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ISSN 2229 – 5046
MATHEMATICAL INEQUALITIES ON FRACTIONAL CALCULUS: A SURVEY
*a
Chinchane V. L. &
bPachpatte D. B.
a
Department of Mathematics,Deogiri Institute of Engineering and Management Studies Aurangabad-431005, INDIA
b
Department of Mathematics,Dr. Babasaheb Ambedkar MarathwadaUniversity, Aurangabad-431 004, INDIA
E-mail:
(Received on: 12-01-12; Accepted on: 22-04-12)
________________________________________________________________________________________________
ABSTRACT
T
he study of mathematical inequalities play very important role in classical differential and integral equations which has applications in many fields. Fractional inequalities are important in studding the existence, uniqueness and other properties of fractional differential equations. In this paper, we present some fractional inequalities, Generalized Gronwall inequalities, Hilbert-Pachpatte inequalities, Minkowski fractional inequalities, Harmite-Hadamard integral inequalities and Opial-Type inequalities.Keywords: Generalized Gronwall inequalities, Hilbert-Pachpatte inequalities, Minkowski fractional inequalities,
Harmite-Hadamard integral inequalities, Opial-Type inequalities, fractional derivative and fractional integration.
________________________________________________________________________________________________
1. INTRODUCTION AND PRELIMINARIES
The main objective of the paper is to survey of some recent development in the field of fractional inequalities which is currently receiving considerable attention. We prepare some material which will be needed later since we will be dealing with different result and application in different paper, it will be also task here in this section to unify the notations. We denote
C
m[(0,
x
)]
the space of all functions of all[0,
x
]
which have continuous derivative up to order m,AC
([0,
x
])
is the space of all absolutely continuous function on[0,
x
]
.AC
m([0,
x
])
denote the space of all])
([0,
1
x
C
g
∈
m− withg
(m−1)∈
A
([0,
x
])
for anyα
∈
R
we denoted by[
α
]
the integral part of α (the integerK
satisfyingK
≤
α
≤
K
+
1
). letα
>
0
, andm
=
[
α
]
+
1
, the spaceJ
α(
L
1)
consist of all functions off
on]
[0,
x
of the formf
∈
J
αψ
for someψ
∈
L
[0,
x
]
.We shall introduce the following definitions and properties which are used through our this paper.
Definition 1.1: A function
f
(
t
)
, is convex function if),
(
)
(1
)
(
)
)
(1
(
t
s
f
t
t
f
s
f
α
+
−
α
≤
α
+
−
(1)where
0
≤
α
≤
1
.Definition 1.2: A function
f
(
t
)
, is concave function if),
(
)
(1
)
(
)
)
(1
(
t
s
f
t
t
f
s
f
α
+
−
α
≥
α
+
−
(2)where
0
≤
α
≤
1
.Definition 1.3: A real valued function
f
(
t
)
,t
≥
0
is said to in spaceC
µ,µ
∈
R
if there exist real numberµ
≥
p
such that)
(
=
)
(
t
t
f
1t
f
p , wheref
1(
t
)
∈
C
([0,
∞
[)
.________________________________________________________________________________________________
*Corresponding author: *a
Page: 1472-1481
© 2012, IJMA. All Rights Reserved 1473
Definition 1.4: A function
f
(
t
)
,t
≥
0
is said to in spaceC
µn,µ
∈
R
iff
n(
t
)
∈
C
([0,
∞
[)
.Definition 1.5: The Riemann-Liouville fractional integral operator of order
α
≥
0
, for functionf
∈
C
µ(
µ
≥
−
1)
is defined as:
(
)
(
)
;
>
0,
>
0,
)
(
1
=
)
(
10
t
f
d
t
t
f
J
tτ
τ
τ
α
α
αα
−
−Γ
∫
(3)
J
0f
(
t
)
=
f
(
t
)
,where
e
uu
1du
0=
)
(
∫
∞ − −Γ
α
α.
For our convenience of establishing the result, we give the semigroup property:
0,
0,
),
(
=
)
(
t
J
+f
t
≥
t
≥
f
J
J
α β α βα
(4)Definition 1.6: The Riemann-Liouville fractional derivative of
f
of orderα
, is defined as:
(
)
(
)
,
)
(
1
=
)
(
10
τ
τ
τ
α
αα
d
f
t
dt
d
m
t
f
D
t mm
− −
−
−
Γ
∫
(5)where
m
=
[
α
]
+
1
. In addition, we stipulate1.
