= Solution
{c}
Write $\omega_h=\dot\alpha$, positive on opening. The <no-slip boundary condition> equates the fluid <velocity> with each plate's <velocity>:
$$
\begin{aligned}
v_r(r,0)&=v_\theta(r,0)=0,\\
v_r(r,\alpha)&=0,\qquad v_\theta(r,\alpha)=r\omega_h.
\end{aligned}
$$
The upper condition is a moving-wall <no-penetration boundary condition>, not zero normal <velocity> in the laboratory frame. At the hinge the <velocity> tends to zero. With the arbitrary constant of the <streamfunction> chosen so that $\psi(r,0)=0$, these conditions become
$$
\psi_\theta(r,0)=\psi_\theta(r,\alpha)=0,\qquad
\psi(r,\alpha)=-\frac12\omega_h r^2.
$$
A changing angle cannot give a literally <steady flow> in fixed coordinates. We interpret the stated steadiness as the <quasi-steady Stokes approximation> at each instantaneous $\alpha$.
\b[The printed expression is not biharmonic.] To test it without reproducing the source formula, write its bracket as $G_p(\theta)$ and its <streamfunction> as $\psi_p=-\Omega r^2G_p/2$. For any angular function $G$, the <Laplacian> gives
$$
\nabla^4(r^2G)=\frac{G^{(4)}+4G''}{r^2}.
$$
At the concrete angle $\alpha=\pi/4$, differentiation of the printed bracket gives
$$
G_p^{(4)}+4G_p''
=\frac{-16\pi\cos2\theta+32\sin2\theta}{\pi^2/4-1}.
$$
For example, at the interior point $\theta=\pi/8$,
$$
\boxed{\nabla^4\psi_p
=\frac{4\sqrt2(\pi-2)\Omega}{(\pi^2/4-1)r^2}\ne0\quad(\Omega\ne0).}
$$
The printed bracket does satisfy $G_p(0)=G_p'(0)=G_p'(\alpha)=0$ and $G_p(\alpha)=1$, so its boundary <velocities> would require $\Omega=\omega_h$. Satisfying these <boundary conditions> does not cure its nonzero <biharmonic operator>. The stated series is likewise not a general biharmonic basis: its $r^2\theta\sin2\theta$ term already fails the displayed test. The <separated polar solutions of the biharmonic equation> contain additional radial powers; for the required $r^2$ dependence, the angular basis is $1,\theta,\sin2\theta,\cos2\theta$.
The corrected <slowly opening Stokes wedge> <similarity solution> follows by setting $\psi=-\omega_h r^2G(\theta)/2$. The <biharmonic equation> gives $G^{(4)}+4G''=0$. The two lower-wall conditions reduce its general solution to
$$
G=A(\cos2\theta-1)+B(\sin2\theta-2\theta).
$$
The upper-wall condition $G'(\alpha)=0$ fixes the ratio of $A$ and $B$, and $G(\alpha)=1$ fixes their scale. An expression that also remains finite at $\alpha=\pi/2$ is
$$
\boxed{G(\theta)=
\frac{\sin\alpha(1-\cos2\theta)+\cos\alpha(\sin2\theta-2\theta)}
{2(\sin\alpha-\alpha\cos\alpha)},\qquad
\psi=-\frac12\dot\alpha r^2G(\theta).}
$$
For $0<\alpha<\pi$, its denominator is positive: the <derivative> of $\sin\alpha-\alpha\cos\alpha$ is $\alpha\sin\alpha>0$. Direct differentiation verifies all four <boundary conditions> and the <biharmonic equation>. The resulting <velocity> components are $v_r=-\dot\alpha rG'/2$ and $v_\theta=\dot\alpha rG$.
For a fixed angle of order one, $U(r)\sim|\dot\alpha|r$, so the local <Reynolds number> is $\mathrm{Re}_r\sim|\dot\alpha|r^2/\nu$. One breakdown <length> is therefore
$$
\boxed{l_1\sim\sqrt{\frac{\nu}{|\dot\alpha|}}.}
$$
A second is the geometric <length> $l_2\sim\min(L,W)$, where $L$ is the distance to the radial ends and $W$ the spanwise width. At $r\sim l_2$, matching to the surrounding bath or three-dimensional end flow matters. Thus $r\ll\min(l_1,l_2)$ is required for this local two-dimensional <Stokes flow>, with angle-dependent factors in the inertial estimate. Ideal infinite plates remove the geometric cutoff but not the inertial one.
There is also a temporal restriction: the relevant <viscous diffusion time> must be short compared with the time $T_s$ over which the shape and its <velocity field> change appreciably. Using $r$ as a conservative diffusion <length> gives $r\ll\sqrt{\nu T_s}$. For order-one angles and constant $\dot\alpha$, $T_s\sim|\dot\alpha|^{-1}$ gives the same scale as $l_1$; narrow angles require using their actual gap and angular factors rather than extrapolating this order-one-angle estimate.
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