Research ArticleA Study of Caputo-Hadamard-Type Fractional DifferentialEquations with Nonlocal Boundary Conditions

Wafa Shammakh

Faculty of Science King Abdulaziz University AL Faisaliah Campus Jeddah Saudi Arabia

Correspondence should be addressed to Wafa Shammakh wshammakhkauedusa

Received 17 February 2016 Accepted 16 March 2016

Academic Editor Jozef Banas

Copyright copy 2016 Wafa Shammakh This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

Existence and uniqueness results of positive solutions to nonlinear boundary value problems for Caputo-Hadamard fractionaldifferential equations by using some fixed point theorems are presented The related Greenrsquos function for the boundary valueproblem is given and some useful properties of Greenrsquos function are obtained Example is presented to illustrate the main results

1 Introduction

In this paper we study the existence and uniqueness ofpositive solutions for the 119898-point boundary value problemsfor Caputo-Hadamard fractional differential equations of theform119862

119867119863120572119909 (119905) + 119891 (119905 119909 (119905)) = 0

119905 isin [1 119890] 119899 minus 1 lt 120572 le 119899 119899 gt 2

119862

119867119863119909 (1) =

119898minus2

sum

119894=1

120573119894

119862

119867119863119909 (120578119894)

119862

1198671198632119909 (1) =

119862

1198671198633119909 (1) = sdot sdot sdot =

119862

119867119863(119899minus1)

119909 (1) = 0

119909 (119890) =

119898minus2

sum

119894=1

120574119894119909 (120578119894)

(1)

where 119862119867119863120572 is the Caputo-Hadamard fractional derivative of

order 119899 minus 1 lt 120572 le 119899 sum119898minus2119894=1

120574119894 lt 1 sum119898minus2119894=1

120573119894 lt 1 120578119894 isin (1 119890)

for all 119894 = 1 2 119898 minus 2 1205781lt 1205782lt sdot sdot sdot lt 120578

119898minus2and 119891

[1 119890] times [0infin) rarr [0infin)

Thederivative is a kind of fractional derivatives attributedtoHadamard in 1892 [1] this fractional derivative differs fromthe Riemann-Liouville and Caputo fractional derivatives inthe sense that the kernel of the integral contains a logarithmicfunction of arbitrary exponent The Riemann-Liouville and

Hadamard derivative have their own disadvantages as wellone of which is the fact that the derivative of a constant isnot equal to zero in general The subject of Hadamard-typefractional differential equations has received much attentionby many researchers Some new results on the existence ofsolutions for a fractional boundary value problem involvingHadamard-type fractional differential inclusions and integralboundary conditions can be found in [2]

In [3] Tariboon et al studied the existence and unique-ness of solutions to the boundary value problems consistingof a fractional differential equation of Riemann-Liouvilletype subject to the Hadamard fractional integral equationsThiramanus et al [4] investigated the existence and unique-ness of solutions for a Hadamard fractional differential equa-tions with nonlocal fractional integral boundary conditionsAhmad and Ntouyas [5 6] studied the existence results fora boundary value problem of nonlinear fractional hybriddifferential inclusions of Hadamard type with Dirichletboundary condition and for a coupled system of Hadamardfractional differential equations and Hadamard-type integralboundary conditions respectively

Jarad et al [7] modified the Hadamard fractional deriva-tive into a more suitable one having physically interpretableinitial conditions similar to the ones in the Caputo setting

Basic definitions and properties of fractional calculus andHadamard-type fractional calculus can be found in [8ndash10]

The idea of this paper is to demonstrate sufficient con-ditions on existence and uniqueness of positive solutions

Hindawi Publishing CorporationJournal of Function SpacesVolume 2016 Article ID 7057910 9 pageshttpdxdoiorg10115520167057910

2 Journal of Function Spaces

to nonlinear boundary value problems (1) for modifiedHadamard fractional (Caputo-Hadamard) differential equa-tions by using Banachrsquos fixed point theorem Leray-Schaudernonlinear alternative theorem for single valued maps Kras-noselskiirsquos fixed point theorem and some properties of theGreen function

2 Preliminaries

In this section we introduce some notations and definitionsof Hadamard-type fractional calculus

Definition 1 (see [11 12]) The Hadamard derivative of frac-tional order 120572 for a function 119892 [1infin) rarr R is defined as

119863120572119892 (119905) =

1

Γ (119899 minus 120572)

(119905

119889

119889119905

)

119899

int

119905

1

(log 119905119904

)

119899minus120572minus1119892 (119904)

119904

119889119904

119899 minus 1 lt 120572 lt 119899 119899 = [120572] + 1

(2)

where [120572] denotes the integer part of the real number 120572 andlog(sdot) = log

119890(sdot)

Definition 2 (see [11 12]) The Hadamard fractional integralof order 120572 isin R+ of a function 119892(119909) forall119909 gt 0 is defined as

119868120572

119886+119892 (119909) =

1

Γ (120572)

int

119909

119886

(log 119909119905

)

120572minus1119892 (119905)

119905

119889119905

119868120572

119887minus119892 (119909) =

1

Γ (120572)

int

119887

119909

(log 119905

119909

)

120572minus1119892 (119905)

119905

119889119905

(3)

Definition 3 (see [1 7]) Let R(120572) ge 0 119899 = [R(120572) + 1] and119892 isin 119860119862

119899

120575[119886 119887] 0 lt 119886 lt 119887 lt infinThen

119862119863120572

119886+119892 (119909) = 119863

120572

119886+[119892 (119909) minus

119899minus1

sum

119896=0

120575119896119892 (119886)

119896

(log 119905119886

)

119896

] (119909)

119862119863120572

119887minus119892 (119909)

= 119863120572

119887minus[

[

119892 (119909) minus

119899minus1

sum

119896=0

(minus1)119896120575

119896

119892 (119887)

119896

(log 119887119905

)

119896

]

]

(119909)

(4)

HereR(120572) ge 0 119899 = [R(120572) + 1] 0 lt 119886 lt 119887 lt infin and

119892 isin 119860119862119899

120575[119886 119887] = 119892 [119886 119887] 997888rarr C 120575

(119899minus1)119892 (119909)

isin 119860119862 [119886 119887] 120575 = 119909

119889

119889119909

(5)

In particular if 0 lt R(120572) lt 1 then

119862119863120572

119886+119892 (119909) = 119863

120572

119886+[119892 (119909) minus 119892 (119886)] (119909)

119862119863120572

119887minus119892 (119909) = 119863

120572

119887minus[119892 (119909) minus 119892 (119887)] (119909)

(6)

Theorem4 (see [1 7]) LetR(120572) ge 0 119899 = [R(120572)+1] and 119892 isin119860119862119899

120575[119886 119887] 0 lt 119886 lt 119887 lt infin Then 119862119863120572

119886+119892(119909) and 119862119863120572

119887minus119892(119909)

exist everywhere on [119886 119887] and(a) if 120572 notin N0

119862119863120572

119886+119892 (119909) =

1

Γ (119899 minus 120572)

int

119909

119886

(log 119909119905

)

119899minus120572minus1

120575119899119892 (119905)

119889119905

119905

= 119868119899minus120572

119886+120575119899119892 (119909)

119862119863120572

119887minus119892 (119909) =

(minus1)119899

Γ (119899 minus 120572)

int

119887

119909

(log 119905

119909

)

119899minus120572minus1

120575119899119892 (119905)

119889119905

119905

= (minus1)119899119868119899minus120572

119886+120575119899119892 (119909)

(7)

(b) if 120572 = 119899 isin N0

119862119863120572

119886+119892 (119909) = 120575

119899119892 (119909)

119862119863120572

119887minus119892 (119909) = (minus1)

119899120575119899119892 (119909)

(8)

In particular1198621198630

119886+119892 (119909) =

1198621198630

119887minus119892 (119909) = 119892 (119909) (9)

Lemma 5 (see [1 7]) Let R(120572) ge 0 119899 = [R(120572) + 1] and119892 isin 119862[119886 119887]

IfR(120572) = 0 or 120572 isin N then119862119863120572

119886+(119868120572

119886+119892) (119909) = 119892 (119909)

119862119863120572

119887minus(119868120572

119887minus119892) (119909) = 119892 (119909)

(10)

Lemma 6 (see [1 7]) Let 119892 isin 119860119862119899

120575[119886 119887] or 119862119899

120575[119886 119887] and 120572 isin

119862 and then

119868120572

119886+(119862119863120572

119886+119892) (119909) = 119892 (119909) minus

119899minus1

sum

119896=0

120575119896119892 (119886)

119896

(log 119905119886

)

119896

119868120572

119887minus(119862119863120572

119887minus119892) (119909) = 119892 (119909)

minus

119899minus1

sum

119896=0

(minus1)119896120575

119896

119892 (119886)

119896

(log 119887119905

)

119896

(11)

Lemma 7 (see [1 7]) For 119899 minus 1 lt 120572 le 119899 119899 gt 2 and ℎ isin

119862([1 119890]R) the unique solution of the problem119862

119867119863120572119909 (119905) + ℎ (119905) = 0 119905 isin [1 119890]

119862

119867119863119909 (1) =

119898minus2

sum

119894=1

120573119894

119862

119867119863119909 (120578119894)

119862

1198671198632119909 (1) =

119862

1198671198633119909 (1) = sdot sdot sdot =

119862

119867119863(119899minus1)

119909 (1)

= 0

119909 (119890) =

119898minus2

sum

119894=1

120574119894119909 (120578119894)

(12)

is given by119909 (119905)

= int

119890

1

[119866 (119905 119904) + 119867 (119905 119904 1205781 1205782 120578119898minus2)] ℎ (119904)

119889119904

119904

(13)

Journal of Function Spaces 3

where

119866 (119905 119904) =

1

Γ (120572)

(log 119890119904

)

120572minus1

minus (log 119905119904

)

120572minus1

1 le 119904 le 119905 le 119890

(log 119890119904

)

120572minus1

1 le 119905 le 119904 le 119890

119867 (119905 119904 1205781 1205782 120578

119898minus1)

=

sum119898minus2

119894=1120574119894 [(log (119890119904))

120572minus1minus (log (120578119894119904))

120572minus1]

1205752Γ (120572)

+

(120583 minus 1205752(log 119905))sum119898minus2

119894=1120573119894(log (120578

119894119904))120572minus2

12057511205752Γ (120572 minus 1)

1 le 119904 le 120578119894 119894 = 1 2 119898 minus 1

sum119898minus2

119894=1120574119894(log (119890119904))120572minus1

1205752Γ (120572)

120578119894le 119904 le 119890 119894 = 1 2 119898 minus 1

1205751= 1 minus

119898minus2

sum

119894=1

120573119894 1205752= 1 minus

119898minus2

sum

119894=1

120574119894 120583 = 1 minus

119898minus2

sum

119894=1

120574119894(log 120578

119894)

(14)

Proof In view of Lemma 6 the solution of the Hadamarddifferential equation (12) can be written as

119909 (119905) = minus119867119868120572ℎ (119905) + 1198880 + 1198881 log 119905 + 1198882 (log 119905)

2

+ 1198883(log 119905)3 + sdot sdot sdot + 119888

119899minus1(log 119905)119899minus1

119862

119867119863119909 (119905) = minus

119867119868120572minus1

ℎ (119905) + 1198881+ 21198882log 119905 + 3119888

3(log 119905)2

+ sdot sdot sdot + (119899 minus 1) 119888119899minus1(log 119905)119899minus2

119862

1198671198632119909 (119905) = minus

119867119868120572minus2

ℎ (119905) + 21198882+ 61198883log 119905 + sdot sdot sdot

+ (119899 minus 1) (119899 minus 2) 119888119899minus1(log 119905)119899minus3

119862

119867119863(119899minus1)

119909 (119905) = minus119867119868120572minus119899+1

ℎ (119905) + (119899 minus 1)119888119899minus1

(15)

In view of the boundary conditions we conclude that

1198882 = 1198883 = sdot sdot sdot = 119888119899minus1 = 0

1198881= minus

1

1205751

119898minus2

sum

119894=1

120573119894119868120572minus1

ℎ (120578119894)

1198880 =

1

1205752

[119868120572ℎ (119890) minus

119898minus2

sum

119894=1

120574119894119868120572ℎ (120578119894)]

+

120583

12057511205752

119898minus2

sum

119894=1

120573119894119868120572minus1

ℎ (120578119894)

(16)

Substituting the values of 119888119894 119894 = 0 1 119899 minus 1 we obtain

119909 (119905) = minus119868120572ℎ (119905) +

1

1205752

[119868120572ℎ (119890) minus

119898minus2

sum

119894=1

120574119894119868120572ℎ (120578119894)]

+

(120583 minus 1205752(log 119905))

12057511205752

(

119898minus2

sum

119894=1

120573119894119868120572minus1

ℎ (120578119894)) = minus119868

120572ℎ (119905)

+ 119868120572ℎ (119890) +

1

1205752

[(1 minus 1205752) 119868120572ℎ (119890) minus

119898minus2

sum

119894=1

120574119894119868120572ℎ (120578119894)]

+

(120583 minus 1205752(log 119905))

12057511205752

(

119898minus2

sum

119894=1

120573119894119868120572minus1

ℎ (120578119894)) = minus119868

120572ℎ (119905)

+ 119868120572ℎ (119890) +

1

1205752

[

119898minus2

sum

119894=1

120574119894119868120572ℎ (119890) minus

119898minus2

sum

119894=1

120574119894119868120572ℎ (120578119894)]

+

(120583 minus 1205752(log 119905))

12057511205752

(

119898minus2

sum

119894=1

120573119894119868120572minus1

ℎ (120578119894)) = minus

1

Γ (120572)

sdot int

119905

1

(log 119905119904

)

120572minus1

ℎ (119904)

119889119904

119904

+

1

Γ (120572)

sdot int

119890

1

(log 119890119904

)

120572minus1

ℎ (119904)

119889119904

119904

+

1

1205752Γ (120572)

119898minus2

sum

119894=1

120574119894

sdot [int

119890

1

(log 119890119904

)

120572minus1

ℎ (119904)

119889119904

119904

minus int

120578119894

1

(log120578119894

119904

)

120572minus1

ℎ (119904)

119889119904

119904

] +

(120583 minus 1205752(log 119905))

12057511205752Γ (120572 minus 1)

sdot

119898minus2

sum

119894=1

120573119894 int

120578119894

1

(log120578119894

119904

)

120572minus2

ℎ (119904)

119889119904

119904

(17)

Lemma 8 The functions 119866(119905 119904) 119867(119905 119904 1205781 1205782 120578

119898minus2)

defined by (15) satisfy

(i) 119866(119905 119904) ge 0119867(119905 119904 1205781 1205782 120578

119898minus2) ge 0 forall119905 119904 isin [1 119890]

(ii) min1205911le119905le1205912

119866(119905 119904) ge 1205910max1le119905le119890

119866(119905 119904) = 1205910119866(119904 119904) forall119905

119904 isin (1 119890) 1 lt 1205911lt 1205912lt 119890

4 Journal of Function Spaces

119866 (119904 119904) =

1

Γ (120572)

(log 119890119904

)

120572minus1

1205910= min1205911le119905le1205912

120595 (119905) = (log 119890

1205912

)

(18)

(iii) 1198732119902(119904) le 119867(119905 119904 120578

1 1205782 120578

119898minus2) le 119873

1119902(119904) where

119902 (119904) =

(log (119890119904))120572minus2

12057511205752Γ (120572)

1198731= [1205751

119898minus2

sum

119894=1

120574119894+ (120572 minus 1) 120583

119898minus2

sum

119894=1

120573119894]

1198732= min

119898minus2

sum

119894=1

120574119894(log 119890

120578119894

)

1205751sum119898minus2

119894=1120574119894

(120572 minus 1)

(19)

(iv) min1205911le119905le1205912

119867(119905 119904 1205781 1205782 120578

119898minus2) ge 120591

lowastmax1le119905le119890

119867(119905

119904 1205781 1205782 120578

119898minus2) 119904 isin (1 119890) where

120591lowast=

1

120583

[120583 minus 1205752(log 1205912)] lt 1 1 lt 120591

1lt 1205912lt 119890 (20)

Proof It is clear that (i) holds So we prove that (ii) is true(ii) In view of the expression for 119866(119905 119904) it follows that

119866(119905 119904) le 119866(119904 119904) for all 119904 119905 isin [1 119890]

If 1 le 119904 le 119905 le 119890 we have

119866 (119905 119904)

119866 (119904 119904)

=

[(log (119890119904))120572minus1 minus (log (119905119904))120572minus1]

(log (119890119904))120572minus1

=

[(log (119890119904)) (log (119890119904))120572minus2 minus (log (119905119904)) (log (119905119904))120572minus2]

(log (119890119904))120572minus1

ge

(log (119890119904))120572minus2 [(log (119890119904)) minus (log (119905119904))](log (119890119904))120572minus1

=

(log (119890119905))(log (119890119904))

ge (log 119890119905

) fl 120595 (119905)

(21)

If 1 le 119905 le 119904 le 119890 we have

119866 (119905 119904)

119866 (119904 119904)

= 1 ge 120595 (119905) (22)

Thus

max1le119905le119890

119866 (119905 119904) = 119866 (119904 119904)

120595 (119905) 119866 (119904 119904) le 119866 (119905 119904) le 119866 (119904 119904)

forall119905 119904 isin (1 119890)

(23)

Therefore

min1205911le119905le1205912

119866 (119905 119904) ge 12059101le119905le119890

max119866 (119905 119904) = 1205910119866 (119904 119904)

forall119905 119904 isin (1 119890) 1 lt 1205911 lt 1205912 lt 119890

(24)

(iii) If 1 le 119904 le 120578119894 119894 = 1 2 119898 minus 2

119867(119905 119904 1205781 1205782 120578119898minus2) =

sum119898minus2

119894=1120574119894[(log (119890119904))120572minus1 minus (log (120578119894119904))

120572minus1]

1205752Γ (120572)

+

(120583 minus 1205752(log 119905))sum119898minus2

119894=1120573119894(log (120578

119894119904))120572minus2

12057511205752Γ (120572 minus 1)

le

sum119898minus2

119894=1120574119894 (log (119890119904))

120572minus1

1205752Γ (120572)

+

(120583 minus 1205752 (log 119905))sum

119898minus2

119894=1120573119894 (log (120578119894119904))

120572minus2

12057511205752Γ (120572 minus 1)

=

1

12057511205752Γ (120572)

[1205751

119898minus2

sum

119894=1

120574119894 (log119890

119904

)

120572minus1

+ (120572 minus 1) 120583

119898minus2

sum

119894=1

120573119894 (log120578119894

119904

)

120572minus2

]

le

(log (119890119904))120572minus2

12057511205752Γ (120572)

[1205751(log 119890

119904

)

119898minus2

sum

119894=1

120574119894+ (120572 minus 1) 120583

119898minus2

sum

119894=1

120573119894]

le

(log (119890119904))120572minus2

12057511205752Γ (120572)

[1205751

119898minus2

sum

119894=1

120574119894 + (120572 minus 1) 120583

119898minus2

sum

119894=1

120573119894] fl 1198731119902 (119904)

119867 (119905 119904 1205781 1205782 120578119898minus2) =

sum119898minus2

119894=1120574119894[(log (119890119904))120572minus1 minus (log (120578119894119904))

120572minus1]

1205752Γ (120572)

+

(120583 minus 1205752(log 119905))sum119898minus2

119894=1120573119894(log (120578

119894119904))120572minus2

12057511205752Γ (120572 minus 1)

Journal of Function Spaces 5

ge

sum119898minus2

119894=1120574119894 [(log (119890119904)) (log (119890119904))

120572minus2minus (log (120578119894119904)) (log (120578119894119904))

120572minus2]

1205752Γ (120572)

ge

(log (119890119904))120572minus2

1205752Γ (120572)

119898minus2

sum

119894=1

120574119894[(log 119890

119904

) minus (log120578119894

119904

)] =

(log (119890119904))120572minus2

1205752Γ (120572)

119898minus2

sum

119894=1

120574119894(log 119890

120578119894

) fl 1198732119902 (119904)

(25)

If 120578119894le 119904 le 119890 119894 = 1 2 119898 minus 2 we have

119867(119905 119904 1205781 1205782 120578119898minus2) =

sum119898minus2

119894=1120574119894(log (119890119904))120572minus1

1205752Γ (120572)

le

(log (119890119904))120572minus2

12057511205752Γ (120572)

1205751

119898minus2

sum

119894=1

120574119894 le 1205751

119898minus2

sum

119894=1

120574119894119902 (119904) lt 1198731119902 (119904)

119867 (119905 119904 1205781 1205782 120578

119898minus2) =

sum119898minus2

119894=1120574119894(log (119890119904))120572minus1

1205752Γ (120572)

=

sum119898minus2

119894=1120574119894 (120572 minus 1) (log (119890119904))120572minus1

1205752 (120572 minus 1) Γ (120572)

ge

(log (119890119904))120572minus2

12057511205752Γ (120572)

1205751sum119898minus2

119894=1120574119894

(120572 minus 1)

ge 1198732119902 (119904)

(26)

(iv) Since 120597119867(11990511990412057811205782 120578119898minus2

)120597119905=minussum119898minus2

119894=1120573119894(log(120578

119894119904))120572minus2

1205751Γ(120572minus1)119905 le 0 then119867(119905 119904 120578

1 1205782 120578

119898minus2) is nonincreasing

in 119905 somin1205911le119905le1205912

119867(119905 119904 1205781 1205782 120578119898minus2)

=

sum119898minus2

119894=1120574119894 [(log (119890119904))

120572minus1minus (log (120578119894119904))

120572minus1]

1205752Γ (120572)

+

(120583 minus 1205752(log 1205912))sum119898minus2

119894=1120573119894(log (120578

119894119904))120572minus2

12057511205752Γ (120572 minus 1)

=

sum119898minus2

119894=1120574119894[(log (119890119904))120572minus1 minus (log (120578119894119904))

120572minus1]

1205752Γ (120572)

+ 120591lowast120583sum119898minus2

119894=1120573119894 (log (120578119894119904))

120572minus2

12057511205752Γ (120572 minus 1)

ge 120591lowastmax1le119905le119890

119867(119905 1205781 1205782 120578119898minus2 119904)

(27)

3 Existence Results

Let us denote by 119864 = 119862([1 119890] 119877) the Banach space of allcontinuous functions from [1 119890] toR endowedwith the norm119909 = Sup

119905isin[1119890]|119909(119905)| and let119875 be the cone119875 = 119909 isin 119864 119909(119905) ge

0 119905 isin [1 119890]

Through this paper we assume that the function 119891

[1 119890] times [0infin) rarr [0infin) satisfies the following conditionsof Caratheodory type

(1198671) (i) 119891(119905 119909) is Lebesgue measurable with respect to 119905

on [1 119890](ii) 119891(119905 119909) is continuous with respect to 119909 on [0infin)

By Lemma 6 we obtain an operatorF 119864 rarr 119864 as

(F119909) (119905) = 119909 (119905)

= int

119890

1

[119866 (119905 119904) + 119867 (119905 119904 1205781 1205782 120578119898minus2)] ℎ (119904)

119889119904

119904

119905 isin [1 119890]

(28)

It should be noticed that problem (1) has solutions if and onlyif the operatorF has fixed points

The first existence and uniqueness result is based on theBanach contraction principle

Theorem 9 Assume that the condition (1198671) holds and thereexists a real-valued function 119892(119905) isin 119871[1 119890] such that

1003816100381610038161003816119891 (119905 119909 (119905)) minus 119891 (119905 119910 (119905))

1003816100381610038161003816le 119892 (119905)

1003816100381610038161003816119909 minus 119910

1003816100381610038161003816

forall119905 isin [1 119890] 119909 119910 isin R(29)

Then problem (1) has a unique solution provided 119892Ψ lt 1where

Ψ = [

1

Γ (120572 + 1)

+

1198731

12057511205752(120572 minus 1) Γ (120572)

] (30)

Proof We set Sup119905isin[1119890]

|119891(119904 0)| = 119901 lt infin and choose 120588 ge

Ψ119901(1 minus Ψ119892)

Now we show that FB120588sub B120588 where B

120588= 119909 isin 119864

119909 le 120588 For any 119909 isin B120588 we have

(F119909) (119905)

= Sup119905isin[1119890]

int

119890

1

[119866 (119905 119904) + 119867 (119905 119904 1205781 1205782 120578

119898minus2)]

sdot 119891 (119904 119909 (119904))

119889119904

119904

le int

119890

1

(119866 (119904 119904) + 1198731119902 (119904))

sdot (1003816100381610038161003816119891 (119904 119909 (119904)) minus 119891 (119904 0)

1003816100381610038161003816+1003816100381610038161003816119891 (119904 0)

1003816100381610038161003816)

119889119904

119904

6 Journal of Function Spaces

le int

119890

1

(119866 (119904 119904) + 1198731119902 (119904)) (119892 (119904) 120588 + 119901)

119889119904

119904

le (10038171003817100381710038171198921003817100381710038171003817120588 + 119901) [

1

Γ (120572)

int

119890

1

(log 119890119904

)

120572minus1119889119904

119904

+

1198731

12057511205752Γ (120572)

int

119890

1

(log 119890119904

)

120572minus2119889119904

119904

] le (10038171003817100381710038171198921003817100381710038171003817120588 + 119901)

sdot [

1

Γ (120572 + 1)

+

1198731

12057511205752(120572 minus 1) Γ (120572)

] le Ψ (10038171003817100381710038171198921003817100381710038171003817120588

+ 119901) le 120588

(31)

It follows that FB120588 sub B120588 For 119909 119910 isin 119864 and for each 119905 isin

[1 119890] we have1003816100381610038161003816(F119909) (119905) minus (F119910) (119905)

1003816100381610038161003816

=

10038161003816100381610038161003816100381610038161003816

int

119890

1

[119866 (119905 119904) + 119867 (119905 119904 1205781 1205782 120578

119898minus2)]

sdot (119891 (119904 119909 (119904)) minus 119891 (119904 119910 (119904)))

119889119904

119904

10038161003816100381610038161003816100381610038161003816

le int

119890

1

(119866 (119904 119904)

+ 1198731119902 (119904)) 119892 (119904)

1003816100381610038161003816119909 (119904) minus 119910 (119904)

1003816100381610038161003816

119889119904

119904

le10038171003817100381710038171198921003817100381710038171003817

1003817100381710038171003817119909 minus 119910

1003817100381710038171003817

sdot int

119890

1

(119866 (119904 119904) + 1198731119902 (119904))

119889119904

119904

le Ψ10038171003817100381710038171198921003817100381710038171003817

1003817100381710038171003817119909 minus 119910

1003817100381710038171003817

(32)

Hence it follows that (F119909)(119905) minus (F119910)(119905) le Ψ119892119909 minus 119910where Ψ119892 lt 1ThereforeF is a contraction Hence by thecontraction mapping principle problem (1) has a uniquenesssolution

