Lorentz reciprocal theorem for Stokes flow

For two Stokes velocity-stress fields in the same domain,It follows by integrating the divergence of the corresponding cross-work flux and using symmetry of the Newtonian stress.

For two incompressible Stokes flows with the same dynamic viscosity, $\nabla \cdot {\sigma }^{(a)}+{\mathbf{f}}^{(a)}=0$. The divergence of $u_i^{(2)}{\sigma }_{ij}^{(1)}-u_i^{(1)}{\sigma }_{ij}^{(2)}$ equals $-{\mathbf{u}}^{(2)}\cdot {\mathbf{f}}^{(1)}+{\mathbf{u}}^{(1)}\cdot {\mathbf{f}}^{(2)}$. The cross-strain-rate tensor products cancel by reciprocity. Integrating gives equality of the two surface cross-work terms after adding the corresponding volume body-force work. The outward domain normal must be used consistently, especially at an exterior domain's inner surface.

Use a rotlet as the auxiliary Stokes flow, with a rigid sphere at its center. Its exterior-domain boundary torque is the imposed auxiliary couple. In the Lorentz reciprocal theorem, the actual sphere's zero torque removes the other surface term, while the volume term is the rotlet dotted with the body force. Varying the auxiliary couple gives the displayed angular velocity, assuming the integrals and the far boundary limit converge.

For homogeneous Stokes flow in a smooth bounded volume, use $J_{ij}(r)=({\delta }_{ij}/|r|+r_i{r}_j/|r{|}^3)/(8\pi \mu )$ and $K_{ijk}(r)=-3r_i{r}_j{r}_k/(4\pi |r{|}^5)$. The Lorentz reciprocal theorem with a point-force Stokeslet gives the displayed representation, with $c=1$ inside, $0$ outside and $1/2$ on a smooth boundary. The double-layer integral at the boundary has its limiting principal-value meaning. In an exterior problem the body's outward normal is opposite to the fluid-domain normal, reversing both boundary terms; a nonzero ambient flow adds its far-boundary contribution.

For a drop with internal viscosity $\lambda \mu$, exterior viscosity $\mu$, continuous velocity and a decaying exterior Stokes flow, use the drop-outward normal $n$ and curvature $\kappa ={\nabla }_s\cdot n$, twice the mean curvature. Constant surface tension gives $({\sigma }^{+}-{\sigma }^{-})n=\gamma \kappa n$. Multiply the interior boundary representation by $\lambda$ and add it to the exterior one: the traction difference yields the single-layer term shown, while the viscosity-independent stress kernel yields $(\lambda -1)D[u]$. Here $S,D$ denote the kernels in boundary integral representation of Stokes flow. For equal viscosities the double layer vanishes.