Page 62 - IJOCTA-15-4

MN. Khan et.al. / IJOCTA, Vol.15, No.4, pp.594-609 (2025)

−5
x 10 sharp gradients or discontinuities, which often
1.035
challenge numerical methods. The Haar wavelet
1.0349
basis, with its compact support and localized
1.0348
resolution, is particularly effective in capturing
1.0347
these smoother variations. Additionally, the non-
1.0346
L ∞ 1.0345 local nature of the boundary conditions, com-
bined with negative parameter values, tends to
1.0344
1.0343 suppress high-frequency components, thereby en-
1.0342 hancing the accuracy and stability of the Haar
wavelet collocation method in such scenarios.
1.0341
1.034
0 50 100 150 200 250 300
M Table 4. The numerical results generated using
Haar wavelets for Test Problem 4
Figure 5. Comparison of the L ∞ error norm versus
M using Haar wavelets for Test Problem 3 ϑ = −10 ϑ = 0
α L ∞ κ L ∞ κ
−10 2.6598e − 05 5.3441e + 08 3.8218e − 05 5.3696e + 08
−03 3.3559e − 05 5.3452e + 08 2.7566e − 05 5.3271e + 08
Problem 4. Consider the steady-state problem of
−02 3.5066e − 05 5.3504e + 08 2.5162e − 05 5.3221e + 08
Equation (32), where the functions are specified as 00 3.8785e − 05 5.3660e + 08 2.8558e − 05 5.3133e + 08
follows: 02 4.3805e − 05 5.3893e + 08 6.8019e − 05 5.3597e + 08
03 4.7036e − 05 5.4041e + 08 3.9434e − 04 5.3866e + 08
f(κ, y) = 2e κ+y , 10 2.2494e − 04 5.5670e + 08 2.9504e − 04 5.6261e + 08
1
Z
y
ℑ 1 (y) = e − ϑ(κ)e κ+y dκ, Table 5. The numerical results generated using
0 Haar wavelets for Test Problem 4
1 (60)
Z
ℑ 2 (y) = e y+1 − α(κ)e κ+y dκ, ϑ = 2 ϑ = 10
0 α L ∞ κ L ∞ κ
κ
ℑ 3 (y) = e , −10 4.3238e − 05 5.3938e + 08 2.2439e − 04 5.5741e + 08
−03 3.5767e − 05 5.3567e + 08 2.5779e − 04 5.5929e + 08
ℑ 4 (y) = e κ+1 , −02 3.4646e − 05 5.3548e + 08 2.5701e − 04 5.6030e + 08
00 6.7483e − 04 5.3623e + 08 2.7761e − 04 5.6309e + 08
where the exact solution is: 02 2.5416e − 05 5.3965e + 08 7.5130e − 05 5.6705e + 08
03 8.5762e − 05 5.4213e + 08 8.7342e − 05 5.6948e + 08
s(κ, y) = e κ+y . (61) 10 8.4653e − 05 5.6667e + 08 4.4201e − 05 5.7578e + 08
We select M = 16 and dt = 0.005, using ϑ
values of −10, 0, 2, and 10 to generate numer-
ical results for different values of α, specifically
Problem 5. Consider the time-dependent prob-
α = −10, −3, −2, 0, 2, 3, and 10, as shown in Ta-
lem in Equation (32), along with the functions:
bles 4 and 5. These tables present the numerical 2
errors and condition numbers for various values f(κ, y, t) = −3(−t + 2κ + 2y),
3
3
of α, demonstrating that only slight variations in ℜ(κ, y) = κ + y ,
the error norm occur when the parameters α and Z 1
3
3
3
3
3
ϑ are varied. In Tables 4 and 5, for ϑ = −10 and ℑ 1 (y, t) = y + t − α 1 α(x)(t + y + x )dx
ϑ = 0, the L ∞ error norms and condition num- 0
Z 1
bers show minor fluctuations as α changes, with 3 3 3 3 3
ℑ 2 (y, t) = 1 + y + t − α 2 ϑ(x)(t + y + x )dx,
a slight increase in error as α moves further away 0
from zero. ℑ 3 (κ, t) = κ + t ,
3
3
Tables 4 and 5 emphasize the importance of 3 3
selecting an appropriate parameter ϑ to obtain ℑ 4 (κ, t) = κ + 1 + t ,
accurate solutions when using the Haar wavelet (62)
method for 2D integral problems. Figure 6
where α 1 and α 2 are both set to 1. The exact
presents a comparison between the exact solution
solution is given by:
and the numerical solution obtained with the pro- 3 3 3
s(κ, y, t) = κ + y + t . (63)
posed method.
Negative values of α and ϑ influence the inte- We select ϑ values of -10, 0, 2, and 10, and set
gral boundary conditions in a way that introduces dt = 0.0005 with T = 1 to investigate how the
a smoothing or damping effect on the solution, accuracy of the proposed method depends on ϑ.
particularly near the domain boundaries. This The numerical results for different values of α
regularizing behavior reduces the influence of (i.e., α = −10, −3, −2, 0, 2, 3, 10) are presented in
604