Burgers vortex

Example

Fix a strain rate $\alpha > 0$, a circulation $\Gamma > 0$ and a kinematic viscosity $\nu > 0$. In cylindrical coordinates $(r, \theta, z)$, the Burgers vortex is the steady flow with $$u_r = -\alpha r, \qquad u_z = 2 \alpha z, \qquad u_\theta = \frac{\Gamma}{2 \pi r} \Big( 1 - e^{-\alpha r^2 / (2 \nu)} \Big).$$ It is incompressible: $\frac{1}{r} \partial_r (r u_r) + \partial_z u_z = -2\alpha + 2\alpha = 0$. Its vorticity is axial, $\omega = (0, 0, \zeta(r))$ with $$\zeta(r) = \frac{1}{r} \frac{d}{dr}(r u_\theta) = \frac{\Gamma \alpha}{2 \pi \nu} \, e^{-\alpha r^2 / (2 \nu)}.$$ The steady vorticity equation reads $u_r \zeta' = 2 \alpha \zeta + \nu (\zeta'' + \zeta'/r)$, since the stretching term is $\zeta \, \partial_z u_z = 2 \alpha \zeta$. With $\zeta = C e^{-k r^2}$ and $k = \alpha / (2\nu)$, the left side is $2 \alpha k r^2 \zeta$, and the right side is $2\alpha \zeta + \nu(4 k^2 r^2 - 4k) \zeta = 2 \alpha k r^2 \zeta$, so the equation holds. With a suitable pressure, the flow is an exact steady solution of the incompressible Navier–Stokes equations.