Some new integral inequalities using Hadamard fractional integral operator

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Adv. Inequal. Appl. 2014, 2014:12 ISSN: 2050-7461

SOME NEW INTEGRAL INEQUALITIES USING HADAMARD FRACTIONAL INTEGRAL OPERATOR

VAIJANATH L. CHINCHANE1,∗AND DEEPAK B. PACHPATTE2

1_{Department of Mathematics, Deogiri Institute of Engineering and Management Studies} Aurangabad-431005, INDIA

2_{Department of Mathematics, Dr. Babasaheb Ambedkar Marathwada University} Aurangabad-431 004, INDIA

Copyright c2014 Chinchane and Pachpatte. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract. The main objective of this paper is to develop certain type of fractional integral inequalities based on Hadamard fractional integral operator.

Keywords: fractional integral inequality; Hadamard fractional integral; integral inequality.

2000 AMS Subject Classification:26D10, 26A33.

1. Introduction

In last few decades, considerable interests has been shown in developing various aspect of the fractional differential and integral equations both for their own sake and for their appli-cation in science and technology; for detail, see [2, 13, 14, 15] and the references therein. In recent years, many authors have worked on fractional differential and integral inequalities using

∗_{Corresponding author}

E-mail addresses: [email protected] (V.L. Chinchane), [email protected] (D.B. Pachpatte) Received August 6, 2013

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Riemann-Liouville and Caputo fractional integrals; see [3, 6, 7, 8, 9, 12, 13] and the references therein. Ahmed Anberet al. established fractional integral inequality using Riemann-Liouville fractional integration; for more details, see [1]. In the literature, few results were obtained on some fractional integral inequalities using Hadamard fractional integral; see [4, 5]. Motivated by the results presented in [1], we prove some new results using Hadamard fractional integral operator. The paper is organized as follows. In Section 2, basic definitions and propositions related to Hadamard fractional derivatives and integrals are given. In Section 3, the results on fractional integral inequality using fractional Hadamard integral are presented.

2. Preliminaries

The necessary details of fractional Hadamard calculus are given in Kilbas [14] and Samkoet al. [15]. Here we present some definitions of Hadamard derivative and integral as given in [2].

Definition 2.1.The Hadamard fractional integral of orderα ∈R+of function f(x), for allx>1

is defined as,

HD−₁_,α_x f(x) = 1

Γ(α)

Z x

1 ln(x

t)

α−1_f₍_t₎dt

t , (2.1)

whereΓ(α) =R₀∞e−uuα−1du.

Definition 2.2. The Hadmard fractional derivative of orderα ∈[n−1,n),n∈Z+, of function

f(x)is given as follows

HDα1,xf(x) = 1

Γ(n−α)(x

d dx)

nZ x

1 ln(x

t)

n−α−1_f₍_t₎dt

t . (2.2)

From the above definitions, we can see obviously the difference between Hadamard fraction-al and Riemann-Liouville fractionfraction-al derivative and integrfraction-als, which include two aspects. The kernel in the Hadamard integral has the form of ln(x_t)instead of the form of (x−t), which is involves both in the Riemann-Liouville and Caputo integral. The Hadamard derivative has the operator(xd

dx)

n_{, whose construction is well suited to the case of the half-axis and is invariant} relation to dilation [15, p.330], while the Riemann-Liouville derivative has the operator(_dxd)n.

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Proposition 2.1.[2] If 0<α <1, the following equalities hold:

HD−₁,αx(lnx)

β−1₌ Γ(β)

Γ(β+α)(lnx)

β+α−1_, ₍₂_.₃₎

and

HDα1,x(lnx)β−1=

Γ(β)

Γ(β−α)(lnx)

β−α−1_, ₍₂_.₄₎

respectively.

For the convenience of establishing the result, we give the semigroup property,

(_HD−α

1,x)(HD1−,βx)f(x) =HD −(α+β)

1,x f(x). (2.5)

3. Main results

Now, we are in a position to give the main result.

Theorem 3.1.Letα >0, p>1, 1_p+1_q=1and let f , g be two positive functions on[0,∞[, such

that for all t>0,_HD−α

1,t f(t)<∞,HD −α

1,t g(t)<∞. If0<m≤ f(τ)

g(τ) ≤M<∞,τ ∈[0,t], then we have the following

h

HD−1,αt f(t)

i1_ph

HD−1,αt g(t)

i1_q

≤(M

m)

1 pq

h

HD−1,αt (f(t))

1 p₍_g₍_t₎₎

1 q i

. (3.1)

Proof.Since _gf(₍τ)

τ) ≤M,τ∈[0,t[,t>0, we find that [g(τ)]

1

p ≥_M

−1 q _[_f₍_τ_)]

1

q ₍₃_.₂₎

and

[f(τ)]

1 p_[_g₍_τ_)]

1

q _≥_M

−1 q _[_f₍_τ_)]

1 q_[_f₍_τ_)]

1 p

≥M−q1_[_f₍_τ_)] 1

q+

1 q

≥M−q1_[_f₍_τ_)]_.