Theorem 10 (nonlinear alternative for single valued maps[13]) Let 119864 be a Banach spaceC a closed convex subset of 119864U an open subset ofC and 0 isin U Suppose that 119865 U rarr C isa continuous compact (ie 119865(U) is a relatively compact subsetofC) map Then either

(i) 119865 has a fixed point inU or(ii) There is 119909 isin 120597U (the boundary of U in C) and 120582 isin

(1 119890) with 119909 = 120582119865(119909)

Theorem 11 Assume that (1198671) and the following conditions

hold(1198672) There exist two nonnegative real-valued functions

ℎ1 ℎ2isin [1 119890] such that

119891 (119905 119909) le ℎ1 (119905) + ℎ2 (119905) 119909

for every 119905 isin [1 119890] all 119909 isin [0infin)

(33)

(1198673)There exists a constant119872 gt 0 such that

119872

(1003817100381710038171003817ℎ1

1003817100381710038171003817+1003817100381710038171003817ℎ2

1003817100381710038171003817119872)Ψ

gt 1 (34)

Then the boundary value problem (1) has at least one solutionon [1 119890]

Proof First we show that the operator F 119875 rarr 119875 iscontinuous

For any 119909119899 119909 isin 119875 119899 = 1 2 with lim

119899rarrinfin119909119899(119905) =

119909(119905) 119905 isin [1 119890] Thus by condition (ii) of (1198671) we have

lim119899rarrinfin

119891(119905 119909119899(119905)) = 119891(119905 119909(119905)) 119905 isin [1 119890] So we can

conclude that

Sup119905isin[1119890]

1003816100381610038161003816119891 (119905 119909

119899 (119905)) minus 119891 (119905 119909 (119905))

1003816100381610038161003816997888rarr 0 as 119899 997888rarr infin (35)

On the other hand

1003816100381610038161003816(F119909119899) (119905) minus (F119909) (119905)

1003816100381610038161003816

=

10038161003816100381610038161003816100381610038161003816

int

119890

1

[119866 (119905 119904) + 119867 (119905 119904 1205781 1205782 120578119898minus2)]

sdot (119891 (119904 119909 (119904)) minus 119891 (119904 119910 (119904)))

119889119904

119904

10038161003816100381610038161003816100381610038161003816

le Sup119905isin[1119890]

1003816100381610038161003816119891 (119905 119909

119899 (119905)) minus 119891 (119905 119909 (119905))

1003816100381610038161003816int

119890

1

(119866 (119904 119904)

+ 1198731119902 (119904))

119889119904

119904

le [

1

Γ (120572 + 1)

+

1198731

12057511205752(120572 minus 1) Γ (120572)

] Sup119905isin[1119890]

1003816100381610038161003816119891 (119905 119909

119899 (119905))

minus 119891 (119905 119909 (119905))1003816100381610038161003816

(36)

Hence

1003817100381710038171003817(F119909119899) (119905) minus (F119909) (119905)

1003817100381710038171003817997888rarr 0 as 119899 997888rarr infin (37)

This means thatF is continuousNow we show that F maps bounded sets into bounded

sets in 119875 It suffices to show that for any 120590 gt 0 there exists apositive constant 120588

1gt 0 such that for each 119909 isin B

120588lowast = 119909 isin

119875 119909 le 120588lowast we have F119909 le 120588

1 By (33) for each 119905 isin [1 119890]

we have

|(F119909) (119905)| =

10038161003816100381610038161003816100381610038161003816

int

119890

1

[119866 (119905 119904) + 119867 (119905 119904 1205781 1205782 120578

119898minus2)]

sdot 119891 (119904 119909 (119904))

119889119904

119904

10038161003816100381610038161003816100381610038161003816

le int

119890

1

[119866 (119904 119904) + 1198731119902 (119904)] (ℎ1 (

119904)

+ ℎ2 (119904) 119909 (119904))

119889119904

119904

le (1003817100381710038171003817ℎ1

1003817100381710038171003817+1003817100381710038171003817ℎ2

1003817100381710038171003817120588lowast)

sdot int

119890

1

(119866 (119904 119904) + 1198731119902 (119904))

119889119904

119904

le (1003817100381710038171003817ℎ1

1003817100381710038171003817+1003817100381710038171003817ℎ2

1003817100381710038171003817120588lowast)

sdot [

1

Γ (120572 + 1)

+

1198731

12057511205752(120572 minus 1) Γ (120572)

] le (1003817100381710038171003817ℎ1

1003817100381710038171003817+1003817100381710038171003817ℎ2

1003817100381710038171003817

sdot 120588lowast) Ψ fl 120588

1

(38)

Journal of Function Spaces 7

which implies that (F119909) le 1205881 Further we let 119905

1 1199052isin [1 119890]

with 1199051lt 1199052and 119909 isin B

120588lowast where B

120588lowast is a bounded set of 119875

and then we find that

10038161003816100381610038161003816(119862

119867119863F119909) (119905)

10038161003816100381610038161003816=

1003816100381610038161003816100381610038161003816100381610038161003816

minus

1

Γ (120572 minus 1)

sdot int

119890

1

[(log 119905119904

)

120572minus2

+

1

1205751

119898minus2

sum

119894=1

120573119894(log

120578119894

119904

)

120572minus2

]

sdot 119891 (119904 119909 (119904))

119889119904

119904

1003816100381610038161003816100381610038161003816100381610038161003816

le (1003817100381710038171003817ℎ1

1003817100381710038171003817+1003817100381710038171003817ℎ2

1003817100381710038171003817120588lowast)

sdot

(1205751 + sum119898minus2

119894=1120573119894)

1205751Γ (120572 minus 1)

fl 1205882

(39)

Hence for 1199051 1199052isin [1 119890] we have

1003816100381610038161003816(F119909) (1199052

) minus (F119909) (1199051)1003816100381610038161003816le int

1199052

1199051

(119862

119867119863F119909) (119904) 119889119904

le 1205882 (1199052 minus 1199051)

(40)

This implies that F maps bounded sets into equicontinuoussets of 119875

Thus by the Arzela-Ascoli theorem the operator F

119875 rarr 119875 is completely continuousNext we consider the set 119881 = 119906 isin 119864 | 119909 = 120583F119909 0 lt

120583 lt 1 and show that the set 119881 is bounded let 119909 isin 119881 andthen 119909 = 120583F119909 0 lt 120583 lt 1 For any 119905 isin [1 119890] we have

|119909 (119905)| = 120583 |(F119909) (119905)|

le int

119890

1

[119866 (119905 119904) + 119867 (119905 119904 1205781 1205782 120578119898minus2)]

sdot1003816100381610038161003816119891 (119904 119909 (119904))

1003816100381610038161003816

119889119904

119904

le 1205881

(41)

Thus 119909(119905) le 1205881for any 119905 isin [1 119890] so that set 119881 is bounded

Thus by the conclusion of Theorem 10 the operator F hasat least one fixed point which implies that problem (1) has atleast one solution

Theorem 12 (Krasnoselskii fixed point theorem [14]) Let 119864be a Banach space and 119870 sub 119864 is a cone in 119864 Assume that Ω1and Ω

2are open subsets of 119864 with 0 isin Ω

1and Ω

1sub Ω2 Let

119879 119870 cap (Ω2 Ω1) rarr 119870 be completely continuous operator In

addition suppose that either

(i) 119879119906 le 119906 forall119906 isin 119870 cap 120597Ω1and 119879119906 ge 119906 forall119906 isin

119870 cap 120597Ω2or

(ii) 119879119906 le 119906 forall119906 isin 119870 cap 120597Ω2and 119879119906 ge 119906 forall119906 isin

119870 cap 120597Ω1

holds Then 119879 has a fixed point in 119870 cap (Ω2 Ω1)

To state the last result of this section we set

1198721 = (Ψ)minus1

1198722 = (

1205910

Γ (120572 + 1)

+ (120572 minus 1) 120591lowast[

120583sum119898minus2

119894=1120573119894

12057511205752Γ (120572)

minus

sum119898minus2

119894=1120574119894

1205752Γ (120572 + 1)

])

minus1

(42)

Theorem 13 Suppose that there exist two positive constants1199032 gt 1199031 gt 0 and1198721 isin (01198722)

(i) 119891(119905 119909) le 11987211199032 for (119905 119909) isin [1 119890] times [0 119903

2]

(ii) 119891(119905 119909) ge 11987221199031 for (119905 119909) isin [1 119890] times [0 1199031]

Then (1) has at least a positive solution

Proof Let Ω119894= 119909 isin 119864 | 119909 lt 119903

119894 119894 = 1 2 From the proof

of Theorem 11 we know that the operatorF defined by (28)is completely continuous on 119875

For any 119909 isin 119875 cap 120597Ω1 it follows that

(F119909) (119905)

= Sup119905isin[1119890]

int

119890

1

[119866 (119905 119904) + 119867 (119905 119904 1205781 1205782 120578

119898minus2)]

sdot 119891 (119904 119909 (119904))

119889119904

119904

ge int

119890

1

(1205910119866 (119904 119904)

+ 120591lowast119867(1 119904 120578

1 1205782 120578

119898minus2))11987221199031

119889119904

119904

ge 11987221199031[

1205910

Γ (120572 + 1)

+ (120572 minus 1) 120591lowast[

120583sum119898minus2

119894=1120573119894

12057511205752Γ (120572)

minus

sum119898minus2

119894=1120574119894

1205752Γ (120572 + 1)

]] ge 1199031 = 119909

(43)

that is (F119909)(119905) ge 119909 119909 isin 119875 cap 120597Ω1

On the other hand for any 119909 isin 119875 cap 120597Ω2 it follows that

(F119909) (119905)

= Sup119905isin[1119890]

int

119890

1

[119866 (119905 119904) + 119867 (119905 119904 1205781 1205782 120578

119898minus2)]

sdot 119891 (119904 119909 (119904))

119889119904

119904

le int

119890

1

(119866 (119904 119904) + 1198731119902 (119904))

sdot 11987211199032

119889119904

119904

le 11987211199032Ψ = 1199032 = 119909

(44)

that is (F119909)(119905) le 119909 119909 isin 119875 cap 120597Ω2

In view ofTheorem 12F has a fixed point in119875cap(Ω2Ω1)which is a positive solution to (1)

4 Examples

In this section we exemplify our theoretical results obtainedin Section 3

8 Journal of Function Spaces

Example 1 Consider the following BVP for Hadamard frac-tional differential equation

119862

11986711986373119909 (119905) +119891 (119905 119909 (119905)) = 0 1 le 119905 le 119890

119862

119867119863119909 (1) =

1

2

119862

119867119863119909(

4

3

)

119862

1198671198632119909 (1) = 0

119909 (119890) =

1

4

119909 (

4

3

)

(45)

Here 120572 = 73 120573 = 05 120574 = 025 120578 = 43 and

119891 (119905 119909) =

119890119905

2 (119890119905+ 1)

(

1199092

119909 + 1

+

119909

4 (119909 + 1)

+

3

4

)

(119905 119909) isin [1 119890] times [0infin)

(46)

Using the given data we find that Ψ = 1609 1205751= 05 120575

2=

075 120583 = 0928 and

1003816100381610038161003816119891 (119905 119909) minus 119891 (119905 119910)

1003816100381610038161003816le

119890119905

2 (119890119905+ 1)

1003816100381610038161003816119909 minus 119910

1003816100381610038161003816le

1

2

1003816100381610038161003816119909 minus 119910

1003816100381610038161003816

with 119892 (119905) =

119890119905

2 (119890119905+ 1)

(47)

Hencewe obtain 119892Ψ = 0805 lt 1Therefore byTheorem9problem (45) has a unique solution on [1 119890]

Example 2 Consider the following BVP for Hadamard frac-tional differential equation

119862

11986711986352119909 (119905) +119891 (119905 119909 (119905)) = 0 1 le 119905 le 119890

119862

119867119863119909 (1) =

1

4

119862

119867119863119909(

3

2

)

119862

1198671198632119909 (1) = 0

119909 (119890) =

1

3

119909 (

3

2

)

(48)

Here 120572 = 52 120573 = 025 120574 = 13 120578 = 32 Let

119891 (119905 119909) = 9 sin2119905 + 119890119905119909

12 (119890119905+ 1) (119909

2+ 1)

(119905 119909) isin [1 119890] times [0infin)

(49)

with ℎ1(119905) = 9 sin2119905 ℎ

2(119905) = 119890

11990512(119890119905+ 1) Here 119891(119905 119909) le

ℎ1(119905) + ℎ

2(119905)119909

It is easy to verify that119872(ℎ1 + ℎ

2119872)Ψ = 1147 gt 1

Then by Theorem 11 problem (48) has at least onesolution on [1 119890]

Example 3 Consider the following BVP for Hadamard frac-tional differential equation

119862

11986711986372119909 (119905) +119891 (119905 119909 (119905)) = 0 1 le 119905 le 119890

119862

119867119863119909 (1) =

1

2

119862

119867119863119909(

4

3

) +

1

4

119862

119867119863119909(

5

2

)

119862

1198671198632119909 (1) = 0

119909 (119890) =

1

3

119909 (

4

3

) +

1

5

119909 (

5

2

)

(50)

where 120572 = 72 1205731 = 05 1205732 = 025 1205741 = 13 1205742 = 15

1205781 = 43 1205782 = 52 Let

119891 (119905 119909) =

1

4

119909 +

119890119905

5 (119890119905+ 1)

(119905 119909) isin [1 119890] times [0infin) (51)

It is easy to verify that

1198721= (Ψ)

minus1= 06196

1198722= (

1205910

Γ (120572 + 1)

+ (120572 minus 1) 120591lowast[

120583sum119898minus2

119894=1120573119894

12057511205752Γ (120572)

minus

sum119898minus2

119894=1120574119894

1205752Γ (120572 + 1)

])

minus1

= 0521

(52)

Choosing 1199031= 15 119903

2= 119890 we have

119891 (119905 sdot 119909) le 12 le 11987211199032 (119905 119909) isin [1 119890] times [0 1199032

]

119891 (119905 sdot 119909) gt 05 ge 11987221199031 (119905 119909) isin [1 119890] times [0 1199031

]

(53)

Hence all conditions of Theorem 13 are satisfied then prob-lem (50) has at least one positive solution 119909 such that 15 lt

119909 lt 119890

Competing Interests

The author declares no competing interests

References

[1] Y Y Gambo F Jarad D Baleanu and T Abdeljawad ldquoOnCaputo modification of the Hadamard fractional derivativesrdquoAdvances in Difference Equations vol 2014 article 10 12 pages2014

[2] B Ahmad S K Ntouyas and A Alsaedi ldquoNew results forboundary value problems of Hadamard-type fractional differ-ential inclusions and integral boundary conditionsrdquo BoundaryValue Problems vol 275 p 14 2013

[3] J Tariboon S K Ntouyas and W Sudsutad ldquoNonlo-cal Hadamard fractional integral conditions for nonlinearRiemann-Liouville fractional differential equationsrdquo BoundaryValue Problems vol 2014 article 253 2014

Journal of Function Spaces 9

[4] P Thiramanus S K Ntouyas and J Tariboon ldquoExistence anduniqueness results for Hadamard-type fractional differentialequations with nonlocal fractional integral boundary condi-tionsrdquo Abstract and Applied Analysis vol 2014 Article ID902054 9 pages 2014

[5] B Ahmad and S K Ntouyas ldquoAn existence theorem forfractional hybrid differential inclusions of Hadamard type withDirichlet boundary conditionsrdquo Abstract and Applied Analysisvol 2014 Article ID 705809 7 pages 2014

[6] B Ahmad and S K Ntouyas ldquoA fully Hadamard type integralboundary value problem of a coupled system of fractional dif-ferential equationsrdquo Fractional Calculus and Applied AnalysisAn International Journal forTheory and Applications vol 17 no2 pp 348ndash360 2014

[7] F Jarad T Abdeljawad and D Baleanu ldquoCaputo-type modi-fication of the Hadamard fractional derivativesrdquo Advances inDifference Equations vol 2012 no 1 article 142 8 pages 2012

[8] AA Kilbas ldquoHadamard-type fractional calculusrdquo Journal of theKorean Mathematical Society vol 38 no 6 pp 1191ndash1204 2001

[9] I Podlubny Fractional Differential Equations Academic PressSan Diego Calif USA 1999

[10] S Pooseh RAlmeida andD FMTorres ldquoExpansion formulasin terms of integer-order derivatives for the Hadamard frac-tional integral and derivativerdquo Numerical Functional Analysisand Optimization vol 33 no 3 pp 301ndash319 2012

[11] A A Kilbas H M Srivastava and J J Trujillo Theoryand Applications of Fractional Differential Equations vol 204Elsevier Science Amsterdam The Netherlands 2006

[12] S G Samko A A Kilbas and O I Marichev FractionalIntegrals and Derivatives Gordon and Breach Science YverdonSwitzerland 1993

[13] B Ahmad S K Ntouyas and J Tariboon ldquoExistence resultsformixedHadamard andRiemann-Liouville fractional integro-differential equationsrdquo Advances in Difference Equations vol2015 no 1 article 293 8 pages 2015

[14] R P Agarwal M Meehan and D OrsquoRegan Fixed PointTheory and Applications vol 141 Cambridge University PressCambridge UK 2001

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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Decision SciencesAdvances in

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

2 Journal of Function Spaces

to nonlinear boundary value problems (1) for modifiedHadamard fractional (Caputo-Hadamard) differential equa-tions by using Banachrsquos fixed point theorem Leray-Schaudernonlinear alternative theorem for single valued maps Kras-noselskiirsquos fixed point theorem and some properties of theGreen function

2 Preliminaries

In this section we introduce some notations and definitionsof Hadamard-type fractional calculus

Definition 1 (see [11 12]) The Hadamard derivative of frac-tional order 120572 for a function 119892 [1infin) rarr R is defined as

119863120572119892 (119905) =

1

Γ (119899 minus 120572)

(119905

119889

119889119905

)

119899

int

119905

1

(log 119905119904

)

119899minus120572minus1119892 (119904)

119904

119889119904

119899 minus 1 lt 120572 lt 119899 119899 = [120572] + 1

(2)

where [120572] denotes the integer part of the real number 120572 andlog(sdot) = log

119890(sdot)

Definition 2 (see [11 12]) The Hadamard fractional integralof order 120572 isin R+ of a function 119892(119909) forall119909 gt 0 is defined as

119868120572

119886+119892 (119909) =

1

Γ (120572)

int

119909

119886

(log 119909119905

)

120572minus1119892 (119905)

119905

119889119905

119868120572

119887minus119892 (119909) =

1

Γ (120572)

int

119887

119909

(log 119905

119909

)

120572minus1119892 (119905)

119905

119889119905

(3)

Definition 3 (see [1 7]) Let R(120572) ge 0 119899 = [R(120572) + 1] and119892 isin 119860119862

119899

120575[119886 119887] 0 lt 119886 lt 119887 lt infinThen

119862119863120572

119886+119892 (119909) = 119863

120572

119886+[119892 (119909) minus

119899minus1

sum

119896=0

120575119896119892 (119886)

119896

(log 119905119886

)

119896

] (119909)

119862119863120572

119887minus119892 (119909)

= 119863120572

119887minus[

[

119892 (119909) minus

119899minus1

sum

119896=0

(minus1)119896120575

119896

119892 (119887)

119896

(log 119887119905

)

119896

]

]

(119909)

(4)

HereR(120572) ge 0 119899 = [R(120572) + 1] 0 lt 119886 lt 119887 lt infin and

119892 isin 119860119862119899

120575[119886 119887] = 119892 [119886 119887] 997888rarr C 120575

(119899minus1)119892 (119909)

isin 119860119862 [119886 119887] 120575 = 119909

119889

119889119909

(5)

In particular if 0 lt R(120572) lt 1 then

119862119863120572

119886+119892 (119909) = 119863

120572

119886+[119892 (119909) minus 119892 (119886)] (119909)

119862119863120572

119887minus119892 (119909) = 119863

120572

119887minus[119892 (119909) minus 119892 (119887)] (119909)

(6)

Theorem4 (see [1 7]) LetR(120572) ge 0 119899 = [R(120572)+1] and 119892 isin119860119862119899

120575[119886 119887] 0 lt 119886 lt 119887 lt infin Then 119862119863120572

119886+119892(119909) and 119862119863120572

119887minus119892(119909)

exist everywhere on [119886 119887] and(a) if 120572 notin N0

119862119863120572

119886+119892 (119909) =

1

Γ (119899 minus 120572)

int

119909

119886

(log 119909119905

)

119899minus120572minus1

120575119899119892 (119905)

119889119905

119905

= 119868119899minus120572

119886+120575119899119892 (119909)

119862119863120572

119887minus119892 (119909) =

(minus1)119899

Γ (119899 minus 120572)

int

119887

119909

(log 119905

119909

)

119899minus120572minus1

120575119899119892 (119905)

119889119905

119905

= (minus1)119899119868119899minus120572

119886+120575119899119892 (119909)

(7)

(b) if 120572 = 119899 isin N0

119862119863120572

119886+119892 (119909) = 120575

119899119892 (119909)

119862119863120572

119887minus119892 (119909) = (minus1)

119899120575119899119892 (119909)

(8)

In particular1198621198630

119886+119892 (119909) =

1198621198630

119887minus119892 (119909) = 119892 (119909) (9)

Lemma 5 (see [1 7]) Let R(120572) ge 0 119899 = [R(120572) + 1] and119892 isin 119862[119886 119887]

IfR(120572) = 0 or 120572 isin N then119862119863120572

119886+(119868120572

119886+119892) (119909) = 119892 (119909)

119862119863120572

119887minus(119868120572

119887minus119892) (119909) = 119892 (119909)

(10)

Lemma 6 (see [1 7]) Let 119892 isin 119860119862119899

120575[119886 119887] or 119862119899

120575[119886 119887] and 120572 isin

119862 and then

119868120572

119886+(119862119863120572

119886+119892) (119909) = 119892 (119909) minus

119899minus1

sum

119896=0

120575119896119892 (119886)

119896

(log 119905119886

)

119896

119868120572

119887minus(119862119863120572

119887minus119892) (119909) = 119892 (119909)

minus

119899minus1

sum

119896=0

(minus1)119896120575

119896

119892 (119886)

119896

(log 119887119905

)

119896

(11)

Lemma 7 (see [1 7]) For 119899 minus 1 lt 120572 le 119899 119899 gt 2 and ℎ isin

119862([1 119890]R) the unique solution of the problem119862

119867119863120572119909 (119905) + ℎ (119905) = 0 119905 isin [1 119890]

119862

119867119863119909 (1) =

119898minus2

sum

119894=1

120573119894

119862

119867119863119909 (120578119894)

119862

1198671198632119909 (1) =

119862

1198671198633119909 (1) = sdot sdot sdot =

119862

119867119863(119899minus1)

119909 (1)

= 0

119909 (119890) =

119898minus2

sum

119894=1

120574119894119909 (120578119894)

(12)

is given by119909 (119905)

= int

119890

1

[119866 (119905 119904) + 119867 (119905 119904 1205781 1205782 120578119898minus2)] ℎ (119904)

119889119904

119904

(13)

Journal of Function Spaces 3

where

119866 (119905 119904) =

1

Γ (120572)

(log 119890119904

)

120572minus1

minus (log 119905119904

)

120572minus1

1 le 119904 le 119905 le 119890

(log 119890119904

)

120572minus1

1 le 119905 le 119904 le 119890

119867 (119905 119904 1205781 1205782 120578

119898minus1)

=

sum119898minus2

119894=1120574119894 [(log (119890119904))

120572minus1minus (log (120578119894119904))

120572minus1]

1205752Γ (120572)

+

(120583 minus 1205752(log 119905))sum119898minus2

119894=1120573119894(log (120578

119894119904))120572minus2

12057511205752Γ (120572 minus 1)

1 le 119904 le 120578119894 119894 = 1 2 119898 minus 1

sum119898minus2

119894=1120574119894(log (119890119904))120572minus1

1205752Γ (120572)

120578119894le 119904 le 119890 119894 = 1 2 119898 minus 1

1205751= 1 minus

119898minus2

sum

119894=1

120573119894 1205752= 1 minus

119898minus2

sum

119894=1

120574119894 120583 = 1 minus

119898minus2

sum

119894=1

120574119894(log 120578

119894)

(14)

Proof In view of Lemma 6 the solution of the Hadamarddifferential equation (12) can be written as

119909 (119905) = minus119867119868120572ℎ (119905) + 1198880 + 1198881 log 119905 + 1198882 (log 119905)

2

+ 1198883(log 119905)3 + sdot sdot sdot + 119888

119899minus1(log 119905)119899minus1

119862

119867119863119909 (119905) = minus

119867119868120572minus1

ℎ (119905) + 1198881+ 21198882log 119905 + 3119888

3(log 119905)2

+ sdot sdot sdot + (119899 minus 1) 119888119899minus1(log 119905)119899minus2

119862

1198671198632119909 (119905) = minus

119867119868120572minus2

ℎ (119905) + 21198882+ 61198883log 119905 + sdot sdot sdot

+ (119899 minus 1) (119899 minus 2) 119888119899minus1(log 119905)119899minus3

119862

119867119863(119899minus1)

119909 (119905) = minus119867119868120572minus119899+1

ℎ (119905) + (119899 minus 1)119888119899minus1

(15)

In view of the boundary conditions we conclude that

1198882 = 1198883 = sdot sdot sdot = 119888119899minus1 = 0

1198881= minus

1

1205751

119898minus2

sum

119894=1

120573119894119868120572minus1

ℎ (120578119894)

1198880 =

1

1205752

[119868120572ℎ (119890) minus

119898minus2

sum

119894=1

120574119894119868120572ℎ (120578119894)]

+

120583

12057511205752

119898minus2

sum

119894=1

120573119894119868120572minus1

ℎ (120578119894)

(16)

Substituting the values of 119888119894 119894 = 0 1 119899 minus 1 we obtain

119909 (119905) = minus119868120572ℎ (119905) +

1

1205752

[119868120572ℎ (119890) minus

119898minus2

sum

119894=1

120574119894119868120572ℎ (120578119894)]

+

(120583 minus 1205752(log 119905))

12057511205752

(

119898minus2

sum

119894=1

120573119894119868120572minus1

ℎ (120578119894)) = minus119868

120572ℎ (119905)

+ 119868120572ℎ (119890) +

1

1205752

[(1 minus 1205752) 119868120572ℎ (119890) minus

119898minus2

sum

119894=1

120574119894119868120572ℎ (120578119894)]

+

(120583 minus 1205752(log 119905))

12057511205752

(

119898minus2

sum

119894=1

120573119894119868120572minus1

ℎ (120578119894)) = minus119868

120572ℎ (119905)

