Page 59 - IJOCTA-15-4

Solving parabolic differential equations via Haar wavelets: A focus on integral boundary conditions
From Equations (17), (44), and (43), we have: coarse selection of collocation points, the method

s(L, t) − s(L, t 0 ) achieves satisfactory accuracy, which improves
= γ (H ⊗ 1)σ+ as M increases and dt decreases. Additionally,
∆t
(H ⊗ y)δ + (1 ⊗ H)ζ + (L ⊗ H)ω+ (53) Figures 1-2 depict the method’s performance in
terms of L ∞ . It is evident from these figures that

((ℜ 2 ⊗ H) + (H ⊗ ℜ 2 ))µ + f(L, t). the method’s accuracy remains relatively stable
regardless of the increase in collocation points M
or changes in ϑ.
2
According to Equation (53), a system of 4M
2
equations involving 4M + 8M unknowns is pro- Table 1. The numerical results generated using
duced by Equation (52). By setting Equations Haar wavelets for Test Problem 1 with parameters
(48) equal to (52) and substituting κ = 0, κ = 1, ϑ = 1, T = 1, and γ = 1
y = 0, and y = 1, we derive another system of 8M
2
equations with 4M + 8M unknowns. The Haar dt = 0.0005 dt = 0.005 dt = 0.05
coefficients for the unknowns µ, ζ, ω, σ, and δ M \L ∞
are determined by solving both sets of equations 4 2.4349e − 04 2.1402e − 04 2.1902e − 03
simultaneously. These coefficients subsequently 8 2.1342e − 05 2.6395e − 04 2.3891e − 03
16 2.2141e − 05 2.5393e − 04 2.2890e − 03
substitute the solution of the given problem ei-
32 2.5340e − 05 2.3393e − 04 2.1889e − 03
ther in Equation (48) or Equation (52).
64 2.6340e − 05 2.1393e − 04 2.5889e − 03
Table 2. The numerical results generated using
Haar wavelets for Test Problem 2 with parameters
4. Numerical results and discussion ϑ = −50, γ = 2, and T = 1

In this section, we present numerical experiments
dt = 0.0005 dt = 0.005 dt = 0.05
on various one- and two-dimensional parabolic
M \L ∞
PDEs with integral boundary conditions using an 4 3.5616e − 04 4.2593e − 04 7.5728e − 03
effective Haar wavelet collocation method. These 8 3.7148e − 05 6.8963e − 04 6.8353e − 03
experiments indicate that the proposed method 16 6.2383e − 05 7.7280e − 04 7.8397e − 03
yields accurate solutions, particularly when the 32 7.6387e − 05 8.5852e − 04 7.6956e − 03
parameters of the integral boundary conditions 64 8.5374e − 05 8.7694e − 04 8.7802e − 03
are negative.

Problem 2. Consider the problem presented in
Problem 1. Consider the problem presented in Equation (10), where the functions are defined as
40
Equation (10), where the functions are defined as follows :
follows: π
ℜ(κ) = cos κ ,
f(κ, t) = −e (t+κ) (γ − 1), 2
−π 2 t
κ
ℜ(κ) = e , g(t) = e 4 , (56)
t
g(t) = e , (54) 2 −π 2 t
m(t) = e 4 .
1 π
Z
m(t) = ϑ(x)e x+t dx,
0
with the exact solution: The theoretical solution of this problem is:
s(κ, t) = e κ+t . (55) −π 2 π
s(κ, t) = e 4 t cos κ . (57)
Table 1 presents the numerical results for Test 2
Problem 1 obtained using the proposed Haar Table 2 presents the numerical results of the
wavelets method. The parameter values used Haar wavelet method for Test Problem 2 with
in Table 1 are ϑ = −50, γ = 2, and T = 1. varying values of M and dt. The parameter val-
To evaluate the effect of different parameters on ues used in Table 2 are ϑ = 1, γ = 1, and
the accuracy of the proposed method, various T = 1. Furthermore, different values of M (M =
values of M and dt are considered, specifically 4, 8, 16, 32, 64) and dt (dt = 0.05, 0.005, 0.0005)
M = 4, 8, 16, 32, 64 and dt = 0.05, 0.005, 0.0005. are considered to assess their influence on the ac-
These results were generated with different values curacy of the proposed method. The accuracy
of M and dt. The table indicates that even with a of the method improves as M increases and the
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