15

Received 19/12/15

ON SOME INTEGRAL INEQUALITIES FOR (k; H) RIEMANN-LIOUVILLE FRACTIONAL INTEGRAL ABDULLAH AKKURT, M. ESRA YILDIRIM, AND HÜSEYIN YILDIRIM Abstract. In this study, giving the de…nition of fractional integral, which are with the help of synchronous and monotonic function, some fractional integral inequalities have established.

1. Introduction Integral inequalities play a fundamental role in the theory of di¤erential equations, functional analysis and applied sciences. Important development in this theory has been achieved for the last two decades. For these, see [6]-[11] and the references therein. Moreover, the study of fractional type inequalities is also of vital importance. Also see [1]-[5] for further information and applications. The researchers have studied Fractional Calculus since seventeenth century. From this date, mathematicians as well as biologists, chemists, economists, engineers and physicists have found this new theory very attractive. Many di¤erent derivatives were introduced. 2. Fractional Integrals Now we will give fundamental de…nitions and notations for fractional integrals. De…nition 1. Let a; b 2 R; a < b; and (2.1)

Ja+ f (x) =

1 (x ( )a

Jb f (x) =

1 Rb (t ( )x

and (2.2)

> 0: For f 2 L1 (a; b)

Rx

t)

1

f (t)dt;

x)

1

f (t)dt; b > 0; b > x:

> 0; x > a

These integrals are called right-sided Riemann-Liouville fractional integral and leftsided Riemann-Liouville fractional integral respectively [12]-[17]. This integrals is motivated by the well known Cauchy formula: (2.3)

Zx a

d

1

Z1 a

d 2 :::

Zn

1

f(

n )d n

=

1 (n)

Zx

(x

)n

1

f ( )d :

a

a

Key words and phrases. Riemann-Liouville Fractional Integral, Integral Inequalities. 2010 Mathematics Sub ject Classi…cation. 26A33, 26D15, 41A55. 1

2

ABDULLAH AKKURT, M . ESRA YILDIRIM , AND HÜSEYIN YILDIRIM

De…nition 2. Let (a; b) be a …nite interval of the real line R and < ( ) > 0. Also let h (x) be an increasing and positive monotone function on (a; b], having a continuous 0 derivative h (x) on (a; b). The left- and right-sided fractional integrals of a function f with respect to another function h on [a; b] are de…ned by [17] (2.4) 1 Rx 1 0 Ja+ ;h f (x) := [h (x) h (t)] h (t) f (t) dt; x a; < ( ) > 0 ( ) a and (2.5)

Jb

;h f

(x) :=

For (2.4) and (2.5)

1 Rb [h (t) ( ) x

1

h (x)]

Ja+ ;h f (a) = Jb

0

h (t) f (t) dt; x

;h f

b; < ( ) > 0:

(b) = 0:

If we take h (x) = x in (2.4) and (2.5) integral formulas, we will obtain Ja+ ;h = Ja+ and Jb Also if we choose h(x) =

x

;h

= Jb :

+1

+1

for

0, then the equalities (2.4) and (2.5) will

be (2.6)

(Ja+ ; f )(x) =

( + 1)1 ( )

Rx

(Jb

( + 1)1 ( )

Rb

and (2.7)

;

f )(x) =

(x

+1

t

+1

)

1

t f (t)dt; x > a

(t

+1

x

+1

)

1

t f (t)dt; x < b

a

x

respectively. This kind of generalized fractional integrals are studied in [12], [13], [16], [18]. In [16], Katugampola gave a new fractional integration which generalized RiemannLiouville fractional integrals. This (2.6) and (2.7) generalizations is based on the following equality, (2.8) Zx Z1 Zn 1 Zx ( + 1)1 n +1 n 1 (x +1 ) f ( )d : n f ( n )d n = 1d 1 2 d 2 ::: (n 1)! a

a

a

a

For a = 0 in (2.4), we can write 1 Rx (h (x) J0+ ;h f (x) = (2.9) ( )0

h (t))

1

0

h (t) f (t)dt; x > 0

J00+ ;h f (x) = f (x):

Semi group and commutative properties of (2.9) integral operator is the following Ja+ ;h Ja+ ;h f (x) = Ja++;h f (x);

0;

and Ja+ ;h Ja+ ;h f (x) = Ja+ ;h Ja+ ;h f (x):

0

ON SOM E INTEGRAL INEQUALITIES FOR (k; H) RIEM ANN-LIOUVILLE...

