5.3 Numerical Test Problems
5.3.2 Two-component aggregation
100 105 10−20
10−15 10−10 10−5 100
volume
number density
τ=0 τ=1.0 τ=3.2
10−2 10−1 100 101
10−2 10−1 100 101 102 103
time M i(t)/M i(0)
M0 M1 M2
Figure 5.6: Test problem 4: One-component aggregation with Gaussian-type initial distri-bution forN = 200.
where τ = N0β0x0t, and α = 1−exp(−x0τ). The exact values of moments are given in Table 5.1. Figure 5.6 give the average number density at τ = 0,1.0,3.2 and the moments until τ = 3.2. The numerical results are again in good agreement with the analytical results. Nevertheless, there is an observable under estimation in the front region of the solution at τ = 3.2. The two time instants correspond to a degree of aggregation of 63%
and 96%. If we compare the results of the current scheme with those in [46], the current scheme give comparable results to the improved scheme in [46] and better than the scheme in [47]. The first three numerical moments are also comparable to the analytical ones.
Here τ =β0N0t and I0 is the modified Bessel function of first kind of order zero. For this solution, we have the analytical solution for the following moments
M0,0(τ) = 2N0/(2 +τ), M1,0(τ) = x0N0, M1,1(τ) =x0y0N0(1 +τ). (5.88) For the numerical calculations we take N0 = 1, x0 = 1 and y0 = 1. The numerical results are shown in Figure 5.7. The first plot show the comparison of analytical and numerical results which are plotted along the diagonal for three dimensionless times τ = 5,20,100.
The three time instants correspond to a degree of aggregation of 71, 91, and 98%, respec-tively. In this log-log plot the abscissa axis represents the variablex. The numerical results are in good agreement with the analytical ones. However, one can see the over estimation in the result of τ = 100. The current test problem is an extension of the Test Problem 1 where one-component aggregation was considered. If we compare the results of Figures 5.3 and 5.7, one can see that in both cases the overall behavior of the results are the same.
However, the two-component results seems even better than the one-component case. This shows that, the current finite volume scheme maintains its accuracy when extended to two-component aggregation case. The second plot in Figure 5.7 show that the numerical moments are in excellent agreement with the analytical ones. The last plot is the three-dimensional mesh plot of the number density at τ = 100 which we have obtained from the same finite volume scheme. The CPU time of the current scheme att = 100 is 4.2 minutes.
According to our expectations, it was found found that our scheme in two-component ag-gregation case is also second order accurate and the order of convergence remains around 2.
Test problem 6: The initial data are f(0, x, y) = 16N0
x0y0
x x0
y y0
exp
−2x x0 − 2y
y0
. (5.89)
The analytical solution is given as [25]
f(t, x, y) = 8N0
x0y0
pτ(τ+ 2)3 exp
−2x x0 −2y
y0
[I0(θ)−J0(θ)] , (5.90) where
θ= 4 xy
x0y0
12 τ τ + 2
14
. (5.91)
Hereτ =β0N0tandJ0andI0 are, respectively, the Bessel function and the modified Bessel functions of first kind of order zero. For the numerical calculations we take N0 = 1, x0 = 1 and y0 = 1. The exact moments for this problem are given as [118]:
M0,0(τ) = 2N0/(2 +τ), M1,0(τ) = x0N0, M2,0(τ) = 1
2x20N0(3 + 2τ), (5.92) M1,1(τ) =x0y0N0(1 +τ), M3,0(τ) = 3
2x30N0(1 +τ)(2 +τ), (5.93) M2,1(τ) = 1
2x20y0N0(3 + 7τ + 3τ2). (5.94)
10−1 100 101 102 103 104 10−20
10−15 10−10 10−5 100
x−component
number density
τ=0 τ=5 τ=20 τ=100
100 101 102
10−2 10−1 100 101 102 103
time M i,j(t)/M i,j(0)
M0,0
M1,0 M1,1
100
102 100
102 10−20
10−10
x−component y−component
number density
Figure 5.7: Test problem 5: Results of two-component aggregation.
