= Solution
Let $n$ point out of the drop, and use $-$ for the interior and $+$ for the exterior. The interfacial <velocity> is continuous. Define $\kappa=\nabla_s\cdot n$; for a sphere of radius $a$ it is $2/a$. With constant <surface tension>, the <interfacial stress balance with variable surface tension> reduces to
$$
(\sigma^+-\sigma^-)n=\gamma\kappa n.
$$
Use the exterior-viscosity kernels to define $S[t]=\int J(y-x)t(x)\,dS_x$ and $D[u]=\int u(x)K(y-x)n(x)\,dS_x$. The interior <viscosity> is $\lambda\mu$, so its <velocity> kernel is $J/\lambda$, while its <stress> kernel $K$ is unchanged. The interior boundary equation is
$$
\frac\lambda2u=S[\sigma^-n]+\lambda D[u].
$$
For the decaying exterior flow, the inner boundary's fluid-domain normal is $-n$, so
$$
\frac12u=-S[\sigma^+n]-D[u].
$$
Add these equations and use the <traction> jump. The required <capillary boundary integral equation for a viscous drop> is
$$
\boxed{\frac{1+\lambda}{2}u(y)
=-\gamma\int_{\partial V}J(y-x)\kappa(x)n(x)\,dS_x
+(\lambda-1)\int_{\partial V}u(x)K(y-x)n(x)\,dS_x.}
$$
The formula uses the drop-outward normal consistently in both integrals. If $\lambda=1$, the double-layer term cancels. A spherical drop of constant curvature has zero <velocity>: a constant Laplace-pressure jump is balanced without motion. A prescribed ambient flow would add its known far-boundary contribution; none is present for the surface-tension-driven flow here.
Back to article page