3

To show the being unit operator property of (2.9) integral operator, we choose h function specially as f (x) = h(x) we obtain the following equality 1 Rx 0 J0+ ;h h (x) = (h (x) h (t)) 1 h(t)h (t) dt ( )0 (2.10) (h (x) h (0)) [h(x) + h (0)] : = ( + 2) Let

= 0 in (2.10), then we have J00+ ;h h (x) = h(x): +1

In (2.9), let f (x) = x ; f (x) = 1 and h(x) = x +1 for > 0; then we have ( + 1) ( ++1+1 ) t ( +1)+ J0+ ;h (x ) = ( + ++1+1 )

0;

>

1; t > 0;

and J0+ ;h (1) De…nition 3. Let (2.11)

=

( + 1) t ( + 1)

( +1)

:

> 0 and x > 0; de…ned by [14], [19] Rx 1 (k J f ) (x) = (x t) k 1 f (t)dt: k k( ) 0

Where k gamma function is de…ned by Z1 x (x) = t k 1e k

tk k

dt; x > 0:

0

and 1 Bk (x; y) = k

Z1

x

tk

1

y

(1

t) k

1

dt:

0

Also Bk (x; y) =

k (x) k (y) k (x

+ y)

and Bk (x; y) =

1 x y Bk ( ; ): k k k

De…nition 4. Let (a; b) be a …nite interval of the real line R and < ( ) > 0. Also let h (x) be an increasing and positive monotone function on (a; b], having a continuous 0 derivative h (x) on (a; b). The left- and right-sided fractional integrals of a function f with respect to another function h on [a; b] are de…ned by (2.12) Zx 1 1 0 [h (x) h (t)] k h (t) f (t) dt; k > 0; < ( ) > 0 k Ja+ ;h f (x) := k k( ) a

and (2.13) 1 k Jb ;h f (x) := k k( )

Zb x

[h (t)

h (x)] k

1

0

h (t) f (t) dt; k > 0; < ( ) > 0:

4


If we take h (x) = x in (2.12) and (2.13) integral formulas, we will obtain 1 (k Ja+ f )(x) = k k( ) 1 (k Jb f ) (x) = k k( )

Zx

(x

t) k

1

f (t)dt; x > a

x) k

1

f (t) dt; b > x:

a

Zb

(t

x

Note that when k ! 1; then it reduces to the classical Riemann-Liouville fractional integral. x

Also if we choose h(x) =

+1

+1

for

2 R= f 1g, then the equalities (2.12) and

(2.13) will be (2.14)

(k Ja+ f )(x) =

( + 1)1 k Rx (x k k( ) a

(k Jb f )(x) =

( + 1)1 k Rb k+1 (t k k( ) x

and (2.15)

+1

t

+1

)k

xk+1 )

1

t f (t)dt; x > a

1 k

t f (t)dt; x < b

respectively. This kind of generalized fractional integrals are studied in [20]. For a = 0 in (2.12), we can write (2.16)

k J0+ ;h f

(x) =

Rx 1 (h (x) k k( ) 0 0 k J0+ ;h f

h (t)) k

1

0

h (t) f (t)dt; x > 0

(x) = f (x):

Semi group and commutative properties of (2.16) integral operator is the following i h f (x) = Ja++;h f (x); 0; 0 k Ja+ ;h k Ja+ ;h and

h

k Ja+ ;h

k Ja+ ;h

i

f (x) =

k Ja+ ;h

k Ja+ ;h

f (x):

