## 2 Methodology
### Solutions to the Problem  
Inserting Equation (4) into Equation (8), it can be obtained as  
\[C_{1m}^{(2)}=-C_{1m}^{(1)}\frac{H_{m}^{(1)\prime}(k_{0}b)}{H_{m}^{(2)\prime}( k_{0}b)},\quad D_{1m}^{(2)}=-D_{1m}^{(1)}\frac{H_{m}^{(1)\prime}(k_{0}b)}{H_{m}^{(2) \prime}(k_{0}b)} \tag{13}\]  
By using Equation (12) in Equation (4), Equation (4) can be rewritten as  
\[w_{1}^{f}(r_{1},\theta_{1})=\sum_{m=0}^{\infty}G_{1m}(k_{0}r_{1})(C_{1m}^{(1 )}\cos m\theta_{1}+D_{1m}^{(1)}\sin m\theta_{1}) \tag{14}\]  
Similarly, it can be obtained from Equation (11) as  
\[w_{2}^{f}(r_{2},\theta_{2})=\sum_{m=0}^{\infty}G_{2m}(k_{0}r_{2})(C_{2m}^{(1 )}\cos m\theta_{2}+D_{2m}^{(1)}\sin m\theta_{2}) \tag{15}\]  
where  
\[G_{1m}(k_{0}r_{1})=H_{m}^{(1)}(k_{0}r_{1})-\frac{H_{m}^{(1)\prime}(k_{0}b)}{H _{m}^{(2)\prime}(k_{0}b)}H_{m}^{(2)}(k_{0}r_{1}) \tag{16}\]  
\[G_{2m}(k_{0}r_{2})=H_{m}^{(1)}(k_{0}r_{2})-\frac{H_{m}^{(1)\prime}(k_{0}b)}{H _{m}^{(2)\prime}(k_{0}b)}H_{m}^{(2)}(k_{0}r_{2}) \tag{17}\]  
For solving the response of tunnel 1, the incident wave \(w_{2}^{i}(r_{3},\theta_{3})\) should be transformed into the coordinate (\(r_{1}\), \(\theta_{1}\)). Then, the incident waves \(w_{1}^{i}(r_{1},\theta_{1})\) and \(w_{2}^{i}(r_{1},\theta_{1})\) can be expressed as  
\[w_{1}^{i}(r_{1},\theta_{1})=\sum_{m=0}^{\infty}\varepsilon_{m}(-i)^{m}J_{m}( kr_{1})(\cos m\gamma\cos m\theta_{1}+\sin m\gamma\sin m\theta_{1}) \tag{18}\]  
\[w_{2}^{i}(r_{1},\theta_{1})=\exp(-2 ikD\cos\gamma)\sum_{m=0}^{\infty} \varepsilon_{m}{}^{im}J_{m}(kr_{1})(\cos m\gamma\cos m\theta_{1}-\sin m\gamma \sin m\theta_{1}) \tag{19}\]  
Similarly, all the scattered wavefields should also be transformed into the coordinate (\(r_{1}\), \(\theta_{1}\)) as  
\[w_{2}^{r}(r_{1},\theta_{1})=\sum_{m=0}^{\infty}J_{m}(kr_{1})(G_{1}^{*}cosm\theta _{1}+G_{2}^{*}\ sin\ m\theta_{1});\ r_{1}<s \tag{20}\]  
\[w_{3}^{r}(r_{1},\theta_{1})=\sum_{m=0}^{\infty}J_{m}(kr_{1})(A_{m}^{*}cosm\theta _{1}+B_{m}^{*}\ sin\ m\theta_{1});\ r_{1}<2D \tag{21}\]  
\[w_{4}^{r}(r_{1},\theta_{1})=\sum_{m=0}^{\infty}J_{m}(kr_{1})(G_{3}^{*}cosm\theta _{1}+G_{4}^{*}\ sin\ m\theta_{1});\ r_{1}<e \tag{22}\]where e is the distance between point \(o1\) and \(o4\) (or \(o_{2}\) and \(o_{3}\)) shown in Figure 2. Inserting Equations (17)-(21) and \(w_{1}^{\prime}(r_{1},\theta_{1})\) into Equations (6) and (7), it can be obtained as  
\[\frac{H_{m}^{(2)}(ka)-G_{1m}(ka)R_{1m}}{J_{m}(ka)}A_{1,m}\] \[+\sum\limits_{n=0}^{\infty}\left[P_{mn}^{+}(2 kD)A_{1,n}+(F3C_{mn}^ {+}(ks)+F1C_{mn}^{+}(ke,\beta))A_{2,n}+(-F4C_{mn}^{-}(ks)+F2C_{mn}^{-}(ke,\beta) )B_{2,n}\right]\] (23) \[=-e_{m}i^{m}((-1)^  
\[\begin{split}&\sum\limits_{n=0}^{\infty}\left[(F4C_{mn}^{+}(ks)+F2C_{mn}^{+ }(ke,\beta))A_{1,n}+(F3C_{mn}^{-}(ks)-F1C_{mn}^{-}(ke,\beta))B_{1,n}+P_{mn}^{-}(2 kD)B_{2,n}\right]\\ &+\frac{H_{m}^{(2)}(ka)-G_{mn}(ka)R_{2m}}{\int_{m}(ka)}B_{2,m}\\ &=-\varepsilon_{m}m^{m}(-(-1)^{m}\exp(-iks\sin\gamma)+\exp(-2 ikD \cos\gamma-iks\sin\gamma))\sin m\gamma;\;\;m=1,2,3,\ldots\end{split} \tag{37}\]  
Equations (22), (23), (35), and (36) constitute an infinite algebraic system of equations. Though setting the truncated number N, the four sets of unknown coefficients A\({}_{1}\),m, B\({}_{1}\),m, A\({}_{2}\),m, and B\({}_{2}\),m can be obtained by solving Equations (22), (23), (35), and (36) together. The analytical solution of the problem can be obtained by substituting the coefficients into the corresponding wave fields.