we find $$\begin{aligned}
\label{c11}
&& \frac{d}{d t}\left(\frac{\delta\underline{L}}{\delta
\left(\underline{\dot{A}}^*_i(\vec{k},t)\right)}\right)-\frac{\delta\underline{L}}{\delta\left(\underline{A}^*_i(\vec{k},t)\right)}=0
\hspace{2.00 cm} i=1,2,3\nonumber\\
\Longrightarrow&&\mu_0\varepsilon_0\underline{\ddot{\vec{A}}}(\vec{k},t)+\mu_0\varepsilon_0
\ \imath\vec{k}\ \underline{\dot{\varphi}}(\vec{k},t)-\vec{k}\times\left(\vec{k}\times\underline{\vec{A}}(\vec{k},t)\right)=\nonumber\\
&&\mu_0\ \underline{\dot{\vec{P}}}(\vec{k},t)+\imath\mu_0\
\vec{k}\times\underline{\vec{M}}(\vec{k},t)\end{aligned}$$ $$\begin{aligned}
\label{c12}
&&\frac{d}{d t}\left(\frac{\delta\underline{L}}{\delta
\left(\underline{\dot{\varphi}}^*(\vec{k},t)\right)}\right)-\frac{\delta\underline{L}}{\delta\left(\underline{\varphi}^*(\vec{k},t)\right)}=0\nonumber\\
\Longrightarrow&&-\varepsilon_0\
\imath\vec{k}\cdot\underline{\dot{\vec{A}}}(\vec{k},t)+\varepsilon_0|\vec{k}|^2\underline{\varphi}(\vec{k},t)
=-\imath\vec{k}\cdot\underline{\vec{P}}(\vec{k},t)\end{aligned}$$ for any wave vector $\vec{k}$ in the half space $k_z\geq 0$ where $$\label{c13}
\underline{\vec{P}}(\vec{k},t)=\int_0^\infty d\omega\int d^3p \
\underline{f}( \omega , \vec{k},\vec{p})\cdot\
\underline{X}_\omega(\vec{p},t)$$ $$\label{c13.1}
\underline{\vec{M}}(\vec{k},t)=\int_0^\infty d\omega\int d^3p \
\underline{g}( \omega , \vec{k},\vec{p})\cdot\
\underline{Y}_\omega(\vec{p},t)$$ are respectively the spatial Fourier transforms of the electric and magnetic polarization densities of the medium and it has been used from the relations (\[c6.1\]). Therefore in this method the polarization fields of the medium are naturally concluded in terms of the coupling tensors $f , g$ and the dynamical variables of the $E$ and $M$ fields. Similarly the Euler-Lagrange equations for the fields $ \underline{\vec{X}}_\omega $ and $
\underline{\vec{Y}}_\omega $ for any vector $\vec{k}$ in the half space $k_z\geq 0$ are easily obtained as $$\begin{aligned}
\label{c14}
&& \frac{d}{d t}\left(\frac{\delta\underline{L}}{\delta
\left(\underline{\dot{X}}^*_{\omega
i}(\vec{k},t)\right)}\right)-\frac{\delta\underline{L}}{\delta\left(\underline{X}^*_{\omega
i}(\vec{k},t)\right)}=0
\hspace{2.00 cm} i=1,2,3\nonumber\\
\Longrightarrow &&\underline{\ddot{\vec{X}}}_\omega
(\vec{k},t)+\omega^2\ \underline{\vec{X}}_\omega(\vec{k},t)= -\int
d^3q\  \underline{f}^\dag( \omega , \vec{q}, \vec{k})\cdot
\left(\underline{\dot{\vec{A}}}(\vec{q},t)+\imath\vec{q}\underline{\varphi}(\vec{q},t)\right)\nonumber\\
&&\end{aligned}$$ $$\begin{aligned}
\label{c15}
&& \frac{d}{d t}\left(\frac{\delta\underline{L}}{\delta
\left(\underline{\dot{Y}}^*_{\omega
i}(\vec{k},t)\right)}\right)-\frac{\delta\underline{L}}{\delta\left(\underline{Y}^*_{\omega
i}(\vec{k},t)\right)}=0
\hspace{2.00 cm} i=1,2,3\nonumber\\
\Longrightarrow &&\underline{\ddot{\vec{Y}}}_\omega
(\vec{k},t)+\omega^2\ \underline{\vec{Y}}_\omega(\vec{k},t)=\int
d^3q \ \underline{g}^\dag( \omega , \vec{q}, \vec{k})\cdot
\left(\imath\ \vec{q}\times\underline{\vec{A}}(\vec{q},t)\right)\nonumber\\
&&\end{aligned}$$ where $\underline{f}^\dag and  \underline{g}^\dag$ are the hermitian conjugate 