The Marichev-Saigo-Maeda Fractional Calculus Operators Pertaining to the Generalized K-Struve Function

Seema Kabra; Harish Nagar; Kottakkaran Sooppy Nisar; D.L. Suthar

doi:10.2478/amns.2020.2.00064

Source Title:: Applied Mathematics and Nonlinear Sciences
Volume/Issue:: Volume 5: Issue 2

The Marichev-Saigo-Maeda Fractional Calculus Operators Pertaining to the Generalized K-Struve Function

Seema Kabra¹, Harish Nagar², Kottakkaran Sooppy Nisar³, and D.L. Suthar⁴

¹ Department of Mathematics, Sangam University, Bhilwara, India
² Department of Mathematics, Sangam University, Bhilwara, India
³ Department of Mathematics, College of Arts & Sciences, Wadi Aldawaser, Prince Sattam bin Abdulaziz University, Saudi Arabia
⁴ Department of Mathematics, Wollo University, Dessie, P.O. Box:1145, Amahara Region, Ethiopia

Seema Kabra

Department of Mathematics, Sangam University, Bhilwara, Rajasthan, India
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, Harish Nagar

Department of Mathematics, Sangam University, Bhilwara, Rajasthan, India
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, Kottakkaran Sooppy Nisar

Department of Mathematics, College of Arts & Sciences, Wadi Aldawaser, Prince Sattam bin Abdulaziz University, Saudi Arabia
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and D.L. Suthar

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Department of Mathematics, Wollo University, Dessie, P.O. Box:1145, Amahara Region, Ethiopia
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DOI:: https://doi.org/10.2478/amns.2020.2.00064
Published online:: 30 Dec 2020

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Abstract

In the present paper, we establish some compositions formulas for Marichev-Saigo-Maeda (MSM) fractional calculus operators with k-Struve function $S_{ν, c}^{k}$ as of the kernel. The results are presented in terms of generalized k-Wright function $_{p} Ψ_{q}^{k}$ .

Keywords:: Generalized k-Struve function; Marichev-Saigo-Maeda fractional calculus operators; Generalized k-Wright function; k-Pochhammer symbol; k-Gamma function; 26A33; 33C65; 33C20

1 Introduction and Preliminaries

The Wright function play an important role in the partial differential equation of fractional order which is familiar and extensively treated in papers by a number of authors including Gorenflo et al. [6].

For ς_i, τ_j ∈ ℝ\{0} and a_i,b_j ∈ ℂ,i = (1̅, p); j = (1̅,q) the generalized form of Wright function is defined by Wright ([13,14,15,16,17]) as following:

_{p} Ψ_{q} (z) =_{p} Ψ_{q} [\begin{matrix} {(a_{i}, ς_{i})}_{1, p} \\ {(b_{j}, τ_{j})}_{1, q} \end{matrix} | z] = \sum_{n = 0}^{\infty} \frac{\prod_{i = 1}^{p} Γ (a_{i} + n ς_{i})}{\prod_{j = 1}^{q} Γ (b_{j} + n τ_{j})} \frac{z^{n}}{n!}, z \in ℂ,

(1.1)

where Γ (z) is the well-known Euler gamma function [4].The condition for existence of (1.1) with its depiction in terms of Mellin-Barnes integral and the H-function were obtained by Kilbas et al. [10].

The generalized form of the above Wright function (1.1) was given by Gehlot and Prajapati [5], named as generalized K-Wright function which is defined as

_{p} Ψ_{q}^{k} (z) =_{p} Ψ_{q}^{k} [\begin{matrix} {(a_{i}, ς_{i})}_{1, p} \\ {(b_{j}, τ_{j})}_{1, q} \end{matrix} | z] = \sum_{n = 0}^{\infty} \frac{\prod_{i = 1}^{p} Γ_{k} (a_{i} + n ς_{i})}{\prod_{j = 1}^{q} Γ_{k} (b_{j} + n τ_{j})} \frac{z^{n}}{n!}, z \in ℂ,

(1.2)

where k ∈ ℝ⁺ and (a_i + nς_i), (b_j + bτ_j) ∈ ℂ\kℤ⁻ for all n ∈ ℕ₀. The generalized k-gamma function [3] is defined as

Γ_{k} (z) = \int_{0}^{\infty} e^{- \frac{t^{k}}{k}} t^{z - 1} dt; (ℜ (z) > 0; k \in ℝ^{+})

(1.3)

and

Γ_{k} (z) = lim_{n \to \infty} \frac{n! k^{n} {(nk)}^{\frac{z}{k} - 1}}{{(z)}_{n, k}}, k \in ℝ^{+}, z \in ℂ \ k ℤ^{-}

(1.4)

Also

Γ_{k} (z) = k^{\frac{z}{k} - 1} Γ (\frac{z}{k}),

(1.5)

where (z)_n,k is the k-Pochammer symbol introduced by Diaz and Pariguan [3] defined for complex z ∈ ℂ and k ∈ ℝ as

{(z)}_{n, k} = {\begin{matrix} 1 & if n = 0, \\ z (z + k) (z + 2 k) \dots (z + (n - 1) k) & if n \in ℕ . \end{matrix}}

(1.6)

On taking k = 1, then the generalized K-Wright function (1.2) diminishes to the generalized Wright function (1.1).

1.1 Saigo fractional calculus operators

Saigo [18] defined the fractional integral and differential operators with the Gauss hyergeometric function as kernel, which are remarkable generalizations of the Riemann-Liouville (R-L) and Erdélyi-Kober fractional calculus operators (see; [11]).

