Page 55 - IJOCTA-15-4

Solving parabolic differential equations via Haar wavelets: A focus on integral boundary conditions
Where f(κ, t), ℜ(κ), g(t), ϑ(κ), and m(t) are interval [t 0 , t], Euler’s formula provides an approx-
known functions, and the constant γ is also imation of the time evolution.
known. Let us consider the following approxi- ∂s(κ, t) s(κ, t) − s(κ, t 0 )
mation using Haar wavelets: ∂t ≈ ∆t . (17)
Equations (10), (14), and (17) yield the following:
2M
2
∂ s X
= µ i h i (κ). (14) s(L, t) − s(L, t 0 )
2 = γHµ + f(L, t), (18)
∂κ
i=1 ∆t
After integrating Equation (14) twice from 0 to where
 
∂s(0,t) h 1 (κ 1 ) h 2 (κ 1 ) ... h 2M (κ 1 )
κ, calculate by substituting κ with 1. In-
∂κ
corporate this result back into the expression for  h 1 (κ 2 ) h 2 (κ 2 ) ... h 2M (κ 2 ) 
. . . .  .


s by rearranging its terms, which results in the . . . . .
H = 
 . . . 
following matrix representation of s. h 1 (κ 2M ) h 2 (κ 2M ) ... h 2M (κ 2M )
s = (1 − L) s(0, t) + Ls(1, t) + ℜ 2 − Lℜ T µ, By applying Equation (16) to Equation (18), we
obtain:
(15)
L T
ℜ 2 − ℜ − γ∆tH µ =s(L, t 0 ) + ∆tf(L, t)
ϑ
where B 2

T
T
L = [κ 1 , κ 2 , ..., κ 2M ] , 1 = [1, 1, ..., 1] , L LB 1
− (1 − L) g(t) + m(t) + g(t) .
T
µ = [µ 1 , µ 2 , ..., µ 2M ] , B 2 B 2
(19)
T
ℜ = ℜ 1,2 (1), ℜ 2,2 (1), ... , ℜ 2M,2 (1) , We need to find the parameter µ from this set of
 
ℜ 1,2 (κ 1 ) ℜ 2,2 (κ 1 ) ... ℜ 2M,2 (κ 1 ) equations. In the approximate solution given by
 ℜ 1,2 (κ 2 ) ℜ 2,2 (κ 2 ) ... ℜ 2M,2 (κ 2 )  Equation (16), µ is subsequently introduced, pro-
 
. . .  . viding an approximate solution to problem (10).
. . . .
ℜ 2 =  .
 . . . . 
ℜ 1,2 (κ 2M ) ℜ 2,2 (κ 2M ) ... ℜ 2M,2 (κ 2M )
To determine s(1, t), we multiply ϑ(κ) by the 3.2. 1D problem with two integrals BC
Equation (15) and integrate from 0 to 1 with re-
spect to κ. This process involves incorporating Now, consider the one-dimensional time-
the nonlocal integral boundary condition. Subse- dependent diffusion equation 41 provided below:
2
quently, we substitute the value of s(1, t) obtained ∂s ∂ s
back into Equation (15) and apply the boundary ∂t = γ ∂κ 2 +f(κ, t), 0 < κ < 1, 0 < t ≤ T, (20)
condition given in Equation (12) to derive the having the initial condition defined as:
approximation for s.
s(κ, 0) = ℜ(κ), 0 ≤ κ ≤ 1, (21)
L LB 1 and in accordance with the nonlocal integral

s = (1 − L) g(t) + m(t) + g(t)
B 2 B 2 boundary condition:

L T s(0, t) = D 1 (t) + g 1 (t), 0 < t ≤ T, (22)
+ ℜ 2 − ℜ ϑ µ,
B 2
s(1, t) = D 2 (t) + g 2 (t), 0 < t ≤ T, (23)
(16)
where
where
1
Z
D 1 (t) = α(κ)s(κ, t) dκ,
1 1
Z Z
B 1 = ϑ(κ)(1 − κ) dκ, B 2 = ϑ(κ)κ dκ, 0
Z 1
0 0 D 2 (t) = ϑ(κ)s(ϑ, t) dκ,
and
0
1 and the functions f(κ, t), ℜ(κ), α(κ), ϑ(κ),
Z
ℜ ϑ = ϑ(L)ℜ dκ. g 1 (κ), g 2 (κ), and γ are known. After approxi-
0
mating the solution using Equation (15) and in-
The time increment is denoted by ∆t. The dis-
corporating the boundary conditions from Equa-
crete time instances are represented as t n = t 0 +
tions (22) and (23), we obtain:
n · ∆t, where t 0 marks the initial time and t n de-
notes the time at each discrete step. For the time s = (1 − L) (D 1 (t) + g 1 (t)) + L(D 2 (t) + g 2 (t))+
ℜ 2 − Lℜ T µ.
(24)
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