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The biharmonic equation is ${\nabla }^{4}f=0$, where the biharmonic operator is the square of the Laplacian. In planar Stokes flow, taking the curl of the momentum equation and using a Cartesian streamfunction produces this equation. Its decaying spatial Fourier series modes have form $(A+By)e^{-|k|y}{e}^{ikx}$.

Use a rotlet as the auxiliary Stokes flow, with a rigid sphere at its center. Its exterior-domain boundary torque is the imposed auxiliary couple. In the Lorentz reciprocal theorem, the actual sphere's zero torque removes the other surface term, while the volume term is the rotlet dotted with the body force. Varying the auxiliary couple gives the displayed angular velocity, assuming the integrals and the far boundary limit converge.

For homogeneous Stokes flow in a smooth bounded volume, use $J_{ij}(r)=({\delta }_{ij}/|r|+r_i{r}_j/|r{|}^3)/(8\pi \mu )$ and $K_{ijk}(r)=-3r_i{r}_j{r}_k/(4\pi |r{|}^5)$. The Lorentz reciprocal theorem with a point-force Stokeslet gives the displayed representation, with $c=1$ inside, $0$ outside and $1/2$ on a smooth boundary. The double-layer integral at the boundary has its limiting principal-value meaning. In an exterior problem the body's outward normal is opposite to the fluid-domain normal, reversing both boundary terms; a nonzero ambient flow adds its far-boundary contribution.

The Brinkman equation augments Stokes flow with a local drag against a background matrix of velocity ${\mathbf{u}}_m$. The inverse screening length $\alpha$ has units of inverse length. With equal viscous coefficients the permeability is ${\alpha }^{-2}$. Taking the curl produces a screened vorticity equation.

For a drop with internal viscosity $\lambda \mu$, exterior viscosity $\mu$, continuous velocity and a decaying exterior Stokes flow, use the drop-outward normal $n$ and curvature $\kappa ={\nabla }_s\cdot n$, twice the mean curvature. Constant surface tension gives $({\sigma }^{+}-{\sigma }^{-})n=\gamma \kappa n$. Multiply the interior boundary representation by $\lambda$ and add it to the exterior one: the traction difference yields the single-layer term shown, while the viscosity-independent stress kernel yields $(\lambda -1)D[u]$. Here $S,D$ denote the kernels in boundary integral representation of Stokes flow. For equal viscosities the double layer vanishes.

A control volume is a chosen region over which fluid mass conservation and momentum balance are integrated. It may be fixed or move with a prescribed velocity. Boundary fluxes must use velocity relative to the moving boundary. In Stokes flow, negligible inertia reduces the momentum balance to the sum of body forces and boundary tractions, making a carefully chosen control volume useful for finding drag without integrating directly over a curved body.

A pair of equal and opposite forces acting along distinct lines has zero resultant force but a nonzero torque. If $\mathbf{F}$ acts at ${\mathbf{x}}_1$ and $-\mathbf{F}$ at ${\mathbf{x}}_2$, the total torque about any origin is $({\mathbf{x}}_1-{\mathbf{x}}_2)\times \mathbf{F}$. Shifting that origin changes the moment by the cross product of the shift with the resultant force, which is zero. A prescribed couple therefore specifies a pure turning moment independently of its reference point. In Stokes flow it can drive a rotating sphere in Stokes flow without a directly imposed translational force.

On the symmetric small-$Q/\nu$ branch of radial viscous flow in a wedge, the leading Stokes flow solves $G_0^{(3)}+4G_0^{\prime }=0$, zero wall values and unit integral. HenceFor $\pi /2<\alpha <{\alpha }_c$, where ${\alpha }_c\simeq 2.24670$, this is positive in $|\theta |<\pi -\alpha$ and negative near the walls. The normalization denominator tends to zero at the stated root. At the root the homogeneous angular mode has zero net flux, so it cannot be rescaled to carry a fixed nonzero source flux.

Approaching that angle makes large outward and reverse contributions cancel to leave the prescribed flux. The leading amplitude diverges and the nonlinear inertial correction ceases to be perturbative, however small its coefficient is at fixed angle. This is a failure of a regular expansion, not a proof of infinite speed in the full Navier-Stokes equation. Distinct inner scalings or different steady branches require separate analysis.

A spherical clean bubble with constant gas mass density and prescribed dissolution law $\dot{M}=-4\pi aD{\rho }_a$ obeys $\dot{a}=-D/a$. With negligible gas dynamic viscosity, instantaneous terminal velocity is $\dot{z}=\Delta \rho g{a}^2/(3\mu )$. Eliminating time yields the displayed radius-height relation and maximum rise $z_{\mathrm{diss}}=\Delta \rho g{a}_0^4/(12\mu D)$. This uses an instantaneous force balance; a literally constant speed during shrinkage instead gives $a^2$ linear in height. The Reynolds number must remain small for the Stokes flow drag model to apply.