To show the being unit operator property of (2.16) integral operator, we choose h function specially as f (x) = h(x) we obtain the following equality k J0+ ;h h

(x)

=

(2.17) = For

Rx 1 (h (x) k k( ) 0

h (t)) k

1

0

h(t)h (t) dt

(h (x) h (0)) k [h(x) + h (0)] : ( + k + 1)

= 0 and k = 1 in (2.17), we have J00+ ;h h (x) = h(x):

The main aim of this work is to establish a new fractional integral inequality for (k; h) Riemann-Liouville fractional integral. Using the technique of [20] a key role in our study.


5

3. Main Results Theorem 1. Let f and g are two synchronous functions on [0; 1]: Then for t > 0; > 0; (3.1)

k

k Ja+ ;h (f g)(t)

( + k)

(h (x)

k Ja+ ;h

h (a)) k

f (t)

k Ja+ ;h

g(t):

Proof. For f and g synchronous functions, we have (3.2)

(f ( )

f ( )) (g ( )

g( ))

0:

From (3.2) it can be written as following (3.3)

f ( )g( ) + f ( )g( )

f ( )g( ) + f ( )g( ): (h (t)

If we multiply two sides of the (3.3) with

h ( )) k k k( )

1

0

h ( );

obtain (h (t)

+

h ( )) k k k( )

(h (t)

1

h ( )) k k k( )

0

h ( ) f ( )g( ) 1

0

h ( ) f ( )g( )

(3.4) (h (t)

+

(h (t)

1

h ( )) k k k( )

h ( )) k k k( )

0

h ( ) f ( )g( ) 1

0

h ( ) f ( )g( ):

Integrating (3.4) inequality on (a; t), then Rt 1 (h (t) h ( )) k k k( ) a + (3.5)

Rt 1 (h (t) k k( ) a

k

Therefore

g( )

+f ( )

k

h ( )) k

+ f ( )g( )

1 k k( )

Z

t

(h (t)

0

h ( ) f ( )g( )d 1

1 ( )

Rt a

0

h ( ) f ( )g( )d

1

h ( )) k

Rt 1 (h (t) a ( ) k

k Ja+ ;h (f g)(t)

(3.6)

h ( )) k

Rt 1 (h (t) k k( ) a

+

1

1

0

h ( ) f ( )g( )d 0

h ( ) f ( )g( )d :

(h (t)

h ( )) k

1

h ( )) k

0

h ( ) f ( )d

a

Rt 1 (h (t) a ( ) k

h ( )) k

1

0

h ( ) g( )d

1

0

h ( )d

2 (a; t); we

6


and k Ja+ ;h

(f g)(t) + f ( )g( )

k Ja+ ;h

(1)

(3.7) g( )

k Ja+ ;h

(f )(t) + f ( ) (h (t)

Now multiplying two sides of (3.7) with

k Ja+ ;h

(g)(t): 1

h ( )) k k k( )

0

h ( );

2 (a; t); we

have (h (t)

+

1

h ( )) k k k( )

(h (t)

0

h ( ) Ja+ ;h (f g)(t) 1

h ( )) k k k( )

0

h ( ) f ( )g( )Ja+ ;h (1)

(3.8) (h (t)

+

(h (t)

1

h ( )) k k k( )

1

h ( )) k k k( )

0

h ( ) g( )Ja+ ;h f (t) 0

h ( ) f ( )Ja+ ;h (g)(t):

By integrating to (3.9) on (a; t), then k Ja+ ;h

+

(f g)(t)

k Ja+ ;h (1)

k

(3.9) k

+

k(

k Ja+ ;h

k

Rt a

k( )

k Ja+ ;h

R t (h (t) a

f ( )g( )(h (t)

f (t) R t a

)

g(t) R t

k(

a

)

(h (t)

(h (t)

1

h ( )) k k k( )

