Stokes flow 

An object moving through a gas or liquid experiences a force in direction opposite to its motion. Terminal velocity is achieved when the drag force is equal in magnitude but opposite in direction to the force propelling the object. Shown is a sphere in Stokes flow, at very low Reynolds number.

Stokes flow (named after George Gabriel Stokes) is a type of fluid flow where advective inertial forces are small compared with viscous forces. The Reynolds number is low, i.e. \textit{Re} \ll 1. This is a typical situation in flows where the fluid velocities are very slow, the viscosities are very large, or the length-scales of the flow are very small, such as in MEMS devices or in the flow of viscous polymers.

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Stokes equations

For this type of flow, the inertial forces are assumed to be negligible and the Navier–Stokes equations simplify to give the Stokes equations:

\boldsymbol{\nabla} \cdot \mathbb{P} + \boldsymbol{f} = 0

where \mathbb{P} is the comoving stress tensor, and \boldsymbol{f} an applied body force. There is also an equation for conservation of mass. In the common case of an incompressible Newtonian fluid, the Stokes equations are:

\boldsymbol{\nabla}p = \mu \nabla^2 \boldsymbol{u} + \boldsymbol{f}
\boldsymbol{\nabla}\cdot\boldsymbol{u}=0

Properties

The Stokes equations represent a considerable simplification of the full Navier–Stokes equations, especially in the incompressible Newtonian case.

Instantaneity
A Stokes flow has no dependence on time other than through time-dependent boundary conditions. This means that, given the boundary conditions of a Stokes flow, the flow can be found without knowledge of the flow at any other time.
Time-reversibility
An immediate consequence of instantaneity, time-reversibility means then a time-reversed Stokes flow solves the same equations as the original Stokes flow. This property can sometimes be used (in conjunction with linearity and symmetry in the boundary conditions) to derive results about a flow without solving it fully.

While these properties are true for incompressible Newtonian Stokes flows, the non-linear and sometimes time-dependent nature of non-Newtonian fluids means that they do not hold in the more general case.

Methods of solution

By stream function

It can be shown that in 2-D, the stream function for an incompressible Newtonian Stokes flow satisfies the biharmonic equation \nabla^4 \psi = 0.

In the 3-D axisymmetric case, the Stokes stream function Ψ solves the equation E2Ψ = 0, where E = {\partial^2 \over \partial r^2} + {\sin{\theta} \over r^2} {\partial \over \partial \theta} { 1 \over \sin{\theta}} {\partial \over \partial \theta}.

By Papkovich-Neuber solution

The Papkovich-Neuber Solution represents the velocity and pressure fields of an incompressible Newtonian Stokes flow in terms of two harmonic potentials.

By Boundary element method

Certain problems, such as the evolution of the shape of a bubble in a Stokes flow, are conducive to numerical solution by the Boundary Element method. This technique can be applied in both 2- and 3-dimensional flows.

By Green's function

The linearity of the Stokes equations in the case of an incompressible Newtonian fluid means that a Green's function for the equations can be found. The solution for the pressure p and velocity \boldsymbol{u} due to a point force \boldsymbol{F}\delta(\boldsymbol{x}) acting at the origin with |\boldsymbol{u}|,p\to 0 as |\boldsymbol{x}|\to\infty is given by

\boldsymbol{u}(\boldsymbol{x}) = \boldsymbol{F} \cdot \mathbb{J}(\boldsymbol{x})
p(\boldsymbol{x}) = \frac{\boldsymbol{F}\cdot\boldsymbol{x}}{4 \pi |\boldsymbol{x}|^3}

where

\mathbb{J}(\boldsymbol{x}) = {1 \over 8 \pi \mu} \left( \frac{\mathbb{I}}{|\boldsymbol{x}|} + \frac{\boldsymbol{x}\boldsymbol{x}}{|\boldsymbol{x}|^3} \right).

is a second-rank tensor known as the Oseen Tensor (after Carl Wilhelm Oseen).

The solution for a distributed force density \boldsymbol{f}(\boldsymbol{x}) (again with decay at infinity) can then be constructed by superposition:

\boldsymbol{u}(\boldsymbol{x}) = \int \boldsymbol{f}(\boldsymbol{y}) \cdot \mathbb{J}(\boldsymbol{x} - \boldsymbol{y}) \, \mathrm{d}^3\!y
p(\boldsymbol{x}) = \int \frac{\boldsymbol{f}(\boldsymbol{y})\cdot(\boldsymbol{x}-\boldsymbol{y})}{4 \pi |\boldsymbol{x}-\boldsymbol{y}|^3} \, \mathrm{d}^3\!y

See also

References