Stokes problem

The initial condition is not required because of periodicity. Since both the equation and the boundary conditions are linear, the velocity can be written as the real part of some complex function

${\displaystyle u=U\Re \left[e^{i\omega t}f(y)\right]}$

because ${\displaystyle \cos \omega t=\Re e^{i\omega t}}$.

Substituting this into the partial differential equation reduces it to ordinary differential equation

${\displaystyle f''-{\frac {i\omega }{\nu }}f=0}$

with boundary conditions

${\displaystyle f(0)=1,\quad f(\infty )=0}$

The solution to the above problem is

${\displaystyle f(y)=\exp \left[-{\frac {1+i}{\sqrt {2}}}{\sqrt {\frac {\omega }{\nu }}}y\right]}$

${\displaystyle u(y,t)=Ue^{-{\sqrt {\frac {\omega }{2\nu }}}y}\cos \left(\omega t-{\sqrt {\frac {\omega }{2\nu }}}y\right)}$

The disturbance created by the oscillating plate travels as the transverse wave through the fluid, but it is highly damped by the exponential factor. The depth of penetration ${\displaystyle \delta ={\sqrt {2\nu /\omega }}}$ of this wave decreases with the frequency of the oscillation, but increases with the kinematic viscosity of the fluid.

The force per unit area exerted on the plate by the fluid is

${\displaystyle F=\mu \left({\frac {\partial u}{\partial y}}\right)_{y=0}={\sqrt {\rho \omega \mu }}U\cos \left(\omega t-{\frac {\pi }{4}}\right)}$

There is a phase shift between the oscillation of the plate and the force created.

An important observation from Stokes' solution for the oscillating Stokes flow is that vorticity oscillations are confined to a thin boundary layer and damp exponentially when moving away from the wall.[7] This observation is also valid for the case of a turbulent boundary layer. Outside the Stokes boundary layer – which is often the bulk of the fluid volume – the vorticity oscillations may be neglected. To good approximation, the flow velocity oscillations are irrotational outside the boundary layer, and potential flow theory can be applied to the oscillatory part of the motion. This significantly simplifies the solution of these flow problems, and is often applied in the irrotational flow regions of sound waves and water waves.

If the fluid domain is bounded by an upper, stationary wall, located at a height ${\displaystyle y=h}$, the flow velocity is given by

${\displaystyle u(y,t)={\frac {U}{2(\cosh 2\lambda h-\cos 2\lambda h)}}[e^{-\lambda (y-2h)}\cos(\omega t-\lambda y)+e^{\lambda (y-2h)}\cos(\omega t+\lambda y)-e^{-\lambda y}\cos(\omega t-\lambda y+2\lambda h)-e^{\lambda y}\cos(\omega t+\lambda y-2\lambda h)]}$

where ${\displaystyle \lambda ={\sqrt {\omega /(2\nu )}}}$.

Stokes boundary layer due to the sinusoidal oscillation of the far-field flow velocity. The horizontal velocity is the blue line, and the corresponding horizontal particle excursions are the red dots.

The case for an oscillating far-field flow, with the plate held at rest, can easily be constructed from the previous solution for an oscillating plate by using linear superposition of solutions. Consider a uniform velocity oscillation ${\displaystyle u(\infty ,t)=U_{\infty }\cos \omega t}$ far away from the plate and a vanishing velocity at the plate ${\displaystyle u(0,t)=0}$. Unlike the stationary fluid in the original problem, the pressure gradient here at infinity must be a harmonic function of time. The solution is then given by

${\displaystyle u(y,t)=U_{\infty }\left[\,\cos \omega t-{\text{e}}^{-{\sqrt {\frac {\omega }{2\nu }}}y}\,\cos \left(\omega t-{\sqrt {\frac {\omega }{2\nu }}}y\right)\right],}$

which is zero at the wall z = 0, corresponding with the no-slip condition for a wall at rest. This situation is often encountered in sound waves near a solid wall, or for the fluid motion near the sea bed in water waves. The vorticity, for the oscillating flow near a wall at rest, is equal to the vorticity in case of an oscillating plate but of opposite sign.

Consider an infinitely long cylinder of radius ${\displaystyle a}$ exhibiting torsional oscillation with angular velocity ${\displaystyle \Omega \cos \omega t}$ where ${\displaystyle \omega }$ is the frequency. Then the velocity approaches after the initial transient phase to[8]

${\displaystyle v_{\theta }=a\Omega \ \Re \left[{\frac {K_{1}(r{\sqrt {i\omega /\nu }})}{K_{1}(a{\sqrt {i\omega /\nu }})}}e^{i\omega t}\right]}$

where ${\displaystyle K_{1}}$ is the modified Bessel function of the second kind. This solution can be expressed with real argument[9] as:

${\displaystyle {\begin{aligned}v_{\theta }\left(r,t\right)&=\Psi \left\lbrace \left[{\textrm {kei}}_{1}\left({\sqrt {R_{\omega }}}\right){\textrm {kei}}_{1}\left({\sqrt {R_{\omega }}}r\right)+{\textrm {ker}}_{1}\left({\sqrt {R_{\omega }}}\right){\textrm {ker}}_{1}\left({\sqrt {R_{\omega }}}r\right)\right]\cos \left(t\right)\right.\\&+\left.\left[{\textrm {kei}}_{1}\left({\sqrt {R_{\omega }}}\right){\textrm {ker}}_{1}\left({\sqrt {R_{\omega }}}r\right)-{\textrm {ker}}_{1}\left({\sqrt {R_{\omega }}}\right){\textrm {kei}}_{1}\left({\sqrt {R_{\omega }}}r\right)\right]\sin \left(t\right)\right\rbrace \\\end{aligned}}}$

where

${\displaystyle \Psi =\left[{\textrm {kei}}_{1}^{2}\left({\sqrt {R_{\omega }}}\right)+{\textrm {ker}}_{1}^{2}\left({\sqrt {R_{\omega }}}\right)\right]^{-1},}$

${\displaystyle \mathrm {kei} }$ and ${\displaystyle \mathrm {ker} }$ are Kelvin functions and ${\displaystyle R_{\omega }}$is to the dimensionless oscillatory Reynolds number defined as ${\displaystyle R_{\omega }=\omega a^{2}/\nu }$, being ${\displaystyle \nu }$ the kinematic viscosity.

If the cylinder oscillates in the axial direction with velocity ${\displaystyle U\cos \omega t}$, then the velocity field is

${\displaystyle u=U\ \Re \left[{\frac {K_{0}(r{\sqrt {i\omega /\nu }})}{K_{0}(a{\sqrt {i\omega /\nu }})}}e^{i\omega t}\right]}$

where ${\displaystyle K_{0}}$ is the modified Bessel function of the second kind.