# On the Lagged Diffusivity Method for the Solution of Nonlinear Finite Difference Systems
## 5 Numerical Experiments
### Effect of the Discretization Scheme  
Let us analyze the effect of the space discretization. In Table 1 we report the results obtained for different choices of \(\bar{\sigma}\) with \(u^{*}=u_{1}^{*}:=(1+x-y)^{3}t\). We set \(N=250\); with this choice, Equation (7) is satisfied for \(\bar{\sigma}_{1},\bar{\sigma}_{2}<200\). In Table 1, we see that the two discretization schemes are identical for \(\bar{\sigma}=\mathbf{0}\), also in terms of global errors and number of iterations. Indeed, in this case, the first-order derivative in the differential equation is multiplied by a null term. Thus, \(A(\mathbf{u})\) is symmetric positive definite and the two discretizations are equivalent. In the other cases, both methods always work properly when \(\bar{\sigma}\) respects the spacing condition, although global errors are larger when we use the upwind discretization. As mentioned above, this \begin{table}
\begin{tabular}{c c c c c c c c c} \hline \hline \(\bar{\sigma}\) & **Discretization** & **Lin. Solver** & \(res_{0}\) & \(res\) & \(err_{h}\) & \(err_{2}\) rel. & \(j^{*}\) & \(t_{tot}\) \\ \hline \multirow{4}{*}{\((0,0)^{T}\)} & \multirow{4}{*}{Central FD} & AM & 91,812 & \(1.37\times 10^{-5}\) & \(4.16\times 10^{-5}\) & \(1.97\times 10^{-5}\) & 7581 & 341.9 \\  & & BiCG(1) & 91,812 & \(1.36\times 10^{-5}\) & \(4.16\times 10^{-5}\) & \(1.97\times 10^{-5}\) & 1199 & 57.5 \\  & & AM & 91,812 & \(1.37\times 10^{-5}\) & \(4.16\times 10^{-5}\) & \(1.97\times 10^{-5}\) & 7581 & 356.5 \\  & & BiCG(1) & 91,812 & \(1.36\times 10^{-5}\) & \(4.16\times 10^{-5}\) & \(1.97\times 10^{-5}\) & 1199 & 64.0 \\ \hline \multirow{4}{*}{\((10,10)^{T}\)} & \multirow{4}{*}{Central FD} & AM & 91,814 & \(1.37\times 10^{-5}\) & \(3.63\times 10^{-5}\) & \(1.71\times 10^{-5}\) & 6608 & 311.0 \\  & & BiCG(1) & 91,814 & \(1.33\times 10^{-5}\) & \(3.61\times 10^{-5}\) & \(1.71\times 10^{-5}\) & 979 & 63.6 \\  & & AM & 92,397 & \(1.38\times 10^{-5}\) & \(2.25\times 10^{-3}\) & \(1.07\times 10^{-3}\) & 6710 & 407.9 \\  & & BiCG(1) & 92,397 & \(1.33\times 10^{-5}\) & \(2.25\times 10^{-3}\) & \(1.07\times 10^{-3}\) & 967 & 72.5 \\ \hline \multirow{4}{*}{\((300,300)^{T}\)} & \multirow{4}{*}{Central FD} & AM & 93,532 & \(1.38\times 10^{-5}\) & \(1.21\times 10^{-5}\) & \(5.71\times 10^{-6}\) & 594 & 37.5 \\  & & BiCG(1) & - & - & - & - & - & - \\ \cline{1-1}  & & AM & 110,834 & \(1.62\times 10^{-5}\) & \(9.07\times 10^{-3}\) & \(4.30\times 10^{-3}\) & 748 & 41.7 \\ \cline{1-1}  & & BiCG(1) & 110,834 & \(1.57\times 10^{-5}\) & \(9.08\times 10^{-3}\) & \(4.30\times 10^{-3}\) & 1534 & 79.7 \\ \hline \multirow{4}{*}{\((500,500)^{T}\)} & \multirow{4}{*}{Central FD} & AM & - & - & - & - & - & - \\ \cline{1-1}  & & BiCG(1) & - & - & - & - & - & - \\ \cline{1-1}  & & AM & 124,785 & \(1.71\times 10^{-5}\) & \(9.41\times 10^{-3}\) & \(4.46\times 10^{-3}\) & 552 & 42.2 \\ \cline{1-1}  & & BiCG(1) & 124,785 & \(1.47\times 10^{-5}\) & \(9.41\times 10^{-3}\) & \(4.46\times 10^{-3}\) & 1398 & 94.0 \\ \hline \hline \end{tabular}
\end{table} 
Table 1: Comparison between discretization by central finite differences and by upwind scheme. is to be attributed to the discretization of the first-order derivative by forward or backward finite difference quotients, leading to a larger discretization error. Lastly, when the spacing condition is not respected anymore, we can have problems in the solution of the linear system. Indeed, the convergence of the AM method requires that the coefficient matrix is an M-matrix. Although the AM still works for values of \(\bar{\sigma}\) for which this is not satisfied (e.g., \(\bar{\sigma}=(300,300)^{T}\)), we soon have problems for larger values of the velocity term. For example, \(\bar{\sigma}=(500,500)^{T}\) (and any other larger velocity term) leads the AM to stall. Also the BiCGstab(1) presents some problems and it already stalls for \(\bar{\sigma}=(300,300)^{T}\). Contrary to the AM, however, the BiCGstab(1) does not explicitly require that the coefficient matrix is an M-matrix. Its stalling is, however, not surprising.