(3.3)

Multiplying both sides of (3.3) by (ln(

t τ))

α−1

τΓ(α) , which is positive because τ ∈(0,t), t >0, we

integrate resulting identity with respect toτfrom 1 tot ge that

1

Γ(α)

Z t

1

(ln(t

τ)) α−1_[_f₍

τ)]

1 p_[_g₍_τ_)]

1

qdτ

τ ≥M

−1

q 1

Γ(α)

Z t

1

(ln(t

τ)) α−1_f₍

τ)dτ

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which implies that

HD−1,tα

h

[f(t)]1p_[_g₍_t_)] 1 q i

≤M−q1 h

HD−1,tαf(t)

i

. (3.5)

It follows that

HD−₁_,α_t

h

[f(t)]1p_[_g₍_t_)] 1 q

i1_p

≤M−

1 pq

h

HD−₁_,α_t f(t)

i1_p

. (3.6)

Notice thatmg(τ)≤ f(τ),τ ∈[0,t],t>0. It follows that

[f(τ)]

1

p _≥_m

1 p_[_g₍_τ_)]

1

p_. ₍₃_.₇₎

Multiplying the equation (3.7) by[g(τ)]

1

q_{, we arrive at}

[f(τ)]

1 p_[_g₍_τ_)]

1

q _≥_m

1 p_[_g₍_τ_)]

1 q_[_g₍_τ_)]

1

p ₌_m

1

p_[_g₍_τ_)] ₍₃_.₈₎

Multiplying both sides of (3.8) by (ln(

t τ))

α−1

τΓ(α) , which is positive because τ ∈(0,t), t >0, we

integrate resulting identity with respect toτfrom 1 tot obtaining that

1

Γ(α)

Z t

1

(ln(t

τ)) α−1_[_g₍

τ)]

1 q_[_f₍_τ_)]

1

pdτ

τ ≥

m1p 1

Γ(α)

Z t

1

(ln(t

τ)) α−1_g₍

τ)dτ

τ , (3.9)

that is,

HD−₁,αt

h

[f(t)]1p_[_g₍_t_)] 1 q i

≤m1p h

HD−₁,αt g(t)

i

. (3.10)

Hence we have

HD−1,αt

h

[f(t)]1p_[_g₍_t_)] 1 q

i1_q ≤mpq1

h

HD−1,tαg(t)

i1_q

. (3.11)

Multiplying the equation (3.6) and (3.11), we can draw the desired conclusion easily.

Lemma 3.2.Letα>0, f and g be two positive functions on[0,∞[, such that for allHD−₁_,_tαfp(t)<

∞,HD−₁_,_tαgq(t)<∞, t>0. If0<m≤ f(τ)

p

g(τ)q ≤M<∞,τ ∈[0,t], then we have h

HD−₁_,α_t fp(t)

i1_ph

HD−₁_,α_t gq(t)

i1_q

≤(M

m)

1 pq

h

HD−₁_,α_t (f(t)(g(t)

i

, (3.12)

where p>1, 1

p+ 1 q=1.

Proof.Replacing f(τ)andg(τ)by f(τ)pandg(τ)q,τ∈[0,t],t>0 in theorem 3.1, we find the

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Lemma 3.3. Let α >0 and let f , g be two positive functions on[0,∞[, such that f is

nonde-creasing and g is non-innonde-creasing. Then

HD−₁_,_tαfγ(t)gδ(t)≤

Γ(α+1)

(lnt)α ) α

HD−₁_,α_t fγ(t)HD−₁_,_tαgδ(t). (3.13)

for any t >0γ >0δ >0.

Proof.Letτ,ρ∈[0,t],t >0, for anyδ >0,γ >0. Then we have

(fγ₍

τ)−fγ(ρ))

gδ₍

ρ)−gδ(τ)

≥0 (3.14)

and

fγ₍

τ)gδ(ρ)−fγ(τ)gδ(τ)−fγ(ρ)(gδ(ρ) + fγ(ρ)gδ(τ)≥0. (3.15)

It follows that

fγ(τ)gδ(τ) + fγ(ρ)(gδ(ρ)≤ fγ(τ)gδ(ρ) + fγ(ρ)gδ(τ), (3.16)

Now, multiplying both sides of (3.16) by (ln(

t τ))

α−1

τΓ(α) , which is positive becauseτ∈(0,t),t >0,

we integrate resulting identity with respect toτ from 1 totfind that

HD−₁_,α_t fγ(t)gδ(t) +fγ(ρ)(gδ(ρ)HD₁−_,_tα(1)≤gδ(ρ)HD−₁_,_tαfγ(t) + fγ(ρ)HD−₁_,α_t gδ(t), (3.17) Again, multiplying both sides of (3.17) by(ln(

t ρ))

α−1

ρΓ(α) , which is positive becauseρ∈(0,t),t>0,

we integrate resulting identity with respect toρ from 1 totobtaining that

HD−₁_,_tαfγ(t)gδ(t)HD−₁_,α_t (1) +HD−₁_,α_t fγ(t)(gδ(t)HD−₁_,α_t (1)

≤HD−₁_,_tαgδ(t)HD−₁_,_tαfγ(t) + fγ(t)HD₁−_,α_t gδ(t).

