The (k, s)-Fractional Calculus of Class of a Function

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THE (k, s)-FRACTIONAL CALCULUS OF CLASS OF A FUNCTION

GAUHAR RAHMAN, ABDUL GHAFFAR, K. S. NISAR*, AZEEMA

Abstract. In this present paper, we deal with the generalized (k, s)-fractional integral

and differential operators recently defined by Nisar et al. and obtain some generalized (k, s)-fractional integral and differential formulas involving the class of a function as its

kernels. Also, we investigate a certain number of their consequences containing the said function in their kernels.

1. introduction

Fractional calculus has gained appreciable fame and significance due to its numerous and boundless applications in Science; more particularly in engineering. For the recent development in the field of fractional calculus, one can refer to [5, 13, 15] and [6, 7]. One more direction to such study was proposed by Atangana and Baleanu [1] by introducing derivatives which are based upon the generalized Mittag-Leffler function. Integral inequal-ities are considered to be of highly importance because they are useful in the research of different subjects of differential and integral equations (see [8]). We begin with the work of Diaz and Pariguan [3] which is defined as:

(p)n,k = (

p(p+k), ..., (p+ (n−1)k) (n ∈N, p ∈_C),

1, (n= 0, p∈_C), (1.1)

and

Γk(η) = Z ∞

0

tη−1e−tkk dt, (1.2)

where η∈_C, k >0, <(z)>0.

They also define the following relations:

Γk(η+k) = ηΓk(η), (1.3)

2010Mathematics Subject Classification. 26B25, 33C45, 33B15, 26A33.

Key words and phrases. fractional integral operators; fractional differential operators; generalized (k, s)-fractional integral operators; generalized (k, s)-fractional differential operators.

*Corresponding author

G. Rahman: Department of Mathematics, International Islamic University, Islamabad, Pakistan;

[email protected]

A. Ghaffar: Department of Mathematical Science, BUITEMS Quetta, Pakistan;

[email protected]

K. S. Nisar: Department of Mathematics, College of Arts and Science, Prince Sattam bin Abdulaziz

Uni-versity, Wadi Al dawaser, Riyadh region 11991, Saudi Arabia; [email protected], [email protected] Azeema: Department of Mathematics, SBK Women’s University, Quetta, Pakistan;

[email protected].

1

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and

Γk(η) =k η k−1Γ

η

k

. (1.4)

A new generalization of the k-fractional integral has been defined by Mubeen and Habibul-lah [11] as follows:

I_kµ(f(x)) = 1 kΓk(µ)

Z x

0

(x−τ)µk−1f(τ)dτ. (1.5)

Clearly, it is observed that whenk = 1 (1.5) coincide with the result of Riemann-Liouville (R-L) fractional integration formula (see [5]). In fact, the following particular cases:

Iµ(f(x)) = 1 Γ(µ)

Z x

0

(x−τ)µ−1f(τ)dτ. (1.6)

In the same paper, they also define the following results:

I_kρ(xβk−1) = Γk(β)

Γk(ρ+β)

xkρ+ β

k−1, (1.7)

and

I_kρ((x−u)βk−1) = Γk(β)

Γk(ρ+β)

(x−u)ρk+ β

k−1. (1.8)

Sarikaya et al. [14] have developed the R-L (k, s)-fractional integral of order µ > 0 is defined by:

s kI

µ

af(x) =

(s+ 1)1−µk

kΓk(µ) Z x

0

(xs+1−ts+1)µk−1tsf(t)dt, (1.9)

where s∈ < \ {−1}, k > 0 andx∈[a;b]. Also, they define,

s kI

µ a[(x

s+1₋_as+1₎λ_k−1_{] =} Γk(λ)

(s+ 1)µkΓ_k(λ+µ)

(xs+1−as+1)λ+kµ−1. (1.10)

For more details about (k, s)-fractional integrals interesting readers can refer to [2, 12, 16]. Recently Nisar et al., [9, 10] defined the following R-L left and right sided (k, s)-fractional integral and differential operators of order µas:

s a,kD

−µ

x f(x) = s a,kI

µ

xf(x) = s kI

µ a+f(x)

= (s_kI_aµ₊f)(x) = (s+ 1) 1−µ_k

kΓk(µ) Z x

a

f(t)

(xs+1₋_ts+1₎1−µ_kt

s_dt,_{(x > a),}

(1.11)

s ρ,kD

−µ

β f(x) = s ρ,kI

µ

βf(x) = s k I

µ a−f(x)

= (s_kI_aµ₋f)(x) = (s+ 1) 1−µ

k

kΓk(µ) Z β

x

f(t) (xs+1₋_ts+1₎1−µ

k

tsdt,(x < β),

(1.12)

(s_kD_aµ₊f)(x) = 1 xs

d dx

n

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and

(s_kD_aµ₋f)(x) =− 1

xs

d dx

n

kn s_kI_ank₋−µf(x), (1.14) respectively. Substituting k = 1 and s = 0, then the above relation will coincide to the R-L left and right sided (k, s)-fractional integral and derivatives see ([5],[7]).

