Consider an infinitely long plate which is suddenly made to move with constant velocity ${\displaystyle U}$  in the ${\displaystyle x}$  direction, which is located at ${\displaystyle y=0}$  in an infinite domain of fluid, which is at rest initially everywhere. The incompressible Navier-Stokes equations reduce to^[2][3]

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

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

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

the last condition is due to the fact that the motion at ${\displaystyle y=0}$  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 edit

The problem on the whole is similar to the one dimensional heat conduction problem. Hence a self-similar variable can be introduced^[4]

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

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

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

with boundary conditions

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

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

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

The force per unit area exerted on the plate is

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

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^[5]

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

Rotating cylinder edit

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

${\displaystyle v_{\theta }={\frac {a\Omega }{2\pi i}}\int _{-i\infty }^{i\infty }{\frac {K_{1}(r{\sqrt {s/\nu }})}{K_{1}(a{\sqrt {s/\nu }})}}e^{st}{\frac {ds}{s}}}$ 

where ${\displaystyle K_{1}}$  is the modified Bessel function of the second kind. As ${\displaystyle t\rightarrow \infty }$ , the solution approaches that of a rigid vortex. The force per unit area exerted on the cylinder is

${\displaystyle F=\mu \left({\frac {\partial v_{\theta }}{\partial r}}-{\frac {v_{\theta }}{r}}\right)_{r=a}={\frac {\rho a^{2}\Omega }{t}}e^{-{\frac {a^{2}}{2\nu t}}}I_{0}\left({\frac {a^{2}}{2\nu t}}\right)-2\mu \Omega }$ 

where ${\displaystyle I_{0}}$  is the modified Bessel function of the first kind.

Sliding cylinder edit

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

${\displaystyle u={\frac {U}{2\pi i}}\int _{-i\infty }^{i\infty }{\frac {K_{0}(r{\sqrt {s/\nu }})}{K_{0}(a{\sqrt {s/\nu }})}}e^{st}{\frac {ds}{s}}.}$ 

• Stokes problem