## 3 Iterative Algorithm of Two-Dimensional Central Difference Galerkin Spectral Method
Therefore, it is easy to obtain the following according to the above conditions. The exact solution and approximate solution of Equation (1) on the Chebyshev-Gauss-Lobatto type collocation points can be replaced by sets \(\left\{u_{k}^{i}\right\}_{i=1}^{I}(1\leq k\leq M)\), \(\left\{u_{H_{k}}^{i}\right\}_{i=1}^{I}(1\leq k\leq G)\), and \(u^{i}=\left(u_{1}^{i},u_{2}^{i},\cdots,u_{G}^{i}\right)^{T}\), \(u_{H}^{i}=\left(u_{H_{1}}^{i},u_{H_{2}}^{i},\cdots,u_{H_{G}}^{i}\right)^{T}\), \(S^{i}=\left(S_{1}^{i},S_{2}^{i},\cdots,S_{G}^{i}\right)^{T}\), \(S_{1}=\left(S_{1_{1}},S_{1_{2}},\cdots,S_{1_{G}}\right)^{T}\), \(S_{2}=\left(S_{2_{1}},S_{2_{2}},\cdots,S_{2_{G}}\right)^{T}\). The following matrix relations can be obtained. \[\partial^{2}/\partial x^{2}= \begin{bmatrix}u_{H}^{i}(x_{0},y_{m})\\ u_{H}^{i}(x_{1},y_{m})\\ \vdots\\ u_{H}^{i}(x_{H},y_{m})\end{bmatrix} = \begin{bmatrix}F_{0}^{\prime\prime}(x_{0}) &F_{1}^{\prime\prime}(x_{0})&\cdots&F_{H}^{\prime\prime}(x_{0})\\ F_{0}^{\prime\prime}(x_{1})&F_{1}^{\prime\prime}(x_{1})&\cdots&F_{H}^{\prime \prime}(x_{1})\\ \vdots&\vdots&\ddots&\vdots\\ F_{0}^{\prime\prime}(x_{H})&F_{1}^{\prime\prime}(x_{H})&\cdots&F_{H}^{\prime \prime}(x_{H})\end{bmatrix}\begin{bmatrix}W_{0}^{i}(m)\\ W_{1}^{i}(m)\\ \vdots\\ W_{H}^{i}(m)\end{bmatrix}  \tag{16}\]  
\[\partial^{2}/\partial y^{2}= \begin{bmatrix}u_{H}^{i}(x_{n},y_{0})\\ u_{H}^{i}(x_{n},y_{1})\\ \vdots\\ u_{H}^{i}(x_{n},y_{H})\end{bmatrix} = \begin{bmatrix}F_{0}^{\prime\prime}(y_{0}) &F_{1}^{\prime\prime}(y_{0})&\cdots&F_{H}^{\prime\prime}(y_{0})\\ F_{0}^{\prime\prime}(y_{1})&F_{1}^{\prime\prime}(y_{1})&\cdots&F_{H}^{\prime \prime}(y_{1})\\ \vdots&\vdots&\ddots&\vdots\\ F_{0}^{\prime\prime}(y_{H})&F_{1}^{\prime\prime}(y_{H})&\cdots&F_{H}^{\prime \prime}(y_{H})\end{bmatrix}\begin{bmatrix}V_{0}^{i}(n)\\ V_{1}^{i}(n)\\ \vdots\\ V_{H}^{i}(n)\end{bmatrix}  \tag{17}\]  
\[ \begin{bmatrix}W_{0}^{i}(m)\\ W_{1}^{i}(m)\\ \vdots\\ W_{H}^{i}(m)\end{bmatrix} =Q \begin{bmatrix}u^{i}(x_{0},y_{m})\\ u^{i}(x_{1},y_{m})\\ \vdots\\ u^{i}(x_{H},y_{m})\end{bmatrix} , \begin{bmatrix}V_{0}^{i}(m)\\ V_{1}^{i}(m)\\ \vdots\\ V_{H}^{i}(m)\end{bmatrix} =Q \begin{bmatrix}u^{i}(x_{n},y_{0})\\ u^{i}(x_{n},y_{1})\\ \vdots\\ u^{i}(x_{n},y_{H})\end{bmatrix}  \tag{18}\]  
Written down as:  
\[\left\{F_{f}^{\prime\prime}(x_{n})\right\}_{f,n=0}^{H}=F_{x}^{T}P^{T},\left\{F _{g}^{\prime\prime}(y_{m})\right\}_{g,m=0}^{H}=F_{y}^{T}P^{T}\,P_{k}=\sum_{k=0}^{n-2}\frac{1}{b_{k}}k\left(k^{2}-n^{2}\right)\]  
The element of matrix \(F_{x}\) is \(F_{k}(x_{n})=\cos\frac{kn\pi}{H}\), and Matrix \(F_{x}\) can be expressed as follows:  
\[ \begin{bmatrix}1&1&\cdots&1&\cdots&1&1\\ 1&\cos\frac{\pi}{H}&\cdots&\cos\frac{k\pi}{H}&\cdots&\cos\frac{(H-1)\pi}{H}&-1 \\ \vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\ 1&\cos\frac{i\pi}{H}&\cdots&\cos\frac{i\pi}{H}&\cdots&\cos\frac{i(H-1)\pi}{H}& (-1)^{i}\\ \vdots&\vdots&\cdots&\vdots&\vdots&\vdots\\ 1&\cos\frac{(H-1)\pi}{H}&\cdots&\cos\frac{(H-1)k\pi}{H}&\cdots&\cos\frac{(H-1)^ {2}\pi}{H}&(-1)^{H-1}\\ 1&-1&\cdots&(-1)^{k}&\cdots&(-1)^{H-1}&(-1)^{H}\end{bmatrix} _{(H+1)\times(H+1)}\]  
Similarly, the matrix of \(F_{y}\) can also be expressed by the above formula. \[Q=\frac{2}{H}[Q_{0},Q_{1},Q_{2},\cdots Q_{H}]_{(H+1)\times(H+1)},\,Q_{0}=\frac{1}{2 }[\frac{1}{2},1,\cdots,\frac{1}{2}]^{T}\]  
\[Q_{H}=\frac{1}{2}[\frac{1}{2},-1,\cdots,(-1)^{i},\frac{1}{2}(-1)^{H}]^{T}\]  
\[P=[Q_{0},Q_{1},Q_{2},\cdots Q_{H-2},0,0]_{(H+1)\times(H+1)},\]  
\[P_{0}=[0,0,\frac{2^{3}}{2},0,\frac{4^{3}}{2},\cdots,0,\frac{H^{3}}{2}]^{T}\]  
\[P_{1}=[0,0,0,3^{3}-3,0,\cdots,(H-1)^{3}-(H-1),0]^{T}\]  
\[P_{H-2}=[0,0,0,0,0,\cdots,H^{3}-H(n-2)^{2}]^{T}\]  
From the matrix given above, it is not difficult to find the [PERSON] spectral method iterative format of the following temperature field \(u(x,y,t)\). \[\left\{ \begin{array}{l}\rho cu_{H}^{i+1}=\rho cu_{H}^{i-1}+2\Lambda t\mu Cu_{ H_{x}}^{i}+2\Lambda t\mu Du_{H_{y}}^{i}+2\Lambda tS^{i}\\ u_{H}^{0}=S_{1}\\ u_{H}^{i}(x_{0},y_{m})=u_{H}^{i}(x_{H},y_{m})=0\\ u_{H}^{i}(x_{n},y_{0})=u_{H}^{i}(x_{n},y_{H})=0\end{array} \right.