Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 75 / 1

For two incompressible Stokes flows $(\mathbf{u},\mathbf{\sigma })$ and $(\hat{\mathbf{u}},\hat{\mathbf{\sigma }})$ with the same viscosity and no body forces, the Lorentz reciprocal theorem for Stokes flow statesIndeed the divergence of their difference is zero: stress divergences vanish, pressure contracts with zero velocity divergence, and the two symmetric strain-rate contractions cancel.

Apply it to the decaying disturbance ${\mathbf{u}}^{\prime }=\mathbf{u}-{\mathbf{u}}_{\infty }$ around the sphere and an auxiliary translating sphere of velocity $\mathbf{V}$. On the body surface, ${\mathbf{u}}^{\prime }=\mathbf{U}+{\mathbf{\Omega }}_s\times \mathbf{x}-{\mathbf{u}}_{\infty }$. Using normals from the body into the fluid, the given auxiliary traction is $-3\mu \mathbf{V}/(2a)$; reversing both normal conventions does not change the reciprocal equality. The background traction integrates to zero because it is regular inside the sphere. The disturbance's hydrodynamic force is consequently ${\mathbf{F}}_H=-\mathbf{F}$, where $\mathbf{F}$ is the externally applied force. The rotational term integrates to zero. Thus for arbitrary $\mathbf{V}$,Therefore $\mathbf{U}=\mathbf{F}/(6\pi \mu a)+⟨{\mathbf{u}}_{\infty }{⟩}_{S_a}$.

The background Stokes equations imply ${\nabla }^2{p}_{\infty }=0$ and ${\nabla }^4{\mathbf{u}}_{\infty }=0$. The spherical mean of a biharmonic function gives exactly $⟨{\mathbf{u}}_{\infty }⟩={\mathbf{u}}_{\infty }+(a^2/6){\nabla }^2{\mathbf{u}}_{\infty }$, evaluated at the sphere center. One can obtain it from Taylor expansion and $⟨x_i{x}_j⟩=a^2{\delta }_{ij}/3$; every subsequent spherical average contains ${\nabla }^4$ and vanishes. This requires the background to be regular throughout the enclosed ball. Hence the Faxén relation isThe printed denominator $6\pi \mu$ omits $a$. The given traction integrates to $-6\pi \mu a\mathbf{V}$, fixing that factor; the printed force term also fails dimensional consistency unless its force were redefined, which the question does not do.

Use the Papkovich–Neuber representation in the conventionIt gives $\nabla \cdot \mathbf{u}=0$ and $\mu {\nabla }^2\mathbf{u}=-\mu \nabla (\nabla \cdot \mathbf{\Phi })=\nabla p$. Additive pressure constants are immaterial.

Let $q=\mathbf{x}\cdot \mathbf{Ex}$, with $\mathbf{E}$ symmetric and traceless. The harmonic vector $\mathbf{Ex}/r^3$ is a derivative of $1/r$. The scalar $q/r^5$ is a decaying degree-two harmonic because $q$ is a harmonic homogeneous quadratic. These Papkovich potentials for a strained sphere have precisely the strain symmetry and decay required for the disturbance, so trySubstitution givesAt $r=a$, require ${\mathbf{u}}^{\prime }=-\mathbf{Ex}$. The independent vector structures give $B=a^5$ and $A=-5a^3/3$. The disturbance for a sphere in a uniform straining Stokes flow isIt decays at infinity and satisfies no slip exactly at the sphere. Its leading far field is a stresslet, of velocity scale $E{a}^3/r^2$.

A force-free sphere in a linear background translates with ${\mathbf{U}}_0$, by Faxén's first law, and a couple-free sphere rotates with the background rigid-body rotation $\mathbf{\Omega }$, by Faxén's rotational law. Uniform translation and solid-body rotation can then be matched without disturbance. The remaining relative boundary velocity is exactly $-\mathbf{Ex}$. The Linearity of Stokes flow and Uniqueness of Stokes flow therefore give the same disturbance. The pure strain produces no resultant force or couple, by its symmetric degree-two angular structure.

The shear has $E_{xy}=E_{yx}=C/2$. The separation between centers is $2(X,Y,0)$, so in the leading stresslet from part b, $q=4CXY$, $r=2R$ and the radial vector's $y$ component is $2Y$. Evaluating the other sphere's disturbance at the positive center givesThis is the first method of reflections for Stokes flow approximation. The next multipole in part b and the receiver's Faxén Laplacian contribute at $C{a}^5/R^4$; changes to the incident strain from further reflections are smaller, $C{a}^6/R^5$. The hypothesis $Y_{\infty }\gg a$ ensures well-separated spheres throughout the encounter.

The leading horizontal velocity is $\dot{X}=CY+O(C{a}^3/R^2)$. Dividing and using $Y_{\infty }$ on the right at first order givesIntegrating from negative infinity,The correction increases until $X=0$ and decreases afterward. ThusThe expansion gives $Y\to Y_{\infty }$ as $X\to +\infty$. More generally, reversible scattering of two spheres in simple shear gives no net transverse shift for this smooth, noncontact, deterministic two-sphere encounter: reflection $X\mapsto -X$ reverses the imposed shear, and time reversal restores it. Uniqueness makes the incoming and outgoing branches mirror images. Brownian motion, contact forces or finite inertia would remove the assumptions behind this conclusion.