= Solution
For a <velocity potential>, $\mathbf u=\nabla\phi$. The <incompressible flow> condition $\nabla\cdot\mathbf u=0$ therefore becomes <Laplace equation> $\nabla^2\phi=0$. The <Laplace equation in polar coordinates> gives
$$
\nabla^2\phi
=A(\gamma^2-\lambda^2)r^{\gamma-2}\cos(\lambda\theta),
$$
so positivity of $\gamma$ and $\lambda$ requires
$$
\boxed{\gamma=\lambda}.
$$
The velocity components are
$$
u_r=\frac{\partial\phi}{\partial r}
=A\gamma r^{\gamma-1}\cos(\lambda\theta),
\qquad
u_\theta=\frac1r\frac{\partial\phi}{\partial\theta}
=-A\lambda r^{\gamma-1}\sin(\lambda\theta).
$$
The <no-penetration boundary condition> on both walls says $u_\theta=0$ at $\theta=0$ and $\theta=\alpha$, hence $\lambda\alpha=n\pi$. If $n\ge2$, $\sin(n\pi\theta/\alpha)$ changes sign inside the wedge. The stated sign condition therefore selects $n=1$, and
$$
\boxed{\gamma=\lambda=\frac\pi\alpha}.
$$
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