For ς, τ, γ ∈ ℂ and x ∈ ℝ⁺ with ℜ(ς) > 0, the left-hand and the right-hand sided generalized fractional integral operators connected with Gauss hypergeometric function are defined as below:

(I_{0 +}^{ς, τ, γ} f) (x) = \frac{x^{- ς - τ}}{Γ (ς)} \int_{0}^{x} {(x - t)}^{ς - 1}_{2} F_{1} (ς + τ, - γ; ς; 1 - \frac{t}{x}) f (t) d t

(1.7)

and

(I_{-}^{ς, τ, γ} f) (x) = \frac{1}{Γ (ς)} \int_{x}^{\infty} {\frac{{(t - x)}^{ς - 1}}{t^{ς + τ}}}_{2} F_{1} (ς + τ, - γ; ς; 1 - \frac{x}{t}) f (t) dt

(1.8)

respectively. Here, ₂F₁(ς, τ; γ; z) is the Gauss hypergeometric function [11] defined for z ∈ ℂ, |z| < 1 and ς, τ ∈ ℂ,

γ \in ℂ \ ℤ_{0}^{-}

_{2} F_{1} (ς, τ; γ; z) = \sum_{n = 0}^{\infty} \frac{{(ς)}_{n} {(τ)}_{n}}{{(γ)}_{n}} \frac{z^{n}}{n!},

where (z)_n = (z)_n,₁. The corresponding fractional differential operators are

(D_{0 +}^{ς, τ, γ} f) (x) = {(\frac{d}{dx})}^{l} (I_{0 +}^{- ς + l, - τ - l, ς + γ - l} f) (x)

(1.9)

and

(D_{-}^{ς, τ, γ} f) (x) = {(- \frac{d}{dx})}^{l} (I_{-}^{- ς + l, - τ - l, ς + γ} f) (x)

(1.10)

where l = [ℜ (ς)] + 1 and [ℜ (ς)] is the integer part of ℜ(ς). Substituting τ = −ς and τ = 0 in equation (1.7) – (1.10), we get the corresponding R-L and Erdélyi-Kober fractional operators, respectively.

1.2 Marichev-Saigo-Maeda fractional operators

Marichev [13] was introduced and studied fractional calculus operators which are the generalization of the Saigo operators, later generalized by Saigo and Maeda [19]. For ς, ς′, τ, τ′, γ ∈ ℂ and x ∈ ℝ⁺ with ℜ(γ) > 0, the left-hand and right-hand sided MSM fractional integral and derivative operators associated with third Appell function F₃ are defined as

(I_{0 +}^{ς, ς^{'}, τ, τ^{'}, γ} f) (x) = \frac{x^{- ς}}{Γ (γ)} \int_{0}^{x} \frac{{(x - t)}^{γ - 1}}{t^{ς^{'}}} F_{3} (ς, ς^{'}, τ, τ, γ, 1 - \frac{t}{x}, 1 - \frac{x}{t}) f (t) dt

(1.11)

and

(I_{-}^{ς, ς^{'}, τ, τ^{'}, γ} f) (x) = \frac{x^{- ς^{'}}}{Γ (γ)} \int_{x}^{\infty} \frac{{(t - x)}^{γ - 1}}{t^{ς}} F_{3} (ς, ς^{'}, τ, τ, γ, 1 - \frac{x}{t}, 1 - \frac{t}{x}) f (t) dt

(1.12)

(D_{0 +}^{ς, ς^{'}, τ, τ^{'}, γ} f) (x) = {(\frac{d}{dx})}^{m} (I_{0 +}^{- ς^{'}, - ς, - τ^{'} + m, - τ, - γ + m} f) (x)

(1.13)

and

(D_{-}^{ς, ς^{'}, τ, τ^{'}, γ} f) (x) = {(- \frac{d}{dx})}^{m} (I_{-}^{- ς^{'}, - ς, - τ^{'}, - τ + m, - γ + m} f) (x)

(1.14)

respectively, where m = [ℜ(γ)] + 1 and the third Appell function [17], is defined by

F_{3} (ς, ς^{'}, τ, τ^{'}, γ; x, y) = \sum_{m, n, = 0}^{\infty} \frac{{(ς)}_{m} {(ς^{'})}_{n} {(τ)}_{m} {(τ^{'})}_{n}}{{(γ)}_{m + n}} \frac{x^{m} y^{n}}{m! n!}, max {| x |, | y |} < 1 .

(1.15)

1.3 Generalized k-Struve function

The generalized k-Struve function was defined by Nisar et al. [14] as

\begin{matrix} S_{ν, c}^{k} (t) = \sum_{n = 0}^{\infty} \frac{{(- c)}^{n}}{Γ_{k} (nk + ν + \frac{3 k}{2}) Γ (n + \frac{3}{2}) n!} {(\frac{t}{2})}^{2 n + \frac{ν}{k} + 1} \\ (k \in ℝ^{+}; c \in ℝ; ν > - 1) \end{matrix}

(1.16)

taking k→ 1 and c = 1; (1.15) reduces to yield the well-known Struve function of order ν is defined by [1] as

H_{ν} (t) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{Γ (n + ν + \frac{3}{2}) Γ (n + \frac{3}{2}) n!} {(\frac{t}{2})}^{2 n + ν + 1}

(1.17)

For more details about Struve functions, their generalizations and properties, the esteemed reader is invited to consider references [2, 7, 8, 14, 15, 20,21,22].

The following MSM integral operators are required here [19, p. 394] to obtain the MSM fractional integration of generalized k-Struve function.

