= Solution
The leading disturbance in <sphere> <Stokes flow> has size $u\sim Ua/r$. Consequently viscous <diffusion> has size $\nu u/r^2$, background <advection> has size $Uu/r$, and their ratio is $Ur/\nu$. Even if the <sphere> <Reynolds number> $Ua/\nu$ is small, this ratio becomes order one at $r\sim\nu/U$. Thus \b[the Stokes approximation is nonuniform at distances of order $\nu/U$]. At that outer distance the disturbance itself is small relative to $U$, so self-convection is smaller than background <advection> and the <Oseen approximation> is appropriate:
$$
\boxed{\rho\mathbf U\cdot\nabla\mathbf u=-\nabla p+\mu\nabla^2\mathbf u,\qquad\nabla\cdot\mathbf u=0.}
$$
Write $\mathbf u=\nabla\phi+\nu\nabla\chi-\mathbf U\chi$. It is sufficient outside the origin to impose $\Delta\phi=0$ and $(\nu\Delta-\mathbf U\cdot\nabla)\chi=0$: these give zero <divergence>, and substitution into <momentum> gives $p=-\rho\mathbf U\cdot\nabla\phi$. With $\chi=h e^{\mathbf U\cdot\mathbf x/(2\nu)}$, differentiation yields
$$
(\nu\Delta-\mathbf U\cdot\nabla)\chi=\nu e^{\mathbf U\cdot\mathbf x/(2\nu)}\left(\Delta h-\frac{U^2}{4\nu^2}h\right),\qquad U=|\mathbf U|.
$$
A decaying radial choice is $h=C e^{-Ur/(2\nu)}/r$. Therefore
$$
\chi=\frac C r\exp\left(\frac{\mathbf U\cdot\mathbf x-Ur}{2\nu}\right).
$$
The exponent is nonpositive, so this scalar decays at infinity, including as $1/r$ directly downstream. In the overlap $a\ll r\ll\nu/U$,
$$
\chi=\frac C r+\frac C{2\nu}\frac{\mathbf U\cdot\mathbf x}{r}-\frac{CU}{2\nu}+\cdots.
$$
Choosing $\phi=-\nu C/r$ cancels the first term's <gradient> in $\nabla\phi+\nu\nabla\chi$. The remaining leading <velocity> is
$$
\mathbf u\sim\frac C2\nabla\left(\frac{\mathbf U\cdot\mathbf x}{r}\right)-\frac C r\mathbf U=-\frac C{2r}\left[\mathbf U+\mathbf n(\mathbf U\cdot\mathbf n)\right].
$$
Matching fixes $C=3a/2$. Thus the <potential-source and wake decomposition of sphere Oseen flow> is
$$
\boxed{\phi=-\frac{3a\nu}{2r},\qquad\chi=\frac{3a}{2r}e^{(\mathbf U\cdot\mathbf x-Ur)/(2\nu)},\qquad\mathbf u=\nabla\phi+\nu\nabla\chi-\mathbf U\chi.}
$$
Its <pressure> is $p=-3\mu a(\mathbf U\cdot\mathbf x)/(2r^3)$, the sign appropriate to a fixed <sphere> in an incident flow. The <velocity> decays in every direction and matches the requested leading inner disturbance; the <sphere>'s potential-dipole term enters a higher matching order.
The scalar potential is a <point source>: $\nabla\phi=(3a\nu/2)\mathbf x/r^3$, so <integration> over a centred <sphere> gives
$$
\boxed{Q_{\rm source}=\int\nabla\phi\cdot\mathbf n\,dS=6\pi\nu a,\qquad\dot M_{\rm source}=\rho Q_{\rm source}=6\pi\mu a.}
$$
To see the <fluid wake>, take $z$ downstream along $\mathbf U$ and transverse radius $R_\perp$. For $z\gg\nu/U$, $r-z\simeq R_\perp^2/(2z)$, giving
$$
\chi\simeq\frac{3a}{2z}\exp\left(-\frac{UR_\perp^2}{4\nu z}\right),\qquad u_z\simeq-U\chi.
$$
The <fluid wake> is concentrated within $R_\perp=O(\sqrt{\nu z/U})$. Its integrated mass deficit is $\rho U\int\chi\,dA\simeq6\pi\mu a$, balancing the outward potential-source <mass flux>. The corresponding <momentum> deficit is
$$
\boxed{\rho U^2\int\chi\,dA\simeq6\pi\mu aU,}
$$
which equals the leading <sphere> drag. The <Gaussian integral> used here is $\int_0^\infty e^{-UR_\perp^2/(4\nu z)}2\pi R_\perp dR_\perp=4\pi\nu z/U$.
The PDF labels $6\pi\mu aU$ as the mass-source strength, but this has dimensions of <force>. \b[The actual mass-source strength is $6\pi\mu a$; $6\pi\mu aU$ is the <momentum> deficit or drag.] The decomposition above supplies both quantities and identifies the printed normalization error.
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