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Reflect the configuration in the horizontal plane through the cylinder axis. The wall and cylinder geometry are unchanged. Downward translation becomes upward translation, a possible wall-normal migration velocity is unchanged, and the axial angular velocity changes sign because it is an axial vector.

On the other hand, reversing the original Stokes flow makes all three velocities change sign: translation becomes upward, wall-normal migration reverses, and rotation reverses. The reflected and reversed problems therefore have the same forcing and geometry except that their wall-normal velocities have opposite signs. Uniqueness of Stokes flow forces that velocity to equal its own negative, so it is zero. The rotational velocities already agree under the two operations, so this argument places no restriction on rotation.

The force-free Stokes flow equations arePut $p={\nabla }^2\Pi$. Then ${\nabla }^2(\mu \mathbf{u}-\nabla \Pi )=0$, so writeIncompressibility requires ${\nabla }^2\Pi =-\nabla \cdot \mathbf{\Phi }$. Since ${\nabla }^2(\mathbf{x}\cdot \mathbf{\Phi })=2\nabla \cdot \mathbf{\Phi }$ for harmonic $\mathbf{\Phi }$, takeThis gives the Papkovich–Neuber representation

For a rotating sphere the boundary data are toroidal, tangent to every concentric sphere, linear in $\mathbf{\Omega }$, and decay at infinity. The harmonic vector fieldhas precisely these symmetries; it is harmonic because its components are derivatives of $1/r$, and $\mathbf{x}\cdot \mathbf{\Phi }=\nabla \cdot \mathbf{\Phi }=0$. HenceThe first pressure argument is $p=-\nabla \cdot \mathbf{\Phi }=0$. Independently, this velocity is harmonic, so the Stokes momentum equation gives $\nabla p=0$; matching the ambient pressure sets that constant to zero.

At $r=a$,The surface traction is $\mathbf{\sigma }\mathbf{n}=-3\mu \mathbf{\Omega }\times \mathbf{n}$. Its moment gives the standard rotational resistance

Apply the Lorentz reciprocal theorem for Stokes flow to ${\mathbf{u}}_1$ and the test flow around a sphere rotating with arbitrary $\hat{\mathbf{\Omega }}$. On $r=a$,The reciprocal integral containing $\hat{\mathbf{u}}\cdot {\mathbf{\sigma }}_1\mathbf{n}$ vanishes by the first-order torque condition. Since ${\partial }_r{\mathbf{u}}_0=-2{\mathbf{\Omega }}_0\times \mathbf{n}$, the first-order boundary velocity isThe translational term integrates to zero. Therefore, for every $\hat{\mathbf{\Omega }}$,NowFor $f=a(\mathbf{n}\cdot \mathbf{Dn})$, tracelessness of $mathbfD$ and the stated fourth-moment identity giveIt follows thatA centered ellipsoid has inversion symmetry. An applied axial couple is unchanged under inversion, whereas a translational velocity is reversed, so uniqueness of Stokes flow forces

The shallow-ice approximation makes the pressure hydrostatic pressure,Horizontal Stokes flow balance and the stress-free boundary condition at $z=h$ givesoThe basal shear stress and the stated basal sliding law are thereforeIntegrating once more and imposing $u(0)=u_b$ yieldsFor the right half of an ice cap, $h_x<0$, so both basal sliding and internal deformation carry ice away from the ice divide.

For $\mathbf{u}=\nabla \times (\psi \hat{\mathbf{y}})$,Hence $u_x+w_z=-{\psi }_{zx}+{\psi }_{xz}=0$, so mass conservation holds identically. Removing the hydrostatic part by writingand taking the curl of the Stokes flow equation gives the biharmonic equation

For a Fourier mode proportional to $e^{ikx}$ with $k>0$, decay as $z\to -\infty$ selectsThe velocity and pressure fields are thereforeDirect substitution verifies the momentum equation.

For a unidirectional, $z$-independent velocityincompressibility holds automatically. The axial Stokes flow equation isNo external pressure gradient is imposed, so $p_z=0$ andThus the axial velocity is a harmonic function in the disk.

Stokes flow obeys $-\nabla p+\mu {\nabla }^{2}u=0$, $\nabla \cdot u=0$. Reversing all applied forces reverses $u$ while leaving streamlines unchanged. Reflection in the vertical plane containing the cylinder axis rules out rotation normal to that plane; reflection combined with reversibility rules out rotation within it and spin about the axis. Thus its orientation remains fixed.

