# Dynamic Lateral Response of the Partially-Embedded Single Piles in Layered Soil
## 3 Green's Functions for Dynamic Response  
### Embedded Section  
We can assume that there is a simple harmonic lateral load at the pile top, which can be written as:  
\[P_{0}(t)=\delta(z-z_{0})P_{0}\mathrm{e}^{\mathrm{i}\Omega t} \tag{52}\]  
where \(\delta(\ )\) is the Dirac function, \(\Omega\) is the frequency of the external excitation. We can let \(\Omega=\omega\) when calculating the Green's functions for the dynamic response of the pile. For the pile top, \(z_{0}=0\).  
Substitution of Equation (52) into Equation (35) gives:  
\[u_{i}^{{}^{\prime\prime\prime}}+a_{1}u_{i}^{{}^{\prime\prime}}+a_{2}u_{i}=bP_{ 0}\delta(z-z_{0}),\ (i=2,3,4),\ l_{1}<z\leq L \tag{53}\]  
where \(a_{1}=-\frac{2i_{i}}{E_{p}I_{p}}\), \(a_{2}=\frac{k_{i}-m_{i}\Omega^{2}}{E_{p}I_{p}}\), \(b=\frac{1}{E_{p}I_{p}}\).  
Using the [PERSON]'s function method, the solution of Equation (53) can be written as:  
\[u_{i}(z)=\int_{l_{1}}^{L}P_{0}(z_{0})G(z;z_{0})\mathrm{d}z_{0} \tag{54}\]  
where \(P_{0}(z_{0})\) is the function of the external excitation, \(G(z;z_{0})\) is the Green's function to be determined which is the solution of the following formulation:  
\[u_{i}^{{}^{\prime\prime\prime}}+a_{1}u_{i}^{{}^{\prime\prime}}+a_{2}u_{i}=b \delta(z-z_{0}) \tag{55}\]  
From Equation (55), it is seen that \(G(z;z_{0})=U(z;z_{0})\). Laplace transformation is applied to the variable \(z\) in Equation (55), we can obtain:  
\[\hat{u}_{i}(s;z_{0})=\frac{1}{s^{4}+a_{1}s^{2}+a_{2}}[b\mathrm{e}^{-sz_{0}}+( s^{3}+a_{1}s)u_{i}(0)+(s^{2}+a_{1})u_{i}^{{}^{\prime\prime}}(0)+su_{i}^{{}^{ \prime\prime}}(0)+u_{i}^{{}^{\prime\prime\prime}}(0)] \tag{56}\]  
where \(s=\sigma+\mathrm{i}\tau\), \(u_{i}(0)\), \(u_{i}^{{}^{\prime}}(0)\), \(u_{i}^{{}^{\prime\prime}}(0)\) and \(u_{i}^{{}^{\prime\prime\prime}}(0)\) are constants which can be determined by the boundary conditions. To obtain the inverse transform of \(\hat{u}_{i}(s;z_{0})\), we assume that \(s^{4}+a_{1}s^{2}+a_{2}=\prod\limits_{i=1}^{4}(s-s_{i})\). We can get the results of the inverse transform by referring to the literature [23]:  
\[L^{-1}[\frac{be^{-sz_{0}}}{(s-s_{1})(s-s_{2})(s-s_{3})(s-s_{4})}]=H(z-z_{0})[A _{1}(z-z_{0})b+A_{2}(z-z_{0})b+A_{3}(z-z_{0})b+A_{4}(z-z_{0})b] \tag{57}\]  
\[L^{-1}[\frac{s^{3}+a_{1}s}{(s-s_{1})(s-s_{2})(s-s_{3})(s-s_{4})}]=A_{1}(z  
\[\begin{array}{ll}u_{i}(z)&=\int_{0}^{L}P_{0}(z_{0})G(z;z_{0})dz_{0}=\int_{0}^{L}P_ {0}\phi(z-z_{0})[H(z-z_{0})\phi_{1}(z-z_{0})+\phi_{2}(z)u_{i}(0)+\phi_{3}(z)u_{i} {}^{\prime}(0)+\phi_{4}(z)u_{i}{}^{\prime\prime}(0)+\phi_{5}(z)u_{i}{}^{\prime \prime\prime}(0)]dz_{0}\\ &=P_{0}[H(z-z_{0})\phi_{1}(z-z_{0})+\phi_{2}(z)u_{i}(0)+\phi_{3}(z)u_{i}{}^{ \prime}(0)+\phi_{4}(z)u_{i}{}^{\prime\prime}(0)+\phi_{5}(z)u_{i}{}^{\prime \prime\prime}(0)]\end{array} \tag{82}\]  
Correspondingly, the dynamic response function of the embedded pile is given by  
\[u_{i}(z,t)=P_{0}[H(z-z_{0})\phi_{1}(z-z_{0})+\phi_{2}(z)u_{i}(0)+\phi_{3}(z)u_{ i}{}^{\prime}(0)+\phi_{4}(z)u_{i}{}^{\prime\prime}(0)+\phi_{5}(z)u_{i}{}^  
\[u_{1}(z) =\int_{0}^{L}P_{0}(z_{0})G(z;z_{0})dz_{0}\] \[=\int_{0}^{L}P_{0}\delta(z-z_{0})[H(z-z_{0})\psi_{1}(z-z_{0})+\psi_{ 2}(z)u_{1}(0)+\psi_{3}(z)u_{1}{}^{\prime}(0)+\psi_{4}(z)u_{1}{}^{\prime\prime}(0 )+\psi_{5}(z)u_{1}{}^{\prime\prime\prime}(0)]dz_{0} \tag{94}\] \[=P_{0}[H(z-z_{0})\psi_{1}(z-z_{0})+\psi_{2}(z)u_{1}(0)+\psi_{3}(z)u _{1}{}^{\prime}(0)+\psi_{4}(z)u_{1}{}^{\prime\prime}(0)+\psi_{5}(z)u_{1}{}^{ \prime\prime\prime}(0)]\]  
The dynamic response function of the pile of the non-embedded section is given by  
\[u_{1}(z,t)=P_{0}[H(z-z_{0})\psi_{1}(z-z_{0})+\psi_{2}(z)u_{1}(0)+\psi_{3}(z)u_{ 1}{}^{\prime}(0)+\psi_{4}(z)u_{1}{}^{\prime\prime}(0)+\psi_{5}(z)u_{1}{}^{ \prime\prime\prime}(0)]\cos\Omega t \tag{95}\]