Von Kármán swirling flow

Example

An infinite plane disk $z = 0$ rotates with constant angular velocity $\Omega > 0$ about the $z$-axis. A fluid of kinematic viscosity $\nu > 0$ fills $z > 0$ and is at rest far away. In cylindrical coordinates $(r, \theta, z)$ set $\zeta = z \sqrt{\Omega / \nu}$ and look for a velocity of the form $$u_r = r \Omega F(\zeta), \qquad u_\theta = r \Omega G(\zeta), \qquad u_z = \sqrt{\nu \Omega} \, H(\zeta),$$ with a pressure that depends only on $z$. Von Kármán swirling flow is the solution of the Navier–Stokes equations of this form. The functions $F, G, H$ solve $$2F + H' = 0, \qquad F'' = F^2 - G^2 + H F', \qquad G'' = 2 F G + H G',$$ with $F(0) = 0$, $G(0) = 1$, $H(0) = 0$ and $F, G \to 0$ as $\zeta \to \infty$.

Check: the divergence is $\frac{1}{r} \partial_r (r u_r) + \partial_z u_z = \Omega (2F + H')$, which vanishes. In the radial momentum equation every term is $r \Omega^2$ times a function of $\zeta$. The pressure term drops because $p$ does not depend on $r$, and this gives the equation for $F''$. The azimuthal equation gives the equation for $G''$ in the same way. The conditions at $\zeta = 0$ are the no-slip condition on the rotating disk.