0

h ( )d

h ( )) k

h ( )) k

h ( )) k

1

1

1

0

h ( )d

0

h ( ) g( )d

0

h ( ) f ( )d :

This inequality is can be written as the following at the same time (3.10)

Ja+ ;h (f g)(t)

k

( + k)

(h (x)

h (a)) k

Ja+ ;h f (t)Ja+ ;h g(t):

So the proof is completed. Theorem 2. Let f and g are two synchronous functions on [a; b]: Then for t > a; > 0; > 0 and k > 0; h (a)) k h k ( + k)

(h (x)

k Ja+ ;h (f g) (t) + k Ja+ ;h

f (t)

i J (f g) (t) k a+ ;h

k Ja+ ;h

g(t) +

k Ja+ ;h

g(t)

k Ja+ ;h

f (t):


7

Proof. Since the f and g are two synchronous functions on [a; b] then for all ; (h (t) h ( )) k 1 0 0: If we multiply two sides of (3.7) with h ( ), then we obtain k k( ) (h (t)

+

h ( )) k k k( )

(h (t)

1

0

h ( )

h ( )) k k k( )

1

k Ja+ ;h

(f g)(t)

0

h ( ) f ( )g( )

k Ja+ ;h

(1)

(3.11) (h (t)

+

(h (t)

h ( )) k k k( )

h ( )) k k k( )

1

0

h ( ) g( ) 1

0

h ( ) f( )

k Ja+ ;h

(f )(t)

k Ja+ ;h

(g)(t):

Integrating to (3.11) on (a; t), then Rt (h (t) a

+

h ( )) k k k( )

Rt (h (t) a

(3.12)

h ( )) k k k( )

Rt (h (t) a

+

Rt (h (t) a

1

0

h ( ) 1

k Ja+ ;h

(f g)(t)dt

0

h ( ) f ( )g( )

h ( )) k k k( )

1

h ( )) k k k( )

1

0

h ( ) g( )

0

h ( ) f( )

k Ja+ ;h (1)

dt

k Ja+ ;h

(f )(t)dt

k Ja+ ;h

(g)(t)dt:

This is the proof of the theorem k Ja+ ;h

(1)

k Ja+ ;h

(f g) (t) +

k Ja+ ;h

(1)

k Ja+ ;h

g(t)

k Ja+ ;h

(f g) (t)

(3.13) k Ja+ ;h

f (t)

k Ja+ ;h

g(t) +

Remark 1. It is obvious that if we take Theorem 1.

=

k Ja+ ;h

f (t):

in this theorem we will obtain

Theorem 3. Let f; g and h be tree monotonic functions de…ned on [0; 1) satisfying the following inequality (f ( )

f ( )) (g ( )

g( )) (h ( )

h ( ))

0

8


for all ; 2 [a; t] ; then for all t > a 0; for (k; H) fractional integrals hold: h i J (f gh) (t) k a+ ;h k Ja+ ;h (1) h

(3.14)

h +

h

k Ja+ ;h

> 0;

> 0; the following inequalities

k Ja+ ;h (1)

h

i J (f gh) (t) k a+ ;h

ih i h ih i (f h) (t) k Ja+ ;h (g) (t) + k Ja+ ;h (gh) (t) k Ja+ ;h (f ) (t)

ih i h ih i J (h) (t) J (f g) (t) + J (f g) (t) J (h) (t) + + k a ;h k a+ ;h k a ;h k a+ ;h k Ja+ ;h

ih i (f ) (t) k Ja+ ;h (gh) (t)

h

k Ja+ ;h

ih i (g) (t) k Ja+ ;h (f h) (t) :

Proof. Since the functions f; g and h monotonic functions on [0; 1); then for all ; 0; we have (3.15)

(f ( )

f ( )) (g ( )

g( )) (h ( )

h ( ))

From (3.15) it can be written as following (3.16) f ( )g( )h( ) f ( )g( )h( ) f ( )g( )h( ) +f ( ) g ( ) h ( )

f ( )g( )h( )

