## Appendix A  
This appendix aims to demonstrate the solution process of the two-dimensional Laplace governing equation discussed in Section 3.2. The two-dimensional Laplace governing equation is  
\[\frac{\partial^{2}h}{\partial y^{2}}+\frac{\partial^{2}h}{\partial z^{2}}=0 \tag{1}\]  
Based on the linear nature of the governing equation and boundary conditions, its analytical solution can be decomposed into four parts according to the boundary conditions, namely  
\[h\big{(}z,y\big{)}=h_{1}\big{(}z,y\big{)}+h_{2}\big{(}z,y\big{)}+h_{3}\big{(}z,y\big{)}+h_{4}\big{(}z,y\big{)} \tag{2}\]  
Each equation, while satisfying the two-dimensional Laplace equation, meets the following boundary conditions, respectively:  
\[\left\{ \begin{array}{l}h_{1}(z,h_{0})=f_{1}(z)\\ h_{1}(z,0)=0\\ h_{1}(L,y)=0\\ h_{1}(0,y)=0\end{array} \right. \tag{3}\]  
\[\left\{ \begin{array}{l}h_{2}(z,h_{0})=0\\ h_{2}(z,0)=f_{2}(z)\\ h_{2}(L,y)=0\\ h_{2}(0,y)=0\end{array} \right. \tag{4}\]  
\[\left\{ \begin{array}{l}h_{3}(z,h_{0})=0\\ h_{3}(z,0)=0\\ h_{3}(L,y)=f_{3}(y)\\ h_{3}(0,y)=0\end{array} \right. \tag{5}\]  
\[\left\{ \begin{array}{l}h_{4}(z,h_{0})=0\\ h_{4}(z,0)=0\\ h_{4}(L,y)=0\\ h_{4}(0,y)=f_{4}(y)\end{array} \right. \tag{6}\]For the four equations, the method of separation of variables can be used to solve them. This paper demonstrates the solution process using \(h_{1}(z,y)\) as an example. First, \(h_{1}(z,y)\) is decomposed into univariate functions of \(y\) and \(z\):  
\[h_{1}(z,y)=Z(z)\cdot Y(y) \tag{10}\]  
Substituting Equation (10) into the Laplace equation of Equation (10), we obtain  
\[Z(z)\cdot Y^{\prime\prime}(y)+Y(y)\cdot Z^{\prime\prime}(z)=0 \tag{11}\]  
Let \(Z^{\prime\prime}(z)/Z(z)\) and \(-Y^{\prime\prime}(y)/Y(y)\) be constants (denoted as \(\lambda\)). Thus, we obtain two ordinary differential equations:  
\[ \begin{cases}Z^{\prime\prime}(z)-\lambda Z(z)=0\\ Y^{\prime\prime}(y)+\lambda Y(y)=0\end{cases}  \tag{12}\]  
Given that \(Z(0)=Z(L)=Y(0)=0\), the general solution can only be  
\[\left\{ \begin{array}{l}Z(z)=A\cos(\sqrt{\lambda}y)+B\sin(\sqrt{\lambda}y)\\ Y(y)=C\exp(\sqrt{\lambda}z)+D\exp(-\sqrt{\lambda}z)\end{array} \right. \tag{13}\]  
Substituting the boundary conditions \(Z(0)=Z(L)=0\) gives \(A=0\), thus  
\[Z(z)=B\sin(\sqrt{\lambda}z)=0 \tag{14}\]  
Substituting the boundary condition \(Y(0)=0\) gives \(C=-D\), thus  
\[Y(y)=C(\mathrm{e}^{\sqrt{\lambda}y}-\mathrm{e}^{-\sqrt{\lambda}y})=C\mathrm{ sinh}(\sqrt{\lambda}y) \tag{15}\]  
To ensure the existence of a non-trivial solution, let \(B\
eq 0\); then, we have:  
\[\sqrt{\lambda}L=n\pi\;(n=1,2,3,\dots) \tag{16}\]  
Thus  
\[\lambda=(n\pi/L)^{2} \tag{17}\]  
Substituting the boundary condition \(h_{1}(z,h_{0})=f_{1}(z)\) and using Fourier series expansion, we obtain  
\[K_{A,n}=BD=\frac{2}{L\mathrm{sinh}(h_{0}\sqrt{\lambda})}\int_{0}^{L}f_{1}(z) \sin\left(z\sqrt{\lambda}\right)\mathrm{d}z \tag{18}\]  
Thus, we obtain \(h_{1}(z,y)\)  
\[h_{1}(z,y)=\sum\limits_{n=1}^{\infty}K_{A,n}\sin\left(\sqrt{\lambda_{n}}\cdot z \right)\mathrm{sinh}\left(\sqrt{\lambda_{n}}\cdot y\right) \tag{19}\]  
Similarly, we can obtain  
\[\left\{ \begin{array}{l}h_{2}\big{(}\ z,y\ \big{)}=\sum\limits_{n=1}^{\infty}K_{B,n} \sin\left(\sqrt{\lambda_{n}}z\right)\cdot\mathrm{sinh}\left[\sqrt{\lambda_{n} }(h_{0}-y)\right]\\ h_{3}\big{(}\ z,y\ \big{)}=\sum\limits_{n=1}^{\infty}K_{C,n}\sin\left[\frac{L}{h_{0}}\sqrt{ \lambda_{n}}y\right]\cdot\mathrm{sinh}\left[\frac{L}{h_{0}}\sqrt{\lambda_{n}} z\right]\\ h_{4}\big{(}\ z,y\ \big{)}=\sum\limits_{n=1}^{\infty}K_{D,n}\sin\left[\frac{L}{h_{0}}\sqrt{ \lambda_{n}}y\right]\cdot\mathrm{sinh}\left[\frac{L}{h_{0}}\sqrt{\lambda_{n}} (L-z)\right]\end{array} \right. \tag{20}\]  
where  
\[\left\{