= Potential-source and wake decomposition of sphere Oseen flow
{title2=$\phi=-3a\nu/(2r),\quad\chi=3a e^{(\mathbf U\cdot\mathbf x-Ur)/(2\nu)}/(2r)$}
Let $U=|\mathbf U|$, $\nu=\mu/\rho$, and $r=|\mathbf x|$. A decaying outer disturbance matched to the leading <translating sphere in Stokes flow> is
$$
\phi=-\frac{3a\nu}{2r},\qquad\chi=\frac{3a}{2r}\exp\left[\frac{\mathbf U\cdot\mathbf x-Ur}{2\nu}\right],\qquad\mathbf u=\nabla\phi+\nu\nabla\chi-\mathbf U\chi.
$$
Here $\Delta\phi=0$ and $(\nu\Delta-\mathbf U\cdot\nabla)\chi=0$ away from zero. The substitution $\chi=h e^{\mathbf U\cdot\mathbf x/(2\nu)}$ reduces the latter equation to $(\Delta-U^2/(4\nu^2))h=0$, whose decaying radial solution is proportional to $e^{-Ur/(2\nu)}/r$. In the inner overlap, cancellation of the potential monopole gives $\mathbf u\sim-3a[\mathbf U+\mathbf n(\mathbf U\cdot\mathbf n)]/(4r)$. The potential source's <volume flux> is $6\pi\nu a$, and its <mass flux> is $6\pi\mu a$. Downstream, the <fluid wake> width grows as $\sqrt{\nu x/U}$; its mass deficit balances this source, while its <momentum> deficit is $6\pi\mu aU$, the <Stokes drag law>.
Back to article page