0:

f ( )g( )h( )

f ( )g( )h( ) + f ( )g( )h( )

If we multiply two sides of the (3.16) with

(h (t)

h ( )) k k k( )

1

0

h ( );

obtain Rt (h (t) a

h ( )) k k k( )

f ( )g( )h( )

1

0

h ( ) f ( ) g ( ) h ( ) dt

Rt (h (t) a

g( )

Rt (h (t) a

+f ( )

Rt (h (t) a

(3.17)

f ( )g( )

h ( )) k k k( )

1

h ( )) k k k( )

1

Rt (h (t) a

+h ( )

Rt (h (t) a

+g ( ) h ( )

f ( )h( )

a

0

h ( ) dt

0

0

h ( ) g ( ) h ( ) dt

1

1

0

h ( ) h ( ) dt

0

h ( ) f ( ) g ( ) dt

h ( )) k k k( )

Rt (h (t)

1

h ( ) f ( ) h ( ) dt

h ( )) k k k( )

h ( )) k k k( )

Rt (h (t) a

h ( )) k k k( )

h ( )) k k k( )

1

0

h ( ) f ( ) dt 1

0

h ( ) g ( ) dt:

0:

2 (a; t); we


9

Therefore k Ja+ ;h

(f gh) (t)

f ( )g( )h( )

k Ja+ ;h

(1)

g ( )k Ja+ ;h (f h) (t) + f ( )k Ja+ ;h (gh) (t)

(3.18)

f ( ) g ( )k Ja+ ;h (h) (t) + h ( )k Ja+ ;h (f g) (t) +g ( ) h ( )k Ja+ ;h (f ) (t)

f ( ) h ( )k Ja+ ;h (g) (t):

Now multiplying two sides of (3.18) with

(h (t)

h ( )) k k k( )

1

0

h ( );

2 (a; t); we

have h

i Rt (h (t) J (f gh) (t) k a+ ;h a

k Ja+ ;h

h

(1)

Rt (h (t) a

1

h ( )) k k k( )

h ( )) k k k( )

0

h ( )d

1

0

f ( )g( )h( )h ( )d

i Rt (h (t) (f h) (t)

h ( )) k 1 0 g( )h ( )d k ( ) k a h i Rt (h (t) h ( )) k 1 0 + k Ja+ ;h (gh) (t) f ( )h ( )d k ( ) k a h i Rt (h (t) h ( )) k 1 0 f ( )g( )h ( )d k Ja+ ;h (h) (t) k k( ) a i Rt (h (t) h ( )) k 1 h 0 + k Ja+ ;h (f g) (t) h( )h ( )d k k( ) a h i Rt (h (t) h ( )) k 1 0 g( )h( )h ( )d + k Ja+ ;h (f ) (t) k ( ) k a i Rt (h (t) h ( )) k 1 h 0 f ( )h( )h ( )d : k Ja+ ;h (g) (t) k ( ) k a

(3.19)

k Ja+ ;h

This is the proof of the theorem, h

k Ja+ ;h

h

h +

h

i (f gh) (t)

k Ja+ ;h

k Ja+ ;h

k Ja+ ;h

k Ja+ ;h

(1)

k Ja+ ;h

(1)

h

k Ja+ ;h

i (f gh) (t)

ih i h ih i (f h) (t) k Ja+ ;h (g) (t) + k Ja+ ;h (gh) (t) k Ja+ ;h (f ) (t)

ih i h ih i (h) (t) k Ja+ ;h (f g) (t) + k Ja+ ;h (f g) (t) k Ja+ ;h (h) (t) ih i (f ) (t) k Ja+ ;h (gh) (t)

h

k Ja+ ;h

ih i (g) (t) k Ja+ ;h (f h) (t) :

