Page 47 - IJOCTA-15-1

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

For ST, with the help of the convolution theorem,
r
we get ∥ϑ m − ϑ r ∥ ≤ α + α r+1 + .. + α m−1 ∥ϑ 1 − ϑ 0 ∥ ,
1
1
1
≤ α r 1 1 + α 1 + .. + α m−r−1 ∥ϑ 1 − ϑ 0 ∥ ,
1
h m−r−1 i
∥ϑ m − ϑ r ∥ ≤ max {(ϕ ℓ + ℏ) |ϑ m−1 − ϑ r−1 | ≤ α r 1−α 1 ∥ϑ 1 − ϑ 0 ∥ .
τ∈η 1 1 1−α 1
τ o
2
R ∂(ϑ 2 −ϑ r−1 ) 2
+ℏ 1 m−1 ∂ (ϑ m−1 −ϑ r−1 )
∂µ +b ∂µ 2
2 Similarly,
0 δ
× (τ−ζ) dζ,
Γ(δ+1)
m−r−1
∥ω m − ω r ∥ ≤ max {(ϕ ℓ + ℏ) |ω m−1 − ω r−1 | r 1 − α 2
τ∈η 2 ∥ω m − ω r ∥ ≤ α 2 ∥ω 1 − ω 0 ∥ .
τ 1 − α 2
R ∂ω m−1 ∂ω r−1
+ℏ ϑ − + (ω m−1 − ω r−1 ) ∂ϑ
∂µ ∂µ ∂µ
As 0 < α 1 , α 2 < 1, so, 1 − α m−r−1 < 1 and
2 2 δ 1
o
0
∂ ω m−1 − ∂ ω r−1 (τ−ζ)
∂µ ∂µ 2 Γ(δ+1) 1 − α 2 < 1, then we have
+b 2 dζ, m−r−1
or,
∥ϑ m − ϑ r ∥ ≤ α r 1 ∥ϑ 1 − ϑ 0 ∥ ,
1−α 1
r
∥ϑ m − ϑ r ∥ ≤ max {(ϕ ℓ + ℏ) |ϑ m−1 − ϑ r−1 | ∥ω m − ω r ∥ ≤ α 2 ∥ω 1 − ω 0 ∥ .
τ∈η 1 1−α 2
τ
R
1
+ℏ (c + d) λ 1 |ϑ m−1 − ϑ r−1 |
2 But ∥ϑ 1 − ϑ 0 ∥ < ∞ and ∥ω 1 − ω 0 ∥ < ∞, con-
0 δ sequently as m → ∞ than ∥ϑ m − ϑ r ∥ → 0
+bλ 2 |ϑ m−1 − ϑ r−1 |)} (τ−ζ) dζ,
Γ(δ+1) and ∥ω m − ω r ∥ → 0. Hence, the sequence
∥ω m − ω r ∥ ≤ max {(ϕ ℓ + ℏ) |ω m−1 − ω r−1 | {ϑ r } and {ω r } each one is a Cauchy sequence
τ∈η 2
τ in (J 1 [η 1 , ∥.∥]) and (J 2 [η 2 , ∥.∥]), so these se-
R
+ℏ (P 1 λ 1 |ω m−1 − ω r−1 | + C 1 |ω m−1 − ω r−1 | quences are convergent. It completes our required
0 δ proof. □
+bλ 1 |ω m−1 − ω r−1 |)} (τ−ζ) dζ.
Γ(δ+1)
6. Applications of fractional WBK
Next, by using the integral mean value theorem, equation
we have
1
Example 1 If a = 0 and b = , then fractional
2
WBK equation (22) can be rewritten as
∥ϑ m − ϑ r ∥ ≤ max {(ϕ ℓ + ℏ) |ϑ m−1 − ϑ r−1 |
τ∈η 1
1
+ℏ (c + d) λ 1 |ϑ m−1 − ϑ r−1 | ∂ϑ(µ,ξ) ∂ω(µ,ξ)
δ
2 C D ϑ (µ, ξ) + ϑ (µ, ξ) +
+bλ 2 |ϑ m−1 − ϑ r−1 |) T 1 } , ξ ∂µ 2 ∂µ
∥ω m − ω r ∥ ≤ max {(ϕ ℓ + ℏ) |ω m−1 − ω r−1 | + 1 ∂ ϑ(µ,ξ) = 0,
2
2
∂µ
τ∈η 2
δ
C D ω (µ, ξ) + ω (µ, ξ) ∂ϑ(µ,ξ) + ϑ (µ, ξ) ∂ω(µ,ξ)
+ℏ (P 1 λ 1 |ω m−1 − ω r−1 | + C 1 ξ ∂µ ∂µ
2
× |ω m−1 − ω r−1 | + bλ 2 |ω m−1 − ω r−1 |) T 2 } , − 1 ∂ ω(µ,ξ) = 0,
2 ∂µ 2
or, (41)
∥ϑ m − ϑ r ∥ ≤ α 1 ∥ϑ m−1 − ϑ r−1 ∥ , subject to initial conditions
∥ω m − ω r ∥ ≤ α 2 ∥ω m−1 − ω r−1 ∥ .
ϑ (µ, 0) = θ − κ 1 coth [κ 1 (µ + ι)] ,
2
2
ω (µ, 0) = −κ 1 csch [κ 1 (µ + ι)] . (42)
Taking m = r + 1 1 gives,
According to Eqs. (33)-(35), we get following
results
∥ϑ r+1 − ϑ r ∥ ≤ α 1 ∥ϑ r − ϑ r−1 ∥
r
2
≤ α ∥ϑ r−1 − ϑ r−2 ∥ ≤ ... ≤ α ∥ϑ 1 − ϑ 0 ∥ ,
1
1
∥ω r+1 − ω r ∥ ≤ α 2 ∥ω r − ω r−1 ∥ ϑ 0 (µ, ξ) = θ − κ 1 coth (κ 1 (µ + ι)) ,
2
2
r
2
≤ α ∥ω r−1 − ω r−2 ∥ ≤ ... ≤ α ∥ω 1 − ω 0 ∥ . ω 0 (µ, ξ) = − κ 1 csch (κ 1 (µ + ι)) ,
2
2
δ
ξ
2
2
ϑ 1 (µ, ξ) = ℏ θ κ 1 csch (κ 1 (µ + ι)) Γ(δ+1) ,
On using triangular inequality, we have 3 2
ω 1 (µ, ξ) = 2 ℏ κ 1 θcsch (κ 1 (µ + ι))
ξ δ
× coth (κ 1 (µ + ι)) ,
Γ(δ+1)
2
2
∥ϑ m − ϑ r ∥ ≤ ∥ϑ r+1 − ϑ r ∥ + ∥ϑ r+2 − ϑ r+1 ∥ ϑ 2 (µ, ξ) = ℏ (1 + ℏ) θ κ 1 csch (κ 1 (µ + ι))
2
2 2
3
+... + ∥ϑ m − ϑ m−1 ∥ . × ξ δ − 2 ℏ θ κ 1 csch (κ 1 (µ + ι))
Γ(δ+1)
ξ 2δ
Hence, × coth (κ 1 (µ + ι)) Γ(2δ+1) ,
41