Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 329 / 1

The Papkovich–Neuber representation writes a homogeneous incompressible Stokes flow in terms of a harmonic vector field $\mathbf{\Phi }$ and harmonic scalar $\chi$:with ${\nabla }^2\mathbf{\Phi }=0$ and ${\nabla }^2\chi =0$. The representation satisfies incompressibility because ${\nabla }^2(\mathbf{x}\cdot \mathbf{\Phi })=2\nabla \cdot \mathbf{\Phi }$, and substitution then verifies the Stokes equation.

For a sphere translating with constant vector velocity $\mathbf{U}$, rotational covariance and decay at infinity suggest a harmonic vector monopole and scalar dipole:Substitution gives the translating sphere in Stokes flowAt $r=a$ the radial tensor terms cancel and $\mathbf{u}=\mathbf{U}$, while $\mathbf{u}\to 0$ as $r\to \infty$, so the no-slip boundary condition and far-field condition hold. The resulting traction integrates to the Stokes drag law $6\pi \mu a\mathbf{U}$ in magnitude.

To the requested order, sphere 1 has velocity $\mathbf{V}$ and creates the translating-sphere field from part a. At the centre of sphere 2, a vector distance $\mathbf{R}$ away, this incident field isSphere 2 is force free, so Faxén's first law givesThe source-dipole term is harmonic, whileThereforeThis is the two-sphere Rotne--Prager mobility through order $a^3/R^3$.

The incident strain at sphere 2 is $O(Va/R^2)$. A force-free sphere in this strain creates a stresslet of size $O(\mu a^{3}Va/R^2)=O(\mu V{a}^4/R^2)$, whose velocity back at sphere 1 is $O(V{a}^4/R^4)$. HenceThe nearly uniform returned flow merely advects sphere 1 and does not change its fixed Stokeslet strength, because its applied force remains fixed. The next scattered disturbance is therefore generated by the returned velocity gradient, of order $V{a}^4/R^5$. It induces a stresslet of size $O(\mu V{a}^7/R^5)$ at sphere 1 and hence velocity $O(V{a}^7/R^7)$ at sphere 2. This method of reflections for Stokes flow explains both the absence of an $O(V{a}^5/R^5)$ term and the next order $O(V{a}^7/R^7)$.

At leading order, $X_1=Vt$, $d{X}_1/dt=V$, $Y_1=0$, and the much smaller motion of sphere 2 may be neglected when evaluating the separation:The $y$ component of the leading $a/R$ term in the mobility from part b isThusIntegrating from the initial position $X_1=0$ to infinity gives the hydrodynamic displacement of a force-free sphere

The leading horizontal velocity isConsequently $X_2\sim (3a/2)\log (X_1/Y_0)$ andlogarithmically. The sphere is carried arbitrarily far downstream even though its transverse displacement approaches a finite limit.