0
,
:=
0,
>
,
:=
,
:=
:=
00 −β β
β
−α α≤
α
≤
if
f
J
f
D
if
f
D
f
J
f
J
f
f
D
(6)If
α
is positive integer, thenf
dt
d
f
D
α=
(
)
α .We also need the following result on law of indices for fractional integral and differentiation.
Lemma 1.1: (Chapter 1 Theorem 2.5) The law of indices
)
(
=
)
(
t
J
f
t
f
J
J
u v u+v (7)is valid in the following case:
1.
v
>
0,
u
+
v
>
0
andf
∈
L
[0,
t
]
; 2.v
<
0,
u
>
0
andf
∈
J
v(
L
1)
; 3.u
<
0,
u
+
v
<
0
andf
∈
J
−u−v(
L
1)
.
Lemma 1.2: (5.Lemma 3.1) Let
α
≥
0
,β
≥
α
, letf
∈
L
(0,
x
)
have anL
∞ fractional derivativeD
βf
in [0,x], and letD
β−kf
(0)
=
0
, fork
=
1...[
β
]
+
1
. Then].
[0,
,
)
(
)
(
)
(
1
=
)
(
10
t
D
f
d
t
x
t
f
D
t−
∈
−
Γ
− −
∫
τ
τ
τ
α
β
β α β β
(8)
2. GENERALIZED GRONWALL INEQUALITY
The Gronwall inequality, which play a very important role in classical differential equations. We first recall standard Gronwall inequality which found in [3].
Theorem 2.1: If
x
t
h
t
tk
s
x
s
ds
t(
)
(
)
)
(
)
(
0
∫
+
≤
t
∈
[
t
0,
T
)
,where all functions involved are continuous on
[
t
0,
T
)
,T
≤
+∞
, andk
(
t
)
≥
0
, thenx
(
t
)
satisfiesds
du
u
k
exp
s
k
s
h
t
h
t
x
ts t
t
(
)
(
)
[
(
)
]
)
(
)
(
0
∫
∫
+
Page: 1472-1481
© 2012, IJMA. All Rights Reserved 1474 If, in addition,
h
(
t
)
nondecreasing, then
≤
h
t
exp
∫
k
s
ds
t
x
tt
(
)
)
(
)
(
0
,
t
∈
[
t
0,
T
)
.In [11] author establish A Generalized fractional Gronwall inequality on the base of iteration argument.
Theorem 2.2: Suppose that
β
≥
0
,a
(
t
)
is nonnegative function locally integrable on0
≤
t
≤
T
(someT
≤
+∞
) and g(t) is nonnegative, nondecreasing continuous function defined on0
≤
t
≤
T
,g
(
t
)
≤
M
(
constant
)
, and supposeu
(
t
)
is nonnegative and locally integrable on0
≤
t
≤
T
withds
s
u
s
t
t
g
t
a
t
u
(
)
(
)
(
)
t(
)
1(
)
0
−
−
+
≤
∫
β.
