= Solution
For the two <Stokes flow> solutions take the same incompressible <Newtonian fluid>, with constant viscosity $\mu$, no body forces, and
$$
\nabla\cdot u=\nabla\cdot\widehat u=0,\qquad
\nabla\cdot\sigma=\nabla\cdot\widehat\sigma=0,\qquad
\sigma=-pI+2\mu e,\quad\widehat\sigma=-\widehat pI+2\mu\widehat e.
$$
Here $e$ and $\widehat e$ are the symmetric <rate-of-strain tensors>. The equality of viscosities is part of comparing two solutions in the same fluid; symmetry of an arbitrary constitutive stress alone would not suffice.
In components, with derivatives $\partial_j u_i$, use the <product rule> to compute the divergence of the cross-work flux:
$$
\partial_j(\widehat u_i\sigma_{ij})
=(\partial_j\widehat u_i)\sigma_{ij}
=-p\,\partial_i\widehat u_i+2\mu e_{ij}\partial_j\widehat u_i
=2\mu e_{ij}\widehat e_{ij}.
$$
The pressure term vanishes by incompressibility; the antisymmetric part of the <velocity gradient> has zero contraction with the symmetric stress. Interchanging the hatted and unhatted fields gives exactly the same scalar. Thus
$$
\nabla\cdot(\sigma\cdot\widehat u-\widehat\sigma\cdot u)=0.
$$
Integrate over $V$ and apply the <divergence theorem>. In its usual form the normal is outward from the fluid volume, whereas the printed $n$ points into the fluid. Reversing the normal multiplies both integrals by the same minus sign, leaving the <Lorentz reciprocal theorem for Stokes flow>:
$$
\boxed{\int_S\widehat u\cdot\sigma\cdot n\,dS
=\int_Su\cdot\widehat\sigma\cdot n\,dS.}
$$
This proof applies to smooth fields or the corresponding finite-energy weak formulation. For an exterior domain one first truncates at a large sphere, applies the same identity, and takes the limit when the outer cross-work terms vanish, as for the decaying fields used below.
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