential-source and wake decomposition of sphere Oseen flow isIts pressure is $p=-3\mu a(\mathbf{U}\cdot \mathbf{x})/(2r^3)$, the sign appropriate to a fixed sphere in an incident flow. The velocity decays in every direction and matches the requested leading inner disturbance; the sphere's potential-dipole term enters a higher matching order.

The scalar potential is a point source: $\nabla \varphi =(3a\nu /2)\mathbf{x}/r^3$, so integration over a centred sphere givesTo see the fluid wake, take $z$ downstream along $\mathbf{U}$ and transverse radius $R_{\perp }$. For $z\gg \nu /U$, $r-z\simeq R_{\perp }^2/(2z)$, givingThe fluid wake is concentrated within $R_{\perp }=O(\sqrt{\nu z/U})$. Its integrated mass deficit is $\rho U\int \chi dA\simeq 6\pi \mu a$, balancing the outward potential-source mass flux. The corresponding momentum deficit iswhich equals the leading sphere drag. The Gaussian integral used here is ${\int }_0^{\infty }{e}^{-U{R}_{\perp }^2/(4\nu z)}2\pi R_{\perp }d{R}_{\perp }=4\pi \nu z/U$.

The PDF labels $6\pi \mu aU$ as the mass-source strength, but this has dimensions of force. The actual mass-source strength is $6\pi \mu a$; $6\pi \mu aU$ is the momentum deficit or drag. The decomposition above supplies both quantities and identifies the printed normalization error.

First solve the hinted fixed-gap problem. Let $y$ measure distance across the gap and $H(x)=h+x^2/(2a)$ be its leading local height. The lubrication approximation gives $\mu u_{yy}=p_x$, with $u=0$ at $y=0,H$, soThe constant flux per unit transverse length is $q$. Integrating its pressure gradient gives the pressure-driven flux through a parabolic lubrication gap:With $x=\sqrt{2ah}t$, the integral is $\sqrt{2a}{h}^{-5/2}{\int }_{-\infty }^{\infty }(1+t^2{)}^{-3}dt$. The substitution $t=\tan \chi$ gives ${\int }_{-\pi /2}^{\pi /2}{\cos }^4\chi d\chi =3\pi /8$, and therefore

Now take the sphere speed to be $U>0$ downwards and assume zero net liquid through-flow, so the displaced liquid must pass through the annular gap. Near the equator, the sphere's surface radius is $R-z^2/(2R)+\cdots$, giving $H(z)=d+z^2/(2R)$. In the sphere frame the relative volume flux is $U\pi (R+d{)}^2$; dividing by circumference $2\pi R$ gives $q\simeq UR/2$. The Couette contribution from the moving tube wall is only order $Ud$, smaller by $d/R$, so the fixed-gap pressure result givesThe pressure changes principally within axial distance $\sqrt{Rd}$ around the equator. It is nearly constant on each hemisphere, so its leading force is the pressure difference times projected area, $F_p\simeq \pi R^2\Delta p$. Direct viscous shear contributes a relative order $d/R$ and is smaller. Balance with the buoyant weight:Thus the settling speed of a nearly occluding sphere without through-flow isThe printed speed lacks the factor $1/\pi$. The parabolic-gap integral above fixes the normalization: its $3\pi /8$ cannot be dropped. With standard dynamic viscosity and Stokes drag law conventions, the requested coefficient must be corrected by dividing the printed expression by $\pi$. Also, the bypass result assumes no net through-flow; allowing a piston-like displacement of the entire fluid column would require specifying tube end conditions and a different resistance balance.

A sphere of radius $d$ in an unbounded fluid has sedimentation speed $U_d=2\Delta \rho g{d}^2/(9\mu )$. ThereforeThe large confined sphere falls much more slowly even than this much smaller unconfined sphere.

For radial offset $e<d$, let $ϵ=e/d$. To leading order the narrowest gap varies as $d(\varphi )=d(1+ϵ\cos \varphi )$. All circumferential strips share the same pressure drop and act as parallel channels. Inverting the gap-resistance formula shows that each carries flux proportional to $d(\varphi {)}^{5/2}\Delta p$. At fixed buoyant load, the leading pressure drop is fixed, so the speed ratio isThe eccentricity increases narrow-gap bypass mobility result holds because the gain through the wider side outweighs the loss through the narrower side. Strict convexity of $x^{5/2}$ and the zero mean of cosine prove the inequality. Pairing the positive- and negative-cosine contributions in the derivative proves monotonicity with offset. The broader hint's assertion for every exponent $\alpha >0$ is not correct for $0<\alpha <1$, when concavity reverses the inequality; the required exponent $5/2$ is in the valid convex range.

Let $\rho$ be the liquid density and neglect atmospheric shear. The shallow-channel lubrication approximation makes the surface nearly horizontal in each cross-section, at $z=h(x,t)$, with hydrostatic pressure $p=p_{\mathrm{atm}}+\rho g(h-z)$. At fixed $y$, the floor is $z_b=\alpha |y|$ and the local liquid thickness is $H_y=h-z_b$. Axial momentum and boundary conditions areWriting $\zeta =z-z_b$, integration givesThe wetted interval is $|y|<h/\alpha$. Its cross-sectional area and total axial flux areThe shallow V-channel lubrication flux and volume conservation $A_t+Q_x=0$ therefore giveThe slope $\alpha \ll 1$ makes transverse shear small relative to vertical shear; the longitudinal current must also be slender for this reduction.

For typical height $H$ and axial length $L$, fixed volume gives $H^{2}L\sim \alpha V$, while the equation gives $L^2\sim C{H}^{3}t$. Eliminating either scale yieldsThus the current grows in length like $t^{2/7}$ and thins like $t^{-1/7}$. Equivalently, the porous-medium transformation of V-channel spreading sets $w=h^2$ and gives $w_t=(C/5)(w^{5/2}{)}_{xx}$ with constant mass $\alpha V$.

Seek a symmetric solution $h=t^{-1/7}f(\xi )$ with $\xi =x/t^{2/7}$. Substitution into the conservative depth equation givesThe left side is $-(2/7)(\xi f^2{)}^{\prime }$. Symmetry sets the integration constant to zero, so $C{f}^4{f}^{\prime }=-(2/7)\xi f^2$ within the support. Integrating $f^2{f}^{\prime }=-2\xi /(7C)$ gives $f^3=3({\xi }_0^2-{\xi }^2)/(7C)$. In physical variables the constant-volume similarity in a V-shaped channel iswhere $X(t)$ is the half-length of its support and the positive part makes $h=0$ outside it.

Define the positive normalizationThe first substitution is $s=\cos \theta$; the gamma form follows from the beta function integral. Conservation of volume now givesand consequentlyThe full current length is $2X(t)$. Its flux vanishes at the two noses, since the integrated similarity equation gives $C{h}^4{h}_x=-2x{h}^2/(7t)$ there, so no volume is lost at the support boundary. A shifted origin or a positive shift of time gives the corresponding translated similarity family.

In an intended trigonometric $I(\beta )$ notation, $J=I(7/3)$ requires either cosine integrated over $[-\pi /2,\pi /2]$ or absolute cosine over $[0,\pi ]$. The printed ${\cos }^{\beta }\theta$ on $[0,\pi ]$ is not a positive real normalization for the required fractional exponent: with a real cube-root interpretation at $\beta =7/3$, its two halves cancel, while the principal complex power is not real. The similarity normalization uses the positive sine integral $J$ above.