Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 342 / 4 / a

For the two Stokes flow solutions take the same incompressible Newtonian fluid, with constant viscosity $\mu$, no body forces, andHere $e$ and $\hat{e}$ 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: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. ThusIntegrate 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: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.