Lemma 1

Let ς, ς′, τ, τ′, γ, ρ ∈ ℂ such that ℜ(ς) > 0

(i)ℜ(ρ) > 0 max{0,ℜ(ς′ − τ′),ℜ(ς + ς′ + τ − γ)}, then
$(I_{0 +}^{ς, ς^{'}, τ, τ^{'}, γ} t^{ρ - 1}) (x) = \frac{Γ (ρ) Γ (- ς^{'} + τ^{'} + ρ) Γ (- ς - ς^{'} - τ + γ + ρ)}{Γ (τ^{'} + ρ) Γ (- ς - ς^{'} + γ + ρ) Γ (- ς^{'} - τ + γ + ρ)} x^{- ς - ς^{'} + γ + ρ - 1}$ (1.18)
(ii)If ℜ(ρ) > max{ℜ(τ),ℜ(−ς − ς′ + γ),ℜ(−ς − τ′ + γ}, then
$(I_{-}^{ς, ς^{'}, τ, τ^{'}, γ} t^{- ρ}) (x) = \frac{Γ (- τ + ρ) Γ (ς + ς^{'} - γ + ρ) Γ (ς + τ^{'} - γ + ρ)}{Γ (ρ) Γ (ς - τ + ρ) Γ (ς + ς^{'} + τ^{'} - γ + ρ)} x^{- ς - ς^{'} + γ - ρ}$ (1.19)

Further, to obtain the MSM fractional differentiation of the generalized k-Struve function, following results will be used from [9] as below:

Lemma 2

Let ς, ς′, τ, τ′, γ, ρ ∈ ℂ, such that ℜ(ς) > 0;

(i)If ℜ(ρ) > max{0,ℜ(−ς + τ),ℜ(−ς − ς′ − τ′ + γ)}
$(D_{0 +}^{ς, ς^{'}, τ, τ^{'}, γ} t^{ρ - 1}) (x) = \frac{Γ (ρ) Γ (- τ + ς + ρ) Γ (ς + ς^{'} + τ^{'} - γ + ρ)}{Γ (- τ + ρ) Γ (ς + ς^{'} - γ + ρ) Γ (ς + τ^{'} - γ + ρ)} x^{ς + ς^{'} - γ + ρ - 1}$ (1.20)
(ii)If ℜ(ρ) > max{ℜ(−τ′),ℜ(ς′ + τ − γ),ℜ(ς + ς′ − γ) + [ℜ(γ)] + 1}, then
$(D_{-}^{ς, ς^{'}, τ, τ^{'}, γ} t^{- ρ}) (x) = \frac{Γ (τ^{'} + ρ) Γ (- ς - ς^{'} + γ + ρ) Γ (- ς^{'} - τ + γ + ρ)}{Γ (ρ) Γ (- ς^{'} + τ^{'} + ρ) Γ (- ς - ς^{'} - τ + γ + ρ)} x^{ς + ς^{'} - γ - ρ}$ (1.21)

2 Fractional Calculus Approach

In this section, the following six theorems for k-Struve function concerning to MSM fractional integral and differential operators are established here as main results.

Theorem 1

Let ς, ς′, τ, tau′, γ, ρ ∈ ℂ and k ∈ ℝ⁺ be such that

ℜ (γ) > 0, ℜ (\frac{λ}{k}) > max {0, ℜ (ς^{'} - τ^{'}), ℜ (ς + ς^{'} + τ - γ)}

. Also let c ∈ ℝ; ν > −1, then for t > 0

\begin{matrix} (I_{0 +}^{ς, ς^{'}, τ, τ^{'}, γ} (t^{\frac{λ}{k} - 1} S_{ν, c}^{k} (t))) (x) = \frac{k^{γ + \frac{1}{2}} x^{- ς - ς^{'} + γ + \frac{λ}{k} + \frac{ν}{k}}}{2^{\frac{ν}{k} + 1}} \\ \times_{3} Ψ_{5}^{k} [\begin{matrix} (λ + ν + k, 2 k), & (- k ς^{'} + k τ^{'} + λ + ν + k, 2 k), \\ (k τ^{'} + λ + ν + k, 2 k), & (- k ς - k ς^{'} + k γ + λ + ν + k, 2 k), \end{matrix} \\ \begin{matrix} (- k ς - k ς^{'} - k τ + k γ + λ + ν + k, 2 k) \\ (- k ς^{'} - k τ + k γ + λ + ν + k, 2 k), & (ν + \frac{3 k}{2}, k), (\frac{3 k}{2}, k) \end{matrix} | \frac{- c x^{2} k}{4}] . \end{matrix}

(2.1)

Proof

On using (1.16) and taking the left-hand sided MSM fractional integral operator inside the summation, the left-hand side of (2.1) becomes

= \sum_{n = 0}^{\infty} \frac{{(- c)}^{n}}{Γ_{k} (nk + ν + \frac{3 k}{2}) Γ (n + \frac{3}{2}) n {! 2}^{2 n + \frac{ν}{k} + 1}} (I_{0 +}^{ς, ς^{'}, τ, τ^{'}, γ} {t^{\frac{λ}{k} + \frac{ν}{k} + 2 n}}) (x),

Making use of (1.18), we obtain

\begin{matrix} = \sum_{n = 0}^{\infty} \frac{{(- c)}^{n} x^{- ς - ς^{'} + γ \frac{λ}{k} + \frac{ν}{k} + 2 n}}{Γ_{k} (nk + ν + \frac{3 k}{2}) Γ (n + \frac{3}{2}) n {! 2}^{2 n + \frac{ν}{k} + 1}} \frac{Γ (\frac{λ}{k} + \frac{ν}{k} + 2 n + 1)}{Γ (τ^{'} + \frac{λ}{k} + \frac{ν}{k} + 2 n + 1)} \\ \times \frac{Γ (- ς - ς^{'} - τ + γ + \frac{λ}{k} + \frac{ν}{k} + 2 n + 1) Γ (- ς^{'} + τ^{'} + \frac{λ}{k} + \frac{ν}{k} + 2 n + 1)}{Γ (- ς^{'} - τ + γ + \frac{λ}{k} + \frac{ν}{k} + 2 n + 1) Γ (- ς - ς^{'} + γ + \frac{λ}{k} \frac{ν}{k} + 2 n + 1)}, \end{matrix}

