## 2 Establishment of Prediction Model
### TLM-PML Method
\begin{cases}u_{x}\\ u_{y}\\ u_{z}\end{cases} = \begin{bmatrix}\cos\theta&-\sin\theta&0\\ \sin\theta&\cos\theta&0\\ 0&0&1\end{bmatrix}\begin{cases}\overline{u}_{r}\\ \overline{u}_{\theta}\\ \overline{u}_{z}\end{cases}  \tag{19}\]  
The displacement \(U_{a\beta}^{nm}\) along the \(\alpha\) direction at the \(m\) layer under the unit load along the direction \(\beta\) at the \(n\) layer is illustrated in Figure 4. \(k_{\text{Rj}}\), \(\phi_{xj}^{m}\), and \(\phi_{zj}^{m}\) represent the eigenvalue and eigenvector of the \(j\)-order mode in the \(m\) layer of the Rayleigh wave signature problem. \(k_{\text{Lj}}\) and \(\phi_{yj}^{m}\) represent the eigenvalue and eigenvector of the \(j\)-mode in the \(m\) layer of the Love eigenvalue problem, respectively. The calculation results are shown in Equations (20) and (21). Equation (20) is the fundamental solution of the displacement in the frequency-space domain under the point load. Equation (21) is the kernels of fundamental solutions. \[ \begin{split}& U_{xx}^{mn}=\sum\limits_{j}^{2N}I_{3j}\phi_{xj}^{ m}\phi_{xj}^{n}+\sum\limits_{j}^{N}I_{4j}\phi_{yj}^{m}\phi_{yj}^{n}\\ & U_{yy}^{mn}=\sum\limits_{j}^{2N}I_{4j}\phi_{xj}^{m}\phi_{xj}^{n}+\sum \limits_{j}^{N}I_{3j}\phi_{yj}^{m}\phi_{yj}^{n}\\ & U_{xy}^{mn}=\sum\limits_{j}^{2N}I_{2j}\phi_{xj}^{m}\phi_{xj}^{n}-\sum \limits_{j}^{N}I_{2j}\phi_{yj}^{m}\phi_{yj}^{n}=U_{yx}^{mn}\\ & U_{xz}^{mn}=-i\sum\limits_{j}^{2N}I_{5j}\phi_{xj}^{m}\phi_{zj}^{n},\quad U_{ zx}^{mn}=i\sum\limits_{j}^{2N}I_{5j}\phi_{zj}^{m}\phi_{xj}^{n}=-U_{xz}^{mn}\\ & U_{yz}^{mn}=-i\sum\limits_{j}^{2N}I_{6j}\phi_{xj}^{m}\phi_{zj}^{n},\quad U_{ zy}^{mn}=i\sum\limits_{j}^{2N}I_{5j}\phi_{zj}^{m}\phi_{xj}^{n}=-U_{yz}^{mn}\\ & U_{zz}^{mn}=\sum\limits_{j}^{2N}I_{1j}\phi_{zj}^{m}\phi_{zj}^{n}\end{split}  \tag{20}\]\[ \begin{array}{l}I_{1j}=\frac{1}{4i}H_{0}^{(2)}\left([PERSON]_{j}\right)\\ I_{2j}=\frac{\cos\theta\sin\theta}{4i}\left\{H_{0}^{(2)}\left([PERSON]_{j}\right)-\frac {2}{Z_{j}}\left\{H_{1}^{(2)}\left([PERSON]_{j}\right)-\frac{2i}{\pi Z_{j}}\right\} \right\}\\ I_{3j}=\frac{1}{4i}\left\{\cos^{2}\theta H_{0}^{(2)}\left([PERSON]_{j}\right)-\frac{ \cos^{2}\theta-\sin^{2}\theta}{Z_{j}}\left\{H_{1}^{(2)}\left([PERSON]_{j}\right)- \frac{2i}{\pi Z_{j}}\right\}\right\}\\ I_{4j}=\frac{1}{4i}\left\{\sin^{2}\theta H_{0}^{(2)}\left([PERSON]_{j}\right)+\frac {\cos^{2}\theta-\sin^{2}\theta}{Z_{j}}\left\{H_{1}^{(2)}\left([PERSON]_{j}\right)- \frac{2i}{\pi Z_{j}}\right\}\right\}\\ I_{5j}=-\frac{1}{4}\cos\theta H_{1}^{(2)}\left([PERSON]_{j}\right)\\ I_{6j}=-\frac{1}{4}\sin\theta H_{1}^{(2)}\left(Z_{j}\right)\\ Z_{j}=k_{j}r\ \ \ \text{r}=\sqrt{x^{2}+y^{2}}\ \ \cos\theta=x/r\ \ \sin\theta=y/r\end{array}  \tag{21}\]  
In addition, when \(r\) approaches 0, the points along the vertical axis at the excitation point are divergent. In order to eliminate this singularity in the calculation, a circular uniform distribution load with radius \(r_{0}\) is considered to replace the concentrated point load. In this research, the fundamental solution of displacement in the frequency-space domain is calculated when a uniform load with radius \(r_{0}\) acts on the axis of the center of the circle (\(r=0\) m), as shown in Equation (22). \[