Picard’s iterative method for nonlinear multicomponent transport equations

Figures & data

Figure 1. The two-level Picard’s method compared with the analytical solution.

Table 1. Comparative results of the two-level method with k iterative steps and the analytical solution for $\mathrm{\Delta }t={10}^{-2}$ and $T=1.0$

Number of iterations	$\mathrm{\Delta }t$	$er{r}_{L1}$	comp. Time
K = 1	${10}^{-1}$	$3.33\times {10}^{-2}$	$2.3611\times {10}^{-4}$
K = 2	${10}^{-1}$	$9.0912\times {10}^{-4}$	$3.1310\times {10}^{-4}$
K = 3	${10}^{-1}$	$1.1806\times {10}^{-5}$	$3.7982\times {10}^{-4}$
K = 4	${10}^{-1}$	$1.5768\times {10}^{-5}$	$4.3628\times {10}^{-4}$
K = 5	${10}^{-1}$	$1.4943\times {10}^{-5}$	$4.8248\times {10}^{-4}$
K = 6	${10}^{-1}$	$1.4968\times {10}^{-5}$	$5.2867\times {10}^{-4}$
K = 1	${10}^{-2}$	$3.33\times {10}^{-2}$	$2.3\times {10}^{-3}$
K = 2	${10}^{-2}$	$8.7588\times {10}^{-5}$	$3.1\times {10}^{-3}$
K = 3	${10}^{-2}$	$1.0439\times {10}^{-7}$	$3.8\times {10}^{-3}$
K = 4	${10}^{-2}$	$1.5461\times {10}^{-7}$	$4.5\times {10}^{-3}$
K = 5	${10}^{-2}$	$1.5385\times {10}^{-7}$	$4.8\times {10}^{-3}$
K = 6	${10}^{-2}$	$1.5385\times {10}^{-7}$	$5.3\times {10}^{-3}$
K = 1	${10}^{-3}$	$3.33\times {10}^{-2}$	$2.3611\times {10}^{-4}$
K = 2	${10}^{-3}$	$8.7256\times {10}^{-6}$	$3.1310\times {10}^{-4}$
K = 3	${10}^{-3}$	$1.0303\times {10}^{-9}$	$3.7982\times {10}^{-4}$
K = 4	${10}^{-3}$	$1.5435\times {10}^{-9}$	$4.3628\times {10}^{-4}$
K = 5	${10}^{-3}$	$1.5427\times {10}^{-9}$	$4.8248\times {10}^{-4}$
K = 6	${10}^{-3}$	$1.5427\times {10}^{-9}$	$5.2867\times {10}^{-4}$

Figure 2. The three-level Picard’s method compared with the analytical solution.

Table 2. Comparative results of the three-level method with k iterative steps and the analytical solution for $\mathrm{\Delta }t={10}^{-2}$ and $T=1.0$

Three-Level method	Accuracy $er{r}_{L_1}$	CPU-time (sec)
$k=1$	${10}^{-4}$	$8\times {10}^{-2}$
$k=2$	${10}^{-1}$	$3\times {10}^{-2}$
$k=3$	${10}^{-1}$	$3\times {10}^{-2}$
$k=4$	${10}^{-1}$	$3\times {10}^{-2}$
$k=5$	${10}^{-1}$	$3\times {10}^{-2}$

Figure 3. The two-level Picard’s method compared with the analytical solution.

Figure 4. The three-level Picard’s method compared with the analytical solution.

Figure 5. The three-level Picard’s method compared with the analytical solution (left side: solutions, right side: errors between analytical and numerical methods).

Figure 6. The numerical error of general Picard’s fix-point method and a reference solution (fourth-order RK-method with fine time steps).

Figure 7. The numerical solution of the three- and four-level Picard’s fix-point methods with different iterative steps.

Figure 8. The figures present the numerical $L_1$-error between the four-level Picard’s fix-point scheme and the reference solution for different iterative steps (left figure: high resolution of the error, right figure: low resolution based on averaging the high oscillating errors).

Figure 9. The figures present the numerical solutions of the uphill experiment (left side) and the concentrations at spatial point $x=0.72$ with a four-level Picard’s fix-point scheme computed (right side).