= Solution
For a homogeneous <Newtonian fluid> with constant <kinematic viscosity> $\nu>0$, assume the same steady purely azimuthal flow. In <cylindrical coordinates>, the azimuthal <Navier-Stokes equation> has no pressure forcing or convective term in this state, and becomes
$$
0=\nu\left[\frac{d^2v_\phi}{dr^2}+\frac1r\frac{dv_\phi}{dr}-\frac{v_\phi}{r^2}\right],\qquad v_\phi=r\Omega.
$$
Equivalently, $\Omega''+3\Omega'/r=0$, or $(r^3\Omega')'=0$. Two integrations give the <steady circular Couette profile>,
$$
\boxed{\Omega(r)=A+\frac B{r^2},\qquad v_\phi(r)=Ar+\frac Br.}
$$
This is the basic <Taylor-Couette flow> between rotating cylinders. The constants depend on boundary <angular velocities>; for radii $R_<,R_>$ and prescribed <angular velocities> $\Omega_<,\Omega_>$,
$$
A=\frac{\Omega_>R_>^2-\Omega_<R_<^2}{R_>^2-R_<^2},\qquad
B=\frac{R_<^2R_>^2(\Omega_<-\Omega_>)}{R_>^2-R_<^2}.
$$
If the domain includes $r=0$, regularity requires $B=0$, leaving <solid-body rotation>. On an annulus a nonzero $B$ is allowed and wall torques maintain the shear.
The restriction concerns a steady profile. An arbitrary initial <angular velocity> does not cease to exist when <viscosity> is introduced; within the purely azimuthal ansatz it evolves according to $\partial_t\Omega=\nu(\Omega''+3\Omega'/r)$ until compatible boundary conditions and angular-momentum constraints determine its long-time state.
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