## 3 Boundary Conditions and Modal Coefficients  
On the mapping plane, we can represent the boundary conditions as follows  
\[\tau_{\eta_{2}}^{(t)}\Big{|}_{\zeta_{0}}=0,\,B_{\eta}^{(t)}\Big{|}_{\zeta_{0}}=\,B _{\eta}^{(I)}\Big{|}_{\zeta_{0}},\,\phi^{(t)}\Big{|}_{\zeta_{0}}=\,\phi^{(I)} \Big{|}_{\zeta_{0}}. \tag{16}\]  
By utilizing Equation (5) and substituting the third equation of Equation (10) into the second equation of Equation (16), we can obtain  
\[h_{15}\Bigg{(}\frac{\partial f^{(t)}}{\partial\zeta}\mathrm{e}^{\mathrm{i} \theta}+\frac{\partial f^{(t)}}{\partial\overline{\zeta}}\mathrm{e}^{-\mathrm{ i}\theta}\Bigg{)}=\mu_{0}\Bigg{(}\frac{\partial\phi^{(I)}}{\partial\overline{ \zeta}}\mathrm{e}^{\mathrm{i}\theta}+\frac{\partial\phi^{(I)}}{\partial \overline{\zeta}}\mathrm{e}^{-\mathrm{i}\theta}\Bigg{)}. \tag{17}\]  
where \(\mu_{0}\) is the vacuum permeability. By utilizing the second formula of Equations (14) and (15), we can express Equation (17) as follows  
\[-\mu_{11}\sum_{n=1}^{\infty}n\Big{(}B_{n}\zeta^{-n-1}\mathrm{e}^{\mathrm{i} \theta}+C_{n}\overline{\zeta}^{-n-1}\mathrm{e}^{-\mathrm{i}\theta}\Big{)}=\mu _{0}\sum_{n=1}^{\infty}n\Big{(}D_{n}\zeta^{n-1}\mathrm{e}^{\mathrm{i}\theta}+E _{n}\overline{\zeta}^{n-1}\mathrm{e}^{-\mathrm{i}\theta}\Big{)}. \tag{18}\]  
Assuming that the mapping plane of the opening is the unit circle, we can express Equations \(\zeta=\eta\mathrm{e}^{\mathrm{i}\theta}\) and \(\eta=1\), as well as Equation (18), as follows  
\[\sum_{n=1}^{\infty}n\Bigg{[}\bigg{(}\frac{\mu_{11}}{\mu_{0}}B_{n}+D_{n}\bigg{)} \mathrm{e}^{-\mathrm{i}n\theta}+\bigg{(}\frac{\mu_{11}}{\mu_{0}}C_{n}+E_{n} \bigg{)}\mathrm{e}^{\mathrm{i}n\theta}\Bigg{]}=0. \tag{19}\]  
As \(\mathrm{e}^{-\mathrm{i}n\theta}\) and \(\mathrm{e}^{\mathrm{i}n\theta}\) are both non-zero and independent of each other, we can conclude that  
\[\left\{ \begin{array}{l}D_{n}=-\frac{\mu_{11}}{\mu_{0}}B_{n}\\ E_{n}=-\frac{\mu_{11}}{\mu_{0}}C_{n}\end{array} ,n\geq 1.\right. \tag{20}\]  
By substituting the third equation of Equations (14), (15), and (20) into the third equation of Equation (16), we obtain  
\[ \begin{array}{l}0=\sum_{n=-\infty}^{\infty}\Big{[}\mathrm{i}^{n}J_{n}(k|h( \zeta)|)\mathrm{e}^{-\mathrm{i}n\varphi}+A_{n}H_{n}^{(1)}(k|h(\zeta)|)\Big{]} \bigg{\{}\frac{\hbar(\zeta)}{|\hbar(\zeta)|}\bigg{\}}^{n}\\ \quad+\Big{(}1+\frac{\mu_{11}}{\mu_{0}}\Big{)}\sum_{n=1}^{\infty}\big{(}B_{n} \mathrm{e}^{-\mathrm{i}n\theta}+C_{n}\mathrm{e}^{\mathrm{i}n\theta}\big{)}-D_ {0}.\end{array}  \tag{21}\]  
Substituting the first equation of Equation (10) into the first equation of Equation (16), we obtain  
\[\frac{c^{*}}{\big{|}h^{\prime}(\zeta)\big{|}}\Bigg{(}\frac{\partial w^{(t)}}{ \partial\zeta}\mathrm{e}^{\mathrm{i}\theta}+\frac{\partial w^{(t)}}{\partial \overline{\zeta}}\mathrm{e}^{-\mathrm{i}\theta}\Bigg{)}+\frac{h_{15}^{2}/\mu_ {11}}{\big{|}h^{\prime}(\zeta)\big{|}}\Bigg{(}\frac{\partial f^{(t)}}{ \partial\overline{\zeta}}\mathrm{e}^{\mathrm{i}\theta}+\frac{\partial f^{(t)} }{\partial\overline{\zeta}}\mathrm{e}^{-\mathrm{i}\theta}\Bigg{)}=0. \tag{22}\]  
We can obtain the following relation  
\[ \begin{array}{l}\frac{\partial}{\partial\overline{\zeta}}\bigg{[}H_{n}(k|h( \zeta)|)\Big{\{}\frac{\hbar(\zeta)}{|\hbar(\zeta)|}\Big{\}}^{n}\bigg{]}=\frac{ k}{2}H_{n-1}(k|h(\zeta)|)\Big{\{}\frac{\hbar(\zeta)}{|\hbar(\zeta)|}\Big{\}}^{n-1}h^{ \prime}(\zeta),\\ \frac{\partial}{\partial\overline{\zeta}}\bigg{[}H_{n}(k|h(\zeta)|)\Big{\{} \frac{\hbar(\zeta)}{|\hbar(\zeta)|}\Big{\}}^{n}\bigg{]}=-\frac{k}{2}H_{n+1}(k|h( \zeta)|)\Big{\{}\frac{\hbar(\zeta)}{|\hbar(\zeta)|}\Big{\}}^{n+1}\overline{h^{ \prime}(\zeta)}.\end{array}  \tag{23}\]By utilizing Equation (23), Equation (22) can be expressed as follows  
\[