The R-L left and right sided (k, s)-fractional derivative operators kD

µ

a+ defined in (1.12) is generalized by (k, s)-fractional derivative operator is denoted by s

kD µ,ν

a+, whereµ is the order such that 0 < ν < 1, we define as

(s_kD_aµ,ν₊f)(x) =

s kI

ν(k−µ)

ρ+

1 xs

d dx

(kn s_kI_a(1₊−ν)(k−µ)f)(x), (1.15)

Obviously, when v = 0 then (1.15) approaches to the R-L (k, s)-fractional derivatives operator s

kD µ

a+ (1.13).

They [9] also defined the following lemma as:

Lemma 1.1. For k > 0, with x > ρ, 0 < ν < 1 and <(λ) > 0, then the following result for (k, s)-fractional derivatives operator s_kD_aµ₊ in (1.15) hold true:

(s_kD_aµ,ν₊(τs+1−as+1)λk−1)(x) = Γk(λ)

(s+ 1)−µkΓ_k(λ−µ)

(xs+1−as+1)λ−kµ−1. (1.16)

For the present investigation, we consider the following class of a function recently defined by Tunc [17]

zσ,kρ,λ(x) =

∞ X

n=0

σ(m) kΓk(ρkm+λ)

xm,(ρ, λ >0;|x|<_R), (1.17)

where the coefficients σ(m)(m ∈ N0 = N ∪ {0} is a bounded sequence of positive real numbers _R is the set of real numbers.

2. Generalized (k, s)-fractional integrals and differentials formulas

In this section, we present the generalized (k, s)-fractional integrals and differentials formulas involving a class of a function _zσ,k_ρ,λ(x) as defined in (1.17). In this continuation of the study of generalizedk-fractional calculus, we define the following fractional integral operator.

Definition 2.1. If k >0, ρ, δ, ω ∈ _C; with R(ρ)>0, R(β)> 0, R(δ)> 0, and x > ρ, then

(s_kεω_a_+;;σ_ρ,λf)(x) = 1 k

Z x

0

(xs+1−τs+1)λk−1_zσ,k ρ,λ(w(x

s+1₋_τs+1₎ρ_)τs_f(τ_)dτ, _(2.1)

By putting s= 0, then (2.1) can be written as

(kεωa+;;σρ,λf)(x) =

1 k

Z x

0

(x−τ)λk−1_zσ,k ρ,λ(w(x

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see [17]. Similarly, when ω = 0 andk = 1 then (2.2) turns to:

(I_aλ₊f)(x) = 1 Γ(λ)

Z x

0

f(τ)

(x−τ)1−λdτ,(R(µ)>0). (2.3)

To prove the generalized (k, s)-fractional integral and differential formulas of a class of a function, we first prove the following result.

Lemma 2.1. For k >0, the following result holds true:

1 xs

d dx

m n

(τs+1−as+1)λk−1_zα,k ρ,λ[w(x

s+1₋

as+1)]ρ

o

=k−m(s+ 1)m(xs+1−as+1)λk−m−1_zα,k

ρ,λ−mk[w(x

s+1₋_as+1_)]ρ _(2.4) where s ∈ _R\ {−1}; ω, α, λ, ρ∈_C; R(ω)>0, R(λ)>0, R(ρ)>0 and R(α)>0. Proof. LetS1 be the L. H. S. of (2.4) then

S1 =

1 xs

d dx

m n

(τs+1−as+1)λk−1_zα,k ρ,λ[w(x

s+1₋_as+1_)]ρo

=

1 xs

d dx

m n

(τs+1−as+1)λk−1

∞ X

n=0

σ(n) kΓk(ρkn+λ)

[w(xs+1−as+1)ρ]no.

Changing the order of summation and differentiation, we have

S1 =

∞ X

n=0

σ(n) kΓk(ρkn+λ)

[w]n

1 xs

d dx

m

(τs+1−as+1)λk+ρn−1.

