## 2 The Simplified Calculation Model and Control Equation of an Underwater Immersed Tunnel
\begin{array}{c}u_{x}=\frac{\partial\phi}{\partial x}-\frac{\partial\phi}{ \partial z};u_{z}=\frac{\partial\phi}{\partial z}+\frac{\partial\psi}{ \partial x}\\ w_{x}=\frac{\partial X}{\partial x}-\frac{\partial\sigma}{\partial z};w_{z}= \frac{\partial X}{\partial z}+\frac{\partial\sigma}{\partial x}\end{array} \] (14a,b)  
By substituting the substitution of Equation (14) into Equation (10), the expressions for displacement, stress, and pore pressure in the Fourier transform domain as follows:  
\[ \begin{array}{c}\left[\widetilde{u}\widetilde{x}_{x}\ \widetilde{\overline{u}}_{z}\ \widetilde{\overline{w}}_{z}\ \widetilde{\overline{w}}_{z}\ \widetilde{\overline{w}}_{z}\ \widetilde{\overline{w}}_{z}\ \widetilde{\overline{w}}_{zz}\ \widetilde{\overline{w}}_{zz}\ \widetilde{\overline{p}}\right]^{T}=\left[\begin{array}{cc} \mathbf{D}_{d}&\mathbf{D}_{u}\\ \mathbf{S}_{d}&\mathbf{S}_{u}\end{array}\right]\left[\mathbf{W}_{d}(z)^{T} \ \mathbf{W}_{u}(z)^{T}\right]\end{array} \] (15a) In formula: \[ \begin{array}{c}\mathbf{W}_{d}(\xi,z,\omega)=\left[be^{-\gamma_{12}}\ de^{-\gamma_{2}z}\ fe^{-\gamma_{3}z}\right]\\ \mathbf{W}_{u}(\xi,z,\omega)=\left[ae^{\gamma_{12}}\ ce^{\gamma_{2}z}\ ce^{ \gamma_{3}z}\right]\end{array} \]  
\[ \begin{array}{c}\mathbf{D}_{u}=\left[\begin{array}{ccc}-\xi&-\xi&i\gamma_{ 3}\\ \gamma_{1}&\gamma_{2}&-i\xi\\ \gamma_{1}\chi_{1}&\gamma_{2}\chi_{2}&-i\xi\gamma_{3}\end{array}\right];\ \mathbf{D}_{d}=\left[\begin{array}{ccc}-\xi&-\xi&-i\gamma_{3}\\ \gamma_{1}&-\gamma_{2}&-i\xi\\ -\gamma_{1}\chi_{1}&-\gamma_{2}\chi_{2}&-i\xi\gamma_{3}\end{array}\right];\\ \mathbf{S}_{u}=\left[\begin{array}{ccc}-2\mu\xi\gamma_{1}&-2\mu\gamma_{2} \xi&i\chi_{3}\\ c_{1}&c_{2}&-2i\mu\xi\gamma_{3}\\ a_{1}&a_{2}&0\end{array}\right];\ \mathbf{S}_{d}=\left[\begin{array}{ccc}2\mu\xi\gamma_{1}&2\mu \gamma_{2}\xi&i\chi_{3}\\ c_{1}&c_{2}&2i\mu\xi\gamma_{3}\\ a_{1}&a_{2}&0\end{array}\right];\\ \chi_{k}=(\alpha+\kappa_{k})M(L_{k})^{2};\ \gamma_{k}=\sqrt{\xi^{2}-L_{k} ^{2}}\ (k=1,2);\\ c_{k}=2\mu\gamma_{1}^{2}-\lambda(L_{k})^{2}-\alpha\chi_{k}(k=1,2);\\ \kappa_{k}=[\left(\lambda+\alpha^{2}M+2\mu\right)(L_{k})^{2}-\rho\omega^{2}]/ [\rho_{t}\omega^{2}-\alpha M(L_{k})^{2}]\ (k=1,2);\\ \chi_{3}=\mu[\xi^{2}+(\gamma_{3})^{2}];\ \kappa_{3}=\rho_{t}\omega^{2}/(ib \omega-m\omega^{2});\\ (S)^{2}=(\rho_{t}\kappa_{3}+\rho)\omega^{2}/\mu;\ \gamma_{3}=\sqrt{\xi^{2}-(S)^{2}};\\ (L_{1})^{2}=\frac{1}{2}(\beta_{1}-\sqrt{\left(\beta_{1}\right)^{2}-4\beta_{2} });\ (L_{2})^{2}=\frac{1}{2}(\beta_{1}+\sqrt{\left(\beta_{1}\right)^{2}-4\beta_{2} });\\ \beta_{1}=[(m\omega^{2}-ib\omega)(\lambda+\alpha^{2}M+2\mu)+\rho\omega^{2}M-2 \alpha M\rho_{t}\omega^{2}]/[(\lambda+2\mu)M];\\ \beta_{2}=[(m\omega^{2}-ib\omega)\rho\omega^{2}M-\rho_{t}^{2}\omega^{4}]/[( \lambda+2\mu)M];\end{array} \]  
For Equation (1), using the Fourier transformation of time and space domains, the pore pressure of the riverbed fluid in the frequency domain and wavenumber domain is:  
\[\frac{\partial^{2}p_{fw}}{\partial z^{2}}+\gamma_{w}^{2}p_{fw}=-2\delta(z-z_{ s}) \tag{16}\]  
In formula: \(\gamma_{w}=\sqrt{(k_{x}^{2}-k_{w}^{2})}\), \(k_{w}=\omega/v_{w}\), and to ensure that the compression wave travels vertically along the fluid, Re(\(k_{w}\)) \(\geq 0\). This can be obtained:  
\[p_{fw}=A_{w}e^{\gamma=z}+B_{w}e^{-\gamma_{w}z} \tag{17}\]  
\[u_{zw}=\frac{1}{\rho_{w}\omega^{2}}\frac{\partial p_{fw}}{\partial z}=\frac{ \gamma_{w}}{\rho_{w}\omega^{2}}(A_{w}e^{\gamma_{w}z}-B_{w}e^{-\gamma_{w}z}) \tag{18}\]  
The vibration equation of the Euler beam in the frequency and wave fields is as follows:  
\[(E_{b}I_{k}k_{x}^{4}-\rho_{B}\omega^{2})\widetilde{\overline{W}}_{b}=2\pi Fz \delta(\omega-k_{x}v_{c})+a(\widetilde{\overline{\sigma}}_{zz}(k_{x},h^{-}, \omega)-\widetilde{\overline{\sigma}}_{zz}(k_{x},h^{+},\omega)) \tag{19}\]