Boundary integral representation of Stokes flow ($c(y)u(y)=\int J(y-x)\sigma ndS+\int uK(y-x)ndS$)

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.