## 3 Analytical Solutions  
Firstly, Equation (1) needs to be adopted to transform \(u_{\text{a}}\) and \(u_{\text{w}}\) into \(\overline{u}_{\text{a}}\) and \(\overline{u}_{\text{w}}\) respectively, to introduce the equal-strain theorem into the solution. Rearranging Equation (1) produces:  
\[\frac{\partial}{\partial r}(r\mathbf{u}_{r})=-r\mathbf{C_{v}}^{-1}\big{(} \mathbf{C}\overline{\mathbf{u}}_{i}-\mathbf{C_{q}}\mathbf{q}_{i}\big{)} \tag{11}\]  
Incorporated with Equations (5), (7) and (8), the following expressions of excess pore pressures in the smeared and influenced areas can be obtained by integrating Equation (11) upon \(r\) in intervals \([r_{\text{w}},\,r]\) and \([r_{\text{s}},\,r]\):  
\[\mathbf{u}_{\text{s}}=\frac{1}{2}\bigg{[}r_{e}^{2}\ln\left(\frac{r}{r_{w}} \right)-\frac{r^{2}-r_{w}^{2}}{2}\bigg{]}\mathbf{C_{v_{s}}}^{-1}\big{(} \mathbf{C}\overline{\mathbf{u}}_{,t}-\mathbf{C_{q}}\mathbf{q}_{,t}\big{)}+ \mathbf{u}_{\text{d}} \tag{12}\]  
\[\mathbf{u}=\frac{1}{2}\bigg{[}r_{e}^{2}\ln\left(\frac{r}{r_{s}}\right)-\frac{ r^{2}-r_{\text{s}}^{2}}{2}\bigg{]}\mathbf{C_{v_{r}}}^{-1}\big{(}\mathbf{C} \overline{\mathbf{u}}_{,t}-\mathbf{C_{q}}\mathbf{q}_{,t}\big{)}+\left.\mathbf{ u}_{\text{s}}\right|_{r=r_{s}} \tag{13}\]  
By substituting Equations (12) and (13) into Equations (1) and (2), one can produce the expression of excess pore pressure under the equal-strain assumption [15]:  
\[\overline{\mathbf{u}}=\frac{r_{e}^{2}}{2}\mathbf{F}\mathbf{C_{v}}^{-1}\big{(} \mathbf{C}\overline{\mathbf{u}}_{t}-\mathbf{C_{q}}\mathbf{q}_{t}\big{)}+ \mathbf{u}_{\text{d}} \tag{14}\]  
where, \(\mathbf{F}=\left[ \begin{array}{cc}F_{\text{a}}&0\\ 0&F_{\text{w}}\end{array} \right]\). \(F_{\text{a}}\) and \(F_{\text{w}}\) are the intermediate variables concerning smear effects:  
\[F_{\text{a}}=\frac{N^{2}}{N^{2}-1}\bigg{[}\left(\ln\frac{N}{S}-\frac{3}{4}- \frac{S^{4}}{4N^{4}}+\frac{S^{2}}{N^{2}}\right)+\alpha_{\text{a}}\bigg{(} \frac{S^{4}-1}{4N^{4}}-\frac{S^{2}-1}{N^{2}}+\ln S\bigg{)}\bigg{]} \tag{15}\]  
\[F_{\text{w}}=\frac{N^{2}}{N^{2}-1}\bigg{[}\left(\ln\frac{N}{S}-\frac{3}{4}- \frac{S^{4}}{4N^{4}}+\frac{S^{2}}{N^{2}}\right)+\alpha_{\text{w}}\bigg{(} \frac{S^{4}-1}{4N^{4}}-\frac{S^{2}-1}{N^{2}}+\ln S\bigg{)}\bigg{]} \tag{16}\]  
\(\alpha_{\text{a}}\) and \(\alpha_{\text{w}}\) are the smear coefficients regarding air and water phases respectively, that is \(\alpha_{\text{a}}\) = \(k_{\text{a}}/k_{\text{as}}\) and \(\alpha_{\text{w}}\) = \(k_{\text{w}}/k_{\text{ws}}\), and \(S\) = \(r_{\text{s}}/r_{\text{w}}\) and \(N\) = \(r_{\text{e}}/r_{\text{w}}\). It is obvious that the derivation of item \(\mathbf{u}_{\text{d}}\) in Equation (14) is the key to obtaining the final solutions of \(\overline{\mathbf{u}}\). Therefore, substituting Equation (12) into Equation (6) affords:  
\[\mathbf{u}_{\text{d},zz}=\Big{(}1-N^{2}\Big{)}\mathbf{K}^{-1}\mathbf{C_{v_{s} }}^{-1}\big{(}\mathbf{C}\overline{\mathbf{u}}_{i}-\mathbf{C_{q}}\mathbf{q}_{i} \big{)} \tag{17}\]  
Combining Equations (14) and (17) produces the following equation:  
\[\mathbf{u}_{\text{d},zz}=\rho(\mathbf{u}_{\text{d}}-\overline{\mathbf{u}}) \tag{18}\]  
where, \(\rho=\left[ \begin{array}{cc}\rho_{\text{a}}&0\\ 0&\rho_{\text{w}}\end{array} \right]\), \(\rho_{\text{a}}=\frac{8\left(N^{2}-1\right)}{N^{2}H^{2}}\frac{G_{\text{a}}}{F_{ \text{a}}}\), \(\rho_{\text{w}}=\frac{8\left(N^{2}-1\right)}{N^{2}H^{2}}\frac{G_{\text{w}}}{F_{ \text{w}}}\). \(G_{\text{a}}\) and \(G_{\text{w}}\) are the factors of drain resistance of air and water phases respectively, and \(G_{\text{a}}=\left(\frac{H}{d_{\text{w}}}\right)^{2}\frac{k_{\text{a}}}{F_{ \text{a}_{\text{d}}}}\), \(G_{\text{w}}=\left(\frac{H}{d_{\text{w}}}\right)^{2}\frac{k_{\text{w}}}{F_{ \text{w}_{\text{d}}}}\), \(d_{\text{w}}=2r_{\text{w}}\).