Page 56 - IJOCTA-15-4

MN. Khan et.al. / IJOCTA, Vol.15, No.4, pp.594-609 (2025)
1 − L
After multiplying α(L) by Equation (24) and in- s(L, t) =(1 − L)g 1 (t) + Lg 2 (t) +
tegrating between 0 and 1, we obtain: m 1 (1 − B 1 )

B 2 E 1
1 + g 2 (t) (B 2
Z g 1 (t) B 1 +
D 1 (t) =(D 1 (t) + g 1 (t)) α(L) (1 − L) dκ+ 1 − E 2

0 B 2 E 2 L
1
Z + + E 1 g 1 (t)
(D 2 (t) + g 2 (t)) α(L)L dκ+ 1 − E 2 m 1 (1 − E 2 )

0 B 1 B 2 E 1
1
Z 1 + + g 2 (t) E 2 +
α(L) ℜ 2 − Lℜ T dκµ. 1 − B 1 1 − B 1

0 T 1 − L
(25) + ℜ − Lℜ + ℜ ϑ +
m 1 (1 − B 1 )
T

B 2 E 2
Similarly, ℜ α − B 2 ℜ 1 + µ
1 − E 2 1 − E 2

1 L E 1
Z
D 2 (t) =(D 1 (t) + g 1 (t)) ϑ(L) (1 − L) dκ+ + m 1 (1 − E 2 ) ℜ α + ℜ ϑ
0 1 − B 1
T
1 E 1 B 2
Z
(D 2 (t) + g 2 (t)) ϑ(L)L dκ − ℜ E 2 + µ.
0 1 − B 1
1
Z (29)
+ ϑ(L) ℜ 2 − Lℜ T dκµ.
0
(26) From Equations (17) and (20), we obtain:
From Equations (25) and (26), the values of D 1 (t) s(L, t) − s(L, t 0 )
= γHµ + f(L, t). (30)
and D 2 (t) are calculated: ∆t
Inserting from Equation (29) the value of s(L, t)
in Equation (30).

1 B 2 E 1
D 1 (t) = g 1 (t) B 1 + + T 1 − L B 2
m 1 (1 − B 1 ) 1 − E 2 ℜ − Lℜ − γ∆tH + ℜ ϑ ℜ α
m 1 (1 − B 1 ) 1 − E 2
T
E 2 L
B 2 E 2 B 2 − B 2 ℜ 1 + µ + (ℜ α
g 2 (t) B 2 + + ℜ ϑ + ℜ α − 1 − E 2 m 1 (1 − E 2 )
1 − E 2 1 − E 2
T

E 1 E 1 B 2
E 2 + ℜ ϑ − ℜ E 2 + µ,
B 2 ℜ 1 + µ,
1 − B 1 1 − B 1
1 − E 2

1 B 1 1 − L .
D 2 (t) = E 1 g 1 (t) 1 + + = − (1 − L)g 1 (t) + Lg 2 (t) + g 1 (t)
m 1 (1 − E 2 ) 1 − B 1 m 1 (1 − B 1 )

B 2 E 1 B 2 E 2
B 1 + + g 2 (t) B 2 +

1 − E 2 1 − E 2
B 2 E 1 E 1
g 2 (t) E 2 + + ℜ α + ℜ ϑ − L B 1
1 − B 1 1 − B 1 − E 1 g 1 (t) 1 + +
m 1 (1 − E 2 )
1 − B 1
E 1 B 2
ℜ E 2 + µ,
B 2 E 1
1 − B 1 g 2 (t) E 2 + + s(L, t 0 ) + ∆tf(L, t).
(27) 1 − B 1
(31)
Computing µ by solving this system of equations.
where
Next, substitute µ in Equation (29), the approx-
Z 1 Z 1
B 1 = ϑ(κ)(1 − κ) dκ, B 2 = ϑ(κ)κ dκ, imate solution.
0 0
Z 1 Z 1
E 1 = α(κ)(1 − κ) dκ, E 2 = α(κ)κ dκ,
0 0 3.3. 2D problem with integral BC
1 1
Z Z
ρ α = α(κ)ℜ dκ, ρ ϑ = ϑ(κ)ℜ dκ, Consider, 42
0 0
2 2
∂s ∂ s ∂ s
B 2 E 1
m 1 = 1 − . =γ 2 + 2 + f(κ, y, t),
(1 − B 1 )(1 − E 1 ) ∂t ∂κ ∂y (32)
(28) 0 < κ < 1, 0 < y < 1,
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