on the interval. Then
ds
s
a
s
t
n
t
g
t
a
t
u
n n t
−
Γ
Γ
+
≤
−∫
)(
)
(
)
)
(
))
(
)
(
(
)
(
)
(
1 0 ββ
β
,
0
≤
t
≤
T
.Proof: Let
B
(
t
)
=
g
(
t
)
t(
t
s
)
1u
(
s
)
ds
0−
−
∫
βφ
,t
≥
0
, for locally integrable functionφ
. Then)
(
)
(
)
(
t
a
t
Bu
t
u
≤
+
implies
)
(
)
(
=
)
(
t
=10B
a
t
u
t
u
∑
nk− k+
.Let us prove
,
))
(
)
)(
)
(
))
(
)
(
(
(
)
(
10
n
t
s
u
s
ds
t
g
t
u
B
n n tn
−
−Γ
Γ
≤
∫
ββ
β
(9)
and
B
nu
(
t
)
→
0
asn
→
∞
for each t in0
≤
t
≤
T
. We know this relation (2.1) is true forn
=
1
. Assume that it is true for somen
=
k
. Ifn
=
k
+
1
then the induction hypothesis impliesds
d
u
s
k
s
g
s
t
t
g
t
u
B
B
t
u
B
k k s t k k
−
Γ
Γ
−
≤
− − +∫
∫
τ
τ
τ
β
β
β β)
(
)
)(
)
(
))
(
)
(
(
(
)
(
)
(
))
(
(
=
)
(
1 0 1 0 1since
g
(
t
)
is nondecreasing , it follows thatds
d
u
s
k
s
t
t
g
t
u
B
k k s t k k
−
Γ
Γ
−
≤
+ − − +∫
∫
τ
τ
τ
β
β
ββ
)(
)
(
)
)
(
))
(
(
(
)
(
))
(
(
)
(
1 0 1 0 1 1 .By interchanging the order of integration, we have
τ
τ
τ
β
β
β βd
u
ds
s
s
t
k
t
g
t
u
B
k k t t k k)
(
)
(
)
)(
)
(
))
(
(
(
))
(
(
)
(
1 10 0 1 1
−
−
Γ
Γ
≤
+ − − +∫
∫
ds
s
u
s
t
k
k k t)
(
)
)(
)
1)
((
))
(
(
(
=
( 1) 11 = 0 − +
−
+
Γ
Γ
∫
ββ
β
,where the integral
dz
z
z
t
ds
s
s
t
k k kt 1 1 0 1 1 1
)
(1
)
(
=
)
(
)
(
−
−−
−−
+ −∫
−
−∫
β β β β βτ
τ
τ
)
,
(
)
(
=
t
−
τ
(k+1)β−1β
k
β
β
1 1) (
)
(
1)
((
)
(
)
(
=
−
+ −+
Γ
Γ
Γ
τ
ββ
β
β
kt
k
k
,Page: 1472-1481
© 2012, IJMA. All Rights Reserved 1475 Hence the relation (2.1) is proved, since
0
))
(
)
)(
)
(
))
(
(
(
)
(
10
Γ
−
→
Γ
≤
−∫
t
s
u
s
ds
n
M
T
u
B
nn t
n β
β
β
as
n
→
+∞
fort
∈
[0,
T
)
,hence theorem is proved.
3. HILBERT-PACHPATTE TYPE INTEGRAL INEQUALITY FOR FRACTIONAL DERIVATIVE
Recall the original Hilbert double integral inequality see [6.Theorem 316]
Theorem 3.1: If
p
>
1
,q
=
p
/
p
−
1
andf
p(
t
)
dt
<
F
0
∫
∞ ,g
q(
s
)
ds
<
G
0∫
∞ ,then
q p
G
F
p
sin
ds
dt
s
t
s
g
t
f
1/ 1/0
0
<
/
)
(
)
(
π
π
+
∫
∫
∞ ∞ ,where
f
,g
are nonmeasurable function not identically zero.In [7. theorem 1] Pachpatte obtained analogues of Hilbert inequality
Theorem 3.2: If
n
≥
1
, and0
≤
k
≤
n
−
1
be integers. Letu
∈
C
n([0,
x
])
and
v
∈
C
n([0,
y
])
, wherex
>
0
,0
>
y
and letu
(i)(0)
=
v
(j)(0)
=
0
forj
∈
{0,...
n
−
1}
. Then(
) (
)
( ) ( ) 1/ 2 1/ 2
2 2
2 2 1 2 2 1
0 0 0
|
( ) ||
( ) |
( . , , )
(
) |
( ) |
(
) |
( ) |
k k
y x n y n
n k n k
u
t
v
s
dt ds
M n k x y
x t
u t
dt
y
s
v s
ds
t
− −+
s
− −≤
−
×
−
∫
∫
∫
(10)where
.
1]
2
[2
]
1)!
[(
2
1
=
)
,
,
.
(
2−
−
−
−
k
n
k
n
xy
y
x
k
n
M
(11)
In [5] to drive the theorem for Hilbert-Pachpatte Type Integral Inequality for Fractional Derivative they used some facts about fractional derivative in[9. Chapter 1].
