## 3 Numerical Results
In Table 2, we repeat the same process but in this case we set the number of space nodes and try different amount of time nodes. So \(t_{max}=1\), \(n=100\) and \(\upvarepsilon=\frac{1}{Re}=0.1\). As we can observe, at first it seems that the more time subintervals we take, the less is GE (similarly as it happened with the previous table, where we tried different amount of space subintervals) however there is a limit for this which if we exceed the error starts to increase again. In this case, the minimum error is obtained for \(m=20\). In Table 3, we repeat the process again, however now we study the effect of choosing a small, regular or big value of the Reynolds constant (\(Re\)), although we actually use \(\varepsilon=\frac{1}{Re}\) for simplicity. For doing \begin{table}
\begin{tabular}{c c c c c c} \hline  & \(\mathbf{n=10}\) & \(\mathbf{n=20}\) & \(\mathbf{n=40}\) & \(\mathbf{n=80}\) & \(\mathbf{n=160}\) \\ \hline Greatest error & 0.09033 & 0.029932 & 0.0070658 & 0.0017149 & 0.00039376 \\ Avg. iter. [PERSON] & 4 & 4 & 4 & 4 \\ Avg. iter. [PERSON] & 3 & 3 & 3 & 3 \\ Avg. iter. M5 & 3 & 3 & 3 & 3 \\ \hline \end{tabular}
\end{table} 
Table 1: Comparison of the GE (greatest error) and methods iterations for different amount of space subintervals. \begin{table}
\begin{tabular}{c c c c c c} \hline  & \(\mathbf{m=10}\) & \(\mathbf{m=20}\) & \(\mathbf{m=40}\) & \(\mathbf{m=80}\) & \(\mathbf{m=160}\) \\ \hline Greatest error & 0.0043069 & 0.00015291 & 0.00083542 & 0.0010565 & 0.0011128 \\ Avg. iter. Newton & 4.4 & 4.15 & 4 & 4 & 4 \\ Avg. iter. [PERSON] & 3.4 & 3.1 & 3 & 3 & 3 \\ Avg. iter. M5 & 3 & 3 & 3 & 3 & 3 \\ \hline \end{tabular}
\end{table} 
Table 2: Comparison of the GE and methods iterations for different amount of time subintervals. Figure 1: Solution (**a–c**) and error progression curve (**d–f**) for different times and \(\varepsilon\) values, with \(n=m=40\) and \(t_{max}=1\). this, we set the number of space and time nodes and the maximum time of study, so \(t_{max}=1\), \(n=40\) and \(m=40\). First of all, it has to be said that the maximum value of \(\varepsilon\) that can be used to keep the method working properly will depend of the amount of time subintervals we take, it happens because the bigger is the attenuation, more abrupt are the curves on the time axis. For keeping the error low, a finer mesh will be needed, what can be obtained by increasing the amount of time subintervals. From Table 3 we can numerically see this phenomenon, for a fast attenuation (\(\varepsilon=1\)) the error is quite big due to the curve is very abrupt (as we said we could decrease the error of this case by taking more time subintervals), however the slower is the attenuation (lower values of \(\varepsilon\)), lower is GE. If we keep on taking lower values of attenuation (lower \(\varepsilon\)), the peak of the GE characteristic curve will not be reached and consequently the GE will be even lower (for this example it happens for \(\varepsilon=0.01\) and even more for \(\varepsilon=0.005\)). See Figure 1 for a better understanding of these different situations. Some conclusions can be obtained from Tables 1 to 3: to use a higher order method (like M5) does not represent a bigger advantage compared with a typical third order method (like Traub), however there is a significant difference between using a third order method compared to a typical second order method (like Newton). The more space subintervals we take, lower is GE; however, it does not happen with the amount of time subintervals, it means that there is an optimum amount of time subintervals that reduces the GE to its minimum but after it the error starts to increase again and the computational cost of having to solve so many systems does not make any sense (due to the increasing number of floating points operations, execution time,...).