+ 119868120572ℎ (119890) +

1

1205752

[

119898minus2

sum

119894=1

120574119894119868120572ℎ (119890) minus

119898minus2

sum

119894=1

120574119894119868120572ℎ (120578119894)]

+

(120583 minus 1205752(log 119905))

12057511205752

(

119898minus2

sum

119894=1

120573119894119868120572minus1

ℎ (120578119894)) = minus

1

Γ (120572)

sdot int

119905

1

(log 119905119904

)

120572minus1

ℎ (119904)

119889119904

119904

+

1

Γ (120572)

sdot int

119890

1

(log 119890119904

)

120572minus1

ℎ (119904)

119889119904

119904

+

1

1205752Γ (120572)

119898minus2

sum

119894=1

120574119894

sdot [int

119890

1

(log 119890119904

)

120572minus1

ℎ (119904)

119889119904

119904

minus int

120578119894

1

(log120578119894

119904

)

120572minus1

ℎ (119904)

119889119904

119904

] +

(120583 minus 1205752(log 119905))

12057511205752Γ (120572 minus 1)

sdot

119898minus2

sum

119894=1

120573119894 int

120578119894

1

(log120578119894

119904

)

120572minus2

ℎ (119904)

119889119904

119904

(17)

Lemma 8 The functions 119866(119905 119904) 119867(119905 119904 1205781 1205782 120578

119898minus2)

defined by (15) satisfy

(i) 119866(119905 119904) ge 0119867(119905 119904 1205781 1205782 120578

119898minus2) ge 0 forall119905 119904 isin [1 119890]

(ii) min1205911le119905le1205912

119866(119905 119904) ge 1205910max1le119905le119890

119866(119905 119904) = 1205910119866(119904 119904) forall119905

119904 isin (1 119890) 1 lt 1205911lt 1205912lt 119890

4 Journal of Function Spaces

119866 (119904 119904) =

1

Γ (120572)

(log 119890119904

)

120572minus1

1205910= min1205911le119905le1205912

120595 (119905) = (log 119890

1205912

)

(18)

(iii) 1198732119902(119904) le 119867(119905 119904 120578

1 1205782 120578

119898minus2) le 119873

1119902(119904) where

119902 (119904) =

(log (119890119904))120572minus2

12057511205752Γ (120572)

1198731= [1205751

119898minus2

sum

119894=1

120574119894+ (120572 minus 1) 120583

119898minus2

sum

119894=1

120573119894]

1198732= min

119898minus2

sum

119894=1

120574119894(log 119890

120578119894

)

1205751sum119898minus2

119894=1120574119894

(120572 minus 1)

(19)

(iv) min1205911le119905le1205912

119867(119905 119904 1205781 1205782 120578

119898minus2) ge 120591

lowastmax1le119905le119890

119867(119905

119904 1205781 1205782 120578

119898minus2) 119904 isin (1 119890) where

120591lowast=

1

120583

[120583 minus 1205752(log 1205912)] lt 1 1 lt 120591

1lt 1205912lt 119890 (20)

Proof It is clear that (i) holds So we prove that (ii) is true(ii) In view of the expression for 119866(119905 119904) it follows that

119866(119905 119904) le 119866(119904 119904) for all 119904 119905 isin [1 119890]

If 1 le 119904 le 119905 le 119890 we have

119866 (119905 119904)

119866 (119904 119904)

=

[(log (119890119904))120572minus1 minus (log (119905119904))120572minus1]

(log (119890119904))120572minus1

=

[(log (119890119904)) (log (119890119904))120572minus2 minus (log (119905119904)) (log (119905119904))120572minus2]

(log (119890119904))120572minus1

ge

(log (119890119904))120572minus2 [(log (119890119904)) minus (log (119905119904))](log (119890119904))120572minus1

=

(log (119890119905))(log (119890119904))

ge (log 119890119905

) fl 120595 (119905)

(21)

If 1 le 119905 le 119904 le 119890 we have

119866 (119905 119904)

119866 (119904 119904)

= 1 ge 120595 (119905) (22)

Thus

max1le119905le119890

119866 (119905 119904) = 119866 (119904 119904)

120595 (119905) 119866 (119904 119904) le 119866 (119905 119904) le 119866 (119904 119904)

forall119905 119904 isin (1 119890)

(23)

Therefore

min1205911le119905le1205912

119866 (119905 119904) ge 12059101le119905le119890

max119866 (119905 119904) = 1205910119866 (119904 119904)

forall119905 119904 isin (1 119890) 1 lt 1205911 lt 1205912 lt 119890

(24)

(iii) If 1 le 119904 le 120578119894 119894 = 1 2 119898 minus 2

119867(119905 119904 1205781 1205782 120578119898minus2) =

sum119898minus2

119894=1120574119894[(log (119890119904))120572minus1 minus (log (120578119894119904))

120572minus1]

1205752Γ (120572)

+

(120583 minus 1205752(log 119905))sum119898minus2

119894=1120573119894(log (120578

119894119904))120572minus2

12057511205752Γ (120572 minus 1)

le

sum119898minus2

119894=1120574119894 (log (119890119904))

120572minus1

1205752Γ (120572)

+

(120583 minus 1205752 (log 119905))sum

119898minus2

119894=1120573119894 (log (120578119894119904))

120572minus2

12057511205752Γ (120572 minus 1)

=

1

12057511205752Γ (120572)

[1205751

119898minus2

sum

119894=1

120574119894 (log119890

119904

)

120572minus1

+ (120572 minus 1) 120583

119898minus2

sum

119894=1

120573119894 (log120578119894

119904

)

120572minus2

]

le

(log (119890119904))120572minus2

12057511205752Γ (120572)

[1205751(log 119890

119904

)

119898minus2

sum

119894=1

120574119894+ (120572 minus 1) 120583

119898minus2

sum

119894=1

120573119894]

le

(log (119890119904))120572minus2

12057511205752Γ (120572)

[1205751

119898minus2

sum

119894=1

120574119894 + (120572 minus 1) 120583

119898minus2

sum

119894=1

120573119894] fl 1198731119902 (119904)

119867 (119905 119904 1205781 1205782 120578119898minus2) =

sum119898minus2

119894=1120574119894[(log (119890119904))120572minus1 minus (log (120578119894119904))

120572minus1]

1205752Γ (120572)

+

(120583 minus 1205752(log 119905))sum119898minus2

119894=1120573119894(log (120578

119894119904))120572minus2

12057511205752Γ (120572 minus 1)

Journal of Function Spaces 5

ge

sum119898minus2

119894=1120574119894 [(log (119890119904)) (log (119890119904))

120572minus2minus (log (120578119894119904)) (log (120578119894119904))

120572minus2]

1205752Γ (120572)

ge

(log (119890119904))120572minus2

1205752Γ (120572)

119898minus2

sum

119894=1

120574119894[(log 119890

119904

) minus (log120578119894

119904

)] =

(log (119890119904))120572minus2

1205752Γ (120572)

119898minus2

sum

119894=1

120574119894(log 119890

120578119894

) fl 1198732119902 (119904)

(25)

If 120578119894le 119904 le 119890 119894 = 1 2 119898 minus 2 we have

119867(119905 119904 1205781 1205782 120578119898minus2) =

sum119898minus2

119894=1120574119894(log (119890119904))120572minus1

1205752Γ (120572)

le

(log (119890119904))120572minus2

12057511205752Γ (120572)

1205751

119898minus2

sum

119894=1

120574119894 le 1205751

119898minus2

sum

119894=1

120574119894119902 (119904) lt 1198731119902 (119904)

119867 (119905 119904 1205781 1205782 120578

119898minus2) =

sum119898minus2

119894=1120574119894(log (119890119904))120572minus1

1205752Γ (120572)

=

sum119898minus2

119894=1120574119894 (120572 minus 1) (log (119890119904))120572minus1

1205752 (120572 minus 1) Γ (120572)

ge

(log (119890119904))120572minus2

12057511205752Γ (120572)

1205751sum119898minus2

119894=1120574119894

(120572 minus 1)

ge 1198732119902 (119904)

(26)

(iv) Since 120597119867(11990511990412057811205782 120578119898minus2

)120597119905=minussum119898minus2

119894=1120573119894(log(120578

119894119904))120572minus2

1205751Γ(120572minus1)119905 le 0 then119867(119905 119904 120578

1 1205782 120578

119898minus2) is nonincreasing

in 119905 somin1205911le119905le1205912

119867(119905 119904 1205781 1205782 120578119898minus2)

=

sum119898minus2

119894=1120574119894 [(log (119890119904))

120572minus1minus (log (120578119894119904))

120572minus1]

1205752Γ (120572)

+

(120583 minus 1205752(log 1205912))sum119898minus2

119894=1120573119894(log (120578

119894119904))120572minus2

12057511205752Γ (120572 minus 1)

=

sum119898minus2

119894=1120574119894[(log (119890119904))120572minus1 minus (log (120578119894119904))

120572minus1]

1205752Γ (120572)

+ 120591lowast120583sum119898minus2

119894=1120573119894 (log (120578119894119904))

120572minus2

12057511205752Γ (120572 minus 1)

ge 120591lowastmax1le119905le119890

119867(119905 1205781 1205782 120578119898minus2 119904)

(27)

3 Existence Results

Let us denote by 119864 = 119862([1 119890] 119877) the Banach space of allcontinuous functions from [1 119890] toR endowedwith the norm119909 = Sup

119905isin[1119890]|119909(119905)| and let119875 be the cone119875 = 119909 isin 119864 119909(119905) ge

0 119905 isin [1 119890]

Through this paper we assume that the function 119891

[1 119890] times [0infin) rarr [0infin) satisfies the following conditionsof Caratheodory type

(1198671) (i) 119891(119905 119909) is Lebesgue measurable with respect to 119905

on [1 119890](ii) 119891(119905 119909) is continuous with respect to 119909 on [0infin)

By Lemma 6 we obtain an operatorF 119864 rarr 119864 as

(F119909) (119905) = 119909 (119905)

= int

119890

1

[119866 (119905 119904) + 119867 (119905 119904 1205781 1205782 120578119898minus2)] ℎ (119904)

119889119904

119904

119905 isin [1 119890]

(28)

It should be noticed that problem (1) has solutions if and onlyif the operatorF has fixed points

The first existence and uniqueness result is based on theBanach contraction principle

Theorem 9 Assume that the condition (1198671) holds and thereexists a real-valued function 119892(119905) isin 119871[1 119890] such that

1003816100381610038161003816119891 (119905 119909 (119905)) minus 119891 (119905 119910 (119905))

1003816100381610038161003816le 119892 (119905)

1003816100381610038161003816119909 minus 119910

1003816100381610038161003816

forall119905 isin [1 119890] 119909 119910 isin R(29)

Then problem (1) has a unique solution provided 119892Ψ lt 1where

Ψ = [

1

Γ (120572 + 1)

+

1198731

12057511205752(120572 minus 1) Γ (120572)

] (30)

Proof We set Sup119905isin[1119890]

|119891(119904 0)| = 119901 lt infin and choose 120588 ge

Ψ119901(1 minus Ψ119892)

Now we show that FB120588sub B120588 where B

120588= 119909 isin 119864

119909 le 120588 For any 119909 isin B120588 we have

(F119909) (119905)

= Sup119905isin[1119890]

int

119890

1

[119866 (119905 119904) + 119867 (119905 119904 1205781 1205782 120578

119898minus2)]

sdot 119891 (119904 119909 (119904))

119889119904

119904

le int

119890

1

(119866 (119904 119904) + 1198731119902 (119904))

sdot (1003816100381610038161003816119891 (119904 119909 (119904)) minus 119891 (119904 0)

1003816100381610038161003816+1003816100381610038161003816119891 (119904 0)

1003816100381610038161003816)

119889119904

119904

6 Journal of Function Spaces

le int

119890

1

(119866 (119904 119904) + 1198731119902 (119904)) (119892 (119904) 120588 + 119901)

119889119904

119904

le (10038171003817100381710038171198921003817100381710038171003817120588 + 119901) [

1

Γ (120572)

int

119890

1

(log 119890119904

)

120572minus1119889119904

119904

+

1198731

12057511205752Γ (120572)

int

119890

1

(log 119890119904

)

120572minus2119889119904

119904

] le (10038171003817100381710038171198921003817100381710038171003817120588 + 119901)

sdot [

1

Γ (120572 + 1)

+

1198731

12057511205752(120572 minus 1) Γ (120572)

] le Ψ (10038171003817100381710038171198921003817100381710038171003817120588

+ 119901) le 120588

(31)

It follows that FB120588 sub B120588 For 119909 119910 isin 119864 and for each 119905 isin

[1 119890] we have1003816100381610038161003816(F119909) (119905) minus (F119910) (119905)

1003816100381610038161003816

=

10038161003816100381610038161003816100381610038161003816

int

119890

1

[119866 (119905 119904) + 119867 (119905 119904 1205781 1205782 120578

119898minus2)]

sdot (119891 (119904 119909 (119904)) minus 119891 (119904 119910 (119904)))

119889119904

119904

10038161003816100381610038161003816100381610038161003816

le int

119890

1

(119866 (119904 119904)

+ 1198731119902 (119904)) 119892 (119904)

1003816100381610038161003816119909 (119904) minus 119910 (119904)

1003816100381610038161003816

119889119904

119904

le10038171003817100381710038171198921003817100381710038171003817

1003817100381710038171003817119909 minus 119910

1003817100381710038171003817

sdot int

119890

1

(119866 (119904 119904) + 1198731119902 (119904))

119889119904

119904

le Ψ10038171003817100381710038171198921003817100381710038171003817

1003817100381710038171003817119909 minus 119910

1003817100381710038171003817

(32)

Hence it follows that (F119909)(119905) minus (F119910)(119905) le Ψ119892119909 minus 119910where Ψ119892 lt 1ThereforeF is a contraction Hence by thecontraction mapping principle problem (1) has a uniquenesssolution

Theorem 10 (nonlinear alternative for single valued maps[13]) Let 119864 be a Banach spaceC a closed convex subset of 119864U an open subset ofC and 0 isin U Suppose that 119865 U rarr C isa continuous compact (ie 119865(U) is a relatively compact subsetofC) map Then either

(i) 119865 has a fixed point inU or(ii) There is 119909 isin 120597U (the boundary of U in C) and 120582 isin

(1 119890) with 119909 = 120582119865(119909)

Theorem 11 Assume that (1198671) and the following conditions

hold(1198672) There exist two nonnegative real-valued functions

ℎ1 ℎ2isin [1 119890] such that

119891 (119905 119909) le ℎ1 (119905) + ℎ2 (119905) 119909

for every 119905 isin [1 119890] all 119909 isin [0infin)

(33)

(1198673)There exists a constant119872 gt 0 such that

119872

(1003817100381710038171003817ℎ1

1003817100381710038171003817+1003817100381710038171003817ℎ2

1003817100381710038171003817119872)Ψ

gt 1 (34)

Then the boundary value problem (1) has at least one solutionon [1 119890]

Proof First we show that the operator F 119875 rarr 119875 iscontinuous

For any 119909119899 119909 isin 119875 119899 = 1 2 with lim

119899rarrinfin119909119899(119905) =

119909(119905) 119905 isin [1 119890] Thus by condition (ii) of (1198671) we have

lim119899rarrinfin

119891(119905 119909119899(119905)) = 119891(119905 119909(119905)) 119905 isin [1 119890] So we can

conclude that

Sup119905isin[1119890]

1003816100381610038161003816119891 (119905 119909

119899 (119905)) minus 119891 (119905 119909 (119905))

1003816100381610038161003816997888rarr 0 as 119899 997888rarr infin (35)

On the other hand

1003816100381610038161003816(F119909119899) (119905) minus (F119909) (119905)

1003816100381610038161003816

=

10038161003816100381610038161003816100381610038161003816

int

119890

1

[119866 (119905 119904) + 119867 (119905 119904 1205781 1205782 120578119898minus2)]

sdot (119891 (119904 119909 (119904)) minus 119891 (119904 119910 (119904)))

119889119904

119904

10038161003816100381610038161003816100381610038161003816

le Sup119905isin[1119890]

1003816100381610038161003816119891 (119905 119909

119899 (119905)) minus 119891 (119905 119909 (119905))

1003816100381610038161003816int

119890

1

(119866 (119904 119904)

+ 1198731119902 (119904))

119889119904

119904

le [

1

Γ (120572 + 1)

+

1198731

12057511205752(120572 minus 1) Γ (120572)

] Sup119905isin[1119890]

1003816100381610038161003816119891 (119905 119909

119899 (119905))

minus 119891 (119905 119909 (119905))1003816100381610038161003816

(36)

Hence

1003817100381710038171003817(F119909119899) (119905) minus (F119909) (119905)

1003817100381710038171003817997888rarr 0 as 119899 997888rarr infin (37)

This means thatF is continuousNow we show that F maps bounded sets into bounded

sets in 119875 It suffices to show that for any 120590 gt 0 there exists apositive constant 120588

1gt 0 such that for each 119909 isin B

120588lowast = 119909 isin

119875 119909 le 120588lowast we have F119909 le 120588

1 By (33) for each 119905 isin [1 119890]

we have

|(F119909) (119905)| =

10038161003816100381610038161003816100381610038161003816

int

119890

1

[119866 (119905 119904) + 119867 (119905 119904 1205781 1205782 120578

119898minus2)]

sdot 119891 (119904 119909 (119904))

119889119904

119904

10038161003816100381610038161003816100381610038161003816

le int

119890

1

[119866 (119904 119904) + 1198731119902 (119904)] (ℎ1 (

119904)

+ ℎ2 (119904) 119909 (119904))

119889119904

119904

le (1003817100381710038171003817ℎ1

1003817100381710038171003817+1003817100381710038171003817ℎ2

1003817100381710038171003817120588lowast)

sdot int

119890

1

(119866 (119904 119904) + 1198731119902 (119904))

119889119904

119904

le (1003817100381710038171003817ℎ1

1003817100381710038171003817+1003817100381710038171003817ℎ2

1003817100381710038171003817120588lowast)

sdot [

1

Γ (120572 + 1)

+

1198731

12057511205752(120572 minus 1) Γ (120572)

] le (1003817100381710038171003817ℎ1

1003817100381710038171003817+1003817100381710038171003817ℎ2

1003817100381710038171003817

sdot 120588lowast) Ψ fl 120588

1

(38)

Journal of Function Spaces 7

which implies that (F119909) le 1205881 Further we let 119905

1 1199052isin [1 119890]

with 1199051lt 1199052and 119909 isin B

120588lowast where B

120588lowast is a bounded set of 119875

and then we find that

10038161003816100381610038161003816(119862

119867119863F119909) (119905)

10038161003816100381610038161003816=

1003816100381610038161003816100381610038161003816100381610038161003816

minus

1

Γ (120572 minus 1)

sdot int

119890

1

[(log 119905119904

)

120572minus2

+

1

1205751

119898minus2

sum

119894=1

120573119894(log

120578119894

119904

)

120572minus2

]

sdot 119891 (119904 119909 (119904))

119889119904

119904

1003816100381610038161003816100381610038161003816100381610038161003816

le (1003817100381710038171003817ℎ1

1003817100381710038171003817+1003817100381710038171003817ℎ2

1003817100381710038171003817120588lowast)

sdot

(1205751 + sum119898minus2

119894=1120573119894)

1205751Γ (120572 minus 1)

fl 1205882

(39)

Hence for 1199051 1199052isin [1 119890] we have

1003816100381610038161003816(F119909) (1199052

) minus (F119909) (1199051)1003816100381610038161003816le int

1199052

1199051

(119862

119867119863F119909) (119904) 119889119904

le 1205882 (1199052 minus 1199051)

(40)

This implies that F maps bounded sets into equicontinuoussets of 119875

Thus by the Arzela-Ascoli theorem the operator F

119875 rarr 119875 is completely continuousNext we consider the set 119881 = 119906 isin 119864 | 119909 = 120583F119909 0 lt

120583 lt 1 and show that the set 119881 is bounded let 119909 isin 119881 andthen 119909 = 120583F119909 0 lt 120583 lt 1 For any 119905 isin [1 119890] we have

|119909 (119905)| = 120583 |(F119909) (119905)|

le int

119890

1

[119866 (119905 119904) + 119867 (119905 119904 1205781 1205782 120578119898minus2)]

sdot1003816100381610038161003816119891 (119904 119909 (119904))

1003816100381610038161003816

119889119904

119904

le 1205881

(41)

Thus 119909(119905) le 1205881for any 119905 isin [1 119890] so that set 119881 is bounded

Thus by the conclusion of Theorem 10 the operator F hasat least one fixed point which implies that problem (1) has atleast one solution

Theorem 12 (Krasnoselskii fixed point theorem [14]) Let 119864be a Banach space and 119870 sub 119864 is a cone in 119864 Assume that Ω1and Ω

2are open subsets of 119864 with 0 isin Ω

1and Ω

1sub Ω2 Let

119879 119870 cap (Ω2 Ω1) rarr 119870 be completely continuous operator In

addition suppose that either

(i) 119879119906 le 119906 forall119906 isin 119870 cap 120597Ω1and 119879119906 ge 119906 forall119906 isin

119870 cap 120597Ω2or

(ii) 119879119906 le 119906 forall119906 isin 119870 cap 120597Ω2and 119879119906 ge 119906 forall119906 isin

119870 cap 120597Ω1

holds Then 119879 has a fixed point in 119870 cap (Ω2 Ω1)

To state the last result of this section we set

1198721 = (Ψ)minus1

1198722 = (

1205910

Γ (120572 + 1)

+ (120572 minus 1) 120591lowast[

120583sum119898minus2

119894=1120573119894

12057511205752Γ (120572)

minus

sum119898minus2

119894=1120574119894

1205752Γ (120572 + 1)

])

minus1

(42)

Theorem 13 Suppose that there exist two positive constants1199032 gt 1199031 gt 0 and1198721 isin (01198722)

(i) 119891(119905 119909) le 11987211199032 for (119905 119909) isin [1 119890] times [0 119903

2]

(ii) 119891(119905 119909) ge 11987221199031 for (119905 119909) isin [1 119890] times [0 1199031]

Then (1) has at least a positive solution

Proof Let Ω119894= 119909 isin 119864 | 119909 lt 119903

119894 119894 = 1 2 From the proof

of Theorem 11 we know that the operatorF defined by (28)is completely continuous on 119875

For any 119909 isin 119875 cap 120597Ω1 it follows that

(F119909) (119905)

= Sup119905isin[1119890]

int

119890

1

[119866 (119905 119904) + 119867 (119905 119904 1205781 1205782 120578

119898minus2)]

sdot 119891 (119904 119909 (119904))

119889119904

119904

ge int

119890

1

(1205910119866 (119904 119904)

+ 120591lowast119867(1 119904 120578

1 1205782 120578

119898minus2))11987221199031

119889119904

119904

ge 11987221199031[

1205910

Γ (120572 + 1)

+ (120572 minus 1) 120591lowast[

120583sum119898minus2

119894=1120573119894

12057511205752Γ (120572)

minus

sum119898minus2

119894=1120574119894

1205752Γ (120572 + 1)

]] ge 1199031 = 119909

(43)

that is (F119909)(119905) ge 119909 119909 isin 119875 cap 120597Ω1

On the other hand for any 119909 isin 119875 cap 120597Ω2 it follows that

(F119909) (119905)

= Sup119905isin[1119890]

int

119890

1

[119866 (119905 119904) + 119867 (119905 119904 1205781 1205782 120578

119898minus2)]

sdot 119891 (119904 119909 (119904))

119889119904

119904

le int

119890

1

(119866 (119904 119904) + 1198731119902 (119904))

sdot 11987211199032

119889119904

119904

le 11987211199032Ψ = 1199032 = 119909

(44)

that is (F119909)(119905) le 119909 119909 isin 119875 cap 120597Ω2

In view ofTheorem 12F has a fixed point in119875cap(Ω2Ω1)which is a positive solution to (1)

4 Examples

In this section we exemplify our theoretical results obtainedin Section 3

8 Journal of Function Spaces

Example 1 Consider the following BVP for Hadamard frac-tional differential equation

119862

11986711986373119909 (119905) +119891 (119905 119909 (119905)) = 0 1 le 119905 le 119890

119862

119867119863119909 (1) =

1

2

119862

119867119863119909(

4

3

)

119862

1198671198632119909 (1) = 0

119909 (119890) =

1

4

119909 (

4

3

)

(45)

Here 120572 = 73 120573 = 05 120574 = 025 120578 = 43 and

119891 (119905 119909) =

119890119905

2 (119890119905+ 1)

(

1199092

119909 + 1

+

119909

4 (119909 + 1)

+

3

4

)

(119905 119909) isin [1 119890] times [0infin)

(46)

Using the given data we find that Ψ = 1609 1205751= 05 120575

2=

075 120583 = 0928 and

1003816100381610038161003816119891 (119905 119909) minus 119891 (119905 119910)

1003816100381610038161003816le

119890119905

2 (119890119905+ 1)

1003816100381610038161003816119909 minus 119910

1003816100381610038161003816le

1

2

1003816100381610038161003816119909 minus 119910

1003816100381610038161003816

with 119892 (119905) =

119890119905

2 (119890119905+ 1)

(47)

Hencewe obtain 119892Ψ = 0805 lt 1Therefore byTheorem9problem (45) has a unique solution on [1 119890]

Example 2 Consider the following BVP for Hadamard frac-tional differential equation

119862

11986711986352119909 (119905) +119891 (119905 119909 (119905)) = 0 1 le 119905 le 119890

119862

119867119863119909 (1) =

1

4

119862

119867119863119909(

3

2

)

119862

1198671198632119909 (1) = 0

119909 (119890) =

1

3

119909 (

3

2

)

(48)

Here 120572 = 52 120573 = 025 120574 = 13 120578 = 32 Let

119891 (119905 119909) = 9 sin2119905 + 119890119905119909

12 (119890119905+ 1) (119909

2+ 1)

(119905 119909) isin [1 119890] times [0infin)

(49)

with ℎ1(119905) = 9 sin2119905 ℎ

2(119905) = 119890

11990512(119890119905+ 1) Here 119891(119905 119909) le

ℎ1(119905) + ℎ

2(119905)119909

It is easy to verify that119872(ℎ1 + ℎ

2119872)Ψ = 1147 gt 1

Then by Theorem 11 problem (48) has at least onesolution on [1 119890]

Example 3 Consider the following BVP for Hadamard frac-tional differential equation

119862

11986711986372119909 (119905) +119891 (119905 119909 (119905)) = 0 1 le 119905 le 119890

119862

119867119863119909 (1) =

1

2

119862

119867119863119909(

4

3

) +

1

4

119862

119867119863119909(

5

2

)

119862

1198671198632119909 (1) = 0

119909 (119890) =

1

3

119909 (

4

3

) +

1

5

119909 (

5

2

)

