Vorticity equation of axi symmetric flow with swirl We consider in cylindrical coordinates $u=(u^{r}, u^{\theta},u^{z})$ of axi symmetric flow with swirl. And we have the following vorticity equation in cylindrical form.
$$\omega^r = \frac{1}{r}\frac{\partial u^z}{\partial \theta} - \frac{\partial u^\theta}{\partial z} = - \frac{\partial u^\theta}{\partial z}, \\ \omega^\theta = \frac{\partial u^r}{\partial z} - \frac{\partial u^z}{\partial r}, \\ \omega^z = \frac{1}{r}\frac{\partial}{\partial r}(r u^\theta) - \frac{1}{r} \frac{\partial u^r}{\partial \theta} = \frac{1}{r}\frac{\partial}{\partial r}(r u^\theta).$$
My question is: how to derive $$u^{r}e_{r}+u^{z}e_{z}=\nabla \times(-\Delta)^{-1}(\omega_{\theta}e_{\theta})$$
It seems true to me since the $\theta$ term on the RHS is related to $r$ and $z$ term, just like in cartesian coordinates but I have no idea how it's derived.
 A: In an axisymmetric flow without swirl, $u^\theta = 0$. Thus from your formulas, $\omega^r = 0 = \omega^z$, and hence $\omega = \omega^\theta e_\theta$. Now apply the Biot-Savart law Wikipedia link
$$ u = \nabla\times(-\Delta)^{-1} \omega.$$
This comes from the fact that $\omega = \nabla\times u$ and the double curl identity
$$ \nabla\times(\nabla\times u) = \nabla(\nabla\cdot u) - \Delta u,$$
i.e. for $u$ incompressible, $\nabla\times \omega = -\Delta u$. 
For an axisymmetric flow with swirl, (under some unstated assumptions) there's only one solution $u$ to the Poisson equation $\nabla\times (\omega^\theta e_\theta ) = -\Delta u$, which is the axisymmetric flow without swirl $u = u^r e_r + u^z e_z$.  
In a more roundabout way: you're asking equivalently (by linearity and Biot-Savart) if
$$ \nabla\times (u^\theta e_\theta) = \omega^r e_r + \omega^ze_z =-\partial_z u^\theta e_r + \frac1r\partial_r(ru^\theta)e_z$$
Reminding (Wikipedia link) ourselves of the curl formula in cyllindrical coordinates,
$$\nabla \times A = \begin{array}{r}{\left(\frac{1}{r} \frac{\partial A_{z}}{\partial \theta}-\frac{\partial A_{\theta}}{\partial z}\right) e_r} \\ {+\left(\frac{\partial A_{r}}{\partial z}-\frac{\partial A_{z}}{\partial r}\right) e_\theta} \\ {+\frac{1}{r}\left(\frac{\partial\left(r A_{\theta}\right)}{\partial r}-\frac{\partial A_{r}}{\partial \theta}\right) e_z}\end{array} $$ this is indeed true.