Theorem 3.3: For each
i
∈
{1,...
n
}
letx
i>
0,
u
i∈
L
(0,
x
i)
andΦ
i∈
L
∞(0,
x
i)
be non negative,r
i>
−
1
, let1
>
,
i iq
p
satisfy1/
p
i+
1/
q
i=
1
,w
i>
0
satisfy =1 i=
1
n iw
∑
, anda
i,
b
i∈
[0,1]
satisfya
i+
b
i=
1
; in addition, ii i
i
r
q
r
b
>
(
+
1)/(1
−
)
for thosei
for for whichr
i<
0
. If,
1,...
=
],
[0,
,
)
(
)
(
|
)
(
|
0
t
d
t
x
i
n
t
u
ri i i i ii i i t i
i
≤
∫
−
τ
Φ
τ
τ
∈
(12)then
(
)
1/1/ 1
1 =1
1
0 0 (( ) 1)/ 0
=1 =1 =1
|
( ) |
...
...
(
)
( )
n
i i n n p
x xn q xi a r i
i i i i
n i i i i i i
n
ai b q ri i i q wi i i i i i
i
u t
dt
dt
x
x
t
t dt
w t
+
+ +
≤ Ω
−
Φ
∏
∏ ∏
∫ ∫
∫
∑
(13)Where
[
pi]
i i i q i i i i n
i
r
a
r
q
b
a
1/ 1/1 =
1)
(
1)
)
((
1
=
+
+
+
Ω
∏
(14)
Page: 1472-1481
© 2012, IJMA. All Rights Reserved 1476
Theorem 3.4: Let
n
≥
N
. For eachi
∈
{1,...
n
}
letx
i>
0
,α
i>
0
,β
i>
α
i. Letp
i,q
i satisfy1/
p
i+
1/
q
i=
1
,0
>
i
w
satisfy=
1
1 = i
n i
w
∑
, anda
i,b
i∈
[0,1]
satisfya
i+
b
i=
1
; ifβ
i<
α
i+
1
, let in addition)
1)(1
)/(
(
>
i i i i ii
q
b
β
−
α
−
−
β
+
α
. Writer
i=
β
i+
α
i−
1
. If, for eachi
∈
{1,...
n
}
,f
i∈
L
[0,
x
i]
has anL
∞fractional derivative
D
βif
i andD
βi−jf
i(1)
=
0
for=
1,...[
]
+
1
j
j
β
, then(
)
1/1/ 1
1 =1
1
0 0 (( ) 1)/ 0
=1 =1 =1
|
( ) |
...
...
(
)
|
( ) |
n
i
i i n n p
x xn q xi a r p i
i i i i i i
n i i i i i i
n
ai b q ri i i q wi i i i i i
i
D f t
dt
dt
x
x
t
D f t
dt
w t
α
β
+
+ +
≤ Ω
−
∏
∏ ∏
∫ ∫
∫
∑
(15)Where
[
]
.
1)
(
1)
)
1)((
(
1
=
1/ 1/
1 =
i p i i i q i i i i i
n
i
r
a
r
q
b
a
r
+
+
+
+
Γ
Ω
∏
(16)Proof: By lemma(1.2),
(
)
(
)
,
[0,
].
1)
(
1
=
)
(
0 i i i i i
i i r i i i t
i i
i
i
t
D
f
d
t
x
r
t
f
D
−
∈
+
Γ
∫
τ
τ
τ
β α
(17)
Set
1)
(
|
)
(
|
=
)
(
+
Γ
Φ
i i i i
i i
r
t
f
D
t
β
.
Then theorem [3.5] applies with
r
i=
β
i+
α
i−
1
>
−
1
.4. MINKOWSKI INTEGRAL INEQUALITY
The well-know Minkowski Integral Inequality is given in [2].
Theorem 4.1: Let
p
≥
1
,∫
bf
pt
dt
≤
∞
a
(
)
<
0
and∫
bg
qt
dt
≤
∞
a
(
)
<
0
. Then
(
(
)
(
))
(
)
(
)
.
1/ 1/
1/ p
q b
a p p
b
a p p b
a
f
t
g
t
dt
f
t
dt
g
t
dt
+
≤
∫
+
∫
∫
(18)
and using this in author established the reverse Minkowski integral inequality as:
Theorem 4.2: Let
f
andg
be positive function satisfying].
