Rayleigh problem

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., ${\displaystyle U=f(t)}$. Then the solution is given by

${\displaystyle u(y,t)={\int }_0^t\frac{f(\tau )}{2\sqrt{\pi \nu }}\frac{y}{(t-\tau {)}^{3/2}}{e}^{-\frac{y^2}{4\nu (t-\tau )}}d\tau .}$

Rayleigh's problem in cylindrical geometry

Rotating cylinder

Consider an infinitely long cylinder of radius ${\displaystyle a}$ starts rotating suddenly at time ${\displaystyle t=0}$ with an angular velocity ${\displaystyle \mathrm{\Omega }}$. Then the velocity in the ${\displaystyle \theta }$ direction is given by

${\displaystyle v_{\theta }=\frac{a\mathrm{\Omega }}{2\pi i}{\int }_{-i\mathrm{\infty }}^{i\mathrm{\infty }}\frac{K_1(r\sqrt{s/\nu })}{K_1(a\sqrt{s/\nu })}{e}^{st}\frac{ds}{s}}$

where ${\displaystyle K_1}$ is the modified Bessel function of the second kind. As ${\displaystyle t\to \mathrm{\infty }}$, the solution approaches that of a rigid vortex. The force per unit area exerted on the cylinder is

${\displaystyle F=\mu {\left(\frac{\mathrm{\partial }{v}_{\theta }}{\mathrm{\partial }r}-\frac{v_{\theta }}{r}\right)}_{r=a}=\frac{\rho a^2\mathrm{\Omega }}{t}{e}^{-\frac{a^2}{2\nu t}}{I}_0\left(\frac{a^2}{2\nu t}\right)-2\mu \mathrm{\Omega }}$

where ${\displaystyle I_0}$ is the modified Bessel function of the first kind.

Sliding cylinder

Exact solution is also available when the cylinder starts to slide in the axial direction with constant velocity ${\displaystyle U}$. If we consider the cylinder axis to be in ${\displaystyle x}$ direction, then the solution is given by

${\displaystyle u=\frac{U}{2\pi i}{\int }_{-i\mathrm{\infty }}^{i\mathrm{\infty }}\frac{K_0(r\sqrt{s/\nu })}{K_0(a\sqrt{s/\nu })}{e}^{st}\frac{ds}{s}.}$