(50)

where 120572 = 72 1205731 = 05 1205732 = 025 1205741 = 13 1205742 = 15

1205781 = 43 1205782 = 52 Let

119891 (119905 119909) =

1

4

119909 +

119890119905

5 (119890119905+ 1)

(119905 119909) isin [1 119890] times [0infin) (51)

It is easy to verify that

1198721= (Ψ)

minus1= 06196

1198722= (

1205910

Γ (120572 + 1)

+ (120572 minus 1) 120591lowast[

120583sum119898minus2

119894=1120573119894

12057511205752Γ (120572)

minus

sum119898minus2

119894=1120574119894

1205752Γ (120572 + 1)

])

minus1

= 0521

(52)

Choosing 1199031= 15 119903

2= 119890 we have

119891 (119905 sdot 119909) le 12 le 11987211199032 (119905 119909) isin [1 119890] times [0 1199032

]

119891 (119905 sdot 119909) gt 05 ge 11987221199031 (119905 119909) isin [1 119890] times [0 1199031

]

(53)

Hence all conditions of Theorem 13 are satisfied then prob-lem (50) has at least one positive solution 119909 such that 15 lt

119909 lt 119890

Competing Interests

The author declares no competing interests

References

[1] Y Y Gambo F Jarad D Baleanu and T Abdeljawad ldquoOnCaputo modification of the Hadamard fractional derivativesrdquoAdvances in Difference Equations vol 2014 article 10 12 pages2014

[2] B Ahmad S K Ntouyas and A Alsaedi ldquoNew results forboundary value problems of Hadamard-type fractional differ-ential inclusions and integral boundary conditionsrdquo BoundaryValue Problems vol 275 p 14 2013

[3] J Tariboon S K Ntouyas and W Sudsutad ldquoNonlo-cal Hadamard fractional integral conditions for nonlinearRiemann-Liouville fractional differential equationsrdquo BoundaryValue Problems vol 2014 article 253 2014

Journal of Function Spaces 9

[4] P Thiramanus S K Ntouyas and J Tariboon ldquoExistence anduniqueness results for Hadamard-type fractional differentialequations with nonlocal fractional integral boundary condi-tionsrdquo Abstract and Applied Analysis vol 2014 Article ID902054 9 pages 2014

[5] B Ahmad and S K Ntouyas ldquoAn existence theorem forfractional hybrid differential inclusions of Hadamard type withDirichlet boundary conditionsrdquo Abstract and Applied Analysisvol 2014 Article ID 705809 7 pages 2014

[6] B Ahmad and S K Ntouyas ldquoA fully Hadamard type integralboundary value problem of a coupled system of fractional dif-ferential equationsrdquo Fractional Calculus and Applied AnalysisAn International Journal forTheory and Applications vol 17 no2 pp 348ndash360 2014

[7] F Jarad T Abdeljawad and D Baleanu ldquoCaputo-type modi-fication of the Hadamard fractional derivativesrdquo Advances inDifference Equations vol 2012 no 1 article 142 8 pages 2012

[8] AA Kilbas ldquoHadamard-type fractional calculusrdquo Journal of theKorean Mathematical Society vol 38 no 6 pp 1191ndash1204 2001

[9] I Podlubny Fractional Differential Equations Academic PressSan Diego Calif USA 1999

[10] S Pooseh RAlmeida andD FMTorres ldquoExpansion formulasin terms of integer-order derivatives for the Hadamard frac-tional integral and derivativerdquo Numerical Functional Analysisand Optimization vol 33 no 3 pp 301ndash319 2012

[11] A A Kilbas H M Srivastava and J J Trujillo Theoryand Applications of Fractional Differential Equations vol 204Elsevier Science Amsterdam The Netherlands 2006

[12] S G Samko A A Kilbas and O I Marichev FractionalIntegrals and Derivatives Gordon and Breach Science YverdonSwitzerland 1993

[13] B Ahmad S K Ntouyas and J Tariboon ldquoExistence resultsformixedHadamard andRiemann-Liouville fractional integro-differential equationsrdquo Advances in Difference Equations vol2015 no 1 article 293 8 pages 2015

[14] R P Agarwal M Meehan and D OrsquoRegan Fixed PointTheory and Applications vol 141 Cambridge University PressCambridge UK 2001

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

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Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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Algebra

Discrete Dynamics in Nature and Society

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Decision SciencesAdvances in

Discrete MathematicsJournal of

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Journal of Function Spaces 3

where

119866 (119905 119904) =

1

Γ (120572)

(log 119890119904

)

120572minus1

minus (log 119905119904

)

120572minus1

1 le 119904 le 119905 le 119890

(log 119890119904

)

120572minus1

1 le 119905 le 119904 le 119890

119867 (119905 119904 1205781 1205782 120578

119898minus1)

=

sum119898minus2

119894=1120574119894 [(log (119890119904))

120572minus1minus (log (120578119894119904))

120572minus1]

1205752Γ (120572)

+

(120583 minus 1205752(log 119905))sum119898minus2

119894=1120573119894(log (120578

119894119904))120572minus2

12057511205752Γ (120572 minus 1)

1 le 119904 le 120578119894 119894 = 1 2 119898 minus 1

sum119898minus2

119894=1120574119894(log (119890119904))120572minus1

1205752Γ (120572)

120578119894le 119904 le 119890 119894 = 1 2 119898 minus 1

1205751= 1 minus

119898minus2

sum

119894=1

120573119894 1205752= 1 minus

119898minus2

sum

119894=1

120574119894 120583 = 1 minus

119898minus2

sum

119894=1

120574119894(log 120578

119894)

(14)

Proof In view of Lemma 6 the solution of the Hadamarddifferential equation (12) can be written as

119909 (119905) = minus119867119868120572ℎ (119905) + 1198880 + 1198881 log 119905 + 1198882 (log 119905)

2

+ 1198883(log 119905)3 + sdot sdot sdot + 119888

119899minus1(log 119905)119899minus1

119862

119867119863119909 (119905) = minus

119867119868120572minus1

ℎ (119905) + 1198881+ 21198882log 119905 + 3119888

3(log 119905)2

+ sdot sdot sdot + (119899 minus 1) 119888119899minus1(log 119905)119899minus2

119862

1198671198632119909 (119905) = minus

119867119868120572minus2

ℎ (119905) + 21198882+ 61198883log 119905 + sdot sdot sdot

+ (119899 minus 1) (119899 minus 2) 119888119899minus1(log 119905)119899minus3

119862

119867119863(119899minus1)

119909 (119905) = minus119867119868120572minus119899+1

ℎ (119905) + (119899 minus 1)119888119899minus1

(15)

In view of the boundary conditions we conclude that

1198882 = 1198883 = sdot sdot sdot = 119888119899minus1 = 0

1198881= minus

1

1205751

119898minus2

sum

119894=1

120573119894119868120572minus1

ℎ (120578119894)

1198880 =

1

1205752

[119868120572ℎ (119890) minus

119898minus2

sum

119894=1

120574119894119868120572ℎ (120578119894)]

+

120583

12057511205752

119898minus2

sum

119894=1

120573119894119868120572minus1

ℎ (120578119894)

(16)

Substituting the values of 119888119894 119894 = 0 1 119899 minus 1 we obtain

119909 (119905) = minus119868120572ℎ (119905) +

1

1205752

[119868120572ℎ (119890) minus

119898minus2

sum

119894=1

120574119894119868120572ℎ (120578119894)]

+

(120583 minus 1205752(log 119905))

12057511205752

(

119898minus2

sum

119894=1

120573119894119868120572minus1

ℎ (120578119894)) = minus119868

120572ℎ (119905)

+ 119868120572ℎ (119890) +

1

1205752

[(1 minus 1205752) 119868120572ℎ (119890) minus

119898minus2

sum

119894=1

120574119894119868120572ℎ (120578119894)]

+

(120583 minus 1205752(log 119905))

12057511205752

(

119898minus2

sum

119894=1

120573119894119868120572minus1

ℎ (120578119894)) = minus119868

120572ℎ (119905)

+ 119868120572ℎ (119890) +

1

1205752

[

119898minus2

sum

119894=1

120574119894119868120572ℎ (119890) minus

119898minus2

sum

119894=1

120574119894119868120572ℎ (120578119894)]

+

(120583 minus 1205752(log 119905))

12057511205752

(

119898minus2

sum

119894=1

120573119894119868120572minus1

ℎ (120578119894)) = minus

1

Γ (120572)

sdot int

119905

1

(log 119905119904

)

120572minus1

ℎ (119904)

119889119904

119904

+

1

Γ (120572)

sdot int

119890

1

(log 119890119904

)

120572minus1

ℎ (119904)

119889119904

119904

+

1

1205752Γ (120572)

119898minus2

sum

119894=1

120574119894

sdot [int

119890

1

(log 119890119904

)

120572minus1

ℎ (119904)

119889119904

119904

minus int

120578119894

1

(log120578119894

119904

)

120572minus1

ℎ (119904)

119889119904

119904

] +

(120583 minus 1205752(log 119905))

12057511205752Γ (120572 minus 1)

sdot

119898minus2

sum

119894=1

120573119894 int

120578119894

1

(log120578119894

119904

)

120572minus2

ℎ (119904)

119889119904

119904

(17)

Lemma 8 The functions 119866(119905 119904) 119867(119905 119904 1205781 1205782 120578

119898minus2)

defined by (15) satisfy

(i) 119866(119905 119904) ge 0119867(119905 119904 1205781 1205782 120578

119898minus2) ge 0 forall119905 119904 isin [1 119890]

(ii) min1205911le119905le1205912

119866(119905 119904) ge 1205910max1le119905le119890

119866(119905 119904) = 1205910119866(119904 119904) forall119905

119904 isin (1 119890) 1 lt 1205911lt 1205912lt 119890

4 Journal of Function Spaces

119866 (119904 119904) =

1

Γ (120572)

(log 119890119904

)

120572minus1

1205910= min1205911le119905le1205912

120595 (119905) = (log 119890

1205912

)

(18)

(iii) 1198732119902(119904) le 119867(119905 119904 120578

1 1205782 120578

119898minus2) le 119873

1119902(119904) where

119902 (119904) =

(log (119890119904))120572minus2

12057511205752Γ (120572)

1198731= [1205751

119898minus2

sum

119894=1

120574119894+ (120572 minus 1) 120583

119898minus2

sum

119894=1

120573119894]

1198732= min

119898minus2

sum

119894=1

120574119894(log 119890

120578119894

)

1205751sum119898minus2

119894=1120574119894

(120572 minus 1)

(19)

(iv) min1205911le119905le1205912

119867(119905 119904 1205781 1205782 120578

119898minus2) ge 120591

lowastmax1le119905le119890

119867(119905

119904 1205781 1205782 120578

119898minus2) 119904 isin (1 119890) where

120591lowast=

1

120583

[120583 minus 1205752(log 1205912)] lt 1 1 lt 120591

1lt 1205912lt 119890 (20)

Proof It is clear that (i) holds So we prove that (ii) is true(ii) In view of the expression for 119866(119905 119904) it follows that

119866(119905 119904) le 119866(119904 119904) for all 119904 119905 isin [1 119890]

If 1 le 119904 le 119905 le 119890 we have

119866 (119905 119904)

119866 (119904 119904)

=

[(log (119890119904))120572minus1 minus (log (119905119904))120572minus1]

(log (119890119904))120572minus1

=

[(log (119890119904)) (log (119890119904))120572minus2 minus (log (119905119904)) (log (119905119904))120572minus2]

(log (119890119904))120572minus1

ge

(log (119890119904))120572minus2 [(log (119890119904)) minus (log (119905119904))](log (119890119904))120572minus1

=

(log (119890119905))(log (119890119904))

ge (log 119890119905

) fl 120595 (119905)

(21)

If 1 le 119905 le 119904 le 119890 we have

119866 (119905 119904)

119866 (119904 119904)

= 1 ge 120595 (119905) (22)

Thus

max1le119905le119890

119866 (119905 119904) = 119866 (119904 119904)

120595 (119905) 119866 (119904 119904) le 119866 (119905 119904) le 119866 (119904 119904)

forall119905 119904 isin (1 119890)

(23)

Therefore

min1205911le119905le1205912

119866 (119905 119904) ge 12059101le119905le119890

max119866 (119905 119904) = 1205910119866 (119904 119904)

forall119905 119904 isin (1 119890) 1 lt 1205911 lt 1205912 lt 119890

(24)

(iii) If 1 le 119904 le 120578119894 119894 = 1 2 119898 minus 2

119867(119905 119904 1205781 1205782 120578119898minus2) =

sum119898minus2

119894=1120574119894[(log (119890119904))120572minus1 minus (log (120578119894119904))

120572minus1]

1205752Γ (120572)

+

(120583 minus 1205752(log 119905))sum119898minus2

119894=1120573119894(log (120578

119894119904))120572minus2

12057511205752Γ (120572 minus 1)

le

sum119898minus2

119894=1120574119894 (log (119890119904))

120572minus1

1205752Γ (120572)

+

(120583 minus 1205752 (log 119905))sum

119898minus2

119894=1120573119894 (log (120578119894119904))

120572minus2

12057511205752Γ (120572 minus 1)

=

1

12057511205752Γ (120572)

[1205751

119898minus2

sum

119894=1

120574119894 (log119890

119904

)

120572minus1

+ (120572 minus 1) 120583

119898minus2

sum

119894=1

120573119894 (log120578119894

119904

)

120572minus2

]

le

(log (119890119904))120572minus2

12057511205752Γ (120572)

[1205751(log 119890

119904

)

119898minus2

sum

119894=1

120574119894+ (120572 minus 1) 120583

119898minus2

sum

119894=1

120573119894]

le

(log (119890119904))120572minus2

12057511205752Γ (120572)

[1205751

119898minus2

sum

119894=1

120574119894 + (120572 minus 1) 120583

119898minus2

sum

119894=1

120573119894] fl 1198731119902 (119904)

119867 (119905 119904 1205781 1205782 120578119898minus2) =

sum119898minus2

119894=1120574119894[(log (119890119904))120572minus1 minus (log (120578119894119904))

120572minus1]

1205752Γ (120572)

+

(120583 minus 1205752(log 119905))sum119898minus2

119894=1120573119894(log (120578

119894119904))120572minus2

12057511205752Γ (120572 minus 1)

Journal of Function Spaces 5

ge

sum119898minus2

119894=1120574119894 [(log (119890119904)) (log (119890119904))

120572minus2minus (log (120578119894119904)) (log (120578119894119904))

120572minus2]

1205752Γ (120572)

ge

(log (119890119904))120572minus2

1205752Γ (120572)

119898minus2

sum

119894=1

120574119894[(log 119890

119904

) minus (log120578119894

119904

)] =

(log (119890119904))120572minus2

1205752Γ (120572)

119898minus2

sum

119894=1

120574119894(log 119890

120578119894

) fl 1198732119902 (119904)

(25)

If 120578119894le 119904 le 119890 119894 = 1 2 119898 minus 2 we have

119867(119905 119904 1205781 1205782 120578119898minus2) =

sum119898minus2

119894=1120574119894(log (119890119904))120572minus1

1205752Γ (120572)

le

(log (119890119904))120572minus2

12057511205752Γ (120572)

1205751

119898minus2

sum

119894=1

120574119894 le 1205751

119898minus2

sum

119894=1

120574119894119902 (119904) lt 1198731119902 (119904)

119867 (119905 119904 1205781 1205782 120578

119898minus2) =

sum119898minus2

119894=1120574119894(log (119890119904))120572minus1

1205752Γ (120572)

=

sum119898minus2

119894=1120574119894 (120572 minus 1) (log (119890119904))120572minus1

1205752 (120572 minus 1) Γ (120572)

ge

(log (119890119904))120572minus2

12057511205752Γ (120572)

1205751sum119898minus2

119894=1120574119894

(120572 minus 1)

ge 1198732119902 (119904)

(26)

(iv) Since 120597119867(11990511990412057811205782 120578119898minus2

)120597119905=minussum119898minus2

119894=1120573119894(log(120578

119894119904))120572minus2

1205751Γ(120572minus1)119905 le 0 then119867(119905 119904 120578

1 1205782 120578

119898minus2) is nonincreasing

in 119905 somin1205911le119905le1205912

119867(119905 119904 1205781 1205782 120578119898minus2)

=

sum119898minus2

119894=1120574119894 [(log (119890119904))

120572minus1minus (log (120578119894119904))

120572minus1]

1205752Γ (120572)

+

(120583 minus 1205752(log 1205912))sum119898minus2

119894=1120573119894(log (120578

119894119904))120572minus2

12057511205752Γ (120572 minus 1)

=

sum119898minus2

119894=1120574119894[(log (119890119904))120572minus1 minus (log (120578119894119904))

120572minus1]

1205752Γ (120572)

+ 120591lowast120583sum119898minus2

119894=1120573119894 (log (120578119894119904))

120572minus2

12057511205752Γ (120572 minus 1)

ge 120591lowastmax1le119905le119890

119867(119905 1205781 1205782 120578119898minus2 119904)

(27)

3 Existence Results

Let us denote by 119864 = 119862([1 119890] 119877) the Banach space of allcontinuous functions from [1 119890] toR endowedwith the norm119909 = Sup

119905isin[1119890]|119909(119905)| and let119875 be the cone119875 = 119909 isin 119864 119909(119905) ge

0 119905 isin [1 119890]

Through this paper we assume that the function 119891

[1 119890] times [0infin) rarr [0infin) satisfies the following conditionsof Caratheodory type

(1198671) (i) 119891(119905 119909) is Lebesgue measurable with respect to 119905

on [1 119890](ii) 119891(119905 119909) is continuous with respect to 119909 on [0infin)

By Lemma 6 we obtain an operatorF 119864 rarr 119864 as

(F119909) (119905) = 119909 (119905)

= int

119890

1

[119866 (119905 119904) + 119867 (119905 119904 1205781 1205782 120578119898minus2)] ℎ (119904)

119889119904

119904

119905 isin [1 119890]

(28)

It should be noticed that problem (1) has solutions if and onlyif the operatorF has fixed points

The first existence and uniqueness result is based on theBanach contraction principle

Theorem 9 Assume that the condition (1198671) holds and thereexists a real-valued function 119892(119905) isin 119871[1 119890] such that

1003816100381610038161003816119891 (119905 119909 (119905)) minus 119891 (119905 119910 (119905))

1003816100381610038161003816le 119892 (119905)

1003816100381610038161003816119909 minus 119910

1003816100381610038161003816

forall119905 isin [1 119890] 119909 119910 isin R(29)

Then problem (1) has a unique solution provided 119892Ψ lt 1where

Ψ = [

1

Γ (120572 + 1)

+

1198731

12057511205752(120572 minus 1) Γ (120572)

] (30)

Proof We set Sup119905isin[1119890]

|119891(119904 0)| = 119901 lt infin and choose 120588 ge

Ψ119901(1 minus Ψ119892)

Now we show that FB120588sub B120588 where B

120588= 119909 isin 119864

119909 le 120588 For any 119909 isin B120588 we have

(F119909) (119905)

= Sup119905isin[1119890]

int

119890

1

[119866 (119905 119904) + 119867 (119905 119904 1205781 1205782 120578

119898minus2)]

sdot 119891 (119904 119909 (119904))

119889119904

119904

le int

119890

1

(119866 (119904 119904) + 1198731119902 (119904))

sdot (1003816100381610038161003816119891 (119904 119909 (119904)) minus 119891 (119904 0)

1003816100381610038161003816+1003816100381610038161003816119891 (119904 0)

1003816100381610038161003816)

119889119904

119904

6 Journal of Function Spaces

le int

119890

1

(119866 (119904 119904) + 1198731119902 (119904)) (119892 (119904) 120588 + 119901)

119889119904

119904

le (10038171003817100381710038171198921003817100381710038171003817120588 + 119901) [

1

Γ (120572)

int

119890

1

(log 119890119904

)

120572minus1119889119904

119904

+

1198731

12057511205752Γ (120572)

int

119890

1

(log 119890119904

)

120572minus2119889119904

119904

] le (10038171003817100381710038171198921003817100381710038171003817120588 + 119901)

sdot [

1

Γ (120572 + 1)

+

1198731

12057511205752(120572 minus 1) Γ (120572)

] le Ψ (10038171003817100381710038171198921003817100381710038171003817120588

+ 119901) le 120588

(31)

It follows that FB120588 sub B120588 For 119909 119910 isin 119864 and for each 119905 isin

[1 119890] we have1003816100381610038161003816(F119909) (119905) minus (F119910) (119905)

1003816100381610038161003816

=

10038161003816100381610038161003816100381610038161003816

int

119890

1

[119866 (119905 119904) + 119867 (119905 119904 1205781 1205782 120578

119898minus2)]

sdot (119891 (119904 119909 (119904)) minus 119891 (119904 119910 (119904)))

119889119904

119904

10038161003816100381610038161003816100381610038161003816

le int

119890

1

(119866 (119904 119904)

+ 1198731119902 (119904)) 119892 (119904)

1003816100381610038161003816119909 (119904) minus 119910 (119904)

1003816100381610038161003816

119889119904

119904

le10038171003817100381710038171198921003817100381710038171003817

1003817100381710038171003817119909 minus 119910

1003817100381710038171003817

sdot int

119890

1

(119866 (119904 119904) + 1198731119902 (119904))

119889119904

119904

le Ψ10038171003817100381710038171198921003817100381710038171003817

1003817100381710038171003817119909 minus 119910

1003817100381710038171003817

(32)

Hence it follows that (F119909)(119905) minus (F119910)(119905) le Ψ119892119909 minus 119910where Ψ119892 lt 1ThereforeF is a contraction Hence by thecontraction mapping principle problem (1) has a uniquenesssolution

Theorem 10 (nonlinear alternative for single valued maps[13]) Let 119864 be a Banach spaceC a closed convex subset of 119864U an open subset ofC and 0 isin U Suppose that 119865 U rarr C isa continuous compact (ie 119865(U) is a relatively compact subsetofC) map Then either

(i) 119865 has a fixed point inU or(ii) There is 119909 isin 120597U (the boundary of U in C) and 120582 isin

(1 119890) with 119909 = 120582119865(119909)

Theorem 11 Assume that (1198671) and the following conditions

hold(1198672) There exist two nonnegative real-valued functions

ℎ1 ℎ2isin [1 119890] such that

119891 (119905 119909) le ℎ1 (119905) + ℎ2 (119905) 119909

for every 119905 isin [1 119890] all 119909 isin [0infin)

(33)

(1198673)There exists a constant119872 gt 0 such that

119872

(1003817100381710038171003817ℎ1

1003817100381710038171003817+1003817100381710038171003817ℎ2

1003817100381710038171003817119872)Ψ

gt 1 (34)

Then the boundary value problem (1) has at least one solutionon [1 119890]

Proof First we show that the operator F 119875 rarr 119875 iscontinuous

For any 119909119899 119909 isin 119875 119899 = 1 2 with lim

119899rarrinfin119909119899(119905) =

119909(119905) 119905 isin [1 119890] Thus by condition (ii) of (1198671) we have

lim119899rarrinfin

119891(119905 119909119899(119905)) = 119891(119905 119909(119905)) 119905 isin [1 119890] So we can

conclude that

Sup119905isin[1119890]

1003816100381610038161003816119891 (119905 119909

119899 (119905)) minus 119891 (119905 119909 (119905))

1003816100381610038161003816997888rarr 0 as 119899 997888rarr infin (35)

On the other hand

1003816100381610038161003816(F119909119899) (119905) minus (F119909) (119905)

1003816100381610038161003816

=

10038161003816100381610038161003816100381610038161003816

int

119890

1

[119866 (119905 119904) + 119867 (119905 119904 1205781 1205782 120578119898minus2)]

sdot (119891 (119904 119909 (119904)) minus 119891 (119904 119910 (119904)))

119889119904

119904

10038161003816100381610038161003816100381610038161003816

le Sup119905isin[1119890]

1003816100381610038161003816119891 (119905 119909

119899 (119905)) minus 119891 (119905 119909 (119905))

1003816100381610038161003816int

119890

1

(119866 (119904 119904)

+ 1198731119902 (119904))

119889119904

119904

le [

1

Γ (120572 + 1)

+

1198731

12057511205752(120572 minus 1) Γ (120572)

] Sup119905isin[1119890]

1003816100381610038161003816119891 (119905 119909

119899 (119905))

minus 119891 (119905 119909 (119905))1003816100381610038161003816

(36)

Hence

1003817100381710038171003817(F119909119899) (119905) minus (F119909) (119905)

1003817100381710038171003817997888rarr 0 as 119899 997888rarr infin (37)

This means thatF is continuousNow we show that F maps bounded sets into bounded

sets in 119875 It suffices to show that for any 120590 gt 0 there exists apositive constant 120588

1gt 0 such that for each 119909 isin B

120588lowast = 119909 isin

119875 119909 le 120588lowast we have F119909 le 120588

1 By (33) for each 119905 isin [1 119890]

we have

|(F119909) (119905)| =

10038161003816100381610038161003816100381610038161003816

int

119890

1

[119866 (119905 119904) + 119867 (119905 119904 1205781 1205782 120578

119898minus2)]

sdot 119891 (119904 119909 (119904))

119889119904

119904

10038161003816100381610038161003816100381610038161003816

le int

119890

1

[119866 (119904 119904) + 1198731119902 (119904)] (ℎ1 (

119904)

+ ℎ2 (119904) 119909 (119904))

119889119904

119904

le (1003817100381710038171003817ℎ1

1003817100381710038171003817+1003817100381710038171003817ℎ2

1003817100381710038171003817120588lowast)

sdot int

119890

1

(119866 (119904 119904) + 1198731119902 (119904))

119889119904

119904

le (1003817100381710038171003817ℎ1

1003817100381710038171003817+1003817100381710038171003817ℎ2

1003817100381710038171003817120588lowast)

sdot [

1

Γ (120572 + 1)

+

1198731

12057511205752(120572 minus 1) Γ (120572)

] le (1003817100381710038171003817ℎ1

1003817100381710038171003817+1003817100381710038171003817ℎ2

1003817100381710038171003817

sdot 120588lowast) Ψ fl 120588

1

(38)

Journal of Function Spaces 7

which implies that (F119909) le 1205881 Further we let 119905

1 1199052isin [1 119890]

with 1199051lt 1199052and 119909 isin B

120588lowast where B

120588lowast is a bounded set of 119875

and then we find that

10038161003816100381610038161003816(119862

119867119863F119909) (119905)

10038161003816100381610038161003816=

1003816100381610038161003816100381610038161003816100381610038161003816

minus

1

Γ (120572 minus 1)

sdot int

119890

1

[(log 119905119904

)

120572minus2

+

1

1205751

119898minus2

sum

119894=1

120573119894(log

120578119894

119904

)

120572minus2

]

sdot 119891 (119904 119909 (119904))