Now, using equation (1.5) on above term, then we get

\begin{matrix} = \frac{x^{- ς - ς^{'} + γ + \frac{λ}{k} + \frac{ν}{k}}}{2^{\frac{ν}{k} + 1} k^{- γ - \frac{1}{2}}} \sum_{n = 0}^{\infty} \frac{Γ_{k} (λ + ν + k + 2 nk) Γ_{k} (- k ς^{'} + k τ^{'} + λ + ν + k + 2 nk)}{Γ_{k} (k τ^{'} + λ + ν + k + 2 nk) Γ_{k} (- k ς - k ς^{'} + k γ + λ + ν + k + 2 nk)} \\ \times \frac{Γ_{k} (- k ς - k ς^{'} - k τ + k γ + λ + ν + k + 2 nk)}{Γ_{k} (- k ς^{'} - k τ + k γ + λ + ν + k + 2 nk) Γ_{k} (nk + ν + \frac{3 k}{2}) Γ_{k} (\frac{3 k}{2} + nk) n!} {(\frac{- c x^{2} k}{2^{2}})}^{n} . \end{matrix}

Using the definition of (1.2) in the above term, we arrive at the result (2.1).

Next theorem gives the right-hand MSM fractional integration of $S_{ν, c}^{k} (.)$ .

Theorem 2

Let ς, ς′, τ, τ′, γ, ρ ∈ ℂ and k ∈ ℝ⁺ be such that ℜ(γ) > 0,

ℜ (γ) > 0, ℜ (\frac{λ}{k}) > max {ℜ (τ), ℜ (- ς - ς^{'} + γ), ℜ (- ς - τ^{'} + γ)}

. Also let c ∈ ℝ; ν > −1, then for t > 0

\begin{matrix} (I_{-}^{ς, ς^{'}, τ, τ^{'}, γ} (t^{\frac{λ}{k} - 1} S_{ν, c}^{k} (t))) (x) = \frac{k^{γ + \frac{1}{2}} x^{- ς - ς^{'} + γ + \frac{λ}{k} + \frac{ν}{k}}}{2^{\frac{ν}{k} + 1}} \\ \times_{3} Ψ_{5}^{k} [\begin{matrix} (- k τ - λ - ν, - 2 k), & (k ς + k ς^{'} - k γ - nu, - 2 k), \\ (- λ - ν, - 2 k), & (k ς - k τ - λ - ν, - 2 k), \end{matrix} \\ \begin{matrix} (k ς + k τ^{'} - k γ - λ - ν, - 2 k) \\ (k ς + k ς^{'} + k τ^{'} - k γ - λ - ν, - 2 k), & (ν + \frac{3 k}{2}, k), (\frac{3 k}{2}, k) \end{matrix} | \frac{- c x^{2} k}{4}] . \end{matrix}

(2.2)

Proof

On using (1.16) and taking the right-hand sided MSM fractional integral operator inside the summation, the left hand side of (2.2) becomes

= \sum_{n = 0}^{\infty} \frac{{(- c)}^{n}}{Γ_{k} (nk + ν + \frac{3 k}{2}) Γ (n + \frac{3}{2}) n {! 2}^{2 n + \frac{ν}{k} + 1}} (I_{-}^{ς, ς^{'}, τ, τ^{'}, γ} {t^{\frac{λ}{k} + \frac{ν}{k} + 2 n}}) (x)

On using (1.19), we get

\begin{matrix} = \sum_{n = 0}^{\infty} \frac{{(- c)}^{n} x^{- ς - ς^{'} + γ + \frac{λ}{k} + \frac{ν}{k} + 2 n}}{Γ_{k} (nk + ν + \frac{3 k}{2}) Γ (n + \frac{3}{2}) n {! 2}^{2 n + \frac{ν}{k}} + 1} \frac{Γ (- τ - \frac{λ}{k} - \frac{ν}{k} - 2 n)}{Γ (- \frac{λ}{k} - \frac{ν}{k} - 2 n)} \\ \times \frac{Γ (ς + τ^{'} - γ - \frac{λ}{k} - \frac{ν}{k} - 2 n) Γ (ς + ς^{'} - γ - \frac{λ}{k} - \frac{ν}{k} - 2 n)}{Γ (ς + ς^{'} + τ^{'} - γ - \frac{λ}{k} - \frac{ν}{k} - 2 n) Γ (ς - τ - \frac{λ}{k} - \frac{ν}{k} - 2 n)} \\ = \sum_{n = 0}^{\infty} \frac{{(- c x^{2})}^{n} x^{- ς - ς^{'} + γ + \frac{λ}{k} + \frac{ν}{k}}}{Γ_{k} (nk + ν + \frac{3 k}{2}) n {! 2}^{2 n + \frac{ν}{k} + 1}} \frac{Γ (- τ - \frac{λ}{k} - \frac{ν}{k} - 2 n) Γ (ς + ς^{'} - γ - \frac{λ}{k} - \frac{ν}{k} - 2 n)}{Γ (n + \frac{3}{2}) Γ (- \frac{λ}{k} - \frac{ν}{k} - 2 n) Γ (ς - τ - \frac{λ}{k} - \frac{ν}{k} - 2 n)} \\ \times \frac{Γ (ς + τ^{'} - γ - \frac{λ}{k} - \frac{ν}{k} - 2 n)}{Γ (ς + ς^{'} + τ^{'} - γ - \frac{λ}{k} - \frac{ν}{k} - 2 n)} \\ = \frac{x^{- ς - ς^{'} + γ + \frac{λ}{k} + \frac{ν}{k}}}{2^{\frac{ν}{k} + 1} k^{- γ - \frac{1}{2}}} \sum_{n = 0}^{\infty} {(\frac{- ck x^{2}}{4})}^{n} \frac{1}{n!} \frac{Γ_{k} (- k τ - λ - ν - 2 nk)}{Γ_{k} (- λ - ν - 2 nk) Γ_{k} (k ς - k τ - λ - ν - 2 nk)} \\ \times \frac{Γ_{k} (k ς + k ς^{'} - k γ - λ - ν - 2 nk) Γ_{k} (k ς + k τ^{'} - k γ - λ - ν - 2 nk)}{Γ_{k} (k ς + k ς^{'} + k τ^{'} - k γ - λ - ν - 2 nk) Γ_{k} (nk + ν + \frac{3 k}{2}) Γ_{k} (\frac{3 k}{2} + nk)} \end{matrix}

