I have the following incompressible axisymmetric velocity field with a general swirl:
$u(x,t) = u_r(r,z,t)e_r + u_{\theta}(r, z, t)e_{\theta} + u_z(r,z,t)e_z$.
The corresponding vorticity is given by $\nabla \times u = \omega$, such that we can write for each component of the vorticity in cylindrical coordinates:
$\omega_r = \frac{1}{r} \frac{\partial u_z}{\partial \theta} - \frac{\partial u_\theta}{\partial z} = -(\frac{\partial u_\theta}{\partial z})e_r$
$\omega_{\theta} = (\frac{\partial u_r} {\partial z} - \frac{\partial u_z}{\partial r})e_{\theta}$
$\omega_z = \frac{1}{r} \frac{\partial}{\partial r}(r u_{\theta}) -\frac{1}{r} \frac{\partial u_r}{\partial_{\theta}} = \frac{1}{r} \frac{\partial}{\partial r}(r u_{\theta}) =\frac{1}{r}(\frac{u_{\theta}}{r} + \frac{\partial u_{\theta}}{\partial r})$, where all partial derivatives with respect to $\theta$ vanish (since its axisymmetric flow).
Also, we have the Euler equations and vorticity equation given by:
$\frac{\partial u}{\partial t} + (\nabla \cdot u)u = -\nabla p, \nabla\cdot {u} = 0,u_{|_{t=0}} = u_0 $,
as well as
$\frac{\partial \omega}{\partial t} + \omega \cdot \nabla u = u \cdot \nabla \omega$.
Now, we find that for axissymetric flows, the $\theta$-component of the Euler equation and vorticity equation in cylindrical coordinates are given by:
$\frac{{D}}{Dt} u_{\theta} = -\frac{u_r u_{\theta}}{r}$
$\frac{{D}}{Dt} \omega_{\theta} = \frac{1}{r}(\omega_{\theta}u_r - u_{\theta}\omega_r) +\omega_r\frac{\partial u_{\theta}}{\partial r} + \omega_z \frac{\partial u_{\theta}}{\partial z}$,
where $\frac{D}{Dt} = \frac{\partial}{\partial t} + u_r\frac{\partial}{\partial r} + u_z\frac{\partial}{\partial z}$..
I am mostly interested in the $\theta$- component of the vorticity equation, when dividing both sides by r, i.e, which yields:
$\frac{ D}{Dt}(\frac{\omega_{\theta}}{r}) = \frac{1}{r^2}(u_r \omega_{\theta} - \omega_r u_{\theta}) + \frac{1}{r} \omega_r \frac{\partial u_{\theta}}{\partial r} + \frac{1}{r}\omega_z \frac{\partial u_{\theta}}{\partial z} + u_r \frac{\omega_{\theta}}{r^2}$
Then when we plug in $\omega_r, \omega_{\theta}, \omega_z$, this gives:
$-\frac{1}{r} (\frac{\partial u_{\theta}}{\partial z})e_r (\frac{\partial u_{\theta}}{\partial r})e_r + \frac{1}{r}(\frac{1}{r} u_{\theta} +\frac{\partial u_{\theta}}{\partial r})e_z (\frac{\partial u_{\theta}}{\partial z})e_z + \frac{1}{r^2} (u_r(\frac{\partial u_r}{\partial uz} - \frac{\partial u_z}{\partial r})e_{\theta} + u_{\theta} \frac{\partial u_{\theta}}{\partial z} e_r) + \frac{u_r}{r^2}((\frac{\partial u_r}{\partial uz} - \frac{\partial u_z}{\partial r})e_{\theta}$.
Apparently, after cancelling some terms and collecting terms with the same derivative, this should give us the PDE: $\frac{D}{Dt}(\frac{\omega_{\theta}}{r}) = -\frac{1}{r^4}\frac{\partial}{\partial z} (ru_{\theta})^2$
I have tried to arrive at this equation, but all of my attempts failed. Could maybe someone help me?
Maybe its also good to note that the unit vectors in cylindrical coordinates are given by:
$e_r = \cos \theta \hat{x}+\sin \theta \hat{y}$
$e_{\theta} = \hat{z} \times \hat{r} = -\sin \theta \hat{x} + \cos \theta \hat{y}$
$e_z = \hat{z}$