Suppose a non-zero a surface charge density $\sigma$ to be present on the interface between two different dielectrics materials (indicated by the subscripts 1 and 2).
The boundary condition $D_{2_{\perp}}-D_{1_{\perp}}=\sigma$ for the electric displacement field (let's suppose $D=\varepsilon E$ with $\varepsilon \in \mathbb{R}$) says that for an incident electromagnetic wave on the surface:
$$ E_{\perp}e^{i(\mathbf{k}\cdot \mathbf{x}-\omega t)}+ E_{\perp}''e^{i(\mathbf{k}''\cdot \mathbf{x}-\omega'' t +\varphi')}=\frac{\sigma}{\varepsilon_2} +\frac{\varepsilon_1}{\varepsilon_2}E_{\perp}'e^{i(\mathbf{k}'\cdot \mathbf{x}-\omega' t+\varphi'')} $$
where $E$ is the incident electric field, $E''$ the reflected and $E'$ the transmitted.
I can write $\sigma=\sigma e^{0}$. Because the exponential are linearly indipendent the following equations must be true:
$$ (\mathbf{k}\cdot \mathbf{x})_{z=0}-\omega t=(\mathbf{k}'\cdot \mathbf{x})_{z=0}-\omega' t+\varphi'=(\mathbf{k}''\cdot \mathbf{x})_{z=0}-\omega'' t+\varphi''=0 $$ where the condition $z=0$ means that the phase is evaluated on the incident plane.
Since this conditions must be true for every position $(x,y)$ and for every time $t$, follows that:
$$0=\varphi'=\varphi''$$ $$\omega=\omega'=\omega''=0$$ $$\mathbf{k}=\mathbf{k}'=\mathbf{k}''=\mathbf{0}$$
i.e. after the wave has encountered the interface, it has become a constant and no longer a wave. This result is obviously unphysical, so there is a problem. Can someone please show me how to get out of this problem?
Answer
You have gotten yourself in trouble by inserting a constant surface charge density into a field that was not designed to describe it. There is nothing that forces the field to look like that combination of incident-reflected-transmitted plane waves; rather, it is one possible solution of the Maxwell equations which is interesting and which is useful to model certain situations - namely, the transmission of a plane wave at the interface between two linear dielectrics. If you change the situation, then there is no guarantee the fields will remain the same.
In your case, you do obtain a contradiction, but the lesson from it is "you have chosen the wrong Ansatz to describe your field". Luckily, it's pretty easy to describe the fields, using the superposition principle: if you want to describe the field caused by an incoming plane wave plus a static surface charge density, simply solve separately for both and then add the solutions. This will add constant fields pointing away from the boundary, which will solve all your concerns:
The Gauss' Theorem applied to the charged plane grants that:
$$ D_{\sigma_1}+D_{\sigma_2}=\sigma \tag{1}$$
Where $D_{\sigma_i}$ is the electric displacement field in the $i$th dielectric generated by the charged plane itself. Using the boundary condition $D_{2_{\perp}}-D_{1_{\perp}}=\sigma$ for the total field (static + wave) you get:
$$ D_{\perp}e^{i(\mathbf{k}\cdot \mathbf{x}-\omega t)}+ D_{\perp}''e^{i(\mathbf{k}''\cdot \mathbf{x}-\omega'' t +\varphi')}+D_{\sigma_2}=\sigma -D_{\sigma_1}+D_{\perp}'e^{i(\mathbf{k}'\cdot \mathbf{x}-\omega' t+\varphi'')} \tag{2}$$ where $D_{\sigma_1}$ is subtracted because is pointing in the opposite direction of $D_{\sigma_2}$.
Putting $(1)$ in $(2)$ you get:
$$ D_{\perp}e^{i(\mathbf{k}\cdot \mathbf{x}-\omega t)}+ D_{\perp}''e^{i(\mathbf{k}''\cdot \mathbf{x}-\omega'' t +\varphi')}=D_{\perp}'e^{i(\mathbf{k}'\cdot \mathbf{x}-\omega' t+\varphi'')}$$
from that follow the know phase relationships:
$$0=\varphi'=\varphi''$$ $$\omega=\omega'=\omega''$$ $$\mathbf{k}=\mathbf{k}'=\mathbf{k}''$$
...unless, that is, the incoming plane wave were to mess with the surface charge and make it non-constant, but you haven't specified enough of the physics to tell what would happen. In the simplest case, you would simply get an oscillatory component to the surface charge, and it then follows from the linear dependence argument (as well as from the physical consideration that the dynamics of the charge are driven by the plane wave) that it will oscillate with the same exponential as the fields. In fact, if you look carefully, this is already happening in the standard configuration, if you rephrase the Gauss law in terms of the $\mathbf E$ fields instead of $\mathbf D$.
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