and the result follows on making use of (1.5) and definition of generalized k-Wright function.

Theorem 3

Let ς, ς′, τ, τ′, γ, ρ ∈ ℂ and k ∈ ℝ⁺ be such that ℜ(γ) > 0,

ℜ (γ) > 0, ℜ (\frac{λ}{k}) > max {ℜ (τ), ℜ (- ς - ς^{'} + γ), ℜ (- ς - τ^{'} + γ)}

. Also let c ∈ ℝ; ν > −1, then for t > 0

\begin{matrix} (I_{-}^{ς, ς^{'}, τ, τ^{'}, γ} (t^{\frac{- λ}{k}} S_{ν, c}^{k} (t))) (x) = \frac{k^{γ - \frac{1}{2}} x^{- ς - ς^{'} + γ + \frac{ν}{k} - \frac{λ}{k} + 1}}{2^{\frac{ν}{k} + 1}} \\ \times_{3} Ψ_{5}^{k} [\begin{matrix} (- k τ + λ - ν, - 2 k), & (k ς + k ς^{'} - k γ + λ - nu - k, - 2 k), \\ (- λ - ν - k, - 2 k), & (k ς - k τ + λ - ν - k, - 2 k), \end{matrix} \\ \begin{matrix} (k ς + k τ^{'} - k γ + λ - ν - k, - 2 k) \\ (k ς + k ς^{'} + k τ^{'} - k γ + λ - ν - k, - 2 k), & (ν + \frac{3 k}{2}, k), (\frac{3 k}{2}, k) \end{matrix} | \frac{- c x^{2} k}{4}] . \end{matrix}

(2.3)

Proof

On using (1.16) and taking the right-hand sided MSM fractional integral operator inside the summation, the left hand side of (2.3) becomes

= \sum_{n = 0}^{\infty} \frac{{(- c)}^{n}}{Γ_{k} (nk + ν + \frac{3 k}{2}) Γ (n + \frac{3}{2}) n {! 2}^{2 n + \frac{ν}{k} + 1}} (I_{-}^{ς, ς^{'}, τ, τ^{'}, γ} {t^{\frac{ν - λ}{k} + 2 n + 1}})

On using (1.19), we obtain

\begin{matrix} = \frac{{(- c)}^{n} t^{- ς - ς^{'} + γ + \frac{ν}{k} - \frac{λ}{k} + 2 n + 1}}{Γ_{k} (nk + ν + \frac{3 k}{2}) Γ (n + \frac{3}{2}) n {! 2}^{2 n + \frac{ν}{k} + 1}} \frac{Γ (- τ + \frac{λ - ν}{k} - 2 n - 1)}{Γ (\frac{λ - nu}{k} - 2 n - 1)} \\ \times \frac{Γ (ς + ς^{'} - γ + \frac{λ - ν}{k} - 2 n - 1) Γ (ς + τ^{'} - γ + \frac{λ - ν}{k} - 2 n - 1)}{γ (ς - τ + \frac{λ - ν}{k} - 2 n - 1) Γ (ς + ς^{'} + τ^{'} - γ + \frac{λ - ν}{k} - 2 n - 1)} \end{matrix}

Making use of (1.5), we get

\begin{matrix} = \frac{x^{- ς - ς^{'} + γ + \frac{ν}{k} - \frac{λ}{k} + 1}}{k^{- γ + \frac{1}{2}} 2^{\frac{ν}{k} + 1}} \sum_{n = 0}^{\infty} \frac{{(- ck x^{2})}^{n}}{4^{n} n!} \frac{Γ_{k} (- k τ + λ - ν - k - 2 nk)}{Γ_{k} (λ - ν - k - 2 nk) Γ_{k} (k ς - k τ + λ - ν - k - 2 nk)} \\ \times \frac{Γ_{k} (k ς + k ς^{'} - k γ + λ - ν - k - 2 nk) Γ_{k} (k ς + k τ^{'} - k γ + λ - ν - k - 2 nk)}{Γ_{k} (k ς + k ς^{'} + k τ^{'} - k γ + λ - ν - k - 2 nk) Γ_{k} (nk + ν + \frac{3 k}{2}) Γ_{k} (nk + \frac{3 k}{2})} \end{matrix}

This on expressing in terms of k-Wright function

_{p} Ψ_{q}^{k}

using (1.2) leads to the right-hand side of (2.3). This completes the proof of theorem.

