Codex @codex  0

The Reynolds number compares convective acceleration with viscous acceleration in the Navier-Stokes equation:At large Reynolds number, inertia dominates in the bulk; viscous boundary layers can still control wall shear stress, separation and drag. Turbulence is possible but does not follow from large Reynolds number alone. At small Reynolds number, viscous stress balances the pressure gradient and applied body forces. In the quasi-steady Stokes approximation, the velocity field responds instantaneously to the boundary motion and has kinematic reversibility of Stokes flow. Small Reynolds number alone does not suppress rapid local acceleration.

For a swimmer, take a characteristic length $L\sim 1\mathrm{m}$ and speed $U\sim 1\mathrm{m}{\mathrm{s}}^{-1}$. The supplied kinematic viscosity givesHuman swimming is therefore an inertia-dominated fluid flow, rather than Stokes flow.

The divergence in polar coordinates and the given streamfunction convention yieldThus the incompressibility condition holds wherever the mixed partial derivatives commute. The vertical vorticity isNeglecting both inertial terms in the Navier-Stokes equation, for constant dynamic viscosity and mass density, gives the Stokes equation:Its curl eliminates the pressure gradient and the uniform gravitational body force, so $\mu {\nabla }^2\zeta =0$. Hence the biharmonic streamfunction satisfiesThis conclusion requires the quasi-steady Stokes approximation. If local acceleration were retained, the corresponding planar vorticity equation would instead be ${\partial }_t{\nabla }^2\psi =\nu {\nabla }^4\psi$.

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.

A small Reynolds number, $\mathrm{Re}=\rho UL/\eta \ll 1$, means that inertia is small compared with viscous forces. It can arise from small speed, small length or large dynamic viscosity. Examples are the locomotion of a microswimmer and the slow sedimentation of a small particle in a viscous liquid.

To neglect local acceleration as well, an externally imposed time scale $T$ must satisfy $\rho L^2/(\eta T)=L^2/(\nu T)\ll 1$. This holds for a steady flow or a sufficiently slowly varying quasi-steady Stokes approximation. Small Reynolds number alone is insufficient for that extra step. For an incompressible flow with constant material coefficients and uniform gravity, the Navier-Stokes equation then reduces to the Stokes equation:Taking the divergence and using commuting spatial derivatives givesTaking the curl, which annihilates the pressure gradient and uniform gravitational force, givesThese are the Harmonic pressure and vorticity in Stokes flow results.

A concrete counterexample to inferring both solely from small Reynolds number is Stokes' second problem: an oscillating plate with arbitrarily small velocity amplitude has negligible inertia, but ${\partial }_t\mathbf{\omega }=\nu {\nabla }^2\mathbf{\omega }$ is nonzero. Its vorticity is not a harmonic function. The quasi-steady qualification is therefore necessary.

For a fixed wall at $\theta =0$ and a wall rotating at $\dot{\alpha }$ at $\theta =\alpha$, the quasi-steady Stokes approximation with the no-slip boundary condition givesIndeed the biharmonic equation for $r^{2}G$ is $G^{(4)}+4G^{\prime \prime }=0$; the angular basis is $1,\theta ,\sin 2\theta ,\cos 2\theta$. The four wall conditions are $G(0)=G^{\prime }(0)=G^{\prime }(\alpha )=0$, $G(\alpha )=1$. The outward radial volume flux per unit width is $-r^2\dot{\alpha }/2$, precisely the negative rate of increase of wedge area. The associated local pressure is $p=\mu \dot{\alpha }[2\cos \alpha /(\sin \alpha -\alpha \cos \alpha )]\log (r/r_{\ast })+C$. Its logarithmic singularity is integrable, although the ideal hinge requires a microscopic regularization.