Consider electromagnetic cylindrical waves. Cylindrical waves can be derived from the plane waves using energy conservation consideration: since the power must be a constant the amplitude of a cylindrical wave must decrease with $\sqrt{r}$. Therefore a cylindrical wave expression must be
$$\mathbf{E}(r,t)=\frac{\mathbf{E}_0}{\sqrt{r}} \mathrm{sin}(kr-\omega t)$$
The function $\sqrt{r} \mathbf{E}(r,t)$ satisfies one dimensional wave equation
$$\frac{\partial^2\xi}{\partial r^2}-\frac{1}{c^2}\frac{\partial^2\xi}{\partial t^2}=0$$
In complex notation the cylindrical wave becomes $$\mathbf{E}(r,t)=\frac{\mathbf{E}_0}{\sqrt{r}} e^{j(kr-\omega t)}\tag{1}$$
If we call $\xi$ a generic component of $\mathbf{E}$, the three dimensional wave equation is
$$\nabla^2\xi-\frac{1}{c^2}\frac{\partial^2\xi}{\partial t^2}=\square \xi=0$$
The solution in cylindrical coordinates is
$$\xi (r,\phi , z,t) =\sum_{\omega,n,h} R^{0}_{\omega, n, h } H_n\Bigg(r \sqrt{\frac{\omega^2}{c^2}-h^2}\Bigg) e^{j(n\phi +hz-\omega t)} \tag{2}$$
Where $R^{0}_{\omega, n, h }$ is a (complex) constant and $H_n$ is the Hankel function of order $n$.
Under the assumption of cylindrical symmetry of the wave, that is
$$\frac{\partial \xi}{\partial \phi}=0 \,\,\,\,\,\, \mathrm{and} \,\,\,\,\,\, \frac{\partial \xi}{\partial z}=0$$ the asymptotic approximation of $(2)$ (for $r >> \frac{c}{\omega}$) lead to a field that is the same as $(1)$.
My question is: why (under cylindrical symmetry) is $(2)$ equal to $(1)$ only at large distances?
I always thought that $(1)$ gives the expression of a cylindrical wave in all the circumstances. So is $(1)$ "wrong" for small $r$? Or are $(1)$ and $(2)$ describing two different things? If so, what are the differences?
(I have an identical doubt for spherical waves).