 Solution

For steady unidirectional pressure-driven channel flow, write $G=-dp/dz>0$. The axial momentum equation and no-slip boundary condition areNear an ideal wedge, the constant forcing has an $r^2$ particular solution by scaling. Substitution gives $f^{\prime \prime }+4f=-1$, with $f(\pm \alpha )=0$. For $\cos 2\alpha \ne 0$, the symmetric particular solution isThis corner expansion of pressure-driven sector flow is a particular solution, not necessarily the dominant corner term; a homogeneous corner mode can be larger, as the requested cases below show.

Separation of the symmetric harmonic correction with zero values on the straight walls gives modes $r^{{\lambda }_n}\cos ({\lambda }_n\theta )$, whereRegularity at the vertex excludes negative powers. Orthogonality on $[-\alpha ,\alpha ]$ has norm $\alpha$. To cancel the particular solution on $r=a$, use its cosine coefficientsThe integral follows by product-to-sum and $\sin ({\lambda }_n\alpha )=(-1{)}^n$. ThusThe arc data are satisfied by the cosine expansion, and the Poisson equation by construction. At the omitted thresholds $2\alpha =\pi /2,3\pi /2$, a mode has ${\lambda }_n=2$, so the pure $r^{2}f$ ansatz is resonant. The corresponding finite-limit term in the full sector solution is $(G/\mu )(-1{)}^n{r}^2\log (a/r)\cos 2\theta /(4\alpha )$; the other modes remain regular. This explains rather than hides the division by $\cos 2\alpha$.