Page 58 - IJOCTA-15-4

MN. Khan et.al. / IJOCTA, Vol.15, No.4, pp.594-609 (2025)
1 1
Z Z
Equation (45) makes it simple to get the value ϑs dκ = ϑℑ 3 (L, t) dκ ⊗ 1 +
of s y | y=0 . To do this, simply substitute L for 1 0 0
in Equation (45), find s y | y=c , and then put this Z 1
value back in Equation (45). We get: ϑ(ℑ 4 (L, t) − ℑ 3 (L, t)) dκ ⊗ y +
0
Z 1
ϑ dκ ⊗ ℜ 2 − yℜ T ζ+
s =(s(L, 0) ⊗ 1) + ((s(L, d) − s(L, 0)) ⊗ y) 0
+ 1 ⊗ ℜ 2 − yℜ T ζ + L ⊗ ℜ 2 − yℜ T ω Z 1 T
ϑL dκ ⊗ ℜ 2 − yℜ ω+
+ ℜ 2 ⊗ ℜ 2 − yℜ T µ, 0
Z 1
(46) T
ϑℜ 2 dκ ⊗ ℜ 2 − yℜ µ,
0
(50)
where
where
T
T α = [α(κ 1 ), α(κ 2 ), α(κ 3 ), ..., , α(κ 2M )] ,
ℜ = ℜ 1,2 (1) ℜ 2,2 (1) ... ℜ 2M,2 (1) . (51)
T
ϑ = [ϑ(κ 1 ), ϑ(κ 2 ), ϑ(κ 3 ), ..., , ϑ(κ 2M )] .
The integral boundary conditions, denoted as
In the same way, from Equation (44), we de- Equations (33) and (34), result from multiplying
rive: α 1 by Equation (49) and α 2 by Equation (50).
After substituting these derived expressions into
s =(1 ⊗ s(0, y)) + (L ⊗ (s(1, y) − s(0, y)))
Equation (47) and rearranging, we derive an al-

+ ℜ 2 − Lℜ T ⊗ 1 σ + ℜ 2 − Lℜ T ⊗ y δ ternative expression for s.

+ ℜ 2 − Lℜ T ⊗ ℜ 2 µ. s = ((1 − L) ⊗ ℑ 1 (y)) + (L ⊗ ℑ 2 (y)) +
(47) Z 1
α 1 (1 − L) αℑ 3 (L, t) dκ ⊗ 1 +
0
The expression below results from applying the Z 1
boundary conditions described by Equation (33) Lα 2 ϑℑ 3 (L, t) dκ ⊗ 1 +
0
to Equation (34) within Equation (46). Z 1
(1 − L) α 1 α(ℑ 4 (L, t) − ℑ 3 (L, t)) dκ ⊗ y +
0
s =(ℑ 3 (L, t) ⊗ 1) + ((ℑ 4 (L, t) − ℑ 3 (L, t)) ⊗ y) Z 1
+ 1 ⊗ ℜ 2 − yℜ T ζ + L ⊗ ℜ 2 − yℜ T ω Lα 2 0 ϑ(ℑ 4 (L, t) − ℑ 3 (L, t)) dκ ⊗ y +
Z 1

+ ℜ 2 ⊗ ℜ 2 − yℜ T µ,
(1 − L) α 1 α dκ+
(48) 0
R 1 Z 1
The expressions α(κ)s(κ, y, t) dκ and T
0 (L) α 2 ϑ dκ ⊗ ℜ 2 − yℜ ζ +
R 1 ϑ(κ)s(κ, y, t) dκ can be derived from Equa- 0
0 Z
tion (48) and are presented as follows: 1
(1 − L) α 1 Lα dκ+
Z 1 Z 1
0
αs dκ = αℑ 3 (L, t) dκ ⊗ 1 + Z 1
0 0 Lϑ dκ ⊗ ℜ 2 − yℜ T ω +
(L) α 2
1 0
Z
α(ℑ 4 (L, t) − ℑ 3 (L, t)) dκ ⊗ y + T
1

Z
0
(1 − L) α 1 αℜ 2 dκ +
1 0
Z
α dκ ⊗ ℜ 2 − yℜ T ζ+
0 Z 1 T
1 (L) α 2 ϑℜ 2 dκ +
Z ⊗ ℜ 2 − yℜ T
αL dκ ⊗ ℜ 2 − yℜ T ω+ 0

0
1 ⊗ ℜ 2
Z ℜ 2 − Lℜ T µ.
αℜ 2 dκ ⊗ ℜ 2 − yℜ T µ,
0 (52)
(49)
600