Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 78 / 1 / Solution

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.