# Study on the Time Domain Semi Analytical Method for Horizontal Vibration of Pile in Saturated Clay
## 3 Equation Solving  
### Solving the Control Equation of Saturated Clay Movement  
By introducing potential functions \(\phi_{s}\), \(\psi_{s}\), \(\phi_{f}\), \(\psi_{f}\) and Helmholtz decomposition of the soil skeleton displacement and water displacement, we obtain  
\[\left. \begin{array}{l}u_{r}=\frac{\partial\phi_{s}}{\partial r}+\frac{1}{r} \frac{\partial\psi_{s}}{\partial\theta},\quad u_{\theta}=\frac{1}{r}\frac{ \partial\phi_{s}}{\partial\theta}-\frac{\partial\psi_{s}}{\partial r}\\ w_{r}=\frac{\partial\phi_{f}}{\partial r}+\frac{1}{r}\frac{\partial\psi_{f}}{ \partial\theta},\quad w_{\theta}=\frac{1}{r}\frac{\partial\phi_{f}}{\partial \theta}-\frac{\partial\psi_{f}}{\partial r}\end{array} \right\} \tag{23}\]  
Substituting Equation (23) into Equations (11) and (12) and organizing it, we obtain:  
\[\left. \begin{array}{l}L\
abla^{2}\phi_{s}-\rho s^{2}\phi_{s}+\mu\frac{ \partial^{2}\phi_{s}}{\partial z^{2}}+\alpha M\
abla^{2}\phi_{f}-\rho_{f}s^{2 }\phi_{f}=0\\ \
abla^{2}\psi_{s}-\frac{\rho s^{2}}{\mu}\psi_{s}+\frac{\partial^{2}\psi_{s} }{\partial z^{2}}-\frac{\rho_{f}s^{2}}{\mu}\psi_{f}=0\\ \alpha M\
abla^{2}\phi_{s}-\frac{\rho_{f}s^{2}}{\alpha M}\phi_{s}+\frac{1}{ \alpha}\
abla^{2}\phi_{f}-N\phi_{f}=0\\ s\psi_{s}+\Big{(}g/k_{f}+s/n\Big{)}\psi_{f}=0\end{array} \right\} \tag{24}\]  
where \(L=\lambda+2\mu+\alpha^{2}M\), \(N=\Big{(}gn+sk_{f}\Big{)}\rho_{f}s/\Big{(}\alpha Mnk_{f}\Big{)}\). Using the method of separating variables to solve Equation (24),  
\[\left. \begin{array}{l}\phi_{s}=\Phi_{s}(r,\theta,s)T_{1}(z),\quad\phi_{f}= \Phi_{f}(r,\theta,s)T_{1}(z)\\ \psi_{s}=\Psi_{s}(r,\theta,s)T_{2}(z),\quad\psi_{f}=\Psi_{f}(r,\theta,s)T_{2} (z)\end{array} \right\} \tag{25}\]  
Substituting Equation (25) into Equation (24) and using boundary condition Equations (16)-(18), we obtain the general solutions of \(T_{1}\)(z) and \(T_{2}\)(z) as follows [16]:  
\[\left. \begin{array}{l}T_{1}(z)=A_{1}(e^{\gamma_{k}z}+e^{-\gamma_{k}z})\\ T_{2}(z)=A_{2}(e^{\gamma_{k}z}+e^{-\gamma_{k}z})\\ \Phi_{s}(r,\theta,s)=\left[A_{3}K_{1}(\beta_{1}r)+A_{4}K_{1}(\beta_{2}r) \right]\cos\theta\\ \Phi_{f}(r,\theta,s)=\left[A_{3}\varepsilon_{1}K_{1}(\beta_{1}r)+A_{4} \varepsilon_{2}K_{1}(\beta_{2}r)\right]\cos\theta\\ \Psi_{s}(r,\theta,s)=A_{5}K_{1}(\beta_{3}r)\sin\theta\\ \Psi_{f}(r,\theta,s)=A_{5}\varepsilon_{3}K_{1}(\beta_{3}r)\sin\theta\end{array} \right\} \tag{26}\]  
In the equation, \(\gamma_{k}=i\pi(2k-1)/(2H)\), \(k=1\), \(2\), \(3\); \(K_{1}\)(0) represents the Bessel function of the second type of first-order imaginary argument, with \(A_{1}\)-\(A_{5}\) as undetermined coefficients; \(\beta_{1}\), \(\beta_{2}\), \(\beta_{3}\) and \(\varepsilon_{1}\), \(\varepsilon_{2}\), \(\varepsilon_{3}\) are all known functions related to \(s\), and the specific form is shown in Appendix A. Substituting Equation (26) into Equation (25) yields:  
\[\left. \begin{array}{l}\phi_{s}=\left[B_{1}K_{1}(\beta_{1}r)+B_{2}K_{1}(\beta_ {2}r)\right]\cos\theta\text{ch}(\gamma_{k}z)\\ \phi_{f}=\left[B_{1}\varepsilon_{1}K_{1}(\beta_{1}r)+B_{2}\varepsilon_{2}K_{1 }(\beta_{2}r)\right]\cos\theta\text{ch}(\gamma_{k}z)\\ \psi_{s}=B_{3}K_{1}(\beta_{3}r)\sin\theta\text{ch}(\gamma_{k}z),\ \psi_{f}= \varepsilon_{3}\psi_{s}\end{array} \right\} \tag{27}\]  
where ch(0) represents the hyperbolic cosine function, and \(B_{1}\), \(B_{2}\), and \(B_{3}\) are the reorganized undetermined coefficients. Substituting Equation (27) into Equation (23) yields the displacement solution of the soil skeleton and water in the Laplace transform domain in saturated soil:  
\[ \begin{array}{l}\widetilde{u}_{r}=\left[B_{1}\frac{\partial K_{1}(\beta_{1}r)}{ \partial r}+B_{2}\frac{\partial K_{1}(\beta_{2}r)}{\partial r}+\frac{1}{r}B_{3} K_{1}(\beta_{3}r)\right]\cos\theta\text{ch}(\gamma_{k}z)\\ \widetilde{u}_{\theta}=\left[-\frac{B_{1}}{r}K_{1}(\beta_{1}r)-\frac{B_{2}}{ r}K_{1}(\beta_{2}r)-B_{3}\frac{\partial K_{1}(\beta_{3}r)}{\partial r}\right]\sin \theta\text{ch}(\gamma_{k}z)\end{array}  \tag{28}\]  
\[