,
[
,
)
(
)
(
<
0
M
for
all
t
a
b
t
g
t
f
m
≤
≤
∈
(19)Then
.
))
(
)
(
(
)
(
)
(
1/ 1/
1/ p
p b
a p q
b
a p p
b
a
f
t
dt
g
t
dt
c
f
t
g
t
dt
+
≤
+
∫
∫
∫
(20)where
1)
1)(
(
1)
(
1)
(
=
+
+
+
+
+
M
m
M
m
M
c
.In [4] author established reverse Minkowski fractional integral inequality.
Page: 1472-1481
© 2012, IJMA. All Rights Reserved 1477
∞
<
)
(
t
f
J
α p ,J
αg
p(
t
)
<
∞
. IfM
g
f
m
≤
≤
)
(
)
(
<
0
τ
τ
,
τ
∈
[0,
t
]
, then we have[
] [
]
[
(
)
(
)
]
.
1)
1)(
(
2)
(
1
)
(
)
(
1/p p 1/p p 1/pp
t
g
f
J
M
m
m
M
t
g
J
t
f
J
+
+
+
+
+
≤
+
α αα
(21)
Proof: Using the condition
M
g
f
≤
)
(
)
(
τ
τ
,
τ
∈
[0,
t
]
,t
>
0
, we can write(
M
+
1)
pf
(
τ
)
≤
M
p(
f
+
g
)
p(
t
).
(22)Multiplying both side of [4.5] by
)
(
)
(
1α
τ
αΓ
−
−t
,
τ
∈
(0,
t
)
, then integrating the resulting inequality with respect to τ ,over
(0,
t
)
, we obtain,τ
τ
τ
α
τ
τ
τ
α
αt
αf
g
d
M
d
f
t
M
t pp p
t p
)
(
)
(
)
(
)
(
)
(
)
(
)
(
1)
(
10 1
0
−
≤
Γ
−
+
Γ
+
− −∫
∫
,which is equivalent to
)
(
)
(
1)
(
)
(
J
f
g
t
M
M
t
f
J
p pp
p
≤
+
α+
α
.
Hence we can write
[
]
[
(
)
(
)
]
.
1)
(
)
(
1/p p 1/pp
t
g
f
J
M
M
t
f
J
+
+
≤
αα
(23)
On other hand, using the condition
mg
(
t
)
≤
f
(
τ
)
, we can write(
(
)
(
)
)
1
)
(
1
1
τ
f
τ
g
τ
m
g
m
≤
+
+
.Therefore,
(
(
)
(
)
)
.
)
1
(
)
(
1
1
p p pp
g
f
m
g
m
τ
≤
τ
+
τ
+
(24)Now, multiplying both side of [4.7] by
)
(
)
(
1α
τ
αΓ
−
−t
,
τ
∈
(0,
t
)
, then integrating the resulting inequality with respect toτ
, over(0,
t
)
, we obtain[
]
[
(
)
(
)
]
.
1
1
)
(
1/p p 1/pp
t
g
f
J
m
t
g
J
+
+
≤
αα
(25)
Adding the inequality [4.6] and [4.8], we obtain the inequality
[
p] [
p p]
p[
p]
pt
g
f
J
M
m
m
M
t
g
J
t
f
J
1/ 1/(
)
(
)
1/1)
1)(
(
2)
(
1
)
(
)
(
+
+
+
+
+
≤
+
α αα
.
5. HERMIT-HADAMARD'S INEQUALITY
We recall Hermit-Hadamard inequality in [10] as follows:
Hermit-Hadamard Inequality: Let
f
:
I
⊂
R
→
R
be convex function on intervalI
of real numbers anda
,
b
∈
I
with
a
<
b
. The following Hermit-Hadamard inequality for convex function holds:.
2
)
(
)
(
)
(
1
2
b
f
a
f
dt
t
f
a
b
b
a
f
ba
+
≤
−
≤
+
Page: 1472-1481
© 2012, IJMA. All Rights Reserved 1478 If the function
f
is concave, the inequality [5.1] can be written as follows:.
