**Appendix A. Dirichlet-to-Neumann Operators**

By projecting the terms of Equation (5a) on the orthonormal basis, spanned by the normalized eigenfunctions *<sup>Z</sup>*1(*i*) *<sup>n</sup>* (*z*) <sup>=</sup> *<sup>Z</sup>*(*i*) *<sup>n</sup>* (*z*)/ 5 5 5 *Z*(*i*) *n* 5 5 <sup>5</sup>, with 5 5 5 *Z*(*i*) *n* 5 5 <sup>5</sup> standing for the *<sup>L</sup>*2<sup>−</sup> norm of each vertical function:

$$\left\| Z\_n^{(i)} \right\| = \left\{ \int\_{z=-h\_i}^{0} \left[ Z\_n^{(i)}(\mathbf{x}\_3) \right]^2 d\mathbf{x}\_3 \right\}^{1/2}, i = L, R,\tag{A1}$$

we obtain

$$\left\langle \boldsymbol{\varrho}\_{p}^{(\boldsymbol{L})}(\mathbf{x}) \cdot \widetilde{\boldsymbol{Z}}\_{n}^{(\boldsymbol{L})}(\mathbf{x}\_{3}) \right\rangle = \begin{cases} \exp\left(ik\_{0}^{(\boldsymbol{L})}\mathbf{x}\_{2}\right) + \mathbb{C}\_{0}^{(\boldsymbol{L})}\exp\left(-ik\_{0}^{(\boldsymbol{L})}\mathbf{x}\_{2}\right), & n=0\\\ \mathbb{C}\_{n}^{(\boldsymbol{L})}\exp\left[k\_{n}^{(\boldsymbol{L})}(\mathbf{x}\_{2} - a)\right], & n \ge 1 \end{cases} \tag{A2}$$

where *f*(*x*3), *g*(*x*3) = *<sup>x</sup>*3=<sup>0</sup> *x*3=−*ha* [ *f*(*x*3) · *g*(*x*3)] *dx*3. Therefore, the reflection coefficient in the left half-strip *DL* is equal to

$$\mathbf{C}\_{0}^{(L)} = \frac{-\exp\left(ik\_{0}^{(L)}\mathbf{x}\_{2}\right) + \left<\boldsymbol{\varphi}\_{p}^{(L)}(\mathbf{x}) \cdot \bar{\mathbf{Z}}\_{0}^{(L)}(\mathbf{x}\_{3})\right>}{\exp\left(-ik\_{0}^{(L)}\mathbf{x}\_{2}\right)}, \ (\mathbf{x}\_{2}, \mathbf{x}\_{3}) \in D\_{L} \tag{A3}$$

Moreover, by calculating the derivative of Equation (5a) with respect to the unit normal vector *n* (which is directed opposite to the *x*<sup>2</sup> direction on *∂D*5) and replacing in Equation (6e)

$$\begin{split} -\frac{\partial \boldsymbol{\varrho}\_{p}^{(L)}(\mathbf{x})}{\partial \boldsymbol{n}} &= i \boldsymbol{k}\_{0}^{(L)} \Big[ \exp \Big( \boldsymbol{k}\_{0}^{(L)} \mathbf{x}\_{2} \Big) - \mathsf{C}\_{0}^{(L)} \exp \Big( -i \boldsymbol{k}\_{0}^{(L)} \mathbf{x}\_{2} \Big) \Big] \boldsymbol{Z}\_{0}^{(L)}(\mathbf{x}\_{3}) + \\ + \sum\_{n=1}^{\infty} \boldsymbol{k}\_{n}^{(L)} \mathsf{C}\_{n}^{(L)} \exp \Big[ \boldsymbol{k}\_{n}^{(L)}(\mathbf{x}\_{2} - a) \Big] \boldsymbol{Z}\_{n}^{(L)}(\mathbf{x}\_{3}) &= -\boldsymbol{T}\_{L} \Big[ \boldsymbol{\varrho}\_{p}^{(L)}(\mathbf{x}) \Big] - \boldsymbol{Q}\_{p\prime} \cdot \mathbf{x} \in \boldsymbol{D}\_{L\prime} \end{split} \tag{A4}$$

