## 2 The Calculation Theory of Deep Roadway Surrounding Rock Temperature Field
### Derivation of Finite Difference Equations  
#### 2.2.1 Derivation Based on the Principle of Taylor Series Expansion  
Polar coordinate transformation of the heat conduction differential equation. In polar coordinates, the direction along the centerline of the roadway is defined as the \(Z\)-axis, the direction perpendicular to the roadway axis pointing towards the deeper part of the surrounding rock mass is designated as the \(r\)-axis, and the circumferential direction of the roadway is defined as the \(\varphi\)-axis. Therefore:  
\[\frac{\partial T}{\partial\tau}=a\left(\frac{\partial^{2}T}{\partial r^{2}}+ \frac{1}{r}\frac{\partial T}{\partial r}+\frac{1}{r^{2}}\frac{\partial^{2}T} {\partial\varphi^{2}}+\frac{\partial^{2}T}{\partial\varepsilon^{2}}\right) \tag{3}\]  
Assuming the surrounding rock of the roadway is a circular axisymmetric figure in the surrounding rock mass, the axial temperature is the same at the same radius. Therefore, the temperature \(T\) is not a function of \(\varphi\), and the derivative of \(T\) with respect to \(\varphi\) is zero. \[\frac{\partial T}{\partial\tau}=a\left(\frac{\partial^{2}T}{\partial r^{2}}+ \frac{1}{r}\frac{\partial T}{\partial r}+\frac{\partial^{2}T}{\partial \varepsilon^{2}}\right) \tag{4}\]  
Substitute the boundary conditions:  
\[ \begin{array}{l}1\leq r\leq\infty\;T=1\\ 0\leq z\leq\Theta\;\Theta=0\end{array} \left(\text{At}\;\text{t}=0\right)\]  
\[ \begin{array}{l}z=0,\;\Theta=\Theta_{in}=0\;\frac{\partial T}{\partial z}=0 \\ r=\infty,\;T=1\\ r=1,\;\frac{\partial T}{\partial r}=Bi(T-\Theta)+q\end{array} \left(\text{At}\; \text{t}>0\right)\]  
where \(B_{i}=\alpha r_{0}/\lambda\); and \(B_{i}\) is the Biot number; \(\alpha\) is the convective heat transfer coefficient of the fluid; \(r_{0}\) is the characteristic length of the solid. Establish the difference equation. Translate the heat conduction differential equation into a difference equation. The one-dimensional unsteady heat conduction equation is as follows:  
\[\frac{\partial T}{\partial t}=a\frac{\partial^{2}T}{\partial y^{2}} \tag{5}\]  
Expand the temperature \(T\) at node \(i-1\) and \(i+1\) in the y-direction at time t using a Taylor series:  
\[T_{i+1,t}=T_{i,t}+\Delta y\left(\frac{\partial T}{\partial y}\right)_{i}+ \frac{\Delta y^{2}}{2!}\left(\frac{\partial^{2}T}{\partial y^{2}}\right)_{i}+ \frac{h^{3}}{3!}\left(\frac{\partial^{3}T}{\partial y^{3}}\right)_{i}+\dots\dots. \tag{6}\]  
\[T_{i-1,t}=T_{i,t}-\Delta y\left(\frac{\partial T}{\partial y}\right)_{i}+ \frac{\Delta y^{2}}{2!}\left(\frac{\partial^{2}T}{\partial y^{2}}\right)_{i}- \frac{h^{3}}{3!}\left(\frac{\partial^{3}T}{\partial y^{3}}\right)_{i}+\dots\dots. \tag{7}\]  
Since \(\Delta y\) is very small in the difference equation, we neglect terms of the third order and higher. Adding Equations (6) and (7):  
\[\left(\frac{\partial^{2}T}{\partial y^{2}}\right)_{i}=\frac{T_{i+1,t}+T_{i-1,t }-2T_{i,t}}{\Delta y^{2}} \tag{8}\]For the time difference:  
\[\left(\frac{\partial T}{\partial t}\right)_{i}=\frac{T_{i,t+\Delta t}-T_{i,t}}{ \Delta t} \tag{9}\]  
Therefore, we can obtain:  
\[T_{i,t+\Delta t}=T_{i,t}+a\cdot\Delta t\cdot\frac{T_{i+1,t}+T_{i-1,t}-2T_{i,t} }{\Delta y^{2}} \tag{10}\]  
To solve for the temperature field in the surrounding rock, we can use Equations (3), (9), and (10):  
\[T_{i,j}^{t+1}=T_{i,j}^{t}+\Delta\tau\left(\frac{\partial^{2}T}{\partial r^{2} }+\frac{1}{r}\frac{\partial T}{\partial r}+\frac{\partial^{2}T}{\partial z^{2 }}\right) \tag{11}\]  
Further, by differencing Equation (8), we obtain:  
\[