2
)
(
1
2
)
(
)
(
+
≤
−
≤
+
∫
f
t
dt
f
a
b
a
b
b
f
a
f
ba (27)
In [4] author develop some new result related to Hermit-Hadamard Integral Inequality by using Riemann-Liouville fractional integral as follows
Lemma 5.1: Let
h
be a concave function on[
a
,
b
]
. Then we have.
2
2
)
(
)
(
)
(
)
(
+
≤
+
−
+
≤
−
h
b
h
b
a
t
h
t
h
a
b
a
h
(28)Theorem 5.1: Let
α
>
0,
β
>
0,
p
>
1,
q
>
1
and letf
,g
be two positive function on[0,
∞
[
. Iff
p,
g
q are concave function on[0,
∞
[
, then we have(
) (
)
[
]
.
))
(
(
))
(
(
)
(
)
(
))
(
(
))
(
(
)
(
)
(
)
(
)
(
(0)
)
(
(0)
2
1 1
1 1
2 1 2
+
Γ
Γ
+
Γ
Γ
≤
+
+
− −
− −
− −
−
t
g
t
J
t
g
t
J
t
f
t
J
t
f
t
J
t
J
t
g
g
t
f
f
q q
p p
q p
q p
β α α
β β
α α
β
β α
α
β
α
β
(29)
Proof: Since
f
p andg
q are concave function on[0,
∞
[
, then by Lemma[5.1] for anyt
>
0
, we have).
2
(
2
)
(
))
(
)
(
(0)
f
t
f
t
f
f
t
f
p+
p≤
p−
τ
+
pτ
≤
p (30)).
2
(
2
)
(
))
(
)
(
(0)
g
t
g
t
g
g
t
g
q+
q≤
q−
τ
+
qτ
≤
q (31)Multiplying both side on [5.5] and [5.6] by
)
(
)
(
1 1α
τ
α βΓ
−
− −J
t
,
τ
∈
(0,1)
, then integrating the resulting inequality w.r.tτ
over(0,
t
)
, we obtain1 1 1 1 1 1
0 0 0
(0)
( )
1
1
(
)
(
)
(
)
(
)
( )
( )
( )
( )
p p
t t t
p p
f
f
t
t
τ
ατ
βd
τ
t
τ
ατ
βf
t
τ τ
d
t
τ
ατ
βf
τ τ
d
α
α
α
− − − − − −
+
−
≤
−
−
+
−
Γ
∫
Γ
∫
Γ
∫
1 1
0
2
( )
2
(
)
( )
p t
t
f
t
τ
ατ
βd
τ
α
− −≤
−
Γ
∫
(32)and
1 1 1 1 1 1
0 0 0
(0)
( )
1
1
(
)
(
)
(
)
(
)
( )
( )
( )
( )
q q
t t t
q q
g
g t
t
τ
ατ
βd
τ
t
τ
ατ
βg t
τ τ
d
t
τ
ατ
βg
τ τ
d
α
α
α
− − − − − −
+
−
≤
−
−
+
−
Γ
∫
Γ
∫
Γ
∫
1 1
0
2
( )
2
(
)
.
( )
q t
t
g
t
τ
ατ
βd
τ
α
− −
≤
−
Γ
∫
(33)Using change of variable
t
−
τ
=
y
, we obtain,)).
(
(
)
(
)
(
=
)
(
)
(
)
(
)
(
)
(
1 1 10
t
f
t
d
J
t
f
t
p p
t − − −
Γ
Γ
−
−
Γ
Γ
Γ
∫
α β β αα
β
τ
τ
τ
τ
α
β
β
(34)
and
(
(
)).
)
(
)
(
=
)
(
)
(
)
(
)
(
)
(
1 1 10
t
g
t
d
J
t
g
t
q q
t − − −
Γ
Γ
−
−
Γ
Γ
Γ
∫
α β β αα
β
τ
τ
τ
τ
α
β
β
Page: 1472-1481
© 2012, IJMA. All Rights Reserved 1479 By the relation [5.7] and [5.9], we can state that
(
)
)(
(
)).
2
(
2
))
(
(
))
(
(
)
(
)
(
))
(
(
)
(
(0)
−1 −1+
−1≤
−1Γ
Γ
≤
+
α β β α α β α βα
β
t
J
t
f
t
f
t
J
t
f
t
J
t
J
t
f
f
p p p p p (36)and [5.8] and [5.10], we can write
(
)
)(
(
)).