119889119904

119904

1003816100381610038161003816100381610038161003816100381610038161003816

le (1003817100381710038171003817ℎ1

1003817100381710038171003817+1003817100381710038171003817ℎ2

1003817100381710038171003817120588lowast)

sdot

(1205751 + sum119898minus2

119894=1120573119894)

1205751Γ (120572 minus 1)

fl 1205882

(39)

Hence for 1199051 1199052isin [1 119890] we have

1003816100381610038161003816(F119909) (1199052

) minus (F119909) (1199051)1003816100381610038161003816le int

1199052

1199051

(119862

119867119863F119909) (119904) 119889119904

le 1205882 (1199052 minus 1199051)

(40)

This implies that F maps bounded sets into equicontinuoussets of 119875

Thus by the Arzela-Ascoli theorem the operator F

119875 rarr 119875 is completely continuousNext we consider the set 119881 = 119906 isin 119864 | 119909 = 120583F119909 0 lt

120583 lt 1 and show that the set 119881 is bounded let 119909 isin 119881 andthen 119909 = 120583F119909 0 lt 120583 lt 1 For any 119905 isin [1 119890] we have

|119909 (119905)| = 120583 |(F119909) (119905)|

le int

119890

1

[119866 (119905 119904) + 119867 (119905 119904 1205781 1205782 120578119898minus2)]

sdot1003816100381610038161003816119891 (119904 119909 (119904))

1003816100381610038161003816

119889119904

119904

le 1205881

(41)

Thus 119909(119905) le 1205881for any 119905 isin [1 119890] so that set 119881 is bounded

Thus by the conclusion of Theorem 10 the operator F hasat least one fixed point which implies that problem (1) has atleast one solution

Theorem 12 (Krasnoselskii fixed point theorem [14]) Let 119864be a Banach space and 119870 sub 119864 is a cone in 119864 Assume that Ω1and Ω

2are open subsets of 119864 with 0 isin Ω

1and Ω

1sub Ω2 Let

119879 119870 cap (Ω2 Ω1) rarr 119870 be completely continuous operator In

addition suppose that either

(i) 119879119906 le 119906 forall119906 isin 119870 cap 120597Ω1and 119879119906 ge 119906 forall119906 isin

119870 cap 120597Ω2or

(ii) 119879119906 le 119906 forall119906 isin 119870 cap 120597Ω2and 119879119906 ge 119906 forall119906 isin

119870 cap 120597Ω1

holds Then 119879 has a fixed point in 119870 cap (Ω2 Ω1)

To state the last result of this section we set

1198721 = (Ψ)minus1

1198722 = (

1205910

Γ (120572 + 1)

+ (120572 minus 1) 120591lowast[

120583sum119898minus2

119894=1120573119894

12057511205752Γ (120572)

minus

sum119898minus2

119894=1120574119894

1205752Γ (120572 + 1)

])

minus1

(42)

Theorem 13 Suppose that there exist two positive constants1199032 gt 1199031 gt 0 and1198721 isin (01198722)

(i) 119891(119905 119909) le 11987211199032 for (119905 119909) isin [1 119890] times [0 119903

2]

(ii) 119891(119905 119909) ge 11987221199031 for (119905 119909) isin [1 119890] times [0 1199031]

Then (1) has at least a positive solution

Proof Let Ω119894= 119909 isin 119864 | 119909 lt 119903

119894 119894 = 1 2 From the proof

of Theorem 11 we know that the operatorF defined by (28)is completely continuous on 119875

For any 119909 isin 119875 cap 120597Ω1 it follows that

(F119909) (119905)

= Sup119905isin[1119890]

int

119890

1

[119866 (119905 119904) + 119867 (119905 119904 1205781 1205782 120578

119898minus2)]

sdot 119891 (119904 119909 (119904))

119889119904

119904

ge int

119890

1

(1205910119866 (119904 119904)

+ 120591lowast119867(1 119904 120578

1 1205782 120578

119898minus2))11987221199031

119889119904

119904

ge 11987221199031[

1205910

Γ (120572 + 1)

+ (120572 minus 1) 120591lowast[

120583sum119898minus2

119894=1120573119894

12057511205752Γ (120572)

minus

sum119898minus2

119894=1120574119894

1205752Γ (120572 + 1)

]] ge 1199031 = 119909

(43)

that is (F119909)(119905) ge 119909 119909 isin 119875 cap 120597Ω1

On the other hand for any 119909 isin 119875 cap 120597Ω2 it follows that

(F119909) (119905)

= Sup119905isin[1119890]

int

119890

1

[119866 (119905 119904) + 119867 (119905 119904 1205781 1205782 120578

119898minus2)]

sdot 119891 (119904 119909 (119904))

119889119904

119904

le int

119890

1

(119866 (119904 119904) + 1198731119902 (119904))

sdot 11987211199032

119889119904

119904

le 11987211199032Ψ = 1199032 = 119909

(44)

that is (F119909)(119905) le 119909 119909 isin 119875 cap 120597Ω2

In view ofTheorem 12F has a fixed point in119875cap(Ω2Ω1)which is a positive solution to (1)

4 Examples

In this section we exemplify our theoretical results obtainedin Section 3

8 Journal of Function Spaces

Example 1 Consider the following BVP for Hadamard frac-tional differential equation

119862

11986711986373119909 (119905) +119891 (119905 119909 (119905)) = 0 1 le 119905 le 119890

119862

119867119863119909 (1) =

1

2

119862

119867119863119909(

4

3

)

119862

1198671198632119909 (1) = 0

119909 (119890) =

1

4

119909 (

4

3

)

(45)

Here 120572 = 73 120573 = 05 120574 = 025 120578 = 43 and

119891 (119905 119909) =

119890119905

2 (119890119905+ 1)

(

1199092

119909 + 1

+

119909

4 (119909 + 1)

+

3

4

)

(119905 119909) isin [1 119890] times [0infin)

(46)

Using the given data we find that Ψ = 1609 1205751= 05 120575

2=

075 120583 = 0928 and

1003816100381610038161003816119891 (119905 119909) minus 119891 (119905 119910)

1003816100381610038161003816le

119890119905

2 (119890119905+ 1)

1003816100381610038161003816119909 minus 119910

1003816100381610038161003816le

1

2

1003816100381610038161003816119909 minus 119910

1003816100381610038161003816

with 119892 (119905) =

119890119905

2 (119890119905+ 1)

(47)

Hencewe obtain 119892Ψ = 0805 lt 1Therefore byTheorem9problem (45) has a unique solution on [1 119890]

Example 2 Consider the following BVP for Hadamard frac-tional differential equation

119862

11986711986352119909 (119905) +119891 (119905 119909 (119905)) = 0 1 le 119905 le 119890

119862

119867119863119909 (1) =

1

4

119862

119867119863119909(

3

2

)

119862

1198671198632119909 (1) = 0

119909 (119890) =

1

3

119909 (

3

2

)

(48)

Here 120572 = 52 120573 = 025 120574 = 13 120578 = 32 Let

119891 (119905 119909) = 9 sin2119905 + 119890119905119909

12 (119890119905+ 1) (119909

2+ 1)

(119905 119909) isin [1 119890] times [0infin)

(49)

with ℎ1(119905) = 9 sin2119905 ℎ

2(119905) = 119890

11990512(119890119905+ 1) Here 119891(119905 119909) le

ℎ1(119905) + ℎ

2(119905)119909

It is easy to verify that119872(ℎ1 + ℎ

2119872)Ψ = 1147 gt 1

Then by Theorem 11 problem (48) has at least onesolution on [1 119890]

Example 3 Consider the following BVP for Hadamard frac-tional differential equation

119862

11986711986372119909 (119905) +119891 (119905 119909 (119905)) = 0 1 le 119905 le 119890

119862

119867119863119909 (1) =

1

2

119862

119867119863119909(

4

3

) +

1

4

119862

119867119863119909(

5

2

)

119862

1198671198632119909 (1) = 0

119909 (119890) =

1

3

119909 (

4

3

) +

1

5

119909 (

5

2

)

(50)

where 120572 = 72 1205731 = 05 1205732 = 025 1205741 = 13 1205742 = 15

1205781 = 43 1205782 = 52 Let

119891 (119905 119909) =

1

4

119909 +

119890119905

5 (119890119905+ 1)

(119905 119909) isin [1 119890] times [0infin) (51)

It is easy to verify that

1198721= (Ψ)

minus1= 06196

1198722= (

1205910

Γ (120572 + 1)

+ (120572 minus 1) 120591lowast[

120583sum119898minus2

119894=1120573119894

12057511205752Γ (120572)

minus

sum119898minus2

119894=1120574119894

1205752Γ (120572 + 1)

])

minus1

= 0521

(52)

Choosing 1199031= 15 119903

2= 119890 we have

119891 (119905 sdot 119909) le 12 le 11987211199032 (119905 119909) isin [1 119890] times [0 1199032

]

119891 (119905 sdot 119909) gt 05 ge 11987221199031 (119905 119909) isin [1 119890] times [0 1199031

]

(53)

Hence all conditions of Theorem 13 are satisfied then prob-lem (50) has at least one positive solution 119909 such that 15 lt

119909 lt 119890

Competing Interests

The author declares no competing interests

References

[1] Y Y Gambo F Jarad D Baleanu and T Abdeljawad ldquoOnCaputo modification of the Hadamard fractional derivativesrdquoAdvances in Difference Equations vol 2014 article 10 12 pages2014

[2] B Ahmad S K Ntouyas and A Alsaedi ldquoNew results forboundary value problems of Hadamard-type fractional differ-ential inclusions and integral boundary conditionsrdquo BoundaryValue Problems vol 275 p 14 2013

[3] J Tariboon S K Ntouyas and W Sudsutad ldquoNonlo-cal Hadamard fractional integral conditions for nonlinearRiemann-Liouville fractional differential equationsrdquo BoundaryValue Problems vol 2014 article 253 2014

Journal of Function Spaces 9

[4] P Thiramanus S K Ntouyas and J Tariboon ldquoExistence anduniqueness results for Hadamard-type fractional differentialequations with nonlocal fractional integral boundary condi-tionsrdquo Abstract and Applied Analysis vol 2014 Article ID902054 9 pages 2014

[5] B Ahmad and S K Ntouyas ldquoAn existence theorem forfractional hybrid differential inclusions of Hadamard type withDirichlet boundary conditionsrdquo Abstract and Applied Analysisvol 2014 Article ID 705809 7 pages 2014

[6] B Ahmad and S K Ntouyas ldquoA fully Hadamard type integralboundary value problem of a coupled system of fractional dif-ferential equationsrdquo Fractional Calculus and Applied AnalysisAn International Journal forTheory and Applications vol 17 no2 pp 348ndash360 2014

[7] F Jarad T Abdeljawad and D Baleanu ldquoCaputo-type modi-fication of the Hadamard fractional derivativesrdquo Advances inDifference Equations vol 2012 no 1 article 142 8 pages 2012

[8] AA Kilbas ldquoHadamard-type fractional calculusrdquo Journal of theKorean Mathematical Society vol 38 no 6 pp 1191ndash1204 2001

[9] I Podlubny Fractional Differential Equations Academic PressSan Diego Calif USA 1999

[10] S Pooseh RAlmeida andD FMTorres ldquoExpansion formulasin terms of integer-order derivatives for the Hadamard frac-tional integral and derivativerdquo Numerical Functional Analysisand Optimization vol 33 no 3 pp 301ndash319 2012

[11] A A Kilbas H M Srivastava and J J Trujillo Theoryand Applications of Fractional Differential Equations vol 204Elsevier Science Amsterdam The Netherlands 2006

[12] S G Samko A A Kilbas and O I Marichev FractionalIntegrals and Derivatives Gordon and Breach Science YverdonSwitzerland 1993

[13] B Ahmad S K Ntouyas and J Tariboon ldquoExistence resultsformixedHadamard andRiemann-Liouville fractional integro-differential equationsrdquo Advances in Difference Equations vol2015 no 1 article 293 8 pages 2015

[14] R P Agarwal M Meehan and D OrsquoRegan Fixed PointTheory and Applications vol 141 Cambridge University PressCambridge UK 2001

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

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Algebra

Discrete Dynamics in Nature and Society

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Decision SciencesAdvances in

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

4 Journal of Function Spaces

119866 (119904 119904) =

1

Γ (120572)

(log 119890119904

)

120572minus1

1205910= min1205911le119905le1205912

120595 (119905) = (log 119890

1205912

)

(18)

(iii) 1198732119902(119904) le 119867(119905 119904 120578

1 1205782 120578

119898minus2) le 119873

1119902(119904) where

119902 (119904) =

(log (119890119904))120572minus2

12057511205752Γ (120572)

1198731= [1205751

119898minus2

sum

119894=1

120574119894+ (120572 minus 1) 120583

119898minus2

sum

119894=1

120573119894]

1198732= min

119898minus2

sum

119894=1

120574119894(log 119890

120578119894

)

1205751sum119898minus2

119894=1120574119894

(120572 minus 1)

(19)

(iv) min1205911le119905le1205912

119867(119905 119904 1205781 1205782 120578

119898minus2) ge 120591

lowastmax1le119905le119890

119867(119905

119904 1205781 1205782 120578

119898minus2) 119904 isin (1 119890) where

120591lowast=

1

120583

[120583 minus 1205752(log 1205912)] lt 1 1 lt 120591

1lt 1205912lt 119890 (20)

Proof It is clear that (i) holds So we prove that (ii) is true(ii) In view of the expression for 119866(119905 119904) it follows that

119866(119905 119904) le 119866(119904 119904) for all 119904 119905 isin [1 119890]

If 1 le 119904 le 119905 le 119890 we have

119866 (119905 119904)

119866 (119904 119904)

=

[(log (119890119904))120572minus1 minus (log (119905119904))120572minus1]

(log (119890119904))120572minus1

=

[(log (119890119904)) (log (119890119904))120572minus2 minus (log (119905119904)) (log (119905119904))120572minus2]

(log (119890119904))120572minus1

ge

(log (119890119904))120572minus2 [(log (119890119904)) minus (log (119905119904))](log (119890119904))120572minus1

=

(log (119890119905))(log (119890119904))

ge (log 119890119905

) fl 120595 (119905)

(21)

If 1 le 119905 le 119904 le 119890 we have

119866 (119905 119904)

119866 (119904 119904)

= 1 ge 120595 (119905) (22)

Thus

max1le119905le119890

119866 (119905 119904) = 119866 (119904 119904)

120595 (119905) 119866 (119904 119904) le 119866 (119905 119904) le 119866 (119904 119904)

forall119905 119904 isin (1 119890)

(23)

Therefore

min1205911le119905le1205912

119866 (119905 119904) ge 12059101le119905le119890

max119866 (119905 119904) = 1205910119866 (119904 119904)

forall119905 119904 isin (1 119890) 1 lt 1205911 lt 1205912 lt 119890

(24)

(iii) If 1 le 119904 le 120578119894 119894 = 1 2 119898 minus 2

119867(119905 119904 1205781 1205782 120578119898minus2) =

sum119898minus2

119894=1120574119894[(log (119890119904))120572minus1 minus (log (120578119894119904))

120572minus1]

1205752Γ (120572)

+

(120583 minus 1205752(log 119905))sum119898minus2

119894=1120573119894(log (120578

119894119904))120572minus2

12057511205752Γ (120572 minus 1)

le

sum119898minus2

119894=1120574119894 (log (119890119904))

120572minus1

1205752Γ (120572)

+

(120583 minus 1205752 (log 119905))sum

119898minus2

119894=1120573119894 (log (120578119894119904))

120572minus2

12057511205752Γ (120572 minus 1)

=

1

12057511205752Γ (120572)

[1205751

119898minus2

sum

119894=1

120574119894 (log119890

119904

)

120572minus1

+ (120572 minus 1) 120583

119898minus2

sum

119894=1

120573119894 (log120578119894

119904

)

120572minus2

]

le

(log (119890119904))120572minus2

12057511205752Γ (120572)

[1205751(log 119890

119904

)

119898minus2

sum

119894=1

120574119894+ (120572 minus 1) 120583

119898minus2

sum

119894=1

120573119894]

le

(log (119890119904))120572minus2

12057511205752Γ (120572)

[1205751

119898minus2

sum

119894=1

120574119894 + (120572 minus 1) 120583

119898minus2

sum

119894=1

120573119894] fl 1198731119902 (119904)

119867 (119905 119904 1205781 1205782 120578119898minus2) =

sum119898minus2

119894=1120574119894[(log (119890119904))120572minus1 minus (log (120578119894119904))

120572minus1]

1205752Γ (120572)

+

(120583 minus 1205752(log 119905))sum119898minus2

119894=1120573119894(log (120578

119894119904))120572minus2

12057511205752Γ (120572 minus 1)

Journal of Function Spaces 5

ge

sum119898minus2

119894=1120574119894 [(log (119890119904)) (log (119890119904))

120572minus2minus (log (120578119894119904)) (log (120578119894119904))

120572minus2]

1205752Γ (120572)

ge

(log (119890119904))120572minus2

1205752Γ (120572)

119898minus2

sum

119894=1

120574119894[(log 119890

119904

) minus (log120578119894

119904

)] =

(log (119890119904))120572minus2

1205752Γ (120572)

119898minus2

sum

119894=1

120574119894(log 119890

120578119894

) fl 1198732119902 (119904)

(25)

If 120578119894le 119904 le 119890 119894 = 1 2 119898 minus 2 we have

119867(119905 119904 1205781 1205782 120578119898minus2) =

sum119898minus2

119894=1120574119894(log (119890119904))120572minus1

1205752Γ (120572)

le

(log (119890119904))120572minus2

12057511205752Γ (120572)

1205751

119898minus2

sum

119894=1

120574119894 le 1205751

119898minus2

sum

119894=1

120574119894119902 (119904) lt 1198731119902 (119904)

119867 (119905 119904 1205781 1205782 120578

119898minus2) =

sum119898minus2

119894=1120574119894(log (119890119904))120572minus1

1205752Γ (120572)

=

sum119898minus2

119894=1120574119894 (120572 minus 1) (log (119890119904))120572minus1

1205752 (120572 minus 1) Γ (120572)

ge

(log (119890119904))120572minus2

12057511205752Γ (120572)

1205751sum119898minus2

119894=1120574119894

(120572 minus 1)

ge 1198732119902 (119904)

(26)

(iv) Since 120597119867(11990511990412057811205782 120578119898minus2

)120597119905=minussum119898minus2

119894=1120573119894(log(120578

119894119904))120572minus2

1205751Γ(120572minus1)119905 le 0 then119867(119905 119904 120578

1 1205782 120578

119898minus2) is nonincreasing

in 119905 somin1205911le119905le1205912

119867(119905 119904 1205781 1205782 120578119898minus2)

=

sum119898minus2

119894=1120574119894 [(log (119890119904))

120572minus1minus (log (120578119894119904))

120572minus1]

1205752Γ (120572)

+

(120583 minus 1205752(log 1205912))sum119898minus2

119894=1120573119894(log (120578

119894119904))120572minus2

12057511205752Γ (120572 minus 1)

=

sum119898minus2

119894=1120574119894[(log (119890119904))120572minus1 minus (log (120578119894119904))

120572minus1]

1205752Γ (120572)

+ 120591lowast120583sum119898minus2

119894=1120573119894 (log (120578119894119904))

120572minus2

12057511205752Γ (120572 minus 1)

ge 120591lowastmax1le119905le119890

119867(119905 1205781 1205782 120578119898minus2 119904)

(27)

3 Existence Results

Let us denote by 119864 = 119862([1 119890] 119877) the Banach space of allcontinuous functions from [1 119890] toR endowedwith the norm119909 = Sup

119905isin[1119890]|119909(119905)| and let119875 be the cone119875 = 119909 isin 119864 119909(119905) ge

0 119905 isin [1 119890]

Through this paper we assume that the function 119891

[1 119890] times [0infin) rarr [0infin) satisfies the following conditionsof Caratheodory type

(1198671) (i) 119891(119905 119909) is Lebesgue measurable with respect to 119905

on [1 119890](ii) 119891(119905 119909) is continuous with respect to 119909 on [0infin)

By Lemma 6 we obtain an operatorF 119864 rarr 119864 as

(F119909) (119905) = 119909 (119905)

= int

119890

1

[119866 (119905 119904) + 119867 (119905 119904 1205781 1205782 120578119898minus2)] ℎ (119904)

119889119904

119904

119905 isin [1 119890]

(28)

It should be noticed that problem (1) has solutions if and onlyif the operatorF has fixed points

The first existence and uniqueness result is based on theBanach contraction principle

Theorem 9 Assume that the condition (1198671) holds and thereexists a real-valued function 119892(119905) isin 119871[1 119890] such that

1003816100381610038161003816119891 (119905 119909 (119905)) minus 119891 (119905 119910 (119905))

1003816100381610038161003816le 119892 (119905)

1003816100381610038161003816119909 minus 119910

1003816100381610038161003816

forall119905 isin [1 119890] 119909 119910 isin R(29)

Then problem (1) has a unique solution provided 119892Ψ lt 1where

Ψ = [

1

Γ (120572 + 1)

+

1198731

12057511205752(120572 minus 1) Γ (120572)

] (30)

Proof We set Sup119905isin[1119890]

|119891(119904 0)| = 119901 lt infin and choose 120588 ge

Ψ119901(1 minus Ψ119892)

Now we show that FB120588sub B120588 where B

120588= 119909 isin 119864

119909 le 120588 For any 119909 isin B120588 we have

(F119909) (119905)

= Sup119905isin[1119890]

int

119890

1

[119866 (119905 119904) + 119867 (119905 119904 1205781 1205782 120578

119898minus2)]

sdot 119891 (119904 119909 (119904))

119889119904

119904

le int

119890

1

(119866 (119904 119904) + 1198731119902 (119904))

sdot (1003816100381610038161003816119891 (119904 119909 (119904)) minus 119891 (119904 0)

1003816100381610038161003816+1003816100381610038161003816119891 (119904 0)

1003816100381610038161003816)

119889119904

119904

6 Journal of Function Spaces

le int

119890

1

(119866 (119904 119904) + 1198731119902 (119904)) (119892 (119904) 120588 + 119901)

119889119904

119904

le (10038171003817100381710038171198921003817100381710038171003817120588 + 119901) [

1

Γ (120572)

int

119890

1

(log 119890119904

)

120572minus1119889119904

119904

+

1198731

12057511205752Γ (120572)

int

119890

1

(log 119890119904

)

120572minus2119889119904

119904

] le (10038171003817100381710038171198921003817100381710038171003817120588 + 119901)

sdot [

1

Γ (120572 + 1)

+

1198731

12057511205752(120572 minus 1) Γ (120572)

] le Ψ (10038171003817100381710038171198921003817100381710038171003817120588

+ 119901) le 120588

(31)

It follows that FB120588 sub B120588 For 119909 119910 isin 119864 and for each 119905 isin

[1 119890] we have1003816100381610038161003816(F119909) (119905) minus (F119910) (119905)

1003816100381610038161003816

=

10038161003816100381610038161003816100381610038161003816

int

119890

1

[119866 (119905 119904) + 119867 (119905 119904 1205781 1205782 120578

119898minus2)]

sdot (119891 (119904 119909 (119904)) minus 119891 (119904 119910 (119904)))

119889119904

119904

10038161003816100381610038161003816100381610038161003816

le int

119890

1

(119866 (119904 119904)

+ 1198731119902 (119904)) 119892 (119904)

1003816100381610038161003816119909 (119904) minus 119910 (119904)

1003816100381610038161003816

119889119904

119904

le10038171003817100381710038171198921003817100381710038171003817

1003817100381710038171003817119909 minus 119910

1003817100381710038171003817

sdot int

119890

1

(119866 (119904 119904) + 1198731119902 (119904))

119889119904

119904

le Ψ10038171003817100381710038171198921003817100381710038171003817

1003817100381710038171003817119909 minus 119910

1003817100381710038171003817

(32)

Hence it follows that (F119909)(119905) minus (F119910)(119905) le Ψ119892119909 minus 119910where Ψ119892 lt 1ThereforeF is a contraction Hence by thecontraction mapping principle problem (1) has a uniquenesssolution

Theorem 10 (nonlinear alternative for single valued maps[13]) Let 119864 be a Banach spaceC a closed convex subset of 119864U an open subset ofC and 0 isin U Suppose that 119865 U rarr C isa continuous compact (ie 119865(U) is a relatively compact subsetofC) map Then either

(i) 119865 has a fixed point inU or(ii) There is 119909 isin 120597U (the boundary of U in C) and 120582 isin

(1 119890) with 119909 = 120582119865(119909)

Theorem 11 Assume that (1198671) and the following conditions

hold(1198672) There exist two nonnegative real-valued functions

ℎ1 ℎ2isin [1 119890] such that

119891 (119905 119909) le ℎ1 (119905) + ℎ2 (119905) 119909

for every 119905 isin [1 119890] all 119909 isin [0infin)

(33)

(1198673)There exists a constant119872 gt 0 such that

119872

(1003817100381710038171003817ℎ1

1003817100381710038171003817+1003817100381710038171003817ℎ2

1003817100381710038171003817119872)Ψ

gt 1 (34)

Then the boundary value problem (1) has at least one solutionon [1 119890]

Proof First we show that the operator F 119875 rarr 119875 iscontinuous

For any 119909119899 119909 isin 119875 119899 = 1 2 with lim

119899rarrinfin119909119899(119905) =

119909(119905) 119905 isin [1 119890] Thus by condition (ii) of (1198671) we have

lim119899rarrinfin

119891(119905 119909119899(119905)) = 119891(119905 119909(119905)) 119905 isin [1 119890] So we can

conclude that

Sup119905isin[1119890]

1003816100381610038161003816119891 (119905 119909

119899 (119905)) minus 119891 (119905 119909 (119905))

1003816100381610038161003816997888rarr 0 as 119899 997888rarr infin (35)

On the other hand

1003816100381610038161003816(F119909119899) (119905) minus (F119909) (119905)

1003816100381610038161003816

=

10038161003816100381610038161003816100381610038161003816

int

119890

1

[119866 (119905 119904) + 119867 (119905 119904 1205781 1205782 120578119898minus2)]

sdot (119891 (119904 119909 (119904)) minus 119891 (119904 119910 (119904)))

119889119904

119904

10038161003816100381610038161003816100381610038161003816

le Sup119905isin[1119890]

1003816100381610038161003816119891 (119905 119909

119899 (119905)) minus 119891 (119905 119909 (119905))

1003816100381610038161003816int

119890

1

(119866 (119904 119904)

+ 1198731119902 (119904))

119889119904

119904

le [

1

Γ (120572 + 1)

+

1198731

12057511205752(120572 minus 1) Γ (120572)

] Sup119905isin[1119890]

