 Solution

No slip on the two fixed radial walls gives $g(\pm \alpha )=0$. The outward flux through a circular arc is ${\int }_{-\alpha }^{\alpha }{u}_{r}rd\theta ={\int }_{-\alpha }^{\alpha }gd\theta$, so source strength $Q$ requires that integral to equal $Q$. Set $g=QG$ and $ϵ=Q/\nu$. ThenAt leading order $G_0^{(3)}+4G_0^{\prime }=0$, 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. ThusIndeed its integral is $C(\sin 2\alpha -2\alpha \cos 2\alpha )=1$. At isolated angles where an odd zero-flux homogeneous mode satisfies both wall conditions, symmetry selects the stated branch.