10−2 100 102 104 10−20
10−15 10−10 10−5 100
x−component
number density
τ=0 τ=5 τ=20 τ=100
100 101 102
10−2 10−1 100 101 102 103 104
time M i,j(t)/M i,j(0)
M0,0 M1,0 M2,0 M1,1 M3,0 M1,2
100
102 100
102 10−20
10−10
x−component y−component
number density
Figure 5.8: Test problem 6: Results of two-component aggregation.
This test problem was considered in [118]. The numerical results are shown in Figure 5.8.
The first plot shows the comparison of analytical and numerical results which are plotted along the diagonal for three dimensionless times τ = 5,20,100. The three time instants correspond to a degree of aggregation of 71, 91, and 98%, respectively. In this log-log plot the abscissa axis represents the variable x. The numerical results are again comparable with the analytical ones. If we compare our numerical results with those in [118], it is seems that our results are better for simulation timesτ = 5,20. For these simulation times there is a visible over estimation in the results of the scheme in [118], while our results are in excellent agreement with analytical results. Although, there is a very slight under estimation in our result for τ = 100, but this under estimation seems to be less than the over estimation in the results of the scheme in [118] for the same time. The second plot in Figure 5.8 show that the numerical moments are in good agreement with the analytical solutions. However, the momentM30 has a visible mismatch with analytical results which is also visible in the results given in [118]. The last plot is the three-dimensional mesh plot of the number density from our scheme at τ = 100. The CPU time of the current scheme att = 100 is 4.5 minutes.
Test problem 7: The initial data are f(0, x, y) = 4N0
x0y0
x x0
exp
−2x x0 − y
y0
. (5.95)
The analytical solution is given as [25]
f(t, x, y) = 16N0
x0y0(τ+ 2)2exp
−2x x0 − y
y0
X∞
k=0
4τ y (τ+2)y0
k
x x0
2k+1
k!(2k+ 1)! . (5.96) Hereτ =β0N0t. For the numerical calculations we take N0 = 1, x0 = 1 and y0 = 1. The exact moments for this problem are also the same as given in (5.92)-(5.94).
This test problem was also considered in [118]. The numerical results are shown in Fig-ure 5.9. The first plot show the comparison of analytical and numerical results which are plotted along the diagonal for three dimensionless times τ = 5,20,100. The three time instants correspond to a degree of aggregation of 71, 91, and 98%, respectively. In this log-log plot the abscissa axis represents the variable x. The numerical results are again quite comparable with the analytical ones.
The comparison of our numerical results with those in [118] show that, both schemes give comparable solutions for this test problem. However, our results shows a small under es-timation while the results in [118] show over eses-timation and that is the main difference in both results.
The second plot in Figure 5.9 show that the numerical moments are in good agreement with the analytical solutions. The moments plot of our scheme and the scheme in [118]
10−2 100 102 104 10−20
10−15 10−10 10−5 100
x−component
number density
τ=0 τ=5 τ=20 τ=100
100 101 102
10−2 10−1 100 101 102 103 104
time M i,j(t)/M i,j(0)
M0,0 M1,0 M2,0 M1,1 M3,0 M1,2
100
102 100
102 10−20 10−10
x−component y−component
number density
Figure 5.9: Test problem 7: Results of two-component aggregation.
give similar trends. Again, the main difference is the under estimation in our results and over estimation in the results of the scheme in [118]. The last plot is the three-dimensional mesh plot of the number density from our scheme atτ = 100. The CPU time of the current scheme att = 100 is 13.5 minutes.
Test problem 8: The initial data for this problem is the same as in Test problem 5.
However instead of the constant kernel, here we consider sum kernel β(t, x, y, x′, y′) = β0(x+y+x′ +y′). As mentioned before, in the two-component case we have no exact solutions for other than a constant kernel. Here, we want to show that the current scheme can also be used for other kernels as well. The first plot in Figure 5.10 give the average number density from our scheme at τ = 0.2,0.4,1.0 which are plotted along the diagonal.
The second plot is a three dimensional plot of number density at t= 1.0 from our scheme.
10−1 100 101 102 103 104
10−20 10−15 10−10 10−5 100
x−component
number density
τ=0 τ=0.2 τ=0.4 τ=1.0
100
102
100 102
10−20 10−10
x−component y−component
nuumber density
Figure 5.10: Test problem 8: Results of two-component aggregation.