1003816100381610038161003816119891 (119905 119909

119899 (119905))

minus 119891 (119905 119909 (119905))1003816100381610038161003816

(36)

Hence

1003817100381710038171003817(F119909119899) (119905) minus (F119909) (119905)

1003817100381710038171003817997888rarr 0 as 119899 997888rarr infin (37)

This means thatF is continuousNow we show that F maps bounded sets into bounded

sets in 119875 It suffices to show that for any 120590 gt 0 there exists apositive constant 120588

1gt 0 such that for each 119909 isin B

120588lowast = 119909 isin

119875 119909 le 120588lowast we have F119909 le 120588

1 By (33) for each 119905 isin [1 119890]

we have

|(F119909) (119905)| =

10038161003816100381610038161003816100381610038161003816

int

119890

1

[119866 (119905 119904) + 119867 (119905 119904 1205781 1205782 120578

119898minus2)]

sdot 119891 (119904 119909 (119904))

119889119904

119904

10038161003816100381610038161003816100381610038161003816

le int

119890

1

[119866 (119904 119904) + 1198731119902 (119904)] (ℎ1 (

119904)

+ ℎ2 (119904) 119909 (119904))

119889119904

119904

le (1003817100381710038171003817ℎ1

1003817100381710038171003817+1003817100381710038171003817ℎ2

1003817100381710038171003817120588lowast)

sdot int

119890

1

(119866 (119904 119904) + 1198731119902 (119904))

119889119904

119904

le (1003817100381710038171003817ℎ1

1003817100381710038171003817+1003817100381710038171003817ℎ2

1003817100381710038171003817120588lowast)

sdot [

1

Γ (120572 + 1)

+

1198731

12057511205752(120572 minus 1) Γ (120572)

] le (1003817100381710038171003817ℎ1

1003817100381710038171003817+1003817100381710038171003817ℎ2

1003817100381710038171003817

sdot 120588lowast) Ψ fl 120588

1

(38)

Journal of Function Spaces 7

which implies that (F119909) le 1205881 Further we let 119905

1 1199052isin [1 119890]

with 1199051lt 1199052and 119909 isin B

120588lowast where B

120588lowast is a bounded set of 119875

and then we find that

10038161003816100381610038161003816(119862

119867119863F119909) (119905)

10038161003816100381610038161003816=

1003816100381610038161003816100381610038161003816100381610038161003816

minus

1

Γ (120572 minus 1)

sdot int

119890

1

[(log 119905119904

)

120572minus2

+

1

1205751

119898minus2

sum

119894=1

120573119894(log

120578119894

119904

)

120572minus2

]

sdot 119891 (119904 119909 (119904))

119889119904

119904

1003816100381610038161003816100381610038161003816100381610038161003816

le (1003817100381710038171003817ℎ1

1003817100381710038171003817+1003817100381710038171003817ℎ2

1003817100381710038171003817120588lowast)

sdot

(1205751 + sum119898minus2

119894=1120573119894)

1205751Γ (120572 minus 1)

fl 1205882

(39)

Hence for 1199051 1199052isin [1 119890] we have

1003816100381610038161003816(F119909) (1199052

) minus (F119909) (1199051)1003816100381610038161003816le int

1199052

1199051

(119862

119867119863F119909) (119904) 119889119904

le 1205882 (1199052 minus 1199051)

(40)

This implies that F maps bounded sets into equicontinuoussets of 119875

Thus by the Arzela-Ascoli theorem the operator F

119875 rarr 119875 is completely continuousNext we consider the set 119881 = 119906 isin 119864 | 119909 = 120583F119909 0 lt

120583 lt 1 and show that the set 119881 is bounded let 119909 isin 119881 andthen 119909 = 120583F119909 0 lt 120583 lt 1 For any 119905 isin [1 119890] we have

|119909 (119905)| = 120583 |(F119909) (119905)|

le int

119890

1

[119866 (119905 119904) + 119867 (119905 119904 1205781 1205782 120578119898minus2)]

sdot1003816100381610038161003816119891 (119904 119909 (119904))

1003816100381610038161003816

119889119904

119904

le 1205881

(41)

Thus 119909(119905) le 1205881for any 119905 isin [1 119890] so that set 119881 is bounded

Thus by the conclusion of Theorem 10 the operator F hasat least one fixed point which implies that problem (1) has atleast one solution

Theorem 12 (Krasnoselskii fixed point theorem [14]) Let 119864be a Banach space and 119870 sub 119864 is a cone in 119864 Assume that Ω1and Ω

2are open subsets of 119864 with 0 isin Ω

1and Ω

1sub Ω2 Let

119879 119870 cap (Ω2 Ω1) rarr 119870 be completely continuous operator In

addition suppose that either

(i) 119879119906 le 119906 forall119906 isin 119870 cap 120597Ω1and 119879119906 ge 119906 forall119906 isin

119870 cap 120597Ω2or

(ii) 119879119906 le 119906 forall119906 isin 119870 cap 120597Ω2and 119879119906 ge 119906 forall119906 isin

119870 cap 120597Ω1

holds Then 119879 has a fixed point in 119870 cap (Ω2 Ω1)

To state the last result of this section we set

1198721 = (Ψ)minus1

1198722 = (

1205910

Γ (120572 + 1)

+ (120572 minus 1) 120591lowast[

120583sum119898minus2

119894=1120573119894

12057511205752Γ (120572)

minus

sum119898minus2

119894=1120574119894

1205752Γ (120572 + 1)

])

minus1

(42)

Theorem 13 Suppose that there exist two positive constants1199032 gt 1199031 gt 0 and1198721 isin (01198722)

(i) 119891(119905 119909) le 11987211199032 for (119905 119909) isin [1 119890] times [0 119903

2]

(ii) 119891(119905 119909) ge 11987221199031 for (119905 119909) isin [1 119890] times [0 1199031]

Then (1) has at least a positive solution

Proof Let Ω119894= 119909 isin 119864 | 119909 lt 119903

119894 119894 = 1 2 From the proof

of Theorem 11 we know that the operatorF defined by (28)is completely continuous on 119875

For any 119909 isin 119875 cap 120597Ω1 it follows that

(F119909) (119905)

= Sup119905isin[1119890]

int

119890

1

[119866 (119905 119904) + 119867 (119905 119904 1205781 1205782 120578

119898minus2)]

sdot 119891 (119904 119909 (119904))

119889119904

119904

ge int

119890

1

(1205910119866 (119904 119904)

+ 120591lowast119867(1 119904 120578

1 1205782 120578

119898minus2))11987221199031

119889119904

119904

ge 11987221199031[

1205910

Γ (120572 + 1)

+ (120572 minus 1) 120591lowast[

120583sum119898minus2

119894=1120573119894

12057511205752Γ (120572)

minus

sum119898minus2

119894=1120574119894

1205752Γ (120572 + 1)

]] ge 1199031 = 119909

(43)

that is (F119909)(119905) ge 119909 119909 isin 119875 cap 120597Ω1

On the other hand for any 119909 isin 119875 cap 120597Ω2 it follows that

(F119909) (119905)

= Sup119905isin[1119890]

int

119890

1

[119866 (119905 119904) + 119867 (119905 119904 1205781 1205782 120578

119898minus2)]

sdot 119891 (119904 119909 (119904))

119889119904

119904

le int

119890

1

(119866 (119904 119904) + 1198731119902 (119904))

sdot 11987211199032

119889119904

119904

le 11987211199032Ψ = 1199032 = 119909

(44)

that is (F119909)(119905) le 119909 119909 isin 119875 cap 120597Ω2

In view ofTheorem 12F has a fixed point in119875cap(Ω2Ω1)which is a positive solution to (1)

4 Examples

In this section we exemplify our theoretical results obtainedin Section 3

8 Journal of Function Spaces

Example 1 Consider the following BVP for Hadamard frac-tional differential equation

119862

11986711986373119909 (119905) +119891 (119905 119909 (119905)) = 0 1 le 119905 le 119890

119862

119867119863119909 (1) =

1

2

119862

119867119863119909(

4

3

)

119862

1198671198632119909 (1) = 0

119909 (119890) =

1

4

119909 (

4

3

)

(45)

Here 120572 = 73 120573 = 05 120574 = 025 120578 = 43 and

119891 (119905 119909) =

119890119905

2 (119890119905+ 1)

(

1199092

119909 + 1

+

119909

4 (119909 + 1)

+

3

4

)

(119905 119909) isin [1 119890] times [0infin)

(46)

Using the given data we find that Ψ = 1609 1205751= 05 120575

2=

075 120583 = 0928 and

1003816100381610038161003816119891 (119905 119909) minus 119891 (119905 119910)

1003816100381610038161003816le

119890119905

2 (119890119905+ 1)

1003816100381610038161003816119909 minus 119910

1003816100381610038161003816le

1

2

1003816100381610038161003816119909 minus 119910

1003816100381610038161003816

with 119892 (119905) =

119890119905

2 (119890119905+ 1)

(47)

Hencewe obtain 119892Ψ = 0805 lt 1Therefore byTheorem9problem (45) has a unique solution on [1 119890]

Example 2 Consider the following BVP for Hadamard frac-tional differential equation

119862

11986711986352119909 (119905) +119891 (119905 119909 (119905)) = 0 1 le 119905 le 119890

119862

119867119863119909 (1) =

1

4

119862

119867119863119909(

3

2

)

119862

1198671198632119909 (1) = 0

119909 (119890) =

1

3

119909 (

3

2

)

(48)

Here 120572 = 52 120573 = 025 120574 = 13 120578 = 32 Let

119891 (119905 119909) = 9 sin2119905 + 119890119905119909

12 (119890119905+ 1) (119909

2+ 1)

(119905 119909) isin [1 119890] times [0infin)

(49)

with ℎ1(119905) = 9 sin2119905 ℎ

2(119905) = 119890

11990512(119890119905+ 1) Here 119891(119905 119909) le

ℎ1(119905) + ℎ

2(119905)119909

It is easy to verify that119872(ℎ1 + ℎ

2119872)Ψ = 1147 gt 1

Then by Theorem 11 problem (48) has at least onesolution on [1 119890]

Example 3 Consider the following BVP for Hadamard frac-tional differential equation

119862

11986711986372119909 (119905) +119891 (119905 119909 (119905)) = 0 1 le 119905 le 119890

119862

119867119863119909 (1) =

1

2

119862

119867119863119909(

4

3

) +

1

4

119862

119867119863119909(

5

2

)

119862

1198671198632119909 (1) = 0

119909 (119890) =

1

3

119909 (

4

3

) +

1

5

119909 (

5

2

)

(50)

where 120572 = 72 1205731 = 05 1205732 = 025 1205741 = 13 1205742 = 15

1205781 = 43 1205782 = 52 Let

119891 (119905 119909) =

1

4

119909 +

119890119905

5 (119890119905+ 1)

(119905 119909) isin [1 119890] times [0infin) (51)

It is easy to verify that

1198721= (Ψ)

minus1= 06196

1198722= (

1205910

Γ (120572 + 1)

+ (120572 minus 1) 120591lowast[

120583sum119898minus2

119894=1120573119894

12057511205752Γ (120572)

minus

sum119898minus2

119894=1120574119894

1205752Γ (120572 + 1)

])

minus1

= 0521

(52)

Choosing 1199031= 15 119903

2= 119890 we have

119891 (119905 sdot 119909) le 12 le 11987211199032 (119905 119909) isin [1 119890] times [0 1199032

]

119891 (119905 sdot 119909) gt 05 ge 11987221199031 (119905 119909) isin [1 119890] times [0 1199031

]

(53)

Hence all conditions of Theorem 13 are satisfied then prob-lem (50) has at least one positive solution 119909 such that 15 lt

119909 lt 119890

Competing Interests

The author declares no competing interests

References

[1] Y Y Gambo F Jarad D Baleanu and T Abdeljawad ldquoOnCaputo modification of the Hadamard fractional derivativesrdquoAdvances in Difference Equations vol 2014 article 10 12 pages2014

[2] B Ahmad S K Ntouyas and A Alsaedi ldquoNew results forboundary value problems of Hadamard-type fractional differ-ential inclusions and integral boundary conditionsrdquo BoundaryValue Problems vol 275 p 14 2013

[3] J Tariboon S K Ntouyas and W Sudsutad ldquoNonlo-cal Hadamard fractional integral conditions for nonlinearRiemann-Liouville fractional differential equationsrdquo BoundaryValue Problems vol 2014 article 253 2014

Journal of Function Spaces 9

[4] P Thiramanus S K Ntouyas and J Tariboon ldquoExistence anduniqueness results for Hadamard-type fractional differentialequations with nonlocal fractional integral boundary condi-tionsrdquo Abstract and Applied Analysis vol 2014 Article ID902054 9 pages 2014

[5] B Ahmad and S K Ntouyas ldquoAn existence theorem forfractional hybrid differential inclusions of Hadamard type withDirichlet boundary conditionsrdquo Abstract and Applied Analysisvol 2014 Article ID 705809 7 pages 2014

[6] B Ahmad and S K Ntouyas ldquoA fully Hadamard type integralboundary value problem of a coupled system of fractional dif-ferential equationsrdquo Fractional Calculus and Applied AnalysisAn International Journal forTheory and Applications vol 17 no2 pp 348ndash360 2014

[7] F Jarad T Abdeljawad and D Baleanu ldquoCaputo-type modi-fication of the Hadamard fractional derivativesrdquo Advances inDifference Equations vol 2012 no 1 article 142 8 pages 2012

[8] AA Kilbas ldquoHadamard-type fractional calculusrdquo Journal of theKorean Mathematical Society vol 38 no 6 pp 1191ndash1204 2001

[9] I Podlubny Fractional Differential Equations Academic PressSan Diego Calif USA 1999

[10] S Pooseh RAlmeida andD FMTorres ldquoExpansion formulasin terms of integer-order derivatives for the Hadamard frac-tional integral and derivativerdquo Numerical Functional Analysisand Optimization vol 33 no 3 pp 301ndash319 2012

[11] A A Kilbas H M Srivastava and J J Trujillo Theoryand Applications of Fractional Differential Equations vol 204Elsevier Science Amsterdam The Netherlands 2006

[12] S G Samko A A Kilbas and O I Marichev FractionalIntegrals and Derivatives Gordon and Breach Science YverdonSwitzerland 1993

[13] B Ahmad S K Ntouyas and J Tariboon ldquoExistence resultsformixedHadamard andRiemann-Liouville fractional integro-differential equationsrdquo Advances in Difference Equations vol2015 no 1 article 293 8 pages 2015

[14] R P Agarwal M Meehan and D OrsquoRegan Fixed PointTheory and Applications vol 141 Cambridge University PressCambridge UK 2001

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Journal of Function Spaces 5

ge

sum119898minus2

119894=1120574119894 [(log (119890119904)) (log (119890119904))

120572minus2minus (log (120578119894119904)) (log (120578119894119904))

120572minus2]

1205752Γ (120572)

ge

(log (119890119904))120572minus2

1205752Γ (120572)

119898minus2

sum

119894=1

120574119894[(log 119890

119904

) minus (log120578119894

119904

)] =

(log (119890119904))120572minus2

1205752Γ (120572)

119898minus2

sum

119894=1

120574119894(log 119890

120578119894

) fl 1198732119902 (119904)

(25)

If 120578119894le 119904 le 119890 119894 = 1 2 119898 minus 2 we have

119867(119905 119904 1205781 1205782 120578119898minus2) =

sum119898minus2

119894=1120574119894(log (119890119904))120572minus1

1205752Γ (120572)

le

(log (119890119904))120572minus2

12057511205752Γ (120572)

1205751

119898minus2

sum

119894=1

120574119894 le 1205751

119898minus2

sum

119894=1

120574119894119902 (119904) lt 1198731119902 (119904)

119867 (119905 119904 1205781 1205782 120578

119898minus2) =

sum119898minus2

119894=1120574119894(log (119890119904))120572minus1

1205752Γ (120572)

=

sum119898minus2

119894=1120574119894 (120572 minus 1) (log (119890119904))120572minus1

1205752 (120572 minus 1) Γ (120572)

ge

(log (119890119904))120572minus2

12057511205752Γ (120572)

1205751sum119898minus2

119894=1120574119894

(120572 minus 1)

ge 1198732119902 (119904)

(26)

(iv) Since 120597119867(11990511990412057811205782 120578119898minus2

)120597119905=minussum119898minus2

119894=1120573119894(log(120578

119894119904))120572minus2

1205751Γ(120572minus1)119905 le 0 then119867(119905 119904 120578

1 1205782 120578

119898minus2) is nonincreasing

in 119905 somin1205911le119905le1205912

119867(119905 119904 1205781 1205782 120578119898minus2)

=

sum119898minus2

119894=1120574119894 [(log (119890119904))

120572minus1minus (log (120578119894119904))

120572minus1]

1205752Γ (120572)

+

(120583 minus 1205752(log 1205912))sum119898minus2

119894=1120573119894(log (120578

119894119904))120572minus2

12057511205752Γ (120572 minus 1)

=

sum119898minus2

119894=1120574119894[(log (119890119904))120572minus1 minus (log (120578119894119904))

120572minus1]

1205752Γ (120572)

+ 120591lowast120583sum119898minus2

119894=1120573119894 (log (120578119894119904))

120572minus2

12057511205752Γ (120572 minus 1)

ge 120591lowastmax1le119905le119890

119867(119905 1205781 1205782 120578119898minus2 119904)

(27)

3 Existence Results

Let us denote by 119864 = 119862([1 119890] 119877) the Banach space of allcontinuous functions from [1 119890] toR endowedwith the norm119909 = Sup

119905isin[1119890]|119909(119905)| and let119875 be the cone119875 = 119909 isin 119864 119909(119905) ge

0 119905 isin [1 119890]

Through this paper we assume that the function 119891

[1 119890] times [0infin) rarr [0infin) satisfies the following conditionsof Caratheodory type

(1198671) (i) 119891(119905 119909) is Lebesgue measurable with respect to 119905

on [1 119890](ii) 119891(119905 119909) is continuous with respect to 119909 on [0infin)

By Lemma 6 we obtain an operatorF 119864 rarr 119864 as

(F119909) (119905) = 119909 (119905)

= int

119890

1

[119866 (119905 119904) + 119867 (119905 119904 1205781 1205782 120578119898minus2)] ℎ (119904)

119889119904

119904

119905 isin [1 119890]

(28)

It should be noticed that problem (1) has solutions if and onlyif the operatorF has fixed points

The first existence and uniqueness result is based on theBanach contraction principle

Theorem 9 Assume that the condition (1198671) holds and thereexists a real-valued function 119892(119905) isin 119871[1 119890] such that

1003816100381610038161003816119891 (119905 119909 (119905)) minus 119891 (119905 119910 (119905))

1003816100381610038161003816le 119892 (119905)

1003816100381610038161003816119909 minus 119910

1003816100381610038161003816

forall119905 isin [1 119890] 119909 119910 isin R(29)

Then problem (1) has a unique solution provided 119892Ψ lt 1where

Ψ = [

1

Γ (120572 + 1)

+

1198731

12057511205752(120572 minus 1) Γ (120572)

] (30)

Proof We set Sup119905isin[1119890]

|119891(119904 0)| = 119901 lt infin and choose 120588 ge

Ψ119901(1 minus Ψ119892)

Now we show that FB120588sub B120588 where B

120588= 119909 isin 119864

119909 le 120588 For any 119909 isin B120588 we have

(F119909) (119905)

= Sup119905isin[1119890]

int

119890

1

[119866 (119905 119904) + 119867 (119905 119904 1205781 1205782 120578

119898minus2)]

sdot 119891 (119904 119909 (119904))

119889119904

119904

le int

119890

1

(119866 (119904 119904) + 1198731119902 (119904))

sdot (1003816100381610038161003816119891 (119904 119909 (119904)) minus 119891 (119904 0)

1003816100381610038161003816+1003816100381610038161003816119891 (119904 0)

1003816100381610038161003816)

119889119904

119904

6 Journal of Function Spaces

le int

119890

1

(119866 (119904 119904) + 1198731119902 (119904)) (119892 (119904) 120588 + 119901)

119889119904

119904

le (10038171003817100381710038171198921003817100381710038171003817120588 + 119901) [

1

Γ (120572)

int

119890

1

(log 119890119904

)

120572minus1119889119904

119904

+

1198731

12057511205752Γ (120572)

int

119890

1

(log 119890119904

)

120572minus2119889119904

119904

] le (10038171003817100381710038171198921003817100381710038171003817120588 + 119901)

sdot [

1

Γ (120572 + 1)

+

1198731

12057511205752(120572 minus 1) Γ (120572)

] le Ψ (10038171003817100381710038171198921003817100381710038171003817120588

+ 119901) le 120588

(31)

It follows that FB120588 sub B120588 For 119909 119910 isin 119864 and for each 119905 isin

[1 119890] we have1003816100381610038161003816(F119909) (119905) minus (F119910) (119905)

1003816100381610038161003816

=

10038161003816100381610038161003816100381610038161003816

int

119890

1

[119866 (119905 119904) + 119867 (119905 119904 1205781 1205782 120578

119898minus2)]

sdot (119891 (119904 119909 (119904)) minus 119891 (119904 119910 (119904)))

119889119904

119904

10038161003816100381610038161003816100381610038161003816

le int

119890

1

(119866 (119904 119904)

+ 1198731119902 (119904)) 119892 (119904)

1003816100381610038161003816119909 (119904) minus 119910 (119904)

1003816100381610038161003816

119889119904

119904

le10038171003817100381710038171198921003817100381710038171003817

1003817100381710038171003817119909 minus 119910

1003817100381710038171003817

sdot int

119890

1

(119866 (119904 119904) + 1198731119902 (119904))

119889119904

119904

le Ψ10038171003817100381710038171198921003817100381710038171003817

1003817100381710038171003817119909 minus 119910

1003817100381710038171003817

(32)

Hence it follows that (F119909)(119905) minus (F119910)(119905) le Ψ119892119909 minus 119910where Ψ119892 lt 1ThereforeF is a contraction Hence by thecontraction mapping principle problem (1) has a uniquenesssolution

Theorem 10 (nonlinear alternative for single valued maps[13]) Let 119864 be a Banach spaceC a closed convex subset of 119864U an open subset ofC and 0 isin U Suppose that 119865 U rarr C isa continuous compact (ie 119865(U) is a relatively compact subsetofC) map Then either

(i) 119865 has a fixed point inU or(ii) There is 119909 isin 120597U (the boundary of U in C) and 120582 isin

(1 119890) with 119909 = 120582119865(119909)

Theorem 11 Assume that (1198671) and the following conditions

hold(1198672) There exist two nonnegative real-valued functions

ℎ1 ℎ2isin [1 119890] such that

119891 (119905 119909) le ℎ1 (119905) + ℎ2 (119905) 119909

for every 119905 isin [1 119890] all 119909 isin [0infin)

(33)

(1198673)There exists a constant119872 gt 0 such that

119872

(1003817100381710038171003817ℎ1

1003817100381710038171003817+1003817100381710038171003817ℎ2

1003817100381710038171003817119872)Ψ

gt 1 (34)

Then the boundary value problem (1) has at least one solutionon [1 119890]

Proof First we show that the operator F 119875 rarr 119875 iscontinuous

For any 119909119899 119909 isin 119875 119899 = 1 2 with lim

119899rarrinfin119909119899(119905) =

119909(119905) 119905 isin [1 119890] Thus by condition (ii) of (1198671) we have

lim119899rarrinfin

119891(119905 119909119899(119905)) = 119891(119905 119909(119905)) 119905 isin [1 119890] So we can

conclude that

Sup119905isin[1119890]

1003816100381610038161003816119891 (119905 119909

119899 (119905)) minus 119891 (119905 119909 (119905))

1003816100381610038161003816997888rarr 0 as 119899 997888rarr infin (35)

On the other hand

1003816100381610038161003816(F119909119899) (119905) minus (F119909) (119905)

1003816100381610038161003816

=

10038161003816100381610038161003816100381610038161003816

int

119890

1

[119866 (119905 119904) + 119867 (119905 119904 1205781 1205782 120578119898minus2)]

sdot (119891 (119904 119909 (119904)) minus 119891 (119904 119910 (119904)))

119889119904

119904

10038161003816100381610038161003816100381610038161003816

le Sup119905isin[1119890]

1003816100381610038161003816119891 (119905 119909

119899 (119905)) minus 119891 (119905 119909 (119905))

1003816100381610038161003816int

119890

1

(119866 (119904 119904)

+ 1198731119902 (119904))

119889119904

119904

le [

1

Γ (120572 + 1)

+

1198731

12057511205752(120572 minus 1) Γ (120572)

] Sup119905isin[1119890]

1003816100381610038161003816119891 (119905 119909

119899 (119905))

minus 119891 (119905 119909 (119905))1003816100381610038161003816

(36)

Hence

1003817100381710038171003817(F119909119899) (119905) minus (F119909) (119905)

1003817100381710038171003817997888rarr 0 as 119899 997888rarr infin (37)

This means thatF is continuousNow we show that F maps bounded sets into bounded

sets in 119875 It suffices to show that for any 120590 gt 0 there exists apositive constant 120588

1gt 0 such that for each 119909 isin B

120588lowast = 119909 isin

119875 119909 le 120588lowast we have F119909 le 120588

1 By (33) for each 119905 isin [1 119890]

we have

|(F119909) (119905)| =

10038161003816100381610038161003816100381610038161003816

int

119890

1

[119866 (119905 119904) + 119867 (119905 119904 1205781 1205782 120578

119898minus2)]

sdot 119891 (119904 119909 (119904))

119889119904

119904

10038161003816100381610038161003816100381610038161003816

le int

119890

1

[119866 (119904 119904) + 1198731119902 (119904)] (ℎ1 (

119904)

+ ℎ2 (119904) 119909 (119904))

119889119904

119904

le (1003817100381710038171003817ℎ1

1003817100381710038171003817+1003817100381710038171003817ℎ2

1003817100381710038171003817120588lowast)

sdot int

119890

1

(119866 (119904 119904) + 1198731119902 (119904))

119889119904

119904

le (1003817100381710038171003817ℎ1

1003817100381710038171003817+1003817100381710038171003817ℎ2

1003817100381710038171003817120588lowast)

sdot [

1

Γ (120572 + 1)

+

1198731

12057511205752(120572 minus 1) Γ (120572)

] le (1003817100381710038171003817ℎ1

1003817100381710038171003817+1003817100381710038171003817ℎ2

1003817100381710038171003817

sdot 120588lowast) Ψ fl 120588

1

(38)

Journal of Function Spaces 7

which implies that (F119909) le 1205881 Further we let 119905

1 1199052isin [1 119890]

with 1199051lt 1199052and 119909 isin B

120588lowast where B

120588lowast is a bounded set of 119875

and then we find that

10038161003816100381610038161003816(119862

119867119863F119909) (119905)

