International Journal of Analysis and Applications
ISSN 2291-8639
Volume 14, Number 2 (2017), 162-166
http://www.etamaths.com

SOME IMPROVEMENTS OF CONFORMABLE FRACTIONAL INTEGRAL

INEQUALITIES

FUAT USTA∗ AND MEHMET ZEKI SARIKAYA

Abstract. In this study, we wish to set up and present some new conformable fractional integral

inequalities of the Gronwall type which have a great variety of implementation area in differential

and integral equations.

1. Introduction & Preliminaries

In light of recent events in theory of differential and integral equations, it is becoming extremely
difficult to ignore the existence of integral inequalities which help to determine of bounds on unknown
functions. For example, Gronwall and Pachpatte have great contribution in the literature [19], [20], [5],
[6]. Together with this contributions, Gronwall inequality has been extended and applied in a number
of context. However, in non-integer order of models the bound provided by the above authors are not
feasible.

Additionally non-integer order calculus called fractional calculus has a number of fields of application
such as control theory, computational analysis and engineering [12], see also [13]. Thus a number of
new definitions have been introduced in academia to provide the best method for fractional calculus.
For instance in more recent times a new local, limit-based definition of a conformable derivative has
been introduced in [1], [4], [10], with several follow-up papers [2], [3], [7]- [9], [11], [14]- [18].

In this research, we presented conformable fractional version of some significant integral inequalities
with the help of the Katugampola conformable fractional calculus. In detail, Katugampola conformable
derivatives for α ∈ (0, 1] and t ∈ [0,∞) given by

Dα (f) (t) = lim
ε→0

f
(
teεt

−α
)
−f (t)

ε
, Dα (f) (0) = lim

t→0
Dα (f) (t) , (1.1)

provided the limits exist (for detail see, [10]). If f is fully differentiable at t, then

Dα (f) (t) = t1−α
df

dt
(t) . (1.2)

A function f is α−differentiable at a point t ≥ 0 if the limit in (1.1) exists and is finite. This definition
yields the following results;

Theorem 1.1. Let α ∈ (0, 1] and f,g be α−differentiable at a point t > 0. Then
i. Dα (af + bg) = aDα (f) + bDα (g) , for all a,b ∈ R,
ii. Dα (λ) = 0, for all constant functions f (t) = λ,
iii. Dα (fg) = fDα (g) + gDα (f) ,

iv. Dα
(
f

g

)
=
fDα (g) −gDα (f)

g2

v. Dα (tn) = ntn−α for all n ∈ R
vi. Dα (f ◦g) (t) = f′ (g (t)) Dα (g) (t) for f is differentiable at g(t).

2010 Mathematics Subject Classification. 26D15, 26A51, 26A33, 26A42.

Key words and phrases. Gronwall integral inequality; conformable fractional differential equation; global existence.

c©2017 Authors retain the copyrights of
their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License.

162



SOME IMPROVEMENTS OF CONFORMABLE FRACTIONAL INTEGRAL INEQUALITIES 163

Definition 1.1 (Conformable fractional integral). Let α ∈ (0, 1] and 0 ≤ a < b. A function f : [a,b] →
R is α-fractional integrable on [a,b] if the integral∫ b

a

f (x) dαx :=

∫ b
a

f (x) xα−1dx

exists and is finite. All α-fractional integrable on [a,b] is indicated by L1α ([a,b])

Remark 1.1.

Iaα (f) (t) = I
a
1

(
tα−1f

)
=

∫ t
a

f (x)

x1−α
dx,

where the integral is the usual Riemann improper integral, and α ∈ (0, 1].

We will also use the following important results, which can be derived from the results above.

Lemma 1.1. Let the conformable differential operator Dα be given as in (1.1), where α ∈ (0, 1] and
t ≥ 0, and assume the functions f and g are α-differentiable as needed. Then

i. Dα (ln t) = t−α for t > 0

ii. Dα
[∫ t
a
f (t,s) dαs

]
= f(t,t) +

∫ t
a
Dα [f (t,s)] dαs

iii.
∫ b
a
f (x) Dα (g) (x) dαx = fg|

b
a −

∫ b
a
g (x) Dα (f) (x) dαx.

