I am going through some lecture notes and I have been absolutely stumped by the derivation of Stokes' drag (it is very short). The sphere is of radius $a$ and is instantaneously located at $\bf{x} = 0$ with velocity $\bf{U}$.
We have Oseem's tensor $$\textbf{G} = \frac{\textbf{I}}{r}+\frac{\textbf{x}\textbf{x}}{r^3},$$and we may as such write the velocity everywhere in the fluid as $$\textbf{u}(\textbf{x}) = 6\pi\mu a\textbf{U}\cdot\left(1+\frac{a^2}{6}\nabla^2\right)\frac{\textbf{G}}{6\pi\mu},$$ which is quite suggestive of Stokes' drag. The derivation of this I have followed. The text then states:
Moreover, since we know that $\textbf{G}$ solves $$\mu\nabla^2\left(\frac{\textbf{G}}{8\pi\mu}\right)+\nabla \textbf{p} = \textbf{I}\delta(\textbf{x}),\space \nabla\cdot\textbf{G} =0,$$ we get the effective force on the fluid due to the sphere as $6 \pi \mu a\textbf{U}$ with no integration required. (The pressure here is a vector $\bf{p}$ as the other terms are tensors).
I assume this is some trick with the green's function, but I can't figure this part out, so if someone could put in a few extra steps to guide me from here to the end then that would be appreciated.
