= Solution
The similarity form gives
$$
u=-{A\over r^2}F',\qquad
w={(Aν)^{1/2}\over2r^{3/2}}(F-3ηF').
$$
Using the outer pressure gradient $ρ^{-1}dP/dr=2A^2/r^5$ in the correctly written radial boundary-layer equation $u u_r+w u_z=-ρ^{-1}P_r+νu_{zz}$ gives
$$
F^{(3)}+2(1-F'^2)-{1\over2}FF''=0.
$$
No slip and matching to the sink require $F(0)=F'(0)=0$ and $F'(∞)=1$. The boundary-layer flux is of order $rUδ$, relative to the sink flux $A$ it is
$$
{rUδ\over A}\sim(νr/A)^{1/2}.
$$
Mass conservation therefore requires an outer correction of this relative size, which is small in the assumed region.
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