= 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 dissipation>
$$
\mathcal D[u]=2\mu\int e(u):e(u)\,dV.
$$
Assume 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 $e(w)$,
$$
2\mu\int e(u):e(w)\,dV=\int\sigma:\nabla w\,dV=\int_{\partial V}w\cdot\sigma n\,dS-\int_Vw\cdot(\nabla\cdot\sigma)\,dV=0.
$$
Expanding the square proves
$$
\boxed{\mathcal D[v]=\mathcal D[u]+2\mu\int e(w):e(w)\,dV\geq\mathcal D[u].}
$$
Equality 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 $u=F(r)\boldsymbol\Omega\times x$. 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 $F(a)=1$, $F(b)=0$ give
$$
B=\frac{a^3b^3}{b^3-a^3},\qquad A=-\frac{a^3}{b^3-a^3}.
$$
Thus the <Rotational Stokes flow between concentric spheres> is
$$
\boxed{u(x)=\frac{a^3}{b^3-a^3}\left(\frac{b^3}{r^3}-1\right)\boldsymbol\Omega\times x.}
$$
It 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 $u=-2\Phi$. Also
$$
\nabla\cdot\Phi=-\tfrac12\{\nabla F\cdot(\boldsymbol\Omega\times x)+F\nabla\cdot(\boldsymbol\Omega\times x)\}=0,
$$
and therefore \b[$p=0$], 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 $\boldsymbol\Omega$, and isotropy leaves only a possible term proportional to $\boldsymbol\Omega\cdot x$. 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 $v=\boldsymbol\Omega\times x$ in the following stress calculation. The symmetric <velocity> gradient is
$$
e_{ij}=\frac{F'(r)}{2r}(x_jv_i+x_iv_j),\qquad F'=-\frac{3B}{r^4}.
$$
The full <stress tensor>, in the zero-<pressure> gauge, is therefore
$$
\boxed{\sigma_{ij}=-\frac{3\mu B}{r^5}(x_jv_i+x_iv_j).}
$$
For $n=x/r$, the traction on a radial surface is $\sigma n=-3\mu B(\boldsymbol\Omega\times n)/r^3$. This completely specifies the stress field; with the polar axis along $\boldsymbol\Omega$, its only nonzero spherical shear component is $\sigma_{r\phi}=-3\mu B\Omega\sin\theta/r^3$.
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 $\int n_i n_j\,d\omega=(4\pi/3)\delta_{ij}$,
$$
\int_{r=a}x\times\sigma n\,dS=-3\mu B\int\{\boldsymbol\Omega-n(n\cdot\boldsymbol\Omega)\}\,d\omega=-8\pi\mu B\boldsymbol\Omega.
$$
Hence the sustaining applied <Torque in rotational Stokes flow between concentric spheres> is
$$
\boxed{\boldsymbol G=\frac{8\pi\mu a^3b^3}{b^3-a^3}\boldsymbol\Omega.}
$$
For $a\ll b$, it approaches the isolated <rotating sphere in Stokes flow> result $8\pi\mu a^3\boldsymbol\Omega$. For a narrow gap $d=b-a\ll a$, it becomes $8\pi\mu a^4\boldsymbol\Omega/(3d)$: 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 $\mathcal D_p\geq\mathcal D_0$.
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 $\mathcal D_p=\boldsymbol G_p\cdot\boldsymbol\Omega$, while $\mathcal D_0=\boldsymbol G_0\cdot\boldsymbol\Omega$. Therefore
$$
\boxed{(\boldsymbol G_p-\boldsymbol G_0)\cdot\boldsymbol\Omega\geq0.}
$$
For $\boldsymbol\Omega\ne0$ 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 \b[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.
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