In this paper, by using the Katugampola type conformable fractional calculus, we introduced re-
tarded Gronwall-Bellman and Bihari like conformable fractional integrals inequalities.

2. Main Findings & Cumulative Results

In this article, all the functions which appear in the inequalities are assumed to be real-valued and all
the integrals involved exist on the respective domains of their definitions, and C (M,S) and C1 (M,S)
denote the class of all continuous functions and the first order conformable derivative, respectively,
defined on set M with range in the set S. Additionally, R denotes the set of real numbers such that
R+ = [0,∞), R1 = [1,∞) and Q = [0,T) are the given subset of R.

Theorem 2.1. [14] Let k,y,x ∈ C (R+,R+) , r ∈ C1 (R+,R+) and assume that r is nondecreasing
with r(t) ≤ t for t ≥ 0. If u ∈ C (R+,R+) satisfies

u(t) ≤ k(t) + y(t)
∫ r(t)
0

x(s)u(s)dαs, t ≥ 0, (2.1)

then

u(t) ≤ k(t) + y(t)
∫ t
0

e
∫ r(t)
r(τ)

x(s)y(s)dαsx (r(τ)) k(r(τ))Dαr(τ)dατ, t ≥ 0. (2.2)

Theorem 2.2. Let u,c,x,h,y ∈ C(R+,R+), r ∈ C1(R+,R+) and assume that r is non-decreasing
with r(t) ≤ t for t ≥ 0. Let w(t,u) be a positive , continuous, monotonic, non-decreasing, sub-additive
and sub-multiplicative function for u > 0 for each fixed t. Let the function k(t) > 0 and Ψ(t) ≥ 0 be a
non-decreasing in t and continuous on [0,∞). Ψ(0) = 0 and suppose further that the inequality

u(t) ≤ k(t) + c(t)
∫ r(t)
0

x(s)u(s)dαs + h(t)Ψ

[∫ r(t)
0

y(s)w(s,u(s))dαs

]
(2.3)

is satisfied for all t > 0. Then

u(t) ≤
[
k(t) + h(t)Ψ

(
G−1

(
G
[∫ t

0

y(s)w(s,k(s)m(s))Dαr(s)dαs

]
+

∫ t
0

[y(s)w(s,h(s)m(s))Dαr(s)]dαs

))]
m(t) (2.4)

where

m(t) = 1 + c(t)

∫ t
0

e
∫ r(t)
r(τ)

x(s)c(s)dαsx(r(τ))Dαr(τ)dατ (2.5)



164 USTA AND SARIKAYA

and G−1 is inverse of G such that

G(ξ) =:
∫ ξ
1

1

w(s, Ψ(s))
dαs, ξ ≥ 0,

and

G
[∫ t

0

y(s)w(s,k(s)m(s))Dαr(s)dαs

]
+

∫ t
0

[y(s)w(s,h(s)m(s))Dαr(s)]dαs ∈ Dom(G−1), ∀t ≥ 0.

Proof. Let define

z(t) =

∫ r(t)
0

y(s)w(s,u(s))dαs. (2.6)

So z(0) = 0, then

u(t) ≤ [k(t) + h(t)Ψ(z(t))] + c(t)
∫ r(t)
0

x(s)u(s)dαs. (2.7)

As [k(t) + h(t)Ψ(z(t))] is positive, monotonic, non-decreasing, continuous function over [0,∞), we can
apply the Theorem 2.1, that is

u(t) ≤ [k(t) + h(t)Ψ(z(t))] m(t) (2.8)

where m(t) defined in 2.5. Then if we take the conformable fractional derivative of equation 2.6, we
obtain

Dαz(t) = y(r(t))w(r(t),u(r(t)))Dαr(t)

≤ y(t)w(t,u(t))Dαr(t)
≤ y(t)w(t, [k(t) + h(t)Ψ(z(t))] m(t))Dαr(t)
≤ y(t)w(t,k(t)m(t))Dαr(t) + y(t)w(t,h(t)m(t))w(t, Ψ(z(t)))Dαr(t)

hence
Dαz(t)

w(t, Ψ(z(t)))
≤
y(t)w(t,k(t)m(t))

w(t, Ψ(z(t)))
Dαr(t) + y(t)w(t,h(t)m(t))Dαr(t). (2.9)