For a sphere of radius $a$ rotating with angular velocity $\Omega$, the primary Stokes flow has $v_{\varphi }=K\sin \theta /r^2$, $K=\Omega a^3$. In the weak, inertialess second-order fluid correction, put $c={\psi }_1+2{\psi }_2$. The stream function in the title hasBoth meridional components vanish at $r=a$ and decay at infinity. Indeed, $E_{r\varphi }=-3K\sin \theta /(2r^3)$ gives $[\nabla \times \nabla \cdot (2c{E}^2){]}_{\varphi }=72c{K}^2\sin \theta \cos \theta /r^8$, and the curl of the momentum equation reduces to $(d^2/d{r}^2-6/r^2{)}^{2}H=72c{K}^2/(\mu r^7)$. For $c>0$ the secondary flow enters towards the equator and leaves near the poles; $c<0$ reverses it and $c=0$ removes this correction. The meridional circulation is even in the sign of sphere rotation. Inertial secondary flow is excluded from this calculation.

A slender axisymmetric column of a Newtonian fluid, with radius $a(z,t)$, axial velocity $w$, external pressure $p_e$ and axial body-force density $f$, obeys $(a^2{)}_t+(a^{2}w{)}_z=0$ and $3\mu (a^2{w}_z{)}_z/a^2=p_{e,z}-f$ when inertia, surface tension and leading external shear are negligible. The factor three is the Trouton ratio. Radial mass conservation gives $u_r=-r{w}_z/2$; the Newtonian fluid stress tensor gives $p=p_e-\mu w_z$ and axial excess stress $3\mu w_z$. This reduces a three-dimensional Stokes flow to one-dimensional evolving geometry.

For a sphere of radius $a$ in a slowly varying ambient Stokes flow of vorticity ${\mathbf{\omega }}_{\infty }$, its angular velocity and applied torque obeyat the sphere centre.

In a three-dimensional exterior Stokes flow with no body force, $\nabla \cdot \mathbf{\sigma }=0$. The divergence theorem on the fluid between a rigid body and a large sphere $S_R$ gives the displayed force on the body, where $\mathbf{n}$ points outward from the sphere. The body's outward normal points oppositely to the inner normal of this fluid domain. A nonzero resultant force produces a leading Stokeslet, so velocity is $O(R^{-1})$, gauge-fixed pressure is $O(R^{-2})$ and rate-of-strain tensor is $O(R^{-2})$. The finite force is therefore compatible with pointwise decay of the stress tensor. Force-free bodies may have faster decay.

At fixed sphere rotation in an annular Stokes flow, add rigid particles with zero net force and torque. Extend their actual rigid velocities through their interiors. The extension is an admissible incompressible competitor in the particle-free annulus, with zero interior rate-of-strain tensor. The Minimum-dissipation theorem for Stokes flow gives $D\ge D_0$. The boundary-work identity gives $D=\mathbf{\Omega }\cdot \mathbf{G}$, because the stationary outer wall and the freely moving particles contribute no power. Equality would require the extension to equal the unique particle-free flow, whose strain cannot vanish on a finite particle volume when $\Omega \ne 0$. Thus the resistance component along the imposed rotation strictly increases.

For a two-dimensional decaying Stokes flow in $y>0$, prescribe boundary Fourier mode $(U,V)e^{ikx}$. The boundary shear stress and normal stress are the displayed diagonal map, with the common exponential understood. For $k>0$ and $V=0$, the Unscaled Papkovich–Neuber representation gives $u=U(1-ky)e^{ikx-ky}$, $v=-ikUy{e}^{ikx-ky}$ and $p=-2i\mu kU{e}^{ikx-ky}$. Differentiation proves the tangential traction and vanishing normal stress. The normal-velocity calculation supplies the second diagonal entry. The map converts a viscous exterior into a nonlocal boundary resistance proportional to $|k|$.

A physicist and mathematician associated with Stokes flow, Stokes' second problem and Stokes theorem.

A fixed sphere in a slowly varying incident Stokes flow requires an external force $\mathbf{F}=-6\pi \mu a{\mathbf{u}}_{\infty }$ and external couple $\mathbf{G}=-4\pi \mu a^3{\mathbf{\omega }}_{\infty }$ to leading order. These follow from Faxén translation law and Faxén rotation law; the incident Stokeslet supplies ${\mathbf{u}}_{\infty }$ and ${\mathbf{\omega }}_{\infty }$. Finite-radius Laplacian terms enter the translational force at the next nonzero order. The signs here refer to applied holding loads, not hydrodynamic loads.

Short-wavelength corrugations on the underside of a highly viscous ice shelf relax through local Stokes flow. For a wavelength much smaller than the shelf thickness, the deformation occupies a depth of order the inverse wavenumber and the ice may be treated as a viscous half-space.

In thermal-time units, taking the Prandtl number large removes momentum inertia at leading order. The Stokes flow momentum equation then determines velocity instantaneously from the temperature perturbation, while the nonlinear heat equation retains its time derivative. Small Prandtl number does not generally give this model. An exact contrasting shear in a free-slip layer is $u=U_0{e}^{-\mathrm{Pr}{\pi }^{2}t}\cos \pi z$, $w=\theta =0$: it decays slowly as the Prandtl number tends to zero but is excluded by the instantaneous Stokes equation.

A viscous current with density between those of two ambient fluids spreads along their interface. Vertical buoyancy divides its thickness above and below the undisturbed interface. When its internal shear and in-plane extension are both negligible, resistance from the two ambient Stokes flows controls its spreading.