The next theorem obtains the left-hand sided MSM fractional differentiation of k-Struve function.

Theorem 4

Let ς, ς′, τ, τ′, γ, ρ ∈ ℂ and k ∈ ℝ⁺ be such that

ℜ (\frac{λ}{k}) > max {0, ℜ (- ς + τ), ℜ (- ς - ς^{'} - τ^{'} + γ)}

. Also let c ∈ ℝ; ν > −1, then for t > 0

\begin{matrix} (D_{0 +}^{ς, ς^{'}, τ, τ^{'}, γ} (t^{\frac{λ}{k} - 1} S_{ν, c}^{k} (t))) (x) = \frac{k^{- γ + \frac{1}{2}} x^{ς + ς^{'} - γ + \frac{λ}{k} + \frac{ν}{k}}}{2^{\frac{ν}{k} + 1}} \\ \times_{3} Ψ_{5}^{k} [\begin{matrix} (λ + ν + k, 2 k), & (- k τ + k ς + λ + ν + k, 2 k), \\ (- k τ + λ + ν + k, 2 k), (k ς + k ς^{'} - k γ + λ + ν + k, 2 k), \end{matrix} \\ \begin{matrix} (k ς + k ς^{'} + k τ - k γ + λ + ν + k, 2 k) \\ (k ς + k τ^{'} - k γ + λ + ν + k, 2 k), & (ν + \frac{3 k}{2}, k), (\frac{3 k}{2}, k) \end{matrix} | \frac{- c x^{2} k}{4}] . \end{matrix}

(2.4)

Proof

On using (1.16) and taking the left-hand sided MSM fractional derivative inside the summation, the left-hand side of (2.4) becomes

= \sum_{n = 0}^{\infty} \frac{{(- c)}^{n}}{Γ_{k} (nk + ν + \frac{3 k}{2}) Γ (n + \frac{3}{2}) n {! 2}^{\frac{ν}{k} + 2 n + 1}} (D_{0 +}^{ς, ς^{'}, τ, τ^{'}, γ} (t^{\frac{λ}{k} + \frac{ν}{k} + 2 n}))

Using (1.20) in above term, we obtain

\begin{matrix} = \sum_{n = 0}^{\infty} \frac{{(- c)}^{n} Γ (\frac{λ}{k} + \frac{ν}{k} + 2 n + 1) Γ (- τ + ς + \frac{λ}{k} + \frac{ν}{k} + 2 n + 1)}{Γ_{k} (nk + ν + \frac{3 k}{2}) Γ (n + \frac{3}{2}) n {! 2}^{\frac{ν}{k} + 2 n + 1} Γ (- τ + \frac{λ}{k} + \frac{ν}{k} + 2 n + 1)} \\ \times \frac{Γ (ς + ς^{'} + τ^{'} - γ + \frac{λ}{k} + \frac{ν}{k} + 2 n + 1)}{Γ (ς + ς^{'} - γ + \frac{λ}{k} + \frac{ν}{k} + 2 n + 1) Γ (ς + τ^{'} - γ + \frac{λ}{k} + \frac{ν}{k} + 2 n + 1)} x^{ς + ς^{'} - γ + \frac{λ}{k} + \frac{ν}{k} + 2 n} \\ = \frac{x^{ς + ς^{'} - γ \frac{λ}{k} + \frac{ν}{k}}}{2^{\frac{ν}{k} + 1}} \sum_{n = 0}^{\infty} \frac{{(- c x^{2})}^{n}}{n {! 4}^{n} Γ_{k} (nk + ν + \frac{3 k}{2})} \frac{Γ (\frac{λ}{k} + \frac{ν}{k} + 2 n + 1)}{Γ (n + \frac{3}{2}) Γ (- τ + \frac{λ}{k} + \frac{ν}{k} + 2 n + 1)} \\ \times \frac{Γ (- τ + ς + \frac{λ}{k} + \frac{ν}{k} + 2 n + 1) Γ (ς + ς^{'} + τ^{'} - γ + \frac{λ}{k} + \frac{ν}{k} + 2 n + 1)}{Γ (ς + ς^{'} - γ + \frac{λ}{k} + \frac{ν}{k} + 2 n + 1) Γ (ς + τ^{'} - γ + \frac{λ}{k} + \frac{ν}{k} + 2 n + 1)} \\ = \frac{k^{- γ + \frac{1}{2}} x^{ς + ς^{'} - γ + \frac{λ}{k} + \frac{ν}{k}}}{2^{\frac{ν}{k} + 1}} \sum_{n = 0}^{\infty} \frac{{(- ck x^{2})}^{n}}{n {! 4}^{n}} \\ \times \frac{Γ_{k} (λ + ν + k + 2 nk) Γ_{k} (- k τ + k ς + λ + ν + k + 2 nk)}{Γ_{k} (nk + ν + \frac{3 k}{2}) Γ_{k} (nk + \frac{3 k}{2}) Γ_{k} (- k τ + λ + ν + k + 2 nk)} \\ \times \frac{Γ_{k} (k ς + k ς^{'} + k τ^{'} - k γ + λ + ν + k + 2 nk)}{Γ_{k} (k ς + k ς^{'} - k γ + λ + ν + k + 2 nk) Γ_{k} (k ς + k τ^{'} - k γ + λ + ν + k + 2 nk)} \end{matrix}

In above term, we use equation (1.5), and the result follows by using (1.2), then we arrive at (2.4).

The next theorem gives the right-hand sided MSM fractional derivative of k-Struve function.

