## 6 Numerical Experiments
By using the L1 formula on non-uniform mesh, we improvde the order of accuracy in time of the proposed method. In fact, we observe that the solutions are more accurate on the non-uniform mesh and the convergence order on the non-uniform mesh is greater than the convergence order obtained with the uniform mesh. The numerical results agree well with the theoretical results. **Example 2**. : _We consider the following TFADE_  
\[ \begin{array}{llll}\frac{\partial^{\alpha}}{\partial t^{\alpha}}u(x,t)+ \frac{\partial}{\partial x}u(x,t)-\frac{\partial^{2}}{\partial x^{2}}u(x,t)=f (x,t),&0<x<1,&0<t\leq T,\\ \\ u(x,0)=0,&0\leq x\leq 1,\\ u(0,t)=0,&u(1,t)=t^{3},&0<t\leq T,\end{array}  \tag{24}\]  
_where the source term is given by_  
\[f(x,t)=\frac{6}{\Gamma(4-\alpha)}x^{2}t^{3-\alpha}+2t^{3}(x-1)\]  
_and the exact solution is_  
\[u(x,t)=x^{2}t^{3}.\]  
We took \(K_{1}=K_{2}=1\) and set \(\alpha=0.1\). Figure 3 shows the comparison between the numerical solution, \(U_{j}^{n}\), and the exact solution, \(u(x_{j},t^{n})\), at different times computed with \(N=J=20\). The exact \begin{table}
\begin{tabular}{c c c c c} \hline \(\alpha\) & \(N\) & \(\varepsilon_{\infty}\) & \(Order\) & \(\varepsilon_{\infty}\) & \(Order_{Q}\) \\ \hline \multirow{4}{*}{0.1} & 10 & \(3.6363\times 10^{-4}\) & & \(1.3741\times 10^{-3}\) & \\  & 20 & \(9.1021\times 10^{-5}\) & \(1.9982\) & \(4.3575\times 10^{-4}\) & \(1.6570\) \\  & 40 & \(2.2054\times 10^{-5}\) & \(2.0452\) & \(1.3222\times 10^{-4}\) & \(1.7206\) \\  & 80 & \(4.4649\times 10^{-6}\) & \(2.3043\) & \(3.8355\times 10^{-5}\) & \(1.7854\) \\ \hline \multirow{4}{*}{0.5} & 10 & \(5.5793\times 10^{-3}\) & & \(1.9875\times 10^{-2}\) & \\  & 20 & \(1.7121\times 10^{-3}\) & \(1.7044\) & \(7.7106\times 10^{-3}\) & \(1.3660\) \\  & 40 & \(5.4236\times 10^{-4}\) & \(1.6584\) & \(2.8895\times 10^{-3}\) & \(1.4160\) \\  & 80 & \(1.7544\times 10^{-4}\) & \(1.6283\) & \(1.0602\times 10^{-3}\) & \(1.4465\) \\ \hline \multirow{4}{*}{0.9} & 10 & \(4.6556\times 10^{-2}\) & & \(1.0115\times 10^{-1}\) & \\  & 20 & \(2.1358\times 10^{-2}\) & \(1.1242\) & \(4.9512\times 10^{-2}\) & \(1.0307\) \\ \cline{1-1}  & 40 & \(9.8735\times 10^{-3}\) & \(1.1131\) & \(2.3691\times 10^{-2}\) & \(1.0634\) \\ \cline{1-1}  & 80 & \(4.5790\times 10^{-3}\) & \(1.1085\) & \(1.1201\times 10^{-2}\) & \(1.0807\) \\ \hline \end{tabular}
\end{table} 
Table 1: \(e_{\infty}\), \(\varepsilon_{\infty}\) and convergence orders, related to the numerical solutions \(U_{j}^{n}\) and \(\bar{U}_{j}^{n}\) respectively, for different values of \(N\) and \(\alpha\), with \(J=100\). solution is reported with the solid line. The numerical solution \(U_{j}^{m}\) is in good agreement with the exact solution \(u(x_{j},t^{m})\). We report only the numerical results obtained for the value of the parameter, \(\alpha=0.1\). Analogous results were obtained for \(0.1<\alpha\leq 0.9\). The values of \(e_{\infty}\) and \(\bar{e}_{\infty}\) and the corresponding numerical convergence orders obtained for \(\alpha=0.1,0.5\) and \(0.9\) are reported in Table 2. The results confirm that the method is stable and convergent when solving problems (24) on both the computational grids. The numerical solutions computed on non-uniform mesh were more accurate and the convergence order on the non-uniform mesh was greater than the convergence order obtained with the uniform mesh. The errors \(e_{\infty}\) and \(\bar{e}_{\infty}\) satisfied the relationships (22). The convergence orders, \(Order\) and \(Order_{\alpha}\), satisfied the relationships (23). **Example 3**. : _In this test, in order to show the efficiency of the method, we solve the following TFADE_  
\[ \begin{array}{l}\frac{\partial^{2}}{\partial x^{2}}u(x,t)+K_{1}\frac{ \partial}{\partial x}u(x,t)-K_{2}\frac{\partial^{2}}{\partial x^{2}}u(x,t)=f(u (x,t)),\\ \\ u(x,0)=x^{2}(5-x)^{2},\qquad a\leq x\leq b,\\ u(a,t)=u(b,t)=0,\qquad 0<t\leq T,\end{array}  \tag{25}\]