Rayleigh problem

In fluid dynamics, Rayleigh problem also known as Stokes first problem is a problem of determining the flow created by a sudden movement of a plane from rest, named after Lord Rayleigh and Sir George Stokes. This is considered as one of the simplest unsteady problem that have exact solution for the Navier-Stokes equations.

Flow description

Consider an infinitely long plate which suddenly made to move with constant velocity $U$ in the $x$ direction, which is located at $y=0$ in an infinite domain of fluid, which is at rest initially everywhere. The incompressible Navier-Stokes equations reduce to

${\frac {\partial u}{\partial t}}=\nu {\frac {\partial ^{2}u}{\partial y^{2}}}$

where $\nu$ is the kinematic viscosity. The initial and the no-slip condition on the wall are

$u(y,0)=0,\quad u(0,t>0)=U,\quad u(\infty ,t>0)=0,$

the last condition is due to the fact that the motion at $y=0$ is not felt at infinity. The flow is only due to the motion of the plate, there is no imposed pressure gradient.

Self-Similar solution

The problem on the whole is similar to the one dimensional heat conduction problem. Hence a self-similar variable can be introduced

$\eta ={\frac {y}{\sqrt {\nu t}}},\quad f(\eta )={\frac {u}{U}}$

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

$f''+{\frac {1}{2}}\eta f'=0$

with boundary conditions

$f(0)=1,\quad f(\infty )=0$

The solution to the above problem can be written in terms of complementary error function

$u=U\mathrm {erfc} \left({\frac {y}{\sqrt {4\nu t}}}\right)$

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

$F=\mu \left({\frac {\partial u}{\partial y}}\right)_{y=0}=\rho {\sqrt {\frac {\nu U^{2}}{\pi t}}}$

Arbitrary wall motion

Instead of using a step boundary condition for the wall movement, the velocity of the wall can be prescribed as an arbitrary function of time, i.e., $U=f(t)$. Then the solution is given by

$u(y,t)=\int _{0}^{t}{\frac {f(\tau )}{2{\sqrt {\pi \nu }}}}{\frac {y}{(t-\tau )^{3/2}}}e^{-{\frac {y^{2}}{4\nu (t-\tau )}}}d\tau .$

See also

• Stokes problem