Skin effect fully explained by plane wave attenuation?

I went over the explanations of the skin effect in multiple sources. However, I still don't understand how the fact that this equation:

$$(\Delta - \mu_0\sigma\partial_t - \frac{1}{c^2}\partial_t ^2)\vec{E} = 0$$

has solutions in the form of plane waves with complex wave vector, i.e. with amplitude attenuated in the direction of propagation ($\hat{z}$),

$$\vec{E} = E_0 e^{-\kappa z}\cos{(kz-\omega t)}\hat{x}, \quad \kappa = \frac{1}{\delta}$$

explains the AC density distribution.

Can you please help me understand how the AC distribution in a cylindrical wire is related to this fact?

$$......$$

Comment

I like Olin Lathrop's answer here (1st and 2nd paragraph),

When you first apply a potential to a wire, the electric field is a thin shell only on the outside of the wire. This causes current to flow in a thin layer along the outer surface of the wire. Since the conducting material of the wire has some non-zero resistance, this current in the outer shell causes a voltage drop, which means a layer a little further into the wire will "see" the electric field, which causes current deeper into the wire, which causes a electric field deeper into the wire, etc. In steady state, the electric field inside the wire is uniform along the crossection of the wire.

The fact that it takes time before most of the material in a wire is conducting current means that the effective resistance of the wire starts out high and exponentially converges to its DC (steady state) resistance over time. Taking this further, it also means that the wire looks like a different resistance to different frequencies.

but I don't trust the statements:

1)

When you first apply a potential to a wire, the electric field is a thin shell only on the outside of the wire.

Is this a consequence of the condcutor "trying to kill the field inside"? Why are the inner layers "shielded"? Doesn't the force due to hopping of electrons at the electrodes affect the core as well?

2)

Since the conducting material of the wire has some non-zero resistance, this current in the outer shell causes a voltage drop, which means a layer a little further into the wire will "see" the electric field...

How does that work? According to Ohm's law $\vec{j} = \sigma \vec{E}$, the direction of the current is locally the same as of the voltage drop.