# Stokes boundary layer

Stokes boundary layer in a viscous fluid due to the harmonic oscillation of a plane rigid plate (bottom black edge). Velocity (blue line) and particle excursion (red dots) as a function of the distance to the wall.

In fluid dynamics, the Stokes boundary layer, or oscillatory boundary layer, refers to the boundary layer close to a solid wall in oscillatory flow of a viscous fluid. Or, it refers to the similar case of an oscillating plate in a viscous fluid at rest, with the oscillation direction(s) parallel to the plate. For the case of laminar flow at low Reynolds numbers over a smooth solid wall, George Gabriel Stokes – after whom this boundary layer is called – derived an analytic solution, one of the few exact solutions for the Navier–Stokes equations.[1][2] In turbulent flow, this is still named a Stokes boundary layer, but now one has to rely on experiments, numerical simulations or approximate methods in order to obtain useful information on the flow.

The thickness of the oscillatory boundary layer is called the Stokes boundary-layer thickness.

## Vorticity oscillations near the boundary

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.[3] 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.

## Stokes boundary layer for laminar flow near a wall

The oscillating flow is assumed to be uni-directional and parallel to the plane wall. The only non-zero velocity component is called u (SI measure in meter/second, or m/s) and is in the x-direction parallel to the oscillation direction. Moreover, since the flow is taken to be incompressible, the velocity component u is only a function of time t (in seconds) and distance from the wall z (in meter). The Reynolds number is taken small enough for the flow to be laminar. Then the Navier–Stokes equations, without additional forcing, reduce to:[4]

$\frac{\partial u}{\partial t} = -\frac{1}{\rho} \frac{\partial p}{\partial x} + \nu \frac{\partial^2 u}{\partial z^2},$

with:

and

• u the velocity of the fluid along the plate (m/s)
• x the position along the plate (m)
• z the distance from the plate (m)
• t the time (s)

Because the velocity u is not a function of position x along the plate, the pressure gradient ∂p/∂x is also independent of x (but the pressure p varies linearly with x). Moreover, the Navier–Stokes equation for the velocity component perpendicular to the wall reduces to ∂p/∂z = 0, so the pressure p and pressure gradient ∂p/∂x are also independent of the distance to the wall z. In conclusion, the pressure forcing ∂p/∂x can only be a function of time t.[4]

The only non-zero component of the vorticity vector is the one in the direction perpendicular to x and z, called ω (in s-1) and equal to:[3]

$\omega = \frac{\partial u}{\partial z}.$

Taking the z-derivative of the above equation, ω has to satisfy[3]

$\frac{\partial \omega}{\partial t} = \nu \frac{\partial^2 \omega}{\partial z^2}.$

As usual for the vorticity dynamics, the pressure drops out of the vorticity equation.[5]

### Oscillation of a plane rigid plate

Harmonic motion, parallel to a plane rigid plate, will result in the fluid near the plate being dragged with the plate, due to the viscous shear stresses. Suppose the motion of the plate is

$u_0(t) = U_0\, \cos\left( \Omega\, t \right),\,$

with

The plate, located at z = 0, forces the viscous fluid adjacent to have the same velocity u1zt ) resulting in the no-slip condition:

$u_1(0,t) = u_0(t) = U_0\, \cos\left( \Omega\, t \right) \quad \text{ at }\; z = 0.$

Far away from the plate, for z → ∞, the velocity u1 approaches zero. Consequently, the pressure gradient ∂p/∂x is zero at infinity and, since it is only a function of time t and not of z, has to be zero everywhere:[6]

$\frac{\partial u_1}{\partial t} = \nu \frac{\partial^2 u_1}{\partial z^2}.$

Such an equation is called a one-dimensional heat equation or diffusion equation.

As a result, the solution for the flow velocity is[7]

$u_1(z,t) = U_0\, \text{e}^{-\kappa\, z}\, \cos\left( \Omega\, t\, -\, \kappa\, z\right) \quad \text{ with }\; \kappa\, =\, \sqrt{\frac{\Omega}{2\nu}}.$

Here, κ is a kind of wavenumber in the z-direction, associated with a length[7]

$\delta = \frac{2\pi}{\kappa} = 2\pi\, \sqrt{\frac{2\nu}{\Omega}}$

which is called the Stokes boundary-layer thickness. At a distance δ from the plate, the velocity amplitude has been reduced to e-2π ≈ 0.002 times its value U0 at the plate surface. Further, as can be seen from the phase changes Ω t - κ z in the solution u1, the velocity oscillations propagate as a damped wave away from the wall, with wavelength δ and phase speed Ω / κ.

The vorticity ω1 is equal to

$\omega_1(z,t) = \frac{\partial u_1}{\partial z} = -\kappa\, U_0\, \text{e}^{-\kappa\, z}\, \Bigl[\, \cos\left( \Omega\, t\, -\, \kappa\, z \right)\, -\, \sin\left( \Omega\, t\, -\, \kappa\, z \right)\, \Bigr]$

and, as u1, dampens exponentially in amplitude when moving away from the plate surface.

### Flow due to an oscillating pressure gradient near a plane rigid plate

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 u:

$u_\infty(z,t) = U_0\, \cos\left( \Omega\, t \right), \,$

which satisfies the flow equations for the Stokes boundary layer, provided it is driven by a pressure gradient

$\frac{\partial p_2}{\partial x} = \rho\, \Omega\, U_0\, \sin\left( \Omega\, t \right).$

Subtracting the solution u1zt ) from uzt ) gives the desired solution for an oscillating flow near a rigid wall at rest:[3]

$u_2(z,t) = U_0\, \Bigl[\, \cos\left( \Omega\, t \right)\, -\, \text{e}^{-\kappa\, z}\, \cos\left( \Omega\, t\, -\, \kappa\, z \right)\, \Bigr],$

which is zero at the wall z = 0, corresponding with the no-slip condition for a wall at rest. Further the velocity u2 oscillates with amplitude U0 far away from the wall, z → ∞. 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: ω2 = - ω1.