## 4 Planar Dielectric Layer: Reflection-Transmission  
The problem's geometry is shown in Figure 3a. The system is excited by the external current \(f(x)\) distributed inside the dielectric layer along the \(x\)-axis and vanishing outside the layer. Therefore, the field is dependent only on one variable, \(x\). The equation for the total field becomes  
\[\frac{d^{2}U_{k}}{dx^{2}}+k^{2}\varepsilon(x)U_{k}=f(x), \tag{40}\]  
where the permittivity \(\varepsilon(x)=\varepsilon\) inside the layer and \(\varepsilon(x)=1\) outside it. The boundary conditions for the unknown field stem from the boundary conditions for the tangential component of the electric field at the boundaries of the dielectric layer [19; 46]. The problem should be supplemented by the radiation condition at \(x<0\) and \(x>l\). The total field may be represented as a sum of incident and reflected-transmitted fields: \(U_{k}=U_{k0}+\bar{U}_{k}\). The incident field \(U_{k0}\) is produced by the same source in the  
Figure 3: Examples of the dielectric structures for resolvent analysis: (**a**) dielectric layer; (**b**) dielectric cylinder of circular cross section; the sources (colored grey) placed inside the dielectric body. absence of the dielectric layer. Therefore, it satisfies equation \(\frac{\partial^{2}U_{00}}{dx^{2}}+k^{2}U_{R0}=f(x)\). For the reflected-transmitted field inside the layer, we have equation \(\frac{\partial^{2}U_{R}}{dx^{2}}+k^{2}\epsilon\widetilde{U}_{k}=vU_{k0}\), where \(v=-k^{2}(\epsilon-1)\), and \(\frac{\partial^{2}\widetilde{U}_{k}}{dx^{2}}+k^{2}\widetilde{U}_{k}=0\) outside. Our goal is to calculate the resolvent of the integral equation  
\[\widetilde{U}_{k}(x)=w_{k0}(x)+v\int\limits_{0}^{l}g\big{(}x,x^{\prime}\big{)} \widetilde{U}_{k}\big{(}x^{\prime}\big{)}dx^{\prime}, \tag{41}\]  
where \(w_{k0}(x)=v\int_{0}^{l}g(x,x^{\prime})U_{k0}(x^{\prime})dx^{\prime}\). It is an analog of the general integral Equation (A1) for the 1D case (where \(g(x,x^{\prime})=e^{ik|x-x^{\prime}|}/2 ik\)). We can represent the unknown field as  
\[\widetilde{U}_{k}(x)=\left\{ \begin{array}{c}\frac{v}{2 ik\sqrt{ \epsilon}}\int\limits_{0}^{l}e^{ik\sqrt{\epsilon}|x-x^{\prime}|}U_{k0}(x^{ \prime})dx^{\prime}+Ae^{ik\sqrt{\epsilon}x}+Be^{-ik\sqrt{\epsilon}x},0\leq x \leq l,\\ \qquad\qquad Ce^{ikx},x>l,\\ \qquad\qquad De^{-ikx},x<0,\end{array} \right. \tag{42}\]  
where \(A\), \(B\), \(C\), and \(D\) are unknown constant coefficients. Using boundary conditions, we obtain a system of four linear equations  
\[ \begin{array}{l}A+B-D=-\frac{1}{\sqrt{\epsilon}}f_{+},\\ \sqrt{\epsilon}(A-B)+D=f_{+},\\ Ae^{i\Phi_{r}}+Be^{-i\Phi_{r}}-Ce^{i\Phi}=-\frac{e^{i\Phi_{r}}}{\sqrt{ \epsilon}}f_{-},\\ \sqrt{\epsilon}\big{(}Ae^{i\Phi_{r}}-Be^{-i\Phi_{r}}\big{)}-Ce^{i\Phi}=-e^{i \Phi_{r}}f_{-},\end{array}  \tag{43}\]  
where \(\Phi=kl,\Phi_{z}=k\sqrt{\epsilon}l\), \(f_{\pm}=\frac{v}{2 ik}\int_{0}^{l}e^{\pm ik\sqrt{\epsilon}x^{\prime}}U_{k0}(x^ {\prime})dx^{\prime}\). System (43) is easily solvable. We need only two of unknowns, for which we have  
\[ \begin{array}{l}A=-\frac{\xi}{\sqrt{\epsilon}}\big{(}f_{+}e^{-2i\Phi_{r}}- rf_{-}\big{)},\\ B=\frac{\xi}{\sqrt{\epsilon}}(rf_{+}-f_{-}),\end{array}  \tag{44}\]  
where \(r=\big{(}1-\sqrt{\epsilon}\big{)}\cdot\big{(}1+\sqrt{\epsilon}\big{)}^{-1}\) is the Fresnel coefficient for normal incidence and \(\xi=re^{2i\Phi_{r}}\big{(}1-r^{2}e^{2i\Phi_{r}}\big{)}^{-1}\). Using Solution (44) together with Relation (42), we obtain the solution inside the layer in the canonical form  
\[\widetilde{U}_{k}(x)=v\int\limits_{0}^{l}\Gamma\big{(}x,x^{\prime};k\big{)}U_{ k0}\big{(}x^{\prime}\big{)}dx^{\prime}, \tag{45}\]  
where the function  
\[ \begin{array}{l}\Gamma(x,x^{\prime};k)=\frac{1}{2 ik\sqrt{\epsilon}}e^{ik \sqrt{\epsilon}|x-x^{\prime}|}+\\ \frac{1}{2 ik\sqrt{\epsilon}\big{(}1-r^{2}e^{2 ik\sqrt{\epsilon}l}\big{)}} \Big{(}-re^{ik\sqrt{\epsilon}(x+x^{\prime})}+r^{2}e^{ik\sqrt{\epsilon}(x-x^{ \prime}+2l)}+r^{2}e^{-ik\sqrt{\epsilon}(x-x^{\prime}-2l)}-re^{-ik\sqrt{ \epsilon}(x+x^{\prime}-2l)}\Big{)}\end{array}