The Question:
Suppose we have an axisymmetric, steady, incompressible flow whose velocity $$\vec u (r,\theta ,z)=u_r(r,\theta ,z)\vec e_r+u_\theta(r,\theta ,z) \vec e_\theta+u_z(r,\theta ,z) \vec e_z$$ in cylindrical coordinates only has $\vec e_\theta$ component (i.e. the transverse component).
If the fluid in $r<a$ rotates rigidly about the $z$-axis with angular velocity $\Omega$, whereas the fluid in $r≥a$ is irrotational, show that the velocity in $r≥a$ is given by
$$\vec u=\Omega \frac{a^2}{r}\vec e_\theta$$
My Attempt:
In cylindrical coordinates the curl of $\vec u$ is
$$\vec \nabla \times \vec u=\bigg(\frac 1r \frac{\partial u_z}{\partial \theta}-\frac{\partial u_\theta}{\partial z} \bigg) \vec e_r+\bigg(\frac{\partial u_r}{\partial z}-\frac{\partial u_z}{\partial r} \bigg)\vec e_\theta+\frac 1r \bigg(\frac{\partial (ru_\theta)}{\partial r}-\frac{\partial u_r}{\partial \theta} \bigg)\vec e_z$$
In the region $r≥a$, the fluid is irrotational, so $\vec \nabla \times \vec u=\vec 0$. Moreover, we are given that $u_r=u_z=0$, so if we plug this in we get
\begin{align} & \vec 0 = \vec \nabla \times \vec u=\bigg(-\frac{\partial u_\theta}{\partial z} \bigg) \vec e_r+\frac 1r \bigg(\frac{\partial (ru_\theta)}{\partial r}\bigg) \vec e_z \\ \implies & \frac{\partial u_\theta}{\partial z}=\frac{\partial (ru_\theta)}{\partial r}=0 \\ \implies & ru_\theta = f(\theta) \\ \implies & u_\theta = \frac 1r f(\theta) \end{align}
And then I am stuck. How should I proceed from here?
$$\frac{1}{r}\frac{\partial u_z}{\partial\theta}=\frac{\partial u_\theta}{\partial z}\implies r\frac{\partial u_\theta}{\partial z}=0$$
and
$$\frac{1}{r}\frac{\partial u_r}{\partial\theta}=\frac{\partial u_\theta}{\partial r}\implies r\frac{\partial u_\theta}{\partial r}=0$$
$r\neq0$
$$ru_\theta(\theta)=f(\theta)$$
Use a matching condition at $r=a$
$$u_\theta(\theta)=\frac{f(\theta)}{a}=\Omega a$$
$$\therefore f(\theta)=\Omega a^2$$