## 2 Solution of Foundation Pit Seepage Field
### Establishment and Solution of Theoretical Analysis Model
For the seepage in the phreatic aquifer, the governing equation is [28] as follows:  
\[\frac{\partial^{2}h}{\partial x^{2}}+\frac{\partial^{2}h}{\partial z^{2}}=0, \tag{1}\]  
The boundary conditions are as follows:  
\[\left\{\begin{array}{l}\vspace{0.2 cm}\v  
* The horizontal seepage gradient gradually decreases from the bottom of the foundation pit curtain to the aquiclude.
* The gradient at the top of the aquiclude approaches zero, which is considered here to be zero.
* The integral of the horizontal velocity on the AB section is equal to the water inflow per unit width on the side of the foundation pit, which is recorded as \(Q\), \(\text{m}^{3}/\text{m}\cdot\text{d}\).  
In order to ensure the successful solution of the governing equation, it is assumed that the horizontal seepage gradient decreases linearly from point A to point B, and the horizontal seepage gradient at point B is zero; it can be assumed that the horizontal seepage gradient distribution function \(J_{x}(z)\) of the AB section is as follows:  
\[J_{x}(z)=-k(z-h_{n}), \tag{3}\]  
where \(k\) is the gradient decreasing rate. According to [PERSON]'s law, \(V_{x}(z)=KJ_{x}(z)\), \(K\) is the permeability coefficient of the formation, \(\text{m}/\text{d}\). It can be obtained from the integral of the horizontal flow velocity over the AB section equal to the water inflow per unit width on one side of the foundation pit:  
\[\int_{L}^{h_{n}}V_{x}(z)dz=\int_{L}^{h_{n}}KJ_{x}(z)dz=Q, \tag{4}\]  
Substituting Equation (3) into Equation (4), the horizontal seepage gradient distribution function \(J_{x}(z)\) of the AB section can be obtained after solving the following:  
\[J_{x}(z)=-\frac{2Q}{KM^{2}}(z-h_{n}), \tag{5}\]  
The second-order linear partial differential Equation (1) is solved by using the separated variable method [29]. Assuming that the solution of Equation (1) can be expressed as \(h(x,z)=X(x)Z(z)\), substituting it into Equation (1) results in the following:  
\[\frac{\frac{d}{dx}(\frac{dX}{dx})}{x}=-\frac{\frac{d}{dz}(\frac{dZ}{dz})}{z}, \tag{6}\]  
Equation (6) is denoted by \(\frac{X^{\prime\prime}}{x}=-\frac{Z^{\prime\prime}}{z}=\lambda^{2}\), and the general solution of Equation (1) is the superposition of positive, negative, and zero cases of \(\lambda^{2}\):  
\[ \begin{array}{ll}h(x,z)=A_{1}&xz+B_{1}x+C_{1}z+D_{1}\\ &+(A_{2}e^{\lambda x}+B_{2}e^{-\lambda x})cos\lambda z+(C_{2}e^{\lambda x}+D_{ 2}e^{-\lambda x})sin\lambda z\\ &+(A_{3}e^{\lambda z}+B_{3}e^{-\lambda z})cos\lambda x+(C_{3}e^{\lambda z}+D_{ 3}e^{-\lambda z})sin\lambda x\end{array} , \tag{7}\]  
where \(A_{i}\), \(B_{i}\), \(C_{i}\), and \(D_{i}\) are constant coefficients, and \(i\) is 1, 2, and 3. Substituting the boundary conditions 1, 2, and 3 in Equation (2) into Equation (7), the following can be obtained:  
\[\frac{\partial h}{\partial x}=\sum_{p=1}^{\infty}A_{p}C_{2}[e^{A_{p}x}-e^{A_{p }(2b-x)}]\sin A_{p}z, \tag{8}\]  
where \(A_{p}=\frac{(2p-1)\pi}{2h_{n}}\), \(p=1\), 2, 3,..., \(\infty\). Substituting the boundary conditions 4a, 3b into Equation (2) into Equation (8), the following can be obtained:  
\[\sum_{p=1}^{\infty}A_{p}C_{2}\Big{(}1-e^{2A_{p}b}\Big{)}\sin A_{p}z= \begin{cases} 0;&0\leq z\leq L.\\ -\frac{2Q}{KM^{2}}(z-h_{n});&L\leq z\leq h_{n}.\end{cases} ^{\prime}. \tag{9}\]Let the right side of the equal sign of Equation (9) be regarded as a function \(f(x)\), then the left side of the equal sign of Equation (9) can be regarded as the Fourier sine series of \(f(\text{x})\) expansion [29], and from \(\cos A_{p}h_{n}=0,\sin A_{p}h_{n}=1\), the following can be obtained:  
\[ \begin{array}{ll}A_{p}C_{2}\Big{(}1-e^{2}{A_{p}}^{b}\Big{)}&=\frac{2}{h_{n}} \left(\begin{matrix}1\\ 0\end{matrix}\,0\,\cos\,A_{p}zd_{z}+\int\limits_{L}^{h_{n}}-\frac{2Q}{KM^{2}} \left(z-h_{n}\right)\times\sin A_{p}zd_{z}\right)\\ &=\frac{2}{h_{n}}[-\frac{2Q}{KM^{2}{A_{p}}^{2}}(1-\sin A_{p}L)]\end{array} , \tag{10}\]  
The coefficient \(C_{2}=-\frac{4Q(1-\sin A_{p}L)}{h_{n}KM^{2}A_{p}^{3}(1-e^{2A_{p}B})}\) can be obtained. At this stage, the unknown coefficients have been solved.