Page 57 - IJOCTA-15-4

Solving parabolic differential equations via Haar wavelets: A focus on integral boundary conditions
which is accompanied by the nonlocal integral
boundary conditions: It is possible to write Equation (40) with the ad-
Z 1 ditions of Equation (39):
s(0, y, t) =α 1 α(κ)s(κ, y, t) dκ + ℑ 1 (y, t), 2M 2M
2
0 ∂ s(κ k , y l ) X X
0 < y < 1, 0 < t ≤ T, ∂y 2 = ζ j h j (y l ) + κ k ω j h j (y l )
j=1 j=1
(33)
2M 2M
Z 1 X X
+ µ i,j ρ i,2 (κ k )h j (y l ),
s(1, y, t) =α 2 ϑ(κ)s(κ, y, t) dκ + ℑ 2 (y, t),
0 i=1 j=1
0 < y < 1, 0 < t ≤ T. (41)
(34) where the collocation points are:
k − 1/2
κ k = , k = 1, 2, . . . , 2M,
Dirichlet boundary conditions and the initial con- 2M (42)
dition are expressed as: l − 1/2
y l = , l = 1, 2, . . . , 2M.
2M
The Equation (41) in matrix form can be repre-
s(κ, 0, t) = ℑ 3 (κ, t), 0 < κ < 1, 0 < t ≤ T, (35) sented as:
s(κ, 1, t) = ℑ 4 (κ, t), 0 < κ < 1, 0 < t ≤ T, (36) s yy = (1 ⊗ H)ζ + (L ⊗ H)ω + (ℜ 2 ⊗ H)µ, (43)
s(κ, y, 0) = ϕ(κ, y), 0 ≤ κ ≤ 1, 0 ≤ y ≤ 1. (37)
where as:
2
2
2
∂ s ∂ s ∂ s
where f(κ, y, t), α(κ), ϑ(κ), ℑ 1 (y), ℑ 2 (κ), ℑ 3 (κ), s yy = [ (κ 1 , y 1 ), (κ 1 , y 2 ), ..., (κ 1 , y n ),
∂y 2 ∂y 2 ∂y 2
ℑ 4 (y) are known smooth functions, parameters α 1 2 2 2
and α 2 are given, while the function s(κ, y, t) is ∂ s (κ 2 , y 1 ), ∂ s (κ 2 , y 2 ), ..., ∂ s (κ 2 , y n ), ...,
unknown. It is assumed that there is mutual com- ∂y 2 ∂y 2 ∂y 2
2
2
2
patibility between the Dirichlet boundary condi- ∂ s ∂ s ∂ s T
tions (Equations (35) and (36)) and the nonlocal ∂y 2 (κ n , y 1 ), ∂y 2 (κ 2M , y 2 ), ..., ∂y 2 (κ 2M , y 2M )] ,
integral boundary conditions (Equations (33) and T T
ζ =[ζ 1 , ζ 2 , ..., ζ 2M ] , ω = [ω 1 , ω 2 , ..., ω 2M ] ,
(34)).
µ =[µ 1,1 , µ 1,2 , ..., µ 1,2M , µ 2,1 , µ 2,2 , ..., µ 2,2M , ...,
For the mixed fourth-order derivative, let us in- µ 2M,1 , µ 2M,2 , ..., µ 2M,2M ] ,
T
vestigate the Haar wavelet approximation:
Following a similar approach, integrating Equa-
tion (38) twice with respect to y over the interval
2M 2M
4
∂ s X X from 0 to y:
= µ i,j h i (κ)h j (y). (38)
2
∂κ ∂y 2 s κκ = (H ⊗ 1)σ + (H ⊗ y)δ + (H ⊗ ℜ 2 )µ, (44)
i=1 j=1
where
We derive this by performing two successive par- y = [y 1 , y 2 , ..., y 2M ] , σ = [σ 1 , σ 2 , ..., σ 2M ] ,
T
T
tial integrations of Equation (38) with respect to
T
κ, from 0 to κ. δ = [δ 1 , δ 2 , ..., δ 2M ] .
2
2
3
∂ s ∂ s(0, y) ∂ s(0, y) By integrating Equation (43) twice over the inter-
= + κ +
∂y 2 ∂y 2 ∂κ∂y 2 val from 0 to y, we obtain:
2M 2M Z Z (39) s =(s(L, 0) ⊗ 1) + (s y | y=0 ⊗ y) + (1 ⊗ ℜ 2 )ζ
X X κ κ
µ i,j h j (y) h i (κ) dκ dκ. + (L ⊗ ℜ 2 ) ω + (ℜ 2 ⊗ ℜ 2 )µ,
i=1 j=1 0 0 (45)
In light of that, we consider:
where
2M
3
∂ s(0, y) X ∂s ∂s ∂s
= ω j h j (y), s y | y=0 =[ (κ 1 , 0), (κ 1 , 0), ..., (κ 1 , 0),
∂κ∂y 2 ∂y ∂y ∂y
j=1
∂s ∂s ∂s
2M
2
∂ s(0, y) X (40) ∂y (κ 2 , 0), ∂y (κ 2 , 0), ..., ∂y (κ 2 , 0), ...,
= ζ j h j (y),
∂y 2 ∂s ∂s ∂s T
j=1 (κ n , 0), (κ 2M , 0), ..., (κ 2M , 0)] .
Z Z ∂y ∂y ∂y
κ κ
ρ i,2 (κ) = h i (κ) dκ dκ.
0 0
599