2
(
2
))
(
(
))
(
(
)
(
)
(
))
(
(
)
(
(0)
−1 −1+
−1≤
−1Γ
Γ
≤
+
α β β α α β α βα
β
t
J
t
g
t
g
t
J
t
g
t
J
t
J
t
g
g
q q q q q (37)The inequality [5.11] and [5.12] implies that
(
)(
)
[
]
.
))
(
(
))
(
(
)
(
)
(
))
(
(
))
(
(
)
(
)
(
)
(
)
(
(0)
)
(
(0)
1 1
1 1
2 1
+
Γ
Γ
+
Γ
Γ
≤
+
+
− −
− −
−
t
g
t
J
t
g
t
J
t
f
t
J
t
f
t
J
t
J
t
g
g
t
f
f
q q
p p
q q
p p
β α α
β β
α α
β
β α
α
β
α
β
(38)As before, since
f
andg
are positive function, then for anyt
>
0
,p
≥
1
,q
≥
1
we have),
(
))
(
(0)
(
2
)
(
2
)
(
(0)
(
+
α β−1≥
−+
α β−1t
J
t
f
f
t
J
t
f
f
p pp p
(39)
and
).
(
))
(
(0)
(
2
)
(
2
)
(
(0)
(
+
α β−1≥
−+
α β−1t
J
t
q
q
t
J
t
g
g
q qq q
(40)
The inequalities [5.14] and [5.15] implies that
2 2
1 1
(
(0)
( ))(
(0)
( ))
(
)
2
( (0)
( )) ( (0)
( ))
(
)
.
4
p p q q
p q p q
f
f
t
g
g t
J
αt
β− − −f
f t
q
q t
J
αt
β−+
+
≥
+
+
(41)Combining [5.13]and [5.16], we obtain required inequality [5.4].
6. OPIAL-TYPE INEQUALITIES FOR FRACTIONAL DERIVATIVE
We know that the original Opial-Type inequality see [1. Chapter 1 Theorem 1.1].
Theorem 6.1: Let
f
∈
C
1(
[0,
h
]
)
such that0
=
)
(
=
(0)
f
h
f
,f
(
t
)
>
0
on(0,
h
)
. Then
(
)
,
4
|
)
(
'
)
(
|
'20
0
f
t
dt
h
dt
t
f
t
f
hh
∫
∫
≤
(42)where
4
h
the best constant.
In [1. Chapter 2] author established Opial-Type Inequality For Fractional Derivative as:
Theorem 6.2: Let
(
[
,
]
)
0
a
b
C
f
∈
tυ ,υ
≥
0
andf
i(
t
0)
=
0,
i
=
0,1,2...
n
−
1,
n
:=
[
υ
]
. Heret
,
t
0∈
[
a
,
b
];
t
≥
t
0, Letp
,
q
>
1
such that1/
p
+
1/
q
=
1
. Then(
)
2/
1/ 2/
0
1/
0 0
0 0
2
(
)
| ( ) ||
( ) |
| (
) ( ) |
( ) (
1)(
2)
q
p p p p
t t q
t p t
t t
t
t
f w
D f w dw
D f
w
dw
p
p
p
p
υ
υ υ
υ
υ
υ
− − +
−
≤
×
Γ
− +
− +
Page: 1472-1481
© 2012, IJMA. All Rights Reserved 1480
Proof: Let
(
[
,
]
)
0
a
b
C
f
∈
tυυ
≥
0
andf
i(
t
0)
=
0,
i
=
0,1,2...
n
−
1,
n
:=
[
υ
]
. Then by fractional Taylor formula [1. Chapter 2 Theorem 2.1 ] as:.
)
)(
(
)
(
)
(
1
=
)
(
0 1 0
τ
τ
τ
υ
t
υD
υf
d
t
f
t tt
−
−
Γ
∫
(44)hence from [6.3] and Holder's inequity we get
(
t
≥
t
0)
..
