## 3 Analytical Solution of Stress and Displacement of Frozen Wall
### Calculation of Elastic Stress of Frozen Wall
#### 3.2.2 Frozen Wall  
When the frozen wall was in the elastic plane strain state, the following basic equations are satisfied:  
Equation of equilibrium [24]:  
\[\frac{d\sigma_{r}}{dr}+\frac{\sigma_{r}-\sigma_{\theta}}{r}=0 \tag{12}\]  
Geometric equation [24]:  
\[\left\{\begin{array}{ll}\varepsilon_{r}=du_{r}/dr\\ \varepsilon_{\theta}=u_{r}/r\end{array}\right. \tag{13}\]Considering the condition of non-uniform frost heaving, the physical equation under plane strain in the elastic region is:  
\[\begin{cases}\varepsilon_{r}^{e}=\frac{1-{\mu_{2}}^{2}}{E(r)}\sigma_{r}-\frac{{ \mu_{2}}(1+{\mu_{2}})}{E(r)}\sigma_{\theta}+(\alpha_{r}+\mu_{2}\alpha_{\theta}) \\ \varepsilon_{\theta}^{e}=\frac{1-{\mu_{2}}^{2}}{E(r)}\sigma_{\theta}-\frac{{ \mu_{2}}(1+{\mu_{2}})}{E(r)}\sigma_{r}+(\alpha_{\theta}+\mu_{2}\alpha_{r}) \end{cases} \tag{14}\]  
Introducing the stress function \(\varphi\), \(\sigma_{r}\) and \(\sigma_{\theta}\) are represented by \(\varphi\):  
\[\begin{cases}\sigma_{r}=\varphi/r\\ \sigma_{\theta}=\varphi\end{cases} \tag{15}\]  
Combining (12)-(15) yields the following formula:  
\[r^{2}\varphi^{\prime\prime}+r\bigg{(}1-\frac{E^{\prime}(r)}{E(r)}r\bigg{)} \varphi^{\prime}+\bigg{(}\frac{E^{\prime}(r)}{E(r)}r-1\bigg{)}\varphi=\frac{r( \alpha_{r}-\alpha_{\theta})E(r)}{1+\mu_{2}} \tag{16}\]  
As mentioned in Section 2.3, the elastic modulus changes with the relative radius \(r\), and the expression of the elastic modulus \(E(r)\) is substituted into Equation (16) to obtain  
\[r^{2}\varphi^{\prime\prime}-r\frac{ar^{2}-\mathrm{c}}{ar^{2}+br+c}\varphi^{ \prime}+\frac{ar^{2}-c}{ar^{2}+br+c}\varphi=\frac{(k-1)\alpha_{v}(ar^{2}+br+c )r}{(k+2)(1+\mu_{2})} \tag{17}\]  
Equation (17) is a second-order differential equation with variable coefficients, and the general solution is  
\[\varphi=r\mathrm{C}_{0}\bigg{(}a\ln r-b\frac{1}{r}-\frac{\mathrm{c}}{2}\frac {1}{r^{2}}\bigg{)}+B_{0}r+rA_{0}(ar^{2}+2 br+2c\ln r-c) \tag{18}\]  
where \(A_{0}=\frac{\alpha_{v}(k-1)}{4(k+2)(\mu_{2}+1)}\).  
By substituting the boundary condition in Equation (8) into Equation (18), the following is obtained:  
\[C_{0}=\frac{-P_{1}-A_{0}(a\big{(}r_{1}^{2}-r_{2}^{2}\big{)}+2b(r_{1}-r_{2})+2 c\ln(r_{1}/r_{2}))}{a\ln(r_{1}/r_{2})+b\Big{(}\frac{1}{r_{2}}-\frac{1}{r_{1}} \Big{)}+\frac{\zeta}{2}\bigg{(}\frac{1}{r_{2}^{2}}-\frac{1}{r_{1}^{2}}\bigg{)}} \tag{19}\]  
\[B_{0}=-A_{0}(ar_{1}^{2}+2 br_{1}+2c\ln r_{1}-c)-C_{0}\bigg{(}a\ln r_{1}-b\frac {1}{r_{1}}-\frac{\mathrm{c}}{2}\frac{1}{r_{1}^{2}}\bigg{)} \tag{20}\]  
\[\begin{cases}\sigma_{r}^{e}=\frac{A_{0}A_{2}(A_{5}-A_{3})+(P_{1}+A_{0}A_{1})(A _{4}-A_{6})}{A_{2}}\\ \sigma_{\theta}^{e}=\frac{A_{0}A_{2}(A_{8}-A_{3})+(P_{1}+A_{0}A_{1})(A_{4}-A_{ 7})}{A_{2}}\end{cases}r\in[r_{1},r_{2}] \tag{21}\]  
where \(A_{1}=a\big{(}r_{1}^{2}-r_{2}^{2}\big{)}+2b(r_{1}-r_{2})+2c\ln(r_{1}/r_{2})\), \(A_{2}=a\ln(r_{1}/r_{2})+b\big{(}\frac{1}{r_{2}}-\frac{1}{r_{1}}\big{)}+\frac{ \zeta}{2}\bigg{(}\frac{1}{r_{2}^{2}}-\frac{1}{r_{1}^{2}}\bigg{)}\), \(A_{3}=ar_{1}^{2}+2 br_{1}+2c\ln r_{1}-c\), \(A_{4}=a\ln r_{1}-b\frac{1}{r_{1}}-\frac{\zeta}{2}\frac{1}{r_{1}^{2}}\), \(A_{5}=ar^{2}+2 br+2c\ln r-c\), \(A_{6}=a\ln r-\frac{b}{r}-\frac{c}{2}\frac{1}{r_{2}^{2}}\), \(A_{7}=a\ln r+\frac{\zeta}{2}\frac{1}{r_{2}}+a\), \(A_{8}=3 ar^{2}+4 br+2c\ln r+c\).  
By substituting Equation (21) into physical Equation (14) and geometric Equation (13), the analytical solution of the elastic displacement of the heterogeneous frozen wall considering frost heave characteristics is obtained:  
\[U^{e}=\frac{r\cdot(1-\mu_{2}^{2})}{E(r)}\cdot\sigma_{\theta}^{e}-\frac{r\cdot \mu_{2}(1+\mu_{2})}{E(r)}\cdot\sigma_{r}^{e}+\frac{(1+\mu_{2}k)\alpha_{v}\cdot r }{k+2} \tag{22}\]By substituting Equations (11) and (22) into the fourth expression in Equation (8) of interface displacement continuity, the expression of \(P_{1}\) is obtained.