## 5 Scattering by the Dielectric Cylinder  
The purpose of this section is to find the resolvent for the dielectric cylinder of the circular cross section. To this end, we will solve the Helmholtz equation:  
\[\left(\frac{\partial^{2}}{\partial\rho^{2}}+\frac{1}{\rho}\frac{\partial}{ \partial\rho}+\frac{1}{\rho^{2}}\frac{\partial^{2}}{\partial\rho^{2}}\right) U_{k}+k^{2}\varepsilon(\rho)U_{k}=f(\rho) \tag{48}\]  
with permittivity \(\varepsilon(\rho)=\varepsilon\) inside the layer and \(\varepsilon(\rho)=1\) outside. The right-hand side in (48) is associated with the external force arbitrarily distributed inside the cylinder. The unknown function satisfies the boundary conditions for the longitudinal component of the electric field at the boundary of the cylinder [19; 47] and radiation condition outside. The scattered field may be represented similarly to Equation (42) by  
\[ \begin{array}{c}\widetilde{U}_{k}(\rho)=-\frac{i\
u}{4}\int\limits_{S}H_{0} ^{(1)}\big{(}k\sqrt{\varepsilon}|\rho-\rho^{\prime}|\big{)}\,\widetilde{U}_{ k0}(\rho^{\prime})d\rho^{\prime}+\\ \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\
u \sum\limits_{n}A_{n}J_{n}\big{(}k\sqrt{\varepsilon}\rho\big{)}e^{in\varphi}, 0\leq\rho\leq a\end{array}  \tag{49}\]  
and  
\[\widetilde{U}_{k}(\rho)=\
u\sum\limits_{n}B_{n}H_{n}^{(1)}(k\rho)e^{in\varphi},\rho\geq a, \tag{50}\]  
where \(A_{n},B_{n}\) are unknown coefficients which can be founded from boundary conditions. We will omit their calculations here (see Appendix D). The final result for the field inside the cylinder is  
\[\widetilde{U}_{k}(\rho)=\
u\int\limits_{S}\Gamma\big{(}\rho,\rho^{\prime};k \big{)}U_{k0}\big{(}\rho^{\prime}\big{)}d\rho^{\prime}, \tag{51}\]  
where  
\[\Gamma\big{(}\rho,\rho^{\prime};k\big{)}=-\frac{i}{4}H_{0}^{(1)}\big{(}k\sqrt {\varepsilon}|\rho-\rho^{\prime}|\big{)}+\delta g\big{(}\rho,\rho^{\prime};k \big{)}, \tag{52}\]  
\[\delta g\big{(}\rho,\rho^{\prime};k\big{)}=\frac{i}{4}\sum\limits_{n}W_{n}(k )J_{n}\big{(}k\sqrt{\varepsilon}\rho\big{)}J_{n}\big{(}k\sqrt{\varepsilon} \rho^{\prime}\big{)}e^{in(\varphi-\varphi^{\prime})}, \tag{53}\]  
And  
\[W_{n}(k)=\frac{H_{n}^{(1)}(ka)H_{n}^{(1)}\big{(}k\sqrt{\varepsilon}a\big{)}- \sqrt{\varepsilon}H_{n}^{(1)}\big{(}ka)H_{n}^{(1)}{}^{\prime}\big{(}k\sqrt{ \varepsilon}a\big{)}}{H_{n}^{(1)}{}^{\prime}\big{(}ka\big{)}J_{n}\big{(}k \sqrt{\varepsilon}a\big{)}-\sqrt{\varepsilon}H_{n}^{(1)}\big{(}ka\big{)}J^{ \prime}{}_{n}\big{(}k\sqrt{\varepsilon}a\big{)}}. \tag{54}\]  
Relation (51) expresses the scattered field through the incident field inside the cylinder in the canonical integral form. Therefore, Relation (52) may be associated with the resolvent kernel of the integral operator and used for the analysis of commutation Relation (18) and the existence of noise. Let us transform Relation (18) for the dielectric cylinder to the form, and similarly transform Equation (47) for the 1D dielectric layer (integration over the whole axis instead of the positive semi-axis with the subsequent integration in the complex plane). It is necessary to note that here the situation dramatically changes compared with the 1D case. The difficulty lies in the non-analyticity of the function \(W_{n}(k)\) given by Relation (54) in the complex \(k\)-plane (this function is multi-valued with a branch point at \(k=0\) in contrast to the dielectric layer case). One can see from well-known identities \(J_{-n}(x)=(-1)^{n}J_{n}(x)\), \(H_{n}^{(1)}\big{(}e^{j\pi}x\big{)}=-H_{-n}^{(2)}(x)\), and \(H_{n}^{(2)}\big{(}e^{-j\pi}x\big{)}=-H_{-n}^{(1)}(x)\) that it is natural to place the cut in the complex \(k\)-plane along the negative semi-axis (gagged line in Figure 4). It is easy to show that \(W_{-n}(x)=-W_{n}^{*}\big{(}e^{i\pi}x\big{)}\) and \(W_{n}(k)\) goes to zero if \(|k|\to\infty\) in the upper half-plane.