10038161003816100381610038161003816=

1003816100381610038161003816100381610038161003816100381610038161003816

minus

1

Γ (120572 minus 1)

sdot int

119890

1

[(log 119905119904

)

120572minus2

+

1

1205751

119898minus2

sum

119894=1

120573119894(log

120578119894

119904

)

120572minus2

]

sdot 119891 (119904 119909 (119904))

119889119904

119904

1003816100381610038161003816100381610038161003816100381610038161003816

le (1003817100381710038171003817ℎ1

1003817100381710038171003817+1003817100381710038171003817ℎ2

1003817100381710038171003817120588lowast)

sdot

(1205751 + sum119898minus2

119894=1120573119894)

1205751Γ (120572 minus 1)

fl 1205882

(39)

Hence for 1199051 1199052isin [1 119890] we have

1003816100381610038161003816(F119909) (1199052

) minus (F119909) (1199051)1003816100381610038161003816le int

1199052

1199051

(119862

119867119863F119909) (119904) 119889119904

le 1205882 (1199052 minus 1199051)

(40)

This implies that F maps bounded sets into equicontinuoussets of 119875

Thus by the Arzela-Ascoli theorem the operator F

119875 rarr 119875 is completely continuousNext we consider the set 119881 = 119906 isin 119864 | 119909 = 120583F119909 0 lt

120583 lt 1 and show that the set 119881 is bounded let 119909 isin 119881 andthen 119909 = 120583F119909 0 lt 120583 lt 1 For any 119905 isin [1 119890] we have

|119909 (119905)| = 120583 |(F119909) (119905)|

le int

119890

1

[119866 (119905 119904) + 119867 (119905 119904 1205781 1205782 120578119898minus2)]

sdot1003816100381610038161003816119891 (119904 119909 (119904))

1003816100381610038161003816

119889119904

119904

le 1205881

(41)

Thus 119909(119905) le 1205881for any 119905 isin [1 119890] so that set 119881 is bounded

Thus by the conclusion of Theorem 10 the operator F hasat least one fixed point which implies that problem (1) has atleast one solution

Theorem 12 (Krasnoselskii fixed point theorem [14]) Let 119864be a Banach space and 119870 sub 119864 is a cone in 119864 Assume that Ω1and Ω

2are open subsets of 119864 with 0 isin Ω

1and Ω

1sub Ω2 Let

119879 119870 cap (Ω2 Ω1) rarr 119870 be completely continuous operator In

addition suppose that either

(i) 119879119906 le 119906 forall119906 isin 119870 cap 120597Ω1and 119879119906 ge 119906 forall119906 isin

119870 cap 120597Ω2or

(ii) 119879119906 le 119906 forall119906 isin 119870 cap 120597Ω2and 119879119906 ge 119906 forall119906 isin

119870 cap 120597Ω1

holds Then 119879 has a fixed point in 119870 cap (Ω2 Ω1)

To state the last result of this section we set

1198721 = (Ψ)minus1

1198722 = (

1205910

Γ (120572 + 1)

+ (120572 minus 1) 120591lowast[

120583sum119898minus2

119894=1120573119894

12057511205752Γ (120572)

minus

sum119898minus2

119894=1120574119894

1205752Γ (120572 + 1)

])

minus1

(42)

Theorem 13 Suppose that there exist two positive constants1199032 gt 1199031 gt 0 and1198721 isin (01198722)

(i) 119891(119905 119909) le 11987211199032 for (119905 119909) isin [1 119890] times [0 119903

2]

(ii) 119891(119905 119909) ge 11987221199031 for (119905 119909) isin [1 119890] times [0 1199031]

Then (1) has at least a positive solution

Proof Let Ω119894= 119909 isin 119864 | 119909 lt 119903

119894 119894 = 1 2 From the proof

of Theorem 11 we know that the operatorF defined by (28)is completely continuous on 119875

For any 119909 isin 119875 cap 120597Ω1 it follows that

(F119909) (119905)

= Sup119905isin[1119890]

int

119890

1

[119866 (119905 119904) + 119867 (119905 119904 1205781 1205782 120578

119898minus2)]

sdot 119891 (119904 119909 (119904))

119889119904

119904

ge int

119890

1

(1205910119866 (119904 119904)

+ 120591lowast119867(1 119904 120578

1 1205782 120578

119898minus2))11987221199031

119889119904

119904

ge 11987221199031[

1205910

Γ (120572 + 1)

+ (120572 minus 1) 120591lowast[

120583sum119898minus2

119894=1120573119894

12057511205752Γ (120572)

minus

sum119898minus2

119894=1120574119894

1205752Γ (120572 + 1)

]] ge 1199031 = 119909

(43)

that is (F119909)(119905) ge 119909 119909 isin 119875 cap 120597Ω1

On the other hand for any 119909 isin 119875 cap 120597Ω2 it follows that

(F119909) (119905)

= Sup119905isin[1119890]

int

119890

1

[119866 (119905 119904) + 119867 (119905 119904 1205781 1205782 120578

119898minus2)]

sdot 119891 (119904 119909 (119904))

119889119904

119904

le int

119890

1

(119866 (119904 119904) + 1198731119902 (119904))

sdot 11987211199032

119889119904

119904

le 11987211199032Ψ = 1199032 = 119909

(44)

that is (F119909)(119905) le 119909 119909 isin 119875 cap 120597Ω2

In view ofTheorem 12F has a fixed point in119875cap(Ω2Ω1)which is a positive solution to (1)

4 Examples

In this section we exemplify our theoretical results obtainedin Section 3

8 Journal of Function Spaces

Example 1 Consider the following BVP for Hadamard frac-tional differential equation

119862

11986711986373119909 (119905) +119891 (119905 119909 (119905)) = 0 1 le 119905 le 119890

119862

119867119863119909 (1) =

1

2

119862

119867119863119909(

4

3

)

119862

1198671198632119909 (1) = 0

119909 (119890) =

1

4

119909 (

4

3

)

(45)

Here 120572 = 73 120573 = 05 120574 = 025 120578 = 43 and

119891 (119905 119909) =

119890119905

2 (119890119905+ 1)

(

1199092

119909 + 1

+

119909

4 (119909 + 1)

+

3

4

)

(119905 119909) isin [1 119890] times [0infin)

(46)

Using the given data we find that Ψ = 1609 1205751= 05 120575

2=

075 120583 = 0928 and

1003816100381610038161003816119891 (119905 119909) minus 119891 (119905 119910)

1003816100381610038161003816le

119890119905

2 (119890119905+ 1)

1003816100381610038161003816119909 minus 119910

1003816100381610038161003816le

1

2

1003816100381610038161003816119909 minus 119910

1003816100381610038161003816

with 119892 (119905) =

119890119905

2 (119890119905+ 1)

(47)

Hencewe obtain 119892Ψ = 0805 lt 1Therefore byTheorem9problem (45) has a unique solution on [1 119890]

Example 2 Consider the following BVP for Hadamard frac-tional differential equation

119862

11986711986352119909 (119905) +119891 (119905 119909 (119905)) = 0 1 le 119905 le 119890

119862

119867119863119909 (1) =

1

4

119862

119867119863119909(

3

2

)

119862

1198671198632119909 (1) = 0

119909 (119890) =

1

3

119909 (

3

2

)

(48)

Here 120572 = 52 120573 = 025 120574 = 13 120578 = 32 Let

119891 (119905 119909) = 9 sin2119905 + 119890119905119909

12 (119890119905+ 1) (119909

2+ 1)

(119905 119909) isin [1 119890] times [0infin)

(49)

with ℎ1(119905) = 9 sin2119905 ℎ

2(119905) = 119890

11990512(119890119905+ 1) Here 119891(119905 119909) le

ℎ1(119905) + ℎ

2(119905)119909

It is easy to verify that119872(ℎ1 + ℎ

2119872)Ψ = 1147 gt 1

Then by Theorem 11 problem (48) has at least onesolution on [1 119890]

Example 3 Consider the following BVP for Hadamard frac-tional differential equation

119862

11986711986372119909 (119905) +119891 (119905 119909 (119905)) = 0 1 le 119905 le 119890

119862

119867119863119909 (1) =

1

2

119862

119867119863119909(

4

3

) +

1

4

119862

119867119863119909(

5

2

)

119862

1198671198632119909 (1) = 0

119909 (119890) =

1

3

119909 (

4

3

) +

1

5

119909 (

5

2

)

(50)

where 120572 = 72 1205731 = 05 1205732 = 025 1205741 = 13 1205742 = 15

1205781 = 43 1205782 = 52 Let

119891 (119905 119909) =

1

4

119909 +

119890119905

5 (119890119905+ 1)

(119905 119909) isin [1 119890] times [0infin) (51)

It is easy to verify that

1198721= (Ψ)

minus1= 06196

1198722= (

1205910

Γ (120572 + 1)

+ (120572 minus 1) 120591lowast[

120583sum119898minus2

119894=1120573119894

12057511205752Γ (120572)

minus

sum119898minus2

119894=1120574119894

1205752Γ (120572 + 1)

])

minus1

= 0521

(52)

Choosing 1199031= 15 119903

2= 119890 we have

119891 (119905 sdot 119909) le 12 le 11987211199032 (119905 119909) isin [1 119890] times [0 1199032

]

119891 (119905 sdot 119909) gt 05 ge 11987221199031 (119905 119909) isin [1 119890] times [0 1199031

]

(53)

Hence all conditions of Theorem 13 are satisfied then prob-lem (50) has at least one positive solution 119909 such that 15 lt

119909 lt 119890

Competing Interests

The author declares no competing interests

References

[1] Y Y Gambo F Jarad D Baleanu and T Abdeljawad ldquoOnCaputo modification of the Hadamard fractional derivativesrdquoAdvances in Difference Equations vol 2014 article 10 12 pages2014

[2] B Ahmad S K Ntouyas and A Alsaedi ldquoNew results forboundary value problems of Hadamard-type fractional differ-ential inclusions and integral boundary conditionsrdquo BoundaryValue Problems vol 275 p 14 2013

[3] J Tariboon S K Ntouyas and W Sudsutad ldquoNonlo-cal Hadamard fractional integral conditions for nonlinearRiemann-Liouville fractional differential equationsrdquo BoundaryValue Problems vol 2014 article 253 2014

Journal of Function Spaces 9

[4] P Thiramanus S K Ntouyas and J Tariboon ldquoExistence anduniqueness results for Hadamard-type fractional differentialequations with nonlocal fractional integral boundary condi-tionsrdquo Abstract and Applied Analysis vol 2014 Article ID902054 9 pages 2014

[5] B Ahmad and S K Ntouyas ldquoAn existence theorem forfractional hybrid differential inclusions of Hadamard type withDirichlet boundary conditionsrdquo Abstract and Applied Analysisvol 2014 Article ID 705809 7 pages 2014

[6] B Ahmad and S K Ntouyas ldquoA fully Hadamard type integralboundary value problem of a coupled system of fractional dif-ferential equationsrdquo Fractional Calculus and Applied AnalysisAn International Journal forTheory and Applications vol 17 no2 pp 348ndash360 2014

[7] F Jarad T Abdeljawad and D Baleanu ldquoCaputo-type modi-fication of the Hadamard fractional derivativesrdquo Advances inDifference Equations vol 2012 no 1 article 142 8 pages 2012

[8] AA Kilbas ldquoHadamard-type fractional calculusrdquo Journal of theKorean Mathematical Society vol 38 no 6 pp 1191ndash1204 2001

[9] I Podlubny Fractional Differential Equations Academic PressSan Diego Calif USA 1999

[10] S Pooseh RAlmeida andD FMTorres ldquoExpansion formulasin terms of integer-order derivatives for the Hadamard frac-tional integral and derivativerdquo Numerical Functional Analysisand Optimization vol 33 no 3 pp 301ndash319 2012

[11] A A Kilbas H M Srivastava and J J Trujillo Theoryand Applications of Fractional Differential Equations vol 204Elsevier Science Amsterdam The Netherlands 2006

[12] S G Samko A A Kilbas and O I Marichev FractionalIntegrals and Derivatives Gordon and Breach Science YverdonSwitzerland 1993

[13] B Ahmad S K Ntouyas and J Tariboon ldquoExistence resultsformixedHadamard andRiemann-Liouville fractional integro-differential equationsrdquo Advances in Difference Equations vol2015 no 1 article 293 8 pages 2015

[14] R P Agarwal M Meehan and D OrsquoRegan Fixed PointTheory and Applications vol 141 Cambridge University PressCambridge UK 2001

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

6 Journal of Function Spaces

le int

119890

1

(119866 (119904 119904) + 1198731119902 (119904)) (119892 (119904) 120588 + 119901)

119889119904

119904

le (10038171003817100381710038171198921003817100381710038171003817120588 + 119901) [

1

Γ (120572)

int

119890

1

(log 119890119904

)

120572minus1119889119904

119904

+

1198731

12057511205752Γ (120572)

int

119890

1

(log 119890119904

)

120572minus2119889119904

119904

] le (10038171003817100381710038171198921003817100381710038171003817120588 + 119901)

sdot [

1

Γ (120572 + 1)

+

1198731

12057511205752(120572 minus 1) Γ (120572)

] le Ψ (10038171003817100381710038171198921003817100381710038171003817120588

+ 119901) le 120588

(31)

It follows that FB120588 sub B120588 For 119909 119910 isin 119864 and for each 119905 isin

[1 119890] we have1003816100381610038161003816(F119909) (119905) minus (F119910) (119905)

1003816100381610038161003816

=

10038161003816100381610038161003816100381610038161003816

int

119890

1

[119866 (119905 119904) + 119867 (119905 119904 1205781 1205782 120578

119898minus2)]

sdot (119891 (119904 119909 (119904)) minus 119891 (119904 119910 (119904)))

119889119904

119904

10038161003816100381610038161003816100381610038161003816

le int

119890

1

(119866 (119904 119904)

+ 1198731119902 (119904)) 119892 (119904)

1003816100381610038161003816119909 (119904) minus 119910 (119904)

1003816100381610038161003816

119889119904

119904

le10038171003817100381710038171198921003817100381710038171003817

1003817100381710038171003817119909 minus 119910

1003817100381710038171003817

sdot int

119890

1

(119866 (119904 119904) + 1198731119902 (119904))

119889119904

119904

le Ψ10038171003817100381710038171198921003817100381710038171003817

1003817100381710038171003817119909 minus 119910

1003817100381710038171003817

(32)

Hence it follows that (F119909)(119905) minus (F119910)(119905) le Ψ119892119909 minus 119910where Ψ119892 lt 1ThereforeF is a contraction Hence by thecontraction mapping principle problem (1) has a uniquenesssolution

Theorem 10 (nonlinear alternative for single valued maps[13]) Let 119864 be a Banach spaceC a closed convex subset of 119864U an open subset ofC and 0 isin U Suppose that 119865 U rarr C isa continuous compact (ie 119865(U) is a relatively compact subsetofC) map Then either

(i) 119865 has a fixed point inU or(ii) There is 119909 isin 120597U (the boundary of U in C) and 120582 isin

(1 119890) with 119909 = 120582119865(119909)

Theorem 11 Assume that (1198671) and the following conditions

hold(1198672) There exist two nonnegative real-valued functions

ℎ1 ℎ2isin [1 119890] such that

119891 (119905 119909) le ℎ1 (119905) + ℎ2 (119905) 119909

for every 119905 isin [1 119890] all 119909 isin [0infin)

(33)

(1198673)There exists a constant119872 gt 0 such that

119872

(1003817100381710038171003817ℎ1

1003817100381710038171003817+1003817100381710038171003817ℎ2

1003817100381710038171003817119872)Ψ

gt 1 (34)

Then the boundary value problem (1) has at least one solutionon [1 119890]

Proof First we show that the operator F 119875 rarr 119875 iscontinuous

For any 119909119899 119909 isin 119875 119899 = 1 2 with lim

119899rarrinfin119909119899(119905) =

119909(119905) 119905 isin [1 119890] Thus by condition (ii) of (1198671) we have

lim119899rarrinfin

119891(119905 119909119899(119905)) = 119891(119905 119909(119905)) 119905 isin [1 119890] So we can

conclude that

Sup119905isin[1119890]

1003816100381610038161003816119891 (119905 119909

119899 (119905)) minus 119891 (119905 119909 (119905))

1003816100381610038161003816997888rarr 0 as 119899 997888rarr infin (35)

On the other hand

1003816100381610038161003816(F119909119899) (119905) minus (F119909) (119905)

1003816100381610038161003816

=

10038161003816100381610038161003816100381610038161003816

int

119890

1

[119866 (119905 119904) + 119867 (119905 119904 1205781 1205782 120578119898minus2)]

sdot (119891 (119904 119909 (119904)) minus 119891 (119904 119910 (119904)))

119889119904

119904

10038161003816100381610038161003816100381610038161003816

le Sup119905isin[1119890]

1003816100381610038161003816119891 (119905 119909

119899 (119905)) minus 119891 (119905 119909 (119905))

1003816100381610038161003816int

119890

1

(119866 (119904 119904)

+ 1198731119902 (119904))

119889119904

119904

le [

1

Γ (120572 + 1)

+

1198731

12057511205752(120572 minus 1) Γ (120572)

] Sup119905isin[1119890]

1003816100381610038161003816119891 (119905 119909

119899 (119905))

minus 119891 (119905 119909 (119905))1003816100381610038161003816

(36)

Hence

1003817100381710038171003817(F119909119899) (119905) minus (F119909) (119905)

1003817100381710038171003817997888rarr 0 as 119899 997888rarr infin (37)

This means thatF is continuousNow we show that F maps bounded sets into bounded

sets in 119875 It suffices to show that for any 120590 gt 0 there exists apositive constant 120588

1gt 0 such that for each 119909 isin B

120588lowast = 119909 isin

119875 119909 le 120588lowast we have F119909 le 120588

1 By (33) for each 119905 isin [1 119890]

we have

|(F119909) (119905)| =

10038161003816100381610038161003816100381610038161003816

int

119890

1

[119866 (119905 119904) + 119867 (119905 119904 1205781 1205782 120578

119898minus2)]

sdot 119891 (119904 119909 (119904))

119889119904

119904

10038161003816100381610038161003816100381610038161003816

le int

119890

1

[119866 (119904 119904) + 1198731119902 (119904)] (ℎ1 (

119904)

+ ℎ2 (119904) 119909 (119904))

119889119904

119904

le (1003817100381710038171003817ℎ1

1003817100381710038171003817+1003817100381710038171003817ℎ2

1003817100381710038171003817120588lowast)

sdot int

119890

1

(119866 (119904 119904) + 1198731119902 (119904))

119889119904

119904

le (1003817100381710038171003817ℎ1

1003817100381710038171003817+1003817100381710038171003817ℎ2

1003817100381710038171003817120588lowast)

sdot [

1

Γ (120572 + 1)

+

1198731

12057511205752(120572 minus 1) Γ (120572)

] le (1003817100381710038171003817ℎ1

1003817100381710038171003817+1003817100381710038171003817ℎ2

1003817100381710038171003817

sdot 120588lowast) Ψ fl 120588

1

(38)

Journal of Function Spaces 7

which implies that (F119909) le 1205881 Further we let 119905

1 1199052isin [1 119890]

with 1199051lt 1199052and 119909 isin B

120588lowast where B

120588lowast is a bounded set of 119875

and then we find that

10038161003816100381610038161003816(119862

119867119863F119909) (119905)

10038161003816100381610038161003816=

1003816100381610038161003816100381610038161003816100381610038161003816

minus

1

Γ (120572 minus 1)

sdot int

119890

1

[(log 119905119904

)

120572minus2

+

1

1205751

119898minus2

sum

119894=1

120573119894(log

120578119894

119904

)

120572minus2

]

sdot 119891 (119904 119909 (119904))

119889119904

119904

1003816100381610038161003816100381610038161003816100381610038161003816

le (1003817100381710038171003817ℎ1

1003817100381710038171003817+1003817100381710038171003817ℎ2

1003817100381710038171003817120588lowast)

sdot

(1205751 + sum119898minus2

119894=1120573119894)

1205751Γ (120572 minus 1)

fl 1205882

(39)

Hence for 1199051 1199052isin [1 119890] we have

1003816100381610038161003816(F119909) (1199052

) minus (F119909) (1199051)1003816100381610038161003816le int

1199052

1199051

(119862

119867119863F119909) (119904) 119889119904

le 1205882 (1199052 minus 1199051)

(40)

This implies that F maps bounded sets into equicontinuoussets of 119875

Thus by the Arzela-Ascoli theorem the operator F

119875 rarr 119875 is completely continuousNext we consider the set 119881 = 119906 isin 119864 | 119909 = 120583F119909 0 lt

120583 lt 1 and show that the set 119881 is bounded let 119909 isin 119881 andthen 119909 = 120583F119909 0 lt 120583 lt 1 For any 119905 isin [1 119890] we have

|119909 (119905)| = 120583 |(F119909) (119905)|

le int

119890

1

[119866 (119905 119904) + 119867 (119905 119904 1205781 1205782 120578119898minus2)]

sdot1003816100381610038161003816119891 (119904 119909 (119904))

1003816100381610038161003816

119889119904

119904

le 1205881

(41)

Thus 119909(119905) le 1205881for any 119905 isin [1 119890] so that set 119881 is bounded

Thus by the conclusion of Theorem 10 the operator F hasat least one fixed point which implies that problem (1) has atleast one solution

Theorem 12 (Krasnoselskii fixed point theorem [14]) Let 119864be a Banach space and 119870 sub 119864 is a cone in 119864 Assume that Ω1and Ω

2are open subsets of 119864 with 0 isin Ω

1and Ω

1sub Ω2 Let

119879 119870 cap (Ω2 Ω1) rarr 119870 be completely continuous operator In

addition suppose that either

(i) 119879119906 le 119906 forall119906 isin 119870 cap 120597Ω1and 119879119906 ge 119906 forall119906 isin

119870 cap 120597Ω2or

(ii) 119879119906 le 119906 forall119906 isin 119870 cap 120597Ω2and 119879119906 ge 119906 forall119906 isin

119870 cap 120597Ω1

holds Then 119879 has a fixed point in 119870 cap (Ω2 Ω1)

To state the last result of this section we set

1198721 = (Ψ)minus1

1198722 = (

1205910

Γ (120572 + 1)

+ (120572 minus 1) 120591lowast[

120583sum119898minus2

119894=1120573119894

12057511205752Γ (120572)

minus

sum119898minus2

119894=1120574119894

1205752Γ (120572 + 1)

])

minus1

(42)

Theorem 13 Suppose that there exist two positive constants1199032 gt 1199031 gt 0 and1198721 isin (01198722)

(i) 119891(119905 119909) le 11987211199032 for (119905 119909) isin [1 119890] times [0 119903

2]

(ii) 119891(119905 119909) ge 11987221199031 for (119905 119909) isin [1 119890] times [0 1199031]

Then (1) has at least a positive solution

Proof Let Ω119894= 119909 isin 119864 | 119909 lt 119903

119894 119894 = 1 2 From the proof

of Theorem 11 we know that the operatorF defined by (28)is completely continuous on 119875

For any 119909 isin 119875 cap 120597Ω1 it follows that

(F119909) (119905)

= Sup119905isin[1119890]

int

119890

1

[119866 (119905 119904) + 119867 (119905 119904 1205781 1205782 120578

119898minus2)]

sdot 119891 (119904 119909 (119904))

119889119904

119904

ge int

119890

1

(1205910119866 (119904 119904)

+ 120591lowast119867(1 119904 120578

1 1205782 120578

119898minus2))11987221199031

119889119904

119904

ge 11987221199031[

1205910

Γ (120572 + 1)

+ (120572 minus 1) 120591lowast[

120583sum119898minus2

119894=1120573119894

12057511205752Γ (120572)

minus

sum119898minus2

119894=1120574119894

1205752Γ (120572 + 1)

]] ge 1199031 = 119909

(43)

that is (F119909)(119905) ge 119909 119909 isin 119875 cap 120597Ω1

On the other hand for any 119909 isin 119875 cap 120597Ω2 it follows that

(F119909) (119905)

= Sup119905isin[1119890]

int

119890

1

[119866 (119905 119904) + 119867 (119905 119904 1205781 1205782 120578

119898minus2)]

sdot 119891 (119904 119909 (119904))

119889119904

119904

le int

119890

1

(119866 (119904 119904) + 1198731119902 (119904))

sdot 11987211199032

119889119904

119904

le 11987211199032Ψ = 1199032 = 119909

(44)

that is (F119909)(119905) le 119909 119909 isin 119875 cap 120597Ω2

In view ofTheorem 12F has a fixed point in119875cap(Ω2Ω1)which is a positive solution to (1)

4 Examples

In this section we exemplify our theoretical results obtainedin Section 3

8 Journal of Function Spaces

Example 1 Consider the following BVP for Hadamard frac-tional differential equation

119862

11986711986373119909 (119905) +119891 (119905 119909 (119905)) = 0 1 le 119905 le 119890

119862

119867119863119909 (1) =

1

2

119862

119867119863119909(

4

3

)

119862

1198671198632119909 (1) = 0

119909 (119890) =

1

4

119909 (

4

3

)

(45)

Here 120572 = 73 120573 = 05 120574 = 025 120578 = 43 and

119891 (119905 119909) =

119890119905

2 (119890119905+ 1)

(

1199092

119909 + 1

+

119909

4 (119909 + 1)

+

3

4

)

(119905 119909) isin [1 119890] times [0infin)

(46)

Using the given data we find that Ψ = 1609 1205751= 05 120575

2=

075 120583 = 0928 and

1003816100381610038161003816119891 (119905 119909) minus 119891 (119905 119910)

1003816100381610038161003816le

119890119905

2 (119890119905+ 1)

1003816100381610038161003816119909 minus 119910

1003816100381610038161003816le

1

2

1003816100381610038161003816119909 minus 119910

1003816100381610038161003816

with 119892 (119905) =

119890119905

2 (119890119905+ 1)

(47)

Hencewe obtain 119892Ψ = 0805 lt 1Therefore byTheorem9problem (45) has a unique solution on [1 119890]

Example 2 Consider the following BVP for Hadamard frac-tional differential equation

119862

11986711986352119909 (119905) +119891 (119905 119909 (119905)) = 0 1 le 119905 le 119890

119862

119867119863119909 (1) =

1

4

119862

119867119863119909(

3

2

)

119862

1198671198632119909 (1) = 0

119909 (119890) =

1

3

119909 (

3

2

)

(48)

Here 120572 = 52 120573 = 025 120574 = 13 120578 = 32 Let

119891 (119905 119909) = 9 sin2119905 + 119890119905119909

12 (119890119905+ 1) (119909

2+ 1)

(119905 119909) isin [1 119890] times [0infin)