and by using Equation (A2), we conclude to

$$\begin{split} T\_L \left[ \boldsymbol{\varrho}\_p^{(L)}(\mathbf{x}) \right] &= i \boldsymbol{k}\_0^{(L)} \widetilde{\boldsymbol{Z}}\_0^{(L)}(\boldsymbol{\chi}\_3) \left\langle \boldsymbol{\varrho}\_p^{(L)}(\mathbf{x}) \cdot \widetilde{\boldsymbol{Z}}\_0^{(L)}(\boldsymbol{\chi}\_3) \right\rangle + \\ &- \sum\_{n=1}^{\infty} \boldsymbol{k}\_n^{(L)} \widetilde{\boldsymbol{Z}}\_n^{(L)}(\boldsymbol{\chi}\_3) \left\langle \boldsymbol{\varrho}\_p^{(L)}(\mathbf{x}) \cdot \widetilde{\boldsymbol{Z}}\_n^{(L)}(\boldsymbol{\chi}\_3) \right\rangle, \end{split} \tag{A5}$$

where

$$Q\_{\mathcal{P}} = -2ik\_0^{(L)} \exp\left(ik\_0^{(L)} \mathbf{x}\_2\right) \tilde{Z}\_0^{(L)}(\mathbf{x}\_3). \tag{A6}$$

Similarly, for *k* = 2, 3, 4, we obtain

$$\begin{split} -\frac{\partial \boldsymbol{\varrho}\_{k}^{(L)}(\mathbf{x})}{\partial \mathbf{n}} &= -i\mathbf{k}\_{0}^{(L)} \mathbf{C}\_{0}^{(L)} \exp\left(-i\mathbf{k}\_{0}^{(L)} \mathbf{x}\_{2}\right) \mathbf{Z}\_{0}^{(L)}(\mathbf{x}\_{3}) + \\ &+ \sum\_{n=1}^{\infty} \boldsymbol{k}\_{n}^{(L)} \mathbf{C}\_{n}^{(L)} \exp\left[\boldsymbol{k}\_{n}^{(L)}(\mathbf{x}\_{2} - a)\right] \mathbf{Z}\_{n}^{(L)}(\mathbf{x}\_{3}) = -T\_{L} \left[\boldsymbol{\varrho}\_{k}^{(L)}(\mathbf{x})\right] - Q\_{k\prime} \mathbf{x} \in D\_{L}. \end{split} \tag{A7}$$

Using the above results, we obtain *TL ϕ*(*L*) *<sup>k</sup>* (**x**) = *TL ϕ*(*L*) *<sup>p</sup>* (**x**) , *k* = 2, 3, 4, and *Qk* = 0, *k* = 2, 3, 4. Similarly, for the wavefield in the domain *DR*:

$$\begin{split} \frac{\partial \boldsymbol{\varrho}\_{p}^{(R)}(\mathbf{x})}{\partial \mathbf{n}} &= \boldsymbol{k}\_{0}^{(R)} \mathbf{C}\_{0}^{(R)} \exp\left(\boldsymbol{k}\_{0}^{(R)} \mathbf{x}\_{2}\right) \mathbf{Z}\_{0}^{(R)}(\mathbf{x}\_{3}) + \\ &+ \sum\_{n=1}^{\infty} \boldsymbol{k}\_{n}^{(R)} \mathbf{C}\_{n}^{(R)} \exp\left[\boldsymbol{k}\_{n}^{(R)}(\boldsymbol{b} - \mathbf{x}\_{2})\right] \mathbf{Z}\_{n}^{(R)}(\mathbf{x}\_{3}) = \boldsymbol{T}\_{\mathcal{R}} \Big[\boldsymbol{\varrho}\_{p}^{(R)}(\mathbf{x})\Big], \mathbf{x} \in D\_{\mathcal{R}}. \end{split} \tag{A8}$$

and thus

$$\begin{split} T\_{\mathbb{R}}\left[\boldsymbol{\varphi}\_{p}^{(\mathcal{R})}(\mathbf{x})\right] &= ik\_{0}^{(\mathcal{R})}Z\_{0}^{(\mathcal{R})}(\boldsymbol{\chi}\_{3})\Big\langle\boldsymbol{\varphi}\_{k}^{(\mathcal{R})}(\mathbf{x})\cdot\tilde{Z}\_{0}^{(\mathcal{R})}(\boldsymbol{\chi}\_{3})\Big\rangle+ \\ &+ \sum\_{n=1}^{\infty}k\_{n}^{(\mathcal{R})}Z\_{n}^{(\mathcal{R})}(\boldsymbol{\chi}\_{3})\Big\langle\boldsymbol{\varphi}\_{k}^{(\mathcal{R})}(\mathbf{x})\cdot\tilde{Z}\_{n}^{(\mathcal{R})}(\boldsymbol{\chi}\_{3})\Big\rangle. \end{split} \tag{A9}$$
