## 3 Results
### In the Two-Dimensional Case  
In this section, a two-dimensional forward-facing step case is used to demonstrate the performance of the optimized TDG method. According to the demonstration of time consuming in Section 3.1, the optimized TDG inside form method obtains better efficiency than the outside form. In two-dimensional test case, the optimized TDG inside form is employed. The main equations defining the shock problem are the Euler equations, which are expressed as follows:  
\[\frac{\partial\mathbf{U}}{\partial t}+\frac{\partial\mathbf{F}}{\partial x}+\frac{ \partial\mathbf{G}}{\partial y}=\mathbf{0}. \tag{39}\]  
where \(\mathbf{U}\) is the solution vector, and \(\mathbf{F},\mathbf{G}\) are the flux vectors. The pressure \(p\) can be defined using the two above vectors as follows:  
\[\mathbf{U}=[\rho,u,v,e]^{T}, \tag{40}\]  
\[\mathbf{F}=\mathbf{U}u+p[0,1,0,u]^{T};\mathbf{G}=\mathbf{U}v+p[0,0,1,v]^{T}, \tag{41}\]  
\[p=(\gamma-1)\bigg{(}e-\frac{u^{2}+v^{2}}{2\rho}\bigg{)}. \tag{42}\]  
where \(\rho\) represents density, \(u\) is velocity, and \(e\) denotes energy; \(\gamma\) is the heat capacity ratio, and in this study, it was set to 1.4. According to the definitions of \(\mathbf{U}\) and \(\mathbf{f}\), a Jacobian Matrix \(\mathbf{A}\) and \(\mathbf{B}\) can be expressed by:  
\[\mathbf{A}= \begin{bmatrix}0&1&0&0\\ (\gamma-1)\frac{u^{2}+v^{2}}{2}-u^{2}&(3-\gamma)u&(1-\gamma)v&\gamma-1\\ -uv&v&u&0\\ \Big{(}\frac{\gamma-1}{2}\Big{)}\big{(}u^{2}+v^{2}\big{)}u-Hu&H+(1-\gamma)u^ {2}&(1-\gamma)uv&\gamma u\end{bmatrix} , \tag{43}\]  
\[\mathbf{B}= \begin{bmatrix}0&0&1&0\\ -uv&v&u&0\\ (\gamma-1)\frac{u^{2}+v^{2}}{2}-v^{2}&(1-\gamma)v&(3-\gamma)v&\gamma-1\\ \Big{(}\frac{\gamma-1}{2}\Big{)}\big{(}u^{2}+v^{2}\big{)}v-Hv&(1-\gamma)uv&H+( 1-\gamma)v^{2}&\gamma v\end{bmatrix} . \tag{44}\]  
Here \(H=\mathrm{e}+p/\rho\), and the temporal derivative of H is:  
\[\frac{dH}{dt}=\frac{de}{dt}+\frac{\frac{dp}{dt}\rho-\frac{d\rho}{dt}p}{\rho^{ 2}}. \tag{45}\]  
Then, the temporal derivative of Jacobian Matrix \(\partial\mathbf{A}/\partial t\) and \(\partial\mathbf{B}/\partial t\) can be calculated as: \begin{table}
\begin{tabular}{c c c c} \hline \hline  & \multicolumn{2}{c}{**Conventional TDG**} & \multicolumn{2}{c}{**Optimized TDG**} \\ \cline{2-4}  & & **Inside Form** & **Outside Form** \\ \hline \(N=3\) & 192.5 s & 18.6 s & 20.3 s \\ \(N=4\) & 346.1 s & 25.5 s & 26.3 s \\ \(N=5\) & 517.7 s & 42.6 s & 43.9 s \\ \(N=6\) & 757.5 s & 65.7 s & 67.8 s \\ \hline \hline \end{tabular}
\end{table} 
Table 2: The time (seconds) required by the three methods to solve the one-dimensional Euler equations at \(K=200\); \(N\) is the order of polynomials. \[\frac{\partial\mathbf{A}}{\partial t}= \begin{bmatrix}0&0&0&0&0\\ (\gamma-1)\Big{(}u\frac{du}{dt}+v\frac{dv}{dt}\Big{)}-2u\frac{du}{dt}&(3-\gamma) \frac{du}{dt}&(1-\gamma)\frac{dv}{dt}&0\\ -\Big{(}\frac{du}{dt}v+u\frac{dv}{dt}\Big{)}&\frac{dv}{dt}&\frac{du}{dt}&0\\ \Big{(}\frac{\gamma-1}{2}\Big{)}2\Big{(}u\frac{du}{dt}+v\frac{dv}{dt}\Big{)}u +(u^{2}+v^{2})\frac{du}{dt}-\frac{dH}{dt}u-H\frac{du}{dt}&\frac{dH}{dt}+2(1- \gamma)\frac{dv}{dt}&(1-\gamma)\Big{(}\frac{du}{dv}v+u\frac{dv}{dt}\Big{)}& \gamma\frac{du}{dt}\\ \end{bmatrix} , \tag{46}\] \[\frac{\partial\mathbf{B}}{\partial t}= \begin{bmatrix}0&0&0&0\\ -\Big{(}\frac{du}{dt}v+u\frac{dv}{dt}\Big{)}&\frac{dv}{dt}&\frac{du}{dt}&0\\ (\gamma-1)\Big{(}u\frac{du}{dt}+v\frac{dv}{dt}\Big{)}-2v\frac{dv}{dt}&(1- \gamma)\frac{dv}{dt}&(3-\gamma)\frac{dv}{dt}&0\\ \Big{(}\frac{\gamma-1}{2}\Big{)}2\Big{(}u\frac{du}{dt}+v\frac{dv}{dt}\Big{)}v +(u^{2}+v^{2})\frac{dv}{dt}-\frac{dH}{dt}v-H\frac{dv}{dt}&(1-\gamma)\Big{(} \frac{du}{dt}v+u\frac{dv}{dt}\Big{)}&\frac{dH}{dt}+2(1-\gamma)\frac{dv}{dt}& \gamma\frac{dv}{dt}\\ \end{bmatrix} . \tag{47}\]  
The computational domain is illustrated in Figure 5. The inflow boundary condition is defined as a uniform Mach 3 flow, while no specific boundary conditions are applied at the outflow.