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L.K. Yadav et.al. / IJOCTA, Vol.15, No.1, pp.35-49 (2025)

            and
                                                              Now, we define a non-linear operator as

                                  0,   ℓ ≤ 1,
                          ϕ ℓ =                        (18)
                                  1,   ℓ > 1.
                                                                 1
                                                               N [Υ 1 (µ, ξ; q) , Υ 2 (µ, ξ; q)] = S [Υ 1 (µ, ξ; q)]
                                                                 v

            Applying the inverse ST to both sides of (16), we  − f (µ) +   v δ S [Υ 1 (µ, ξ; q)  ∂Υ 1 (µ,ξ;q)
                                                                 s         s                  ∂µ
                                                                               2
            obtain:                                              ∂Υ 2 (µ,ξ;q)  ∂ Υ 1 (µ,ξ;q)  i
                                                               +          + b           ,
                                                                    ∂µ           ∂µ 2
               ϑ ℓ (µ, ξ)                                      N [Υ 1 (µ, ξ; q) , Υ 2 (µ, ξ; q)] = S [Υ 2 (µ, ξ; q)]
                                                                 2
                                      h    − →      i          v        v δ              ∂Υ 1 (µ,ξ;q)

               = ϕ ℓ ϑ ℓ−1 (µ, ξ) + ℏS −1  R ℓ  ϑ ℓ−1 (µ, ξ)  ,  − g (µ) +  s  S [Υ 2 (µ, ξ; q)  ∂µ  +
                                                                 s
                                                                                                             i
                                                                                       3
                                                                                                    2
                                                                          ∂Υ 2 (µ,ξ;q)
                                                                                                   ∂ Υ 2 (µ,ξ;q)
                                                                                      ∂ Υ 1 (µ,ξ;q)
                                                       (19)    Υ 1 (µ, ξ; q)  ∂µ  + a    ∂µ 3   − b   ∂µ 2    .
                                                                                                         (25)
            Therefore, the Mth-order approximate solution of
            equation (6) is provided as follows:                            1
                                                              Here q ∈    0,    is an embedding parameter.
                                                                             n
                                                              Liao 32,33  constructed zeroth-order deformation
                                    M
                                   X                          equation such as
                         ϑ(µ, ξ) =    ϑ ℓ (µ, ξ) .     (20)
                                   ℓ=0
                                                                (1 − q) S [Υ 1 (µ, ξ; q) − ϑ (µ, 0)]
            Moreover, for M → ∞, we get                                       1
                                                                = ℏH (µ, ξ) N [Υ 1 (µ, ξ; q) , Υ 2 (µ, ξ; q)] ,
                                    ∞                                                                    (26)
                                   X
                         ϑ(µ, ξ) =    ϑ ℓ (µ, ξ) .     (21)     (1 − q) S [Υ 2 (µ, ξ; q) − ω (µ, 0)]
                                                                              2
                                   ℓ=0                          = ℏH (µ, ξ) N [Υ 1 (µ, ξ; q) , Υ 2 (µ, ξ; q)] .
            4. HASTM solution of fractional WBK
                                                              Here, S represent the Shehu transform, ℏ is
                equation                                      nonzero auxiliary parameter, H (µ, ξ) ̸= 0 denotes
                                                              an auxiliary function, ϑ (µ, 0) and ω (µ, 0) indicate
            The fractional Whitham-Broer-Kaup equations
                                                              initial guesses of ϑ (µ, ξ) and ω (µ, ξ), respectively.
            (1) with Caputo fractional derivative can be
                                                              Let q = 0 and q = 1 in equation (26), we get
            rewritten as
              C   δ                ∂ϑ(µ,ξ)  ∂ω(µ,ξ)            Υ 1 (µ, ξ; q) = ϑ 0 (µ, ξ) , Υ 1 (µ, ξ; 1) = ϑ (µ, ξ) ,
                D ϑ (µ, ξ) + ϑ (µ, ξ)    +
                  ξ                  ∂µ       ∂µ               Υ 2 (µ, ξ; q) = ω 0 (µ, ξ) , Υ 2 (µ, ξ; 1) = ω (µ, ξ) .
                  2
                 ∂ ϑ(µ,ξ)
              +b        = 0,                                                                             (27)
                   ∂µ 2
              C D ω (µ, ξ) + ω (µ, ξ)  ∂ϑ(µ,ξ)  + ϑ (µ, ξ)  ∂ω(µ,ξ)
                  δ
                  ξ
                                     ∂µ
                                                     ∂µ
                  3
                             2
              +a ∂ ϑ(µ,ξ)  − b ∂ ω(µ,ξ)  = 0,                 Thus, if q increases from 0 to 1, then Υ 1 (µ, ξ; q)
                   ∂µ 3       ∂µ 2                            varies from initial guess ϑ 0 (µ, ξ) to the solu-
                                                       (22)
                                                              tion ϑ(µ, ξ), and Υ 2 (µ, ξ; q) varies from initial
                                                              guess ω 0 (µ, ξ) to the solution ω(µ, ξ), respectively.
            subject to initial conditions
                                                              Upon expanding Υ 1 (µ, ξ; q) and Υ 2 (µ, ξ; q) ac-
                                                              cording to Taylor’s series near q, we have
                ϑ (µ, 0) = f (µ) ,  ω (µ, 0) = g (µ) ,  (23)
                                                                                         ∞
                                                                                        P            ℓ
                                                                 Υ 1 (µ, ξ; q) = ϑ 0 (µ, ξ) +  ϑ ℓ (µ, ξ) q ,
                                       δ
            where 0 < δ ≤ 1 and   C D ξ present the Caputo                              ℓ=1
                                                                                         ∞               (28)
            fractional derivative of order δ.                    Υ 2 (µ, ξ; q) = ω 0 (µ, ξ) +  P  ω ℓ (µ, ξ) q ,
                                                                                                     ℓ
            Applying ST to both side of (22) and on simpli-                             ℓ=1
            fication, we get                                  where

                                       nh                                             ℓ
                          v

             S [ϑ (µ, ξ)] − f (µ) +  v δ  ϑ (µ, ξ)  ∂ϑ(µ,ξ)             ϑ ℓ (µ, ξ) =  1 ∂ Υ 1 (µ,ξ;q)  ,
                          s         s              ∂µ                              ℓ!   ∂q ℓ   q=0
                                     2
                          ∂ω(µ,ξ)   ∂ ϑ(µ,ξ)  io                                      ℓ                 (29)
                        +       + b           = 0,                      ω ℓ (µ, ξ) =  1 ∂ Υ 2 (µ,ξ;q)  .
                            ∂µ        ∂µ 2                                         ℓ!   ∂q ℓ
                                       nh        ∂ϑ(µ,ξ)                                       q=0
                          v

             S [ω (µ, ξ)] − g (µ) +  v δ  ω (µ, ξ)
                          s         s              ∂µ
                                                   io         By proper choosing of ϑ 0 (µ, ξ), ω 0 (µ, ξ), ℏ, and
                                  3
                                             2
              +ϑ (µ, ξ)  ∂ω(µ,ξ)  + a ∂ ϑ(µ,ξ)  − b ∂ ω(µ,ξ)  = 0 .
                         ∂µ        ∂µ 3       ∂µ 2            H (µ, ξ), the series in equation (28) converges at
                                                       (24)   q = 1 , we will get
                                                            38
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