(2.5) Now we find

1 xs

d dx

m

(xs+1−as+1)λk+ρn−1

= (s+ 1)m(ρn+λ

k −1)...(ρn+ λ

k −m)(x

s+1₋_as+1₎λ_k+ρn−m−1

= (s+ 1)m Γ(ρn+

λ k)

Γ(ρn+ λ_k −m)(x

s+1₋_as+1₎λ_k+ρn−m−1 _(2.6)

Using (1.4), we have

Γ(ρn+λ_k) Γ(ρn+ λ_k −m) =

Γk(nρk+λ)

km_Γ

k(nρk+λ−mk)

. (2.7)

Using (2.7) in (2.6), we get

1 xs

d dx

m

(xs+1−as+1)λk+ρn−1

= (s+ 1)m Γk(nρk+λ) km_Γ

k(nρk+λ−mk)

(xs+1−as+1)kλ+ρn−m−1. (2.8)

Thus by using (2.8) in (2.5), we get the following required result. S1 = k−m(s+ 1)m(xs+1−as+1)

λ

k−m−1_zα,k

ρ,λ−mk[w(x

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Theorem 2.1. For k >0, the following integral formulas holds true: s

kI µ

a+(τs+1−as+1)

λ

k−1)_zα,k ρ,λ[w(τ

s+1₋_as+1₎ρ_](x)

= (x

s+1₋_as+1₎λ+_kµ−1 (s+ 1)µk z

α,k

ρ,λ+µ[w(x

s+1₋_as+1_)]ρ_.

(2.10)

where x > a(a ∈ _R+ = [0,∞)); ω, α, λ, ρ ∈ _C; R(ω) > 0, R(λ) > 0, R(ρ) > 0, and

R(µ)>0.

Proof. LetS2 be L. H. S. side of (2.10) then S2 = skI

µ

a+(τs+1−as+1)

λ

s+1₋_as+1₎ρ_](x)

= (s+ 1) 1−µ

k

kΓk(µ) Z x

a

(τs+1₋_as+1₎λ_k−1

zα,kρ,λ[w(τs+1−as+1)ρ](x)

(xs+1₋_τs+1₎1−µ k

dτ

= (s+ 1) 1−µ

k

kΓk(µ) Z x

a

(τs+1−as+1)λk−1

(xs+1₋_τs+1₎1−µ_k ∞ X

n=0

σ(n) kΓk(nρk+λ)

[w(τs+1−as+1)ρ]ndτ

Changing the order of summation and integration, we have

S2 =

(s+ 1)1−µk

kΓk(µ)

∞ X

n=0

σ(n) kΓk(nρk+λ)

[w]n

Z x

a

(τs+1−as+1)λk+ρn−1

(xs+1₋_τs+1₎1−µ_k dτ

(2.11) Substitutingτs+1 =as+1+y(xs+1−as+1) in (2.11), then this implies τsdτ = (xs+1_s₊₁−as+1)dy, when τ →a, ⇒ y→0 and τ →x, ⇒ y →1, we have

S2 =

∞ X

n=0

σ(n)[w]n

kΓk(nρk+λ)

(s+ 1)1−µ_k

kΓk(µ)

×

Z 1

0

(αs+1₊_y(xs+1₋_αs+1₎₋_as+1₎λ_k+ρn−1 (xs+1₋_αs+1₋_y(xs+1₋_αs+1₎₎1−µ

k

(xs+1₋_αs+1₎ s+ 1 dy

=

∞ X

n=0

σ(n)[w]n

kΓk(nρk+λ)

(xs+1₋_αs+1₎µ+λ+_kρnk−1 (s+ 1)µkkΓ_k(µ)

Z 1

0

yλk+ρn−1(1−y) µ k−1dy

=

∞ X

n=0

σ(n)[w]n kΓk(nρk+λ)

(xs+1−αs+1)µ+λ+kρnk−1

(s+ 1)µkkΓ_k(µ)

Γk(λ+ρnk)Γk(µ)

Γk(λ+ρnk+µ)

= (x

s+1₋_αs+1₎µ+_kλ−1 (s+ 1)µk

∞ X

n=0

σ(n)

kΓk(nρk+λ+µ)

[w(τs+1−as+1)ρ]n

= (x

s+1₋_as+1₎λ+_kµ−1 (s+ 1)µk z

α,k

ρ,λ+µ[w(x

s+1₋_as+1_)]ρ_.

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Theorem 2.2. For k >0, the following result holds true: s

kD µ a+(τ

s+1₋

as+1)λk−1)_zα,k ρ,λ[w(τ

s+1₋

as+1)ρ](x)

= (x

s+1₋_as+1₎λ−_kµ−1 (s+ 1)−µk z

α,k

ρ,λ−µ[w(x

s+1₋_as+1_)]ρ_.

