Skin depth of current density in magnetic conductor at boundary between two different materials Imagine a magnetic conductor with a cylindrical cross section, surrounded by a coil with a time varying current of
$$I = I_0\cdot \cos (2\pi f t)$$
The conductor is split into two parts, the first with a conductivity and a relative permeability of $\kappa, \mu$, the second with $4\kappa, \mu$. There is a magnetic field $B$ through the conductor, which is caused by the current and therefore time varying as well:
$$B = B_0\cdot \cos (2\pi f t)$$

The change of this magnetic field induces a voltage inside the material and causes a current density $J$. This current density has the value $J_1$ on the surface of the left conductor and $J_2$ on the right side. 
The skin depth $\delta$ is defined by the distance from the surface where $J = 0.37 \cdot J_1$, respectively $J = 0.37 \cdot J_2$ , with $0.37 = 1/e$ and also:
$$\delta = \frac{1}{\sqrt{\pi f\kappa\mu}} = \frac{\sqrt{2j}}{\alpha}$$
where a $\alpha$ is the propagation constant. I found out by simulation, that at the boundary between both materials, the blue one, and the orange one, applies:
$$\frac{1}{\delta_{12}} = \frac{1}{2}(\frac{1}{\delta_{1}}+\frac{1}{\delta_{2}})$$
and therefore
$$\alpha_{12} = \frac{1}{2}(\alpha_1 + \alpha_2)$$
But I'm really struggling to prove that. Can someone give me some hints, how I could get these relations analytically?

Here another plot:
The upper one shows the current density at the surface. The second one shows the contour line where the current density decreased about $63\% = skin depth$. At $z=0$ is the boundary between both materials. Though the current density is a step function, the skin depth is continuous and has the value $\delta_{12}=\frac{2}{\frac{1}{\delta_{1}}+\frac{1}{\delta_{2}}}$ at $z=0$.

 A: Well, I hope I didn't go too much off track. I'm open for discussion.(possible solution path is on the bottom)
Looking at the formula for $\delta$ I see that it is actually related to the speed of EM waves in the medium. The speed of propagation for a medium with the properties $\kappa_1$ and $\mu_1$ is:
 $$c_1=\frac{1}{\sqrt{\kappa_1\mu_1}}$$
Therefore we can rewrite it as:
$$\delta=\frac{1}{c\sqrt{\pi f}}$$
Looking at it, I would like to have the $\omega$ in it rather than $f$:
$$\delta=\frac{\sqrt{2}}{c\sqrt{\omega}}$$
Lets square it:
$$\delta^2=\frac{2}{c^2\omega}$$
What stays same, I guess its $\omega$:
$$\omega=\frac{2}{c^2\delta^2} $$
The frequency is the same everywhere, so it is:
$$\frac{1}{c_1^2\delta_1^2}=\frac{1}{c_2^2\delta_2^2}  $$
or:
$$\frac{c_2}{c_1}=\frac{\delta_1}{\delta_2}  $$
So now we actually reduced it to a problem of finding the speed of propagation of EM-waves at the interface. That will not help directly but it gives me another idea.
I would now suggest to look at it similar to a semiconductor PN-junction problem, and calculate the change of $\kappa$ on the interface due to the differences in charged particle density. The drift currents will do the charge compensation on the interface.
So the diffusion current is:
$$I_{diff}=qD\frac{dn}{dx}$$
Using this formula you get some current, this affects the local $\kappa$ and creates a $\kappa'$. Using this value you should get the $\delta$ on the boundary(and even the functional dependence on the distance from the interface).