It follows that

2HD−₁,tαf

γ₍_t₎_gδ₍_t₎_≤ 1

[HD−₁_,α_t (1)]−1

2HD−₁,αt f

γ₍_t₎

HD−₁,tαg

δ₍_t₎

This completes the proof.

Theorem 3.4. Let f , g be two positive functions on[0,∞[, such that f is nondecreasing and g

is non-increasing. Then for all t>0γ >0δ >0, we have

(lnt)β Γ(β+1)_HD

−α

1,t f

γ₍_t₎_gδ₍_t_{) +} (lnt) α

Γ(α+1)_HD

−α

1,t f

γ₍_t₎_gδ₍_t₎

≤HD−₁,αt f

γ₍_t₎

HD−₁_,β_t gδ(t)

+HD−₁,tαg

δ₍_t₎

HD−₁_,_tβfγ(t)

.

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Proof.Multiplying both sides of the equation (3.17)(ln(

t ρ))

β−1

ρΓ(β) ,ρ∈(0,t),t>owhich is positive,

we integrate resulting identity with respective toρ from 1 totfinding that

HD−1,αt fγ(t)gδ(t) 1

Γ(β)

Z t

1

(ln(t

ρ)) β−1dρ

ρ +

1

Γ(β)

Z t

1

(ln(t

ρ))

β−1_fγ₍

ρ)gδ(ρ)dρ

ρ _HD

−α

1,t (1)

≤ 1

Γ(β)

Z t

1

(ln(t

ρ))

β−1_gδ₍ ρ)dρ

ρ H

D−α

1,t f

γ₍_t_{) +} 1 Γ(β)

Z t

1

(ln(t

ρ))

β−1_fγ₍ ρ)dρ

ρ H

D−α

1,t g

δ₍_t₎_.

(3.19)

which implies (3.18). This completes proof.

Remark 3.1. Applying theorem 3.4 forα =β, we obtain theorem 3.3 immediately.

Theorem 3.5. Let f ≥0, g≥0be two functions defined on[0,∞[, such that g is non-decreasing. If

HD−1,αt f(t)≥HD1−,αt g(t),t>o. (3.20)

then, for allα>0,γ >0,δ >0andγ−δ >0,

HD−1,tαf

γ−δ₍_t₎_≤

HD−1,tαf

γ₍_t₎_g−δ₍_t₎_. ₍₃_.₂₁₎

Proof.In view of the arithmetic-geometric inequality, forγ >0,δ >0, we have γ

γ−δ f γ−δ₍

τ)− δ

γ−δg γ−δ₍

τ)≤ fγ(τ)g−δ(τ), τ∈(0,t),t>0. (3.22)

Now, multiplying both sides of (3.22) by (ln(

t τ))

α−1

τΓ(α) , which is positive, we have

γ

γ−δ

(ln(t

τ)) α−1

τΓ(α) f γ−δ₍

τ)− δ

γ−δ

(ln(t

τ)) α−1

τΓ(α) g γ−δ₍

τ)≤

(ln(t

τ)) α−1

τΓ(α) f γ₍

τ)g−δ(τ). (3.23)

Integrating (3.23) with respective toτ from 1 tot, we obtain γ γ−δ 1 Γ(α) Z t 1 ln(t

τ)

α−1_fγ−δ₍ τ)dτ

τ − δ γ−δ 1 Γ(α) Z t 1 ln(t

τ)

α−1_gγ−δ₍ τ)dτ

τ

≤ 1

Γ(α)

Z t

1 ln(t

τ)

α−1_fγ₍

τ)g−δ(τ)dτ τ .

(3.24)

It follows that

γ

γ−δH

D−α

1,t f

γ−δ₍_t₎₋ δ γ−δ H

D−α

1,t g

γ−δ₍_t₎_≤

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which implies that

γ

γ−δ HD

−α

1,t f

γ−δ₍_t₎_≤

HD−₁,tαf

γ₍_t₎_g−δ₍_t_{) +} δ γ−δ HD

−α

1,t g

γ−δ₍_t₎ ₍₃_.₂₆₎

and

HD−₁,αt f

γ−δ₍_t₎_≤ γ−δ

γ HD

−α

1,t f

γ₍_t₎_g−δ₍_t_{) +}δ γ HD

−α

1,t f

γ−δ₍_t₎_.

This completes the proof.

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