(2.12)

R(µ)>0.

Proof. LetS3 be L. H. S. of (2.12) then

S3 =

1 xs

d dx

n n

kn s_kI_ank₊−µ(τs+1−as+1)λk−1_zα,k ρ,λ[w(τ

s+1₋_as+1₎ρ_](x)o

Applying (2.10), this takes the following form

s kD

µ a+(τ

s+1₋_as+1₎λ_k−1₎

zα,kρ,λ[w(τ

s+1₋_as+1₎ρ_](x)

=kn

1 xs

d dx

n

(xs+1₋_as+1₎λ−_kµ+n−1 (s+ 1)−µk+n z

α,k

ρ,λ−µ+nk[w(x

s+1₋_as+1_)]ρ_.

Applying Lemma 2.4, we have

S3 =

(xs+1−as+1)λ−kµ−1

(s+ 1)−µk z α,k

ρ,λ−µ[w(x s+1₋

as+1)]ρ.

Theorem 2.3. For k >0, the following result holds true: s

kD µ,ν

a+(τs+1−as+1)

λ

s+1₋_as+1₎ρ_](x)

= (x

s+1₋_as+1₎λ−_kµ−1 (s+ 1)−kµ

zα,kρ,λ−µ[w(x s+1₋

as+1)]ρ.

(2.13)

R(µ)>0. Proof.

s kD

µ,ν

a+(τs+1−as+1)

λ

s+1₋_as+1₎ρ_](x)

=s_k Dµ,ν_a₊

" _∞ X

n=0

σ(n) kΓk(nρk+λ)

[w(τs+1−as+1)λnk−1+ρ]n #

.

This can be written as

s kD

µ,ν

a+(τs+1−as+1)

λ

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=

∞ X

n=0

σ(n)[w]n

kΓk(nρk+λ) h

s kD

µ,ν

a+(τs+1−as+1)

λ k+ρn−1

i

.

By applying (1.16), we get

s kD

µ,ν

a+(τs+1−as+1)

λ

s+1₋_as+1₎ρ_](x)

=

∞ X

n=0

Γk(λ+ρnk)

Γk(λ+ρnk−µ)

(τs+1−as+1)λk+ρn−1

(s+ 1)−µk

= (x

s+1₋_as+1₎λ−_kµ−1 (s+ 1)−kµ

zα,kρ,λ−µ[w(x s+1₋

as+1)]ρ.

Which complete the desired proof.

3. Properties of the operator skε ω,σ a+;ρ,λf

(x)

Theorem 3.1. For k > 0, ω, σ, λ, ρ ∈ _C; R(ω) > 0, R(λ) > 0, R(ρ) > 0, and

R(µ)>0, then the following result holds:

s kε

ω,σ a+;ρ,λ τ

s+1₋_as+1µ_k−1

(x) = (x

s+1₋_as+1₎λ+kµ−1_Γ k(µ)

(s+ 1) z

σ,k

ρ,λ+µ[w x

s+1₋_as+1ρ

]. (3.1)

Proof. From (2.1), we have

s kε

ω,σ a+;ρ,λ τ

s+1₋_as+1µ_k−1

(x)

= 1 k

Z x

a

τs+1−as+1

µ

k−1_(xs+1₋_τs+1₎λ_k−1

zσ,kρ,λ[ω x s+1₋

τs+1ρ]τsf(τ)dτ.

This can be written as

s kε

ω,σ a+;ρ,λ τ

s+1₋_as+1µ_k−1

(x)

=

∞ X

n=0

1 k

Z x

a

xs+1−τs+1

λ+ρkn

k −1 _τs+1₋_as+1 µ

k−1_τs_dτ

= (x

s+1₋_αs+1₎λ+kµ−1_Γ k(µ)

(s+ 1)

∞ X

n=0

σ(n)[ω(xs+1₋_τs+1₎ρ_]n

kΓk(nρk+λ)

Γk(nρk+λ)

Γk(nρk+λ+µ)

= (x

s+1₋_as+1₎λ+kµ−1_Γ k(µ)

(s+ 1) z

σ,k

ρ,λ+µ[w x

s+1₋_as+1ρ

].