(49)

with ℎ1(119905) = 9 sin2119905 ℎ

2(119905) = 119890

11990512(119890119905+ 1) Here 119891(119905 119909) le

ℎ1(119905) + ℎ

2(119905)119909

It is easy to verify that119872(ℎ1 + ℎ

2119872)Ψ = 1147 gt 1

Then by Theorem 11 problem (48) has at least onesolution on [1 119890]

Example 3 Consider the following BVP for Hadamard frac-tional differential equation

119862

11986711986372119909 (119905) +119891 (119905 119909 (119905)) = 0 1 le 119905 le 119890

119862

119867119863119909 (1) =

1

2

119862

119867119863119909(

4

3

) +

1

4

119862

119867119863119909(

5

2

)

119862

1198671198632119909 (1) = 0

119909 (119890) =

1

3

119909 (

4

3

) +

1

5

119909 (

5

2

)

(50)

where 120572 = 72 1205731 = 05 1205732 = 025 1205741 = 13 1205742 = 15

1205781 = 43 1205782 = 52 Let

119891 (119905 119909) =

1

4

119909 +

119890119905

5 (119890119905+ 1)

(119905 119909) isin [1 119890] times [0infin) (51)

It is easy to verify that

1198721= (Ψ)

minus1= 06196

1198722= (

1205910

Γ (120572 + 1)

+ (120572 minus 1) 120591lowast[

120583sum119898minus2

119894=1120573119894

12057511205752Γ (120572)

minus

sum119898minus2

119894=1120574119894

1205752Γ (120572 + 1)

])

minus1

= 0521

(52)

Choosing 1199031= 15 119903

2= 119890 we have

119891 (119905 sdot 119909) le 12 le 11987211199032 (119905 119909) isin [1 119890] times [0 1199032

]

119891 (119905 sdot 119909) gt 05 ge 11987221199031 (119905 119909) isin [1 119890] times [0 1199031

]

(53)

Hence all conditions of Theorem 13 are satisfied then prob-lem (50) has at least one positive solution 119909 such that 15 lt

119909 lt 119890

Competing Interests

The author declares no competing interests

References

[1] Y Y Gambo F Jarad D Baleanu and T Abdeljawad ldquoOnCaputo modification of the Hadamard fractional derivativesrdquoAdvances in Difference Equations vol 2014 article 10 12 pages2014

[2] B Ahmad S K Ntouyas and A Alsaedi ldquoNew results forboundary value problems of Hadamard-type fractional differ-ential inclusions and integral boundary conditionsrdquo BoundaryValue Problems vol 275 p 14 2013

[3] J Tariboon S K Ntouyas and W Sudsutad ldquoNonlo-cal Hadamard fractional integral conditions for nonlinearRiemann-Liouville fractional differential equationsrdquo BoundaryValue Problems vol 2014 article 253 2014

Journal of Function Spaces 9

[4] P Thiramanus S K Ntouyas and J Tariboon ldquoExistence anduniqueness results for Hadamard-type fractional differentialequations with nonlocal fractional integral boundary condi-tionsrdquo Abstract and Applied Analysis vol 2014 Article ID902054 9 pages 2014

[5] B Ahmad and S K Ntouyas ldquoAn existence theorem forfractional hybrid differential inclusions of Hadamard type withDirichlet boundary conditionsrdquo Abstract and Applied Analysisvol 2014 Article ID 705809 7 pages 2014

[6] B Ahmad and S K Ntouyas ldquoA fully Hadamard type integralboundary value problem of a coupled system of fractional dif-ferential equationsrdquo Fractional Calculus and Applied AnalysisAn International Journal forTheory and Applications vol 17 no2 pp 348ndash360 2014

[7] F Jarad T Abdeljawad and D Baleanu ldquoCaputo-type modi-fication of the Hadamard fractional derivativesrdquo Advances inDifference Equations vol 2012 no 1 article 142 8 pages 2012

[8] AA Kilbas ldquoHadamard-type fractional calculusrdquo Journal of theKorean Mathematical Society vol 38 no 6 pp 1191ndash1204 2001

[9] I Podlubny Fractional Differential Equations Academic PressSan Diego Calif USA 1999

[10] S Pooseh RAlmeida andD FMTorres ldquoExpansion formulasin terms of integer-order derivatives for the Hadamard frac-tional integral and derivativerdquo Numerical Functional Analysisand Optimization vol 33 no 3 pp 301ndash319 2012

[11] A A Kilbas H M Srivastava and J J Trujillo Theoryand Applications of Fractional Differential Equations vol 204Elsevier Science Amsterdam The Netherlands 2006

[12] S G Samko A A Kilbas and O I Marichev FractionalIntegrals and Derivatives Gordon and Breach Science YverdonSwitzerland 1993

[13] B Ahmad S K Ntouyas and J Tariboon ldquoExistence resultsformixedHadamard andRiemann-Liouville fractional integro-differential equationsrdquo Advances in Difference Equations vol2015 no 1 article 293 8 pages 2015

[14] R P Agarwal M Meehan and D OrsquoRegan Fixed PointTheory and Applications vol 141 Cambridge University PressCambridge UK 2001

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Journal of Function Spaces 7

which implies that (F119909) le 1205881 Further we let 119905

1 1199052isin [1 119890]

with 1199051lt 1199052and 119909 isin B

120588lowast where B

120588lowast is a bounded set of 119875

and then we find that

10038161003816100381610038161003816(119862

119867119863F119909) (119905)

10038161003816100381610038161003816=

1003816100381610038161003816100381610038161003816100381610038161003816

minus

1

Γ (120572 minus 1)

sdot int

119890

1

[(log 119905119904

)

120572minus2

+

1

1205751

119898minus2

sum

119894=1

120573119894(log

120578119894

119904

)

120572minus2

]

sdot 119891 (119904 119909 (119904))

119889119904

119904

1003816100381610038161003816100381610038161003816100381610038161003816

le (1003817100381710038171003817ℎ1

1003817100381710038171003817+1003817100381710038171003817ℎ2

1003817100381710038171003817120588lowast)

sdot

(1205751 + sum119898minus2

119894=1120573119894)

1205751Γ (120572 minus 1)

fl 1205882

(39)

Hence for 1199051 1199052isin [1 119890] we have

1003816100381610038161003816(F119909) (1199052

) minus (F119909) (1199051)1003816100381610038161003816le int

1199052

1199051

(119862

119867119863F119909) (119904) 119889119904

le 1205882 (1199052 minus 1199051)

(40)

This implies that F maps bounded sets into equicontinuoussets of 119875

Thus by the Arzela-Ascoli theorem the operator F

119875 rarr 119875 is completely continuousNext we consider the set 119881 = 119906 isin 119864 | 119909 = 120583F119909 0 lt

120583 lt 1 and show that the set 119881 is bounded let 119909 isin 119881 andthen 119909 = 120583F119909 0 lt 120583 lt 1 For any 119905 isin [1 119890] we have

|119909 (119905)| = 120583 |(F119909) (119905)|

le int

119890

1

[119866 (119905 119904) + 119867 (119905 119904 1205781 1205782 120578119898minus2)]

sdot1003816100381610038161003816119891 (119904 119909 (119904))

1003816100381610038161003816

119889119904

119904

le 1205881

(41)

Thus 119909(119905) le 1205881for any 119905 isin [1 119890] so that set 119881 is bounded

Thus by the conclusion of Theorem 10 the operator F hasat least one fixed point which implies that problem (1) has atleast one solution

Theorem 12 (Krasnoselskii fixed point theorem [14]) Let 119864be a Banach space and 119870 sub 119864 is a cone in 119864 Assume that Ω1and Ω

2are open subsets of 119864 with 0 isin Ω

1and Ω

1sub Ω2 Let

119879 119870 cap (Ω2 Ω1) rarr 119870 be completely continuous operator In

addition suppose that either

(i) 119879119906 le 119906 forall119906 isin 119870 cap 120597Ω1and 119879119906 ge 119906 forall119906 isin

119870 cap 120597Ω2or

(ii) 119879119906 le 119906 forall119906 isin 119870 cap 120597Ω2and 119879119906 ge 119906 forall119906 isin

119870 cap 120597Ω1

holds Then 119879 has a fixed point in 119870 cap (Ω2 Ω1)

To state the last result of this section we set

1198721 = (Ψ)minus1

1198722 = (

1205910

Γ (120572 + 1)

+ (120572 minus 1) 120591lowast[

120583sum119898minus2

119894=1120573119894

12057511205752Γ (120572)

minus

sum119898minus2

119894=1120574119894

1205752Γ (120572 + 1)

])

minus1

(42)

Theorem 13 Suppose that there exist two positive constants1199032 gt 1199031 gt 0 and1198721 isin (01198722)

(i) 119891(119905 119909) le 11987211199032 for (119905 119909) isin [1 119890] times [0 119903

2]

(ii) 119891(119905 119909) ge 11987221199031 for (119905 119909) isin [1 119890] times [0 1199031]

Then (1) has at least a positive solution

Proof Let Ω119894= 119909 isin 119864 | 119909 lt 119903

119894 119894 = 1 2 From the proof

of Theorem 11 we know that the operatorF defined by (28)is completely continuous on 119875

For any 119909 isin 119875 cap 120597Ω1 it follows that

(F119909) (119905)

= Sup119905isin[1119890]

int

119890

1

[119866 (119905 119904) + 119867 (119905 119904 1205781 1205782 120578

119898minus2)]

sdot 119891 (119904 119909 (119904))

119889119904

119904

ge int

119890

1

(1205910119866 (119904 119904)

+ 120591lowast119867(1 119904 120578

1 1205782 120578

119898minus2))11987221199031

119889119904

119904

ge 11987221199031[

1205910

Γ (120572 + 1)

+ (120572 minus 1) 120591lowast[

120583sum119898minus2

119894=1120573119894

12057511205752Γ (120572)

minus

sum119898minus2

119894=1120574119894

1205752Γ (120572 + 1)

]] ge 1199031 = 119909

(43)

that is (F119909)(119905) ge 119909 119909 isin 119875 cap 120597Ω1

On the other hand for any 119909 isin 119875 cap 120597Ω2 it follows that

(F119909) (119905)

= Sup119905isin[1119890]

int

119890

1

[119866 (119905 119904) + 119867 (119905 119904 1205781 1205782 120578

119898minus2)]

sdot 119891 (119904 119909 (119904))

119889119904

119904

le int

119890

1

(119866 (119904 119904) + 1198731119902 (119904))

sdot 11987211199032

119889119904

119904

le 11987211199032Ψ = 1199032 = 119909

(44)

that is (F119909)(119905) le 119909 119909 isin 119875 cap 120597Ω2

In view ofTheorem 12F has a fixed point in119875cap(Ω2Ω1)which is a positive solution to (1)

4 Examples

In this section we exemplify our theoretical results obtainedin Section 3

8 Journal of Function Spaces

Example 1 Consider the following BVP for Hadamard frac-tional differential equation

119862

11986711986373119909 (119905) +119891 (119905 119909 (119905)) = 0 1 le 119905 le 119890

119862

119867119863119909 (1) =

1

2

119862

119867119863119909(

4

3

)

119862

1198671198632119909 (1) = 0

119909 (119890) =

1

4

119909 (

4

3

)

(45)

Here 120572 = 73 120573 = 05 120574 = 025 120578 = 43 and

119891 (119905 119909) =

119890119905

2 (119890119905+ 1)

(

1199092

119909 + 1

+

119909

4 (119909 + 1)

+

3

4

)

(119905 119909) isin [1 119890] times [0infin)

(46)

Using the given data we find that Ψ = 1609 1205751= 05 120575

2=

075 120583 = 0928 and

1003816100381610038161003816119891 (119905 119909) minus 119891 (119905 119910)

1003816100381610038161003816le

119890119905

2 (119890119905+ 1)

1003816100381610038161003816119909 minus 119910

1003816100381610038161003816le

1

2

1003816100381610038161003816119909 minus 119910

1003816100381610038161003816

with 119892 (119905) =

119890119905

2 (119890119905+ 1)

(47)

Hencewe obtain 119892Ψ = 0805 lt 1Therefore byTheorem9problem (45) has a unique solution on [1 119890]

Example 2 Consider the following BVP for Hadamard frac-tional differential equation

119862

11986711986352119909 (119905) +119891 (119905 119909 (119905)) = 0 1 le 119905 le 119890

119862

119867119863119909 (1) =

1

4

119862

119867119863119909(

3

2

)

119862

1198671198632119909 (1) = 0

119909 (119890) =

1

3

119909 (

3

2

)

(48)

Here 120572 = 52 120573 = 025 120574 = 13 120578 = 32 Let

119891 (119905 119909) = 9 sin2119905 + 119890119905119909

12 (119890119905+ 1) (119909

2+ 1)

(119905 119909) isin [1 119890] times [0infin)

(49)

with ℎ1(119905) = 9 sin2119905 ℎ

2(119905) = 119890

11990512(119890119905+ 1) Here 119891(119905 119909) le

ℎ1(119905) + ℎ

2(119905)119909

It is easy to verify that119872(ℎ1 + ℎ

2119872)Ψ = 1147 gt 1

Then by Theorem 11 problem (48) has at least onesolution on [1 119890]

Example 3 Consider the following BVP for Hadamard frac-tional differential equation

119862

11986711986372119909 (119905) +119891 (119905 119909 (119905)) = 0 1 le 119905 le 119890

119862

119867119863119909 (1) =

1

2

119862

119867119863119909(

4

3

) +

1

4

119862

119867119863119909(

5

2

)

119862

1198671198632119909 (1) = 0

119909 (119890) =

1

3

119909 (

4

3

) +

1

5

119909 (

5

2

)

(50)

where 120572 = 72 1205731 = 05 1205732 = 025 1205741 = 13 1205742 = 15

1205781 = 43 1205782 = 52 Let

119891 (119905 119909) =

1

4

119909 +

119890119905

5 (119890119905+ 1)

(119905 119909) isin [1 119890] times [0infin) (51)

It is easy to verify that

1198721= (Ψ)

minus1= 06196

1198722= (

1205910

Γ (120572 + 1)

+ (120572 minus 1) 120591lowast[

120583sum119898minus2

119894=1120573119894

12057511205752Γ (120572)

minus

sum119898minus2

119894=1120574119894

1205752Γ (120572 + 1)

])

minus1

= 0521

(52)

Choosing 1199031= 15 119903

2= 119890 we have

119891 (119905 sdot 119909) le 12 le 11987211199032 (119905 119909) isin [1 119890] times [0 1199032

]

119891 (119905 sdot 119909) gt 05 ge 11987221199031 (119905 119909) isin [1 119890] times [0 1199031

]

(53)

Hence all conditions of Theorem 13 are satisfied then prob-lem (50) has at least one positive solution 119909 such that 15 lt

119909 lt 119890

Competing Interests

The author declares no competing interests

References

[1] Y Y Gambo F Jarad D Baleanu and T Abdeljawad ldquoOnCaputo modification of the Hadamard fractional derivativesrdquoAdvances in Difference Equations vol 2014 article 10 12 pages2014

[2] B Ahmad S K Ntouyas and A Alsaedi ldquoNew results forboundary value problems of Hadamard-type fractional differ-ential inclusions and integral boundary conditionsrdquo BoundaryValue Problems vol 275 p 14 2013

[3] J Tariboon S K Ntouyas and W Sudsutad ldquoNonlo-cal Hadamard fractional integral conditions for nonlinearRiemann-Liouville fractional differential equationsrdquo BoundaryValue Problems vol 2014 article 253 2014

Journal of Function Spaces 9

[4] P Thiramanus S K Ntouyas and J Tariboon ldquoExistence anduniqueness results for Hadamard-type fractional differentialequations with nonlocal fractional integral boundary condi-tionsrdquo Abstract and Applied Analysis vol 2014 Article ID902054 9 pages 2014

[5] B Ahmad and S K Ntouyas ldquoAn existence theorem forfractional hybrid differential inclusions of Hadamard type withDirichlet boundary conditionsrdquo Abstract and Applied Analysisvol 2014 Article ID 705809 7 pages 2014

[6] B Ahmad and S K Ntouyas ldquoA fully Hadamard type integralboundary value problem of a coupled system of fractional dif-ferential equationsrdquo Fractional Calculus and Applied AnalysisAn International Journal forTheory and Applications vol 17 no2 pp 348ndash360 2014

[7] F Jarad T Abdeljawad and D Baleanu ldquoCaputo-type modi-fication of the Hadamard fractional derivativesrdquo Advances inDifference Equations vol 2012 no 1 article 142 8 pages 2012

[8] AA Kilbas ldquoHadamard-type fractional calculusrdquo Journal of theKorean Mathematical Society vol 38 no 6 pp 1191ndash1204 2001

[9] I Podlubny Fractional Differential Equations Academic PressSan Diego Calif USA 1999

[10] S Pooseh RAlmeida andD FMTorres ldquoExpansion formulasin terms of integer-order derivatives for the Hadamard frac-tional integral and derivativerdquo Numerical Functional Analysisand Optimization vol 33 no 3 pp 301ndash319 2012

[11] A A Kilbas H M Srivastava and J J Trujillo Theoryand Applications of Fractional Differential Equations vol 204Elsevier Science Amsterdam The Netherlands 2006

[12] S G Samko A A Kilbas and O I Marichev FractionalIntegrals and Derivatives Gordon and Breach Science YverdonSwitzerland 1993

[13] B Ahmad S K Ntouyas and J Tariboon ldquoExistence resultsformixedHadamard andRiemann-Liouville fractional integro-differential equationsrdquo Advances in Difference Equations vol2015 no 1 article 293 8 pages 2015

[14] R P Agarwal M Meehan and D OrsquoRegan Fixed PointTheory and Applications vol 141 Cambridge University PressCambridge UK 2001

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

8 Journal of Function Spaces

Example 1 Consider the following BVP for Hadamard frac-tional differential equation

119862

11986711986373119909 (119905) +119891 (119905 119909 (119905)) = 0 1 le 119905 le 119890

119862

119867119863119909 (1) =

1

2

119862

119867119863119909(

4

3

)

119862

1198671198632119909 (1) = 0

119909 (119890) =

1

4

119909 (

4

3

)

(45)

Here 120572 = 73 120573 = 05 120574 = 025 120578 = 43 and

119891 (119905 119909) =

119890119905

2 (119890119905+ 1)

(

1199092

119909 + 1

+

119909

4 (119909 + 1)

+

3

4

)

(119905 119909) isin [1 119890] times [0infin)

(46)

Using the given data we find that Ψ = 1609 1205751= 05 120575

2=

075 120583 = 0928 and

1003816100381610038161003816119891 (119905 119909) minus 119891 (119905 119910)

1003816100381610038161003816le

119890119905

2 (119890119905+ 1)

1003816100381610038161003816119909 minus 119910

1003816100381610038161003816le

1

2

1003816100381610038161003816119909 minus 119910

1003816100381610038161003816

with 119892 (119905) =

119890119905

2 (119890119905+ 1)

(47)

Hencewe obtain 119892Ψ = 0805 lt 1Therefore byTheorem9problem (45) has a unique solution on [1 119890]

Example 2 Consider the following BVP for Hadamard frac-tional differential equation

119862

11986711986352119909 (119905) +119891 (119905 119909 (119905)) = 0 1 le 119905 le 119890

119862

119867119863119909 (1) =

1

4

119862

119867119863119909(

3

2

)

119862

1198671198632119909 (1) = 0

119909 (119890) =

1

3

119909 (

3

2

)

(48)

Here 120572 = 52 120573 = 025 120574 = 13 120578 = 32 Let

119891 (119905 119909) = 9 sin2119905 + 119890119905119909

12 (119890119905+ 1) (119909

2+ 1)

(119905 119909) isin [1 119890] times [0infin)

(49)

with ℎ1(119905) = 9 sin2119905 ℎ

2(119905) = 119890

11990512(119890119905+ 1) Here 119891(119905 119909) le

ℎ1(119905) + ℎ

2(119905)119909

It is easy to verify that119872(ℎ1 + ℎ

2119872)Ψ = 1147 gt 1

Then by Theorem 11 problem (48) has at least onesolution on [1 119890]

Example 3 Consider the following BVP for Hadamard frac-tional differential equation

119862

11986711986372119909 (119905) +119891 (119905 119909 (119905)) = 0 1 le 119905 le 119890

119862

119867119863119909 (1) =

1

2

119862

119867119863119909(

4

3

) +

1

4

119862

119867119863119909(

5

2

)

119862

1198671198632119909 (1) = 0

119909 (119890) =

1

3

119909 (

4

3

) +

1

5

119909 (

5

2

)

(50)

where 120572 = 72 1205731 = 05 1205732 = 025 1205741 = 13 1205742 = 15

1205781 = 43 1205782 = 52 Let

119891 (119905 119909) =

1

4

119909 +

119890119905

5 (119890119905+ 1)

(119905 119909) isin [1 119890] times [0infin) (51)

It is easy to verify that

1198721= (Ψ)

minus1= 06196

1198722= (

1205910

Γ (120572 + 1)

+ (120572 minus 1) 120591lowast[

120583sum119898minus2

119894=1120573119894

12057511205752Γ (120572)

minus

sum119898minus2

119894=1120574119894

1205752Γ (120572 + 1)

])

minus1

= 0521

(52)

Choosing 1199031= 15 119903

2= 119890 we have

119891 (119905 sdot 119909) le 12 le 11987211199032 (119905 119909) isin [1 119890] times [0 1199032

]

119891 (119905 sdot 119909) gt 05 ge 11987221199031 (119905 119909) isin [1 119890] times [0 1199031

]

(53)

Hence all conditions of Theorem 13 are satisfied then prob-lem (50) has at least one positive solution 119909 such that 15 lt

119909 lt 119890

Competing Interests

The author declares no competing interests

References

[1] Y Y Gambo F Jarad D Baleanu and T Abdeljawad ldquoOnCaputo modification of the Hadamard fractional derivativesrdquoAdvances in Difference Equations vol 2014 article 10 12 pages2014

[2] B Ahmad S K Ntouyas and A Alsaedi ldquoNew results forboundary value problems of Hadamard-type fractional differ-ential inclusions and integral boundary conditionsrdquo BoundaryValue Problems vol 275 p 14 2013

[3] J Tariboon S K Ntouyas and W Sudsutad ldquoNonlo-cal Hadamard fractional integral conditions for nonlinearRiemann-Liouville fractional differential equationsrdquo BoundaryValue Problems vol 2014 article 253 2014

Journal of Function Spaces 9

[4] P Thiramanus S K Ntouyas and J Tariboon ldquoExistence anduniqueness results for Hadamard-type fractional differentialequations with nonlocal fractional integral boundary condi-tionsrdquo Abstract and Applied Analysis vol 2014 Article ID902054 9 pages 2014

[5] B Ahmad and S K Ntouyas ldquoAn existence theorem forfractional hybrid differential inclusions of Hadamard type withDirichlet boundary conditionsrdquo Abstract and Applied Analysisvol 2014 Article ID 705809 7 pages 2014

[6] B Ahmad and S K Ntouyas ldquoA fully Hadamard type integralboundary value problem of a coupled system of fractional dif-ferential equationsrdquo Fractional Calculus and Applied AnalysisAn International Journal forTheory and Applications vol 17 no2 pp 348ndash360 2014

[7] F Jarad T Abdeljawad and D Baleanu ldquoCaputo-type modi-fication of the Hadamard fractional derivativesrdquo Advances inDifference Equations vol 2012 no 1 article 142 8 pages 2012

[8] AA Kilbas ldquoHadamard-type fractional calculusrdquo Journal of theKorean Mathematical Society vol 38 no 6 pp 1191ndash1204 2001

[9] I Podlubny Fractional Differential Equations Academic PressSan Diego Calif USA 1999

[10] S Pooseh RAlmeida andD FMTorres ldquoExpansion formulasin terms of integer-order derivatives for the Hadamard frac-tional integral and derivativerdquo Numerical Functional Analysisand Optimization vol 33 no 3 pp 301ndash319 2012

[11] A A Kilbas H M Srivastava and J J Trujillo Theoryand Applications of Fractional Differential Equations vol 204Elsevier Science Amsterdam The Netherlands 2006

[12] S G Samko A A Kilbas and O I Marichev FractionalIntegrals and Derivatives Gordon and Breach Science YverdonSwitzerland 1993

[13] B Ahmad S K Ntouyas and J Tariboon ldquoExistence resultsformixedHadamard andRiemann-Liouville fractional integro-differential equationsrdquo Advances in Difference Equations vol2015 no 1 article 293 8 pages 2015

[14] R P Agarwal M Meehan and D OrsquoRegan Fixed PointTheory and Applications vol 141 Cambridge University PressCambridge UK 2001

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Journal of Function Spaces 9

[4] P Thiramanus S K Ntouyas and J Tariboon ldquoExistence anduniqueness results for Hadamard-type fractional differentialequations with nonlocal fractional integral boundary condi-tionsrdquo Abstract and Applied Analysis vol 2014 Article ID902054 9 pages 2014

[5] B Ahmad and S K Ntouyas ldquoAn existence theorem forfractional hybrid differential inclusions of Hadamard type withDirichlet boundary conditionsrdquo Abstract and Applied Analysisvol 2014 Article ID 705809 7 pages 2014

[6] B Ahmad and S K Ntouyas ldquoA fully Hadamard type integralboundary value problem of a coupled system of fractional dif-ferential equationsrdquo Fractional Calculus and Applied AnalysisAn International Journal forTheory and Applications vol 17 no2 pp 348ndash360 2014

[7] F Jarad T Abdeljawad and D Baleanu ldquoCaputo-type modi-fication of the Hadamard fractional derivativesrdquo Advances inDifference Equations vol 2012 no 1 article 142 8 pages 2012

[8] AA Kilbas ldquoHadamard-type fractional calculusrdquo Journal of theKorean Mathematical Society vol 38 no 6 pp 1191ndash1204 2001

[9] I Podlubny Fractional Differential Equations Academic PressSan Diego Calif USA 1999

[10] S Pooseh RAlmeida andD FMTorres ldquoExpansion formulasin terms of integer-order derivatives for the Hadamard frac-tional integral and derivativerdquo Numerical Functional Analysisand Optimization vol 33 no 3 pp 301ndash319 2012

[11] A A Kilbas H M Srivastava and J J Trujillo Theoryand Applications of Fractional Differential Equations vol 204Elsevier Science Amsterdam The Netherlands 2006

[12] S G Samko A A Kilbas and O I Marichev FractionalIntegrals and Derivatives Gordon and Breach Science YverdonSwitzerland 1993

[13] B Ahmad S K Ntouyas and J Tariboon ldquoExistence resultsformixedHadamard andRiemann-Liouville fractional integro-differential equationsrdquo Advances in Difference Equations vol2015 no 1 article 293 8 pages 2015

[14] R P Agarwal M Meehan and D OrsquoRegan Fixed PointTheory and Applications vol 141 Cambridge University PressCambridge UK 2001

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of