Page 45 - IJOCTA-15-1

Approximate analytical solutions of fractional coupled Whitham-Broer-Kaup equations . . .

∞ M
P P
ϑ (µ, ξ) = ϑ 0 (µ, ξ) + ϑ ℓ (µ, ξ) , ϑ (µ, ξ) = ϑ 0 (µ, ξ) + ϑ ℓ (µ, ξ) ,
ℓ=1 ℓ=1
∞ (30) (36)
P M
ω (µ, ξ) = ω 0 (µ, ξ) + ω ℓ (µ, ξ) . ω (µ, ξ) = ω 0 (µ, ξ) + P ω ℓ (µ, ξ) .
ℓ=1 ℓ=1
− → − →
We define the vectors ϑ ℓ (µ, ξ) , and ω ℓ (µ, ξ) as
Moreover, for M → ∞ , we get
− →
ϑ ℓ (µ, ξ) = {ϑ 0 (µ, ξ), ϑ 1 (µ, ξ), ........, ϑ ℓ (µ, ξ)} , ∞
− → P
ω ℓ (µ, ξ) = {ω 0 (µ, ξ), ω 1 (µ, ξ), ........, ω ℓ (µ, ξ)} . ϑ (µ, ξ) = ϑ 0 (µ, ξ) + ϑ ℓ (µ, ξ) ,
(31) ℓ=1 (37)
∞
P
ω (µ, ξ) = ω 0 (µ, ξ) + ω ℓ (µ, ξ).
ℓ=1
First, differentiating equation (26) ℓ- times with
respect to q, then evaluate at q = 0 and finally
dividing by Γ (ℓ + 1), we get

S [ϑ ℓ (µ, ξ) − ϕ ℓ ϑ ℓ−1 (µ, ξ)]
h − → − → i 5. Convergence analysis of the
= ℏH (µ, ξ) R 1,ℓ ϑ , ω ,
ℓ−1 ℓ−1
proposed method
S [ω ℓ (µ, ξ) − ϕ ℓ ω ℓ−1 (µ, ξ)]
− → − →
h i Theorem 1. (Uniqueness Theorem) For
= ℏH (µ, ξ) R 2,ℓ ϑ , ω ,
ℓ−1 ℓ−1 equation (22), the HASTM solution is unique
(32) wherever 0 < α 1 < 1 and 0 < α 2 < 1,
1
where, where, α 1 = (ϕ ℓ + ℏ) + ℏ 2 (c + d) λ 1 + bλ 2 T 1
and
− → − →
h i α 2 = (ϕ ℓ + ℏ) + ℏ (P 1 λ + C 1 + bλ 2 ) T 2 .
R 1,ℓ ϑ , ω = S [ϑ ℓ−1 (µ, ξ)] − (1 − ϕ ℓ )
ℓ−1 ℓ−1

ℓ−1
v

× [f (µ)] + v δ S P ϑ k ∂ϑ ℓ−1−k + ∂ω ℓ−1
s s ∂µ ∂µ
k=0
2 i Proof. For fractional WBK equation (22), the
+ b ∂ ϑ ℓ−1 ,
∂µ 2 solution is illustrated as following
− → − →
h i
R 2,ℓ ϑ , ω = S [ω ℓ−1 (µ, ξ)] − (1 − ϕ ℓ ) ∞
ℓ−1 ℓ−1 ϑ (µ, ξ) = P ϑ ℓ (µ, ξ) ,
ℓ−1
v

× [g (µ)] + v δ S P ϑ k ∂ω ℓ−1−k ℓ=0 (38)
∞
s s ∂µ P
k=0 ω (µ, ξ) = ω ℓ (µ, ξ) ,
ℓ−1 i ℓ=0
3
2
P ∂ϑ ℓ−1−k ∂ ϑ ℓ−1 ∂ ω ℓ−1
+ ω k + a 3 − b ,
∂µ ∂µ ∂µ 2
k=0 where,
(33)
and
ϑ ℓ (µ, ξ) = (ϕ ℓ + ℏ) ϑ ℓ−1 (µ, ξ) − (1 − ϕ ℓ )

ℓ−1
0, ℓ ≤ 1 −1 v −1 v δ P

ϕ ℓ = (34) ×S s [f (µ)] + ℏS s S ϑ k
1, otherwise . k=0
2 ii
∂ϑ ℓ−1−k
∂ ϑ ℓ−1
∂ω ℓ−1
Next, applying inverse ST to both sides of equa- × ∂µ + ∂µ + b ∂µ 2 ,
tion (32) and H (µ, ξ) = 1, we get ω ℓ (µ, ξ) = (ϕ ℓ + ℏ) ω ℓ−1 (µ, ξ) − (1 − ϕ ℓ )

ℓ−1

×S −1 v [g (µ)] + ℏS −1 P ϑ k ∂ω ℓ−1−k
ϑ ℓ (µ, ξ) = ϕ ℓ ϑ ℓ−1 (µ, ξ) s k=0 ∂µ
− → − → ℓ−1
h h ii
3
2
+S −1 ℏR 1,ℓ ϑ , ω , P ∂ϑ ℓ−1−k ∂ ϑ ℓ−1 ∂ ω ℓ−1
ℓ−1 ℓ−1 + ω k + a − b .
3
∂µ ∂ µ ∂µ 2
(35) k=0
(39)
ω ℓ (µ, ξ) = ϕ ℓ ω ℓ−1 (µ, ξ)
h h − → − → ii
+S −1 ℏR 2,ℓ ϑ , ω .
ℓ−1 ℓ−1
In order to prove the required result, we consider
∗
Finally, we compute ϑ ℓ (µ, ξ) by using equation ϑ and ϑ are two different solutions for equation
(35) for ℓ ≥ 1. Hence the M th order approximate (22) such that |ϑ| ≤ P 1 and |ω| ≤ P 2 , then using
solution of equation (22) can be represented as the above relation, it is obtained as following
39