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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.

Use the Newtonian fluid stress tensor $\sigma =-pI+2\mu e$, where $e=(\nabla u+\nabla u^T)/2$. The Stokes equations for a point force on the fluid are $\nabla \cdot u=0$ and $\nabla \cdot \sigma +F\delta (x)=0$. Put $r=|x|$. Away from the origin, a harmonic vector potential proportional to $F/r$ in the Unscaled Papkovich–Neuber representation gives the rotationally covariant decaying solution. Specifically, take $\Phi =-F/(8\pi \mu r)$ and $\chi =0$. Differentiation yieldsThis satisfies the homogeneous Stokes equations for $r>0$. Its stress isTo verify the point-force normalization, on a sphere $r=\epsilon$ the traction is $\sigma n=-3(F\cdot n)n/(4\pi {\epsilon }^2)$. Since ${\int }_{S^2}{n}_i{n}_{j}d\Omega =4\pi {\delta }_{ij}/3$, its surface integral is $-F$. Thus integrating momentum balance over the small ball gives exactly the applied force $F$, fixing the normalization of the Stokeslet. Its radial velocity is $(F\cdot n)/(4\pi \mu r)$, soThere is no net volume or mass source at the singularity. The kernels have the parity $J(-x)=J(x)$ and $K(-x)=-K(x)$.

For two Stokes flows with the same viscosity and body forces $f^{(1)},f^{(2)}$, the Lorentz reciprocal theorem including body forces isIndeed, apply the divergence theorem to $u_i^{(1)}{\sigma }_{ij}^{(2)}-u_i^{(2)}{\sigma }_{ij}^{(1)}$. The cross-gradient terms cancel: incompressibility removes each pressure contraction and the remaining terms are both $2\mu e^{(1)}:e^{(2)}$. The two stress divergences give the body-force terms shown.

Choose flow 1 to be the desired body-force-free field and flow 2 to be a Stokeslet with arbitrary force vector $q$ at $y$. For $y\in V$, the body-force term is $-q\cdot u(y)$. Rearrange the reciprocal identity and use the parity of $J,K$; equality for every $q$ givesFor $y$ outside the volume there is no point force in $V$, so the same two integrals sum to zero. To obtain the smooth boundary value, exclude a small hemisphere about $y$, use continuity of $u$ to replace it by $u(y)$ on that hemisphere, and take the radius to zero. A full small sphere contributes $u(y)$; a hemisphere contributes half of it. Equivalently, the local solid-angle factor is $2\pi /(4\pi )$. The result is the boundary integral representation of Stokes flow with right side $\frac{1}{2}u(y)$ on a smooth boundary, and with the double layer interpreted by its boundary principal value. A nonsmooth corner would have a different solid-angle coefficient.

With body-outward normal $n$, put $F=-\int \sigma ndS$, ${\Sigma }_{ij}=\int x_i(\sigma n{)}_{j}dS$, $U_{ij}=\int u_i{n}_{j}dS$, and $M=\Sigma -2\mu U$. Taylor expansion of the exterior boundary integral representation of Stokes flow gives a Stokeslet from $F$, a rotlet from $G_j=-{ϵ}_{jkl}{\Sigma }_{kl}$, a source from $Q=\mathrm{Tr}U$, and a stresslet from $S=(M+M^T)/2-I\mathrm{Tr}M/3$. Thus $F$ and $G$ are the force and torque exerted on the fluid, while $Q$ is the instantaneous displaced-volume rate. A volume-preserving deforming body has $Q=0$.