= Solution
Curling Navier-Stokes and using incompressibility gives $\omega_t+(u\cdot\nabla)\omega=(\omega\cdot\nabla)u+\nu\nabla^2\omega$. For pure swirl, $\omega=r^{-1}(rv)_r$, so regularity gives $v(r,t)=r^{-1}\int_0^rs\omega(s,t)ds$. Adding strain $(-\alpha r,v,2\alpha z)$ yields
$$
\omega_t-\alpha r\omega_r=2\alpha\omega+\nu(\omega_{rr}+r^{-1}\omega_r).
$$
Multiplication by $2\pi r$ and integration proves $\dot\Gamma=0$. The decaying steady <Burgers vortex> is
$$
\boxed{\omega_s(r)=\frac{\Gamma\alpha}{2\pi\nu}e^{-\alpha r^2/(2\nu)}.}
$$
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