Theorem 5

Let ς, ς′, τ, τ′, γ, ρ ∈ ℂ and k ∈ ℝ⁺ be such that

ℜ (\frac{λ}{k}) > max {ℜ (- τ^{'}), ℜ (ς^{'} + τ - γ), ℜ (ς + ς^{'} - γ) + [ℜ (γ)] + 1}

. Also let c ∈ ℝ; ν > −1, then for t > 0

\begin{matrix} (D_{-}^{ς, ς^{'}, τ, τ^{'}, γ} (t^{\frac{λ}{k} - 1} S_{ν, c}^{k} (t))) (x) = \frac{k^{- γ + \frac{1}{2}} x^{ς + ς^{'} - γ + \frac{λ}{k} + \frac{ν}{k}}}{2^{\frac{ν}{k} + 1}} \\ \times_{3} Ψ_{5}^{k} [\begin{matrix} (k τ^{'} - λ - ν, - 2 k), & (- k ς - k ς^{'} + k γ - λ - ν, - 2 k), \\ (- λ - ν, - 2 k), & (- k ς^{'} + k τ^{'} - λ - ν, - 2 k), \end{matrix} \\ \begin{matrix} (- k ς^{'} + k τ + k γ - λ - ν, - 2 k) \\ (- k ς - k ς^{'} - k τ + k γ - λ - ν, - 2 k), & (ν + \frac{3 k}{2}, k), (\frac{3 k}{2}, k) \end{matrix} | \frac{- c x^{2} k}{4}] . \end{matrix}

(2.5)

Proof

On using (1.16) and taking the left-hand sided MSM fractional derivative inside the summation, the left-hand side of (2.5) becomes

= \sum_{n = 0}^{\infty} \frac{{(- c)}^{n}}{Γ_{k} (nk + ν + \frac{3 k}{2}) Γ (n + \frac{3}{2}) n {! 2}^{\frac{ν}{k} + 2 n + 1}} (D_{-}^{ς, ς^{'}, τ, τ^{'}, γ} (t^{\frac{λ}{k} + \frac{ν}{k} + 2 n}))

Using (1.21) in above term, we obtain

\begin{matrix} = \sum_{n = 0}^{\infty} \frac{{(- c)}^{n} Γ (τ^{'} - \frac{λ}{k} - \frac{ν}{k} - 2 n)}{Γ_{k} (nk + ν + \frac{3 k}{2}) Γ (n + \frac{3}{2}) n {! 2}^{\frac{ν}{k} + 2 n + 1} Γ (- \frac{λ}{k} - \frac{ν}{k} - 2 n)} \\ \times \frac{Γ (- ς - ς^{'} + γ - \frac{λ}{k} - \frac{ν}{k} - 2 n) Γ (- ς^{'} - τ + γ - \frac{λ}{k} - \frac{ν}{k} - 2 n)}{Γ (- ς - ς^{'} - τ + γ - \frac{λ}{k} - \frac{ν}{k} - 2 n) Γ (- ς^{'} + τ^{'} - \frac{λ}{k} - \frac{ν}{k} - 2 n)} x^{ς + ς^{'} - γ + \frac{λ}{k} + \frac{ν}{k} + 2 n} \\ = \frac{x^{ς + ς^{'} - γ \frac{λ}{k} + \frac{ν}{k}}}{2^{\frac{ν}{k} + 1}} \sum_{n = 0}^{\infty} \frac{{(- c x^{2})}^{n}}{n {! 4}^{n} Γ_{k} (nk + ν + \frac{3 k}{2})} \frac{Γ (τ^{'} - \frac{λ}{k} - \frac{ν}{k} - 2 n)}{Γ (- \frac{λ}{k} - \frac{ν}{k} - 2 n)} \\ \times \frac{Γ (- ς - ς^{'} + γ - \frac{λ}{k} - \frac{ν}{k} - 2 n) Γ (- ς^{'} - τ + γ - \frac{λ}{k} - \frac{ν}{k} - 2 n)}{Γ (- ς^{'} + τ^{'} - \frac{λ}{k} - \frac{ν}{k} - 2 n) Γ (- ς - ς^{'} - τ + γ - \frac{λ}{k} - \frac{ν}{k} - 2 n)} \\ = \frac{k^{- γ + \frac{1}{2}} x^{ς + ς^{'} - γ + \frac{λ}{k} + \frac{ν}{k}}}{2^{\frac{ν}{k} + 1}} \sum_{n = 0}^{\infty} \frac{{(- ck x^{2})}^{n}}{n {! 4}^{n}} \\ \times \frac{Γ_{k} (k τ^{'} - λ - ν - 2 nk) Γ_{k} (- k ς - k ς^{'} + k γ - λ - ν - 2 nk)}{Γ_{k} (- λ - ν - 2 nk) Γ_{k} (- k ς^{'} + k τ^{'} - λ - ν - 2 nk)} \\ \times \frac{Γ_{k} (- k ς^{'} - k τ + k γ - λ - ν - 2 nk)}{Γ_{k} (- k ς - k ς^{'} - k τ + k γ - λ - ν - 2 nk) t Γ_{k} (ν + \frac{3 k}{2} + nk) Γ_{k} (\frac{3 k}{2} + nk)} \end{matrix}

Thus, in accordance with (1.2), we get the required result (2.5).