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Theorem 3.2. Let s ∈ _R\ {−1}; k > 0; ω, λ, ρ ∈ _C; R(ω) > 0, R(ω) > 0, R(λ) > 0, R(ρ)>0, R(µ)>0 and x > a, then

(s_kI_aµ₊) _ksεω,σ_a_+;_ρ,λf(x) = 1 (s+ 1)µk

s kε

ω,σ a+;ρ,λ+µf

(x) = s_kεω,σ_a_+;_ρ,λ[s_kI_aµ₊f](x),

(3.2)

hold for any f ∈L[ρ, λ].

Proof. From equations (1.11) and (2.1), we have (s_kI_aµ₊) s_kεω,σ_a_+;_ρ,λf

(x)

= (s+ 1) 1−µ_k

kΓk(µ) Z x

a s kε

ω,σ a+;ρ,λf

(τ) (xs+1₋_τs+1₎1−kµ

τsdτ

= (s+ 1) 1−µ

k

k2_Γ

k(µ) Z x

a

xs+1−τs+1µ_k−1

×

Z τ

a

τs+1−us+1

λ k−1

zσ,kρ,λ w τ

s+1₋_us+1ρ

f(u)usdu

τsdτ.

By changing the order of integration, we obtain (s_kI_aµ₊) _ksεω,σ_a_+;_ρ,λf(x)

= 1 k

Z x

a

(s+ 1)1−µk

kΓk(µ)

Z x

u

xs+1−τs+1

µ

k−1 _τs+1₋_us+1 λ k−1

zσ,kρ,λ w τ

s+1₋_us+1ρ

τsdτif(u)usdu. (3.3)

Substitutingτs+1₋_us+1 ₌_ts+1 _{, this implies} _τs_dτ ₌_ts_{dt. Therefore (3.3) can be written}

as

(s_kI_aµ₊) _ksεω,σ_a_+;_ρ,λf(x)

= 1 k

Z x

a

(s+ 1)1−µk

kΓk(µ) "

Z xs+1−us+1

0

xs+1−ts+1−us+1

µ

k−1 _ts+1 λ k−1

zσ,kρ,λ w t s+1ρ

tsdt

i

f(u)usdu.

By the use of (1.11) and applying (2.10), we have (s_kI_aµ₊) s_kεω,σ_a_+;_ρ,λf

(x)

= 1

k(s+ 1)µk Z x

a

xs+1−us+1µ+_kλ−1

zσ,kρ,λ+µ w x

s+1₋_us+1ρ

f(u)usdu.

thus, we get

(s_kI_aµ₊) _ksεω,σ_a_+;_ρ,λf(x) = 1 (s+ 1)µk

s kε

ω,σ a+;ρ,λ+µf

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To prove the second part, consider the right part of (3.2), we have

s kε

ω,σ a+;ρ,λ[

s kI

µ a+f]

(x) = 1

k

Z x

a

xs+1−τs+1λ_k−1

zσ,kρ,λ w x

s+1₋_τs+1ρ

[s_kI_aµ₊f]τsdτ

= 1 k

Z x

a

xs+1−τs+1

λ k−1

zσ,kρ,λ

w xs+1−τs+1

ρ k

(s+ 1)1−µ k

kΓk(µ) Z τ

a

f(u)

(τs+1₋_us+1₎1−µk

usdu

!

dτ

By changing the order of integration, we obtain

s kε

ω,σ a+;ρ,λ[

s kI

µ a+f]

(x)

= 1 k

Z x

a

(s+ 1)1−µk

kΓk(µ)

Z x

u

xs+1−τs+1

µ

k−1 _τs+1₋_us+1 λ k−1

zσ,kρ,λ w x

s+1₋_τs+1ρ

τsdτif(u)usdu. Again making the use of (1.11) and applying (2.10), we have

s kε

ω,σ a+;ρ,λ[

s kI

µ a+f]

(x)

= 1

k(s+ 1)µk Z x

a

xs+1−us+1

µ+λ k −1

zσ,kρ,λ+µ w x

s+1₋_us+1ρ

f(u)us.

thus, we get

s kε

ω,σ a+;ρ,λ[

s kI

µ a+f]

(x) = 1

(s+ 1)µk s kε

ω,σ a+;ρ,λ+µf

(x).

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G. Rahman: Department of Mathematics, International Islamic University, Islamabad, Pakistan

E-mail address: [email protected]

A. Ghaffar: Department of Mathematical Science, BUITEMS Quetta, Pakistan

E-mail address: [email protected]

K. S. Nisar: Department of Mathematics, College of Arts and Science, Prince Sattam bin Abdulaziz University, Wadi Al dawaser, Riyadh region 11991, Saudi Arabia

E-mail address: [email protected], [email protected]

Azeema: Department of Mathematics, SBK Women’s University, Quetta, Pakistan