# Full Information Numerical Simulation of Two-Dimensional Steady Ground Temperature Field
## 2 Principle of Methods
### Holographic Fourier Transform Principle  
One-dimensional Fourier transforms can be expressed as [13]  
\[\left\{\begin{array}{l}F(k_{x})=\int_{-\infty}^{\infty}f(x)e^{-ik_{x}x}dx\\ f(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}F(k_{x})e^{ik_{x}x}dk_{x}\end{array} \right.. \tag{4}\]  
where \(k_{x}\) represents wavenumber in the \(x\) direction, \(f(x)\) is spatial domain function, and \(F(k_{x})\) is wavenumber spectrum. The positive transform integral in Equation (4) is discretized to obtain  
\[F(k_{x})=\sum_{j=1}^{M}\int\limits_{x_{1}}^{x_{j+2}}f(x)e^{-ik_{x}x_{j}}dx. \tag{5}\]  
where \(M\) represents the number of units, and \(j\) represents the unit \(j\).  
When the quadratic interpolation form function is used in the unit to fit, the coordinates of the three nodes inside any unit are \(x_{1},x_{2},x_{3}\), respectively, \(x_{1}+x_{3}=2x_{2}\), the midpoint, satisfied, the nodes in the unit are shown in Figure 1[14]:  
Figure 1: Schematic diagram of element and nodes in the element.  
The values on each node are respectively \(f(x_{1}),f(x_{2}),f(x_{3})\), and \(f(x)\) can be obtained by using a quadratic function  
\[f(x)=N_{1}f(x_{1})+N_{2}f(x_{2})+N_{3}f(x_{3}). \tag{6}\]  
where,  
\[N_{1}=\frac{(x-x_{2})(x-x_{3})}{(x_{1}-x_{2})(x_{1}-x_{3})},\ N_{2}=\frac{(x-x_ {1})(x-x_{3})}{(x_{2}-x_{1})(x_{2}-x_{3})},\ N_{3}=\frac{(x-x_{1})(x-x_{2})}{(x _{3}-x_{1})(x_{3}-x_{2})}. \tag{7}\]  
Thus, Formula (7) can be written as  
\[F(k_{x})=\sum_{j=1}^{M}\big{[}\int_{x_{1}}^{x_{3}}N_{1}f(x_{1})e^{-ik_{x}x}dx+ \int_{x_{1}}^{x_{3}}N_{2}f(x_{2})e^{-ik_{x}x}dx+\int_{x_{1}}^{x_{3}}N_{3}f(x_{3} )e^{-ik_{x}x}dx\big{]}. \tag{8}\]  
Let \(W_{1}=\int_{x_{1}}^{x_{3}}N_{1}e^{-ik_{x}x}dx\), \(W_{2}=\int_{x_{1}}^{x_{3}}N_{2}e^{-ik_{x}x}dx\), \(W_{3}=\int_{x_{1}}^{x_{3}}N_{3}e^{-ik_{x}x}dx\) is the Fourier transform node coefficient within the unit, then Equation (8) can be abbreviated as  
\[F(k_{x})=\sum_{j=1}^{M}[f(x_{1})W_{1}+f(x_{2})W_{2}+f(x_{3})W_{3}]. \tag{9}\]  
When the wavenumber is not 0, the Formula (9) is substituted into \(W_{1},W_{2},W_{3}\), the internal Fourier transform node coefficient of the unit can be obtained as  
\[\begin{split} W_{1}&=\frac{1}{(x_{1}-x_{3})^{2}}[2 \int_{x_{1}}^{x_{3}}x^{2}e^{-ik_{x}x}dx-(x_{1}+3x_{3})\int_{x_{1}}^{x_{3}}xe^{ -ik_{x}x}dx+(x_{1}+x_{3})x_{3}\int_{x_{1}}^{x_{3}}e^{-ik_{x}x}dx]\\ W_{2}&=-\frac{4}{(x_{1}-x_{3})^{2}}[\int_{x_{1}}^{x _{3}}x^{2}e^{-ik_{x}x}dx-(x_{1}+x_{3})\int_{x_{1}}^{x_{3}}xe^{-ik_{x}x}dx+x_{1 }x_{3}\int_{x_{1}}^{x_{3}}e^{-ik_{x}x}dx]\\ W_{3}&=\frac{1}{(x_{1}-x_{3})^{2}}[2\int_{x_{1}}^{x _{3}}x^{2}e^{-ik_{x}x}dx-(3x_{1}+x_{3})\int_{x_{1}}^{x_{3}}xe^{-ik_{x}x}dx+(x_{ 1}+x_{3})x_{1}\int_{x_{1}}^{x_{3}}e^{-ik_{x}x}dx]\\ \end{split} \tag{10}\]  
When the wavenumber is 0, \(W_{1}^{0}=\int_{x_{1}}^{x_{3}}N_{1}dx\), \(W_{2}^{0}=\int_{x_{1}}^{x_{3}}N_{2}dx\), \(W_{3}^{0}=\int_{x_{1}}^{x_{3}}N_{3}dx\) can be integrated to obtain the Fourier transform node coefficient under zero wavenumber as  
\[W_{1}=\frac{1}{6}(x_{3}-x_{1}),W_{2}=\frac{2}{3}(x_{3}-x_{1}),W_{3}=\frac{1}{6 }(x_{3}-x_{1}). \tag{11}\]  
The final one-dimensional positive Fourier transform result can be obtained by summating the analytic expressions of different elements. In addition, when the division of spatial domain and frequency domain is unchanged, the Fourier transform node coefficients \(W_{1}\) and \(W_{2}\) also remain unchanged. Calculating and storing the Fourier transform coefficients in advance can reduce repeated calculation and improve the efficiency of the algorithm, which is also one of the advantages of this algorithm.  
In addition, since the form of the forward and inverse transform is the same, the final expression result of the Fourier transform is similar to the expression of the forward transform, so it is not necessary to go into details.