## 3 Numerical Results
As we have seen from Table 3, if we want to study a situation with a big attenuation (low \(Re\)), we will need a higher amount of time subintervals compared to those situations where the attenuation is lower (it means that the computational cost for having to solve more nonlinear systems will be higher), however taking more space subintervals does not seem to help. **Example 2. ** Let us consider now the following initial and boundary conditions for Burgers' equation  
\[u(x,0)=\left\{ \begin{array}{ll}\sin(\pi x),&0<x\leq 1\\ -\frac{1}{2}\sin(\pi x),&1<x\leq 2\\ 0&2<x\leq 5\end{array} \right. \tag{15}\]  
and \(u(0,t)=u(5,t)=0,t\geq 0\). In Figure 2 we see a plot of the approximated solution by our method at different times for \(0\leq x\leq 5\) and for two different values of \(\varepsilon\), and in Table 4 we show the numerical results of this solution for some values of \(x,t\) and \(\varepsilon\). \begin{table}
\begin{tabular}{c c c c c c c c} \hline \hline  & \(\varepsilon=1\) & \(\varepsilon=0.1\) & \(\varepsilon=0.05\) & \(\varepsilon=0.03\) & \(\varepsilon=0.02\) & \(\varepsilon=0.01\) & \(\varepsilon=0.005\) \\ \hline Greatest error & 0.22216 & 0.0068572 & 0.0035207 & 0.0021242 & 0.0014188 & 0.000708 & 0.000296 \\ Avg. iter. [PERSON] & 3.9 & 4 & 4 & 4 & 4 & 3.85 & 3 \\ Avg. iter. [PERSON] & 3.15 & 3 & 3 & 3 & 3 & 3 & 3 \\ Avg. iter. M5 & 2.8 & 3 & 3 & 3 & 3 & 3 & 2 \\ \hline \hline \end{tabular}
\end{table} 
Table 3: Comparison of the GE and methods iterations for different values of \(\varepsilon\). This results are very similar to those obtained for [PERSON] and [PERSON] on their paper (see [12]), where they transform [PERSON] equation into a system of nonlinear ordinary differential equations and solve it by Runge-Kutta-Chebyshev second order method. Although the techniques used to solve the problem are different, the results obtained by [PERSON] and ours, showed in Table 4 and Figure 2 have the same order of magnitude. As we can see in Figure 2 it happens indeed that the lower is \(\varepsilon\), slower is the attenuation. Although in this example we have used a piecewise initial condition, and it usually brings more problems, we still have obtained satisfactory results. Since we have considered a quite high upper bound of the studied time (\(t=10\)) we have had to take a bigger value of \(m\) in order to keep a fine \begin{table}
\begin{tabular}{c c c c} \hline \hline \(x\) & \(t\) & \(\varepsilon=0.1\) & \(\varepsilon=0.01\) \\ \hline
1.5 & 2 & 0.44533 & 0.63548 \\  & 4 & 0.28961 & 0.34573 \\  & 6 & 0.20701 & 0.23678 \\  & 8 & 0.16020 & 0.17999 \\  & 10 & 0.13041 & 0.14516 \\ \hline
3 & 2 & 0.02972 & 0.00000 \\  & 4 & 0.14900 & 0.00470 \\  & 6 & 0.22312 & 0.47269 \\  & 8 & 0.22501 & 0.35971 \\  & 10 & 0.20562 & 0.29021 \\ \hline
4.5 & 2 & 0.00001 & 0.00000 \\  & 4 & 0.00145 & 0.00000 \\  & 6 & 0.01171 & 0.00000 \\  & 8 & 0.03557 & 0.00000 \\  & 10 & 0.06242 & 0.02066 \\ \hline \hline \end{tabular}
\end{table} 
Table 4: Numerical results of the approximated solution for different values of \(x,t\) and \(\varepsilon\). Figure 2: Solution for different times and \(\varepsilon\) values, both of them with \(n=m=200\) and \(t_{max}=10\). (**a**) \(\varepsilon=0.1\); (**b**) \(\varepsilon=0.01\). mesh size. The results obtained in both Figure 2 and Table 4 are the same regardless if we use Newton', [PERSON]'s or M5 method.