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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 toUse the exterior-viscosity kernels to define $S[t]=\int J(y-x)t(x)d{S}_x$ and $D[u]=\int u(x)K(y-x)n(x)d{S}_x$. The interior viscosity is $\lambda \mu$, so its velocity kernel is $J/\lambda$, while its stress kernel $K$ is unchanged. The interior boundary equation isFor the decaying exterior flow, the inner boundary's fluid-domain normal is $-n$, soAdd these equations and use the traction jump. The required capillary boundary integral equation for a viscous drop isThe 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.