))
(
|
|
(
1)
)(
(
)
(
=
))
(
|
|
(
)
)
((
)
(
1
|
)
)(
(
|
)
(
)
(
1
|
)
(
|
1/ 0
0 1/ 1/ 0
1/ 0
0 1/ 1 0
0 1 0
q q t
t
t p p p p
q q t
t
t p p t
t
t t
t
d
f
D
p
p
t
t
dt
f
D
dt
t
d
f
D
t
t
f
+
−
Γ
−
−
Γ
≤
−
Γ
≤
∫
∫
∫
∫
+ −
− −
τ
τ
υ
υ
τ
τ
υ
τ
τ
τ
υ
υ υ
υ υ
υ υ
(45)
Set
z
t
D
tf
t
qdt
tt
(
|
|
(
))
:=
)
(
0 0
υ
∫
,(
z
(
t
0)
=
0)
.Then
z
t
D
tf
t
1/q0 '
))
(
|
|
(
=
))
(
(
υ andD
tf
z
't
1/q0
|
=
(
(
))
|
υ ,∀
t
0≤
t
≤
b
.Therefore [6.4], we have
.
)
(
))
(
|
|
(
1)
)(
(
)
(
)
(
|
||
)
(
|
1/ ' 0
0 1/ 1 0 0
q q
t w
t p p p
t
D
f
d
z
w
p
p
t
w
w
f
D
w
f
+
−
Γ
−
≤
− +∫
τ
τ
υ
υ
υυ υ
(46)
all
t
0≤
w
≤
t
. Next we integrate [6.5] over[
t
0,
t
]
1/ ' 1/
0 1/
0
0 0
1
| ( ) ||
| ( )
(
)
( ( ) ( ))
( )(
1)
t t p p p q
t p
t
f w
D f
w dw
p
p
tw t
z w z w
dw
υ υ
υ
υ
− +≤
−
Γ
− +
∫
∫
(
) (
)
[ ]
1/ 1/
1 '
0 1/
0
2/ 2/
0
1/ 1/ 1/
1
(
)
( )( ( ))
( )(
1)
( )
(
)
=
( )(
1)
(
2)
2
t p q
p p p t
q p p p
p p q
w t
dw
z w z w dw
p
p
z t
t
t
p
p
p
p
υ
υ
υ
υ
υ
υ
υ
− +
− +
≤
−
Γ
− +
−
Γ
− +
− +
∫
(
)
2/
1/ 2/
0
1/ 0
1/ 0
2
(
)
=
(|
( ) |)
.
( ) (
1)(
2)
q
q p p p
t q
t p t p
t
t
D f w
dw
p
p
p
p
υ
υ
υ
υ
υ
−
−
− +
Γ
− +
− +
∫
Thus the proof.
REFERENCES
[1] Anastassiou George A. Fractional Differentiation Inequalities, Springer Publishing Company, Incorporated, New York, NY, 2009.
[2] L.Baugoffa, On Minkowski And Hardy Integral Inequalities, Jurnal Of Inequalities In Pure and Appl.Math,vol.7, Issue 2, Aritcle 60, 2006.
[3] C.Carduneanu, Principal Of Differential And Integral Equation, Allynand Bacon, Boston 1971.
[4] Dahamni Zoubir, On Minkowski And Hermit-Hadamard Integral Inequalities Via Fractional Integration, Ann. Funct.(2010), no 1, 51-58.
Page: 1472-1481
© 2012, IJMA. All Rights Reserved 1481 [6] G.H.Hardley, J.E.Littlewood, G.Polya, Inequalities, Cambridge University Press 1934.
[7] B.G.Pachpatte, Inequalities Similar To The Integral Anglague Of Hilbert Inequality, Tamiking. J. Math 30(1999), 139-146.
[8] I.Podlubny, Fractional Differentioal Equation, Academic Press, New York,1999.
[9] S.G.Somko, A.A.Kilbas, O.I.Marichev, Fractional Integral And Derivative Theory And Application, Gordon And Breach, Yverdon, Switzerland, 1993.
[10] Erhan Set, M.Emin Ozdemir, Serer Dragomir On Hermit-Hadamard Inequality And Other Integral Inequalities Involving Two Function, Journal of Inequalities And Applictions. Volume 2010 (2010),Article ID 148102,9 pages.
[11] Haping Ye, Jianming Gao, Yongs Ding, Generalized Grownwall Inequality And It's Application To Fractional Differential Equation, J. Math. Annal. 328(2007)1075-1081.
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