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Write ${\omega }_h=\dot{\alpha }$, positive on opening. The no-slip boundary condition equates the fluid velocity with each plate's velocity: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 becomeA 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 $\alpha$.

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 ${\psi }_p=-\Omega r^2{G}_p/2$. For any angular function $G$, the Laplacian givesAt the concrete angle $\alpha =\pi /4$, differentiation of the printed bracket givesFor example, at the interior point $\theta =\pi /8$,The printed bracket does satisfy $G_p(0)=G_p^{\prime }(0)=G_p^{\prime }(\alpha )=0$ 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 $r^2\theta \sin 2\theta$ 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 $1,\theta ,\sin 2\theta ,\cos 2\theta$.

The corrected slowly opening Stokes wedge similarity solution follows by setting $\psi =-{\omega }_h{r}^{2}G(\theta )/2$. The biharmonic equation gives $G^{(4)}+4G^{\prime \prime }=0$. The two lower-wall conditions reduce its general solution toThe upper-wall condition $G^{\prime }(\alpha )=0$ fixes the ratio of $A$ and $B$, and $G(\alpha )=1$ fixes their scale. An expression that also remains finite at $\alpha =\pi /2$ isFor $0<\alpha <\pi$, 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 $v_r=-\dot{\alpha }r{G}^{\prime }/2$ 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 thereforeA 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.