A potential dipole in three dimensions is the potential flow $\mathbf{u}=\nabla \varphi$ with $\varphi =\mathbf{d}\cdot \mathbf{x}/r^3$. Away from the origin its potential is a harmonic function, so it is also a constant-pressure Stokes flow. Its velocity decays as $r^{-3}$ and its vorticity is zero. It is distinct from a stresslet, whose leading velocity decays as $r^{-2}$.

The Prandtl number is the ratio of kinematic viscosity to thermal diffusivity. It compares the rates at which momentum and temperature variations diffuse. In thermal-diffusion units the Boussinesq equations have momentum inertia multiplied by ${\mathrm{Pr}}^{-1}$; large, rather than small, Prandtl number supplies the usual instantaneous Stokes flow limit.

A rotlet is the singular Stokes flow generated by a point torque $\mathbf{G}$:It is the far field of a rotating sphere in Stokes flow.

Attach weights $mg$ at the joint and $\lambda mg$ at each outer endpoint of the right-angle two-rod resistance matrix geometry. With the first rod initially horizontal and the second vertically upwards, quasistatic Stokes flow gives $\dot{\theta }=(1-\lambda )mg(\cos \theta -\sin \theta )/(4C{L}^2)$. For $\lambda <1$ the orientation tends to $\pi /4$; for $\lambda >1$ it tends to $-3\pi /4$. Eliminating time gives $x-x_0=(2L/3)(1+\sin \theta -\cos \theta )$ and the same rightward net displacement $2L/3$ in both cases. For $\lambda >1$ the initial drift is leftwards before reversing. At $\lambda =1$ there is no rotation or lateral drift, illustrating noncommuting infinite-time and parameter limits.

An inner cylinder of radius $a$ centered at $(0,a)$ and an outer cylinder of radius $2a$ centered at $(0,2a)$ have polar boundaries $r=2a\sin \theta$ and $r=4a\sin \theta$. If their angular velocities are $\Omega$ and $-\Omega /4$, respectively, the no-slip Stokes flow between them has streamfunctionIts unique interior stagnation point is $(x,y)=(0,2\sqrt{2}a)$, and its nonzero streamlines are closed curves around that point.

The Taylor swimming sheet is an infinite two-dimensional microswimmer whose prescribed traveling deformation drives Stokes flow. A transverse wave $y_s=ϵ\sin (x-t)$ of small amplitude swims opposite to its direction of propagation, with speed $U={ϵ}^2/2+O({ϵ}^4)$ when lengths are scaled by inverse wavenumber and velocities by wave speed. Its biharmonic stream function for planar Stokes flow has first-order part ${\psi }_1=(1+y)e^{-y}\sin (x-t)$.

A spherical squirmer with tangential surface slip velocity $(I-nn)(A+Bn)$, where $B$ is a symmetric second-rank tensor and a traceless second-rank tensor, translates with $U=-2A/3$ and does not rotate in an unbounded quiescent fluid. With $S=x\cdot Bx$ its exterior Stokes flow isThis follows from the Unscaled Papkovich–Neuber representation using harmonic functions as potentials $\Phi =-a^{2}Bx/(2r^3)$ and $\chi =a^{3}A\cdot x/(3r^3)+a^{4}S/(2r^5)$. Matching the surface coefficients proves the formula. The two modes generate a potential dipole and a stresslet, respectively.

With prescribed velocity on the boundary and suitable decay or far-field data, a Stokes flow is unique up to an additive pressure constant. Applying the viscous dissipation identity to the difference of two solutions makes its rate-of-strain tensor vanish; the homogeneous boundary data then eliminate the remaining rigid motion.

In the free-slip Stokes flow temperature model, let the critical temperature be $g=\sin \pi z$ and vertical velocity $f=sg$, $s=k_c^2+{\pi }^2$. A forced positive second harmonic $(W,\Theta )e^{2i{k}_{c}x}$ couples to the negative critical harmonic. Its temperature solvability condition has integrandThe identity follows by substituting $f=sg$ and differentiating. Its integral is zero because $g$ vanishes on both plates, regardless of the forced boundary value of $\Theta$. Thus the first resonant forcing coefficient vanishes. A generic nonzero conjugate-amplitude term cannot be inferred from wave-number matching alone.