## 2 Materials and Methods
### A Useful 'Equivalent' Problem: A Forced Infinite Elastic Plate
The general solutions to these equations are given by  
\[u =\frac{\partial\phi}{\partial x}+\frac{\partial\psi}{\partial y}, \tag{1}\]1 \[u =\frac{\partial\phi}{\partial x}+\frac{\partial\psi}{\partial y}, \tag{1}\]2  
where we have defined  
\[u =\frac{\partial\phi}{\partial x}+\frac{\partial\psi}{\partial y}, \tag{1}\]3 \[u =\frac{\partial\phi}{\partial x}+\frac{\partial\psi}{\partial y}, \tag{1}\]4the stresses in Fourier space are denoted by \(T\), and are given by  
\[u =\frac{\partial\phi}{\partial x}+\frac{\partial\psi}{\partial y}, \tag{1}\]5 \[u =\frac{\partial\phi}{\partial x}+\frac{\partial\psi}{\partial y}, \tag{1}\]6 (15b) \[u =\frac{\partial\phi}{\partial x}+\frac{\partial\psi}{\partial y}, \tag{1}\]7  
the Fourier transformed boundary conditions are \(T_{yy}(\alpha,1/2)=P^{+}(\alpha)\), \(T_{xy}(\alpha,1/2)=S^{+}(\alpha)\), \(T_{yy}(\alpha,-1/2)=R^{+}(\alpha)\), and \(T_{xy}(\alpha,-1/2)=Q^{+}(\alpha)\), where  
\[u =\frac{\partial\phi}{\partial x}+\frac{\partial\psi}{\partial y}, \tag{1}\]8  
etc., and the superscript '\(+\)' ('\(-\)') notation denotes a 'plus' ('minus') function, that is, a function that is analytic in the upper (lower) half of the complex \(\alpha\)-plane. Note that, from the theory of half-range Fourier transforms [24], we know that such well-behaved functions are analytic in the upper or lower half space of the \(\alpha\)-plane. We may use the symmetry of the general solutions to simplify the boundary conditions. Using the linear combination of boundary conditions given by \(T_{xy}(\alpha,1/2)-T_{xy}(\alpha,-1/2)=S^{+}(\alpha)-Q^{+}(\alpha)\) and \(T_{yy}(\alpha,1/2)+T_{yy}(\alpha,-1/2)=P^{+}(\alpha)+R^{+}(\alpha)\), we find that the unknown coefficients \(A\) and \(D\) can be found by solving a system decoupled from the system governing \(B\) and \(C\). The solutions for \(A\) and \(D\) are known as the symmetric part and must satisfy the linear system given by  
\[ \begin{bmatrix}(\alpha^{2}-\beta_{t}^{2})\cos\left(\frac{\beta_{t}}{2}\right)& -2\mathrm{i}\alpha\beta_{t}\cos\left(\frac{\beta_{t}}{2}\right)\\ -2\mathrm{i}\alpha\beta_{t}\sin\left(\frac{\beta_{t}}{2}\right)&(\alpha^{2}- \beta_{t}^{2})\sin\left(\frac{\beta_{t}}{2}\right)\end{bmatrix}\begin{bmatrix} A\\ D\end{bmatrix} =\frac{1}{2} \begin{bmatrix}P^{+}(\alpha)+R^{+}(\alpha)\\ S^{+}(\alpha)-Q^{+}(\alpha)\end{bmatrix} . \tag{17}\]  
for the antisymmetric part, we use the combination of conditions given by \(T_{xy}(\alpha,1/2)+T_{xy}(\alpha,-1/2)=S^{+}(\alpha)+Q^{+}(\alpha)\) and \(T_{yy}(\alpha,1/2)-T_{yy}(\alpha,-1/2)=P^{+}(\alpha)-R^{+}(\alpha)\). This generates the linear system given by  
\[ \begin{bmatrix}(\alpha^{2}-\beta_{t}^{2})\sin\left(\frac{\beta_{t}}{2}\right)& 2\mathrm{i}\alpha\beta_{t}\sin\left(\frac{\beta_{t}}{2}\right)\\ 2\mathrm{i}\alpha\beta_{t}\cos\left(\frac{\beta_{t}}{2}\right)&(\alpha^{2}- \beta_{t}^{2})\cos\left(\frac{\beta_{t}}{2}\right)\end{bmatrix}\begin{bmatrix} B\\ C\end{bmatrix} =\frac{1}{2} \begin{bmatrix}P^{+}(\alpha)-R^{+}(\alpha)\\ S^{+}(\alpha)+Q^{+}(\alpha)\end{bmatrix} . \tag{18}\]  
In linear theory, the symmetric and antisymmetric solutions do not interact at a symmetric boundary. Therefore, if we assume a symmetric incoming wave, the contributions from the antisymmetric Lamb waves must be zero. This implies that \(Q^{+}(\alpha)=-S^{+}(\alpha)\) and \(R^{+}(\alpha)=P^{+}(\alpha)\). Similarly for the antisymmetric case, the symmetric waves must be zero, which requires \(Q^{+}(\alpha)=S^{+}(\alpha)\) and \(R^{+}(\alpha)=-P^{+}(\alpha)\). We will present the method for symmetric waves here; the antisymmetric case follows similarly.