= Solution
The <divergence> of this axisymmetric <velocity field> is $r^{-1}\partial_r(-\alpha r^2)+\partial_z(2\alpha z)=-2\alpha+2\alpha=0$. Thus it satisfies <incompressibility>. The apparent axis singularity in the azimuthal component is removable, since $u_\theta\sim\gamma\beta r$ as $r\to0$.
Only the axial <vorticity> component is nonzero:
$$
\boxed{\boldsymbol\omega=\nabla\times\mathbf u=(0,0,w(r)),\qquad w(r)=2\beta\gamma e^{-\beta r^2}.}
$$
The <cross product> is $\mathbf u\times\boldsymbol\omega=(u_\theta w,\alpha r w,0)$. Its components are independent of $z$ and $\theta$, so its <curl> also has only an axial component:
$$
\bigl[\nabla\times(\mathbf u\times\boldsymbol\omega)\bigr]_z=\frac1r\frac d{dr}(\alpha r^2w)=\alpha(2w+rw')=2\alpha(1-\beta r^2)w.
$$
Meanwhile $w'=-2\beta rw$ and $w''=(-2\beta+4\beta^2r^2)w$, giving
$$
\nabla^2\boldsymbol\omega=(0,0,w''+w'/r)=(0,0,-4\beta(1-\beta r^2)w).
$$
Thus the required steady <vorticity equation> holds with \b[$\nu=\alpha/(2\beta)$]. This is the balance between vortex stretching and viscous diffusion in a <Burgers vortex>.
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