Taylor–Couette flow

Example

Consider a fluid of density $\rho > 0$ and kinematic viscosity $\nu > 0$ in the gap $R_1 < r < R_2$ between two coaxial cylinders, in cylindrical coordinates $(r, \theta, z)$. The inner cylinder rotates with angular velocity $\Omega_1$ and the outer one with $\Omega_2$. Circular Couette flow is the azimuthal flow $u = v(r) \, e_\theta$ with $$v(r) = A r + \frac{B}{r}, \qquad A = \frac{\Omega_2 R_2^2 - \Omega_1 R_1^2}{R_2^2 - R_1^2}, \qquad B = \frac{(\Omega_1 - \Omega_2) R_1^2 R_2^2}{R_2^2 - R_1^2},$$ and a pressure $p(r)$ with $p'(r) = \rho \, v(r)^2 / r$.

Check: the flow has no radial or axial component and does not depend on $\theta$ or $z$, so it is divergence-free. The azimuthal Navier–Stokes equation reduces to $v'' + v'/r - v/r^2 = 0$, and both $r$ and $1/r$ solve it. The radial equation reduces to $-\rho v^2 / r = -p'(r)$, which is the choice of $p$. At $r = R_1$ one gets $v(R_1) = \Omega_1 R_1$ after simplifying, and similarly $v(R_2) = \Omega_2 R_2$, so the no-slip condition holds.