= Solution
<No-slip boundary condition>[No slip] on the two fixed radial walls gives $g(\pm\alpha)=0$. The outward <flux> through a circular arc is $\int_{-\alpha}^\alpha u_r r\,d\theta=\int_{-\alpha}^\alpha g\,d\theta$, so <volume flux>[source strength] $Q$ requires that integral to equal $Q$. Set $g=QG$ and $\epsilon=Q/\nu$. Then
$$
G^{(3)}+4G'+2\epsilon GG'=0,\quad G(\pm\alpha)=0,\quad\int_{-\alpha}^\alpha G\,d\theta=1.
$$
At leading order $G_0^{(3)}+4G_0'=0$, whose general solution is a constant plus sine and cosine of $2\theta$. The symmetric source-driven branch has no sine term; for generic $\alpha$ this also follows directly by subtracting its two wall conditions. Thus
$$
\boxed{G_0=C(\cos2\theta-\cos2\alpha),\qquad C=\frac1{\sin2\alpha-2\alpha\cos2\alpha}.}
$$
Indeed its integral is $C(\sin2\alpha-2\alpha\cos2\alpha)=1$. At isolated angles where an odd zero-<flux> homogeneous mode satisfies both wall conditions, symmetry selects the stated branch.
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