Theorem 6

Let ς, ς′, τ, τ′, γ, ρ ∈ ℂ and k ∈ ℝ⁺ be such that

ℜ (\frac{λ}{k}) > max {ℜ (- τ^{'}), ℜ (ς^{'} + τ - γ), ℜ (ς + ς^{'} - γ) + [ℜ (γ)] + 1}

. Also let c ∈ ℝ; ν > −1, then for t > 0

\begin{matrix} (D_{-}^{ς, ς^{'}, τ, τ^{'}, γ} (t^{- \frac{λ}{k}} S_{ν, c}^{k} (t))) (x) = \frac{k^{- γ + \frac{1}{2}} x^{ς + ς^{'} - γ + \frac{λ}{k} + \frac{ν}{k} + 1}}{2^{\frac{ν}{k} + 1}} \\ \times_{3} Ψ_{5}^{k} [\begin{matrix} (k τ^{'} + λ - ν - k, - 2 k), & (- k ς - k ς^{'} + k γ - λ - ν - k, - 2 k), \\ (- λ - ν - k, - 2 k), & (- k ς^{'} + k τ^{'} + λ - ν - k, - 2 k), \end{matrix} \\ \begin{matrix} (- k ς^{'} - k τ + k γ + λ - ν - k, - 2 k) \\ (- k ς - k ς^{'} - k τ + k γ + λ - ν - k, - 2 k), & (ν + \frac{3 k}{2}, k), (\frac{3 k}{2}, k) \end{matrix} | \frac{- c x^{2} k}{4}] . \end{matrix}

(2.6)

Proof

On using (1.16) and taking the right-hand sided MSM fractional derivative inside the summation, the left-hand side of (2.6) becomes

= \sum_{n = 0}^{\infty} \frac{{(- c)}^{n}}{Γ_{k} (nk + ν + \frac{3 k}{2}) Γ (n + \frac{3}{2}) n {! 2}^{\frac{ν}{k} + 2 n + 1}} (D_{-}^{ς, ς^{'}, τ, τ^{'}, γ} (t^{\frac{ν}{k} - \frac{λ}{k} + 2 n + 1}))

Using (1.21), we have

\begin{matrix} = \sum_{n = 0}^{\infty} \frac{{(- c)}^{n}}{Γ_{k} (nk + ν + \frac{3 k}{2}) Γ (n + \frac{3}{2}) n {! 2}^{\frac{ν}{k} + 2 n + 1}} \frac{Γ (τ^{'} + \frac{λ - ν}{k} - 2 n - 1)}{Γ (\frac{λ - ν}{k} - 2 n - 1)} \\ \times \frac{Γ (- ς - ς^{'} + γ + \frac{λ - ν}{k} - 2 n - 1) Γ (- ς^{'} - τ + γ + \frac{λ - ν}{k} - 2 n - 1)}{Γ (- ς^{'} + τ^{'} + \frac{λ - ν}{k} - 2 n - 1) Γ (- ς - ς^{'} - τ + γ + \frac{λ - ν}{k} - 2 n - 1)} x^{ς + ς^{'} - γ + \frac{ν}{k} - \frac{λ}{k} + 2 n + 1} \end{matrix}

Making use of (1.5), we obtain

\begin{matrix} = \frac{t^{ς + ς^{'} - γ + \frac{ν}{k} - \frac{λ}{k} + 1}}{2^{\frac{ν}{k} + 1}} \sum_{n = 0}^{\infty} \frac{{(- ck t^{2})}^{n}}{n {! 4}^{n} k^{γ - \frac{1}{2}}} \\ \times \frac{Γ_{k} (k τ^{'} + λ - ν - k - 2 nk)}{Γ_{k} (λ - ν - k - 2 nk) Γ_{k} (- k ς^{'} + k τ^{'} + λ - ν - k - 2 nk)} \\ \times \frac{Γ_{k} (- k ς - k ς^{'} + k γ + λ - ν - k - 2 nk) Γ_{k} (- k ς^{'} - k τ + k γ + λ - ν - k - 2 nk)}{Γ_{k} (- k ς - k ς^{'} - k τ + k γ + λ - ν - k - 2 nk) Γ_{k} (nk + ν + \frac{3 k}{2}) Γ_{k} (nk + \frac{3 k}{2})} \end{matrix}

This on expressing in terms of k-Wright function

_{p} Ψ_{q}^{k}

using (1.2) leads to the right-hand side of (2.6). This completes the proof.

3 Concluding Remark

MSM fractional calculus operators have more advantage due to the generalize of Riemann-Liouville, Weyl, Erdélyi-Kober, and Saigo's fractional calculus operators; therefore, many authors are called as general operator. Now we are going to conclude of this paper by emphasizing that our leading results (Theorems 1 – 6) can be derived as the specific cases involving familiar fractional calculus operators as above said. On other hand, the k Struve function defined in (1.16) possesses the lead that a number of special functions occur to be the particular cases. Some of special cases respect to the integrals relating with k Struve function have been discovered in the earlier research works by various authors with not the same arguments.

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Article information

Received:: 2019-09-20

Accepted:: 2019-11-11

Published Online:: 2020-12-30

Citation Information:: Applied Mathematics and Nonlinear Sciences, Volume 5, Issue 2, Pages 593–602, eISSN 2444-8656, DOI: https://doi.org/10.2478/amns.2020.2.00064.

OPEN ACCESS

Journal + Issues

Applied Mathematics and Nonlinear Sciences

Online ISSN:: 2444-8656

First Published:: 01 Jan 2016

Language:: English

Publisher:: Sciendo

The Marichev-Saigo-Maeda Fractional Calculus Operators Pertaining to the Generalized K-Struve Function

Abstract

1 Introduction and Preliminaries

1.1 Saigo fractional calculus operators

1.2 Marichev-Saigo-Maeda fractional operators

1.3 Generalized k-Struve function

2 Fractional Calculus Approach

3 Concluding Remark

References

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Applied Mathematics and Nonlinear Sciences

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