10


References [1] Anastassiou GA, Hooshmandasl MR, Ghasemi A, Moftakharzadeh F. Montgomery identities for fractional integrals and related fractional inequalities, J. Inequal. Pure Appl. Math., 10(4)(2009), 1-6. [2] Anastassiou GA. Fractional Di¤erentiation Inequalities, Springer Science, LLC, 2009. [3] Belarbi S, Dahmani Z. On some new fractional integral inequalities, J. Inequal. Pure Appl. Math., 10(3)(2009), 1-12. [4] Dahmani Z. New inequalities in fractional integrals, International Journal of Nonlinear Sciences, 9(4)(2010), 493-497. [5] Dahmani Z. On Minkowski and Hermite-Hadamad integral inequalities via fractional integration, Ann. Funct. Anal., 1(1)(2010, 51-58. [6] Dragomir SS. A generalization of Gruss’s inequality in inner product spaces and applications, J. Math. Annal. Appl., 237(1)(1999), 74-82. [7] Mitrinovic DS, Pecaric JE, Fink AM. Classical and New Inequalities in Analysis, Kluwer Academic Publishers, Dordrecht, 1993. [8] Pachpatte BG. On multidimensional Gruss type integral inequalities, J. Inequal. Pure Appl. Math., 32 (2002), 1-15. [9] Qi F, Li AJ, Zhao WZ, Niu DW, Cao J. Extensions of several integral inequalities, J. Inequal. Pure Appl. Math., 7(3)(2006), 1-6. [10] Qi F. Several integral inequalities, J. Inequal. Pure Appl. Math., 1(2)(2000), 1-9. [11] Sarikaya MZ, Aktan N, Yildirim H. On weighted Chebyshev-Gruss like inequalities on time scales, J. Math. Inequal., 2(2)(2008), 185-195. [12] Samko SG, Kilbas AA, Marichev OI. Fractional Integrals and Derivatives - Theory and Applications, Gordon and Breach, Linghorne, 1993. [13] Akkurt A, Kaçar Z, Yildirim H. Generalized Fractional Integrals Inequalities for Continuous Random Variables, Journal of Probability Statistics, Volume 2015, http://dx.doi.org/10.1155/2015/958980, (2015). [14] Diaz, R. and Pariguan, E., On hypergeometric functions and Pochhammer k symbol, Divulg.Math, 15.(2007),179-192. [15] P. L. Butzer, A. A. Kilbas and J.J. Trujillo, Fractional calculus in the Mellin setting and Hadamard-type fractional integrals, Journal of Mathematical Analysis and Applications, 269, (2002), 1-27. [16] U.N. Katugampola, New Approach to a Generalized Fractional Fntegral, Appl. Math. Comput. 218(3), (2011), 860-865. [17] A. A. Kilbas, H.M. Srivastava and J.J. Trujillo, Theory and Applications of Fractional Diferential Equations, Elsevier B.V., Amsterdam, Netherlands, 2006. [18] Akkurt, A., & Y¬ld¬r¬m, H. (2014). Genelle¸stirilmi¸s Fractional I·ntegraller I·çin Feng Qi Tipli I·ntegral E¸sitsizlikleri Üzerine. Fen Bilimleri Dergisi, 1(2). [19] Mubeen, S. and Habibullah, G.M., k fractional integrals and application, Int. J. Contemp. Math. Sciences, 7(2), 2012, 89-94. [20] M.Z. Sarikaya, Z. Dahmani, M.E. Kiris and F. Ahmad, (k; s) Riemann-Liouville fractional integral and applications, Hacettepe Journal of Mathematics and Statistics, Accepted. [Department of Mathematics, Faculty of Science and Arts, University of KahramanmaraS¸ Sütçü I·mam, 46000, KahramanmaraS¸, Turkey E-mail address : [email protected] [Department of Mathematics, Faculty of Science, University of Cumhuriyet, 58140, Sivas, Turkey E-mail address : [email protected] [Department of Mathematics, Faculty of Science and Arts, University of KahramanmaraS¸ Sütçü I·mam, 46000, KahramanmaraS¸, Turkey E-mail address : [email protected]