Then using the definition of G, we get

G(z(t)) ≤G
[∫ t

0

y(s)w(s,k(s)m(s))Dαr(s)dαs

]
+

∫ t
0

[y(s)w(s,h(s)m(s))Dαr(s)]dαs. (2.10)

Hence

z(t) ≤G−1
(
G
[∫ t

0

y(s)w(s,k(s)m(s))Dαr(s)dαs

]
+

∫ t
0

[y(s)w(s,h(s)m(s))Dαr(s)]dαs

)
. (2.11)

If we combine the equation 2.8 and 2.11, we get the desired bound. �

Theorem 2.3. Let u,x ∈ C(R+,R+), r ∈ C1(R+,R+) and assume that r is non-decreasing with
r(t) ≤ t for t ≥ 0, for which the inequality

Dαu(t) ≤ p +
∫ r(t)
0

x(s)Dαuq(s)[u(s) + Dαu(s)]dαs (2.12)

holds, where p is a positive constant and 0 < q < 1. If

[1 −q(p + u(0))q
∫ r(t)
0

x(s)eqsdαs] > 0, t ≥ 0, (2.13)

then

Dαu(t) ≤ (pβ + β
∫ t
0

x(s)Ω(s)dαs)
1/β, t ≥ 0. (2.14)

where q + β = 1,

Ω(t) =
(u(0) + p)et

[1 −q(u(0) + p)q
∫ t
0
x(s)eqsdαs]1/q

. (2.15)



SOME IMPROVEMENTS OF CONFORMABLE FRACTIONAL INTEGRAL INEQUALITIES 165

Proof. Let denote the right hand side of equation 2.12 by z(t), that is

z(t) = p +

∫ r(t)
0

x(s)Dαuq(s)[u(s) + Dαu(s)]dαs. (2.16)

Here z(0) = p and Dαu(t) ≤ z(t). If we integrate both sides of Dαu(t) ≤ z(t) according to rules of
conformable fractional calculus, we get

u(t) ≤ u(0) +
∫ t
0

z(s)dαs. (2.17)

Then if we take conformable fractional derivative of equation 2.16, we obtain

Dαz(t) ≤ x(r(t))Dαuq(r(t))[u(r(t)) + Dαu(r(t))]Dαr(t). (2.18)
After simple manipulation, we get

Dαz(t) ≤ x(t)zq(t)[u(0) + z(t) +
∫ t
0

z(s)dαs]D
αr(t) (2.19)

Let define

w(t) = u(0) + z(t) +

∫ t
0

z(s)dαs. (2.20)

Here w(0) = u(0) + p. Then by taking both sides of conformable fractional derivative of above
expression and using Dαz(t) ≤ x(t)zq(t)w(t)Dαr(t) and z(t) < w(t), we get

Dαw(t) = Dαz(t) + z(t)

≤ x(t)zq(t)w(t)Dαr(t) + w(t)
≤ x(t)wq+1(t)Dαr(t) + w(t).

So we have

w(t) ≤ Ω(t), t ≥ 0, (2.21)
where

Ω(t) =
(u(0) + p)et

[1 −q(u(0) + p)q
∫ t
0
x(s)eqsdαs]1/q

(2.22)

If we substitute 2.22 into Dαz(t) ≤ x(t)zq(t)w(t)Dαr(t), we get

Dαz(t) ≤ x(t)zq(t)Ω(t)Dαr(t). (2.23)

which implies the estimation for z(t) such that,

z(t) ≤ (pβ + β
∫ t
0

x(s)Ω(s)dαs)
1/β, t ≥ 0. (2.24)

If we combine the equation 2.24 and Dαu(t) ≤ z(t), we get the desired result. �

3. Concluding Remark

In this study we established the explicit bounds on retarded integral inequalities with the help of
conformable fractional calculus. We take the advantage of Katugampola type conformable fractional
derivatives and integrals.

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166 USTA AND SARIKAYA

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Department of Mathematics, Faculty of Science and Arts, Düzce University, Düzce-Turkey

∗Corresponding author: fuatusta@duzce.edu.tr


	1. Introduction & Preliminaries
	2. Main Findings & Cumulative Results
	3. Concluding Remark
	References